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Given a diagram of n infinitely long straight lines in the plane, let them intersect in the points p_i, let the angles at $p_i$ be $v_i^j$, such that $360=\sum_j v_i^j$.

Given the suggestive diagram, and some angles, if one is given the task of calculating some $v_m^n$, and the diagram and angles determine this angle uniquely, is it always sufficient for finding this angle to solve the linear system $360=\sum_j v_i^j$ for every i, together with $a+b+c=180$ for every triangle?

One cant read any distances or other things off the diagram, only the order in which line a intersects line b, for every a, for every b.

What if one is given some additional information, that some distances are equal? Is it necessary to draw any more straight lines? Does one never need any more advanced formula/geomtry to find $v_n^m$?

(Why) does the algorithm fail for http://www.cut-the-knot.org/triangle/80-80-20/60-70Sol1.shtml#solution

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    How would your question be answered in the case of two lines (one point of intersection)?2012-05-10
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    Yes, we need this rule too, angle on opposite side=equal2012-05-10
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    In mathematics (as opposed to art), lines are by definition infinitely long and straight. You could thus simplify the premise: "Given $n$ lines in the plane, ...".2012-05-10
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    You may have missed the point of the @hardmath comment. If all you know is that there are 2 lines, you can write down all the equations for angles you like, but you'll never be able to solve them, since the angle between the two lines could be anything (between zero and ninety degrees).2012-05-11
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    @GerryMyerson Sure, but Im asking, if its possible to deduce the angle in some way, ie the problem has sufficient parameters, then is my algorithm always going to work, or does it require use of other trigonometric theorems?2012-05-11
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    The question is a bit ill-defined. If I read it correctly, then the answer is 'No'. For each new line (starting from the second line), you need to be given the angle it makes with one (anyone will do) of the earlier lines, as long as the "sign" of that angle is also given. Then your algorithm should work. This is the same thing that hardmath and Gerry Myerson have been explaining. Those extra angles will specify the compass bearing of each and every line relative to the first one. That knowledge will determine all the angles, and your algorithm will find them recursively.2012-05-11
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    ...(cont.) So if you have other ways of finding the compass bearings of the other lines relative to the first one, then the answer is "Yes".2012-05-11
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    Ok, but this doesnt seem to work for this triangle, here they have to add additional lines, why? http://www.cut-the-knot.org/triangle/80-80-20/60-70Sol1.shtml#solution2012-05-11
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    No direct information about the compass bearing of the line $ED$ was given. That line was defined using points of intersection of earlier lines. Therefore an additional mechanism for converting that kind of a specification of a line to information about its direction relative to the earlier lines was required.2012-05-11
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    Just think about the case of two lines. Of course it's not possible to deduce the angle in some way; the problem has *too many* parameters, and you don't have enough information; no algorithm is going to work; all the trigonometric theorems in all the libraries in the world aren't going to help. If all you know is you have two lines, how in the name of all that's mathematical are you going to calculate the angle between them? You need more information about the lines. Maybe you have more information about the lines, but if so you are keeping it to yourself.2012-05-11

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