You probably worked with the point $\displaystyle{\left(\frac{\sqrt{{3}}}{{2}},-\frac{{1}}{{2}}\right)}\usepackage{cmbright}\Huge\$ , which corresponds to the angle $\displaystyle\frac{{{11}\pi}}{{6}}\usepackage{cmbright}\Huge\$ and gives you this information. However, the point $\displaystyle{\left(-\frac{{1}}{{2}},\frac{\sqrt{{3}}}{{2}}\right)}\usepackage{cmbright}\Huge\$ is where the angle $\displaystyle\frac{{{2}\pi}}{{3}}\usepackage{cmbright}\Huge\$ radians or $\displaystyle{{120}}^{\circ}\usepackage{cmbright}\Huge\$ intersects the unit circle.