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0580/42 Oct/Nov 2022, Question 7: Bearings with Sine and Cosine Rules

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 18 marks

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The question, in short

Use triangle PQR with sides PR = 39.4 cm, RQ = 38.2 cm, PQ = 46.5 cm, and a cuboid with AB = 29 cm, BC = 21 cm, CG = 20 cm, plus a lighthouse problem with KL = 112 km, to work out angles, lengths and a bearing.

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

Part (ai) · 4 marks

Work out the size of angle QPR in triangle PQR.

  1. 1cos QPR = (39.4² + 46.5² − 38.2²)/(2 × 39.4 × 46.5)
  2. 2= 0.6155…
  3. 3Angle QPR = 52.01… ≈ 52.0°

Answer52.0 or 52.01…

Part (aii) · 3 marks

Work out the perpendicular distance from Q to side PR.

  1. 1The shortest distance is the perpendicular from Q to PR
  2. 2d = 46.5 × sin 52.01…
  3. 3= 36.64… ≈ 36.6 cm

Answer36.6 or 36.64 to 36.65

Part (bi) · 3 marks

Work out the length of the space diagonal AG of the cuboid.

  1. 1AG² = 29² + 21² + 20² = 1682
  2. 2AG = √1682 = 41.01… ≈ 41.0 cm

Answer41.0 or 41.01…

Part (bii) · 3 marks

Work out the angle that AG makes with the base ABCD.

  1. 1The angle between AG and the base is angle GAC, with CG = 20 cm
  2. 2AC² = 29² + 21² = 1282
  3. 3AG = √(1282 + 20²) = √1682 = 41.01… cm
  4. 4sin GAC = 20 ÷ 41.01…
  5. 5Angle GAC = 29.18… ≈ 29.2°

Answer29.2 or 29.18 to 29.2

Part (c) · 5 marks

Using bearing 155° of L from K, bearing 010° of K from M and angle KML = 96°, work out the bearing and distance of M from L.

  1. 1Bearing of M from K = 010 + 180 = 190°, so angle MKL = 190 − 155 = 35°
  2. 2Angle MLK = 180 − 96 − 35 = 49°
  3. 3ML/sin 35 = 112/sin 96, so ML = 112 × sin 35 ÷ sin 96 = 64.59… km
  4. 4Bearing of K from L = 155 + 180 = 335°
  5. 5Bearing of M from L = 335 − 49 = 286°, distance ≈ 64.6 km

AnswerBearing 286, distance 64.6 or 64.59… km

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Also searched as: 0580/42/O/N/22 · 0580_w22_qp_42 · 0580 w22 42 · question 7.
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