CEMC Banner

Problem of the Week

Problem C and Solution

Angled

Problem

In \(\triangle PQS\), \(R\) lies on \(PQ\) such that \(PR=RQ=RS\) and \(\angle QRS =70^{\circ}\).
Determine the measure of \(\angle PSQ\).

Solution

Solution 1
In \(\triangle PRS\), since \(PR=RS\), \(\triangle PRS\) is isosceles and \(\angle RPS=\angle RSP=x^{\circ}\).
Similarly, in \(\triangle QRS\), since \(RQ=RS\), \(\triangle QRS\) is isosceles and \(\angle RQS=\angle RSQ=y^{\circ}\).

Since \(PRQ\) is a straight line, \(\angle PRS + \angle QRS=180^{\circ}\). Since \(\angle QRS =70^{\circ}\), we have \(\angle PRS =110^{\circ}\).
The angles in a triangle sum to \(180^{\circ}\), so in \(\triangle PRS\) \[\begin{aligned} \angle RPS +\angle RSP + \angle PRS&=180^{\circ}\\[-1mm] x^{\circ} + x^{\circ} + 110^{\circ}&=180^{\circ}\\[-1mm] 2x&=70\\[-1mm] x&=35\\[-10mm]\end{aligned}\]

The angles in a triangle sum to \(180^{\circ}\), so in \(\triangle QRS\) \[\begin{aligned} \angle RQS +\angle RSQ + \angle QRS&=180^{\circ}\\[-1mm] y^{\circ} + y^{\circ} + 70^{\circ}&=180^{\circ}\\[-1mm] 2y&=110\\[-1mm] y&=55\\[-10mm]\end{aligned}\] Then \(\angle PSQ=\angle RSP + \angle RSQ=x^{\circ}+y^{\circ}=35^{\circ}+55^{\circ}=90^{\circ}\).
Therefore, the measure of \(\angle PSQ\) is \(90^{\circ}\).

See Solution 2 for a more general approach to the solution of this problem.

It turns out that it is not necessary to determine the values of \(x\) and \(y\) to solve this problem.
Solution 2
In \(\triangle PRS\), since \(PR=RS\), \(\triangle PRS\) is isosceles and \(\angle RPS=\angle RSP=x^{\circ}\).
Similarly, in \(\triangle QRS\), since \(RQ=RS\), \(\triangle QRS\) is isosceles and \(\angle RQS=\angle RSQ=y^{\circ}\).

The angles in a triangle sum to \(180^{\circ}\), so in \(\triangle PQS\) \[\begin{aligned} \angle QPS +\angle PSQ + \angle PQS&=180^{\circ}\\[-1mm] x^{\circ} + (x^{\circ}+y^{\circ}) + y^{\circ}&=180^{\circ}\\[-1mm] (x^{\circ}+y^{\circ}) + (x^{\circ}+y^{\circ}) &=180^{\circ}\\[-1mm] 2(x^{\circ}+y^{\circ}) &=180^{\circ}\\[-1mm] x^{\circ}+y^{\circ}&=90^{\circ}\\[-1mm]\end{aligned}\] But \(\angle PSQ=\angle RSP + \angle RSQ=x^{\circ}+y^{\circ}=90^{\circ}\).
Therefore, the measure of \(\angle PSQ\) is \(90^{\circ}\).