12.074. A sphere is inscribed in a cone. The ratio of the radius of the circle of contact between the spherical and conical surfaces to the radius of the base of the cone is $k$. Find the cosine of the angle between the slant height of the cone and the plane of the base.
Solution.

From the problem statement, the radius of the circle of tangency between the spherical and conical surfaces will be $K O_{1}$, and the radius of the base of the cone will be $L O_{2}$ (Fig. 12.76). $\angle S L O_{2}=\alpha$. From the diagram, it is clear that $\triangle S K O \sim \Delta S O_{2} L$, and $\triangle S K O \sim \Delta K O_{1} O$, so $\angle K O O_{1}=\angle S L O_{2}=\alpha$. Then from $\Delta K O_{1} O$, $\mathrm{KO}=\frac{\mathrm{KO}_{1}}{\sin \alpha}$. From $\Delta \mathrm{LOO}_{2}$, $\mathrm{OO}_{2}=\mathrm{LO}_{2} \cdot \operatorname{tg} \frac{\alpha}{2}$. Since $\mathrm{KO}$ and $O \mathrm{O}_{2}$ are equal, then $\frac{K O_{1}}{\sin \alpha}=L O_{2} \operatorname{tg} \frac{\alpha}{2} \cdot \frac{K O_{1}}{L O_{2}}=\sin \alpha \operatorname{tg} \frac{\alpha}{2}=2 \sin ^{2} \frac{\alpha}{2}=k$. Therefore, $\sin \frac{\alpha}{2}=\sqrt{\frac{k}{2}}$. Hence, $\cos \alpha=1-2 \sin ^{2} \frac{\alpha}{2}=1-2 \cdot \frac{k}{2}=1-k$.

Answer: $1-k$.