How smart are you, /b7?

How smart are you, /b7?

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60?

Nevermind that's wrong.

It's 50.

It's impossiblecrayton esye

GOD FUCKING DAMMIT IM A RETARD. ITS FUCKING 70. HOLY SHIT I SHOULD KILL MYSELF

Nope. I miscalculated. It's 60. Final fucking answer

There you go.

no

Prove that it's wrong.

The sum of any angle in any triangle you can make out of that adds up to 180 degrees.

sum of all angles in any triangle adds up to 180 degrees* or pi

X = 30
I AM INTERNET SEARCH GOD!!!

Where the fuck did you get this 50 from?

Just use a protractor, dumbass.

youtube.com/watch?v=HQc-54hQ8kw

You had it

40°

x and the red angle could be any 2 fucking numbers that add up to 110, how the fuck is this even a discussion?

This can't be solved. The immediate triangle the x is surrounded by has another angle that can't be solved for. Thus, you have another missing angle. The only useful equation relevant here is that all angles in a triangle add to 180. Trig cant be used, and the nonparallel opposing sides of the quadrilateral further prevents a resolution. Two missing variables and one equation makes this unsolvable.

there is no way to get it.
x plus that other bottom angle will equal 110, but that is the closest you can get.

40 degrees

Wrong. All fucking wrong, there is no answer you dumbasses

Brought to you by SketchUp.

Well this guy solved it

Fucking retards

It's not even the same fucking problem

Well then fucking prove me wrong and stop spewing bullshit

Yes it is... now I completely understand why you can't get the answer

A^2 +b^2= c^2
X=350°

Do you not see the shape of the 2 fucking problems? Do you not notice how they're different?

Or, you know, just pull out your fucking protractor, slap it on the screen and get your X.

X will always be 30.

360-160= 200

What the fuck where did you get this answer

Did you watch the video at all, do you see the numbers, do you see what the problem is asking to be solved. Do you see that the quadrilateral here is the same one in the video except flipped. Do you see the user that posted a protractor on the same quadrilateral mocked up in SketchUp. It's the same problem

Has already been solved on youtube.
youtube.com/watch?v=HQc-54hQ8kw
The video is titled "Can You Solve The Hardest Easy Geometry Problem?"

Not flipped, sorry, rotated

Here it is again, genius.

>implying geometry problems are drawn to scale

you are implying that the shape corresponds to the actual angles written there.
also, we cannot draw the shape because we dont have lengths of any of the lines.

lol google "langley's adventitious angles" you will see this EXACT same problem lol

X= 60.

Also, use a fucking protractor.

>How smart are you Sup Forums?
Smart enough to watch a youtube video I guess, good on you Sup Forums

Of course, because all geometry problems are to scale

told you I AM INTERNET SEARCH GOD!! lol i can find anything you could think of

if you use a protractor it is 30 degrees. i know because i just did it in response to your post.
also the given angles are close, but not exact.

>"use a fucking protractor"
>angle is clearly below 45 but gives says 60 anyways

How'd you find this?

i cant believe i actually took a picture.
as you can see i slipped a little and the protractor isn't line up perfectly, but here. also i did measure some of the other angles and they are very close to what it says they are.

dedictive reasoning

>china

you have to convert from metric to freedom.

x is 30 degrees, its more acute than the 60 listed and more obtuse than the 20

did it take you a long time to think of something to say?

fucking retards

you don't know if it's 2d or 3d.

therefore, you cannot solve for x.

Yeah you can actually.

Since retards here think you can't solve this because they are retarded, try this one instead.

It can't be done. Listing all potential solutions with a function is the furthest you can go

you cant solve this one with the given information.
nobody is falling for it except people who are as retarded as you are.

Check it out
youtu.be/HQc-54hQ8kw
Ok retard, solution for OP. If you can't solve the one I just posted there is no helping you
youtu.be/HQc-54hQ8kw

it's 60 you retards

Since you know all the angles of several triangles, can you just set the side of some triangle to 1 and use the sin/cos/tan formulas to get the lengths of the rest of the sides (relative to the side of 1) and calculate the rest of the angles?

Would you have to make any assumptions? Can't be bothered to do the trig myself

you are the retard.
just because a=b you cannot assume that c=d (c and d for those symbols you used)
that is actually explicitly untrue because the angles in corners a and b are different.

Hahahahaha...

Me either. I don't think you would because all of those trig calculations are basically defining ratios. So you're just saying "some side is definitely 100% of itself (thus 1) and so now I can use all of these angles to calculate how big every side is relative to that first side". Once you fill all of those in, getting those last angles should be easy. With these trig problems it's never really about the actual length of sides in actual length values. It's more or less saying "if I get the ratio of side A and B, then I know angle N on this triangle". Just a thought.

cant you only do that with right triangles?

If you do the law of sines a bunch of times you could do it eventually

i never said angles A and B are the same as C and D I stated that since all quadrilaterals are equal to 360 if the top two angles are equal to each other, then the bottom two must be as well

80

Yeah. The Law of Sines will let you figure it out because you have two angles and can get a side equal to 1 for purposes of finding ratios. I think this is actually the answer for solving the initial problem once you have figured out all the implicit angles. I'm still not going to do it myself, though. I don't feel like doing the law of sines that many times.

nigger what? can you read?
i said in that YOU claim that A=B and therefore C=D.
I said that was a logical fallacy. I never said that A=C or A=D or B=C or B=D or A=B=C=D at all.
So no. the bottom two sets of angles may all add up to be 200 degrees, but you cannot say that it is split evenly between the two corners C and D.
also omfg god captcha

That does not necessarily apply in an irregular quadrilateral. You're making assumptions.

dude, no.
i said that you cannot use sine, cosine, or tangent to calculate it because you don't have any 90 degree angles. how do you think the math even works? you need a 90 degree angle to find the ratio in the other two angles.

You can use the Law of Sines. It's not limited by right triangles. Give it a quick Google search to see how it works. All it needs is two angles and the length of one side to start calculating the rest of the triangle. It technically needs three angles, but if you know two then you also know the third by sum of angles.

no, you cannot actually.

oops. you are right.
i havent seen that shit since highschool so i forgot about it.
is still a nigger though

No worries, man. At least we have the answer in abstract. I'm not doing all those sines calculations, but it turns out you can solve for x.

You can do this without the law of sines as well. Have to draw in some isosceles triangles, but it can be solved using simple methods.

45

no

you can solve this one because you are given information about the lengths of the sides

in OP's, you aren't, so you cannot definitively solve it

only fuck it’s 0
wholesome of retards in this thread

pi is fucking circumference in a circle based off diameter your "or pi" just pissed me off

pi is the ratio of circumference to diameter

180 degrees is equivalent to pi radians

damn, that's a great video. and it's a great problem.

it only works for these particular angles (maybe there are other ones too...not sure).

i thought the angles were arbitrary, and therefore thought there was no solution.

thanks for posting the vid :)

you actually don't need to know the lengths of the lines.

watch the video.

it actually is solvable.

holy shit, i didn't think it was.

x+the angle next to it do not equal a right angle