geometry-and-trigonometry

[heavy snow]

sure hold on

area = 15 and perimeter = 16,4

I think thats how you say it

[dark sparrow]

yes

[heavy snow]

O = 5 + 4 + 5,38 + 2 = 16,38 A = 5*((2+4)/2) = 15

[dark sparrow]

so then why did you say that the book said perimeter = 15

[heavy snow]

mixed up the words there, sorry

[hearty rain]

@forest drift ...

[forest drift]

what

you want that proof?

[dark sparrow]

you've been all promises and no delivery, anish bhat

you know this makes you look very suspicious at best, right?

[earnest basin]

Help

[forest drift]

lol

[heavy snow]

if the task is to solve what coordinates say B has in the line AB if what is given to you is only A = (5,2) and the midpoint M = (-1, 1) how would you solve it?

[dark sparrow]

the midpoint, as in the midpoint of AB?

[heavy snow]

I assume the midpoint here is between A and B

my first thought is to find out the length of A and M and then double it to find B?

[dark sparrow]

you will only find the length of AB

which won't help you much

[heavy snow]

B is supposed to be (-5, 4)

[dark sparrow]

it's better to remember how you would find the midpoint given A and B

$M = \paren{\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}}$

[somber coyote]

Kanga Gang Annihilator Ann

[heavy snow]

right and that gives me an equation

[dark sparrow]

two equations

one for x, one for y

[heavy snow]

the Y equation is quite difficult to solve

[dark sparrow]

is it?

it should not be any more difficult than the x equation

they have the exact same structure just with different numbers

[heavy snow]

2 * (3 + x) / 2 = -1 * 2
3 - 3 + x = -2 - 3
x = -5

[dark sparrow]

and you had difficulty going through the exact same motions with (y+2)/2 = 1?

[heavy snow]

actually I was doing something wrong

got it

2 * (-2 + y) / 2 = 1 * 2
-2 + y = 2
2 + -2 + y = 2 + 2
y = 4

[dark sparrow]

you said A = (5, 2) not A = (5, -2)

but ok

[heavy snow]

sorry I was looking at the wrong task it should be A = (3, -2) B = (?,?) and M = (-1, 1)

[dark sparrow]

ah.

[heavy snow]

thats why I got Y wrong there

[hearty rain]

What do mean, 'lol'?

[upper karma]

how do we visualize the graph by reflection of a function f(x) at a given line like y=x. Is there any way we could visualize this in a nice easier way rather than plotting all the points?

[heavy snow]

is the midpoint normal between a line AB just another name for a midpoint?

[upper karma]

Like how do we visualize the reflection of the graph of a function f(x) = cosx along y=x line. Please give me some idea on what exactly you do to know this?

[heavy snow]

visualize as in using your imagination or an actual graph tool?

[upper karma]

imagination and drawing it on copy

[heavy snow]

https://www.geogebra.org/classic

put in the function here and practice visualization?

would be my take on it

if you actually know what it looks like then you can approximate a similar function

[upper karma]

actually I was tryin' to learn the graphs of inverse functions which is in fact graphs of a function reflected along y=x but I was having hard time understanding it . so

[upper karma]

I mean how do you get to know that this part of line will be concave here , convex here, will point this side after reflection?

[heavy snow]

stuff like this?

otherwise I don't know because it sounds beyond my knowledge

[upper karma]

yeah, exactly like this.

[forest drift]

ummmmmmmmmm I never proved ramanujan wrong.

[heavy snow]

no you didnt

[forest drift]

I just proved another case of ramanujan correct

[dark sparrow]

......

okay so you misled us all

that's not very nice

[forest drift]

nope

[forest drift]

nope

k

nice

[heavy snow]

stop spamming these boards with nonsense

[upper karma]

suppose u're given a function and how do u know it's graph after reflection along y=x will be like you've drawn in the ss u've sent.. plz tell me. that's the thing i'm havin' dilemma over

[heavy snow]

well put in the function you have into that graph there and it'll draw it for you

[upper karma]

okay.. sounds cool... ty so much, king

[heavy snow]

what does the function you are trying to solve look like?

[forest drift]

I just hope that you guys hate me

[dark sparrow]

you hope that we hate you?

weird but ok

[forest drift]

you hate me right?

[dark sparrow]

no

[wise pawn]

stop spamming about non math in here please

[forest drift]

ok

I'll send my proof

[heavy snow]

you can send it down the toilet

[forest drift]

ah

[heavy snow]

how do you find the midpoint normal of a line AB if A = (1,0) and B = (5,4) ?

erm the task is to write it as a linear equation, y=kx+m

[nocturne remnant]

midpointnormal?

you sure you didn't mean perpendicular bisector?

[heavy snow]

yes thats the correct term

I didn't know what it was called

[nocturne remnant]

welp i think i have one way to do it

[forest drift]

basically find the mid term using the formula and then just get the equation of a line passing through the point

[heavy snow]

(1 + 5) / 2, (0 + 4) / 2 = M(3, 2)

[forest drift]

mid point is (3,2)

so use basic graphing to find the equation of line

[nocturne remnant]

alternatively, any point T(m,n) lies on the desired line if TA = TB
so you might be able to just plug the distance formula or sth

[heavy snow]

y = 3x + 2

but for some reason the facts wants it to be y = -x + 5

[nocturne remnant]

how did you get

[nocturne remnant]

...

the line y = 3x + 2 very clearly does not pass through (3,2)

[forest drift]

ooh geogebra is funn

[heavy snow]

how do you find it then without using the graph tool?

[forest drift]

you cannot

or maybe

just find the locus

[nocturne remnant]

does the question tell you to use a graph?

[heavy snow]

no

[nocturne remnant]

well you cant use a graph in the exams

[heavy snow]

it gives you the A and B points and then you are supposed to determine the equation to the perpendicular bisector

[nocturne remnant]

was there a similar-type problem in the lesson / handouts / textbooks?

[heavy snow]

surprisingly nothing

and I looked it up and other people have being asking the same question when it comes to this problem

[dark sparrow]

well i would suggest writing out (x-1)^2 + y^2 = (x-5)^2 + (y-4)^2, expanding it out and letting the algebraic dust settle

but i'm almost certain that wouldn't go well

[nocturne remnant]

(I'll assume that means you can approach the problem whichever way you like)

[dark sparrow]

this is just TA^2 = TB^2 in elon mask's notation

[nocturne remnant]

so yeah, either plug the distance formula
or find the mid-pt, and then the desired slope in order to construct the eqn

[heavy snow]

well I solved the mid-pt (1 + 5) / 2, (0 + 4) / 2 = M(3, 2)

[dark sparrow]

sure, that's going to be your midpoint

[nocturne remnant]

desired line should be perpendicular to the original line AB

[dark sparrow]

if you want to do it that way, the next step would be to find the slope of the line AB and take its negative reciprocal

[heavy snow]

so you inverse AB ?

