geometry-and-trigonometry

[mild birch]

ah well

now you know

i like these types of questions

please ping me the next time you do these

[river marsh]

i've gotl ike

another one after this

will probably work on it tomorrow but i can ping

[mild birch]

ooh sweet

[river marsh]

though i may need some help right now

[mild birch]

thanks man

for resolving vectors?

[river marsh]

sum of forces in x and y are 0, which makes sense

[mild birch]

yeah

[river marsh]

so basically my equation is this 'T1z+T2z+T3z = mg'

[mild birch]

mg is just weight

526N

i think

[river marsh]

okay, so i don't have any of the tension forces, but the height to A from origin from trig is 0.086603m

[mild birch]

do we need the height

they’re all the same height i’d presume

[river marsh]

yes, they are

[mild birch]

slant height of a cone

yeah, just find the angle

you don’t need the height, just the angle really

so that you plug it in the trig thing

they all form the same angle right?

[river marsh]

well the angle is 30%

30 deg.*

unless we're talking bout different angles

[mild birch]

(T1+T2+T3)sin/cos(30) =526

angle from base of the cone

or from between z axis and a string

it makes sense right ?

[river marsh]

so it expands out as T1 sin/cos(30) + T2 sin/cos(30) + T3 sin/cos(30) = 526?

[mild birch]

don’t expand i

idk if it’s sin or cos because i’ve not done it myself

you can just divide by the constant

to get your last eqn

bc both sin and cos30 are constants

[river marsh]

okay so

(T1+T2+T3)sin/cos(30) =526

divide 526 by the constant gives you T1+T2+T3 = 526/sin or cos 30

[mild birch]

yep

[river marsh]

okay, so then you don't know what the ratio for each cable is

[mild birch]

that’s what your simultaneous eqns are gonna be

you’ve got 3 eqns

[river marsh]

so that's one equation, the others are sum of forces x and sum of forces y?

[mild birch]

yep

precisely

[river marsh]

aight, i'll attempt that now

[mild birch]

nice one

don’t forget to call one direction as positive and the other as negative

when you do that, you’ll subtract a few times

but that’s exactly what we want

@river marsh

[river marsh]

yep, the sum of forces in x and y are the forces on the plane of the disk, there's an x and y axis so i know what's positive or negative

[mild birch]

yep

[river marsh]

@mild birch

I've done everything except the last line there, wouldn't it be tan 30?

wait, you said slant height of the cone

so yeah it would be sin

plugging that into wolfram alpha gave me these answers:

x = -0.977559 and y = 526.322 and z = 526.656

x is a weird number

well, all those numbers don't seem right

[mild birch]

i think you’ve missed out something

what’s the 0.05

[mild birch]

T is the slant height

[river marsh]

radius of the plate is 0.05m

[river marsh]

yep, i used sin 30

[mild birch]

that’s Tcos30

Tsin30 would be resultant of x and y

[river marsh]

ah, of course

[mild birch]

did you get these two so far

oh no, it’s 60 degrees

one sec

there

fixed it

see what you get now

[river marsh]

not sure what the three bottom lines are representing

[mild birch]

which line

[river marsh]

i'm from australia so the style we use to lay these out are different

[mild birch]

oh, yeah no, i tried to do it quickly, baso each of them will have sin30 as a factor when you try to make them as components of x and y.

literally every single term in 2 and 3

since the RHS is zero, i factored out this sin30 and divided by it

did you get any nice answer this time tho

[river marsh]

I’ll check now

[mild birch]

looks about right ey?

seems reasonable to me

[mild birch]

i didn’t know wolphram alpha did that so thank you

i’m not taking the piss, i genuinely didn’t

[river marsh]

haha no problem. yeah it looks good, it's just that for other similar examples in the textbook there's way more involved calculations

like let me show you

[mild birch]

sure

i’ll have a look

[river marsh]

though i guess this question we just did is SIMPLER in that all the cables sit at the same distance away from the center of the plate

so it eliminates the need to use this more involved cartesian method

[mild birch]

yeah, because it’s round plate

[river marsh]

thank you heaps for that. since you mentioned it, i've got another trig question, up to you whether to work on it later or have a look now or do now

[echo ibex]

@upper merlinis ans 70degree

[mild birch]

if it wasn’t equidistant, i would have still resorted to trig vectors ,

[river marsh]

though i probably need to catch up on some textbook stuff lol

[mild birch]

it’s upto you

just dm me if i’m not here

[river marsh]

yep, gotcha. i think the question is probably there to trick you

[upper merlin]

no its 40

[river marsh]

i'll add u as a friend

[mild birch]

nice one

[river marsh]

i'll grab an image so you can look over it

[mild birch]

cheers

[river marsh]

eh, this one has moments though

so a bit more involved

[mild birch]

doesn’t block and tackle pulley system reduce the required force ?

[echo ibex]

@upper merlinwhat is the angle DBC in the question

[upper merlin]

20 deg

[river marsh]

i'd say so, but i think the question is ignoring that

[mild birch]

moments are just forces that produce a rotational effect

[river marsh]

yeah

[mild birch]

oh, thank god

makes life easier

[river marsh]

lol, it does

it shouldn't be too hard to solve

[mild birch]

shouldn’t no

it’s just formulaic

[river marsh]

the other question i have i had to email the teacher about because they gave me a stupid angle

0 = 90 + 75 sin alpha - 239 cos 33

sum of forces in y direction

since resultant is along x axis

simplifies to sine alpha = -1.473

which is impossible lol

[echo ibex]

@mild birchi am getting 70 degree have u done it

[mild birch]

okie dokes wait

[mild birch]

which one?

[mild birch]

f1 is perpendicular and of no use

oh wait

[echo ibex]

@mild birchangle of ECA

[mild birch]

question please

all perpendicular forces are equal

[echo ibex]

@mild birchthat was ur question

[mild birch]

where ?

