12.131. In an acute-angled triangle $A B C$, the altitude $A D=a$, the altitude $C E=b$, and the acute angle between $A D$ and $C E$ is $\alpha$. Find $A C$.
Solution.

Since $\triangle A B C$ is an acute-angled triangle, the point $O$ of intersection of its altitudes lies inside the triangle, $\angle A O E=\alpha$ (Fig. 12.4).

$\triangle A E O$ and $\triangle A D B$ are right triangles with a common angle $\angle D A B$. Therefore, $\angle B=\angle A O E=\alpha$.

From $\triangle C E B\left(\angle C E B=90^{\circ}\right): C B=\frac{C E}{\sin \angle B}=\frac{b}{\sin \alpha}$.

From $\triangle A D B\left(\angle A D B=90^{\circ}\right): A B=\frac{A D}{\sin \angle B}=\frac{a}{\sin \alpha}$:

From $\triangle A B C$:

$A C^{2}=A B^{2}+B C^{2}-2 \cdot A B \cdot B C \cdot \cos \angle B=\frac{a^{2}}{\sin ^{2} \alpha}+\frac{b^{2}}{\sin ^{2} \alpha}-\frac{2 a b \cos \alpha}{\sin ^{2} \alpha} ;$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0894.jpg?height=421&width=581&top_left_y=64&top_left_x=97)

Fig. 12.4

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0894.jpg?height=301&width=558&top_left_y=193&top_left_x=704)

Fig. 12.5

$$
A C=\frac{\sqrt{a^{2}+b^{2}-2 a b \cos \alpha}}{\sin \alpha}
$$

Answer: $\frac{\sqrt{a^{2}+b^{2}-2 a b \cos \alpha}}{\sin \alpha}$.