Geometry - Isosceles triangle made from intersection of circles 
Please view the picture above. 


*

*It is an isosceles triangle

*It is mirror symmetric, all circles are EQUAL in radius.

*Let the green and purple angles be P and G. Let's assume that Green = 75 deg, Purple = 30 deg (could be different, is it possible to have these as variable?)

*I am looking to find the BLUE and RED angles, given the values of P and G.

*The value of P and G are determined by how the circles are placed, basically it depends on radius of the circles as well as arc sector length. (check below)

*Note that, the angles are formed by the tangent of the circles, at the intersection point (Not necessarily equal to P or G for example)


The following picture is a better example of my last point

The only difference in those picture, is that the circles are closer together, same size, different arc length
(different view, to show the interior angles)

If you look at this picture, I consider arc length is more like "arc sector", from point G to point C.

Since it's an isosceles triangle and symmetrical, two of the circles will have the same arc length (let's call it L1), and other one, the bottom circle, will be L2.
Radius is R
Is there a way to determine a relationship between the angles, L1, L2 and R? If a simple "plug and chug" formula is not possible, then could there be a non-linear relationship that I could use?
Or one for each angle.
Thank you all!
Note: I may have asked a similar question before, but I had the wrong picture in my mind when I drew it, therefore I doubt it could have been solved.
Some clarification:

There are 3 circles of equal radius (known as fact). The "triangle" is for visual purpose to simplify the problem, its vertices are the intersection of the circles. My problem has 3 circles, intersecting like that, with one being a bit lower which forms an isosceles triangle when joining points. I want to find the angle formed by the tangent lines at the intersection point, as a function of arc length (sector) and radius. Again, the triangle came to be AFTER I laid down the circles, circles are defined arbitrarily on a plane (keeping symmetry of course)

 A: Here is the partial solution to the requested if my guess to your description is correct. 
Let the circles $C_A$ and $C_B$ intersect at D and P. PS and PT are tangents drawn at P w.r.t. $C_A$ and $C_B$ respectively. Join CB and also BP such that $\angle PBC = \theta$; and $\angle BPT = π/2$.

I assume the arc PDC is L (given). Then, L = rθ; meaning θ is therefore a known quantity (since L and r are known).
By (base angles isosceles triangle + angle sum of triangle + adj. angles on a st. line), we have
$(0.5)Blue + (π/2) + (0.5)(π – θ) + (0.5)purple = π$
Then, … blue = θ – purple
I think the orange one can be handled similarly.
A: It turns out that you need neither the symmetry nor the equal radii. Consider the following generic situation:

Here I assume that you know the outer arc lengths $L_1,L_2,L_3$ as well as the radii $R_1,R_2,R_3$. From these you can compute the angles
$$
\angle B_2A_1B_3=\frac{L_1}{R_1} \qquad
\angle B_3A_2B_1=\frac{L_2}{R_2} \qquad
\angle B_1A_3B_2=\frac{L_3}{R_3}
$$
You can use these to compute the edges of the triangle, since these are chords:
$$
\lvert B_2B_3\rvert=2R_1\sin\frac{L_1}{2R_1} \qquad
\lvert B_3B_1\rvert=2R_2\sin\frac{L_2}{2R_2} \qquad
\lvert B_1B_2\rvert=2R_3\sin\frac{L_3}{2R_3}
$$
Using the edge lengths, you can use the cosine law to compute the angles of the triangle $B_1B_2B_3$:
$$
\angle B_2B_1B_3 = \arccos\frac{\lvert B_1B_2\rvert^2-\lvert B_2B_3\rvert^2+\lvert B_3B_1\rvert^2}{2\cdot\lvert B_1B_2\rvert\cdot\lvert B_3B_1\rvert}
$$
and likewise for the other two corners with cyclically shifted indices.
Now consider triangle $A_1B_2B_3$. It is isosceles, so from the angle sum in a triangle you can conclude
$$
\angle B_3B_2A_1=\angle A_1B_3B_2=\frac{\pi-\angle B_2A_1B_3}{2}
$$
which will be negative if $\frac{L_1}{R_1}>\pi$ but that's just what I want. Again you get similar equations by shifting indices. Now you can combine them to compute
$$
\angle A_3B_1A_2=\angle B_2B_1B_3-\angle B_2B_1A_3-\angle A_2B_1B_3
$$
The angle of intersection between circles $C_2$ and $C_3$ is supplementary to this, because the tangents are orthogonal to the radii:
$$\angle C_2C_3 = \pi-\angle A_3B_1A_2$$
By setting $R_1=R_2=R_3=R$ and $L_1=L_3$ in all of the above, you obtain your symmetric situation.
