12.151. The hypotenuse of a right-angled triangle is divided by the point of tangency of the inscribed circle into segments, the ratio of which is $k$. Find the angles of the triangle.
## Solution.

Let $O$ be the center of the circle inscribed in triangle $ABC$, $\angle BAC = \pi / 2$, and $D$ be the point of tangency of this circle with the hypotenuse $BC$ (Fig. 12.24), $BD: DC = k$, $\angle BAC = \alpha$. Then $\angle OBD = \frac{\alpha}{2}$, $\angle OCD = \frac{\pi}{4} - \frac{\alpha}{2}$.

$$
\text{In } \triangle ODC \left(\angle ODC = \frac{\pi}{2}\right): OD = DC \tan\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)
$$

In $\triangle ODB \left(\angle ODB = \frac{\pi}{2}\right): OD = DB \tan \frac{\alpha}{2} = k DC \tan \frac{\alpha}{2} \Rightarrow$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0909.jpg?height=520&width=339&top_left_y=77&top_left_x=194)

Fig. 12.24

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0909.jpg?height=436&width=555&top_left_y=161&top_left_x=643)

Fig. 12.25

$$
\begin{aligned}
& \Rightarrow k \tan \frac{\alpha}{2} = \tan\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \Leftrightarrow \frac{k \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} = \frac{\sin \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)}{\cos \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)} \Leftrightarrow \\
& \Leftrightarrow k \sin \frac{\alpha}{2} \cos \left(\frac{\pi}{2} - \frac{\alpha}{2}\right) = \sin \left(\frac{\pi}{4} - \frac{\alpha}{2}\right) \cos \frac{\alpha}{2} \Leftrightarrow \\
& \Leftrightarrow k \frac{1}{2}\left(\sin \left(\alpha - \frac{\pi}{4}\right) + \sin \frac{\pi}{4}\right) = \frac{1}{2}\left(\sin \frac{\pi}{4} + \sin \left(\frac{\pi}{4} - \alpha\right)\right) \Leftrightarrow \\
& \Leftrightarrow \sin \left(\frac{\pi}{4} - \alpha\right) + k \sin \left(\frac{\pi}{4} - \alpha\right) = k \sin \frac{\pi}{4} - \sin \frac{\pi}{4} \Leftrightarrow \\
& \Leftrightarrow \sin \left(\frac{\pi}{4} - \alpha\right) \cdot (k + 1) = \frac{\sqrt{2}(k - 1)}{2} \Leftrightarrow \\
& \Leftrightarrow \sin \left(\frac{\pi}{4} - \alpha\right) = \frac{\sqrt{2}(k - 1)}{2(k + 1)} \Leftrightarrow \alpha = \frac{\pi}{4} - \arcsin \frac{\sqrt{2}(k - 1)}{2(k + 1)} \\
& \angle ACB = \frac{\pi}{2} - \alpha = \frac{\pi}{4} + \arcsin \frac{\sqrt{2}(k - 1)}{2(k + 1)} \\
& \text{Answer: } \frac{\pi}{4} \pm \arcsin \frac{\sqrt{2}(k - 1)}{2(k + 1)}
\end{aligned}
$$