Question

9. (a) An aircraft took off from an airstrip at an average speed of 35km/h on a bearing of 015 degrees for 2 hours, then changed course to a bearing of 100 degrees at 22km/h for 2.5 hours. Find its: i. Distance from the starting point (2 d.p.) ii. Bearing from the airstrip to the nearest degree. (b) Using a ruler and compasses only, construct: i. Triangle PQR with |PQ|=8cm, angle RPQ=120 degrees, angle PQR=30 degrees ii. The locus L1 of points equidistant from P and Q iii. The locus L2 of points equidistant from PQ and PR passing through the triangle. Label the intersection of L1 and L2 as X and measure |QX|. (12 marks)

Accepted Answer

(a)(i) First leg = 35x2=70km on bearing 015. Second leg = 22x2.5=55km on bearing 100. Using the cosine rule with the included angle between the two legs (96 degrees): distance^2 = 70^2+55^2-2(70)(55)cos(96) ~= 8596.099, so distance ~= 92.72km.
(ii) Using the sine rule: sin(A)/55 = sin(96)/92.72, giving A ~= 36.22 degrees. Bearing from the airstrip ~= 15+36.22 ~= 51 degrees (to the nearest degree).
(b) Draw PQ=8cm. Construct a 120 degree angle at P and a 30 degree angle at Q; extend both lines to locate R, forming triangle PQR. Construct the perpendicular bisector of PQ (locus L1: points equidistant from P and Q). Construct the bisector of angle RPQ (locus L2: points equidistant from lines PQ and PR). Mark the intersection of L1 and L2 as X, then measure |QX| directly with a ruler.

NECO Mathematics 2025 Theory — Question 9

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