12.279. The cosine of the angle between two adjacent lateral faces of a regular quadrilateral pyramid is $k$. Find the cosine of the angle between a lateral face and the plane of the base and the permissible values of $k$.
## Solution.

Let $S O$ be the height of the regular pyramid $S A B C D$ (Fig. 12.142), $\angle B E D$ be the angle between the lateral faces $B S C$ and $D S C$, $F$ be the midpoint of $C D$, $\angle S F O$ be the angle between the lateral face and the base plane, $\angle B E D=\alpha, \angle S C D=\beta, \angle S F O=\gamma$.

By the cosine theorem for the trihedral angle with vertex $C$:

$$
\cos \alpha=\frac{\cos 90^{\circ}-\cos \beta \cos \gamma}{\sin \beta \sin \beta}=-\operatorname{ctg}^{2} \beta
$$

In $\triangle S F C\left(\angle S F C=90^{\circ}\right): \operatorname{ctg} \beta=\frac{C F}{S F}$.

In $\triangle \operatorname{SOF}\left(\angle S O F=90^{\circ}\right): \cos \gamma=\frac{O F}{S F}$.

Since $C F=O F$, then $\cos \gamma=\operatorname{ctg} \beta \Rightarrow \cos \alpha=-\cos ^{2} \gamma$, $\cos \gamma=\sqrt{-\cos \alpha}=\sqrt{-k}$.

Since $0<\operatorname{cosg}<1$, then $0<\sqrt{-k}<1,0<-k<1,-1<k<0$.

Answer: $\sqrt{-k} ;-1<k<0$.