[ Law of Cosines ]

In triangle $ABC$, it is known that $AC=13, AB=14, BC=15$. A point $M$ is taken on side $BC$, such that $CM: MB=1: 2$. Find $AM$.

#
According to the Law of Cosines, find $\cos \angle C$ from triangle $A B C$.

## Solution

From the Law of Cosines, it follows that

$$
\cos \angle C=\frac{A C^{2}+B C^{2}-A B^{2}}{2 A C \cdot B C}=\frac{196+225-196}{2 \cdot 13 \cdot 15}=\frac{33}{13 \cdot 5}
$$

Thus,

$$
\begin{gathered}
A M^{2}=A C^{2}+C M^{2}-2 A C \cdot C M \cos \angle C= \\
=169+25-2 \cdot 13 \cdot 5 \cdot \frac{33}{13 \cdot 5}=194-66=128
\end{gathered}
$$

Therefore, $A M=8 \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_04ea0aec75eb28164af1g-05.jpg?height=397&width=577&top_left_y=1259&top_left_x=749)

Answer

$8 \sqrt{2}$.