text: What is the difference in radii of two concentric circles given an angle and length of a triangle that is inscribed in the annulus?
  
  In relation to this geometric construction:
  
  where D is the center of both circles, if the inner radius (x = length of line segments DA and DE), the angle φ = ∠CAB, and the length Δg of line segment AB are given, what is the length Δx of line segment AC?
  
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  @muad: The inner radius is non-zero, but if it were, then Δx—the length that I seek—would be equal to the given value Δg. I cannot assign a value to the inner radius, Δg, or φ because these are all given quantities. – Daniel Trebbien Sep 26 '10 at 15:19
  
  Consider the triangle DAB. You know two sides and an angle. The length of the third side is $x + \Delta x$.
  Notice that we have, $\angle DAB=\pi-\phi$, $AD=x$, $AB=\Delta g$ & $BD=x+\Delta x$
  Now, applying Cosine rule in $\Delta ABD$ as follows $$\cos \angle DAB=\frac{(AD)^2+(AB)^2-(BD)^2}{2(AD)(AB)}$$ $$\cos (\pi-\phi)=\frac{(x)^2+(\Delta g)^2-(x+\Delta x)^2}{2(x)(\Delta g)}$$ $$\implies (x+\Delta x)^2=x^2+(\Delta g)^2+2x(\Delta g)\cos\phi$$ $$\implies (x+\Delta x)=\pm \sqrt{x^2+(\Delta g)^2+2x(\Delta g)\cos\phi}$$ $$\implies \Delta x=\pm \sqrt{x^2+(\Delta g)^2+2x(\Delta g)\cos\phi}-x$$ But $\Delta x>0$, Hence, we get $$\color{blue}{\Delta x=\sqrt{x^2+(\Delta g)^2+2x(\Delta g)\cos\phi}-x}$$
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