Question

How do you find the polar coordinate of `(2sqrt3, 2)`?

Answer 1

`(4, (pi)/6)` (radians) or `(4, 60^@)` (degrees)

Explanation:

Rectangular `->` Polar: `(x, y) -> (r, theta)`

• Find `r` (radius) using `r = sqrt(x^2 + y^2)`
• Find `theta` by finding the reference angle: `tantheta = y/x` and use this to find the angle in the correct quadrant

`r = sqrt((2sqrt3)^2 + (2)^2)`

`r = sqrt((4*3)+4)`

`r = sqrt(12+4)`

`r = sqrt(16)`

`r = 4`

Now we find the value of `theta` using `tantheta = y/x`.

`tantheta = 2/(2sqrt3)`

`tantheta = 1/sqrt3`

`theta = tan^-1(1/sqrt3)`

`theta = (pi)/6` or `(7pi)/6`

To determine which one it is, we have to look at our coordinate `(2sqrt3, 2)`. First, let's graph it:

As you can see, it is in the first quadrant. Our `theta` has to match that quadrant, meaning that `theta = (pi)/6`.

From `r` and `theta`, we can write our polar coordinate:
`(4, (pi)/6)` (radians) or `(4, 60^@)` (degrees)

Hope this helps!