Two forces are applied to a car in an effort to move it, as shown in the following figure, where F1 = 401 N and F2 = 363 N. (Assume up and to the right as positive directions.)

What is the direction? I am having the world of trouble.
Here is the link to the image:
http://www.webassign.net/sercp9/4-p-012-alt.gif

Question

Two forces are applied to a car in an effort to move it, as shown in the following figure, where F1 = 401 N and F2 = 363 N. (Assume up and to the right as positive directions.) 

What is the direction? I am having the world of trouble. 
Here is the link to the image:
http://www.webassign.net/sercp9/4-p-012-alt.gif

anonymous · Answer

I already figured out the magnitude.

anonymous · Answer

I already know the resultant vector. I added the x and y components and tried taking the inverse tan but I got it wrong.

shane_b · Answer

I recommend posting your work so that someone can try to figure out what's wrong

anonymous · Answer

so okay you say 401sin10 = 69.63   These added are 251.13
                          363sin30 = 181.5
401cos10 = 394.91     These added together are 709.28
363cos30 = 314.37

shane_b · Answer

My equations look like this:
$$\Large F1 = [401sin(-10), 401cos(10)]=(-69.6,394.9)$$$$\Large F2 = [363sin(30), 363cos(30)]=(181.5,314.4)$$

shane_b · Answer

Note that in the first equation it should be -10 in both x and y....but you end up with the 394.9 either way..

shane_b · Answer

I'm not sure if you're still there but you should end up with a magnitude of about 718N and a direction of about +9 degrees.