/sci/ - Science & Math » Thread #13996025

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Anonymous Wed 22 Dec 16:15:34 2021 No.13996025 View Reply Original Report

Quoted By: >>13996037 >>13997063 >>13997072 >>13998044 >>13999541 >>14002661 >>14003897 >>14004049 >>14004188

You should be able to solve this.

Anonymous

Anonymous Wed 22 Dec 2021 16:17:22 No.13996037 Report

Quoted By:

>>13996025
You can also use Times text in paint

Anonymous

Anonymous Wed 22 Dec 2021 16:36:48 No.13996126 Report

Quoted By: >>13996130 >>13996137 >>13997063

l = sqrt(a^2-d^2) also
l = sqrt(b^2-e^2) also
l = sqrt(c^2-f^2)

Anonymous

Anonymous Wed 22 Dec 2021 16:37:50 No.13996130 Report

Quoted By:

>>13996126
oh umm maybe l doesn't form right angles with d, e, or f... umm...

Anonymous

Anonymous Wed 22 Dec 2021 16:38:58 No.13996137 Report

Quoted By:

>>13996126

OP here

forgot to say that the red line is not orthogonal to the basis

either waat it would be too easy

OP Wed 22 Dec 2021 19:38:13 No.13997044 Report

Quoted By:

OP here again
im trans btw

Anonymous

Anonymous Wed 22 Dec 2021 19:41:12 No.13997063 Report

Quoted By:

>>13996025
>>13996126
You can't solve this shit like this because the angles are three dimensional which means normal trigonometry rules do no apply

Anonymous

Anonymous Wed 22 Dec 2021 19:43:05 No.13997072 Report

Quoted By:

>>13996025
l=?
a=!
b=*
c=&
d=)
f=(
e=[]

Anonymous

Anonymous Wed 22 Dec 2021 23:19:07 No.13998044 Report

Quoted By:

>>13996025
thank you for your service.

Anonymous

Anonymous Thu 23 Dec 2021 07:58:59 No.13999541 Report

Quoted By: >>13999615 >>14004041

>>13996025
No general solution.
When yall do test or hw questions like this, you gotta consider whether or not the question has enough info to have an exact answer, and mentally go to the most extreme cases. It's like asking given 3 random side lengths, what are the angles of the triangle? Can you consider a case where there is no solution?

Anonymous

Anonymous Thu 23 Dec 2021 08:57:14 No.13999615 Report

Quoted By: >>14001850 >>14004035

>>13999541
>It's like asking given 3 random side lengths, what are the angles of the triangle? Can you consider a case where there is no solution?

A triangle is completely determined by three lengths.

Anonymous

Anonymous Thu 23 Dec 2021 19:19:46 No.14001474 Report

Quoted By:

yee haw

Anonymous

Anonymous Thu 23 Dec 2021 20:50:24 No.14001850 Report

Quoted By:

>>13999615
Ackchuwally, He clearly means values not given

Anonymous

Anonymous Fri 24 Dec 2021 00:04:48 No.14002517 Report

Quoted By:

has anyone actually tried to solve it ?

Anonymous

Anonymous Fri 24 Dec 2021 00:36:19 No.14002607 Report

Quoted By: >>14002612

Is it a regular tetrahedron?

Anonymous

Anonymous Fri 24 Dec 2021 00:37:42 No.14002612 Report

Quoted By:

>>14002607

no it's not, a,b,c,d,e,f can be any value

Anonymous

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Anonymous Fri 24 Dec 2021 00:57:37 No.14002661 Report

Quoted By: >>14002677

>>13996025
Well there’s 3 polyhedrons, all sharing 2 faces with others.
All faces have angles adding up to 180 degrees.
Just write down all the equations and solve.

Anonymous

Anonymous Fri 24 Dec 2021 01:03:57 No.14002677 Report

Quoted By: >>14002770

>>14002661

Two problems :

You can't use trigonometry to compute angles in terms of lengths as there are no right angles.

You don't know the base triangle's side lengths.

Anonymous

Anonymous Fri 24 Dec 2021 01:25:39 No.14002770 Report

Quoted By: >>14002776

>>14002677
Very well.
I can make L the hypotenuse of the 3 right triangles where the angle goes from the bottom of L to the edges a, b, and c. This doesn’t add any unknowns - Pythagoras gives the new edge lengths formed by the right triangles.
This adds a bunch of trig equations.
I have 3 equations for sum of angles equals 180.

Shouldn’t that be enough equations to solve?

Anonymous

Anonymous Fri 24 Dec 2021 01:27:21 No.14002776 Report

Quoted By:

>>14002770
It actually forms 6 right triangles - two for each triangle.
L is the hypotenuse of 3 of them.

Anonymous

Anonymous Fri 24 Dec 2021 07:08:48 No.14003897 Report

Quoted By:

>>13996025
Case 1:
Let the point where d, e, and f meet be (0, 0, 0).
Let a and d meet at (0, 2, 0).
Let b and e meet at (1, -1, 0).
Let c and f meet at (-1, -1, 0).
Let a, b, and c meet at (0, 1, 1).
a = 2
b = c = d = sqrt(2)
e = f = sqrt(6)
l = sqrt(2)

Case 2:
Let the point where d, e, and f meet be (0, 0, 0).
Let a and d meet at (0, 2, 0).
Let b and e meet at (sqrt(2), 0, 0).
Let c and f meet at (-sqrt(2), 0, 0).
Let a, b, and c meet at (0, 3/2, sqrt(7)/2).
a = 2
b = c = d = sqrt(2)
e = f = sqrt(6)
l = 2

Anonymous

Anonymous Fri 24 Dec 2021 08:17:52 No.14004035 Report

Quoted By: >>14004144

>>13999615
The third side of a triangle is not determined by the other two sides.

Anonymous

Anonymous Fri 24 Dec 2021 08:22:11 No.14004041 Report

Quoted By:

>>13999541
I have an intuition that this has an exact solution. Look at the picture and think about it.

Anonymous

Anonymous Fri 24 Dec 2021 08:25:24 No.14004049 Report

Quoted By:

>>13996025
your 3d drawing skills are pure shit, either way, use the law of sines

Anonymous

Anonymous Fri 24 Dec 2021 09:38:42 No.14004144 Report

Quoted By:

>>14004035

But the third side is determined by the lengths d, e, f and the fact that these line segments are coplanar.

You can't arbitrarily choose the third length.

Anonymous

Anonymous Fri 24 Dec 2021 10:12:14 No.14004188 Report

Quoted By: >>14004191 >>14004203

>>13996025
I've added the side lengths g, h, i, which we also don't know.
>If we are given a,b,d,e is it possible to figure out g?
>If we are given a,c,d,f is it possible to figure out i?
>If we are given b,c,e,f is it possible to figure out h?

Anonymous

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Anonymous Fri 24 Dec 2021 10:13:17 No.14004191 Report

Quoted By:

>>14004188
forgot pic

Anonymous

Anonymous Fri 24 Dec 2021 10:27:02 No.14004203 Report

Quoted By:

>>14004188
no, no and no because adl, bel and cfl can all rotate freely around l without changing their lengths. You need to include d, e and f, and the fact that they are coplanar, to get anywhere.

Anonymous

Anonymous Fri 24 Dec 2021 19:05:17 No.14005686 Report

Quoted By:

ho ho ho

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