[ The Sum of Angles in a Triangle. The Exterior Angle Theorem. ] [ Criteria and Properties of an Isosceles Triangle. ]

In an isosceles triangle $A B C$, the heights $A D$ and $C E$, dropped to the lateral sides, form an angle $A M C$ equal to $48^{\circ}$. Find the angles of triangle $A B C$.
Angle $A M C$ together with angle $B$ sums up to $180^{\circ}$.

## Solution

Since angle $A M C$ together with angle $B$ sums up to $180^{\circ}$, then $\angle B=180^{\circ}-48^{\circ}=132^{\circ}$. Therefore, $\angle A=\angle C=\left(180^{\circ}-\right.$ $\left.132^{\circ}\right): 2=24^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4fce0705b02fa9e106d4g-12.jpg?height=329&width=548&top_left_y=950&top_left_x=755)

Answer

$24^{\circ}, 24^{\circ}, 132^{\circ}$.