Answer
$${\text{17;}}\,\,{331.9^ \circ }$$
Work Step by Step
$$\eqalign{
& \left\langle {15, - 8} \right\rangle \cr
& {\text{The magnitude of the vector is }} \cr
& \left| {\left\langle {15, - 8} \right\rangle } \right| = \sqrt {{{\left( {15} \right)}^2} + {{\left( { - 8} \right)}^2}} \cr
& \left| {\left\langle {15, - 8} \right\rangle } \right| = \sqrt {289} \cr
& \left| {\left\langle {15, - 8} \right\rangle } \right| = 17 \cr
& {\text{The direction of the vector is }} \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{{ - 8}}{{15}}} \right) + {360^ \circ } \cr
& \theta \approx {331.9^ \circ } \cr} $$
