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.
Step-by-step explanation:
The vectors can be represented in the diagram as shown below
The x-component of the resultant will be
![\begin{gathered} R_x=70cos(50^(\circ))+40cos(80^(\circ)) \\ =51.941\text{ m} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/vqdrqfcqjcl0rpatvoz8myjzf17fcqsnvh.png)
The y-component of the resultant will be
![\begin{gathered} R_y=70sin(50^(\circ))+40sin(80^(\circ)) \\ =93.015\text{ m} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/cv0aoqyhznpu60b8b7sbh9os9p11lsnhs1.png)
The magnitude of the resultant can be calculated as
![\begin{gathered} R=√(R_x+R_y) \\ =√((51.941)^2+(93.015)^2) \\ =106.535\text{ m} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/8y0l2dcrjels296p9oavcukou9bdkfre88.png)
The direction can be calculated as
![\begin{gathered} \theta\text{ =tan}^(-1)((R_y)/(R_x)) \\ =\text{ tan}^(-1)((93.015)/(51.941)) \\ =\text{ 60.82}^(\circ) \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/a0ln85m97c1nd53cuakqatjj2itwi1aeu1.png)
Thus, option D is the correct option.