ECZ 2024 · O-Level Paper 2 · Question 12 — Trigonometry
Question from ECZ 2024 · O-Level Paper 2 · Question 12
(a) In triangle PQR, PQ = 16.1 m, angle QPR = 42° and angle PQR = 53°. Calculate the
(a(i)) length of QR,[4]
Verified working — step 1
The third angle: angle PRQ = 180 - 42 - 53 = 85°.(a(ii)) area of triangle PQR,[2]
Verified working — step 1
Two sides and the included angle at Q: Area = (12)(PQ)(QR) sin(PQR)(a(iii)) shortest distance from R to PQ.[2]
Verified working — step 1
The shortest distance is the perpendicular height h from R onto PQ.(b) Solve the equation 6 sin θ = -3, for 180° ≤ θ ≤ 360°.[2]
Verified working — step 1
sin θ = -36 = -12, with related acute angle 30°.(c) Simplify x3 - 9xx + 3.[2]
Verified working — step 1
Factor x out first: x3 - 9x = x(x2 - 9)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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