Question

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Answer 1

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∠COD=65 o

[ Given ]

⇒ ∠COD=∠AOB [ Vertically opposite angles ]

∴ ∠AOB=65 o .

In △AOB,

⇒ ∠BAC+∠AOB+∠ABO=180 o

[ Sum of angles of a triangle is 180 o. ]

⇒ 35 o+65 o+∠ABO=180 o .

⇒ 100 o

+∠ABO=180 o

⇒ ∠ABO=80 o

∴ ∠ABD=80 o

(ii) AB∥CD abd BD is transversal.

∴ ∠ABD=∠BDC [ Alternate angles ]

∴ ∠BDC=80 o

(iii)⇒ ∠AOB+∠BOC=180 o

[ Linear pair ]

⇒ 65 o

+∠BOC=180 o

∴ ∠BOC=115 o

⇒ ∠DAO=∠OCB [ Alternate angles ]

∴ ∠OCB=40 o

In △OCB

⇒ ∠BOC+∠OCB+∠CBO=180 o .

⇒ 115 o +40 o+∠CBO=180 o

⇒ 155 o +∠CBD=180 o

⇒ ∠CBD=25 o

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Answer 2

Step-by-step explanation:

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