Let's begin by identifying key information given to us:
We have one known angle & one side
[tex]\begin{gathered} \theta=60^{\circ} \\ adjacent(a)=2\sqrt[]{6} \\ opposite(b)=\text{?} \\ hypotenuse(c)=\text{?} \end{gathered}[/tex]
Based on the information we have been provided, we will solve for the missing sides using Trigonometric Ratio (SOHCAHTOA). This is shown below:
[tex]\begin{gathered} TOA\Rightarrow\tan \theta=\frac{opposite}{adjacent} \\ tan\theta=\frac{opposite}{adjacent} \\ tan60^{\circ}=\frac{opposite}{2\sqrt[]{6}} \\ But,tan60^{\circ}=\sqrt[]{3} \\ opposite=2\sqrt[]{6}\text{ x }\sqrt[]{3} \\ opposite=2\sqrt[]{18} \\ \\ CAH\Rightarrow cos\theta=\frac{adjacent}{hypotenuse} \\ cos\theta=\frac{adjacent}{hypotenuse} \\ cos60^{\circ}=\frac{2\sqrt[]{6}}{hypotenuse} \\ But,\cos 60^{\circ}=\frac{1}{2} \\ hypotenuse\text{ x }\cos 60^{\circ}=2\sqrt[]{6} \\ hypotenuse=\frac{2\sqrt[]{6}}{\frac{1}{2}} \\ hypotenuse=\frac{2\cdot2\sqrt[]{6}}{1}=4\sqrt[]{6} \\ hypotenuse=4\sqrt[]{6} \end{gathered}[/tex]
