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?

-
@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 $$\color{blue}{\Delta x=\sqrt{x^2+(\Delta g)^2+2x(\Delta g)\cos\phi}-x}$$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}