Question 1184155
By hypothesis, {{{CosA/a = CosB/b = CosC/c}}}.

Since the triangle also obeys the Sine Law, we also have {{{sinA/a = SinB/b = SinC/c}}}.


Now {{{CosA/a = CosB/b}}} and {{{sinA/a = SinB/b}}}  ===>  {{{Cos^2A/a^2 = Cos^2B/b^2}}} and {{{Sin^2A/a^2 = Sin^2B/b^2}}}

By adding corresponding sides of the last two equations, we get


{{{1/a^2 = 1/b^2}}}, which implies that {{{a = b}}}.


Similarly, {{{CosA/a = CosC/c}}} and {{{sinA/a = SinC/c}}}  ===> {{{a = c}}}.


Therefore, the triangle is actually an isosceles triangle, each side having a measure of a = 2 units.


Hence its area is {{{A = (1/2)*a*((sqrt(3)/2)*a) = (sqrt(3)/4)*a^2 = (sqrt(3)/4)*2^2 = highlight(sqrt(3))}}} square units.