The following problem comes from "Vector Mechanics for Engineers: Statics" from F.Beer
Solution:
$60$lb Force:
$F_x =(60\text{lb})\cos20°=56.38\text{lb}$
$F_y=(60\text{lb})\sin20°=20.52\text{lb}$
$80$lb Force:
$F_x=(80\text{lb})\cos60°=40.00\text{lb}$
$F_y=(80\text{lb})\sin60°=69.28\text{lb}$
$120$lb Force:
$F_x=(120\text{lb})\cos30°=103.92\text{lb}$
$F_y=-(120\text{lb})\sin30°=-60.00\text{lb}$
$R_x=\sum F_x=200.30\text{lb}$
$R_y=\sum F_y=29.80\text{lb}$
What I'm struggling to understand here it's how to find these angles, my guess is that the used angles are between the forces and the inclined plane, so since the $60$lb force is parallel to the plane it's angle is $20°$, for the $80$lb force we should have $20° + \alpha$ following the same logic, but for the $120$lb force I don't know how to find this $30°$ angle. Can you help me visualize this?
They have calculated the components of the individual “forces” in the horizontal and vertical directions, not parallel and perpendicular to the plane.
The $120$lb “force” makes an angle of $\alpha+20=60$ to the vertical and hence an angle $90-60=30$ to the horizontal.
BTW lbs are a measure of mass, not force IMHO, but I guess F. Beer has some explanation for this terminology somewhere…