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 (Some Guy) 258 More: Interesting
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57763 clicks; posted to Main » on 30 Mar 2008 at 7:07 PM (9 years ago)   |   Favorite    |   share:    more»

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Manderhozen: If you use a loose interpretation of "elementary geometry" and click on "very small hint" you get the simplest solution. it's a diagram drawn to scale. measure it with a protractor.

technically first grade is elementary school!

I've taught geometry. I had to explain, on numerous occasions, how to use a protractor to my ninth and tenth graders.

...and they wonder why I quit after a semester.

Ferserious: Quantumbunny: I remember SOHCAHTOA... Sine = Adjacent Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, and Tangent = Opposite/Angle Adjacent, what does that have to do with my numbers... when we have no lengths? I was working purely in angles. FAIL

S'weird when SOHCAHTOA has a built in fail-safe RIGHT in it...and yet, fail... Unless of course, my secret decoder ring is wrong and O now stands for Adjacent.

Sorry, I copy and pasted a few words into the wrong places trying to format it. However, I maintain that without side lengths, it's farking worthless to know SOHCAHTOA. For this problem then, it doesn't help.

I fail at teh maff but I do know the value of the square root of a Flexnard.

Quantumbunny: So an algebraic solution is still totally possible, and far preferably in my humble opinion to the purely geometric one.

Where's subby when the English police are needed?

Oooooh, Poppycock!

S'weird when SOHCAHTOA has a built in fail-safe RIGHT in it...and yet, fail... Unless of course, my secret decoder ring is wrong and O now stands for Adjacent.

Sorry, I copy and pasted a few words into the wrong places trying to format it. However, I maintain that without side lengths, it's farking worthless to know SOHCAHTOA. For this problem then, it doesn't help.

But more importantly, the point was that some chilren were struggling with their algebra. ;-)

I copied off an asian guy too, therefore, 20.

thumbsucker: I copied off an asian guy too, therefore, 20.

NOW we're talkin'!!

Ferserious: But more importantly, the point was that some chilren were struggling with their algebra. ;-)

That's fair, I admit to my misdeeds; but where the heck did SOHCAHTOA come from? It has no bearing on this problem as far as I can tell.

\Also I apologize in full to the thread for my high quality English skills today.
\\Has a doctor's excuse
\\\Is on meds.

C is 20 degrees

I can tell that drunk!

i cheated, i drew it out in a parametric cad software and then measured the results

the results? X= 20

Oddly enough, works out fine using 70 as the answer.

ADB = 180 - DAB - ABD
= 180 - (80) - (60)
= 40

calling the point where line AE and line BD intersect F:
DFA = 180 - FAD - FDA
= 180 - (40) - (10)
= 130

AEB = 180 - ABE - EAB
= 180 - (80) - (70)
= 30

BFE = 180 - FEB - FBE
= 180 - (30) - (20)
= 130

DFE = 360 - DFA - AFB - BFE
= 360 - () - () - ()
= 50

ACB = 180 - CAB - CBA
= 180 - (80) - (80)
= 20

Those are all the angles I can prove without drawing lines.
That leaves x (DEF, DEA) unknown, and also leaves CED, CDE, and EDB unknown.
x can be either 20 or 70.
I call bullshiat on line drawing being part of elementary geometry.

anthonix: Jeez, all you have to do is use AutoCAD.

//I love my job!
///Except I'm working on a Sunday. :(

out of the stone ages man, I used inventor, drew it badly, then changed the numbers to fit,

....and also without SOHCAHTOA, which the link said we are not allowed to use.
Suck it.

Hint: Draw more lines and create right triangles.

Quantumbunny: I can say with confidence, x = 40, y = 110, z = 50, w = 90. If F is the intersection of DB and AE, then Triangle DEF is a 40, 50, 90 right triangle.

x=20.

That algebra is a bunch of equivalent equations which will take you nowhere.

Drew one perpendicular, and X is 20.

This is the moment when I chime in with how easy that problem was without having clicked on the link.

Drathus: No one who writes anything in Comic Sans is believable when it comes to difficulty. =P

Off topic, but your comment reminded me of an acquaintance's website. It's kind of lighthearted, but funny:

http://bancomicsans.com/home.html

BarstoolPhilosopher: ....and also without SOHCAHTOA, which the link said we are not allowed to use.
Suck it.

C'maaannnn...the ACTUAL message in there is that "...Having Trouble Over Algebra" Little message in there to clue the bunny in. Subliminal? Yes, I agree. Totally farking nerdy? I concede.

Is it just my imagination or has my class grown considerably? Well, by no stretch of my imagination do I believe you've all come here to hear me lecture. But rather to ascertain the identity of the mystery math magician. So, without further ado, come forward silent rogue and receive thy prize...Well, I'm sorry to disappoint my spectators, but it seems there will be no unmasking here today. However, um...my colleagues and I have conferred, and there is a problem on the board right now that took us more than two years to prove. So, let this be said: the gauntlet has been thrown down, but the faculty have answered, and answered with vigor.
~ Gerald Lambeau

Sique: Here's a step-by-step solution:

I've found the easiest way to solve this puzzle is simply to keep drawing right triangles inside the first triangles, then labelling each angle as you go while continuing to draw right triangles.

