1. ABCD is a cyclic quadrilateral whose vertices are equidistant from the point 0 (centre of the circle). If ∠COD = 120° and ∠BAC = 30°, then the measure of ∠BCD is
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By: anil on 05 May 2019 01.58 am
Given : OA = OB = OC = OD To find : $$angle$$BCD = ? Solution : $$angle$$COD + $$angle$$BOC = 180° [Linear Pair] => $$angle$$BOC = 180° - 120° = 60° Also, $$angle$$OBC = $$angle$$OCB [$$ecause$$ OB = OC] In $$ riangle$$BOC => $$angle$$BOC + $$angle$$OCB + $$angle$$OBC = 180° => $$angle$$OCB = frac{120°}{2} = 60° --------Eqn(1) Also, $$angle$$OAB = $$angle$$OCD [Alternate interior angles] => $$angle$$OCD = 30° ---------------Eqn(2) Adding eqn (1) & (2), we get : => $$angle$$OCB + $$angle$$OCD = 60° + 30° => $$angle$$BCD = 90°
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Given : OA = OB = OC = OD To find : $$angle$$BCD = ? Solution : $$angle$$COD + $$angle$$BOC = 180° [Linear Pair] => $$angle$$BOC = 180° - 120° = 60° Also, $$angle$$OBC = $$angle$$OCB [$$ecause$$ OB = OC] In $$ riangle$$BOC => $$angle$$BOC + $$angle$$OCB + $$angle$$OBC = 180° => $$angle$$OCB = frac{120°}{2} = 60° --------Eqn(1) Also, $$angle$$OAB = $$angle$$OCD [Alternate interior angles] => $$angle$$OCD = 30° ---------------Eqn(2) Adding eqn (1) & (2), we get : => $$angle$$OCB + $$angle$$OCD = 60° + 30° => $$angle$$BCD = 90°