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0580/43 May/Jun 2024, Question 6: The Sine Rule

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 15 marks

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The question, in short

Right-angled and general triangles: a ladder length by Pythagoras, a bearing from north and east distances, a kite-shaped perimeter, and an angle across a quadrilateral's diagonal.

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

Part (a) · 2 marks

Ladder with foot 1.5 m from the wall reaching 4 m up: find its length.

  1. 1GH² = 4² + 1.5² = 16 + 2.25 = 18.25
  2. 2GH = √18.25 = 4.27 m

Answer4.27 or 4.272…

Part (b) · 3 marks

W is 120 m north and 50 m east of V: bearing of V from W.

  1. 1Angle at W: tan θ = 50/120
  2. 2θ = 22.6°
  3. 3V is south-west of W: bearing = 180 + 22.6 = 202.6 ≈ 203°

Answer203 or 202.6…

Part (c) · 5 marks

AD = DC = 5 cm, AB = BC, angles 25° and 15°: find the perimeter.

  1. 1Angle ADB = 180 − 25 − 15 = 140°
  2. 2AB/sin 140 = 5/sin 25
  3. 3AB = 5 sin 140/sin 25 = 7.604… cm
  4. 4Perimeter = 2 × 7.604… + 5 + 5 = 25.2 cm

Answer25.2 or 25.20 to 25.21

Part (d) · 5 marks

Use triangle PSR (8 cm, 11 cm, 110°) to get PR, then find angle PQR with PQ 10 and QR 14 cm.

  1. 1PR² = 11² + 8² − 2 × 11 × 8 × cos 110° = 245.19…
  2. 2cos PQR = (10² + 14² − 245.19…)/(2 × 10 × 14) = 0.1814…
  3. 3Angle PQR = 79.5°

Answer79.5 or 79.6 or 79.54 to 79.55…

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Also searched as: 0580/43/M/J/24 · 0580_s24_qp_43 · 0580 s24 43 · question 6.
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