12.095. The lateral face of a regular truncated triangular pyramid forms an acute angle $\alpha$ with the base plane. Find the angle between the height and the lateral edge of the pyramid.
Solution.

From $\triangle S O D\left(\angle S O D=90^{\circ}\right): S O=D O \operatorname{tg} \alpha$, and $D O=\frac{x \sqrt{3}}{6}$ (radius of the inscribed circle) (Fig. 12.96). Therefore, $S O=\frac{x \sqrt{3}}{6} \operatorname{tg} \alpha$. From $\triangle S O A$ $\left(\angle S O A=90^{\circ}\right) \operatorname{tg} \beta=\frac{A O}{S O} ; A O=\frac{x \sqrt{3}}{3}$ (radius of the circumscribed circle). Then $\operatorname{tg} \beta=\frac{x \sqrt{3} \cdot 6}{3 x \sqrt{3} \operatorname{tg} \alpha}=2 \operatorname{ctg} \alpha$. Therefore, $\beta=\operatorname{arctg}(2 \operatorname{ctg} \alpha)$.

Answer: $\quad \operatorname{arctg}(2 \operatorname{ctg} \alpha)$.