In the diagram, $P Q$ and $R S$ intersect at $T$. If $\angle S T Q=140^{\circ}$ and $\angle P T U=90^{\circ}$, what is the measure of $\angle R T U$ ?
(A) $30^{\circ}$
(B) $90^{\circ}$
(C) $50^{\circ}$
(D) $40^{\circ}$
(E) $140^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-012.jpg?height=358&width=282&top_left_y=225&top_left_x=1344)
Solution 1

$P Q$ and $R S$ intersect at $T$. Thus, $\angle P T R$ and $\angle S T Q$ are vertically opposite angles and so $\angle P T R=\angle S T Q=140^{\circ}$.

Since $\angle P T R=\angle P T U+\angle R T U$, then

$$
\begin{aligned}
\angle R T U & =\angle P T R-\angle P T U \\
& =140^{\circ}-90^{\circ} \\
& =50^{\circ}
\end{aligned}
$$

and so the measure of $\angle R T U$ is $50^{\circ}$.

## Solution 2

$R S$ is a line segment and so $\angle R T Q+\angle S T Q=180^{\circ}$ or $\angle R T Q=180^{\circ}-140^{\circ}=40^{\circ}$.

$P Q$ is a line segment and so $\angle R T Q+\angle R T U+\angle P T U=180^{\circ}$ or $\angle R T U=180^{\circ}-40^{\circ}-90^{\circ}=50^{\circ}$.