# Problem of the Week Problem C and Solution The Missing Pieces

## Problem

The following information is known about $$\triangle PQR$$.

• The point $$S$$ is on side $$PR$$ and the point $$T$$ is on side $$PQ$$.

• The distance from $$P$$ to $$S$$ is equal to the distance from $$T$$ to $$Q$$.

• The distance from $$S$$ to $$R$$ is equal to the distance from $$P$$ to $$T$$.

• $$\angle PRQ=40^{\circ}$$ and $$\angle PTS=20^{\circ}$$.

Determine the value of each of the five other interior angles. That is, determine the values of $$\angle RPQ$$, $$\angle STQ$$, $$\angle TQR$$, $$\angle RST$$, and $$\angle PST$$.

## Solution

First, we let $$\angle RPQ$$ measure $$a^{\circ}$$, $$\angle STQ$$ measure $$b^{\circ}$$, $$\angle TQR$$ measure $$c^{\circ}$$, $$\angle RST$$ measure $$d^{\circ}$$, and $$\angle PST$$ measure $$e^{\circ}$$.

Since $$\angle PTQ$$ is a straight angle, $$20+b=180$$, and so $$b=160$$.

Since $$PS=TQ$$ and $$SR=PT$$, it follows that $$PS+PR=PT+TQ$$, and so $$PR=PQ$$ and $$\triangle PQR$$ is isosceles. Therefore $$\angle PRQ=\angle PQR$$, and so $$c=40$$.

Since the angles in a triangle sum to $$180^{\circ}$$, in $$\triangle PQR$$, \begin{align} a+40+c&=180\\ a+40+40&=180\\ a+80&=180\\ a&=100 \end{align}

Similarly, in $$\triangle PST$$, \begin{align} a+e+20&=180\\ 100+e+20&=180\\ 120+e&=180\\ e&=60 \end{align}

Since $$\angle PSR$$ is a straight angle, \begin{align} e+d&=180\\ 60+d&=180\\ d&=120 \end{align}

We have determined the value of all the other five interior angles. $$\angle RPQ=a^{\circ}=100^{\circ}$$, $$\angle STQ=b^{\circ}=160^{\circ}$$, $$\angle TQR=c^{\circ}=40^{\circ}$$, $$\angle RST=d^{\circ}=120^{\circ}$$, and $$\angle PST=e^{\circ}=60^{\circ}$$.