#1**+5 **

The polar representation of cartesian point (x,y) is (r,θ) where

$$\\

r=\sqrt{x^2+y^2}\\ \theta=\tan^{-1}(\frac{y}{x})$$

So, with x = 4 and y = 30

$${\mathtt{r}} = {\sqrt{{{\mathtt{4}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{30}}}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{r}} = {\mathtt{30.265\: \!491\: \!900\: \!843\: \!111\: \!9}}$$

$${\mathtt{theta}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{30}}}{{\mathtt{4}}}}\right)} \Rightarrow {\mathtt{theta}} = {\mathtt{82.405\: \!356\: \!631\: \!409^{\circ}}}$$

.Alan May 13, 2014

#1**+5 **

Best Answer

The polar representation of cartesian point (x,y) is (r,θ) where

$$\\

r=\sqrt{x^2+y^2}\\ \theta=\tan^{-1}(\frac{y}{x})$$

So, with x = 4 and y = 30

$${\mathtt{r}} = {\sqrt{{{\mathtt{4}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{30}}}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{r}} = {\mathtt{30.265\: \!491\: \!900\: \!843\: \!111\: \!9}}$$

$${\mathtt{theta}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{30}}}{{\mathtt{4}}}}\right)} \Rightarrow {\mathtt{theta}} = {\mathtt{82.405\: \!356\: \!631\: \!409^{\circ}}}$$

Alan May 13, 2014