Line segments $P Q$ and $R S$ are parallel. Points $T, U, V$ are placed so that $\angle Q T V=30^{\circ}, \angle S U V=40^{\circ}$, and $\angle T V U=x^{\circ}$, as shown. What is the value of $x$ ?
(A) 80
(B) 85
(C) 65
(D) 70
(E) 75

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-024.jpg?height=287&width=461&top_left_y=2231&top_left_x=1274)

## Part C: Each correct answer is worth 8.
Solution 1

We begin by constructing a line segment $W X$ perpendicular to $P Q$ and passing through $V$, as shown.

Since $R S$ is parallel to $P Q$, then $W X$ is also perpendicular to $R S$. In $\triangle T W V, \angle T W V=90^{\circ}$ and $\angle W T V=30^{\circ}$.

Since the sum of the angles in a triangle is $180^{\circ}$, then $\angle T V W=180^{\circ}-30^{\circ}-90^{\circ}=60^{\circ}$.

In $\triangle U X V, \angle U X V=90^{\circ}$ and $\angle V U X=40^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-119.jpg?height=284&width=442&top_left_y=281&top_left_x=1514)

Similarly, $\angle U V X=180^{\circ}-40^{\circ}-90^{\circ}=50^{\circ}$.

Since $W X$ is a straight line segment, then $\angle T V W+\angle T V U+\angle U V X=180^{\circ}$.

That is, $60^{\circ}+\angle T V U+50^{\circ}=180^{\circ}$ or $\angle T V U=180^{\circ}-60^{\circ}-50^{\circ}$ or $\angle T V U=70^{\circ}$, and so $x=70$.

## Solution 2

We begin by extending line segment $U V$ to meet $P Q$ at $Y$, as shown.

Since $R S$ is parallel to $P Q$, then $\angle T Y V$ and $\angle V U S$ are alternate angles, and so $\angle T Y V=\angle V U S=40^{\circ}$.

In $\triangle T Y V, \angle T Y V=40^{\circ}$ and $\angle Y T V=30^{\circ}$.

Since the sum of the angles in a triangle is $180^{\circ}$, then $\angle T V Y=180^{\circ}-40^{\circ}-30^{\circ}=110^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-119.jpg?height=296&width=509&top_left_y=882&top_left_x=1445)

Since $U Y$ is a straight line segment, then $\angle T V U+\angle T V Y=180^{\circ}$.

That is, $\angle T V U+110^{\circ}=180^{\circ}$ or $\angle T V U=180^{\circ}-110^{\circ}$ or $\angle T V U=70^{\circ}$, and so $x=70$.

## Solution 3

We begin by constructing a line segment $C D$ parallel to both $P Q$ and $R S$, and passing through $V$, as shown.

Since $C D$ is parallel to $P Q$, then $\angle Q T V$ and $\angle T V C$ are alternate angles, and so $\angle T V C=\angle Q T V=30^{\circ}$.

Similarly, since $C D$ is parallel to $R S$, then $\angle C V U$ and $\angle V U S$ are alternate angles, and so $\angle C V U=\angle V U S=40^{\circ}$.

Since $\angle T V U=\angle T V C+\angle C V U$, then $\angle T V U=30^{\circ}+40^{\circ}=70^{\circ}$,

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-119.jpg?height=287&width=461&top_left_y=1407&top_left_x=1447)
and so $x=70$.

ANSWER: (D)

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