Answer: The correct option is tan(F).
Explanation:
In the given figure the triangle is a right angle triangle because [tex]\angle E=90^{\circ}[/tex].
It is also given that [tex]DE=EF=5[/tex] and [tex]DF=5\sqrt{2}[/tex].
Since it is an isosceles right angle triangle, therefore the value of perpendicular and base is same for both angles D and F, which is 5.
[tex]\cos (F)=\frac{Base}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}[/tex]
The value of cos(F) is [tex]\frac{1}{\sqrt{2}}[/tex].
[tex]\sin (F)=\frac{Perpendicular}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}[/tex]
The value of sin(F) is [tex]\frac{1}{\sqrt{2}}[/tex].
[tex]\sin (D)=\frac{Perpendicular}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}[/tex]
The value of sin(D) is [tex]\frac{1}{\sqrt{2}}[/tex].
[tex]\tan (F)=\frac{Perpendicular}{Base}=\frac{5}{5} =1[/tex]
The value of tan(F) is 1. Which is not equal to [tex]\frac{1}{\sqrt{2}}[/tex].
[tex]\cos (D)=\frac{Base}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}[/tex]
The value of cos(D) is [tex]\frac{1}{\sqrt{2}}[/tex].
Therefore, the value of tan(F) is not equal to the value of cos(F), so the correct option is third.
