Cho tam giác ABC có AB= 2a, BC =a, B= 30 độ. Tính độ dài AC và diện tích tam giác ABC

Ta có:
$\displaystyle S_{ABC} =\frac{1}{2} AB.BC.\sin\hat{B} =\frac{1}{2} .a.2a.\sin 30^{o} =\frac{1}{2} .a.2a.\frac{1}{2} =\frac{a^{2}}{4}$

Nửa chu vi $\displaystyle \vartriangle $ABC là $\displaystyle p=\frac{AB+BC+CA}{2} =\frac{a+2a+AC}{2} =\frac{3a+AC}{2} =\frac{3a}{2} +\frac{AC}{2}$
Lại có $\displaystyle S_{ABC} =\sqrt{p( p-AB)( p-AC)( p-BC)}$.
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Rightarrow \sqrt{\left(\frac{3a}{2} +\frac{AC}{2}\right)\left(\frac{3a}{2} +\frac{AC}{2} -a\right)\left(\frac{3a}{2} +\frac{AC}{2} -2a\right)\left(\frac{3a}{2} +\frac{AC}{2} -AC\right)} =\frac{a^{2}}{4}\\
\Leftrightarrow \left(\frac{3a}{2} +\frac{AC}{2}\right)\left(\frac{a}{2} +\frac{AC}{2}\right)\left(\frac{AC}{2} -\frac{a}{2}\right)\left(\frac{3a}{2} -\frac{AC}{2}\right) =\frac{a^{4}}{16}\\
\Leftrightarrow \left(\frac{9a^{2}}{4} -\frac{AC^{2}}{4}\right)\left(\frac{AC^{2}}{4} -\frac{a^{2}}{4}\right) =\frac{a^{4}}{16}\\
\Leftrightarrow \left( 9a^{2} -AC^{2}\right)\left( AC^{2} -a^{2}\right) =a^{4}\\
\Leftrightarrow AC^{4} -10a^{2} .AC^{2} +9a^{4} +a^{4} =0\\
\Leftrightarrow AC^{4} -10a^{2} .AC^{2} +25a^{4} =16a^{4}\\
\Leftrightarrow \left( AC^{2} -5a^{2}\right)^{2} =16a^{4}\\
\Leftrightarrow AC^{2} -5a^{2} =4a^{2}\\
\Leftrightarrow AC=9a^{2}\\
\Leftrightarrow AC=3a\\
\end{array}$