geometry-and-trigonometry

[worn scroll]

in this case, DE = kAB

CD = kAC

CE = kBC

these k's are all the same

it makes each side k times bigger

the area k^2 times

[azure helm]

ye

[solar slate]

Can any1 do this ?

any1 need solution DM me

[deep pivot]

would this be the same answer as this

[solar slate]

this is easy

[tiny sundial]

yes

[solar slate]

try this

[tiny sundial]

i mean no

noooo absolutely nor

haha

[tiny sundial]

have you solved this?

[solar slate]

trying

almost half way till solution

[tiny sundial]

No it would not. your expressiom simplifies down to $-\frac{2}{3}\sqrt{3}-\frac{1}{3}$. These are not the same

[somber coyote]

joel

[solar slate]

it would not u r right

btw igot the solution

[deep pivot]

Ah, thank you

[azure helm]

I need help

[solar slate]

hey hey

can any1 solve a question made by me

if need solution DM me in person

[silk pollen]

We got DG/BD = GF/AB = 3/10
Rewrite it, we have DG/3 = BD/10 = (DG + BD)/(3+10) = 10/13
From there we have DG => area of DGF => area of CDEF

[azure helm]

oh my badd I misread I'm stupid

[worldly juniper]

what's the answer?

i tried lel

well i could try verifying when i have the time so just ignore me

[frosty rose]

i might genuinely be losing it but can someone draw out this triange please

[solar slate]

wait

[frosty rose]

it's probably something basic sry but im on 2 hours of sleep 😭

[solar slate]

alr

[frosty rose]

how do i draw that

[solar slate]

were'nt u on sleep bud ?

[frosty rose]

wym

[solar slate]

nvm

[regal phoenix]

is that it? considering right at a

[frosty rose]

oh so thats a height

thanks vro 🙏

[frosty rose]

so the area of ABC here is (BC x AH)/2 ?

[regal phoenix]

yep

you can also use x' and x'' depending on what the exercise wants

[frosty rose]

this is the question\

[regal phoenix]

damn

[frosty rose]

it looks so simplie but goddamn

im losing sanity

the whole worksheet took me two days

im still at half

[regal phoenix]

does it give any other value/info?

[frosty rose]

this is the whole exercise

but idk if anything before 4 is relevant

is there missing given ?

[flint marlin]

Nmtc questions?

[slender rain]

Can anyone explain why does on line 5 the a(x + b/[2a])² + c - b²/(4a) is not a(x + b/[2a])² + c - b²/(4a²) ??

[trail tendon]

a(x^2 + (b/a)x) + c
= a(x^2 + (b/a) x + (b^2)/(4a^2) - (b^2)/(4a^2)) + c
= a(x^2 + (b/a) x + (b^2)/(4a^2)) - a(b^2)/(4a^2) + c
= a(x + b/(2a))^2 - (b^2)/(4a) + c
the a gets distributed to the (b^2)/(4a^2) to give (b^2)/(4a)

[azure helm]

help please

[maiden brook]

Find the area of AED

[regal phoenix]

since AF:FD=3:1, the area of AEF and DEF also have the same ratio

[frosty rose]

can anyone help with part 4 in here?

[gritty topaz]

17:51 = 1:3

(51+17):126 = 1:2

(51+17+126):194 = 1:1

if you think you don't understand, don't bother with the problems and learn the theorems first

[elfin gorge]

which class is it

[gritty topaz]

in turkish i know but in english 😂

1 sec

[elfin gorge]

its way too ezzzzzzzzzzzzz

[gritty topaz]

does this even have a name

[frosty rose]

triangle(s).

[earnest yarrow]

Area lemma

[maiden brook]

A = 1/2 bh

[brisk brook]

Hi there, i am new in this group! Is this group for student in High Schools?

And can this dc group help me with my GCSE'S?

[trail tendon]

highschool and beyond yea

[trail tendon]

if you ask questions people usually try to help yuh

[brisk brook]

Oh tysm bro

the discussion room is crazy

here

https://qph.cf2.quoracdn.net/main-qimg-b2a7b1481a32095c04cf58272361a9b0-lq

[slender rain]

Now I'm confused why does the h = - b/(2a) in a(x - h)² is in negative sign, while the equation literally showing + b/(2a)

[trail tendon]

(x-h)^2
h = -b/(2a)
(x-h)^2 = (x- (-b/(2a)) )^2 = (x + b/(2a) )^2
double negative makes a positive 😎

[slender rain]

Where did the negative h comes from anyway?

A formula created by some scientist?

[slender rain]

Isn't this physics....

[trail tendon]

f(x) = a(x-h)^2 + b
is designed such that if you plug in x=h, you get f(h) = b where (h,b) is the vertex of the parabola f(x)

so if h = -b/(2a), you know that -b/(2a) is the vertex of f(x) = ax^2 + bx + c

[brisk brook]

Yup guess it right lol

[limpid light]

Cot(4x-40°)=cot(2x-20°)
roots between [0-2π]
how do i do this?

[slender rain]

Makes sense. Last question, why does on my book it showing that to find the y vertex the formula is -D/(4a) wich I expand it will be -(b² - 4ac)/(4a) while the picture above showing that b² - c/(4a) is the k. Where did a and 4 comes from in -(b² - 4ac)/(4a)?

[slender rain]

This is maths server bro

[trail tendon]

what "picture above"?

[slender rain]

In this one

[trail tendon]

-(b^2 - 4ac)/(4a) is what you get when you plug x=-b/(2a) into f(x), so its f(-b/(2a)) which is the y-coordinate of the vertex

i dont think thats ur question

[slender rain]

OOOO

[trail tendon]

oh you mean this?

that looks like a typo

oh wait

nah

-(b^2 - 4ac)/(4a) = -(b^2)/(4a) + (4ac)/(4a) = c - (b^2)/(4a)

i was assuming they forgot parenthesis but they didnt intentionally

[slender rain]

Ikr

Alright thanks, took me 3 days just to understand this formula. F for my slow brain

[onyx forum]

Ahh just solved this qn few days ago

[brisk brook]

Lol

is it hard

[solar slate]

lol what

what are you a weaksauce

where is your brain man

or cant you see whats the server abt

it math

not physics

and also showing these easy questions wont make you look cool

[worn scroll]

ay man you should calm down

[solar slate]

still uk who's the wrong one here

[cinder raven]

Does anyone have ideas on how to do it? (top f_

[obsidian harness]

use the identities $z - 1/z = 2i \sin \theta$ and hence $z^n - 1/z^n = 2i \sin n \theta$ given that $z = \cos \theta + i \sin \theta$

so you just need to expand $(z - 1/z)^5$ and group them into terms of the form $z^n - 1/z^n$

[somber coyote]

south

south

[brisk brook]

lol mb 😢

[shrewd fog]

i didn't expect this community to be so toxic

[obsidian harness]

how so?

[shrewd fog]

😕

[obsidian harness]

these users aren't representative of our server

[shrewd fog]

well that's reassuring

[obsidian harness]

like we let people join without phone verification and so on

this is just what happens, especially in discussy

either they calm down or they get moderated

[shrewd fog]

ok

[brisk brook]

like bro i am here to learn new thing not to make fun of or get ganged my the rooms

by*

[worldly juniper]

rooms??

[charred grove]

Hey, can anyone that's avaliable in Geometry let me know if I can dm them i need some help!! For some reason claiming a help channel doesn't work for me

[storm zephyr]

just send it here, if someone knows they might help

[onyx forum]

If it is a difficult / moderate question i am more than happy to solve..

[charred grove]

Okayyy thank youu

I've been trying to solve these problems, and I was wondering if they are correct or if I messed up somewhere. The yellows are my answers.

This one i don't understand how to do

[trail tendon]

are you familiar with the law of sines?

[brisk brook]

Aren't they meant to be Circle Theorem Problem?

[slender rain]

I hv science helper server if u want to join

[brisk brook]

do u?

Like for Triple Science Students

[slender rain]

Yes, they cover all aspects of science. I don't really know the detail since I just randomly bumped into the server

[upper karma]

Basic trig

[rotund granite]

This is simple trig. You know the three ratios, sin, cos and tan, you use those to find the other sides.

[charred grove]

Okay, are the others correct?

And tyy

[charred grove]

Is the answer 8√5mm and 16 feet?

[trail tendon]

naw, actually are you familiar with the side ratio of 30-60-90 triangles and 45-45-90 triangles?

[charred grove]

Idk what u mean by that 😭

Like they all add up to 180 that's what I know

[trail tendon]

if you have a triangle and the inside angles are 30, 60, and 90

then the sides have ratio 1:sqrt(3):2

so like uhh

[trail tendon]

as long as the angles are 30, 60, and 90, the sides will always be these for some number t

so then if you plug in t=8mm, do u know what a and b are?

[slender rain]

Where did you get 1:√3:2 ?

