The first thing to do, even if you do it crudely, is to draw a diagram.
I have labelled these vectors #ulA# and #ulB#
We can write these vectors in terms of #uli# and #ulj# notation, where #uli# is the horizontal unit vector and #ulj# is the vertical unit vector. This will make it easier to find the resultant force.
First, lets find #ulB# since this is easier than #ulA#.
#ulB=-8ulj#;
it is not going anywhere horizontally and goes 8 down, or -8 up.
For #ulA#, we will use trigonometry to find the components.
#"vertical"_ulA=5cos30# from SOHCAHTOA.
#=5*sqrt3/2#
#=5/2sqrt3#
#"horizontal"_ulA=5sin30#
#=5/2#
Its unusual in mechanics that we leave answers as fractions or surds, but it helps us to remain accurate until the end.
#:. ulA=5/2sqrt3uli+5/2ulj#
Let the resultant force be #ulR#.
#ulR=ulA+ulB#
#=5/2sqrt3uli+5/2ulj-8j#
#=5/2sqrt3uli-11/2ulj#
Let #x# be the angle that the resultant force makes with the positive x-axis, or with the unit vector #uli#.
First, magnitude. By Pythagoras' Theorem:
#abs(ulR)=sqrt(((5sqrt3)/2)^2+(11/2)^2#
#=sqrt(75/4+121/4)#
#=sqrt(196/4)#
#=sqrt49#
#=7#
#:. absulR=7N#
#tanx=(11/2)/((5sqrt3)/2)#
#=11/(5sqrt3)#
No point trying to rationalize this, chuck it into your calculator.
#x=51.79^@# (4sf)
