WAEC 2021 · Paper 2 · Q3

The points $X$ , $Y$ and $Z$ are located such that $Y$ is $15\text{ km}$ south of $X$ , and $Z$ is $20\text{ km}$ from $X$ on a bearing of $270^\circ$ . Calculate, correct to:

(a)
two significant figures, $|YZ|$ ;
(b)
the nearest degree, the bearing of $Y$ from $Z$ .

Worked solution (try it first)

Draw north at $X$ .
$Y$ is 15 km due south of $X$ .
$Z$ is 20 km from $X$ on $270^\circ$ , which is due west.
South and west are $90^\circ$ apart, so $\angle YXZ = 90^\circ$ .

(a)

$|YZ| = \sqrt{15^2 + 20^2}$
$= \sqrt{625}$
$= 25$ km.

(b)

At $Z$ : the opposite side is $XY = 15$ and the adjacent side is $ZX = 20$ , so $\tan\angle XZY = \frac{15}{20}$ and $\angle XZY \approx 36.87^\circ$ .
At $Z$ , $X$ is due east ( $090^\circ$ ), and $Y$ is $36.87^\circ$ further round clockwise.
Bearing of $Y$ from $Z$ $= 090^\circ + 36.87^\circ$
$\approx 127^\circ$ .

Report a problem with this question