Sphere in cone

A sphere is inscribed in the cone (the intersection of their boundaries consists of a circle and one point). The ratio of the ball's surface and the area of the base is 4:3. A plane passing through the axis of a cone cuts the cone in an isosceles triangle. Find the size of the angle for the base of this triangle.

Correct answer:

A =  60 °

Step-by-step explanation:

$$\begin{gathered} G=1 \\ G:S_1=4:3 \\ {S}_1=\frac{3}{4}\cdot G=\frac{3}{4}\cdot 1=\frac{3\cdot 1}{4}=\frac{3}{4}=0.75 \\ {S}_1=\pi r_1^2 \\ {r}_1=\sqrt{S_1/\pi }=\sqrt{0.75/3.1416}\doteq 0.4886 \\ G=4\pi r_2^2 \\ {r}_2=\sqrt{\frac{G}{4\pi }}=\sqrt{\frac{1}{4\cdot 3.1416}}\doteq 0.2821 \\ \tan \alpha =r_2:r_1 \\ \alpha =\arctan (r_2/r_1)=\arctan (0.2821/0.4886)\doteq 0.5236\text{rad} \\ {\alpha }_2=2\cdot \alpha =2\cdot 0.5236\doteq 1.0472\text{rad} \\ A={\alpha }_2\to \text{°}={\alpha }_2\cdot \frac{180}{\pi }\text{°}=1.0472\cdot \frac{180}{\pi }\text{°}=60\text{°}=60^{\circ } \end{gathered}$$

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