[echo ibex]

@mild birchgeometrical problem

[mild birch]

could you please sc and send the question

i’m a bit lost

[echo ibex]

@upper merlinhave u done this problem

[upper merlin]

yes

read this

[echo ibex]

@upper merlinsend me

[upper merlin]

someone sent a trig solution too above that

[mild birch]

@river marsh

[river marsh]

okay, let me have a look

[echo ibex]

@upper merlinwhere are u from sir

[upper merlin]

india

[mild birch]

sir is a bit too much tho innit

[echo ibex]

@upper merlinis this server for only indians

[upper merlin]

ikr someone said that to me the first time

[mild birch]

huh

nah, i’m from england

[dark sparrow]

no, this server is far from being for indians only.

very far.

[river marsh]

@mild birch yeah, there's no answer

if you plug that in to the calculator

[mild birch]

oh oops

[river marsh]

i think the actual angle is wrong

[mild birch]

but yeah

[river marsh]

well the teacher said

[mild birch]

if there was an answer

that’s your answer 😂

[river marsh]

yep, well the teacher said i can change F3 to another value so that it becomes possible to calculate

[echo ibex]

@mild bircho i got it what is ur major

[mild birch]

yeah 239 is a bit much

[mild birch]

maths aha

[river marsh]

so i guess it's the block question now

[mild birch]

i have a tragic backstory and all that

[dark sparrow]

can i see the original problem

[mild birch]

which one

[dark sparrow]

the one y'all are talking about rn

[river marsh]

there's like two being talked about haha

[mild birch]

this ? or the triangle construction one

[dark sparrow]

the one with forces

okay yeah that one

let's see

[mild birch]

a more challenging one if you like challenges

[mild birch]

i got that

[mild birch]

well it’s still formulaic

[dark sparrow]

oh yeah that one is impossible as stated y'all are right

[mild birch]

yeah

[river marsh]

so just make it a smaller value?

[dark sparrow]

F1 + F3 has too big of a negative y component

[river marsh]

^ teacher email

[dark sparrow]

alright so then yeah just decrease the magnitude of F3 to like 50 or something

[mild birch]

yeah. because even if F2 was fully perpendicular to x, it’s not gonna equal to y

[mild birch]

not 50, that’s too small,

[river marsh]

at least it's a rather simple question

100?

[mild birch]

100

yeah

[dark sparrow]

okay sure 100

[river marsh]

kk, easy done then

[mild birch]

otherwise you’ll get negative alpha

120 even

[river marsh]

ok

i'll go 150

[mild birch]

baso this formula is true

[river marsh]

yep, so just need to replace alpha

[mild birch]

you need to find alpha

by replacing F3

[river marsh]

yeah, meant F3 sorry

[mild birch]

plug in all the other values and see

[river marsh]

ok

28.512

deg

[mild birch]

sweet

there you go

[river marsh]

thanks!

you seem to enjoy trig

[mild birch]

now for the last one

[river marsh]

😄

a bit harder

[mild birch]

the diagram is a bit confusing

what are the reference points

[river marsh]

that's all the info i get

[mild birch]

a for example is from. the hanging place to the axis

[river marsh]

yes

[mild birch]

d is the height at which it’s held

isb the height till centre of gravity of the object

[river marsh]

hm

[mild birch]

?

[river marsh]

i don't think this question will involve center of gravity

[mild birch]

no, cuz weight acts from the centre of gravity

any object can be modelled as a particle from its centre of gravity with a mass m

[river marsh]

ah right

[mild birch]

don’t worry about the terms

i just wanted to know what the variables meant

we’re mathematicians not physicists

[river marsh]

hahahha

yeah, i think it's just denoting the max height we need to worry about on that axis

[echo ibex]

anyone have a try

[mild birch]

d

those are the lengths in the respective axes

anything on y axis is parallel

[echo ibex]

@echo ibexanyone do this

[river marsh]

yep, that makes sense

[mild birch]

so it’s a vector product thing

[mild birch]

i did it, it’s d

[echo ibex]

@mild birchans is b

[mild birch]

constant dv/dt

and it’s negative

ohh, it’s it’s aceleration to displacement

apologies, my bad

i thought it was acceleration time graph

[mild birch]

back to this

[mild birch]

you’re right. it’s b

[echo ibex]

@mild birchcan u explain

[mild birch]

process of elimination, not a because negative acceleration and positive displacement, not c because constant acceleration but displacement increases so it should slope the other way, towards the axes, same reason for why it’s not d, b slopes the right way

[echo ibex]

@mild birchis it correct

[mild birch]

acceleration is a negative constant and displacement is a variable that increases

[echo ibex]

have u seen it

[mild birch]

yeah well, just by using this fact

[echo ibex]

yeah

[mild birch]

tends to zero

you cannot strictly have 0

cuz asymptote

[echo ibex]

are u indian

[mild birch]

that’s how it should look

[mild birch]

i’m from england why?

but realistically, not possible

but mathematically is

but anyways,

[echo ibex]

@mild birchjust asking

[mild birch]

ah

nws

[mild birch]

you seemed to complicate this one a bit too much i reckon

when you’ve got mcq, try eliminate obviously wrong answers

[mild birch]

it says most suitable

[echo ibex]

@mild birchyeah i know i am preparing IIT thats why i am learning the process

[mild birch]

you know there is no way dv/dtds is equal to zero

[mild birch]

sweet, wish you the very best my friend

i applied to oxford and i got rejected even though the disruption in my test centre wasn’t my fault

bit unfortunate really

but i hope you have some luck with your dream uni

[echo ibex]

@mild birchbe strong there is many more to acheive

[river marsh]

that sounds fucked

[mild birch]

true indeed

[mild birch]

it’s so broken because one of the lead tutors in oxford literally scouted me kinda

she was like, just do the mat well and i’ll take care of the rest kinda thing

mat is the admissions test for oxford

i have emails to prove it

[echo ibex]

yeah it happens dont worry you will do something is good

[mild birch]

it’s not even hard, it’s just you bleed marks in mcqs and you have 60 marks to salvage

