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0580/42 Oct/Nov 2021, Question 3: The Cosine Rule

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 12 marks

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The question, in short

Solve problems on pentagon ABCDE using given side lengths and angles, and find a distance in triangle PQR with area equal to triangle LMN.

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

Part (ai) · 2 marks

Use triangle ADE with DE = 46.1 m, angle DEA = 64° and a right angle at D to calculate AD, showing it rounds to 94.5 m.

  1. 1Triangle ADE is right-angled at D, so tan 64° = AD/46.1
  2. 2AD = 46.1 × tan 64° = 94.519… ≈ 94.5 m

Answer94.51 to 94.52

Part (aii) · 3 marks

In triangle ABC with BC = 56.5 m, AC = 78.4 m and angle ABC = 94°, calculate angle BAC.

  1. 1sin BAC/56.5 = sin 94°/78.4
  2. 2sin BAC = 56.5 × sin 94° ÷ 78.4 = 0.7189…
  3. 3Angle BAC = 45.96… ≈ 46.0°

Answer46.0 or 45.96…

Part (aiii) · 4 marks

Using AD from part (i), CD = 38.6 m and AC = 78.4 m, find the largest angle in triangle CAD.

  1. 1In right-angled triangle ADE: AD = 46.1 × tan 64° = 94.51… m
  2. 2AD is the longest side, so the largest angle is ACD, opposite AD
  3. 3cos ACD = (38.6² + 78.4² − 94.51…²) ÷ (2 × 38.6 × 78.4) = −0.2143…
  4. 4Angle ACD = 102.37… ≈ 102.4°

Answer102.3 or 102.4 or 102.34 to 102.38

Part (b) · 3 marks

Find the area of triangle LMN from LM = 21.5 cm, MN = 27.6 cm and angle LMN = 111°, then calculate the shortest distance from R to PQ given PQ = 34.3 cm.

  1. 1Area LMN = 1/2 × 21.5 × 27.6 × sin 111° = 276.99…
  2. 21/2 × 34.3 × d = 276.99…
  3. 3d = 2 × 276.99… ÷ 34.3 = 16.15… ≈ 16.2 cm

Answer16.2 or 16.15…

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Also searched as: 0580/42/O/N/21 · 0580_w21_qp_42 · 0580 w21 42 · question 3.
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