as it is given that
[tex]r_x = 16 m[/tex]
[tex]r_y = -8.5 m[/tex]
now we will have
[tex]\vec r = 16 \hat i - 8.5 \hat j[/tex]
now the magnitude of this vector is given as
[tex]|r| = \sqrt{16^2 + 8.5^2}[/tex]
[tex]|r| = 18 m[/tex]
now to find the direction we can use
[tex]tan\theta = \frac{r_y}{r_x}[/tex]
[tex]tan\theta = \frac{-8.5}{16}[/tex]
[tex]\theta = tan^{-1}(-0.53)[/tex]
[tex]\theta = - 28^0[/tex]
