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A bearing is not just an angle written beside a line. In the standard school convention, it is a clockwise angle from north at the point you are looking from. Change the starting point, the reference direction or the direction of measurement, and you may be describing a different angle.
For example, the bearing of B from A is measured at A, from A’s north line towards the direction from A to B. A bearing of 060° means sixty degrees clockwise from north. It does not mean sixty degrees anticlockwise from east, and it does not mean an angle measured at B.
The official 2027 G3 Mathematics syllabus includes problems involving bearings within trigonometry. This guide builds the directional interpretation before introducing selected triangle applications. All journeys are fictional flat-plan classroom models, not operational navigation instructions. Your school determines the teaching sequence and required techniques.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–25
Sections 26–28
Read the phrase of B from A as an instruction to stand at A and look towards B. Draw north at A, mark the ray from A to B and measure clockwise from north. Write the result using three digits before the degree sign, such as 025°, 090° or 315°.
The cardinal directions provide immediate checks: north is 000°, east 090°, south 180° and west 270°. A northeast direction lies between 000° and 090°; a northwest direction lies between 270° and 360°. Three-digit notation records the angle clearly but does not by itself prove the correct viewpoint was used.
For opposite directions along the same straight segment in the usual flat school diagram, add or subtract 180° and express the result in the standard one-turn range. For a triangle problem, first convert bearings into the relevant interior angles. Do not insert every printed bearing directly into a sine or cosine formula.
The bearing of B from A belongs at A. Think of A as the observation point and B as the point being sighted. The wording describes a directed relationship, not merely the undirected line joining the two points.
If the bearing of B from A is 040°, draw a ray from A leaning towards the northeast, forty degrees clockwise from A’s north line. To find the bearing of A from B, move your viewpoint to B and look back along the opposite direction. That second angle is not measured from the original north arrow at A.
The distinction resembles giving instructions from different starting positions. The same segment is involved, but which way you face along it changes. Writing A → B beside the first bearing and B → A beside the second can prevent the two descriptions from merging in your working.
Before calculating, underline the word from and circle its point name. This is a reading step with a geometric consequence. If the angle is placed at the wrong point, even flawless angle arithmetic may solve a different problem from the one asked.
In an ordinary school plan, the north arrows establish the reference direction. They are usually drawn vertically, but it is the labelled north direction that matters. A page can be rotated without changing the mathematical bearings if the north reference rotates with the diagram.
A line that looks horizontal on a tilted page is not automatically east-west. Likewise, the top of an illustration is not automatically north unless the diagram or standard setup establishes it. Read the north arrow rather than relying on how you are holding the paper.
At each relevant point, imagine a north ray parallel to the others in the flat-plan model. This gives you a local reference for the bearing measured there. It also creates the parallel-line angle relationships used later in triangle questions.
If no north direction is supplied and none is established by the context, a numerical bearing cannot generally be determined from the segment alone. A line has an orientation relative to a reference, so the reference is essential information. Do not invent one merely because the diagram has a convenient vertical edge.
Starting at north, clockwise movement goes through east, then south, then west, before returning to north. A direction thirty degrees east of north therefore has bearing 030°. A direction thirty degrees west of north has bearing 330°, not 030°.
The smaller visible angle can be misleading. For a northwest ray, the small angle between the ray and north may be 30°, but the required clockwise turn from north is 360 − 30 = 330°. The bearing convention chooses a particular directed turn, not always the smallest angle.
Use a small curved arrow when sketching the bearing. It need not be artistic; its purpose is to record which way the turn is measured. This is especially helpful for bearings greater than 180°, where the required clockwise turn may be reflex rather than a triangle’s interior angle.
The degree symbol alone does not encode clockwise or anticlockwise direction. The convention and your working must do that. A correct-looking angle value with the wrong turn direction can point to the other side of north and produce a different location.
Standard three-figure bearings use a leading zero when needed. Write 007° rather than 7°, 045° rather than 45°, and 090° rather than 90°. Values already containing three digits, such as 225° or 310°, need no extra zero.