[trail tendon]

its just a commonly known ratio that people memorize

you can see it works because sin(30 deg) = 1/2 and sin(60 deg) = sqrt(3)/2, which it follows from

[charred grove]

Can someone please tell me if this is correct?

[charred grove]

HUHH😭😭

[worn scroll]

confused about what its asking. is it asking about how they are represented in the circle?

[regal phoenix]

i think it’s asking for the values of sine, cosine and tangent of angle alpha

cosine seems to be 1/3

which exercises do you need answers to?

[slender iris]

Please help

.

AN=NS

[regal phoenix]

!show

[lime crown]

Show your work, and if possible, explain where you are stuck.

[slender iris]

I need a way to calculate the value of x or prove that it is worth 30 degrees.

[regal phoenix]

to find 124 you just do 180-32-24, since all angles in a triangle add up to 180º

you cand find 56 since both angles are in the same line, so they add up to 180

now do the same with the blue triangle, to get angle "a"

a = 180 - (x + 24) - 32

can you solve it from here?

[slender iris]

Nou😔

I understand that, but how do I show that x is 30 degrees?

[slender iris]

I'm stuck in that

[regal phoenix]

harder than i thought

[upper echo]

But the answer is not 30 degrees. It is about 31.208...

[slender iris]

Aprox. But how?

[upper echo]

you can use the law of sines.

The exact construction shows this picture and tan(x)=sin(2Pi/9)/(2cos(2Pi/15) - cos(2Pi/9)).

[upper karma]

Wanna see my problem

[upper karma]

Its drawing

[gritty topaz]

cool

[crude nebula]

how old are you?
do you know sin and cos?

[stone nimbus]

Looks alot like etherium

[hybrid belfry]

Guys, i have a question which is kinda related to geometry but it's not 100% about math:

What gave human the sense of straightness and perpendicularity?
(i'm not sure if these words exist, i'm not a native English speaker)

What about the other animals?
Is there any kind of animal which can't sense straightness or perpendicularity?

[slender iris]

17, yes I know them, but I would like a way through geometric constructions calculate the value of x

[slender iris]

In nature, it is quite possibly the positioning of some objects over others. Example when stacking rocks, perhaps even see the positioning of trees with respect to the ground and the particularity of some. Things like that xd. Animals should be investigated

[quiet bloom]

This may be a dumb question (proving a postulate), but is there any way to show Euclid's 5th postulate is true exclusively for flat space?

I have been looking through some of the proofs from The Elements, and in the first (prop. 27) there is a proof that equal alternate angles implies parallelism (implication that parallel lines can be constructed), but the converse (29) (only exist if a transverse between them forms 2 internal angles = 180deg/2pi) implies their uniqueness- that parallel lines exist only when internal angles equal two right angles.

I have been struggling to understand why it exists as a postulate from looking online (other than various 2D geometric spaces such as hyperbolic/spherical can exist with altered forms of the postulate, and it can't be proved from the first 4)

[woeful canopy]

Can anyone explain to me what the heck my teacher is trying to show me here? She was out of school and she gave us this same worksheet but without the asnwers so I am trying to fill it out. I have no clue what I am doing and what the "in the line y=-1" and "in the line x=-2" means and what it does.

[hidden grotto]

Hi can someone help me with my cofunctions

I dont understand c when it is 180 instead of 90

Ik im looking csc but how does the value change and does it switch quadrants

[terse hull]

well, if you don't know the formula for $\sec(180^\circ - \theta)$ then just substetute $\sec (\theta) = \frac{1}{\cos(\theta)}$

[somber coyote]

Umatriz

[terse hull]

so

$\sec (180^\circ - \theta) = \frac{1}{\cos(180^\circ - \theta)}$ then $\cos (180^\circ - \theta) = -\cos(\theta)$

[somber coyote]

Umatriz

[terse hull]

$\sec(180^\circ - \theta) = \frac{1}{-\cos(\theta)} = -\sec(x)$

[somber coyote]

Umatriz

[hidden grotto]

Thank you!! 😼

[sage maple]

all ik is that your cooked

[woeful canopy]

Dawg 😭

Idk what she is trying to show me

[sage maple]

srry g

its a mess

thats all i know

tbh all i know is that ur cooked

maybe someone else knows different

[rough violet]

Reflect across those lines

[pallid smelt]

pls help

[vapid totem]

help >> #help-33 message

[wooden widget]

hi

i do not have motivation to learn trigonomtrical formulas

how do i memorise

[silent plank]

go through the tedious process of deriving them
to disincentivise yourself from forgetting

and on the off chance, you do forget, you'll know how to get them

[gritty hemlock]

s/t = tan b ~ tan(0.693) as given @pallid smelt

[dense anvil]

ABC is a triangle in which angle B=2 of angle C. D is a point on BC such that Ad bisects angle BAC and AB=CD.prove that angle BAC=72 degree

[flint marlin]

It's a INMO question, you can google the solution

Very old though

[dense anvil]

@flint marlin but that won't help me though

'searching on google

[flint marlin]

Do you want a hint?

[dense anvil]

ye

plz

[flint marlin]

Let's see

Try joining some line which would give you congurant triangle

Also keep the fact that angle b is 2 times of c close to your heart

[dense anvil]

but if i wanted to join these lines which you have said it should have been given in the question if i am not wrong

like how it is given for others

[flint marlin]

Lol geometry doesn't work like that

[flint marlin]

I don't know which book you are using but if the construction is already given

What's the use of our brain and that question

[dense anvil]

bro i am in 9th grade

[flint marlin]

So am o

I*

[dense anvil]

and construction is not there

[flint marlin]

Are you Indian?

[dense anvil]

yeah

ofc

[flint marlin]

Ah rd sharma

Lnfao

[dense anvil]

you indian?

[flint marlin]

Yes

[dense anvil]

wanna do group study?>

i mean ur wish

[flint marlin]

Brother I can be your teacher (sorry might sound rude)

Are you aware about rmo?

[dense anvil]

nah teach me like a friend

[flint marlin]

Sure

[dense anvil]

i don't wanna hear the same crap from everyone plz

everyone said the same thing

when i reached them

[flint marlin]

Ok lol but we all want to inspire ppl to do maths

[dense anvil]

i love math

[flint marlin]

School maths I shit

Is*

Don't think that's maths

[dense anvil]

k

[flint marlin]

Enter in the realm of non routine maths

Ok back to your question

[dense anvil]

you if you are comfortable can we talk like in vc

for study

[flint marlin]

Nah vc is too much

Come to dms

[dense anvil]

k

[sharp void]

what are "postulates" ?

you mean euclids postulates? or something different?

[merry cloud]

This

[sharp void]

oh alright thats what I was thinking

you need SAS for level 2

[obsidian harness]

murica spotted

[drifting zealot]

Could someone help with this?

[craggy wave]

What is spin 1/2?

[young helm]

im taking this course in like 2 weeks can someone help me like prepare

taking trig

[wise pawn]

electron is spin 1/2

wait this question doesn't belong here

[vital fulcrum]

true

that some quantum mechanics

you should to physics server

electrons are dimensionless so what the spin means is a change in the state of the electron, basically this a super mega oversimplification

[wind shard]

https://en.m.wikipedia.org/wiki/Weitzenböck's_inequality

w.r.t the geometric proof, why exactly do we choose the Fermat point?

[orchid glen]

The Fermat point is chosen in geometric proofs because it provides an optimal solution for minimizing the sum of distances from a point to the vertices of a triangle. Specifically, the Fermat point is the point inside the triangle where the total distance to the three vertices is as small as possible, making it highly useful in optimization problems, such as minimizing travel distances or energy consumption. Its geometric properties allow for elegant constructions, and it can be derived using simple principles, such as drawing equilateral triangles on the sides of the given triangle. Additionally, the Fermat point is closely related to Fermat’s principle of least time, which asserts that light travels along paths that minimize the total travel distance. This principle has broader applications in optimization problems, making the Fermat point an important concept in both theoretical and practical geometry.

[wind shard]

Yeah ... I also asked AI before

[orchid glen]

ok . good . i also get some help from ai to make it more easier

[wind shard]

But this is not the answer I'm looking for.

[orchid glen]

ok .what are you looking for then ?

[wind shard]

The Wikipedia article mentions that it is used to partition the initial triangle into three obtuse subtriangles with 120 Degrees. I know that this is a property of the Fermat point. But why should we care about the subtriangles having this angle. Why couldn't we use the centroid for example. What's the reasoning behind choosing the Fermat point, or, a level deeper, to have the 120 Degree angles in subtriangles.

And how exactly do we know that these fit into the equilateral triangles three times?

[orchid glen]

This 120° property is significant because it represents the most "efficient" geometric configuration for minimizing the total distance between the point and the vertices of the triangle. This characteristic is related to optimization and minimal distance, which is why it's chosen in many geometric proofs and problems

[wind shard]

What is the optimization here?

