Respuesta :
[tex]\bf \begin{array}{rlclll}
(2&,&-2)\\
\uparrow &&\uparrow \\
x&&y
\end{array}\qquad
\begin{cases}
r=\sqrt{x^2+y^2}\\
\theta =tan^{-1}\left( \frac{y}{x}\right)\\
----------\\
r=\sqrt{(2)^2+(-2)^2}\\
\qquad \sqrt{8}\\
\qquad 2\sqrt{2}\\
\theta =tan^{-1}\left( \frac{-2}{2} \right)\\
\qquad tan^{-1}(-1)\\\\
\qquad \cfrac{3\pi }{4}\ ,\ \boxed{\frac{7\pi }{4}}
\end{cases}\implies \left(2\sqrt{2}\ ,\ \frac{7\pi }{4} \right)[/tex]
now, notice, there are two feasible angles, they both have a tangent of -1, however, our x,y pair is +,-, the x is positive and the y is negative, that means the rectangular point is in the IV quadrant, thus, the angle is in the IV quadrant, not the II quadrant.
now, notice, there are two feasible angles, they both have a tangent of -1, however, our x,y pair is +,-, the x is positive and the y is negative, that means the rectangular point is in the IV quadrant, thus, the angle is in the IV quadrant, not the II quadrant.
Answer:
([tex]2\sqrt{2}[/tex], 315°), ([tex]-2\sqrt{2}[/tex], 135°)
Hope this helps!
