The question being asked of me is the following:

What is the sum of a polygon's exterior angles?

Assuming, again, that the polygon is simple and convex, the answer I see repeatedly given is 360°. I'm wondering if there are some other criteria for defining an exterior angle that I'm ignorant of, because if we say only that an exterior angle is the angle around a vertex that is not the interior angle, or put another way, that an exterior angle is 360°-$\measuredangle interior$, centered at a vertex, how does the sum of all vertices' exterior angles = 360°?

An example

Let there be a drawing! ...or two:
Exterior angle: exterior angle

Am I defining exterior angle incorrectly? The most common illustrations employed essentially lob off 180° from every exterior angle so that they fit nicely together into a kind of pie chart, but no reasoning is given for ignoring 180° from each vertex. (I can't link to an example because I'm limited to two links.) The illustrations suppose that the polygon is shrunk to a size where the polygon becomes only a one-dimensional point, i.e. where the polygon is no longer a polygon and doesn't have even interior angles. Is there justification for this? My essential question is, if an interior angle = 30°, why isn't the exterior angle = 330°?

I'll use the following polygon as an example:


$\sum\measuredangle interior=a+b+c$
$\text{ }=(360-A)+(360-B)+(360-C)$
$\text{ }=3*360-(A+B+C)$
$\text{ }=1080-\sum\measuredangle exterior$
$\text{ }=180^\circ$ (a given, since it's a triangle)

$\Longrightarrow A+B+C=1080-180=900^\circ$

$\sum\measuredangle exterior=900^\circ$

General solution

For any n-sided polygon, with interior angles a, b, ... n:

$\sum\measuredangle interior=a+b+...+n$
$\text{ }=(360-A)+(360-B)+...+(360-N)$
$\text{ }=360n-(A+B+...+N)$
$\text{ }=360n-\sum\measuredangle exterior$

Using the equation for the sum of interior angles of an n-sided polygon, $(n-2)180$,

$(n-2)180=360n-\sum\measuredangle exterior$

$\mathbf{\Longrightarrow\sum\measuredangle exterior=(n+2)180}$

For the triangle, this also works out to 900°.
For a hexagon, for example:
$\sum\measuredangle exterior=6*180+360=1440^\circ$

Granted, these answers have nothing to do with whether my definition of exterior angle is correct, but they're simply the answers I would get.

Also, for n=0, such as when the polygon is shrunk down to a point, as in the illustrations previously mentioned, the equation gives the expected 360°.


Well, the intuition that "exterior angle" captures is how much one would have to turn at each vertex if they were traversing the edge of the polygon. For instance, if you traverse a triangle counterclockwise and consider these angles, you get a picture like this:

enter image description here

where all the edges are extended to rays in the counterclockwise direction, representing what happens if you don't turn at the vertex. It happens that this equals $180^{\circ}$ minus the interior angle.

This is to say that "exterior angle" is meant to capture a different intuition than the one you are pursuing - it isn't literally "the part of the angle which is exterior" so much as "an angle, which happens to be represented on the exterior." All the work you've done is consistent with your definition (and your definition would be a reasonable interpretation of the phrase "exterior angle" were it not defined otherwise)

  • $\begingroup$ Ahh, that makes sense. Wouldn't you get the same answer (360°), then, for the sum of interior angles? $\endgroup$ – benJephunneh Jan 18 '16 at 6:04
  • $\begingroup$ @benJephunneh Yes, one's body would turn $360^{\circ}$ if they walked around a polygon. The interior angle's don't follow the same logic - consider that for interior angles near $180^{\circ}$, one barely needs to turn, since the directions are nearly parallel. The difference is that exterior angles rotate the ray $AB$ into the ray $BC$ and then $BC$ into $CD$, meaning the rotations are somehow "consistent", where as interior angles rotate the ray $BA$ into the ray $BC$ then $CB$ into $CD$ - so we can't "chain" the transformations together, since $BC$ and $CB$ are different rays. $\endgroup$ – Milo Brandt Jan 18 '16 at 6:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.