# What is the sum of the exterior angles of a polygon with 4 sides? 5 sides? 6 sides? n sides?

What is the sum of the exterior angles of a polygon with 4 sides? 5 sides? 6 sides? n sides?
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lelestalis80d
Step 1
The interior angle of a polygon can be found by:
$\frac{180\left(n-2\right)}{n}$ where n is the number of sides
a) 4 sides
Interior angle is equal to:
$\frac{180\left(4-2\right)}{4}=90$
Therefore, exterior angle is equal to $180-90={90}^{\circ }$
The sum $=4×90°=360°$
b) 5 sides
Interior angle is equal to:
$\frac{180\left(5-2\right)}{5}=108$
Therefore, exterior angle is equal to $180-108={72}^{\circ }$
The sum $=5×72°=360°$
c) 6 sides
Interior angle is equal to:
$\frac{180\left(6-2\right)}{6}=120$
Therefore, exterior angle is equal to $180-120={60}^{\circ }$
The sum $=6×60°=360°$
d) n sides
Interior angle is equal to:
$\frac{180\left(n-2\right)}{n}$
So exterior angle is equal to $180-\frac{180\left(n-2\right)}{n}$ which can be simplified
$180-\frac{180\left(n-2\right)}{n}$
$=\frac{180n-\left(180\left(n-2\right)\right)}{n}$
$=\frac{180n-180n+360}{n}$
exterior angle $=\frac{360}{n}$
The sum $=\frac{360}{n}×n=360°$
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Aleah Avery
$\text{The sum of the exterior angles of any polygon is}\phantom{\rule{1ex}{0ex}}{360}^{\circ }$