[orchid glen]

The reason we care about the 120° angles in the subtriangles is that it directly relates to the minimal total distance property of the Fermat point. The Fermat point, with its 120° angles, minimizes the sum of distances from the point to the vertices of the triangle, and this is what makes it so important in many optimization and geometric problems. While the centroid is a central point with its own important properties, it does not minimize the distance in the same way, and thus doesn't lead to the same optimal solution or the desired geometric configuration of 120° angles.

[wind shard]

Again, this is not helpful. I know what the Fermat point is and some of its properties. These AI generated answers are not enlighting. I tried to use ChatGPT aswell. It wasn't helpful.

[orchid glen]

bro that was not ai generated . i am literally looking my books

i read it almost a year ago

can you text this again . i will try to text this to you through direct message

[wind shard]

It is obviously AI-generated. You're writing is very different. Even your interpunction is different.

[orchid glen]

I guess I need to think it for a while

[wind shard]

Anyways, thanks for trying to help. But I don't think this is getting at anything valueable.

[orchid glen]

I am from a bengali speaking country . I cant speak or write english like a native one '

[wind shard]

No problem

[orchid glen]

I guess I can solve it . please text this question again and your doubts .

I will revise it and try to elaborate it .

[wind shard]

I understand it up to the point where they choose the Fermat point. I know what the Fermat point is (it minimizes the distances to A, B and C) and the subtriangles will have 120 degree angles.

But why should one care about them having 120 degree angles? Why is that important for the proof. Also, how do we know (not just by looking at the picture), that these subtriangles fit into the equilateral triangles 3 times?

[orchid glen]

is that all?

[wind shard]

yes

I understand the other proofs of the theorem, but they don't really shed light on why this theorem is reasonable. They just show that it holds.

[orchid glen]

can i text you through direct message?

[wind shard]

I'd prefer it here

[orchid glen]

I need some time . there is chance to lose this chat and you . now thats become my confusion too

[twilit girder]

I know this should be easy for people but I haven't done quadratics in a while and I'm studying on my exam could anyone help me with this?

[wise pawn]

for starters, do you know the quadratic formula?

[twilit girder]

yeah

[wise pawn]

cool, so you can use that as a trick to help remember

the roots are equally spaced away from the vertex +-

[twilit girder]

oh wait so we use the quadratic formula to find the 2 x intercepts and choose the one that's positive?

[wise pawn]

no we don't

[twilit girder]

oh

[wise pawn]

but x=-b/2a is where the line of symmetry is

[twilit girder]

ohhhh thank you i just remembered

[wise pawn]

see what I mean by that? it's just the piece of the quadratic formula with the +- part under the sqrt thrown out

[twilit girder]

yeah I understand now thank you!!

[wise pawn]

cool, yeah you're welcome

[orchid glen]

For an acute angle the Fermat point is the point when the angles formed by joining each vertex to the Fermat point is all 120 degree. This is not valid for obtuse triangles*. To find out The Fermat point you have to draw equilateral triangles. You know the proof. When you construct an equilateral triangle on one side of the original triangle, the interior angles of the equilateral triangle are 60. Since the exterior angle is supplementary to the interior angle (they add up to 180), the exterior angle of the equilateral triangle is 120.

Now, when you connect the vertices of these equilateral triangles to the opposite vertices of the original triangle, the lines meet at the Fermat point, and the angles formed between them are 120. This 120 angle configuration minimizes the total distance from the Fermat point to the vertices of the original triangle, which is why it’s an optimal solution. you can think through another way . This happens because the total angle around the Fermat point is 360 . Dividing the angle equally among the vertices gives you 120 . The balance ensures that the Fermat point as optimized as possible. That why we equilateral triangles are need to find or distribute angles equally. the minimum summation of minimum distances is fixed . like fixed sums or products, the sum is minimized when the numbers are as equal as possible . but the distance is proportional to the Sin ratio of the angle between two another distances . Sin ratio is proportional to angle . so it is divided into 3 parts equally . 120+120+120=360 will give the minimum sum. Let me know if there is anything wrong ?

[wind shard]

Here's what I got:

The 120 degree angle guarantees that the side opposite is the largest side of the corresponding sub-triangle, and since the outward-facing equilateral triangles each use this largest side, the sub-triangles each fit into them three times. Furthermore, because of the sum of the angles in the triangle, it follows from the 120 degree angle that the other two angles of the sub-triangle together add up to 60 degrees, i.e. exactly the angle size of the equilateral triangles.

[gritty topaz]

theres also 2 fermat points

one in the triangle and one outside

[tepid hedge]

hi question, if you know three corner points of a triangle, how can you know wheither a random point is inside the triangle, outside it, or on the border?

[orchid glen]

To determine the position of a random point relative to a triangle defined by three corner points, one must first compute the area of the triangle. Subsequently, the random point is connected to each of the triangle's vertices, resulting in the formation of three smaller triangles. The next step involves calculating the total area of these three smaller triangles. If the total area matches the area of the original triangle, it indicates that the random point is located within the triangle. Conversely, if the total area exceeds that of the original triangle, it confirms that the random point lies outside. Additionally, it is important to verify whether the area of any of the smaller triangles is zero, as this would suggest that the random point is situated on the boundary of the original triangle.

[hybrid belfry]

Hey guys, i have a fundamental question.

I'm working from the most fundamental definitions/axioms of math,
so let's imagine that i'm a survivor in a library in the Mad Max world, i can read and right very well,
but i know absolutely nothing about math at all.
I have never seen anyone else but my parents and they both passed away.

(I know this is a bit dark but i just want a suitable context :))

Then i found a math book which says a lot of things about perpendicular straight lines, right angles
and i saw them a lot in the book,
but the book never told me their definitions.

I wanna know what their RIGOROUS definitions are.
I know how they look like, but i can't define them.

How should i define them?
(Let's assume that i knew the ground is not flat, so gravity is not an option to define them)

[trail tendon]

you want the axioms of euclidean geometry? or wut

[hybrid belfry]

What do you mean?

I didn't see them in Euclid's axioms

[trail tendon]

there aren't one set of axioms for the entirety of math...

[hybrid belfry]

How is that relevant to my question?

[trail tendon]

maybe i don't understand your question...

[hybrid belfry]

define:

• What is a right angle?
• What does 2 perpendicular straight lines mean?

Define them in any order you want.

[exotic yarrow]

https://mathcs.clarku.edu/~djoyce/java/elements/bookI/defI10.html

Google is your friend

[trail tendon]

hmm

I think I see what you mean

they don't explicitly define every term

[hybrid belfry]

It's very difficult for me to Google this because i'm not very good at English and the more fundamental things are the more difficult to define them.

I tried to search for an answer but couldn't find one

[exotic yarrow]

I just typed in "definition of a right angle"

The first result was this: https://math.stackexchange.com/questions/3623361/definition-of-a-right-angle

[hybrid belfry]

hm... 🤔...

[exotic yarrow]

(side note: stack exchange is very useful. However, if you ask a question on there, read their rules so that you don't ask "garbage" questions, aka questions with no context/low-effort questions/etc.)

and the first answer contains the link I sent earlier

In general, if you need the definition of something, searching

definition of [...] usually suffices

oh on the topic of searching math

if you wanna search equations

[trail tendon]

not me searching up "definition of perpendicular" and them talking about right angles and then me searching up "definition of right angles" and them talking about perpendicular

[exotic yarrow]

don't use google

https://approach0.xyz/search/

use this

note that you do need to type the equation in latex

(If you don't know latex, then just put the equation in chat gpt and tell it to write it for you in latex. Or you can always learn latex too )

there's probably more stuff I can spill later if you ask me at a different time

but this is all I can think of rn

[hybrid belfry]

This is a pretty good definition, 😄
now i'll go search for "the definition of equal angles",
i'll come back if something else comes up. 👌

[exotic yarrow]

(btw, "spill" is a slang term that I'm roughly using to mean "dumping information")

it's not the most appropriate meaning but

such is English

[trail tendon]

spill the beans :D

[exotic yarrow]

idt Euclid ever defined what it means for angles to be equal lol

he kinda just went off of vibe

💀

(idt = I don't think)

[orchid glen]

Should I try ?

[hybrid belfry]

Well i'm working from the fundamentals
so i don't care who defined what at when.

I just wanna have good definitions/proofs
(meaning the most intuitive and rigorous definitions/proofs).

A better context than the "Mad Max world" context i said earlier is:
Let's assume that i'm an immortal and i was borned alone on a far far away planet,
now i'm working on my eternal science exploration from scratch.

That would be a better context, i guess. 😄

[hybrid belfry]

Try to define them?
Sure, please do.

