Urgent SUMMER SCHOOL CALCULUS
posted by Derek .
Add these two vectors: I used component law, and I got an answer, but I am unsure as to how to do the direction at the end. Add: 9m/s [N30E] and 2m/s [N60E] I wound up with 10.8m/s and I found an angle of 55 but I am unsure as to what do with the 55: thanks for the help, I appreciate this...this class is online, so you all are acting as my teacher! Thanks. I think the answer would be 10.8m/s [N35E]. Am I correct?

Urgent SUMMER SCHOOL CALCULUS 
drwls
The direction, measured counterclockwise relative to east (+x), is the arctangent of the ratio of the sum of the y components to the sum of the x components. In other words.
theta = arctan (Ry/Rx),
where Ry and Rx are the y and x components of the resultant, R.
Rx = 9 sin 30 + 2 sin 60 = 6.232
Ry = 9 cos 30 + 2 cos 60 = 8.793
R = 10.78 m/s
theta = arctan 8.793/6.232 = 54.7 degrees N of E, or 25.3 degrees E of N
No guarantees; check my numbers. 
Urgent SUMMER SCHOOL CALCULUS 
Derek
I did it a little differently:
Horizontal Components: 9cos 60=4.5 and 2cos30=1.7 and added they are 6.2 which gives the horizontal resultant
Vertical: 9sin60=7.8 and 2sin30=1 and added they are 8.8 which is the vertical resultant
Then using an equation I found magnitude, and got 10.8 and then did tan to find the angles in which I got 55....I am unsure however, as to why the 55 is N55E should the 55 not be subtracted from 90? An example was given in which the angle the example was found was 40 and in the answer they stated 50...this is online so I am not sure if it was just a typo or not...any thoughts...thanks a bunch!
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