The leading zero does not change the numerical angle. It makes the notation consistent and signals that the number is being used as a bearing. It is a presentation convention that supports precise communication, not a separate mathematical operation.
When checking a copied value, read the digits aloud as an angle: zero four five means forty-five degrees in this notation. A missing leading zero is a presentation issue; writing 450° instead of 045° changes the turn. Separate a formatting correction from a numerical or directional correction so that the learner knows which part actually needs attention.
For north, use 000° in the ordinary one-turn bearing range. A complete 360° turn returns to the same direction, so 360° is directionally equivalent, but standard school answers generally represent that direction as 000° rather than as the end of the turn.
Do not confuse a leading zero with decimal notation. A bearing of 025° means twenty-five degrees, not 0.25° and not two and a half degrees. If the task requires a decimal angle, retain the stated precision while following its formatting instructions. The basic directional meaning remains a clockwise angle from north.
Point B lies 35° east of north from A. Find the bearing of B from A and the bearing of A from B in the standard flat-plan model. Begin with a sketch showing A, its north ray and the ray from A to B.
The phrase east of north starts at north and turns towards east. That is clockwise by 35°, so the first bearing is 035°. For the reverse direction, the ray from B to A points southwest. Adding 180° gives 35 + 180 = 215°, so the return bearing is 215°.
Check the quadrants independently. The outward ray is northeast, so its bearing should lie between 000° and 090°. The return ray is southwest, so its bearing should lie between 180° and 270°. Both answers match these directional expectations.
Finally label the two answers by direction. A bare pair 035°, 215° leaves the reader to infer which belongs to which question. Write B from A: 035°; A from B: 215°. The labels preserve the viewpoint through to the final line.
The cardinal directions anchor the full clockwise turn. Use them to locate an answer before worrying about the last few degrees. A value in the wrong quadrant cannot represent the intended ray, even if the arithmetic that produced it was performed correctly.
| Bearing | Direction | Clockwise interpretation |
|---|---|---|
| 000° | North | Starting reference |
| 045° | Northeast | Halfway from north to east |
| 090° | East | Quarter-turn |
| 180° | South | Half-turn |
| 225° | Southwest | Halfway from south to west |
| 270° | West | Three-quarter-turn |
| 315° | Northwest | Halfway from west back to north |
The diagonal values in this table describe exact halfway directions. Not every northeast ray has bearing 045°. Northeast as a broad quadrant includes other bearings between north and east, while the exact compass direction northeast usually identifies the halfway ray.
When a problem gives a precise angle, use that angle rather than replacing it with a memorised diagonal. For example, 020° is still northeast of the starting point, but it is much closer to north than 045°. The table provides orientation checks; it does not replace the supplied data.
Direction phrases have a starting reference and a turn towards another direction. Twenty degrees east of south starts at south and turns towards east. Its bearing is 180 − 20 = 160°. Twenty degrees west of south gives 180 + 20 = 200°.
Similarly, forty degrees west of north gives 360 − 40 = 320°. Forty degrees north of west starts at west and turns towards north, giving 270 + 40 = 310°. The words may involve the same compass directions but not the same reference ray.
Sketch before applying a remembered addition or subtraction. The phrase after of identifies the starting direction in these compass descriptions. Then locate the ray in its quadrant and determine its clockwise distance from north.
A quick comparison helps: 20° east of north is 020°, whereas 20° north of east is 070°. Their rays both lie in the northeast quadrant, but the angles are measured from different reference directions. Swapping the words changes the ray unless the angle is the special halfway value of 45°.
A bearing may be greater than 180°, but a non-degenerate triangle’s interior angle is less than 180°. Therefore a printed bearing such as 250° cannot simply be copied as a triangle’s interior angle. You must identify the two rays bounding the interior angle and compare their directions.
Suppose B and C are viewed from A on bearings 040° and 110°. The smaller angle BAC between the rays is 110 − 40 = 70°. That angle may be an interior angle of triangle ABC if those are its sides from A.
If the bearings are 350° and 020°, subtracting mechanically gives a difference of 330°, but the smaller angle between the rays is 30°, crossing north. The clockwise turns from north are labels for the rays; the angle between the rays needs its own calculation.
