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'Bisector Intersection' printed from https://nrich.maths.org/
$110^{\circ}$
As $\angle QPR=40^{\circ}$, $\angle PQR+\angle
PRQ=180^{\circ}-40^{\circ} =140^{\circ}$.
So $\angle SQR+\angle SRQ=140^{\circ}/2=70^{\circ}$. Therefore
$\angle QSR =180^{\circ}-70^{\circ}=110^{\circ}$.