Answer:
[tex]sin(t)=\frac{\sqrt{99}}{10}[/tex]
Step-by-step explanation:
Remember that
If angle t belong to the First Quadrant
then
The value of sin(t) and cos(t) are positive values
we know that
[tex]sin^2(t)+cos^2(t)=1[/tex] ---> by trigonometric identity
we have
[tex]cos(t)=0.1=\frac{1}{10}[/tex]
substitute
[tex]sin^2(t)+(\frac{1}{10})^2=1[/tex]
[tex]sin^2(t)+\frac{1}{100}=1[/tex]
[tex]sin^2(t)=1-\frac{1}{100}[/tex]
[tex]sin^2(t)=\frac{99}{100}[/tex]
[tex]sin(t)=\frac{\sqrt{99}}{10}[/tex]