MHB Finding the Counter Clockwise Angle of Vector Difference B-A with the +x Axis

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To find the counterclockwise angle of the vector difference B-A with the +x axis, the correct approach involves using the tangent function with the components 2.77 and -5.95. The calculation should be tan(θ) = 5.95/2.77, leading to θ ≈ 65 degrees when using the inverse tangent. It's important to ensure that the calculator is set to degrees, not radians, to avoid confusion. The discussion clarifies that the signs of the components can be adjusted based on the quadrant, confirming that the vector indeed lies in the first quadrant after correcting the direction of vector A.
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I have a question on finding the counter clockwise angle the vector difference B-A makes with the +x axis
I already have the components of vector difference for B-A and I have checked that they are correct they are 2.77,-5.95
I started with doing the tan function (opp/adj) which was -5.95/2.77
I got -2.148 which I found the tan(-2.148) to be -1.536
I then found the arctan to be 56.934
I then subtracted 360-56.93 and got an answer to be 303.07
This proved to be wrong and I am wondering if someone could help find where my mistake is.
 
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I find that your question is a bit unclear, do you have the original question?

You got $2.77$ and $-5.95$; are those the lengths of the $x$ and $y$ components of the vector respectively?

$$\tan\left({\theta}\right)=\frac{5.95}{2.77}$$

I do not get $\theta$ the be $57$ degrees.
 
Sorry for the confusing question, the question word for word is
"Find the counterclockwise angle the vector difference B⃗ −A⃗ makes with the +x axis."
2.77 is for B and -5.95 is for A
So my math must be off, I am assuming I went wrong getting
tan(θ)=-5.95/2.77
First I took what I got from dividing the two which was -2.148, and plugged it in my calculator and like this (tan -2.148) and got 1.536
then I do arctan(1.536) am I right?
 
I see your mistake now.

For all intents and purposes, the signs don't don't matter since we're dealing with magnitudes here. We can just adjust accordingly depending on what quadrant we are on.

$$\tan\left({\theta}\right)=\frac{5.95}{2.77}$$

Theta is computed by the inverse tangent:

$$\tan^{-1}\left({\frac{5.95}{2.77}}\right)\approx65 \text{ degrees}$$

If your calculator is on radians, use the conversion factor $\frac{180}{\pi}$

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Notice, however, that your question wants $\vec{B-A}$, so we need to flip the vector $\vec{A}$ in the opposite directions. Therefore, the vector lies on the first quadrant.
 
I understand my mistake now!
Thank you so much!
 
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