Polar points in rectangular coordinates help

[lakitu]

My teacher has told me I will need to be able to do these in my up and comping exam later this month. He covered the topic a while ago but I didn't quite understand. Here is an example I have.

Express the polar point (6, 120 °) in rectangular coordinates.

I have no idea how to start doing these? Guidence of any kind would be wonderful.

Regards Lakitu

 

[assyrian_77]

Surely you know the relation between cartesian coordinates and polar coordinates? It is two very simple equations that express x and y in r and $\theta$.

 

[lakitu]

Sorry to disappoint, I have no idea how to do these

Lakitu

 

[topsquark]

Sorry to disappoint, I have no idea how to do these

Lakitu

I fail to see how your teacher would expect you to be able to do this problem if (s)he didn't give you the above mentioned equations. Look in your book or notes under "polar coordinates."

-Dan

 

[VietDao29]

My teacher has told me I will need to be able to do these in my up and comping exam later this month. He covered the topic a while ago but I didn't quite understand. Here is an example I have.

Express the polar point (6, 120 °) in rectangular coordinates.

I have no idea how to start doing these? Guidence of any kind would be wonderful.

Regards Lakitu

You may want to give this page a glance: Polar coordinate.
The relation between Rectangular coordinate, and Polar coordinate is:
$$\begin{array}{l} x = r \cos \theta \\ y = r \sin \theta \end{array}$$
Now what's r, and what's $$\theta$$? From there, can you find x, y?
Can you go from here? :)

 

[lakitu]

Thank you for your help VietDao. Big Help. ok i have come up with the answers x = -3 and y = 5.196. I have drawn a triangle also and this seems to be the same.

Thanks again

 

[VietDao29]

Thank you for your help VietDao. Big Help. ok i have come up with the answers x = -3 and y = 5.196. I have drawn a triangle also and this seems to be the same.

Thanks again

Yes, it looks good. However, I'd prefer the y to be:
$$y = 3 \sqrt{3}$$ instead of an approximation. Other than that, everything seems correct. :)