[ Doubling the Median ] [ Law of Cosines ]

Determine the angle $A$ between the sides of lengths 2 and 4, if the median drawn from vertex $A$ is $\sqrt{7}$.
On the extension of the median $A M$ of the given triangle, lay off the segment $M D$, equal to the segment $A M$.

## Solution

On the extension of the median $A M$ of the given triangle $A B C$ with sides $A B=2$ and $A C=4$, lay off the segment $M D$, equal to the segment $A M$. Then the quadrilateral $A B D C$ is a parallelogram, so $C D=A B=2$. Applying the cosine rule to triangle $A C D$, we find that

$$
\cos \angle A C D=\frac{A C^{2}+C D^{2}-A D^{2}}{2 \cdot A C \cdot C D}=\frac{16+4-4 \cdot 7}{2 \cdot 4 \cdot 2}=-\frac{1}{2}
$$

so $\angle A C D=120^{\circ}$. Therefore,

$$
\angle B A C=180^{\circ}-\angle A C D=180^{\circ}-120^{\circ}=60^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-17.jpg?height=623&width=594&top_left_y=0&top_left_x=732)

## Answer

$60^{\circ}$.