I have attached the figure :
Based on the figure:
Net force in vertical and horizontal direction on kneecap will be 0.
∑F(horiz) = 0
\(F_h\) = 60 cos 42 = 44.58
∑F(vert) = 0,
\(F_v\) = 60 - 60 sin 42 = 60 - 40.148 = 19.85
F = \(\sqrt{F_h^2 + F_v^2}\) = 48.8
