Answer
$\vec {F_1}=(9.64\hat i+11.49\hat j)\;kN$
$\vec {F_2}=(-24\hat i+10\hat j)\;kN$
$\vec {F_3}=(31.18\hat i-18\hat j)\;kN$
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
$F_{3x}=36\cos30^\circ\;kN\approx 31.18\;kN$
$F_{3y}=-36\sin30^\circ\;kN\approx -18\;kN$
Having determined the magnitudes and directions of the components of each force, we can express each force as a Cartesian vector.
$\vec {F_1}=(9.64\hat i+11.49\hat j)\;kN$
$\vec {F_2}=(-24\hat i+10\hat j)\;kN$
$\vec {F_3}=(31.18\hat i-18\hat j)\;kN$
where $\hat i$ and $\hat j$ are the unit vectors along positive $x$ and $y$ axes.
