Đáp án:
$\begin{array}{l}
1)Theo\,Pytago:\\
A{C^2} + B{C^2} = A{B^2}\\
\Rightarrow A{B^2} = {2^2} + {5^2} = 29\\
\Rightarrow AB = \sqrt {29} \\
+ )sin\widehat A = \dfrac{{BC}}{{AB}} = \dfrac{5}{{\sqrt {29} }}\\
\Rightarrow \widehat A = {68^0}\\
\Rightarrow \widehat B = {90^0} - \widehat A = {22^0}\\
2){S_{ABC}} = \dfrac{1}{2}.AC.BC = \dfrac{1}{2}.2.5 = 5\left( {c{m^2}} \right)\\
3)\\
{S_{ABC}} = \dfrac{1}{2}.AC.BC = \dfrac{1}{2}CH.AB\\
\Rightarrow CH = \dfrac{{5.2}}{{AB}} = \dfrac{{10}}{{\sqrt {29} }} = \dfrac{{10\sqrt {29} }}{{29}}\left( {cm} \right)\\
\Rightarrow AH = \sqrt {A{C^2} - C{H^2}} = \sqrt {{2^2} - \dfrac{{100}}{{29}}} = \dfrac{{4\sqrt {29} }}{{29}}\\
\Rightarrow {S_{ACH}} = \dfrac{1}{2}.AH.CH = \dfrac{1}{2}.\dfrac{{4\sqrt {29} }}{{29}}.\dfrac{{10\sqrt {29} }}{{29}} = \dfrac{{20}}{{29}}\left( {c{m^2}} \right)
\end{array}$
