Sec 2 G1 Maths Tutor Bukit Timah | Sine Cosine Tangent

Did you know? A right triangle can be made three times larger without changing any of its angles. Its sides all triple, but the ratios of opposite to hypotenuse, adjacent to hypotenuse and opposite to adjacent stay the same. That simple observation explains why the three trigonometric functions—sine, cosine and tangent—can connect an angle with unknown side lengths.

Quick answer: sine, cosine and tangent

sinθ = opposite / hypotenuse (SOH).
cosθ = adjacent / hypotenuse (CAH).
tanθ = opposite / adjacent (TOA).

For these formulas, θ is one acute angle in a right triangle. The hypotenuse is opposite the 90° angle; opposite and adjacent are chosen relative to θ. If two sides are known, an appropriate inverse function (sin⁻¹, cos⁻¹ or tan⁻¹) can recover an acute angle. When angles are given in degrees, use degree mode or an explicit unit conversion. Show the relevant mathematical steps and follow the accuracy required by the question.

This guide is for families searching for a Secondary 2 G1 Mathematics tutor in Bukit Timah. The published 2027 SEC G1 Mathematics K110 graduating syllabus includes acute-angle sine, cosine and tangent ratios, with angle-of-elevation and depression applications. That does not establish that these methods are on every 2026 Secondary 2 G1 school test. Use this article as a later-stage G1 bridge when right-angle properties, Pythagoras and signed-number calculator skills are secure, and follow current teacher-issued scope first.

Choose a trigonometry problem

One-minute readiness diagnostic
How to label a right triangle
Why the three ratios work
Identify opposite, adjacent and hypotenuse
Compute sine, cosine and tangent from sides
Find an unknown side with a known angle
Find an acute angle from two side lengths
Compare trig ratios when triangles are resized
Decide which trig ratio the question needs
Check calculator mode, accuracy and feasibility
A three-pupil teaching example
Questions and answers
Official syllabus references

One-minute readiness diagnostic

Ask a pupil to identify the hypotenuse in a 3–4–5 right triangle, verify 3²+4²=5², and calculate 3/5 and 4/5. Then mark an acute angle opposite the side of length three and ask which leg is adjacent. If the child cannot identify the opposite side and the right angle, adding a calculator mnemonic now may compound that problem. Start with labelled triangle geometry before extending to trigonometric function input.

The hypotenuse never changes when the viewpoint changes

The right angle is marked 90°. Its opposite side is the longest side—the hypotenuse. Pick one of the remaining two angles and label it θ. The shorter side facing θ across the triangle is opposite, and the other leg touching θ is adjacent. If θ moves to the other acute corner, opposite and adjacent swap, but the hypotenuse remains the same. This is a definition-based decision, not a rule that ‘the horizontal side is always adjacent’.

A ratio is a relationship, not a length

Sine, cosine and tangent are quotients of lengths expressed in the same unit, so their units cancel. For a 3–4–5 triangle viewed from the angle opposite the side of length three, sinθ=3/5, cosθ=4/5 and tanθ=3/4. The angle remains the same if the triangle is uniformly resized. The mnemonic SOHCAHTOA helps remember the pairings only after the diagram has been labelled correctly.

238 original right-triangle trigonometry clinics

The worked bank contains exact side ratios, calculated unknown sides, acute angles, similarity checks and ratio-selection questions. The scenarios are original mathematical illustrations, not copies of official examination papers. Pupils should choose a small selection based on readiness, then attempt changed-number tasks with the worked solution hidden. Every drawing or side relationship is described explicitly so no missing photograph is required.

Identify opposite, adjacent and hypotenuse

In this identify opposite, adjacent and hypotenuse section, start from the 90° mark and identify which side is being compared with which. Write one appropriate ratio before calculating; do not use every function simply because it appears on a calculator. Exact fractions are retained when possible and approximate lengths or angles are labelled accordingly. The important skill is explaining the side relationship on a fresh triangle.

1. Label a right triangle 1

Question: A right triangle has perpendicular legs of 3 cm and 4 cm, and hypotenuse 5 cm. Focus on the acute angle θ whose opposite leg has length 4 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 5 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=4 cm, adjacent=3 cm and hypotenuse=5 cm. The right angle lies between the two legs 3 and 4; θ is one of the other two triangle angles.

First tempting mistake: A pupil who labels the longest leg as the hypotenuse without identifying the side opposite 90° may choose the wrong ratio. Calling both shorter sides ‘adjacent’ ignores the selected angle.

Independent check: Pythagoras gives 3²+4²=9+16=25=5². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 3 and 4; the hypotenuse stays 5. Ask which labels change and which never do.

2. Label a right triangle 2

Question: A right triangle has perpendicular legs of 5 cm and 12 cm, and hypotenuse 13 cm. Focus on the acute angle θ whose opposite leg has length 5 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 13 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=5 cm, adjacent=12 cm and hypotenuse=13 cm. The right angle lies between the two legs 5 and 12; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 5²+12²=25+144=169=13². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 12 and 5; the hypotenuse stays 13. Ask which labels change and which never do.

3. Label a right triangle 3

Question: A right triangle has perpendicular legs of 8 cm and 15 cm, and hypotenuse 17 cm. Focus on the acute angle θ whose opposite leg has length 15 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 17 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=15 cm, adjacent=8 cm and hypotenuse=17 cm. The right angle lies between the two legs 8 and 15; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 8²+15²=64+225=289=17². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 8 and 15; the hypotenuse stays 17. Ask which labels change and which never do.

4. Label a right triangle 4

Question: A right triangle has perpendicular legs of 7 cm and 24 cm, and hypotenuse 25 cm. Focus on the acute angle θ whose opposite leg has length 7 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 25 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=7 cm, adjacent=24 cm and hypotenuse=25 cm. The right angle lies between the two legs 7 and 24; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 7²+24²=49+576=625=25². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 24 and 7; the hypotenuse stays 25. Ask which labels change and which never do.

5. Label a right triangle 5

Question: A right triangle has perpendicular legs of 9 cm and 40 cm, and hypotenuse 41 cm. Focus on the acute angle θ whose opposite leg has length 40 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 41 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=40 cm, adjacent=9 cm and hypotenuse=41 cm. The right angle lies between the two legs 9 and 40; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 9²+40²=81+1600=1681=41². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 9 and 40; the hypotenuse stays 41. Ask which labels change and which never do.

6. Label a right triangle 6

Question: A right triangle has perpendicular legs of 20 cm and 21 cm, and hypotenuse 29 cm. Focus on the acute angle θ whose opposite leg has length 20 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 29 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=20 cm, adjacent=21 cm and hypotenuse=29 cm. The right angle lies between the two legs 20 and 21; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 20²+21²=400+441=841=29². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 21 and 20; the hypotenuse stays 29. Ask which labels change and which never do.

7. Label a right triangle 7

Question: A right triangle has perpendicular legs of 12 cm and 35 cm, and hypotenuse 37 cm. Focus on the acute angle θ whose opposite leg has length 35 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 37 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=35 cm, adjacent=12 cm and hypotenuse=37 cm. The right angle lies between the two legs 12 and 35; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 12²+35²=144+1225=1369=37². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 12 and 35; the hypotenuse stays 37. Ask which labels change and which never do.

8. Label a right triangle 8

Question: A right triangle has perpendicular legs of 11 cm and 60 cm, and hypotenuse 61 cm. Focus on the acute angle θ whose opposite leg has length 11 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 61 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=11 cm, adjacent=60 cm and hypotenuse=61 cm. The right angle lies between the two legs 11 and 60; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 11²+60²=121+3600=3721=61². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 60 and 11; the hypotenuse stays 61. Ask which labels change and which never do.

9. Label a right triangle 9

Question: A right triangle has perpendicular legs of 28 cm and 45 cm, and hypotenuse 53 cm. Focus on the acute angle θ whose opposite leg has length 45 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 53 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=45 cm, adjacent=28 cm and hypotenuse=53 cm. The right angle lies between the two legs 28 and 45; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 28²+45²=784+2025=2809=53². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 28 and 45; the hypotenuse stays 53. Ask which labels change and which never do.

10. Label a right triangle 10

Question: A right triangle has perpendicular legs of 33 cm and 56 cm, and hypotenuse 65 cm. Focus on the acute angle θ whose opposite leg has length 33 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 65 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=33 cm, adjacent=56 cm and hypotenuse=65 cm. The right angle lies between the two legs 33 and 56; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 33²+56²=1089+3136=4225=65². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 56 and 33; the hypotenuse stays 65. Ask which labels change and which never do.

11. Label a right triangle 11

Question: A right triangle has perpendicular legs of 16 cm and 63 cm, and hypotenuse 65 cm. Focus on the acute angle θ whose opposite leg has length 63 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

Worked answer: For θ, opposite=63 cm, adjacent=16 cm and hypotenuse=65 cm. The right angle lies between the two legs 16 and 63; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 16²+63²=256+3969=4225=65². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 16 and 63; the hypotenuse stays 65. Ask which labels change and which never do.

12. Label a right triangle 12

Question: A right triangle has perpendicular legs of 13 cm and 84 cm, and hypotenuse 85 cm. Focus on the acute angle θ whose opposite leg has length 13 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 85 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=13 cm, adjacent=84 cm and hypotenuse=85 cm. The right angle lies between the two legs 13 and 84; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 13²+84²=169+7056=7225=85². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 84 and 13; the hypotenuse stays 85. Ask which labels change and which never do.

13. Label a right triangle 13

Question: A right triangle has perpendicular legs of 36 cm and 77 cm, and hypotenuse 85 cm. Focus on the acute angle θ whose opposite leg has length 77 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

Worked answer: For θ, opposite=77 cm, adjacent=36 cm and hypotenuse=85 cm. The right angle lies between the two legs 36 and 77; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 36²+77²=1296+5929=7225=85². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 36 and 77; the hypotenuse stays 85. Ask which labels change and which never do.

14. Label a right triangle 14

Question: A right triangle has perpendicular legs of 39 cm and 80 cm, and hypotenuse 89 cm. Focus on the acute angle θ whose opposite leg has length 39 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 89 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=39 cm, adjacent=80 cm and hypotenuse=89 cm. The right angle lies between the two legs 39 and 80; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 39²+80²=1521+6400=7921=89². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 80 and 39; the hypotenuse stays 89. Ask which labels change and which never do.

15. Label a right triangle 15

Question: A right triangle has perpendicular legs of 48 cm and 55 cm, and hypotenuse 73 cm. Focus on the acute angle θ whose opposite leg has length 55 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 73 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=55 cm, adjacent=48 cm and hypotenuse=73 cm. The right angle lies between the two legs 48 and 55; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 48²+55²=2304+3025=5329=73². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 48 and 55; the hypotenuse stays 73. Ask which labels change and which never do.

16. Label a right triangle 16

Question: A right triangle has perpendicular legs of 15 cm and 112 cm, and hypotenuse 113 cm. Focus on the acute angle θ whose opposite leg has length 15 cm. Name its opposite, adjacent and hypotenuse and identify the right angle.

How to read the triangle: The hypotenuse is always opposite the 90° angle, so it remains 113 cm regardless of which acute angle is being considered. Opposite and adjacent are relative words: a leg may be opposite one acute angle and adjacent to the other. State which angle has been selected before labelling.

Worked answer: For θ, opposite=15 cm, adjacent=112 cm and hypotenuse=113 cm. The right angle lies between the two legs 15 and 112; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 15²+112²=225+12544=12769=113². The right angle and selected viewpoint are consistent with these side lengths.

Changed-number question: Focus instead on the other acute angle. Its opposite and adjacent sides exchange roles, becoming 112 and 15; the hypotenuse stays 113. Ask which labels change and which never do.

17. Label a right triangle 17

Worked answer: For θ, opposite=4 cm, adjacent=3 cm and hypotenuse=5 cm. The right angle lies between the two legs 3 and 4; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 3²+4²=9+16=25=5². The right angle and selected viewpoint are consistent with these side lengths.

18. Label a right triangle 18

Worked answer: For θ, opposite=5 cm, adjacent=12 cm and hypotenuse=13 cm. The right angle lies between the two legs 5 and 12; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 5²+12²=25+144=169=13². The right angle and selected viewpoint are consistent with these side lengths.

19. Label a right triangle 19

Worked answer: For θ, opposite=15 cm, adjacent=8 cm and hypotenuse=17 cm. The right angle lies between the two legs 8 and 15; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 8²+15²=64+225=289=17². The right angle and selected viewpoint are consistent with these side lengths.

20. Label a right triangle 20

Worked answer: For θ, opposite=7 cm, adjacent=24 cm and hypotenuse=25 cm. The right angle lies between the two legs 7 and 24; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 7²+24²=49+576=625=25². The right angle and selected viewpoint are consistent with these side lengths.

21. Label a right triangle 21

Worked answer: For θ, opposite=40 cm, adjacent=9 cm and hypotenuse=41 cm. The right angle lies between the two legs 9 and 40; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 9²+40²=81+1600=1681=41². The right angle and selected viewpoint are consistent with these side lengths.

22. Label a right triangle 22

Worked answer: For θ, opposite=20 cm, adjacent=21 cm and hypotenuse=29 cm. The right angle lies between the two legs 20 and 21; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 20²+21²=400+441=841=29². The right angle and selected viewpoint are consistent with these side lengths.

23. Label a right triangle 23

Worked answer: For θ, opposite=35 cm, adjacent=12 cm and hypotenuse=37 cm. The right angle lies between the two legs 12 and 35; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 12²+35²=144+1225=1369=37². The right angle and selected viewpoint are consistent with these side lengths.

