Respuesta :

For this problem we first use the pythagorean theorem to find QH

[tex]\begin{gathered} QH^2+HE^2=QE^2 \\ QH^2=QE^2-HE^2=101^2-99^2=400 \\ QH=20 \end{gathered}[/tex]

Then

[tex]\sin (E)\text{ =}\frac{QH}{QE}=\frac{20}{101}[/tex]