JAMB 1993 · UME · Q30

In the diagram, $QP \parallel ST$ , $\angle PQR = 34^\circ$ , $\angle QRS = 73^\circ$ and $RS = RT$ . Find $\angle SRT$ .

Worked solution (try it first)

Draw a line through $R$ parallel to $QP$ .
By alternate angles, $RQ$ makes $34^\circ$ with it and $RS$ makes $\angle RST$ with it, so $73^\circ = 34^\circ + \angle RST$ .
So $\angle RST = 39^\circ$ .
$RS = RT$ , so triangle $RST$ is isosceles and $\angle RTS = \angle RST = 39^\circ$ .
The angles of the triangle add up to $180^\circ$ : $\angle SRT = 180^\circ - 39^\circ - 39^\circ$
$= 102^\circ$ , option B.