Answer
$$90^\circ ;{\text{orthogonal}}$$
Work Step by Step
$$\eqalign{
& {\text{Let the vectors }}\left\langle {5, - 3} \right\rangle {\text{ and }}\left\langle {3,5} \right\rangle \cr
& {\text{Geometric interpretation of the dot product}} \cr
& \cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{u}} \right|\left| {\bf{b}} \right|}} \cr
& {\text{Substitute the given vectors}} \cr
& \cos \theta = \frac{{\left\langle {5, - 3} \right\rangle \cdot \left\langle {3,5} \right\rangle }}{{\left| {\left\langle {5, - 3} \right\rangle } \right|\left| {\left\langle {3,5} \right\rangle } \right|}} \cr
& {\text{Simplify}} \cr
& \cos \theta = \frac{{\left( 5 \right)\left( 3 \right) + \left( { - 3} \right)\left( 5 \right)}}{{\sqrt {{{\left( 5 \right)}^2} + {{\left( { - 3} \right)}^2}} \cdot \sqrt {{{\left( 3 \right)}^2} + {{\left( 5 \right)}^2}} }} \cr
& \cos \theta = \frac{0}{{\sqrt {25 + 9} \cdot \sqrt {9 + 25} }} \cr
& \cos \theta = 0 \cr
& {\text{Therefore,}} \cr
& \theta = 90^\circ \cr
& {\text{The vectors are orthogonal}} \cr} $$
