12.046. The base of the pyramid is an equilateral triangle. One lateral edge is perpendicular to the base plane and equals $l$, while the other two form an angle $\alpha$ with the base plane. A right prism is inscribed in the pyramid; three of its vertices lie on the lateral edges of the pyramid, and the other three lie on the base of the pyramid. The diagonal of the lateral face of the prism forms an angle $\beta$ with the base plane. Find the height of the prism.
## Solution.

Given $A B=B C=A C, S A \perp A B, S A \perp A C, \angle S B A=\alpha=\angle S C A$, $\angle O A E=\beta, S A=l$ (Fig. 12.48). From $\triangle S A B\left(\angle S A B=90^{\circ}\right): A B=\dot{B C}=A C=$ $=l \cdot \operatorname{ctg} \alpha, S B=\frac{l}{\sin \alpha}$. Let $A E=F E=A F=x$, then from $\triangle A E O$ we have $O E=x \cdot \operatorname{tg} \beta=M A$. From $\triangle S M O\left(\angle S M O=90^{\circ}\right): S M=x \cdot \operatorname{tg} \alpha, S A=x \cdot \operatorname{tg} \alpha+$ $+x \cdot \operatorname{tg} \beta, x(\operatorname{tg} \alpha+\operatorname{tg} \beta)=l, x=\frac{l}{\operatorname{tg} \alpha+\operatorname{tg} \beta}, O E=\frac{l \operatorname{tg} \beta}{\operatorname{tg} \alpha+\operatorname{tg} \beta}=\frac{l \sin \beta}{\cos \beta\left(\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}\right)}=$ $=\frac{l \cos \alpha \sin \beta}{\sin (\alpha+\beta)}$.

Answer: $\frac{l \cos \alpha \sin \beta}{\sin (\alpha+\beta)}$.