You should be able to solve this

Solve for all the angles you can.
Then notice x depends on two triangles, which will yield two simultaneous equations in two variables. Solve the system. x=40

Sorry, x=50

40

Literally high school geometry
Triangle sum theorem

I measured it with a protractor, it came out as 30 degrees.

wrong angle brainlet

BE cuts CA in half, angles are (1:3,5), then

x = (180-40)/3,5
x = 140/3,5
x = 40

Prove me wrong, brainlets.

...

>BE cuts CA in half
prove it

x = x .
Checkmate.

Oops. Haha..yeah, that makes more sense. I was beginning to think this whole board is retarded.

Anyway, this is as far as I got. I am pretty shifty at math, admittedly. Is this system of equations solvable ?

>Is this system of equations solvable ?

even better, x ≡ x

Nevermind. I just realized that this provlem is not well defined. There are infinitely many solutions. Let the angle FEA be y. Then as long as x = y - 20, the system of angles remains stable.

I originally "found" x=50 and y=70.

Check that if you plug those angles in, the system is filled without contradictions. SHIT PROBLEM

Jesus christ I meant let y be the angle EFA

I admitted I suck at math. I haven't done geometry since high school. I am a med-fag, so I never use math really.

Anyway, instead of simply pointing out that I'm a brainlet, can you please explain what exactly was so stupid about my question regarding the system of equations? I vaguely remember there being some rule saying you need a certain number of equations to be able to solve a system with a certain number of variables, but I don't remember the specifics.

Bullshit, that's only if we didn't have the angles of 20 and 30. They 'fix' A in place, which 'fixes' others in place. Draw the diagram to scale with the angles in the diagram, and try and insert your angles. It won't work.

>Draw the diagram to scale
>Draw
Are you retarded? We are not given any measurement to think that the way the diagram is presented is how the system looks. I showed that the system of just the angles (what is given) is stable with infinitely many solutions.

Obviously if you literally measure one of the sides and fix a configuration then you collapse the set of solutions into a unique one, but that is because you inserted new information. BRAINLET

All I am saying is that if you change x and y so that x = y - 20 then the system of angles is still not contradictory.

>I showed that the system of just the angles (what is given) is stable with infinitely many solutions.

No, you didn't.

>All I am saying is that if you change x and y so that x = y - 20 then the system of angles is still not contradictory.

Maybe that's because you haven't considered all the information.

Case in point: youtube.com/watch?v=HQc-54hQ8kw

Yes I did. Insert y and x= y-20 in the system and see it is not contradictory.

I am considering all the information except the drawing, which is not information. Drawings are not meant to say anything, they just give you a picture. The problem's information is just the system of angles. That's all there is.

wow it was so simple, I would've got it if I only cared and drew it out

>The problem's information is just the system of angles. That's all there is.
I wasn't disputing that, all I was saying was to draw it out with some x and y pulled out of your arse with the equation, and to see that it wouldn't work.

it's 30, try drawing a bunch of lines to make it a bunch of isoceles triangles

to early in the morning for this shit.

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

...

Yeah, number of equations >= number of variables. He's a dick. I can't solve it although it should be solveable

>waited all day to get home to solve this
>an hour late
Geometry is the only math I'm not a complete embarrassment at.

Only one question remained: where's point D?

got to this point and just guessed and checked. took me ~5 minutes
x = 30

Just do basis transformation

(Or Gaussian elimination)

>where's point D
Inside your mom... like always.

Found the mechanic.

These niggas are the future.

waitwaiwait, how did you find out the angles in the complementary triangle (the one above BF)?

Looks like BFD reflected.

That's just a coincidence. If the construction was made on the other side, no reflection.

Did he just assume its angles?

Triangle above BF is reflected so angles are reflected too.

You cant prove it's reflected, the triangle constructed the same way on the other side won't be reflected.

Solving for this, you get 1 number equals itself. There's infinitely many solutions; if you use Gaussian elimination, one equation will cancel out.

Fuck off man. Get the solutions before posting this shit.

This is as far as I've gotten. If I was able to figure out either of either of the angles I've marked with red question marks, I would have been completely able to figure out out, using the equation 180 - (70 + (180 - (50 + n)), with n being one of the red angles.

Or is my potential solution completely incorrect?

Try looking for isoceles triangles, or creating some.

Update: By drawing a rectangle where the unknown angles are I was able to deduce the unknown angle and in turn calculate the angle of x.

The angle of x is 30.

As it turns out, calculating the angles of A, B, and C, was useless, along with attempting to calculate x directly, by making a circle at the bottom.

The actual image, which i forgot to attach the image because I'm a dumbass.

Apart from you getting the correct result, which you did, how do you know that the triangle in that rect is Isosceles? As in, how do you know it's 50 50 up there and not idk, 60 40.

Final update: Some of the things I had drawn without much thought lead to contradictions, so I removed them and cleaned up the image, to make it easier to figure out what's going on.

Call me retarded but I still don't get where you got those angles from. How do you know the triangle written into the rectangle is Isosceles? Is that relevant? Can you scale it up and number the angles that shows your progression? This just looks like you assumed some shit and said "solved" to me.

