Answer
$$\eqalign{
& \left( {\bf{a}} \right){\text{ In radians}} \cr
& \theta = \frac{\pi }{{12}}{\text{rad}} \cr
& \left( {\bf{b}} \right){\text{ In degrees}} \cr
& \theta = 15^\circ \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let the vectors be }} \cr
& {\bf{u}} = 5\left[ {\cos \left( {\frac{{3\pi }}{4}} \right){\bf{i}} + \sin \left( {\frac{{3\pi }}{4}} \right){\bf{j}}} \right] = - \frac{{5\sqrt 2 }}{2}{\bf{i}} + \frac{{5\sqrt 2 }}{2}{\bf{j}} \cr
& {\bf{v}} = 2\left[ {\cos \left( {\frac{{2\pi }}{3}} \right){\bf{i}} + \sin \left( {\frac{{2\pi }}{3}} \right){\bf{j}}} \right] = - {\bf{i}} + \sqrt 3 {\bf{j}} \cr
& {\bf{u}} \cdot {\bf{v}} = \left( { - \frac{{5\sqrt 2 }}{2}{\bf{i}} + \frac{{5\sqrt 2 }}{2}{\bf{j}}} \right) \cdot \left( { - {\bf{i}} + \sqrt 3 {\bf{j}}} \right) \cr
& {\bf{u}} \cdot {\bf{v}} = \frac{{5\sqrt 2 }}{2} + \frac{{5\sqrt 6 }}{2} \cr
& \left\| {\bf{u}} \right\| = \left\| { - \frac{{5\sqrt 2 }}{2}{\bf{i}} + \frac{{5\sqrt 2 }}{2}{\bf{j}}} \right\| = \sqrt {\frac{{25}}{2} + \frac{{25}}{2}} = 5 \cr
& \left\| {\bf{v}} \right\| = \left\| { - {\bf{i}} + \sqrt 3 {\bf{j}}} \right\| = \sqrt {1 + 3} = 2 \cr
& {\text{Find the Angle Between Two Vectors}} \cr
& \cos \theta = \frac{{{\bf{u}} \cdot {\bf{v}}}}{{\left\| {\bf{u}} \right\|\left\| {\bf{v}} \right\|}} \cr
& \cos \theta = \frac{{\frac{{5\sqrt 2 }}{2} + \frac{{5\sqrt 6 }}{2}}}{{\left( 5 \right)\left( 2 \right)}} \cr
& \theta = {\cos ^{ - 1}}\left( {\frac{{\sqrt 6 + \sqrt 2 }}{4}} \right) = \frac{\pi }{{12}}{\text{rad}} = 15^\circ \cr
& \cr
& \left( {\bf{a}} \right){\text{ In radians}} \cr
& \theta = \frac{\pi }{{12}}{\text{rad}} \cr
& \left( {\bf{b}} \right){\text{ In degrees}} \cr
& \theta = 15^\circ \cr} $$