24. Label a right triangle 24

Worked answer: For θ, opposite=11 cm, adjacent=60 cm and hypotenuse=61 cm. The right angle lies between the two legs 11 and 60; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 11²+60²=121+3600=3721=61². The right angle and selected viewpoint are consistent with these side lengths.

25. Label a right triangle 25

Worked answer: For θ, opposite=45 cm, adjacent=28 cm and hypotenuse=53 cm. The right angle lies between the two legs 28 and 45; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 28²+45²=784+2025=2809=53². The right angle and selected viewpoint are consistent with these side lengths.

26. Label a right triangle 26

Worked answer: For θ, opposite=33 cm, adjacent=56 cm and hypotenuse=65 cm. The right angle lies between the two legs 33 and 56; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 33²+56²=1089+3136=4225=65². The right angle and selected viewpoint are consistent with these side lengths.

27. Label a right triangle 27

Worked answer: For θ, opposite=63 cm, adjacent=16 cm and hypotenuse=65 cm. The right angle lies between the two legs 16 and 63; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 16²+63²=256+3969=4225=65². The right angle and selected viewpoint are consistent with these side lengths.

28. Label a right triangle 28

Worked answer: For θ, opposite=13 cm, adjacent=84 cm and hypotenuse=85 cm. The right angle lies between the two legs 13 and 84; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 13²+84²=169+7056=7225=85². The right angle and selected viewpoint are consistent with these side lengths.

29. Label a right triangle 29

Worked answer: For θ, opposite=77 cm, adjacent=36 cm and hypotenuse=85 cm. The right angle lies between the two legs 36 and 77; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 36²+77²=1296+5929=7225=85². The right angle and selected viewpoint are consistent with these side lengths.

30. Label a right triangle 30

Worked answer: For θ, opposite=39 cm, adjacent=80 cm and hypotenuse=89 cm. The right angle lies between the two legs 39 and 80; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 39²+80²=1521+6400=7921=89². The right angle and selected viewpoint are consistent with these side lengths.

31. Label a right triangle 31

Worked answer: For θ, opposite=55 cm, adjacent=48 cm and hypotenuse=73 cm. The right angle lies between the two legs 48 and 55; θ is one of the other two triangle angles.

Independent check: Pythagoras gives 48²+55²=2304+3025=5329=73². The right angle and selected viewpoint are consistent with these side lengths.

Compute sine, cosine and tangent from sides

In this compute sine, cosine and tangent from sides section, start from the 90° mark and identify which side is being compared with which. Write one appropriate ratio before calculating; do not use every function simply because it appears on a calculator. Exact fractions are retained when possible and approximate lengths or angles are labelled accordingly. The important skill is explaining the side relationship on a fresh triangle.

32. Ratio from exact side lengths 1

Question: In a right triangle, relative to acute angle θ, opposite side=4, adjacent=3, hypotenuse=5. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

How to read the triangle: The three ratios compare corresponding sides of the same right triangle. Sine uses opposite/hypotenuse; cosine adjacent/hypotenuse; tangent opposite/adjacent. Each quantity is a dimensionless ratio of two lengths using the same unit.

Worked answer: sinθ=4/5, cosθ=3/5 and tanθ=4/3. Their decimal values are about 0.8, 0.6 and 1.33, respectively, when approximation is useful.

First tempting mistake: A common error is to call opposite/adjacent the sine or divide by the hypotenuse for tangent. Writing the three named side pairs before the fractions is quicker than trying to recall them from a picture without labels.

Independent check: Because 4²+3²=5², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=4/3, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 4/5, 3/5 and 4/3. Explain why the common scale cancels.

33. Ratio from exact side lengths 2

Question: In a right triangle, relative to acute angle θ, opposite side=7, adjacent=24, hypotenuse=25. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=7/25, cosθ=24/25 and tanθ=7/24. Their decimal values are about 0.28, 0.96 and 0.292, respectively, when approximation is useful.

Independent check: Because 7²+24²=25², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=7/24, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 7/25, 24/25 and 7/24. Explain why the common scale cancels.

34. Ratio from exact side lengths 3

Question: In a right triangle, relative to acute angle θ, opposite side=35, adjacent=12, hypotenuse=37. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=35/37, cosθ=12/37 and tanθ=35/12. Their decimal values are about 0.946, 0.324 and 2.92, respectively, when approximation is useful.

Independent check: Because 35²+12²=37², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=35/12, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 35/37, 12/37 and 35/12. Explain why the common scale cancels.

35. Ratio from exact side lengths 4

Question: In a right triangle, relative to acute angle θ, opposite side=33, adjacent=56, hypotenuse=65. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=33/65, cosθ=56/65 and tanθ=33/56. Their decimal values are about 0.508, 0.862 and 0.589, respectively, when approximation is useful.

Independent check: Because 33²+56²=65², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=33/56, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 33/65, 56/65 and 33/56. Explain why the common scale cancels.

36. Ratio from exact side lengths 5

Question: In a right triangle, relative to acute angle θ, opposite side=77, adjacent=36, hypotenuse=85. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=77/85, cosθ=36/85 and tanθ=77/36. Their decimal values are about 0.906, 0.424 and 2.14, respectively, when approximation is useful.

Independent check: Because 77²+36²=85², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=77/36, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 77/85, 36/85 and 77/36. Explain why the common scale cancels.

37. Ratio from exact side lengths 6

Question: In a right triangle, relative to acute angle θ, opposite side=15, adjacent=112, hypotenuse=113. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=15/113, cosθ=112/113 and tanθ=15/112. Their decimal values are about 0.133, 0.991 and 0.134, respectively, when approximation is useful.

Independent check: Because 15²+112²=113², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=15/112, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 15/113, 112/113 and 15/112. Explain why the common scale cancels.

38. Ratio from exact side lengths 7

Question: In a right triangle, relative to acute angle θ, opposite side=15, adjacent=8, hypotenuse=17. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=15/17, cosθ=8/17 and tanθ=15/8. Their decimal values are about 0.882, 0.471 and 1.88, respectively, when approximation is useful.

Independent check: Because 15²+8²=17², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=15/8, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 15/17, 8/17 and 15/8. Explain why the common scale cancels.

39. Ratio from exact side lengths 8

Question: In a right triangle, relative to acute angle θ, opposite side=20, adjacent=21, hypotenuse=29. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=20/29, cosθ=21/29 and tanθ=20/21. Their decimal values are about 0.69, 0.724 and 0.952, respectively, when approximation is useful.

Independent check: Because 20²+21²=29², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=20/21, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 20/29, 21/29 and 20/21. Explain why the common scale cancels.

40. Ratio from exact side lengths 9

Question: In a right triangle, relative to acute angle θ, opposite side=45, adjacent=28, hypotenuse=53. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=45/53, cosθ=28/53 and tanθ=45/28. Their decimal values are about 0.849, 0.528 and 1.61, respectively, when approximation is useful.

Independent check: Because 45²+28²=53², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=45/28, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 45/53, 28/53 and 45/28. Explain why the common scale cancels.

41. Ratio from exact side lengths 10

Question: In a right triangle, relative to acute angle θ, opposite side=13, adjacent=84, hypotenuse=85. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=13/85, cosθ=84/85 and tanθ=13/84. Their decimal values are about 0.153, 0.988 and 0.155, respectively, when approximation is useful.

Independent check: Because 13²+84²=85², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=13/84, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 13/85, 84/85 and 13/84. Explain why the common scale cancels.

42. Ratio from exact side lengths 11

Question: In a right triangle, relative to acute angle θ, opposite side=55, adjacent=48, hypotenuse=73. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=55/73, cosθ=48/73 and tanθ=55/48. Their decimal values are about 0.753, 0.658 and 1.15, respectively, when approximation is useful.

Independent check: Because 55²+48²=73², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=55/48, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 55/73, 48/73 and 55/48. Explain why the common scale cancels.

43. Ratio from exact side lengths 12

Question: In a right triangle, relative to acute angle θ, opposite side=5, adjacent=12, hypotenuse=13. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=5/13, cosθ=12/13 and tanθ=5/12. Their decimal values are about 0.385, 0.923 and 0.417, respectively, when approximation is useful.

Independent check: Because 5²+12²=13², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=5/12, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 5/13, 12/13 and 5/12. Explain why the common scale cancels.

44. Ratio from exact side lengths 13

Question: In a right triangle, relative to acute angle θ, opposite side=40, adjacent=9, hypotenuse=41. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=40/41, cosθ=9/41 and tanθ=40/9. Their decimal values are about 0.976, 0.22 and 4.44, respectively, when approximation is useful.

Independent check: Because 40²+9²=41², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=40/9, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 40/41, 9/41 and 40/9. Explain why the common scale cancels.

45. Ratio from exact side lengths 14

Question: In a right triangle, relative to acute angle θ, opposite side=11, adjacent=60, hypotenuse=61. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=11/61, cosθ=60/61 and tanθ=11/60. Their decimal values are about 0.18, 0.984 and 0.183, respectively, when approximation is useful.

Independent check: Because 11²+60²=61², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=11/60, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 11/61, 60/61 and 11/60. Explain why the common scale cancels.

46. Ratio from exact side lengths 15

Question: In a right triangle, relative to acute angle θ, opposite side=63, adjacent=16, hypotenuse=65. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=63/65, cosθ=16/65 and tanθ=63/16. Their decimal values are about 0.969, 0.246 and 3.94, respectively, when approximation is useful.

Independent check: Because 63²+16²=65², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=63/16, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 63/65, 16/65 and 63/16. Explain why the common scale cancels.

47. Ratio from exact side lengths 16

Question: In a right triangle, relative to acute angle θ, opposite side=39, adjacent=80, hypotenuse=89. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=39/89, cosθ=80/89 and tanθ=39/80. Their decimal values are about 0.438, 0.899 and 0.487, respectively, when approximation is useful.

Independent check: Because 39²+80²=89², the exact side ratios satisfy sin²θ+cos²θ=1 as an optional consistency check. Also tanθ=(sinθ)/(cosθ)=39/80, which agrees with the direct side quotient.

Changed-number question: If the same triangle were enlarged uniformly by factor three, all lengths would triple but sinθ, cosθ and tanθ would stay exactly 39/89, 80/89 and 39/80. Explain why the common scale cancels.

48. Ratio from exact side lengths 17

Question: In a right triangle, relative to acute angle θ, opposite side=4, adjacent=3, hypotenuse=5. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=4/5, cosθ=3/5 and tanθ=4/3. Their decimal values are about 0.8, 0.6 and 1.33, respectively, when approximation is useful.

49. Ratio from exact side lengths 18

Question: In a right triangle, relative to acute angle θ, opposite side=7, adjacent=24, hypotenuse=25. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=7/25, cosθ=24/25 and tanθ=7/24. Their decimal values are about 0.28, 0.96 and 0.292, respectively, when approximation is useful.

50. Ratio from exact side lengths 19

Question: In a right triangle, relative to acute angle θ, opposite side=35, adjacent=12, hypotenuse=37. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=35/37, cosθ=12/37 and tanθ=35/12. Their decimal values are about 0.946, 0.324 and 2.92, respectively, when approximation is useful.

51. Ratio from exact side lengths 20

Question: In a right triangle, relative to acute angle θ, opposite side=33, adjacent=56, hypotenuse=65. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=33/65, cosθ=56/65 and tanθ=33/56. Their decimal values are about 0.508, 0.862 and 0.589, respectively, when approximation is useful.

52. Ratio from exact side lengths 21

Question: In a right triangle, relative to acute angle θ, opposite side=77, adjacent=36, hypotenuse=85. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=77/85, cosθ=36/85 and tanθ=77/36. Their decimal values are about 0.906, 0.424 and 2.14, respectively, when approximation is useful.

53. Ratio from exact side lengths 22

Question: In a right triangle, relative to acute angle θ, opposite side=15, adjacent=112, hypotenuse=113. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=15/113, cosθ=112/113 and tanθ=15/112. Their decimal values are about 0.133, 0.991 and 0.134, respectively, when approximation is useful.

54. Ratio from exact side lengths 23

Question: In a right triangle, relative to acute angle θ, opposite side=15, adjacent=8, hypotenuse=17. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=15/17, cosθ=8/17 and tanθ=15/8. Their decimal values are about 0.882, 0.471 and 1.88, respectively, when approximation is useful.

55. Ratio from exact side lengths 24

Question: In a right triangle, relative to acute angle θ, opposite side=20, adjacent=21, hypotenuse=29. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=20/29, cosθ=21/29 and tanθ=20/21. Their decimal values are about 0.69, 0.724 and 0.952, respectively, when approximation is useful.

56. Ratio from exact side lengths 25

Question: In a right triangle, relative to acute angle θ, opposite side=45, adjacent=28, hypotenuse=53. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=45/53, cosθ=28/53 and tanθ=45/28. Their decimal values are about 0.849, 0.528 and 1.61, respectively, when approximation is useful.

57. Ratio from exact side lengths 26

Question: In a right triangle, relative to acute angle θ, opposite side=13, adjacent=84, hypotenuse=85. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=13/85, cosθ=84/85 and tanθ=13/84. Their decimal values are about 0.153, 0.988 and 0.155, respectively, when approximation is useful.

58. Ratio from exact side lengths 27

Question: In a right triangle, relative to acute angle θ, opposite side=55, adjacent=48, hypotenuse=73. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=55/73, cosθ=48/73 and tanθ=55/48. Their decimal values are about 0.753, 0.658 and 1.15, respectively, when approximation is useful.

