A polygon is an enclosed 2 dimensional figure made of three or more line segments. The line segments are called the sides of the polygon and a point where two sides meet is called a vertex. An angle formed between two adjacent sides at any of the vertices is called an interior angle. An exterior angle is an angle formed outside the polygon’s enclosure by one of its sides and the extension of its adjacent side. The sum of the exterior angles of a polygon is 360 degrees. Let us learn in detail the concept of exterior angles of polygons.

Sum of the Exterior Angles of a Polygon

Exterior angles of a polygon are the angles formed between one of the sides of the polygon and outward extension of the adjacent side. The sum of all the exterior angles of a polygon is 360 degrees. We will prove this statement in the following sections and look at some examples. But before that let us first learn some basic definitions about polygons and exterior angles.

What is a Polygon?

The word polygon comes from the Greek word poly, meaning many, and gon, meaning angle. In other words, a polygon is a simple closed two-dimensional shape formed by joining the straight line segments. Examples of polygons are equilateral triangle, square, scalene triangle, rectangle, pentagon etc.

Some of the common polygons are shown in the image below:

Exterior Angles of a Polygon

A polygon has two types of angles:

(i) Interior angles are those angles formed inside the polygon at the vertices.

(ii) Exterior angle is formed by one of the sides of a polygon and the extension of the adjacent side. See the image below, which shows a pentagon with five vertices. The pentagon’s exterior angles are produced by extending the length of its adjacent sides. Look at the below figures to understand exterior angles in detail.

Important Points to Remember about Exterior Angles

1. They are formed on the outside or exterior of the polygon.
2. The sum of an interior angle and its corresponding exterior angle is always \({\rm{18}}{{\rm{0}}^{\rm{o}}}\) since they lie on the same straight line.

Note: Exterior angles of a regular polygon are equal in measure.

Sum of the Exterior Angles of a Polygon

Assume that you start your journey from the vertex at angle \(1.\) You turn to the same vertex in the right clockwise direction, turning at angles \(2,3,4\) and \(5.\) You covered the entire polygon perimeter and finished the whole cycle. One complete turn covers \({\rm{36}}{{\rm{0}}^{\rm{o}}}.\) As a result, the total of \(1, 2, 3, 4,\) and \(5\) equals \({\rm{36}}{{\rm{0}}^{\rm{o}}}.\)

As a result, regardless of the number of sides in the polygons, the sum of the measures of the exterior angles equals \({\rm{36}}{{\rm{0}}^{\rm{o}}}.\)

Polygon Exterior Angle Sum Theorem

Let us prove that if a polygon is a convex polygon, then the sum of its exterior angles (one at each vertex) is equal to \({\rm{36}}{{\rm{0}}^{\rm{o}}}.\)

Proof: 

Let us consider a regular polygon with \(n\) number of sides. Let the sum of its exterior angles is \(S.\)

For any polygon, the sum of interior angles\( = (n – 2){\rm{18}}{{\rm{0}}^{\rm{o}}}\)

We know that for a regular polygon, all the sides are equal. So, the measure of each angle will be\( = \frac{{(n – 2){\rm{18}}{{\rm{0}}^{\rm{o}}}}}{n}\)

For any polygon, the sum of the interior and exterior angles are always supplementary.

So, the measure of each exterior angle will be\( = {\rm{18}}{{\rm{0}}^{\rm{o}}} – \frac{{(n – 2){\rm{18}}{{\rm{0}}^{\rm{o}}}}}{n}\)
\( = \frac{{{\rm{18}}{{\rm{0}}^{\rm{o}}} \times n – (n – 2){\rm{18}}{{\rm{0}}^{\rm{o}}}}}{n}\) 
\( = \frac{{{\rm{18}}{{\rm{0}}^{\rm{o}}} \times n – {\rm{18}}{{\rm{0}}^{\rm{o}}} \times n + {\rm{36}}{{\rm{0}}^{\rm{o}}}}}{n}\) 
\( = \frac{{{\rm{36}}{{\rm{0}}^{\rm{o}}}}}{n}\) 

Now the sum of exterior angles\( = \frac{{{\rm{36}}{{\rm{0}}^{\rm{o}}}}}{n} \times n = {\rm{36}}{{\rm{0}}^{\rm{o}}}\) 

Hence, the sum of exterior angles of a polygon with \(n\) number of sides is equal to \({\rm{36}}{{\rm{0}}^{\rm{o}}}.\)

Measure of Each Exterior Angle of a Regular Polygon

The measure of each exterior angle of a regular polygon\( = \frac{{{\rm{36}}{{\rm{0}}^{\rm{o}}}}}{n},\)
where \(n=\)the number of sides of a polygon.

