A problem on a triangle's inradius and circumradius . I'm trying to solve the following problem :  
In $△ABC$, $AB = AC, BC = 48$ and inradius $r = 12$. Find the
circumradius $R$.
Here is a figure that I drew : ( note : it was not given in the question so there may be some mistakes ) 

I don't know how to solve it , am I missing any relation between inradius , circumradius and sides of a isosceles triangle?
EDIT: Is there a simple solution without using trigonometry ?

Let $M$ be the midpoint of $BC$, let $P$ be the point where the perpendicular from $O$ meets the side $AB$, and let $|PA|=:x$. Since the two tangent segments from $B$ to the incircle have equal length it follows that  $|PB|=24$; therefore $|AB|=24+x$, and $|AO|^2= 12^2+x^2$. It follows that
$$(24+x)^2=24^2+\bigl(12+\sqrt{12^2+x^2}\bigr)^2\ .$$
Solving for $x$ gives $x=16$, whence $|AB|=40$, $|AO|=20$, $|AM|=32$.
Now let $|MK|=:y$. Then $\sqrt{24^2+y^2}=32-y$, which enforces $y=7$. It follows that $R=32-7=25$.
