Đáp án:
$\begin{array}{l}
b1:\widehat B = {30^0} \Rightarrow \widehat C = {60^0}\\
a)cos\left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right) + \sin \left( {\overrightarrow {BA} ,\overrightarrow {BC} } \right) + \tan \frac{{\left( {\overrightarrow {AC} ,\overrightarrow {CB} } \right)}}{2}\\
= cos{90^0} + \sin \,{30^0} + \tan \frac{{{{180}^0} - {{60}^0}}}{2}\\
= 0 + \frac{1}{2} + \tan {60^0}\\
= \frac{1}{2} + \sqrt 3
\end{array}$
$\begin{array}{l}
b2:AC = AB.\tan B = a.\tan {60^0} = \sqrt 3 .a \Rightarrow BC = 2a\\
\overrightarrow {AB} .\overrightarrow {AC} = AB.AC.c{\rm{os9}}{{\rm{0}}^0} = 0\\
\overrightarrow {CA} .\overrightarrow {CB} = CA.CB.c{\rm{osC = }}\sqrt 3 .a.2a.c{\rm{os3}}{{\rm{0}}^0} = 2\sqrt 3 {a^2}.\frac{{\sqrt 3 }}{2} = 3{a^2}
\end{array}$
các ý còn ại tương tự nhé
