Riddles
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- Chronically Blathering
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Re: Riddles
It should be able to be calculated, as there's only one triangle that could work. However, my math knowledge seems to be failing me. 150 is a close estimate, but it may be a few degrees short of it based on actually building such a triangle. Looked to me to be around 147 degrees.
- gamecreator
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Re: Riddles
If people will have troubles solving this, I will provide a hint.
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- Chronically Blathering
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Re: Riddles
Perhaps, the key may be to define the lengths of the sides and apply the laws of sines, cosines, and tangents to get lengths of the legs of the triangles, and from there the measurement of the angles involved. Further, there is no reason not to create points E, F, and G by elongating line segments until they cross through the outer lines of the outer triangle.
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- Game Master
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Re: Riddles
Yes, something like that came into my mind too, but I just did not have the time.Davecom3 wrote:Perhaps, the key may be to define the lengths of the sides and apply the laws of sines, cosines, and tangents to get lengths of the legs of the triangles, and from there the measurement of the angles involved. Further, there is no reason not to create points E, F, and G by elongating line segments until they cross through the outer lines of the outer triangle.
Since right and left line are equal, we might be able to calculate the other lines to provide the line length as the missing 5th formula I need.
I tried with E-F-G but had not enough information to get anything out of them (without line length).
- Zathyr
- Smiths Silly Smiles
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Re: Riddles
Well if you extend the line CD it creates another isosceles triangle with sides equal to CB and AB. If the point of intersection along AB is labelled E, then CB = CE = AB.
Furthermore, the triangle AED is also isosceles, with AE = ED. Meaning also that BE = CD.
But it's been ages since I've done any real trigonometry so turning this into something more useful is going to take some time, when I have it. Right now the best I have is that the angle we're looking for is equal to the angle ABD + 80 degrees.
Furthermore, the triangle AED is also isosceles, with AE = ED. Meaning also that BE = CD.
But it's been ages since I've done any real trigonometry so turning this into something more useful is going to take some time, when I have it. Right now the best I have is that the angle we're looking for is equal to the angle ABD + 80 degrees.
And always make sure your dragons are happy little dragons.
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- Chronically Blathering
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Re: Riddles
Well, the sine of angle BCD is equal to the sine of 20 degrees divided by the square root of ((the sine of 80 degrees multiplied by the sine of 30 degrees divided by the product of the sine of 50 degrees and the sine of 100 degrees) squared + (1 - (the sine of 80 degrees multiplied by the sine of 30 degrees divided by the product of the sine of 50 degrees and the sine of 100 degrees)) squared - (2 * (the sine of 80 degrees multiplied by the sine of 30 degrees divided by the product of the sine of 50 degrees and the sine of 100 degrees) * (1 - (the sine of 80 degrees multiplied by the sine of 30 degrees divided by the product of the sine of 50 degrees and the sine of 100 degrees)) * the cosine of 80 degrees)). What that means the angle equals I leave to somebody who has a functioning graphic calculator.
- gamecreator
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- Chronically Blathering
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Re: Riddles
I think i did something wrong there. Going to recheck my work. Sine somehow wasn't positive when I managed to find batteries for my graphing calculator, so obviously I must have done something wrong.
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- Chronically Blathering
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Re: Riddles
Given that the law of sines starts us off.by, if we define the lengths of the side we know as 1, stating the length of the side opposite the 80 degree angle, which I shall define as x:
sin(50)/1=sin(80)/x
sin(50)*x=sin(80)
x=sin(80)/sin(50)
Now, given this we endeavor to prove the point where CD would bisect AB, which we will define as point E, which happens at a 100 degree angle for angle AEC.
We again use the law of sines to measure the length between A and this point, which I shall define as y:
x/sin(100)=y/sin(30)
x*sin(30)=y*sin(100)
x*sin(30)/sin(100)=y
(sin(80)/sin(50)*sin(30)/sin(100))=y
Now, BE measures 1-y, and given we know that 100 degrees+40degrees=140 degrees and that's 40 degrees short of completing a triangle, and given properties of isoscoles triangles we know DE is length y. We can next proceed by using the 80 degrees of BED to determine the length of BD, which I shall define as length z, using the law of cosines.
z=(y2+(1-y)2-2*(y)*(1-y)*cos(80))1/2
z=(((sin(80)/sin(50)*sin(30)/sin(100))2+(1-(sin(80)/sin(50)*sin(30)/sin(100)))2-2*((sin(80)/sin(50)*sin(30)/sin(100)))*(1-(sin(80)/sin(50)*sin(30)/sin(100)))*cos(80))1/2)
From this measure z, we can calculate, using the measure the angle of BDC by plugging values into the law of sines.
z/sin(20)=1/sin(BDC)
sin(20)=z*sin(BDC)
sin(20)/z=sin(BDC)
sin-1(sin(20)/z)=BDC
sin-1(sin(20)/((((sin(80)/sin(50)*sin(30)/sin(100))2+(1-(sin(80)/sin(50)*sin(30)/sin(100)))2-2*((sin(80)/sin(50)*sin(30)/sin(100)))*(1-(sin(80)/sin(50)*sin(30)/sin(100)))*cos(80))1/2)))=BDC
This reveals that the angle we're looking for is 28.334490435743 according to my calculator, but this is obviously incorrect. That said, if we subtract this number from 180, we get 151.665509564257 degrees, which is rather close to the correct angle.
sin(50)/1=sin(80)/x
sin(50)*x=sin(80)
x=sin(80)/sin(50)
Now, given this we endeavor to prove the point where CD would bisect AB, which we will define as point E, which happens at a 100 degree angle for angle AEC.
