Here, we have a special right triangle.
Let's solve for the variables, x and y.
Given:
common side = x
Hypotenuse of the larger triangle = 8
Let's find x using trigonometric ratio.
We have:
[tex]\begin{gathered} \sin \theta=\frac{opposite}{hypotenuse} \\ \\ \sin 30=\frac{x}{8} \\ \\ x=8\sin 30 \\ \\ x=8(0.5) \\ \\ x=4 \end{gathered}[/tex]To solve for y, we have:
[tex]\begin{gathered} \tan \theta=\frac{opposite}{adjacent} \\ \\ \tan 60=\frac{x}{y} \\ \\ \tan 60=\frac{4}{y} \\ \\ \text{Multiply both sid}es\text{ by y:} \\ y\tan 60=\frac{4}{y}\ast y \\ \\ y\tan 60=4 \\ \\ \text{Divide both sides by tan60} \\ \\ \frac{y\tan 60}{\tan 60}=\frac{4}{\tan60} \\ \\ \\ y=\frac{4}{\tan 60} \end{gathered}[/tex]Solving further:
[tex]\begin{gathered} y=\frac{4}{\sqrt[]{3}} \\ \\ \end{gathered}[/tex]Multiply both numerator and denominator by √3:
[tex]\begin{gathered} y=\frac{4}{\sqrt[]{3}}\ast\frac{\sqrt[]{3}}{\sqrt[]{3}} \\ \\ y=\frac{4\sqrt[]{3}}{3} \end{gathered}[/tex]ANSWER:
[tex]\begin{gathered} x=4 \\ \\ y=\frac{4\sqrt[]{3}}{3} \end{gathered}[/tex]