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0580/43 May/Jun 2026, Question 13: The Sine Rule

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 7 marks

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The question, in short

Triangle ABC has BC = 15 cm, AC = 15√3 cm and angle ABC = 120°. Show the triangle is isosceles, then find the shortest distance from B to AC.

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

Part (a) · 4 marks

Show that triangle ABC is isosceles using the given side lengths and the 120° angle.

  1. 1sin BAC/15 = sin 120°/(15√3)
  2. 2sin BAC = 15 × sin 120°/(15√3) = (√3/2)/√3 = 1/2
  3. 3Angle BAC = 30°
  4. 4Angle BCA = 180° − 120° − 30° = 30°
  5. 5Angle BAC = angle BCA, so triangle ABC is isosceles

Answersin BAC = 15 sin 120/(15√3) = 1/2, so angle BAC = 30°, angle BCA = 180 − 120 − 30 = 30°, two equal angles so isosceles

Part (b) · 3 marks

Calculate the shortest distance from B to the base AC.

  1. 1Shortest distance is the perpendicular from B to AC
  2. 2It meets AC at its midpoint, 15√3/2 from C
  3. 3Angle BCA = 30°, so d = 15 sin 30°
  4. 4d = 7.5 cm

Answer7.5

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