Question

What is the polar form of `( 13,1 )`?

Answer 1

`(sqrt(170),tan^-1(1/13))-=(13.0,0.0768^c)`

Explanation:

For a given set of coordinates `(x,y)`, `(x,y)->(rcostheta,rsintheta)`

`r=sqrt(x^2+y^2)`
`theta=tan^-1(y/x)`

`r=sqrt(13^2+1^2)=sqrt(169+1)=sqrt(170)=13.0`
`theta=tan^-1(1/13)=0.0768^c`

`(13,1)->(sqrt(170),tan^-1(1/13))-=(13.0,0.0768^c)`