To solve the question we need to draw a sketch for the given triangle
Triangle EFG is a right triangle at F
FG = 6
EG = 8
Let us revise the trigonometry ratios
[tex]\begin{gathered} sin(x)=\frac{opposite}{hypotenuse} \\ cos(x)=\frac{adjacent}{hypotenuse} \\ tan(x)=\frac{opposite}{adjacent} \end{gathered}[/tex][tex]\begin{gathered} sec(x)=\frac{1}{cos(x)}=\frac{hypotenuse}{adjacent} \\ csc(x)=\frac{1}{sin(x)}=\frac{hypotenuse}{opposite} \\ cot(x)=\frac{1}{tan(x)}=\frac{adjacent}{opposite} \end{gathered}[/tex]
We want to find cot G, then we need to find the adjacent side and the opposite side of Since the opposite side of Since the adjacent side of
Then we need to find EF using Pythagoras' relation
[tex]\begin{gathered} EG^2=FG^2+EF^2 \\ 8^2=6^2+EF^2 \\ 64=36+EF^2 \\ 64-36=EF^2 \\ 28=EF^2 \\ \sqrt{28}=EF \\ 2\sqrt{7}=EF \end{gathered}[/tex]
Now, we can find cot(G)
[tex]cot(G)=\frac{FG}{EF}=\frac{6}{2\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}=\frac{3\sqrt{7}}{7}[/tex]
It is the 4th choice from left to right
[tex]sin(E)=\frac{FG}{EG}=\frac{6}{8}=\frac{\frac{6}{2}}{\frac{8}{2}}=\frac{3}{4}[/tex]
It is the last choice from left to right
[tex]sec(G)=\frac{EG}{FG}=\frac{8}{6}=\frac{\frac{8}{2}}{\frac{6}{2}}=\frac{4}{3}[/tex]
It is the first choice from left to right
