Question

The diagonals AC and BD of a
parallelogram ABCD intersect each other at
the point o. If CAD = 32° and AOB = 70°,
find DBC.​

Answer 1

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

∘

Answer 2

Step-by-step explanation:

In given figure,

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

∘