In the diagram, $\triangle P Q R$ is isosceles with $P Q=P R$, and $Q R S T$ is a rectangle. If $\angle Q P R=70^{\circ}, \angle P Q R=x^{\circ}$, and $\angle R Q T=y^{\circ}$, the value of $x+y$ is
(A) 70
(B) 90
(C) 160
(D) 145
(E) 60

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-008.jpg?height=350&width=334&top_left_y=1950&top_left_x=1302)
$\triangle P Q R$ is isosceles with $P Q=P R$, and so $\angle P R Q=\angle P Q R$.

The sum of the angles in $\triangle P Q R$ is $180^{\circ}$.

Since $\angle Q P R=70^{\circ}$, then the other two angles in this triangle add to $180^{\circ}-70^{\circ}=110^{\circ}$.

Since these two angles are equal, they each measure $110^{\circ} \div 2=55^{\circ}$, and so $x=55$.

Since $Q R S T$ is a rectangle, each of its interior angles is a right angle, and so $y=90$.

The value of $x+y$ is $55+90=145$.

ANSWER: (D)