We can use the sine function and tangent function in order to solve for x and y:
[tex]\begin{gathered} sin(\theta)=\frac{opposite}{hypotenuse} \\ sin(60)=\frac{15}{x} \\ x=\frac{15}{sin(60)} \\ x=10\sqrt{3} \end{gathered}[/tex][tex]\begin{gathered} tan(\theta)=\frac{opposite}{adjacent} \\ tan(60)=\frac{15}{y} \\ y=\frac{15}{tan(60)} \\ y=5\sqrt{3} \end{gathered}[/tex]
