Convert a Point from Rectangular to Polar Coordinates
-$( 6 , - 6 \sqrt { 3 } )$
A) $\left( 12 , \frac { 5 \pi } { 3 } \right)$
B) $\left( 6 , \frac { 5 \pi } { 3 } \right)$
C) $\left( 12 , \frac { 11 \pi } { 6 } \right)$
D) $\left( 6 , \frac { 11 \pi } { 6 } \right)$

Question

Quizplus · Accepted Answer

The polar coordinates $(r, \theta)$ of a point can be found from its rectangular coordinates $(x, y)$ using the formulas $r = \sqrt{x^2 + y^2}$ and $\theta = \tan^{-1}\left(\frac{y}{x}\right)$. For the point $(6, -6\sqrt{3})$, $r = \sqrt{6^2 + (-6\sqrt{3})^2} = 12$, and $\theta$ falls in the fourth quadrant, where $\tan^{-1}\left(\frac{-6\sqrt{3}}{6}\right) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$. However, since polar coordinates are usually given in the range $0$ to $2\pi$ for angles in the fourth quadrant, we add $2\pi$ to $-\frac{\pi}{3}$, resulting in $\frac{5\pi}{3}$.

Convert a Point from Rectangular to Polar Coordinates - (6,−63)( 6 , - 6 \sqrt { 3 } )(6,−63​)

Convert a Point from Rectangular to Polar Coordinates
- $( 6 , - 6 \sqrt { 3 } )$