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)