Question

In the given figure, O is the centre of the circle angle AOB is equal to 140 degree Find angle ACB ​

Answer 1

Answer:

(1)∠ACB=

2

1

∠AOBbisectors

=

2

1

×(360

0

−140

0

)=

2

1

×220=110

0

(2)∠OBC=360

0

−(∠AOB+∠OAC+∠ACB)

=360

0

−(140

0

+50

o

+110

0

)

=360

0

−300

0

=60

0

(3)since,OA&OBisradius

∴ΔOABisisoscelesΔ

∴∠OAB=

2

1

(180

0

−140

0

)=

2

1

×40

0

=20

0

(4)∠CBA−∠OBC−∠OBA

=∠OBC−∠OAB[ΔOABisisoscelesΔ]

=60

0

−20

0

−40

0