Answer
$F_R\approx 1254.40\;lb$
$\theta\approx 258.68^\circ$
Work Step by Step
First we resolve each force into its $x$ and $y$ components using the parallelogram law,
Let $F_{1x}$ and $F_{1y}$ are the $x$ and $y$ components of $\vec {F_1}$ force in which $F_1=700\;lb$. Thus we have
$F_{1x}=-700\cos30^\circ\;lb\approx -606.2\;lb$
$F_{1y}=-700\sin30^\circ\;lb=-350\;lb$
Let $F_{2x}$ and $F_{2y}$ are the $x$ and $y$ components of $\vec {F_2}$ force, in which $F_2=400\;lb$. Thus, we have
$F_{2x}=400\cos90^\circ\;lb=0\;lb$
$F_{2y}=-400\sin90^\circ\;lb=-400\;lb$
Let $F_{3x}$ and $F_{3y}$ are the $x$ and $y$ components of $\vec {F_3}$ force, in which $F_3=600\;lb$. Using proportional parts of similar triangles and considering the directions we have
$\frac{F_{3x}}{600\;lb}=\frac{3}{5}$
or, $F_{3x}=600\;lb\Big(\frac{3}{5}\Big)=360\;lb$
Similarly,
$\frac{F_{3y}}{600\;lb}=-\frac{4}{5}$
or, $F_{3y}=-600\;lb\Big(\frac{4}{5}\Big)=-480\;lb$
Second, 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, $(F_R)_x=(-606.2+0+360)\;lb=-246.2\;lb$
Summing the y components yields
$+\uparrow (F_R)_y=\sum F_y$
or, $(F_R)_y=(-350-400-480)\;lb=-1230\;lb$
$\therefore$ The resultant force has a magnitude of
$F_R=\sqrt {[(F_R)_x]^2+[(F_R)_y]^2}$
or, $F_R=\sqrt {(-246.2)^2+(-1230)^2}\;lb$
or, $F_R\approx 1254.40\;lb$
The direction $\theta$ of resultant force $\vec F_R$, measured counterclockwise from the $x$ axis, is
$\theta=180^\circ+\tan^{-1}\Big(\frac{-1230\;lb}{-246.2\;lb}\Big)\approx 258.68^\circ$
