Question

In Fig. 12, O is the centre of the circle and BA = AC, If ZABC = 50., find ZBOC and CBDC

Answer 1

Answer:

BA=AC

∠ABC=50

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Since BA = AC

∠ABC=∠ACB=50

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so ∠A=180

∘

−(50

∘

+50

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)

=80

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We know ∠BOC=2(∠BAC)

∠BOC=2×80=160

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ABCD is a cyclic Quadrilateral

So ∠A+∠D=180

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80

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+∠D=180

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⇒∠BDC=100