Question

ABCD is a parallelogram whose diagonals intersect at point O. If angle ADO = 61°, angle DOC = 34° and angle OAB = 53°. Find angle BDC, DBC, DOA and DCB​

Answer 1

Answer:

ABCD is a parallelogram .

∴ AD | | BC ⇒ ∠ACB = ∠DAC = 34°

Now, ∠AOB is an exterior angle of △BOC

∴ ∠OBC + OCB = ∠AOB [∵ ext ∠ = sum of two int. opp. ∠S]

⇒ ∠OBC + 34° = 75°

⇒ ∠OBC = 75° - 34° = 41°

or ∠DBC = 41°

Step-by-step explanation:

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