CEMC Gauss (Grade 7), 2018

Consider the following diagram.

Since $S,T,U$ lie on a straight line, $\angle STU$ measures $180^{\circ}$.
Therefore, $\angle RTU=180^{\circ}-\angle STR=180^{\circ}-120^{\circ}=60^{\circ}$.
Similarly, $Q,U,R$ lie on a straight line, and so $\angle QUR$

measures $180^{\circ}$.

Therefore, $\angle TUR=180^{\circ}-\angle TUQ=180^{\circ}-95^{\circ}=85^{\circ}$.

The sum of the angles in $\triangle TUR$ is $180^{\circ}$.

Thus, $\angle TRU=180^{\circ}-\angle RTU-\angle TUR=180^{\circ}-60^{\circ}-85^{\circ}=35^{\circ}$.

Since $\triangle PQR$ is isosceles with $PQ=PR$, then $\angle PQR=\angle PRQ=35^{\circ}$.

Finally, the sum of the angles in $\triangle PQR$ is $180^{\circ}$, and so $x^{\circ}=180^{\circ}-\angle PQR-\angle PRQ$ or $x^{\circ}=180^{\circ}-35^{\circ}-35^{\circ}$ and so $x=110$.