WAEC 2013 · Paper 2 · Q10

Town $Q$ is $20\text{ km}$ due north of $P$ . The bearing of town $R$ from $Q$ is $140^\circ$ . If $R$ is $8\text{ km}$ from $Q$ , calculate:

Worked solution (try it first)

Sketch it: $Q$ is 20 km due north of $P$ , and $R$ is 8 km from $Q$ on a bearing of $140^\circ$ (to the south-east of $Q$ ).
At $Q$ , the direction back to $P$ is due south ( $180^\circ$ ), so $\angle PQR = 180^\circ - 140^\circ$
$= 40^\circ$ .
Cosine rule: $|PR|^2 = 20^2 + 8^2 - 2(20)(8)\cos40^\circ$
$= 218.87$ , so $|PR| = 14.79\text{ km}$ .
Sine rule for $\theta = \angle QPR$ : $\frac{\sin\theta}{8} = \frac{\sin40^\circ}{14.79}$ , so $\sin\theta = 0.3476$ and $\theta = 20.34^\circ$ .
$\theta$ is measured clockwise from north at $P$ , so the bearing of $R$ from $P$ is $020^\circ$ to the nearest degree.

$R$ is $8\cos40^\circ = 6.13\text{ km}$ south of $Q$ .
So $R$ is $20 - 6.13 = 13.87\text{ km}$ north of $P$ (the same as $|PR|\cos20.34^\circ$ ).
To 2 significant figures, $R$ is $14\text{ km}$ north of $P$ .