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 $$90^{\circ}$$.