A Hard Geometry Problem on circle

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°