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The calculator’s inverse-sine function returns a principal angle. In degree mode, sin⁻¹(0.5) = 30°. But sin 150° is also 0.5, so thirty degrees is not the only angle with that sine value. In a triangle, the other information decides whether an acute or an obtuse candidate is possible.
The quick reason is symmetry: supplementary angles have equal sines. For an acute angle α, sin(180° − α) = sin α. Thus thirty and one hundred and fifty degrees share the same sine even though they are different angles.
This guide focuses on that calculator interpretation and its use in triangle problems. It is written for G3 readers studying trigonometry beyond right triangles. Optional comments on wider angle intervals are identified separately. Use the current school course for the required scope and precision; a general explanation of inverse functions is not a claim that every advanced trigonometric topic belongs in the same syllabus.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–25
Sections 26–28
If sin B = 0.5 and 0° < B < 180°, then B can be 30° or 150°. The inverse-sine key gives the principal value 30°. The second candidate is 180° − 30° = 150°. Check both against any additional triangle conditions.
If B is an acute angle in a right triangle, only thirty degrees fits that condition. If the question states that B is obtuse, only one hundred and fifty degrees fits. If neither condition is supplied, use the remaining side and angle information rather than assuming the displayed value finishes the problem.
The notation sin⁻¹ here means inverse sine, also written arcsin. It does not mean 1/sin. Inverse sine returns an angle from a ratio; the reciprocal of sine returns a number from an angle. Keep the degree setting, interval and meaning of the notation clear before interpreting the display.
For an acute angle in a right triangle, sine is opposite side divided by hypotenuse. Both lengths are positive and the opposite side is shorter than the hypotenuse, so the ratio lies between zero and one.
If the opposite side is half the hypotenuse, the acute angle is thirty degrees. In that setting, the calculator result sin⁻¹(0.5) = 30° is the required angle because the geometric context already restricts it to an acute value.
The other non-right angle in a nondegenerate right triangle is also acute. An obtuse angle cannot appear alongside a ninety-degree angle because their sum would already exceed one hundred and eighty degrees. That geometric restriction removes the supplementary candidate.
The important habit is to identify why the single answer is justified. It is not because inverse sine magically lists every possible angle. It is because the right-triangle condition narrows the allowed range to one where the principal acute answer fits uniquely.
When the lesson later moves to non-right triangles, that earlier range restriction no longer applies automatically. The sine ratio still supplies useful information, but the solver must now reconsider which angles the context allows.
Sine is not limited to acute right-triangle angles. It can be extended consistently to obtuse angles, allowing relationships such as the sine rule and the triangle-area formula to work for non-right triangles.
A simple unit-circle interpretation is that sine records the vertical coordinate of a point at the chosen angle from the positive horizontal direction. Angles thirty and one hundred and fifty degrees reach points with the same vertical coordinate, although their horizontal coordinates have opposite signs.
That shared height explains why their sines agree. The angles are reflections across the vertical axis of the circle. Their sum is one hundred and eighty degrees, so they are supplementary.
You do not need a full course in trigonometric functions to use the resulting triangle fact: sin B = sin(180° − B). A sketch or altitude construction can also help explain the symmetry for triangle work.
Keep the model’s purpose clear. The unit circle explains the extended function, while the triangle’s angle sum decides whether a particular supplementary value can actually occur in the given triangle. Equal sine values alone do not establish that both candidates satisfy every part of the problem.
An inverse function must assign one output to each permitted input. But sine across all angles repeats values. If inverse sine tried to return every angle with a given sine as a single ordinary function value, it would not have a unique output.
The conventional inverse-sine function therefore uses a restricted output range. In degrees, its principal values lie from −90° to 90°, including the endpoints. For an input of 0.5, it selects thirty degrees within that range.
This convention makes the function well defined. It is not an error or a limitation that the calculator should somehow fix by displaying every other angle at once. Solving a trigonometric equation over a stated interval is a broader task than evaluating the inverse function.
For triangle angles, we usually work within 0° < B < 180°. A positive sine value below one can correspond to one acute and one obtuse candidate in that interval. The principal inverse value supplies the acute candidate, and supplementary symmetry supplies the other.
The output range and the problem’s allowed angle range are different pieces of information. Keeping both visible makes the apparent conflict between thirty and one hundred and fifty degrees disappear.
