this is a question that involves angle of elevation and right angled triangles.
first, a diagram deoicting the scenerio is drawn below;
using the SOHCAHTOA rule for right angled triangles, we would name he distance of the ramp to the platform; x
y is length of the ramp up to the platform (the side opposite the right angle), H
8ft is the height from the ground to the platform ( the distance of the side opposite the angle), O
x is the length of the end of the ramp to the base of the platform( is the adjacent) A
THEREFORE, we will be applying TOA
[tex]\begin{gathered} \tan \theta=\text{ opposite/adjacent} \\ \cos 30=\frac{O}{A} \\ \cos 30=\frac{8}{X} \\ 0.8660=\frac{8}{x} \\ x=\frac{8}{0.8660} \\ x=9.24ft \end{gathered}[/tex]the end of the ramp is 9.24ft from the base of the platform
[tex]\begin{gathered} \sin 30=\frac{posite}{\text{hypotenuse}} \\ \sin 30=\frac{8}{y} \\ 0.5=\frac{8}{y} \\ y=\frac{8}{0.5} \\ y=\text{ 16ft} \end{gathered}[/tex]the ramp is 16ft long up to the top of the platform