Question 981278
I assume that the answer wanted is the direction and magnitude of the resultant of those forces.
I also assume that the angles are measured counter-clockwise from the positive x-axis, as shown below.
{{{drawing(300,200,-0.9,0.9,-0.21,0.99,
grid(0),arrow(0,0,-0.174,0.985),
arrow(0,0,0.821,0.143),
locate(-0.4,0.99,30),locate(-0.28,0.99,lb),
locate(0.72,0.25,25),locate(0.85,0.25,lb),
red(arc(0,0,0.5,0.5,-100,0)),red(arc(0,0,1.1,1.1,-10,0)),
locate(0.2,0.25,red(100^o)),locate(0.57,0.15,red(10^o))
)}}} The angle between the forces is {{{100^o-10^o=90^o}}} .
Based on that angle we can find the magnitude of the resultant force,
and the angles between the resultant force and the two given forces.
Leaving out the x- and y-axes, the two forces and their resultant force can be drawn as a rectangle with its diagonal, like this
{{{drawing(350,400,-5,30,-5,35,
rectangle(0,0,25,30),
locate(-3.7,20,30),locate(-1.7,20,lb),
locate(18,0,25),locate(20,0,lb),
arrow(15,0,25,0),line(0,0,25,0),
arrow(0,20,0,30),line(0,0,0,30),
rectangle(22.5,0,25,2.5),locate(15,18,red(R)),
red(arrow(15,18,25,30)),red(line(0,0,25,30)),
red(arc(0,0,25,25,-50.2,0)),locate(12,1,red(theta))
)}}} The magnitude (in lb) of the resultant, {{{red(R)}}} , is
the lengths of the hypotenuse of a right triangle with legs measuring {{{25}}} and {{{30}}} :
{{{sqrt(30^3+25^2)=sqrt(900+325)=sqrt(1525)=61sqrt(5)}}}= about {{{39.05}}} (rounded).
The angle {{{red(R)}}} makes with the 25 pound force is {{{red(theta)}}} such that
{{{tan(red(theta))=30/25=1.2}}} ---> {{{red(theta)=about 50.19^o}}} (rounded)
So the angle between the positive x-axis and the resultant measures about
{{{10^o+50.19^o=highlight(60.19^o)}}} (rounded),
and the magnitude of the resultant is about {{{highlight(39.05)}}} pounds (rounded).