Supplementary angles add to 180°. Complementary angles add to 90°. The second triangle-angle candidate with the same positive sine is supplementary to the acute candidate, not generally complementary to it.
For an acute candidate of thirty degrees, the supplementary angle is one hundred and fifty degrees. The complementary angle is sixty degrees, whose sine is √3/2 rather than one half. Confusing these relationships generates a wrong second candidate.
The identity sin(180° − α) = sin α is the one used here. Another relationship, sin α = cos(90° − α), connects complementary angles but changes the function from sine to cosine. These are different statements.
Write the target condition before choosing a relationship. If the task is to find another angle with the same sine in a triangle-angle range, use the supplementary relation. If the task is to express a sine as a cosine of a complementary acute angle, use the second relation.
Similar-looking angle subtractions should not be treated as interchangeable shortcuts. Each formula states a particular equality, and its function names are as important as the number subtracted from.
The question is to solve sin B = 0.5 for 0° < B < 180°. First use degree mode and find the principal inverse value: B₁ = sin⁻¹(0.5) = 30°.
Because sine is positive for both acute and obtuse angles within this interval, consider the supplementary candidate B₂ = 180° − 30° = 150°. Both candidates lie strictly between zero and one hundred and eighty degrees.
Check the original equation: sin 30° = 0.5 and sin 150° = 0.5. Therefore the solutions within the stated interval are thirty degrees and one hundred and fifty degrees. There is no further triangle information in this isolated question to reject either.
Do not report only the principal value unless the question asks specifically for sin⁻¹(0.5) or supplies a condition restricting B to the acute range. Conversely, do not add extra turns such as 390° when the interval ends at 180°.
The complete answer combines a calculator evaluation, a symmetry relationship and an interval check. Each has a different role: the key finds a principal value, the identity generates the supplementary possibility, and the stated range selects which generated values belong.
The supplementary relationship applies to every acute candidate, not only the special value thirty degrees. The pairs below show the structure using exact angle descriptions.
| Acute candidate | Supplementary candidate | Relationship | Triangle-angle range |
|---|---|---|---|
| 20° | 160° | Same sine | Both between 0° and 180° |
| 30° | 150° | Same sine | Both between 0° and 180° |
| 45° | 135° | Same sine | Both between 0° and 180° |
| 60° | 120° | Same sine | Both between 0° and 180° |
At ninety degrees, the supplementary angle is ninety degrees again. The two candidates coincide, so there is one distinct angle rather than two different answers. Avoid duplicating the same value in a final list.
The table gives possible equal-sine pairs, not a promise that every pair forms two valid solutions in every triangle problem. A given triangle may already contain another angle that excludes its obtuse candidate.
Use the table to recognise the symmetry, then return to the actual data. The additional conditions are what turn possible angle values into valid solutions for a particular geometric situation.
In a triangle, the sine rule relates each side to its opposite angle: a/sin A = b/sin B = c/sin C. The side letters must stay paired with the angles opposite them.
If A and its opposite side a are known, along with another side b, rearrange to sin B = b sin A/a. A calculator then gives a principal candidate for B, but the sine equation may also allow its supplement.
This is the familiar ambiguity that can arise from certain side-side-angle data. The phrase does not mean every sine-rule question has two triangles. Depending on the lengths and angle, there may be no valid triangle, one valid triangle or two.
The first task is accurate correspondence. If side b is not actually opposite B in the supplied diagram, the equation itself is wrong before any inverse-sine issue arises. Mark the side-angle pairings before substituting numbers.
After calculating candidates, test their compatibility with the triangle’s angle sum and any other given conditions. The sine rule supplies a relation; it does not remove the need to interpret the geometry that relation describes.
Suppose A = 30°, side a = 7 units and side b = 10 units. These labels mean that the seven-unit side lies opposite A and the ten-unit side lies opposite B. The sine rule gives sin B = 10 × sin 30°/7 = 5/7.
The principal candidate is B₁ = sin⁻¹(5/7), approximately 45.5847°. The supplementary candidate is B₂ = 180° − B₁, approximately 134.4153°. Both must be checked with the known thirty-degree angle.
For the first, C₁ = 180° − 30° − 45.5847° ≈ 104.4153°, which is positive. For the second, C₂ ≈ 15.5847°, also positive. Thus both candidates can form nondegenerate triangles with the supplied data.
