Answer:
[tex]\begin{gathered} \sin A=\frac{\sqrt[]{6}}{3} \\ \cos A=\frac{\sqrt[]{3}}{3} \\ \sin B=\frac{\sqrt[]{3}}{3} \\ \cos B=\frac{\sqrt[]{6}}{3} \end{gathered}[/tex]Explanation:
Let x represent unknown side length
We can go ahead and find x using the Pythagorean Theorem as seen below;
[tex]\begin{gathered} (5\sqrt[]{3})^2=5^2+x^2 \\ (25\times3)=25+x^2 \\ 75-25=x^2 \\ 50=x^2 \\ x=\sqrt[]{50}=\sqrt[]{25\times2}=\sqrt[]{25}\times\sqrt[]{2}=5\sqrt[]{2} \\ x=5\sqrt[]{2} \end{gathered}[/tex]Let's find sin A as seen below;
[tex]\begin{gathered} \sin A=\frac{opposite}{\text{hypotenuse}}=\frac{5\sqrt[]{2}}{5\sqrt[]{3}} \\ \sin A=\frac{\sqrt[]{2}}{\sqrt[]{3}}=\frac{\sqrt[]{2}\times\sqrt[]{3}}{\sqrt[]{3}\times\sqrt[]{3}}=\frac{\sqrt[]{6}}{3} \\ \sin A=\frac{\sqrt[]{6}}{3} \end{gathered}[/tex]Let's find cos A as seen below;
[tex]\begin{gathered} \cos A=\frac{adjacent}{\text{hypotenuse}}=\frac{5}{5\sqrt[]{3}} \\ \cos A=\frac{1}{\sqrt[]{3}}=\frac{\sqrt[]{3}}{\sqrt[]{3}\times\sqrt[]{3}}=\frac{\sqrt[]{3}}{3} \\ \cos A=\frac{\sqrt[]{3}}{3} \end{gathered}[/tex]Let's find sin B as seen below;
[tex]\begin{gathered} \sin B=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{5}{5\sqrt[]{3}} \\ \sin B=\frac{1}{\sqrt[]{3}}=\frac{\sqrt[]{3}}{\sqrt[]{3}\times\sqrt[]{3}}=\frac{\sqrt[]{3}}{3} \\ \sin B=\frac{\sqrt[]{3}}{3} \end{gathered}[/tex]Let's find cos B as seen below;
[tex]\begin{gathered} \cos B=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{5\sqrt[]{2}}{5\sqrt[]{3}} \\ \cos B=\frac{\sqrt[]{2}}{\sqrt[]{3}}=\frac{\sqrt[]{2}\times\sqrt[]{3}}{\sqrt[]{3}\times\sqrt[]{3}}=\frac{\sqrt[]{6}}{3} \\ \cos B=\frac{\sqrt[]{6}}{3} \end{gathered}[/tex]