The angle a vector [tex](a\hat{i}+b\hat{j})[/tex] makes with positive direction of x axis is given by
[tex]\theta = tan^{-1} (\frac{b}{a})[/tex]
Here a and b are the components of vector [tex]\vec{F}[/tex] along x and y axis and [tex]\theta[/tex] be the angle made by vector with x axis.
Also the magnitude resultant of a vector of the form [tex]a\hat{i}+b\hat{j}[/tex] is
[tex]|\vec{F}| = \sqrt{a^2+b^2}[/tex]
Replacing with our values we have that
[tex]|\vec{F}| = \sqrt{(-2980)^2+(8200)^2}[/tex]
[tex]|\vec{F}| = 8724.7N[/tex]
Therefore the magnitude of the given force vector is 8724.7N
For the given vector substitute -2980N for a and 8200 N for b in the equation of angle of vector from positive direction of x axis
[tex]\theta = tan^{-1} (\frac{8200}{-2980})[/tex]
[tex]\theta = -70.01\°[/tex]
Therefore the angle the force vector makes is at -70.01° from east
