ECZ 2018 · GCE Paper 2 · Question 8 — Trigonometry
Question from ECZ 2018 · GCE Paper 2 · Question 8
(a) Three villages A, B and C are connected by straight paths as shown in the diagram below. Given that AB = 15 km, angle ABC = 79° and angle ACB = 40°, calculate the
(a(i)) distance AC,[4]
Verified working — step 1
Third angle A = 180 - 79 - 40 = 61°.(a(ii)) area of triangle ABC,[2]
Verified working — step 1
Two sides AB and AC enclose angle A (61°): Area = (12)(AB)(AC) sin A.(a(iii)) shortest distance from B to AC.[2]
Verified working — step 1
The shortest distance is the perpendicular height h from B to AC.(b) Solve the equation cos θ = 0.937 for 0° ≤ θ ≤ 360°.[2]
Verified working — step 1
cos θ = 0.937, so θ = cos-1(0.937) = 20.4°.(c) Sketch the graph of y = sin θ for 0° ≤ θ ≤ 360°.[2]
Verified working — step 1
Amplitude 1, one full period over 0 to 360°.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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