# Does the sum of exterior angles of a simple, convex polygon truly = 360°?

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:

triangle

$$\sum\measuredangle interior=a+b+c$$
$$\text{ }=(360-A)+(360-B)+(360-C)$$
$$\text{ }=3*360-(A+B+C)$$
$$\text{ }=1080-\sum\measuredangle exterior$$
$$\text{ }=180^\circ$$ (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°. 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.
• @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. – Milo Brandt Jan 18 '16 at 6:08