59. Ratio from exact side lengths 28

Question: In a right triangle, relative to acute angle θ, opposite side=5, adjacent=12, hypotenuse=13. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=5/13, cosθ=12/13 and tanθ=5/12. Their decimal values are about 0.385, 0.923 and 0.417, respectively, when approximation is useful.

60. Ratio from exact side lengths 29

Question: In a right triangle, relative to acute angle θ, opposite side=40, adjacent=9, hypotenuse=41. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=40/41, cosθ=9/41 and tanθ=40/9. Their decimal values are about 0.976, 0.22 and 4.44, respectively, when approximation is useful.

61. Ratio from exact side lengths 30

Question: In a right triangle, relative to acute angle θ, opposite side=11, adjacent=60, hypotenuse=61. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=11/61, cosθ=60/61 and tanθ=11/60. Their decimal values are about 0.18, 0.984 and 0.183, respectively, when approximation is useful.

62. Ratio from exact side lengths 31

Question: In a right triangle, relative to acute angle θ, opposite side=63, adjacent=16, hypotenuse=65. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=63/65, cosθ=16/65 and tanθ=63/16. Their decimal values are about 0.969, 0.246 and 3.94, respectively, when approximation is useful.

63. Ratio from exact side lengths 32

Question: In a right triangle, relative to acute angle θ, opposite side=39, adjacent=80, hypotenuse=89. Calculate sinθ, cosθ and tanθ as exact simplified fractions.

Worked answer: sinθ=39/89, cosθ=80/89 and tanθ=39/80. Their decimal values are about 0.438, 0.899 and 0.487, respectively, when approximation is useful.

Find an unknown side with a known angle

In this find an unknown side with a known angle section, start from the 90° mark and identify which side is being compared with which. Write one appropriate ratio before calculating; do not use every function simply because it appears on a calculator. Exact fractions are retained when possible and approximate lengths or angles are labelled accordingly. The important skill is explaining the side relationship on a fresh triangle.

64. Find an unknown side 1: 20°

Question: A right triangle has acute angle 20° and hypotenuse 10 cm. Find the opposite leg to three significant figures.

How to read the triangle: Sketch and mark the angle θ=20° next to the known leg where relevant. The right-angle symbol identifies the hypotenuse. Choose a ratio containing the requested side and the known side, then write the formula with symbols before entering a calculator.

Worked answer: Here Opposite/hypotenuse is sine. Hence opposite=10×sin20°≈3.42 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

First tempting mistake: The first risk is treating the adjacent leg as the hypotenuse, or choosing cos when the question actually gives opposite and adjacent. Another is using radian mode without conversion, which can produce an implausible output.

Independent check: For a right triangle with known hypotenuse 10 cm, a leg must be shorter than 10 cm. The calculated 3.42 cm is between zero and 10, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 25° while preserving the same known side. The new opposite length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

65. Find an unknown side 2: 25°

Question: A right triangle has acute angle 25° and hypotenuse 17 cm. Find the adjacent leg to three significant figures.

How to read the triangle: Sketch and mark the angle θ=25° next to the known leg where relevant. The right-angle symbol identifies the hypotenuse. Choose a ratio containing the requested side and the known side, then write the formula with symbols before entering a calculator.

Worked answer: Here Adjacent/hypotenuse is cosine. Hence adjacent=17×cos25°≈15.4 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 17 cm, a leg must be shorter than 17 cm. The calculated 15.4 cm is between zero and 17, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 30° while preserving the same known side. The new adjacent length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

66. Find an unknown side 3: 30°

Question: A right triangle has acute angle 30° and adjacent leg 24 cm. Find the opposite leg to three significant figures.

How to read the triangle: Sketch and mark the angle θ=30° next to the known leg where relevant. The right-angle symbol identifies the hypotenuse. Choose a ratio containing the requested side and the known side, then write the formula with symbols before entering a calculator.

Worked answer: Here Opposite/adjacent is tangent. Hence opposite=24×tan30°≈13.9 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: Here 24 cm is the known adjacent leg, not the hypotenuse. The hypotenuse would be 27.7 cm, longer than the adjacent leg and the calculated opposite leg. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 35° while preserving the same known side. The new adjacent length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

67. Find an unknown side 4: 35°

Question: A right triangle has acute angle 35° and hypotenuse 31 cm. Find the opposite leg to three significant figures.

How to read the triangle: Sketch and mark the angle θ=35° next to the known leg where relevant. The right-angle symbol identifies the hypotenuse. Choose a ratio containing the requested side and the known side, then write the formula with symbols before entering a calculator.

Worked answer: Here Opposite/hypotenuse is sine. Hence opposite=31×sin35°≈17.8 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 31 cm, a leg must be shorter than 31 cm. The calculated 17.8 cm is between zero and 31, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 40° while preserving the same known side. The new opposite length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

68. Find an unknown side 5: 40°

Question: A right triangle has acute angle 40° and hypotenuse 38 cm. Find the adjacent leg to three significant figures.

How to read the triangle: Sketch and mark the angle θ=40° next to the known leg where relevant. The right-angle symbol identifies the hypotenuse. Choose a ratio containing the requested side and the known side, then write the formula with symbols before entering a calculator.

Worked answer: Here Adjacent/hypotenuse is cosine. Hence adjacent=38×cos40°≈29.1 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 38 cm, a leg must be shorter than 38 cm. The calculated 29.1 cm is between zero and 38, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 45° while preserving the same known side. The new adjacent length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

69. Find an unknown side 6: 45°

Question: A right triangle has acute angle 45° and adjacent leg 45 cm. Find the opposite leg to three significant figures.

How to read the triangle: Sketch and mark the angle θ=45° next to the known leg where relevant. The right-angle symbol identifies the hypotenuse. Choose a ratio containing the requested side and the known side, then write the formula with symbols before entering a calculator.

Worked answer: Here Opposite/adjacent is tangent. Hence opposite=45×tan45°≈45 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: Here 45 cm is the known adjacent leg, not the hypotenuse. The hypotenuse would be 63.6 cm, longer than the adjacent leg and the calculated opposite leg. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 50° while preserving the same known side. The new adjacent length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

70. Find an unknown side 7: 50°

Question: A right triangle has acute angle 50° and hypotenuse 52 cm. Find the opposite leg to three significant figures.

How to read the triangle: Sketch and mark the angle θ=50° next to the known leg where relevant. The right-angle symbol identifies the hypotenuse. Choose a ratio containing the requested side and the known side, then write the formula with symbols before entering a calculator.

Worked answer: Here Opposite/hypotenuse is sine. Hence opposite=52×sin50°≈39.8 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 52 cm, a leg must be shorter than 52 cm. The calculated 39.8 cm is between zero and 52, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 55° while preserving the same known side. The new opposite length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

71. Find an unknown side 8: 55°

Question: A right triangle has acute angle 55° and hypotenuse 59 cm. Find the adjacent leg to three significant figures.

How to read the triangle: Sketch and mark the angle θ=55° next to the known leg where relevant. The right-angle symbol identifies the hypotenuse. Choose a ratio containing the requested side and the known side, then write the formula with symbols before entering a calculator.

Worked answer: Here Adjacent/hypotenuse is cosine. Hence adjacent=59×cos55°≈33.8 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 59 cm, a leg must be shorter than 59 cm. The calculated 33.8 cm is between zero and 59, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 60° while preserving the same known side. The new adjacent length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

72. Find an unknown side 9: 60°

Question: A right triangle has acute angle 60° and adjacent leg 66 cm. Find the opposite leg to three significant figures.

How to read the triangle: Sketch and mark the angle θ=60° next to the known leg where relevant. The right-angle symbol identifies the hypotenuse. Choose a ratio containing the requested side and the known side, then write the formula with symbols before entering a calculator.

Worked answer: Here Opposite/adjacent is tangent. Hence opposite=66×tan60°≈114 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: Here 66 cm is the known adjacent leg, not the hypotenuse. The hypotenuse would be 132 cm, longer than the adjacent leg and the calculated opposite leg. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 65° while preserving the same known side. The new adjacent length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

73. Find an unknown side 10: 65°

Question: A right triangle has acute angle 65° and hypotenuse 73 cm. Find the opposite leg to three significant figures.

How to read the triangle: Sketch and mark the angle θ=65° next to the known leg where relevant. The right-angle symbol identifies the hypotenuse. Choose a ratio containing the requested side and the known side, then write the formula with symbols before entering a calculator.

Worked answer: Here Opposite/hypotenuse is sine. Hence opposite=73×sin65°≈66.2 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 73 cm, a leg must be shorter than 73 cm. The calculated 66.2 cm is between zero and 73, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 70° while preserving the same known side. The new opposite length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

74. Find an unknown side 11: 70°

Question: A right triangle has acute angle 70° and hypotenuse 80 cm. Find the adjacent leg to three significant figures.

How to read the triangle: Sketch and mark the angle θ=70° next to the known leg where relevant. The right-angle symbol identifies the hypotenuse. Choose a ratio containing the requested side and the known side, then write the formula with symbols before entering a calculator.

Worked answer: Here Adjacent/hypotenuse is cosine. Hence adjacent=80×cos70°≈27.4 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 80 cm, a leg must be shorter than 80 cm. The calculated 27.4 cm is between zero and 80, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 75° while preserving the same known side. The new adjacent length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

75. Find an unknown side 12: 20°

Question: A right triangle has acute angle 20° and adjacent leg 12 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/adjacent is tangent. Hence opposite=12×tan20°≈4.37 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: Here 12 cm is the known adjacent leg, not the hypotenuse. The hypotenuse would be 12.8 cm, longer than the adjacent leg and the calculated opposite leg. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 25° while preserving the same known side. The new adjacent length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

76. Find an unknown side 13: 25°

Question: A right triangle has acute angle 25° and hypotenuse 19 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/hypotenuse is sine. Hence opposite=19×sin25°≈8.03 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 19 cm, a leg must be shorter than 19 cm. The calculated 8.03 cm is between zero and 19, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 30° while preserving the same known side. The new opposite length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

77. Find an unknown side 14: 30°

Question: A right triangle has acute angle 30° and hypotenuse 26 cm. Find the adjacent leg to three significant figures.

Worked answer: Here Adjacent/hypotenuse is cosine. Hence adjacent=26×cos30°≈22.5 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 26 cm, a leg must be shorter than 26 cm. The calculated 22.5 cm is between zero and 26, as expected. The side definitions and bounds can reveal a wrong input.

78. Find an unknown side 15: 35°

Question: A right triangle has acute angle 35° and adjacent leg 33 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/adjacent is tangent. Hence opposite=33×tan35°≈23.1 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: Here 33 cm is the known adjacent leg, not the hypotenuse. The hypotenuse would be 40.3 cm, longer than the adjacent leg and the calculated opposite leg. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 40° while preserving the same known side. The new adjacent length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

79. Find an unknown side 16: 40°

Question: A right triangle has acute angle 40° and hypotenuse 40 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/hypotenuse is sine. Hence opposite=40×sin40°≈25.7 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 40 cm, a leg must be shorter than 40 cm. The calculated 25.7 cm is between zero and 40, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 45° while preserving the same known side. The new opposite length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

80. Find an unknown side 17: 45°

Question: A right triangle has acute angle 45° and hypotenuse 47 cm. Find the adjacent leg to three significant figures.

Worked answer: Here Adjacent/hypotenuse is cosine. Hence adjacent=47×cos45°≈33.2 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 47 cm, a leg must be shorter than 47 cm. The calculated 33.2 cm is between zero and 47, as expected. The side definitions and bounds can reveal a wrong input.

81. Find an unknown side 18: 50°

Question: A right triangle has acute angle 50° and adjacent leg 54 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/adjacent is tangent. Hence opposite=54×tan50°≈64.4 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: Here 54 cm is the known adjacent leg, not the hypotenuse. The hypotenuse would be 84 cm, longer than the adjacent leg and the calculated opposite leg. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 55° while preserving the same known side. The new adjacent length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

82. Find an unknown side 19: 55°

Question: A right triangle has acute angle 55° and hypotenuse 61 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/hypotenuse is sine. Hence opposite=61×sin55°≈50 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 61 cm, a leg must be shorter than 61 cm. The calculated 50 cm is between zero and 61, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 60° while preserving the same known side. The new opposite length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

83. Find an unknown side 20: 60°

Question: A right triangle has acute angle 60° and hypotenuse 68 cm. Find the adjacent leg to three significant figures.

Worked answer: Here Adjacent/hypotenuse is cosine. Hence adjacent=68×cos60°≈34 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 68 cm, a leg must be shorter than 68 cm. The calculated 34 cm is between zero and 68, as expected. The side definitions and bounds can reveal a wrong input.

84. Find an unknown side 21: 65°

Question: A right triangle has acute angle 65° and adjacent leg 75 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/adjacent is tangent. Hence opposite=75×tan65°≈161 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: Here 75 cm is the known adjacent leg, not the hypotenuse. The hypotenuse would be 177 cm, longer than the adjacent leg and the calculated opposite leg. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 70° while preserving the same known side. The new adjacent length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

85. Find an unknown side 22: 70°

Question: A right triangle has acute angle 70° and hypotenuse 82 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/hypotenuse is sine. Hence opposite=82×sin70°≈77.1 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 82 cm, a leg must be shorter than 82 cm. The calculated 77.1 cm is between zero and 82, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 75° while preserving the same known side. The new opposite length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

86. Find an unknown side 23: 20°

Question: A right triangle has acute angle 20° and hypotenuse 14 cm. Find the adjacent leg to three significant figures.

Worked answer: Here Adjacent/hypotenuse is cosine. Hence adjacent=14×cos20°≈13.2 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 14 cm, a leg must be shorter than 14 cm. The calculated 13.2 cm is between zero and 14, as expected. The side definitions and bounds can reveal a wrong input.

