In the diagram, $\triangle P Q R$ is isosceles with $P Q=P R$. What is the value of $x$ ?
(A) 110
(B) 90
(C) 95
(D) 100
(E) 105

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-026.jpg?height=301&width=647&top_left_y=175&top_left_x=1086)

## Part C: Each correct answer is worth 8.
We begin by labelling points $S, T$ and $U$, as shown.

Since $S, T, U$ lie on a straight line, $\angle S T U$ measures $180^{\circ}$. Therefore, $\angle R T U=180^{\circ}-\angle S T R=180^{\circ}-120^{\circ}=60^{\circ}$. Similarly, $Q, U, R$ lie on a straight line, and so $\angle Q U R$ measures $180^{\circ}$.

Therefore, $\angle T U R=180^{\circ}-\angle T U Q=180^{\circ}-95^{\circ}=85^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-110.jpg?height=298&width=596&top_left_y=1995&top_left_x=1339)

The sum of the angles in $\triangle T U R$ is $180^{\circ}$.

Thus, $\angle T R U=180^{\circ}-\angle R T U-\angle T U R=180^{\circ}-60^{\circ}-85^{\circ}=35^{\circ}$.

Since $\triangle P Q R$ is isosceles with $P Q=P R$, then $\angle P Q R=\angle P R Q=35^{\circ}$.

Finally, the sum of the angles in $\triangle P Q R$ is $180^{\circ}$, and so $x^{\circ}=180^{\circ}-\angle P Q R-\angle P R Q$ or $x^{\circ}=180^{\circ}-35^{\circ}-35^{\circ}$ and so $x=110$.