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Right △EFG has its right angle at F, EG=6, and FG=4.

What is the value of the trigonometric ratio of an angle of the triangle?

Drag a value to each box to match the trigonometric ratio with its value.
cscE
cosG
cotG

Answer :

Please find the attachment.

We have been given that △EFG has a right angle at F, EG=6, and FG=4. We are asked to find trigonometric ratio for given angles of the triangle.

First of all, we will draw a right triangle using our given information.

Now we will find length of side EF using Pythagoras theorem.  

[tex]EF^2=EG^2-FG^2[/tex]

[tex]EF^2=6^2-4^2[/tex]

[tex]EF^2=36-16[/tex]

[tex]EF^2=20[/tex]

[tex]EF=\sqrt{20}=2\sqrt{5}[/tex]

We know that cosecant relates hypotenuse with opposite side of right triangle.

[tex]\csc=\frac{\text{Hypotenuse}}{\text{Opposite}}[/tex]

We can see that opposite side to angle E is FG and hypotenuse is EG.

[tex]\csc(E)=\frac{6}{4}[/tex]

[tex]\csc(E)=\frac{3}{2}[/tex]

We know that cosine relates adjacent side with hypotenuse.

[tex]\cos=\frac{\text{Adjacent}}{\text{Hypotenuse}}[/tex]

We can see that adjacent side to angle G is FG and hypotenuse is EG.

[tex]\cos(G)=\frac{4}{6}[/tex]

[tex]\cos(G)=\frac{2}{3}[/tex]

We know that cotangent relates adjacent side with opposite side of right triangle.

[tex]\cot=\frac{\text{Adjacent}}{\text{Opposite}}[/tex]

We can see that adjacent side to angle G is FG and opposite side is EG.

[tex]\cot(G)=\frac{4}{2\sqrt{5}}[/tex]

[tex]\cot(G)=\frac{2}{\sqrt{5}}[/tex]

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