If the lengths of the sides of a triangle are a - b , a + b and `sqrt(3a^(2)+b^(2)),(a,bgt,0)`, then the largest angle of the triangle , is

Options

A. `(2pi)/3`
B. `(3pi)/4`
C. `(pi)/2`
D. `(7pi)/8`

Explanation

Evidently,`sqrt(3a^(2)+b^(2))` is the largest side. Therefore, cosine of the largest angle `theta` is, given by
`costheta=((a-b)^(2)+(a+b)^(2)-(3a^(2)+b^(2)))/(2(a-b)(a+b))=-1/2rArrtheta=(2pi)/3`