Once you've filled in the triangle completely, you no longer have to stress out about X, as it is no longer visible.

Jen thinks math is for suckers. Who are you to argue?

You know... i kinda figured that out just by looking at the angles... and it worries me.

You damn libs and your math. You think your so smart but
your not. Try using that stuff againts the terrorist when
they come over here for you!!!1

BarstoolPhilosopher: Those are all the angles I can prove without drawing lines.
That leaves x (DEF, DEA) unknown, and also leaves CED, CDE, and EDB unknown.
x can be either 20 or 70.
I call bullshiat on line drawing being part of elementary geometry.

Where exactly do you get the either 20 or 70? Due to how you define them, the drawing changes, but those angles could be pretty much anything and the math will work out since they are all unknowns, you can fill in any value for one, and fit the others to that one.

whatshisname: That algebra is a bunch of equivalent equations which will take you nowhere.

Uh, you don't need them all, but one way to ensure you have all the needed info is to just add them all up:

2w + 3x +2y + 3z = 670
2x = 670 - 2w - x - 2y -3z
2x = 670 - (w + x) - (w + z) - 2(y + z)
2x = 670 - 130 - 140 - 2(160)
2x = 80
x = 40

The second step is the tricky one -- you need to notice that you need an even number of variables on the right to swap out.

i eyeballed it and got it right.

whatshisname: Quantumbunny: I can say with confidence, x = 40, y = 110, z = 50, w = 90. If F is the intersection of DB and AE, then Triangle DEF is a 40, 50, 90 right triangle.

x=20.

That algebra is a bunch of equivalent equations which will take you nowhere.

I concede that I lucked out, if you change the existing angles, the same math fails.

\Damn these meds.

Where's Max Fischer when you need him?

If the plan involves drawing the correctly-scaled diagram out using a protractor (as suggested by the hint), geometry becomes pretty irrelevant. Just measure the angle and forget about it, since you're going to all that trouble.

I got an answer different than 20 by just figuring out every angle based on the angles given. My solution is here if anyone wants to check my work for me. I don't see anything wrong with it.

um... X does not equal 50, it's 10 or twenty.

AEB 70, BDA 60
(Naming the other X shape Z now.
ZAB = 50, so it's inverse ZDE = 50
X is more acute than ZDE so it is not 50
DCB = 140, DBA = 40
Now we know all but two angles, draw them and use a bloody protractor to find X.

This is not hard, I never took Geometry.

What a farking crock of steaming stagnant shiat.

I always hated math and this still proves it.

ZoeNekros: whatshisname: That algebra is a bunch of equivalent equations which will take you nowhere.

Uh, you don't need them all, but one way to ensure you have all the needed info is to just add them all up:

2w + 3x +2y + 3z = 670
2x = 670 - 2w - x - 2y -3z
2x = 670 - (w + x) - (w + z) - 2(y + z)
2x = 670 - 130 - 140 - 2(160)
2x = 80
x = 40

The second step is the tricky one -- you need to notice that you need an even number of variables on the right to swap out.

Ha ha, you used my initial bad equation set up for this.

The real way from there would be
2w + 2z + 2y + 2x = 580
Aka 2x + 2w = 260 or x + w = 130, back to the initial equation set.

\My vodoo in the problems was legendary, I could solve them and get the right answer (if I set them up properly, unlike this one originally) in totally invalid ways.

I met this guy.
He looked like he might have been a hat check clerk at an ice rink.
Which, in fact, he turned out to be.
And I said, "Oh boy. Right again."

Quantumbunny: Ha ha, you used my initial bad equation set up for this.

Heh, oops. I fail.

doh my statements start angles with the vertex instead of vertex in the middle, that will teach me not to take a math class. I guess I didn't list all angles, but all the ones you need to find everything else.

I_Hate_Iowa: I got an answer different than 20 by just figuring out every angle based on the angles given. My solution is here if anyone wants to check my work for me. I don't see anything wrong with it.

I'm at work and not paying attention, but I don't see how you decided that that angle DA[your imaginary extension off from DE] was 40.

Also, I suck terribly at math. So, yeah.

I look at a problem like that and know that I have enough education, deduction and practical application to figure out the answer. But instead I go "Meh. Don't want to." and come into the thread to read the funny responses.

/Lazy nerd

I did it the easy way.

I figured out all the angles that I could without drawing extra lines, then saw that triangle CED and EBF were pretty much the same shape. I substituted the angles from EBF into CED and used them to get the angle x. From there I was able to get every other missing angle. I made sure they all added up properly and proved that my guess was correct. Much faster than the solution page posted earlier.

Hey cool, I got it right, and all I did was go "Well x looks the same size as C."

/Always got dinged for not showing work in math class
//Fark all that extra work, though.

I think it can be 30, but I'm sure I'm wrong. The very top triangle would be 20-120-40, and the triangle with x in it would be 30-50-100.

Relatively Obscure: I'm at work and not paying attention, but I don't see how you decided that that angle DA[your imaginary extension off from DE] was 40.

I don't remember how I did that now either. I thought I had solid reasoning behind every value I plugged in, but now I can't find that one and a lot of other values flow from that one. Even with that assumption though, I can't find a flaw in my math, so I think my answer is possible still.

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