Respuesta :

DeanR

[tex]\sin \angle FHG = \dfrac{FG}{FH}[/tex]


[tex]FH = \dfrac{FG}{\sin \angle FHG}[/tex]


[tex]EF^2 + FH^2 = EH^2[/tex]


[tex]\cos \angle FHE = \dfrac{FH}{EH}[/tex]


We have all the pieces we need.


[tex]FH = \dfrac{FG}{\sin \angle FHG} = \dfrac{\sqrt{8}}{1/\sqrt{2}}=\sqrt{16}=4[/tex]


[tex]EH= \sqrt{ EF^2 + FH^2} = \sqrt{3^2 + 4^2}=5[/tex]


[tex]\cos \angle FHE = \dfrac{FH}{EH} = \dfrac 4 5[/tex]


It's nice when the numbers work out.



Answer:

\cos \angle FHE = \dfrac{FH}{EH} = \dfrac 4 5

It's nice when the numbers work out.

Step-by-step explanation: