Question

find an answer its urgent....​

Answer 1

In ΔBDC

BD = DC

Therefore, ∠DCB = ∠DBC (angle opp. to equal side)

∠DCB = 40°

Now,

∠DCB + ∠DBC +∠BDC = 180° ( ASP)

40°+40°+∠BDC = 180°

∠BDC = 180 - 80

∠BDC = 100°

Now,

ABCD is cyclic quadrilateral

⇒∠BDC + ∠BAC = 180 (sm of opp sides of cyclic quad)

100 + ∠BAC = 180°

∠BAC = 80°

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