If angles are requested to one decimal place, report approximately 45.6° or 134.4° for B, with the corresponding C values approximately 104.4° or 15.6°. Keep unrounded values during intermediate calculations.
The two triangles have different shapes but satisfy the same stated side-side-angle information. A diagram drawn in one orientation or with one apparent angle size does not erase the second possibility unless it supplies an actual condition that distinguishes them.
Suppose A = 40°, a = 10 units and b = 6 units. The sine rule gives sin B = 6 sin 40°/10, approximately 0.385673. The principal candidate is approximately 22.6855°.
The supplementary candidate is approximately 157.3145°. But adding that candidate to A = 40° already exceeds 180°. It would leave a negative third angle, so it cannot belong to the triangle.
The acute candidate leaves C ≈ 117.3145°, which is positive. Thus there is one valid triangle from these data, and B is approximately 22.7° if one decimal place is required.
The side comparison supports the conclusion: side b is shorter than side a, so its opposite angle B must be smaller than A in a nondegenerate triangle. The acute candidate fits that relation; the obtuse candidate does not.
This illustrates two useful geometric checks pointing to the same result. Neither requires inventing a rule that inverse sine always gives the only answer. Instead, the other supplied conditions explain why one of the mathematically generated sine candidates is rejected.
Suppose A = 30°, a = 5 units and b = 12 units. The sine-rule rearrangement gives sin B = 12 × 0.5/5 = 1.2. No real angle has sine 1.2 because real sine values lie between negative one and one.
The supplied triangle data are therefore inconsistent under the stated labels. There is no real triangle satisfying them. A calculator error on inverse sine of 1.2 is a meaningful signal to inspect the model or the data.
Do not subtract 1.2 from 180 or try another quadrant to rescue the calculation. The supplementary relationship applies to angles with valid sine values; it cannot create a real angle for an input outside the function’s domain.
First check correspondence, units and arithmetic. If those are correct and the data really imply a sine above one, report that no such triangle exists. A question can legitimately have an impossibility conclusion.
The point is not to distrust every calculator error. It is to connect the error to the mathematical condition it expresses. Here the inverse-sine input is outside its real domain, so the absence of an angle follows from the function’s range.
Suppose A = 30°, a = 5 units and b = 10 units. Then sin B = 10 × 0.5/5 = 1. The principal angle is B = 90°, and its supplement is also ninety degrees.
There is therefore only one distinct B candidate. The third angle is C = 180° − 30° − 90° = 60°, so the triangle is valid. Listing ninety degrees twice would not describe two different triangles.
This boundary example helps prevent the mechanical instruction “always write two answers after inverse sine”. For a positive ratio strictly between zero and one, there are two distinct candidates in the open triangle-angle interval. At one, they merge.
At zero, the angles zero and one hundred and eighty degrees do not form a nondegenerate triangle angle, so the geometric setting excludes them. A separate trigonometric equation with inclusive endpoints might treat them differently.
Always combine the function fact with the domain of the actual problem. Boundary values are where a memorised general pattern often needs adjustment, and a quick interval check is more reliable than counting answers by habit.
Suppose A = 30°, a = 10 units and b = 10 units. The sine rule gives sin B = 0.5, with candidates thirty and one hundred and fifty degrees. The equal side lengths also suggest equal opposite angles in an ordinary triangle.
With B = 30°, the third angle is 120°, so the triangle is valid. With B = 150°, the third angle would be zero degrees. That is a degenerate configuration, not a nondegenerate triangle of the kind assumed in these questions.
Rejecting the second candidate requires the strict condition C > 0°, not merely C ≥ 0°. This is why the triangle-angle interval was written with strict inequalities at zero and one hundred and eighty degrees.
The example differs from the preceding one-angle case because the second candidate reaches the boundary rather than exceeding it. Both must be rejected for an ordinary triangle, but the arithmetic reason should be stated accurately.
Attention to endpoints is a general Mathematics habit. A candidate can satisfy a sine equation while failing a strict geometric condition. The final solution must satisfy both parts of the problem, including boundaries that are easy to overlook when the numbers otherwise seem neat.
If a question states that B is obtuse and sin B = 0.6, first find α = sin⁻¹(0.6) ≈ 36.8699°. The required obtuse angle is B = 180° − α ≈ 143.1301°, or 143.1° to one decimal place.
If it states that B is acute, the required value is approximately 36.9°. The same sine ratio supports different answers because the question supplies different range restrictions.