We again use the law of sines to measure the length between A and this point, which I shall define as y:
x/sin(100)=y/sin(30)
x*sin(30)=y*sin(100)
x*sin(30)/sin(100)=y
(sin(80)/sin(50)*sin(30)/sin(100))=y
Now, BE measures 1-y, and given we know that 100 degrees+40degrees=140 degrees and that's 40 degrees short of completing a triangle, and given properties of isoscoles triangles we know DE is length y. We can next proceed by using the 80 degrees of BED to determine the length of BD, which I shall define as length z, using the law of cosines.
z=(y2+(1-y)2-2*(y)*(1-y)*cos(80))1/2
z=(((sin(80)/sin(50)*sin(30)/sin(100))2+(1-(sin(80)/sin(50)*sin(30)/sin(100)))2-2*((sin(80)/sin(50)*sin(30)/sin(100)))*(1-(sin(80)/sin(50)*sin(30)/sin(100)))*cos(80))1/2)
From this measure z, we can calculate, using the measure the angle of BDC by plugging values into the law of sines.
z/sin(20)=1/sin(BDC)
sin(20)=z*sin(BDC)
sin(20)/z=sin(BDC)
sin-1(sin(20)/z)=BDC
sin-1(sin(20)/((((sin(80)/sin(50)*sin(30)/sin(100))2+(1-(sin(80)/sin(50)*sin(30)/sin(100)))2-2*((sin(80)/sin(50)*sin(30)/sin(100)))*(1-(sin(80)/sin(50)*sin(30)/sin(100)))*cos(80))1/2)))=BDC
This reveals that the angle we're looking for is 28.334490435743 according to my calculator, but this is obviously incorrect. That said, if we subtract this number from 180, we get 151.665509564257 degrees, which is rather close to the correct angle.
Last edited by Davecom3 on Wed Apr 24, 2013 2:09 pm, edited 1 time in total.
- Zathyr
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Re: Riddles
Did you switch the labels of some of the points? It seems like you start referring to point B as A halfway through there. ÔêáBDC is what we're looking for in the original problem. ÔêáADC is 140. ÔêáAED is ÔêáAEC, 100. ÔêáBED is 80.
And always make sure your dragons are happy little dragons.
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- Chronically Blathering
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- Location: Gainesville, Florida
Re: Riddles
Yes, I admit I did start calling it by the wrong name somewhere there, but nowhere that affects the calculation of the angle. Kept thinking of A as the top point for most of that.
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- Chronically Blathering
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Re: Riddles
And, yes, that was the answer I'm going to give, unless you require we be more precise.
- gamecreator
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Re: Riddles
FYI sin(80)/sin(100)=1 (for obvious reasons). And you calculated your expression wrong.
- Guus
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Re: Riddles
Can I use super threadromancy skills to revive this topic with a new riddle?
I won't block you but your sight,
You see me most in morning light.
Quite easy, I'd say ;-)
I won't block you but your sight,
You see me most in morning light.
Quite easy, I'd say ;-)
Last edited by Guus on Fri Mar 21, 2014 3:58 pm, edited 1 time in total.
I feel smart, but I'm pretty sure I'm an idiot.
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- Game Master
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Re: Riddles
Go ahead I would say. I studied math, and some seem to have done the same or are at least on the tops of school math, but still I see no solution or further attemps for nearly a year.
- Zathyr
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Re: Riddles
For a while I meant to get back to the triangle problem and really give it some attention, but never quite had the time/inclination and then I just forgot about it entirely.
As for Guus' new riddle:
Eyelids? They block my sight and my eyes are generally at least half-closed for most of the morning.
As for Guus' new riddle:
Eyelids? They block my sight and my eyes are generally at least half-closed for most of the morning.
And always make sure your dragons are happy little dragons.
- Guus
- Floods your Ears
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Re: Riddles
Fog, yes!
Was it a good riddle? I made it up myself, like most other riddles I use for D&D sessions
Was it a good riddle? I made it up myself, like most other riddles I use for D&D sessions
I feel smart, but I'm pretty sure I'm an idiot.
- gamecreator
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Re: Riddles
Good enough for me.
I don't have any riddles, so whoever wants to post one is welcome to do so.
Or you can try and solve one that killed the thread. Here is a hint: you need to plot an additional regular polygon.
I don't have any riddles, so whoever wants to post one is welcome to do so.
Or you can try and solve one that killed the thread. Here is a hint: you need to plot an additional regular polygon.
- Guus
- Floods your Ears
- Posts: 2131
- Location: Beneath sea level
Re: Riddles
But they're all number riddles And I suck at those
A new normal riddle!
I'll be with you for hours on end,
Which could be bad, most at the end.
People still come back to me,
For need or joy the case may be.
What am I?
A new normal riddle!
I'll be with you for hours on end,
Which could be bad, most at the end.
People still come back to me,
For need or joy the case may be.
What am I?
I feel smart, but I'm pretty sure I'm an idiot.
- Guus
- Floods your Ears
- Posts: 2131
- Location: Beneath sea level
Re: Riddles
Nope, but it fits the riddle ridiculously well
I feel smart, but I'm pretty sure I'm an idiot.
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- Chronically Blathering
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- Location: Gainesville, Florida
Re: Riddles
alky, *urp*, alcohol