12.118. Find the cosine of the angle between adjacent lateral faces of a regular quadrilateral pyramid, where the lateral edge is equal to the side of the base.
Solution.

From $\triangle A O C$ by the cosine theorem $A C^{2}=2 A O^{2}-2 A O^{2} \cos \alpha$ (Fig. 12.118). From this, $\cos \alpha=\frac{2 A O^{2}-A C^{2}}{2 A O^{2}}$. From $\triangle A B C\left(\angle A B C=90^{\circ}\right)$: $A C=\sqrt{2 a^{2}}=a \sqrt{2}$. From $\triangle A O B \quad\left(\angle A O B=90^{\circ}\right): A O=\sqrt{a^{2}-\frac{a^{2}}{4}}=\frac{a \sqrt{3}}{2}$.

Then we finally get: $\cos \alpha=\frac{\frac{2 \cdot 3 a^{2}}{4}-2 a^{2}}{\frac{3 a^{2}}{2}}=-\frac{1}{3}$.

Answer: $-\frac{1}{3}$.