3.38. The base of the pyramid $ABCD$ is a right triangle $ABC (\angle C=90^{\circ})$. The lateral edge $AD$ is perpendicular to the base. Find the acute angles of triangle $ABC$ if $\angle DBA=\alpha$ and $\angle DBC=\beta (\alpha<\beta)$.
3.38. By the condition, $DABC$ is a pyramid, $\angle ACB=90^{\circ}, AD \perp(ABC)$, $\angle DBA=\alpha, \angle DBC=\beta(\alpha<\beta)$ (Fig. 3.40). Let $AD=h$; then from $\triangle ADB$ we find $AB=h \operatorname{ctg} \alpha, DB=\frac{h}{\sin \alpha}$. Since $BC \perp AC$, then $DC \perp BC$ and from $\triangle DCB$ we get $BC=DB \cos \beta=\frac{h \cos \beta}{\sin \alpha}$. Thus, in $\triangle ABC$ we have

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-523.jpg?height=569&width=510&top_left_y=102&top_left_x=43)

Fig. 3.40

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-523.jpg?height=573&width=500&top_left_y=106&top_left_x=656)

Fig. 3.41

$$
\sin \angle A=\frac{BC}{AB}=\frac{h \cos \beta}{\sin \alpha \cdot h \operatorname{ctg} \alpha}=\frac{\cos \beta}{\cos \alpha}
$$

from which

$$
\angle A=\arcsin \frac{\cos \beta}{\cos \alpha}, \angle B=\frac{\pi}{2}-\arcsin \frac{\cos \beta}{\cos \alpha} .
$$