As $QS=SR$, triangle $SQR$ is isosceles, so $\angle SRQ= \angle SQR =x^{\circ}$.

So by the exterior angle theorem $\angle QST=2x^{\circ}$.

Also, $\angle TQS = 2x^{\circ}$ since $QT=TS$.

As $PT=QT$, $\angle TPQ=\angle TQP =20^{\circ}$.

Since the interior angles of triangle $PQR$ must sum to $180^{\circ}$ we obtain $$20+(20+2x+x)+x=180$$ $$40+4x=180$$ $$4x=140$$ $$x=35.$$

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.