3.70. 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.
3.70. By the condition, in $\triangle A B C$ we have: $B C=a, \angle C=\alpha$ radians, $\angle B=\beta$ radians, $A D \perp B C, A D-$ is the radius of the circle intersecting $A C$ and $A B$ at points $K$ and $L$ (Fig. 3.71); we need to find $l_{\cup K L}$. From $\triangle A D C$ and $\triangle A D B$ we find $C D=$ $=A D \operatorname{ctg} \alpha, B D=A D \operatorname{ctg} \beta$. Then $a=$ $=C D+D B=A D(\operatorname{ctg} \alpha+\operatorname{ctg} \beta)$, i.e.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-544.jpg?height=396&width=525&top_left_y=415&top_left_x=628)

Fig. 3.71

$$
A D=\frac{a}{\operatorname{ctg} \alpha+\operatorname{ctg} \beta}=\frac{a \sin \alpha \sin \beta}{\sin (\alpha+\beta)}
$$

Considering that $\angle K A L=\pi-(\alpha+\beta)$, we finally get

$$
l_{\cup K L}=A D(\pi-\alpha-\beta)=\frac{a(\pi-\alpha-\beta) \sin \alpha \sin \beta}{\sin (\alpha+\beta)}
$$