[echo ibex]

you know what india ranks first in sucide worldwide students cant handle so much pressure bcoz of exams

[mild birch]

ooh, i’ve got a response from dr. Bowles from ucl for my approach on collatz conjecture

ah, it’s a bit of shame really

[echo ibex]

hey have u heard about JEE

[mild birch]

nope

[echo ibex]

@mild birchit's considered one of the toughest exam in the world for engineering

[mild birch]

tried not to dox myself

that’s from ucl

you said you’re from india right ?

gandhi went to ucl

[echo ibex]

yeah i am indian

[echo ibex]

yup

how do u come to know about mahatma gandhi

[mild birch]

everyone does

wdym

[river marsh]

yeah, everyone knows who he is

[mild birch]

look at the address on the last one

[echo ibex]

@river marshthat's pretty awesome

[mild birch]

she literally made one of her colleagues persuade me to switch to maths and cs

[mild birch]

you didn’t know people outside india knew about gandhi ?

[river marsh]

you'd be hard pressed finding someone in the west who doesn't know the name

[mild birch]

ye

[echo ibex]

@mild birchi am not a big fan of him but u will find his picture on our currency

[mild birch]

fair plays

[mild birch]

everyone knows him tho

history my friend is a bunch of lies written by the victors

[river marsh]

does that include the allies in ww2?

😉

[echo ibex]

@mild birchhow many county in west follow his principle

[mild birch]

you know it

[mild birch]

who does what

[echo ibex]

@mild birchyou know very well he was against violence have violence stopped there

[mild birch]

i’m not a fan of violence either

would much rather curl in a ball and do my maths

do mat papers and see how you find them, i averaged 80s on them under time pressure.

any maths is good, right ?

step is prolly harder tho

step is the cambridge entrance test

[echo ibex]

most precious gift we got from british were trains

[mild birch]

fun fact. most trains and train lines were built by indian companies

in colonial india

you can blame oli ball if that’s not true cuz he told that to me

[echo ibex]

@mild birchbut technology was given by brits

[mild birch]

tech exchange happens nevertheless

you didn’t need a war or colonialism for people to share tech now, do you

[echo ibex]

@mild birchcan u remember what we gave most precious thing to british

[mild birch]

kohinoor?

there are two such priceless diamonds the queen owns

[echo ibex]

@mild birchgreat u know that but there is many more things we gave

i can say the everything what is in london museum

[mild birch]

ah, ye, they stole a fuckton of shit from a lot of places

apologies for the language

[echo ibex]

data says they looted a worth of 45trillion dollar from india

[mild birch]

perhaps even more

[echo ibex]

@mild birchno problem

[mild birch]

you never know

the country is in a trillion dollar debt tho

uk i mean

the great recession changed it all

[echo ibex]

we lost our 20million peoples bcoz of winsten churchill

he was the reason for food shortage in india

[mild birch]

ah dw, not many people like churchill

[echo ibex]

and still he is considered great in uk

[mild birch]

says who

[mild birch]

see, history wise , he is quite great, cuz he inspired many people even in our country to fight

right?

but most of the other things he did

were pretty questionable

[echo ibex]

we are still suffering bcoz of uk

[mild birch]

germany had it worse, they could turn it around, it’s more to do with how corrupt is your system

dubai started with nothing but oil reserves

uk is pretty corrupt

we have corrupt governments that are keeping us from progressing my friend

[echo ibex]

@mild birchi agree but what about illogical partition done by brits

[mild birch]

not done by brits?

[echo ibex]

@mild birchwho was radcliffe indian?

[mild birch]

which partition ?

[echo ibex]

leave it i don't want to talk about these all

will u do it my friend

[river marsh]

@mild birch how did u go on the block question?

[iron spoke]

ok well

look at the graph

[mild birch]

i’m back

[mild birch]

oh yeah sorry

forgot to send this to you

[mild birch]

when a is 0

but that gives me two values

is a max and a min not the same value ?

[grizzled kindle]

@echo ibex well there are 2 ways to go about it

1. a function is either maximum or minimum when it's derivative is 0
   So , possible answers are 0 , 2 , 4
   Then you'd realise acceleration derivative is +ve in [1,2] so 2 is the answer

The other way would be ... Using familiar graphs to approximate

[mild birch]

yeah, to me, it looks like a = -sin(kt)

[grizzled kindle]

This graph looks awfully close to - sin(x)

So it's integral would look something like cos(x)

[mild birch]

so v = kcos(kt) +c

[grizzled kindle]

Ya ...

[mild birch]

yeah

the only problem is 2 and 4 are both valid answers

unless the graph looks more like a harmonic motion graph

[mild birch]

@river marsh

[river marsh]

hmm, you will have to explain it to me

[grizzled kindle]

Noo

[river marsh]

i haven't done moments for a bit

[mild birch]

force x perpendicular distance

[grizzled kindle]

It's not

[river marsh]

i'll have to look at similar examples in my textbook to understand it

[mild birch]

you’ve got two perpendicular distances

[river marsh]

yep, force x perp. distance

[grizzled kindle]

Using second derivative test 4 is the answer

[mild birch]

so i took the hypotenuse, i’m not sure if we’re allowed to do that, but for some reason it made sense in my head

[upper karma]

Hi there.

What are you doing?

[mild birch]

because one is negative and the other is positive

oh yeah right

you’re right

but no, it says speed

not velocity right ?

[mild birch]

both x and y are perpendicular to z and i’ve got both x and y

lengths

[river marsh]

hmmmm okay

i'll save that working out and do it on my own, see if i can come up with the same answer

[mild birch]

and the effective length of x and y is using one string in x and y

is the hyp

just see the answer, i wanna know if it works too

[mild birch]

because you’re right in saying coordinate geometry would work

like, very formulaic and monotonous

but it’ll defo work

i just tried to find the answer quicker

[river marsh]

okay, give me one minute. why don't you try using coordinate geometry?

