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0580/43 Oct/Nov 2023, Question 8: The Cosine Rule

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 17 marks

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The question, in short

For a shape built from major sector AOB plus triangles OBC and AOD, with OB = 6 cm, BC = 2 cm, angle AOC = 135° and right angle at C, work out angles, areas and a length.

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

Part (ai) · 2 marks

Demonstrate that the angle BOC equals 19.5° when rounded to one decimal place.

  1. 1sin BOC = BC/OB = 2/6
  2. 2Angle BOC = sin⁻¹(1/3) = 19.47… ≈ 19.5°

Answersin BOC = 2/6 leading to 19.47…

Part (aii) · 3 marks

Work out the area of the major sector AOB.

  1. 1In triangle OBC: sin BOC = 2/6, so angle BOC = 19.47…°
  2. 2Major sector angle = 360 − 135 − 19.47… = 205.52…°
  3. 3Area = 205.52…/360 × π × 6²
  4. 4= 64.56… ≈ 64.6 cm²

Answer64.6 or 64.55 to 64.58

Part (aiii) · 5 marks

Given C is midway along OD, work out the length AD.

  1. 1OC = √(6² − 2²) = √32 = 5.656…
  2. 2OD = 2 × 5.656… = 11.31…
  3. 3AD² = 6² + 11.31…² − 2 × 6 × 11.31… × cos 135° = 36 + 128 + 96 = 260
  4. 4AD = √260 = 16.12… ≈ 16.1 cm

Answer16.1 or 16.10 to 16.13

Part (aiv) · 4 marks

Work out the combined area of the whole shape.

  1. 1Triangle AOD = 1/2 × 6 × 11.31… × sin 135° = 24
  2. 2Triangle OBC = 1/2 × 5.656… × 2 = 5.656…
  3. 3Total = 64.56… + 24 + 5.656… = 94.22… ≈ 94.2 cm²

Answer94.2 or 94.3 or 94.15 to 94.27…

Part (b) · 3 marks

A sector with radius 8 cm has area 160 cm²; find the area of a similar sector with radius 20 cm.

  1. 1Length scale factor = 20 ÷ 8 = 2.5
  2. 2Area scale factor = 2.5² = 6.25
  3. 3Area = 160 × 6.25 = 1000 cm²

Answer1000

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Also searched as: 0580/43/O/N/23 · 0580_w23_qp_43 · 0580 w23 43 · question 8.
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