geometry problem for sci

[Anonymous No.15052796]

interesting problem. gonna be nice to see different approachs

[Anonymous No.15052807]

is it suppose to be a square or any wrecktango?

[Anonymous No.15052811]

>>15052807
square

[Anonymous No.15052834]

>>15052796
70?

[Anonymous No.15052843]

>>15052834
no, the inscribed triangle is not isosceles

[Anonymous No.15052846]

>>15052811
It's clearly a 10x9 rectangle. OP is lying.

Problem is unsolvable.

[Anonymous No.15052860]

>>15052846
its a square and its solvable bro what's wrong with you

[sage No.15052862]

>>15052860
>what's wrong with you
YOU ACCUSER!

SEED OF SATAN YE BE.

LORD[clap]OF[clap]LIES[clap]

[Anonymous No.15052863]

>>15052846
you realize that the grid in the background page doesnt mean anything right? If OP says it's a square, then it's a square

[sage No.15052879]

>>15052863
>If OP says it's a square, then it's a square
If she identifies as a rectangle who are you to deny her lived truth?

Her documents even say she is; >>15052811

Deal with it, racist asshole.

[Anonymous No.15052883]

>>15052796
Not enough information, stop wasting our time

[Anonymous No.15052888]

>>15052879
(you)

[Anonymous No.15052895]

>>15052883
here is a similar task that has been answered

https://mindyourdecisions.com/blog/2020/06/30/a-magical-triangle-think-outside-the-box/

[sage No.15052897]

>>15052888
>your face
>no u r
>pthhghgh

Got'em.

[Anonymous No.15052935]

>>15052796
100?

[Anonymous No.15052939]

>>15052935
don't think so but can you prove it?

[Anonymous No.15052992]

Assume side length 1. In the middle triangle call the top side (between the 40 and 80) a and the bottom (between 40 and x) b. Don't care about the other one.

Law of sines on top triangle
a/sin(90) = 1/sin(80)
a = 1/sin(80)

On left triangle
b/sin(90) = 1/sin(50)
b = 1/sin(50)

Middle triangle
b/sin(140-x) = a/sin(x)
sin(80)/(sin(140)cos(x) − cos(140)sin(x)) = sin(50)/sin(x)
sin(80)/(sin(140) - cos(140) tan(x)) = sin(50)/tan(x)
tan(x) = sin(50)sin(140)/(sin(80) + sin(50)cos(140))
x = atan(all that shit) = 51 degrees

[Anonymous No.15053029]

>>15052992
this is right but how can you do it without trig?

[Anonymous No.15053038]

>>15053029
You use your grade school education to earn $10 and buy a trig book.

[sage No.15053063]

>>15052992
>Assume side length 1
...and .9.

Do it again and do it right this time.

[Anonymous No.15053069]

>>15052796
Use:
>Use angles around a straight lime add up to 180
>Angles the trisect the top left corner add up to 90
>Angles in a triangle add up to 180
To work around the triangles making up the square. Eventually you'll be left with 4 angles
>x + y on the inner triangle
>a + b on the bottom right
Use the fact angles on a straight line add up to 180 and angles in a triangle add up to 180 to set up a system of equations.
>Solve for x
Think that works.

[Anonymous No.15053073]

>hey why hasn't anyone tried the completely fucking obvious thing yet? I can't be bothered to check but I'm definitely sure it will work.

[Anonymous No.15053080]

>>15052796
Undefined, but dependent on side. Is it a square?

[Anonymous No.15053084]

>>15053080
yes

[sage No.15053100]

>>15053080
No.

[Anonymous No.15053105]

>>15053069
Doesn't work.
Essentially, you aren't using the fact that it's a square, so the problem is missing a constraint.

Yes, I did write the entire system out and throw it in wolfram.
>>15053029
The answer is irrational, hence you can't solve it without trig.

[Anonymous No.15053116]

>>15053105
>The answer is irrational, hence you can't solve it without trig.
The fuck are to you taking about?

[sage No.15053118]

>>15053105
So I was right? OP is a liar?

>>15052846
>Problem is unsolvable.

[Anonymous No.15053127]

>>15053038
Patrician's Response.
I applaud you sir.

[Anonymous No.15053201]

I find this approach interesting. I figure it should work, but it's not working. Could be arithmetic errors tho

[Anonymous No.15053802]

>>15052846
based greek geometer

[Anonymous No.15054780]

>>15052796
it was 85

[Anonymous No.15054783]

>>15054780
adjusted

[Anonymous No.15054790]

>>15053029
this

[Anonymous No.15054862]

>>15053029
>>15054790
The answer is irrational while all the inputs are rational. The only non-trig formulas that you could use to solve for angles are the "the 3 angles of a triangle add to 180" types, and none of them will ever give an irrational answer.

[Anonymous No.15054867]

>>15054783
Following you until the lower right rectangle. Why is the 90 degrees angle above x split into 45 and 45?

