# 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.

Question:

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. Determine the number of sides of two polygons.

Solution:

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