For a triangle, \(n=3.\) The equilateral triangle is the regular \(3-\)sided polygon. The measure of each exterior angle of an equilateral triangle\( = \frac{{{\rm{36}}{{\rm{0}}^{\rm{o}}}}}{3} = {120^{\rm{o}}}\)

For a quadrilateral, \(n=4.\) A square is a regular polygon.

The measure of each exterior angle of a square\( = \frac{{{\rm{36}}{{\rm{0}}^{\rm{o}}}}}{4} = {90^{\rm{o}}}\)

For a pentagon, \(n=5\)

The measure of each exterior angle of a pentagon\( = \frac{{{\rm{36}}{{\rm{0}}^{\rm{o}}}}}{5} = {72^{\rm{o}}}\)

NOTE: Irregular polygons have different measures of interior and exterior angles.

Solved Examples

Q.1. Determine the measure of each exterior angle of a regular hexagon.

Ans: We need to find the measure of each exterior angle of a regular hexagon.
We know that the number of sides of a hexagon is, \(n=6.\)
The measure of each exterior angle of a \(n-\)sided polygon\( = \frac{{{\rm{36}}{{\rm{0}}^{\rm{o}}}}}{n}\)
Therefore, each exterior angle of a hexagon\( = \frac{{{\rm{36}}{{\rm{0}}^{\rm{o}}}}}{6} = {60^{\rm{o}}}\)

Q.2. By using the sum of exterior angles formula, prove that each interior angle and its corresponding exterior angle in any polygon are supplementary.
Ans: To prove: The sum of an interior angle and its corresponding exterior angle is \({180^{\rm{o}}}.\)
Let us consider a \(n-\)sided polygon.
By using the sum of exterior angles formula,
Sum of exterior angles of any polygon\( = {360^{\rm{o}}} \ldots \ldots {\rm{(i)}}\)
Sum of interior angles of any polygon\( = (n – 2){180^{\rm{o}}} \ldots \ldots {\rm{(ii)}}\)
By adding equations \({\rm{(i)}}\) and \({\rm{(ii)}},\) we get the sum of all n interior angles and the sum of all \(n\) exterior angles:
\({360^{\rm{o}}} + (n – 2){180^{\rm{o}}} = {360^{\rm{o}}} + {180^{\rm{o}}}n – {360^{\rm{o}}} = 180n\)
Therefore, the sum of one interior angle and its corresponding exterior angle is:
\( = \frac{{180n}}{n} = {180^{\rm{o}}}\)
Hence, the sum of an interior angle and its corresponding exterior angle in any polygon are supplementary.

Q.3. Calculate the exterior angle of a regular polygon with \(12\) sides.
Ans: Given, \(n=12\)
We know that the measure of each exterior angle\( = \frac{{{\rm{36}}{{\rm{0}}^{\rm{o}}}}}{n}\)
So, the measure of each exterior angle\( = \frac{{{\rm{36}}{{\rm{0}}^{\rm{o}}}}}{{12}} = {30^{\rm{o}}}\)

