How smart are you, Sup Forums?

How smart are you, Sup Forums?

30

smart enough to see that it doesnt add up

What do you mean it doesn't add up?

solve your homework yourself

it's 30

another one

solved

It has to be 360, you can never have more degrees than that.

pic related

A circle has 360 degrees, a triangle has 180 degrees within it.
Did you drop out of middle school or something?

HAHAHAHAHHAH

It DOES add up - interior angles of a quadrilateral = 360 and interior angles of a triangles = 180

wtf are you talking about?

lmao braindead

That's what I said, 360 is the max degrees you can have, what is your problem?

GTFO weeb

omfg

solved another one 2 ez

This is a triangle, user

HAHAHAHHAHAH

Nowhere is it noted that the angles are equal, so those 110 and 70 degrees are graphically deducted, not mathematically. There's not enough info to solve this, it's a troll.

There's obviously enough info to solve it, it just requires you not to be retarded.

Vertically Opposite angles aren't equal? WTF

hurr durr but I can just measure the angle... let's give up math alltogether

When it's stated... they are, they are not because they look that way

ikr

I dont know if I'm right. Im 16

No, faggot. We're not going to show you how to show your work on your math homework. user is right, now go cry and pray you can pass 10th grade.

Why not dude?

good job, now find the value of X

For the same reason that the left one is a right triangle, the right one it not. It looks that way, but there's not geometric proof. Get it?

You would have to assume the the intersecting lines are truly straight. They may not be, and then the distribution of angles will be lopsided.

If they are not straight then there is no point then.

Obviously the intersecting lines are straight. If they aren't, the puzzle is impossible to solve.

Yea that's what I thought

How the hell did you determine the 50/80 and 100/x side?

I got everything else but there's no way to determine the rest.

But if the lines are straight, there is no need to figure out the intersection at all.

You can do it with just the larger triangles.

My attempt...

It doesn't say to find the value, just to find x.

A straight line is 180 deg, so to speak. If the point of a triangle touches it, the sum of those angles will be 180.

was going to show it to him
thx

There is a way, it's just difficult. No one in this thread has demonstrated a way that works yet, though.

How did you determine 60 &100? I don't see how you can determine the angle of the line with no information?

So it could be anything who adds up to 130 and not specifically 50/80

for example this user choose 60/70 instead of 50/80 and he gets a different result

exactly all triangles = 180 degree's not hard to work backwards from that

he's wrong, it should be 50 and 110

Knowing that the internal angles of a triangle add up to 180 degrees is not enough to find x.

Not in a triangle...

This much is obvious.. find the remaining ones

I don't think there is a single solution

you can get from here these equation : ( and i don't think i've missed some)


180=y+(x+40)+30
180=x+(50+y)+20

180=x+y+70
180=20+z+t

180=50+y+z
180=x+40+t

and when combined it boils down to :


x+y=110
z+t=160
y+z=130
x+t=140


which has multiple solution

40 + X + C = 180
A + B + 50 = 180
70 + B + X = 180
All equations have two unknowns

Yes it is, you can 100% solve this with only that knowlege.

you can add the top triangle and the inner triangle as i did there but it doesn't help you

OHHHHH!!!!!
(A+B) + 20 + (50 + 30) = 180

>A straight line is 180 deg

I know that but there should be no way to get the 80/50 or 110/x after getting everything else first.

I assume you guessed the 50/80 and 110/x.

...

Just kidding

lmao

Show such a solution, then.

A + C +20 =180
20+50+x+b

5 equations and 4 unknown this can now be solved. I'm in mobile, so I ain't doing the work.

It work with all value of X who doesn't generate negative value of y,z,t in this set of equation.


x+y=110
z+t=160
y+z=130
x+t=140

( or :

y=110-x
z=x+20
t=140-x

because "y+z=130" is redundant with the 2 first line.

So there is only 3 equation and 4 unknown

See the following and solve for the variables

try solving them you'll see that some of them are redundant. I've found 6 equation here : but when you work on them you see that it boils down to only 3


All these solution for example would work

x= 10 and y = 100 z = 30 t = 130

x= 20 and y = 90 z =40 t = 120 etc...

there is not an unique solution for these set of equation

Tony thinks ur dumb!

> but when you work on them you see that it boils down to only 3

dude, no. There's a much much wider range since we don't have enough facts.

for example

x = 1, b=109, a=21, c=139
x = 2, b=108, a=22, c=138
And so forth.
This is for the pic in There's no real answer because we don't have enough info.

linked to wrong pic. i meant

There is obviously enough info and a real answer to the puzzle. You just haven't incorporated all the available info into your equations.

Same result. Took me a while bc I was solving writing on a bag.

it was an example.

As i said any value of x is an answer as long as it doesn't generate value of y,z, and t that are negative based on this set of 3 equation :

y=110-x
z=x+20
t=140-x

No user. Once you figured out the rest, those four are all that's left over that you cannot figure out.

Take a good look at it again and see for yourself. a,b,c and x are unsolveable.

They are unsolvable from the equations posted. They are not unsolvable from the original puzzle, you just have to go about solving it in a different way.

Use trigonometry instead you useless piece of shit

Couldn't you use these equations at the bottom to substitute and swap around, find values for X,y,t, and z?

no because when you do so you see that these equation are redundant

that can be a solution, show us.

>that can be a solution, show us.

however i'm pretty sure it will lead to the same dead end

Fixed it

in that sense yes you are right, it's anything within that certain range, tho it's a really wide range.

There's really no concrete way of figuring out the exact numbers. All we got are ranges of possible solutions.

Anyone who thinks there's one absolute solution to this is a fucking idiot.

>They are not unsolvable from the original puzzle, you just have to go about solving it in a different way.

And what way is that? Guessing games until you find one that works?

We're not looking for the lengths of each side here.

It makes sense now

Holyshit i did not think about that

gg

>And what way is that? Guessing games until you find one that works?

Obviously not, finding out a way to solve it that works is the whole challenge of this puzzle. You can straightforwardly use the laws of sines and cosines to figure it out, but it can also be solved purely with elementary methods, though finding that solution is quite difficult.

how do you infer that the small triangle is isosceles from knowing the big one is ?

>And what way is that? Guessing games until you find one that works?

You could use Trigonometry to figure out the sides, which would let you figure out the angle.

>You can straightforwardly use the laws of sines and cosines to figure it out, but it can also be solved purely with elementary methods, though finding that solution is quite difficult.

user, you can't because you don't know two variables in every possible equation here.

This guy got it

i don't see how trigonometry will add any information. But you can show us wrong.

answer this :

stop being baited by him you dumbass.

Never learned trig. But, If you make an assumption about the length of a side, calculate all the other lengths, and then find the angle, would that angle be the same regardless of what you initial assumption was?

Yes you can, you just need to set a length for one side.

Top is 20, not 40.

The biggest, outer triangle is:
> (20 + 60) + (50 + 30) + X = 180
> (80) + (80) + X = 180
> 160 + X = 180
> X = 20

Recalc your work accordingly.

are you fucking retarded? A triangle adds up to 180, a square adds up to 360.

except you'll end up with length expressed as a variable of x and W ( where W is the "assumption" you've made about one length), and that in the end you'll end up having a circular equation like x+w= x + w

we're talking about triangles here, user.

I.e the unknown angles are all constrained by the lengths of the two long intersecting lines. It's those lengths that are missing.

Clearly you don't know shit about trigonometry.

the do the math, show me

then*