Two tugboats are pulling a freighter. Find the net force and direction of the acceleration for the following cases. Assume that each tugboar pulls with a force of 1.2 x 10^5 N. Assume no resistive force at this time. a) Tugboat 1 pulls in a direction given by [N 60degrees E] and tugboat 2 pulls in a direction given by [S60 degrees E].

I drew the diagram, and I found that the contained angle was 60 degrees, so I used the cosine law to solve for Fnet, but the answer at the back was different.

Thanks in advance!

First, remember to always draw a picture of your problem.
Next, resolve the two vectors into their respective
x and y components. After that, the respective
x and y components of the two vectors are simply
added to together.

For vector F1:

sin 60 = y1 / F1 ; y1 = F1 sin 60
cos 60 = x1 / F1 ; x1 = F1 cos 60

For vector F2:

sin -60 = y2 / F2; y2 = F2 sin -60
cos -60 = x2 / F2; x2 = F2 sin -60

For the resultant vector F3:

F3 = F1 + F2
F3 = (x1 + x2)i + (y1 + y2)j
F3 = [(F1 cos 60)+(F2 cos -60)]I + [(F1 sin 60) + (F2 sin -60)]j

Since we know the value of F1 and F2, we simply need to know
the values of sin 60, cos 60, sin -60, and cos -60.
It is left as an exercise for you to evaluate the last equation
for the resultant vector F3 above.

Thank you!

Indeed, but if you make a picture and fully understand it, you can draw a parallelogram of forces to get the resultant force (in this case, also a rhombus).

This gives you directly another pair of forces that you use with the cosine rule to get the resultant.

F_{net} = \sqrt{F_1^2 + F_2^2 - 2(F_1)(F_2)\cos(120)}