Problem involving triangle. Find $x$ in the figure 
I need to find $x$ in the triangle above.
I tried to do basic things, like sum of a triangle's internal angles $= 180^\circ$ but I only found $2$ equations for $3$ variables
Any help is appreciated
 A: In triangle $ABC$ you have the following angles $\angle A=120^\circ$ (top corner), $\angle B=20^\circ$ (left corner) and $\angle C=40^\circ$ (right corner). Denote the central point with $D$ and introduce lengths $AD=a, BD=b,CD=c$.
By law of sines applied to triangles $ABD,ACD,BCD$:
$$a\sin20^\circ=b\sin10^\circ\tag{1}$$
$$a\sin100^\circ=c\sin(40^\circ-x)\tag{2}$$
$$b\sin10^\circ=c\sin x\tag{3}$$
From (1) and (2):
$$b=\frac{a\sin20^\circ}{\sin10^\circ}$$
$$c=\frac{a\sin100^\circ}{\sin(40^\circ-x)}$$
Replace that into (3):
$$\frac{a\sin20^\circ}{\sin10^\circ}\sin10^\circ=\frac{a\sin100^\circ}{\sin(40^\circ-x)}\sin x$$
$$\sin20^\circ \sin(40^\circ-x)=\sin100^\circ\sin x$$
$$\sin20^\circ \sin(40^\circ-x)=\cos10^\circ\sin x$$
$$2\sin10^\circ \cos10^\circ \sin(40^\circ-x)=\cos10^\circ\sin x$$
$$2\sin10^\circ \sin(40^\circ-x)=\sin x$$
Sometimes you have to make things more complicated before your are able to jump over the last hurdle: multiply the right side with $1=2\sin30^\circ$.
$$2\sin10^\circ \sin(40^\circ-x)=2\sin x\sin30^\circ$$
$$\cos(-30^\circ+x)-\cos(50^\circ-x)=\cos(x-30^\circ)-\cos(x+30^\circ)$$
$$\cos(50^\circ-x)=\cos(x+30^\circ)$$
For obviously acute angle $x$
$$50^\circ-x=x+30^\circ$$
$$x=10^\circ$$
No calculator needed.
A: Using the same labeling as the author of the trig proof:
Cut off $BE$ on $BC$, such that $BE=BA$. Then, using the fact that $BD$ is the bisector of  $\angle B$, we can easily show that  $\triangle ADE$ is equilateral, so that $EA=ED$. Similarly, we can show that $\triangle AEC$ is isosceles (with base angles of $40^\circ$), so that $EA=EC$. Therefore $ED=EC$, so that, in $\triangle EDC, \angle EDC = \angle ECD = x$. 
But the external angle $BED (=\angle BAD) = 20^\circ$ , therefore $2x = 20^\circ$ ,
yielding that $x=10^\circ$.
