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In concurrent forces, forces lies on the same point, since the resultant is on the y axis, obviously its direction is 90 degrees..so we need to find out their forces, obviously in the third and 4th quadrant at 45 degrees, and its angle are parallel to first and second quadrant, respectively by components.

@2nd quad;
Fx=-f (cos 45);=-f(1/squaroot of2)=-0.707f;
Fy=f (sin45); =f(1/squaroot of 2)=0.707f;

@1st quadrant

since they are in same direction, let us add all fx  components and fy components:
fx(total)= 0.707f-0.707f=0
fy(total)= 0.707f+0.707f=1.414f;

Resultant) R^2= fx^2+fy^2
                  300^2=0 +[1.414f]^2
                   f=212.13 lbs..@ 45 degrees south east
                   f=212.13 lbs..@45 degrees  south west

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