Answer
$F_R\approx 31.24\;kN$
$\theta\approx 39.8^\circ$
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 $F_1=15\;kN$. Using proportional parts of similar triangles and considering the directions, we have
$\frac{F_{1x}}{15\;kN}=\frac{4}{5}$
or, $F_{1x}=15\;kN\Big(\frac{4}{5}\Big)=12\;kN$
Similarly,
$\frac{F_{1y}}{15\;kN}=-\frac{3}{5}$
or, $F_{1y}=-15\;kN\Big(\frac{3}{5}\Big)=-9\;kN$
Let $F_{2x}$ and $F_{2y}$ are the $x$ and $y$ components of $\vec {F_2}$ force, in which $F_2=20\;kN$. Thus we have
$F_{2x}=20\cos90^\circ\;kN=0\;kN$
$F_{2y}=20\sin90^\circ\;kN=20\;kN$
Let $F_{3x}$ and $F_{3y}$ are the $x$ and $y$ components of $\vec {F_3}$ force, in which $F_3=15\;kN$. Using proportional parts of similar triangles and considering the directions, we have
$\frac{F_{3x}}{15\;kN}=\frac{4}{5}$
or, $F_{3x}=15\;kN\Big(\frac{4}{5}\Big)=12\;kN$
Similarly,
$\frac{F_{3y}}{15\;kN}=\frac{3}{5}$
or, $F_{3y}=15\;kN\Big(\frac{3}{5}\Big)=9\;kN$
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, $(F_R)_x=(12+0+12)\;kN=24\;kN$
Summing the y components yields
$+\uparrow (F_R)_y=\sum F_y$
or, $(F_R)_y=(-9+20+9)\;kN=20\;kN$
$\therefore$ The resultant force has a magnitude of
$F_R=\sqrt {[(F_R)_x]^2+[(F_R)_y]^2}$
or, $F_R=\sqrt {(24)^2+(20)^2}\;kN$
or, $F_R\approx 31.24\;kN$
The direction $\theta$ of resultant force $\vec F_R$, measured counterclockwise from the positive $x$ axis, is
$\theta=\tan^{-1}\Big(\frac{20\;kN}{24\;kN}\Big)\approx 39.8^\circ$
