## Task A-3.6.

Given is a right triangle $A B C$. On the bisector of the right angle $\varangle A C B$ a point $M$ is chosen. If $\sin \varangle M A C=\frac{8}{17}$ and $\cos \varangle M B C=\frac{3}{5}$, find the ratio $|A B|:|C M|$.
## First Solution.

Let $D$ and $E$ be the feet of the perpendiculars from point $M$ to lines $A C$ and $B C$, respectively.

Then $C D M E$ is a rectangle whose diagonal $C M$ is the angle bisector of $\varangle E C D$. Therefore, $C D M E$ is a square.

![](https://cdn.mathpix.com/cropped/2024_05_30_acf83088931d085246bdg-18.jpg?height=603&width=1003&top_left_y=1983&top_left_x=458)

Let $\alpha = \varangle M A C$, $\beta = \varangle M B C$, $a = |B C|$, $b = |A C|$, and $d = |C D|$.

Since $C D M E$ is a square, we have $|C M| = d \sqrt{2}$.

Since $|A D| = b - d$, from the right triangle $A M D$ we see that

$$
\operatorname{ctg} \alpha = \frac{b - d}{d}
$$

1 point

Similarly, $|B E| = a - d$, so from the right triangle $A M D$ we see that

$$
\operatorname{ctg} \beta = \frac{a - d}{d}
$$

From $\sin \alpha = \frac{8}{17}$, we get $\operatorname{ctg} \alpha = \frac{15}{8}$, so from $\frac{b - d}{d} = \frac{15}{8}$ we get $b = \frac{23}{8} d$.

From $\cos \beta = \frac{3}{5}$, we get $\operatorname{ctg} \beta = \frac{3}{4}$, so from $\frac{a - d}{d} = \frac{3}{4}$ we get $a = \frac{7}{4} d$.

1 point

By the Pythagorean theorem, we have

$$
|A B|^{2} = a^{2} + b^{2} = \left(\frac{7}{4} d\right)^{2} + \left(\frac{23}{8} d\right)^{2} = \left(\frac{7^{2}}{4^{2}} + \frac{23^{2}}{8^{2}}\right) d^{2} = \frac{49 \cdot 4 + 529}{8^{2}} d^{2} = \frac{725}{64} d^{2}
$$

Finally, since $|C M| = d \sqrt{2}$, the desired ratio is

$$
\frac{|A B|}{|C M|} = \sqrt{\frac{725}{2 \cdot 64}} = \frac{5 \sqrt{29}}{8 \sqrt{2}} = \frac{5 \sqrt{58}}{16}
$$