Answer : (b) 60º
Given,
∠OBD – ∠CDB = ∠CBD – ∠ODB
⇒ ∠OBD + ∠ODB = ∠CBD + ∠CDB ...(i)
∵ OB = OD (radii)
∠OBD = ∠ODB = θ (say)
Let ∠CBD = θ1, ∠CDB = θ2
Then putting these value in eqn (i), we have
θ + θ = θ1 + θ2 ⇒ 2θ = θ1 + θ2 ...(ii)
Also, ∠BOD = 180° – 2θ
⇒ Reflex ∠BOD = 360° – (180° – 2θ)
⇒ ∠BCD = \(\frac{1}{2}\) × Reflex × ∠BOD
= \(\frac{1}{2}\)× [360° – (180° – 2θ)] (Angle subtended at centre by an arc = 2 × Angle subtended at any point on remaining part of the circle)
Also ∠BCD = 180° – (θ1 + θ2)
180° – (θ1 + θ2 ) = \(\frac{360°- (180° -2\theta)}{2}\)
⇒ 180° – 2θ = 90° + θ (∵ θ1 + θ2 = 2θ)
⇒ 3θ = 90°
⇒ θ = 30°
∴ ∠BOD = 180° – 60° = 120°
⇒ ∠BAD = \(\frac{1}{2}\) × ∠BOD = 60°.