Question

in the given figure, o is the centre of the circle. find angle BDC, if angle BOC 80°
pls solve this question.​

Answer 1

$\large{ \underline{ \pmb{ \sf{answer}}}}$

$\large \displaystyle \implies b \: a = a \: c$

angel ABC = 50°

Since BA = AC

∠ABC= ∠ACB = 50°

so ∠A=180° - (50 + 50)

= 80°

We know,

∠BOC=2(∠BAC)

∠BOC=2×80=160°

ABCD is a cyclic Quadrilateral

So ∠A+∠D=180°

80° +∠ D = 80°

⇒∠BDC=100°

hope its help u

Answer 2

Answer:

⟹ba=ac

angel ABC = 50°

Since BA = AC

∠ABC= ∠ACB = 50°

so ∠A=180° - (50 + 50)

= 80°

We know,

∠BOC=2(∠BAC)

∠BOC=2×80=160°

ABCD is a cyclic Quadrilateral

So ∠A+∠D=180°

80° +∠ D = 80°

⇒∠BDC=100°

Step-by-step explanation:

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