Huge Rally Next Wednesday

Next Wednesday there's going to be a huge fucking pump for both BTC and ETH. Absolutely huge.

Just letting you all know. That's when the volume is going to pick up again.

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geogebra.org/geometry
twitter.com/AnonBabble

The outside triangle is 60 on all sides, but like, the top left one, if that's 60+70 the big side of it is 50 yet it's clearly > 90, is that triangle to scale at all? Unless her noted angles are part of the larger triangle and it's 80+80 so the far side is 20... those numbers are wrong I think.

I can only do math based on logic and gut.

The numbers might be wrong but that doesn't seem to matter. But from what I can tell, it's only possible to solve if you assume that lines E-D and B-A are parallel.

If you assume that, then it becomes really easy. But if not, then I don't think it's possible, and I've been trying to do it without that.

>NO TRIGONOMETRY
Because you can totally use trigonometric functions without knowing any of the lengths.

Ok so I've now shown that it's not possible to solve.

I've shown that the line E-D is not parallel to B-A, and if it's not parallel it can't be solved.

bump

Those a hooks not bait you retard.

Fuck you OP

Holy fucking shit...it's just a stock picture from Veeky Forums.

I'm saving your picture though.. Wow.

actually, we know that angle a is larger than 10 deg, but smaller than the third angle. we know the largest triangle is 80-80-20 - D, E and an unnamed point make a tiny isosceles, BDC has an angle of 20 deg, angle at point C is once again 20deg, meaning angle at point D must be 140 deg. since angle a is part of a small scalene, angle a has to be equal to the angle at point d, the upper left triangle is 60-70-30, meaning that means the smaller triangles' reflecting side is also 30deg. angle a would have to be 75 deg and angle At point D would = 105 deg

easy
you can literally solve this by finding the corresponding angles in the correct order
>BEA - 30
>BCA - 20
>BDC - 140
>BDA - 40
Let the intersection point in the center be X
BXA = 50
Therefore EXD = 50
Now the issue is finding BDC
50 - Alpha = BDE, and BDE + DEC = BDC = 140
Since
BCA = 20
DEC = 180 - 20 - EDC
And DEC = 180 - Alpha - 30
If you solve for Alpha you get a fucking non-answer because none of hte fucking corresponding angles are dfeined so there are 2 unknowns i'm so fucking mad I spent 30 minutes on this bullshit holy fuck I'm a mathematician and this is literally impossible tos olve without trig fuck fuck fuck fuck ufck

Woops, meant scalene twice not isosceles, and math is wrong for BAD, should be 60, 80, 40.since we know point C has an angle of 20, point Ds lower triangle has an angle of 140. Meaning the upper triangle at D has an angle of 40. meaning angle a, is 10-40-a, angle a is 130. The scaling makes no fucking sense, i know.

I...I am too. I got that far but that was kind of my point.

i forgot to edit out that part after spending 30 minutes trying to work through this bullshit

who /brainlet/ here

nevermind again i'm retarded, yeah so we still dont know angle a triangles' angle of point D, even with the other inferences we've made. it isnt 40 deg. it's (40+angle a)-10- angle a. dont really know or care enough to continue

There's no way to solve it without knowing at least 1 other angle, since there are 2 unknowns that are unsolveable with trig unfortunately.

See for the answer

I only did the drawing, my guess is to work out the answer you have to find out how long BC and AC are compared to AB (by making use of the angles), then find the lengths of the other segments as a function of AB and finally make use of that to find alpha. Using the angles alone is not enough

if the angle at the center is 30 based on upper lefts proportions, that means the angle under it is reflecting and also 30. meaning that angle a is apparent twice in the small scalene triangle. 30+a+a. a = 75, if a = 75, and point A = 10 deg for our triangle, that means that the last angle will have to be 75 as well.

its a 75-75-10 triangle.

Engineers waste at least 15 minutes on this, mathematicians waste at least 30, thanks for sticking it in another 15, mathbros, and confirming my hunch that it's a garbage troll problem. This was gonna bother me all day.

The picture is misleading because if you look at the large angles forming the larger triangle A and B angles add up to 80 + 80 meaning C is only 20 degrees. Pictorially this doesn't seem the case but mathematically it must be the case. The top left triangle formed by A B and Nameless (now b) must be 50 at b. By filling this in, we can find the angle of c in the triangle BEc as it is a long a straight line meaning it must add up to 180 as well. c ends up being 130 degrees making the E angle in BEc 30 degrees. Likewise, we can use the value of 50 for b to determine the unknown angle in ADd (top right triangle) as this must also equal 180 degrees, doing the same as we did for c this means c=d and is also 130 degrees making angle D 40 degrees. Now we're inside the triangle in which degree a is located and can safely say that it's top angle is 50 degrees as angles c and d in the circle that is created by the crossing of the two lines are both 130 and angle b being 50 makes this new angle also only 50. Here, knowing that angles add up to 180 in a straight line, the value of C comes in to play. So far, we have the following angles and triangles

ABb b = 50
BEc c = 130 E = 30
ADd d = 130
and angles C = 20 and e (top of mystery inside triangle) at 50 degrees.

Now we deal with the bottom triangle. The value of D being 40 gives us a clue to the value of the inner triangle as the 3 angles being drawn from it need to be = 180 and one of them is already determined to be 40, leaving 140 degrees between the inner triangle and the bottom triangle. The same can be done with the E value, leaving 150 degrees.


