It was reported that although this question was not attempted by many candidates, those who attempted it displayed a good understanding of the topic.  They were able to resolve the forces into the horizontal and vertical components of the resultant force thus:                                          
  8 cos 55°   +   6 cos0o  +   4 cos 25   +  -5sin25°  +  -4cos25o     which
  8 sin 55°         6 sin0o       -4sin 25o        -5cos25°        4sin25o
after a little computation gave  6.0571   . With the result, they were able
                                                        4.4401
to obtain the magnitude of the resultant force

/F/ =  = 7.51 N to 2 decimal places. The direction of the resultant force is gotten by: 090° - tan-1(4.4401 / 6.0571)  = 054° to the nearest degree.