Find this angle, in terms of variables I have a geometry question.
Take a look at this figure:

It is 3 circles, symmetrically placed so that the arc length that is outside is equal for all sides. I tried to determine the angle that is tangent to two sides as they intersect (Theta in the drawing) in terms of $L$ (arc length) and $R$ (radius). Obviously, all $L_1=L_2=L_3$ and $R_1=R_2=R_3$ since the circles are the same.
I figured the angle to be $\dfrac{5\pi}{3} - \dfrac{L}{R}$.
Anyways, so my problem is: If I were to make a polygon, made of symmetrical circles where the vertices are intersection, e.q.

Is there a way to find the angle in term of the number of sides.
I am assuming it would look like N( ) - L/R, where N is the number of sides and ( ) is some angle expression.
I appreciate all helps!
 A: 
Consider the regular $N$-gon formed by the centers of the circles:
The outer angle of that polygon, the orange one, is
$$\alpha=2\pi-\frac{(N-2)\pi}{N}=\pi+\frac{2\pi}N$$
The green angle $\beta$ is related to the angle you're asking for. Due to a mistake in my formulation of this answer, I originally assumed $\beta=\frac\theta2$, but that is not the case even though it looks that way in the figure. Instead, the tangents are perpendicular to the radii, so we get
$$\beta=\frac\pi2-\frac\theta2=\frac{\pi-\theta}2$$
Therefore the purple angle is
$$\gamma=\frac\pi2-\beta=\frac{\theta}2$$
The angle corresponding to the outer arc, colored in cyan, is
$$\delta=\alpha-2\gamma=\pi+\frac{2\pi}N-\theta=\frac LR$$
So you can solve this for $\theta$ and obtain
$$\theta=\alpha-\frac LR=\pi+\frac{2\pi}N-\frac LR$$
This isn't exactly the shape you suggested, particularly since $N$ occurs in the denominator, not in the numerator. But for $N=3$ it now agrees with your result, thanks to the fix your comment triggered.
