we know that
[tex]sin(60\°)=\frac{\sqrt{3}}{2}[/tex]-----> equation A
[tex]cos(60\°)=\frac{1}{2}[/tex] ------> equation B
In the right triangle of the figure
[tex]sin(60\°)=\frac{y}{x}[/tex] -----> equate to the equation A
[tex]\frac{y}{x}=\frac{\sqrt{3}}{2}[/tex] -----> equation C
[tex]cos(60\°)=\frac{8}{x}[/tex] ------> equate to the equation B
[tex]\frac{8}{x}=\frac{1}{2}\\\\x=16[/tex]
Substitute the value of x in the equation C
[tex]\frac{y}{16}=\frac{\sqrt{3}}{2}\\ \\y=16 \sqrt{3}/2 \\ \\y=8 \sqrt{3}\ units[/tex]
therefore
the answer is the option B
8 square root 3
