Lemmara logoLemmaraWorked solution

0580/42 May/Jun 2026, Question 20: Shortest Distance from a Point to a Line

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 10 marks

اقرأ هذا الحل مشروحاً بالعربي ←

The question, in short

Use the field diagram with footpath AC and given lengths and angles to find a bearing, a distance CD, and a perpendicular distance from B to AC.

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

Part (a) · 1 mark

Find the bearing of A from B using the North line at B and the given angle.

  1. 1The 103° is measured anticlockwise from North to BA
  2. 2Bearing = 360° − 103° = 257°

Answer257

Part (b) · 3 marks

Calculate the length CD using the given sides and angle in triangle ACD.

  1. 1CD² = 38.2² + 86.8² − 2 × 38.2 × 86.8 × cos 79°
  2. 2= 1459.24 + 7534.24 − 1265.36… = 7728.1…
  3. 3CD = 87.9 m

Answer87.9 or 87.90 to 87.91

Part (c) · 6 marks

Calculate the shortest distance from B to the footpath AC.

  1. 1sin ACB = 50.3 × sin 112° ÷ 86.8 = 0.5372…
  2. 2angle ACB = 32.49…°
  3. 3angle BAC = 180° − 112° − 32.49…° = 35.51…°
  4. 4Distance = 50.3 × sin 35.51…°
  5. 5= 29.2 m

Answer29.2 or 29.20 to 29.21

Stuck on a step? Read the free lesson: Shortest Distance from a Point to a Line

Practise more shortest distance from a point to a line 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 shortest distance from a point to a line

Also searched as: 0580/42/M/J/26 · 0580_s26_qp_42 · 0580 s26 42 · question 20.
Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education. The worked solutions are Lemmara’s own.