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# Find the number of sides of a regular polygon whose each exterior angle is half of it's interior angle

Given :

The exterior angle of a regular polygon is half of its interior angle.

To do :

We have to find the number of sides of the polygon.

Solution :

Let the number of sides of the regular polygon be 'n'.

The exterior angle of a regular polygon with n sides $= \frac{360}{n}$

The interior angle of a regular polygon with n sides $=180 - \frac{360}{n}$

Here, the exterior angle is half of its interior angle.

$\frac{360}{n} = \frac{1}{2}(180 - \frac{360}{n})$

$\frac{360\times 2}{n} = 180 - \frac{360}{n}$

$\frac{360\times 2}{n} = \frac{180 n - 360}{n}$

$360\times 2 =180 n - 360 $ [$n$ on both sides get cancelled]

$360\times 2 + 360=180 n $

Take 360 as common in LHS,

$360(2 + 1) = 180 n $

$360 \times 3 = 180 n$

Rewrite,

$180 n = 360 \times 3$

$n = \frac{360 \times 3}{180}$

$n = 2 \times 3$ $[\frac{360}{180} = 2]$

$n = 6$

Therefore, the number of sides of the regular polygon is 6.