87. Find an unknown side 24: 25°

Question: A right triangle has acute angle 25° and adjacent leg 21 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/adjacent is tangent. Hence opposite=21×tan25°≈9.79 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: Here 21 cm is the known adjacent leg, not the hypotenuse. The hypotenuse would be 23.2 cm, longer than the adjacent leg and the calculated opposite leg. The side definitions and bounds can reveal a wrong input.

88. Find an unknown side 25: 30°

Question: A right triangle has acute angle 30° and hypotenuse 28 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/hypotenuse is sine. Hence opposite=28×sin30°≈14 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 28 cm, a leg must be shorter than 28 cm. The calculated 14 cm is between zero and 28, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 35° while preserving the same known side. The new opposite length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

89. Find an unknown side 26: 35°

Question: A right triangle has acute angle 35° and hypotenuse 35 cm. Find the adjacent leg to three significant figures.

Worked answer: Here Adjacent/hypotenuse is cosine. Hence adjacent=35×cos35°≈28.7 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 35 cm, a leg must be shorter than 35 cm. The calculated 28.7 cm is between zero and 35, as expected. The side definitions and bounds can reveal a wrong input.

90. Find an unknown side 27: 40°

Question: A right triangle has acute angle 40° and adjacent leg 42 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/adjacent is tangent. Hence opposite=42×tan40°≈35.2 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: Here 42 cm is the known adjacent leg, not the hypotenuse. The hypotenuse would be 54.8 cm, longer than the adjacent leg and the calculated opposite leg. The side definitions and bounds can reveal a wrong input.

91. Find an unknown side 28: 45°

Question: A right triangle has acute angle 45° and hypotenuse 49 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/hypotenuse is sine. Hence opposite=49×sin45°≈34.6 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 49 cm, a leg must be shorter than 49 cm. The calculated 34.6 cm is between zero and 49, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 50° while preserving the same known side. The new opposite length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

92. Find an unknown side 29: 50°

Question: A right triangle has acute angle 50° and hypotenuse 56 cm. Find the adjacent leg to three significant figures.

Worked answer: Here Adjacent/hypotenuse is cosine. Hence adjacent=56×cos50°≈36 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 56 cm, a leg must be shorter than 56 cm. The calculated 36 cm is between zero and 56, as expected. The side definitions and bounds can reveal a wrong input.

93. Find an unknown side 30: 55°

Question: A right triangle has acute angle 55° and adjacent leg 63 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/adjacent is tangent. Hence opposite=63×tan55°≈90 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: Here 63 cm is the known adjacent leg, not the hypotenuse. The hypotenuse would be 110 cm, longer than the adjacent leg and the calculated opposite leg. The side definitions and bounds can reveal a wrong input.

94. Find an unknown side 31: 60°

Question: A right triangle has acute angle 60° and hypotenuse 70 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/hypotenuse is sine. Hence opposite=70×sin60°≈60.6 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 70 cm, a leg must be shorter than 70 cm. The calculated 60.6 cm is between zero and 70, as expected. The side definitions and bounds can reveal a wrong input.

Changed-number question: Change the stated angle to 65° while preserving the same known side. The new opposite length is determined by the same named ratio, but its direction of change should be predicted before entering numbers.

95. Find an unknown side 32: 65°

Question: A right triangle has acute angle 65° and hypotenuse 77 cm. Find the adjacent leg to three significant figures.

Worked answer: Here Adjacent/hypotenuse is cosine. Hence adjacent=77×cos65°≈32.5 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 77 cm, a leg must be shorter than 77 cm. The calculated 32.5 cm is between zero and 77, as expected. The side definitions and bounds can reveal a wrong input.

96. Find an unknown side 33: 70°

Question: A right triangle has acute angle 70° and adjacent leg 84 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/adjacent is tangent. Hence opposite=84×tan70°≈231 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: Here 84 cm is the known adjacent leg, not the hypotenuse. The hypotenuse would be 246 cm, longer than the adjacent leg and the calculated opposite leg. The side definitions and bounds can reveal a wrong input.

97. Find an unknown side 34: 20°

Question: A right triangle has acute angle 20° and hypotenuse 16 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/hypotenuse is sine. Hence opposite=16×sin20°≈5.47 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 16 cm, a leg must be shorter than 16 cm. The calculated 5.47 cm is between zero and 16, as expected. The side definitions and bounds can reveal a wrong input.

98. Find an unknown side 35: 25°

Question: A right triangle has acute angle 25° and hypotenuse 23 cm. Find the adjacent leg to three significant figures.

Worked answer: Here Adjacent/hypotenuse is cosine. Hence adjacent=23×cos25°≈20.8 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 23 cm, a leg must be shorter than 23 cm. The calculated 20.8 cm is between zero and 23, as expected. The side definitions and bounds can reveal a wrong input.

99. Find an unknown side 36: 30°

Question: A right triangle has acute angle 30° and adjacent leg 30 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/adjacent is tangent. Hence opposite=30×tan30°≈17.3 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: Here 30 cm is the known adjacent leg, not the hypotenuse. The hypotenuse would be 34.6 cm, longer than the adjacent leg and the calculated opposite leg. The side definitions and bounds can reveal a wrong input.

100. Find an unknown side 37: 35°

Question: A right triangle has acute angle 35° and hypotenuse 37 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/hypotenuse is sine. Hence opposite=37×sin35°≈21.2 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 37 cm, a leg must be shorter than 37 cm. The calculated 21.2 cm is between zero and 37, as expected. The side definitions and bounds can reveal a wrong input.

101. Find an unknown side 38: 40°

Question: A right triangle has acute angle 40° and hypotenuse 44 cm. Find the adjacent leg to three significant figures.

Worked answer: Here Adjacent/hypotenuse is cosine. Hence adjacent=44×cos40°≈33.7 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: For a right triangle with known hypotenuse 44 cm, a leg must be shorter than 44 cm. The calculated 33.7 cm is between zero and 44, as expected. The side definitions and bounds can reveal a wrong input.

102. Find an unknown side 39: 45°

Question: A right triangle has acute angle 45° and adjacent leg 51 cm. Find the opposite leg to three significant figures.

Worked answer: Here Opposite/adjacent is tangent. Hence opposite=51×tan45°≈51 cm to three significant figures. Keep full precision until the final answer; the degree mark means the trig function uses degrees.

Independent check: Here 51 cm is the known adjacent leg, not the hypotenuse. The hypotenuse would be 72.1 cm, longer than the adjacent leg and the calculated opposite leg. The side definitions and bounds can reveal a wrong input.

Find an acute angle from two side lengths

In this find an acute angle from two side lengths section, start from the 90° mark and identify which side is being compared with which. Write one appropriate ratio before calculating; do not use every function simply because it appears on a calculator. Exact fractions are retained when possible and approximate lengths or angles are labelled accordingly. The important skill is explaining the side relationship on a fresh triangle.

103. Recover the acute angle 1 from two legs

Question: A right triangle is known to have opposite leg 4 cm and adjacent leg 3 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

How to read the triangle: When opposite and adjacent legs are given, tangent is the simplest direct ratio. Write tanθ=opposite/adjacent and use inverse tangent to retrieve the acute angle. On a scientific calculator set to degrees, the inverse gives an angle measured in degrees, not a length.

Worked answer: tanθ=4/3=4/3, so θ=tan⁻¹(4/3)≈53.1° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

First tempting mistake: Using ordinary tangent on the quotient rather than inverse tangent is a function-selection error. Another misunderstanding is interpreting tan⁻¹ as the reciprocal 1/tan instead of the inverse function that returns an angle.

Independent check: The recovered angle 53.1° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.8 and 0.6.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 36.9°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

104. Recover the acute angle 2 from two legs

Question: A right triangle is known to have opposite leg 20 cm and adjacent leg 21 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=20/21=20/21, so θ=tan⁻¹(20/21)≈43.6° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 43.6° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.69 and 0.724.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 46.4°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

105. Recover the acute angle 3 from two legs

Question: A right triangle is known to have opposite leg 63 cm and adjacent leg 16 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=63/16=63/16, so θ=tan⁻¹(63/16)≈75.7° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 75.7° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.969 and 0.246.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 14.3°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

106. Recover the acute angle 4 from two legs

Question: A right triangle is known to have opposite leg 15 cm and adjacent leg 112 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=15/112=15/112, so θ=tan⁻¹(15/112)≈7.6° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 7.6° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.133 and 0.991.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 82.4°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

107. Recover the acute angle 5 from two legs

Question: A right triangle is known to have opposite leg 40 cm and adjacent leg 9 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=40/9=40/9, so θ=tan⁻¹(40/9)≈77.3° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 77.3° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.976 and 0.22.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 12.7°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

108. Recover the acute angle 6 from two legs

Question: A right triangle is known to have opposite leg 33 cm and adjacent leg 56 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=33/56=33/56, so θ=tan⁻¹(33/56)≈30.5° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 30.5° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.508 and 0.862.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 59.5°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

109. Recover the acute angle 7 from two legs

Question: A right triangle is known to have opposite leg 55 cm and adjacent leg 48 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=55/48=55/48, so θ=tan⁻¹(55/48)≈48.9° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 48.9° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.753 and 0.658.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 41.1°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

110. Recover the acute angle 8 from two legs

Question: A right triangle is known to have opposite leg 7 cm and adjacent leg 24 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=7/24=7/24, so θ=tan⁻¹(7/24)≈16.3° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 16.3° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.28 and 0.96.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 73.7°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

111. Recover the acute angle 9 from two legs

Question: A right triangle is known to have opposite leg 45 cm and adjacent leg 28 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=45/28=45/28, so θ=tan⁻¹(45/28)≈58.1° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 58.1° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.849 and 0.528.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 31.9°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

112. Recover the acute angle 10 from two legs

Question: A right triangle is known to have opposite leg 39 cm and adjacent leg 80 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=39/80=39/80, so θ=tan⁻¹(39/80)≈26° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 26° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.438 and 0.899.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 64°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

113. Recover the acute angle 11 from two legs

Question: A right triangle is known to have opposite leg 15 cm and adjacent leg 8 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=15/8=15/8, so θ=tan⁻¹(15/8)≈61.9° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 61.9° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.882 and 0.471.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 28.1°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

114. Recover the acute angle 12 from two legs

Question: A right triangle is known to have opposite leg 11 cm and adjacent leg 60 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=11/60=11/60, so θ=tan⁻¹(11/60)≈10.4° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 10.4° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.18 and 0.984.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 79.6°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

115. Recover the acute angle 13 from two legs

Question: A right triangle is known to have opposite leg 77 cm and adjacent leg 36 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=77/36=77/36, so θ=tan⁻¹(77/36)≈64.9° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 64.9° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.906 and 0.424.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 25.1°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

116. Recover the acute angle 14 from two legs

Question: A right triangle is known to have opposite leg 5 cm and adjacent leg 12 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=5/12=5/12, so θ=tan⁻¹(5/12)≈22.6° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 22.6° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.385 and 0.923.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 67.4°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

117. Recover the acute angle 15 from two legs

Question: A right triangle is known to have opposite leg 35 cm and adjacent leg 12 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=35/12=35/12, so θ=tan⁻¹(35/12)≈71.1° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 71.1° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.946 and 0.324.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 18.9°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

118. Recover the acute angle 16 from two legs

Question: A right triangle is known to have opposite leg 13 cm and adjacent leg 84 cm relative to an unknown acute angle θ. Determine θ to one decimal place and name the inverse function used.

Worked answer: tanθ=13/84=13/84, so θ=tan⁻¹(13/84)≈8.8° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

Independent check: The recovered angle 8.8° lies strictly between zero and 90°, consistent with an acute angle in a right triangle. Its sine and cosine from the same side triple would be approximately 0.153 and 0.988.

Changed-number question: Swap the opposite and adjacent legs. The other acute angle is approximately 81.2°, and the two acute angles add to 90°. This independent geometric check does not require guessing from the diagram.

119. Recover the acute angle 17 from two legs

Worked answer: tanθ=4/3=4/3, so θ=tan⁻¹(4/3)≈53.1° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

120. Recover the acute angle 18 from two legs

121. Recover the acute angle 19 from two legs

122. Recover the acute angle 20 from two legs

123. Recover the acute angle 21 from two legs

Worked answer: tanθ=40/9=40/9, so θ=tan⁻¹(40/9)≈77.3° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

124. Recover the acute angle 22 from two legs

125. Recover the acute angle 23 from two legs

126. Recover the acute angle 24 from two legs

Worked answer: tanθ=7/24=7/24, so θ=tan⁻¹(7/24)≈16.3° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

127. Recover the acute angle 25 from two legs

128. Recover the acute angle 26 from two legs

Worked answer: tanθ=39/80=39/80, so θ=tan⁻¹(39/80)≈26° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

129. Recover the acute angle 27 from two legs

Worked answer: tanθ=15/8=15/8, so θ=tan⁻¹(15/8)≈61.9° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

130. Recover the acute angle 28 from two legs

131. Recover the acute angle 29 from two legs

132. Recover the acute angle 30 from two legs

Worked answer: tanθ=5/12=5/12, so θ=tan⁻¹(5/12)≈22.6° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

133. Recover the acute angle 31 from two legs

134. Recover the acute angle 32 from two legs

Worked answer: tanθ=13/84=13/84, so θ=tan⁻¹(13/84)≈8.8° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

135. Recover the acute angle 33 from two legs

Worked answer: tanθ=4/3=4/3, so θ=tan⁻¹(4/3)≈53.1° to one decimal place. This is a rounded display of an acute angle; it should be checked before any early rounding of the ratio.

