, 2012

Since $\triangle PRS$ is isosceles with $PR = PS$, then $\angle PRS = \angle PSR$. Since the angles in $\triangle PRS$ add to $180^\circ$, then $\angle PRS + \angle PSR + \angle RPS = 180^\circ$. Therefore, $2(\angle PRS) + 34^\circ = 180^\circ$ or $2(\angle PRS) = 146^\circ$ or $\angle PRS = 73^\circ$. Since $\triangle PQT$ is isosceles with $PQ = PT$, then $\angle PQT = \angle PTQ = 62^\circ$. Since $\angle PRS$ is an exterior angle to $\triangle PQR$, then $\angle PRS = \angle PQR + \angle QPR$ or $73^\circ = 62^\circ + x^\circ$. Therefore, $x = 73 - 62 = 11$. (Instead, we could have determined that $\angle PRQ = 180^\circ - \angle PRS = 180^\circ - 73^\circ = 107^\circ$, and then looked at the sum of the angles in $\triangle PQR$ to get $62^\circ + x^\circ + 107^\circ = 180^\circ$ or $x = 180 - 169 = 11$.)