Question

9(a) Two ships leave a harbour Q at the same time: One sails at a speed of 5kmh^-1 on a bearing of 049 deg; the other sails at a speed of 9kmh^-1 on a bearing of 319 deg. The two ships stop at points X and Y respectively, on the sea, after 2 hours. (i) Draw a diagram to show the bearings of X and Y from Q. (b) Calculate the: (i) Distance between X and Y, correct to the nearest kilometre; (ii) Bearing of X from Y, correct to the nearest degree; (c) If the ship at X changes its course and sails for 4 hours to reach Y, calculate, correct to three significant figures, its average speed.

Accepted Answer

(a)(i) [Diagram showing point Q with two bearings drawn: QX on a bearing of 049 deg and QY on a bearing of 319 deg, forming an angle of 90 deg between them (since 319-49-180=90, confirming XQY is a right angle).]
(b)(i) Distance travelled by ship A (to X) = speed x time = 5 x 2 = 10 km. Distance travelled by ship B (to Y) = 9 x 2 = 18 km. Since angle XQY=90 deg (right angle, as derived from the bearings), using Pythagoras' theorem: XY^2 = QX^2+QY^2 = 10^2+18^2 = 100+324 = 424, so XY = sqrt424 = 20.59, approximately 21 km (to the nearest km).
(ii) To find the bearing of X from Y, we use the main diagram and trigonometric ratios. Using tan(theta) = opposite/adjacent = 10/18 = 0.5556, theta = tan^-1(0.5556) = 29.06 deg. Given the geometry (angle theta, angle alpha (bearing of X from Y) and 41 deg are all on a straight line, summing to 180 deg): alpha + 41 + 29.06 = 180, so alpha = 180-(41+29.06) = 109.94, approximately 110 deg (to the nearest degree) as the bearing of X from Y.
(c) Distance travelled by ship from X to Y = |XY| = 21 km (from part (b)(i), more precisely 20.59km). Time of travel = 4 hours. Average speed = distance/time = 20.59/4 = 5.25 km/hr (to 3 s.f).

WAEC Mathematics 2020 Theory — Question 9

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