**Introduction**

In the previous post, we have shown that the sum of the interior angles of a polygon with $latex n$ sides is $latex 180(n-2)$ 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 $latex 360^\circ$. For instance, in the pentagon, the sum of the interior angles is $latex 5(72^\circ) = 360^\circ$. 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 $latex 180(n-2)^\circ$. Let us say, if $latex a_1, a_2$ and $latex a_3$ are measures of the angles of a 3-sided polygon (a triangle), then $latex a_1 + a_2 + a_3 = 180(3-2) = 180^\circ$. Similarly, if we have a quadrilateral, $latex a_1 + a_2 + a_3 + a_4 = 180(4-2) = 360^\circ$. In general, if we have a polygon with $latex n$ sides, the equation

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

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 *$latex n$* 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 *$latex 1 + 2 + 3 + \cdots + 100$, *we mean the sum of all the positive integers from* $latex 1$* to* $latex 100$.

**Theorem**

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

**Proof**

We add all the exterior angles of a polygon with $latex n$ 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 $latex S$ of a polygon is

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

Regrouping we have

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

Since there are $latex n$ $latex 180$’s, we have

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

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

$latex S = 180n – (180n – 360)$

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

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