Solution:
Given:
The sketch of the two vectors can be made as shown below;
Resolving the forces into the horizontal and vertical components,
[tex]\begin{gathered} \text{For vector u,} \\ 30\cos 70i+30\sin 70j \\ \\ \text{For vector v,} \\ -40\cos 40i-40\sin 40j \end{gathered}[/tex]
Hence, the resultant vector (u + v) is;
[tex]\begin{gathered} u+v=(30\cos 70-40\cos 40)i+(30\sin 70-40\sin 40)j \\ u+v=-20.38i+2.4793j \\ \\ \text{where,} \\ F_x=-20.38 \\ F_y=2.4793 \end{gathered}[/tex]
The direction of u + v is given by;
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{F_y}{F_x})^{}_{} \\ \theta=\tan ^{-1}(\frac{2.4793}{-20.38}) \\ \theta=-6.936 \end{gathered}[/tex]
Since the tangent of the angle is negative, then it falls in the second or fourth quadrant.
In the second quadrant,
[tex]\begin{gathered} \theta=-6.936+180 \\ \theta=173.064^0 \\ \\ To\text{ the nearest degre}e,\text{ the direction of u + v is 173 degr}ees \end{gathered}[/tex]
Therefore, the direction of u + v to the nearest degree is 173 degrees.
