CEMC Pascal, 2012

Geometry Difficulty 2.6 Junior Find the answer Canada

In the diagram, points $R$ and $S$ lie on $QT$ . Also, $\angle PTQ=62^\circ$ , $\angle RPS = 34^\circ$ , and $\angle QPR = x^\circ$ .
What is the value of $x$ ?

Pick one

Solution

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 $x = 180 - 169 = 11$ .)

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