Answer
$$45^\circ $$
Work Step by Step
$$\eqalign{
& {\text{Let the vectors }}\left\langle {0,4} \right\rangle {\text{ and }}\left\langle { - 4,4} \right\rangle \cr
& {\text{Geometric interpretation of the dot product}} \cr
& \cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left| {\bf{u}} \right|\left| {\bf{b}} \right|}} \cr
& {\text{Substitute the given vectors}} \cr
& \cos \theta = \frac{{\left\langle {0,4} \right\rangle \cdot \left\langle { - 4,4} \right\rangle }}{{\left| {\left\langle {0,4} \right\rangle } \right|\left| {\left\langle { - 4,4} \right\rangle } \right|}} \cr
& {\text{Simplify}} \cr
& \cos \theta = \frac{{\left( 0 \right)\left( { - 4} \right) + \left( 4 \right)\left( 4 \right)}}{{\sqrt {{{\left( 0 \right)}^2} + {{\left( 4 \right)}^2}} \cdot \sqrt {{{\left( { - 4} \right)}^2} + {{\left( 4 \right)}^2}} }} \cr
& \cos \theta = \frac{{16}}{{\sqrt {16} \cdot \sqrt {16 + 16} }} \cr
& \cos \theta = \frac{4}{{\sqrt {32} }} \cr
& \cos \theta = \frac{1}{{\sqrt 2 }} \cr
& {\text{Therefore,}} \cr
& \theta = 45^\circ \cr} $$
