In a `!ABC` , if `1/(b+c)+1/(c+a)=3/(a+b+c)`, then `angleC`=

Options

A. `90^(@)`
B. `60^(@)`
C. `45^(@)`
D. `30^(@)`

Explanation

We have,
`1/(b+c)+1/(c+a)=3/(a+b+c)`
`rArr(a+b+c)/(b+c)+(a+b+c)/(c+a)=3`
`rArra/(b+c)+1+b/(c+a)+1=3`
`rArr` a (c +a)+ b (b + c) = (b + c) (c + a)
`rArra^(2)+b^(2)-c^(2)=ab`
`rArr(a^(2)+b^(2)-c^(2))/(2ab)=1/2rArrcosC=1/2rArrC=60^(@)`