Q.4. The exterior angles of an octagon are \({x^{\rm{o}}},2{x^{\rm{o}}},3{x^{\rm{o}}},4{x^{\rm{o}}},5{x^{\rm{o}}},6{x^{\rm{o}}},7{x^{\rm{o}}},\) and \(8{x^{\rm{o}}}.\)
Find the measure of the largest exterior angle of this octagon?
Ans: Given, the exterior angles of an octagon are \({x^{\rm{o}}},2{x^{\rm{o}}},3{x^{\rm{o}}},4{x^{\rm{o}}},5{x^{\rm{o}}},6{x^{\rm{o}}},7{x^{\rm{o}}},\) and \(8{x^{\rm{o}}}.\)
For any polygon, the sum of the exterior angles\( = {360^{\rm{o}}}\)
Therefore \({x^{\rm{o}}} + 2{x^{\rm{o}}} + 3{x^{\rm{o}}} + 4{x^{\rm{o}}} + 5{x^{\rm{o}}} + 6{x^{\rm{o}}} + 7{x^{\rm{o}}} + 8{x^{\rm{o}}} = {360^{\rm{o}}} \Rightarrow 36{x^{\rm{o}}} = {360^{\rm{o}}}\)
\( \Rightarrow {x^{\rm{o}}} = \frac{{{{360}^{\rm{o}}}}}{{36}} = {10^{\rm{o}}}\)
Therefore the size of the largest exterior angle\( = 8{x^{\rm{o}}} = 8 \times {10^{\rm{o}}} = {80^{\rm{o}}}\)

Q.5. The exterior angles of a heptagon are \({y^{\rm{o}}},2{y^{\rm{o}}},3{y^{\rm{o}}},4{y^{\rm{o}}},5{y^{\rm{o}}}\) and \(6{y^{\rm{o}}}\)
What is the value of \(y\)?
Ans: Given, the exterior angles of a heptagon are \({y^{\rm{o}}},2{y^{\rm{o}}},3{y^{\rm{o}}},4{y^{\rm{o}}},5{y^{\rm{o}}}\) and \(6{y^{\rm{o}}}.\)
For any polygon, the sum of the exterior angles\( = {360^{\rm{o}}}\)
Therefore \({y^{\rm{o}}} + 2{y^{\rm{o}}} + 3{y^{\rm{o}}} + 4{y^{\rm{o}}} + 5{y^{\rm{o}}} + 6{y^{\rm{o}}} = {360^{\rm{o}}} \Rightarrow 24{y^{\rm{o}}} = {360^{\rm{o}}}\)
\( \Rightarrow {y^{\rm{o}}} = \frac{{{{360}^{\rm{o}}}}}{{24}} = {15^{\rm{o}}}\)
Hence, the value of \(y\) is \(15.\)

Summary: Sum of the Exterior Angles of a Polygon

The main definition of a polygon, its sides, the adjacent side and opposites and its vertices was discussed briefly in this article. We also discussed two kinds of inside and outside angles of a polygon. Then we discussed in detail the external angles of a polygon, the sum of the exterior angles of a polygon, the sum of the external angle theorem, and the solved examples.

Learn About Angle Sum Property of Quadrilateral

FAQs on Sum of the Exterior Angles of a Polygon

The most frequently asked questions about angle sum property of a quadrilateral are answered here:

Q.1. Why is the sum of exterior angles of a polygon 360? Ans: The exterior angles and interior angles of a polygon makes a linear pain, and hence they are supplementary. The interior angles add up to \(180{(n – 2)^{\rm{o}}},\) and the sum of the exterior angles is supplementary to this interior angle sum. So, the sum of any polygon’s exterior equals \({360^{\rm{o}}}.\) Remember that here that we are just talking about convex polygons here.

Q.2. What is the sum of exterior angles of a regular polygon? Ans: The sum of exterior angles of any regular polygon or any polygon is \({360^{\rm{o}}}.\)

Q.3. What is the formula of measure of each exterior angle of a polygon? Ans: We know that the sum of exterior angles of any polygon is \({360^{\rm{o}}}.\) So, the measure of each exterior angle of a regular polygon of \(n\) sides\( = \frac{{{{360}^{\rm{o}}}}}{n}.\)

Q.4. What is the sum of all exterior angles of a triangle? Ans: The sum of all the exterior angles of a triangle is \({360^{\rm{o}}}.\)

Q.5. What is the measure of each exterior angle of a decagon? Ans: For a decagon, \(n=10\) The measure of each exterior angle of a regular polygon\( = \frac{{{{360}^{\rm{o}}}}}{n}\) \( = \frac{{{{360}^{\rm{o}}}}}{{10}} = {36^{\rm{o}}}\) Hence, the measure of each exterior angle of a decagon is \({36^{\rm{o}}}.\)

Learn All the Concepts on Polygons

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