Finding the area of a triangle, given the distance between center of incircle and circumscribed circle Consider the following depiction:

$ABC$ is an isosceles triangle ($AB=AC$), where the two angles opposite the equal sides are equal $\beta$ ($\beta>60$), and $AD$ perpendicular to $BC$.
$O$ is the center of the circumscribed circle of $ABC$, and $M$ is the center of the incircle of $ABC$.
I want to express the area of the triangle $BOM$ and the radius of the incircle of $ABC$, in terms of $k$ and $\beta$.  
This is what I tried: I denoted the radius of the circumscribed circle as $R$, and the radius of the incircle as $r$.
The area of  $BOM$ can be obtained by: $\frac{k\cdot BD}{2}$, and $BD=\frac{BC}{2}$, I get that the area is actually:  $\frac{k\cdot BC}{4}$.
Now from Law of sines I get $BC=\sin(180-2\beta)\cdot 2 R$.
So this reduces to problem to expressing $R$ (which is $OA$) in terms of $\beta$ and $k$. This is where I'm stuck. I couldn't find a way to do that.
I know that $OA=OB=R$, and that $AOB$ is also isosceles triangle, and I have a feeling it is important, but I can't figure out what am I missing...
Hints and guidelines would be very appreciated. 

Hint:-
Let the equal sides be $x$,
Distance of circumcenter from A, OA = circumradius, $R=\dfrac{abc}{4\triangle}$
Distance of incenter from A, MA = $r\csc{\dfrac{A}{2}}$
You will get MA - OA in terms of $x$ and $\beta$, but MA - OA = k
Then you can substitute the value of $x$ in terms of $k$ and $\beta$ in the formula for the area, i.e.
$\dfrac{1}{2}\cdot2x\cos\beta\cdot x\sin\beta$
