Finding the interior angles of an irregular polygon inscribed on a circle

Question

Is there any way to calculate the interior angles of an irregular N-sided polygon inscribed on a circle?

I only have a list of edge lengths (in order). I don't know any of the interior angles nor the radius of the circle the polygon is inscribed upon.

Here is an example of what I'm trying to figure out:

Irregular polygon points inscribed on a circle

The polygon can have any number of sides, but I'll always know the lengths of each side (for example, in the picture above I know what the lengths are for AB, BC, CD, DE, EF, and FA) and the polygon is always guaranteed to be inscribed on a circle.

I have found numerous solutions for solving triangles (which is easy, that's trig 101) and quadrangles (or rather, "cyclic quadrilaterals"). I'm wondering if a similar solution exists for N-sided irregular polygons, but so far I've not been able to find anything.

Answer 1

[Improved answer, based on excellent comment from David]

We have two different cases: one with the center inside the polygon and one with the center outside of it.

WLOG, we can assume that $a_1$ is the length of the longest edge. If the length of the $i^{th}$ edge of the polygon is $a_i$ and the radius of the circumscribed circle is $R$, the corresponding central angle would be (in both cases):

$$\alpha_i=2\arcsin \frac{a_i}{2R}$$

Case on the left: Sum of all central angles is $2\pi$:

$$2\pi=\sum_{i=1}^{N}\alpha_i=2\arcsin \frac{a_1}{2R}+\sum_{i=2}^{N}2\arcsin \frac{a_i}{2R}$$

$$\pi=\arcsin \frac{a_1}{2R}+\sum_{i=2}^{N}\arcsin \frac{a_i}{2R}$$

$$\pi-\arcsin \frac{a_1}{2R}=\sum_{i=2}^{N}\arcsin \frac{a_i}{2R}$$

$$\frac{a_1}{2R}=\sin(\sum_{i=2}^{N}\arcsin \frac{a_i}{2R})\tag{1}$$

Case on the rigth: $$2\pi=(2\pi-2\arcsin \frac{a_1}{2R})+\sum_{i=2}^{N}2\arcsin \frac{a_i}{2R}$$

$$\arcsin \frac{a_1}{2R}=\sum_{i=2}^{N}\arcsin \frac{a_i}{2R}$$

$$\frac{a_1}{2R}=\sin(\sum_{i=2}^{N}\arcsin \frac{a_i}{2R})\tag{2}$$

Actually, both equations (1),(2) are the same. The equation (1,2) has only one unknown ($R$) but I doubt that it is possible to solve it in a closed form for an arbitrary $N$. So numerical approach seems to be your only option. Once you have $R$ you can calculate pretty much anything, all internal angles included.

Example for the case on the left with $a_1=a_2=a_3=6$, $a_4=3$:

a1 = 6
a2 = 6
a3 = 6
a4 = 3
Solve[a1/(2 R) == Sin[ArcSin[a2/(2 R)] + ArcSin[a3/(2 R)] + ArcSin[a4/(2 R)]], R]

The result is $R=6\sqrt\frac25$.

Example for the case on the right with $a1=4$, $a2=3$, $a3=2$:

a1 = 4
a2 = 3
a3 = 2
Solve[a1/(2 R) == Sin[ArcSin[a2/(2 R)] + ArcSin[a3/(2 R)]], R]

This gives one solution: $R={8\over\sqrt{15}}$

Answer 2

If I'm understanding your quesiton correctly, can you get each angle as the angle of the triangle of the two edges. If you want the angle ABC, then look only at the triangle ABC on the circle that you know. You don't have length AC at first, but given that points ABc define the circle, you should be able to find it easily. You then end up having to solve N triangle problems, but it seems very tractable.