Let's start by sketching the triangle:
For the trignometric rates, we have:
[tex]\begin{gathered} \sin \alpha=\frac{\text{opposite leg}}{\text{hypotenuse}} \\ \cos \alpha=\frac{\text{adjacent leg}}{\text{hypotenuse}} \\ \tan \alpha=\frac{\text{opposite leg}}{\text{adjacent leg}} \\ \sec \alpha=\frac{\text{hypotenuse}}{\text{adjacent leg}} \\ \csc \alpha=\frac{\text{hypotenuse}}{\text{opposite leg}} \\ \cot \alpha=\frac{\text{adjacent leg}}{\text{opposite leg}} \end{gathered}[/tex]
So firts, let's find the missing side using Pythagora's Theorem:
[tex]\begin{gathered} FG^2+EF^2=EG^2 \\ 4^2+EF^2=10^2 \\ EF^2=100-16 \\ EF=\sqrt[]{84}=\sqrt[]{4\cdot21}=2\sqrt[]{21} \end{gathered}[/tex]
For E, the opposite leg is FG and the adjancet leg is EF, so:
[tex]\cos E=\frac{EF}{EG}=\frac{2\sqrt[]{21}}{10}=\frac{\sqrt[]{21}}{5}[/tex]
For G, the opposite leg is EF and the adjacent leg is FG, so:
[tex]\sec G=\frac{EG}{FG}=\frac{10}{4}=\frac{5}{2}[/tex]
And:
[tex]\tan G=\frac{EF}{FG}=\frac{2\sqrt[]{21}}{4}=\frac{\sqrt[]{21}}{2}[/tex]