[Anonymous No.15054904]

x = I HATE THE ANTICHRIST

[Anonymous No.15054967]

>>15054867
to find the angle x-40

[Anonymous No.15054996]

50
X+y=140
Y=90-a (a is adjacent x)
A=130-x
X+90-a=140
X+90-130+x=140
2x-40=140
2x=100
Dumbest board by far

[Anonymous No.15055020]

>>15054996

[Anonymous No.15055201]

>>15052796
X is 50

Bros, can one improve iq by seriously working hard every single day solving puzzles; reading high brow lit, philosophy, history, etc., ; exercising & improving endocrine profile; meditating; and have some sort of epigenetic Lamarckian change in mys lf and and my progeny. Intelligence after all is a polygenic trait, so is it unreasonable to have such expectations?

[Anonymous No.15055205]

>>15055201
>>15055020

[Anonymous No.15055210]

>>15055205
Why?

[Anonymous No.15055214]

>>15052796
Ignore bottom right triangle.
Let R = |line between 40 and 80|, L = |line between 40 and x|, W the width of rectangle, H the height of rectangle.
Consider the altitudes from the top left point.
Altitude of top triangle is Wsin(90) = Rsin(80)
Altitude of middle triangle is Rsin(140-x) = Lsin(x)
Altitude of bottom triangle is Lsin(50) = Hsin(90)
Multiply these three together (keeping the lefts on the left and the rights on the right).
WRLsin(140-x)sin(50) = HRLsin(80)sin(x)
giving
Wsin(140-x)sin(50) = Hsin(80)sin(x)
giving
Wsin(50)[sin(140)cos(x) - sin(x)cos(140)] = Hsin(80)sin(x)
giving
sin(50)[sin(50)cos(x) + cos(50)sin(x)] = (H/W)sin(80)sin(x)
giving
cos(x)sin(50)^2 = sin(x)[(H/W)sin(80) - sin(50)cos(50)]
giving
cos(x)sin(50)^2 = sin(x)[(H/W)sin(80) - sin(100)/2]
giving
cos(x)sin(50)^2 = sin(x)[(H/W)sin(100) - sin(100)/2]
giving
cos(x)sin(50)^2 = sin(x)sin(100)[(H/W) - 1/2]
giving
cos(x)sin(50)^2 = 2sin(x)sin(50)cos(50[(H/W) - 1/2]
giving
cos(x)sin(50) = sin(x)cos(50)[(2H/W) - 1]

In the case of a square, tan(50) = tan(x) giving x = 50
In the case of the 10x9, tan(50)*(5/4) = tan(x) giving x = 56.12733964970903

[Anonymous No.15055225]

>>15055210
FCEB isn't a square

[Anonymous No.15055260]

>>15052796
Ignore bottom right triangle.
Let R = |line between 40 and 80|, L = |line between 40 and x|, W the width of rectangle, H the height of rectangle.
Consider the altitudes from the top left point.
Altitude of top triangle is Wsin(90) = Rsin(80)
Altitude of middle triangle is Rsin(140-x) = Lsin(x)
Altitude of bottom triangle is Lsin(50) = Hsin(90)
Multiply these three together (keeping the lefts on the left and the rights on the right).
WRLsin(140-x)sin(50) = HRLsin(80)sin(x)
giving
Wsin(140-x)sin(50) = Hsin(80)sin(x)
giving
Wsin(140-x)sin(50) = 2Hsin(40)cos(40)sin(x)
giving
Wsin(140-x)sin(50) = 2Hsin(40)sin(50)sin(x)
giving
Wsin(140-x) = 2Hsin(40)sin(x)
giving
W[sin(140)cos(x) - sin(x)cos(140)] = 2Hsin(40)sin(x)
giving
W[sin(40)cos(x) + sin(x)cos(40)] = 2Hsin(40)sin(x)
giving
Wsin(40)cos(x) = sin(x)[2Hsin(40)-Wcos(40)]
giving
1/[2H/W - cot(40)] = tan(x)

In the case of a square, x = 51.05324821679765
In the case of the 10x9, x = 58.69009207912

[Anonymous No.15055279]

>>15055260
Yeah, u're right. Is X, 51 or 50?

[Anonymous No.15055281]

>>15055279
Meant for>>15055225

[Anonymous No.15055296]

>>15055225
Alright, join FE. Feb is congruent to bed. Then z=10 & y=80

[Anonymous No.15055299]

>>15055225
And yes, fceb is a rectangle

[Anonymous No.15055339]

>>15055279
x=51.stuff t. >>15052992 >>15055260

[Anonymous No.15055352]

[Anonymous No.15055357]

>>15055352
Why is there 10%?

[Anonymous No.15055362]

>>15055357
Easier to tell it to put a 10 degree angle at a known place than an 80 degree angle at an unknown place.

