The angle in one regular polygon is to that in another as 3 : 2 and the number of sides in first is twice that in the second.

Let the number of sides in the first polygon be 2x and the number of sides in the second polygon is x.

We know:

Angle of an $n$-sided regular polygon $=\left(\frac{n-2}{n}\right) \pi$ radian

$\therefore$ Angle of the first polygon $=\left(\frac{2 x-2}{2 x}\right) \pi=\left(\frac{x-1}{x}\right) \pi$ radian

Angle of the second polygon $=\left(\frac{x-2}{x}\right) \pi$ radian

Thus, we have:

$\frac{\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right) \pi}{\left(\frac{\mathrm{x}-2}{\mathrm{x}}\right) \pi}=\frac{3}{2}$

$\Rightarrow \frac{x-1}{x-2}=\frac{3}{2}$

$\Rightarrow 2 x-2=3 x-6$

$\Rightarrow x=4$

Thus,

Number of sides in the first polygon = 2x = 8

Number of sides in the first polygon = x = 4