[mild birch]

bit long to write down?

[river marsh]

it is

[mild birch]

yeah, i have a bad habit which is actually good while doing maths, try find shortcuts or alternative approaches

[mild birch]

i shall try your method in a bit tho,

[river marsh]

haha, good luck

[mild birch]

tell me if you’ve already done it tho

[river marsh]

no i haven't myself

i'm going to bed soon, 1 am! but i will be tomorrow for sure

[mild birch]

ah

nws

it’s half 3 here

[grizzled kindle]

Which one?

@storm heath

[storm heath]

Example.

[grizzled kindle]

Okay , first tell me what you understood after reading the question?

[steady olive]

If the only information given about a point is that it is "inside" a circle, could the point be on the very edge of the circle(i.e. its distance from the center is equal to the radius of the circle)?

[dark sparrow]

probably depends on context & exact wording

do you have the problem statement in front of you? @steady olive

[steady olive]

the points are vertices of a triangle

[dark sparrow]

do you have the problem statement in front of you?

i want you to show me the problem exactly as it is stated. maybe there is something in there that will clarify whether points exactly on the circle count or not.

[steady olive]

I do but the issue is im only allowed to ask very general questions about the problem and I'm worried that sharing the problem could count as asking a specific question

[dark sparrow]

well the devil is in the details here

so unfortunately i do not see a way around this

maybe you could solve two versions of the problem, one where points exactly on the circle count as inside and one where they don't

maybe it won't make a difference. maybe it will. whatever it is, make it clear you are considering both cases as it was not made clear to you which one applies.

[steady olive]

Ok thank you

[rain solar]

does anyone know the answer for this

[alpine cliff]

you should drag the vertices to match the congruencies correctly

[upper karma]

How do i do this

[alpine cliff]

that square indicates right angle

on angle GDF

its also less of a square

and more of a smaller pointy angle symbol

instead of a round one

[echo ibex]

@upper karmalength of arc HG is 77degree into radius i guess

[alpine cliff]

it is not

[echo ibex]

angle EDG 113DEGREE

[alpine cliff]

you shouldnt just give the answers thats not what this server is for

[echo ibex]

arc HEG is piR

[alpine cliff]

its to help and guide through problem solving

[echo ibex]

@alpine clifftell me whats the correct ans

[alpine cliff]

first, the first three questions are asking for arc measure not length

and im not going to put the answer in chat, again that is not what this server is meant for

[echo ibex]

@alpine cliffthen how will i get to know

[alpine cliff]

its not your question*, you will get to know when erravi answers or just message privately

[echo ibex]

@alpine cliffokkkay where are u from man?

[alpine cliff]

the US why

[echo ibex]

@alpine cliffjust asking

[quaint cradle]

if x = cos(theta), y = sin(theta), r = hypotenuse how do we solve for theta?

i got arctan(y/x) but they said to add pi but i dont really get why

[spiral mesa]

You mean to do any multiple of pi?

[silent plank]

you're not representing theta properly on your diagram

[iron spoke]

arctan's range is -pi/2 < x < pi/2

and since that angle is outside the range you need to add pi

to get the correct angle

[quaint cradle]

hm

i dont think im visualizing it properly

pi/2 is the top and -pi/2 is the bottom

are we talking about this angle being outside the range?

[iron spoke]

yep

[quaint cradle]

if we add pi its something like this?

[iron spoke]

yep

[quaint cradle]

oh i see

thansk

[iron spoke]

<@&268886789983436800>

[clear haven]

band

[iron spoke]

epic

[echo ibex]

hello

[upper karma]

hi

[upper karma]

what would be the answer of this

I can't understand it cuz its for 12th grade and I am not in 12th grade

[heavy snow]

not sure what they mean by focus but I would guess its the vertex point at (9.8, -0.5)

[upper karma]

I had this question for a competition

[upper karma]

hmmm

[heavy snow]

hmm no I was wrong

https://images.saymedia-content.com/.image/t_share/MTc0NjQ1MDYzMzA0Njg0OTIy/how-to-understand-the-equation-of-a-parabola-directrix-and-focus.gif

[dark sparrow]

@upper karma have you worked with parabolas before

cause if you havent it would probably involve like, learning about them for the first time

no real way around doing that

[upper karma]

oh

[prisma star]

[Properties of the euclidean parabola]
Hello people, I pick hilbert's book "Geometry and the imagination" as a summer read. Yet I came upon this sentence (marked in yellow) which I am unable to make sense of. Could some explain it? maybe with a drawing?.
What I am thinking is that, if we let F be the fixed point and A, B be two points on the parabola and L be a straight line, then FA + AA' = FB + FB' = const (where A', B' are the respective distances from the point into the line), but since this doesn't hold I must be misunderstanding the sentence.

[grizzled kindle]

@prisma star dude it's the above animation in words

[prisma star]

I don't think so, the one below of the yellow mark says that. I'm trying to understand the first one (marked in yellow), which I think is an stronger sentence.

[grizzled kindle]

Look at fig 5

It says BF1 + BL' = constant

That statement is this equation in words

[prisma star]

Indeed, but that's an special of the construction (deduced from the ellipse property/definition since F1 is a focii), what the statesment says, to my understanding is that this holds for any B, F1, L (where F1, L are fixed and B on the parabola), and i'm saying that this doesn't hold, take for instance F1 near the intersection of L with the parabola, and B near F1 on the parabola, now take B2 on the other intersection of L with the parabola, then clearly both sums are not equal.

so, could you point it out what am I misunderstanding?

[grizzled kindle]

Yes Actually

But not in same parabola

See focus and directrix uniquely define a parabola

So if you interpret the statement that you can choose any fixed point or line of your liking

Then you can't choose a parabola

But if you have already decided your parabola then fixed point and directrix is also decided

@prisma star

It says " Parabola is a curve with the property that the sum of the distances of any point on it from a fixed point and a fixed line is a constant "

It doesn't say anything about it being true for all points

[prisma star]

mmh isn't "any point" the same as "all points (on the parabola)"?

