In the diagram, $P T$ is parallel to $Q R$. What is the size of $\angle P Q R$ ?
(A) $116^{\circ}$
(B) $168^{\circ}$
(C) $138^{\circ}$
(D) $144^{\circ}$
(E) $122^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-111.jpg?height=244&width=496&top_left_y=672&top_left_x=1194)

## Part B: Each correct answer is worth 6.
Solution 1

Since $P T$ and $R Q$ are parallel, then $2 x^{\circ}=128^{\circ}$, so $x=64$, so $\angle T P Q=64^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-202.jpg?height=249&width=504&top_left_y=2166&top_left_x=867)

Since $P T$ and $Q R$ are parallel, then $\angle T P Q$ and $\angle P Q R$ are supplementary. Thus, $\angle P Q R+64^{\circ}=180^{\circ}$, so $\angle P Q R=116^{\circ}$.

## Solution 2

Since the two angles at $R$ add to $180^{\circ}$, then $\angle Q R T+128^{\circ}=180^{\circ}$, so $\angle Q R T=52^{\circ}$.

Since $P T$ and $Q R$ are parallel, then $\angle P T R$ and $\angle Q R T$ are supplementary, so $2 x^{\circ}+52^{\circ}=180^{\circ}$ or $2 x^{\circ}=128^{\circ}$ or $x=64$.

Therefore, three of the angles of quadrilateral $P Q R T$ are $64^{\circ}, 128^{\circ}$ and $52^{\circ}$.

Since the angles in a quadrilateral add to $360^{\circ}$, then $\angle P Q R=360^{\circ}-64^{\circ}-128^{\circ}-52^{\circ}=116^{\circ}$.

ANSWER: (A)

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