[orchid glen]

To begin with, it is essential to grasp the concept of a point. A point is defined as a distinct position in space that possesses no dimensions. Next, we consider a line, which can be visualized as a series of points arranged in a specific direction, thereby giving it a measurable length. The intersection of two lines creates angles, which represent a measure of space or rotation. When a line rotates and returns to its initial position, it has completed a full rotation. In contrast, when a line has undergone one-fourth of its total rotational movement, or when two intersecting lines create an angle that corresponds to one-fourth of a complete rotation, this angle is recognized as a right angle, indicating that the lines are perpendicular to one another.

I am not a native English speaker so I got help from ai re writer . hope that you can understand what am i trying to say .

[hybrid belfry]

I do understand what you said.
But i'm trying to avoid using terms that i couldn't/haven't defined or at least explain them in a clear and rigorous way.

For example i haven't defined "direction", "measure of space", "rotation" (should i use a circle to define rotation or vice versa?), etc.

[hybrid belfry]

I know that i might ask very intuitive things like
what is a point,
what is space,
what is a straight line,
what is a circle,
what is an angle,
how to prove a straight line which goes throught a circle's center will spilt the circle into 2 identical halves,
etc.

But i'm just starting my exploration, so... i gotta ask ((:

My goal is NOT defining every single thing or prove every single thing.

My goal is to REDUCE THE TOTAL AMOUNT of axioms as much as i can
(when i say "axioms" i mean all things that we usually just say
"it's obvious so we just accept that it's true and we understand it".
I don't know how to shorten this in English 😂)

[orchid glen]

I guess you cant skip space , negligible space or point , set of point or lie etc. They comes manually .

Human started geometry from realizing them and giving them a random name

I know students memorize it sometimes without understanding but you have to define them or you have to go practical measures .

Like you can define a right angle through rope

Let a rope be knotted at regular intervals so that there are 12

equal segments separated by knots.

Tie the rope in a loop consisting of those 12

segments.

Fix one of the knots to the ground at the point you want the right angle to be placed.

Stretch the rope tightly in the direction of one of the legs of the right angle you want to create and fix the fourth knot to the ground.

Stretch the rope tightly in the direction of the other leg of the right angle you want to create and fix the third knot in the ground, at the same time making sure the remaining segment of 5

knots is also tight.

The point where the 3
-knot section and the 4

-knot section meet, the rope will be bent at a right angle.

RightAngleRopeConstruction.png

are you looking for such kind of practical thing for right angels ?

[hybrid belfry]

I think these definitions rely too much on human's intuition.

It's will be difficult to build more definitions/statements on top of these. 😅

[orchid glen]

i guess thats why we have to go through from basic things . practically any axioms can be defined . but it cant be used for more definition . I mean realizing axioms after axioms.

[hybrid belfry]

Sorry but i can't understand what you said. 😂

I couldn't understand it after you say "any axioms can be defined"

[orchid glen]

my english or the concept

?

[hybrid belfry]

I guess it's your English that i don't understand

[orchid glen]

ok . sorry for that . i mean axioms are foundational but also inherently limited in their scope. They’re the basic building blocks, but they don’t explain themselves further, which is why every new theory must be grounded in new axioms or assumptions. This step-by-step approach seems necessary to construct more complex ideas from simple starting points

I wanna say that new definition needs previous defined concept .

[hybrid belfry]

I don't think creating new axioms or assumptions is a good things,
we should try to avoid that as much as possible.

It's like rebuilding a house over and over again,
it's very expensive 😄

[orchid glen]

thats why practical definition cant work in that case .

what do you think can we avoid axioms ?

[hybrid belfry]

avoid using axioms or avoid creating new axioms?

[orchid glen]

what are you trying to avoid by your imaginative fundamental exploration ?

[hybrid belfry]

By the way, i don't know any theory that require new axioms/assumptions.

Please share more informations about this? 😃

[orchid glen]

I can explain through physics

[hybrid belfry]

I know something like that in physics,
it's the fundamental difference between relativity and quantum physics.

But i was talking about math.

[orchid glen]

ok . before inventing complex number we cant find many solutions of equation .

like x^3 =1

before inventing integral we couldnt compute are of curvy bodies

before inventing derivatives integrals used to very complex

Leibniz calculus made more simple than newtonian calculus .

So , new axioms are also important to make previous axioms easier

[willow falcon]

is this the chat for ???

[charred magnet]

yes

[obsidian harness]

all users must have completed Clubstep before being allowed to type in this channel

[obsidian harness]

@trail tendon go complete Clubstep

you've been reduced to emoting in this channel I can tell

[worn scroll]

yes

[earnest yarrow]

Can anyone help me with latex

[hybrid belfry]

I have a few questions:

1. Why is the name of this channel geometry and trigonometry?
   Doesn't geometry include trigonometry already?
   Or i'm having some kind of misconception here?

2. How did people discover the Pythagorean Theorem?
   Maybe it was discovered at many different places around the world,
   i wanna know a few stories about how people discovered the theorem
   (including Pythagos' story if there is one).
   I wanna know the STORY, not just the proof, i can prove them alright.

3. What source would you suggest to find stories that explain how old science knowledges were discovered?
   (again, the STORY, not just the proof)
   The new stories are pretty easy to find so i don't have a problem with it.

[obsidian harness]

yes, but Euclidean geometry and trigonometry are very different

a lot of people will not accept a solution using trigonometry to a Euclidean geometry question

not "elegant"

https://hsm.stackexchange.com/ - this should be your go-to

[obsidian harness]

the problem is that our historical sources are just pieces of paper, some numbers on some clay tablet and so on

you really need a high-quality production and research team to put together this story if you want it in video

Veritasium's videos on mathematics have been excellent so far

I recommend looking into those even if they aren't an exact match

define new, cause there's a lot of stuff from the Industrial Revolution and that time that has been covered by Veritasium

[hybrid belfry]

Well, the older it is the harder to find its origin story.

So... as old as possible?

[obsidian harness]

fair enough then

you'd really need to find historians then who specialise in this

the next best are books on the history of mathematics, there definitely should be some for Babylonian and ancient Egyptian mathematics

https://archive.org/details/howmathematicsha0000rudm_m4o0

[hybrid belfry]

Veritasium is cool,
but i couldn't find many stories like
How was the eclipse discovered?
How were the theorems of similar triangles discovered?
How...
Why...

A lot of questions, i haven't seen them yet on Veritasium and they're not specifically focused on just the science history and their details.

I was hoping for some kind of channel or website something to search for these science origin stories

[obsidian harness]

not sure about the quality of these but here

[obsidian harness]

honestly Wikipedia is your best bet for a lot of these

but also it's hard to answer, like you can find examples of when different cultures used ellipses and similar triangles

I think you're asking for something that is unknowable

all of these things were discovered by ancient humans playing around

[hybrid belfry]

There's a problem with history books/websites/videos is that
most of them just tell which event happened around what time and who involved,
they usually don't focus on the progress of discovering the solutions from the moment people encounter the problems for the first time until they solve them.

What i wanna know is the way people think and do or encountered that led them to the solutions.
I wanna learn the way people solved problems as human,
the way they thought and things they did to find the solution,
not as a machine that requires a teacher to feed them a bunch data to memorize.
I don't care who those people are or where they are or at what time it happened, etc.

As you said,
it seems like these kinds of story are pretty rare to find if they're old. 😐

[mortal estuary]

hey, so I got an practise example with a rhumbus, it got me the height(h) and one diagonal(f) am I able to get other values out of those 2 values like side a and stuff, I can't figure it out

no but you could just take a sketch for it

should I draw one real quick? (could look super bad since I would do it on paint)
but its overall a more theoretical question

like I am not sure if its possible

the a's are all the same size

[orchid glen]

When provided with the height ( h ) and one diagonal ( f ) of a rhombus, it is possible to determine additional parameters such as the side length ( a ) and the other diagonal ( g ). Initially, it is important to recognize that the diagonals of a rhombus intersect at right angles and bisect each other. This characteristic allows the formation of right triangles, where half the lengths of the diagonals serve as the legs and the side of the rhombus acts as the hypotenuse. To calculate the side length ( a ), one can utilize the Pythagorean theorem: ( a^2 = \left(\frac{f}{2}\right)^2 + h^2 ), where ( \frac{f}{2} ) represents half of the given diagonal and ( h ) denotes the height. Upon solving this equation, the side length ( a ) will be obtained. Subsequently, to determine the second diagonal ( g ), one can employ the area formula of the rhombus in two distinct manners: ( \text{Area} = a \times h ) (where ( a ) is the side length and ( h ) is the height) and ( \text{Area} = \frac{1}{2} \times f \times g ) (which involves the diagonals). By equating these two area expressions, it becomes feasible to solve for ( g ) using the equation ( g = \frac{2 \times a \times h}{f} ). This calculation will yield the length of the other diagonal.

