# 6.10: Sum of the Exterior Angles of a Polygon

**At Grade**Created by: CK-12

## Learning Objectives

- Identify the exterior angles of convex polygons.
- Find the sums of exterior angles in convex polygons.

This lesson focuses on the **exterior angles** in a polygon. There is a surprising feature of the *sum of the exterior angles* in a polygon that will help you solve problems about **regular** polygons.

## Exterior Angles in Convex Polygons

Recall that **interior** means *inside* and that **exterior** means *outside*. So, an **exterior angle** is an angle on the *outside* of a polygon. An **exterior angle** is formed by extending a side of the polygon:

There are two possible **exterior angles** for any given vertex on a polygon. In the figure above, one set of exterior angles is shown, the set in the *counter-clockwise* direction. The other set of **exterior angles** would be formed by extending each side of the polygon in the opposite *(clockwise)* direction. However, it does not matter which exterior angles you use because their measurement will be the *same* on each vertex. Look closely at one vertex below, where we draw both of the *possible* **exterior angles:**

In the above diagrams, both **exterior angles** are drawn separately. On the next page, both exterior angles on a single vertex are drawn *together*:

As you can see, the two **exterior angles** at the same vertex are **vertical angles**.

Since *vertical angles are congruent*, the two exterior angles possible around a single vertex are *congruent*.

The clockwise exterior angle and the counter-clockwise exterior angle at the *same* vertex are _______________________________ .

Additionally, because the exterior angle will be a **linear pair** with its *adjacent interior angle*, it will always be **supplementary** to that interior angle.

As a reminder, **supplementary** angles have a sum of \begin{align*}180^\circ\end{align*}.

This means the exterior angle and interior angle at the same vertex will sum to ______\begin{align*}^\circ\end{align*}.

**Reading Check**

1. *True or False: Vertical angles are supplementary.*

2. *True or False:*

*Exterior angles are on the outside of a polygon and are formed when you extend one side of the polygon.*

3. *Name the 2 sets of exterior angles:*

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

4. *Fill in the blank: Angles that form a linear pair add up to* ________\begin{align*}^\circ\end{align*}.

5. *Complete the sentence:*

*Since the interior angle and the exterior angle at the same vertex of a polygon form a linear pair,* ...

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

**Example 1**

*What is the measure of the exterior angle \begin{align*}\angle OKL\end{align*} in the diagram below?*

The **interior angle** \begin{align*}\angle JKL\end{align*} is labeled as \begin{align*}45^\circ\end{align*}.

Notice that the **interior angle** and the **exterior angle** form a **linear pair**, meaning the two angles are **supplementary** – they add up to \begin{align*}180^\circ\end{align*}.

So, to find the measure of the **exterior angle**, subtract \begin{align*}45^\circ\end{align*} from \begin{align*}180^\circ\end{align*}:

\begin{align*}180^\circ - 45^\circ = 135^\circ\end{align*}

The measure of \begin{align*}\angle OKL\end{align*} is \begin{align*}135^\circ\end{align*}.

## Summing Exterior Angles in Convex Polygons

By now you might expect that if you add up various angles in polygons, there will be some sort of pattern or rule.

For example, you already know that the sum of the **interior angles** in a triangle will always be \begin{align*}180^\circ\end{align*}.

From that fact, you learned that you can find the *sums* of the interior angles of any polygon with \begin{align*}n\end{align*} sides using the expression \begin{align*}180^\circ(n - 2)\end{align*}.

There is also a rule for **exterior angles** in a polygon.

Let’s begin by looking at a triangle:

To find the **exterior angles** at each vertex, extend the segments and find angles supplementary to the interior angles:

The *sum* of these three **exterior angles** is:

\begin{align*}150^\circ + 120^\circ + 90^\circ = 360^\circ\end{align*}

So, the **exterior angles** in this triangle will sum to \begin{align*}360^\circ\end{align*}.

Let’s see what happens with another shape.

To compare, examine the exterior angles of a rectangle:

In a rectangle, each **interior angle** measures ______ \begin{align*}^\circ\end{align*}.

Since **exterior angles** are **supplementary** to **interior angles**, all exterior angles in a rectangle will *also* measure ______ \begin{align*}^\circ\end{align*}.

Find the sum of the 4 **exterior angles** in a rectangle:

\begin{align*}90^\circ + 90^\circ + 90^\circ + 90^\circ = 360^\circ\end{align*}

So, the *sum* of the **exterior angles** in a rectangle is *also* \begin{align*}360^\circ\end{align*}.