[nocturne remnant]

find the slope of AB first

[heavy snow]

so rise/run = 4/5 = 0,8

[nocturne remnant]

hmm

the slope should be || 1 ||

rise and run are both 4

[heavy snow]

ah I see you start from where the point is at

I was counting from 0

[nocturne remnant]

so, the slope of AB is 1

what should the slope of our answer line be?

[heavy snow]

not sure maybe (2, 1) ?

[nocturne remnant]

slope!

[heavy snow]

I don't know how you do that

[nocturne remnant]

a slope is a number

(Any two perpendicular lines have their slopes multiply to -1)

[heavy snow]

1 ?

[nocturne remnant]

well AB is 1

[heavy snow]

1 * -1 = -1

[nocturne remnant]

so the required slope is?

[heavy snow]

-1 I guess?

[nocturne remnant]

Yep!

So, we already have the slope, and that the line passes through (3,2)

now, you should be able to find the y=mx+c

[heavy snow]

my mind is blank, I have no idea

slope is the x? so y = -x + something

[nocturne remnant]

yes

just sub x = 3, y = 2 to find the "something"

[heavy snow]

you mean add? 3 +2 = 5

3 -2 = 1

[nocturne remnant]

y = mx + c
sub x = 3, y = 2, m = -1

what is c?

[heavy snow]

what do you mean sub x = 3, y = 2, m = -1?. 3 - 2 + -1 ?

[nocturne remnant]

substitution

juse replace the letters in the equation with the corresponding numbers

[heavy snow]

2 = -3x + -1 ?

[nocturne remnant]

?

be careful

[heavy snow]

2 = 3 - 1 unless you count out the -x slope

I'm just guessing because I don't know how you'd approach this

[nocturne remnant]

???

this is basic algebra

substituting

y = mx+c
2 = (-1)(3) + c

[heavy snow]

weeeell that explains it

[nocturne remnant]

yeah c is 5

problem solved.

bye

[heavy snow]

wouldn't have been easier just to show me that to begin with?

[nocturne remnant]

well i didnt think you'd get stuck on such a thing

since you know about midpoints and stuff

[heavy snow]

how am I supposed to know you are supposed to solve it this way?

its the reason I asked for help to begin with

[nocturne remnant]

i reckon this is quite clear already

but whatever

bye

[heavy snow]

for you it is, not me, okey, thanks, bye

its a common pattern I've noticed when you ask for help about a math problem. People just assume you know everything already and they are too lazy to imagine why the person who is asking for help is thinking in a particular way about a problem. If you can imagine what the person who is asking for help is thinking which leads to the wrong answer, you can also guide and show that person the correct way of thinking.

[hearty rain]

I am not angry, I am disappointed

[molten delta]

Do you still need help?

[abstract fog]

We can see here this is supplementary angles

In the supplementry formula we use =180

So we will have
90 + 3x + 2x-20= 180

70 + 5x = 180 we combine like terms

5x= 110

x= 22

We know x is 22 but we are not done yet

It wants to know what angle c is

So we now go (2*22 - 20)

44-20

24

So we get our answer which is m<c is equal to 24

Hopefully I got it right and you understand

@hexed tiger

Did I do the work correctly

I’m not that good at geometry and idk if I am leading in wrong path

[loud vale]

hey guys does anyobdy want to join help 1 with me

its on solving equations

[upper karma]

What does x positive mean over here as highlighted in the blue text? Can x ever be considered negative? How could distance be negative? Plz tell

[humble pulsar]

x could be negative once you introduce the unit circle for trig

but distance is strictly non-negative, yes

Because you use right angle trig, you restrict yourself to only considering positive x

[upper karma]

what if we want to consider negative x in above example?

[lunar heart]

excuse me can anyone help me with geometry

[humble pulsar]

Then the argument doesn't hold

Cause you wouldn't have a triangle

Or.. if you impose it on the unit circle then you'd just get the triangle the other way, so it should actually work in technicality?

[upper karma]

ty so much, king

[earnest basin]

Circle M has a central angel of 100 degrees with radius 12 cm. Find the area of the segment.

54.75 sq. cm
4.56 sq. m
10.47 sq. cm
76.46 sq. cm

Helppppp

<@&286206848099549185>

[signal swallow]

also multiple choice leads me to believe it's a test/quiz

[sacred wyvern]

I really don't understand trigonometry much could anyone help me what exactly it is I tried going through some websites for them but I still didn't understand it?

[dark sparrow]

somewhat loosely, trigonometry is the study of the relationship between lengths and angles in triangles (and in other shapes that can be built up from triangles)

[fallen sleet]

how to prove that a circle has a finite perimeter?

[dark sparrow]

i guess you would need to work with the definition of a rectifiable curve

ie the definition of what it means for a curve to have a length in the first place

[fallen sleet]

does this mean changing the definition of a rectifiable curve [for example making it x2] would not cause any issues? [even if it is not useful]

[dark sparrow]

i think you misunderstand

the definition of a rectifiable curve concerns what curves have a length in the first place, not the assignment of lengths to curves

[fallen sleet]

oh my bad

[silk jacinth]

I know this solution I saw is wrong, but my sleep- and glucose-deprived brain figure out why:

tan x = sin x
sin x / cos x = sin x
sin x / cos x = sin x / 1
cos x = 1
x = 2πn

oh wait now I get it, if sin x = 0, then cos x might not be 1

so yeah nevermind 😄

[humble pulsar]

tan(x)-sin(x)=0 then factor

[silk jacinth]

yeah that's how I did it

but someone gave that solution and I couldn't immediately find the error in their reasoning

[empty thicket]

Hi anyone know how to do proofs

[humble pulsar]

nah, no one's ever done proofs here

[twin axle]

Can someone help me with a very easy trig identities q

[upper karma]

2SINxCOSx = SIN(2x)

1 = cos²x+ sin²x

you know these formulas?

or not?

