Đáp án:
$\begin{array}{l}
a)Theo\,Pytago:\\
B{C^2} = A{B^2} + A{C^2} = {6^2} + 4,{5^2}\\
\Rightarrow BC = 7,5\left( {cm} \right)\\
\sin \widehat B = \dfrac{{AC}}{{BC}} = \dfrac{{4,5}}{{7,5}} = \dfrac{3}{5}\\
\Rightarrow \widehat B = {37^0}\\
\Rightarrow \widehat C = {90^0} - {37^0} = {53^0}\\
\text{Vậy}\,BC = 7,5cm;\widehat B = {37^0};\widehat C = {53^0}\\
b)A{B^2} = HB.BC\\
\Rightarrow HB = \dfrac{{A{B^2}}}{{BC}} = \dfrac{{{6^2}}}{{7,5}} = 4,8\left( {cm} \right)\\
\Rightarrow HC = BC - HB = 7,5 - 4,8 = 2,7\left( {cm} \right)\\
{S_{ABC}} = \dfrac{1}{2}.AB.AC = \dfrac{1}{2}.AH.BC\\
\Rightarrow AH = \dfrac{{6.4,5}}{{7,5}} = 3,6\left( {cm} \right)\\
\text{Vậy}\,HB = 4,8;HC = 2,7;AH = 3,6cm\\
c)\\
AC.\sin B + AB.\sin C\\
= AC.\dfrac{{AC}}{{BC}} + AB.\dfrac{{AB}}{{BC}}\\
= \dfrac{{A{C^2} + A{B^2}}}{{BC}}\\
= \dfrac{{B{C^2}}}{{BC}}\left( {do:A{C^2} + A{B^2} = B{C^2}} \right)\\
= BC\\
\text{Vậy}\,AC.\sin B + AB.\sin C = BC
\end{array}$
