Ramp Angle for Wheelchair Access
A building must install a wheelchair ramp. Accessibility guidelines require the ramp's angle of incline to be no more than 5 degrees. The entrance door is 0.9 meters above the sidewalk. The contractor proposes a ramp with a horizontal run of 9 meters. A second contractor proposes a run of 6 meters. The building manager wants to know which proposal complies and how much horizontal run is needed if the angle is set exactly at the 5-degree limit. A surveyor also notes the ramp length (the sloped surface) differs from the horizontal run.
1. For each proposed run, compute the incline angle and determine compliance.
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The angle satisfies $\tan\theta = \frac{\text{rise}}{\text{run}} = \frac{0.9}{\text{run}}$. For 9 m: $\tan\theta = 0.9/9 = 0.10$, so $\theta = \arctan(0.10) \approx 5.71°$, which exceeds 5° and does not comply. For 6 m: $\tan\theta = 0.9/6 = 0.15$, $\theta \approx 8.53°$, also noncompliant. Both proposals fail.
2. What minimum horizontal run meets the 5-degree limit?
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Require $\tan\theta \le \tan 5° \approx 0.0875$. So $\frac{0.9}{\text{run}} \le 0.0875 \Rightarrow \text{run} \ge \frac{0.9}{0.0875} \approx 10.29$ m. The run must be at least about 10.3 meters. The 9 m proposal is short by roughly 1.3 m.
3. At exactly the 5-degree limit, what is the sloped ramp length, and how does it compare to the run?
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Use $\sin\theta = \frac{\text{rise}}{\text{length}}$: length $= \frac{0.9}{\sin 5°} \approx \frac{0.9}{0.0872} \approx 10.32$ m. Alternatively from the run: length $= \sqrt{10.29^2 + 0.9^2} \approx \sqrt{105.9 + 0.81} \approx \sqrt{106.7} \approx 10.33$ m. The sloped length slightly exceeds the horizontal run (by about 0.04 m), a small but real difference; using run for construction and length for material ordering gives nearly the same figure here because the angle is small.