Convert the rectangular coordinate (-3V3, -3),to a polar coordinate (r, 0).[ Select ]0[ Select ]

Step 1: Write out the formula for finding r and Θ
[tex]\begin{gathered} r=\sqrt[]{y^2+x^2} \\ \emptyset=\tan ^{-1}\frac{y}{x} \\ \text{Where x and y are the given cartesian coordinates and r and }\emptyset\text{ are the equivilent polar coordinates} \\ x=-3\sqrt[]{3} \\ y=-3 \end{gathered}[/tex]Step2: Evaluate r and Θ
[tex]\begin{gathered} r=\text{ }\sqrt[]{(-3)^2}+(-3\sqrt[]{3})^2 \\ =\sqrt[]{9+9(3)} \\ =\sqrt[]{9+27} \\ =\sqrt[]{36}=6 \\ r=6 \end{gathered}[/tex][tex]\begin{gathered} \varnothing=tan^{-1}(\frac{-3}{-3\sqrt[]{3}}) \\ tan^{-1}(\frac{3}{3\sqrt[]{3}}) \\ By\text{ rationalization, we have} \\ tan^{-1}(\frac{3}{3\sqrt[]{3}}\text{ =}\frac{3}{3\sqrt[]{3}}\times\frac{\sqrt[]{3}}{\sqrt[]{3}}=\frac{3\sqrt[]{3}}{9}=\frac{\sqrt[]{3}}{3}) \end{gathered}[/tex][tex]\varnothing=\tan ^{-1}\frac{\sqrt[]{3}}{3}[/tex][tex]\varnothing=30^0[/tex]Hence, r=6, Θ =30°