In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB.

Question:

In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If ACD = 25°, then ∠AOD = ?
(a) 50°
(b) 75°
(c) 90°
(d) 100°

 

Solution:

(b) 75°
OB = BC (Given)
⇒ ∠BOC = ∠BCO = 25°
Exterior ∠OBA = ∠BOC + ∠BCO = (25° + 25°) = 50°
OA = OB (Radius of a circle)
⇒ ∠OAB = ∠OBA  = 50°
In Δ AOC, side CO has been produced to D.
∴ Exterior ∠AOD = ∠OAC  + ∠ACO

= ∠OAB + ∠BCO

= (50° + 25°) = 75°

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