In the diagram, points $Q$ and $R$ lie on $P S$ and $\angle Q W R=38^{\circ}$. If $\angle T Q P=\angle T Q W=x^{\circ}$, $\angle V R S=\angle V R W=y^{\circ}$, and $U$ is the point of intersection of $T Q$ extended and $V R$ extended, then the measure of $\angle Q U R$ is
(A) $71^{\circ}$
(B) $45^{\circ}$
(C) $76^{\circ}$
(D) $81^{\circ}$
(E) $60^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-046.jpg?height=515&width=542&top_left_y=1605&top_left_x=1171)

Part C: Each correct answer is worth 8.
Since $\angle T Q P$ and $\angle R Q U$ are opposite angles, then $\angle R Q U=\angle T Q P=x^{\circ}$.

Similarly, $\angle Q R U=\angle V R S=y^{\circ}$.

Since the angles in a triangle add to $180^{\circ}$, then

$$
\angle Q U R=180^{\circ}-\angle R Q U-\angle Q R U=180^{\circ}-x^{\circ}-y^{\circ}
$$

Now $\angle W Q P$ and $\angle W Q R$ are supplementary, as they lie along a line.

Thus, $\angle W Q R=180^{\circ}-\angle W Q P=180^{\circ}-2 x^{\circ}$.

Similarly, $\angle W R Q=180^{\circ}-\angle W R S=180^{\circ}-2 y^{\circ}$.

Since the angles in $\triangle W Q R$ add to $180^{\circ}$, then

$$
\begin{aligned}
38^{\circ}+\left(180^{\circ}-2 x^{\circ}\right)+\left(180^{\circ}-2 y^{\circ}\right) & =180^{\circ} \\
218^{\circ} & =2 x^{\circ}+2 y^{\circ} \\
x^{\circ}+y^{\circ} & =109^{\circ}
\end{aligned}
$$

Finally, $\angle Q U R=180^{\circ}-x^{\circ}-y^{\circ}=180^{\circ}-\left(x^{\circ}+y^{\circ}\right)=180^{\circ}-109^{\circ}=71^{\circ}$.

ANSWER: (A)