The triangle given in the exercise is a Right Triangle because the square inside it indicates that it has an angle that measures 90 degrees.
Then, you can use the following Trigonometric Function:
[tex]\sin \alpha=\frac{opposite}{hypotenuse}[/tex]
In this case:
[tex]\begin{gathered} \alpha=31\degree_{} \\ opposite=11 \\ hypotenuse=x \end{gathered}[/tex]
Then, by substituting values and solving for "x", you get:
[tex]\begin{gathered} \sin (31\degree)=\frac{11}{x} \\ \\ x\cdot\sin (31\degree)=11 \end{gathered}[/tex][tex]\begin{gathered} x=\frac{11}{\sin (31\degree)} \\ \\ x\approx21.4 \end{gathered}[/tex]
Therefore, the answer is:
[tex]x\approx21.4[/tex]
