ECZ 2018 · O-Level Paper 2 · Question 8 — Trigonometry
In the diagram, K, N, B and R are places on a horizontal surface. KN = 80 m, NB = 50 m, angle KNR = 60 degrees and angle KRN = 52 degrees.
(a(i)(a)) In the diagram, K, N, B and R are places on a horizontal surface with KN = 80 m, NB = 50 m, angle KNR = 60 degrees and angle KRN = 52 degrees. Calculate KR.[4]
Verified working — step 1
In triangle KNR, angle NKR = 180 - 60 - 52 = 68 degrees.(a(i)(b)) Calculate the area of triangle KNB.[2]
Verified working — step 1
B lies on NR, so angle KNB = angle KNR = …(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(theta) for 0 <= theta <= 360 degrees.[2]
Verified working — step 1
The curve has amplitude 1 and one full period over 0 to 3…(c) Simplify 12dn^315cd^3divided by 9c^3n10c^2d^2.[2]
Verified working — step 1
Multiply 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.
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