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?
 A: By law of sines we obtain:
$$\frac{BG}{GD}=\frac{\frac{BG}{AG}}{\frac{GD}{AG}}=\frac{\sin{x}}{\sin(100^{\circ}-x)}.$$
In another hand,
$$\frac{BG}{GD}=\frac{\frac{BG}{CG}}{\frac{GD}{CG}}=\frac{\frac{\sin30^{\circ}}{\sin50^{\circ}}}{\frac{\sin70^{\circ}}{\sin30^{\circ}}}=\frac{1}{4\sin50^{\circ}\sin70^{\circ}},$$
which gives
$$\frac{\sin{x}}{\sin(100^{\circ}-x)}=\frac{1}{4\sin50^{\circ}\sin70^{\circ}}$$ or
$$\sin100^{\circ}\cot{x}-\cos100^{\circ}=4\sin50^{\circ}\sin70^{\circ}$$ or
$$\cot{x}=\frac{\cos100^{\circ}+4\sin50^{\circ}\sin70^{\circ}}{\sin100^{\circ}},$$
which gives $x=20^{\circ}.$
Indeed,
$$\cot{x}-\cot20^{\circ}=\frac{\cos100^{\circ}+4\sin50^{\circ}\sin70^{\circ}}{\sin100^{\circ}}-\cot20^{\circ}=$$
$$=\frac{\cos100^{\circ}+2(\cos20^{\circ}-\cos120^{\circ})}{\cos10^{\circ}}-\frac{\cos20^{\circ}}{\sin20^{\circ}}=$$
$$=\frac{2\sin10^{\circ}(\cos100^{\circ}+2\cos20^{\circ}+1)-\cos20^{\circ}}{\sin20^{\circ}}=$$
$$=\frac{2\sin10^{\circ}(\cos40^{\circ}+\cos20^{\circ}+1)-\cos20^{\circ}}{\sin20^{\circ}}=$$
$$=\frac{\sin50^{\circ}-\sin30^{\circ}+\sin30^{\circ}-\sin10^{\circ}+2\sin10^{\circ}-\cos20^{\circ}}{\sin20^{\circ}}=$$
$$=\frac{\sin50^{\circ}+\sin10^{\circ}-\cos20^{\circ}}{\sin20^{\circ}}=0$$ and we are done!
A: 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 $x=20^\circ$.
