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# Exterior Angles Theorems

## Exterior angles equal the sum of the remote interiors.

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Exterior Angles Theorems

### Exterior Angle Theorems

An exterior angle is the angle formed by one side of a polygon and the extension of the adjacent side. In all polygons, there are two sets of exterior angles, one going around the polygon clockwise and the other goes around the polygon counterclockwise. By the definition, the interior angle and its adjacent exterior angle form a linear pair.

The Exterior Angle Sum Theorem states that each set of exterior angles of a polygon add up to 360\begin{align*}360^\circ\end{align*}.

m1+m2+m3m4+m5+m6=360=360\begin{align*}m \angle 1+m \angle 2+m \angle 3 &= 360^\circ\\ m \angle 4+m \angle 5+m \angle 6 &= 360^\circ\end{align*}

Remote interior angles are the two angles in a triangle that are not adjacent to the indicated exterior angle. A\begin{align*}\angle A\end{align*} and B\begin{align*}\angle B\end{align*} are the remote interior angles for exterior angle ACD\begin{align*}\angle ACD\end{align*}.

The Exterior Angle Theorem states that the sum of the remote interior angles is equal to the non-adjacent exterior angle. From the picture above, this means that mA+mB=mACD\begin{align*}m \angle A+m \angle B=m \angle ACD\end{align*}. Here is the proof of the Exterior Angle Theorem. From the proof, you can see that this theorem is a combination of the Triangle Sum Theorem and the Linear Pair Postulate.

Given: ABC\begin{align*}\triangle ABC\end{align*} with exterior angle ACD\begin{align*}\angle ACD\end{align*}

Prove: mA+mB=mACD\begin{align*}m \angle A+m \angle B=m \angle ACD\end{align*}

Statement Reason
1. ABC\begin{align*}\triangle ABC\end{align*} with exterior angle ACD\begin{align*}\angle ACD\end{align*} Given
2. mA+mB+mACB=180\begin{align*}m \angle A+m \angle B+m \angle ACB=180^\circ\end{align*} Triangle Sum Theorem
3. mACB+mACD=180\begin{align*}m \angle ACB+m \angle ACD=180^\circ\end{align*} Linear Pair Postulate
4. mA+mB+mACB=mACB+mACD\begin{align*}m \angle A+m \angle B+m \angle ACB=m \angle ACB+m \angle ACD\end{align*} Transitive PoE
5. mA+mB=mACD\begin{align*}m \angle A+m \angle B=m \angle ACD\end{align*} Subtraction PoE

#### Measuring Angles in a Triangle

1. Find the measure of RQS\begin{align*}\angle RQS\end{align*}.

112\begin{align*}112^\circ\end{align*} is an exterior angle of RQS\begin{align*}\triangle RQS\end{align*}. Therefore, it is supplementary to RQS\begin{align*}\angle RQS\end{align*} because they are a linear pair.

112+mRQSmRQS=180=68\begin{align*}112^\circ + m \angle RQS &= 180^\circ\\ m \angle RQS &= 68^\circ\end{align*}

If we draw both sets of exterior angles on the same triangle, we have the following figure:

Notice, at each vertex, the exterior angles are also vertical angles, therefore they are congruent.

456789\begin{align*}\angle 4 & \cong \angle 7\\ \angle 5 & \cong \angle 8\\ \angle 6 & \cong \angle 9\end{align*}

2. Find the measure of the numbered interior and exterior angles in the triangle.

m1+92=180\begin{align*}m \angle 1 + 92^\circ = 180^\circ\end{align*} by the Linear Pair Postulate, so m1=88\begin{align*}m \angle 1 = 88^\circ\end{align*}.

m2+123=180\begin{align*}m \angle 2 + 123^\circ = 180^\circ\end{align*} by the Linear Pair Postulate, so m2=57\begin{align*}m \angle 2 = 57^\circ\end{align*}.

m1+m2+m3=180\begin{align*}m \angle 1+m \angle 2+m \angle 3=180^\circ\end{align*} by the Triangle Sum Theorem, so 88+57+m3=180\begin{align*}88^\circ + 57^\circ + m \angle 3 = 180^\circ\end{align*} and m3=35\begin{align*}m \angle 3 = 35^\circ\end{align*}.

m3+m4=180\begin{align*}m \angle 3 + m \angle 4 = 180^\circ\end{align*} by the Linear Pair Postulate, so m4=145\begin{align*}m \angle 4 = 145^\circ\end{align*}.