[opaque mauve]

Hello I need some help with geometry

My teacher gave me this and idk where to start

This is a example of what I need to do

[tribal tendon]

@nocturne pebble

[forest drift]

yes I know

[forest drift]

hey

can you come to DMs

I can help you out

for the first question

[opaque mauve]

It’s ok I finished it already

[forest drift]

okk

[forest drift]

it's very easy

[opaque mauve]

@trail tree helped me

🙏👍

[heady adder]

Hello

@forest drift pranam ramanujan ji

I was playing with this simulation and i am unable to understand how 230° is found here.

[dark sparrow]

@heady adder

[heady adder]

This picture proves im dumb

Thanks

[forest drift]

pranam

ramanujan ka pranam

[potent orbit]

Anyone know about this?

[heavy snow]

direction is the length of p2-p1

the line started at point p1(2.5) and ended at point p2(11,13)

you then take the origin or p1 * direction * force which is 2,8

[potent orbit]

Is it fine with you if I DM you about it?

[heavy snow]

sure

[dense badger]

can somone help me with this

its like a summative but i am confused on a and c

[dense badger]

nvm got c can somone help with a

[upper karma]

The first one

Sec=1/cos

And cot=cos/sin

So 1/cot= sin/cos

And sinsec= sin/cos

[dense badger]

oh

what about a

i solved c

[upper karma]

This is a

[dense badger]

oh

yea

like i was confused when us aid taht was b

[upper karma]

Sorry i wrote it wrong

[dense badger]

nah all good

any idea on b

[upper karma]

Yes

[dense badger]

how

[upper karma]

1-cos²x=sin²x

[dense badger]

wait how sin 2 x

how did u get that

[upper karma]

That is thé basic formula of goniometri

Cos²a+sin²a=1

[dense badger]

oh wiat

i see

yea

[upper karma]

Is your problem solved?

[dense badger]

yea

i got it

ik what to do

just needed these hints

thanks a lot

[upper karma]

Ok

Done with pleasur

[upper karma]

Give every Point a letter first

[potent orbit]

I figured it out, sorta, but i do have a more complex one

Here's the one I'm talking about

[upper karma]

So and than you look first for side AD

You know the formules?

[upper karma]

Draw the situation and it will be much easier

It doesnt matter if the picture is bad it is vetter than No picture

Better

[potent orbit]

Seriously? You don't know how to help if there isn't a picture involved?

[viscid ginkgo]

https://i.imgur.com/FMB3cdQ.png am I crazy or does khan academy not explain how to get to a maximum point of (2.5pi, 6) here at all??

https://i.imgur.com/VVyJKK2.png

this is often my problem with my classes where I feel like people don't explain things properly

I look all over for the answer and spend so much time confused

[upper karma]

A picture make things simplier

You put 2.5π in your function

And 2.5π-π=1.5π

And than (1/3)•1.5π=(π/2)

π/2=90°

Sin 90°= 1

So (2•1)+4=6

[upper karma]

Thé answer is correct but you have to work with radian and degrees

[upper blaze]

Guys I have a question

Does anyone know how much 3D geogebra calculations are precise?

[onyx mesa]

im stuck

i dont know how to prove the last part

[viscid ginkgo]

yes I finally understood this when someone explained ( [] - C) = + horizontal shift

thank you sir

[upper blaze]

I dont know what you dont understand

You have all 3 angles and 1 common side

[onyx mesa]

yeah i figured it out after a bit

you have to get the shortest proof possible so stating it's right triangles wouldnt count

[onyx mesa]

can someone help me

[upper blaze]

Hmmm but thats shortest? I dont see what shorter you want xD

[forest drift]

I learnt this in grade 4 lmao

[dark sparrow]

great, who cares

[upper blaze]

Anish, I think you should be more respectfull to others and if you cant help them dont comment. If you have something usefull to say please do otherwise dont say anything. You alredy trolled here before now please stop

[forest drift]

I never trolled

[upper blaze]

About ramanujan proof?

I have never saw your work

[forest drift]

first of all correct your english

[forest drift]

I never trolled

I have done it but I am hesitant to show it to others cuz you guys might steal

[upper blaze]

yea yea

dont argue here

[forest drift]

then come to DMs

guys doubt

pls help

nvm i got it

[dark sparrow]

I have done it but I am hesitant to show it to others cuz you guys might steal
lmfao

wow

[upper blaze]

Fast Geogebra question. If I have a regular triangle in some plane in space how can I make tetraheadron over that triangle so that 4th point of tetraheadron is on wanted side of a plane?

[wise pawn]

you could use the cross product

[spiral gulch]

hi i am confused in one of my mock exam questions i dont really get it can someone help me?

[upper karma]

use the basic formula

sin²x+cos²x=1

1-cos²x=(4/7)²

1-cos²x= 16/49

1-16/49=cos²x

so cosx = 0.8206518066

cotx = (cosx/sinx)

[dark sparrow]

so cosx = 0.8206518066
ew

[upper karma]

yea i know

[dark sparrow]

there is no need whatsoever to shove these ugly decimals in

[upper karma]

it is squerooth of (33/49)

[dark sparrow]

and also do not do other ppl's work for them

and also it's square root

[upper karma]

bro i am dutch it is my phone his keyboard

i try to help

[dark sparrow]

please do not call me bro

[upper karma]

boy or girl?

you dont have to answer

[dark sparrow]

if you really care so much about a stranger's gender, then i'm a girl.

[upper karma]

ok tha i take my words bro and so back

sorry there is u shorter way

so we now secx= (1/cosx)

so secx= square root of (49/33)

but tha is cot > 0 and it has to be <0

[upper karma]

0<cosx>1 so the corner is in quadrant 1

and if -1<cosx>0 than is cotx<0

so secx have to be -square root (49/33)

[upper karma]

so secx have to be -square root (49/33)

[lost gale]

does Diagonal in a rectangle across an angle?

[upper karma]

yes?

what is your question?

[radiant fractal]

hey quick question, is it no triangle because angle c gets cancelled or what exactly?

[upper blaze]

What is ssa?

[iron spoke]

side side angle

[upper karma]

Ooooh yea

There can be only one with ssa so 1 and 2 are false

[silent plank]

There can be only one with ssa so 1 and 2 are false
bad reasoning

[upper karma]

you know something else

[upper karma]

and if you know it so much better give than your opinion/answer ok

[silent plank]

i try not to give the whole answer away.

There can be only one with ssa
is false or incomplete

[vagrant quail]

How do you motivate yourself to learn euclidean geometry at the Olympiad level? It just seems so unmotivated and dry

[silk patio]

Motivation by beauty. What is the main motivation for you for any mathematics?

