Question

The municipal building code specifies that public buildings have a handicapped access ramp with an incline of no more than 9°. What length of ramp than 9°. What length of ramp is needed if the ramp climbs 4 ft 8 in.? Express the answer in feet and inches to the nearest inch. (Hint: Work the problem in inches only, then convert the answer to feet and inches.)(Round to the nearest inch, as needed.)

Answer 1

Given:

Angle of the ramp = no more than 9 degrees.

height that needed be climb by the ramp = 4ft and 8 inches

Find: length of the ramp

Solution:

Let's convert the height first into inches. Know that in every 1 ft, there are 12 inches.

`\begin{gathered} 1ft=12inches\text{ } \\ 4ft=4*12=48inches\text{ } \\ 4ft+8inches=48inches+8inches=56inches\text{ } \end{gathered}`

Hence, the height that needs to be climbed is 56 inches.

Let's illustrate this problem. (Assuming the wall and the base of the incline is perpendicular to each other)

To solve the length of the ramp or the hypotenuse of the triangle, we can use the sine function.

`\sin \theta=\frac{oppositeside}{\text{hypotenuse}}`

Our θ = 9 degrees and our opposite side is 56 inches, let's plug it in to the function above.

`\sin 9=(56)/(x)`

Then, solve for "x" or the hypotenuse.

`\begin{gathered} x\sin 9=56 \\ x=(56)/(\sin 9) \\ x\approx358\text{inches} \end{gathered}`

Hence, the length of the ramp must be 358 inches. To convert that to feet and inches, let's divide it by 12.

`(358)/(12)=29ft\text{ and 10 inches}`

Hence, the length of the ramp must be 29ft and 10 inches.