Giải thích các bước giải:
$\begin{array}{l} a)\Delta ABC \bot A\\ \Rightarrow \widehat B + \widehat C = {90^0}\\ \Rightarrow \widehat C = {90^0} - {30^0} = {60^0}\\ + \sin \widehat C = \sin {60^0} = \dfrac{{AB}}{{BC}}\\ \Rightarrow AB = BC.\sin {60^0} = 10.\dfrac{{\sqrt 3 }}{2} = 5\sqrt 3 \left( {cm} \right)\\ \Rightarrow AC = BC.cos{60^0} = 10.\dfrac{1}{2} = 5\left( {cm} \right)\\ Vay\,\widehat C = {60^0};AB = 5\sqrt 3 cm;AC = 5cm\\ b){S_{ABC}} = \dfrac{1}{2}.AB.AC\\ = \dfrac{1}{2}.5\sqrt 3 .5\\ = \dfrac{{25\sqrt 3 }}{2}\left( {c{m^2}} \right) \end{array}$
