you get these shit-tier posts go viral all the time on the shit-tier

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this is a perfect fit.

>>8519414
>pear = 17/13
>apple = 20/13
>bonono = 32/13

fuk dis

>>8519541
The first one has infinitetly many solutions.I stopped there.

>>8519549
m8

>>8519541
I got 13/17, 10/17, 28/17

That integral comes down to
[math]\frac{1}{x^{4}+1}[/math]
So it's not too bad.

>>8519433
REVEAL YOUR SECRETS TO ME

>>8519600
haven't done it, but consider trying to draw a line parallel to AC that passes through B, and then using parallel line rules to find the inner angles of that triangle

Did the integration on the trusty Ti-84 and got 1.758

>>8519646

Italians do It Better

>>8519646
"Wrong"
-Donald J. Trump

>>8519646
>copied from wolfram and forgot to replace log

Obviously, if you can not do the math, you say that others are wrong

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>>8519681
And who are you motherlicker?

>>8519681
>copied from wolfram and forgot to replace log
yes yes sure , replace log with ?

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>>8519686
Image flip + substitution method as well

>>8519689
Fuck it, no image flip, sick

the result of an integral is written in algebraic form, where is the log?

>>8519646
this is right

>>8519711
Is this just a separate solution
Or is >>8519686 correct as well?

>>8519433
im bad at geometry, i don't know why am i even browsing this board, but isn't answer 10?

>>8519433
What the fuck

>>8519414
Let the intersection in the middle be X

180=BCX+XBC+BXC, BXC=20
180=FXB+BXC, FXB=160
180=FBX+FXB+BFX, BFX=0

Am I retarded or was this a troll image?

>>8520237
Replied to wrong post
>>8519433

>>8520237
The 80 shown is ABC, not EBC.

>>8519433
Right-angle triangles total 180 degrees

>>8519433
20
tell me what i did wrong

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>>8519689
>he bars his augmented matrix

biggest sign of a FUCKING BRAINLET

>>8519433
i spent an entire evening to solve this
feels good mang

2.125296257155888

>>8520392
u fucking wot m89?

>>8520591
Us chosen ones just leave the matrix without the bar that separates what is on the sides of the equation.

[spoiler]I got the same answer except I didn't feel like putting the simplified integral into mathematica.[/spoiler]
[spoiler]Also I used my calculator to row reduce[/spoiler]

>>8519941
this a well-known geometry problem. There's no way a run-of-the-mill high-schooler or even math undergraduate can solve it.
It can be bruteforced using the law of sines, and there are quite elegant solutions that involve extra constructions.

See https://en.wikipedia.org/wiki/Langley%E2%80%99s_Adventitious_Angles

>>8519687
natural log

>>8519433
10
wasn't even hard

>>8519414
These are always embarrassing because whoever spent the time making them was evidently trying to be witty but they were too dumb to understand that the thing that made the simpler versions effective bait was that they included undefined and ambiguous notation such as stacking two symbols, which could be interpreted in various ways. This one on the other hand is just a standard problem with fruit symbols for variables.

>>8521765
wrong

>>8521742
you mean

ln

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>>8519433
There must be an easier wat, my answer is 10 degrees (used sin and cosine rule and have set the base of the largest triangle to 1 to aid the calculation)

>>8519433
I got 58.93 degrees, kek.

>>8521923
In what?

>>8522387
Right answer is 30°...

>>8522387
lol kys

>>8519686
>>8519689
Congrats, that is the correct answer

you could totes have used Wolfram Alpha or some shit but I don't blame you, I designed it to be as annoying as possible to solve while still being doable with basic calculus and linear algebra knowledge

>>8519433
If I'm correct that's one of the "world's hardest simple geometry problems" http://thinkzone.wlonk.com/MathFun/Triangle.htm

I remember solving this particular one, or a very similar one back in highschool,

you had to construct a bunch of annoying triangles and the answer was like 20 degrees or sth

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>>8523227

>>8523231
autism

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>>8519433
Are all of you retarded? Triangles total 180 degrees, no matter what. Look again.
20+20+80+80+30=230

>>8523422
It's just 80 + 80 + 20.

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>>8523444
Fuck me I get it now, is that a common way to write in the angles?

>>8523453
>is that a common way to write in the angles?
no not really.

>>8519414
I'm confused by this. Does a set of 3 numbers exist that could satisfy all of those equations? I tried solving for banana and apple based on the pear but I get different amounts for apple in equations 1 and 3. I did this: (variables are first letters of respective fruits)

2b - 3p = 1
2b = 3p + 1
b = 1.5p + 0.5

substitute into eq. 1:
2p - 2a + 1.5p + 0.5 = 2
3.5p - 2a = 1.5
-2a = -3.5p + 1.5
a = 1.75p - 0.75

substitute with eq. 3:
3p + a - (1.5p + 0.5) = 3
3p + a - 1.5p - 0.5 = 3
1.5p + a - 0.5 = 3
1.5p + a = 3.5
a = 3.5 - 1.5p

What did I do incorrectly? Is the question flawed?

