Question

Find polar coordinates of the point that rectangular coordinates (-3, -Square root of three

Answer 1

`\bf \begin{array}{clclll} (-3&,&√(3))\\ x&&y \end{array}\qquad \begin{cases} r=√(x^2+y^2)\\\\ \theta=tan^(-1)\left( (y)/(x) \right) \end{cases}\\\\ -----------------------------\\\\`


`\bf r=\sqrt{(-3)^2+(√(3))^2}\implies r=√(9+3)\implies r=√(12)\implies r=2√(3) \\\\\\ \theta=tan^(-1)\left( (√(3))/(-3) \right) \impliedby \textit{now, let's rationalize the numerator} \\\\\\ \cfrac{√(3)}{-3}\cdot \cfrac{√(3)}{√(3)}\implies \cfrac{3}{-3√(3)}\implies -\cfrac{1}{√(3)}`

now, let's take a peek at our terminal point, -3, √(3).... the "x" is negative, the "y" is positive, and that means, the 2nd quadrant


`\bf \textit{now, take a look in your Unit Circle at }(5\pi )/(6)\quad \begin{cases} x=-(√(3))/(2)\\\\ y=(1)/(2) \end{cases} \\\\\\ \cfrac{y}{x}\implies \cfrac{(1)/(2)}{-(√(3))/(2)}\implies \cfrac{1}{2}\cdot -\cfrac{2}{√(3)}\implies -\cfrac{1}{√(3)} \\\\\\ \theta=\cfrac{5\pi }{6} \\\\\\ thus\qquad \qquad \left( 2√(3)\ ,\ (5\pi )/(6) \right)`