ECZ 2019 · O-Level Paper 2 · Question 12 — Trigonometry

Question from ECZ 2019 · O-Level Paper 2 · Question 12

(a) The diagram below shows a triangle KMN in which KM = 8 km, MN = 10 km and angle KMN = 92°. Calculate

(a(i)) KN,[5]

Verified working — step 1

Two sides and the included angle: cosine rule.

(a(ii)) the area of triangle KMN,[2]

Verified working — step 1

Two sides and the included angle at M: Area = (12)(KM)(MN) sin(KMN)

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

Verified working — step 1

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

(b) Solve the equation 2 tan θ = -3 for 0° ≤ θ ≤ 180°.[1]

Verified working — step 1

tan θ = -32 = -1.5, related acute angle 56.31°.

(c) Simplify 25p⁴7q² ÷ 5p⁶21q⁴ × p15q.[2]

Verified working — step 1

Turn each division into multiplication by the reciprocal:

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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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