136. Recover the acute angle 34 from two legs

Compare trig ratios when triangles are resized

In this compare trig ratios when triangles are resized section, start from the 90° mark and identify which side is being compared with which. Write one appropriate ratio before calculating; do not use every function simply because it appears on a calculator. Exact fractions are retained when possible and approximate lengths or angles are labelled accordingly. The important skill is explaining the side relationship on a fresh triangle.

137. Enlarge the same triangle 1: does θ change?

Question: Triangle A is a 3–4–5 right triangle. Triangle B has corresponding sides multiplied by 2, giving 6, 8, 10. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 4 in A.

How to read the triangle: Similarity preserves corresponding angles. A uniform scale factor multiplies opposite and hypotenuse by the same amount, so the ratio between them is unchanged. This explains why trigonometry uses ratios instead of fixed side lengths.

Worked answer: For A, sinθ=4/5=4/5. For B, sinθ=8/10=4/5. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 2>1.

First tempting mistake: If a pupil multiplies sine itself by 2, they confuse a dimensionless ratio with a physical length. A second error is scaling only one side; that distorts shape and no longer describes the same similarity transformation.

Independent check: Compare tangent as well: 4/3=8/6; both simplify to 4/3. Equal corresponding ratios match the claim that the acute angle remains unchanged.

Changed-number question: Now change only one leg in Triangle B while holding its other two printed lengths unchanged; check whether the right-triangle relation or similarity survives. The answer is not guaranteed by the old uniform scale argument.

138. Enlarge the same triangle 2: does θ change?

Question: Triangle A is a 11–60–61 right triangle. Triangle B has corresponding sides multiplied by 3, giving 33, 180, 183. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 11 in A.

Worked answer: For A, sinθ=11/61=11/61. For B, sinθ=33/183=11/61. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 3>1.

First tempting mistake: If a pupil multiplies sine itself by 3, they confuse a dimensionless ratio with a physical length. A second error is scaling only one side; that distorts shape and no longer describes the same similarity transformation.

Independent check: Compare tangent as well: 11/60=33/180; both simplify to 11/60. Equal corresponding ratios match the claim that the acute angle remains unchanged.

139. Enlarge the same triangle 3: does θ change?

Question: Triangle A is a 48–55–73 right triangle. Triangle B has corresponding sides multiplied by 4, giving 192, 220, 292. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 55 in A.

Worked answer: For A, sinθ=55/73=55/73. For B, sinθ=220/292=55/73. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 4>1.

First tempting mistake: If a pupil multiplies sine itself by 4, they confuse a dimensionless ratio with a physical length. A second error is scaling only one side; that distorts shape and no longer describes the same similarity transformation.

Independent check: Compare tangent as well: 55/48=220/192; both simplify to 55/48. Equal corresponding ratios match the claim that the acute angle remains unchanged.

140. Enlarge the same triangle 4: does θ change?

Question: Triangle A is a 20–21–29 right triangle. Triangle B has corresponding sides multiplied by 5, giving 100, 105, 145. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 20 in A.

Worked answer: For A, sinθ=20/29=20/29. For B, sinθ=100/145=20/29. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 5>1.

First tempting mistake: If a pupil multiplies sine itself by 5, they confuse a dimensionless ratio with a physical length. A second error is scaling only one side; that distorts shape and no longer describes the same similarity transformation.

Independent check: Compare tangent as well: 20/21=100/105; both simplify to 20/21. Equal corresponding ratios match the claim that the acute angle remains unchanged.

141. Enlarge the same triangle 5: does θ change?

Question: Triangle A is a 36–77–85 right triangle. Triangle B has corresponding sides multiplied by 6, giving 216, 462, 510. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 77 in A.

Worked answer: For A, sinθ=77/85=77/85. For B, sinθ=462/510=77/85. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 6>1.

First tempting mistake: If a pupil multiplies sine itself by 6, they confuse a dimensionless ratio with a physical length. A second error is scaling only one side; that distorts shape and no longer describes the same similarity transformation.

Independent check: Compare tangent as well: 77/36=462/216; both simplify to 77/36. Equal corresponding ratios match the claim that the acute angle remains unchanged.

142. Enlarge the same triangle 6: does θ change?

Question: Triangle A is a 7–24–25 right triangle. Triangle B has corresponding sides multiplied by 7, giving 49, 168, 175. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 7 in A.

Worked answer: For A, sinθ=7/25=7/25. For B, sinθ=49/175=7/25. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 7>1.

First tempting mistake: If a pupil multiplies sine itself by 7, they confuse a dimensionless ratio with a physical length. A second error is scaling only one side; that distorts shape and no longer describes the same similarity transformation.

Independent check: Compare tangent as well: 7/24=49/168; both simplify to 7/24. Equal corresponding ratios match the claim that the acute angle remains unchanged.

143. Enlarge the same triangle 7: does θ change?

Question: Triangle A is a 16–63–65 right triangle. Triangle B has corresponding sides multiplied by 2, giving 32, 126, 130. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 63 in A.

Worked answer: For A, sinθ=63/65=63/65. For B, sinθ=126/130=63/65. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 2>1.

Independent check: Compare tangent as well: 63/16=126/32; both simplify to 63/16. Equal corresponding ratios match the claim that the acute angle remains unchanged.

144. Enlarge the same triangle 8: does θ change?

Question: Triangle A is a 5–12–13 right triangle. Triangle B has corresponding sides multiplied by 3, giving 15, 36, 39. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 5 in A.

Worked answer: For A, sinθ=5/13=5/13. For B, sinθ=15/39=5/13. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 3>1.

Independent check: Compare tangent as well: 5/12=15/36; both simplify to 5/12. Equal corresponding ratios match the claim that the acute angle remains unchanged.

145. Enlarge the same triangle 9: does θ change?

Question: Triangle A is a 28–45–53 right triangle. Triangle B has corresponding sides multiplied by 4, giving 112, 180, 212. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 45 in A.

Worked answer: For A, sinθ=45/53=45/53. For B, sinθ=180/212=45/53. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 4>1.

Independent check: Compare tangent as well: 45/28=180/112; both simplify to 45/28. Equal corresponding ratios match the claim that the acute angle remains unchanged.

146. Enlarge the same triangle 10: does θ change?

Question: Triangle A is a 15–112–113 right triangle. Triangle B has corresponding sides multiplied by 5, giving 75, 560, 565. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 15 in A.

Worked answer: For A, sinθ=15/113=15/113. For B, sinθ=75/565=15/113. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 5>1.

Independent check: Compare tangent as well: 15/112=75/560; both simplify to 15/112. Equal corresponding ratios match the claim that the acute angle remains unchanged.

147. Enlarge the same triangle 11: does θ change?

Question: Triangle A is a 12–35–37 right triangle. Triangle B has corresponding sides multiplied by 6, giving 72, 210, 222. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 35 in A.

Worked answer: For A, sinθ=35/37=35/37. For B, sinθ=210/222=35/37. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 6>1.

Independent check: Compare tangent as well: 35/12=210/72; both simplify to 35/12. Equal corresponding ratios match the claim that the acute angle remains unchanged.

148. Enlarge the same triangle 12: does θ change?

Question: Triangle A is a 39–80–89 right triangle. Triangle B has corresponding sides multiplied by 7, giving 273, 560, 623. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 39 in A.

Worked answer: For A, sinθ=39/89=39/89. For B, sinθ=273/623=39/89. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 7>1.

Independent check: Compare tangent as well: 39/80=273/560; both simplify to 39/80. Equal corresponding ratios match the claim that the acute angle remains unchanged.

149. Enlarge the same triangle 13: does θ change?

Question: Triangle A is a 9–40–41 right triangle. Triangle B has corresponding sides multiplied by 2, giving 18, 80, 82. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 40 in A.

Worked answer: For A, sinθ=40/41=40/41. For B, sinθ=80/82=40/41. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 2>1.

Independent check: Compare tangent as well: 40/9=80/18; both simplify to 40/9. Equal corresponding ratios match the claim that the acute angle remains unchanged.

150. Enlarge the same triangle 14: does θ change?

Question: Triangle A is a 13–84–85 right triangle. Triangle B has corresponding sides multiplied by 3, giving 39, 252, 255. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 13 in A.

Worked answer: For A, sinθ=13/85=13/85. For B, sinθ=39/255=13/85. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 3>1.

Independent check: Compare tangent as well: 13/84=39/252; both simplify to 13/84. Equal corresponding ratios match the claim that the acute angle remains unchanged.

151. Enlarge the same triangle 15: does θ change?

Question: Triangle A is a 8–15–17 right triangle. Triangle B has corresponding sides multiplied by 4, giving 32, 60, 68. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 15 in A.

Worked answer: For A, sinθ=15/17=15/17. For B, sinθ=60/68=15/17. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 4>1.

Independent check: Compare tangent as well: 15/8=60/32; both simplify to 15/8. Equal corresponding ratios match the claim that the acute angle remains unchanged.

152. Enlarge the same triangle 16: does θ change?

Question: Triangle A is a 33–56–65 right triangle. Triangle B has corresponding sides multiplied by 5, giving 165, 280, 325. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 33 in A.

Worked answer: For A, sinθ=33/65=33/65. For B, sinθ=165/325=33/65. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 5>1.

Independent check: Compare tangent as well: 33/56=165/280; both simplify to 33/56. Equal corresponding ratios match the claim that the acute angle remains unchanged.

153. Enlarge the same triangle 17: does θ change?

Question: Triangle A is a 3–4–5 right triangle. Triangle B has corresponding sides multiplied by 6, giving 18, 24, 30. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 4 in A.

Worked answer: For A, sinθ=4/5=4/5. For B, sinθ=24/30=4/5. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 6>1.

Independent check: Compare tangent as well: 4/3=24/18; both simplify to 4/3. Equal corresponding ratios match the claim that the acute angle remains unchanged.

154. Enlarge the same triangle 18: does θ change?

Question: Triangle A is a 11–60–61 right triangle. Triangle B has corresponding sides multiplied by 7, giving 77, 420, 427. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 11 in A.

Worked answer: For A, sinθ=11/61=11/61. For B, sinθ=77/427=11/61. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 7>1.

Independent check: Compare tangent as well: 11/60=77/420; both simplify to 11/60. Equal corresponding ratios match the claim that the acute angle remains unchanged.

155. Enlarge the same triangle 19: does θ change?

Question: Triangle A is a 48–55–73 right triangle. Triangle B has corresponding sides multiplied by 2, giving 96, 110, 146. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 55 in A.

Worked answer: For A, sinθ=55/73=55/73. For B, sinθ=110/146=55/73. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 2>1.

Independent check: Compare tangent as well: 55/48=110/96; both simplify to 55/48. Equal corresponding ratios match the claim that the acute angle remains unchanged.

156. Enlarge the same triangle 20: does θ change?

Question: Triangle A is a 20–21–29 right triangle. Triangle B has corresponding sides multiplied by 3, giving 60, 63, 87. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 20 in A.

Worked answer: For A, sinθ=20/29=20/29. For B, sinθ=60/87=20/29. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 3>1.

Independent check: Compare tangent as well: 20/21=60/63; both simplify to 20/21. Equal corresponding ratios match the claim that the acute angle remains unchanged.

157. Enlarge the same triangle 21: does θ change?

Question: Triangle A is a 36–77–85 right triangle. Triangle B has corresponding sides multiplied by 4, giving 144, 308, 340. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 77 in A.

Worked answer: For A, sinθ=77/85=77/85. For B, sinθ=308/340=77/85. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 4>1.

Independent check: Compare tangent as well: 77/36=308/144; both simplify to 77/36. Equal corresponding ratios match the claim that the acute angle remains unchanged.

158. Enlarge the same triangle 22: does θ change?

Question: Triangle A is a 7–24–25 right triangle. Triangle B has corresponding sides multiplied by 5, giving 35, 120, 125. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 7 in A.

Worked answer: For A, sinθ=7/25=7/25. For B, sinθ=35/125=7/25. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 5>1.

Independent check: Compare tangent as well: 7/24=35/120; both simplify to 7/24. Equal corresponding ratios match the claim that the acute angle remains unchanged.

159. Enlarge the same triangle 23: does θ change?

Question: Triangle A is a 16–63–65 right triangle. Triangle B has corresponding sides multiplied by 6, giving 96, 378, 390. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 63 in A.

Worked answer: For A, sinθ=63/65=63/65. For B, sinθ=378/390=63/65. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 6>1.

Independent check: Compare tangent as well: 63/16=378/96; both simplify to 63/16. Equal corresponding ratios match the claim that the acute angle remains unchanged.

160. Enlarge the same triangle 24: does θ change?

Question: Triangle A is a 5–12–13 right triangle. Triangle B has corresponding sides multiplied by 7, giving 35, 84, 91. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 5 in A.

Worked answer: For A, sinθ=5/13=5/13. For B, sinθ=35/91=5/13. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 7>1.

Independent check: Compare tangent as well: 5/12=35/84; both simplify to 5/12. Equal corresponding ratios match the claim that the acute angle remains unchanged.

161. Enlarge the same triangle 25: does θ change?

Question: Triangle A is a 28–45–53 right triangle. Triangle B has corresponding sides multiplied by 2, giving 56, 90, 106. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 45 in A.

Worked answer: For A, sinθ=45/53=45/53. For B, sinθ=90/106=45/53. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 2>1.

Independent check: Compare tangent as well: 45/28=90/56; both simplify to 45/28. Equal corresponding ratios match the claim that the acute angle remains unchanged.

