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Lesson overview

Geometry — navigation and trigonometry

A bearing is the angle measured clockwise from North. It is always written as a three-figure number (e.g., 120° or 030°). To solve bearing problems: (1) sketch the diagram carefully, showing North, the relevant points and the angles; (2) identify the triangle(s) formed; (3) use trigonometry (sine rule, cosine rule, or basic SOHCAHTOA) to find unknown distances or angles; (4) work back to find the bearing if needed, remembering that bearing is always measured clockwise from North. Multi-step problems require patience with diagram work and careful tracking of known information.

Understanding bearings

Town B is on a bearing of 120° from town A and 8 km away. Describe the position of B relative to A.

Bearings range from 000° (North) to 360°. 090° is East, 180° is South, 270° is West.

Drawing a bearing diagram for a triangle problem

Ship A is at the origin. Ship B is 5 km away on a bearing of 060°. Ship C is 7 km from A on a bearing of 150°. Find the distance from B to C.

Draw North arrows at A. Measure angles clockwise from North. The angle between two bearings is their difference (or 360° minus the difference if needed).

Finding a bearing from an angle

In the triangle ABC (from above), find the bearing of C from B.

The bearing from B back to A is opposite (180° different). Adjust by the angle at B to find the bearing to C.

Worked example — two-stage navigation problem

A ship travels from A to B, a distance of 12 km on a bearing of 040°. From B, it travels to C, a distance of 15 km on a bearing of 130°. Find the direct distance from A to C.

Draw the diagram with North arrows at both A and B.
The bearing from A to B is 040°, so the bearing from B back to A is 040° + 180° = 220°.
The bearing from B to C is 130°.
The angle at B is 220° − 130° = 90° (the difference between these two bearings).
We now have a right-angled triangle with AB = 12 km, BC = 15 km, and angle B = 90°.
Use Pythagoras: ${AC}^{2}$ = ${12}^{2}$ + ${15}^{2}$ = 144 + 225 = 369.
$AC = \sqrt{369} ≈ 19.2 km.$
The direct distance from A to C is approximately 19.2 km.

Try it

Draw the diagram carefully with North arrows. Identify the angles between bearings. Use trigonometry (right-angled or general triangles) to solve.

Question 1

Town B is 10 km away from town A on a bearing of 050°. What is the bearing of A from B?

💡 The reverse bearing is 180° away: 050° + 180° = 230°.

Question 2

From a point, location X is 8 km away on a bearing of 090° (due East). What is the bearing of the starting point from X?

💡 The reverse bearing: 090° + 180° = 270° (due West).

Question 3

Ship P is at the origin. Q is 6 km away on bearing 070°. R is 8 km away on bearing 160°. Find angle QPR.

💡 Angle = 160° − 070° = 90°.

Question 4

A boat travels 5 km on bearing 045°, then 7 km on bearing 315°. Which trigonometric approach is most suitable?

💡 Bearings 045° and 315° differ by 90° − 45° = 90°. This creates a right angle at the turn point.

Question 5

A plane flies 200 km from A to B on bearing 120°. From B, it flies 150 km to C on bearing 040°. Approximately how far is C from A (use cosine rule)?

💡 Angle at B = |120° + 180° − 040°| = 260° or 100°. Use the smaller angle (100°) with cosine rule.

Common mistakes

Watch for these when working through the lesson.

Forgetting that bearing is measured clockwise from North, not from the previous direction.
Confusing the angle in the triangle with the bearings. The angle at a vertex is the difference between two reverse bearings, adjusted carefully.
Not drawing the diagram. Bearing problems are nearly impossible without a clear sketch showing North at each location.
Using the wrong bearing for the reverse journey. The bearing from B to A is NOT the same as the bearing from A to B.

Bearings and trigonometric problem solving

Lesson overview

Worked example — two-stage navigation problem

Try it

Common mistakes

Related topics

Independent practice will plug in here