JAMB 1991 · UME · Q30

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

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 $180^\circ - 110^\circ = 70^\circ$ .
$\angle RST$ and the angle between $RS$ and that line are co-interior too, so the lower part of $x$ is $180^\circ - 120^\circ = 60^\circ$ .
So $x = 70^\circ + 60^\circ = 130^\circ$ , option A.