The central angle of a regular polygon is formed by two lines from consecutive vertices to the centre point or two radii of consecutive vertices of the circumsribed circle.
I can plot each point on the hexagon by using the same length of radii and rotating $60$ degrees from the center.
I know I can work out the exterior angle by $(n−2) \cdot 180^\circ$.
That gives me $720 / 6 = 120$, and $180 - 120 = 60$.
Is there any connection between the central angle and the exterior angle?
On
For a regular $n$-gon the central angle is $\frac {360}n$.
The central angles cut the $n$-gon into $n$ isoceles triangles. So that base of these triangles are $\frac {180 - \frac {360}n}2 = 90 - \frac{180}n$. The interior angles are two of these base angles so the interior angles $180 - \frac {360}n$.
And there for the exterior angles are $180 - (180 - \frac {360}n) = \frac {360}n$.
So this is true no matter what regular $n$-gon you do.
Hexagons and $60$ degrees are particularly important number as represent the sides and radius being equal and tesselates the plane.
As an answer to your question on comments, notice that when we draw a line from a vertex to center of a regular polygon, that line is angle bisector of an interior angle, say interior angle is $2\alpha$. In a triangle constructed this way, there are two such $\alpha$ angles, so central angle is $180 - 2\alpha$. But notice that this is as same as the exterior angle. Therefore this is not special to hexagon. Here is a sketch for a general result: