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0580/42 May/Jun 2024, Question 6: The Cosine Rule

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 10 marks

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The question, in short

Quadrilateral ABCD with AB = 10.4 cm, AD = 6.5 cm, angle DAB = 64°, angle BDC = 26°, angle DBC = 42°. Find BD, angle BCD, CD, and the shortest distance from D to AB.

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

Part (a) · 3 marks

Use triangle ABD with sides AB = 10.4 cm, AD = 6.5 cm and included angle 64° to show BD = 9.55 cm to 2 decimal places.

  1. 1BD² = 10.4² + 6.5² − 2 × 10.4 × 6.5 × cos 64
  2. 2= 91.14…
  3. 3BD = 9.546… = 9.55 cm (2 d.p.)

Answer9.546 to 9.547

Part (bi) · 1 mark

Use the angles in triangle BDC to show that angle BCD = 112°.

  1. 1Angles in a triangle add to 180°
  2. 2Angle BCD = 180 − (26 + 42) = 112°

Answer180 − (26 + 42)

Part (bii) · 3 marks

Calculate the length CD in triangle BDC using BD = 9.55 cm and the given angles.

  1. 1CD/sin 42 = 9.55/sin 112
  2. 2CD = 9.55 × sin 42 / sin 112
  3. 3= 6.892… ≈ 6.89 cm

Answer6.89 or 6.888 to 6.892...

Part (c) · 3 marks

Find the shortest distance from D to line AB, using triangle ABD.

  1. 1Shortest distance is the perpendicular from D to AB
  2. 2h = 6.5 × sin 64
  3. 3= 5.842… cm

Answer5.842…

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Also searched as: 0580/42/M/J/24 · 0580_s24_qp_42 · 0580 s24 42 · question 6.
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