Question

in the figure O is the centre of the circle central of aAXB is 60° arc CYD is 80° then find the all the angle of angle APD​

Answer 1

Answer:

Ref. Image.

BA=AC

∠ABC=50

∘

Since BA = AC

∠ABC=∠ACB=50

∘

so ∠A=180

∘

−(50

∘

+50

∘

)

=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

∘

80

∘

+∠D=180

∘

⇒∠BDC=100