Find the angle in a quadrilateral

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!