Solution
D

Let $\angle S R Q$ be $x^{\circ}$. Since sides $P R$ and $P Q$ are equal, triangle $P Q R$ is isosceles and hence $\angle P R Q=\angle P Q R=\left(180^{\circ}-40^{\circ}\right) / 2=70^{\circ}$. Therefore, since we are given that $\angle T Q P=\angle S R Q$, we have $\angle U Q R$ is $70^{\circ}-x^{\circ}$. Hence, since the exterior angle of a triangle is equal to the sum of the interior opposite angles, $\angle T U R$ is $70^{\circ}-x^{\circ}+x^{\circ}=70^{\circ}$.
13. In the diagram shown, sides $P Q$ and $P R$ are equal. Also $\angle Q P R=40^{\circ}$ and $\angle T Q P=\angle S R Q$. What is the size of $\angle T U R$ ?
A $55^{\circ}$
B $60^{\circ}$
C $65^{\circ}$
D $70^{\circ}$
E $75^{\circ}$