Probability of triangle to be obtuse 
Two points $A,B$ fixed on a plane(distance = 2).
C - random choosen point inside circle with radius $R$ with center at the center of $AB$
Find probability of triangle $ABC$ to be obtuse

My thoughts: 


*

*If $C$ lies in the circle - $ABC$ will be rectangular

*Opposite the larger angle is large side, so using cosine theorem: 
$$
b^2+c^2<a^2=4R^2,
$$
where $a$ - the diagonal, $b=AC, c=BC$ 


And then I don't know...

Not a solution, this image should make it clear that the solution is a function of $R$
Any point $C$ within the purple region will form an obtuse triangle $ABC$, while points within the blue region will form an acute triangle, and points on the border of the two regions will form a right triangle.

On the left, $R=2$, and on the right, $R=14$
