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0580/42 May/Jun 2021, Question 6: The Sine Rule

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

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The question, in short

Using the given side lengths and angles in quadrilateral ABCD, work with triangles ABD and BCD to find BD, an angle, the total area, and a perpendicular distance.

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

Part (a) · 3 marks

Use triangle ABD with AB = 16 m, AD = 19 m and angle DAB = 57° to verify that BD rounds to 16.9 m.

  1. 1BD² = 16² + 19² − 2 × 16 × 19 × cos 57°
  2. 2BD² = 285.8…
  3. 3BD = 16.90… = 16.9 m (1 d.p.)

Answer√(16² + 19² − 2 × 16 × 19 cos 57) = 16.90 to 16.91

Part (b) · 4 marks

In triangle BCD, with BC = 32 m, BD from part (a) and angle BDC = 75°, determine angle CBD.

  1. 1sin BCD = 16.9 × sin 75° ÷ 32
  2. 2Angle BCD = 30.67…°
  3. 3Angle CBD = 180 − 75 − 30.67… = 74.32… ≈ 74.3°

Answer74.3 or 74.30 to 74.33

Part (c) · 3 marks

Combine the areas of triangles ABD and BCD to obtain the area of quadrilateral ABCD.

  1. 1Triangle ABD = 1/2 × 16 × 19 × sin 57° = 127.47…
  2. 2Triangle BCD = 1/2 × 16.9 × 32 × sin 74.32…° = 260.3…
  3. 3Area = 127.47… + 260.3… = 387.8… ≈ 388 m²

Answer388 or 387.7 to 387.9…

Part (d) · 3 marks

Determine the perpendicular distance from B to side AD.

  1. 1Shortest distance is perpendicular to AD: d/16 = sin 57°
  2. 2d = 16 × sin 57° = 13.41… ≈ 13.4 m

Answer13.4 or 13.41 to 13.42

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