Foundation
Mathematics
A Circle
Question
ABCD is a cyclic quadrilateral inscribed in a circle with the centre O. Then ∠OAD is equal to
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Solution
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