Manga VS Sup Forums

Well, Sup Forums, are you smart enough to beat this manga using no trigonometry and only using elementary geometry?

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181

My dick + Kuro's womb + my sperm = a

70°

Don't see how these numbers are correct, considering that the top triangle's missing angle is 50 but visually it's clearly above 90 degrees.

I suppose that's the point of these threads tho

That's math for you, bad drawing but good thinking.

Can I use geology? How about Meteorology?

How exactly does solving stupid math problems prove that you are smart? I never did understand that.

the only "knowledge" you need for this is that all angles in a triangle add up to 180.

Math is the language of the universe.

It isn't.

Math is just a language of logic. It is nothing by itself.

If you need to ask that question you are not ready to know.

Thank you, I thought i was the only one thinking it. After doing that i had to google the amount of degrees in a Triangle when I got 50 but knew it was over 90. I thought I had messed up.

It's the drawing that messed up

the drawing is not supposed to be accurate you fuck, because else you could just measure the correct angle.

No shit, I'm not that stupid. But that angle isn't 50 degrees when it's well over 90

Here's a version drawn to scale.

You should be able to solve this. Stop bitching about scale and do the work.

If you're trying to make me feel stupid like these guys:
And these guys who couldn't grasp the fact that the diagram itself is intentionally misleading to dissuade solving by intuition:
Well... then... you're doing a bloody good job! ;_;

How'd I do teach.

lets see.. after solving the outer triangle by adding up the inner angles of A (70+10=80) and B (60+20=80) giving me C via subtracting from the sum of a triangle which is given as per it's definition (180-80-80=20) and solving the inner triangle formed using AB (180-70-60=50).. i have everything i need to solve the inner angle 'a' by solving it's other 2 angles using the vertical angles theorem, the definition of an isosceles triangle, and the definition of an angle bisector giving me 50 for that left angle, 70 for the right one, and finally 60 for angle 'a' itself

was this supposed to be hard..

We have a 60 degree and a 70 degree answer.

At least one of you is wrong.

Are correct.

You forgot my 181 answer.

You are retarded.

I'm sorry. I have down syndrome.

you didnt show your work so you got an F

>got it right

Haven't done triangle math in forever, awesome.

>F

Here's an hint.

just filling in the numbers doesnt prove you didnt just use trig

Double retard.

I know downs people who can solve integers.

While we're on the subject of math, how about some probability too?

50% yeah?

Do you homework already.

Reported for underage.

Some integrations are so simple even a down could do it.

CDE is not an Isosceles triangle no matter how you look at which is drawn to scale.

I thought this problem didnt have an answer though. Like, it has more variables than equations or something.

Isn't the proper way to write the probability like 0.5 instead of 50%? Like, there is 50% of smth to happen, but the probability of it is 0.5.
It's been a while since I actually learned that, though.

It can be solved, but you need to draw a few auxiliary lines.

More of a style thing, since 0.5 is just 50% of 1 anyway.

Both triangle ABC and triangle BCD are Isosceles triangles as per the given values of the inner angles of A and B which gives you all of the information you need to solve every other angle inside this triangle.

>50% yeah?
Nope.
P(HH) / (P(HH) + P(HT) + P(TH)) = 1/3

This is why you always confirm your sample space.

You can solve it in a straightforward manner using trig (law of sines and/or law of cosines). The hard part is to solve it using only elementary geometry.

H = Heads = 1/2
T = Tails = 1/2
P(H | H) = P(H and H) divided by P(H)
=(1/2*1/2)/(1/2)=1/2=
0.5

Good.

This image makes me mad.
Not because it's hard or something but because of the retarded limitation
>Solve this problem
>Oh, but you are not allowed to use this efficient method to do it
Fuck the guy who made it.

>misleading diagram
>no right angles
how the fuck am i supposed to solve this without redrawing it with a protractor and then using said protractor again to solve it if i'm only able to use elementary geometry?

this is correct

Don't ask this, you'll start another IQ vs multiple intelligence semantic fight and it will be the most pointless thing on Sup Forums.

>how am I supposed to solve it without the only acceptable way to solve it

>Not because it's hard or something but because of the retarded limitation
OK, here's a variant to suit you: Prove that $\alpha$ is exactly 20°. Now trig is a lot less useful, since you can't just plug numbers into a calculator and be satisified with an answer that's accurate to 6 decimal places.

what a waste of time

No unique solution exists.
1/3.

No, there's multiple interpretations for this one.

Depends on how you interpret "given at least one is heads", it's either 1/2 or 1/3. (1/3 is more correct in my opinion).
See en.wikipedia.org/wiki/Boy_or_Girl_paradox

Your mistake is in failing to distinguish between the two distinct coins.
Let H1 mean "Coin 1 is heads"
Let H2 mean "Coin 2 is heads"
Now try solving the problem again.

>there's multiple interpretations for this one.
How about this: "You flip two fair coins. Kuro will kiss you if and only if at least one is heads. What is the probability that both are heads given that Kuro kisses you?"

0%, she is not real

a = 75

>80° 80° 20° triangle
Why teachers never draw geometric figures proportional to their degrees?

>No unique solution exists.
That's not true. You have enough information to draw the diagram from scratch. Line DE's placement is given from lines AE and BD which have given angles applied to them thus allowing you to draw this diagram only one single way. You basically draw the diagram leaving line DE for last, simply drawing a line from point D to point E.

The only way to get a different answer for "a" would be to reconstruct this diagram incorrectly.

Because they don't need to.

I can't into maths, isn't it just 50% because the previous flip is irrelevant ?

See and

My bad, didn't open the thread before responding, thanks

1/3.
"Given that at least one is heads" means the sample space is {HH, HT, TH}. P(HH) = 1/3

>Why teachers never draw geometric figures proportional to their degrees?
Why would you need to?
Are you bad at visualization?

they don't want you to get a correct guess by measuring

answer is 20

the trick is it can be solved with a protractor if you bother to draw the whole thing out to scale. It can't be solved with math, and "elementary geometry" includes using a protractor, and the problem gives you more then enough information to diagram the problem to scale

1/3

yep, this is the right answer

Both 1/2 and 1/3 are correct because the question is written ambiguously.

Alright Sup Forums, I'm stuck. I've got everything except for the important bits, the angles at E and D. BDC has 140 degrees, AEC has 150. Not sure on how to move forward from there. Tried the hint from , but I don't know how to use that to work out anything else outside of the new triangle that I didn't already have.

You literally cannot have a triangle that dosen't have 180 degrees. You fucked up hard.

Sorry, miscommunicated. I meant that the total angle from B to D to C was 140, and A to E to C was 150.