There is a relation between sine and cosine an angle
[tex]sin^2x+cos^2x=1[/tex]
We will use this rule to solve the question
Since cos(t) = 2/9, then substitute it in the rule above
[tex]\begin{gathered} sin^2\left(t\right)+\left(\frac{2}{9}\right)^2=1 \\ sin^2\left(t\right)+\frac{4}{81}=1 \end{gathered}[/tex]
Subtract 4/81 from each side
[tex]\begin{gathered} sin^2\left(t\right)+\frac{4}{81}-\frac{4}{81}=1-\frac{4}{81} \\ sin^2\left(t\right)=\frac{81}{81}-\frac{4}{81} \\ sin^2\left(t\right)=\frac{77}{81} \end{gathered}[/tex]
Take a square root for each side
[tex]\begin{gathered} \sqrt{sin^2\left(t\right)}=\sqrt{\frac{77}{81}} \\ sin\left(t\right)=\frac{\sqrt{77}}{9} \end{gathered}[/tex]
Change it to decimal and round the answer to the nearest hundredth
[tex]sin\left(t\right)=0.97[/tex]
The answer is sin(t) = 0.97
