The following forces are acting at a point.

1)450N force towards north-east.
2)350N force towards east.
3)250N force at 30º west of north.
4)300N force at 60º south of west.
Find magnitude and direction of the resultant.

1 answer

EF = 450 cos 45 + 350 - 250 sin 30 - 300 cos 60
NF = 450 sin 45 + 0 + 250 cos 30 - 300 sin 60

magnitude = sqrt (EF^2 + NF^2)
tan direction East of North = EF/NF
or if you do math not navigation then
tan angle North of East = NF/EF