JAMB 1983 · UME · Q37

A ship $H$ leaves a port $P$ and sails 30 km due south. Then it sails 60 km due west. What is the bearing of $H$ from $P$ ?

Worked solution (try it first)

Sketch it: $H$ ends up 30 km south and 60 km west of $P$ , making a right-angled triangle.
The angle west of due south at $P$ is $\tan^{-1}\frac{60}{30} = \tan^{-1}2$
$= 63^\circ26'$ .
Bearings are measured clockwise from north, and due south is $180^\circ$ .
So the bearing is $180^\circ + 63^\circ26' = 243^\circ26'$ , option B.