149. $n$ equal cones have a common vertex. Each touches two others along a generatrix, and all touch one plane. Find the angle at the vertex of the axial section of these cones.
149. Consider the pyramid $S A B C$ (Fig. 28), in which $|C A| = |A B|, \quad \widehat{B A C} = \frac{2 \pi}{n}, S A$ is perpendicular to the plane $A B C$,

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-089.jpg?height=421&width=326&top_left_y=723&top_left_x=166)

Fig. -8. II such that the vertex $A$ projects onto the plane $S B C$ at point $O$ - the center of the circle inscribed in $S B C$.

We inscribe a cone in this pyramid such that its vertex coincides with $A$, and the circle of its base is the circle inscribed in $S B C$. Clearly, if we take $n$ such pyramids, the bases of which are located on the plane $A B C$ such that their bases, equal to $\triangle A B C$, form a regular $n$-gon with center at $A$, then the cones inscribed in these pyramids form the system of cones you need.

Next, let $D$ be the midpoint of $B C, |O D| = r, |A D| = l$. Then $|S D| = \frac{l^{2}}{r}, |B D| = l \operatorname{tg} \frac{\pi}{n}$. Since $\widehat{S B D} = 2 \widehat{O B D}, \operatorname{tg} \widehat{S B D} = \frac{|S D|}{|B D|} = \frac{l}{r \operatorname{tg} \frac{-\pi}{n}}, \operatorname{tg} \widehat{O B D} = \frac{r}{l \operatorname{tg} \frac{\pi}{n}}$, we can form the equation

$$
\frac{l}{r \operatorname{tg} \frac{\pi}{n}} = \frac{2 \frac{r}{l \operatorname{tg} \frac{\pi}{n}}}{1 - \frac{r^{2}}{l^{2} \operatorname{tg}^{2} \frac{\pi}{n}}}
$$

from which $\frac{r}{l} = \frac{\operatorname{tg} \frac{\pi}{n}}{\sqrt{1 + 2 \operatorname{tg}^{2} \frac{\pi}{n}}}$.

$$
\text { Answer: } \quad 2 \arcsin \frac{\operatorname{tg} \frac{\pi}{n}}{\sqrt{1 + 2 \operatorname{tg}^{2} \frac{\pi}{n}}}
$$