[vagrant quail]

Well exactly, beauty. But there is barely any beauty in Olympiad geometry, cause all the problems always seem contrived or too niche to be relevant

At least that is my impression of it

[dark sparrow]

you're not alone in having that impression, if it's any consolation

[vagrant quail]

Almost all the teachers that have taught me geometry just dump a load of theorems with no proof and then 2-3h of problems

But I still believe that if I just learn it in the right way, I can see something in it

Like Ceva's theorem

[silk patio]

I wouldn’t let uninspiring teachers trick you into thinking the results aren’t beautiful or have no conceptual motivations

[vagrant quail]

I do believe that the teachers are to blame in part for just lazily explaining the subject

But is there an approach to this that doesn't feel like directionless gruntwork?

[wise pawn]

what are you currently doing to learn euclidean geometry, what's your process?

[twilit pagoda]

I started learning these trig functions a few days ago and I'm happy with my progression, I tried solving this problem myself but I'm getting a different answer than the lecturer I'm following.
So this is the question: https://snipboard.io/QCeA3g.jpg
And this is my solution: https://snipboard.io/KSB4tA.jpg
What wrong did I do there?
The lecturer used this cancelling technique which I'm not really aware of: https://snipboard.io/dcXQrW.jpg
And got this answer: sin x - cos x
However I'm getting cos x - sin x which is completely changing my final answer.
So whats wrong there?

[twilit pagoda]

Edit: okay I've just crossed division and I got the same answer as my teacher: https://snipboard.io/ygc3Ax.jpg. Now my question is when do we do cross division? Or we also do cross division and thats something I've never noticed (i started learning math 1 month ago) so...

[upper karma]

your fold is that (sinx/(sinx*cosx))= 1/cosx and not to cosx

[runic dock]

@faint dagger

[upper karma]

Is cot(-x) = -cot X?

Or is that only in the case of sine?

But sin(-x) = -sin X everywhere?

And is that the case in any other trig fns

here you can change this a in your x

if you know the radials than you can use this

pi is the same as 180°

[wise pawn]

if you know cos(-x)=cos(x) and sin(-x)=-sin(x) and how to write cot in terms of sin and cos, then you can work it out

[upper karma]

So like
cot(-x) = [sin(-x)/cos(-x)]
= [-sin(x)/cos(x)]
= -cot(x)
Alright thanks

[silk patio]

It’s cos/sin but doesn’t matter

[upper breach]

can someone help me with t this

[cosmic smelt]

trad: calculate the volume of a parallelepiped with a square base and a diagonal of 26 cm and make an angle with the base of 53 °

how do i graph that angle?

[upper karma]

What is it?

[dark sparrow]

based on it being spammed all over, probably a scam.

[upper karma]

I think it

[smoky palm]

Hey guys, look at this trig I learned

Sin^2(x)+Cos^2(x)=cot^2(45)

[humble pulsar]

that's incorrect

[smoky palm]

Eh

[humble pulsar]

since sin^2(x)cos^2(x) is a function dependent on x

and cot^2(45) isnt

[smoky palm]

What is cot*2(45)

[humble pulsar]

1

[smoky palm]

I rest my case

Shit

[humble pulsar]

You're still wrong

[smoky palm]

I missed the plus sign

There we go

[humble pulsar]

yeah, that's just pythagorean identity

[smoky palm]

That is

On an not unrelated

Which of the types of trig identities should be memorized?

[humble pulsar]

I mean... hard to say b/c idk what your teacher would give you on a formula sheet

[smoky palm]

I mean, for like

Calc

And

Just, most of the general ones

[humble pulsar]

pythagorean identity, double angle

[smoky palm]

What about half angle

[humble pulsar]

derived from double angle.

[smoky palm]

Or the power something?

I forgot the name of that one

[humble pulsar]

then idk

[smoky palm]

What about the sums and differences, or the ones derived from that

[humble pulsar]

compound angle you should know, to which double and subsequently half angle come from them

[smoky palm]

Thank you

[humble pulsar]

as well as the "definition identities"

like tan(x)=sin(x)/cos(x) and sec(x)=1/cos(x)

[smoky palm]

The basic ones

I didn’t know those were identities

But, it makes sense

Do I need to know how to use polar coordinates?

[humble pulsar]

I mean for multi-var yes, but I never used polar in 1st year.

[smoky palm]

When is multi var taught?

Like, after what

[humble pulsar]

2nd year.

[smoky palm]

Do they do a review or something?

[humble pulsar]

no fucking clue what the university you will go to will do

[smoky palm]

That’s fair

Sorry

[wintry tundra]

@smoky palm the ones I use the most in derivatives is pythag and double angle

[smoky palm]

tyty

[forest drift]

guys

I proved ramanujan wrong again

and that too 2 days before Indian Mathematics Day

[upper karma]

How

[forest drift]

basic number theory

sum upto n natural numbers = n(n+1)/2

[upper karma]

are you trying to prove his statements?

he is one of the biggest 'mathers' from the 20th century

good luck with proving

[forest drift]

I am done proving

[silent plank]

sum upto n natural numbers = n(n+1)/2
how does that prove anyone wrong about anything

[turbid star]

I suppose that is not from Ramanujan?

that was from Gauss as far as I can remember

[upper karma]

Ig u mean sum upto infinity = -1/12 one? If so then yes i agree he was so wrong there

[turbid star]

[forest drift]

yeah gauss

[forest drift]

sum upto infinity = -1/8

he told -1/12

[turbid star]

Ramanujan wasn't wrong there, I think that's analytic continuation

[forest drift]

I proved it is -1/8

[turbid star]

also, why is this in geo and trig

[forest drift]

no clue

come to prealg

[upper karma]

I dont think that still makes sense

[upper karma]

Have you watched mathologer's video about that?

[turbid star]

I haven't actually

but I heard that Mathologer got a lot of "hate" from that video

or it was something like that, I dont remember if it was Mathologer

[upper karma]

He might have idk either

But he proved very efficiently every blunder ramanujan made

In sum upto inf = -1/12

[wise pawn]

this channel is for talking about math, not for bragging about math without showing it

you can only brag about it if you show it, otherwise you're trolling

[forest drift]

honestly I never have received hate from people who have done more than me

[wise pawn]

are you trying to imply this is some form of hate?

