EST. 2024 · LONDON·MMXXVI SPECIFICATION
AQA·Edexcel·OCR|Foundation + Higher
GeometryFoundation & HigherTopic 166 of 245

Bearings and Trigonometry –

GCSEMathsAI Team·8 min read·23 May 2026

Bearings combine angle work with trigonometry and are a favourite topic on GCSE papers. You need to know how to measure, calculate, and describe bearings, and at Higher tier you must use trigonometry to find distances and angles in bearings problems.

What Are Bearings?

A bearing is a direction measured as an angle clockwise from north. Bearings are always written as three figures — for example, due east is 090°, due south is 180°, and northwest is 315°. If the angle is less than 100°, a leading zero is added (e.g., 045° rather than 45°).

Bearings are used in navigation, map reading, and real-world distance calculations. GCSE questions typically involve two or three locations and ask you to find a bearing between them or use trigonometry to calculate a distance.

The three rules for bearings are: always measure from north, always measure clockwise, and always give three figures.

In exam diagrams, north is always straight up the page unless otherwise stated. All north lines are parallel to each other.

Key Formulas

Return bearing = original bearing + 180° (if result > 360°, subtract 360°)
SOHCAHTOA for right-angled triangle problems: sin = opp/hyp, cos = adj/hyp, tan = opp/adj

Step-by-Step Method

  1. Draw a clear north line at the starting point.
  2. Measure the angle clockwise from the north line to the line joining the two points.
  3. Write the bearing as three figures. For calculation problems, sketch the triangle formed and use trigonometry to find missing sides or angles.

Worked Example 1 — Foundation Level

Question: Town B is due east of Town A. What is the bearing of B from A?

Working: North is straight up. Due east is a 90° clockwise rotation from north.

Answer: 090°

Worked Example 2 — Higher Level

Question: A ship sails from port P on a bearing of 060° for 80 km to reach point Q. It then sails due south for 50 km to reach point R. Find the direct distance from P to R to 1 decimal place.

Working: From the diagram, the angle at Q between the north line and PQ is 060°. Since QR is due south, the angle PQR (inside the triangle at Q) is 180° − 60° = 120°. Using the cosine rule: PR² = 80² + 50² − 2(80)(50)cos 120° PR² = 6400 + 2500 − 8000 × (−0.5) PR² = 8900 + 4000 = 12900 PR = √12900 = 113.6 (1 d.p.)

Answer: 113.6 km

Worked Example 3 — Exam Style

Question: The bearing of B from A is 125°. Find the bearing of A from B.

Working: The return bearing = 125° + 180° = 305°. Check: 305° is between 0° and 360°, so this is the final answer.

Answer: 305°

Common Mistakes

  • Forgetting three figures. A bearing of 70° must be written as 070°. Losing the leading zero costs marks.
  • Measuring anticlockwise. Bearings are always measured clockwise from north, never anticlockwise.
  • Drawing north in the wrong place. You must draw a north line at the point you are measuring from, not the point you are measuring to.
  • Incorrect return bearings. To find the return bearing, add 180° to the original bearing. If the result exceeds 360°, subtract 360°. Do not simply subtract from 360°.

Exam Tips

  • Always draw a clear north arrow at each point in the problem — this makes angle identification much easier.
  • Label all known angles and sides before starting any calculation.
  • For Higher tier, be prepared to use the sine rule or cosine rule in non-right-angled bearing triangles.
  • State the three-figure bearing clearly in your final answer.
  • Use co-interior angles (add to 180°) and alternate angles to find unknown angles in bearing diagrams with parallel north lines.
  • A return journey always involves the return bearing — practise calculating these quickly.

Practice Questions

Q1 (Foundation): The bearing of Y from X is 210°. Find the bearing of X from Y.

Answer: 210° + 180° = 390°. 390° − 360° = 030°.

Q2 (Foundation): Town B is directly southwest of Town A. What is the bearing of B from A?

Answer: Southwest is halfway between south (180°) and west (270°), so the bearing is 225°.

Q3 (Higher): A boat sails from A on a bearing of 040° for 12 km to B. From B it sails on a bearing of 130° for 9 km to C. The angle ABC is 90°. Find the distance AC.

Answer: AC² = 12² + 9² = 144 + 81 = 225. AC = √225 = 15 km.

Practise bearings and trigonometry questions with instant feedback — completely free on GCSEMathsAI.

Summary

  • A bearing is measured clockwise from north and always given as three figures.
  • Return bearing = original bearing + 180° (subtract 360° if the result exceeds 360°).
  • Always draw a north line at the point you are measuring from.
  • Use SOHCAHTOA for right-angled bearing problems and the sine/cosine rules for non-right-angled ones.
  • Label angles and distances clearly on your diagram before calculating.
  • North lines at different points are parallel — use parallel line angle rules to find angles within triangles.

Test your understanding

5 quick MCQs to identify any misconceptions on this topic.

Take Diagnostic Quiz
§Academic References

Further reading from leading academic institutions — free and open-access.

N
TrigonometryNRICH

Cambridge problems on trigonometric ratios and applications.

University of Cambridge · Free · Open Access
C
TrigonometryCorbett Maths

SOHCAHTOA, sine rule, cosine rule — full GCSE coverage.

Corbett Maths · Free · Open Access
M
TrigonometryMIT OpenCourseWare

MIT trigonometric functions and their applications.

Massachusetts Institute of Technology · Free · Open Access
C
Bearings & ConstructionsCorbett Maths

Three-figure bearings, constructions, and loci.

Corbett Maths · Free · Open Access
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