In the figure given below, AB is the chord of a circle with centre O. AB is extended to C such that BC = OB. The straight line CO is produced to meet the circle at D. If ∠ACD = y° and ∠AOD = x° such that x = ky, then the value of k is
Explanation:
Consider the figure shown above.
In ∆OBC, BC = OB
∴ ∠BOC = ∠BCO = y … (i)
Also, OB = OA = Radius of the circle
∠OBA is an exterior angle of ∆OBC.
∴ ∠OBA = ∠BOC + ∠BCO = 2y
∴ ∠OAB = ∠OBA = 2y …(ii)
∴ In ∆AOB, ∠AOB = 180 − 4y
Now, ∠AOD + ∠AOB + ∠BOC = 180°
∴ x + 180 − 4y + y = 180°
∴ x = 3y
∴ The value of k = 3
Hence, option (a).
Alternatively,
m∠ACD = 1/2 × [m(arc AD) – m(arc BM)]
∴ 2y = x – y
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