[Anonymous No.15055546]

>>15054967
But if you assume it is 45 degrees, that doesn't ensure that angle bisector is perpendicular to the other line

[Anonymous No.15055552]

>>15055352
And what happens when you drag points D and C an equal diatance to the right?

[Anonymous No.15055696]

post solution im done trying

[Anonymous No.15055764]

>>15055352
>t. physicist

[Anonymous No.15055777]

>>15053201
How do I get mathy handwriting?

[Anonymous No.15055798]

>>15055777
I started a thread for more input on this question:
>>15055796

[Anonymous No.15055844]

>>15052796
Sorry I'm too lazy to write this in TeX.

b = 180 - (90 + 80) = 10
=> a = 90 - (40 + 10) = 40
=> f = 180 - (90 + 40) = 50

So we have f=50. Furthermore, we have

c + x = 180 - 40 = 140
c + d = 180 - 80 = 100
d + e = 180 - 90 = 90
e + x = 180 - 50 = 130

This is a system

1c + 0d + 0e + 1x = 140
1c + 1d + 0e + 0x = 100
0c + 1d + 1e + 0x = 90
0c + 0d + 1e + 1x = 130

of four linear equations in four unknowns, of which the corresponding matrix is

1 0 0 1 | 140
1 1 0 0 | 100
0 1 1 0 | 90
0 0 1 1 | 130.

The rref of this matrix is

1 0 0 1 | 140
0 1 0 -1 | -40
0 0 1 1 | 130
0 0 0 0 | 0,

and so there are infinitely many solutions x (depending on d, c, and e) that only need to satisfy

d + x = 140
c - x = -40
e + x = 130.

[Anonymous No.15055846]

>>15055844
forgot the pic, whoops

[Anonymous No.15055875]

>>15055696
Why man, even I haven't given in inspite of posting 4 or 5 wrong answers.

[Anonymous No.15055908]

>>15052796
lmao angular momentum isn't conserved above 80° so your triangle is impossible, fuck off

[Anonymous No.15056600]

>>15055844
>>15055846
I took a similar approach and found the same result. Sneaky sneaky, op.

[Anonymous No.15056613]

>thread posted
>someone posts correct solution in a couple hours
>two days later idiots are still claiming it's impossible

[Anonymous No.15056647]

>>15052796
Upper left upper: 90-80=10
Upper left lower: 90-50=40
Lower mid left: 90-40=50
Lower mid right: ??
Right mid mid: 140-x
Right mid lower: x-40
Lower mid right: 90-x+40=130-x
From triangle: x+40+140-x=180
I'm retarded

[Anonymous No.15056680]

>>15052796
ladies and gentlemen... we gottem

[Anonymous No.15056820]

>>15052796
I'm not gonna do your homework for you kid. You have to be 18 or older to post here.

[Anonymous No.15056844]

>>15053029
Only if the angles are multiples of 15°, not the case here.

[Anonymous No.15056857]

>>15053029
Try doing geometry. You can extend line segments as much as needed for example.

[Anonymous No.15056872]

>>15055764
You mean engineer

[Anonymous No.15056878]

>>15054904

[Anonymous No.15057001]

>sci can’t even solve highschool trigonometry problems
The absolute state of 4chud

[Anonymous No.15057810]

>>15055844
>and so there are infinitely many solutions x (depending on d, c, and e)

Not true at all. The problem with your reasoning is that c, d, e and x all depend on each other, meaning that the equations you found are not linearly independent.

i.e. you got 2 variables and 1 true equation.

[Anonymous No.15057871]

>>15052796
Op, I surrender

[Anonymous No.15058042]

>>15052796
Nice problem anon.
I can only see solution of it through trigonometry... but is there a way to solve this problem without using trigno (simply using angle properties) ?
i think its not possible with it

[Anonymous No.15058640]

>>15057810
>c, d, e, and x all depend on each other
Correct
>meaning the equations aren't linearly independent
Incorrect. Their dependence is given by the equations. They don't depend on each other as scalar multiples, and so their dependence is not "linear."

[Anonymous No.15058643]

>>15058042
see >>15055844 and >>15056600.

[Anonymous No.15058698]

>>15055844
I appreciate the effort, but this answer is wrong. The solution is unique, which can be proven through a trig and coordinate bash, like in >>15052992

An easy way to see your solution is wrong is that if x-c=40, but x+c=140, then x=90 and c=50.

[Anonymous No.15058767]

>>15058698
The solution using trigonometry assumes that the figure is a square and not a rectangle. Perhaps I missed where you said that somewhere in the thread. If you want the solution where the figure is a square, it can be obtained rather easily by finding the d which makes the length of the bottom side equal to the length of the left side.

>An easy way to see your solution is wrong is that if x-c=40, but x+c=140, then x=90 and c=50
Sorry I'm not following. Why can't x be a right angle? Is there anything preventing this other than the way you have drawn the picture?

[Anonymous No.15058768]

>>15058767
Nvm just scrolled up lol. Fair enough.