Đáp án:
$\begin{array}{l}
B3)\\
\sin \widehat B = \dfrac{{AC}}{{BC}}\\
\Rightarrow \sin {60^0} = \dfrac{{\sqrt 3 }}{2} = \dfrac{{\sqrt 3 }}{{BC}}\\
\Rightarrow BC = 2\\
\Rightarrow AB = \sqrt {B{C^2} - A{C^2}} = \sqrt {4 - 3} = 1\\
Do:\widehat {HAB} + \widehat B = {90^0}\\
\widehat C + \widehat B = {90^0}\\
\Rightarrow \widehat {HAB} = \widehat C\\
\Rightarrow \tan \widehat {HAB} = \tan {30^0} = \dfrac{1}{{\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3}\\
B4)\\
AC = 2\sqrt 5 \\
\sin \widehat C = \dfrac{2}{3} = \dfrac{{AB}}{{BC}}\\
\Rightarrow BC = \dfrac{3}{2}AB\\
Theo\,Pytago:\\
A{B^2} + A{C^2} = B{C^2}\\
\Rightarrow A{B^2} + {\left( {2\sqrt 5 } \right)^2} = \dfrac{9}{4}A{B^2}\\
\Rightarrow \dfrac{5}{4}A{B^2} = 20\\
\Rightarrow A{B^2} = 25\\
\Rightarrow AB = 5\\
\Rightarrow {S_{ABC}} = \dfrac{1}{2}.AB.AC = \dfrac{1}{2}.2\sqrt 5 .5 = 5\sqrt 5
\end{array}$
