Given that, side of triangles are
$$
\mathrm{a}=6+2 \sqrt{3}, \mathrm{~b}=4 \sqrt{3} \text { and } \mathrm{c}=\sqrt{24}
$$
Here, we observe that the side ' $c$ ' is small, hence the angle $C$ is also small.
$$
\text { Then, } \begin{aligned}
& \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \\
=& \frac{(6+2 \sqrt{3})^{2}+(4 \sqrt{3})^{2}-(\sqrt{24})^{2}}{2(6+2 \sqrt{3})(4 \sqrt{3})}
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \quad \cos C=\frac{36+12+48-24+24 \sqrt{3}}{16 \sqrt{3}(3+\sqrt{3})} \\
&\Rightarrow \quad \cos C=\frac{72+24 \sqrt{3}}{16 \sqrt{3}(3+\sqrt{3})}=\frac{24(3+\sqrt{3})}{16 \sqrt{3}(3+\sqrt{3})} \\
&\Rightarrow \quad \cos C=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2} \\
&\Rightarrow \quad \cos C=\cos 30^{\circ} \Rightarrow \angle C=30^{\circ} \\
&\text { The smallest angle } C=30^{\circ} \\
&\text { Hence, the tangent of smallest angle is } \\
&\text { tan } C=\tan 30^{\circ} \\
&=\frac{1}{\sqrt{3}}
\end{aligned}
$$