91Ó°ÊÓ

The theorem for the sum of interior angles of a polygon states that the sum of measures of the interior angles of a polygon with sides is $\left(n-2\right)\mathbf{180}$.

The number of sides in a pentagon is 5.

Therefore, the value of $n$is 5.

Therefore, the sum of interior angles of a pentagon is:

$\begin{array}{c}\left(5−2\right)180=3×180\\ =540\end{array}$

Therefore, the sum of interior angles of a pentagon is $540°$.

As, the sum of interior angles of a pentagon is$540°$ and the$∠S$and$∠T$are each three times as large as $∠R$.

$$\begin{gathered} m∠P+m∠Q+m∠R+m∠S+m∠T=540 \\ 60+130+m∠R+3m∠R+3m∠R=540 \\ 190+7m∠R=540 \\ 7m∠R=540−190 \\ 7m∠R=350 \\ m∠R=\frac{350}{7} \\ m∠R=50 \end{gathered}$$

Therefore, the measure of$∠R$ is 50.

$m∠S=3m∠R=3×50=150$

Therefore, the measure of$∠S$ is $150°$.

$m∠T=3m∠R=3×50=150$

Therefore, the measure of $∠T$is $150°$.

The measures of the angles of are:

$m∠P=60,m∠Q=130,m∠R=50,m∠S=150$and$m∠T=150$

The sum of the angles$∠Q$ and$∠R$ is:

$\begin{array}{c}m∠Q+m∠R=130+50\\ =180\end{array}$

As, the sum of the angles $∠Q$and$∠R$ is $180°$, therefore the pair of sides of$PQRST$ that must be parallel are$PQ$ and $RS$.