Given:
The vector A is 70 m 50 deg north of east
The vector B is 40 m 80 deg north of east
To find the magnitude and direction(with respect to the positive x-axis) of the resultant.
Explanation:
The vectors can be represented in the diagram as shown below
The x-component of the resultant will be
[tex]\begin{gathered} R_x=70cos(50^{\circ})+40cos(80^{\circ}) \\ =51.941\text{ m} \end{gathered}[/tex]
The y-component of the resultant will be
[tex]\begin{gathered} R_y=70sin(50^{\circ})+40sin(80^{\circ}) \\ =93.015\text{ m} \end{gathered}[/tex]
The magnitude of the resultant can be calculated as
[tex]\begin{gathered} R=\sqrt{R_x+R_y} \\ =\sqrt{(51.941)^2+(93.015)^2} \\ =106.535\text{ m} \end{gathered}[/tex]
The direction can be calculated as
[tex]\begin{gathered} \theta\text{ =tan}^{-1}(\frac{R_y}{R_x}) \\ =\text{ tan}^{-1}(\frac{93.015}{51.941}) \\ =\text{ 60.82}^{\circ} \end{gathered}[/tex]
Thus, option D is the correct option.
