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!
 A: 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 $\Delta OCD: \frac{OC}{\sin{x}}=\frac{CD}{\sin{(x+20)}}\Rightarrow \sin x=\sin{(x+20)}\Rightarrow x=80.$
A: 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° 
A: Hint...with a bit of angle-chasing you should be able to establish $\angle ADC=70$.
You can then use the sine rule in triangles $ODB$ and $ODC$ (assume the radius is $1$)
The final answer seems to be $x=80$ which would suggest there must be a better way.
A: $$/angle(ABC)=30°//
/angle(AOC)=2*angle(ABC)=60°//$$
So AOC is equilateral triangle (AO=AC=CO=R)
$$/angle(CAB)=/angle(COB)/2=70°
//angle(ACD)=70°$$
So CD=CB=R and CDO is isosceles triangle so
$$/Angle(CDO)=angle(DOC)
Angle(CDO)=90°-Angle(DCO)/2=160°/2=80°$$
A: Produce CD and meet the circle at E
Join OA and AC
angle APC = 60 deg (centre and circumference angles)
OA = AC = OC
Join BE
angle CAB = angle CEB = 70 deg (CE = CB and angle BCE = 40deg)
Therefore AD = CD = CD
x = 80 deg
A: 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.).
A: BCD is equal to 40 (OCD plus BCO). So the unknown corner of the triangle is 180 - (40 + 30) which is 110. ADC, CDB and ODB are supplementary, so ADC is 70. COB is  180 - (20 + 15) which is 155. I'm pretty sure DOB is equal to COB, so 155 + 155 equals 310 therefore COD should be 50? 180 - (20 + 50) is 110 so x is 110. I'm pretty sure that's the answer, however I'm only a Year Seven, that happens to love Geometry and I've just started using this. Basically I'm saying, don't think my word is final. Hope it helps anyways. :) please tell me if I'm not right, I would like to know to correct answer if I've gotten it wrong.
