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.


2 Answers 2


[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.

enter image description here

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: $R={8\over\sqrt{15}}$

  • $\begingroup$ A further point: it's clear that $2R$ is greater than every $a_i$. Therefore the right hand side of your second-last equation is a decreasing function of $R$, and there is at most one solution. $\endgroup$
    – David
    Apr 15, 2019 at 0:05
  • 1
    $\begingroup$ There is a subtle mistake here: the central angle is $2\arcsin(a_i/2R)$ only if the centre is on the "correct" side of the chord, otherwise it is $2\pi-\alpha_i$. In effect, your solution assumes that the centre of the circle is inside the polygon. If for example the sides are $4,3,2$ then your method gives no solution even though there clearly is a solution (every triangle can be inscribed in a circle). $\endgroup$
    – David
    Apr 16, 2019 at 0:24
  • $\begingroup$ @David You are 100% right, thanks! There are actually two cases and I have covered only the first one. In the second case (your one), the central angle of the longest chord is equal to the sum of central angles of all shorter chords. Honestly, I stopped thinking about the probelm when I saw that there was no simple solution.I'll update the solution as soon as possible. And I have upvoted your comment too :) $\endgroup$
    – Saša
    Apr 16, 2019 at 6:25
  • $\begingroup$ @David It looks better now? $\endgroup$
    – Saša
    Apr 16, 2019 at 7:22
  • $\begingroup$ Looks pretty good to me. $\endgroup$
    – David
    Apr 16, 2019 at 7:24

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.

  • $\begingroup$ Wouldn't I need to know the radius of the circle first for this to work? I only have the side lengths of the polygon, so I'm not sure how I could ever figure out AC with just that information. $\endgroup$
    Apr 12, 2019 at 8:12
  • $\begingroup$ I'm sorry I thought you had the radius. I misread. $\endgroup$
    – EMP
    Apr 12, 2019 at 15:28

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