Draw \(CH\perp AB\)

Connected AC

\(\left\{{}\begin{matrix}DA\perp AB\\CH\perp AB\end{matrix}\right.\)=> DA // CH

=> \(\angle HCD+\angle ADC=180^o\) [2 angles in the same side]

But \(\angle ADC=120^o\Rightarrow\angle HCD=60^o\Leftrightarrow\angle BCH=120^o-60^o=60^o\)

\(\angle BCH=60^o\Rightarrow\angle HBC=90^o-60^o=30^o\)

We have:

tan \(\angle HBC=\dfrac{HC}{BC}\)

\(\Rightarrow HC=tan30^o\cdot8=\dfrac{\sqrt{3}}{3}\cdot8=\dfrac{8\sqrt{3}}{3}\left(units\right)\)

The same, we have:

tan \(\angle HCB=\dfrac{HB}{BC}\)

\(\Rightarrow HB=tan60^o\cdot8=\sqrt{3}\cdot8=8\sqrt{3}\left(units\right)\)

\(\Rightarrow S_{BHC}=\dfrac{\dfrac{8\sqrt{3}}{3}\cdot8\sqrt{3}}{2}=32\left(units^2\right)\)