1. Aug 26, 2011

### hyde2042

1. The problem statement, all variables and given/known data

http://i.imgur.com/fiUPD.png

2. Relevant equations
Asin(Theta)
Acos(Theta)
Ax+Bx=Cx, Ay+By=Cy

Cx^2+Cy^2=C^2

3. The attempt at a solution

My problem is with the answer I got for part a). I used 3cos120 and got 2.44. Then put in 3sin120 and got 1.74.

I added these to the Bx and By quantities and got 2.44=Cx and 4.74 for By. (Since B is on the origin I used (0 and 3 for Bx and By respectively).

After squaring and finding the square root of C, I got 5.33. Which is close to the answer of 5.2 in the back of the book, but I just wanna make sure I'm doing everything corrct, but perhaps the number I got is a bit off due to some rounding up. THank you for your time.

Edit: I also got 62.76 degrees for the arctan of Cy/Cx (4.74/2.44), which is still close to the 60 degrees the book says. Thank you once again for your help.

Last edited: Aug 26, 2011
2. Aug 26, 2011

### cepheid

Staff Emeritus
Where did you get 120 from? Theta is 30 degrees.

3cos(30o) = 2.59807621 (I'm keeping all the digits until the end, to avoid rounding error)

3sin(30o) = 1.5

Cx = Ax + Bx = 2.59807621 + 0 = 2.59807621

Cy = Ay + By = 4.5

|C| = (Cx2 + Cy2)1/2

= 5.19615242

= 5.20 (rounding to 3 sig figs, since that is the precision of the magnitudes of 3.00 m that you were given).

By the way, the question says solve graphically. Are you sure you weren't supposed to do vector addition using the parallelogram rule or something (rather than resolving into components)?

3. Aug 26, 2011

### hyde2042

Oh. Thanks. I got 120 from putting the point of B to the tail of A and getting the angle of that since it was A+B.

And thanks for pointing out the "graphically" part. I read through questions too quickly.

4. Aug 26, 2011

### hyde2042

Ah crap... I was also in Radians.

5. Aug 26, 2011

### cepheid

Staff Emeritus
No, you are mixing up two different methods. Either you resolve each of the vectors into components and add them component-wise, or you do the tip-to-tail thing.

Can you see why, when resolving just A into components, the x-component of A has to be equal to |A|cos(theta) and the y-component has to be equal to |A|sin(theta)? Draw it out.