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Question
In the given figure O is the center of the circle, ∠ BAD = 75° and chord BC = chord CD. Find:
(i) ∠BOC (ii) ∠OBD (iii) ∠BCD.
Solution
(i) ∠ BOD = 2 x ∠ BAD = 2 x 75° = 150°
∠ BOC = ∠ COD
∵ BC = CD
∴ ∠ BOD = 2 ∠ BOC
∴ ∠ BOC = `1/2` ∠ BOD = 75°
(ii) ∠ OBD = `1/2`( 180° - ∠ BOD)
∠ OBD = `1/2`( 180° - 150°) = 15°
(iii) ∠ BCD = 180° - ∠ BAD ....(Opposite ∠S of a cyclic quadrilateral is supplementary.)
∠ BCD = 180° - 75°
∠ BCD = 105°
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