Foundation
Mathematics Foundation
Tangent to Circle
Question

ABCD is a cyclic quadrilateral inscribed in a circle with the centre O. Then ∠OAD is equal to

30º    

40º    

50º    

60º    

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Solution
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 ABCD is a cyclic quadrilaterial .
    In triangle BOC
    ∠OCB = ∠OBC = 30º 
    { ∴OB = OC radius of circle}
    Now ∠BCD = ∠BCO + ∠OCD  = 30º + 50º
    ∠BCD = 80            --------(1)
    In cyclic quadrilateral ABCD 
    ∠BAD + ∠BCD = 180º
    ∠BAO + ∠OAD + ∠BCD = 180º
    ∠OAD = 180º – (∠BCD + ∠BAO)
    = 180º – (80º + 40º) = 60º

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