# How to add two vectors?

1. May 6, 2013

### Matriculator

Hello, I was wondering if anyone could explain this to me. Although I know the answer, I'm confused as to how it's obtain. I know that it's the magnitude times the sines and cosines then you add the sines and the cosines for the two Fs. What I'm having a hard time understanding is the finding the angles used for sine and cosine. Sometimes my teacher used what appeared to be the reference angle, sometimes the angles themselves. I have no idea of how to obtain the angles. Are the angles of the F1 and F2 related? Thank you in advance.

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2. May 6, 2013

### SteamKing

Staff Emeritus
In order to find the components of a vector, you use the angle between the vector and the positive x-axis.

For example, the vector (4,0) is also expressed as a vector with magnitude 4, angle 0. Positive angles are measured counterclockwise with respect to the positive x-axis, negative angles are measured clockwise.

For the vectors shown in your attachment, the angles shown would need conversion in order to calculate the x and y components, according to the method discussed above.

The angles for F1 and F2 do not appear to be related; they are just angles chosen for this particular problem.

3. May 6, 2013

### Matriculator

Thank you for responding. I was told that in order to find the resultant I would do <35cos(270+23)+19cos(180-105), 35sin(270+23)+19sin(180-105)>. My question is that, when adding 23 to 270 we seemed to be only finding the angle itself, but at 180-105 it's as if we're finding the reference angle, why is this? Sorry for sounding ignorant. I've taken pre-calc before but it was never this complicated. Thank you.

4. May 6, 2013

### HallsofIvy

There are two very different ways to add vectors. You can, as Steam King says, write the components in some coordinate system. Here if you use "right" as x coordinate and "up" as y coordinate, F1, which has length 35 and makes angle 23 degrees with the negative y axis, has x component 35 sin(23) and y component - 35 cos(23). F2, which has length 19 and makes angle 105 degrees with the negative x-axis, and so angle 180- 105= 75 degrees with the positive x-axis and so has x component 19 cos(75), y component 19 sin(75).

The sum of the two vectors has x component 35 sin(23)+ 19 cos(75), y component -35 cos(23)+ 19 sin(75).

Another way to do this is to use the "cosine law" and "sine law". Attaching vector F1 the tip of F2 gives a triangle with adjoining sides of lenth 35 and 19 and angle between those two side 15+ 23= 38 degrees.

Now, if a triangle has two sides of length a and b and angle between them C, the length of the third side, c, which here is the length of the resultant vector, is given by $$c^2= a^2+ b^2- 2ab cos(C)$$. And once you have that length, you can get the other angles in the triangle, and so the direction of the resultant vector using the sine law, $$\frac{sin(A)}{a}= \frac{sin(B)}{b}= \frac{sin(C)}{c}$$.

5. May 6, 2013

### Matriculator

I think that I'm beginning to understand. Thank you. But may I ask where the 15 is coming from? The one that you added to 23 to get 38 degrees?

6. May 7, 2013

### HallsofIvy

The 15 is 105- 90. In your picture, the 105 degree angle is measured from the negative x-axis. The angle in the triangle I am using is measured from the positive y-axis.

7. May 7, 2013

### Matriculator

Thank you sir, I think that I am starting to see. But why do we use 15? Is this always the case.

8. May 7, 2013

### HallsofIvy

Go back and look at your picture in your first post. Do you see where your 105 degree angle is? Do you see why it is "90+ 15" degrees and where those two angles are in your picture?

But, no, you don't have to do it that way. You should also be able to see, in your picture, that F2 makes angle 180- 105= 75 degrees with the positive x-axis. If you draw a vertical line down from the tip of F2 the angel that vertical makes with F2 (since the two acute angles of a right triangle add to 90 degrees) is 90- 75= 15 degrees.

9. May 7, 2013

### Matriculator

Oh wow, I now get it. Thank you very much for all of your help sir.