यदि त्रिभुज ABC में `(1)/(a+c)+(1)/(b+c)=(3)/(a+b+c)` हो, तो सिद्ध कीजिए कि `angle C = 60^(@)`.

Explanation

यहाँ `(1)/(a+c)+(1)/(b+c)=(3)/(a+b+c)rArr (a+b+c)/(a+c)+(a+b+c)/(b+c)=3`
`rArr " "(1+(b)/(a+c))+(1+(a)/(b+c))=3 rArr (b)/(a+c)+(a)/(b+c)=1`
`rArr " " b(b+c)+a(a+c)=(a+c)(b+c)`
`rArr " " b^(2)+bc+a^(2)+ac=ab+ac+bc+c^(2)`
`rArr " " a^(2)+b^(2)-c^(2)=ab`
`rArr " "(a^(2)+b^(2)-c^(2))/(2ab)=(1)/(2)`
`rArr cos C=(1)/(2)=cos 60^(@)rArr angle C = 60^(@). " "` इति सिद्धम्