>spend ages figuring out the angle of x
>while drawing the rectangle figure that both of the angles are 50 degrees because it's mirrored and shit
>gets asked how exactly i knew it was 50 degrees
>writes response
>realizes that they're not actually mirrored
>mfw when my entire answer was only right by accident and i didn't have an actual reason for a crucial piece of evidence

Call O the intersection of the diagonals of BFCE.
Then COB = 70, hence BOF = 110 and thus COE = BOF = 110 and EOF = COB = 70.
Moreover, BAC = 20, hence CFA = 130
Now we have enough angles that we may solve everything using the law of sines several times:
sin(x) = sin(70)*OF/OE = sin(70)*(sin(20)/sin(50) )*(sin(40)/sin(30))*OB/OC = sin(70)*sin(20)*sin(40)*sin(50)/(sin(50)*sin(30)*sin(60)) = sin(20)*cos(20)*sin(40)/(sin(30)*cos(30)) = sin2(40)/sin(60)
Then, Wolfram gives x = 28.5°

>OF/OE
But we don't know the angle opposite OE (CFE). You implied after that OF/OE = sin20/sin50, but if CFE were 50 and FOE 70, that would make x 60...

That's not what I said, I used the law of sines twice, first in OFB to get OF/OB = sin(20)/sin(50), then in OEC to get OE/OC = sin(30)/sin(40).
But as a general rule, problems like this are always completely solvable because as long as the angles you are given are sufficient to draw a unique picture (up to similarity), then you can use the law of sines to express every other angle as a function of the original angles. Might not be pretty, but it will work

See pic rotated

x=60

I assume by using the 140+40=180 on the bottom and filling up angles until it came out

...

This is the only correct solution ITT
You end up with 4 equations and 4 variables. Solve algebraically

Let's call the third point on the reflected triangle (above BF) D. How do you know that D, F and E lie on the same line?

Dangit I just realized where I screwed up, ignore this I can't be fucked to do it again on mobile

In the real world you just draw and measure. You would be laughed at if you tried these magic tricks.

They degenerate to three equations. What exactly is wrong with ?

Nothing, actually. Quite a nice solution, how did you find it?

a brainlet that realized he's a brainlet gets rewarded with a piece of working brain

if you have the possibility of measuring angles, wouldn't you directly measure X instead of doing that?

it works for other angles too

yeah, show me the passages

I think that's Indeterminate, becouse any number assigned to x satisfies the conditions of the triangle (180°//360° rules).
It si obvious it works with 30°, because it works with every single x value

sorry for my english guys

No, it just means that the equations you get by the condition that the sum of the angles in a triangle be 180° aren't enough to deduce the value of x. But there are other relations (you obviously can't make x equal 90° while preserving the drawing as it is, just look at it)

Oh shit I wasn't reading, you were right of course... My bad.
So, we can just redo it all with the law of sines and the law of cosines...
EF^2 = OF^2 + OE^2 - 2*OF*OE*cos(70)
OF = EF*sin(x)/sin(70)
OF^2 *(sin(x)/sin(70))^2 = OF^2 + OE^2 - 2*OF*OE*cos(70)
Expressing OE/OF as before with 3 sine laws, we get OE/OF = sin(60)/(2*sin(40)*sin(20))
Plugging back, we get sin(x) = 1/2 and x=30°

down with euclid
death to triangles

This site has 13 solutions to the problem. This one is probably my favorite.

cut-the-knot.org/triangle/80-80-20/Classical5.shtml

nice one

Why does this same thread appear once per month?

I didn't measure angles, I was just making a system of equations, but in my hasty notation I set F equal to two different angles so my result is invalid. Sorry for any confusion

thank me later fags

>that filename
You're Italian, aren't you?
You can persuade yourself there's only one solution this way: construct a 80 80 20 triangle, choose point E on AC such that BE=AE, choose point F on the interior of AB such that BC=BF

As for the problem, idk I did it with trig but can't find a noice synthetic solution :(

It's beautiful

that's not how it works

x+y=5
x+y=6

Tell me how to solve this

Find AEB, then EBD, then ACB, then BEC, then DEC, then ECD and then x

T R I G O N O M E T R I C C E V A

Wolfram Alpha says it's an ugly number btw

Couldn't be bothered to write a rigorous proof follow the constructions in rainbow order:
Triangle BCD isosceles (just from the picture)

Point M on side CA such that angle CMB = 80º
Triangles BCM and BME isosceles because of this construction.

Draw segment MF, since Sides BM and BC are congruent and triangle BCF is isosceles we can conclude triangle BMF is an isosceles triangle with different angle 60º, thus becoming an equilateral triangle.

Side MF congruent with ME (because of the congruence shown on the step before this one) thus triangle MFE is isosceles.
We know angles BEC and FME from good ol' 180º in a triangle, thus x+40=70.

Therefore, x=30º
TL:DR follow the rainbow, all colored sides are congruent

Just fill it with trigonometric functions, you'll get an ugly expression.
wolframalpha.com/input/?i=(cos(33º)sin(15º)cos(xº+15º))/(cos(15º)sin(xº+15º))=sin(33º)cot(51º)
Hope I got it right

>then DEC
impossible

the answer is 11.something

That's not a valid solution, you can't just solve it by the sum of angles = 180°
Might as well use a protractor if you're gonna guess