JAMB 1991 · UME · Q42

From a point $Z$ , 60 m north of $X$ , a man walks $60\sqrt3$ m eastwards to another point $Y$ . Find the bearing of $Y$ from $X$ .

Worked solution (try it first)

$Y$ is 60 m north and $60\sqrt3$ m east of $X$ , so there is a right angle at $Z$ .
The angle at $X$ , east of north, has $\tan\theta = \frac{60\sqrt3}{60} = \sqrt3$ , so $\theta = 60^\circ$ .
Measured clockwise from north, the bearing of $Y$ from $X$ is $060^\circ$ , option C.