[ARCHIVED THREAD] - Can you solve this math problem??? (Page 1 of 3)
Posted: 9/11/2011 8:27:10 PM EDT
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You have a deck that is approximately 30 inches high. You need to build a ramp that can only rise 1 inch every 18 inches.
How long does the ramp need to be? |
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The first two posts.
[Edna Mode]Is this a question?[/Edna Mode] Is the rub in the "approximately?" If so, there is no answer, since "approximately" is undefined. EDIT: Is the "18 inches" horizontal distance, or actual ramp length? I see some people are assuming that the question is for the ramp length, which calls for the hypotenuse. The question does not indicate which it is. |
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Quoted: Quoted: 30 x 18 = 540540.83269131959839396788319012057 inches That would be the length of the ramp. 540 inch distance from base of ramp to bottom of deck. 30 inches up. Solve for hypotenuse of said triangle 540^2 + 30^2 = length of ramp^2 You don't want the horizontal distance from the base of the ramp to the deck, you want the length of the ramp. |
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Quoted: You have a deck that is approximately 30 inches high. You need to build a ramp that can only rise 1 inch every 18 inches. How long does the ramp need to be? OK, so I have a deck. And I need to build a ramp to said specs. Are the two connected somehow??? (sorry)... |
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Quoted: I think Chapman covered that. Then he answered the OP's question.Quoted: 30 x 18 = 540540.83269131959839396788319012057 inches That would be the length of the ramp. 540 inch distance from base of ramp to bottom of deck. 30 inches up. Solve for hypotenuse of said triangle 540^2 + 30^2 = length of ramp^2 |
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Quoted: darnit Quoted: Quoted: 30 x 18 = 540540.83269131959839396788319012057 inches That would be the length of the ramp. 540 inch distance from base of ramp to bottom of deck. 30 inches up. Solve for hypotenuse of said triangle 540^2 + 30^2 = length of ramp^2 You don't want the horizontal distance from the base of the ramp to the deck, you want the length of the ramp. |
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Quoted: Quoted: Math can go to hell. Anyone who cannot cope with mathematics is not fully human. At best he is a tolerable subhuman who has learned to wear shoes, bathe and not make messes in the house. ––Lazarus Long, "Time Enough for Love", (Robert A. Heinlein) A quote for a book doesn't address the fact that math can in fact go to hell and burn for all eternity.  |
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Quoted:
You have a deck that is approximately 30 inches high. You need to build a ramp that can only rise 1 inch every 18 inches. How long does the ramp need to be? The vertical distance is 30 inches or 2.5 feet, the horizontal distance is 540 inches or 45 feet, the distance of the hypotenuse is 540.8327 inches or 45.0694 feet. |
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Quoted: Quoted: 30 x 18 = 540540.83269131959839396788319012057 inches That would be the length of the ramp. 540 inch distance from base of ramp to bottom of deck. 30 inches up. Solve for hypotenuse of said triangle 540^2 + 30^2 = length of ramp^2 Yup...pretty easy actually. |
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Quoted: 540 inches (45 ft)is the length of the rampQuoted: Quoted: 30 x 18 = 540540.83269131959839396788319012057 inches That would be the length of the ramp. 540 inch distance from base of ramp to bottom of deck. 30 inches up. Solve for hypotenuse of said triangle 540^2 + 30^2 = length of ramp^2 You don't want the horizontal distance from the base of the ramp to the deck, you want the length of the ramp. |
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Quoted: No, it's not.Quoted: 540 inches (45 ft)is the length of the rampQuoted: Quoted: 30 x 18 = 540540.83269131959839396788319012057 inches That would be the length of the ramp. 540 inch distance from base of ramp to bottom of deck. 30 inches up. Solve for hypotenuse of said triangle 540^2 + 30^2 = length of ramp^2 You don't want the horizontal distance from the base of the ramp to the deck, you want the length of the ramp. |
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It's a simple trig problem yo!!
