$\left.\begin{array}{l}{\left[\begin{array}{l}\text { Auxiliary circle }\end{array}\right]} \\ {[\text { Equilateral (equiangular) triangle }]}\end{array}\right]$

Outside the equilateral triangle $A B C$, but inside the angle $B A C$, a point $M$ is taken such that the angle $C M A$ is $30^{\circ}$ and the angle $B M A$ is $\alpha$. Find the angle $A B M$.
Draw a circle with center at point $B$ and radius $B A$. Point $M$ belongs to this circle.

## Solution

With center at point $B$ and radius $B A$, draw a circle. Then point $M$ belongs to this circle. Therefore, triangle $A B M$ is isosceles. Consequently,

$$
\angle A B M=180^{\circ}-2 \angle A M B=180^{\circ}-2 \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-54.jpg?height=406&width=780&top_left_y=926&top_left_x=639)

Answer

$180^{\circ}-2 \alpha$.