find angle x in triangle

Label the figure as above. We note that

$\begin{aligned} x & = \angle BEG - \angle BEF \\ & = \tan^{-1} \left(\frac {BG}{EB} \right) - \tan^{-1} \left(\frac {BF}{EB} \right) \\ & = \tan^{-1} \left(\frac {BC\tan 30^\circ \tan 40^\circ}{BC\tan 10^\circ} \right) - \tan^{-1} \left(\frac {BC\tan 20^\circ \tan 40^\circ}{BC\tan 10^\circ} \right) \\ & = \tan^{-1} \left(\frac {\tan 30^\circ \tan 40^\circ}{\tan 10^\circ} \right) - \tan^{-1} \left(\frac {\tan 20^\circ \tan 40^\circ}{\tan 10^\circ} \right) \\ & = \tan^{-1} \left(\frac {\tan 30^\circ \tan (30^\circ+10^\circ)}{\tan 10^\circ} \right) - \tan^{-1} \left(\frac {\tan (30^\circ - 10^\circ) \tan (30^\circ+10^\circ)}{\tan 10^\circ} \right) & \small \blue{\text{Let }t = \tan 10^\circ} \\ & = \tan^{-1} \left(\frac {\frac 1{\sqrt 3} + t}{\blue{\sqrt 3 t} \left(1-\frac t{\sqrt 3} \right)} \right) - \tan^{-1} \left(\frac {\left(\frac 1{\sqrt 3} - t\right)\left(\frac 1{\sqrt 3} + t\right)}{t \left(1+\frac t{\sqrt 3} \right)\left(1-\frac t{\sqrt 3} \right)} \right) & \small \blue{\text{then } \frac {3t-t^3}{1-3t^2} = \frac 1{\sqrt 3}} \\ & = \tan^{-1} \left(\frac {\blue{(3-t^2)}(1+\sqrt 3t)}{\blue{(1-3t^2)}(\sqrt 3 - t)} \right) - \tan^{-1} \left(\frac {\frac 13 - t^2}{t \left(1-\frac {t^2} 3 \right)} \right) & \small \blue{\implies \frac {3-t^2}{1-3t^2} = \frac 1{\sqrt 3t}} \\ & = \tan^{-1} \left(\frac {\sqrt 3+t}{1-\sqrt 3t} \right) - \tan^{-1} \left(\frac {1 - 3t^2}{3t-t^3} \right) \\ & = \tan^{-1} \left(\frac {\tan 60^\circ + \tan 10^\circ}{1-\tan 60^\circ \tan 10^\circ} \right) - \tan^{-1} (\cot 30^\circ) \\ & = 70^\circ - 60^\circ = \boxed {10}^\circ \end{aligned}$