JAMB 1986 · UME · Q40✱✱

Angles, triangles & polygons

In the figure, $\triangle PQT$ is isosceles with $PQ = QT$ . $S$ lies on $PT$ , $\angle SRQ = 35^\circ$ , $\angle TPQ = 20^\circ$ and $PQR$ is a straight line. Calculate $\angle TSR$ .

Worked solution (try it first)

Look at triangle $PSR$ : its angle at $P$ is $20^\circ$ and its angle at $R$ is $35^\circ$ .
$\angle TSR$ is the exterior angle of triangle $PSR$ at $S$ , because $PST$ is a straight line.
An exterior angle equals the sum of the two interior opposite angles: $\angle TSR = 20^\circ + 35^\circ$
$= 55^\circ$ , option B.

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