IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~6 min read

Bearings & Constructions

Bearings describe direction using angles. Three rules: measure from North, go clockwise, write in three figures (e.g., 045°). The maths is just the sine rule, cosine rule, and SOH CAH TOA — applied to triangles between locations.

📘 What you need to know

Three bearing rules: from North, clockwise, three figures.
Reverse bearing: from B back to A = bearing from A to B ± 180°.
Compass directions: N = 000° (or 360°), E = 090°, S = 180°, W = 270°.
Sketch first: draw a North line at every point, then mark the angles clockwise from it.
Convert to triangle angles using the geometry of the diagram — often via reverse bearings or angle-sum rules.
Apply sine/cosine rule to find missing distances; use sine rule to find missing internal angles, then convert back to a bearing.

Compass directions and reverse bearings

Direction	Bearing
Due North	000° (or 360°)
Due East	090°
Due South	180°
Due West	270°

Reverse bearing rule: if the bearing from A to B is less than 180°, add 180° to get the bearing from B to A. If it’s 180° or more, subtract 180°. Same line, opposite direction.

🧭 Recipe — every bearing problem

Sketch: draw all locations and the North line at each.
Mark known bearings and distances on the diagram.
Find the angle inside the triangle using reverse bearings, alternate angles, or angle-sum tricks.
Apply sine rule, cosine rule, or SOH CAH TOA to find the unknown distance or angle.
Convert back to a bearing if asked: measure clockwise from North.

Worked examples

WE 1

Find a reverse bearing

The bearing from town A to town B is 040°. Find the bearing from B to A.

Step 1: Same line, opposite direction → add 180° 040° + 180° = 220° bearing from B to A = 220° if the original bearing is < 180°, add 180°; if ≥ 180°, subtract 180°

WE 2

Two-leg journey — find total distance

A ship sails from port P for 10 km on a bearing of 070° to point Q, then for 6 km on a bearing of 160° to point R. Find the distance from P to R, correct to 3 s.f.

Step 1: Find the angle at Q inside triangle PQR reverse bearing Q→P = 070° + 180° = 250° forward bearing Q→R = 160° angle PQR = 250° − 160° = 90° Step 2: Right triangle → use Pythagoras PR² = 10² + 6² = 100 + 36 = 136 PR = √136 = 11.66… PR ≈ 11.7 km (3 s.f.) when bearings differ by exactly 90°, the triangle has a right angle

WE 3

Find a bearing from N–E displacements

A point Q is 5 km east and 3 km north of point P. Find the bearing of Q from P, correct to the nearest degree.

Step 1: Sketch — N is up, E is right Q is up-and-right of P → bearing is between 000° and 090° Step 2: Angle from North = arctan(east / north) tan θ = 5/3 = 1.667 θ = tan⁻¹(1.667) = 59.04° bearing of Q from P = 059° always write as 3 figures — leading zero is required

WE 4

Cosine rule with bearings

A boat sails from harbour H for 12 km on a bearing of 030° to buoy A. It then sails for 9 km on a bearing of 110° to buoy B. Find the distance HB, correct to 3 s.f.

Step 1: Find the angle at A inside triangle HAB reverse bearing A→H = 030° + 180° = 210° forward bearing A→B = 110° angle HAB = 210° − 110° = 100° Step 2: Apply cosine rule HB² = 12² + 9² − 2(12)(9) cos 100° HB² = 144 + 81 − 216 × (−0.1736…) HB² = 225 + 37.50… = 262.50… HB = √262.50… = 16.20… HB ≈ 16.2 km (3 s.f.)

WE 5

Two ships from a port

Two ships leave port P. Ship A travels at 20 km/h on a bearing of 050°. Ship B travels at 12 km/h on a bearing of 170°. Find the distance between the two ships after 1.5 hours.

Step 1: Distances after 1.5 hours PA = 20 × 1.5 = 30 km; PB = 12 × 1.5 = 18 km Step 2: Angle at P (between the two bearings) angle APB = 170° − 50° = 120° Step 3: Cosine rule AB² = 30² + 18² − 2(30)(18) cos 120° AB² = 900 + 324 − 1080 × (−0.5) AB² = 1224 + 540 = 1764 AB = √1764 = 42 distance = 42 km cos 120° = −½ — clean integer pops out

WE 6

Combined — find a distance and a bearing

Town A is due south of town B, with AB = 25 km. Town C is 40 km from B on a bearing of 110°. Find (a) the distance AC, and (b) the bearing of C from A, both correct to 3 s.f.

Step 1: Find the angle at B inside the triangle A is due south of B → bearing B→A = 180° bearing B→C = 110° angle ABC = 180° − 110° = 70° (a) Cosine rule for AC AC² = 25² + 40² − 2(25)(40) cos 70° AC² = 625 + 1600 − 2000 × 0.342… AC² = 2225 − 684.04… = 1540.96… AC = √1540.96… = 39.25… (a) AC ≈ 39.3 km (3 s.f.) (b) Find angle BAC using sine rule sin(BAC)/40 = sin 70°/39.25 sin(BAC) = 40 × 0.9397…/39.25 = 0.9577… angle BAC = sin⁻¹(0.9577…) = 73.2° Step 4: Convert to a bearing from A B is due north of A (bearing 000°) C is east of north → bearing of C from A = 000° + 73.2° (b) bearing of C from A = 073° cosine rule for the distance, sine rule for the angle, then translate the angle into a bearing using the North reference

💡 Top tips

Always draw a North line at each location. This makes converting between bearings and triangle angles much easier.
Three figures, always: 045° not 45°. The leading zero is part of the answer.
Big diagram — small ones get cluttered with bearings, distances, and North lines.
If bearings differ by exactly 90°, the triangle has a right angle — Pythagoras saves you the cosine rule.
Sine rule for angles inside the triangle, then add or subtract from a known bearing to get the final answer.

⚠ Common mistakes

Measuring anti-clockwise. Bearings always go clockwise from North.
Forgetting the leading zero. 45° on its own is not a valid bearing — write 045°.
Confusing “from A to B” with “from B to A”. They differ by 180°.
Not converting bearings to triangle angles. The sine and cosine rules need internal angles, not raw bearings.
Mixing up east/north when finding a bearing from displacements. East is the horizontal component (opposite to North), so tan(bearing) = east/north.

And that closes Section 3.2 — Trigonometry. You’ve now got the full real-world toolkit: Pythagoras & SOH CAH TOA, sine/cosine rules, area formula, elevation/depression, and bearings. The next sections of Topic 3 leave the triangle behind and move into the unit circle, trig identities, equations, and graphs — where trig functions are studied as functions in their own right rather than tools for measuring shapes.

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