Given,
The magnitude of vector A, A=4.00 units
The magnitude of vector B, B=5.40 units
The angle between A and B, θ=60.0°
The magnitude of the sum of two vectors is given by,
[tex]|\vec{A}+\vec{B}|=\sqrt[]{A^2+B^2+2AB\cos \theta}[/tex]
On substituting the known values,
[tex]\begin{gathered} |\vec{A}+\vec{B}|=\sqrt[]{4.00^2+5.40^2+2\times4.00\times5.40\times\cos 60^{\circ}} \\ =8.17\text{ units} \end{gathered}[/tex]
Therefore, the magnitude of the sum of the two given vectors is 8.17 units
