ECZ 2025 · O-Level Paper 2 · Question 8 — Trigonometry

Question from ECZ 2025 · O-Level Paper 2 · Question 8

(a) The diagram shows a triangular piece of land LMN in which LM = 23.7 m, MN = 16.6 m and angle LMN = 106°. Calculate the

(a(i)) distance LN,[5]

Verified working — step 1

Two sides and the INCLUDED angle are known, so use the cosine rule:

(a(ii)) area of triangle LMN,[2]

Verified working — step 1

With two sides and the included angle: Area = (12)(LM)(MN) sin(LMN)

(a(iii)) shortest distance from M to LN.[2]

Verified working — step 1

The shortest distance from M to LN is the perpendicular height h onto LN.

(b) Solve the equation 7 cos θ = 4 for 180° ≤ θ ≤ 360°.[1]

Verified working — step 1

cos θ = 47 = 0.5714

(c) Simplify 216 - 6x²6 + x.[2]

Verified working — step 1

Take out the common factor first: 216 - 6x² = 6(36 - x²)

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Original examination question © Examinations Council of Zambia. Worked solution and commentary © G12 Titan — not to be reproduced without permission.

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