$\underline{\text { Folklore }}$

In triangle $ABC$, the median drawn from vertex $A$ to side $BC$ is four times smaller than side $AB$ and forms an angle of $60^{\circ}$ with it. Find angle $A$.
Extending median $A M$ to its length, we obtain parallelogram $A B D C$ (see figure). In triangle $A B D$, we draw median $D E$, then $A E=1 / 2 A B=A D$. Thus, triangle $A D E$ is isosceles with an angle of $60^{\circ}$, meaning it is equilateral. Therefore, the median $D E$ of triangle $A B D$ is equal to half the side $A B$ to which it is drawn; hence, this triangle is a right triangle. Then $\angle B A C=\angle B A D+\angle C A D=\angle B A D+\angle A D B=150^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0e468145bb05e34a17aag-09.jpg?height=289&width=900&top_left_y=1496&top_left_x=585)

## Answer

$150^{\circ}$.

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