Cartesian to polar conversions

[rcmango]

Homework Statement

find polar coordinates of the points whose cartesian coordinates are given.

Homework Equations

heres the point: (3sqrt(3), 3)

The Attempt at a Solution

well i know that r^2 = (sqrt(a^2 + b^2))
so the answer here is : 6

and if we use tan(theta) = o/a = 3/(3sqrt(3)
so the answer here is: 1/sqrt(3)

so theta is pi/6

so how do i know its pi/6?
how do i convert to get this answer with pi? radians?

help please.

 

[benorin]

theta = arctan(1/sqrt3) = pi/6 (in radians, always radions for polar coordinates).

 

[HallsofIvy]

Homework Statement

find polar coordinates of the points whose cartesian coordinates are given.

Homework Equations

heres the point: (3sqrt(3), 3)

The Attempt at a Solution

well i know that r^2 = (sqrt(a^2 + b^2))
so the answer here is : 6

and if we use tan(theta) = o/a = 3/(3sqrt(3)
so the answer here is: 1/sqrt(3)

The right hand side of that equation, not the "answer" (to what question?!), is \frac{1}{\sqrt{3}}. In general, if you know what tan(\theta) is, you can find \theta by using the arctan function- perhaps on a calculator. Here, you are probably expected to know that sin(\pi/6)= \frac{1}{2} and that cos(\pi/6)= \frac{\sqrt{3}}{2} so that tan(\pi/6)= \frac{1}{\sqrt{3}}.

so theta is pi/6

so how do i know its pi/6?
how do i convert to get this answer with pi? radians?

help please.

As benorin said- in polar coordinates the angle is always in radians. As a rule, the only time you use degrees is when the problem specifically involves angle that are given in degrees.