Find the direction of u=-3i+8j (round to the nearest tenth)

[tex]\bf u=-3i+8j\implies u=\ \textless \ \stackrel{a}{-3}~~,~~\stackrel{b}{8}\ \textgreater \ \qquad \theta =tan^{-1}\left( \frac{b}{a} \right) \\\\\\ \theta =tan^{-1}\left( \frac{8}{-3} \right)\implies \theta \approx -69.4^o[/tex]

keeping in mind that the range of the inverse tangent function is from π/2 to -π/2, that gives us an angle in the IV Quadrant, however, notice "a" and "b", "a" is negative and "b" is positive, that means the II Quadrant.

so we're looking for a reference angle of 69.4° on the II Quadrant, and that'd be 180° - 69.4°, or 110.6°.