Answer
Using scalar notion:
$F_{1x}=0$
$F_{1y}=300\uparrow$
$F_{2x}=225\sqrt 2\;N\leftarrow$
$F_{2y}=225\sqrt 2\;N\uparrow$
$F_{3x}=360\;N\rightarrow$
$F_{3y}=480\;N\uparrow$
Work Step by Step
By the parallelogram law, we can resolve each force into its $x$ and $y$ components.
Let $F_{1x}$ and $F_{1y}$ are the $x$ and $y$ components of $\vec {F_1}$ force. Using scalar notation to represent these components, we have
$F_{1x}=300\cos90^\circ\;N=0$
$F_{1y}=300\sin90^\circ\;N=300\;N=300\uparrow$
Let $F_{2x}$ and $F_{2y}$ are the $x$ and $y$ components of $\vec {F_2}$ force. Using scalar notation to represent these components, we have
$F_{2x}=-450\cos45^\circ\;N=-225\sqrt 2\;N=225\sqrt 2\;N\leftarrow$
$F_{2y}=450\sin45^\circ\;N=225\sqrt 2\;N=225\sqrt 2\;N\uparrow$
Let $F_{3x}$ and $F_{3y}$ are the $x$ and $y$ components of $\vec {F_3}$ force. Using proportional parts of similar triangles, we have
$\frac{F_{3x}}{600\;N}=\frac{3}{5}$
or, $F_{3x}=600\;N\Big(\frac{3}{5}\Big)=360\;N$
Similarly,
$\frac{F_{3y}}{600\;N}=\frac{4}{5}$
or, $F_{3y}=600\;N\Big(\frac{4}{5}\Big)=480\;N$
Using scalar notation to represent these components, we have
$F_{3x}=480\;N=360\;N\rightarrow$
$F_{3y}=480\;N=480\;N\uparrow$
