JAMB 1998 · UME · Q26

In the diagram, $PQ \parallel ST$ , $\angle PQR = 120^\circ$ and $\angle RST = 130^\circ$ . Find the angle marked $x$ .

Worked solution (try it first)

Draw a line through $R$ parallel to $PQ$ and $ST$ .
$\angle PQR$ and the angle between $RQ$ and that line are co-interior: $180^\circ - 120^\circ = 60^\circ$ .
$\angle RST$ and the angle between $RS$ and that line are co-interior: $180^\circ - 130^\circ = 50^\circ$ .
The $60^\circ$ , $x$ and $50^\circ$ make a straight line at $R$ , so $x = 180^\circ - 60^\circ - 50^\circ$
$= 70^\circ$ , option C.