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 =?
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In △OBC, OB = BC
⇒ ∠BOC = ∠BCO = y ...[angles opp. to equal sides are equal]
∠OBA is the exterior angle of △BOC
So, ∠ABO = 2y ...[ext. angle is equal to the sum of int. opp. angles]
Similarly, ∠AOD is the exterior angle of △AOC
∴ x = 2y + y = 3y
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