Question

The rectangular coordinates of a point are given. Find polar coordinates (r,theta) of this point with expressed in radians. Let r>0 and -2pi

(-2(square root)3,2)

PLEASE HELP URGENT

Answer 1

The answer is "The Polar coordinates (r,
`\theta)` =
`(4,(5 \pi)/(6))`"

Explanation:

Given point:

`\to (-2√(3),2)`

Formula:

`r=√(x^2+y^2)\\`

`=\sqrt{(-2√(3))^2+(2)^2}\\\\=√(12+4)\\\\=√(16)\\\\=√(4^2)\\\\=4`

`\therefore \\\\ \tan \theta= (y)/(x)\\\\`

`\tan \theta=(2)/(-2√(3))\\\\ \tan \theta= -(1)/(√(3))\\\\ \tan \theta= (5\pi)/(6)\\\\ \theta= (5\pi)/(6)`

The Polar coordinates (r,
`\theta)` =
`(4,(5 \pi)/(6))`.