In triangle $ABC$, angle $C$ measures $120°$ and side $AC$ is greater than side $BC$. Knowing that the area of the equilateral triangle with side $AB$ is $31$, the area of the equilateral triangle with side $AC-BC$ is $19$, calculate the area of triangle $ABC$.(Answer: $4$)
I try:
$\dfrac{AB^2\sqrt3}{4}=31 \implies AB = \sqrt{\frac{124\sqrt3}{3}}$
$\dfrac{(AC-BC)^2\sqrt3}{4} = 19 \implies AC – BC = \sqrt{\frac{76\sqrt3}{3}}$
$S\triangle ABC = \frac{1}{2}.sen 120^o .AC.BC=\frac{\sqrt3}{2}.AC.BC$
$AB^2=AC^2+BC^2-2.AC.BC.cos 120^o = AC^2+BC^2+2AC.BC$
$\frac{124\sqrt3}{3} = AC^2+BC^2+\frac{2S \triangle ABC \sqrt3}{3}$
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Best Answer
You've made a good start, but there are a couple of small mistakes. As WW1's comment indicates, $\sin(120^{\circ}) = \frac{\sqrt{3}}{2}$ and $\cos(120^{\circ}) = -\frac{1}{2}$. In addition, I'll use its suggestion of single lower-case letters for the triangle side lengths, i.e.,
$$a = \lvert BC\rvert, \;\; b = \lvert AC\rvert, \;\; c = \lvert AB\rvert \tag{1}\label{eq1A}$$
Thus, with the trigonometry formula for triangle area, your final denominator of $2$ should be $4$ instead, i.e.,
$$S\triangle ABC = \frac{1}{2}\cdot \sin 120^{\circ}\cdot ab = \frac{\sqrt3}{\color{blue}{4}}\cdot ab \tag{2}\label{eq2A}$$
Also, using $c^2 = \frac{124}{\sqrt{3}}$ and changing your factor of $2$ to $1$ in the final law of cosines formula, we get
$$c^2 = a^2 + b^2 - 2ab \cdot \cos 120^{\circ} = a^2 + b^2 + \color{blue}{1}\cdot ab = \frac{124}{\sqrt{3}} \tag{3}\label{eq3A}$$
In addition, we have
$$(a - b)^2 = \frac{76}{\sqrt{3}} \;\;\to\;\; a^2 + b^2 - 2ab = \frac{76}{\sqrt{3}} \tag{4}\label{eq4A}$$
Next, \eqref{eq3A} minus \eqref{eq4A} gives
$$3ab = \frac{48}{\sqrt{3}} \;\;\to\;\; ab = \frac{16}{\sqrt{3}} \tag{5}\label{eq5A}$$
Substituting this into \eqref{eq2A} results in
$$S\triangle ABC = \frac{\sqrt{3}}{4}\cdot \frac{16}{\sqrt{3}} = 4 \tag{6}\label{eq6A}$$