The dimensions given are;
Horizontal = 19
Length (diagonal) = 20
Vertical = ???
The description would end up as a right angled triangle with one of the legs as 19 and the hypotenuse as 20. To calculate the vertical;
![\begin{gathered} 20^2=19^2+v^2 \\ 400=361+v^2 \\ \text{Subtract 361 from both sides} \\ 400-361=v^2 \\ 39=v^2 \\ \sqrt[]{39}=v \\ 6.2449=v \\ v\approx6.25\text{ (to 2 decimal places)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zmalwkt1icx2s6i60ep3qr70flvj7lc6k1.png)
The vertical height of the ramp is 6.25 ft