JAMB 1988 · UME · Q46

Construction & loci

$PQR$ is a triangle in which $PQ = 10$ cm and $\angle QPR = 60^\circ$ . $S$ is a point equidistant from $P$ and $Q$ , and also equidistant from $PQ$ and $PR$ . If $U$ is the foot of the perpendicular from $S$ to $PR$ , find $SU$ to one decimal place.

Worked solution (try it first)

$S$ is equidistant from $P$ and $Q$ , so it lies on the perpendicular bisector of $PQ$ .
Its foot on $PQ$ is 5 cm from $P$ .
$S$ is equidistant from $PQ$ and $PR$ , so it lies on the bisector of $\angle QPR$ .
That line makes $60^\circ \div 2 = 30^\circ$ with $PQ$ .
In the right-angled triangle at $P$ , the distance from $S$ to $PQ$ is $5\tan30^\circ \approx 2.887$ cm.
On the angle bisector, the distances to $PQ$ and $PR$ are equal, so $SU \approx 2.9$ cm, option B.

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