in tbe given figure the diagonals of parallelogram ABCD intersect at O If Angle AOB=70 degree and angleDAC=30 degree find angle BDA About the author Caroline

Answer: ans. is 40 Step-by-step explanation: AD ∣∣ BC ∴ ∠DAC = ∠ACB — ( Alternate angle) ∴ ∠ACB = 30 ∘ ∠AOB + ∠BOC = 180 ∘ — (straight angle) ⇒70 ∘ + ∠BOC = 180 ∘ ∴ ∠BOC = 110 ∘ In △BOC, ∠OBC + ∠BOC + ∠OCB = 180 ∘ ⇒∠OBC + 110 ∘ + 30 ∘ = 180 ∘ ⇒ ∠OBC = 40 ∘ ∴ ∠DBC = 40 ∘ Reply

Answer:ans. is 40

Step-by-step explanation:AD ∣∣ BC

∴ ∠DAC = ∠ACB — ( Alternate angle)

∴ ∠ACB = 30

∘

∠AOB + ∠BOC = 180

∘

— (straight angle)

⇒70

∘

+ ∠BOC = 180

∘

∴ ∠BOC = 110

∘

In △BOC,

∠OBC + ∠BOC + ∠OCB = 180

∘

⇒∠OBC + 110

∘

+ 30

∘

= 180

∘

⇒ ∠OBC = 40

∘

∴ ∠DBC = 40

∘