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4024/22 Oct/Nov 2025, Question 22: The Sine Rule

Cambridge O Level Mathematics 4024 · Paper 22 (Calculator) · Calculator · 10 marks

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The question, in short

For quadrilateral field ABCD with right angle at A and diagonal path BD, use given side lengths and angle ABD to find AB, verify an angle in triangle BCD, then find Lara's average speed along the perpendicular path CX in km/h.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 2 marks

Using right-angled triangle ABD with BD = 495 m and angle ABD = 42°, work out the length AB.

  1. 1cos 42° = AB/495
  2. 2AB = 495 × cos 42° = 367.85… ≈ 368 m

Answer368 or 367.8 to 367.9

Part (b) · 3 marks

In triangle BCD with DC = 620 m, BC = 580 m and BD = 495 m, prove that angle BDC equals 61.5° to one decimal place.

  1. 1580² = 620² + 495² − 2 × 620 × 495 × cos BDC
  2. 2cos BDC = (620² + 495² − 580²)/(2 × 620 × 495) = 293 025/613 800 = 0.4773…
  3. 3Angle BDC = 61.48… = 61.5° (1 d.p.)

Answer61.48…

Part (c) · 5 marks

CX is the perpendicular distance from C to BD; given Lara's time of 6 minutes 24 seconds along CX, calculate her average speed in kilometres per hour.

  1. 1CX is perpendicular to BD, so sin 61.5° = CX/620
  2. 2CX = 620 × sin 61.5° = 544.86… m
  3. 3Time = 6 min 24 s = 6.4 minutes = 6.4/60 hours
  4. 4Speed = 0.54486… km ÷ (6.4/60) h = 5.108…
  5. 5≈ 5.11 km/h

Answer5.11 or 5.105 to 5.108…

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