12.081. The diagonals of the lateral faces of a rectangular parallelepiped form angles $\alpha$ and $\beta$ with the base plane. Find the angle between the diagonal of the parallelepiped and the base plane.
Solution.

Let the sides of the parallelepiped be $\boldsymbol{a}$ and $\boldsymbol{b}$, and the height $h$ (Fig. 12.83). Then $h=a \operatorname{tg} \alpha=b \operatorname{tg} \beta$, from which $\frac{b}{a}=\frac{\operatorname{tg} \alpha}{\operatorname{tg} \beta}$. The diagonal of the base

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-775.jpg?height=500&width=442&top_left_y=124&top_left_x=167)

Fig. 12.83

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-775.jpg?height=530&width=474&top_left_y=95&top_left_x=715)

Fig. 12.84

$$
\begin{aligned}
& d=\sqrt{a^{2}+b^{2}} \cdot \operatorname{tg} \gamma=\frac{h}{d}=\frac{a \operatorname{tg} \alpha}{\sqrt{a^{2}+b^{2}}}=\frac{\operatorname{tg} \alpha}{\sqrt{1+\frac{b^{2}}{a^{2}}}}=\frac{\operatorname{tg} \alpha}{\sqrt{1+\frac{\operatorname{tg}^{2} \alpha}{\operatorname{tg}^{2} \beta}}}=\frac{\operatorname{tg} \alpha \operatorname{tg} \beta}{\sqrt{\operatorname{tg}^{2} \alpha+\operatorname{tg}^{2} \beta}}= \\
& =\frac{\frac{1}{\operatorname{ctg} \alpha} \cdot \frac{1}{\operatorname{ctg} \beta}}{\sqrt{\frac{1}{\operatorname{ctg}^{2} \alpha}+\frac{1}{\operatorname{ctg}^{2} \beta}}}=\frac{1}{\sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta}}, \text { or } \operatorname{ctg} \gamma=\sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta}
\end{aligned}
$$

Answer: $\quad \operatorname{arcctg} \sqrt{\operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta}$.