Find w

>>8829954
Your mom's weight divided by two, so that is 2↑↑↑↑↑3

>>8829971
that would've made it like 60 pounds

>>8829954
I solved it but I dont wanna tell u the answer

4.

sqrt( 122 - sqrt(1109) )

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I didn't know which one you meant so I circled both. also there's three M's up there that could also be W's if flipped upside down.

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Took four and a half, no calculator.

>>8830087
Are you retarded? That can't be w you dumb nigger because it's obviously longer than the base.

>>8830087
nice try brainlet but that's completely wrong

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I have solved it but the proof of my solution is too dank to fit in the margin of this shitpost

>>8829954
>If you can't solve that in under 5 minutes
I hate sniveling competition, isn't it good enough that I can solve it?

>>8830087
there are so many things wrong with this, it has to be bait

>>8830109
prove it

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It's about 6cm

There is no "w".

>>8830134
ayy

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>>8829954
Say we call the height of the left triangle l, and that of the right triangle, r. Since we have the lengths of both hypotenuses, we can solve for l in terms of r, or r in terms of l.

Let's express the longer height L in terms of the shorter S and flip the picture so that the vertical leg of the triangle of shorter height is on the left. Lets call the bottom left corner the origin, so that the point where the 5m touches the x-axis is called (x, 0). So, now we have two linear equations in x. We'll call the height of this perpendicular line in the middle h.

1. h = L/w * x
2. h = -S/w * x + S

=> L/w * x = -S/w * x + S
=> (L + S)/w * x = S
=> x = w * S/(L + S)

Because L is expressed in terms of S, S/(L + S) reduces to some value a, 0 < a < 1.


We now know the position of the vertical leg as a percentage of w, and we know the lengths of the hypotenuses of the two big right triangles, we can now calculate the the values of the hypotenuses of the two little right triangles. This gives us the lengths of both the height and the hypotenuse for both of the little triangles. We simply have to calculate the bottom of both triangles and sum the lengths of the bottoms together to get w.

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>>8830157
Here's a picture to illustrate what I mean.

>>8830157
what's the answer then

>>8829954
there is no possible solution
die and be damned

>>8829977
>2^3^3^3^3^3
>about 60

Anon, I...

>>8830164
Wrong

arccos(w/10) = arctan(5/x)
arccos(w/12) = arctan(5/(w-x))

solve

>>8829954
got a 4th degree polynomial

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>>8829954

Simple Geometry
t. Hanzo

>>8830248
was it w^4 - 244w^2 + 13775?

>>8830134
Yes.

>>8829954

w can't exist.

such a configuration isn't possible.

any positive height < 5m is possible though.

>>8830283
Substitute w' = w2 and you'll have a nice quadratic to solve.

>>8830134
heh.

>>8830263
>7' 9 3/16''
My god, what is this mess?

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>>8829954
Well done OP, I was about to call you a Nigger and a Jew, because I didn't think your pic was possible because you obviously shooped the 5m in.

w = 19.5256m

I've solved it numerically since I don't think there is an analytical solution, but I'm not sure since my math is a little rusty. You are still a faggot though.

>>8830619
w < 10

>>8830630
er, maybe..
are the full line lengths 10 and 12, or just the top parts?

>>8830630
>>8830636
Blue circle is 10.
Green circle is 12.
(0,0) is the center of both circles.
Red line is 5.

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>>8830640
With grid. This seems to be the only solution for h=5 btw.

[eqn]w \approx 4.297328005 [/eqn]

workable

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>>8830662
forgot puzzle lol

[eqn] 12^2 = a^2 + w^2 \\
10^2 = b^2 + w^2 \\
\frac{5}{a} + \frac{5}{b} = 1 [/eqn]

Solving this system gives >>8830652

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>>8830666
Forgot picture

>>8830664
This solution involves taking a new measurement that is conducive to calculating w

>>8830666
how'd you get the 5/a + 5/b = 1

[eqn] -w^8 + 388w^6 - 51736w^4 + 2610400w^2 = 32890000 [/eqn]

>>8830672
https://en.wikipedia.org/wiki/Intercept_theorem

Once used for both of the triangles.

>>8830666
>>8830667
I thought the upper segments were 12 and 10m, not the full lines

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>doing someone's homework

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>>8830676
nice

>>8830697
I don't think it can be correct with 5m there though just looking at it

ah nvm, it's probably ok

so we get wa/wb = sa/sb = xa/xb = a/b = 12/10

5/a = wb / w = sb / sb+12 = 10 / 10+sa

>>8830697
>>8830727
[eqn] \frac{5}{a} = \frac{w_a}{w} [/eqn]
[eqn] \frac{5}{b} = \frac{w_b}{w} = \frac{w - w_a}{w} [/eqn]
Add them together
[eqn] \frac{5}{a} + \frac{5}{b} = 1 [/eqn]

>>8829954
Do the measurements "12m" and "10m" refer to the larger line segments, or the smaller line segments?

