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If in triangle ABC, angle C = 60^(@), th...

If in triangle `ABC, angle C = 60^(@)`, then prove that `(1)/(a + c) + (1)/(b + c) = (3)/(a + b + c)`

Text Solution

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Let `1/(a+b)+1/(b+c)=3/(a+b+c)`
`rArr (b+c+a+b)/((a+b)(b+c))=3/(a+b+c)`
`rArr (a+2b+c)(a+b+c)=3(a+b)(b+c)`
`rArr a^(2)+2b^(2)+c^(2)+3ab+3bc+2ac = 3ab + 3b^(2)+3ac+3bc`
`rArr a^(2)+c^(2)-b^(2)=ac`
`rArr (a^(2)+c^(2)-b^(2))/(2ac)=1/2`
`rArr cosB=cos60^(@)`
`rArr B=60^(2)`
`therefore` In `DeltaABC, angleB=60^(@)`
`rArr 1/(a+b)+1/(b+c)=3/(a+b+c)`. Hence Proved.
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