[somber coyote]

Arjuna

[mortal estuary]

wait tho, when I use the pythagorean theorem I don't really end up with a I end up with a and some other part which I will just call x which I marked green (its the dotted line)

[orchid glen]

sorry my bad i missed it

[mortal estuary]

was that a response from chat gpt?

cuz when I asked it I got a similar response

[orchid glen]

can chat gpt understand pictures . i thought it like a Orthographical projection.

I guess chatgpt can give the right answer

[mortal estuary]

?

well it can understand pictures

but not that good yet

well this is the shape

just 2d

idk what u mean with Orthographical projection

[orchid glen]

which diagonal you have ?

smaller one or the longer?

[mortal estuary]

well f. goes from the top left to the bottom right

[orchid glen]

do you know this theorem?

In an obtuse-angled triangle, the square on the side opposite to the obtuse angle is equal to the sum of the squares on the sides containing the obtuse angle together with twice the rectangle contained by one of the sides, and the projection on it of the other.​

Based on the threom$ f^2 = a^2 + a^2 + 2ax$, it can be simplified to$ f^2 = 2a^2 + 2ax$. Utilizing the Pythagorean theorem, we have $x^2 = a^2 - h^2$. Consequently, we can express$ f^2 as (a + x)^2 + h^2$, which expands to $f^2 = a^2 + 2ax + x^2 + h^2.$ This can be rearranged to$ f^2 = a^2 + h^2 + (f^2 - 2a^2) + (a^2 - h^2)$. At this point, with the values of f and h determined, one can derive the value of a and subsequently g by applying the Pythagorean theorem.

[somber coyote]

Arjuna

[orchid glen]

let me know if there is anything wrong.

[mortal estuary]

I don‘t get it like where is the part that actually gets side a, I only see f^2

[orchid glen]

which statement seems wrong to you?

[mortal estuary]

Well they‘re right but not relevant, since I already got h and f and urs is just telling me how to get those values

[orchid glen]

well $ f^2 = a^2 + h^2 + (f^2 - 2a^2) + (a^2 - h^2)$ you have the value of f and h . you easily find a from this equation

[mortal estuary]

But this equation is to find f^2 right?

[orchid glen]

dont you have the value of the diagonal f ?

[mortal estuary]

yes I have

[orchid glen]

you have to put the value of f and h . than write some more lines . and get a

ohh i get it now . sorry man again

a^2 got dismissed

[mortal estuary]

Well I need a so it would need to look like this a = …….
Or th

and I can‘t find an answer

This is the example

Its in german

[sleek ridge]

So I picked C so I could get the rationale (It cannot be seen in this screenshot) instantly but I just wanted someone to back me up on this.

So we know that this is a circle correct which means its perfect yes? {Question 1}

If it is perfect and we were given the point (5,0) then we can infer that C is (-5,0) correct? {Question 2}

Is the equation of the circle that is given be utilized to identify the radius of the circle? {Question 3}

If we can do the inferences of question 1 and 2 that means we know the radius of the circle is 5 yes?

Then we know A is just (0,5)

Then after all these questions I dont definitively know how we can say point B is the answer lol...

[twilit girder]

Hi guys, I kind of forgot how to find the "a" in factored form can anyone help me?

[maiden brook]

What are the roots? And then how can u make the rest of the graph work if it doesn’t work already

[orchid glen]

\section*{Given Data}
The height of the rhombus is ( h = 72.5 , \text{cm} ) and one diagonal is ( f = 96 , \text{cm} ). We need to find the other diagonal ( g ), the side length ( a ), and the angles of the rhombus.

\section*{Step 1: Use the Area Formula}
The area of the rhombus can be written in two ways:

[
A = a \times h
]
and
[
A = \frac{1}{2} \times f \times g
]

Equating these two expressions gives:

[
a \times 72.5 = \frac{1}{2} \times 96 \times g
]

Simplifying:

[
a \times 72.5 = 48 \times g
]

This gives the relationship:

[
a = \frac{48g}{72.5}
]

\section*{Step 2: Use the Pythagorean Theorem}
Since the diagonals bisect each other at right angles, we can use the Pythagorean theorem to find the side length ( a ):

[
a = \sqrt{\left( \frac{f}{2} \right)^2 + \left( \frac{g}{2} \right)^2}
]

Substituting ( f = 96 , \text{cm} ):

[
a = \sqrt{48^2 + \left( \frac{g}{2} \right)^2}
]

Simplifying:

[
a = \sqrt{2304 + \frac{g^2}{4}}
]

[somber coyote]

Arjuna

[orchid glen]

\section*{Step 3: Solve the System of Equations}
We now have two equations for ( a ):

[
a = \frac{48g}{72.5}
]

and

[
a = \sqrt{2304 + \frac{g^2}{4}}
]

Set the two expressions for ( a ) equal:

[
\frac{48g}{72.5} = \sqrt{2304 + \frac{g^2}{4}}
]

Square both sides:

[
\left( \frac{48g}{72.5} \right)^2 = 2304 + \frac{g^2}{4}
]

Simplifying:

[
\frac{2304g^2}{5256.25} = 2304 + \frac{g^2}{4}
]

Multiply both sides by 5256.25 to clear the denominator:

[
2304g^2 = 5256.25 \times 2304 + \frac{5256.25g^2}{4}
]

Now simplify and solve for ( g ) using numerical methods or a calculator.

\section*{Step 4: Find the Side Length ( a )}
Once we find ( g \approx 110.5 , \text{cm} ), we substitute this into the equation for ( a ):

[
a = \frac{48 \times 110.5}{72.5} \approx 73.2 , \text{cm}
]

\section*{Step 5: Find the Angles}
To find the angles, we use the fact that the diagonals bisect each other at right angles. The acute angle ( \theta ) can be found using:

[
\cos \left( \frac{\theta}{2} \right) = \frac{\frac{f}{2}}{a}
]

Substitute ( f = 96 , \text{cm} ) and ( a = 73.2 , \text{cm} ):

[
\cos \left( \frac{\theta}{2} \right) = \frac{48}{73.2} \approx 0.656
]

Now take the inverse cosine to find ( \frac{\theta}{2} ):

[
\frac{\theta}{2} \approx \cos^{-1}(0.656) \approx 49.5^\circ
]

So the full angle ( \theta ) is approximately:

[
\theta = 2 \times 49.5^\circ = 99^\circ
]

Thus, the acute angle is approximately ( 99^\circ ), and the obtuse angle is:

[
180^\circ - 99^\circ = 81^\circ
]

[somber coyote]

Arjuna

[orchid glen]

This time the answers matches . i used chat gpt to make this latex code understandable . let me know if i missed anything again.

[hybrid belfry]

1. Can we somehow prove that a straight line which goes through a circle's center always splits the circle into 2 identical halves?

2. Can we prove the diagonal of a square or a rectangle always splits the shape into 2 identical halves?

I know these questions might sound a bit crazy and maybe there's no explaination/proof for this at all,
i just wanna make sure that there is a way to prove it RIGOROUSLY or not. 😄

[mortal estuary]

Well we got an answer for an angle now but its not correct checking up with my solutions, I think its impossible to calculate

There has to be a mistake made by my teacher

[orchid glen]

I found the diagonal 110.5cm , the side length 73.2cm, the angels 99 and 81 . Its really close to the answer given in the question which was written in german. i used approx values . if you use more accurate values you will find the answer. Its not impossible to calculate. check my solutions .

[mortal estuary]

Thats actually accurate somewhat what formula did u use

[orchid glen]

didn,t you check my solutions ? i have sent them .mentioning you . you have to made equations and solve it

[lyric halo]

What are the geometric implications of the last form?

[obsidian harness]

https://en.wikipedia.org/wiki/Circles_of_Apollonius#Apollonius'_definition_of_a_circle

[lyric halo]

yippee, ty!!

[orchid glen]

I guess its a circle .

[lyric halo]

yes

[orchid glen]

is it related to complex number or something like argand diagram?

[lyric halo]

yea

[coral cobalt]

any and all help is appreciated

[signal field]

90 degrees to revolutions is 1/4 rev?

[obsidian harness]

correct

[fluid narwhal]

is there somone who understand frensh

[worn stag]

You don’t need to copy paste this in like 7 channels

[fluid narwhal]

hhhhhhh

yes

i do

but

why ur just

folloowing me

broooooooooo

[worn stag]

Also don’t type one word every message

I have eyes everywhere

[upper karma]

I need help
<@&286206848099549185>

[toxic jacinth]

<@&286206848099549185>

how to solve this

[trail bison]

I got a question like if I want help with a math assignment I’m i able to stream it so someone that knows how to do that specific math can watch and tell me if I’m wrong and what I’m doing wrong?

