Finding the angle given positive x component and negative y component

[hey123a]

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?

 

[collinsmark]

Hello hey123a,

Welcome to Physics Forums!

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

 

[hey123a]

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

 

[collinsmark]

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!

 

[collinsmark]

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 A_x was -3 m instead of positive 3 m.

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

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 180^o from the result. So in that case the angle would be 53^o - 180^o = -127^o.

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

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

The angle you calculate with the ATAN function gives you -53^o 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 180^o to it, making it the correct +127^o.

 

[hey123a]

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 A_x was -3 m instead of positive 3 m.

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

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 180^o from the result. So in that case the angle would be 53^o - 180^o = -127^o.

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

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

The angle you calculate with the ATAN function gives you -53^o 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 180^o to it, making it the correct +127^o.

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