a, \(cosB=\frac{AB}{BC}\Rightarrow BC=\frac{AB}{cosB}=\frac{6}{cos40}\approx7,8\left(cm\right)\)
\(\Rightarrow AC=\sqrt{BC^2-AB^2}=\sqrt{7,8^2-6^2}\approx4,98\left(cm\right)\)
\(\widehat{C}=90^o-\widehat{B}=90^o-40^o=50^o\)
b, \(AC=BC.cosC=10.cos35\approx8,19\left(cm\right)\)
\(AB=BC.sinC=10.sin35\approx5,74\left(cm\right)\)
\(\widehat{B}=90^o-35^o=55^o\)
c, \(AB=BC.cos58\approx10,6\left(cm\right)\)
\(AC=BC.sin58\approx19,96\left(cm\right)\)
\(\widehat{C}=90^o-58^o=32^o\)
d, \(AB=\sqrt{BC^2-AC^2}=\sqrt{32^2-20^2}\approx24,98\left(cm\right)\)
\(\widehat{B}=sin^{-1}\frac{5}{8}=38,68^o\)
\(\widehat{C}=90^o-38,68^o=51,32^o\)
e, \(AC=BC.cos42\approx60,94\)
\(AB=BC.sin42\approx54,87\)
\(\widehat{B}=90^o-42^o=48^o\)
f, \(BC=\sqrt{AB^2+AC^2}=3\sqrt{85}\left(cm\right)\)
\(\widehat{B}=sin^{-1}\frac{7\sqrt{85}}{85}=49,4^o\)
\(\widehat{C}=90^o-49,4^o=40,6^o\)
