Sine theorem is: a/sina = b/sinb = c/sinc = 2/(√ 3/2) = 4/√ 3.

Therefore, b=(4/√3)sinB and c=(4/√3)sinC.

s△ABC =( 1/2)bcsinA =( 1/2)(4/√3)？ sinBsinC (√3/2)

=(4√3/3)sinBsinC

=(-2√3/3) [cos(B+C)-cos(B-C)]

=(-2√3/3)[(- 1/2)-cos(B-C)]

=(2√3/3) [cos(B-C)+( 1/2)]

Cos(B-C)∈(- 1/2, 1] because B-C∈(-2π/3, 2π/3).

Therefore, S△ABC∈(0, √3)