Do not treat the word obtuse as decorative context. It is mathematical information that selects the correct inverse branch. Likewise, a right-angle marker elsewhere in the triangle can force the unknown remaining angle to be acute.
When no such information is explicit, inspect the remaining side and angle conditions. A drawing that merely looks obtuse is weaker evidence than a label or a logical consequence of the data, especially when the diagram is not drawn to scale.
A clear solution records the branch decision in a sentence: “B is obtuse, so use 180° minus the acute inverse-sine value.” This makes the reasoning visible and shows that the principal display has been interpreted in the problem’s stated range.
For two sides a and b enclosing an angle C, triangle area is 1/2 ab sin C. Supplementary angles C and 180° − C have the same sine, so the same two side lengths with either included angle produce the same area.
For example, sides six and eight units with included angle thirty degrees give area 1/2 × 6 × 8 × 0.5 = 12 square units. An included angle of one hundred and fifty degrees gives the same area because its sine is also one half.
The triangles are generally different shapes, but the perpendicular height relative to a chosen base can have the same magnitude. Equal areas do not force equal included angles or congruent triangles.
If area and two sides determine sin C = 0.5, the possible included angle may therefore be acute or obtuse unless another condition chooses one. This is the same inverse-sine issue appearing through an area relationship rather than the sine rule.
The supplied data still need to describe a valid triangle. Use the conditions in the question, the angle interval and any additional side or angle information. A matching area calculation alone is not permission to ignore those other constraints.
Distinguish these tasks. First, evaluate sin⁻¹(0.5) in degrees. Second, solve sin B = 0.5 for 0° < B < 180°. Third, find the obtuse angle B with sin B = 0.5.
The first answer is thirty degrees because it asks for the inverse function’s principal value. The second has two answers, thirty and one hundred and fifty degrees, because it asks for all solutions in a wider interval. The third has only one answer, one hundred and fifty degrees, because it supplies an obtuse restriction.
The input ratio is identical in all three tasks. The requested output and allowed range differ. This is why the wording should be read before the calculator is used, rather than interpreted after a number appears on the display.
For another practice ratio, use sin B = √2/2. Its principal degree value is forty-five degrees, and the supplementary triangle-angle candidate is one hundred and thirty-five degrees. If a triangle’s other known angle is sixty degrees, only forty-five degrees survives because sixty plus one hundred and thirty-five exceeds one hundred and eighty.
A triangle has A = 20° and sin B = 0.6. Find the possible B and C values to one decimal place. The acute inverse value is approximately 36.8699°, with supplement approximately 143.1301°.
For the acute B, C = 180° − 20° − 36.8699° ≈ 123.1301°. For the obtuse B, C ≈ 16.8699°. Both third angles are positive, so both angle triples are possible. Rounded pairs are B ≈ 36.9°, C ≈ 123.1°, or B ≈ 143.1°, C ≈ 16.9°.
Now change A to forty degrees while keeping sin B = 0.6. The obtuse B leaves C ≈ −3.1301°, which is impossible. The acute B leaves C ≈ 103.1301°, so only that branch survives.
This practice changes just one condition. The sine candidates stay the same, but the admissible triangle changes because the remaining angle space changes. It is a useful way to see why generating a candidate and validating it are separate mathematical actions.
Suppose A = 30°, a = 8 units and b = 8 units. The sine rule gives sin B = 8 × 0.5/8 = 0.5. Candidates are thirty and one hundred and fifty degrees. The second leaves a zero third angle and is rejected, so B = 30° and C = 120°.
Now suppose A = 30°, a = 8 units and b = 4 units. Then sin B = 4 × 0.5/8 = 0.25. The acute candidate is approximately 14.4775°, with supplement approximately 165.5225°.
Adding the obtuse candidate to thirty degrees exceeds one hundred and eighty degrees, so only the acute candidate survives. The third angle is approximately 135.5225°. The shorter side b correctly lies opposite the smaller angle B.
If the formula was accidentally inverted to sin B = a sin A/b, this second example would give one instead of one quarter. The resulting right-angle answer would contradict the side-angle comparison. Correct correspondence and a geometric reasonableness check support each other; both should be part of reliable sine-rule working.
In degree mode, sin⁻¹(0.5) displays thirty. In radian mode, it displays π/6 or approximately 0.523599. These represent the same principal angle in different units, but using one as though it were the other produces incorrect calculations.
