12.191. The base of the triangle is equal to $a$, and the angles at the base are $\alpha$ and $\beta$ radians. From the opposite vertex of the triangle, a circle is drawn with a radius equal to the height of the triangle. Find the length of the arc of this circle that is enclosed within the triangle.
## Solution.

Let in $\triangle A B C$ (Fig.12.58) $B C=a, \angle C=\alpha, \angle B=\beta, A K$-height, $A K=H$.

In $\triangle A K C\left(\angle A K C=\frac{\pi}{2}\right): C K=H \operatorname{ctg} \alpha$.

From $\triangle B K A\left(\angle B K A=\frac{\pi}{2}\right): B K=H \operatorname{ctg} \beta$.

Since $C K+B K=B C$, then $H \operatorname{ctg} \alpha+H \operatorname{ctg} \beta=a$,

$$
\frac{H \sin (\alpha+\beta)}{\sin \alpha \sin \beta}=a, H=\frac{a \sin \alpha \sin \beta}{\sin (\alpha+\beta)}
$$

The length of the arc $l=\gamma R$, where $\gamma$ is the radian measure of the arc, $\mathrm{R}$ is its radius; $\gamma=$ $\angle B A C=\pi-\alpha-\beta, R=H$,

$$
l=a(\pi-\alpha-\beta) \cdot \frac{\sin \alpha \sin \beta}{\sin (\alpha+\beta)}
$$

Answer: $a(\pi-\alpha-\beta) \cdot \frac{\sin \alpha \sin \beta}{\sin (\alpha+\beta)}$.