Question

The diagonals AC and BD of a parallelogram ABCD intersect each other at point O. If Angle DAC=32° and then Angle AOB=70° then DBC is equal to

Answer 1

Hence ∠DBC = 41°.

Step-by-step explanation:

Given:

∠DAC=32° and ∠AOB=70°

Here ABCD is a parallelogram.

Therefore, AD | | BC

So, ∠ACB = ∠DAC = 32°        [1]    

Now, ∠AOB is an exterior angle of △BOC,  

∴ ∠OBC +∠ OCB = ∠AOB  [∵ exterior ∠ = sum of two interior opposite angles]

⇒ ∠OBC + 34° = 75°         [from (1) ∠ACB=∠AOB=32°  ]

⇒  ∠OBC  = 75° - 34° = 41°        

or ∠DBC = 41°  

Hence ∠DBC = 41°.

Answer 2

Answer:

Hence ∠DBC = 41°.

Step-by-step explanation:

Given:

∠DAC=32° and ∠AOB=70°

Here ABCD is a parallelogram.

Therefore, AD | | BC

So, ∠ACB = ∠DAC = 32°        [1]    

Now, ∠AOB is an exterior angle of △BOC,  

∴ ∠OBC +∠ OCB = ∠AOB  [∵ exterior ∠ = sum of two interior opposite angles]

⇒ ∠OBC + 34° = 75°         [from (1) ∠ACB=∠AOB=32°  ]

⇒  ∠OBC  = 75° - 34° = 41°        

or ∠DBC = 41°  

Hence ∠DBC = 41°.