Question

In figure ABCD is a parallelogram in which angle CAD = 30° , angle DCA = 45° and angle COD = 65° . Find (i) angle ABD (ii) angle BDC (iii) angle ACB (iv) angle CBD


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Answer 1

Answer:

don't know sorry

Answer 2

Answer:

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Step-by-step explanation:

(i)

∠COD=65o                             [ Given ]

⇒  ∠COD=∠AOB           [ Vertically opposite angles ]

∴  ∠AOB=65o.

In △AOB,

⇒  ∠BAC+∠AOB+∠ABO=180o            [ Sum of angles of a triangle is 180o. ]

⇒  35o+65o+∠ABO=180o.

⇒  100o+∠ABO=180o

⇒  ∠ABO=80o

∴  ∠ABD=80o

(ii)

AB∥CD abd BD is transversal.

∴  ∠ABD=∠BDC                [ Alternate angles ]

∴  ∠BDC=80o

(iii)

⇒  ∠AOB+∠BOC=180o       [ Linear pair ]

⇒  65o+∠BOC=180o

∴  ∠BOC=115o

⇒  ∠DAO=∠OCB           [ Alternate angles ]

∴  ∠