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 $\angle ACD={25}^{\circ }$ then $\angle AOD$ = ?

(a) ${50}^{\circ }$

(b) ${75}^{\circ }$

(c) ${90}^{\circ }$

(d) ${100}^{\circ }$

Answer 1

ANSWER:
(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°