However this is all nonsensical and does not add up as you can make macrotriangles between lines A E and C and by knowing the values of A and C can come up with the value of a being 150 even though it would not add up with the other previously determined values as angles A and C are 10 and 20 respectively.

Proportions don't work lol
The angles aren't to scale, if the center point is X than the and BXA is 50 degrees, which is 100% incorrect in terms of proportions since it's alot smaller than AXD

the proportions have to purposely be fucked up on this triangle, or the numbers or some shit, then again im just a retarded NEET who stopped doing math after calc 1 so

It's a garbage troll problem and I went through and tried to find every unknown angle and determined that these meme numbers would just not add up with all triangles in place there will be at least one pair that ends up over 180 if all the degrees given are to be believed.

>mfw I'm deep in modern algebraic geometry, but I can't solve simple high school stuff.

Discussing memelines like it's a fact

...

BXA would have to be 30 deg dawg, 60+70+angle X = 180 ;3

alpha should be 60

Find BCA then CDB then ADC, then BDC then AEB then DEC and finally alpha

Yes but you can't solve for Alpha with BAX lol
Look at my long reply, there are 2 unknowns and BAX doesn't help you solve it
You can't solve for DEC because you don't know alpha stoopid

BDC and CDB
Oh god...

Again, the only triangles that matter are the large macro triangle and the right side macro triangle.
ABC can easily be determined as those angles are given, so C = 20
This makes the given (10) and C equal to 30, meaning that a has to equal 150. However, using the other given triangles will determine a 50 degree angle for the inside of the middle triangle at it's top as well, meaning that the given angles add up to nonsense and triangles with over 180 degrees as well as multiple values for angles in the interior. It's a troll problem. It's fucked.

You can if you know DCE and CDE

ABC is an isoceles, which means AD:BE is 6:7.
so alpha should be 60 degrees

>meaning that a has to equal 150.
Where do you solve this from?
Ok brainlet so solve CDE for me then

Construct point P such that ABP is equilateral. P lies on BD since

It's easy, 65 degrees baka

Well we do know ADC is a segment right? So the angle is 180
And we can figure out ACB since we know CAB and ABC
When you figure out ACB you can figure out BDC
Then knowing ADB and BDC you can figure out...
No wait I DUN GOOFED
I believe we should use BADE to solve this...

AC and BC are the same length (drawing not to scale), because angles ABC and BAC are the same (80deg) (look at to see the properly scaled drawing)

So how long are AC and BC compared to AB?

Let's call M the middle of AB, then ACM is a right rectangle, with angles 80 10 90. Then AC = AM/sin(10deg) = AB/2*sin(10deg) ~ 2.879*AB

So if you set AB = 1, then AC = BC ~ 2.879, then from that you can probably calculate the other lengths and eventually be able to find alpha but I'm too lazy

...

something like that works too i guess

>right rectangle
right triangle

>When you figure out ACB you can figure out BDC
Dude, the angle BDE is unknown lol, you can't solve EDC without knowing BDE
There are 2 unknowns and BDE is the primary unknown
No trig you fucking cheater

Triangle BDA has two known angles
60 and 70
They add to 130 leaving the final angle 50
That matches with the top angle in the triangle containing alpha
Subtract 50 from 180 then divide by 2

There's alpha. No trig, just angles
It's 65

Never mentioned BDE
Im done with this, on a phone and O have to scroll uo everytime which is a pain in the ass

I swear I'll kill myself if 65 isn't the answer

Why would you assume the angles alpha and BDE are equal?

Dude what you're missing is that you need BDE to solve for Alpha, and you can't solve for BDE without knowing EDC, and you can't solve for EDC without knowing DEC, and you can't solve for DEC without knowing alpha. It's a logical loop and there are 2 unknowns in the same triangle that make it impossible to solve. All the other complementary angles don't allow you to solve this.

Got bad news for you user...

Yeah thats what I assumed, hence the "I dun goofed"

Anyone can give us the answer, using trig so I

The answer is 50. The opposite is 80. You can figure out the rest. All triangles angles add up to 180 and all square angles equal 360.

Nevermind. Here is the work.

Yeah that's fair, using Trig this shouldn't be too hard

How did you find 100 and 60 in the CED triangle?

Nvm.. I see how you did it.

>mfw opening this thread thinking it's going to have tons of valuable, pertinent information due to post count and it's just a bunch of community college students discussing triangle angles

Literally every single post lol. Not one person has asked me why there's going to be a rally next Wednesday.

Tell us why. I don't think it's happening.

EVERYONE ELSE:
This question is bullshit. If you use geometry you get a different answer than if you use trig.

check that, still don't know how to solve it...

Korean holiday ends and volume will pick up a ton again. Plenty of bullish news and people will all be buying back in to start preparing for forks.

This is the correct answer I still have a hard time understanding some steps though, like why APEC is cyclic

Because it's how it is. The top angle is fo sho 50. Angles add up to 180 so subtract 50 and divide by 2

No....the world is wrong

I was wrong. I just plugged two numbers in that added up to 130 because a + b = 130 according to the 80 + 80 + (30 + a) + (40 + b) = 360 square. You can plug any number in and the whole thing will work so 0 < a < 130. There is no real answer. Plug in any number and the whole thing will always work.

Maybe someone can draw that solution on geogebra.org/geometry to make it clearer

my monitor is too small i can't see shit after drawing too many lines