>>8523401
prove it wrong faggot

>>8522699
Its the notation of natural log "el"+"en"

>>8522387
>degrees
kys

>>8523629
>Not creating a matrix and reducing to row echelon form

Get off my board you filthy pleb

>>8519433
>20+20+30+80+80=230
>230 degrees in a triangle

Common core strikes again

>>8524057
The 80 degree angles are ABC and ACB of the larger triangle, so it's 80+80+20 which equals go fuck yourself for not reading the thread

>>8519414
Not sure it's a linear combination. Could be an arbitrary group. Then it's probably unsolveable.

>>8519433
Isn't this triangle impossible?

>>8519414
I actually did something like this when that thing was circulating. But, I wasn't smart enough to come up with a good one.

>>8524294
>probably unsolvable
Do you even read the threads you're on?

>>8519433
this took an extremely long time but I got 30 degrees

>>8524741
Show me your proof (or give an outline of it)

>>8519433
>110 and 100 degree acute angles

I see what you did there.

>>8524830
Basically I brute-forced the problem.

I drew up the triangle and calculated the length of all of the lines involved. Relative to the length of BC:

BC = 1
AB = AC = 1/(2cos(80)) ~= 2.879385
height (distance from midpoint of BC to A) = tan(80)/2 ~= 2.835641
etc. To do this for FE, I had to calculate their coordinates relative to B = (0,0). FE ~= .684040.

Lastly, I calculated the coordinates of the intersection of BE and a line perpendicular to it, passing through F: the corner of a right triangle with FE as its hypotenuse. A simple distance formula calculation later, I had the lengths of three sides of a right triangle where the angle FEB could be calculated using inverse sine, cosine, or tangent.

>>8519433
Really not that hard desu I'll post the answer

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>>8519433
Here it is
The answer I got is 30 with simple geometry

>>8524066
Falling for this b8

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>>8525139
this

>>8525185
Mfw this

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>>8519433
Why always make it so complicated if you can do it so easily.
From that picture I get 30.0°.

>>8525185
>>8525202
>I spent 40 min finding the answer and find out the entire thing is drawn to scale

>>8525235
guessing 5*pi/2 based on partial integrals t. TI-83

definitely not prepared to prove it though

>>8519433
That is completely wrong because the ABC triangle has angles which have a sum greter than 180°

>>8525831
I can't tell if this is b8 or if you're a woman

>>8519414
First, we'll replace pears with "a", apples with "b", bananas with "c" and lemons with "x"
[math]2a - 2b + c = 2, \quad 2c - 3a = 1, \quad 3a + b - c = 3 \\ 3a = 2c -1 = 3 - b - c \implies 3c = 4 - b \implies b = 4 - 3c \\ (2c - 3a) + (2a - 3b + c) = 1 + 2 = 3 \implies 3c -2b - a = 3a + b + c = 3a + 4 - 2c \\ 5c - 2b - 4a = 4 \implies 5c - 2b - 4c - (2c - 3a) = 3 = 3c - 2b - a = 4 - b -2b - a \\ 4 - (3b + a) = 3 \implies 3b + a = 1 = 2c - 3a \implies 3b + 4a = 2c \implies 3b + 4a + c = 4 - b \\ 4b + 4a + c = 4 \implies a + b + \frac 14 c = 1 \ implies 2a + 2b +\frac 12 c = 2a - 2b + c \implies \frac 12 c = 4b \implies c = 8b \\ 2a - 2b + 8b = 2a + 6b = 2 \implies a + 3b = 1 \implies a = 1- 3b \\ 2c - 3a = 16b - (3 - 9b) = 1 \implies 25b = 4 \implies b = \frac{4}{25} \\ a = 1 - \frac{12}{25} , \quad b = \frac{4}{25}, \quad c = \frac{32}{25} \\ \left[ \int \frac{2\sin^2 x - cos(2x)}{x^4 + 1}\, dx \right]_{b-c}^{c-a} [/math]
Ahora nos enfocamos en la parte de arriba de la integral:
[math]\require{cancel} 2\sin^2 x + cos(2x) = \cancel{\sin^2 x} + \sin^2 x + \cos^2 x \cancel{- \sin^2 x} = \sin^2 x + \cos^2 x = 1 [/math]
And now we solve the integral that gives you:
[math] \left[ \dfrac{\ln\left(\left|x\left(x+\sqrt{2}\right)+1\right|\right)-\ln\left(\left|x\left(x-\sqrt{2}\right)+1\right|\right)+2\left(\arctan\left(\frac{2x+\sqrt{2}}{\sqrt{2}}\right)+\arctan\left(\frac{2x-\sqrt{2}}{\sqrt{2}}\right)\right)}{2^\frac{5}{2}} \right]_{b-c}^{c-a} \approx 1.64 [/math]

>>8526239
>Ahora nos enfocamos en la parte de arriba de la integral
Sorry about that, I was focused in a spanish forum and i forgot to write it in english, that part means "Now we focus on the top of that integral"

>>8526239
Cringe

>>8526239
boi he did it

>>8526239
no mostraste como resolver la integral. sos un puto

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>>8526239
"now we solve the integral"

don't die just yet