Respuesta :

[tex]\\ \rm\rightarrowtail sin45=\dfrac{x}{12}[/tex]

[tex]\\ \rm\rightarrowtail \dfrac{1}{\sqrt{2}}=\dfrac{x}{12}[/tex]

[tex]\\ \rm\rightarrowtail x=\dfrac{12}{\sqrt{2}}[/tex]

[tex]\\ \rm\rightarrowtail x=6\sqrt{2}ft[/tex]

Now

[tex]\\ \rm\rightarrowtail tan60=\dfrac{y}{6\sqrt{2}}[/tex]

[tex]\\ \rm\rightarrowtail y=6\sqrt{6}[/tex]