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
∘