I don't know if I have my phrasing correct here. I am basically trying to get the radius around the parameter of a regular polygon. From here on I will just use a pentagon for a concrete example.
This problem is most easily seen in polar coordinates but does not need to be as such.
So we have a pentagon inscribed in a circle with radius R = 1. Therefore from the origin, O to any of the corners R = 1. Then the distance from the origin to a midpoint (the distance from O to A) is simply cos(36). Now I just need to know all the other points between the two.
These basics and my question are more easily explained in this diagram: 
Now I want to rotate all around the pentagon in polar space sweeping theta. What is the distance, P from the origin to the perimeter?
My approach was to look at that important segment from O to A because this forma a nice right angle with the outside edge. Then after finding φ (which is my biggest concern) I can get what that distance is because I made a right triangle with segment OA segment P and the perimeter segment. I have the length of OA, |OA| = cos(36) and I have the angle of the corner at the origin. This is 36-φ. Then with a little trig voila $P = \frac{\cos(36)}{\cos(36-φ)}$.
Something I haven't covered yet is how to get φ. φ is the angle distance from the closest corner where phi should never get over 36 in this example. Right now I am getting phi by,
$$φ = |(\frac{\theta}{72}) - 36|$$ Wolfram Equation
$$P = \frac{\cos(36)}{\cos(36 - φ)}$$
I feel like this is a bit of overkill to try to solve this problem. Is there a better way to go about this problem? I can also list the n-sided version for these equations
$$a = \frac{360}{n}$$
$$b = \frac{360}{2n}$$
$$φ = |(\frac{\theta}{a}) - b|$$
$$P = \frac{\cos(b)}{\cos(b - φ)}$$
In your image suppose other vertices for the triangle on the right side are B, C because you have already drawn O vertex there. Now some mathematical calculations.
$AB=\sin(36)$
$OB=r$
$PA=P\cos(36-\phi)$
$OA=\cos(36)$
Now apply: $a^2+b^2=c^2$
$OA^2+PA^2=PO^2$
$P^2\cos^2(36-\phi)+\cos^2(36)=P^2$
Solving you get:
$P=\sqrt{\frac{\cos^2(36)}{\sin^2(36-\phi)}}$
Is this the correct solution? Please let me know.