Since the angle is in standard position we can, we draw a right triangle for this point.
See diagram.
Using Pythagoras Theorem, we find the hypotenuse
[tex]h^{2} =3^{2} +4^{2} [/tex]
[tex]\Rightarrow h^{2} =9 +16 [/tex]
[tex]\Rightarrow h^{2} =25 [/tex]
[tex]\Rightarrow h=\sqrt{25} [/tex]
[tex]\Rightarrow h= 5 \ units[/tex]
Now recall the mnemonics SOH CAH TOA
[tex]\sin(\theta)=\frac{opposite}{hypotenuse} [/tex]
[tex]\Rightarrow \sin(\theta)=\frac{4}{5} [/tex]
[tex]\cos(\theta)=\frac{adjacent}{hypotenuse} [/tex]
[tex]\Rightarrow \cos(\theta)=\frac{3}{5} [/tex]
[tex]
\tan(\theta)=\frac{opposite}{adjacent} [/tex]
[tex]
\Rightarrow \tan(\theta)=\frac{4}{3}[/tex]