Write the angle name, such as BAC, beside the derived value. This prevents a bearing at one point from being confused with an interior angle at another. The labels are part of keeping the geometric relationship intact while changing representations.
For two bearings at the same point, calculate their absolute numerical difference d. Since the bearings lie within one full turn, the smaller angle between their rays is the smaller of d and 360 − d. The same-point condition matters: you are comparing rays from one vertex.
For 075° and 205°, d = 130°, so the smaller angle is 130°. For 010° and 300°, d = 290°, so the smaller angle is 70°. In the second case, the short turn passes across north rather than taking the long way around the circle.
If the question asks for a specified clockwise angle rather than the smaller angle, follow that directed instruction instead. Smaller angle and clockwise turn are not interchangeable requests. Read what is being measured before applying the minimum rule.
Degenerate cases also deserve attention. Equal bearings give a zero angle between identical rays. Bearings differing by 180° give opposite rays and a straight angle. Those configurations do not form a non-degenerate triangle with two distinct sides from the same vertex, so a triangle interpretation would need reconsideration.
In a flat school diagram with parallel north references, reversing a straight directed segment rotates its direction by half a turn. If the outward bearing is below 180°, add 180°. If it is at least 180°, subtract 180°. Express north as 000°.
An outward bearing of 128° gives return bearing 308°. An outward bearing of 305° gives return bearing 125°. An outward bearing of 180° gives return bearing 000°. All three pairs describe opposite directions along the same segment.
The rule is not subtract the bearing from 180 in every case. For 128°, that incorrect calculation gives 52°, which points northeast rather than northwest. Draw the return ray and use its quadrant to reject the error.
For a focused repair of this particular viewpoint change, see why return bearings go wrong. This guide uses the return calculation as one part of the wider starting-point and clockwise-reference system, rather than treating it as an isolated shortcut.
School bearing diagrams normally treat north rays at different points as parallel. A segment crossing those reference lines creates corresponding, alternate or supplementary angle relationships. This is the geometric reason you can move directional information between vertices.
If AB has bearing 060° from A, then the line towards A from B points in the opposite direction. At B, the clockwise bearing is 240°. The smaller angle between north at B and the ray BA is 120°. These values describe different directed or undirected angles built around the same line.
Mark the exact angle required instead of copying 60 everywhere near the segment. Parallel-line reasoning preserves particular angle relationships; it does not make all angles around two points equal. Some are equal, some sum to 180°, and some complete a full turn.
When you use a reason in your working, name it accurately. A statement such as north lines are parallel, so these alternate angles are equal is more informative than north lines equal. The equality applies to specific angles, not to every numerical bearing displayed in the diagram.
If B and C lie on the same ray from A, their bearings from A are equal even if they are different distances away. A bearing specifies direction, not distance. A point two kilometres along a ray and a point five kilometres along it share the same bearing from its starting point.
This means one bearing alone generally does not determine a unique location. It identifies a ray containing many possible points. Adding a distance from the same origin can locate a point within the ideal flat-plan model, while adding another suitable bearing from a different known position can constrain it by intersection.
Do not infer distances from how long a line looks in a diagram described as not to scale. The drawing helps organise the geometry, but the numerical distances must come from supplied data or valid calculation.
The distinction is useful when checking answers. If the question asks for a bearing, a distance in kilometres cannot be the final answer. If it asks for a distance, a correct direction alone is incomplete. Both quantities may appear in the solution, but they perform different jobs.
Suppose AB = 8 km and AC = 6 km. The bearing of B from A is 030°, and the bearing of C from A is 120°. The angle BAC is 120 − 30 = 90°, so the two known sides meet at a right angle.
Therefore BC² = 8² + 6² = 100, giving BC = 10 km. The right angle was not supplied as a square marker; it was derived from the two bearings at the same point. That derivation justifies using Pythagoras’ theorem.
A wrong approach might use 30° or 120° as the included angle between AB and AC. Those are measured from north, not from one triangle side to the other. Subtracting the bearings is the bridge between the directional data and the triangle.