162. Enlarge the same triangle 26: does θ change?

Question: Triangle A is a 15–112–113 right triangle. Triangle B has corresponding sides multiplied by 3, giving 45, 336, 339. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 15 in A.

Worked answer: For A, sinθ=15/113=15/113. For B, sinθ=45/339=15/113. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 3>1.

Independent check: Compare tangent as well: 15/112=45/336; both simplify to 15/112. Equal corresponding ratios match the claim that the acute angle remains unchanged.

163. Enlarge the same triangle 27: does θ change?

Question: Triangle A is a 12–35–37 right triangle. Triangle B has corresponding sides multiplied by 4, giving 48, 140, 148. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 35 in A.

Worked answer: For A, sinθ=35/37=35/37. For B, sinθ=140/148=35/37. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 4>1.

Independent check: Compare tangent as well: 35/12=140/48; both simplify to 35/12. Equal corresponding ratios match the claim that the acute angle remains unchanged.

164. Enlarge the same triangle 28: does θ change?

Question: Triangle A is a 39–80–89 right triangle. Triangle B has corresponding sides multiplied by 5, giving 195, 400, 445. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 39 in A.

Worked answer: For A, sinθ=39/89=39/89. For B, sinθ=195/445=39/89. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 5>1.

Independent check: Compare tangent as well: 39/80=195/400; both simplify to 39/80. Equal corresponding ratios match the claim that the acute angle remains unchanged.

165. Enlarge the same triangle 29: does θ change?

Question: Triangle A is a 9–40–41 right triangle. Triangle B has corresponding sides multiplied by 6, giving 54, 240, 246. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 40 in A.

Worked answer: For A, sinθ=40/41=40/41. For B, sinθ=240/246=40/41. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 6>1.

Independent check: Compare tangent as well: 40/9=240/54; both simplify to 40/9. Equal corresponding ratios match the claim that the acute angle remains unchanged.

166. Enlarge the same triangle 30: does θ change?

Question: Triangle A is a 13–84–85 right triangle. Triangle B has corresponding sides multiplied by 7, giving 91, 588, 595. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 13 in A.

Worked answer: For A, sinθ=13/85=13/85. For B, sinθ=91/595=13/85. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 7>1.

Independent check: Compare tangent as well: 13/84=91/588; both simplify to 13/84. Equal corresponding ratios match the claim that the acute angle remains unchanged.

167. Enlarge the same triangle 31: does θ change?

Question: Triangle A is a 8–15–17 right triangle. Triangle B has corresponding sides multiplied by 2, giving 16, 30, 34. Are their corresponding acute angles equal? Compare the sine ratio for the angle opposite side 15 in A.

Worked answer: For A, sinθ=15/17=15/17. For B, sinθ=30/34=15/17. They are exactly equal. Thus the corresponding angle θ is unchanged, and the triangles are similar but not congruent because 2>1.

Independent check: Compare tangent as well: 15/8=30/16; both simplify to 15/8. Equal corresponding ratios match the claim that the acute angle remains unchanged.

Decide which trig ratio the question needs

In this decide which trig ratio the question needs section, start from the 90° mark and identify which side is being compared with which. Write one appropriate ratio before calculating; do not use every function simply because it appears on a calculator. Exact fractions are retained when possible and approximate lengths or angles are labelled accordingly. The important skill is explaining the side relationship on a fresh triangle.

168. Choose the ratio by the given sides 1

Question: A right-triangle exercise states an acute angle of 20°, a hypotenuse side of length 12 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

How to read the triangle: First identify opposite, adjacent and hypotenuse relative to the angle given. The correct relationship must include both the known hypotenuse and requested opposite; selecting a formula from memory without reading these labels is unreliable.

Worked answer: Use sin because it connects opposite and hypotenuse. The rule is opposite=hypotenuse×sin20°. Substituting the full-precision given side and angle gives approximately 4.1 cm. This value respects the geometry of the right triangle.

First tempting mistake: Choosing the function merely because ‘SOHCAHTOA starts with sine’ ignores the side pair. Another risk is reversing known and unknown, dividing instead of multiplying. Write the defining fraction before rearranging.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, sin20° is approximately 0.342, and the recovered quotient agrees with the stated side relationship.

Changed-number question: Hide the ratio name and re-read the same statement with the unknown changed from opposite to the third side. A different side pair might require a different ratio or Pythagoras. Explain the choice rather than reusing the previous formula.

169. Choose the ratio by the given sides 2

Question: A right-triangle exercise states an acute angle of 35°, a hypotenuse side of length 17 cm, and asks for the adjacent side. Which trig ratio is appropriate and why? Find the side to three significant figures.

How to read the triangle: First identify opposite, adjacent and hypotenuse relative to the angle given. The correct relationship must include both the known hypotenuse and requested adjacent; selecting a formula from memory without reading these labels is unreliable.

Worked answer: Use cos because it connects adjacent and hypotenuse. The rule is adjacent=hypotenuse×cos35°. Substituting the full-precision given side and angle gives approximately 13.9 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, cos35° is approximately 0.819, and the recovered quotient agrees with the stated side relationship.

Changed-number question: Hide the ratio name and re-read the same statement with the unknown changed from adjacent to the third side. A different side pair might require a different ratio or Pythagoras. Explain the choice rather than reusing the previous formula.

170. Choose the ratio by the given sides 3

Question: A right-triangle exercise states an acute angle of 50°, a adjacent side of length 14.1413 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

How to read the triangle: First identify opposite, adjacent and hypotenuse relative to the angle given. The correct relationship must include both the known adjacent and requested opposite; selecting a formula from memory without reading these labels is unreliable.

Worked answer: Use tan because it connects opposite and adjacent. The rule is opposite=adjacent×tan50°. Substituting the six-significant-figure given adjacent length and the stated angle gives approximately 16.9 cm to three significant figures. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, using the unrounded calculated opposite, the quotient opposite/14.1413 is tan50°≈1.19. The displayed final length is rounded to three significant figures, so its quotient can differ slightly.

171. Choose the ratio by the given sides 4

Question: A right-triangle exercise states an acute angle of 65°, a hypotenuse side of length 27 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use sin because it connects opposite and hypotenuse. The rule is opposite=hypotenuse×sin65°. Substituting the full-precision given side and angle gives approximately 24.5 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, sin65° is approximately 0.906, and the recovered quotient agrees with the stated side relationship.

172. Choose the ratio by the given sides 5

Question: A right-triangle exercise states an acute angle of 25°, a hypotenuse side of length 32 cm, and asks for the adjacent side. Which trig ratio is appropriate and why? Find the side to three significant figures.

How to read the triangle: First identify opposite, adjacent and hypotenuse relative to the angle given. The correct relationship must include both the known hypotenuse and requested adjacent; selecting a formula from memory without reading these labels is unreliable.

Worked answer: Use cos because it connects adjacent and hypotenuse. The rule is adjacent=hypotenuse×cos25°. Substituting the full-precision given side and angle gives approximately 29 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, cos25° is approximately 0.906, and the recovered quotient agrees with the stated side relationship.

173. Choose the ratio by the given sides 6

Question: A right-triangle exercise states an acute angle of 40°, a adjacent side of length 28.3436 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use tan because it connects opposite and adjacent. The rule is opposite=adjacent×tan40°. Substituting the six-significant-figure given adjacent length and the stated angle gives approximately 23.8 cm to three significant figures. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, using the unrounded calculated opposite, the quotient opposite/28.3436 is tan40°≈0.839. The displayed final length is rounded to three significant figures, so its quotient can differ slightly.

174. Choose the ratio by the given sides 7

Question: A right-triangle exercise states an acute angle of 55°, a hypotenuse side of length 42 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use sin because it connects opposite and hypotenuse. The rule is opposite=hypotenuse×sin55°. Substituting the full-precision given side and angle gives approximately 34.4 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, sin55° is approximately 0.819, and the recovered quotient agrees with the stated side relationship.

175. Choose the ratio by the given sides 8

Question: A right-triangle exercise states an acute angle of 70°, a hypotenuse side of length 47 cm, and asks for the adjacent side. Which trig ratio is appropriate and why? Find the side to three significant figures.

How to read the triangle: First identify opposite, adjacent and hypotenuse relative to the angle given. The correct relationship must include both the known hypotenuse and requested adjacent; selecting a formula from memory without reading these labels is unreliable.

Worked answer: Use cos because it connects adjacent and hypotenuse. The rule is adjacent=hypotenuse×cos70°. Substituting the full-precision given side and angle gives approximately 16.1 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, cos70° is approximately 0.342, and the recovered quotient agrees with the stated side relationship.

176. Choose the ratio by the given sides 9

Question: A right-triangle exercise states an acute angle of 30°, a adjacent side of length 45.0333 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use tan because it connects opposite and adjacent. The rule is opposite=adjacent×tan30°. Substituting the six-significant-figure given adjacent length and the stated angle gives approximately 26 cm to three significant figures. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, using the unrounded calculated opposite, the quotient opposite/45.0333 is tan30°≈0.577. The displayed final length is rounded to three significant figures, so its quotient can differ slightly.

177. Choose the ratio by the given sides 10

Question: A right-triangle exercise states an acute angle of 45°, a hypotenuse side of length 57 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use sin because it connects opposite and hypotenuse. The rule is opposite=hypotenuse×sin45°. Substituting the full-precision given side and angle gives approximately 40.3 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, sin45° is approximately 0.707, and the recovered quotient agrees with the stated side relationship.

178. Choose the ratio by the given sides 11

Question: A right-triangle exercise states an acute angle of 60°, a hypotenuse side of length 62 cm, and asks for the adjacent side. Which trig ratio is appropriate and why? Find the side to three significant figures.

How to read the triangle: First identify opposite, adjacent and hypotenuse relative to the angle given. The correct relationship must include both the known hypotenuse and requested adjacent; selecting a formula from memory without reading these labels is unreliable.

Worked answer: Use cos because it connects adjacent and hypotenuse. The rule is adjacent=hypotenuse×cos60°. Substituting the full-precision given side and angle gives approximately 31 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, cos60° is approximately 0.5, and the recovered quotient agrees with the stated side relationship.

179. Choose the ratio by the given sides 12

Question: A right-triangle exercise states an acute angle of 20°, a adjacent side of length 62.9594 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use tan because it connects opposite and adjacent. The rule is opposite=adjacent×tan20°. Substituting the six-significant-figure given adjacent length and the stated angle gives approximately 22.9 cm to three significant figures. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, using the unrounded calculated opposite, the quotient opposite/62.9594 is tan20°≈0.364. The displayed final length is rounded to three significant figures, so its quotient can differ slightly.

180. Choose the ratio by the given sides 13

Question: A right-triangle exercise states an acute angle of 35°, a hypotenuse side of length 72 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use sin because it connects opposite and hypotenuse. The rule is opposite=hypotenuse×sin35°. Substituting the full-precision given side and angle gives approximately 41.3 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, sin35° is approximately 0.574, and the recovered quotient agrees with the stated side relationship.

181. Choose the ratio by the given sides 14

Question: A right-triangle exercise states an acute angle of 50°, a hypotenuse side of length 77 cm, and asks for the adjacent side. Which trig ratio is appropriate and why? Find the side to three significant figures.

How to read the triangle: First identify opposite, adjacent and hypotenuse relative to the angle given. The correct relationship must include both the known hypotenuse and requested adjacent; selecting a formula from memory without reading these labels is unreliable.

Worked answer: Use cos because it connects adjacent and hypotenuse. The rule is adjacent=hypotenuse×cos50°. Substituting the full-precision given side and angle gives approximately 49.5 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, cos50° is approximately 0.643, and the recovered quotient agrees with the stated side relationship.

182. Choose the ratio by the given sides 15

Question: A right-triangle exercise states an acute angle of 65°, a adjacent side of length 34.6547 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use tan because it connects opposite and adjacent. The rule is opposite=adjacent×tan65°. Substituting the six-significant-figure given adjacent length and the stated angle gives approximately 74.3 cm to three significant figures. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, using the unrounded calculated opposite, the quotient opposite/34.6547 is tan65°≈2.14. The displayed final length is rounded to three significant figures, so its quotient can differ slightly.

183. Choose the ratio by the given sides 16

Question: A right-triangle exercise states an acute angle of 25°, a hypotenuse side of length 16 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use sin because it connects opposite and hypotenuse. The rule is opposite=hypotenuse×sin25°. Substituting the full-precision given side and angle gives approximately 6.76 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, sin25° is approximately 0.423, and the recovered quotient agrees with the stated side relationship.

184. Choose the ratio by the given sides 17

Question: A right-triangle exercise states an acute angle of 40°, a hypotenuse side of length 21 cm, and asks for the adjacent side. Which trig ratio is appropriate and why? Find the side to three significant figures.

How to read the triangle: First identify opposite, adjacent and hypotenuse relative to the angle given. The correct relationship must include both the known hypotenuse and requested adjacent; selecting a formula from memory without reading these labels is unreliable.

Worked answer: Use cos because it connects adjacent and hypotenuse. The rule is adjacent=hypotenuse×cos40°. Substituting the full-precision given side and angle gives approximately 16.1 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, cos40° is approximately 0.766, and the recovered quotient agrees with the stated side relationship.

185. Choose the ratio by the given sides 18

Question: A right-triangle exercise states an acute angle of 55°, a adjacent side of length 14.913 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use tan because it connects opposite and adjacent. The rule is opposite=adjacent×tan55°. Substituting the six-significant-figure given adjacent length and the stated angle gives approximately 21.3 cm to three significant figures. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, using the unrounded calculated opposite, the quotient opposite/14.913 is tan55°≈1.43. The displayed final length is rounded to three significant figures, so its quotient can differ slightly.

