# The Sum of the Exterior Angles of a Polygon

Introduction

In the previous post, we have shown that the sum of the interior angles of a polygon with $n$ sides is $180(n-2)$ degrees. In this post, we explore the exterior angles of a polygon. The exterior angle is formed by extending the side of the polygon as shown in the following figure.

One observation about the exterior angles is that their angle sum is always $360^\circ$. For instance, in the pentagon, the sum of the interior angles is $5(72^\circ) = 360^\circ$. The next question is, is this observation always true? Is this always true for all polygons, even non-regular ones? Try a few more examples by drawing or by using a geometry software.

Scratch Work

First we note that the measure of the exterior angle is

180 – measure of adjacent interior angle.

Second, the sum of the interior angles of a polygon is $180(n-2)^\circ$. Let us say, if $a_1, a_2$ and $a_3$ are measures of the angles of a 3-sided polygon (a triangle), then $a_1 + a_2 + a_3 = 180(3-2) = 180^\circ$. Similarly, if we have a  quadrilateral,  $a_1 + a_2 + a_3 + a_4 = 180(4-2) = 360^\circ$. In general, if we have a polygon with $n$ sides, the equation

$a_1 + a_2 + a_3 + \cdots + a_n = 180 (n-2)$.

We will use  these notations and equations so we can prove the theorems later, but before that…

Don’t Panic!

The previous equation only shows the sum of all the interior angles of a polygon with $n$ sides. The 3-dot symbol is used to indicate that there are terms in between that were not written. The 3-dot symbol is used to shorten the equation.  For instance, if we write $1 + 2 + 3 + \cdots + 100$, we mean the sum of all the positive integers from $1$ to $100$.

Theorem

The sum of the exterior angles of a polygon is $360^\circ$.

Proof

We add all the exterior angles of a polygon with $n$ sides.  From above, we have learned that the angle measure of a regular polygon is 180 minus the measure of its adjacent interior angle. Therefore, the exterior angle sum $S$ of a polygon is

$S = (180 - a_1) + (180 - a_2) + (180 - a_3) + \cdots + (180 - a_n)$.

Regrouping we have

$S = 180 + 180 + 180 + \cdots + 180 - (a_1 + a_2 + a_3 + \cdots + a_n)$

Since there are $n$ $180$‘s, we have

$S = 180n - (a_1 + a_2 + a_3 + \cdots + a_n)$.

But from above, $a_1 + a_2 + a_3 + \cdots a_n = 180(n-2)$, so

$S = 180n - (180n - 360)$

Simplifying, we have $S = 360^\circ$.

Therefore, the sum of the exterior angles of a polygon with $n$ sides is $360^\circ$.