Answer: [tex]-\dfrac{2\sqrt{14}}{9}[/tex]
Step-by-step explanation:
Given
[tex]\cos a=\dfrac{5}{9}[/tex]
[tex]a[/tex] lies in the fourth quadrant
So, sine must be negative in the fourth quadrant
Using identity [tex]\sin ^2 x+\cos^2 x=1[/tex] to find sine value
[tex]\Rightarrow \sin^2 a=1-\dfrac{5^2}{9^2}[/tex]
[tex]\\\\\Rightarrow \sin^2 a=1-\dfrac{25}{81}\\\\\\\Rightarrow \sin^2 a=\dfrac{56}{81}\\\\\\\Rightarrow \sin a=-\sqrt{\dfrac{56}{81}}\\\\\\\Rightarrow \sin a=-\dfrac{2\sqrt{14}}{9}[/tex]