we have
[tex]\sin (t)=\frac{\sqrt[]{5}}{6}[/tex]Remember that
[tex]\sin ^2(t)+\cos ^2(t)=1[/tex]substitute the given value
[tex](\frac{\sqrt[]{5}}{6})^2+\cos ^2(t)=1[/tex][tex]\begin{gathered} \frac{5}{36}+\cos ^2(t)=1 \\ \cos ^2(t)=1-\frac{5}{36} \\ \cos ^2(t)=\frac{31}{36} \\ cos(t)=\pm\frac{\sqrt[]{31}}{6} \end{gathered}[/tex]the angle t is an acute angle and the sine is positive, that means, the angle t lies in the first quadrant
so
the cosine is positive
therefore
[tex]cos(t)=\frac{\sqrt[]{31}}{6}[/tex]