\(\begin{array}{l}
a)\,\,\,Xet\,\,\,\Delta AHC\,\,\,ta\,\,co:\\
AH = AC.\sin {60^0} = 8.\frac{{\sqrt 3 }}{2} = 4\sqrt 3 \,\,cm.\\
HC = AC.\cos {60^0} = 8.\frac{1}{2} = 4\,\,cm.\\
\angle HAC = {90^0} - \angle ACH = {90^0} - {60^0} = {30^0}.\\
b)\,\,Vi\,\,CI\,\,\,la\,\,phan\,\,\,giac\,\,\,cua\,\,\angle ACH\\
\Rightarrow \angle HCI = {30^0}.\\
\Rightarrow HI = HC.\tan {30^0} = 4.\frac{1}{{\sqrt 3 }} = \frac{{4\sqrt 3 }}{3}\,\,cm.\\
\Rightarrow {S_{IHC}} = \frac{1}{2}IH.HC = \frac{1}{2}.\frac{{4\sqrt 3 }}{3}.4 = \frac{{8\sqrt 3 }}{3}\,\,c{m^2}.\\
{S_{AHC}} = \frac{1}{2}AH.HC = \frac{1}{2}.4\sqrt 3 .4 = 8\sqrt 3 \,\,c{m^2}.\\
\Rightarrow {S_{AIC}} = {S_{AHC}} - {S_{IHC}} = 8\sqrt 3 - \frac{{8\sqrt 3 }}{3} = \frac{{16\sqrt 3 }}{3}\,\,c{m^2}.
\end{array}\)
