Answer
$\vec {F_1}=(-15\hat i-15\sqrt 3\hat j)\;kN$
$\vec {F_2}=(-10\hat i+24\hat j)\;kN$
Work Step by Step
Having determined the magnitudes and directions of the components of each force, we can express each force as a Cartesian vector.
$\vec{F_1}=(ā30\sin30^\circ\hat iā30\cos30^\circ\hat j)\;kN$
or, $\vec {F_1}=(-15\hat i-15\sqrt 3\hat j)\;kN$
and
$\vec{F_2}=\Big\{-26\Big(\frac{5}{13}\Big)\hat i+26\Big(\frac{12}{13}\Big)\hat j\Big\}\;kN$
or, $\vec {F_2}=(-10\hat i+24\hat j)\;kN$
where $\hat i$ and $\hat j$ are the unit vectors along positive $x$ and $y$ axes respectively.
