12.070. The lateral edges of a regular triangular pyramid are pairwise perpendicular. Find the angle between a lateral face and the plane of the base.
## Solution.

By the condition $\angle A S C=\angle C S B=\angle A S B=90^{\circ}$ (Fig. 12.72). Moreover, $A S=S B=S C, \triangle A B C$ is equilateral; $S K \perp B C, H K \perp B C$, where $H$ is the foot of the altitude. Since $\triangle S C B$ is a right and isosceles triangle, then $\angle S C B=45^{\circ}$. Let $S C$ be $a$, then $S K=S C \sin \angle S C B=a \sin 45^{\circ}=a \frac{\sqrt{2}}{2}$, and $B C=a \sqrt{2}$ (as the hypotenuse of $\triangle S C B$); $H K$ is the radius of the circle inscribed in $\triangle A B C$, so $H K=r=\frac{a \sqrt{2} \cdot \sqrt{3}}{6}$. Then $\cos \alpha=\frac{H K}{S K}=\frac{a \sqrt{2} \cdot \sqrt{3} \cdot 2}{6 a \sqrt{2}}=\frac{\sqrt{3}}{3}$. Therefore, $\alpha=\arccos \frac{\sqrt{3}}{3}$.

Answer: $\arccos \frac{\sqrt{3}}{3}$.