How Do Vector Angles Behave with Different Quadrant Coordinates?

[CathyCat]

Homework Statement

I had a problem on my physics hw, and I keep on getting it wrong, so I was wondering:
in a vector, if the x direction is negative, and the y direction is positive, would the angle direction also be negative? What happens if the x direction is positive, and the y direction is negative? A degree can't be negative right?

also,

2) what does it means when the angle direction is "from the positive x axis" or "counterclockwise from the positive x-axis is positive"

Homework Equations

The Attempt at a Solution

In my hw, I figured out the magnitude already, which is 81.2, and the x direction is -81.22 and y direction is 1.95, so I found the angle direction by tan-1(1.93/-81.22) and got -1.37, but i just ignore the negative sign, since i assume that a degree can't be negative.

 

[PeterO]

Homework Statement

I had a problem on my physics hw, and I keep on getting it wrong, so I was wondering:
in a vector, if the x direction is negative, and the y direction is positive, would the angle direction also be negative? What happens if the x direction is positive, and the y direction is negative? A degree can't be negative right?

also,

2) what does it means when the angle direction is "from the positive x axis" or "counterclockwise from the positive x-axis is positive"

Homework Equations

The Attempt at a Solution

In my hw, I figured out the magnitude already, which is 81.2, and the x direction is -81.22 and y direction is 1.95, so I found the angle direction by tan-1(1.93/-81.22) and got -1.37, but i just ignore the negative sign, since i assume that a degree can't be negative.

You clearly need to draw your self a picture.

With a vertical y-axis [positive above the x axis, negative below, as per normal] and a horizontal x-axis [positive to the right of the y-axis and negative to the left of the y axis, as per normal]

You should then locate the point (-81.2, 1.95) which would be just above the point -81.2 on the x-axis.

One common convention for angles - mimicking polar co-ordinates - is to define angles relative to the positive x-axis, traveling anti clockwise [travelling clockwise means a negative angle.
There is nothing more mysterious about a negative angle than there is with the negative x-axis. It merely means "going the other way"

The angle you are looking for is just a little less than 180 degrees,

Lets imagine the angle was 177^o. The same point is at angle -183^o - but it seems you were asked for the positive answer.

You will also need Pythagoras to find the actual magnitude of the vector.

Note: If you are familiar with swapping from rectangular to polar co-ordinates on your calculator, you could just use that transform. Most people are not familiar with that use of a calculator however.

 

[PeterO]

Homework Statement

I had a problem on my physics hw, and I keep on getting it wrong, so I was wondering:
in a vector, if the x direction is negative, and the y direction is positive, would the angle direction also be negative? What happens if the x direction is positive, and the y direction is negative? A degree can't be negative right?

also,

2) what does it means when the angle direction is "from the positive x axis" or "counterclockwise from the positive x-axis is positive"

Homework Equations

The Attempt at a Solution

In my hw, I figured out the magnitude already, which is 81.2, and the x direction is -81.22 and y direction is 1.95, so I found the angle direction by tan-1(1.93/-81.22) and got -1.37, but i just ignore the negative sign, since i assume that a degree can't be negative.

Note: your answer of -1.37 is not "entirely" wrong. It means the angle is related to 1.37 degrees, but in a quadrant where the tangent is negative [2nd and 4th]. SO the answer you really want is (180 - 1.37) or (360 - 1.37); meaning 178.63 or 358.63.

A diagram of the situation will show that the 178.63 is what you are after.

 

[CathyCat]

Thank you for helping me out, but i am still confused on why 180-1.37?

 

[Saitama]

PeterO answered the question in his post.

Thank you for helping me out, but i am still confused on why 180-1.37?

but in a quadrant where the tangent is negative [2nd and 4th].

 

[PeterO]

Thank you for helping me out, but i am still confused on why 180-1.37?

If you recall your trigonometry: we have expressions like:

sin(180-θ) = sin(θ) 2nd quadrant
sin(180+θ) = -sin(θ) 3rd quadrant
sin(360-θ) = -sin(θ) 4th quadrant

also
tan(180-θ) = -tan(θ)
tan(180+θ) = tan(θ)
tan(360-θ) = -tan(θ)

and a set for cosine.

since the tangent was negative in your problem, it meant you should use either
tan(180-θ) = -tan(θ)
or
tan(360-θ) = -tan(θ)

A diagram would show your angle was in the 2nd quadrant [between 90 and 180 degrees] so tan(180-θ) was used.

 

[CathyCat]

What would happen if the x direction is +81.22, and the y direction is -1.95?

 

[PeterO]

What would happen if the x direction is +81.22, and the y direction is -1.95?

Your diagram would show you it is in the 4th quadrant, so would use (360 - θ) to get 358.63.

That answer could also be expressed as -1.37