Áp dụng định lý hàm số cos ta được:

$\begin{array}{l}
A{B^2} = A{C^2} + B{C^2} - 2AC.BC\cos \widehat {ACB}\\
 = {7^2} + {8^2} - 2.7.8.\cos {120^o}\\
 = {7^2} + {8^2} + 7.8 = 169 \Rightarrow AB = 13\\
{S_{ABC}} = \dfrac{1}{2}AC.BC.\sin \widehat {ACB} = \dfrac{1}{2}.7.8\sin {120^o}\\
 = \dfrac{1}{2}.7.8.\dfrac{{\sqrt 3 }}{2} = 14\sqrt 3 \left( {dvdt} \right)
\end{array}$

$AB^{2}$ =$AC^{2}$ +$BC^{2}$ -2AC.BC.COS($120^{o}$
=72+82−2.7.8.COS($120^{o}$
=72+82+7.8=169
⇒AB=13
$S_{ABC}$ =$\frac{1}{2}$ .AC.BC.Sin($120^{o}$
=$\frac{1}{2}$.7.8.Sin($120^{o}$
=14$\sqrt{3}$