The area of the region bounded by three mutually-tangent circles

Hints have been given in the comments. To make it more complete, I add a few more.

(1) $AB = R_1 + R_2$, and similar results for $BC$ and $CA$.

(2) Because all sides of $\triangle ABC$ are known, all three angles (in radians) can be found by applying cosine law (three times) (or cosine law then sine law and then " angle sum of triangle").

(3) Need to apply the conversion ratio ($\pi$ radian $= 180$ degrees) if not already done so in (2).

(4) Area of the yellow sector $= \frac {1}{2}(R_2)^2 \theta $. Again, $\theta$ should be in radian.

(5) Area of $\triangle ABC$ can be found by Heron’s formula or by the area formula $A = \frac {1}{2}ab. \sin C$.

The answer is quite complex(I think, assuming only the three radii are given). First, apply Sine Law in the triangle with vertices as centers of the three circles.( Also, assuming their radii to be $R_1,R_2,R_3$ and accordingly $\theta_1,\theta_2,\theta_3$ the angles of the triangle. You have $$\frac{R_1+R_2}{\sin\theta_3}=\frac{R_2+R_3}{\sin\theta_1}=\frac{R_1+R_3}{\sin\theta_2}$$ Now for the question, the blue area is the area of the triangle minus the area of the three sectors. Mathematically it is, $$\frac{1}{2}(R_1+R_3)(R_1+R_2)\sin\theta_1-\frac{\theta_1}{360°}\pi R_1^2-\frac{\theta_2}{360°}\pi R_2^2-\frac{\theta_3}{360°}\pi R_3^2$$ Now replace each of the angle from the Sine Law.(though it will be in $\sin^{-1}$)

Edit: Now I see the angles cannot just be replaced(they are interdependent on each other), there must be some other way of replacing the angles.