Given to find the side of right angle triangle.
use the formula,
[tex]\sin \theta=\frac{opposite\text{ }}{\text{hypotenouse}}[/tex][tex]\begin{gathered} \sin 30^{\circ}=\frac{\sqrt[]{2}}{y} \\ \frac{1}{2}=\frac{\sqrt[]{2}}{y} \\ y=2\sqrt[]{2} \end{gathered}[/tex][tex]\begin{gathered} \sin 60^{\circ}=\frac{x}{2\sqrt[]{2}} \\ \frac{\sqrt[]{3}}{2}=\frac{x}{2\sqrt[]{2}} \\ 2x=2\sqrt[]{2}\cdot\sqrt[]{3} \\ x=\sqrt[]{6} \end{gathered}[/tex][tex]\begin{gathered} \tan 60=\frac{x}{\sqrt[]{2}} \\ \sqrt[]{3}=\frac{x}{\sqrt[]{2}} \\ x=\sqrt[]{6} \end{gathered}[/tex]
so the answer is,
[tex]x=\sqrt[]{6}[/tex]
