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Topic: Trig Problem (Read 3977 times) |
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warriors837
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A twenty foot ladder leans up against a perpendicular wall and just touches the outer top edge of a 6'x6'x6' cube packing case pushed up close to the wall. How far up the wall is the top of the ladder?
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« Last Edit: May 26th, 2009, 11:46am by warriors837 » |
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towr
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Re: Trig Problem
« Reply #1 on: May 26th, 2009, 12:01pm » |
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Let's call the length from the cube to the foot of the ladder a, then the height h, is (6+a) * 6/a, and we have the further constraint that h2+(6+a)2 = 202. I think that should be enough.
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pex
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Re: Trig Problem
« Reply #2 on: May 26th, 2009, 2:31pm » |
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on May 26th, 2009, 12:01pm, towr wrote:Let's call the length from the cube to the foot of the ladder a, then the height h, is (6+a) * 6/a, and we have the further constraint that h2+(6+a)2 = 202. I think that should be enough. |
| Yes it is, but I think the solution outlined here (with different numbers) is aesthetically preferable. (Although it uses the same general ideas.)
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Wiles_euler
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Work and answers in attachment
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« Last Edit: May 29th, 2009, 7:44am by warriors837 » |
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pex
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Re: Trig Problem
« Reply #5 on: May 29th, 2009, 12:26am » |
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on May 28th, 2009, 4:02pm, Wiles_euler wrote:Full proof in attachment... |
| I wouldn't call that a full proof... What it's basically saying is, we have h = (6+a)*6/a and h2+(6+a)2=202, so let's numerically solve ((6+a)*6/a)2+(6+a)2=202. Have you considered reading ThudanBlunder's and my links? Edit: Besides, why would you go through all the simplification if you're going to solve the equation numerically anyway? 202=h2+[6+(36/(h-6)]2 isn't really any harder for a numerical solver than 400=(h4-12h3+72h2)/(h2-12h+36)...
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« Last Edit: May 29th, 2009, 12:31am by pex » |
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towr
wu::riddles Moderator Uberpuzzler
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Re: Trig Problem
« Reply #6 on: May 29th, 2009, 12:33am » |
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It's not only not a full proof, but I'm fairly sure the answers aren't even quite correct (even besides the fact they're both approximations). They're not very far off, but still too far to be accurate. Just put them back into the equation, you won't get 400, but 401 and 413.
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« Last Edit: May 29th, 2009, 12:38am by towr » |
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Wiles_euler
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When you plug in the given values from the attachment, they do work out with the us of the Pythagorean theorem.
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« Last Edit: May 29th, 2009, 7:46am by warriors837 » |
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Ronno
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Re: Trig Problem
« Reply #8 on: May 29th, 2009, 8:45pm » |
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I don't think that statement makes any sense and the correct values are around 9.040515497 and 17.84009752.
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towr
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Re: Trig Problem
« Reply #9 on: May 30th, 2009, 2:14am » |
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on May 29th, 2009, 8:45pm, Ronno wrote:I don't think that statement makes any sense and the correct values are around 9.040515497 and 17.84009752. |
| Yup, that's what I get. Or to be exact: 3 + (109) +/- (82 - 6 (109)) But hey, it's his homework; so by all means he should go with his own results
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