Question

A, B, C, D are four distinct points on a circle whose center is at O. If ∠OBD – ∠CDB = ∠CBD – ∠ODB, then what is ∠A equal to?

(a) 45° 

(b) 60° 

(c) 120° 

(d) 135°

Answer 1

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°.