Draw $\displaystyle{{240}}^{\circ}\usepackage{cmbright}\Huge\$ in standard position. The terminal side is in the third quadrant and the reference angle is $\displaystyle{{60}}^{\circ}\usepackage{cmbright}\Huge\$ . Use the $\displaystyle{{30}}^{\circ}-{{60}}^{\circ}-{{90}}^{\circ}\usepackage{cmbright}\Huge\$ triangle to find the value of this function at the reference angle:

$\displaystyle{{\tan{{60}}}}^{\circ}=\frac{\sqrt{{3}}}{{1}}=\sqrt{{3}}\usepackage{cmbright}\Huge\$

Use the signs of x and y in the third quadrant to determine the sign of tangent:

$\displaystyle{{\tan{{240}}}}^{\circ}=\frac{{y}}{{x}}=\frac{{{\left(-\right)}}}{{{\left(-\right)}}}={\left(+\right)}\usepackage{cmbright}\Huge\$

So: $\displaystyle{{\tan{{240}}}}^{\circ}=\sqrt{{3}}\usepackage{cmbright}\Huge\$