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 surface of the ball and the contents of the base is 4: 3. A plane passing through the axis of a cone cuts the cone in an isosceles triangle. Determine the size of the angle with respect to the base of this triangle.

Correct answer:

A =  60 °

Step-by-step explanation:

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

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