In Figure, OAB = 30° and OCB = 57°. Find BOC , AOC and ADB
Answers
Answer:
Step-by-step explanation:
According to question ∠BOC and ∠AOC.
Given−
O is the centre of the given circle.
AB&BC are chords.
InΔOAB&ΔOCB we have
∠OAB=30 Given−
O is the centre of the given circle.
AB&BC are chords.
InΔOAB&ΔOCB we have
∠OAB=30
o
&∠OCB=57
o
.
To find out−
∠BOC&∠AOC=?
Solution−
InΔOAB we have
OA=OB(Radii of the same circle).
∴ΔOAB is isosceles.
i.e∠OBA=∠OAB=30
o
........(i)
Similarly
inΔOBC we have
OC=OB(Radii of the same circle).
∴ΔOCB is isosceles.
i.e∠OBC=∠OCB=57
o
.......(ii).
∴∠BOC=180
o
−(∠OBC+∠OCB)=180
o
−(57
o
+57
o
)=66
o
.(fromii)
(anglesumpropertyoftriangles.)
SimilarlyinΔAOB
∴∠AOB=180
o
−(∠OAB+∠OBA)=180
o
−(30
o
+30
o
)=120
o
.(fromi)
(anglesumpropertyoftriangles.)
So∠AOC=∠AOB−∠BOC=120
o
−66
o
=54
o
.
∴∠AOC=54
o
&∠BOC=66
o
.
Given−
Oisthecentreofthegivencircle.
AB&BCarechords.
InΔOAB&ΔOCBwehave
∠OAB=30
o
&∠OCB=57
o
.
Tofindout−
∠BOC&∠AOC=?
Solution−
InΔOABwehave
OA=OB(Radiiofthesamecircle).
∴ΔOABisisosceles.
i.e∠OBA=∠OAB=30
o
........(i)
Similarly
inΔOBCwehave
OC=OB(Radiiofthesamecircle).
∴ΔOCBisisosceles.
i.e∠OBC=∠OCB=57
o
.......(ii).
∴∠BOC=180
o
−(∠OBC+∠OCB)=180
o
−(57
o
+57
o
)=66
o
.(fromii)
(anglesumpropertyoftriangles.)
SimilarlyinΔAOB
∴∠AOB=180
o
−(∠OAB+∠OBA)=180
o
−(30
o
+30
o
)=120
o
.(fromi)
(anglesumpropertyoftriangles.)
So∠AOC=∠AOB−∠BOC=120
o
−66
o
=54
o
.
∴∠AOC=54
o
&∠BOC=66
o
.
&∠OCB=57
o
.
Tofindout−
∠BOC&∠AOC=?
Solution−
InΔOABwehave
OA=OB(Radiiofthesamecircle).
∴ΔOABisisosceles.
i.e∠OBA=∠OAB=30
o
........(i)
Similarly
inΔOBCwehave
OC=OB(Radiiofthesamecircle).
∴ΔOCBisisosceles.
i.e∠OBC=∠OCB=57
o
.......(ii).
∴∠BOC=180
o
−(∠OBC+∠OCB)=180
o
−(57
o
+57
o
)=66
o
.(fromii)
(anglesumpropertyoftriangles.)
SimilarlyinΔAOB
∴∠AOB=180
o
−(∠OAB+∠OBA)=180
o
−(30
o
+30
o
)=120
o
.(fromi)
(anglesumpropertyoftriangles.)
So∠AOC=∠AOB−∠BOC=120
o
−66
o
=54
o
.
∴∠AOC=54
o
&∠BOC=66
o
.