[analog cargo]

Highschol

Number 11 is confusing

Given: AT|EY||SB. Complete each proportion with the appropriate measure. AB/UA=?/TU

Is TS/or ST correct according to the image and given

[trim escarp]

Can someone explain how to find an asymptotes when comes to graphing hyperbolas

[zealous pivot]

$\angle{AUT} = \angle{BUS}
and \angle{UTA} = \angle{USB}$ then the other angle is also equal hence by AAA similarity ∆AUT= ∆BUS.
Then$ \frac{AU}{BU} = \frac{AT}{BS} = \frac{UT}{US}$

NOW put UB = UA + AB and solve accordingly.

[somber coyote]

Commander Woojin
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rough matrix]

Use a math-help channel.

[polar ermine]

high eyes reading thsi

[old pilot]

finding x is possible here right?

[silent plank]

the diagram itself doesn't make sense

[old pilot]

thats what im saying

I've been trying to answer this one for like an hour now

im just curious whether there's a scenario where you can get the value of x here

[silent plank]

if you bring complex numbers into this, i suppose

[old pilot]

[cosmic lichen]

x is possible

let me show u how

oh, i think your value for mesurement is wrong

[old pilot]

where

[upper karma]

If angles (A, B) are 45° then x would equal 10

[hybrid belfry]

Is there a RIGOROUS way to prove that a right triangle has the two edges'
(i'm too lazy to search for their name 🙃)
length are 1 and 2 then the angle must be 30° ?

The sin 30° = 1/2 thing?

I'm keeping it simple so let's just talk about a specific right triangle with 3 edges are 1, √3 and 2.

If you don't need to use trigonometry to prove it then that's even better. 😄

[cunning lion]

you can bisect an equilateral triangle to show that right triangles with a 30° angle have that side ratio

and it also shouldn't be too hard to show that any right triangle with side ratio 1/2 is similar to that original triangle

[hybrid belfry]

So you suggest that we should draw an equilateral triangle first then prove the angle is 30° later?
I don't think that's a rigorous way to prove it, that flipped the proof
(i can't think of a good way to say this in English 😂).
I mean i wanna prove that THE triangle has a 30° angle,
I wasn't talking about a right triangle which is created from spliting an equilateral in half,
that's the other way around.

Anyway, before that,
we still need to prove that each angle of an/the equilateral triangle is exactly 60°.

[proud cairn]

guys can someone explain to me why u can only touch one side of the equation when verifying trig identities

no im pretty sure she just means dont apply something to both sides

and not that i can only simplify one side

since that would become needlessly painful

[maiden brook]

u need to start from something true

u cant start from something false/something u need to prove and end with a true statement to conclude what u want

the classic simple example is like: lets prove -1 = 1. square both sides and 1 = 1, and this is true so the proof is done

[robust crane]

if you have two directional arcs with radius,origin,start angle and delta angle is there a simple way to test if the arcs end up close enough to each other by some distance ? the only way i currently think of doing is iterating the arcs and measure distance between each point but thats not computationally performant

[faint pasture]

You can do anything

Tho the tchr might deduct some points

But its not wrong

[unreal storm]

guys i have got 2 equations 3x-y=40
4x-2y=50
Can someone help me solve them graphically?

[obsidian harness]

https://www.desmos.com/calculator

[unreal storm]

?

[obsidian harness]

you can type those equations in here

[unreal storm]

on the left tab?

[obsidian harness]

then click on the intersection to find its coordinates

yeah

[unreal storm]

ohh thanks

also

can i get more cordinates or if it has given 3 its the maximum for that question./

?

[obsidian harness]

not sure what you mean

I get the intersection as (x, y) = (15, 5)

[unreal storm]

leave it;-;

[unreal storm]

how do i explain in my notebook how i got that answer

;-;

[obsidian harness]

I think you might need to copy the graph

pick two points on one of the lines, then draw a straight line through them

do the same for the 2nd line

I'm not sure if you're allowed to just say you did it on Desmos

[unreal storm]

nonono

like we create a table in first row x value and next row y

and we wrtie a random value of x

and write what y will be if that value of x is inputte

[wise zodiac]

guys

i need some help showing that between [2npi, 2(n+1)pi] there's only two intersection points of sin(x) and x/100

for 0 <= n <= 14

take any n for that matter; just make sure whatever n you select is 😭 in that range

so like idk if i'm being silly but i can't fully justify

that there must only be 2 solutions; no more no less

consider n = 1 for simplicity i guess

like uhhh i defined f(x) = x/100 - sin(x) and then by IVT f(2pi) > 0 and f(5pi/2) < 0 so there's a root between them

in a similar fashion f(4pi) > 0 and f(5pi/2) <0 so there's a root between them too

but how can u say for sure there's not more than these two roots?

like there's no guarantee that you can't have 10 roots between 5pi/2 and 4pi, right? all we know is there must exist at least a root?

[worn scroll]

i'd try taking it's derivative and see what happens or find the values for which the function is positive or negative

[nova jungle]

i found this question and i think its wrong

am not sure but

should the answer be 90

the 50 and 30 are irrelevent

and they don't make sense

[wise zodiac]

didn't help yet 😭

[wise zodiac]

we don't know if it passes thru the center

[silent plank]

don't assume things to be diameters when there's nothing indicating it

[nova jungle]

ohhh

then how do isolve this

[silent plank]

the given angles are very relevant

[wise zodiac]

inscribed angle theorem

[silent plank]

or triangle sum
then cyclic quad property

[wise zodiac]

just checked what a cyclic quad is

so a quadrilateral that has a circumcircle?

so i'm guessing the assumption is that the above figure isn't like a parallelogram or something

is that smth you'd check when trying to use cyclic quad property or smth

or is it only "okay the vertices touch the circle thus it is a cyclic quad"?

[wise zodiac]

also bump 😭

[storm zephyr]

yes

[storm zephyr]

squares, rectangles are parallelograms, and they are always cyclic quadrilaterals

but not every parallelogram is a cyclic quadrilateral

[wise zodiac]

well besides squares and rectangles ofc

[wise zodiac]

i think it’s just this

[storm zephyr]

isosceles trapezoids are also always cyclic quadrilaterals

[storm zephyr]

There's Ptolemy's Theorem too

[wise zodiac]

yes i read about it on wiki

i got it, thanks tho

[storm zephyr]

ok

[wise zodiac]

what i do need help with is this 😭

[storm zephyr]

yeah idk that stuff honestly

[flint marlin]

..

[dull bear]

x=10

[mystic shadow]

Hey quick question. When talking about a sinusoid, does it apply to both sin and cos or only sin?

[trail tendon]

both

[mystic shadow]

cool

[sinful schooner]

does anyone know how to do circle therom my maths teacher sucks and i have a quiz on this T-T

[proud cairn]

wym circle theorem

[smoky bridge]

Yroo

I need sum help much appreciated

[proud cairn]

they are similar triangles

[sinful schooner]

these things

[proud cairn]

o i see

what do you need help with

[sinful schooner]

i dont understand them at all

[maiden fox]

Does anyone know how to do angles of depression and elevation?

[covert quarry]

I can help with that

[covert quarry]

Have you learnt properties of angles in a circle?

[sinful schooner]

no

[covert quarry]

Angle C is 90 degrees as angles in a semicircle is 90 degrees

Then you can take sum of angles in triangle - 90 - 29

[sinful schooner]

so b would be 61?

[covert quarry]

Yes

But make sure for each step you state the reason for doing so

For e.g, angle C = 90 degrees ( angles in a semicircle)

[maiden fox]

Please do! I’m so conused, on it

[covert quarry]

Ok let me explain in a way that you might understand

So imagine your looking straight to the wall with invisible laser beams shooting out from your eyes. As you look up a bit higher than you were looking before, an angle of elevation would be form

When you look lower than you were looking before, an angle of depression is form

[maiden fox]

yeah, over/under the horizontal lazer i get that portion of it at least, but its direct scenarios that are messing me up.

i made a help-forum discussion post with like 2 problems i have where i have to draw them out

my teacher doesn’t give us the diagrams and i often dont know how to draw rhem out myself

[covert quarry]

I've seen your forum

[maiden fox]

yeah very confusing stuff

[covert quarry]

Ok I'll try to solve give me some time

[sinful schooner]

but b is also touching the circle so why is it 61 instead of 90?

[median plaza]

someone help me i have a D+ and a dream

[maiden fox]

you solving for x here right

well because its a rectangle, the long sides and short sides should b the same? should x not be radical 28 or do you have to use phythagorean theorem?

[slender rain]

Can we use sincostan for the upper triangle?

[covert quarry]

Cus C is opposite to the centre of the circle

[median plaza]

yea i got to use pyth

i solved it tho

[median plaza]

im doing 9th grade advanced geometry isnt that above my level

[sinful schooner]

oh thanks man

[slender rain]

Ups my bad, how do you solve it tho? It seems hard to me who even in highschool

[covert quarry]

Good to try new things 😂

[median plaza]

man im just screwed

[median plaza]

cant do that when i have a D+ and a test tommorow 😭

[trail tendon]

just using the pythagorean theorem

[covert quarry]

The length of the straw is the same length as the diagonal length of the base

[slender rain]

Ahh this one seems easy, use the 2D rectangle diagonal by searching the phyta, then use phyta again for the space diagonal

[slender rain]

How?

