You should be able to solve this

You should be able to solve this.

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Give me a moment to check my notes.

>tfw being a NEET for so long has destroyed your ability to do HS-level work

Never gets old.

Recent college grad studying for the GRE and I'm actually amazed at how bad I am at high school math. Shit's pedantic as fuck

Well math is naturally pedantic, user. Good luck on the GRE, and enjoy your NEETdom afterwards.

You should be able to solve this.

Wtf

At some point I did remember how to do this but then I realized I had no idea what I was solving other than a math puzzle that served no purpose what so ever. And so is the general state of higher math in school.

BEHOLD, THE THREAD DESTROYER.

What was Sup Forums's reaction to that?

What's elementary geometry?

isn't this one literally impossible?

I can't read squiggly
Z is North, X is East, Y is North-East

120 degrees

It's possible but the angle values given make it impossible. A triangle's interior angles have to add up to 180 degrees. However the numbers given throw that requirement out the window. Once that requirement is gone; the formulas no longer work. You'd get two or three answers where there's meant to be one.

Only a fool couldn't handle such simple math.

Assuming that AB and DE are parallel lines, it would be 70°, otherwise I have no idea.

>Assuming that AB and DE are parallel lines
Why would they?

Dunno, that's why you assume things.

...

You don't just assume things to solve questions like that.

Okay, for a moment they looked parallel to me, and got a flash.

No bully please.

The third angle with 60 and 70 is 50 degrees yet looks more than 90. The angle to either side of it on the same line are 130 degrees but look like 70. This is why you don't assume

110

What's the first question even asking?
If we're supposed to solve for x, we can't because that's not a fully formed equation. It needs to be equal to something.

Simplify the expression.
I know that you really wanted to impress people by showing that you know an equation needs equal signs but you should have thought about it a tiny bit more.

the question and answer are both readily available on the first page of google results, you fucking retards

but there's no answer

nice try at being cool, fag

>no one solved these simple problems
>tfw Sup Forums is brainlet board

It's been a while but I think this is right.

alpha and angle BDE sum to 130 is as far as you can get, there are infinite solutions

I have no clue how get the missing two angles

It's been tried over and over again. You have to guess the answer and try to get the remaining angles that way.

Why did you think any different?

What is with this child level math?

Enlighten us, then, O wise one.

That's not even HS level. That's some primary school shit. Kids that are 10 years old can do it.

This is for 13-14 year olds

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That's because the picture lies to you.
Look at them fucking angles.

Lots and lots of shit posting. Some math posting. General bemusement. That was before the malaise of the actual show set in.

>implying parallel lines exist
t.Gauss

It actually is mathematically provable to be impossible
look at check the value of what should be γ (the angle at C). user here gave it the value 20 because he constructed a rectangle with E,D, C and the intersection between EA and BD. However, when you construct a large triangle ABC, the angle should be 80° to make the sum of all angles 180°.
The question can't be answered with the values given here.

Should have stopped here. Fuck that image.

No, it is not. It is possible to get the answer and prove it mathematically. You just can't solve it by relying on systems of linear equations only.

all you have to do is assign any length to a side and go from there

the answer is 20

It's common practice to not have the image make sense for the angles given so that
a) You can reuse the same image and just stick in different values
b) Some lazy students can't just measure the answer

Are you literally retarded? This has nothing to do with the solving method, if the angles don't add up, it's completely impossible for the given values to add up to something.
Also this has absolutely nothing to do with linear equations, this is geometry.

Pfft.
Literal amerifat common core shit spreading nonsense to the more intelligent youth.

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Even if you didn't have the idea to solve it by elementary geometry (and you are too lazy to find the solution on internet), you can still solve it by trigonometry. So, it is still not "is mathematically provable to be impossible".

If you prefer you can do it without actually calculating the values and end up with a long equation. I heard bitches love long equations.

>bitches
Yep, c/p americuck learning new words.
What a wonderful society we live into.

duckware.com/tech/worldshardesteasygeometryproblem.html
Why didn't you just like, google it? You're baiting me aren't you.

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the "drawing parallel lines" argument is fallacious because he, without a solution doesn't know where (relative to other points) the new point (F) in his examples will fall. If F falls on either side of E, the equations used will change. By linear systems the problem has 4 unknown variables and a system of rank 3. Any solution will work so long as it's orthogonal to the systems nullspace, ie that (\alpha + CDE) - (BDE + CED) = 0
Essentially, he's using a not-to-scale image to wrongly impose extra restrictions on the solution space.

laughed more than I should have

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Quaternion?

Back in high shcool used to get some projects or questions applicable to a familiar application to maintain some semblance of sanity.
Like figuring out the optimal path a butterfly would travel in the wind, or angles required by a cannon to hit a moving target, etc...
I am shit at maths myself but do find it fun on occasion when working on problems you might encounter in practice.

There is only one way to construct the triangle fulfilling all the conditions required (in Euclidean geometry). If any solution will work, the easiest way to show it is to show alternative diagram. But, I really doubt you can do it.

I want to believe you because I can't immediately discredit trig-based (arbitrary fixed length) approaches. The trig-less approach is infinite solutions though.

If the trig-based approach shows that there is only one solution, then there is only one solution even if one goes trig-less approach.

The actual solution.

Iteration is a shit tier method that I had to use to much in chemistry

>when you construct a large triangle ABC, the angle should be 80° to make the sum of all angles 180°.
I think you made a mistake somewhere because γ = 20° is correct for both cases

I finished civil engineer like 4 years ago and I'm amazed at how useless I am without a calculator

>civil
>useless
>amazed

You shouldn't be able to solve this.

checkmate Langley fags

This artist should be shot Jesus fucking Christ.

>small difference in size between 70° and 10°
>130° angles drawn as acute angles
>50° drawn as obtuse
>30° angle close to a right angle in size

Everything is so fucking off visually

At least no one posted any Kurisu board problems

underrated

70deg is literally smaller than the 60deg.

If it's a math problem, the illustration is just an abstraction.

The problem has nothing to do with the artist

>The angles of a triangle add up to 180
>60+60+70=180

Are you dumb?

Yes, user, I am dumb for pointing out the flawed logic in someone else's solution

There is no triangle with angles 60, 60 and 70?

You should be able to solve this

Not on a flat plane.

There isn't one in the image I meant. That 70 obviously refers to the angle between the black lines.

I can, is calc hard now?

I don't even see a shaded area

stop posting your high school cal 1 homework on Sup Forums.