Check the scale of the result. The side opposite a right angle is the longest side, and ten is larger than both eight and six. This does not replace the theorem, but it gives a useful independent warning if a mistaken calculation produces a side shorter than six.
Let AB = 7 km and AC = 5 km, with bearings 020° and 080° respectively from A. The included angle BAC is 60°. Since the known sides surround that angle, the cosine rule gives BC² = 7² + 5² − 2 × 7 × 5 × cos 60°.
Using cos 60° = 1/2 gives BC² = 49 + 25 − 35 = 39. Hence BC = √39 km, approximately 6.245 km. Round only as required by the question, keeping the exact form or full calculator value during subsequent work.
The printed bearings do not go directly into the cosine rule. The relevant angle is between the two known sides. This is why a clear sketch and an angle name are more reliable than choosing whichever angle happens to be next to the side lengths in the text.
Check the triangle inequality: the third side must be greater than 7 − 5 = 2 and less than 7 + 5 = 12. The answer passes. A result outside that interval would be impossible for the stated non-degenerate triangle, even if the calculator accepted the entered expression.
The bearing of Q from P is 070°. State where the angle is measured, which ray represents the direction of travel, and the bearing of P from Q. Then explain why 070° cannot simply be reused as the return bearing.
The angle is measured at P, from north at P clockwise to the ray PQ. The reverse ray QP points in the opposite direction, so its bearing is 70 + 180 = 250°. The first direction is northeast; the reverse direction is southwest.
The same segment is involved, but the directed rays and starting points differ. That is the reason for the half-turn, not a formatting rule about changing a number whenever a question contains two point names.
As a follow-up, suppose Q is moved twice as far from P along the same ray. The bearing of Q from P remains 070°, and the return bearing remains 250°. The distance change does not rotate either ray. This variation checks that the learner distinguishes directional information from the size of the journey.
Find the three-figure bearings for these four directions: 25° east of north, 25° north of east, 25° west of south, and 25° south of west. Make a small quadrant sketch before reading the answers.
The answers are 025°, 065°, 205° and 245°. The first starts from north and turns towards east. The second starts from east and turns back towards north, so its clockwise distance from north is 90 − 25. The third is 180 + 25, while the fourth is 270 − 25.
Notice that the same numerical 25° appears in all four descriptions but produces four different bearings. The reference direction and turn destination are essential parts of the information. Copying the angle alone discards them.
If you obtained 295° for the fourth direction, you turned from west towards north rather than south. Repair that specific directional step. Recalculate only after drawing the ray in the southwest quadrant, where its bearing must be between 180° and 270°. The quadrant check can expose the mistake before any detailed geometry is attempted.
From A, B has bearing 335° and C has bearing 025°. Find the smaller angle BAC. The numerical difference is 310°, but the smaller angle crosses north and is 360 − 310 = 50°. Equivalently, add the 25° gap from 335° to north to the 25° gap from north to 025°.
Now use bearings 280° and 040°. Their difference is 240°, so the smaller angle is 120°. The same method works even though the gaps on either side of north are not equal: 80° to north and 40° after north.
State that both bearings are measured from A. If the values were measured from different origins, subtracting them would compare directions but would not automatically produce an interior angle at A without transferring the relevant rays through the geometry.
Finally check whether the question asks for the smaller angle or a specified clockwise turn. From bearing 335° clockwise to 025° the turn is 50°, whereas from 025° clockwise to 335° it is 310°. The pair of rays is the same, but the directed question changes the answer.
A traveller moves 9 km from A to B on bearing 090°, then 12 km from B to C on bearing 000°. Find the straight-line distance AC and describe the quadrant of C from A. These are ideal flat-plan displacements, not instructions for an actual journey.
The first leg is east and the second is north. The two legs are perpendicular, so AC = √(9² + 12²) = 15 km. From A, C is northeast because it lies nine kilometres east and twelve kilometres north.
If the bearing of C from A is required, measure the angle from north, not from the eastward leg. With northward side twelve and eastward side nine, tan θ = 9/12, so θ is approximately 36.87°. To the nearest degree, the bearing is 037°.
