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0580/42 May/Jun 2025, Question 20: The Cosine Rule

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 13 marks

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The question, in short

Quadrilateral ABCD split by diagonal DB (12 m) with CD 10 m, angles 56° and 34° and a right angle at B: find a side, the area, the perimeter and a shortest distance.

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

Part (a) · 2 marks

Right-angled triangle ABD: find AB.

  1. 1tan 34° = AB/12
  2. 2AB = 12 × tan 34° = 8.094… ≈ 8.09 m

Answer8.09 or 8.094…

Part (b) · 3 marks

Add the areas of the two triangles.

  1. 1Triangle BCD = 1/2 × 10 × 12 × sin 56° = 49.74…
  2. 2In right triangle ABD, AB = 12 × tan 34° = 8.094…
  3. 3Triangle ABD = 1/2 × 12 × 8.094… = 48.56…
  4. 4Area = 49.74… + 48.56… = 98.30… ≈ 98.3 m²

Answer98.3 or 98.28 to 98.31

Part (c) · 5 marks

Find the missing sides, then add all four.

  1. 1BC² = 10² + 12² − 2 × 10 × 12 × cos 56° = 109.79…, so BC = 10.478…
  2. 2AB = 12 × tan 34° = 8.094…
  3. 3AD = 12 ÷ cos 34° = 14.474…
  4. 4Perimeter = 10 + 10.478… + 8.094… + 14.474… = 43.04…
  5. 5Perimeter ≈ 43.0 m

Answer43.0 to 43.1

Part (d) · 3 marks

Perpendicular distance from B to AD.

  1. 1The shortest distance is the perpendicular from B to AD
  2. 2sin 34° = distance/12
  3. 3Distance = 12 × sin 34° = 6.710… ≈ 6.71 m

Answer6.71 or 6.706 to 6.710…

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Also searched as: 0580/42/M/J/25 · 0580_s25_qp_42 · 0580 s25 42 · question 20.
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