IB Maths AI HLTrigonometryPaper 1 & 2Bearings & diagrams~9 min read
Bearings & Constructions
A bearing is a direction measured clockwise from north, always written as three digits (e.g. 075°, 230°). Most bearings questions reduce to a triangle problem: draw a sketch, work out the interior angle at the turning point, then apply the cosine or sine rule. The construction step (turning bearings into a labelled diagram) is half the work.
📘 What you need to know
Three rules for bearings: (1) measured from north, (2) measured clockwise, (3) always written as three digits (e.g. 030°, not 30°).
Compass bearings: north = 000°/360°, east = 090°, south = 180°, west = 270°.
Reverse bearing: the bearing of A from B is the bearing of B from A ± 180°.
Always draw a north arrow at every turning point — the new bearing is measured from the new north line, not the original one.
Interior triangle angle: subtract bearings (or use 180° − difference) to find the angle the journey turns through at each vertex.
Combine with sine/cosine rule: once you have the triangle’s sides and one interior angle, the rest is standard non-right trigonometry.
From bearings to triangle angles
The trick to a bearings question is to convert the compass information into an interior triangle angle. At a turning point Q with an incoming leg PQ and an outgoing leg QR, the angle inside the triangle is the angle PQR formed at Q. Find it by taking the reverse bearing of PQ (= bearing of PQ + 180°, mod 360°) and subtracting the bearing of QR (or vice versa — take the positive answer). Once the interior angle is in hand, the question is just a sine/cosine/area-of-triangle problem.
At every turning point, draw a fresh north arrow and measure the new bearing from it. The interior triangle angle at the turn is found by subtracting the new bearing from the reverse of the incoming one.
Interior angle at a turn
∠PQR = (bearing of PQ + 180°) − (bearing of QR) (or take 360° − result if negative)
reverse bearing: bearing of A from B = (bearing of B from A) ± 180°
Constructing the diagram
Most bearing questions start with a description (e.g. “B is 12 km from A on a bearing of 075°; C is 9 km from B on a bearing of 200°”). Translate the description step-by-step: place the first point, draw a north arrow there, then construct each leg using its stated bearing. Add a fresh north arrow at every new point because each bearing is measured from that point’s own north line. Sketches don’t need to be to scale, but distances and angles roughly proportional make labelling much easier.
Convert bearing to compass direction quickly: 000°–090° is in the NE quadrant; 090°–180° is SE; 180°–270° is SW; 270°–360° is NW. This sanity-check confirms your sketch is in the right region of the plane.
🧭 Recipe — bearings & triangle solving
Sketch each leg with a fresh north arrow at every turning point.
Compute the interior angle at the turning point: reverse the incoming bearing (add 180°), subtract the outgoing one.
Pick the rule: two sides + included angle → cosine rule; opposite pair → sine rule; area? use ½absinC.
Solve the triangle for the unknown distance or angle.
Convert back to a bearing if the question asks for one: add or subtract from a known reference bearing using your diagram.
Worked examples
WE 1
Components & reverse bearing
A ship sails 18 km from port P on a bearing of 060° to point Q. Find: (a) how far north of P the ship is at Q; (b) how far east of P the ship is at Q; (c) the bearing of P from Q. Give distances to 3 s.f.
(a) north componentN = 18 · cos(60°) = 18 × 0.5N = 9.00 km(b) east componentE = 18 · sin(60°) = 18 × 0.8660 = 15.588…E ≈ 15.6 km (3 s.f.)(c) reverse bearing060° + 180° = 240°bearing of P from Q = 240°bearing measured clockwise from N, so cos × distance gives the N-leg, sin × distance gives the E-leg.
WE 2
Cosine rule — two-leg sailing journey
A yacht sails 14 km from port P on a bearing of 075° to point Q. From Q it sails a further 20 km on a bearing of 145° to point R. Find the direct distance PR, to 3 s.f.
Two coastguard stations A and B lie on a coastline running east-west, with B 25 km due east of A. A boat C is observed on a bearing of 040° from A and on a bearing of 305° from B. Find: (a) the distance AC, (b) the distance BC. Give answers to 3 s.f.
interior angles inside triangle ABCat A: AB is east (090°), AC is 040° → ∠A = 090° − 040° = 50°at B: BA is west (270°), BC is 305° → ∠B = 305° − 270° = 35°∠C = 180° − 50° − 35° = 95°(a) sine rule for AC (opp B)AC / sin(35°) = 25 / sin(95°)AC = 25 · sin(35°) / sin(95°) = 14.39…AC ≈ 14.4 km (3 s.f.)(b) sine rule for BC (opp A)BC = 25 · sin(50°) / sin(95°) = 19.22…BC ≈ 19.2 km (3 s.f.)
WE 4
Aircraft — cosine rule
An aircraft flies from airfield A on a bearing of 080° for 45 km to checkpoint B. It then turns and flies on a bearing of 210° for 60 km to landing site C. Find the direct distance AC, to 3 s.f.
From port P, a ship sails 16 km on a bearing of 030° to point Q. From Q, the ship turns and sails 22 km due east to point R. Find: (a) the direct distance PR, to 3 s.f.; (b) the bearing of R from P, to the nearest degree.
(a) interior angle at Q∠PQR = (30° + 180°) − 90° = 120°PR² = 16² + 22² − 2(16)(22)cos(120°) = 256 + 484 − 704 × (−0.5) = 740 + 352 = 1092PR = √1092 = 33.04…PR ≈ 33.0 km (3 s.f.)(b) angle QPR via sine rulesin(∠QPR) / 22 = sin(120°) / 33.04sin(∠QPR) = 22 · sin(120°) / 33.04 = 0.5766∠QPR = sin⁻¹(0.5766) = 35.21°bearing from Pbearing of PQ = 030°; PR sits 35.21° clockwise of PQbearing of R from P = 030° + 35.21° = 65.21°≈ 065° (nearest degree)add the interior angle ∠QPR to the bearing of the first leg to get the bearing of R.
WE 6
Distance & area — observation towers
Three observation towers L, M and N lie on flat ground. M is 12 km from L on a bearing of 075°. N is 9 km from M on a bearing of 200°. Find: (a) the distance LN; (b) the area of triangle LMN. Give answers to 3 s.f.
Three digits always — a bearing of 30° must be written 030°.
New north at every turn — the new bearing is measured from the fresh north arrow, not from the previous leg.
Reverse-bearing trick: bearing of A from B = (bearing of B from A) + 180°. If the result exceeds 360°, subtract 360°.
Interior angle: for a turn at Q, ∠PQR = (incoming reverse bearing) − (outgoing bearing). If negative, add 360°.
Sketch first, calculate second — even a rough diagram avoids most quadrant mistakes.
⚠ Common mistakes
Measuring bearings anticlockwise or from south — always clockwise from north.
Using the original north at a turning point — each new bearing is measured from a fresh north line at that point.
Writing two-digit bearings like “30°” instead of “030°” — can cost a mark in the exam.
Confusing bearing of A from B with bearing of B from A — they differ by 180°.
Forgetting to convert the answer back to a bearing when the question asks for one (e.g. computing an interior angle and stopping there).
Chapter complete — you now have all four Trigonometry sub-topics: Pythagoras & right-angled trig, Sine/Cosine/Area rules, Angles of Elevation & Depression, and Bearings & Constructions. Together they cover every triangle-based problem in AI HL.
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