12.444 The base of the pyramid is an equilateral triangle. One of the lateral faces of the pyramid is perpendicular to the plane of the base. Find the cosine of the angle between the other two lateral faces, if they form an angle \(\alpha\) with the plane of the base.
Solution. Let $FABC$ be the given pyramid, $FAB \perp ABC$. Then the height $FK$ of the pyramid lies in the plane $FAB$. Let $FM$ be the height of $\triangle FBC$. Then $BC \perp KM$ (Theorem 12) $\Rightarrow$

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-315.jpg?height=370&width=371&top_left_y=72&top_left_x=220)

$\Rightarrow \angle FMK = \alpha, KM = FK \cdot \operatorname{ctg} \alpha$, i.e., the distances from point $K$ to the sides $AC$ and $BC$ of the equilateral $\triangle ABC$ are equal (the face $FAC$ also forms an angle $\alpha$ with the base $ABC) \Rightarrow K$ is the midpoint of $AB \Rightarrow FA = FB \Rightarrow$

$\Rightarrow \triangle FAC = \triangle FBC \Rightarrow$ the perpendiculars dropped from $A$ and $B$ to the edge $FC$ intersect at one point $O \Rightarrow$ $\Rightarrow \angle AOB$ is the desired angle.

Let $a$ be the side of $\triangle ABC$. Then $KM = \frac{a \sqrt{3}}{4}; CM = \frac{3}{4}a$; $FM = \frac{KM}{\cos \alpha} \Rightarrow FC = \sqrt{CM^2 + FM^2} = \frac{a}{4 \cos \alpha} \sqrt{3 + 9 \cos^2 \alpha} \Rightarrow$ $\Rightarrow \frac{OB}{a} = \frac{FM}{FC} = \frac{1}{\sqrt{1 + 3 \cos^2 \alpha}} (\triangle FMC \sim \triangle OBC)$. From here, by the cosine rule for $\triangle AOB$, we have: $\cos \angle AOB = \frac{2 \cdot OB^2 - a^2}{2 \cdot OB^2} = 1 - \frac{1}{2} \left( \frac{a}{OB} \right)^2 = 1 - \frac{1 + 3 \cos^2 \alpha}{2}$.

Answer: $-\frac{1 + 3 \cos 2\alpha}{4}$.