A Hard Geometry Problem on circle $$\angle(ABC) = 30°\\
\angle(BCO) = 20°\\
\angle(OCD) = 20°$$

How do i find $\angle(ODC)$? so i wanted to show my teacher this but he is not available yet. Can someone help me to solve? geometry problems on circle seems hard to me. Thanks!
Without loss of generality, we can consider $R=OC=OB=1$.

From $\Delta OBC: \frac{OB}{\sin20}=\frac{BC}{\sin140}\Rightarrow BC=\frac{\sin140}{\sin20}=2\cos20.$

From $\Delta BCD: \frac{BC}{\sin110}=\frac{CD}{\sin30}\Rightarrow CD=\frac{\cos20}{\sin110}=1.$

Finally from $\Delta OCD: \frac{OC}{\sin{x}}=\frac{CD}{\sin{(x+20)}}\Rightarrow \sin x=\sin{(x+20)}\Rightarrow x=80.$
First join $AC$ and $AO$. Now, $\angle ABC$ is $30^\circ$,so AOC is 60*.AO=OC.so,OAC=OCA=(180*-60*)÷2=60*. Triangle AOC is equilateral Triangle.so,AC=OC=OA.angleACD=(ACO-DCO)=(60*-20*)=40*.In triangle ACB ,CAB=70*,ACD=40*.So,ADC=70*.So,AC=CD. Now AC=CO=CD,So,OC=CD.in triangle CDO,DCO=20*,and OC=CD.So,CDO=COD=(180*-20*)÷2=80*.HENCE,CDO=80*.(ANS.).