I'm just telling you your messages are not math and off topic here

[turbid star]

^^^

[upper karma]

Give proof now

Im 10 billion Percent sure sum of all natural numbers ≠ negative

[forest drift]

i dont care

[forest drift]

have you ever heard of mathematical marvels

[upper karma]

No

[forest drift]

Then just don't comment on anything

[weary drift]

@forest drift ur whole history is trolling w/ poor attitude toward others. pls stop or face a ban

[upper karma]

I might appreciate you to explain that to me rather than telling me to stop

[forest drift]

Ok I don't care

[weary drift]

bye

[upper karma]

Pls dont!!

Shizz

@weary drift did you ban him?

[weary drift]

yes

[upper karma]

You shouldn't have

Can i dm because thats off topic?

aw man he was an entertainer

[weary drift]

i suggest not contacting him

[turbid star]

I dont think it would have been fruitful anyways
#geometry-and-trigonometry message

[turbid star]

xD yeah had a good laugh

[weary drift]

whole history was this. entertaining to some but in the end we dont want garbage contributors here

[upper karma]

He's a 9th grader i was like that as well when i was in 9th
That was the exact time when i was the most excited to dive deep into the advanced maths

[turbid star]

then later the cringe strikes back

[upper karma]

He was insecure to show prove because he was afraid they might steal it so he will only send it to the indian maths society (im concluding that because of his earliest chat here)

Agony of 9th graders are cringe indeed

[turbid star]

I'd be really surprised if they even accept it

ANYHOW

off topic chat

[upper karma]

Yep they didn't respond and he's expecting them to publish that in their annual journal which obviously aint gonna happen anyway yes off topic lets get back to trig

Oh btw are functions like sinh, cosh, considered trig?

[turbid star]

hyperbolic functions haha

i guess so?

not sure myself

[weary drift]

i think the enthusiasm for math is genuine. its just the approach to math & trash attitude

[turbid star]

yeah I agree

[upper karma]

Trigonometry
Measure of 3 sides (triangles)

Deals with circular function lol

And then we have hyperbolic function

[turbid star]

deals with hyperbola thingies

[upper karma]

Trig changed a lot haha

[turbid star]

wait that makes me wonder

do sech cosech etc also exist?

[upper karma]

We learnt it as triangles
Changed to circles
Evolved into have others

[turbid star]

im pretty sure they do

[upper karma]

Yes

[turbid star]

but how would they be defined

[upper karma]

Just like it is defined in normal trig

1/sinh = cosech

All that

[turbid star]

oh hmm

but geometrically?

[upper karma]

,w 1/sinh(x) === cosech(x)

Lol

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Insert nerdy name#7892

[upper karma]

Not sure tbh

I have yet to learn conic sections

Im not even sure how they have defined sinh geometrically

I just know it as an exponential function

[turbid star]

something to do with the area around hyperbola

like this

[upper karma]

Ahhhhh i see

But wait area is not dependent of theta i see

[silk patio]

It depends on it, just not linearly

[bright cairn]

A tangent to the ellipse x^2 + 4y^2 = 4 meets the ellipse x^2 + 2y^2 = 6 at P and Q. Prove that the tangents at P and Q will be at right angle

I tried using various method
lets there be a tangent at ellipse 1 - parametric
intersect it with e2
solve quadratic
but it got complicated
tried in reverse
and failed

oh

i am sorry

[dull harness]

Is anyone here

[gleaming nova]

we all just randomly pop into existence once someone has asked a question

so no

[weak fjord]

@gleaming nova

[gleaming nova]

why did you ping me

don’t ping specific people for help

[weak fjord]

cause u love math

[gleaming nova]

did i ask for you to ping me

[weak fjord]

my fault g 🚔 🚶‍♂️

aint know it was illegal

u dont gotta ask, its a feature for a reason, lmfao.

[weary drift]

@weak fjord pls dont ping randos

[weak fjord]

sure

I just seen him in every chat so I figured why not

[gleaming nova]

[dark sparrow]

[gleaming nova]

i feel like that same thing has happened to you as well ann lol

[dark sparrow]

sure has

many times at that

[gleaming nova]

the burden of being way too active

[smoky palm]

When do you start to use hyperbolic trig functions?

[earnest basin]

Help

[wintry tundra]

Calc 2 sometimes

Bc the derivatives and integrals for the basic h-trig stuff are the same

So its weird

[silver willow]

Should I talk about Analytic geometry here or in Pre-calculus?

[smoky palm]

For the half angle identities, are there third angle identities? Fourth angle? And if so, what are they?

[dark sparrow]

you can get something like a 1/4 angle identity by applying the half angle identity to itself

for 1/3 angles you would have to engage in cubic equation fuckery

[smoky palm]

Ooh! Thank you

[dry basalt]

They ask for the distance between the "source" (the red sun) passing through B to the "capteurs = sensors" (big blue arrow)
And they find this result:

I don't see how to get it, if a sympathetic soul can explain it to me. Thank you

[humble pulsar]

pythagorean theorem

the ray follows the 2 equal sides of an isoceles triangle

[dry basalt]

I don't see how you get this result with the pythagorean theorem, can you expand a little more please? How long are the sides of the triangle?

[humble pulsar]

The horizontal side is x

So drop a perpendicular from the reflection point to the horizontal side, you'll form a right triangle with legs e and x/2

Hence the hypotenuse cones from pythagorean

[dry basalt]

This triangle ?

[humble pulsar]

yes.

[vapid terrace]

I need to find 16 different and even distance from each other points on a circumference of a circle it has a radius of 3 (Unity measurements) how do I go about doing this I'm pretty confused?

[covert dew]

someone help me pls

[woven copper]

is there a way to add sinusoidal function without vectors?

[covert dew]

?

i’m not sure

[mortal zealot]

as in sin(x)+cos(x)?

then no

that's the simplest form

[covert dew]

what can i do for an equation

to solve the problem

😣

[woven copper]

like sin(x)+sin(y)=k sin(z)

[covert dew]

??

[mortal zealot]

nope

You might be interested in this though

i hate this png

search up sum to product formulas

[woven copper]

like adding sin waves

[upper karma]

I wanna learn trigonometry

I have exam after two weeks

I need someone truely teach me

[tidal drum]

I'm trying to make a triangular tube 9" long. I'm putting a cylinder inside of it with diameter 4.375". at minimum, what are the lengths of the sides of the equilateral triangle?

I should be able to figure this out, I'm just... failing.

nevermind... google is my friend: Also the radius of Incircle of an equilateral triangle = (side of the equilateral triangle)/ 3. so that gives me the answer I need.

