If in `!ABC,(c+a)/b+(c+b)/a=c/r` then

Options

A. `angleB=pi/2`
B. `angleC=pi/2`
C. `angleA=pi/2`
D. none of these

Explanation

We have,
`(c+a)/b+(c+b)/a=c/r`
`rArr(a^(2)+b^(2)+ac+bc)/(ab)=(cs)/(Delta)`
`rArr(a^(2)+b^(2)+c(a+b))/(ab)=(2cs)/(absinC)`
`rArra^(2)+b^(2)+c(a+b))=(2cs)/(sinC)`
`rArra^(2)+b^(2)+c(a+b+c)-c^(2)=(2cs)/(sinC)`
`rArr(a^(2)+b^(2)-c^(2))+2cs=(2cs)/(sinC)rArr2abcosC+2cs=(2cs)/(sinC)`
Clearly, `angleC=pi/2` satisfies the relataion.