Lemmara logoLemmaraWorked solution

0580/43 May/Jun 2023, Question 5: The Sine Rule

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 12 marks

اقرأ هذا الحل مشروحاً بالعربي ←

The question, in short

Use a 3D setup with a vertical post and two wires to find lengths and a triangle area, then solve a triangle with a midpoint.

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

Part (ai) · 3 marks

Calculate side AB of triangle ABC using AC = 4.8 m, BC = 5.6 m and angle ACB = 20.4°.

  1. 1AB² = 4.8² + 5.6² − 2 × 4.8 × 5.6 × cos 20.4°
  2. 2AB² = 54.4 − 50.388… = 4.0117…
  3. 3AB = √4.0117… = 2.003 m

Answer2.00 or 2.002 to 2.003

Part (aii) · 2 marks

Find the post height AX using angle XBA = 64° and the length AB.

  1. 1Triangle XAB is right-angled at A
  2. 2tan 64° = AX/AB
  3. 3AX = 2.003 × tan 64° = 4.107 m

Answer4.10 or 4.11 or 4.100 to 4.107

Part (aiii) · 2 marks

Work out the area of triangle YAC given AY = 2.9 m and AC = 4.8 m.

  1. 1Triangle YAC is right-angled at A
  2. 2Area = 1/2 × 4.8 × 2.9
  3. 3= 6.96 m²

Answer6.96

Part (b) · 5 marks

Calculate PQ in triangle PQR where M is the midpoint, RM = 8 cm, angle PRM = 30° and angle RMQ = 75°.

  1. 1Exterior angle of triangle PRM: angle RPM = 75 − 30 = 45°
  2. 2Sine rule: PM/sin 30° = 8/sin 45°
  3. 3PM = 8 × sin 30°/sin 45° = 5.656…
  4. 4PQ = 2 × PM = 11.31… ≈ 11.3 cm

Answer11.3 or 11.31…

Stuck on a step? Read the free lesson: The Sine Rule

Practise more the sine rule questions

Real 0580 past-paper questions, marked on your working like the exam, with a worked solution for every part.

Practise free on Lemmara

More past-paper questions on the sine rule

Also searched as: 0580/43/M/J/23 · 0580_s23_qp_43 · 0580 s23 43 · question 5.
Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education. The worked solutions are Lemmara’s own.