Answer:
[tex]sec\ E = \frac{4\sqrt{7} }{7}[/tex]
[tex]Cos\ F = \frac{3}{4}[/tex]
[tex]Tan\ F =\frac{\sqrt{7}}{3}[/tex]
Step-by-step explanation:
Given
EF = 8
FG = 6
We need to find the trigonometric ratios.
Solution:
First we will find the length of the third side.
Now we know that;
△EFG is a right angled triangle with right angle at G.
Now applying Pythagoras theorem which states.
"The sum of square of the two legs of the triangle is equal to square of the hypotenuse."
so we can say that;
[tex]FG^2=EF^2+EG^2\\\\EG^2=FG^2-EF^2[/tex]
Substituting the given values we get;
[tex]EG^2=8^2-6^2=64-36=28[/tex]
Taking square roots on both side we get;
[tex]\sqrt{EG^2} =\sqrt{28}=\sqrt{4\times7}\\ \\EG = 2\sqrt{7}[/tex]
Now we will find the trigonometric values.
[tex]secE=\frac{Hypotenuse}{Adjacent\ side}[/tex]
Here Hypotenuse = EF = 8
Adjacent side of E = EG = [tex]2\sqrt{7}[/tex]
[tex]secE=\frac{8}{2\sqrt{7}} =\frac{4}{\sqrt{7}}[/tex]
Now rationalizing the denominator by multiplying numerator and denominator by [tex]\sqrt{7}[/tex] we get;
[tex]secE=\frac{4\times \sqrt{7} }{\sqrt{7}\times \sqrt{7} }\\\\secE = \frac{4\sqrt{7} }{7}[/tex]
Now,
[tex]Cos F = \frac{Adjacent \ side}{Hypotenuse}[/tex]
Adjacent side to F =GF = 6
Hypotenuse = EF = 8
[tex]Cos\ F = \frac{6}{8}\\\\Cos\ F = \frac{3}{4}[/tex]
Now,
[tex]Tan F = \frac{opposite \ side}{adjacent\ side}[/tex]
Here Opposite side of F = EG = [tex]2\sqrt{7}[/tex]
Adjacent side of F = GF = 6
[tex]Tan\ F= \frac{2\sqrt{7}}{6}\\\\Tan\ F =\frac{\sqrt{7}}{3}[/tex]
Hence Below are required details.
[tex]sec\ E = \frac{4\sqrt{7} }{7}[/tex]
[tex]Cos\ F = \frac{3}{4}[/tex]
[tex]Tan\ F =\frac{\sqrt{7}}{3}[/tex]
