Question

In a triangle ABC with usual notations, if `(cos "A")/"a" = (cos "B")/"b" = (cos "C")/"c"`, then area of triangle ABC with a = `sqrt6` is ____________.

Options

• `2/sqrt3` sq units

• `(3sqrt3)/2` sq units

• `(5sqrt3)/2` sq units

• `sqrt3/2` sq units

Answer 1

In a triangle ABC with usual notations, if `(cos "A")/"a" = (cos "B")/"b" = (cos "C")/"c"`, then area of triangle ABC with a = `sqrt6` is `underline((3sqrt3)/2)` sq units.

Explanation:

Given, `(cos "A")/"a" = (cos "B")/"b" = (cos "C")/"c"` .............(i)

`("b"^2 + "c"^2 - "a"^2)/(2 "abc") = ("a"^2 + "c"^2 - "b"^2)/(2 "abc") = ("a"^2 + "b"^2 - "c"^2)/(2 "abc")`

b² + c² − a² = a² + c² − b² = a² + b² − c²

a = b = c

∴ ΔABC is an equilateral triangle

∴ Area of ΔABC = `sqrt3/4 "a"^2`

= `sqrt3/4 (sqrt6)^2`

= `(sqrt3 xx 6)/4`

= `(3sqrt3)/2` sq units