Bạn vẽ hình để tiện theo dõi nhé:
\(\begin{array}{l}
a)\,\,\angle B = {40^0},\,\,AB = 7cm\\
Xet\,\,\Delta ABC\,\,co:\\
\angle C = {90^0} - \angle B = {90^0} - {40^0} = {50^0}.\\
AC = AB.\tan \angle B = 7.\tan {40^0} \approx 5,87\,\,cm.\\
BC = \frac{{AB}}{{\cos {{40}^0}}} = \frac{7}{{\cos {{40}^0}}} \approx 9,13\,cm.\\
b)\,\,\angle C = {30^0},\,\,\,BC = 16cm\\
Xet\,\,\Delta ABC\,\,co:\\
\angle B = {90^0} - \angle C = {90^0} - {30^0} = {60^0}.\\
AC = BC.\cos \angle C = 16.\cos {30^0} = 8\sqrt 3 \,\,cm.\\
AB = BC.\sin \angle C = 16.\sin {30^0} = 8\,cm.\\
c)\,\,AB = 18cm,\,\,AC = 21\,cm.\\
Xet\,\,\Delta ABC\,\,co:\\
BC = \sqrt {A{B^2} + A{C^2}} = \sqrt {{{18}^2} + {{21}^2}} = 3\sqrt {85} cm.\\
\sin \angle B = \frac{{AC}}{{BC}} = \frac{{18}}{{3\sqrt {85} }} \Rightarrow \angle B \approx {40^0}36'\\
\Rightarrow \angle C = {90^0} - \angle B = {49^0}24'.\\
d)\,\,AC = 12cm,\,\,BC = 13\,cm.\\
Xet\,\,\Delta ABC\,\,co:\\
AB = \sqrt {B{C^2} - A{C^2}} = \sqrt {{{13}^2} - {{12}^2}} = 5cm.\\
\sin \angle B = \frac{{AC}}{{BC}} = \frac{{12}}{{13}} \Rightarrow \angle B \approx {67^0}23'\\
\Rightarrow \angle C = {90^0} - \angle B = {22^0}37'.
\end{array}\)
