Try calculating the magnitude of the forces (individually) acting on the specified object and determine their nature (i.e. attractive/repulsive) , then find the resultant vector of force using trigonometry.
I did these, what did I do wrong?
Multiply it by sin(30) and multiply 7.875 again by cos(30) = (6.82 N, 3.94 N)
Turn the 6uc into vector components. = (-9.02 N,0)
Add the two vectors = (-2.2, 3.94)
Use pythagorean theorem on it = 3.27 N
Arctangent to find direction= -60.809688481761 degrees
3.37 N, 60 degrees N of W
The vector triangle will not be right angled - do you have a justification for assuming one of the angles between the vectors to be 90 degrees? The resultant won't be on the same line as the 10cm one, so using pythagoras' principle won't work. Try using the cosine rule.
Multiply it by sin(30) and multiply 7.88 again by cos(30) = (6.82 N, -3.94 N)
Turn the 6uc into vector components. = (-9.34 N,0)
Add the two vectors = (-2.52, 3.94)
Use pythagorean theorem on it = 4.68 N
Arctangent to find direction= -57.4 Degrees
4.68 N, 57.4deg S of E