12.050. A plane is drawn through vertex $C$ of the base of a regular triangular pyramid $S A B C$ perpendicular to the lateral edge $S A$. This plane forms an angle with the base plane, the cosine of which is $2 / 3$. Find the cosine of the angle between two lateral faces.
## Solution.

Let $C K B$ be the given section (Fig. 12.52). This means that $K C \perp A S$, $K B \perp A S$, $L$ is the midpoint of $B C$, and $\angle A L K=\alpha$. We need to find $\angle C K B=\beta$. Let $A B=B C=A C=a \cdot \triangle B K C$ be isosceles, with $K B=K C$ (since $C K \perp A S$, $K B \perp A S$, $A B=A C$, and $A K$ is common). Then $\operatorname{tg} \frac{\beta}{2}=\frac{L C}{K L}$. From the right triangle $A K L$ (since $K L \perp A S$, as $K L$ lies in the plane $K B C$), we find $K L$: $K L=A L \cos \alpha$, where $A L$ is the height of the equilateral $\triangle A B C$, $A L=\frac{a \sqrt{3}}{2}$. Then $K L=\frac{a \sqrt{3}}{2} \cos \alpha=\frac{a \sqrt{3}}{2} \cdot \frac{2}{3}=\frac{a \sqrt{3}}{3}$. Therefore, $\operatorname{tg} \frac{\beta}{2}=\frac{a \cdot 3}{2 \cdot a \sqrt{3}}=\frac{\sqrt{3}}{2}$. The value of $\cos \beta=\frac{1-\operatorname{tg}^{2} \frac{\beta}{2}}{1+\operatorname{tg}^{2} \frac{\beta}{2}}=\frac{1-\frac{3}{4}}{1+\frac{3}{4}}=\frac{1}{7}$.

Answer: $\frac{1}{7}$.