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CTK Exchange
Sgt Pepper
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Aug-26-05, 07:41 PM (EST) |
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1. "RE: unsolved angle"
In response to message #0
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I only took a quick look at the drawing, but to me it'seems like some detail that locks the geometry is missing. In other words, this drawing is presenting a claim that the X angle will stay constant no matter how you alter all the unspecified angles and distances. Normally, for such problems, the specified angles would lock everything in one place and prevent any adjustments, but in this case you can! Choosing a value for one of the arbitrary angles and the length of one side would lock the rest of the geometry. Choosing an angle that makes the math easier is probably a good idea. To check if X is indeed constant, you would have to solve for a few different situations and see if they give the same solution. It could also be that X is not constant for the given geometry, and the problem lacks some information. |
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mr_homm
Member since May-22-05
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Aug-29-05, 04:50 PM (EST) |
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3. "RE: unsolved angle"
In response to message #0
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If you assume that the large outer triangle is isoceles, you can calculate the answer using trigonometry. Apply the law of sines to several triangles and use some trigonometric identities to simplify the answer. What you finally get is sin(50-x)/sin(x) = sqrt(3), which gives x = 17.8779871443 degrees. Since this shows no sign of being a simple integer or fraction, I think that this is a case of a problem where old fashioned geometry methods will not yield the solution you want. Most of the tools of geometry are used for construction, disection (for angles, bisection only), or proof of equality. This angle is already constructed (there it is on the diagram), it cannot be built in any simple way by repeated bisection and addition of the other angles (look at its numerical value -- there is no obvious combination of 20, 30, and 180 that will give this), and it is not equal to anything else in the picture. Geometry problems typically are about relationships among the geometric objects in the problem statement. This problem is different, because it is asking you to relate an angle to a number (the numeric value of the angle in degrees) rather than to other angles in the diagram. This goes somewhat outside old fashioned Euclidean geometry. There are other problems that look like this one, in that they ask you to find the value of an unknown angle, but are not really the same, because they are actually about the relationship of the unknown angle to the others in the picture. They are always solved by showing how to build the unknown angle out of the other angles, which you could do WITHOUT knowing their numerical values, if the necessary relationships among the angles are given. Giving the angles numerical values instead of just labels hides the fact that you are not really looking for a way to associate a number to the unknown angle. It LOOKS like you're computing a numerical value, but since you could solve the problem exactly the same way using labels instead of numbers, it can't really be about numbers. Then, after you have done many problems of this kind, someone slips in a problem (such as the current one) that's REALLY about computing the numerical value of an unknown angle, and suddenly nothing works. That's what makes this kind of problem so frustrating. Another way to look at this problem is to notice that angle x is not on any triangle where you can find the other two ANGLES. There are lots of triangles in the diagram for which you can find all three angles, but none of them share enough angles with any triangle containing x for you to solve and get x. All three SIDES of a triangle containing x are shared with other triangles whose angles you know, so x is determined, i.e. there is a definite value for x, and it is the only one that works. But to get from the information you have (other angles) to the information you want (x), you absolutely must use reasoning that connects the information via the side lengths of triangles, since sides are the only things connecting triangles that contain x to triangles whose angles you know. Relating sides to angles essentially involves trigonometry, so you absolutely must use trigonometric reasoning to solve the problem. As a trigonometry problem, it is nothing special, just several applications of the law of sines (or cosines) and a little algebra. It is only when you fall for the trick and think it is a geometry problem that it'seems horribly difficult. Hope this explains some of your difficulties. --Stuart Anderson |
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yoyoma
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Sep-13-05, 12:00 PM (EST) |
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4. "RE: unsolved angle"
In response to message #3
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I wrote down 13 equations with 8 variables and put it into a linear solver and it came back with infinitely many solutions with two independent variables. Not sure if it's right, but I checked it pretty carefully and that's what it came back with. |
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mr_homm
Member since May-22-05
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Sep-19-05, 06:55 PM (EST) |
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6. "RE: unsolved angle"
In response to message #4
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Here is what I did, ASSUMING the large triangle is isoceles: 1) Label the points on the original poster's diagram as follows: Top corner is A, lower left and right are B and C. Point where the unknown angle x sits is D, and the point with the unmarked angle on side BC is E. 2) Angle ABC = angle ACB = 80 since ABC is isoceles and angle BAC is given as 20. Since angle EBC is also given as 20, triangle EBC is similar to triangle ABC, hence it is also isoceles and angle BEC is 80. On triangle DCB, the angles DCB is given as 30, and we know angle DBC is 80, so angle BDC is 70. 3) Since angles are independent of overall scale of the diagram, let the base BC of the large triangle have length = 1. Then by the law of sines on triangle BEC, sin(20)/sin(80) = EC/1, and on triangle DCB, sin(30)/sin(70) = DB/1. Now we have EC and DC. 4) On triangle DEB, angle DEB = 180-60-(70+x) = 50-x. Therefore, the law of sines on triangle DEC gives sin(x)/EC = sin(50)/DE, and on triangle DEB, it gives sin(50-x)/DB = sin(60)/DE. Dividing these equations and rearranging the terms gives sin(50-x)/sin(x) = (DBsin(60))/(ECsin(50) = (sin(30)sin(60)sin(80))/(sin(20)sin(50)sin(70)). 5) Working on the denominator, sin(20)sin(50)sin(70) = sin(50)(sin(20)cos(20)) = sin(50)*1/2*sin(40) = 1/2*(sin(40)cos(40) = 1/4*sin(80). This cancels with the sin(80) in the numerator to give sin(50-x)/sin(x) = 4sin(30)sin(60) = 4*1/2*sqrt(3)/2 = sqrt(3). This was rather long, which is why I didn't post it originally, but since there was a question, here it is. By the way, if you put 13 equations in 8 variables into your solver, a lot of the equations must have been redundant. It might be easier to analyze if you first try to eliminate equations that can be derived from others. Also, and this is very important, these are trig equations, so they are NOT linear. If it is really a linear solver instead of a general function solver, it won't be able to do this problem. It is not the appropriate tool for the job. Hope this helps, --Stuart Anderson |
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pluto
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Sep-17-05, 06:45 AM (EST) |
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5. "RE: unsolved angle"
In response to message #0
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Hello, I bet it'll remain unsolved forever... If you deform the largest triangle continuously so that it becomes isocele "on the left" or "on the right side", the little triangle with X in one angle degenerates to a line. In the 'left version' X tends to zero, while in the 'right version' X tends to 20+30. So it does depend on one further variable. |
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momus
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Mar-03-06, 07:53 PM (EST) |
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7. "RE: unsolved angle"
In response to message #5
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This problem is unsolvable without one more variable. Any additional angle around the perimeter would make this system of equations solvable (it wouldn't need to be one of the angles at one of the apexes). Other than that one cannot make the assumption that this is an isosoles triangle without the problem explicitly stating that fact. |
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