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4024/21 May/Jun 2020, Question 10: Choosing the Right Trigonometric Ratio

Cambridge O Level Mathematics 4024 · Paper 21 (Calculator) · Calculator · 13 marks

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The question, in short

Solve triangle and area problems for field ABCD with AB = 750 m, BC = 600 m, AC = 800 m, right angle DAC, and total area 375 000 m².

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

Part (a) · 3 marks

Use triangle ABC side lengths to show angle ACB is 62.9° to 1 decimal place.

  1. 1750² = 600² + 800² − 2 × 600 × 800 × cos ACB
  2. 2cos ACB = (600² + 800² − 750²)/(2 × 600 × 800) = 437 500/960 000 = 0.45572…
  3. 3Angle ACB = 62.888… = 62.9° (1 d.p.)

Answer62.88 to 62.89

Part (b) · 4 marks

Given field area 375 000 m², calculate length AD.

  1. 1Area ABC = 1/2 × 800 × 600 × sin 62.888… = 213 628…
  2. 2Area ACD = 375 000 − 213 628… = 161 371…
  3. 31/2 × 800 × AD = 161 371…
  4. 4AD = 161 371… ÷ 400 = 403.4 m

Answer403 or 403.3 to 403.5

Part (c) · 2 marks

Calculate angle ACD.

  1. 1Triangle ACD has a right angle at A.
  2. 2AD = 403.4 m (from part b) is opposite angle ACD; AC = 800 m is adjacent.
  3. 3tan ACD = 403.4/800 = 0.504…
  4. 4ACD = tan⁻¹(0.504…) = 26.76… ≈ 26.8°

Answer26.7 or 26.8 or 26.73 to 26.76

Part (d) · 4 marks

With X on DC and AX = 500 m, calculate the obtuse angle AXC.

  1. 1From part (b), AD = 403.4…, so tan ACD = 403.4… ÷ 800, giving angle ACD = 26.76…°
  2. 2Sine rule in triangle AXC: sin AXC/800 = sin 26.76…/500
  3. 3sin AXC = 800 × sin 26.76… ÷ 500 = 0.7204…, so the acute angle is 46.09…°
  4. 4Obtuse angle AXC = 180 − 46.09… = 133.9°

Answer133.8 to 134.0…

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