Determine two pairs of polar coordinates for the point with 0° ≤θ

Question

Determine two pairs of polar coordinates for the point with 0° ≤θ < 360°.
-$( - 15,5 \sqrt { 3 } )$
A) $\left( 10 \sqrt { 3 } , 30 ^ { \circ } \right) , \left( - 10 \sqrt { 3 } , 330 ^ { \circ } \right)$
B) $\left( 10 \sqrt { 3 } , 120 ^ { \circ } \right) , \left( - 15 \sqrt { 3 } , 240 ^ { \circ } \right)$
C) $\left( 10 \sqrt { 3 } , 150 ^ { \circ } \right) , \left( - 10 \sqrt { 3 } , 330 ^ { \circ } \right)$
D) $\left( 10 \sqrt { 3 } , 60 ^ { \circ } \right) , \left( - 10 \sqrt { 3 } , 300 ^ { \circ } \right)$

Quizplus · Accepted Answer

The polar coordinates of a point can be found using the formula $r = \sqrt{x^2 + y^2}$ and $\theta = \tan^{-1}\left(\frac{y}{x}\right)$. For the point $(-15, 5\sqrt{3})$, $r = \sqrt{(-15)^2 + (5\sqrt{3})^2} = 10\sqrt{3}$. The angle $\theta$ can be found as $\tan^{-1}\left(\frac{5\sqrt{3}}{-15}\right) = \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)$, which corresponds to $150^\circ$ (since the point is in the second quadrant). The second set of coordinates can be obtained by adding $180^\circ$ to the angle to find the equivalent negative radius representation, resulting in $(-10\sqrt{3}, 330^\circ)$.

Determine Two Pairs of Polar Coordinates for the Point with 0°