JAMB 2013 · UTME · Q25

In the diagram, $PQ \parallel TS$ , $\angle PQR = 110^\circ$ and $\angle TSR = 120^\circ$ . Find the value of $x$ .

Worked solution (try it first)

Draw a line through $R$ parallel to $PQ$ and $TS$ .
$\angle PQR$ and the angle between $RQ$ and that line are co-interior, so the upper part of $x$ is $180^\circ - 110^\circ = 70^\circ$ .
$\angle TSR$ and the angle between $RS$ and that line are co-interior, so the lower part of $x$ is $180^\circ - 120^\circ = 60^\circ$ .
So $x = 70^\circ + 60^\circ = 130^\circ$ , option B.