ECZ 2019 · GCE Paper 2 · Question 8 — Trigonometry
Question from ECZ 2019 · GCE Paper 2 · Question 8
(a) In triangle ABC below, AC = 275 km, angle BAC = 125° and angle ACB = 40°. Calculate
(a(i)) the distance BC,[4]
Verified working — step 1
Find the third angle: B = 180 - 125 - 40 = 15°.(a(ii)) the area of triangle ABC,[2]
Verified working — step 1
Two sides AC and BC enclose angle C, so Area = (12)(AC)(BC) sin C.(a(iii)) the shortest distance from A to BC.[2]
Verified working — step 1
The shortest distance is the perpendicular height h from A to BC.(b) Solve the equation 13 cos θ = 5 for 0° ≤ θ ≤ 360°.[2]
Verified working — step 1
13 cos θ = 5, so cos θ = 513 = 0.3846.(c) Simplify 2x2 - 18x + 3.[2]
Verified working — step 1
Factor the numerator: 2x2 - 18 = 2(x2 - 9) = 2(x - 3)(x + 3).The full step-by-step working and final answer are included with Premium — checked against ECZ marking standards.
Original examination question © Examinations Council of Zambia. Worked solution and commentary © G12 Titan — not to be reproduced without permission.
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