WAEC 2020 · Paper 2 · Q9

Two cyclists $X$ and $Y$ leave town $Q$ at the same time. Cyclist $X$ travels at the rate of $5\text{ km/h}$ on a bearing of $049^\circ$ and cyclist $Y$ travels at the rate of $9\text{ km/h}$ on a bearing of $319^\circ$ .

(a)
Illustrate the information on a diagram.
Model answer
A clear sketch is enough (it need not be to scale), but it must show every given fact: after 2 hours, $QX = 5 \times 2 = 10$ km on bearing $049^\circ$ and $QY = 9 \times 2 = 18$ km on bearing $319^\circ$ ( $41^\circ$ west of north). The angle between them at $Q$ is $49^\circ + 41^\circ = 90^\circ$ . Draw a north line at $Y$ as well, since part (b) asks for a bearing measured from $Y$ .
(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 $Y$ .
Separate values with commas, e.g. 3, −2
(c)
Find the average speed at which cyclist $X$ would cover the distance to $Y$ in 4 hours.

Worked solution (try it first)

(a)

Draw north at $Q$ .
After 2 hours, $X$ is $5 \times 2 = 10$ km from $Q$ on $049^\circ$ and $Y$ is $9 \times 2 = 18$ km from $Q$ on $319^\circ$ .
Join $X$ and $Y$ .

(b)(i)

The angle between the two directions at $Q$ is $049^\circ + (360^\circ - 319^\circ) = 49^\circ + 41^\circ$
$= 90^\circ$ .
So triangle $XQY$ is right-angled at $Q$ : $|XY| = \sqrt{10^2 + 18^2}$
$= \sqrt{424}$
$\approx 20.59$ km, which is 21 km to the nearest whole number.

(ii)

At $Y$ : $\tan\angle QYX = \frac{10}{18}$ , so $\angle QYX \approx 29.1^\circ$ .
At $Y$ , the direction back to $Q$ is $319^\circ - 180^\circ = 139^\circ$ , and $X$ is $29.1^\circ$ further round anticlockwise.
Bearing of $X$ from $Y$ $= 139^\circ - 29.1^\circ$
$\approx 110^\circ$ .

(c)

Average speed $= \frac{\text{distance}}{\text{time}}$
$= \frac{20.59}{4}$
$\approx 5.15$ km/h.

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