The complementary angle from east is approximately 53.13°, but that is not the bearing. Both acute angles belong to the same right triangle. The reference direction decides which one answers the question. Checking that the direction is closer to north than east also supports the smaller bearing angle.
Where coordinate or vector methods are already familiar, let eastward displacement be E and northward displacement be N. A length d on bearing θ has components E = d sin θ and N = d cos θ in the standard flat-plan convention, using degree mode for degree bearings.
These formulas differ in appearance from the usual angle-from-horizontal arrangement because the reference axis is north rather than east. At θ = 000°, they give E = 0 and N = d. At θ = 090°, they give E = d and N = 0. These boundary checks explain the assignment.
For d = 10 and θ = 030°, the eastward component is five and the northward component is 5√3, approximately 8.660. The point lies northeast and closer to north than east, matching the bearing.
Treat this as an optional connection if your class has not introduced components. It is not necessary for the basic viewpoint and angle work. The important habit is consistent reference directions: changing from north-based bearings to horizontal-axis angles requires an explicit conversion, not a silent reuse of a memorised trigonometric diagram.
An inverse-tangent calculation by itself can hide the quadrant of a direction. If the displacement is four kilometres east and three kilometres south, the point is southeast. The acute angle east of south has tangent 4/3 and is approximately 53.13°, but the bearing is 180 − 53.13 = 126.87°.
Alternatively, the direction is about 36.87° south of east, giving 90 + 36.87 = 126.87°. Both descriptions agree when their reference directions are handled correctly. A bare answer of 53° would point northeast and fail the quadrant check.
Do not treat a negative calculator angle as a finished three-figure bearing either. It may be an intermediate signed angle relative to a different axis. Draw the displacement signs, identify the quadrant and convert to clockwise-from-north notation.
This is a reference-system problem, not a reason to distrust the calculator. The calculator evaluates the function entered. You decide which geometric angle that function represents and whether it answers the original bearing question. Keep that interpretation visible whenever several possible acute angles appear in the same diagram.
If a traveller turns right by 40° from an initial bearing of 300°, the new heading is 340° because right corresponds to clockwise rotation in the flat-plan view. A further right turn of 50° gives 390°, which is represented as 030° after removing one full turn.
Turning left means anticlockwise, so it decreases the clockwise bearing value before conversion to the standard range. Starting at 020° and turning left by 70° gives −50°, equivalent to 310° within one full turn.
These heading changes describe the new travel direction, not the bearing of the final destination from the original starting point. If the traveller completes several legs, the overall displacement depends on their lengths as well as their headings.
This distinction prevents a common overreach. You cannot average two headings to find the bearing of the endpoint without considering the geometry and distances. A sequence of turns describes how the route is travelled; the straight-line bearing from start to finish describes the resultant displacement. They are related, but they are not the same quantity.
A learner calculates the acute angle between a northeast journey and east as 28°, then writes the bearing as 028°. The trigonometry may be correct while the final conversion is wrong. Since the bearing starts from north, the required angle is 90 − 28 = 62°, written 062°.
The repair should preserve the valid triangle calculation and change the reference interpretation. Asking the learner to redo every line can hide the fact that the actual difficulty occurs only when moving from a triangle angle to a bearing.
A useful correction sentence is: my 28° was measured from east; the bearing is measured clockwise from north, so I need the complementary angle in this northeast quadrant. That sentence names the condition that makes subtraction from 90 appropriate.
Then vary the quadrant. If the ray is 28° south of east, the bearing is 118°, not 62°. This follow-up prevents a new shortcut from becoming another blind rule. The operation depends on where the ray lies and which reference direction the calculated angle uses.
Start with four rays in different quadrants and ask only for approximate bearing ranges. This establishes clockwise orientation without arithmetic pressure. Next label exact gaps from north, east, south or west and translate them into three-figure bearings.
Then introduce a viewpoint change on one segment. Ask for B from A and A from B, and require the student to mark each angle at its own starting point. Finish with two rays from one point and calculate their smaller included angle, including one pair that crosses north.
