12.266. The base of the pyramid is an equilateral triangle. Two lateral faces are perpendicular to the plane of the base. The sum of two unequal plane angles at the vertex is $\frac{\pi}{2}$. Find these angles.
Solution.

Let the faces $SBA$ and $SBC$ of the pyramid $SABC$ be perpendicular to the plane $ABC$ of the base (Fig. 12.131), and $\triangle ABC$ is equilateral. Then $\triangle SBA = \triangle SBC$ by two legs, and $\angle ASB = \angle CSB, AS = SC$.

If $\angle ASC = \alpha$, then $\angle ASB = \frac{\pi}{2} - \alpha$.

$D$ is the midpoint of $AC$.

In $\triangle ADS \left(\angle ADS = \frac{\pi}{2}\right): AD = AS \sin \frac{\alpha}{2}$.

In $\triangle ABS \left(\angle ABS = \frac{\pi}{2}\right): AB = AS \sin \left(\frac{\pi}{2} - \alpha\right)$.

Since $AB = 2AD$, then $AS \sin \left(\frac{\pi}{2} - \alpha\right) = 2AS \sin \frac{\alpha}{2}; \cos \alpha = 2 \sin \frac{\alpha}{2}$.

$0 < \alpha < \pi, \sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}} \Rightarrow \cos \alpha = 2 \sqrt{\frac{1 - \cos \alpha}{2}};$
$\cos^2 \alpha = 2(1 - \cos \alpha) \cos^2 \alpha + 2 \cos \alpha - 2 = 0; \cos \alpha = -1 \pm \sqrt{3}$.

Considering that $0 < \cos \alpha < 1$, we have $\cos \alpha = \sqrt{3} - 1, \alpha = \arccos (\sqrt{3} - 1)$, $\angle ASB = \frac{\pi}{2} - \arccos (\sqrt{3} - 1)$

Answer: $\arccos (\sqrt{3} - 1); \frac{\pi}{2} - \arccos (\sqrt{3} - 1)$