3. What is the value of p\begin{align*}p\end{align*} in the triangle below?

First, we need to find the missing exterior angle, we will call it x\begin{align*}x\end{align*}. Set up an equation using the Exterior Angle Sum Theorem.

130+110+xxx=360=360130110=120\begin{align*}130^\circ+110^\circ+x &= 360^\circ\\ x &= 360^\circ-130^\circ-110^\circ\\ x &= 120^\circ\end{align*}

x\begin{align*}x\end{align*} and p\begin{align*}p\end{align*} are supplementary and add up to 180\begin{align*}180^\circ\end{align*}.

x+p120+pp=180=180=60\begin{align*}x + p &= 180^\circ\\ 120^\circ + p &=180^\circ\\ p &= 60^\circ\end{align*}

#### Using the Exterior Angle Sum Theorem

The third exterior angle of the triangle below is 1\begin{align*}\angle 1\end{align*}.

By the Exterior Angle Sum Theorem:

m1+130+130=360m1=100\begin{align*} m \angle 1 + 130^\circ + 130^\circ = 360^\circ\\ m \angle 1 = 100^\circ\end{align*}

### Examples

#### Example 1

Find mA\begin{align*}m \angle A\end{align*}.

#### Example 2

Find mC\begin{align*}m \angle C\end{align*}.

Using the Exterior Angle Theorem, mC+16=121\begin{align*}m \angle C + 16^\circ = 121^\circ\end{align*}. Subtracting 16\begin{align*}16^\circ\end{align*} from both sides, mC=105\begin{align*}m \angle C = 105^\circ\end{align*}.

#### Example 3

Find the value of x\begin{align*}x\end{align*} and the measure of each angle.

Set up an equation using the Exterior Angle Theorem.

(4x+2)+(2x9)=(5x+13) interior anglesexterior angle(6x7)=(5x+13) x=20\begin{align*}&(4x+2)^\circ+(2x-9)^\circ =(5x+13)^\circ\\ & \quad \uparrow \qquad \nearrow \qquad \qquad \qquad \quad \uparrow\\ & \ \text{interior angles} \qquad \qquad \text{exterior angle}\\ & \qquad \qquad \quad (6x-7)^\circ = (5x+13)^\circ\\ &\qquad \qquad \qquad \qquad \ x = 20^\circ\end{align*}

Substituting 20\begin{align*}20^\circ\end{align*} back in for x\begin{align*}x\end{align*}, the two interior angles are (4(20)+2)=82\begin{align*}(4(20)+2)^\circ =82^\circ\end{align*} and (2(20)9)=31\begin{align*}(2(20)-9)^\circ=31^\circ\end{align*}. The exterior angle is \begin{align*}(5(20)+13)^\circ=113^\circ\end{align*}. Double-checking our work, notice that \begin{align*}82^\circ + 31^\circ = 113^\circ\end{align*}. If we had done the problem incorrectly, this check would not have worked.

### Explore More

Determine \begin{align*}m\angle{1}\end{align*}.

Use the following picture for the next three problems:

1. What is \begin{align*}m\angle{1}+m\angle{2}+m\angle{3}\end{align*}?
2. What is \begin{align*}m\angle{4}+m\angle{5}+m\angle{6}\end{align*}?
3. What is \begin{align*}m\angle{7}+m\angle{8}+m\angle{9}\end{align*}?

Solve for \begin{align*}x\end{align*}.

1. Suppose the measures of the three angles of a triangle are x, y, and z. Explain why \begin{align*}x+y+z=180\end{align*}.
2. Suppose the measures of the three angles of a triangle are x, y, and z. Explain why the expression \begin{align*}(180-x)+(180-y)+(180-z)\end{align*} represents the sum of the exterior angles of the triangle.
3. Use your answers to the previous two problems to help justify why the sum of the exterior angles of a triangle is 360 degrees. Hint: Use algebra to show that \begin{align*}(180-x)+(180-y)+(180-z)\end{align*} must equal 360 if \begin{align*}x+y+z=180\end{align*}.

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