Well, no one asked for the answer and to show my work.
Yes, the answer is exactly 30 degrees. You must have had some rounding error in your math calculations. I worked it out geometrically not completely mathematically so I don't have any rounding errors.
I drew it out and deduced the angle. That simple.
Originally posted by GhostintheShellSorry, but if you drew it out then you don't know that the answer is exactly 30 degrees.
Well, no one asked for the answer and to show my work.
Yes, the answer is exactly 30 degrees. You must have had some rounding error in your math calculations. I worked it out geometrically not completely mathematically so I don' ...[text shortened]... errors.
I drew it out and deduced the angle. That simple.
You are assuming that the answer is exactly 30 degrees, not deducing. 😛
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Originally posted by timebombted, snipped for spaceI think that this method, although clumsy, would get to a correct answer.
After working out all possible angles through the 180 degrees in a triangle rule, 180 degrees on a straight line rule, Iso tri rules........
Start by assigning a distance of 10 to BC, using either your Cosine Rule or Sine Rule appropriately ....
c2=a2+b2-2abCosC
Gives you side DE.
Now using the Sine rule.......
a b
_ = _
SinA SinB
You can calculate CDE
When you first posted, I tried working this out without assigning a value, but simply an "x." This resulted in a rather complicated fraction, but all the x's canceled out at the end: a good sign. I don't have my notes with me here at work, but I will post the fraction I came up with and see if the total also equals 30 degrees later tonight (say 10 EST or so).
A solution is a solution, whether it is awkward or not, but one must wonder if there is a more elegant way of doing it....Perhaps the poster could give a hint if there is a second way of doing it.
I'd like you guys to read back through the thread and note...who had the correct answer first and knew that he was right. Me, thats who.
Its a simple geometry problem and I solved it on page 1. I never guessed at it, I never made 3 or 4 attempts to answer it. I figured out the answer, stated the answer and it was correct.
Originally posted by GhostintheShellGood Work Ghost!
I'd like you guys to read back through the thread and note...who had the correct answer first and knew that he was right. Me, thats who.
Its a simple geometry problem and I solved it on page 1. I never guessed at it, I never made 3 or 4 attempts to answer it. I figured out the answer, stated the answer and it was correct.
I don't know why no one is giving you the props you deserve... Geeesh.
P-
Originally posted by GhostintheShellGhost: Congrats. I thought I had asked earlier: Would you mind sharing how you did this? Timebombted's method, while it appears sound (I haven't double checked it for myself), is not what I would call "simple." I mean, it employs simple formulae (Laws of Co/Sines), but it takes several iterations in the process to come up with an answer.
Its a simple geometry problem and I solved it on page 1. I never guessed at it, I never made 3 or 4 attempts to answer it. I figured out the answer, stated the answer and it was correct.
How did you do it? By another means?
Nemesio
BTW, you don't need any trig at all if you extend the lines out far enough and begin filling in all the angles you know. Eventually you will solve the puzzle by additive and subtractive geometry.
Also, another rule I haven't heard anyone mention:
any quadralateral contains 360 degrees of included angles
I'd love to post up my drawing for everyone to see that it is possible.
Anyone want it via email?
Post up your address or whatever and I'll send it on.
Originally posted by THUDandBLUNDERMaybe I am just reading into this, but I think that this sort of response makes people like timebombted so aggitated, tantalizing him with "don't YOU know." It makes it a competition, every mind for itself, rather than a potentially collaborative effort where everyone learns, benefits, and prospers, and in a non-aggressive and at times vitriolic environment. If this forum were a high-pressure think tank in mathematical academia, then I could understand the childish goading, but in the friendly forum of the off-topic puzzles section, is this necessary?
Your previous post, a triumph of form over content, could hardly be called a cogent argument.
And how come, given the iron logic of your solution, that you don't KNOW if your answer is correct or not?
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Anyway, I repeat timebombted's question: is his method and answer correct? And while I think the the method is fundamentally correct, one must ask is there a simpler way that we are just not seeing?
