12.115. A sphere is inscribed in a cone, the surface of which is equal to the area of the base of the cone. Find the cosine of the angle at the vertex in the axial section of the cone.
## Solution.

By the condition $S_{\mathrm{w}}=S_{\text {osn }}$ (Fig. 12.115). Hence, $4 \pi R_{\mathrm{u}}^{2}=\pi R_{\mathrm{k}}^{2} \Rightarrow R_{\mathrm{k}}=2 R_{\mathrm{w}}$. Consider $\triangle S C A$. By the Law of Cosines, $(2 A O)^{2}=2 S A^{2}-2 S A^{2} \cos \alpha=2 S A^{2}(1-\cos \alpha)$. Therefore, $\cos \alpha=\frac{2 S A^{2}-2\left(R_{\mathrm{K}}\right)^{2}}{2 S A^{2}}=1-\frac{2 R_{\mathrm{K}}^{2}}{S A^{2}} \cdot \triangle S O A$ and $\triangle S B O_{1}$ are similar: $\frac{O A}{B O_{1}}=\frac{S A}{S O_{1}} \Rightarrow S A=S O_{1} \cdot \frac{R_{\mathrm{\kappa}}}{R_{\mathrm{w}}}=2 S O_{1} \cdot$ In $\triangle S O A: S A^{2}=O A^{2}+S O^{2}, \mathrm{a}$ and $S O=S O_{1}+O_{1} O=\frac{S A}{2}+O_{1} B$. Substituting: $S A^{2}=O A^{2}+\left(\frac{S A}{2}+O_{1} B\right)^{2}$. $S A^{2}=R_{\mathrm{k}}^{2}+\frac{S A^{2}}{4}+S A \cdot R_{\mathrm{w}}+R_{\mathrm{m}}^{2}=5 R_{\mathrm{m}}^{2}+\frac{S A^{2}}{4}+S A \cdot R_{\mathrm{m}}$. Solving the equation
$\frac{3 S A^{2}}{4}-S A \cdot R_{\mathrm{m}}-5 R_{\mathrm{m}}^{2}=0$, we get: 1) $S A=\frac{10 R_{\mathrm{m}}}{3}$; 2) $S A=\frac{-6 R_{\mathrm{m}}}{3}$ (not suitable). Finally, we obtain: $\cos \alpha=1-\frac{2 \cdot 4 R_{\mathrm{m}}^{2} \cdot 9}{100 R_{\mathrm{m}}^{2}}=1-\frac{18}{25}=\frac{7}{25}$. Answer: $\frac{7}{25}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-800.jpg?height=488&width=470&top_left_y=714&top_left_x=749)

Fig. 12.115