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