Finding the angle given positive x component and negative y component

In summary: Thanks again for the help!In summary, to find the angle with respect to the x axis, you can use the formula Tan = Ay/Ax and plot the vector on a graph to determine the correct quadrant and whether to add or subtract 180 degrees from the calculated angle. Blindly trusting the inverse trigonometric functions on your calculator may result in incorrect answers.
  • #1
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Homework Statement


Ax = 3m
Ay = -4m

Find the angle with respect to the x axis


Homework Equations


Tan = Ay/Ax


The Attempt at a Solution


Tan = Ay/Ax
= -4m/3m
= 1.333333
angle = -53 degrees

Is -53 degrees a correct answer or should it be positive 53?
 
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  • #2
Hello hey123a,

Welcome to Physics Forums!

hey123a said:

Homework Statement


Ax = 3m
Ay = -4m

Find the angle with respect to the x axis

Homework Equations


Tan = Ay/Ax

The Attempt at a Solution


Tan = Ay/Ax
= -4m/3m
= 1.333333
angle = -53 degrees

Is -53 degrees a correct answer or should it be positive 53?

I'd tell you, but I think you might be able to determine this one on your own by first plotting the vector on a graph. Even a piece of scratch paper or a paper napkin should be enough for this.

First plot the x- and y-axes.

Then place the vector on the graph, with its tail at the origin and its head at point (3 m, -4 m).

The angle is with respect to the positive x-axis.

Positive angles (with 0o < θ < 180o) represent a vector that is above the x-axis. Negative angles (with -180o < θ < 0o) represent a vector below the x-axis.
 
  • #3
collinsmark said:
Hello hey123a,

Welcome to Physics Forums!



I'd tell you, but I think you might be able to determine this one on your own by first plotting the vector on a graph. Even a piece of scratch paper or a paper napkin should be enough for this.

First plot the x- and y-axes.

Then place the vector on the graph, with its tail at the origin and its head at point (3 m, -4 m).

The angle is with respect to the positive x-axis.

Positive angles (with 0o < θ < 180o) represent a vector that is above the x-axis. Negative angles (with -180o < θ < 0o) represent a vector below the x-axis.

Hey thank you for the greeting, i'll definitely be around the forums a lot because physics is one of the subjects that i don't feel confident in.

Well, I placed the vector on the graph and it is definitely below the positive x axis, so, my conclusion is that the answer is -53 degrees
 
  • #4
hey123a said:
Hey thank you for the greeting, i'll definitely be around the forums a lot because physics is one of the subjects that i don't feel confident in.

Well, I placed the vector on the graph and it is definitely below the positive x axis, so, my conclusion is that the answer is -53 degrees

Sounds good to me! :smile:
 
  • #5
As a heads up though, for future problems, you might want to get into the habit of actually plotting the vectors out. You can't blindly trust the inverse trigonometric functions on your calculator, as they can mislead you.

For example, suppose Ax was -3 m instead of positive 3 m.

The angle you first calculate is θ = ATAN(-4/-3) = 53o.

The angle you calculate is positive, but you can tell when graphing the vector that not only is the angle negative, but it also is in the third quadrant. In that case you need to subtract 180o from the result. So in that case the angle would be 53o - 180o = -127o.

-------------------------------------------

As another example, suppose that the signs of both were opposite what you were given,
Ax = -3
Ay = 4

The angle you calculate with the ATAN function gives you -53o again (this is the same result as in the original problem). But you can tell by graphing the function that its in the second quadrant. So now you need to add 180o to it, making it the correct +127o.
 
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  • #6
collinsmark said:
As a heads up though, for future problems, you might want to get into the habit of actually plotting the vectors out. You can't blindly trust the inverse trigonometric functions on your calculator, as they can mislead you.

For example, suppose Ax was -3 m instead of positive 3 m.

The angle you first calculate is θ = ATAN(-4/-3) = 53o.

The angle you calculate is positive, but you can tell when graphing the vector that not only is the angle negative, but it also is in the third quadrant. In that case you need to subtract 180o from the result. So in that case the angle would be 53o - 180o = -127o.

-------------------------------------------

As another example, suppose that the signs of both were opposite what you were given,
Ax = -3
Ay = 4

The angle you calculate with the ATAN function gives you -53o again (this is the same result as in the original problem). But you can tell by graphing the function that its in the second quadrant. So now you need to add 180o to it, making it the correct +127o.

Thank you for taking the time out of your schedule to help me out. I really appreciate it! :smile:
I pressed that thank you button on your post which I found pretty neat
 

1. What is an angle given positive x component and negative y component?

An angle given positive x component and negative y component refers to the direction or orientation of a vector with a positive horizontal (x) component and a negative vertical (y) component. This angle can be measured using trigonometric functions such as tangent or arctangent.

2. How do you find the angle given positive x component and negative y component?

To find the angle given positive x component and negative y component, you can use the inverse tangent function (arctan) to calculate the angle using the ratio of the y and x components. The resulting angle will be in radians, so you may need to convert it to degrees if necessary.

3. Can the angle given positive x component and negative y component be negative?

Yes, the angle given positive x component and negative y component can be negative. This will depend on the direction and magnitude of the vector. If the angle is measured counterclockwise from the positive x-axis, then a negative angle would indicate that the vector is pointing in the opposite direction (i.e. to the left).

4. Is the angle given positive x component and negative y component always the same as the angle given negative x component and positive y component?

No, the angle given positive x component and negative y component is not always the same as the angle given negative x component and positive y component. The two angles will be complementary, meaning they add up to 90 degrees (or π/2 radians). This is because the x and y components are switched, resulting in a different ratio for the tangent calculation.

5. Can the angle given positive x component and negative y component be greater than 90 degrees?

Yes, the angle given positive x component and negative y component can be greater than 90 degrees. This would indicate that the vector is pointing below the negative y-axis. In this case, the angle would be greater than 90 degrees but less than 180 degrees (or π radians).

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