In Triangle ABC

AB = AC Therefore Triangle ABC is an isosceles traingle

So,

∠BAC + ∠BCA + ∠ABC = 180° [Angle sum property]

But it is an isosceles traingle so base angles are equal

Therefore,

∠BAC = ∠BCA

Therefore,

∠BAC + ∠BAC + 120 = 180

2∠BAC = 180 - 120

∠BAC = 60/2

∠**BAC = 30**°

Now,

∠ECB = 90° **[Angle subtended by the diameter on the circumference of cirlce is complemantary]**

∠ECA + ∠BCA = 90 **[BAC = BCA = 30]**

∠ECA + 30 = 90

**∠ECA = 60°**

Triangle COB is equilateral epecified below

So,

∠CBO = ∠CBE = 60°

In Traingle ECB

∠ECB + ∠CBE + ∠BEC = 180° [Angle sum property]

90 + 60 + ∠BEC= 180

∠CBE = 180 - 9 - 60

**∠BEC = 30°**

**Traingle EOD,DOC and COB are equilateral traingle as OE = OD = OC = OB all are radii of the circle and ED = DC = CB so all the 3 sides of all the 3 triangles are equal so by SSS Test All the 3 triangles are congruent**

So,

∠EOD = ∠DOC = ∠COB **[By C.P.C.T.C]**

∠EOD + ∠DOC + ∠COB = 180° **[EOB is a straight line]**

∠EOD + ∠EOD + ∠EOD = 180°

**3**∠EOD = 180

∠EOD = 60 = ∠COD = ∠BOD

Therefore,

**∠COD = ∠BOD = 60°**

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