In figure , O is the center of the circle , such that m angel AOC =140° then find m angel ABC .
Answers
Answered by
2
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
Similar questions