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Sum of Interior and Exterior Angles

Date: 02/14/2003 at 20:11:26
From: Student E
Subject: Geometry theorems 

Is there a theorem for concave polygons about the sum of the interior 
and the sum of the exterior angles?  

I know that for convex polygons the sum of the interior angles is 
(n-2)180 and the sum of the exterior angles is 360. However, my 
geometry class believes that we might have proven that for concave 
polygons the sum of the interior angles is still (n-2)180 but the 
exterior angles' sum is 540. Do you know if this can be found in any 
textbook or if it has been proven before?

Date: 02/14/2003 at 22:51:16
From: Doctor Peterson
Subject: Re: Geometry theorems 

Hi, E.

The theorem you refer to for convex polygons applies to concave 
polygons as well, IF you take the direction of the exterior angles 
into account. If you don't, then you may get a variety of sums, 
depending on how many angles turn in each direction.

Take this example, a concave hexagon with two "dents":

     \60    60/
      \      /
    120+    +120
      /      \
     /60    60\

The sum of the interior angles is

  60+60+240+60+60+240 = 720 = (6-2)180

as expected.

If you walk around the perimeter counterclockwise, you will turn 
counterclockwise (making positive exterior angles) at all the acute 
angles, but clockwise (making a negative exterior angle) at the two 
obtuse angles, where you turn outward from the interior. The sum of 
the signed exterior angles is therefore

  120+120-60+120+120-60 = 360

This is always 360 because in going around a polygon you always make 
one complete turn, and the sum of the exterior angles in this sense 
tells how far you have turned. It is this sum that can be used to 
prove the sum of the interior angles, since each interior angle is 
the supplement of its exterior angle:

  sum of A = sum of (180-B) = 180n - sum of B = 180n - 360 = (n-2)180

where A is each interior angle, B is each exterior angle, and n is 
the number of angles. In our example, the supplement of interior 
angle A=240 is B=180-240=-60. That's not how we normally think of 
supplements, but it's the only way that makes sense for reflex angles 
like this.

Now, if we sum the absolute values of the exterior angles, not 
treating them as signed angles, we get a different answer:

  120+120+60+120+120+60 = 600

You might like to play with this and see whether you can get any sum 
at all by changing the number and size of the negative angles, or if 
there is some restriction on possible sums.

Did you have an actual proof that the sum is 540 degrees for some 
class of concave polygons, or did you just try it out for some 
examples and find that to be true for those you tried? If you have a 
proof, I'd like to see it, so we can figure out when it applies 
(since it obviously is not true for all concave polygons).

- Doctor Peterson, The Math Forum 
Associated Topics:
High School Triangles and Other Polygons

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