Only after these steps are stable should a longer trigonometry application be added. Otherwise the learner is trying to manage reference direction, viewpoint, angle transfer and side calculations simultaneously, making the first source of error harder to identify.
The session does not need many questions. It needs deliberate variation: different quadrants, a reversed segment, a crossing-north pair and a triangle where the bearing must be converted before a theorem applies. Ask for a brief explanation at each change so that a memorised numerical pattern cannot conceal an unstable concept.
For a broad overview of the topic’s mathematical purpose, visit The Core Aim of Mathematics Mastery: Bearings. It connects direction with parallel-line geometry, triangle reasoning and related representations. This guide provides the narrower reading discipline that makes those connections usable.
For an application perspective, read Marine Navigation, Bearings and Dead Reckoning. Keep its real-world context separate from the ideal flat diagrams used here. Operational navigation can involve reference systems, magnetic corrections, changing conditions and other requirements beyond a school exercise.
Do not add a fixed magnetic adjustment to these classroom answers unless the question explicitly supplies a reference and conversion. The north used in a mathematical diagram is the stated north reference. Introducing unstated real-world corrections would change the problem rather than improve its solution.
The connected learning move is to keep representations aligned. Bearings label directions, diagrams organise rays, geometry finds angles and trigonometry relates angles to lengths. Each stage must preserve which point and which reference the information belongs to.
Before calculation, identify the point after from and draw north there. Mark the target ray and the clockwise turn. During calculation, distinguish printed bearings from derived interior angles. After calculation, check the quadrant, the three-digit format and the requested precision.
For a return bearing, check that the new ray is opposite the original and that the two values differ by 180° within one turn. For an angle between two rays, check whether the smaller turn crosses north. For a triangle, label the interior angle before selecting a theorem.
For a calculated bearing from components or side lengths, compare the relative east-west and north-south displacements. A ray with much larger northward than eastward displacement should be closer to north than east in the northeast quadrant. This qualitative check can expose a complementary-angle mistake.
Finally reread the question’s endpoint names. A correctly calculated 215° can still be the answer to the reverse direction if the requested viewpoint was overlooked. The last check should return to the original wording, not merely confirm that the result is a plausible angle somewhere on the diagram.
If viewpoint is the difficulty, practise only of B from A with labelled rays before adding trigonometry. If clockwise direction is unstable, use quadrant sketches and a full-turn reference. If the triangle work fails, identify the interior angle separately from the bearing before choosing a sine, cosine or Pythagorean relationship.
If the learner consistently finds an accurate triangle angle but reports the wrong bearing, focus on the final reference conversion. Keep the valid earlier work. A precise repair is more useful than restarting a whole topic that is mostly understood.
Use the G1/G2/G3 Mathematics explainer directory to find related geometry and checking guides. The Clementi Secondary 1 Mathematics reference describes the broader teaching emphasis on clear structure and identifying the exact mathematical move that needs support.
Keep the core instruction short: stand at the point after from, begin at north, turn clockwise towards the target. That sentence preserves the meaning before any formula is used. Once it is dependable, the longer geometry becomes easier to organise and easier to check.
These answers concern the standard flat-plan convention used in school bearing questions.
Is a bearing always the smaller angle?
No. A bearing is the clockwise turn from north and may exceed 180°. The smaller angle between north and the ray may therefore be different. For example, a ray 30° west of north has bearing 330°, not 030°.
Where do I measure the bearing of A from B?
At B. Begin with north at B and turn clockwise towards the ray BA. The word from identifies the viewing point. Reversing the names reverses the direction along the segment and changes the bearing by a half-turn in the usual flat model.
Why is my trigonometric angle not the bearing?
It may be measured from east, south or a triangle side rather than from north. First identify exactly which angle your calculation found. Then use the diagram’s quadrant and reference lines to convert it into the requested clockwise-from-north angle.
Does a bearing tell me how far away a point is?
No. A bearing identifies a direction from a starting point. Many points at different distances can lie on that ray. A distance or other suitable constraint is needed to determine a particular location within the stated geometric model.