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WMA11/01 Jun 2025, Question 6: Triangles and Sectors in Radians

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 9 marks

The question, in short

Helicopter landing pad made of major sector AOB (radius 15 m) and triangle AOC, with BC = 2 m and angle ACO = 0.6 radians. Find angle COA, total area and perimeter.

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

Part (a) · 3 marks

Show that angle COA equals 1.847 radians, correct to 3 decimal places.

  1. 1OC = 15 + 2 = 17 m
  2. 2Sine rule: sin CAO / 17 = sin 0.6 / 15
  3. 3sin CAO = 17 sin 0.6 / 15 = 0.6399…, so CAO = 0.6944…
  4. 4COA = π − 0.6 − 0.6944… = 1.847 radians (3 d.p.)

AnswerShown: sin CAO = 17 sin 0.6 / 15 gives CAO = 0.6944, so COA = π − 0.6 − 0.6944 = 1.847

Part (b) · 3 marks

Calculate the total area of the landing pad in m² to 3 significant figures.

  1. 1Major sector angle = 2π − 1.847 = 4.436 rad
  2. 2Major sector area = ½ × 15² × 4.436 = 499.1 m²
  3. 3Triangle AOC area = ½ × 15 × 17 × sin 1.847 = 122.7 m²
  4. 4Total area = 499.1 + 122.7 ≈ 622 m²

Answer622 m²

Part (c) · 3 marks

Calculate the perimeter of the landing pad in metres to 3 significant figures.

  1. 1Major arc AB = 15 × (2π − 1.847) = 66.54 m
  2. 2AC² = 15² + 17² − 2 × 15 × 17 × cos 1.847 = 653.1…, so AC = 25.56 m
  3. 3Perimeter = 66.54 + 25.56 + BC (2) = 94.1 m

Answer94.1 m

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