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Question

# In given figure, O is the centre of the circle ∠BCO=30∘ Find x and y

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Solution

## O is the center of the circle and ∠BCO=30o. In given figure join OB and AC.In △BOC,⇒ CO=BO [ Radius of a circle ]∴ ∠OBC=∠OCB=30o [ Angles opposite to equal sides are equal ]⇒ ∠BOC+∠OBC+∠OCB=180o [ By angles sum property of a triangle ]⇒ ∠BOC+30o+30o=180o⇒ ∠BOC=180o−60o⇒ ∠BOC=120oWe know that, in a circle, the angle subtended by an arc at the center is twice the angle subtended by it at the remaining part of the circle.⇒ ∠BOC=2∠BAC⇒∠BAC=120o2=60o⇒ ∠BAE=∠CAE=30o [ AE is an angle bisector of ∠A ]⇒ ∠BAE=x=30oIn △ABE,⇒ 30o+∠EBA+90=180o⇒ 120o+∠EBA=180o⇒ ∠EBA=60o⇒ ∠ABD+y=60oIn a circle, the angle subtended by an arc at the center is twice the angle subtended by it at the remaining part of the circle.⇒ 12×∠AOD+y=60o∴ 90o2+y=60o⇒ 45o+y=60o⇒ y=60o−45o∴ y=15o

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