[grizzled kindle]

Look at that animation

Again

Parabola is the locus of P

And that "any points" is referring to P

Not F

Or the points on directrix

[prisma star]

that's what i am saying

[grizzled kindle]

The line in the figure 5 is not directrix

[prisma star]

i am not saying you that can move F to your linking, but that given the parabola, fixed a point F and a line, then you can move P whenever you want and the sum FP + PL is a const

which is what i understand from the sentence

right?

[grizzled kindle]

Yes

Wait Lemme right the equations

[prisma star]

okey

[grizzled kindle]

@prisma star you still there?

[prisma star]

yup

[grizzled kindle]

Okay

So I couldn't prove that for a general case probably do it in morning

But I proved it for that said line parallel to directrix

Let's refer the line as L

Distance between directrix and L is d

Then d = l + q

Where l is the distances of points on parabola and the line L

And q is the distance between points on parabola and directrix

Now by definition of parabola q = distance between focus and points on parabola

There you go

I'll prove it for the general case in morning it's 3 am here and my brain is not functioning

[prisma star]

hehe ok, sorry to keep you late

have a good night

[grizzled kindle]

You got it right?

[prisma star]

yes, but i think using that definition of parabola is kind of assuming a weak version of what you are proving. I think a valid proof should (in this case) come from the definition given by hilbert (which is a degenerate case of an ellipse).

[grizzled kindle]

But the weak version can be proven true right?

[prisma star]

mmh haven't tried, i gonna try that now

[grizzled kindle]

Actually several theorems are proven this way

Say Cosine rule

Is proved using Pythagoras theorem

Which is a weak version of it

[prisma star]

i mean, yes, but they are both proved by common axioms, i was saying that the definition given is tricky in this case because it "assumes to much"

[grizzled kindle]

Dude Greeks came up with parabolas

They had no concept of limits

They defined it as distance from focus = distance from directrix

Hilbert is trying to explain parabolas from his perspective

[prisma star]

it doesn't really matters, what matters is the formalism. Like yeah, in physics everything is differentiable and so on, yet, in mathematics we don't take that for granted just because most natural fenoma were described that way.

[prisma star]

that sir, I think it need a source, could you provide it?

[grizzled kindle]

This is from this article

https://en.wikipedia.org/wiki/Parabola?wprov=sfla1

It's said Pappus proved focus and directrix

Property

@prisma star here is your source

[hollow crypt]

??

[gleaming nova]

#odes-and-pdes

[hollow crypt]

and convergence and continuity?

[gleaming nova]

#calculus

[hollow crypt]

okay

thnc

[prisma star]

I think is state it as a result, not as a definition.

[grizzled kindle]

From the same article

Atleast read the article before arguing

[prisma star]

I think sir, you are not understanding, yes, it can be proven that the definition holds and describe a parabola. Yet it not the same as the natural definition as certainly (by your own source) the one given by greeks, which was the discussion all about

[grizzled kindle]

Define natural definition first

[prisma star]

the one that arises from simpler constructions

[grizzled kindle]

Yes

Then explain why is this not a natural definition

Infact explain why any definition is not a natural definition

[prisma star]

because does not arise from simplest constructions, that's what I say

[grizzled kindle]

What's simplest it's relative

[prisma star]

that's morally true, but not in practice

[grizzled kindle]

Good night dude

[prisma star]

have a good night

[mental crystal]

How to get at trigonometry ?

Inverse especially

I know the regular one, I'm quite capable of calculating sin, cos etc from scratch with constructing shapes etc

But how would one do this inverse?

[tame seal]

how do you know the side lengths of a triangle with just the angle when you have it represented in a unit circle?

for example, why does a right triangle with theta at pi/3 have sqrt 3 as the opposite

[grave pond]

In triangle ABC, let O be the circumcenter. Use the central angle theorem to find angle (say) AOB, then divide the isosceles triangle AOB in two right triangles by an altitude from O and use basic trigonometry to find half of AB.

[vital moat]

Can someone help me with these questions I don’t understand it

[pliant jasper]

These trig functions are periodic, so they have infinitely many asymptotic and min/max values. They want you to write this as a sequence

For example, for 1a, the asymptotes are where sin(x) = 0

I'm assuming n is an integer? I haven't seen that specific notation before

[fleet sentinel]

cot x- tan x= 2 cot(2x)

for the left side i got as far as cosx/sinx-sinx/cosx

but how does it come cos^2x-sin^2x/sinxcosx

[pliant jasper]

$\cos^2(x) -\sin^2(x)$ and $\sin(x)\cos(x)$ are both double angle identities

[gleaming nova]

minus not plus

[pliant jasper]

whops

[somber coyote]

qas

[humble pulsar]

2sin(x)cos(x) is the double angle

[dark sparrow]

none of them are double angle identities

an identity is first and foremost an equality between two things

[gleaming nova]

half double angle identity

[silent plank]

magic(2x) = sin(x)cos(x)

[upper karma]

Can somebody please tell the derivation of the equation of the rotation of axes. in conics??

[grizzled kindle]

@upper karma you know the derivative of cos ( x + r ) in terms of x ?

[upper karma]

-sin(x+r)?

[grizzled kindle]

Yes there you go

Let X0 = kcos(t)

[upper karma]

u mean xnot right?

[grizzled kindle]

Then after rotation X = kcos(t+r)

[grizzled kindle]

Yes

[grizzled kindle]

Expand it , using angle sum property

[upper karma]

alright

[upper karma]

the x0 plane is your rectangular plane or the X one?

[grizzled kindle]

x0 is the x co-ordinate co ordinate before rotation

Similarly y0 is the y co ordinate before rotation

[upper karma]

okk

what do I do after expansion

[grizzled kindle]

After expanding substitute for kcost as x0

And similarly ksint = y0

And that's how you get the formula

[upper karma]

I am talking about that eqn Ax^2+Bxy+Cy^2..........., are we doing this for that derivation??

