12.016. Two heights of a parallelogram, drawn from the vertex of the obtuse angle, are equal to $h_{1}$ and $h_{2}$, and the angle between them is $\alpha$. Find the larger diagonal of the parallelogram.
Solution.

By the condition $B F \perp C D, B F=h_{2}, B E \perp A D, B E=h_{1}, \angle E B F=\alpha$. Then $\angle A B F=\angle C F B=90^{\circ} \quad(A B \| C D), \angle A B E=90^{\circ}-\alpha$. From $\triangle B E A$ $\angle B E A=90^{\circ}, A E=h_{1} \cdot \operatorname{tg}\left(90^{\circ}-\alpha\right)=h_{1} \cdot \operatorname{ctg} \alpha$ (Fig. 12.19). From $\triangle B F C$ $\angle B F C=90^{\circ}, \quad C F=h_{2} \cdot \operatorname{ctg} \alpha, \quad A B=C D=\frac{h_{1}}{\sin \alpha}, \quad B C=A D=\frac{h_{2}}{\sin \alpha}$,

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-727.jpg?height=300&width=596&top_left_y=94&top_left_x=124)

Fig. 12.19

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-727.jpg?height=294&width=482&top_left_y=97&top_left_x=751)

Fig. 12.20

$\angle D=180^{\circ}-\alpha$. By the cosine rule $A C^{2}=\frac{h_{2}^{2}+h_{1}^{2}+2 h_{1} h_{2} \cdot \cos \alpha}{\sin ^{2} \alpha}$, therefore, $A C=\frac{\sqrt{h_{2}^{2}+h_{1}^{2}+2 h_{1} h_{2} \cdot \cos \alpha}}{\sin \alpha}$.

Answer: $\frac{\sqrt{h_{2}^{2}+h_{1}^{2}+2 h_{1} h_{2} \cdot \cos \alpha}}{\sin \alpha}$.