Question

How do you convert `(-sqrt 3, -1)` into polar coordinates?

Answer 1

`(r,theta)->(2,240^0)" using the "360^0" system"`
`(r,theta)->(2,(4pi)/3)" using the radian system"`

Explanation:

Polar coordinates consist of the length if the arm from the origin to the point and the angle of that 'arm' measured from the `color(whit)()^+ x` axis.

`color(blue)("Determine the length of the 'arm'")`

Using Pythagoras

`r^2=x^2+y^2 -> r=sqrt[ (-sqrt(3))^2+(-1)^2)`

`color(green)(=> r=sqrt(4)=2)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the angle to the x-axis")`

The point `P_r ->(x,y)->(-sqrt(3),-1)`
which is the 3rd quadrant

So the angle is `180^0+ tan^(-1)( (-sqrt(3))/(-1)) = 180^(0)+60^0`

Thus the polar coordinate is `(r,theta)->(2,240^0)`

Or if you wish to use radians ( `2pi" radians "= 360^0`)

`240^0 ->240/360xx2pi= 4/3 pi`

giving: `(r,theta)->(2,(4pi)/3)`