Let ABC be a triangle such that AB=BC. Let F be midpoint of AB and X be a poin on BC such that FX is perpendicular to AB. If BX =3XC then the ratio BC/AC equals

Options

A. `sqrt(3)`
B. `sqrt(2)`
C. `sqrt((3)/(2))`
D. 1

Answer

C

Explanation

`:. Cos B=(2)/(3)`
Also `cosB=(16y^(2)+16y^(2)-AC^(2))/(2.4y.4y)`
`(2)/(3)=(32y^(2)-AC^(2))/(32y^(2))`
`64y^(2)=96y^(2)-3Ac^(2)`
`3AC^(2)=32y^(2)`
`AC=(4sqrt(2))/(sqrt(3))y`
`:. (BC)/(AC)=(4y)/(((4sqrt(2)y)/(sqrt(3))))=(sqrt(3))/(sqrt(2))`