Math, asked by rishi1309, 7 months ago

In Figure, OAB = 30° and OCB = 57°. Find BOC , AOC and ADB

Answers

Answered by chethan67
0

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

.

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