find x in triangle

Drop a perpendicular from $A$ to $BC$ at $E$ . Let $AB=CD=1$ . Then $|AE|=\sin 20^\circ$ , also

$\begin{aligned} |DE|+|EC| & = |DC| \\ \frac {|AE|}{\tan \angle ADC} + \frac {|AE|}{\tan \angle ACD} & = 1 \\ \frac {\sin 20^\circ}{\tan 30^\circ} + \frac {\sin 20^\circ}{\tan x^\circ} & = 1 \end{aligned}$

$\begin{aligned} \implies \frac 1{\tan x^\circ} & = \frac 1{\sin 20^\circ} - \frac 1{\tan 30^\circ} \\ & = \frac 1{\sin 20^\circ} - \frac {\cos 30^\circ}{\sin 30^\circ} \\ & = \frac {\sin 30^\circ - \blue{\sin 20^\circ} \cos 30^\circ}{\sin 20^\circ \sin 30^\circ} \\ & = \frac {\sin 30^\circ - \blue{\cos 70^\circ}\cos 30^\circ}{\sin 20^\circ \sin 30^\circ} \\ & = \frac {\frac 12 - \frac 12 (\cos 40^\circ + \cos 100^\circ)}{\frac 12 (\cos 10^\circ - \cos 50^\circ)} \\ & = \frac {1 - \cos 40^\circ + \cos 80^\circ}{\sin 80^\circ - \sin 40^\circ} \\ & = \frac {1 - \cos 40^\circ + 2\cos^2 40^\circ - 1}{2\sin 40^\circ \cos 40^\circ - \sin 40^\circ} \\ & = \frac {\cos 40^\circ \cancel{(2\cos 40^\circ - 1)}}{\sin 40^\circ \cancel{(2\cos 40^\circ - 1)}} \\ \implies x & = \boxed{40} \end{aligned}$

take a compass put it on the screen at the angle 'x' you have your answer 40. i know it seems idiotic but try it, no use of math only a tool called compass and boom you got your answer.

"smart people never do hard works, and i do it smarty." -trupal panchal

$\tan x=\dfrac{2\sin 20\degree\sin 30\degree}{1-2\sin 20\degree\cos 30\degree}=\tan 40\degree\implies x=\boxed {40\degree}$ .