For maximum accessibility a wheelchair ramp should have a slope between 1/16 and 1/20. What is the range of angle measures that a ramp should make with the ground? Round to the nearest hundredth of a degree.
+23252 When you graphed, you were told that the slope was the "change in y / change in x" or "rise over run".
Same here.
For a slope to be 1/16, it has a vertical rise of 1 for every horizontal run of 16 .
If, on a piece of graph paper, you place a dot at the origin (0,0) (call this dot A); move over to the point (16,0) and put a dot there (call this dot C); and move up to the point (16,1) and put a dot there (call this dot B), you have drawn a right triangle (triangle ACB, with right angle at C) which represents a ramp with slope 1/16.
You need to find the size of the angle at A. Since you know the opposite side (1) and the adjacent side (16), use tangent (because tan = opp / adj). ---> tan ∠A = 1 /16 ---> ∠A = invtan (1/16) ---> ∠A = 3.576°.
Similarly for the slope of 1/20: tan ∠A = 1 /20 ---> ∠A = invtan (1/20) ---> ∠A = 2.862°.
+23252 When you graphed, you were told that the slope was the "change in y / change in x" or "rise over run".
Same here.
For a slope to be 1/16, it has a vertical rise of 1 for every horizontal run of 16 .
If, on a piece of graph paper, you place a dot at the origin (0,0) (call this dot A); move over to the point (16,0) and put a dot there (call this dot C); and move up to the point (16,1) and put a dot there (call this dot B), you have drawn a right triangle (triangle ACB, with right angle at C) which represents a ramp with slope 1/16.
You need to find the size of the angle at A. Since you know the opposite side (1) and the adjacent side (16), use tangent (because tan = opp / adj). ---> tan ∠A = 1 /16 ---> ∠A = invtan (1/16) ---> ∠A = 3.576°.
Similarly for the slope of 1/20: tan ∠A = 1 /20 ---> ∠A = invtan (1/20) ---> ∠A = 2.862°.
