# In the following figure, O is the center of the circle such that ∠ABC

| In the following figure, O is the center of the circle such that ∠ABC = 30°, OC is the bisector of ∠BCD and ∠BCO = 20° What is ∠ODC?

A. 70°

B. 80°

C. 50°

D. 60°

### Right Answer is: B

#### SOLUTION

Join AC and AO, then ∠AOC = 2 x 30° = 60° (Angle made by a chord at centre of circle is double of the angle made by the same chord on circumference)

Now, AO = CO ⇒ ∠OAC = ∠OCA = 60°

∠ACB = ∠ACO +∠OCB = 60° + 20° = 80°

In Triangle ABC;

⇒ ∠CAB = 180° – (∠ACB +∠ABC) =180° – (80° + 30°) = 70°

Also, ∠CDA is external angle of triangle BDC;

⇒ ∠CDA = ∠DCB + ∠B = 40° + 30° = 70°

Now, From triangle OAC;

OAC is an equilateral Triangle; so

⇒ CA = CO

also, in triangle CDA;

∠CDA = ∠CAD = 70°

so, CD = CO

this makes-

⇒ ∠CDO = ∠COD (in Triangle COD)

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In the following figure, O is the center of the circle such that ∠ABC