In fact, the sum of the exterior angles in **any** convex polygon will *always* be \begin{align*}360^\circ\end{align*}.

It does not matter how many sides the polygon has, the sum will *always* be \begin{align*}360^\circ\end{align*}.

**Exterior Angle Sum**

The sum of the exterior angles of any convex polygon is \begin{align*}360^\circ\end{align*}.

No matter how many sides a polygon has, the sum of its __________________________ angles is equal to _______ \begin{align*}^\circ\end{align*}.

We can prove this using algebra and the facts that at any vertex the sum of the interior and one of the exterior angles is always \begin{align*}180^\circ\end{align*}, and the sum of all interior angles in a polygon is \begin{align*}180^\circ(n - 2)\end{align*}. The proof is on the next page.

**Proof:**

At any vertex of a polygon the exterior angle and the interior angle sum to \begin{align*}180^\circ\end{align*}. So, adding up all of the exterior angles and interior angles gives a total of \begin{align*}180^\circ\end{align*} times the number of vertices:

\begin{align*}(\text{Sum of Exterior Angles}) + (\text{Sum of Interior Angles}) = 180^\circ n\end{align*}

On the other hand, we already saw that the sum of the interior angles was:

\begin{align*}(\text{Sum of Interior Angles}) & = 180^\circ (n - 2)\\ & = 180^\circ n - 360^\circ \ (\text{using the Distributive Property})\end{align*}

Putting these together we have:

\begin{align*}180^\circ n & = (\text{Sum of Exterior Angles}) + (\text{Sum of Interior Angles})\\ & = (180^\circ n - 360^\circ) + (\text{Sum of Exterior Angles})\end{align*}

Subtract \begin{align*}(180^\circ n - 360^\circ)\end{align*} from both sides and:

\begin{align*}360^\circ = (\text{Sum of Exterior Angles})\end{align*}

**Reading Check**

1. *True or False:*

*In a polygon, an interior angle and one of the exterior angles at the same vertex are supplementary.*

2. *True or False: Supplementary angles add up to* \begin{align*}90^\circ\end{align*}.

3. *Fill in the blanks:*

*The sum of all interior __________________ in a polygon with \begin{align*}n\end{align*} sides is equal to* _______\begin{align*}^\circ \cdot (n - 2)\end{align*}.

4. *What is the rule for the sum of exterior angles in a polygon? Describe.*

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

**Example 2**

*What is \begin{align*}m \angle QXZ\end{align*} in the diagram below?*

\begin{align*}\angle QXZ\end{align*} in the diagram is an **exterior angle**. So, we need to find the measure of one exterior angle on a polygon given the measures of all of the others.

4 of the 5 exterior angles on this polygon have their measurements labeled:

\begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;} ^\circ, \underline{\;\;\;\;\;\;\;\;\;\;\;\;} ^\circ, \underline{\;\;\;\;\;\;\;\;\;\;\;\;} ^\circ,\end{align*} and \begin{align*}\underline{\;\;\;\;\;\;\;\;\;\;\;\;} ^\circ\end{align*}

We know that the *sum* of the exterior angles on a polygon must be equal to \begin{align*}360^\circ\end{align*}, regardless of how many sides the shape has. So, we can set up an equation where we set all of the exterior angles shown (including \begin{align*}m \angle QXZ\end{align*}) summed and equal to \begin{align*}360^\circ\end{align*}:

\begin{align*}70^\circ + 60^\circ + 65^\circ + 40^\circ + m \angle QXZ & \ = \ 360^\circ\\ 235^\circ + m \angle QXZ & \ = \ 360^\circ\\ -235^\circ{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} & {\;\;\;\;}- 235^\circ\\ m \angle QXZ & \ = \ 125^\circ\end{align*}

The measure of the missing exterior angle (\begin{align*}\angle QXZ\end{align*}) is \begin{align*}125^\circ\end{align*}.

We can check that our answer is reasonable by inspecting the diagram and checking whether the angle in question is acute, right, or obtuse. Since the angle should be obtuse, \begin{align*}125^\circ\end{align*} is a reasonable answer (assuming the diagram is accurate).

**Reading Check**

*The sum of the exterior angles of any convex polygon is equal to* ________\begin{align*}^\circ\end{align*}.

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