Check the mode before entering the ratio, and label the returned angle. If the triangle problem is expressed in degrees, either use degree mode throughout or convert deliberately. Do not subtract a radian-valued display from 180 as though both numbers used the same unit.
Keep extra digits during intermediate angle calculations. If α is rounded too early, its supplement and the third angle inherit that rounding error. Use the stored or unrounded inverse value, then round final requested angles to the stated precision.
For special exact values, thirty and one hundred and fifty degrees can be written exactly. For a ratio such as 5/7, use an approximate sign with decimal angles. The distinction helps readers see whether the number is an exact result of known geometry or a rounded numerical evaluation.
The notation sin⁻¹(0.5) conventionally means the angle whose sine is 0.5 within the principal inverse range. It equals thirty degrees in degree mode. The reciprocal 1/sin 30° equals two, which is a ratio, not an angle.
The superscript negative one can be confusing because a numerical power a⁻¹ means 1/a. Function inverse notation uses the same superscript for a different idea: undoing a function on a suitable restricted domain.
Context and parentheses help. Write arcsin(0.5) when you want to make the inverse meaning especially clear. Write 1/sin B when you mean the reciprocal. Do not use the two expressions interchangeably.
A quick output check can help: finding an angle from a sine ratio should give an angle, while taking a reciprocal of a dimensionless ratio gives another dimensionless number. If the calculator output is interpreted with the wrong kind of quantity, the notation may have been misread.
This distinction matters beyond one key sequence. Clear function notation helps the learner understand why the principal value restriction exists, while reciprocal reasoning belongs to multiplication and division of numerical values.
For sin θ = 0.5 on 0° ≤ θ < 360°, the solutions are thirty and one hundred and fifty degrees. There are no solutions with sine positive in the lower half of that full-turn interval. For other signs and intervals, the candidate locations change.
If the interval includes additional full turns, adding multiples of 360° can produce further solutions. That wider periodic-equation work should be used only when it belongs to the learner’s course.
The triangle focus of this guide uses 0° < θ < 180°, so negative angles and extra rotations do not describe an ordinary interior triangle angle. A general trigonometric solution list must be filtered back to that geometric interval when used in a triangle problem.
Do not apply the supplement formula mechanically to every trigonometric function or every interval. Sine symmetry, cosine symmetry and tangent periodicity have different forms. The present shortcut is justified by a specific equal-sine relation and a specific angle range.
The transferable habit is to write the interval before solving. It tells you which branches to consider and when the search is complete. A calculator’s principal answer is the starting information, while the interval defines the full question.
A labelled right angle, a stated obtuse angle or equal-side marks supply mathematical conditions. A triangle that merely looks narrow or wide on the printed page may not. Many diagrams are schematic, and their apparent proportions should not override the given data.
For the two-triangle example with A = 30°, a = 7 and b = 10, a drawing of one possible triangle does not automatically establish uniqueness. The supplied conditions permit both candidates unless an additional constraint excludes one.
If a question explicitly gives a configuration, location or direction that distinguishes the possibilities, use that information. For example, a point’s stated position relative to a perpendicular or a clearly defined obtuse-angle requirement can select a branch.
The solver should explain which condition is doing the selecting. Saying “it looks obtuse” is weaker than saying “the question states B is obtuse” or “the acute candidate contradicts this given angle relation”.
This is a general geometry reading skill. Treat diagram labels, marks and stated relationships as data, then distinguish them from visual impressions. A sound calculation begins by reading the evidence accurately and ends by checking that the answer fits it.
Suppose a learner uses the sine rule correctly, gets sin B = 5/7, presses inverse sine and reports only 45.6°. The arithmetic may be accurate. The missing step is interpreting whether the principal angle is the only admissible one.
Ask for the supplement, approximately 134.4°, and then ask whether each candidate leaves a positive third angle with the known thirty degrees. Both do. The correction identifies a branch-selection gap rather than describing all the working as wrong.
Next give a case with A = 40° and sin B = 0.6. The learner should generate both candidates but reject the obtuse one because the angle sum fails. This tests whether they understand the checking stage rather than simply adding a second answer to every question.
Finally, give sin B = 1 in a triangle context. They should recognise one distinct ninety-degree candidate. Together, the three tasks distinguish two valid triangles, one valid branch and a boundary where the candidates coincide.
The aim is a decision process grounded in the range and geometry. A fixed instruction to always keep one or always write two cannot handle all three situations accurately.
