What is the difference in radii of two concentric circles given an angle and length of a triangle that is inscribed in the annulus?

Solution 1:

Consider the triangle DAB. You know two sides and an angle. The length of the third side is $x + \Delta x$.

Solution 2:

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 $$\color{blue}{\Delta x=\sqrt{x^2+(\Delta g)^2+2x(\Delta g)\cos\phi}-x}$$