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Need A Mathematic Genius


Plaidman

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I remember that the hypotenuse of a right triangle is a2 + b2 = c2

but if you no longer have a right angle and the angle is 110 degrees what formula do you use?

You proceed from a false assumption. You, in fact, do still have a right triangle, precisely as Dru (G-spotter) points out earlier in this thread. So the Pythagorean Theorem works here, and the answer is 72.8 feet. This assumes that you are at the extreme end of the 200-foot rope, and we ignore the effects of rope stretch and the amount of rope consumed by knots.

 

Note that keenwash came up with the same answer using the Law of Sines (as one would expect), but in this instance there is no need for such superfluousness in solving the problem. The more elegant solution is to invoke the Pythagorean Theorem.

 

QED.

 

Now to SnailEye's statement...

kind of depends on how far up the 200 foot rope you are...
This is also true. As you ascend the rope, your distance from the wall reduces. At the halfway point (say) you will be 72.8'/2, or 36.4 feet, away from the wall (still assuming an average straight-line slope of 110 degrees). Ultimately, at the top of the rope, you are in contact with the wall.

 

You can calculate how far you are from the wall at any point on the rope through the Law of Similar Triangles. LST states that for any triangles whose angles are the same, the lengths of the sides are proportional to each of the other triangles. So, at the 200-foot mark, you are 72.8 feet away from the wall. At the 100-foot mark, you are 36.4 feet away from the wall. You can see very quickly that the ratio is 200/72.8, or 2.75:1. That means that for every 2.75 feet you ascend the rope, you will hang 1 foot closer to the wall, until at the top of the rope you are touching the wall.

 

I hope this explanation has not been too :nurd: for you. Please carry on!

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Note that keenwash came up with the same answer using the Law of Sines (as one would expect), but in this instance there is no need for such superfluousness in solving the problem. The more elegant solution is to invoke the Pythagorean Theorem.

 

QED

 

Please explain. I was previously led to believe that only the 30-60 and 45-45 triangles can be calculated from knowledge of the length of 1 side. Otherwise 2 sides were necessary and thus the angle irrelevant.

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Note that keenwash came up with the same answer using the Law of Sines (as one would expect), but in this instance there is no need for such superfluousness in solving the problem. The more elegant solution is to invoke the Pythagorean Theorem.

 

QED

 

Please explain. I was previously led to believe that only the 30-60 and 45-45 triangles can be calculated from knowledge of the length of 1 side. Otherwise 2 sides were necessary and thus the angle irrelevant.

Ken,

Yes, the 30/60/90 and 45/45/90 triangles make it very easy to calculate the length of any side knowing only the length of one side. But that is not the problem statement here. Trigonometry says that you can solve for the length of any side of a right triangle knowing either 1) the lengths of the other two sides (by using Pythagorean Theorem), or 2) the length of one side and the value of one of the other two angles (those that are not the right angle).

 

Since we are given the (overhanging) angle of the subject pitch of the Leaning Tower as 110 degrees, we know the interior of that angle is 70 degrees. And since the 200-foot side is known, and that side must be perpendicular to the ground line, we have all that is needed to solve by trig (one known side and two known angles).

 

LeaningTower.jpg

Edited by sobo
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OK I got one (not multi-variable calculus, but fun nevertheless): show from Newton's laws that the fall factor is length of rope divided by fall distance

 

FF = l/d

 

Be sure to include any assumptions you make (e.g. the rope behaves like an ideal spring where f = kx). Bonus points if you include a free body diagram.

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