Question

What is the polar form of `( -13,14 )`?

Answer 1

`(r,theta)~~(19.1,2.3" [radians]")`

Explanation:

`(x,y)=(-13,14)` is a point in Quadrant II
So `theta` must be `in [0,pi]`

The radius (distance from the origin) for the point `(x,y)`
is `sqrt(x^2+y^2)`
In this case
`color(white)("XXX")r=sqrt((-13)^2+14^2)~~19.10497...` (using a calculator)

The tangent for this angle `(theta)` is `y/x`
In this case
`color(white)("XXX")tan(theta)=14/(-13)`
which implies
`color(white)("XXX")theta=arctan(-14/13)~~2.3191744...` (again using a calculator
Note:
you might need to adjust for the quadrant if your `arctan` function gives a value in the range `[-pi/2,+pi/2]`