Answer
$F_R\approx 17.18\;kN$
$\theta\approx 11.72^\circ$
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 $F_1=15\;kN$. Thus we have
$F_{1x}=15\sin40^\circ\;kN\approx 9.64\;kN$
$F_{1y}=15\cos40^\circ\;kN\approx 11.49\;kN$
Let $F_{2x}$ and $F_{2y}$ are the $x$ and $y$ components of $\vec {F_2}$ force, in which $F_2=26\;kN$. Using proportional parts of similar triangles and considering the directions, we have
$\frac{F_{2x}}{26\;kN}=-\frac{12}{13}$
or, $F_{2x}=-26\;kN\Big(\frac{12}{13}\Big)=-24\;kN$
Similarly,
$\frac{F_{2y}}{26\;kN}=\frac{5}{13}$
or, $F_{2y}=26\;kN\Big(\frac{5}{13}\Big)=10\;kN$
Let $F_{3x}$ and $F_{3y}$ are the $x$ and $y$ components of $\vec {F_3}$ force, in which $F_3=36\;kN$. Thus we have
$F_{3x}=36\cos30^\circ\;kN\approx 31.18\;kN$
$F_{3y}=-36\sin30^\circ\;kN\approx -18\;kN$
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, $(F_R)_x=(9.64-24+31.18)\;kN=16.82\;kN$
Summing the y components yields
$+\uparrow (F_R)_y=\sum F_y$
or, $(F_R)_y=(11.49+10-18)\;kN=3.49\;kN$
$\therefore$ The resultant force has a magnitude of
$F_R=\sqrt {[(F_R)_x]^2+[(F_R)_y]^2}$
or, $F_R=\sqrt {(16.82)^2+(3.49)^2}\;kN$
or, $F_R\approx 17.18\;kN$
The direction $\theta$ of resultant force $\vec F_R$, measured counterclockwise from the positive $x$ axis, is
$\theta=\tan^{-1}\Big(\frac{3.49\;kN}{16.82\;kN}\Big)\approx 11.72^\circ$
