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 ?

The standard formula for the inradius is r = area/s,where s is the semi-perimeter.Letting x = the triangle sides that are equal,we have s=(1/2)(x+x+48)=
x+24.The area is calculated  using Heron"s formula = sqrt(s(s-a)(s_b)(s_c)).In
your problem,this becomes sqrt((x+24)(24)(24)(x-24)= 24sqrt(x^2 - 576).
Substituting,we get:  12 = 24sqrt(x^2 - 576)/(x + 24).Dividing by 12,and cross-
multiplying, x+24 = 2sqrt(x^2 - 576).Squaring,(x+24)^2 = 4(x^2 - 576),or
x^2 + 48x + 576 = 4(x^2 - 576) = 4x^2 - 2304.Expanding and transposing,
3x^2 - 48x - 2880=0. Dividing by 3,x^2 - 16x - 960 =0.Solving for x,
2x = 16 + sqrt(256 + 3840),or 2x=16+sqrt(4096),and x=8 + 32 = 40.Then the
area is 24sqrt(1600 - 576) = 24sqrt(1024)=24*32 = 768.A formula for the
circumradius is R = abc/(4*area).In your problem,this becomes:
R = (40)(40)(48)/((4)(8)(8)(12)) = (5)(%)(4)/(4) = 25.      Ed Gray 
