The ratio of angles in one regular polygon to that in another is $3 : 2$ . The number of sides in the first is twice the second. How many sides do the second polygon have?

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Solution

Suppose the second regular polygon has number of side = x
$∴$ The first regular polygon will have number of side = 2x.
$∴$ Each angle of the first polygon $= \frac{(4 x - 4)}{2 x}$ right angle.
And each angle of the second polygon $= \frac{(2 x - 4)}{x}$ right angle.
$∴ \frac{4 x - 4}{2 x} : \frac{2 x - 4}{x} = 3 : 2$
$\frac{4 x - 4}{4} = \frac{6 x - 12}{x}$
$4 x - 4 = 6 x - 12$
$2 x = 8$
$x = 4$
$∴$ The number of sides in the first and second polygons is respectively is 8 and 4

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