Since it is a right triangle then you can use the trigonometric ratio
[tex]\sin (\theta)=\frac{\text{ opposite leg of }\theta}{\text{ hypotenuse}}[/tex]
In this case, you have
[tex]\begin{gathered} \text{opposite leg of x = 6 } \\ \text{hypotenuse = 13} \end{gathered}[/tex]
Then,
[tex]\begin{gathered} \sin (x)=\frac{6}{13} \\ \text{ Using the inverse function }\sin ^{-1}(\theta)\text{ on both sides of the equation} \\ \sin ^{-1}(\sin (x))=\sin ^{-1}(\frac{6}{13}) \\ x=\sin ^{-1}(\frac{6}{13}) \\ x=27.486\text{\degree} \\ \text{ Rounding} \\ x=28\text{\degree} \end{gathered}[/tex]
