We should analyze the forces in two different directions: the forward-backward direction and the left-right direction, and then calculate the resultant of the net forces acting on both directions.
Forward-backward direction. Here we have three forces acting on the monkey: the force applied by Jack, 97.9 N directly ahead; the force applied by Jill, 72.7 N with an angle [tex]45^{\circ}[/tex] to the left; and the force applied by Jane, 145 N with an angle [tex]45^{\circ}[/tex]. Therefore, the resultant on this axis is
[tex]F_x = 97.9 N + 72.7 N\cdot \cos (45^{\circ}) + 145 N \cos (45^{\circ}) = 251.8 N [/tex]
Left-right direction. In this direction, the force applied by Jack is 0 N, because he is applying his force only ahead. The force applied by Jill is 72.7 N with an angle [tex]45^{\circ}[/tex] to the left, while the force applied by Jane is 145 N with an angle [tex]45^{\circ}[/tex] to the right: this means we should write the two forces with opposite signs, because they have opposite direction in the left-right axis. Therefore,
[tex]F_y = 72.7 N \cdot \sin (45^{\circ})-145 N \cdot \sin (45^{\circ})=-51.1 N[/tex]
The net force acting on the monkey is the resultant of these two forces:
[tex]F= \sqrt{F_x^2+F_y^2}= \sqrt{(251.8N)^2+(-51.1N)^2}=257 N [/tex]
