AB is the chord of the circle with centre O. A line segment DOC originating from a point D on the circumference of the circle in major segment meets AB produced at C such that BC = OD. If angle BCO = 30°, then angle AOD is:
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 90°.
According to the given question,
Given that, OD = BC
Also, OD = OB = radius of circle
⇒ OB = BC
Now, triangle OBC is as isosceles triangle.
⇒ ∠BCO = ∠BOC = 30°
Now, ∠BCO + ∠BOC + ∠OBC = 180°
⇒ ∠OBC = 180° - 30° - 30°
⇒ ∠OBC = 120°
Now, ∠OBA = 180° - ∠OBC
⇒ ∠OBA = 60°
In triangle AOB, AO = BO (radius)
⇒ ∠OBA = OAB = 60°
So, ∠AOB = 60°
Now, ∠AOD = 180° - ∠AOB - ∠BOC
⇒ ∠AOD = 180° - 60° - 30°
⇒ ∠AOD = 90°