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

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

Mathematics
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For r I already got 4 (sqroot3) and -4 (Sqroot3)
and then tan(theta)= -(2(sqroot3))/6
@rvc :) I know you just got on. But if you're free!

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Other answers:

  • rvc
ah bit busy dear :( @JoannaBlackwelder
\[\tan(\theta) = -\frac{ 2\sqrt{3} }{ 6 }\]
The tan equation looks good to me :-) Can you solve for theta?
em theta = arctan (x) but i'm not sure how to keep it not in decimal form
....?
We will need to use the unit circle and the idea that tan is sin/cos
http://www.mathsisfun.com/geometry/unit-circle.html Look down at the very bottom chart.
I'm not finding anywhere that tangent equals that on the unit circle. It is sq3/3 at pi/6
and -sq3/3 at 11pi/6
Right, and -2sqrt3/6 is -sqrt3/3
oh.. We can take the 6 and divide it by the 2 in the numerator?
Yep, those can simplify, since they have a common factor of 2
Do you see another one other than 11pi/6?
5pi/6
so how do I know which to choose? 11pi/6 or 5pi/6 ?
Awesome! We are looking for 2 options, so both.
And I don't get what you got for r. How did you get that?
Give me a moment to type that up :)
Oh, sorry, my mistake. You're rs are correct! :-)
As long as you can combine them correctly.
Into two pairs.
\[r^2=x^2+y^2\] \[r^2=-6^2+(2\sqrt{3})^2\] \[r^2=36+12 = \sqrt{48}\] Simplify and get \[r=4\sqrt{3}\]
Right, how do I know how to pair them?
(4 sq3, 5pi/6) (-4 sq3, 11pi/6) right? but i'm not sure how I got there xD or why
Let's look at each of those graphically to see if they look like they are in the right spots.
|dw:1434646002351:dw|
Here is a rough graph of the rectangular coordinate.
|dw:1434646256752:dw|
Yeah?
Yep, great!
Fantastic, thanks :)
No worries :-)

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