Answer:
The answer is "The Polar coordinates (r,[tex]\theta)[/tex] =[tex](4,\frac{5 \pi}{6})[/tex]"
Step-by-step explanation:
Given point:
[tex]\to (-2\sqrt{3},2)[/tex]
Formula:
[tex]r=\sqrt{x^2+y^2}\\[/tex]
[tex]=\sqrt{(-2\sqrt{3})^2+(2)^2}\\\\=\sqrt{12+4}\\\\=\sqrt{16}\\\\=\sqrt{4^2}\\\\=4[/tex]
[tex]\therefore \\\\ \tan \theta= \frac{y}{x}\\\\[/tex]
[tex]\tan \theta=\frac{2}{-2\sqrt{3}}\\\\ \tan \theta= -\frac{1}{\sqrt{3}}\\\\ \tan \theta= \frac{5\pi}{6}\\\\ \theta= \frac{5\pi}{6}[/tex]
The Polar coordinates (r,[tex]\theta)[/tex] =[tex](4,\frac{5 \pi}{6})[/tex].
