Answer
$F_R\approx567.26\;N$
$\theta\approx 38.1^\circ$
Work Step by Step
First, we resolve each force into its $x$ and $y$ components by the parallelogram law:
Let $F_{1x}$ and $F_{1y}$ are the $x$ and $y$ components of $\vec {F_1}$ force, in which $F_1=300\;N$. Then we have
$F_{1x}=300\cos0^\circ\;N=300\;N$
$F_{1y}=300\sin0^\circ\;N=0$
Let $F_{2x}$ and $F_{2y}$ are the $x$ and $y$ components of $\vec {F_2}$ force, in which $F_2=400\;N$. Then we have
$F_{2x}=400\cos30^\circ\;N=200\sqrt 3\;N$
$F_{2y}=400\sin30^\circ\;N=200\;N$
Let $F_{3x}$ and $F_{3y}$ are the $x$ and $y$ components of $\vec {F_3}$ force, in which $F_3=250\;N$. Then we have Using proportional parts of similar triangles and considering the directions, we have
$\frac{F_{3x}}{250\;N}=-\frac{4}{5}$
or, $F_{3x}=-250\;N\Big(\frac{4}{5}\Big)=-200\;N$
Similarly,
$\frac{F_{3y}}{250\;N}=\frac{3}{5}$
or, $F_{3y}=250\;N\Big(\frac{3}{5}\Big)=150\;N$
Then, using scalar notation and indicating the positive directions of components along the $x$ and $y$ axes with symbolic arrows, we sum these components algebraically:
Summing the x components, we have
$\xrightarrow{+} (F_R)_x=\sum F_x$
or, $(F_R)_x=(300+200\sqrt 3-200)\;N=446.41\;N$
Summing the y components yields
$+\uparrow (F_R)_y=\sum F_y$
or, $(F_R)_y=(0+200+150)\;N=350$
$\therefore$ The resultant force has a magnitude of
$F_R=\sqrt {[(F_R)_x]^2+[(F_R)_y]^2}$
or, $F_R=\sqrt {(446.41)^2+(350)^2}\;N$
or, $F_R\approx567.26\;N$
The direction $\theta$ of resultant force $\vec F_R$, measured from positive $x$ axis, is
$\theta=\tan^{-1}\Big(\frac{350\;N}{446.41\;N}\Big)\approx 38.1^\circ$
