In triangle $K M N$, it is known that $\sin \angle K N M=\frac{\sqrt{3}}{2}$ and $\cos \angle K M N=\frac{1}{3}$. Find the ratio of the lengths of the altitudes dropped from vertex $N$ to side $M$ and from vertex $M$ to side $N K$.

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From the right triangles $M A N$ and $N B M$, express the heights $M A$ and $N B$ of triangle $K M N$ in terms of the side $M N$.

## Solution

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-28.jpg?height=392&width=349&top_left_y=2191&top_left_x=860)

Answer

$\frac{4 \sqrt{6}}{9}$. Since $\cos \angle K M N=\frac{1}{3}$ and $\angle K M N$ is an angle of the triangle, then $\sin \angle K M N=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}$. Let $M A$ and $N B$ be the specified heights of triangle $A B C$. From the right triangles $M A N$ and $N B M$, we find that

$$
A M=M N \cdot \sin \angle A N M=M N \cdot \frac{\sqrt{3}}{2} \text { and } B N=M N \cdot \sin \angle K M N=M N \cdot \frac{2 \sqrt{2}}{3} .
$$

Therefore,

$$
\frac{B N}{A M}=\frac{M N \cdot \frac{2 \sqrt{2}}{3}}{M N \cdot \frac{\sqrt{3}}{2}}=\frac{4 \sqrt{6}}{9}
$$

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