Find ratio of the volume of two cone

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.

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. $$r_a=\dfrac{4r\theta}{\pi}$$, $$r_b=\dfrac{9r\theta}{\pi}$$
$$h_a^2=4r^2-r_a^2=4r^2-\dfrac{16r^2\theta^2}{\pi^2}$$
$$h_b^2=9r^2-r_b^2=9r^2-\dfrac{81r^2\theta^2}{\pi^2}$$
$$Va=\dfrac13{\pi}r_a^2h_a=\dfrac13{\pi}(\dfrac{4r\theta}{\pi})^2\sqrt{4r^2-\dfrac{16r^2\theta^2}{\pi^2}}$$
$$V_a:V_b=32\sqrt{1-\dfrac{4\theta^2}{\pi^2}}:243\sqrt{1-\dfrac{9\theta^2}{\pi^2}}$$