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 result:

A =  46.8264 °

Solution:

$$\begin{gathered} G=1 \\ G:S_1=4:3 \\ {S}_1={\displaystyle \frac{4}{3}}\cdot G={\displaystyle \frac{4}{3}}\cdot 1\doteq 1.3333 \\ {S}_1=\pi r_1^2 \\ {r}_1=\sqrt{S_1\mathrm{/}\pi }=\sqrt{1.3333\mathrm{/}3.1416}\doteq 0.6515 \\ G=4\pi r_2^2 \\ {r}_2=\sqrt{{\displaystyle \frac{G}{4\pi }}}=\sqrt{{\displaystyle \frac{1}{4\cdot 3.1416}}}\doteq 0.2821 \\ \tan \alpha =r_2:r_1 \\ \alpha =\arctan (r_2\mathrm{/}{r}_1)=\arctan (0.2821\mathrm{/}0.6515)\doteq 0.4086\text{ rad } \\ {\alpha }_2=2\cdot \alpha =2\cdot 0.4086\doteq 0.8173\text{ rad } \\ A={\alpha }_2{\to }^{\circ }={\alpha }_2\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=0.8173\cdot {{\displaystyle \frac{180}{\pi }}}^{\circ }=46.826^{\circ }=46.8264^{\circ }=46^{\circ }49^{\mathrm{\prime }}35\mathrm{"} \end{gathered}$$

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