WAEC 2019 · Paper 2 · Q12

A town $J$ is $20\text{ km}$ from a lorry station, $K$ , on a bearing $065^\circ$ . Another town, $T$ , is $8\text{ km}$ from $K$ on a bearing $155^\circ$ . Calculate:

(a)(i)
to the nearest kilometre, the distance of $T$ from $J$ ;
(a)(ii)
to the nearest degree, the bearing of $T$ from $J$ .

Try it on a graph

K is at the origin; north is up. The dashed line is JT.

Worked solution (try it first)

Draw north at $K$ .
$J$ is 20 km from $K$ on $065^\circ$ and $T$ is 8 km from $K$ on $155^\circ$ .
The angle between them is $\angle JKT = 155^\circ - 65^\circ$
$= 90^\circ$ , so triangle $JKT$ is right-angled at $K$ .

(a)(i)

$|TJ| = \sqrt{20^2 + 8^2}$
$= \sqrt{464}$
$\approx 21.54$ km, which is 22 km to the nearest kilometre.

(ii)

At $J$ : $\tan\angle KJT = \frac{8}{20} = 0.4$ , so $\angle KJT \approx 21.8^\circ$ .
At $J$ , the direction back to $K$ is $065^\circ + 180^\circ = 245^\circ$ , and $T$ is $21.8^\circ$ further round anticlockwise.
Bearing of $T$ from $J$ $= 245^\circ - 21.8^\circ$
$= 223.2^\circ$
$\approx 223^\circ$ .

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