JAMB 1999 · UME · Q30

Elevation, depression & bearings

From a point $P$ , the bearings of two points $Q$ and $R$ are N $67^\circ$ W and N $23^\circ$ E respectively. If the bearing of $R$ from $Q$ is N $68^\circ$ E and $PQ = 150$ m, calculate $PR$ .

Worked solution (try it first)

At $P$ , $Q$ is $67^\circ$ west of north and $R$ is $23^\circ$ east of north, so $\angle QPR = 67^\circ + 23^\circ$
$= 90^\circ$ .
The bearing of $Q$ from $P$ is $293^\circ$ , so the bearing of $P$ from $Q$ is $113^\circ$ .
$R$ from $Q$ is on $068^\circ$ , so $\angle PQR = 113^\circ - 68^\circ$
$= 45^\circ$ .
A right-angled triangle with a $45^\circ$ angle is isosceles, so $PR = PQ = 150$ m, option C.

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