ECZ 2018 · O-Level Paper 2 · Question 8 — Trigonometry
Question from ECZ 2018 · O-Level Paper 2 · Question 8
(a) In the diagram below, K, N, B and R are places on horizontal surface. KN = 80 m, NB = 50 m, angle KNR = 60° and angle KRN = 52°.
(a(i)) Calculate
(a(i)(a)) KR,[4]
Verified working — step 1
Third angle: angle NKR = 180 - 60 - 52 = 68°.(a(i)(b)) the area of triangle KNB.[2]
Verified working — step 1
B lies on NR, so angle KNB = angle KNR = 60°.(a(ii)) Given that the area of triangle KNR is equal to 3 260 m2, calculate the shortest distance from R to KN.[2]
Verified working — step 1
The shortest distance is the perpendicular height h from R to KN.(b) Sketch the graph of y = cos θ for 0° ≤ θ ≤ 360°.[2]
Verified working — step 1
Amplitude 1, one full period over 0 to 360°.(c) Simplify 12dn315cd3 ÷ 9c3n10c2d2.[2]
Verified working — step 1
Dividing by a fraction means multiplying by its 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.
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