Apparently hard Polar Coordinate Question

Do you think this question is hard?


  • Total voters
    1

durgethedemonhunter

New member
Joined
Dec 10, 2004
Messages
2
Find the polar coordinates of the point with rectangular coordinates; {}=radical

[-6 , (6/ {3})] i know the first part of the formula r={x^2 + y^2} but ive found conflicting info on the second part which i belive to be either
rtan (y/x) or r tan^-1 (y/x) now ive work out the first part and im preety sure its this

r = {(-6)^2 + (6/{3})^2} => r = {36 + 12} => r = {48} => r = 2{3}

that works out into radians at either 2Pi/3 & csc 120 or 5Pi/6 & sec 150
the book im useing in my class suks and only touchs breifly on this so any help woul be awsome:)
 
polar coordinates

For the first part I came up with r = 4{3}

Now for the second part:
Since cos theta = x/r just plug in your known values and solve.

cos theta = -6/4{3}
cos theta = -.866
theta = 150 degrees

Your answer is (4{3}, 150 degrees)
 
your right jeees that was a stupid mistake on that radical .

so its cos (/) = x/r => (/) = 150 deg. or 5Pi/6

thanks for your help
 
Top