Question

A diagonal AC and BC of parallelogram ABCD intersects at the point O. If DAC=34 degrees and AOB = 75 degrees then what is the measure of OCB and DBC​

Answer 1

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°.

Answer 2

Answer:

Quadrilateral ABCD is a parallelogram.

So, AD ∣∣ BC

∴ ∠DAC = ∠ACB --- ( Alternate angle)

∴ ∠ACB = 32

∘

∠AOB + ∠BOC = 180

∘

--- (straight angle)

⇒70

∘

+ ∠BOC = 180

∘

∴ ∠BOC = 110

∘

In △BOC,

∠OBC + ∠BOC + ∠OCB = 180

∘

⇒∠OBC + 110

∘

+ 32

∘

= 180

∘

⇒ ∠OBC = 38

∘

∴ ∠DBC = 38

∘

Step-by-step explanation:

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