Angle created by three distincts random vertices

Assume you have a regular polygon ( $$n$$-sides). and Let $$A=\{ x_0, x_2, \cdots , x_{n-1} \}$$ be vertices of the polygon.

My Question is:

Are there is any formula that tell us what is the angle between any three distincts vertices let say $$x_i, x_j, x_k$$?

As a special case we know the angle made by $$x_m,x_{m+1},x_{m+2}$$ in $$n$$-sided polygon is $$\frac{180(n-2)}{n}$$.

• Do you mean a regular polygon? – Andrei May 8 at 5:10
• yes. I just fixed it. – henry May 8 at 5:11

Let's assume that $$i. Also note that all points in $$A$$ are on a circle, and the angle from $$x_a$$, center of the circle $$O$$, and $$x_{a+1}$$ is $$\frac{360^\circ}{n}$$. Then the angle between $$x_j$$, $$O$$, $$x_i$$ is $$\frac{360^\circ}{n}(j-i)$$. This is an isosceles triangle, so the angle $$\angle Ox_jx_i$$ is $$\frac 12(180^\circ-\frac{360^\circ}{n}(j-i))$$. Similarly $$\angle x_kx_jO=\frac 12(180^\circ-\frac{360^\circ}{n}(k-j))$$. Then just add them together to get$$\angle x_kx_jx_i=180^\circ-\frac{180^\circ}{n}(k-i)$$ Note that the angle depends only on the end points.