For the given triangle, we have that the trigonometric functions are defined as follows:
[tex]\begin{gathered} sinA=\frac{opposite\text{ side}}{hypotenuse}=\frac{1}{2}=0.5 \\ cosA=\frac{adjacent\text{ side}}{hypotenuse}=\frac{\sqrt{3}}{2}=0.86 \\ tanA=\frac{opposite\text{ side}}{adjacent\text{ side}}=\frac{1}{\sqrt{3}}=0.58 \end{gathered}[/tex]
