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0580/42 Oct/Nov 2024, Question 7: The Cosine Rule

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 20 marks

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The question, in short

Solve triangle and area problems for field ABCD with given sides and angles, then find a rectangle width upper bound from rounded area and length.

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

Part (ai) · 1 mark

Use the given triangle BCD data to show that angle CBD equals 81 degrees.

  1. 1Angles in triangle BCD add to 180°
  2. 2Angle CBD = 180 − 60 − 39 = 81°

Answer180 − 60 − 39 = 81

Part (aii) · 3 marks

Calculate the length CD using triangle BCD with angle BDC 39 degrees and angle BCD 60 degrees.

  1. 1CD/sin 81° = 129/sin 60°
  2. 2CD = 129 × sin 81° ÷ sin 60°
  3. 3CD = 147.1… ≈ 147 m

Answer147 or 147.1…

Part (aiii) · 4 marks

Show that angle ABD is 31.6 degrees to 1 decimal place using triangle ABD.

  1. 172² = 85² + 129² − 2 × 85 × 129 × cos ABD
  2. 2cos ABD = (85² + 129² − 72²)/(2 × 85 × 129) = 18 682/21 930
  3. 3cos ABD = 0.8518…
  4. 4Angle ABD = 31.58… = 31.6° (1 d.p.)

Answer31.58…

Part (aiv) · 3 marks

Find the shortest distance from A to the path BD.

  1. 1The shortest distance is the perpendicular from A to BD
  2. 2d = 85 × sin 31.6°
  3. 3d = 44.5 m

Answer44.5 or 44.51 to 44.54

Part (av) · 2 marks

Find the bearing of B from C.

  1. 1Triangle BCD: angle DBC = 180 − 60 − 39 = 81°
  2. 2Cosine rule: cos ABD = (85² + 129² − 72²)/(2 × 85 × 129), so ABD = 31.6°
  3. 3North at B is straight on from AB, so NBC = 180 − 81 − 31.6 = 67.4°
  4. 4Bearing of B from C = 180 + 67.4 = 247.4… ≈ 247°

Answer247 or 247.4…

Part (avi) · 4 marks

Using 1100 trees per hectare and 1 hectare = 10000 m², calculate total trees in the field.

  1. 1Area ABD = 1/2 × 85 × 129 × sin 31.58° = 2871.3 m²
  2. 2Area BCD = 1/2 × 129 × 147.12 × sin 39° = 5971.9 m²
  3. 3Total area = 8843.2 m² = 0.884 32 hectares
  4. 4Trees = 8843.2/10 000 × 1100 = 972.7…, so 972 or 973 trees

Answer972 or 973

Part (b) · 3 marks

Given area 9400 cm² to nearest 100 and length 80 cm to nearest 10, calculate the upper bound of the width.

  1. 1Upper bound of area = 9400 + 50 = 9450 cm²
  2. 2Lower bound of length = 80 − 5 = 75 cm
  3. 3Upper bound of width = 9450 ÷ 75 = 126 cm

Answer126

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Also searched as: 0580/42/O/N/24 · 0580_w24_qp_42 · 0580 w24 42 · question 7.
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