# Positive direction of the X axis?

1. Feb 12, 2008

### lgen0290

1. The problem statement, all variables and given/known data

What are the x and y components of a vector a in the xy plane if its direction is 230° counterclockwise from the positive direction of the x axis and its magnitude is 2 m?

2. Relevant equations

3. The attempt at a solution
My main issue is I'm not sure where counterclockwise from the positive direction of the X axis is. I'd assume that the x and y components wouldn't form a right triangle with the 2m vector being the hypotenuse if the angle is 230°.

Help, I'm lost here.

2. Feb 12, 2008

### lavalamp

The positive side of the x axis points to the right, and this vector apparently forms an angle of 230 degrees to it.

I don't know if you done anything on polar coordinates, but this is very similar to that. On a polar graph, there's only one axis and instead of the y axis, the second coordinate refers to a rotation about the origin where 0 degrees is flat on the positive x axis.

Sounds like your teacher has picked out the next topic and is leading you into it with a coordinate system you're already familiar with.

3. Feb 12, 2008

### CaptainQuasar

In this drawing the point B is the (0,0) origin, AE is the x axis, and DC is the y axis, right?

Counterclockwise is in the direction.

If you draw the 230°, 2m vector coming out of B and call its endpoint F, it's going to be between BA and BD, right? Can you see that it would make two right triangles where F lines up perpendictularly to the axes? You would use those triangles to calculate the x and y components.

The places you'll be marking off on the axes will be negative but you can ignore that; all you care about is that it's a right triangle, so you can turn it around and do your calculations with positive values and you'll still get the correct answer.

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4. Feb 12, 2008

### lgen0290

Thanks.

That would give me a right triangle in the third quadrant with a 50 degree angle.

My y component would then be sin(50)*2, right? The X would then be sin(40)*2?

5. Feb 12, 2008

### lgen0290

OK, I tried what I concluded above and there is a problem somewhere, I can't figure out where though. The answer is not the correct answer.

6. Feb 12, 2008

### HallsofIvy

Staff Emeritus
You are correct that the values from the right triangle would be 2 sin(50) and 2 cos(50)= 2 sin(40) but both x and y are negative in the third quadrant.

Actually, if you are using a calculator, you would do better just to enter 2 cos(230) and 2 sin(230).

Oh, and be sure your calculator is in degree mode!

7. Feb 12, 2008

### XxBollWeevilx

Consider the x-y coordinate plain. The positive x axis is 0 degrees, the negative x axis is 180 degrees. Likewise, the positive y is 90 and the negative y is 270. If your vector is defined by these polar coordinates, yours would be somewhere in the third quadrant, correct? Use this to form your triangle and your components, and you can then use trig to get the components.