[dark sparrow]

that's not correct, or you may have mis-copied it

the ratio between the side length of an equilateral triangle and the radius of its inscribed circle is not 3

it's 2*sqrt(3)

@tidal drum

[crude zenith]

task is to find side of square

pls help

[upper merlin]

there should be some more info to be told

in words

for example what is that point hanging near the center of the square?

[crude zenith]

it is centre of square

[upper merlin]

coord bash 😅

[tidal drum]

Thank you for correcting me. The site must be wrong because there's nothing else there that might change the above statement to be more correct. I just copy/pasted it.

[shut finch]

does anyone know how to work what is the distance from a to b

[humble pulsar]

are the square and circle concentric?

[shut finch]

sorry do not understand what you are saying can you simplify

[humble pulsar]

do they share a center Y or N?

[shut finch]

yes

[humble pulsar]

so find the radius of the circle

[shut finch]

that is the point how

[humble pulsar]

Hint: diameter of the circle and diagonal of the square are the same

[shut finch]

jesus christ(is king) wow how did i look past that thank you so much

@tidal drum

[humble pulsar]

then getting ab should be just line segment manipulation

[shut finch]

@humble pulsar

ok

[echo spear]

https://tenor.com/view/balls-dragon-balls-juggling-spinning-orb-tricks-jukin-video-gif-15365731

[candid sonnet]

cud anyone please help me with this

[upper karma]

is it a cirkle or an oval???

[upper merlin]

@upper karma he already got the solution

[upper karma]

ow i didnt see that

[smoky palm]

Are there any websites or anything that I can use to work on trig?

Like, the more advanced side to trig, with identities and whatnot, not just the right triangles

[dark sparrow]

<@&268886789983436800>

[silk charm]

thx

[storm heath]

What does produced to D mean?

Does it mean it is in the segment AB itself?

Or is that going away from the line?

[nocturne remnant]

I think D should be on the extended line AB but not segment AB
So from left to right the four points might look like A-C-B-D

[storm heath]

Is it like this?

Or this?

Or this?

[smoky palm]

I’ve got
cos(12)cos(24)cos(48)cos(96)

And I have to simplify it, what I did Is I know cos(18)

So, cos(30-18)=Cos30Cos12+Sin18Sin30
Then, do I just do some stuff with double angle identities to find the rest?

[grave nova]

How do I solve this?

Question 2 btw

[humble pulsar]

similar triangles.

[grave nova]

???

[humble pulsar]

???

[grave nova]

What do you mean by similar triangles?

[gleaming nova]

same side ratios

one triangle is just a scaled up / scaled down version of the other

[grave nova]

Oh ok

[brave needle]

Is there a way to increase the period of sin(x) gradually

Kind of like sin(1/x)

[wise pawn]

is sin(x^2) sort of like you want

[brave needle]

Oh that’s good

But is it possible for the period to increase rather than decrease as x tends to inf

[wise pawn]

yeah, try to think in terms of what you're plugging in as hitting the multiples of pi less often

so like for instance y=x is not changing anything, but y=5ln(x) will be hitting them but less often because it's below y=x

so sin(5 ln(x)) will get something kind of like that

althought just a random choice of function that lies below y=x that's monotonically increasing

[brave needle]

Ok that’s exactly what I was looking for thx!

[upper blaze]

Guys I found a claim on Wikipedia that there are 7 lines connected to tetraheadron that and concurent at its centroid. I know that one of them is generalization of Eulers line but does anyone have any idea what are other 6?

[dark sparrow]

can you link the wikipedia article that makes this claim?

[valid osprey]

\

[upper blaze]

https://en.wikipedia.org/wiki/Euler_line
Under section "Generalization"

I mean this claim make a lot of things different but Idk what lines are they talking about. Though I am not suprised to see this. I myself found few lines that pass through centroid in my research though their importance to me is unknown (For my research they were useless)

And plus they dont seem like something that would be associated to tetrahedron in sentance like this.

[grave pond]

Seven lines that pass through the centroid of a tetrahedron would be the line from each vertex to the centroid of the opposite face, and the three lines that join centers of opposite edges.

(This is obviously the case in a regular tetrahedron, and any tetrahedron can be made regular by an affine transformation. Affine transformations preserve lines, planes, midpoints, centroids ...).

The Euler line analogue must be an eighth line.

[upper blaze]

I was hoping for 7 Euler like lines not those easy ones honestly xD

Becouse when you say associated you just cant mean that becouse it is very well known

[torn quail]

hello, could anyone help me figure out how to calculate the surface area of this pot

[torn quail]

im not sure what shape to consider this part. a trapezoid?¿

[upper blaze]

Honestly, why do you think that you can apply math to concrete problem hahahaaha?

[torn quail]

lmao i have to

i need the surface area of this pot for an experiment im doing. it's relative to the cooling of water with varying s.a

[upper blaze]

Can you use aprox or it has to be accurate?

[torn quail]

as accurate as possible, as long as i can justify what im doing

im in high school so the procedure doesnt need to be very extensive

[upper blaze]

project it onto a plane?

[torn quail]

how would i do that? loggerpro?

[upper blaze]

and assume all are straight lines?

Well projection will be close(if not accurate) circle so you can take paper and wrap around it to get diameter of bigger one

or radius whatever

radius of bottom part(small cirlce) you can measure when you find center using normal geometry

rest of it should be easy assuming all are lines and curvature you throw away and say its an error

[torn quail]

but isnt the curvature important tho

wait so basically what you're saying is to ignore the curvature in this part and consider it to be a large flat circle?

[upper blaze]

Yes, and when you find projection of that distance you are showing on picture then you can find that distance using pithagora 😄

assuming they are straight lines

[torn quail]

oh okay thanks

[upper blaze]

Or

ORRR

[torn quail]

ill have to ask my supervisor if i can ignore the curvature lmao

mhm?

[upper blaze]

well you can use thee meter thats used for measuring cloaths?

to find exact measures

But again some aprox should be made

[torn quail]

wait that's not too bad

a measuring tape could work

[upper blaze]

Hmmm I am not sure actualy maybe you can measure those 2 radis I told you and that height difference of those 2 circles and then

oh wait actually maybe you can do it precisly but it will take some work if its possible at all xD

I am thinking that you can find surface that contains that part in between the circles

I mean its equation

using geogebra or something

But hmmm

I dont know hahahah

really I dont know

[clever fable]

Do vertical asymptotes of a function become a 'hole' when we take the function's reciprocal and graph them?

Or will that depend on the function that we are taking the reciprocal of?

