JAMB 2001 · UME · Q22

Angles, triangles & polygons

In the figure, $PQR$ is a straight line and $PQ = QT$ . Triangle $PQT$ is isosceles, $\angle SRQ = 75^\circ$ and $\angle QPT = 25^\circ$ . Calculate $\angle RST$ .

Worked solution (try it first)

$PQ = QT$ , so the base angles of triangle $PQT$ are equal: $\angle QTP = 25^\circ$ .
The exterior angle at $Q$ equals the sum of the two interior opposite angles: $\angle TQR = 25^\circ + 25^\circ$
$= 50^\circ$ .
$Q$ , $T$ and $S$ are in a straight line, so triangle $QRS$ has angles $50^\circ$ at $Q$ and $75^\circ$ at $R$ .
Angles in a triangle add up to $180^\circ$ : $\angle RST = 180^\circ - 50^\circ - 75^\circ$
$= 55^\circ$ , option D.

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