[grizzled kindle]

Yes

This is the equation for rotation about origin

In general

Then translate to the point you want to rotate about then just simplify

Or the other way to go about it is to adjust the slope of directrix

[upper karma]

I am talking about this

[grizzled kindle]

In general equation of conics

Dude the equation I made you do work for every function

Not just conics

[upper karma]

i know u r right

i've already done that tbh

thanks

but i am looking for this eqn's derivation, the second one specifically

[upper karma]

and can u elaborate more on this, I didn't really understand what u meant by adjusting here

[grizzled kindle]

You know the general equation of conics?

[upper karma]

isn't this the general one that i sent rn??

ok wait maybe i asked the wrong way at first, i should have only written eqn of conics, my bad

[upper karma]

do u know its derivation?

[grizzled kindle]

Yes it is but the one I sent you is more helpful here

Here adjust 'a' and 'b' and it'll rotate

Or you can completely ditch it for polar coordinates

Idk what you need it for but if it were for me I prefer polar coordinates

[upper karma]

Thanks a lot

[grizzled kindle]

@upper karma it's derivation can be done using focus - directrix property

[upper karma]

i am just curious lol

[upper karma]

okk, I'll have to look at this property first, give me a sec

yeah i'm back @grizzled kindle

It's what u sent, thanks

[fleet sentinel]

but how does it become that

[pliant jasper]

Sin(2x) = 2sin(x)cos(x) and cos(2x) =cos^2(x)-sin^2(x)

And 2cot(2x) = 2cos(2x)/sin(2x)

Sorry I didn't use latex, I'm on mobile and have to go soon

[nova rose]

https://cdn.discordapp.com/attachments/785319440648503359/925133407963463810/IMG_1110.jpg

bruh can anyone else solve it or is it just me who can't solve it???

[torpid jungle]

just 30-60-90 triangles

[nova rose]

ik

but for me it does not make sense cuz of the variables

nvm found it

[native fractal]

allied angle formula sin(⁡- 300°) would be?

• makes things harder

[deep linden]

isn't -300 = 60?

not sure but
sin(-300)=-sin(300)=-sin(360-60)=-sin(-60)=sin(60)

[deep linden]

nah this one better

[native fractal]

so, final answer comes down to sin(60)?

[deep linden]

yes

[native fractal]

thanks, also is asking help for assignments not allowed? (this was one)

[deep linden]

i think its #❓how-to-get-help (i am also new here)

[native fractal]

no cheating that's it, chill

i just missed a few classes ;-;

anyone?

[deep linden]

pretty straight application of this formula

[native fractal]

i forgot to reply, thanks!

[fervent dune]

How to solve this question?

[grizzled kindle]

@fervent dune is trigonometry allowed?

[clever aurora]

does Delaunay triangulation work on non-convex hulls?

[fervent dune]

I think try to answer with 2 solution, geometricly or and trigonometricly

[ripe stream]

@fervent dune two questions is angle C 90 degrees and what does the top angle of a i cant see it well looks like a 24 but im not sure

[fervent dune]

it's 2 alpha

[ripe stream]

ok

[fervent dune]

$\angle{CAE} = 2\angle{DAB} = \alpha$

[somber coyote]

Its_TheMathCraft

[ripe stream]

is angle BCA 90?

[fervent dune]

not sure

[ripe stream]

i think i can do it but ill have to assume that its 90

using trig

wait nv,

nvm were supposed to find the area

the area should be 0.933

do you get it?

[silk iris]

I need help solving for Segment BC, I have solved all the angles and segment AB. I don’t know what BC or AC is.

[ripe stream]

can you use trig or only geometry

[silk iris]

?

[storm portal]

Are you allowed to use trigonometry?

[silk iris]

No

[ripe stream]

damn

[silk iris]

I’m in geometry

[storm portal]

Oh okay

[ripe stream]

then im lost

[coarse surge]

hellooo i was wondering if anyone could check to see if these are correct T^T (geometry)

[storm portal]

I think I can cover

@coarse surge Not now - another person has a question posted

@silk iris You can figure out AB from the Pythagorean Theorem

[ripe stream]

wait we can use pythagorean theorum

[silk iris]

Yeah it was 24

[storm portal]

Once Airo is done then you're good

[ripe stream]

they do that in geo right

[storm portal]

Exactly

[silk iris]

But idk Bc

[storm portal]

Yup - pythagorean theorem is learned before geometry

[ripe stream]

ohh

[storm portal]

Once you have the 7-24-25 right ∆, you have to use properties of similar triangles

(reposting the image so I don't have to scroll)

[silk iris]

I forgot the other properties

[ripe stream]

so angle BAT is 56 and BAC is 34

[storm portal]

∆ABT is similar to ∆CBA

Therefore, the ratios of corresponding sides are the same

[silk iris]

Ahhhh yes

[ripe stream]

which means angle BCA is 56

[storm portal]

In other words:

AB/CB = BT/BA = TA/AC

[silk iris]

I forgot about the dilation stuff

[storm portal]

Think you can take it from here?

[silk iris]

Yes

[storm portal]

👍

[silk iris]

Thanks

[storm portal]

np

[storm portal]

This looks good to me

[ripe stream]

damn i was tacking a whole another approach

taking

[storm portal]

I am 90% sure that given the statement If p then q:

1. If q then p is the converse
2. If ~p then ~q is the inverse

[fervent dune]

no

[main isle]

I’m stuck on 14

[void atlas]

take two cases of volume V1 and V2, for V1 use radius r and for V2 use radius r/2, then take ratios

[dire vector]

If I have an equilateral triangle

And I have a random point in that triangle

How would I calculate the closet distance to an edge?

[grizzled kindle]

@dire vector write all edges in form of a line

Then use this formula

x0 and y0 are the co ordinate of the point

Then compare them

[dire vector]

So I'd do that above calculation three times?

