$\displaystyle \begin{array}{{>{\displaystyle}l}} S_{ABC} =\frac{1}{2} AB.AC.sinBAC=\frac{1}{2} .2a.3a.sin60^{o} =\frac{3\sqrt{3}}{2} a^{2}\\ Kẻ\ AH\bot BC,\ H\in BC\\ Theo\ định\ lý\ \ cosin\ ta\ có:\ BC^{2} =AB^{2} +AC^{2} -2AB.AC.cosBAC\\ \Leftrightarrow BC=\sqrt{AB^{2} +AC^{2} -2AB.AC.cosBAC} =\sqrt{( 2a)^{2} +( 3a)^{2} -2.2.3a^{2} .cos60^{o}}\\ \Leftrightarrow BC=a\sqrt{7}\\ Mà\ S_{ABC} =\frac{1}{2} BC.AH=\frac{3\sqrt{3}}{2} a^{2}\\ \Leftrightarrow AH=\frac{2.\frac{3\sqrt{3}}{2} a^{2}}{a\sqrt{7}} =\frac{3a\sqrt{21}}{7} \end{array}$ 

Đáp án:

$h_a=\dfrac{3a}{\sqrt{7}}$

Giải thích các bước giải:

 Ta có :

$S_{ABC}=\dfrac{1}{2}.AB.AC.Cos(\widehat{BAC})=\dfrac{1}{2}.2a.3a.\dfrac{1}{2}=\dfrac{3a^2}{2}$

Áp dụng định lý Cô sin ta có :

$BC^d2=AB^2+AC^2-2AB.AC.Cos(\widehat{BAC})$

$BC^2=(2a)^2+(3a)^2-2.2a.3a.\dfrac{1}{2}$

$BC^2=4a^2+9a^2-6a^2$

$BC=a\sqrt{7}$

Mà :

$S_{ABC}=\dfrac{1}{2}.BC.h_a$

$\dfrac{3a^2}{2}=\dfrac{1}{2}.a\sqrt{7}.h_a$

$h_a=\dfrac{3a^2}{a\sqrt{7}}$
$h_a=\dfrac{3a}{\sqrt{7}}$