Q:

If the total number of angles in a polygon is $180(n-2)$, why are there just $360$ degrees in a circle?

For any polygon, the total degrees in the interior angles equals $180(n-2)$. A three-sided polygon has $180$ degrees. A four-sided one has $360$. Five sides gives $540$. Etc. 
So why is it that $\lim_{n\to\inf}180(n-2)=360$? That is, why does a circle, an infinite-sided polygon, have the same number of degrees as a four-sided polygon?

A:

The interior angles on a polygon are the RED angles in the diagram.  The sum of these approach infinity.  They do not exist in a circle.
The central angles are the BLUE angles.  They are always $360$ 
Actually when you calculate the interior (red) angles of a polygon and they are $180(n−2)=360$... well the $180*(-2)$ represent the blue angles.  You don't count the in calculating the red angles so you subtract them.  But in doing the circle they are the only thing you do.

A:

$180^\circ(n-2)=180^\circ n-360^\circ$ and that is the $360^\circ$ of the circle.
Take a regular polygon and join each vertex to the centre.  You have $n$ triangles, each with $180^\circ$.
The angles at the edge add up to $180^\circ(n-2)$ and the angles at the centre give the remaining $360^\circ$ for a total of $180^\circ n$.
I think it is just a coincidence, because $360^\circ=2\times 180^\circ$ and the soltion of $n-2=2$ is $n=4$.

