3. Find the angle at the vertex of the axial section of a right circular cone, given that there exist three generatrices of the lateral surface of the cone that are pairwise perpendicular to each other.

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# Solution

From the problem statement, it follows that a triangular pyramid $P A B C$ can be inscribed in the given cone, where the lateral edges $P A, P B$, and $P C$ are equal, and all planar angles at vertex $P$ are right angles. Therefore, this pyramid is regular, and its height is the segment $P O$, where $O$ is the center of the base of the cone. The angle $B P D$ is then twice the angle $B P O$. Let $P B = b$, then $B C = b \sqrt{2} ; O B = \frac{B C \sqrt{3}}{3} = \frac{b \sqrt{6}}{3}$. Therefore, $\sin \angle B P O = \frac{O B}{P B} = \frac{\sqrt{6}}{3} ; \angle B P D = 2 \arcsin \frac{\sqrt{6}}{3}$. If the desired angle is calculated using the cosine theorem from triangle $B P D$, the answer can be obtained in another form: $\angle B P D = \arccos \frac{2 P H^2 - B D^2}{2 P B^2} = \arccos \left(-\frac{1}{3}\right) = \pi - \arccos \frac{1}{3}$. In any case, the desired angle is obtuse.

![](https://cdn.mathpix.com/cropped/2024_05_06_b07dcda4069114bfba77g-2.jpg?height=294&width=314&top_left_y=830&top_left_x=948)

Answer: $2 \arcsin \frac{\sqrt{6}}{3}$.