I'm trying to find the radius of the larger of the two circular planes of a frustum of a right circular cone.
I know the volume of the frustum, the radius of the smaller circle, and the angle between the sides and vertical. What's the formula to find the larger circle's radius?
Notice in the given figure, consider right triangles
The normal height of the smaller cone (cap of the frustum) $$=r\cot\alpha$$ Volume of smaller cone with circular base of radius $r$ & normal height $r\cot\alpha$ $$V_1=\frac{1}{3}\pi r^2(r\cot\alpha)=\frac{1}{3}\pi r^3\cot\alpha$$
The normal height of the larger cone (frustum with its cap) $$=R\cot\alpha$$ Volume of larger cone with circular base of radius $R$ & normal height $R\cot\alpha$ $$V_2=\frac{1}{3}\pi R^2(R\cot\alpha)=\frac{1}{3}\pi R^3\cot\alpha$$
Now, the volume of the frustum of cone $$V=\text{(volume of larger cone)} - \text{(volume of smaller cone)}$$ $$V=V_1-V_2$$ $$V=\frac{1}{3}\pi R^3\cot\alpha-\frac{1}{3}\pi r^3\cot\alpha$$ $$R^3=\frac{3}{\pi\cot\alpha}\left(V+\frac{1}{3}\pi r^3\cot\alpha\right)=\frac{3V}{\pi\cot \alpha}+r^3$$ $$\color{red}{R=\left(\frac{3V}{\pi\cot\alpha}+r^3\right)^{1/3}}$$
