Given the right triangle:
Taking the cosine of the angle θ:
[tex]\cos \theta=\frac{2\sqrt[]{6}}{2\sqrt[]{15}}=\frac{\sqrt[]{2}}{\sqrt[]{5}}[/tex]Now, taking the arc cosine to find θ:
[tex]\begin{gathered} \theta=\arccos (\sqrt[]{\frac{2}{5}}) \\ \therefore\theta\approx50.8\degree^{} \end{gathered}[/tex]