Question

How is the number of sides related to the sum of the interior angles in a polygon, and what about the sum of the exterior angles?

Answer 1

For convex polygons with `N` sides the sum of interior angles `S_1=pi(N-2)`
The sum of its exterior angles does not depend on `N` and equals to `S_2=4pi`

Explanation:

If polygon is convex, there is a point inside it, from which all vertices are "visible", that is a segment connecting this point with any vertex does not intersect any side.

Assuming such point `P` exists, let's connect it with all the vertices and consider `N` triangles formed by `N` sides of a polygon and segments connecting point `P` with all the vertices. In each such triangle let's call the side that is also a side of a polygon a base, while two other sides - those connections to point `P` - we will call legs.

The sum of all interior angles of a polygon is the same as a sum of all angles adjacent to bases of all triangles (let's call them base angles).
In its turn, this sum of base angles and the sum of all angles formed by legs of all triangles around point `P` equals to `N` times a sum of angles of one triangle, that is `pi`.

Therefore, the sum of base angles of all triangles equals to a difference between `pi N` and the sum of all angles formed by legs of all triangles around point `P`, which is equal to `2pi`.

Hence, the formula for a sum of interior angles of a convex polygon with `N` sides is
`S_1=pi(N-2)`

The sum of all exterior angles (there are two exterior angles per each vertex of a polygon) can be calculated based on the above formula and the fact that the sum of all exterior angles and double the sum of all interior angles equals to `2piN`. This is clearly visible if you consider lines around each vertex of a polygon.

Therefore, the sum of all exterior angles `S_2` can be found as follows:
`S_2+2S_1 = 2piN`
`S_2 = 2piN - 2S_1 = 2piN - 2pi(N-2) = 4pi`