In `triangle ABC`, if `( a + b + c) (a-b+c)` = 3 ac, then

Options

A. `angle B = 60^(@)`
B. `angle B = 30^(@)`
C. `angleC = 60^(@)`
D. `angleA + angleC = 90^(@)`

Answer

A

Explanation

`(a+c)^(2) - b^(2) = 3ac`
`rArr a^(2) + c^(2) - b^(2) = ac`
But `cos B = (a^(2) + c^(2) -b^(2))/(2ac) = (1)/(2)`
`rArr angleB = (pi)/(3) = 60^(@)`