 Problem of the Week

Problem D and Solution

Angled II

Problem

In $$\triangle PQS$$, $$R$$ lies on $$PQ$$ such that $$PR=RQ=RS$$ and $$\angle QRS =z^{\circ}$$.
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 $$\angle QRS =z^{\circ}$$, we have $$\angle PRS =180 - z^{\circ}$$.
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} + 180 - z^{\circ}&=180^{\circ}\\[-1mm] 2x&=z\\[0mm] x&=\frac{z}{2}\\[-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} + z^{\circ}&=180^{\circ}\\[-1mm] 2y&=180-z\\[-1mm] y&= \frac{180-z}{2} \\[-6mm]\end{aligned}

Then $$\angle PSQ=\angle RSP + \angle RSQ=x^{\circ}+y^{\circ}=\frac{z}{2}^{\circ}+\left( \frac{180-z}{2} \right)^{\circ}=\left(\frac{180}{2}\right)^{\circ}=90^{\circ}$$.
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 expressions for $$x$$ and $$y$$ in terms of $$z$$ 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}\\ x^{\circ} + (x^{\circ}+y^{\circ}) + y^{\circ}&=180^{\circ}\\ (x^{\circ}+ y^{\circ}) + (x^{\circ}+y^{\circ}) &=180^{\circ}\\ 2(x^{\circ}+y^{\circ})&=180^{\circ}\\ x^{\circ}+y^{\circ}&=90^{\circ}\\[-8mm]\end{aligned} But $$\angle PSQ=\angle RSP + \angle RSQ=x^{\circ}+y^{\circ}=90^{\circ}$$.
Therefore, the measure of $$\angle PSQ$$ is $$90^{\circ}$$.