JAMB 2015 · UTME · Q6

In the diagram, $PQ$ is parallel to $RS$ . Calculate the value of $x$ .

Worked solution (try it first)

Let the crossing point be $X$ .
Angles on the straight line $RXQ$ : $\angle PXQ = 180^\circ - 100^\circ$
$= 80^\circ$ .
The angles of triangle $PXQ$ add up to $180^\circ$ : $\angle PQX = 180^\circ - 60^\circ - 80^\circ$
$= 40^\circ$ .
$PQ \parallel RS$ , so $x = \angle SRQ = \angle PQR = 40^\circ$ (alternate angles), option B.