I am so confused about the direction of this vector!?

Question

I found that the x component of the vector is -18.4 and the y component of the vector is 23.2.

So from that, I figured tan θ = (23.2 / -18.4)

Then I got θ = -51.6⁰

But the book I have says that translates to 51.6⁰ above the negative x axis.

I can see from the xy graph that it's in the 2nd quadrant, but I don't understand how -51.6⁰ turns into "51.6⁰ above the negative x axis". How come I didn't just get 128.4⁰?

Thanks!

Answer 1

Yes, you do need to be careful with sin and cos as well - sin is positive in the 1st and 2nd quadrants (and negative in the 3rd and 4th) and cos is positive in the 1st and 4th (therefore negative in the 2nd and 3rd). So you have to pay attention to which quadrant your answer should be in, because the equation gives you two correct answers.

It might be easier to ignore the minus signs to begin with. Just say the x length is 18.4 and the y length is 23.2. You will get 51.6 degrees. Then draw the vector on an x-y graph and include the angle.

It's in the second quadrant and makes an angle of 51.6 with the x-axis. Therefore theta should be 180-51.6.

Answer 2

theta = ArcTan[23.2/(-18.4)] = 128.418 Deg.

Subtract this from 180 to get the angle above the -x axos gives, with rounding, 51.6 Deg

Answer 3

you should leave out the negative sign when you take the tangent, when you do that, the calculator doesn't know if you  mean the second quadrant, or the fourth quadrant.

It will give you the answer for the fourth quadrant, which would be -51.6

(it thinks the negative sign belongs to the Y component, and the positive sign belongs to the X component)

Once you have 51.6 take 180 -51.6, to find the degrees from 0 degrees

technically it is 51.6 degrees above the X axis if you measure from the 180 degree mark.

but the book's answer seems to be poorly worded, it really should say 180- 51.6