Once for every edge?

[grizzled kindle]

Yes

[formal cove]

Hi

Please help

[crude niche]

you know 2 angles, and then (i forgot the theorem) u know the third angle, which means that by AAA they r congruent

i forgot how to do proofs tho lol so idk how exactly u write it down

[void atlas]

common angle X, W=X and YZW=YZX =90, three same angles so it is congruent by AAA

[silent plank]

AAA is not a valid justification for congruence

[knotty stump]

wdym? We all know that we live in a world with negative curvature 😌 @silent plank

rectangles don't exist

[weary drift]

@knotty stump pls dont shitpost

[hoary mango]

nvm I’m kinda used to trigonometry

I just need help on radians and how to find ratios on a right triangle and it’ll be fine

[placid solar]

Hii

Can you please help me with finding the angle of a non-isosceles trapezoid

[dark sparrow]

"the" angle? @placid solar

do you have the problem statement in front of you?

[placid solar]

I need to find b

[dark sparrow]

...i see some measurements marked in pencil

what was given and what did you find yourself?

is 13 meant to be the length of AB?

[placid solar]

Yes

[dark sparrow]

so you are given AB = 13, AD = 6, CD = 4 and angle A = 70° and you are asked to find angle B. do i understand this correctly?

[placid solar]

was given AD ,DC, AB , angle DAB

[dark sparrow]

okay

where did CF = 2.052 come from?

[placid solar]

I found DE with sin and DE and CF are same length

[dark sparrow]

please do not reply-ping me so often.

[placid solar]

Oh ok sorry

[dark sparrow]

did you use a calculator to find DE?

[placid solar]

Yes

[dark sparrow]

and you got 2.052 from calculating 6 * sin(70°)?

[placid solar]

yes

[dark sparrow]

...that's strange.

[placid solar]

Oh wait

[dark sparrow]

do you have the calculator with you?

[placid solar]

Nnno

[dark sparrow]

no to what?

"no, i didn't calculate 6 * sin(70°) at all" or "no, i calculated 6 * sin(70°) but didn't get 2.052"?

[placid solar]

I mixed up the numbers

6*sin70 is 5.638

[dark sparrow]

okay

so you calculated CF

do you understand that if you also calculate FB you will have enough info to find angle B?

[placid solar]

I just found the FB

How do I find the angle

[dark sparrow]

CF and FB are known, and CFB is a right triangle

focus on just that triangle

[placid solar]

I calculated

So now I know all the sides of this triangle

[dark sparrow]

"i calculated" is the most vague response possible

ok, look. do you know your way around trigonometric ratios?

[placid solar]

yes

[dark sparrow]

okay

so you have this triangle

tell me, what ratio is CF/FB for angle B?

[placid solar]

You mean like

Cos?

Did I understand right ?

...

What is ratio

[dark sparrow]

no, CF/FB is not cos(B)

[placid solar]

Tan

[dark sparrow]

yes

CF/FB = tan(B)

do you now understand how to find B?

[placid solar]

Yeaaaaah I found it

thanks

[still laurel]

what would a solution for a question like this look like?

i understand the part where you can solve for x in the equation, but then how do you use the value to determine the number of solutions in an interval

[grizzled kindle]

@still laurel let's say Sint = k then how many solution are there in [0,2π]

There are 2

Now you know about streching and squeezing a function wrt to axis

Is done by multiplying the respective co ordinate by a scaling factor

So therefore it'll strech the graph by a factor of 4

Therefore 2× 1/4 is your answer

[still laurel]

im getting one as my answer @grizzled kindle could you explain in like math symbols, the writing is not transferring to my brain, my apologies

why would you multiply 1/4 by 2

[pine iron]

The answer is one

[pine iron]

I like to think this way: for a given number k between 0 and 1, sin(x) = k will have two solutions in a complete trip around the unit circle.

(Such as sin(x) = 0.8, x could be 53.13° or 126.87°)

[pine iron]

Given the equation sin(1x/4) = 0.8, x in the range of [0, 2pi] the maximum "trip" you could get in the unit circle would be x = 2pi. sin(1(2pi)/4) = (sin(pi/2)). Between 0 and pi/2 there is only one solution to get sin(x) = 0.8, which is 53.13°

[foggy cosmos]

can anyone help me with this pls

[carmine spear]

tan(alpha) = 5/6
alpha = arctan(5/6)
alpha = 13.115

[silent plank]

above work is incorrect

what have you tried so far?

@novel hemlock

[novel hemlock]

my teacher tried to explain it

cant remember what he said

he said something about hypotenuse and sine

[silent plank]

do you know what trig functions do in a right triangle?

[novel hemlock]

no todays the first day of trig 💀

[silent plank]

and/or have you been introduced to the
mnemonic soh-cah-toa

[novel hemlock]

oh nvm

looked back at the notes he put up

[west ferry]

It seems that I lack the knowledge of that

[past bobcat]

the godly soh-cah-toa

life saver

so I get that this might sound like a stupid question but is there an equation that works for reflecting a point across any line that isn't y = 0 and the slope is greater than or less than 0?

[hoary mango]

trig triangles for dummies:
hypotenuse is longest (most common hyp is diagonal)
adjacent line is PARALLEL to hypotenuse
opposite line is connected to BOTH the hyp and adj.

[tardy panther]

the adjacent is parallel to the hypotenuse?

since when

[silent plank]

that attempt at oversimplification of triangles is unhelpful and some of it is even wrong

[wise pawn]

I feel the naming already makes sense on its own, no need to add an extra layer to it. Adjacent means the leg that's literally adjacent to the angle you're making with the hypotenuse. Opposite means it's the leg opposite the angle.

