Given a right angle triangle:
As shown:
[tex]\begin{gathered} SU=\sqrt[]{57} \\ ST=4 \\ TU=\sqrt[]{41} \end{gathered}[/tex]
We will find sin S
[tex]\sin S=\frac{opposite}{hypotenuse}=\frac{TU}{SU}[/tex]
Substitute with TU and SU
So,
[tex]\begin{gathered} \sin S=\frac{\sqrt[]{41}}{\sqrt[]{57}}\times\frac{\sqrt[]{57}}{\sqrt[]{57}}=\frac{\sqrt[]{2337}}{57} \\ \\ \sin S=\frac{\sqrt[]{2337}}{57} \end{gathered}[/tex]
