Question

In Fig., angle OAB = 30º and angle OCB = 57º. Find angle AOC.​

Answer 1

Answer:

⇒ ∠OAB=30

o

and ∠OCB=57

o

.

In △AOB,

⇒ AO=OB [ Radius of a circle ]

⇒ ∠OBA=∠BAO=30

o

[ Angles opposite to equal sides are equal ]

In △AOB,

⇒ ∠AOB+∠OBA+∠BAO=180

o

.

∴ ∠AOB+30

o

+30

o

=180

o

∴ ∠AOB=180

o

−30

o

−30

o

∴ ∠AOB=120

o

----- ( 1 )

Now, in △OCB,

⇒ OC=OB [ Radius of a circle ]

⇒ ∠OBC=∠OCB=57

o

. [ Angles opposite to equal sides are equal ]

In △OCB,

⇒ ∠BOC+∠OCB+∠OBC=180

o

.

⇒ ∠BOC+57

o

+57

o

=180

o

⇒ ∠BOC=180

o

−114

o

∴ ∠BOC=66

o

------ ( 2 )

⇒ ∠AOB=120

o

[ From ( 1 ) ]

⇒ ∠AOC+∠COB=120

o

.

⇒ ∠AOC+66

o

=120

o

⇒ ∠AOC=120

o

−66

o

∴ ∠AOC=54

o

Answer 2

Answer:

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