The parallel component of force acting on person is,
[tex]F_x=F_g\sin \theta[/tex]
Substitute the known values,
[tex]\begin{gathered} F_x=(50\text{ N)s}in25 \\ =(50\text{ N)(}0.423) \\ \approx21.15\text{ N} \end{gathered}[/tex]
The perpendicular component of force acting on person is,
[tex]F_y=F_g\cos \theta[/tex]
Substitute the known values,
[tex]\begin{gathered} F_y=(50\text{ N)}\cos 25 \\ =(50\text{ N)(}0.906) \\ \approx45.3\text{ N} \end{gathered}[/tex]
Thus, the parallel force acting on person is 21.15 N and the perpendicular force is 45.3 N.
