Theory question 109
BearingsTwo cyclists X and Y leave town Q at the same time. Cyclist X travels at the rate of 5 km/h on a bearing of 049° and cyclist Y travels at the rate of 9 km/h on a bearing of 319°.
(a) Illustrate the information on a diagram.
(b) After travelling for two hours, calculate, correct to the nearest whole number, the: (i) distance between cyclists X and Y; (ii) bearing of cyclist X from Y.
(c) Find the average speed at which cyclist X will get to Y in 4 hours.
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(a) Diagram from Q: QX on bearing 049°, QY on bearing 319°; the angle XQY = 360° − 319° + 49° = 90°
(b) After 2 h: QX = 10 km, QY = 18 km
(i) XY = √(102 + 182) = √424 ≈ 20.6 ≈ 21 km
(ii) ∠QYX: tan = 10/18 → 29.1°; bearing of Q from Y = 139°, so bearing of X from Y = 139° − 29.1° ≈ 110°
(c) Speed = 20.6 km ÷ 4 h ≈ 5.1 km/h (about 5 km/h)
The two bearings differ by 90°, so the triangle QXY is right-angled at Q.