Given data:
* The magnitude of the vector C is,
[tex]\begin{gathered} C=2\times2 \\ C=4\text{ N} \end{gathered}[/tex]* The magnitude of the vector B is,
[tex]\begin{gathered} B=4\times2 \\ B=8\text{ N} \end{gathered}[/tex]Solution:
From the given diagram, the magnitude of the vector A is,
[tex]\begin{gathered} A=\sqrt[]{Base^2+Perpendicular^2} \\ A=\sqrt[]{(3\times2)^2+(4\times2)^2} \\ A=\sqrt[]{6^2+8^2} \\ A=\sqrt[]{36+64} \\ A=\sqrt[]{100} \\ A=10\text{ N} \end{gathered}[/tex]The count of square grid in the hypotenuse is,
[tex]\begin{gathered} n=\frac{A}{2} \\ n=\frac{10}{2} \\ n=5 \end{gathered}[/tex]The angle of the vector A with the x-axis is,
[tex]\begin{gathered} cos(\theta)=\frac{Base}{\text{Hypotenuse}} \\ \cos (\theta)=\frac{3}{5} \\ \theta=53.13^{\circ} \end{gathered}[/tex]Thus, the value of vector A is,
[tex]\begin{gathered} \vec{A}=A\cos (53.13^{\circ})+A\sin (53.13^{\circ}) \\ \vec{A}=10\times\cos (53.13^{\circ})i+10\times\sin (53.13^{\circ})j \\ \vec{A}=6\text{ i + 8 j} \end{gathered}[/tex]The value of vector B is,
[tex]\vec{B}=-8\text{ j}[/tex]The value of vector C is,
[tex]\vec{C}=-4\text{ i}[/tex]Thus, the sum of the vectors is,
[tex]\begin{gathered} \vec{A}+\vec{B}+\vec{C}=6\text{ i+8 j-8 j-4 i} \\ \vec{A}+\vec{B}+\vec{C}=2\text{ i} \\ |\vec{A}+\vec{B}+\vec{C}|=\sqrt[]{2^2} \\ |\vec{A}+\vec{B}+\vec{C}|=\text{ 2 N} \end{gathered}[/tex]Thus, the magnitude of the sum of three given vectors is 2 N towards the east (positive of the x-axis).