The speed is the magnitude of the vector given by:
[tex]v=\sqrt[]{v^2_x+v^2_y_{}}_{}[/tex]Plugging the components of the vector we have:
[tex]\begin{gathered} v=\sqrt[]{(5)^2+(-12)^2} \\ v=\sqrt[]{25+144} \\ v=\sqrt[]{169} \\ v=13 \end{gathered}[/tex]Now to find the direction we need to notice that the vector lies on the second quadrant of the coordinate plane, this means that the formula:
[tex]\theta=\tan ^{-1}(\frac{v_y}{v_x})[/tex]does not give the angle we are looking for but the supplementary angle.
Then the direction is given as:
[tex]180-\tan ^{-1}(\frac{v_y}{v_x})[/tex]Plugging the values given we have:
[tex]180-\tan ^{-1}(\frac{12}{5})=113[/tex]Therefore the spped of the plane is 13 and the direction is 113°
