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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.

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

$h_a^2=4r^2-r_a^2=4r^2-\dfrac{16r^2\theta^2}{\pi^2}$

$h_b^2=9r^2-r_b^2=9r^2-\dfrac{81r^2\theta^2}{\pi^2}$

$Va=\dfrac13{\pi}r_a^2h_a=\dfrac13{\pi}(\dfrac{4r\theta}{\pi})^2\sqrt{4r^2-\dfrac{16r^2\theta^2}{\pi^2}}$

$V_a:V_b=32\sqrt{1-\dfrac{4\theta^2}{\pi^2}}:243\sqrt{1-\dfrac{9\theta^2}{\pi^2}}$

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