$\begin{array}{l} a)\,\,\widehat{ B} = {40^0},\,\,AB = 7cm\\ \text{Xét}\,\,\Delta ABC\,\,\text{ta có}:\\ \widehat{ C} = {90^0} - \widehat{ B} = {90^0} - {40^0} = {50^0}\\ AC = AB.\tan \widehat{ B} = 7.\tan {40^0} \approx 5,87\,\,cm\\ BC = \dfrac{{AB}}{{\cos {{40}^0}}} = \dfrac{7}{{\cos {{40}^0}}} \approx 9,13\,cm\\ b)\,\,\widehat{ C} = {30^0},\,\,\,BC = 16cm\\ \text{Xét}\,\,\Delta ABC\,\,\text{ta có}:\\ \widehat{ B} = {90^0} - \widehat{ C} = {90^0} - {30^0} = {60^0}\\ AC = BC.\cos \widehat{ C} = 16.\cos {30^0} = 8\sqrt 3 \,\,cm\\ AB = BC.\sin \widehat{ C} = 16\sin {30^0} = 8\,cm\\ c)\,\,AB = 18cm,\,\,AC = 21\,cm\\ \text{Xét}\,\,\Delta ABC\,\,\text{ta có}:\\ BC = \sqrt {A{B^2} + A{C^2}} = \sqrt {{{18}^2} + {{21}^2}} = 3\sqrt {85}\, cm\\ \sin \widehat{ B} = \dfrac{{AC}}{{BC}} = \dfrac{{18}}{{3\sqrt {85} }} \\\Rightarrow \widehat{ B} \approx {40^0}36'\\ \Rightarrow \widehat{ C} = {90^0} - \widehat{ B} = {49^0}24'\\ d)\,\,AC = 12cm,\,\,BC = 13\,cm\\ \text{Xét}\,\,\Delta ABC\,\,\text{ta có}:\\ AB = \sqrt {B{C^2} - A{C^2}} = \sqrt {{{13}^2} - {{12}^2}} = 5cm\\ \sin \widehat{ B} = \dfrac{{AC}}{{BC}} = \dfrac{{12}}{{13}} \Rightarrow \widehat{ B} \approx {67^0}23'\\ \Rightarrow \widehat{ C} = {90^0} - \widehat{ B} = {22^0}37' \end{array}$
