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0580/43 Oct/Nov 2024, Question 12: The Ambiguous Case of the Sine Rule

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 10 marks

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The question, in short

Quadrilateral ABCD with E on CD; given AB = 8.7 cm, BC = 11.4 cm, CE = 10.9 cm, angle ADE = 90°, angle ABC = 119°, angle CAE = 20°. Find AC, angle ACE, and perimeter of ABCD.

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Part (a) · 3 marks

Use triangle ABC with AB = 8.7 cm, BC = 11.4 cm and angle ABC = 119° to show AC = 17.37 cm to 2 decimal places.

  1. 1AC² = 8.7² + 11.4² − 2 × 8.7 × 11.4 × cos119°
  2. 2AC² = 301.81...
  3. 3AC = 17.372... = 17.37 cm

Answer√(8.7² + 11.4² − 2 × 8.7 × 11.4 cos119) = 17.372 to 17.373

Part (b) · 4 marks

Given angle AEC is obtuse, use triangle ACE with AC, CE = 10.9 cm and angle CAE = 20° to calculate angle ACE.

  1. 1sin AEC = 17.37 × sin20°/10.9 = 0.5450...
  2. 2Acute angle = 33.03°, so obtuse AEC = 180 − 33.03 = 146.97°
  3. 3Angle ACE = 180 − 20 − 146.97
  4. 4= 13.03°

Answer13.0 or 13 or 13.02 to 13.03

Part (c) · 3 marks

Calculate the perimeter of quadrilateral ABCD, using the right angle at D and lengths found.

  1. 1AD = 17.37 × sin13.03° = 3.917... cm
  2. 2CD = 17.37 × cos13.03° = 16.92... cm
  3. 3Perimeter = 8.7 + 11.4 + 16.92... + 3.917...
  4. 4= 40.9 cm

Answer40.9 or 40.91 to 40.94

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Also searched as: 0580/43/O/N/24 · 0580_w24_qp_43 · 0580 w24 43 · question 12.
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