[humble pulsar]

they become roots

since you'd have the reciprocal of (not 0 number)/0 becomes 0/(not 0 number), which is 0

[pliant jasper]

Does it matter how you write it? For example the reciprocal of f(x) = 1/x is x, which has no holes. But if I write it as f(x) = $\frac{1}{\frac{1}{x}}$ does x=0 become a hole because it has division by 0

[somber coyote]

qas

[clever fable]

^^^^

[humble pulsar]

In that case any function has a hole

which is absurd

cause it's not true

the reciprocal of 1/x is x

x is continuous on R.

[pliant jasper]

Ok 👍

[clever fable]

what if the function was a log for example, like log(x+1), when we take its reciprocal, it will be 1/log(x+1), but x=-1, so it would be a hole right?

[humble pulsar]

yes that still has problems

and no, it wouldnt

[clever fable]

oh

[humble pulsar]

log(0) is undefined

[clever fable]

Ok I think I kind of understand, it will just be a root then right?

[humble pulsar]

no.

[clever fable]

how would the graph look like?

[humble pulsar]

x=-1 isn't in the domain of either log(x+1) or 1/log(x+1)

,w graph y=1/log_10(x+1)

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Omegabet_#9771

[clever fable]

Wait so what's happening at x = -1, if it's not a root?

[humble pulsar]

not going into the in-depth analysis rn, cba

my guess is just approaching in the limit

[clever fable]

hmm I see, it's just in my teachers' solutions, there was a hole, and I was just confused because I thought all vertical asymptotes are going to have holes - which i have realised is now not the case

Thanks for bothering anyway XD!

[humble pulsar]

Yeah holes and asymptotes are completely different

it approaches the point (-1,0) from the bottom though

[clever fable]

Yea it makes a bit more sense now (I think). I'll probably just check the solutions with my teach, but it seems more clearer (in terms of what's happening when we take the reciprocal of the functions.)

[hoary mango]

what the fuck is with pythagorean theorem 💀

can't even get sides correctly

[spiral mesa]

Wdym

[hoary mango]

i keep squaring at wrong sides

like

i could be trying to solve a right triangle

and my deadass would square the hypotenuse

to the a or b side

[spiral mesa]

Just beware the location of the right angle and that's it

I was about to type angel instead of angle lol

[hoary mango]

i've bewared a fuck ton of times

but i still get it wrong

turns out hypotenuses arent always diagonal

[gleaming nova]

the hypotenuse is the longest side of a right triangle

or just the side opposite to the right angle

[upper karma]

both

[hoary mango]

i know what a hypotenuse is

im just saying it's not always a diagonal line

[silent plank]

how is a line being diagonal relevant

[hoary mango]

😐

have you not heard of the past messages i made

i said a fucking hypotenuse isn't always diagonal

[silent plank]

wdym by isn't always "diagonal"

[hoary mango]

like

hypotenuses could be anything

but diagonal angles aren't always the longest

[silent plank]

if your trying to identify the hyp by whether a line is "diagonal," then that's the source of your problem

the hyp is the side opposite the right angle of a right triangle

diagonal is also vague in this context and I have nfi what you actually mean by that

[dark sparrow]

it's something that depends on the orientation of the triangle as drawn on the page

[hoary mango]

see

its not that fucking hard to know what i fucking mean

[silent plank]

is the swearing necessary?

if as drawn on the page, no lines are vertical/horizontal meaning you'd consider all your lines to be "diagonal", then you're stuffed

[chilly whale]

This comes as a surprise to you? That a triangle can be rotated?

[hoary mango]

why do people take this so personally

[chilly whale]

Who?

[hoary mango]

the fact i said diagonals arent always the longest

[chilly whale]

Not sure how that has any relevancy for why people should take it personal, which I'm pretty sure no one is doing.. It's more or less just surprising that you're defining diagonals that way.

[silent plank]

tbf i took it a little personally when swearing was involved and the implication that i was illiterate

[chilly whale]

^ that I can understand. People that ask for help are not even trying anymore

[grave pond]

Manners aside, in the diagram in Euclid's proof, the hypotenuse is horizontal.

[glacial dawn]

He’s a mobile Roblox player

[weary drift]

@hoary mango curb ur attitude

[versed valve]

Hey can you help me on how do i prove the reflective faculty of hyperbola and ellipse?

[opaque gull]

LMAO

[vital moat]

I've been struggling in advanced functions a lot and got a 53 and 58 on the first 2 unit tests, does anyone know how I could convince my teacher to assign me mark boosting work?

[viscid crag]

hi !

opps sorry, wrng grou

[royal raptor]

Can anyone solve this problem or is it too Hard?

[humble pulsar]

<@&268886789983436800>

[clear haven]

band

[quasi matrix]

hey

[flint wyvern]

If I have a circle centered at (4,4) with radius 2, what are the functions of the two tangent lines to the circle passing through (0,0)?

Just playing around with this for a LaTeX illustration I'm making... can't seem to get it !!

[dark sparrow]

is 53/58 not a high enough grade for you or is your teacher so draconian in their grading that this works out to an F-?

[flint wyvern]

he means he got a 53% and a 58% as two separate grades

which in the states, if repeated on a sufficient number of exams, will force the student to retake the class

[dark sparrow]

...i could have sworn the message said "53/58" with a slash the forest time i read it lol

[empty kite]

I'm having a bit of trouble w' this question

[silent plank]

what did you discover in Q16

[empty kite]

The thing is

I don't know

There was no q16

Teacher just posted this in teams

+bunch of other questions asked us to solve as review for finals

[silent plank]

you could start by drawing in the centre, radii to the points of tangency and denote the length as r

[empty kite]

Sorry I still don't get it

[silent plank]

which part of that don't you get

[empty kite]

I don't understand what's next...

[silent plank]

did you do what I recommended?

[empty kite]

Am I supposed to see something

[silent plank]

that doesn't answer my question

[empty kite]

Yes

[silent plank]

making those constructions might not lead you directly to the answer you seek, but they will most likely be helpful

[empty kite]

Hm

[dark sparrow]

can you show us what your diagram looks like after doing what ramonov instructed you to do?

[silent plank]

(note these initial steps are intended to help prove the implied result from Q16)

[empty kite]

Hold on

[graceful ginkgo]

You can try drawing squares inside the circle to solve it

[lusty abyss]

Help

[dark sparrow]

what is troubling you here?

@lusty abyss

is it ||that the problem presents an obviously impossible scenario||?

[lusty abyss]

Yes

[lusty abyss]

Should I do Pythagoras?

