let's start by drawing a diagram of the situation
In order to find the magnitud of the resulting force we need to find the forces towards the west direction
Then
[tex]\begin{gathered} F_x=F_{1x}+F_{2x} \\ \end{gathered}[/tex][tex]\begin{gathered} F_{1x}=1610\cdot\sin 30=1610\cdot\frac{1}{2}=805 \\ \end{gathered}[/tex][tex]F_{2x}=1250\cdot\sin 55=1023.94[/tex]
Thus,
[tex]F_x=805+1023.94=1828.94[/tex]
In the y axis we have:
[tex]\begin{gathered} F_y=F_{1y}+F_{2y} \\ F_y=1610\cos 30+1250\cos 55 \\ F_y=1394.30+716.97 \\ F_y=2111.27 \end{gathered}[/tex]
Then, finally, we add both forces as follows
[tex]F=\sqrt[]{(F_x)^2+(F_y)^2_{}}_{}[/tex]
replacing
[tex]\begin{gathered} F=\sqrt[]{1828.94^2+2111.27^2} \\ F=2793.29 \end{gathered}[/tex]
Then the magnitud of the resulting forces in x and y directions is: 2793.29 kg
