3.52. A triangular pyramid is inscribed in a cone, with its lateral edges being pairwise perpendicular to each other. Find the angle between the slant height of the cone and its height.
3.52. Given that $DAB$ is a cone, $DABC$ is a pyramid inscribed in this cone, $AD \perp BD, \quad AD \perp CD, \quad DB \perp DC$, and $DO \perp (ABC)$ (Fig. 3.54). Since $DB = DC = DA$, $\triangle ADB = \triangle ADC = \triangle DBC$ as right triangles with two equal legs; therefore, $\triangle ABC$ is equilateral. Let $AB = a$; then $DA = \frac{a \sqrt{2}}{2}$. From $\triangle AOD$, we find $\sin \angle ADO = \frac{OA}{DA} = \frac{a \sqrt{3}}{3} : \frac{a \sqrt{2}}{2} = \frac{\sqrt{6}}{3}$, i.e., $\angle ADO = \arcsin \frac{\sqrt{6}}{3}$.

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Fig. 3.55