186. Choose the ratio by the given sides 19

Question: A right-triangle exercise states an acute angle of 70°, a hypotenuse side of length 31 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use sin because it connects opposite and hypotenuse. The rule is opposite=hypotenuse×sin70°. Substituting the full-precision given side and angle gives approximately 29.1 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, sin70° is approximately 0.94, and the recovered quotient agrees with the stated side relationship.

187. Choose the ratio by the given sides 20

Question: A right-triangle exercise states an acute angle of 30°, a hypotenuse side of length 36 cm, and asks for the adjacent side. Which trig ratio is appropriate and why? Find the side to three significant figures.

How to read the triangle: First identify opposite, adjacent and hypotenuse relative to the angle given. The correct relationship must include both the known hypotenuse and requested adjacent; selecting a formula from memory without reading these labels is unreliable.

Worked answer: Use cos because it connects adjacent and hypotenuse. The rule is adjacent=hypotenuse×cos30°. Substituting the full-precision given side and angle gives approximately 31.2 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, cos30° is approximately 0.866, and the recovered quotient agrees with the stated side relationship.

188. Choose the ratio by the given sides 21

Question: A right-triangle exercise states an acute angle of 45°, a adjacent side of length 28.9914 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use tan because it connects opposite and adjacent. The rule is opposite=adjacent×tan45°. Substituting the six-significant-figure given adjacent length and the stated angle gives approximately 29 cm to three significant figures. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, using the unrounded calculated opposite, the quotient opposite/28.9914 is tan45°≈1. The displayed final length is rounded to three significant figures, so its quotient can differ slightly.

189. Choose the ratio by the given sides 22

Question: A right-triangle exercise states an acute angle of 60°, a hypotenuse side of length 46 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use sin because it connects opposite and hypotenuse. The rule is opposite=hypotenuse×sin60°. Substituting the full-precision given side and angle gives approximately 39.8 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, sin60° is approximately 0.866, and the recovered quotient agrees with the stated side relationship.

190. Choose the ratio by the given sides 23

Question: A right-triangle exercise states an acute angle of 20°, a hypotenuse side of length 51 cm, and asks for the adjacent side. Which trig ratio is appropriate and why? Find the side to three significant figures.

How to read the triangle: First identify opposite, adjacent and hypotenuse relative to the angle given. The correct relationship must include both the known hypotenuse and requested adjacent; selecting a formula from memory without reading these labels is unreliable.

Worked answer: Use cos because it connects adjacent and hypotenuse. The rule is adjacent=hypotenuse×cos20°. Substituting the full-precision given side and angle gives approximately 47.9 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, cos20° is approximately 0.94, and the recovered quotient agrees with the stated side relationship.

191. Choose the ratio by the given sides 24

Question: A right-triangle exercise states an acute angle of 35°, a adjacent side of length 45.8725 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use tan because it connects opposite and adjacent. The rule is opposite=adjacent×tan35°. Substituting the six-significant-figure given adjacent length and the stated angle gives approximately 32.1 cm to three significant figures. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, using the unrounded calculated opposite, the quotient opposite/45.8725 is tan35°≈0.7. The displayed final length is rounded to three significant figures, so its quotient can differ slightly.

192. Choose the ratio by the given sides 25

Question: A right-triangle exercise states an acute angle of 50°, a hypotenuse side of length 61 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use sin because it connects opposite and hypotenuse. The rule is opposite=hypotenuse×sin50°. Substituting the full-precision given side and angle gives approximately 46.7 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, sin50° is approximately 0.766, and the recovered quotient agrees with the stated side relationship.

193. Choose the ratio by the given sides 26

Question: A right-triangle exercise states an acute angle of 65°, a hypotenuse side of length 66 cm, and asks for the adjacent side. Which trig ratio is appropriate and why? Find the side to three significant figures.

How to read the triangle: First identify opposite, adjacent and hypotenuse relative to the angle given. The correct relationship must include both the known hypotenuse and requested adjacent; selecting a formula from memory without reading these labels is unreliable.

Worked answer: Use cos because it connects adjacent and hypotenuse. The rule is adjacent=hypotenuse×cos65°. Substituting the full-precision given side and angle gives approximately 27.9 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, cos65° is approximately 0.423, and the recovered quotient agrees with the stated side relationship.

194. Choose the ratio by the given sides 27

Question: A right-triangle exercise states an acute angle of 25°, a adjacent side of length 64.3479 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use tan because it connects opposite and adjacent. The rule is opposite=adjacent×tan25°. Substituting the six-significant-figure given adjacent length and the stated angle gives approximately 30 cm to three significant figures. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, using the unrounded calculated opposite, the quotient opposite/64.3479 is tan25°≈0.466. The displayed final length is rounded to three significant figures, so its quotient can differ slightly.

195. Choose the ratio by the given sides 28

Question: A right-triangle exercise states an acute angle of 40°, a hypotenuse side of length 76 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use sin because it connects opposite and hypotenuse. The rule is opposite=hypotenuse×sin40°. Substituting the full-precision given side and angle gives approximately 48.9 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, sin40° is approximately 0.643, and the recovered quotient agrees with the stated side relationship.

196. Choose the ratio by the given sides 29

Question: A right-triangle exercise states an acute angle of 55°, a hypotenuse side of length 81 cm, and asks for the adjacent side. Which trig ratio is appropriate and why? Find the side to three significant figures.

How to read the triangle: First identify opposite, adjacent and hypotenuse relative to the angle given. The correct relationship must include both the known hypotenuse and requested adjacent; selecting a formula from memory without reading these labels is unreliable.

Worked answer: Use cos because it connects adjacent and hypotenuse. The rule is adjacent=hypotenuse×cos55°. Substituting the full-precision given side and angle gives approximately 46.5 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, cos55° is approximately 0.574, and the recovered quotient agrees with the stated side relationship.

197. Choose the ratio by the given sides 30

Question: A right-triangle exercise states an acute angle of 70°, a adjacent side of length 5.1303 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use tan because it connects opposite and adjacent. The rule is opposite=adjacent×tan70°. Substituting the six-significant-figure given adjacent length and the stated angle gives approximately 14.1 cm to three significant figures. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, using the unrounded calculated opposite, the quotient opposite/5.1303 is tan70°≈2.75. The displayed final length is rounded to three significant figures, so its quotient can differ slightly.

198. Choose the ratio by the given sides 31

Question: A right-triangle exercise states an acute angle of 30°, a hypotenuse side of length 20 cm, and asks for the opposite side. Which trig ratio is appropriate and why? Find the side to three significant figures.

Worked answer: Use sin because it connects opposite and hypotenuse. The rule is opposite=hypotenuse×sin30°. Substituting the full-precision given side and angle gives approximately 10 cm. This value respects the geometry of the right triangle.

Independent check: Check by recovering the corresponding named ratio from the calculated lengths. For instance, sin30° is approximately 0.5, and the recovered quotient agrees with the stated side relationship.

Check calculator mode, accuracy and feasibility

In this check calculator mode, accuracy and feasibility section, start from the 90° mark and identify which side is being compared with which. Write one appropriate ratio before calculating; do not use every function simply because it appears on a calculator. Exact fractions are retained when possible and approximate lengths or angles are labelled accordingly. The important skill is explaining the side relationship on a fresh triangle.

199. A right angle comes first

Question: A triangle is described with angles 60°,50°,70°. Can SOHCAHTOA be used as a right-triangle rule?

How to read the triangle: A right-triangle answer must follow the given domain, physical dimensions and permitted calculation conventions. Identify the right angle and selected acute angle first, then match the correct side pair. The full K110 graduating curriculum is a long-term reference; pupils’ current class content should determine whether this is an appropriate extension.

Worked answer: No. None is 90°, so it is not a right triangle. The simple opposite–adjacent–hypotenuse right-triangle method does not apply as written.

First tempting mistake: Applying a memorised right-angle procedure to any triangle is a model-selection error.

Independent check: 60+50+70=180 but still no right angle.

Changed-number question: In a changed example, rotate the diagram, vary one input or replace one ratio. The learner should re-label the opposite/adjacent/hypotenuse relationship rather than copying the previous formula. State whether the result is exact or approximate and verify its angle or length units.

200. Hypotenuse must be longest

Question: Can a right triangle have legs 8 and 12 but hypotenuse 10?

Worked answer: No. 8²+12²=208, not 10²=100; ten is shorter than a leg and cannot be the hypotenuse.

First tempting mistake: Labelling the visually sloping side as hypotenuse without testing the right-angle position can be wrong.

Independent check: Use c=√208≈14.4 if legs are 8 and 12.

201. Sine of an acute angle lies between zero and one

Question: A pupil finds sinθ=1.3 for acute θ in a right triangle. Is that plausible?

Worked answer: No. Opposite leg is shorter than hypotenuse, so 0

First tempting mistake: A ratio over one often means the side labels or denominator were reversed.

Independent check: Reidentify opposite and hypotenuse.

202. Tangent can exceed one

Question: Is tanθ=2 impossible for an acute angle?

Worked answer: No. Tangent is opposite/adjacent and can exceed one when opposite is longer than adjacent.

First tempting mistake: Transferring the sine bound to tangent is not valid.

Independent check: A right triangle with legs 2 and 1 has tanθ=2 relative to the leg of length 2 as opposite.

203. Sine and cosine interchange for complementary angles

Question: If θ and φ are the two acute angles of one right triangle, how are sinθ and cosφ related?

Worked answer: They are equal, because a leg opposite θ is adjacent to φ.

First tempting mistake: Treating opposite and adjacent as permanently fixed names for one side is wrong.

Independent check: θ+φ=90° and sinθ=cos(90°−θ).

204. A negative angle is not the stated acute angle

Question: An inverse tangent result appears −35° for a right-triangle acute angle with positive legs. Is the interpretation correct?

Worked answer: No. Positive opposite and adjacent legs should yield a positive acute angle.

First tempting mistake: A negative input or wrong selected mode/ratio may have been entered.

Independent check: Recheck signed entries and diagram orientation.

205. Radian mode changes display

Question: A calculator is in radians and evaluates sin30 as roughly −0.988 instead of 0.5. What should be checked?

Worked answer: The angle unit. A marked 30° needs degree mode or explicit degree-to-radian conversion.

First tempting mistake: A surprising output is not proof the trigonometric identity is wrong.

Independent check: sin30°=1/2.

206. Pythagoras can check trig sides

Question: An acute angle is 30° and hypotenuse 10 cm. Opposite is 5 cm. Find adjacent.

Worked answer: Pythagoras gives √(100−25)=√75≈8.66 cm.

First tempting mistake: The adjacent side should not be equal to ten or larger.

Independent check: 5²+(√75)²=100.

207. Arcsine and arctangent are different inverse functions

Question: Given sinθ=0.5 for acute θ, which inverse function returns θ?

Worked answer: θ=sin⁻¹(0.5)=30°.

First tempting mistake: Using inverse cosine of 0.5 would return 60°, the complementary angle.

Independent check: sin30°=0.5.

208. A tangent ratio needs opposite and adjacent

Question: A right triangle has opposite 6 cm, adjacent 8 cm. Find tanθ.

Worked answer: 6/8=3/4=0.75.

First tempting mistake: Dividing opposite by hypotenuse would be sine, not tangent.

Independent check: tanθ=0.75.

209. Cosine needs adjacent and hypotenuse

Question: A right triangle has adjacent 8 cm and hypotenuse 10 cm. Find cosθ.

Worked answer: 8/10=0.8.

First tempting mistake: Opposite/hypotenuse would give sine.

Independent check: cosθ=0.8.

210. Sine needs opposite and hypotenuse

Question: A right triangle has opposite 6 cm and hypotenuse 10 cm. Find sinθ.

Worked answer: 6/10=0.6.

First tempting mistake: Opposite/adjacent would give tangent.

Independent check: sinθ=0.6.

211. Finding opposite from adjacent and tangent

Question: Given adjacent 8 cm and tanθ=3/4, find opposite.

Worked answer: opposite=8×3/4=6 cm.

First tempting mistake: Dividing by 3/4 finds another relationship, not opposite from tangent.

Independent check: 6/8=3/4.

212. Finding adjacent from opposite and tangent

Question: Given opposite 6 cm and tanθ=3/4, find adjacent.

Worked answer: adjacent=6÷(3/4)=8 cm.

First tempting mistake: Multiplying opposite by 3/4 is the inverse direction.

Independent check: 6/8=3/4.

213. Unknown hypotenuse from sine

Question: Given opposite 5 cm and sinθ=1/2, find hypotenuse.

Worked answer: hypotenuse=5÷(1/2)=10 cm.

First tempting mistake: Multiplying by 1/2 makes the hypotenuse shorter than a leg.

Independent check: 5/10=1/2.

214. Unknown hypotenuse from cosine

Question: Given adjacent 8 cm and cosθ=4/5, find hypotenuse.

Worked answer: hypotenuse=8÷(4/5)=10 cm.

First tempting mistake: Multiplying adjacent by 4/5 is the wrong inverse operation.

Independent check: 8/10=4/5.

215. Unknown side must have a unit

Question: A calculator shows 7.32 for a leg and input lengths were in cm. How should answer be reported?

Worked answer: About 7.32 cm, with appropriate specified rounding.

First tempting mistake: Leaving off the unit loses the physical meaning.

Independent check: The ratio is dimensionless but the recovered side is a length.

216. The diagram is not to scale

Question: A sketch makes the hypotenuse seem short, but lengths and 90° marker identify it. Which should be used?

Worked answer: Use the given right-angle and stated lengths, not a visual measurement of the sketch.

