12.280. The base of the pyramid is a rectangle $A B C D$ ($A B \| C D$). The lateral edge $O A$ is perpendicular to the base. The edges $O B$ and $O C$ form angles with the base, respectively equal to $\alpha$ and $\beta$. Find the angle between the edge $O D$ and the base.
Solution.

Edge $OA$ is the height of the pyramid $OABCD$ (Fig. 12.143), $\angle OBA=\alpha$, $\angle OCA=\beta$, $\angle ODA$ is the angle between edge $OD$ and the base to be found.

Let $OA=1, \angle ODA=\gamma$, then from $\triangle OAB\left(\angle OAB=90^{\circ}\right): AB=OA$ $\operatorname{ctg} \angle OBA=\operatorname{ctg} \alpha$.

$\mathrm{B} \triangle OAC\left(\angle OAC=90^{\circ}\right): AC=OA \operatorname{ctg} \angle OCA=\operatorname{ctg} \beta$.

$\mathrm{B} \triangle OAD\left(\angle OAD=90^{\circ}\right): AD=OA \operatorname{ctg} \angle ODA=\operatorname{ctg} \gamma$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1035.jpg?height=498&width=713&top_left_y=76&top_left_x=322)

Fig. 12.143

From $\triangle ADC\left(\angle ADC=90^{\circ}\right): AD^{2}=AC^{2}-CD^{2} ;$

$$
\begin{aligned}
& \operatorname{ctg}^{2} \gamma=\operatorname{ctg}^{2} \beta-\operatorname{ctg}^{2} \alpha \Rightarrow \\
& \Rightarrow \operatorname{ctg}^{2} \gamma=(\operatorname{ctg} \beta-\operatorname{ctg} \alpha)(\operatorname{ctg} \beta+\operatorname{ctg} \alpha)=\frac{\sin (\alpha-\beta) \sin (\alpha+\beta)}{\sin ^{2} \alpha \sin ^{2} \beta} \\
& \operatorname{tg} \gamma=\frac{\sin \alpha \sin \beta}{\sqrt{\sin (\alpha-\beta) \sin (\alpha+\beta)}} \\
& \gamma=\operatorname{arctg} \frac{\sin \alpha \sin \beta}{\sqrt{\sin (\alpha-\beta) \sin (\alpha+\beta)}}
\end{aligned}
$$

Answer: $\operatorname{arctg} \frac{\sin \alpha \sin \beta}{\sqrt{\sin (\alpha-\beta) \sin (\alpha+\beta)}}$.