Answer to Question #347990 in Analytic Geometry for Helly

Question

Convert the following rectangular coordinates into polar coordinates (r , theta) so that r<0 and 0<theta <2pi: ( 4, -4 square root of 3)

Answer 1

$$r^2=(4)^2+(-4\sqrt{3})^2=64$$

$$r<0=>r=-\sqrt{64}=-8$$

Quadrant IV

$$\tan\theta=\dfrac{-4\sqrt{3}}{4}=-\sqrt{3}$$

$$\theta=\pi+\tan^{-1}(-\sqrt{3})=\dfrac{2\pi}{3}$$

$$(r, \theta)=(-8, \dfrac{2\pi}{3})$$