[maiden fox]

well not really, we learn trig at my school in geometry (im a freshman) but also its an honors course so idk

[trail tendon]

like they did in the picture
the hypotenuse of the triangle who has both side lengths as length 2 would be sqrt((2)^2 + (2)^2) = sqrt(8) = 2sqrt(2)
then the diagonal of the rectangle divides the rectangle into two triangles
the length of the rectangle's base is sqrt(8) and the hypotenuse is 6, so c^2 = a^2 + b^2 -> b^2 = c^2 - a^2
b^2 = 6^2 - sqrt(8)^2 means b^2 = 36-8 = 28, so b = sqrt(28)

[median plaza]

so what does that mean?

please help me out man

[covert quarry]

Cut the base of the rectangle box in half diagonally

[median plaza]

let me show you an example

[slender rain]

Dayum bruh, didn't notice that

[median plaza]

doesnt make sense to me

[median plaza]

im in high school too

im a freshman

[slender rain]

I suggest you to go in to the help channel

[median plaza]

theres a lot of them

?

[slender rain]

#❓how-to-get-help

[median plaza]

like this?

doubt anyone is gonna go check it out

[floral linden]

hello guys what are the prerequisite before learning geometry ??

[median plaza]

algebra

[floral linden]

ok

[covert quarry]

What doesn't make sense to you

[median plaza]

solved it, turns out i found in easier way

ty for tryna help

now im stuck with stuff like this

[vague trench]

DE=EB and then pythagorean

[median plaza]

my only issue is trying to find the hypotenuse and all the weird marks and lines on the shape

[vague trench]

the hypotenuse are given you are trying to find CE and AE using pythagorean theorem

[median plaza]

alr how do i know where the hypotenuse is tho for next time

DE = 5 correct

[vague trench]

the hypotenuse are the oppisite to the right angle

[vague trench]

yep

[median plaza]

when i got it wrong and it showed me the examples it kept mentioning these and they were supposed to help what do they mean

[median plaza]

aight

[covert quarry]

You good or you still need help?

[vague trench]

those two lines mean equal sometimes they could be on line three lines

[median plaza]

still need

this is the last thing i need help with

[covert quarry]

Ok where are you stuck at?

[median plaza]

idk what to do after this

[covert quarry]

What are you asked to find

[vague acorn]

5^2+7^2 to find EC^2?

[covert quarry]

Find AE using Pythagorean theorem

[median plaza]

my only issue is i have no idea where the hyptoneuse is

oh

wait

i do know

is that it

[covert quarry]

Hypotenuse is 6

[vague acorn]

the hypotonuse is 6

[covert quarry]

Yes

[median plaza]

ok now im here

[covert quarry]

Ok find AE with Pythagoras theorem

Do you know how to do Pythagoras theorem

[median plaza]

is this right

[median plaza]

a^2+b^2=c^2

[vague acorn]

u have to sqr 5 and 6

so 25 + 36

sqr root of 61

not 11

[covert quarry]

No 6 is the hypotenuse

6 is c^2

So it's 36-25

[vague acorn]

oh crap ya didnt see notice

ya what he said

[covert quarry]

Np

[median plaza]

therefore its sqrt 11 right

[vague acorn]

my bad brown boy

[covert quarry]

Yes

[median plaza]

u good

DC is longest side correct

[vague acorn]

can i ask a question here or do i gota do help channel

[median plaza]

ur choice

id make it here

[covert quarry]

Yeah

[covert quarry]

You can ask here

[vague acorn]

alr grabbing my packet one sec

[median plaza]

now what

[covert quarry]

Take Square root 24 + square root 11

[vague acorn]

do i just count the lines it passes?

[covert quarry]

If you don't mind, can you show the instructions?

[vague acorn]

like the nm' is 6 long and nm is 2 does that make the scale factor 3?

[covert quarry]

It's a reduction scale

So it will be lower than 1

[vague acorn]

wdym reduction?

the original is the small one

i just dont know how to find the lengths

im like 95% sure its 3

[median plaza]

so 35

[vague acorn]

is thta how to find side lengt?

[median plaza]

ohhh so its 5.9

[median plaza]

??????

it was incorrect

[covert quarry]

Oh then in that case it's enlargement

So it's more than one

[covert quarry]

Did you add 11 + 24 then square root?

[median plaza]

like this?

[covert quarry]

Yes

[median plaza]

ty bro u saved e

[covert quarry]

Np

[echo geyser]

what's the rules for this channel? and why is it seperated from help

[covert quarry]

Not sure honestly

[median plaza]

i cant tell which line is longer

i think its ED no?

[covert quarry]

Is it a different qn?

[median plaza]

ye

[covert quarry]

Can show the whole qn?

[covert quarry]

To find ED take 10/2

Cus BE and ED are equal

[median plaza]

oh yeah

[covert quarry]

So that makes AD longer than ED

[covert quarry]

Correct

[median plaza]

now ima solve EDC

so 8^2 - 5^2?

[covert quarry]

You wanna solve the angle?

[median plaza]

yea

oh ok i think i got it now

ty

[slender rain]

Use the preoccupied help channel

Is it pre-occupied or just preoccupied, rip languange syntaxis

[flint ingot]

@warm ingot
Example 1: $\Delta ABC$ has side lengths $AB = 8$, $AC = 15$, and $BC = 17$. If $D$, $E$, and $F$ are the circumcenter, centroid, and incenter respectively of $\Delta ABC$, find the area of $\Delta DEF$.
Since $\Delta ABC$ is an $8$-$15$-$17$ right triangle, we can place it on the coordinate plane. Let $A = (0,0)$, $B = (8,0)$, and $C = (0,15)$. (Check for yourself that $AB = 8$, $AC = 15$, $BC = 17$.)

The circumcenter of a triangle is the center of the circle which passes through the vertices of a triangle. In a right triangle, the circumcenter is the midpoint of the hypotenuse. The hypotenuse $BC$ has endpoints $B = (8,0)$ and $C = (0,15)$, the midpoint is $D = (\tfrac{8+0}{2},\tfrac{0+15}{2}) = (4,\tfrac{15}{2})$.

The centroid of a triangle is the intersection of the medians of the triangle. This sounds hard to do with coordinates, but as it turns out, the centroid of a triangle is simply the average of the coordinates of the three vertices. So, $E = (\tfrac{0+8+0}{3},\tfrac{0+0+15}{3}) = (\tfrac{8}{3},5)$.

The incenter of a triangle is the center of the inscribed circle. The area of $\Delta ABC$ is $K := \tfrac{1}{2} \cdot 8 \cdot 15 = 60$, and the semi-perimeter is $s := \tfrac{1}{2}(8+15+17) = 20$. Therefore, the radius of the inscribed circle is $r = \frac{K}{s} = \frac{60}{20} = 3$. So, the distance from $F$ to each of the three sides is $3$ units. Since $AB$ lies on the $x$-axis ($y = 0$), $F$ must lie on one of the lines $y = \pm 3$. Since $AC$ lies on the $y$-axis ($x = 0$), $F$ must lie on one of the lines $x = \pm 3$. Since $F$ is inside $\Delta ABC$, $F = (3,3)$.

Now that we have the coordinates of $D$, $E$, and $F$, we can simply use the Shoelace formula to find that the area of $\Delta DEF$ is $\tfrac{1}{2}\left|4 \cdot 5 + \tfrac{8}{3} \cdot 3 + 3 \cdot \tfrac{15}{2} - \tfrac{8}{3} \cdot \tfrac{15}{2} - 3 \cdot 5 - 4 \cdot 3 \right| = \tfrac{7}{4}$.

[somber coyote]

Kai Cui

[torpid shale]

anyone know how to solve this problem?

[mossy raven]

I need help anyone

[obsidian harness]

can you set up an algebraic equation?

the two angles sum to 90 degrees

[mossy raven]

Correct

So it is complementary

[crisp quail]

@wicked lark for u

[wicked lark]

Thx!!

[crisp quail]

np

[wicked lark]

Is this IXL?

[cedar flower]

Does anybody know is this can further be simplified?

[white lynx]

give me a trigo que

[fallen surge]

somebody ans!?

[cedar flower]

$\arctan(\frac{k \sin(\frac{\pi}{n})}{1 - k \cos(\frac{\pi}{n})})$

[somber coyote]

Siupa

[fallen surge]

uh

[cedar flower]

Where k > 0 is a real number

And n a positive natural

[exotic yarrow]

Hint: recall the Pythagorean identity involving $\sec^2 \theta$

[somber coyote]

Civil Service Pigeon

[white lynx]

wait 5 min

[fallen surge]

okay lemme seee

[fallen surge]

alr

1/m>

???

