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 ?


$AI=k$, $\triangle AIF\sim\triangle ABD$ $\Longrightarrow$ $AB=2k$ , $AD=k+12$    
$AB^{2}=BD^{2}+AD^{2}$ $\Longrightarrow$ $k=20$ , $AF=s-a=16$ , $AT=20$   
$\triangle AOT\sim\triangle AIF$ $\Longrightarrow$ $\dfrac{AO}{AT}=\dfrac{AI}{AF}=\dfrac{5}{4}$ $\Longrightarrow$ $AO=25$
