physics quesitions....vectors ......need help?

Question

i have a few problems left and i need some help with them:
1.) The following forces act on an object resting on a level, frictionless surface. 10 N to the north, 20 N to the east, 10 N at an angle 40 degrees south of east, and 20 N at an angle 50 degrees west of south. Find the magnitude and direction of the resultant force acting on the object.
2.)The following horizontal forces act on an object: A.) has a magnitude of 6.0 N and is in the + y direction; B.) has a magnitude of 10.0 N and is in the - x direction; C.) has a magnitude of 8.0 N and is at an angle of 45 degrees clockwise from hte + x direction. Find the magnitude and direction of A + B - C.
3.) The vector A has a magnitude of 10 cm and points 37 degrees clockwise from the + y direction. The vector B has a magnitude of 10 cm and points 37 degrees clockwise from the + x direction. Find the magnitude and direction of A + B, A - B, and B - A.

Answer 1

1) The following forces act on an object resting on a level, frictionless surface. 10 N to the north, 20 N to the east, 10 N at an angle 40 degrees south of east, and 20 N at an angle 50 degrees west of south. Find the magnitude and direction of the resultant force acting on the object.

10 N north
20 N east
10 N 40° south of east
20 N 50° west of south or 20 N 40° south of west

The horizontal component of the forces is
x = 0 + 20 + 10cos40° - 20cos40° = 20 - 10cos40° ≈ 12.339556

The vertical component of the forces is
y = 10 + 0 - 10sin40° - 20sin40° = 10 - 30sin40° ≈ -9.2836283

The direction is determined
tanθ = y/x = (10 - 30sin40°)/(20 - 10cos40°)
= (1 - 3sin40°)/(2 - cos40°)
θ = arctan{(1 - 3sin40°)/(2 - cos40°)} = -36.955866°

The direction is 36.955866° south of east.

The magnitude is determined
r = √(x² + y²) = √{12.339556² + (-9.2836283)²} = 15.441839 N