Question

please solve it full steps

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Answer 1

(i)

∠COD=65°                            [ Given ]

⇒  ∠COD=∠AOB           [ Vertically opposite angles ]

∴  ∠AOB=65°.

In △AOB,

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

⇒  35o+65o+∠ABO=180°

⇒  100o+∠ABO=180°

⇒  ∠ABO=80°

∴  ∠ABD=80°

(ii)

AB∥CD abd BD is transversal.

∴  ∠ABD=∠BDC                [ Alternate angles ]

∴  ∠BDC=80°

(iii)

⇒  ∠AOB+∠BOC=180°     [ Linear pair ]

⇒  65o+∠BOC=180°

∴  ∠BOC=115°

⇒  ∠DAO=∠∠OCB           [ Alternate angles ]

∴  ∠OCB=40o

In △OCB

⇒  ∠BOC+∠OCB+∠CBO=180°.

⇒  115°+40°+∠CBO=180°

⇒  155°+∠CBD=180°

⇒  ∠CBD=25°.