SOLUTION
Let's illustrate this with a diagram
Now we will use SOHCAHTOA to solve this
We have just the adjacent side and the hypotenuse
[tex]\begin{gathered} \sin 60^o=\frac{opposite}{\text{hypotenuse }} \\ \\ \sin 60^o=\frac{x\sqrt[]{3}}{y} \\ \\ y\sin 60^o=x\sqrt[]{3} \\ y\text{ = }\frac{x\sqrt[]{3}}{\frac{\sqrt[]{3}}{2}} \\ \\ y=x\sqrt[]{3}\times\frac{2}{\sqrt[]{3}} \\ \\ \sqrt[]{3}\text{ cancels out} \\ \\ y=x\times2 \\ y=2x \end{gathered}[/tex]
Therefore, the hypotenuse is 2x
