In the given figure, O is the center of a circle and ZAOC = 120°. Then, LBDC = C 120 A B
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( c ) 30°
∠COB = 180° - 120° = 60° (Linear pair)
Now, arc BC subtends ∠COB at the centre and ∠BDC at the point D of the remaining part of the circle.
∴ ∠COB = 2∠BDC
⇒ ∠BDC=1/2∠COB=1/2×60°=30°
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