Question

angle abc =69° angle acb = 31° find angle dbc​

Answer 1

Answer:

∠ABC = 69°, ∠ACB = 31°   [Given]

In ΔABC

∠ABC + ∠ACB + ∠BAC = 180°

⇒ ∠BAC  = 180° - ∠ABC - ∠ACB

⇒ ∠BAC  = 180° -  69° - 31° = 80°

∠BDC = ∠BAC  ( Angle drawn from same chord) 

∴ ∠BDC = 80°

I hope it helps you ^_^

Answer 2

Answer:

Given :

• A circle with centre O. $\angle$ ABC = 69° and $\angle$ ACB = 31°.

To find :

• $\angle$ BDC

Proof :

$\angle$ BAC = $\angle$ BDC $\qquad$ ( angle in the same segment of the circle)

In ∆ ABC,

$\\ \\ \sf \angle \: BAC + \angle \: ABC + \angle \: ACB = 180 {}^{ \circ} \\ \\ \\ \sf \implies \: \sf \angle \: BAC + 69 {}^{ \circ} + 31 {}^{ \circ} = 180 {}^{ \circ} \\ \\ \\ \implies \sf \: \angle \: BAC = 180 - 100 \\ \\ \\ \implies \sf \: \angle \: BAC = 80 {}^{ \circ} \\ \\ \\ \large{ \boxed{ \sf{thus \: \: \angle \:BAC = 80 {}^{ \circ} }}}$