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 $\Delta OCD: \frac{OC}{\sin{x}}=\frac{CD}{\sin{(x+20)}}\Rightarrow \sin x=\sin{(x+20)}\Rightarrow x=80.$

Join OA and AC

angle AOC = 2xangle ABC=60 deg (center angle and circumference angle)

OC=OA (radii)

Triangle OAC is equilateral

AC=OA=OC

angle CAB =180-30-80=70 deg

angle CDA=40+30=70 deg

Therefore CA=CD=CO

ANGLE ODC=angle BAC+angle ABC

Angle ODC=(180-100)=80°
● ● ● ANGLE ODC IS 80°