$\begin{align}
& \overset{\Delta }{\mathop{ABC}}\,:\sin {{60}^{\circ }}=\frac{AH}{AB}\Rightarrow \frac{\sqrt{3}}{2}=\frac{3}{AB}\Rightarrow AB=\frac{6}{\sqrt{3}} \\
& \Rightarrow AB=\frac{6\sqrt{3}}{3}=2\sqrt{3} \\
& \overset{\Delta }{\mathop{ACH}}\,:\sin {{30}^{\circ }}=\frac{AH}{AC}\Rightarrow \frac{1}{2}=\frac{3}{AC}\Rightarrow AC=6 \\
\end{align}$
و بنابه رابطهٔ فیثاغورس داریم:
$\begin{align}
& BC=\sqrt{A{{B}^{2}}+A{{C}^{2}}}=\sqrt{12+36}=\sqrt{48}=4\sqrt{3} \\
& \Rightarrow ABC\,\,mohit\,mosalas=2\sqrt{3}+6+4\sqrt{3}=6\sqrt{3}+6 \\
\end{align}$