Inverting away

Since $\cos^{-1}(-\frac {\sqrt{3}} {2}) = \theta$ , for some angle $\theta$ , then $-\frac {\sqrt{3}} {2} = \cos(\theta)$ .

We know that $\cos(-x) = -\cos(x + \pi)$ and $\cos(30^{\circ}) = \frac {\sqrt{3}} {2}$ .

So $-\frac {\sqrt{3}} {2} = -\cos(30^{\circ}) = -\cos(-150^{\circ} + \pi) = \cos(150^{\circ})$ .

Therefore $\theta = 150^{\circ}$ .

cos a = x/r

x=-sqrt3 r=2 (the point must be located at 2nd/3rd quadrant 'cause the value of x is minus)

we know that cos 30=sqrt3/2

in 2nd quadrant cos(180-30)=cos 150

in 3rd quadrant cos(180+30)=cos 210