The side AB of an isosceles triangle is along the axis of x with vertices `A (–1, 0)` and `AB = AC`. The equation of the side BC when `angleA=120^(@)` and `BC=4sqrt(3)` is:

Options

A. `x + sqrt3 y - 3 = 0`
B. `sqrt3 x + y = 3 `
C. `x + y = sqrt3`
D. none of these

Explanation

In `Delta`ALB , we have
BL = AB cos `(pi)/(6)`
`implies 2sqrt3 = AB xx (sqrt3)/(2) implies AB = 4 implies OB = 3 " " [because OA = 1]`

So , the coordinates of B are (3,0)
Clearly , `angleXBC = (5pi)/(6)` . So , the equation of BC is
y - 1 = tan `(5pi)/(6) (x - 3) implies y = - (1)/(sqrt3) (x - 3) implies x + sqrt3 y - 3 = 0`