First tempting mistake: Measuring the drawing can override mathematically precise data incorrectly.

Independent check: Locate the 90° symbol.

217. Accuracy should be requested correctly

Question: A trig problem asks for the angle correct to one decimal place. Calculator gives 36.869897°. Report?

Worked answer: 36.9°.

First tempting mistake: Rounding to 37° loses the requested decimal place.

Independent check: Second decimal six rounds first decimal eight up to nine.

218. A right triangle has two acute angles

Question: If one acute angle is 42°, what is the other?

Worked answer: 48°, since the acute angles sum to 90°.

First tempting mistake: Subtracting from 180° alone forgets the right angle already uses ninety.

Independent check: 42+48+90=180.

219. SOHCAHTOA is a memory aid, not proof

Question: Why should a pupil draw and label sides before using a mnemonic?

Worked answer: The mnemonic does not identify which physical side is opposite or adjacent in this specific diagram.

First tempting mistake: Reciting it without a diagram can choose the wrong numerator or denominator.

Independent check: Write sinθ=o/h etc. with labelled parts.

220. An enlargement does not change acute angles

Question: The 3-4-5 triangle is enlarged to 9-12-15. Does its angle opposite side three change?

Worked answer: No. All lengths are tripled and side ratios stay equal.

First tempting mistake: Multiplying the angle itself by three confuses ratio and absolute lengths.

Independent check: 3/5=9/15.

221. An angle is not measured in centimetres

Question: The pupil writes θ=40 cm after inverse tangent. What is wrong?

Worked answer: The unknown is an angle, usually reported in degrees under these questions.

First tempting mistake: The output unit follows the requested quantity, not a side length.

Independent check: Use degree symbol if specified.

222. An angle exactly ninety is not acute

Question: Can θ=90° be used as the acute viewpoint to define tangent opposite/adjacent with adjacent zero?

Worked answer: Not under the ordinary acute-angle right-triangle setup; tangent is undefined at 90°.

First tempting mistake: A denominator zero can appear if the wrong angle is chosen.

Independent check: Use 0°<θ<90° for the typical right-triangle exercises.

223. Sine and cosine are both positive for acute angles

Question: Can sinθ=−0.4 describe an acute angle in an ordinary right triangle with positive side lengths?

Worked answer: No. Opposite and hypotenuse are positive lengths, so their quotient is positive.

First tempting mistake: A negative ratio indicates a mistaken signed entry or different context.

Independent check: Compare the side ratio directly.

224. A 45-degree isosceles right triangle

Question: At θ=45°, what is tanθ?

Worked answer: One, because opposite and adjacent legs are equal.

First tempting mistake: A tangent of one is perfectly valid and does not mean the angle is one degree.

Independent check: tan45°=1.

225. Pythagoras is sometimes easier than trig

Question: A 3-4-5 right triangle has legs three and four. Need third side five. Which method is simplest?

Worked answer: Pythagoras directly gives √(3²+4²)=5. Trigonometry is unnecessary if no angle is involved.

First tempting mistake: Forcing an angle calculation into every right-triangle question wastes steps.

Independent check: Use the givens and requested unknown to choose an efficient method.

226. Graphical perspective may be misleading

Question: A right triangle is drawn rotated. Does that change opposite or adjacent labels relative to a marked θ?

Worked answer: No. Relative position to θ and the 90° angle determines the labels, not page orientation.

First tempting mistake: Calling the bottom horizontal side adjacent in every drawing fails when the triangle rotates.

Independent check: Mark θ before choosing the sides.

227. Trig ratio versus physical unit

Question: If opposite=6 m, adjacent=8 m, what are tangent units?

Worked answer: tanθ=6/8=0.75 with no physical unit because metres cancel.

First tempting mistake: Reporting 0.75 m misstates a dimensionless ratio.

Independent check: 6 m /8 m cancels units.

228. A small-angle intuition

Question: For the same hypotenuse, does opposite increase or decrease as acute angle rises from 20° to 60°?

Worked answer: It increases because sine increases over acute angles.

First tempting mistake: Assuming larger angle means the adjacent side grows too ignores fixed hypotenuse.

Independent check: sin20° is about 0.342, sin60° about 0.866.

229. A larger hypotenuse for fixed angle

Question: If the angle is fixed and the hypotenuse doubles, how does the opposite side change?

Worked answer: It doubles because opposite=hypotenuse×sinθ and sinθ stays fixed.

First tempting mistake: Squaring the scale factor would apply an area rule to a length.

Independent check: Similar triangles scale corresponding lengths linearly.

230. Angle relationships are contextual

Question: A real ladder is not necessarily a perfectly rigid straight segment at one exact angle. Can the triangle formula alone guarantee field safety?

Worked answer: No. The classroom triangle is an idealised mathematical model, not installation or safety guidance.

First tempting mistake: Using abstract formulas as safety certification overstates the model.

Independent check: Follow appropriate practical standards and supervision.

231. An inverse result requires plausible quadrant

Question: For an acute θ, can an inverse cosine angle exceed ninety degrees when its input is adjacent/hypotenuse between zero and one?

Worked answer: No. The principal acute angle for ratio strictly between zero and one is between 0° and 90°.

First tempting mistake: A wrong mode or ratio may create an implausible reading.

Independent check: Check the sketch’s 90° corner and acute viewpoint.

232. Full K110 coverage is not Secondary Two topic sequence

Question: Does the 2027 K110 listing trig automatically mean every 2026 Secondary 2 G1 class has already studied it?

Worked answer: No. It is a graduating-syllabus topic; a current Secondary Two lesson follows the school’s scope and prerequisite readiness.

First tempting mistake: Confusing full-course and year-level content makes tuition advice misleading.

Independent check: Confirm actual school topic before assigning the extension.

233. Trig ratios describe similar triangles

Question: Why does sinθ not depend on whether the triangle is drawn small or large?

Worked answer: For a fixed acute angle in similar right triangles, opposite and hypotenuse change by the same positive scale factor.

First tempting mistake: Assuming sinθ grows with triangle perimeter confuses a ratio with length.

Independent check: Cancel common scale in o/h.

234. A numeric side can be checked by reasonableness

Question: If hypotenuse is 10 cm and calculated opposite is 13 cm, what should the pupil do?

Worked answer: Reject as impossible for a nondegenerate right triangle and revisit the side choice or calculation.

First tempting mistake: A calculator display is not proof of valid geometry.

Independent check: Hypotenuse is longest.

235. Calculator rounding is not exact equality

Question: If tanθ≈0.577, can it be called exactly 0.577 for 30°?

Worked answer: No. tan30°=1/√3, approximately 0.57735…, so the finite decimal is an approximation.

First tempting mistake: Treating a display as an exact identity can distort later results.

Independent check: Keep extra precision until the final requested rounding.

236. Angle addition is a useful inverse check

Question: If one acute angle is 32°, what is the other in a right triangle?

Worked answer: 58°, and both are complementary.

First tempting mistake: Treating two acute angles as supplementary to 180° ignores the right angle.

Independent check: 32+58+90=180.

237. The cosine bound

Question: Can cosθ=2 be valid for an acute angle in a right triangle?

Worked answer: No. Adjacent/hypotenuse must be between zero and one.

First tempting mistake: Dividing hypotenuse by adjacent reverses the ratio.

Independent check: The adjacent leg is shorter than hypotenuse.

238. The zero-angle boundary is not an ordinary triangle angle

Question: Can a nondegenerate right triangle have an acute angle of exactly 0°?

Worked answer: No. Its two non-right angles are strictly between zero and ninety degrees.

First tempting mistake: Using degenerate boundary values as ordinary geometry may cause division and shape problems.

Independent check: The angles of a nondegenerate triangle are positive.

How a three-pupil Maths tutorial could introduce trigonometry

The immutable eduKateSG Clementi Mathematics small-group programme describes 90-minute tutorials for up to three pupils near Sixth Avenue MRT. An appropriate later-stage G1 extension session could have one student rebuild Pythagoras and triangle labels, another practise identifying the relevant SOHCAHTOA ratio, and a third check inverse-angle answers against geometry. This is a differentiated teaching illustration, not confirmation that trigonometry is being taught immediately to every Secondary 2 pupil or that a class vacancy exists.

10 minutes: retrieve the right-angle mark, 3–4–5 triangles and the longest-side rule.
20 minutes: move θ between the two acute corners and relabel opposite/adjacent.
25 minutes: practise ratios and one missing-side problem with individual checks.
15 minutes: demonstrate an inverse-angle calculation where ready and check degree mode.
10 minutes: solve one rotated unseen triangle without model prompts.
10 minutes: explain the first wrong side label and plan a later retest.

A short home practice plan that respects school scope

Day one: label three right triangles with the right angle and hypotenuse.
Day two: move θ between acute corners and swap opposite and adjacent labels.
Day three: write each ratio as a named fraction using simple integer side lengths.
Day four: calculate an unknown leg with a teacher-approved scientific calculator.
Day five: check one inverse angle and explain why it lies between 0° and 90°.

Where the school has not yet introduced trigonometry, this can be treated as a later lesson plan rather than an extra homework demand. Stronger early preparation may instead involve Pythagoras, ratios, diagram interpretation or arithmetic fluency. The aim is to add one connected concept at the right time, not to race beyond school requirements.

Frequently asked questions

What is SOHCAHTOA?

It is a memory aid for sine = opposite/hypotenuse, cosine = adjacent/hypotenuse and tangent = opposite/adjacent in a right triangle.

Does the hypotenuse always have to be the bottom side?

No. It is always opposite the 90° angle, irrespective of how the triangle is rotated.

Why might a calculator give the wrong angle?

The angle unit setting, inverse-function choice, or side ratio may be wrong. Check degree mode for a degree-marked question and verify the angle is acute.

Can tangent be greater than one?

Yes. The opposite leg can be longer than the adjacent leg. Sine and cosine for acute angles remain between zero and one.

Is trigonometry on the complete G1 SEC syllabus?

Yes, in the published 2027 K110 graduating syllabus. That is different from guaranteeing it is in the current Secondary 2 school’s assessment.

What should a tutor check first?

Pythagoras readiness, the ability to locate a marked right angle, correct opposite/adjacent labels, and matching the ratio to the known and requested sides.

Official syllabus and editorial accuracy

Reader note: All triangle dimensions, angles and ratios are original educational examples based on ideal right-triangle geometry. Approximations follow their displayed precision; they are not real safety or construction measurements. Current school topics, actual graduating year, permitted devices, tuition availability and fees must be verified directly. No result or subject-level change is guaranteed.

Bring a labelled right-triangle worksheet to eduKate Singapore and ask whether the best next lesson is to repair side naming and Pythagoras or to introduce the appropriate trigonometric ratio.

Secondary 2 G1 Mathematics Tutor Bukit Timah | Sine, Cosine and Tangent in Right Triangles

Quick answer: sine, cosine and tangent

Choose a trigonometry problem

One-minute readiness diagnostic

The hypotenuse never changes when the viewpoint changes

A ratio is a relationship, not a length

238 original right-triangle trigonometry clinics

Identify opposite, adjacent and hypotenuse

1. Label a right triangle 1

2. Label a right triangle 2

3. Label a right triangle 3

4. Label a right triangle 4

5. Label a right triangle 5

6. Label a right triangle 6

7. Label a right triangle 7

8. Label a right triangle 8

9. Label a right triangle 9

10. Label a right triangle 10

11. Label a right triangle 11

12. Label a right triangle 12

13. Label a right triangle 13

14. Label a right triangle 14

15. Label a right triangle 15

16. Label a right triangle 16

17. Label a right triangle 17

18. Label a right triangle 18

19. Label a right triangle 19

20. Label a right triangle 20

21. Label a right triangle 21

22. Label a right triangle 22

23. Label a right triangle 23

24. Label a right triangle 24

25. Label a right triangle 25

26. Label a right triangle 26

27. Label a right triangle 27

28. Label a right triangle 28

29. Label a right triangle 29

30. Label a right triangle 30

31. Label a right triangle 31

Compute sine, cosine and tangent from sides

32. Ratio from exact side lengths 1

33. Ratio from exact side lengths 2

34. Ratio from exact side lengths 3

35. Ratio from exact side lengths 4

36. Ratio from exact side lengths 5

37. Ratio from exact side lengths 6

38. Ratio from exact side lengths 7

39. Ratio from exact side lengths 8

40. Ratio from exact side lengths 9

41. Ratio from exact side lengths 10

42. Ratio from exact side lengths 11

43. Ratio from exact side lengths 12

44. Ratio from exact side lengths 13

45. Ratio from exact side lengths 14

46. Ratio from exact side lengths 15

47. Ratio from exact side lengths 16

48. Ratio from exact side lengths 17

49. Ratio from exact side lengths 18

50. Ratio from exact side lengths 19

51. Ratio from exact side lengths 20

52. Ratio from exact side lengths 21

53. Ratio from exact side lengths 22

54. Ratio from exact side lengths 23

55. Ratio from exact side lengths 24

56. Ratio from exact side lengths 25

57. Ratio from exact side lengths 26

58. Ratio from exact side lengths 27

59. Ratio from exact side lengths 28

60. Ratio from exact side lengths 29

61. Ratio from exact side lengths 30

62. Ratio from exact side lengths 31

63. Ratio from exact side lengths 32

Find an unknown side with a known angle

64. Find an unknown side 1: 20°

65. Find an unknown side 2: 25°

66. Find an unknown side 3: 30°

67. Find an unknown side 4: 35°

68. Find an unknown side 5: 40°

69. Find an unknown side 6: 45°

70. Find an unknown side 7: 50°