Answer
$$142^\circ $$
Work Step by Step
$$\eqalign{
& {\text{Let the vectors }}\left\langle {3, - 2} \right\rangle {\text{ and }}\left\langle { - 1,3} \right\rangle \cr
& {\text{Geometric interpretation of the dot product}} \cr
& \cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{u}} \right|\left| {\bf{b}} \right|}} \cr
& {\text{Substitute the given vectors}} \cr
& \cos \theta = \frac{{\left\langle {3, - 2} \right\rangle \cdot \left\langle { - 1,3} \right\rangle }}{{\left| {\left\langle {3, - 2} \right\rangle } \right|\left| {\left\langle { - 1,3} \right\rangle } \right|}} \cr
& {\text{Simplify}} \cr
& \cos \theta = \frac{{\left( 3 \right)\left( { - 1} \right) + \left( { - 2} \right)\left( 3 \right)}}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( { - 2} \right)}^2}} \cdot \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 3 \right)}^2}} }} \cr
& \cos \theta = \frac{{ - 9}}{{\sqrt {13} \cdot \sqrt {10} }} \cr
& {\text{Use a calculator}} \cr
& \cos \theta \approx - 0.7893 \cr
& {\text{Use the inverse cosine function}}. \cr
& \theta \approx 142^\circ \cr} $$
