Đáp án:
$\begin{array}{l}
a)Theo\,Pytago:\\
A{B^2} + A{C^2} = B{C^2}\\
\Rightarrow A{C^2} = {10^2} - {6^2} = 64\\
\Rightarrow AC = 8\left( {cm} \right)\\
\sin \widehat B = \dfrac{{AC}}{{BC}} = \dfrac{8}{{10}} = \dfrac{4}{5}\\
\Rightarrow \widehat B = {53^0}\\
\Rightarrow \widehat C = {37^0}\\
b)Xet:\Delta ABH;\Delta AHM:\\
+ \widehat {BAH}\,chung\\
+ \widehat {AHB} = \widehat {AMH} = {90^0}\\
\Rightarrow \Delta ABH \sim \Delta AHM\left( {g - g} \right)\\
\Rightarrow \dfrac{{AH}}{{AM}} = \dfrac{{AB}}{{AH}}\\
\Rightarrow AM.AB = A{H^2}\\
TT:AN.AC = A{H^2}\\
\Rightarrow AM.AB = AN.AC\left( { = A{H^2}} \right)\\
c)Xet:\Delta AMN;\Delta ACB:\\
+ AM.AB = AN.AC\\
+ \widehat A\,chung\\
\Rightarrow \Delta AMN \sim \Delta ACB\left( {c - g - c} \right)\\
\Rightarrow \widehat {AMN} = \widehat {ACB}
\end{array}$