sineX = 1/18 sineX= .0555 Angle = 3.185 degrees sine 3.185 degrees = 30/x .05555 = 30/x solve for X X = 30/.0555 X = 540.54 inches or 540.54/12 = 45.045 feet .......................... depending how you round off. I been drink'in some!!! |
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Quoted:
It's a simple trig problem yo!! sineX = 1/18 sineX= .0555 Angle = 3.185 degrees sine 3.185 degrees = 30/x .05555 = 30/x solve for X X = 30/.0555 X = 540.54 inches or 540.54/12 = 45.045 feet .......................... depending how you round off. I been drink'in some!!! I can tell 
Small mistake in counting the 18 as the hypotenuse measurement, instead of the "adjacent" measurement, therefore using sin instead of tan. 
tan^(-1) (1/18) = 3.17983012* 30 csc (3.17983012) = 540.8326913... For the NON MATHLETES here, allow me to construct a simple experiment for you all to SHOW you why the answer is indeed not 540 inches even. Take two books. Lay one down flat, stand the other up, place the one laying down against the one standing. It should resemble and L shape. 540 inches is the HORIZONTAL distance that the ramp must traverse in order to satisfy known parameters. Raise the book to resemble ramp, which making sure to keep far end in place (since we know the ass end of this ramp must remain at 540 inches to satisfy the slope requirement). Ramp is too short? That's where the extra .83 inches come in on this problem. It's a hypotenuse problem, not multiplication. |
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Quoted:
Quoted:
It's a simple trig problem yo!! sineX = 1/18 sineX= .0555 Angle = 3.185 degrees sine 3.185 degrees = 30/x .05555 = 30/x solve for X X = 30/.0555 X = 540.54 inches or 540.54/12 = 45.045 feet .......................... depending how you round off. I been drink'in some!!! I can tell
For the NON MATHLETES here, allow me to construct a simple experiment for you all to SHOW you why the answer is indeed not 540 inches even. Take two books. Lay one down flat, stand the other up, place the one laying down against the one standing. It should resemble and L shape. 540 inches is the HORIZONTAL distance that the ramp must traverse in order to satisfy known parameters. Raise the book to resemble ramp, which making sure to keep far end in place (since we know the ass end of this ramp must remain at 540 inches to satisfy the slope requirement). Ramp is too short? That's where the extra .83 inches come in on this problem. It's a hypotenuse problem, not multiplication. There ain't no extra .83 inches. Do the math like I did, show me.......... I said the answer can vary by how you round off....... significant digits.......... but in no way do you come up with an extra .83 inches. |
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Quoted:
Quoted:
Quoted:
It's a simple trig problem yo!! sineX = 1/18 sineX= .0555 Angle = 3.185 degrees sine 3.185 degrees = 30/x .05555 = 30/x solve for X X = 30/.0555 X = 540.54 inches or 540.54/12 = 45.045 feet .......................... depending how you round off. I been drink'in some!!! I can tell
For the NON MATHLETES here, allow me to construct a simple experiment for you all to SHOW you why the answer is indeed not 540 inches even. Take two books. Lay one down flat, stand the other up, place the one laying down against the one standing. It should resemble and L shape. 540 inches is the HORIZONTAL distance that the ramp must traverse in order to satisfy known parameters. Raise the book to resemble ramp, which making sure to keep far end in place (since we know the ass end of this ramp must remain at 540 inches to satisfy the slope requirement). Ramp is too short? That's where the extra .83 inches come in on this problem. It's a hypotenuse problem, not multiplication. There ain't no extra .83 inches. Do the math like I did, show me.......... I said the answer can vary by how you round off....... significant digits.......... but in no way do you come up with an extra .83 inches. here's the short version of my edited post above Edit again: I did mine a different way, with less rounding. Yours is correct as well, but rounded a bit more. tan^(-1) (1/18) = 3.17983012* For clarity tan = o/a, csc = 1/sin = h / o 30 csc (3.17983012) = 540.8326913... When a problem is presented as "rise 1 inch for every 18 inches," the convention is to measure said 18 inches along the ground, not along the ramp. |