3.58. In triangle $ABC$, angle $A$ is equal to $\alpha$ and side $BC=a$. Find the length of the bisector $AD$, if the angle between the bisector $AD$ and the altitude $AE$ is $\beta$.
3.58. By the condition, $\angle B A C=\alpha, B C=a, \angle D A B=\angle D A C, A E \perp B C$, $\angle D A E=\beta$ (Fig. 3.60). We have $\angle B A D=\frac{\alpha}{2}, \angle A D E=90^{\circ}-\beta$; then $\angle A B D=90^{\circ}-\beta-\frac{\alpha}{2}$ (according to the property of the exterior angle of a triangle), $\angle A C D=180^{\circ}-\frac{\alpha}{2}-\left(90^{\circ}-\beta\right)=90^{\circ}+\beta-\frac{\alpha}{2}$. Let $A D=l$; then by the sine theorem from $\triangle A D B$ and $\triangle A D C$ we get

$$
\begin{aligned}
& \frac{B D}{\sin \frac{\alpha}{2}}=\frac{l}{\sin \left(90^{\circ}-\left(\beta+\frac{\alpha}{2}\right)\right)}=\frac{l}{\cos \left(\beta+\frac{\alpha}{2}\right)} ; B D=\frac{l \sin \frac{\alpha}{2}}{\cos \left(\beta+\frac{\alpha}{2}\right)} \\
& \frac{D C}{\sin \frac{\alpha}{2}}=\frac{l}{\sin \left(90^{\circ}-\left(\beta-\frac{\alpha}{2}\right)\right)}=\frac{l}{\cos \left(\beta-\frac{\alpha}{2}\right)} ; D C=\frac{l \sin \frac{\alpha}{2}}{\cos \left(\beta-\frac{\alpha}{2}\right)}
\end{aligned}
$$

Therefore,

$$
a=B D+D C=\frac{l \sin \frac{\alpha}{2}\left(\cos \left(\beta-\frac{\alpha}{2}\right)+\cos \left(\beta+\frac{\alpha}{2}\right)\right)}{\cos \left(\beta-\frac{\alpha}{2}\right) \cos \left(\beta+\frac{\alpha}{2}\right)}=
$$

$$
=\frac{l \sin \frac{\alpha}{2} \cdot 2 \cos \beta \cos \frac{\alpha}{2}}{\cos \left(\beta-\frac{\alpha}{2}\right) \cos \left(\beta+\frac{\alpha}{2}\right)}=\frac{l \sin \alpha \cos \beta}{\cos \left(\beta-\frac{\alpha}{2}\right) \cos \left(\beta+\frac{\alpha}{2}\right)}
$$

from which

$$
l=\frac{a \cos \left(\beta-\frac{\alpha}{2}\right) \cos \left(\beta+\frac{\alpha}{2}\right)}{\sin \alpha \cos \beta}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-537.jpg?height=339&width=421&top_left_y=613&top_left_x=45)

Fig. 3.61