Use a short mixed set with simple sine values first. Include an acute restriction, an obtuse restriction, an unrestricted triangle-angle equation and a known second angle that rejects one candidate. Ask for a reason beside every retained or rejected angle.
Then introduce less tidy ratios such as 0.6 or 5/7. The calculator arithmetic changes, but the branch and triangle checks remain the same. This isolates the conceptual decision before adding precision demands.
After a delay, ask the learner to explain why sin⁻¹(0.5) and solving sin B = 0.5 are different tasks. A clear answer should mention the principal output range and the problem’s allowed angle range.
Parents can support this by asking, “What condition decides the angle?” rather than telling the learner to use the other answer whenever the first looks unexpected. Teachers can distinguish correspondence errors, mode errors, missing branches and angle-sum failures.
Those categories suggest different repairs. More calculator practice may help an input error, while a missing obtuse candidate needs work on sine symmetry and range. Identifying the decision that failed makes a small amount of focused practice more useful.
The central application here is triangle trigonometry, including supplementary sine values and selecting angles from the given conditions. Unit-circle language and wider periodic intervals are supporting explanations or optional extensions, not a statement that every part is assessed identically in G3 Mathematics and Additional Mathematics.
For official course context, consult the 2027 G3 Mathematics syllabus. Match practice to the school’s topic sequence and the actual wording of the question.
The existing Sine Rule and Cosine Rule guide is the broader triangle-method owner. The inverse trigonometric functions and principal values guide offers a deeper optional function perspective.
Use the Mathematics explainer directory for related lessons on calculator modes, corresponding sides and diagram evidence. These connections help when the difficulty is not the inverse-sine key itself but the interpretation around it.
Find the obtuse angle B with sin B = √3/2. The acute principal value is sixty degrees, so the required obtuse angle is one hundred and twenty degrees. The word obtuse selects the supplement.
Now solve sin B = √3/2 when B is a triangle angle and another angle is seventy degrees. The candidates are sixty and one hundred and twenty degrees. The second would make the two known angles total one hundred and ninety, so it is rejected. B is sixty degrees and the third angle is fifty degrees.
Finally, a triangle has A = 30°, a = 7 units and b = 10 units. Explain why two B values survive without calculating every side. The sine ratio is 5/7, its acute and obtuse candidates are approximately 45.6° and 134.4°, and each leaves a positive third angle with thirty degrees.
These tasks use the same sine symmetry but different constraints. If you can explain the selection in each, you are interpreting the calculator result within the geometry rather than treating the display as a complete solution by itself.
When inverse sine returns an angle, identify its unit and principal range. Then ask what range the problem allows and whether a supplementary candidate must be considered. In a triangle, test every candidate against the angle sum and the other given conditions.
For sin B = 0.5, thirty degrees is the principal degree value. One hundred and fifty degrees has the same sine and may be a valid triangle angle. Which answer belongs depends on the complete problem.
Keep side-angle correspondence clear when using the sine rule, and keep exact or unrounded values until final rounding. Reject ratios outside the real sine range and reject candidates that leave zero or negative third angles in a nondegenerate triangle.
The essential question is, “What has this calculator value actually answered?” It has evaluated an inverse function on a chosen range. Your geometry and interval checks complete the larger task of finding every admissible angle.
These answers distinguish a calculator’s principal value from the possible angles in a triangle. The same ratio can lead to different final answers when the problem supplies different restrictions.
Is the calculator wrong when it gives thirty degrees?
No. Thirty degrees is the correct principal degree value of inverse sine at 0.5. Finding all angles with sine 0.5 in a wider interval is a separate task requiring symmetry and range checks.
Should I always write the supplementary angle?
Consider it when the allowed interval includes obtuse triangle angles and the sine ratio is positive. Then check the geometry. A known acute restriction or another angle may reject it, and at ninety degrees it duplicates the same candidate.
Can two different triangles fit the same sine-rule data?
Yes, some side-side-angle data permit two nondegenerate triangles. Other data permit one or none. Generate the angle candidates and check the complete conditions rather than assuming a fixed answer count.
Why is inverse sine of 1.2 unavailable over the reals?
Every real sine value lies between negative one and one. An input of 1.2 lies outside the real inverse-sine domain. Check the data and correspondence; if they genuinely imply that ratio, no real angle satisfies the requirement.