The terminal side of $\displaystyle-{{140}}^{\circ}\usepackage{cmbright}\Huge\$ lies in Quadrant III where x and y are both negative, so $\displaystyle{\cos{\theta}}={x}\usepackage{cmbright}\Huge\$ is negative and $\displaystyle{\tan{\theta}}=\frac{{y}}{{x}}=\frac{{{\left(-\right)}}}{{{\left(-\right)}}}={\left(+\right)}\usepackage{cmbright}\Huge\$ is positive.