ECZ 2017 · GCE Paper 2 · Question 10 — Trigonometry
Question from ECZ 2017 · GCE Paper 2 · Question 10
(a) In triangle PQR below, QR = 36.5 m, angle PQR = 36° and angle QPR = 46°. Calculate
(a(i)) PQ,[4]
Verified working — step 1
Third angle R = 180 - 36 - 46 = 98°.(a(ii)) the area of triangle PQR,[2]
Verified working — step 1
Two sides PQ and QR enclose angle Q (36°): Area = (12)(PQ)(QR) sin Q.(a(iii)) the shortest distance from R to PQ.[2]
Verified working — step 1
The shortest distance is the perpendicular height h from R to PQ.(b) Solve the equation sin θ = 0.6792 for 0° ≤ θ ≤ 360°.[2]
Verified working — step 1
sin θ = 0.6792, so θ = sin-1(0.6792) = 42.8°.(c) Simplify p2 q34 × 8pq ÷ 2p2 q.[2]
Verified working — step 1
Work left to right, turning the division into multiplication by the reciprocal.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.
Have a different question?
Search it — if it is from a past paper or a school mock, chances are we have it worked out, or one that uses the same method.
Search any past-paper question