Find the angle in a quadrilateral 
This is the picture, and we are aiming for the angle $x$
It's easy to see that $\angle DGA = \angle CGB = 100°$, $\angle CGD = \angle AGB = 80°$, $\angle CBG = 50°$, but now i'm missing $\angle GBA =   ? $ and $\angle GAB = x$
Any hints?

Let $y$ denote the angle $\angle GBA$. By the law of sines applied to the four triangles with vertex $G$, we obtain the relations
$$\frac{AB}{\sin 80^\circ}=\frac{BG}{\sin x}=\frac{AG}{\sin y}.$$
$$\frac{AD}{\sin 100^\circ}=\frac{AG}{\sin 40^\circ}=\frac{DG}{\sin 40^\circ},$$
$$\frac{BC}{\sin 100^\circ}=\frac{BG}{\sin 30^\circ}=\frac{CG}{\sin 50^\circ},$$
$$\frac{CD}{\sin 50^\circ}=\frac{DG}{\sin 70^\circ}=\frac{CG}{\sin 30^\circ}.$$
Dividing the fourth equation by the third one, we get (cancelling out $CG$)
$$\frac{\sin^2 30^\circ}{\sin 50^\circ\sin 70^\circ}=\frac{BG}{DG}.$$
Now, from the second, we get that $DG=AG$, so 
$$\frac{\sin^2 30^\circ}{\sin 50^\circ\sin 70^\circ}=\frac{BG}{AG}.$$
Now, we apply the law of cosines to get 
$$AB^2=BG^2+AG^2-2BG\cdot AG \cos 80^\circ$$
Hence, by the first equation we gave by the law of sines
$$\frac{1}{\sin^2x} =\frac{1}{\sin^2 80^\circ}\frac{AB^2}{BG^2}=\frac{1}{\sin^2 80^\circ}\left(1+\frac{AG^2}{BG^2}-2\frac{AG}{BG} \cos 80^\circ\right)=\frac{1}{\sin^2 80^\circ}\left(1+\left(\frac{\sin 50^\circ\sin 70^\circ}{\sin^2 30^\circ}\right)^2-2\frac{\sin 50^\circ\sin 70^\circ}{\sin^2 30^\circ} \cos 80^\circ\right)$$
Easily you can check that $x=20^\circ$.
