In $\triangle A B C, \angle B=72^{\circ}$. What is the sum, in degrees, of the other two angles?
(A) 144
(D) 110
(B) 72
(E) 288
(C) 108

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-100.jpg?height=220&width=274&top_left_y=1256&top_left_x=1381)
In $\triangle A B C, \angle B=72^{\circ}$. What is the sum, in degrees, of the other two angles?
(A) 144
(B) 72
(C) 108
(D) 110
(E) 288

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-439.jpg?height=296&width=417&top_left_y=790&top_left_x=1361)

## Solution

There are $180^{\circ}$ in a triangle.

Therefore, $\angle A+\angle C+72^{\circ}=180^{\circ}$ ( $\angle A, \angle B$ are in degrees.)

$$
\angle A+\angle C=108^{\circ}
$$

ANSWER: (C)