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## 'Angle Hunt' printed from http://nrich.maths.org/

Observing that triangle $PQS$ is isosceles, we have $\angle PSQ = \frac {1}{2}(180^{\circ} - 12^{\circ}) = 84^{\circ}$ and hence $\angle PSR = 180^{\circ} - 84^{\circ}=96^{\circ}$.

Since triagle $PRS$ is also isosceles, we have $\angle SPR = \frac {1}{2}(180^{\circ} - 96^{\circ}) = 42^{\circ}$. Hence $\angle QPR = 12^{\circ}+ 42^{\circ} = 54^{\circ}$.

*This problem is taken from the UKMT Mathematical Challenges.*