Question

Find two polar coordinate representations for the rectangular coordinate point
(-6, 2 (sqroot3)), one with r>0, one with r<0 and both with
10 years ago

Answer 1

For r I already got 4 (sqroot3) and -4 (Sqroot3)

Answer 2

and then
tan(theta)= -(2(sqroot3))/6

Answer 3

@rvc :) I know you just got on. But if you're free!

Answer 4

ah
bit busy dear :(
@JoannaBlackwelder

Answer 5

\[\tan(\theta) = -\frac{ 2\sqrt{3} }{ 6 }\]

Answer 6

The tan equation looks good to me :-) Can you solve for theta?

Answer 7

em
theta = arctan (x)
but i'm not sure how to keep it not in decimal form

Answer 8

....?

Answer 9

We will need to use the unit circle and the idea that tan is sin/cos

Answer 10

http://www.mathsisfun.com/geometry/unit-circle.html
Look down at the very bottom chart.

Answer 11

I'm not finding anywhere that tangent equals that on the unit circle.
It is sq3/3 at pi/6

Answer 12

and -sq3/3 at 11pi/6

Answer 13

Right, and -2sqrt3/6 is -sqrt3/3

Answer 14

oh.. We can take the 6 and divide it by the 2 in the numerator?

Answer 15

Yep, those can simplify, since they have a common factor of 2

Answer 16

Do you see another one other than 11pi/6?

Answer 17

5pi/6

Answer 18

so how do I know which to choose? 11pi/6 or 5pi/6 ?

Answer 19

Awesome! We are looking for 2 options, so both.

Answer 20

And I don't get what you got for r. How did you get that?

Answer 21

Give me a moment to type that up :)

Answer 22

Oh, sorry, my mistake. You're rs are correct! :-)

Answer 23

As long as you can combine them correctly.

Answer 24

Into two pairs.

Answer 25

\[r^2=x^2+y^2\] \[r^2=-6^2+(2\sqrt{3})^2\] \[r^2=36+12 = \sqrt{48}\]
Simplify and get \[r=4\sqrt{3}\]

Answer 26

Right, how do I know how to pair them?

Answer 27

(4 sq3, 5pi/6)
(-4 sq3, 11pi/6) right?
but i'm not sure how I got there xD or why

Answer 28

Let's look at each of those graphically to see if they look like they are in the right spots.