In the diagram, $P Q$ is parallel to $R S$. Also, $Z$ is on $P Q$ and $X$ is on $R S$. If $Y$ is located between $P Q$ and $R S$ so that $\angle Y X S=20^{\circ}$ and $\angle Z Y X=50^{\circ}$, what is the measure of $\angle Q Z Y$ ?
(A) $30^{\circ}$
(B) $20^{\circ}$
(C) $40^{\circ}$
(D) $50^{\circ}$
(E) $60^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-064.jpg?height=244&width=499&top_left_y=1583&top_left_x=1214)
Solution 1

Draw a line segment $T Y$ through $Y$ parallel to $P Q$ and $R S$, as shown, and delete line segment $Z X$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-124.jpg?height=231&width=485&top_left_y=1603&top_left_x=880)

Since $T Y$ is parallel to $R S$, then $\angle T Y X=\angle Y X S=20^{\circ}$.

Thus, $\angle Z Y T=\angle Z Y X-\angle T Y X=50^{\circ}-20^{\circ}=30^{\circ}$.

Since $P Q$ is parallel to $T Y$, then $\angle Q Z Y=\angle Z Y T=30^{\circ}$.

Solution 2

Since $Q Z$ and $X S$ are parallel, then $\angle Q Z X+\angle Z X S=180^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-124.jpg?height=241&width=490&top_left_y=2254&top_left_x=877)

Now $\angle Q Z X=\angle Q Z Y+\angle Y Z X$ and $\angle Z X S=\angle Z X Y+\angle Y X S$.

We know that $\angle Y X S=20^{\circ}$.

Also, the sum of the angles in $\triangle X Y Z$ is $180^{\circ}$, so $\angle Y Z X+\angle Z X Y+\angle Z Y X=180^{\circ}$, or $\angle Y Z X+\angle Z X Y=180^{\circ}-\angle Z Y X=180^{\circ}-50^{\circ}=130^{\circ}$.

Combining all of these facts, we obtain $\angle Q Z Y+\angle Y Z X+\angle Z X Y+\angle Y X S=180^{\circ}$ or $\angle Q Z Y+130^{\circ}+20^{\circ}=180^{\circ}$.

From this, we obtain $\angle Q Z Y=180^{\circ}-130^{\circ}-20^{\circ}=30^{\circ}$.

ANSwER: (A)