JAMB 1989 · UME · Q38

Angles, triangles & polygons

In $\triangle PQR$ , the bisector of $\angle QPR$ meets $QR$ at $S$ . $PQ$ is produced to $V$ , and the bisector of $\angle VQS$ meets $PS$ produced at $T$ . If $\angle QPR = 46^\circ$ and $\angle QST = 75^\circ$ , calculate $\angle QTS$ .

Worked solution (try it first)

$PS$ bisects $\angle QPR$ , so $\angle QPS = 46^\circ \div 2 = 23^\circ$ .
$PST$ is a straight line, so $\angle QSP = 180^\circ - 75^\circ$
$= 105^\circ$ .
$\angle VQS$ is an exterior angle of triangle $PQS$ , so $\angle VQS = 23^\circ + 105^\circ$
$= 128^\circ$ .
$QT$ bisects it: $\angle VQT = 64^\circ$ .
$\angle VQT$ is an exterior angle of triangle $PQT$ , so $64^\circ = \angle QPT + \angle QTP$
$= 23^\circ + \angle QTS$ .
So $\angle QTS = 64^\circ - 23^\circ$
$= 41^\circ$ , option A.

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