\(\widehat{C}=180^o-45^o-105^o=30^o\\ \dfrac{AB}{sinC}=\dfrac{BC}{sinA}\\ \Rightarrow BC=\dfrac{AB.sinA}{sinC}=\dfrac{3.sin45^o}{sin30^o}=3\sqrt{2}\\ =>A\)

\(35=5.7\\ 20=2^2.5\\ =>BCNN\left(35;70\right)=5.2^2.7=140\)

\(\dfrac{6}{13}:\left(0,5+x\right)=\dfrac{15}{39}\\ \Rightarrow\dfrac{6}{13}:\left(\dfrac{1}{2}+x\right)=\dfrac{15}{39}\\ \Rightarrow\dfrac{1}{2}+x=\dfrac{6}{13}:\dfrac{15}{39}\\ \Rightarrow\dfrac{1}{2}+x=\dfrac{6}{13}\times\dfrac{39}{15}\\ \Rightarrow\dfrac{1}{2}+x=\dfrac{6}{5}\\ \Rightarrow x=\dfrac{6}{5}-\dfrac{1}{2}=\dfrac{6\times2-5}{10}=\dfrac{7}{10}\)

Vì \(2< 3\)

\(\Rightarrow\dfrac{2}{7}< \dfrac{3}{7}\)

Chiều dài là : \(\dfrac{4}{3}:\dfrac{3}{4}=\dfrac{16}{9}\left(m\right)\)

Chu vi hình chữ nhật là : \(\left(\dfrac{4}{3}+\dfrac{16}{9}\right)\times2=\dfrac{56}{9}\left(m\right)\)

Diện tích : \(\dfrac{4}{3}\times\dfrac{16}{9}=\dfrac{64}{27}\left(m^2\right)\)

Theo định lý Pytago 

\(AB^2+AC^2=BC^2\\ \Rightarrow AB=\sqrt{25^2-20^2}=15\left(cm\right)\)

Tam giác ABC vuông tại A , AH đg cao 

\(AB.AC=AH.BC\\ \Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{20.15}{25}=12\left(cm\right)\)

\(AB^2=BH.BC\\ \Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{15^2}{25}=9\left(cm\right)\)

\(AC^2=CH.BC\\ \Rightarrow HC=\dfrac{AC^2}{BC}=\dfrac{20^2}{25}=16\left(cm\right)\)

Theo định lý Pytago 

\(AB^2+AC^2=BC^2\\ \Rightarrow AC=\sqrt{10^2-8^2}=6\left(cm\right)\)

Tam giác ABC vuông tại A , đg cao AH

\(AB^2=BH.BC\\ \Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{8^2}{10}=\dfrac{32}{5}\left(cm\right)\\ AC^2=HC.BC\\ \Rightarrow HC=\dfrac{AC^2}{BC}=\dfrac{6^2}{10}=\dfrac{18}{5}\left(cm\right)\)

\(AB.AC=AH.BC\\ \Leftrightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{6.8}{10}=\dfrac{24}{5}\left(cm\right)\)

\(=\dfrac{1}{19}+\dfrac{9}{10}.\dfrac{29-19}{19.29}+\dfrac{9}{10}.\dfrac{39-29}{29.39}+.....+\dfrac{9}{10}.\dfrac{2009-1999}{1999.2009}\\ =\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{10}{19.29}+\dfrac{10}{29.39}+....+\dfrac{10}{1999.2009}\right)\\ =\dfrac{1}{19}+\dfrac{9}{10}.\left(\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{39}+....+\dfrac{1}{1999}-\dfrac{1}{2009}\right)\\ =\dfrac{1}{19}+\dfrac{9}{10}.\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)\)

\(=\dfrac{1}{19}+\dfrac{9}{10}\times\dfrac{1990}{38171}=\dfrac{1}{19}+\dfrac{1791}{39171}=\dfrac{200}{2009}\)

Theo định lý Pytago :

\(AB^2+AC^2=BC^2\\ \Rightarrow BC=\sqrt{5^2+12^2}=13\left(cm\right)\)

Tam giác ABC vuông tại A

\(AB^2=BH.BC\\ \Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{5^2}{13}=\dfrac{25}{13}\left(cm\right)\)

\(AB.AC=AH.BC\\ \Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{5.12}{13}=\dfrac{60}{13}\left(cm\right)\)

\(AC^2=HC.BC\\ \Rightarrow HC=\dfrac{AC^2}{BC}=\dfrac{12^2}{13}=\dfrac{144}{13}\left(cm\right)\)

a, Theo định lý Pytago :

\(AB^2+AC^2=BC^2\\ \Rightarrow AB=\sqrt{20^2-16^2}=12\)

\(cosB=\dfrac{AB}{BC}=\dfrac{12}{20}\Rightarrow\widehat{B}=53^o8'\)

\(cosC=\dfrac{AC}{BC}=\dfrac{16}{20}\Rightarrow\widehat{C}=36^o52'\)

b, Vì tam giác ABC vuông tại A

\(\widehat{B}+\widehat{C}=90^o\\ \Rightarrow\widehat{B}=90^o-50^o=40^o\)

\(cosC=\dfrac{AC}{BC}\Rightarrow BC=\dfrac{AC}{cosC}=\dfrac{10}{cos50^o}\approx15,6\)

\(tanC=\dfrac{AB}{AC}\Rightarrow AB=tanC\times AC=tan50^o\times10\approx11,9\)

c, 

\(cosB=\dfrac{AB}{BC}\\ \Rightarrow BC=\dfrac{AB}{cosB}=\dfrac{5}{\dfrac{3}{4}}=\dfrac{20}{3}\)

Theo định lý Pytago :

\(AB^2+AC^2=BC^2\\ \Rightarrow AC=\sqrt{\left(\dfrac{20}{3}\right)^2-5^2}=\dfrac{5\sqrt{7}}{3}\)

\(cosB=\dfrac{3}{4}\Rightarrow\widehat{B}=41^o25'\\ sinC=\dfrac{AB}{BC}=\dfrac{5}{\dfrac{20}{3}}=48^o35'\)