12.296. The diagonals $A B_{1}$ and $C B_{1}$ of two adjacent lateral faces of a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ form angles with the diagonal $A C$ of the base $A B C D$ that are equal to $\alpha$ and $\beta$, respectively. Find the angle between the plane of triangle $A B_{1} C$ and the plane of the base.
## Solution.

In the plane $A B_{1} C$, draw the perpendicular $B_{1} E \perp A C$ (Fig. 12.158) $\angle B_{1} A E=\alpha, \angle B_{1} C E=\beta$.

$B E$ is the projection of $B_{1} E$ on the base plane $A B C D$. Then $B E \perp A C$ and $\angle B_{1} E B$ is the angle between the plane $\triangle \mathrm{A} B_{1} C$ and the base plane.

Let $B_{1} E=1$.

In $\triangle B_{1} E A\left(\angle B_{1} E A=90^{\circ}\right): A E=B_{1} E \operatorname{ctg} \angle B_{1} A E=\operatorname{ctg} \alpha$.

In $\triangle B_{1} E C\left(\angle B_{1} E C=90^{\circ}\right): C E=B_{1} E \operatorname{ctg} \angle B_{1} C E=\operatorname{ctg} \beta$.

$B E$ is the height of the right $\triangle A B C \Rightarrow B E=\sqrt{A E \cdot C E}=\sqrt{\operatorname{ctg} \alpha \operatorname{ctg} \beta}$.

In $\triangle B_{1} B E\left(\angle B_{1} B E=90^{\circ}\right): \cos \angle B_{1} E B=\frac{B E}{B_{1} E}=\sqrt{\operatorname{ctg} \alpha \operatorname{ctg} \beta}$;

$\angle B_{1} E B=\arccos \sqrt{\operatorname{ctg} \alpha \operatorname{ctg} \beta}$.

Answer: $\arccos \sqrt{\operatorname{ctg} \alpha \operatorname{ctg} \beta}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1051.jpg?height=761&width=589&top_left_y=528&top_left_x=93)

Fig. 12.159