$\begin{array}{l}
Bai\,\,1:\\
a)\,\,\,\,\tan \angle C = \frac{{AB}}{{AC}} \Rightarrow AC = AB.\tan C = 21.\tan {40^0} \approx 17,6cm.\\
\cos \angle C = \frac{{AB}}{{BC}} \Rightarrow BC = \frac{{AB}}{{\cos \angle C}} = \frac{{21}}{{\cos {{40}^0}}} \approx 27,41\,\,cm.\\
b)\,\,\,BD\,\,la\,\,\,tia\,\,phan\,\,giac \Rightarrow \angle ABD = \frac{1}{2}\angle ABC = \frac{1}{2}\left( {{{180}^0} - {{40}^0} - {{90}^0}} \right) = {25^0}.\\
\Rightarrow \cos \angle ABD\, = \frac{{AB}}{{BD}} \Rightarrow BD = \frac{{AB}}{{\cos \angle ABD}} = \frac{{21}}{{\cos {{25}^0}}} \approx 23,17\,\,cm.\\
Bai\,\,2:\\
Ap\,\,\,dung\,\,he\,\,\,thuc\,\,\,luong\,\,trong\,\,\,\Delta ABC\,\,\,ta\,\,co:\\
A{H^2} = BH.HC = 25.64 \Rightarrow AH = 40\,\,cm.\\
\tan \angle ABH = \frac{{AH}}{{BH}} = \frac{{40}}{{25}} = \frac{8}{5} \Rightarrow \angle ABC = \angle ABH \approx {58^0}\\
\Rightarrow \angle ACB = {90^0} - \angle ABC = {90^0} - {58^0} = {32^0}.\\
Bai\,\,3:\\
a)\,\,\,\angle B = {30^0} \Rightarrow \angle C = {90^0} - {30^0} = {60^0}.\\
AB = 6cm \Rightarrow AC = AB\tan \angle B = 6.\tan 30 = 2\sqrt 3 \,\,cm.\\
\Rightarrow BC = \sqrt {A{B^2} + A{C^2}} = \sqrt {{6^2} + {{\left( {2\sqrt 3 } \right)}^2}} = 4\sqrt 3 \,\,cm.\\
b)\,\,{S_{ABC}} = \frac{1}{2}AB.AC = \frac{1}{2}.6.2\sqrt 3 = 6\sqrt 3 \,\,c{m^2}.\\
AM\,\,la\,\,\,trung\,\,tuyen \Rightarrow BM = CM\\
CH = AC.\cos \angle C = 2\sqrt 3 .\cos 60 = \sqrt 3 \,\,cm.\\
\Rightarrow AH = \sqrt {A{C^2} - C{H^2}} = \sqrt {{{\left( {2\sqrt 3 } \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} = 3\,\,cm.\\
\Rightarrow HM = CM - CH = \frac{1}{2}BC - CH = 2\sqrt 3 - 3\,cm.\\
{S_{AHM}} = \frac{1}{2}.HM.AH = \frac{1}{2}.3.\left( {2\sqrt 3 - 3} \right)\,\,c{m^2}.
\end{array}$
