<s>[INST] 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
 [/INST] 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.
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