12.113. The base of the pyramid $ABCD$ is a right triangle $ABC\left(\angle C=90^{\circ}\right)$. The lateral edge $AD$ is perpendicular to the base. Find the acute angles of triangle $ABC$, if $\angle DBA=\alpha$ and $\angle DBC=\beta(\alpha<\beta)$.
## Solution.

Let $DABC$ be a pyramid, $\angle ACB=90^{\circ}, AD \perp(ABC), \angle DBA=\alpha$, $\angle DBC=\beta(\alpha<\beta)$ (Fig. 12.113). 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 $\sin \angle A=\frac{BC}{AB}=\frac{h \cos \beta}{\sin \alpha \cdot h \operatorname{tg} \alpha}=\frac{\cos \beta}{\cos \alpha}$, from which $\angle A=\arcsin \frac{\cos \beta}{\cos \alpha}$,

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

Answer: $\arcsin \frac{\cos \beta}{\cos \alpha} ; \arccos \frac{\cos \beta}{\cos \alpha}$.