[white lynx]

done

[fallen surge]

whatrs the ansss

:)

[white lynx]

1/m

lmao

[fallen surge]

thanks

[upper karma]

chat my final is in 11hours and idk how to solve essay questions

am i cooked?

[inner badge]

Not with that pronouns u got

Say Bismillah and do it

[flint ingot]

is this an iscoceles kite

ayo

[maiden brook]

how’s that different than a normal kite

[cursive summit]

i have a question, which side is which? like which is adjacent, which is hypotenuse, and which is opposite?

[cunning lion]

the hypotenuse is the longest side/side opposite the right angle

the "opposite side" is the side which is opposite the angle you're interested in, and the "adjacent side" is the side next to the angle of interest which is not the hypotenuse

[cursive summit]

so lets say the bottom side is unknown, would i use sin cos or tan?

[cunning lion]

depends on which other side is known

[cursive summit]

bottom side unknown and the hypotenuse and adjacent are known (i think)

[cunning lion]

then you have a choice

it also depends on which angle you know

[cursive summit]

in this case i only know of the right angle bottom left

[cunning lion]

you can't use any of the trig functions to find a side if you don't know at least one non-right angle

[cursive summit]

but every problem in my assignment is set up like that

[cunning lion]

well if you know two sides and are trying to find a third side, then the pythagorean theorem seems like a more direct way to do it

[cursive summit]

may i send a screenshot of my assignment because i am very confused right now

[silent plank]

also the lengths of that triangle don't make sense

an ss is fine

[gritty hemlock]

,calc 1414 - 1010 - 12*12

[somber coyote]

Result:

-48

[royal charm]

Hi! Good morning/goodnight I need help on some packets! So what's happening is that this year I'm gonna pre calculus right, but the teacher told me, I didn't think geometry in Puerto Rico, so she gave me packets to do by the end of the year, I already finished 5 and I have 3 more to finish, but 3 of the package are confusing me, if there is anyone that can help me to do this till April I would appreciate this

[royal charm]

So if there's anyone that can help me, pls DM me

[silent plank]

just claim channels in this server when you need help

people usually prefer not to do this through dms

[worn scroll]

you can just ask for help whenever you are stuck or dont understand what is going on when you are going through the exercises

[worldly plank]

Hii does anyone know a good yt vid or website where they cover the rules and different theorems even basics. So for example thales theorem. I feel like I miss a lot of knowledge on the different rulings about geometry while I am making practice questions?

[light fable]

just search theorem by theorem. https://youtu.be/fyA8E-vSICA?si=ckl0tXzorxVkZx9-

[wise pawn]

for what it's worth, Thales' thm is a special case of the Inscribed Angle Thm

[obsidian harness]

Keeps coming up though so it deserves the special name

[wispy shoal]

do i need to study if i feel that i know the stuff thats coming on the test or can i study to be sure i know everything?

[obsidian harness]

feeling isn't enough, practice to check if you're getting the questions right

[worldly plank]

Thanks for all the answers

[undone quest]

Does anyone have idea about st loney

[pine thunder]

In theorem seven how do we get a negative ratio for external division. Won't that make the sides negative

[tiny sundial]

yes, this does not make sence

sense

[atomic prism]

The ratio BX/XC is a ratio of the segments. It does not mean that sides are negative. The ratio is negative only because of the external division happening. When dividing something externally, the divisor is greater than the dividend causing the ratio to always be negative. It is not the sides that are negative, but we are just expressing the ratio negatively. So i can see why it doesn't make sense.

[tardy basin]

Could someone help me figure out a good method to figuring out this proof garbage? I probably wont get it since this math is for lunitics but any help is appreciated

[maiden brook]

what’s wrong with the proof

[tardy basin]

been through like 6 tutors. 3 of them couldnt even get the answer right

[worn scroll]

cos/sin is another way of writing cotangent

1/sin is just cosecant

[flint marlin]

Excursion is boring

Use evan chen

[cunning trellis]

Is 3d trig usually a gr11 topic

[amber dagger]

how do i find the ratio of these angles with respect to sides?

[pine thunder]

I am trying to prepare from an olympiad point of view. Will it help?

All my friends are suggesting excursions

[flint marlin]

Ik , not so useful for geometry though

Maybe other topics are gud

Which is the hardest geo problem y'all have solved and enjoyed?

[pine thunder]

Monges theorem

[compact stirrup]

pls can someone check my solution for a geo problem real quick

[flint marlin]

Homothety should easily do it

[slender rain]

$\text{if angles of a right triangle is} \frac{\pi}{6}, \frac{\pi}{3}, \text{and 90, would the ratio of the side of the triangle would also be
1 to} \frac{1}{2} \text{ to } \frac{1}{2}\sqrt{3}?$

[somber coyote]

カザミ

[obsidian harness]

yes, that is correct

maybe easier to remember the side length ratio as $1: \sqrt{3}: 2$

[somber coyote]

south

[obsidian harness]

in any triangle, the longest side is always opposite the biggest angle

and the same for the smallest side and the smallest angle

[slender rain]

Nice insight

[limber tide]

could anyone tell me if i got it right or wrong? and where if i did get it wrong?

[obsidian harness]

,w (100 sin(60 deg))/(sin(75 deg)

[somber coyote]

Results provided by WolframAlpha

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

Requested by sooouth.

[obsidian harness]

,calc 89.6575 /sqrt(2)

[somber coyote]

Result:

63.397426234233

[obsidian harness]

yep, method is good and answer checks out

well done!

[storm zephyr]

Apparently the answer is option C

But it should be Option D right?

[silent plank]

yes

[storm zephyr]

ok thanks

how do I do this?

[buoyant bobcat]

One thing, im a private teacher (not a professional by any sorts) and my pupil asked me about an exercise her teacher proposed during class of demostrating the following statement

tg^2 x * cos x + cos^2 x = 1

this is not true, right?

or is it just me

[open vapor]

doesnt look right to me

assuminf tg is tangent

[buoyant bobcat]

im left with sin^2x/cos x + cos^2 x = 1

yeah

[slender rain]

If a parabola only has property of ax² + c instead of ax² + bx + c, what is affected to the parabola between those two equation?

[trail tendon]

ax^2 + c is guarenteed even function (symmetric about the x-axis) 👍

[slender rain]

No matter what a, c, x value I put the y will always be even?

Can I intrepreted bx as a horizontal shift to the parabola?

[trail tendon]

yes, because a(-x)^2 + c = ax^2 + c for all a, c

and yes the bx causes horizontal shifts

[slender rain]

This is weird, what I've learned is horizontal shift can be achieved by subtracting or adding the x like (x -1) or (3 + x). Couldn't find any correlation bx to the horizontal shift

[trail tendon]

if y = a(x-b)^2 + c
the b is the horizontal shift
if b=0, y = ax^2 + c and there is no horizontal shift
if you expand y = a(x-b)^2 + c you get y = ax^2 - 2abx + ab^2 + c
y = ax^2 + (-2ab)x + (ab^2 + c)
if A = a, B = -2ab, C = ab^2 + c, then
y = Ax^2 + Bx + C
anyway the point is that if b=0, B=0 no matter what so its even and there no horizontal shift n stuff yea :l

[slender rain]

Ahh so b is being derived from the vertex form, alr thanks

[neat wigeon]

such a cursed question with such a cursed answer

[sharp night]

i need someone to help me with a practice test

[elder mirage]

Pls help cuz I wasn’t in class when my teacher explained this and I have no clue

[elder mirage]

Since the angles of 72 degrees and (g+4) degrees are equal u can solve for g by: g+4= 72 and solving it algebraically and u can do the same thing for the side lengths

[sharp night]

is it 68?

@elder mirage

and h = 11

[maiden brook]

yeah

I think they should state it is a parallelogram but whatever

[elder mirage]

Yes

[sharp night]

@maiden brook

are you good at proofs

[maiden brook]

Decent

Yeah

[maiden brook]

so what is the problem

[trail tendon]

you will never know

thats how they getcha :P

[tropic sierra]

sorry in advance if this is not the right chat box
Is anyone looking for M1 (Cambridge) basic notes? Feel free to knock if interested.

[upper karma]

When you have to prove angles with converse theorems, it uses the converse theorem because you already have the points, but you have to prove if they are congruent or not right?

[vivid dove]

prove angles are what with converse theorems?
prove what is congruent or not?

[upper karma]

yes

[lyric ocean]

Guyss i hav jee today

[subtle tartan]

Hello????? How do you calculate that

[obsidian harness]

the right angle shouldn't be there lol

[obsidian harness]

I also don't understand

[cunning lion]

i think the "right angle" might be intended as kite-themed decoration

[obsidian harness]

do you mean something like this @upper karma

[flint marlin]

Easy exam , crack it