[ Law of Sines [Trigonometric Ratios in a Right Triangle]

The hypotenuse $A B$ of the right triangle $A B C$ is 9, and the leg $B C$ is 3. A point $M$ is taken on the hypotenuse such that $A M: M B=1: 2$. Find $C M$.

#
Use the Law of Cosines.

## Solution

From the right triangle $A B C$, we find that

$$
\cos \angle B=\frac{B C}{A B}=\frac{1}{3}
$$

In triangle $B M C$, the sides $B C=3$ and $B M=\frac{2}{3} A B=6$ and the cosine of the angle between them are known. By the Law of Cosines

$$
C M^{2}=B C^{2}+B M^{2}-2 B C \cdot B M \cos \angle B=9+36-2 \cdot 3 \cdot 6 \cdot \frac{1}{3}=33
$$

Therefore, $C M=\sqrt{33}$.

From the right triangle $A B C$, we find that

$$
\cos \angle B=\frac{B C}{A B}=\frac{1}{3}
$$

In triangle $B M C$, the sides $B C=3$ and $B M=\frac{2}{3} A B=6$ and the cosine of the angle between them are known. By the Law of Cosines

$$
C M^{2}=B C^{2}+B M^{2}-2 B C \cdot B M \cos \angle B=9+36-2 \cdot 3 \cdot 6 \cdot \frac{1}{3}=33
$$

Therefore, $C M=\sqrt{33}$.

From the right triangle $A B C$, we find that

$$
\cos \angle B=\frac{B C}{A B}=\frac{1}{3}
$$

In triangle $B M C$, the sides $B C=3$ and $B M=\frac{2}{3} A B=6$ and the cosine of the angle between them are known. By the Law of Cosines

Therefore, $C M=\sqrt{33}$.

## Answer

$\sqrt{33}$.