[silent plank]

as shown, the scenario as shown is ||impossible|| and it is pointless to continue

[lusty abyss]

Why would someone give something like this to a grade 7 kid

[dark sparrow]

incompetent or inattentive teachers,

or perhaps incompetent board members who overlooked this glaring error

[iron spoke]

how are you even supposed to solve that

with 0 info

[silk patio]

What do you mean it’s impossible

[grave pond]

There are at least two ways.

[silk patio]

Calculate the area in two ways

[grave pond]

The altitude from P is supposed to be longer than either PQ or PR.

The altitude from Q is drawn inside triangle even through 5²+6²<8² so the angle at P must be obtuse.

[silk patio]

This is megahyperbolic-geometry tho

It’s a shut up and calculate moment

[grave pond]

What stuck me first was that they say "QY = x; find the value of x" instead of simply "find QY", but that strange phrasing fades into the background once it turns out that the diagram is impossible anyway.

Even if it is hyperbolic geometry, both of my objections would still apply. (And the Euclidean area formula won't work there anyway).

[dark sparrow]

even in hyperbolic geometry there are no right triangles with legs longer than the hypotenuse, so...

[silk patio]

Megahyperbolic-geometry

[dark sparrow]

what the fuck is that

[silk patio]

A joke that you took seriously

[dark sparrow]

this is no place for jokes

your objection appeared genuine. you cannot accuse me of taking your joke seriously when you present it as if it were serious.

[silk patio]

Don’t blame the fact you didn’t pick up on the social cue on me

[dark sparrow]

what social cue

[silk patio]

Exactly

[dark sparrow]

are you in all seriousness saying that i should have picked up on a social cue that wasn't there?

[silk patio]

It was there, you just missed it. It’s okay

[grave pond]

Just ignore him, he seems ot be a troll.

[dark sparrow]

i asked you

"what social cue"

and you refused to answer

[silk patio]

Yeah you’re right, I’ll ignore them

[junior light]

I request both of you to de-escalate.

[silk patio]

Haha, yeah I think it’s over

All in jest

[dark sparrow]

[chilly whale]

[chilly whale]

Not really, just a badly designed diagram or faulty value assigned to the segment PX

And what the people have mentioned above of course!

[pastel fog]

I have a question, I stumbled upon an interesting post by Andy Sloane, the post is about Rendering a 3D ASCII donut on a 2D plane, his blog post Andy defines the math behind it. But I am quiet stuck somewhere.

The points (x, y, z) is defined by the expression (R2,0,0)+(R1cosθ,R1sinθ,0) I do not get how this was derived.

The blog post link is here: https://www.a1k0n.net/2011/07/20/donut-math.html

[grave pond]

(R2,0,0) is the center of the circle. The added term R1(cos theta,sin theta,0) is the standard parameterization of a circle in the xy-plane.

[pastel fog]

Thank you @grave pond 🙂

[upper merlin]

I found all the trivial angles but it just seems impossible to get to the solution

[hoary mango]

bro how the fuck do i do unit circles

it literally never indicates any point at any time

[hoary mango]

how the fuck do you get 170 on this

[spiral mesa]

Cancel out 180 with 18 and pi with pi, so you get 10*17

[hoary mango]

nah trigonometry is fucking mid

functions are literally like 100x better than this crappy ass course bro

[spiral mesa]

What's so terrible about it

[hoary mango]

the fact it has so much out of context equations

and then it just turns into an AP for no reason at the math part

i feel like precalculus should be a better choice than this since its functions and shit like that

while in trigonometry, you get stupid equations with stupid formulas

like literally

they dont give steps on how they got their first answer

they just skip it off, leaving questions to the fucking void

so overall, trigonometry sucks and it always will

once bad, always bad

[iron spoke]

nah bro they just taught you trig in the worst way possible

[upper merlin]

gotta learn this trig way of solving

I found a pure geometry solution later tho

[short arch]

Hi everyone, I am at 11th grade, and I am learning the cross product, is this solution correct?

[dark sparrow]

you will have to show us the problem you're solving

@short arch

[short arch]

I am still practicing this method, I am asking if I did it correct or not despite the problem, did I apply the right hand rule correctly or not?

[dark sparrow]

,rccw

[dark sparrow]

ok so you're just calculating the cross product of A and B

the sketch looks... adequate i guess

the way you drew C it looks like it's pointing in roughly the right direction

[short arch]

So I did it correct?

[dark sparrow]

i guess so?

[short arch]

Thanks for your help

[dark sparrow]

but you haven't shown what the value of A×B is

and you've also omitted the length symbol in the last line

and also i guess you got this angle of 22° out of nowhere

[short arch]

I just assumed it

[dark sparrow]

oh, you assumed it? why not assume it's 42.069 degrees exactly?!

why not assume that 2+2=5 while you're at it?

[short arch]

I wanted to give some hypothetic values so I can draw it, in fact I wanted to test if I am drawing it right or not

[iron spoke]

i think she was joking lol

[graceful halo]

can you share it because i am also curious on how to solve it using pure geometry

[upper merlin]

draw a line from A to meet BC at F such that angle BAF=70 and join EF; you will be able to prove that triangle AEF is equilateral and CA is a perpendicular bisector of EF which suffices to show angle ECA is 40

I have no clue how I got this

[river marsh]

Just posting this here even though I’ve got a help channel open, just because it’s been open for quite a while without a response. This is a year 1 university problem but none of the early uni channels seem to fit this question

Is the sum of the forces in each cable equal to the weight of the plate? And if so, how do I solve that algebraically? Preferably using Cartesian coordinates

[upper karma]

trig is interesting wdym

[mild birch]

yeah, resolve the Tension into on the plane and perpendicular to the plane T1,T2,T3

all the horizontal vectors/ ones on the plane cancel each other out

vertical ones add up to give the magnitude of weight

you’ve got T1x+T2x+T3x = 0

T1y+T2y+T3y=0

T1z+T2z+T3z = mg

it’s basic trig to see what T1x or wtv is

@river marsh

i hope that made sense

then it’s just simultaneous eqns

if you’ve got a calculator that does matrices, use it, it makes your life easier

[river marsh]

my calc can do simul. equations

[mild birch]

oh, that works too

[river marsh]

i think i may have thought about this too deeply

because i was resolving the forces into cartesian vectors i,j,k and then trying to figure out the tension

[mild birch]

ahh well, i remember doing 2d ones in school, wasn’t too hard to extend it to 3d

you could , but do you need to ?

[river marsh]

well i must have gone wrong there because i got dumb numbers

[mild birch]

t look like bunch of points