The net force is the vector
∑ F = (450 N) (cos(42°) i + sin(42°) j)
and two of the forces provided by the girls are
F₁ = (310 N) (cos(115°) i + sin(115°) j)
F₂ = (250 N) (cos(285°) i + sin(285°) j)
Then the force provided by the third girl is the vector
F₃ = ∑ F - F₁ - F₂
F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i
… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j
F₃ ≈ (400.722 N) i + (261.635 N) j
So, the third girl provided a force of magnitude
||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N
pointing in a direction
arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°
relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.
So, the third girl pushed with force 480 N [N57°E].
