WAEC 2018 · Paper 2 · Q7

(a)
Find the equation of the line passing through the points $(2, 5)$ and $(-4, -7)$ .
(b)
Three ships $P$ , $Q$ and $R$ are at sea. The bearing of $Q$ from $P$ is $030^\circ$ and the bearing of $P$ from $R$ is $300^\circ$ . If $|PQ| = 5\text{ km}$ and $|PR| = 8\text{ km}$ , (i) illustrate the information in a diagram; (ii) calculate, correct to three significant figures, the: (I) distance between $Q$ and $R$ ; (II) bearing of $R$ from $Q$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Draw north at $P$ .
$Q$ is 5 km from $P$ on $030^\circ$ .
The bearing of $P$ from $R$ is $300^\circ$ , so the bearing of $R$ from $P$ is $300^\circ - 180^\circ = 120^\circ$ : $R$ is 8 km from $P$ on $120^\circ$ .
The angle between the lines at $P$ is $120^\circ - 30^\circ = 90^\circ$ .

(I)** Triangle $QPR$ is right-angled at $P$ : $|QR| = \sqrt{5^2 + 8^2}$
$= \sqrt{89}$
$\approx 9.43$ km.
(II) At $Q$ : $\tan\angle PQR = \frac{8}{5} = 1.6$ , so $\angle PQR \approx 58.0^\circ$ .
At $Q$ , the direction back to $P$ is $030^\circ + 180^\circ = 210^\circ$ , and $R$ is $58.0^\circ$ further round anticlockwise.
Bearing of $R$ from $Q$ $= 210^\circ - 58.0^\circ = 152^\circ$ .