[Trigonometric relationships in a right triangle]

Find the angle at the vertex of the axial section of a right circular cone if it is known that there are three generatrices of the lateral surface of the cone, each pair of which are perpendicular to each other.

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From the condition of the problem, 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 plane angles at vertex $P$ are right angles (see Fig. 11.3). Therefore, this pyramid is regular, and its height is the segment $P O$, where $O$ is the center of the base of the cone. Then the required

![](https://cdn.mathpix.com/cropped/2024_05_06_bca77ef9e87e9e47ba11g-23.jpg?height=89&width=2018&top_left_y=2574&top_left_x=0)
$B P D=2 \arcsin \frac{\sqrt{6}}{3}$. If the required angle is calculated using the cosine theorem from triangle $B P D$, the answer can be obtained in another form $\angle B P D=\arccos \frac{2 P^{\prime} B^{2}-B D^{2}}{2 P^{\prime} B^{2}}=\arccos \left(-\frac{1}{3}\right)=\pi-\arccos \frac{\frac{1}{3}}{3}$. In any case, the required angle is obtuse.

![](https://cdn.mathpix.com/cropped/2024_05_06_bca77ef9e87e9e47ba11g-24.jpg?height=292&width=354&top_left_y=0&top_left_x=858)

## Answer

$2 \arcsin ^{\frac{\sqrt{6}}{3}}$.

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