The polar coordinates of the point,whose Cart | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given Cartesian coordinates are $(x, y) = (-2 \sqrt{3}, 2)$.First,calculate the modulus $r$:$r = \sqrt{x^2 + y^2} = \sqrt{(-2 \sqrt{3})^2 + (2)^2}

Vedclass · Accepted Answer

$(4, \frac{5 \pi}{6})$

Vedclass · Answer

(B) Given Cartesian coordinates are $(x, y) = (-2 \sqrt{3}, 2)$.First,calculate the modulus $r$:$r = \sqrt{x^2 + y^2} = \sqrt{(-2 \sqrt{3})^2 + (2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4$.Since the point lies in the second quadrant $(x  0)$,the argument $	heta$ is given by:$	heta = \pi - 	an^{-1} \left| \frac{y}{x} ight| = \pi - 	an^{-1} \left| \frac{2}{-2 \sqrt{3}} ight| = \pi - 	an^{-1} \left( \frac{1}{\sqrt{3}} ight) = \pi - \frac{\pi}{6} = \frac{5 \pi}{6}$.Thus,the polar coordinates are $(r, 	heta) = (4, \frac{5 \pi}{6})$.

The polar coordinates of the point,whose Cartesian coordinates are $(-2 \sqrt{3}, 2)$,are

Similar Questions

If $a$ is a complex number and $b$ is a real number,then the equation $\bar{a}+a+b=0$ represents $a$ as a locus of points in the complex plane,which is a:

The locus of $z=x+iy$, such that $\operatorname{Im}\left(\frac{z-3i}{iz+4}\right)=0$ is

If $P(x)=0$ is a polynomial equation of least degree with integer coefficients and $\sqrt{2}+\sqrt{3} i$ is one of its roots,then that equation is

The complex number $z$ satisfying the equation $|z-i|=|z+1|=1$ is

The number of solutions of the equation $z^{2}+\overline{z}=0$,where $z \in \mathbb{C}$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module