[silent plank]

associating the hyp with orientation is also something you shouldn't be doing

and what kind of triangle doesn't have all their sides connected

[hoary mango]

now i feel stupid

[silk patio]

I think using the words hypotenuse adjacent and opposite is a waste of time too. I directly associate the opposite with sin(x) in my mind for a unit triangle. I don’t have to think of which trig function it corresponds to, I cut out the middleman

[dark sparrow]

@hoary mango you know no two sides in a triangle are ever parallel right

[hoary mango]

I KNOW

I WAS BEING DUMB

[teal goblet]

what does the || and inverted T symbol represent? something about congruency, being similar, etc?

[manic salmon]

$\sum_{k=1}^{13} \frac{1}{sin[\frac{\pi}{4}+(k-1)\frac{\pi}{6}]sin[\frac{\pi}{4}+\frac{k \pi}{6}]}$ I need help with this

[somber coyote]

InertialObservr

[manic salmon]

The problem wants me to simplify it

[silent plank]

is parallel to and is perpendicular to respectively

[astral hull]

Anyone has an idea how to prove point ABCD are concyclic?

[dire marten]

M is the midpoint of AB and N is the midpoint of CD, then you need to draw two perpendicular lines from M and N.

[dire marten]

Sorry for my English

[astral hull]

thanks let me try work on this way

[astral hull]

it is clear enough

[fickle grove]

If I got 2 points A and B, where B determines the direction of the vector AB. How can I scale that vector to a certain bound? I.e. I want to scale the vector AB until either or both coordinates hit the "end" of the coordinate system

[void atlas]

there is no "end" of the coordinate system

[fickle grove]

that is why i put it inside " "

there is for a specific problem I am trying to solve

[void atlas]

do you mean by scaling that the direction remains same but you have to 'extend' its length to a certain extent

then just multiply the magnitude

[fickle grove]

the direction has to stay the same

[void atlas]

just multiply the magnitude to 'scale'

[fickle grove]

got it, ty

[glacial crane]

How to understand trigonometry?

Nobody knows either I guess. Well, that's predictable

[grizzled kindle]

What's it you need help with?

[astral hull]

still unable to figure out

[smoky palm]

Very dumb question

okay so

Do people capitalize the t when writing tan(x)

I mean, For Sin and Cos i do

but not for tan

[gleaming nova]

no

[silent plank]

people don't really capitalise for any of them

[grave pond]

Standard notation uses lowercase letters for all of the trigonometric functions.

[smoky palm]

really

Huh

fight me, I shall continue to capitalize Sin and Cos

and Cot

not sec though

But I will for Csc

[grave pond]

using non-standard notation is your prerogative; you get to pick up the pieces if it confuses your readers. ¯_(ツ)_/¯

[wise pawn]

why use different capitalization for some rather than others

[grizzled kindle]

@astral hull is there any other information ? Like C = 90° ? Or anything else?

[indigo cove]

How do I show there is no isometric from R^3 to R^2

[dark sparrow]

you mean no isometry?

there exists a quadruplet of points in R^3 which are all at distance 1 from each other, but there is no such quadruplet in R^2.

[indigo cove]

Yeah thanks

I thought I had to show using definitions so that makes sense

[glacial crane]

idk, i just cant get it, like, at all

[grizzled kindle]

Happened to me too

It'll take little time to settle in

[astral hull]

yup it is 90 degree

[astral hull]

and cd is along the line passing the centre of the larger circle

[fleet sentinel]

can someone help me with this

[gleaming nova]

first one looks like some sort of sum identity

third one is just sin(2x) multiplied by something

[limber inlet]

hello. can someone please help me turn 2sin^2(x) - 1 = 0 to sin(x) = 1/2?

[silent plank]

that's not possible

[limber inlet]

ah nvm nvm. i read it wrong and that's why i wasnt solving it

i figured it out

[jade ember]

is there a transformation matrix for clockwise 90 degree rotation?

[grave pond]

Yes.

Transform (1,0) according to your specification; that becomes the first column of your matrix.

Then transform (0,1) according to your specification; that becomes the second column of your matrix.

[jade ember]

ah found it thanks

[quick sapphire]

hi, could i pls get help w this?

<@&286206848099549185>

[grave pond]

#❓how-to-get-help

[violet horizon]

Have you gotten help

[hushed arch]

@quick sapphire I would first convert 30 60 90 to 1 2 root 3 and 45 45 90 to 1 1 root2 then do guess and check to verify which triangle each value corrosponds to

[knotty stump]

@fleet sentinel the first one is just $sin(x-\frac{\pi}{6})$

[somber coyote]

Arm

[scenic garnet]

yo i know soh cah toa

[knotty stump]

So... $x-\frac{\pi}{6} = sin^{-1}(\frac{1}{2})+2\pi k$

[somber coyote]

Arm

[humble pulsar]

you have a bunch of similar triangles

You explain it then.

[grave pond]

That's, um, not quite right.

Start by 4/x=x/9

That gives x.

Then just Pythagoras for the two last sides.

An altitude towards the hypotenuse in a right triangle cuts it into two smaller triangles that are both similar to the original one.

(Because two right triangles that share one of the acute angles are always similar).

4/5 is not equal to 5/9.

Right.

y and z become medium-ugly square roots, though.

Why?

There's no such thing as tan 90°.

If x/z=1 and z/13=1 then x=z and z=13 so x=13. Where did you get those relations from???

Can you give more details of your reasoning? Say,, why would z/13 be 1?

How do you know what that angle is?

Ahem. You cannot fight in here, this is a math room.

[gilded thunder]

<@&268886789983436800>

[clear haven]

what

why are the messages vanishing

ok gimme a second

[gilded thunder]

my dude thinks that there are no delete logs

[clear haven]

i think that clears it up

[grim musk]

hello guys

for this pic we have an angle theta

and so (x,y) which is at some point P would be rcos theta and rsin theta respectively

now i had this one doubt

if theta here is some angle( it can be anything but lets take it as say 30 degrees)

and radius is 1

we get (x,y) as cos theta, sin theta

now lets say the initial side which is x also lies on the circle like the point P

then we will get the (x,y) of the side x to be r,0