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Question

In ABC, if 1a+b+1a+c=3a+b+c then prove that A=60o

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Solution

We have,

1a+b + 1a+c = 3a+b+c

Then prove that.

A=600


Now,

1a+b + 1a+c = 3a+b+c

a+c+a+b(a+b)(a+c)=3a+b+c

2a+b+ca2+ac+ab+bc=3a+b+c

(2a+b+c) (a+b + c) = 3 (a2 + ac + ab +bc)

2a2+2ab+2ac+ab+b2+bc+ac+bc+c2=3a2+3ac+3a+3ab+3bc

a2=bc+b2+c2

bc=b2+c2a2

1=b2+c2a2bc


on divide both side by 2.

12=b2+c2a22bc

12=CosA

CosA=Cosπ6

A=π6


Hence, this is the answer.


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