Question

I need help with this practice problem solving Make sure to read the instructions

Answer 1

First, we can find the value of the angle theta using the trigonometric function tangent:

$\begin{gathered} \tan\theta=\frac{y}{x}=\frac{2\sqrt{3}}{-2}=-\sqrt{3} \\ \Rightarrow\theta=\tan^{-1}(-\sqrt{3})=-\frac{\pi}{3}\cong\frac{5}{3}\pi \end{gathered}$

for r, we can use the following expression:

$r=\sqrt{x²+y²}$

then, in this case we have the following:

$r=\sqrt{(-2)²+(2\sqrt{3})}=\sqrt{4+4(3)}=\sqrt{4+12}=\sqrt{16}=4$

therefore, the polar coordinates are (r,theta) = (4,5/3 pi)