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Something like this...??

Yes indeed! How perceptive of you, Dru! But I would posit that Plaidman was actually wondering "how far out from the wall would I be" if I were at the end of a 200-foot rope on a 110 degree overhanging wall. While his actual distance from the wall (at his closest approach) would be 68.4 feet as you so aptly demonstrate, he prolly was looking for the horizontal distance from the wall. Only Plaidman knows... What say ye all to a 4th page? I still have more Fat Tires here...

me! Lest y'all think this pic was not taken moments ago, please note the envelope on my keyboard. Look familiar?? You may now ask the question... "Hey, Sobo! What time is it??"

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I will raise a Fat Tire to the first person who puts this thread to three pages.

Seven more to go... Ken, ya gonna get in on this? You started it, ya know...

You should send it to Joe. It appears from the top of this page that he might need it... Nine more to go...

Only 11 more posts to go...

I'm luv'n this... And Teresa, my surgery's not until the 14th. But I'll spray from the recovery room if that's what the hordes demand...

Hey, I'm just providing answers to the questions being posed. And yes, that really is written on the back of an envelope...

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).

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... 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 for you. Please carry on!

Hey Sobo, what time is it? I don't know what to say... simply at a loss for words here... :shakinghead:

Now that's fuq'n funny, I don't care who ya'are.