In the diagram, points $R$ and $S$ lie on $Q T$. Also, $\angle P T Q=62^{\circ}, \angle R P S=34^{\circ}$, and $\angle Q P R=x^{\circ}$. What is the value of $x$ ?
(A) 11
(B) 28
(C) 17
(D) 31
(E) 34

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-070.jpg?height=448&width=586&top_left_y=768&top_left_x=1168)
Since $\triangle P R S$ is isosceles with $P R=P S$, then $\angle P R S=\angle P S R$.

Since the angles in $\triangle P R S$ add to $180^{\circ}$, then $\angle P R S+\angle P S R+\angle R P S=180^{\circ}$.

Therefore, $2(\angle P R S)+34^{\circ}=180^{\circ}$ or $2(\angle P R S)=146^{\circ}$ or $\angle P R S=73^{\circ}$.

Since $\triangle P Q T$ is isosceles with $P Q=P T$, then $\angle P Q T=\angle P T Q=62^{\circ}$.

Since $\angle P R S$ is an exterior angle to $\triangle P Q R$, then $\angle P R S=\angle P Q R+\angle Q P R$ or $73^{\circ}=62^{\circ}+x^{\circ}$. Therefore, $x=73-62=11$.

(Instead, we could have determined that $\angle P R Q=180^{\circ}-\angle P R S=180^{\circ}-73^{\circ}=107^{\circ}$, and then looked at the sum of the angles in $\triangle P Q R$ to get $62^{\circ}+x^{\circ}+107^{\circ}=180^{\circ}$ or $x=180-169=11$.)