Times you felt retarded

I cant solve this question for the life of me and this is some simple as fuck geometry 101 problem. Now this would not be as bad but considering I'm in calculus 3 this seriously leaves a dent in my ego. This fucking meme of a question is making me think its one of those millennial questions by some faggots at Clay that showed up on my class board to find Will Hunting or some shit.

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>add two numbers on paper
>get wrong answer bc numbers weren't evenly spaced and added up the wrong numbers

user remember angle properties and algebra exists. Everyone has brainless moments.

From the center, assuming the lines are straight, the angle opposite that 50 is another 50, which means the remaining angles to the side of that are 40. Figure it out that same way; 3 angles of any Euclidean triangle add up to 180, total angles from a straight line add up to 180 as well

this one took me way too long but remember that like triangles are a thing

Just chuck it into matlab desu

X=50

Every class period in my physics 1 class. I am nearly a straight A student. ive made 1 B after 3 semesters of college. But my physics professor is so insanely bad at lecturing. She is the anti-lecturer. I come out knowing less after class is over. Our first exam is coming up and I'm screwed. I might drop. I don't know. I swear I'm not a brainlet. I'm in cal 3 and I'm doing fine, I ace quizzes in cal 3 because the professor doesn't suck.

I don't know what to do besides slowly drive myself insane by teaching myself out of the book. Physics is hard enough it doesn't help when you can't read the handwriting on the board and the prof trips over their words for an hour an a half straight and never explains how to solve the problems.

I know most other students don't understand the material at all as well but I don't know if it'll be a good enough curve . I think enough students are taking phys 1 for the 2nd time so theyre not as impacted, so I might be boned on the curve, idk

Looks like there are infinitely many solutions to what x could be. I don't think there's enough information to find the 4 unknown angles. I guess there's a narrower range of possible answers if you want to assume that the unknown angles are acute like they appear to be.

Yeah just looked into it, x can be any angle greater than 10 and less than 130 by the looks of it.

Someone tell me if I'm wrong, good chance I could be missing something.

Defintely, because you could set up this in real life and see that there is no way to adjust x.

The mistake with the linear equation is that it doesn't allow for the fact that inner line are inscribed within the larger triangle, and x lies on the perimeter. The linear solve approach is just chasing angles endlessly around.

If you can get all (or almost all) of your money back then I would recommend dropping the class. I had to deal with something like that and it was hell, it wasn't worth the stress and time.

So I can do the algebra to show there are a range of solutions, but I don't know why that would be the case. I feel like I've written the problem as 4 equations, 4 variables. Are any of those equations not unique?

p sure this is the/a solution

Can you go into more detail?

It's easy, calc the other two angles internal to that triangle first.

Issue is that your equations are effectively just looking at the quadrilateral at the top. Need the extra information about how the sides coverage to fit within the big triangle.

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jk im a huge faggot who is clearly blind plz rape my face

You need to use law of sines and law of cosines

you can solve this without trig btw

for anyone struggling

[spoiler]it actually took me like an hour to figure it out[/spoiler]

X is 50 deg.

Yeah you have missed some equations.

Naming the variables like in pic related.
And using the equations in 's pic.
it solves to give j=X=50

What the hell is going on in that pic my dude?

Also, here are all the solved angles in alphabetical order (as defined by )

OK, how did you do it then?

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Mine is right.
Yours is wrong.

My most sincere apologies to everyone in this thread.
An error of mine has caused me to post incorrect things.

>pic related
Those 2 coefficients need to be swapped around.
When I do that.
The rank of the matrix becomes less than the number of variables.
There are infinitely many solutions.

Both and are correct answers.

But there are posters earlier in the thread who figured this out first.

how the fuck could there be infinite values of x. you obviously cannot increase or decrease x since it is being "restricted" by the 60/20 and 70/10 degree angle lines. just look at the picture and see how x is fixed

all you need for this are triangle properties (angles add to 180) and pons asinorum property of isosceles triangles. It helps to construct one more point as well.

>the first time i was introduced to polish notation

I want to hate your notation but I'm too lazy to pull up Matlab and try to program it myself.

Of the solutions here:

cut-the-knot.org/triangle/80-80-20/IndexToClassical.shtml

I like solution 9 (bisectors everywhere) and solution 5 (isosceles everywhere).

X = 20

The process involves autistic, repeated application of sine law in order to find ratios of the side lengths of the central interior triangle to a common reference (eg. the bottom length). Then apply cosine law for the interior triangle. It will spit out a number for cosX and you will be happy. I might've made a mistake but I don't think so.

And there is probably an easier way but whatever.

Ah t'es le français

x=30

The image is completely retarded and easy the faggot who built this deserves death

Wrong

By the time you're in university you should be able to teach yourself.

This. Self study is key.
A diploma is just an entry ticket to the academic world.

I'm taking high school algebra in college, op. You've got nothing on me.

Speaking of which, why is (-2,0) and (0,4) not apparently the correct x and y intercept of the equation in the picture?

Wait I'm retarded

The x intercept isn't -.5 either what the fuck am I doing wrong

I did it

I'm still retarded I divided the wrong numbers

I might be confused about your labelling, but shouldn't the angle

In fact

Why the FUCK is nobody solving this using analytical geometry? Brainlets, the lot of you!

delet this

why solve something simple in a complicate way just so you can be pretentious about it?

bump for correct answer pls

it was already answered let the thread die

But he was wrong and admitted he was

the matlab guy was wrong, not the triangle guy

Oh, misread. Anyway this guy is correct in pointing out that there aren't infinite solutions

How did you solve it though/

there's not enough information. if X triangle isn't stated to have it's catetii continue in a straight line to the other triangles, X can be a whole range of solutions.

How can there be infinite solutions? If you draw a line from the right bottom corner with an angle of 70 you will hit the left side at a definite angle, same with the left corner, how is there any room for variation here?

The lengths of the lines. You can extend them while still forcing the angles to stay the same. This will change the angles we do not know.

No, it goes from the corner, at an angle, and intersects a side, these determine their lengths.
Pic related, answer is 20 but dunno how to prove it.

you literally just did prove it

Nah, like you can see from the angles, that program is dogshit, I'll need to verify it.

Built this in AutoCAD and got 20

You're right, it is fixed. I went through and worked it through with solution 9.

How, we can't use the BC=CD, and how can T be an excenter when it's one of the corners?

Yeah, I also just realized it's a bullshit solution. They got it from finding AAA similarity, but they used it as if it were congruence.

yeah my bad just realized that, gonna try something else then

Looks like the answer is 25.12. I'm sure there is some way to do this using matrices, but this is much easier

Your bottom right corner is 70 not 80.

Well I'm gonna kill myself

You can prove it with sine and cosine laws like this guy but it sounds ugly

There are actually two congruent pairs of triangles in our problem, 1: GB and RGB and 2: Y and R, dunno if this helps.

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