# 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?

• For an $n$-sided polygon, the sum of the interior angles is equal to $180(n-2)$ (in degrees). I don't think that anyone says that the sum of the "interior angles of a circle" is $360^\circ$. In fact, there is no common notion of an interior angle for a circle. Dec 20, 2018 at 4:30
• The sum of the external angles of a polygon is always 360°; that's what you're looking for.
– MJD
Dec 20, 2018 at 5:17
• If you look at what interior angles are limiting to as a polygon gains more sides, it might make sense to say that a circle has an interior angle of $180$ degrees at every point on its boundary. (The number of degrees swept out by the tangent line.) So in this sense, the sum of interior angles of a circle is infinite, as expected. Dec 20, 2018 at 5:35
• The central angle of a circle doesn't correspond to the interior angle of a polygonal at all. If you pursue this line of thought the interior angles occur at every point of the circumference and measure 180 degrees and there are an infinite number of them. Dec 20, 2018 at 5:56

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.

$$180^\circ(n-2)=180^\circ n-360^\circ$$ 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 $$n=4$$.

As the comment said, it’s hard to define what is meant by “the sum of the interior angles of a circle”. You might define it to mean the limit of this same sum for $$n$$-sided polygons. But then its value is infinite, not 360 degrees.

Things make more sense if we think of exterior angles. At a vertex of a polygon, the exterior angle is the angle you turn through as you move from one polygon edge to the next one. So the sum of these angles is 360 degrees. Of course, traveling around a circle gives you this same total turning angle.