first we will find the missing side with the pythagorean theorem:
[tex]\begin{gathered} a^2+b^2=c^2 \\ a^2+4^2=5^2 \\ a^2+16=25 \\ a^2+16-16=25-16 \\ a^2=9 \\ a=\sqrt[]{9}=3 \end{gathered}[/tex]for sin x
[tex]\sin x=\frac{opposite}{\text{hypotenuse}}=\frac{3}{5}[/tex]for cos x
[tex]\cos x=\frac{adjacent}{hypotenuse}=\frac{4}{5}[/tex]for tan x
[tex]\tan x=\frac{opposite}{adjacent}=\frac{3}{4}[/tex]