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 \(\angle PST\).
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 \(\angle PST=e^{\circ}=60^{\circ}\).
