help with my engineering mechanics problems

Sol. 1
Taking the horizontal axis as X and vertical as Y,
the X- and Y- components of force A are:
Ax = 100cos37o = 100*(0.8) = 80 lbs
Ay = 100sin37o = 100*(0.6) = 60 lbs
the X- and Y- components of force B are:
Bx = (-)100cos30o = (-)100*(0.866) = (-)86.6 lbs
By = 100sin30o = 100*(0.5) = 50 lbs
the X- and Y- components of force C are:
Cx = (-)80sin37o = (-)80*(0.6) = (-)48 lbs
Cy = (-)80cos37o = (-)80*(0.8) = (-)64 lbs
Thus, net forces along X- and Y- axes are:
Fx = Ax + Bx + Cx = 80 + (-)86.6 + 48 = 41.4 lbs
Fy = Ay + By + Cy = 60 + 50 + (-)64 = 46 lbs
The resultant force acts at an angle given by tanθ = Fy/Fx = 46/41.4 =
Thus, θ = tan-1(1.11) = 48.01o
Thus, the magnitude of the resultant force is F = (Fx2 + Fy2)1/2 = 61.88 lbs acting at an angle of 48.01o to
the horizontal as shown in...