Question 1159987
The magnitude of the resultant of the twp forces,  in Newton, is {{{R=320}}} .The magnitude of a known first force, in Newton, is {{{F1=180}}} .
Let the magnitude of the second force be {{{F2}}} Newton.
I would draw my vector diagram as shown below.
{{{drawing(650,240,-10,315,-10,100,
line(0,0,151.2,0),
arrow(100,0,151.2,0),
line(151.2,0,307.08,90),
arrow(255.12,60,307.08,90),
arrow(151.2,0,350,0),
red(line(0,0,307.08,90)),
red(arrow(204.72,60,307.08,90)),
green(arc(151.2,0,60,60,-30,0)),
locate(180,15,green(30^o)),
locate(75,0,F2), locate(150,45,red(R)),
locate(230,45,F1)
))}}}
Depending on instructor preferences, the diagram expected could have
both vectors starting from the same point with parallel segments added forming a parallelogram, like this:
{{{drawing(650,240,-10,315,-10,100,
line(0,0,151.2,0),
arrow(100,0,151.2,0),
triangle(151.2,0,307.08,90,0,0),
triangle(155.88,90,307.08,90,0,0),
arrow(103.92,60,155.88,90),
line(0,0,155.88,90),
red(line(0,0,307.08,90)),
red(arrow(204.72,60,307.08,90)),
green(arc(0,0,60,60,-30,0)),
locate(30,12,green(30^o)),
locate(75,0,F2), locate(150,45,red(R)),
locate(78,45,F1)
))}}} .
I could solve it several different ways,
but here is a simple way, assuming knowing only a little trigonometry.
We are concerned with two right triangles, {{{ABC}}} and {{{ABD}}} below.
{{{drawing(650,240,-10,315,-10,100,
triangle(151.2,0,307.08,90,0,0),
triangle(307.08,0,307.08,90,0,0),
red(line(0,0,307.08,90)),
green(arc(151.2,0,60,60,-30,0)),
locate(180,15,green(30^o)),
locate(75,0,F2), locate(150,45,red(R)),
locate(230,45,F1),locate(-2,0,A),
locate(308,95,B),locate(151,0,C),locate(307,0,D)
))}}}
{{{AB=320}}} (the magnitude of R)
{{{BC=180}}} (the magnitude of F1)
We want to find {{{AC}}} (the magnitude of F2).
From trigonometry, we find that
{{{CD=180*cos(30^o)=155.88}}} and {{{BD=180*sin(30^o)=90}}} .
We can find {{{AD}}} simply using the Pythagorean theorem as
In physics class you could say those are the magnitudes of the horizontal and vertical components of F1.
{{{AD=sqrt(AB^2+BD^2)=sqrt(320^2-90^2)=307.08}}}
You could say that {{{AD is the magnitude of the horizontal component of R.
With {{{AD}}} and {{{CD}}} , we find
{{{AC=AD-CD=307.08-155.88=151.2}}}