Given the information on the right triangle, we can use the cosine function to find the hypotenuse:
[tex]\begin{gathered} \cos (60)=\frac{\text{adjacent side}}{hypotenuse}=\frac{9}{x} \\ \Rightarrow\cos (60)=\frac{9}{x} \end{gathered}[/tex]
solving for x, we get:
[tex]\begin{gathered} \cos (60)=\frac{1}{2} \\ \Rightarrow\cos (60)=\frac{9}{x} \\ \Rightarrow\frac{1}{2}=\frac{9}{x} \\ \Rightarrow\frac{x}{2}=9 \\ \Rightarrow x=9\cdot2=18 \\ x=18 \end{gathered}[/tex]
next, we can use the tangent function to find y:
[tex]\begin{gathered} \tan (60)=\frac{\text{opposite side}}{adjacent\text{ side}}=\frac{y}{9} \\ \Rightarrow\tan (60)=\frac{y}{9} \end{gathered}[/tex]
doing the same for y, we get:
[tex]\begin{gathered} \tan (60)=\sqrt[]{3} \\ \Rightarrow\tan (60)=\frac{y}{9} \\ \Rightarrow\sqrt[]{3}=\frac{y}{9} \\ \Rightarrow y=9\cdot\sqrt[]{3} \end{gathered}[/tex]
therefore, x = 18 and y = 9*sqrt(3)
