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