Given two sector ABC and PQR, $\angle A=2\theta$, $\angle P=3\theta, AC=2r, PR=3r, $ both sectors are folded into a right circular cone, find the ratio of the volume of two cone. tio

I am having trouble doing this question, and I doubt if the result is not a simple ratio. Here is what I have got:

The ratio of the base area = $16 : 81$

The ratio of the height = $\frac{2r}{360^\circ}\sqrt{(360^\circ)^2-4\theta^2}:\frac{3r}{360^\circ}\sqrt{(360^\circ)^2-9\theta^2}$

And it cannot be further simplified.

Any form of help will be appreciated.


I got same ratio of the base area. $r_a=\dfrac{4r\theta}{\pi}$, $r_b=\dfrac{9r\theta}{\pi}$






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