Answer
$F_R\approx 11.08\;kN$
$\theta\approx47.67^\circ$
Work Step by Step
First, we resolve each force into its $x$ and $y$ components by using the parallelogram law:
Let $F_{1x}$ and $F_{1y}$ are the $x$ and $y$ components of $\vec {F_1}$ force, in which $F_1=6\;kN$. Thus we have
$F_{1x}=6\cos0^\circ\;kN=6\;kN$
$F_{1y}=6\sin0^\circ\;kN=0\;kN$
Let $F_{2x}$ and $F_{2y}$ are the $x$ and $y$ components of $\vec {F_2}$ force, in which $F_2=5\;kN$. Thus we have
$F_{2x}=5\sin30^\circ\;kN=2.5\;kN$
$F_{2y}=5\cos30^\circ\;kN\approx 4.33\;kN$
Let $F_{3x}$ and $F_{3y}$ are the $x$ and $y$ components of $\vec {F_3}$ force, in which $F_3=4\;kN$. Thus we have
$F_{3x}=-4\sin15^\circ\;kN\approx-1.04\;kN$
$F_{3y}=4\cos15^\circ\;kN\approx 3.86\;kN$
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, $(F_R)_x=(6+2.5-1.04)\;kN=7.46\;kN$
Summing the y components yields
$+\uparrow (F_R)_y=\sum F_y$
or, $(F_R)_y=(0+4.33+3.86)\;kN=8.19\;kN$
$\therefore$ The resultant force has a magnitude of
$F_R=\sqrt {[(F_R)_x]^2+[(F_R)_y]^2}$
or, $F_R=\sqrt {(7.46)^2+(8.19)^2}\;kN$
or, $F_R\approx 11.08\;kN$
The direction $\theta$ of resultant force $\vec F_R$, measured counterclockwise from the positive $x$ axis, is
$\theta=\tan^{-1}\Big(\frac{8.19\;kN}{7.46\;kN}\Big)\approx 47.67^\circ$
