12.274. A cube is inscribed in a cone (one of the faces of the cube lies in the plane of the base of the cone). The ratio of the height of the cone to the edge of the cube is $k$. Find the angle between the slant height and the height of the cone.
## Solution.

Let the edge of the cube be $a$ (Fig. 12.138).

Then $O_{1} C_{1}=\frac{a \sqrt{2}}{2}, S O=k a, S O_{1}=S O-O O_{1}=a(k-1)$.

In $\triangle S O_{1} C_{1}\left(\angle S O_{1} C_{1}=90^{\circ}\right):$

\[
\begin{aligned}
& \operatorname{ctg} \angle O_{1} S C_{1}=\frac{S O_{1}}{O_{1} C_{1}}=\frac{a(k-1)}{\frac{a \sqrt{2}}{2}}=\sqrt{2}(k-1) \\
& \angle O_{1} S C_{1}=\operatorname{arcctg}(\sqrt{2}(k-1))
\end{aligned}
\]

Answer: $\operatorname{arcctg}(\sqrt{2}(k-1))$.