If it refers to the smaller line segments, I think w has length 21m (approx). I spent way too long on calculating that answer. Label the upper-left acute angle x, and the upper-right acute angle y. By drawing a horizontal line at the top of the 5m segment, we can create some additional triangles, which allows us to find:

You can draw a few more triangles to get:
w = 5 tan y + 5 tan x

With all of that, you can also get:
w = 12 sin x + 10 sin y
12 sin x = 5 tan y
10 sin y = 5 tan x

Through a miserable amount of pain-in-the-ass basic trig and basic algebra, one eventually gets:
720(sin^4)(x) + (-551)(sin^2)(x) = 0

The relevant solution is:
(sin^2) x = 0.76527777777 (approx)

⇒ x = arcsin(sqrt(0.76527777777)) = 61.0214996 degree (approx)
⇒ y = arctan(12 / 5 * sin(x)) = 64.5316102 degree (approx)
⇒ w = 12 sin x + 10 sin y = 19.5258454256 (approx)

Double checking result:
⇒ y = arcsin(5 / 10 * tan(x)) = 64.5316102 degree (approx)

Double checking result:
⇒ w = 5 tan y + 5 tan x = 19.5258453784 (approx)

Looks right to me. I re-checked my work enough, so I think I didn't make a mistake, but goddamnit I haven't done this sort of basic math in years.

>>8829954
If you assume the "12m" and "10m" are the lengths of the larger line segments, then:
w = 12 sin x
w = 10 sin y

Draw a few more triangles, and you still get:
w = 5 tan y + 5 tan x

I know that narrows it down to an exact solution, but I don't know if you can solve that algebriacally. I'm getting a non-trivial 6th degree polynomial, which cannot be solved algebraically. I don't know if I'm missing some other way to solve that. As for numerical solutions, using Wolfram-Alpha, I get:
x = 0.366244 radians (approx)
y = 0.444197 radians (approx)

w = 12 sin x = 4.29733281528 (approx)

double checking work:
w = 10 sin y = 4.29732945233 (approx)
w = 5 tan y + 5 tan x = 4.29733145046 (approx)

>>8830890
>length 21m (approx).
Please ignore typo. I provide correct answer later in same post:
> 19.5258454256 (approx)

>>8830263
Did you make 5m 5 foot?

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>>8829954

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>>8830134
lel

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Here he is!

Why does 5m look about the same size as 10m?

You don't expect these things to be drawn to scale, but come on, make it look somewhat realistic.

>>8829954
I couldn't do it.
What's the bad news?

>>8830461
found the communist

>>8830087
i cant stop laughing at this

>>8832469
What the hell are you talking about?

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>>8829954
There's no unique solution. We need to know at least two angles subtending w and/or at least two other lengths or something. Correct me if I'm wrong

>>8830087
Kek it's impossible for both bottom line segment to be the same length as their top counterpart, unless the thick black lines are not perfectly vertical.

>>8832506
lol

>>8832526
Are you laughing at pic related or some error you think I made? I'm 100% sure this isn't solvable without more information

>>8832529
Well, you're 100% wrong. Look at my solutions above.

>>8829954
God you people are fucking brainlets.

It's trivial that following an analytic continuation, w is -1/12 units in length.

>>8832531
You're wrong though. Just try to manipulate the line segments in your head. Maintain the known measurements but increase the length of each line segment whose length is unknown. You will see that w changes as a result, and so we must conclude that without more information there is no solution

>>8832538
Don't know what you're talking about. The description is somewhat ambiguous whether the lengths refer to the larger or smaller line segments, but both problems are solvable.

>>8832531
Except he's right by your own admission. Why would you ask if they are the smallest or biggest line segments?

>>8832541
?
Because it's unclear if the "12" and "10" are labels on the smaller line segments, or the larger line segments.

>>8829954
I got approx. 19.6 but idk since I have autism.

>>8832548
Which would change the answer. Although not sure why he said angles.

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>>8832540
>>8832548
This is what I mean

>>8832565
Except in w' the lines aren't bounded by the vertical thick black lines.

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>>8832565

>>8832568
Damn, you're right. Guess I'm gonna shoot myself now ;_;

>>8830134
Neat