You may have correctly determined that the reference angle is $\displaystyle{{60}}^{\circ}\usepackage{cmbright}\Huge\$ and found $\displaystyle{{\sin{{60}}}}^{\circ}\usepackage{cmbright}\Huge\$ , but you made a mistake with the sign. Use the signs of x and y in the third quadrant to determine the sign of sine, remembering that: $\displaystyle{\sin{\theta}}=\frac{{y}}{{1}}\usepackage{cmbright}\Huge\$