Question 1102869
Here is a representation of the two forces and their resultant,
 {{{R}}} :
{{{drawing(600,300,-100,900,-100,400,
grid(0),line(0,0,591,0),arrow(0,0,591,0),
line(591,0,889,308.6),arrow(591,0,889,308.6),
line(0,0,889,308.6),arrow(444.5,154.3,889,308.6),
locate(300,30,591N),locate(444.5,154.3,R),
locate(740,154.3,429N),locate(620,50,46^o),
red(arc(591,0,200,200,-46,0)),locate(150,40,theta),
red(arc(0,0,400,400,-19.1,0))
)}}}
The rest is trigonometry.
You could use law of cosines and law of sines,
but I would favor just calculating the horizontal and vertical components.
The 429N force can be seen as the sum of
a horizontal component of magnitude {{{cos(46^o)*429N=about 298.0N}}} , and
a vertical component of magnitude {{{sin(46^o)*429N=about 308.6N}}} .
The 591N force an be seen as the sum of
a horizontal component of magnitude {{{591N}}} , and
a vertical component of magnitude {{{0N}}} .
Adding up components, you get the horizontal and vertical components,
{{{R[x]=298.0N+591N=889.0N}}} and {{{R[y]=308.6N+0N=308.6N}}}
{{{tan(theta)=R[y]/R[x]=308.6N/"740.0 N"=0.3341}}} --> {{{theta=19.1^o}}} .
{{{R=sqrt((889.0N)^2+(308.6N)^2)=sqrt(790336N^2+95232N^2)=sqrt(885568N^2)=about941.0N}}} .