ECZ 2025 · GCE Paper 2 · Question 8 — Trigonometry
Question from ECZ 2025 · GCE Paper 2 · Question 8
(a) A garden is in the form of a quadrilateral PQRS with PQ = 5 m, QR = 8 m, RS = 7 m and PS = 6 m. Given that angle PQR = 140° and angle PRS = 18°, calculate
(a(i)) PR,[5]
Verified working — step 1
Two sides and the included angle: cosine rule.(a(ii)) the area of triangle PRS,[2]
Verified working — step 1
Two sides and the included angle at R: Area = (12)(PR)(RS) sin(PRS)(a(iii)) the shortest distance from S to PR.[2]
Verified working — step 1
The shortest distance from S to PR is the perpendicular height h onto PR.(b) Solve the equation 10 tan θ = 50 for 0° ≤ θ ≤ 90°.[1]
Verified working — step 1
Divide first: tan θ = 5010 = 5(c) Simplify 9a - c81a2 - c2.[2]
Verified working — step 1
81a2 - c2 is a difference of two squares: (9a - c)(9a + c).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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