In the diagram, $A M N, A P Q, Q R M$, and $P R N$ are all straight lines. The value of $a+b$ is
(A) 70
(B) 55
(C) 80
(D) 90
(E) 75

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-124.jpg?height=352&width=369&top_left_y=342&top_left_x=1320)
Since $\angle P R M=125^{\circ}$, then $\angle Q R P=\angle N R M=55^{\circ}$.

Then $\angle A P R=180^{\circ}-\angle Q P R=180^{\circ}-\left[180^{\circ}-a^{\circ}-55^{\circ}\right]=55^{\circ}+a^{\circ}$.

Similarly, $\angle Q P R=55^{\circ}+b^{\circ}$. (This is an external angle in $\triangle Q P R$.)

Since $A Q$ is a straight line,

$$
\begin{aligned}
\left(55^{\circ}+a^{\circ}\right)+\left(55^{\circ}+b^{\circ}\right) & =180^{\circ} \\
a^{\circ}+b^{\circ}+110^{\circ} & =180^{\circ} \\
a+b & =70
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-241.jpg?height=534&width=506&top_left_y=432&top_left_x=1357)