Question

In a rectangle ABCD, the diagonals intersect at o. If OAB =40° , find AOB ​

Answer 1

Answer:

In △ABC,

⇒  ∠CAB+∠ABC+∠ACB=180o.

⇒  30o+90o+∠ACB=180o.

⇒  120o+∠ACB=180o.

∴  ∠ACB=60o

We know that, diagonals of rectangle are equal and bisect each other equally.

∴  AO=OC=BO=OD

In △ABO,

⇒  AO=BO

⇒  ∠OAB=∠ABO        [ Angle opposite to equal side are also equal ]

⇒  ∠OAB=∠ABO=30o

⇒  ∠OAB+∠ABO+∠BOA=180o

⇒  30o+30o+∠BOA=180o.

⇒  ∠BOA=120o.

⇒  ∠BOA=∠COD          [ Vertically opposite angle ]

∴  ∠COD=120o

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

⇒  120o+∠BOC=

Step-by-step explanation:

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Answer 2

Answer:

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