Constructions, Loci and Bearings: Edexcel 4MA1 Past Paper Questions and Worked Solutions
This page collects every constructions, loci and bearings question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 3 questions worth 9 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.
Using ruler and compasses only, construct the bisector of angle ABC. You must show all your construction lines.
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- Put the compass point on B and draw an arc that crosses both arms. Label the crossing points P (on BA) and Q (on BC).
- Keep the same radius (or choose a new one that is more than half of PQ). Put the compass point on P and draw an arc between the arms. Do the same from Q, so the two arcs cross at R.
- Draw a straight line from B through R. This line splits angle ABC into two equal angles.
Answer: The line from B through R, with all arcs left showing
Use ruler and compasses only to construct the perpendicular bisector of line AB. You must show all your construction lines.
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- Open the compasses to more than half the length of AB.
- With the point on A, draw arcs above and below the line. Keep the same radius and repeat from B, so the arcs cross in two places.
- Join the two crossing points with a straight line. This line cuts AB in half at right angles.
Answer: A straight line through the two arc crossings, cutting AB at right angles through its midpoint
The diagram shows the positions of three ships, A, B and C. Ship B is due north of ship A, 150 m away. Ship C is 275 m from ship A on a bearing of 120°. Calculate the bearing of ship C from ship B. Give your answer correct to the nearest degree.
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- In triangle ABC, the angle at A between north (AB) and AC is 120°.
- Cosine rule: BC2 = 1502 + 2752 − 2 × 150 × 275 × cos 120° = 22 500 + 75 625 + 41 250 = 139 375, so BC = 373.32… m
- Sine rule for angle ABC: sin B = 275 × sin 120°373.32… = 0.6379…, so angle ABC = 39.63…° (it must be acute because the angle at A is obtuse)
- The bearing from B is measured clockwise from north. At B, north points away from A, so the line BA points due south (180°). C is 39.63…° round from south towards east, so the bearing = 180° − 39.63…°
Answer: 140°
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