The Exterior Angle Theorem

In the angle sum of a triangle post, we have discussed that the angle sum of a triangle is 180 degrees.  In the angle sum of a polygon post,  we also have discussed that  and that the angle sum of a polygon with n sides is 180(n-2). For example, a pentagon has 5 sides, so the sum of its interior angle is 180(5-2) = 180(3) = 540 degrees.

Figure 1 – The interior and exterior angles a triangle and a quadrilateral.

The angle sums that we have discussed in both blogs refer to the sum of the interior angles. What about the exterior angles?

The exterior angle is formed when we extend a side of a polygon. In the triangle above, \alpha is an exterior angle. The sum of the interior angle and the exterior angle adjacent to it is always 180  degrees (Why?).  Angles whose sum is 180 degrees are called supplementary angles.  If two angles are supplementary, we call them a linear pair.  For example, angles \alpha and a_1 are supplementary angles and at the same time a linear pair, so \alpha + a_1 = 180 degrees. Now this means, that \alpha = 180 - a_1. Therefore, if we want to compute the measure of an exterior angle adjacent to an interior angle, we can always subtract the measure of the interior angle from 180 as shown in Figure 1.

Observe the computation in the two diagrams.  If we let S_t be the angle sum of the exterior angles of a triangle, then S_t = (180 - a_1) + (180 - a_2) + (180 - a_3) = 540. Rearranging the terms, we have S_t = 540 - (a_1 + a_2 + a_3).  But a_1 + a_2 + a_3 is the sum of the interior angles of a triangle which is 180 degrees, so 540 - (a_1 + a_2 + a_3) = 540 - 180 = 360 degrees.

Now, try calculating for the sum of the exterior angles of the quadrilateral above. What is your answer?

To verify our hunch, we will try to compute for the sum of the exterior angles of a pentagon.

Let S_p be the sum of the exterior angles of the pentagon in Figure 2. Then

S_p =(180 - c_1)+ (180 - c_2) + (180 - c_3) +(180 - c_4) +(180 - c_5). Simplifying, we have S_p = 900 - (c_1 + c_2 + c_3 + c_4 + c_5). But according to the angle sum theorem for polygons, c_1 + c_2 + c_3 + c_4 + c_5 = 540. Therefore,900 - (c_1 + c_2 + c_3 + c_4 + c_5) = 900 - 540 = 360 degrees.

We have three polygons – triangle, quadrilateral, pentagon – whose angle sums of exterior angles are always 360 degrees. Now, is this true for all polygons?  Try to compute polygons up to 10 sides and see if the sum is 360 degrees.

Delving Deeper

We know that in a polygon, the number of exterior angles is equal to the number of interior angles.  Furthermore, we know that the angle sum of an interior angle and the exterior angle adjacent to each is always latex 180 degrees. If we have a polygon with 5 sides, then

interior angle sum + exterior angle sum = 180(5)

In general, this means that in a polygon with n sides

interior angle sum* + exterior angle sum = 180n

But the interior angle sum = 180(n – 2). So, substituting in the preceding equation, we have

180(n – 2) + exterior angle sum = 180n

which means that the exterior angle sum = 180n – 180(n – 2)  = 360 degrees. More formal proofs using these arguments are shown below.

Theorem: The sum of the measure of the exterior angles of a polygon with n sides is 360 degrees.

Proof 1:

Let a_1, a_2, \cdots a_n be measures of the interior angles of a polygon with n sides. Letb_1, b_2, \cdots b_n be measures of the exterior angles of the same polygon where all angle names with the same subscripts are adjacent angles from a_1 andb_1 all the way up through a_n and b_n .  We know that adjacent interior and exterior angles are supplementary angles, so this implies that their measures add up to 180 degrees. Hence,

(a1 + b1) + (a2 + b2) + … + (an + bn) = 180 + 180 + … +180 (n of them) = 180n

Regrouping the terms of the preceding equation, we have

(a1 + a2 + … + an) + (b1 + b2 + … + bn) = 180n

But the sum of the interior angles is a1 + a2 + … + an = 180(n – 2)

So,

180(n – 2) + (b1 + b2 + … + bn) = 180n

b1 + b2 + … + bn = 180n – 180(n – 2) = 360

Therefore, the sum of the exterior angles of any polygon is equal to 360 degrees.

Proof 2:

Let a1, a2, …, an be measures of the interior angles of the polygon with n sides. Since each adjacent interior and exterior angle is a linear pair, it follows that the measure of the exterior angles adjacent to them respectively are  180 – a1, 180 – a2, …, 180 – an.

If we let S, be the sum of the measure of the exterior angles, we have

S = (180 – a1) + (180 – a2) + (180 – a3) + … + (180 – an)

= (180 + 180 + 180 + … +180 (n of them)) – a – a2 – a3– … – an

S = 180n – (a1 + a2 + a3 + … + an)

But a1 + a2 + a3 + … + an is the sum of the measures of the interior angles of a polygon  with n sides which equals

180(n – 2), so, S = 180n – 180(n – 2) = 360, which is want we want to show.

Therefore, the sum of the exterior angles of any polygon is equal to 360 degrees.

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