JAMB 1988 · UME · Q31

Angles, triangles & polygons

In the figure, $PQ$ is parallel to $ST$ , $\angle QRS = 40^\circ$ , the angle at $Q$ between $QP$ and $QR$ is $3x$ and $\angle TSR = x$ . Find the value of $x$ .

Worked solution (try it first)

Draw a line through $R$ parallel to $PQ$ .
Angles on a straight line at $Q$ : $QR$ makes $180^\circ - 3x$ with the rightward direction there, so $RQ$ makes $180^\circ - 3x$ with the parallel line at $R$ (alternate angles).
$ST$ is parallel too, so $RS$ makes $x$ with the parallel line at $R$ (alternate angles with $\angle TSR$ ).
Both arms lie on the same side of that line, so $\angle QRS$ is the difference: $x - (180^\circ - 3x) = 40^\circ$ , which gives $4x = 220^\circ$ .
So $x = 55$ , option A.

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