Nemesio
Originally posted by nemesioI was asking a valid question, not having a personal dig at him. I capitalized KNOW, not YOU. One reason why many people like maths is that we can often be SURE when we have the right answer - especially with a simple puzzle like this, which involves no more than a bit of trig.
Maybe I am just reading into this, but I think that this sort of response makes people like timebombted so aggitated, tantalizing him with "don't YOU know."
Anyway, I repeat timebombted's question: is his method and answer correct? ...[text shortened]... t ask is there a simpler way that we are just not seeing?
Nemesio
As for whether his method is correct, I will let you know when he posts one. Until now, all I have seen is a plausible handwaving argument. 😛
Here is how I solved it:
Let the unknown angle CDE = x
EBC = BEC = 50; therefore BC = EC
Now we use the Sine Rule twice.
In triangle CDE, EC/sinx = DC/sin(160-x)
This can be rewritten as BC/sinx = DC/sin(x + 20).......[1]
In triangle BDC, BC/sin40 = DC/sin80.........................[2]
[1] divided by [2] gives sin40/sinx = sin80/sin(x+20)...[3]
The solution of [3] is x = 30
Check: sin40/sin30 = sin80/sin50 = 1.285575219
Of course, some purists would turn their noses up at such an 'inelegant' trig. solution, preferring instead a geometrical approach.
Eight different solutions and some interesting historical background can be found at http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf
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Originally posted by GhostintheShellAlthough you had the right answer, you couldn't prove it.
I'd like you guys to read back through the thread and note...who had the correct answer first and knew that he was right. Me, thats who.
Its a simple geometry problem and I solved it on page 1. I never guessed at it, I never ma ...[text shortened]... I figured out the answer, stated the answer and it was correct.
Perhaps that is why nobody heard you. 😛
If the answer had been x = 29 I doubt if you would have been correct.
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Originally posted by THUDandBLUNDER
I was asking a valid question, etc.
Fair enough. I am interested in understanding the solution as well. But, sometimes, as we have seen, even the suggestion of a dig will result in a digression.
Here is how I solved it:
Let the unknown angle CDE = x
EBC = BEC = 50; therefore BC = EC
Now we use the Sine Rule twice.
In triangle CDE, EC/sinx = DC/sin(160-x)
This can be rewritten as BC/sinx = DC/sin(x + 20).......[1]
In triangle BDC, BC/sin40 = DC/sin80.........................[2]
So far, so good....
[1] divided by [2] gives sin40/sinx = sin80/sin(x+20)...[3]
The solution of [3] is x = 30
Got lost here. Could you fill in the '...' for me. Other than trial and error, how do you arrive at 30? My trig grade in high school was not that stellar...
Nemesio
Originally posted by nemesioI was afraid that you would ask that. 😴
Got lost here. Could you fill in the '...' for me. Other than trial and error, how do you arrive at 30? My trig grade in high school was not that stellar...
Nemesio[/b]
Use the formula
sin(x+y) = sinxcosy + cosxsiny
to get
sin(x+20) = (sinx * cos20) + (sin20 * cosx)
and substitute in [3]
OMG this is ridiculous.
The method I used is exact, and there's enough there to solve it without trig. No guessing, no protractors, no interations.
You extend all the lines, create a perpendicular line and two horizontal lines, and the answer is right there. You'll find two triangles are created at the top left and top right, and there's (barely) enough info that you can solve it with simple subtraction.
I don't see what the resistance is to giving me a little credit for solving the puzzle first. Its laughable.
Originally posted by GhostintheShellAs you may remember from mathematics classes, the answer is not as imporant as the method used to arrive at it. The answer alone will never get you full credit.
OMG this is ridiculous.
The method I used is exact, and there's enough there to solve it without trig. No guessing, no protractors, no interations.
You extend all the lines, create a perpendicular line and two horizontal lines, and the answer is right there. You'll find two triangles are created at the top left and top right, and there's ( ...[text shortened]... istance is to giving me a little credit for solving the puzzle first. Its laughable.