geometry-and-trigonometry

[odd otter]

no, but u can prove the congruent by saying that <ESX=<UCM

[silent plank]

i'd recommend editing those labels, content here should be safe for work

[vestal sky]

Whats been the coolest thing you guys have picked up form trig?

[silent plank]

and also these triangles can't be congruent

[odd otter]

why not?

[silent plank]

pay close attention to which sides are congruent

[vestal sky]

It seems like both marked sides have different sized angles opposite them

[upper karma]

So no one can help with this?

[outer sinew]

how would u use euler's identity to derive the half angle formulas? Or is that possible?

[tepid hatch]

Once you derive double angle formula for cos, you turn it into the half angle formula

[tepid hatch]

.

[outer sinew]

Wait... how?

[tepid hatch]

Once you get cos (2x) = 1 - 2sin^2 (x)

replace x with x/2

done

[tepid hatch]

Although im not sure if you're even able to obtain this using euler's identity, but above you said you did

[tepid hatch]

turns out this is possible

so just replace x with x/2

cos(2 (x/2)) = 1 - 2sin^2 (x/2)
cos (x) = 1 - 2sin^2 (x/2)
then rearrange

half angle formula is just double angle formula in disguise lol

[outer sinew]

ah

[covert wadi]

can someone make me formula for this please? (i am not making torpedo or any ilegal stuff if you wanna know)

@everyone

[civic wasp]

simple question but does sin(2pi x/z) < sin(2pi y/z) follow immediately if x < y < z

[grave pond]

No, consider x=1, y=2, z=4.

[civic wasp]

ah

as part of a problem i have to show the following

For $1 \leq j_1 < j_2 \leq n$, show that $e^{2\pi i \frac{j_1}{n}} \neq e^{2\pi i \frac{j_2}{n}}$

[somber coyote]

not sebbb not stμ₂dying

[grave pond]

I'd observe that e^it is injective on t in (0,2pi].
A rigorous proof could divide into cases on < or > pi (which can be distinguished by the sign of the imaginary part), and then within each of those cases the sine is monotonic.

[civic wasp]

i was gonna argue by desmos

like is it not apparent from the period of cos and sine

[hasty karma]

It is mathcrime if your school never show you this

also the best way to proof

for some reason i see strange proofs with cosine theorem (extended pythagorean) and distances

strange textbooks

[timber bridge]

geometry dash

[last quiver]

This one image taught me more trigonometry than my high school teacher did in an entire year.

[hasty karma]

[mossy ivy]

what do yall call this in english

ive been asking around for a while now but to no avail

[woven shoal]

Wdym in english?

[vestal sky]

Would anyone mind clarifying the ambiguous triangle case? I can do the problems and I’d say I would get it right 90% of the time but I don’t understand it all the way just yet

[mossy ivy]

as in is there an english name to this theorem

[sick hemlock]

Those are just some results obtained via Pythagoras Theorem, don't think you need to know a name for those

Well, Pythagoras Theorem and Similarity

[mossy ivy]

gotcha

thanks a bunch

[lucid shuttle]

Hello. Can someone help me with this exercise? It's Affine Geometry, 1st year.

[dark sparrow]

do you have any progress so far?

[lucid shuttle]

Not really

But a friend will help me. I can send the solution in a couple of hours if you want.
I wanted to solve by then

[dark sparrow]

okay so

wait

do you still want to get help from here or are you relying on your friend

if you want to get help from here: i would start by finding the intersection points of the lines mentioned

[lucid shuttle]

yes. if you can and know how to solve it, you can give me some hints or put the solution here.

Thank you!

[dark sparrow]

i won't give you the solution in full

because that isn't how this server works

however i can give you pointers, and i've already given you one.

now it's your job to follow up on it

[lucid shuttle]

i don t know how to do it

how do i solve S ∩ S′ ?

[lucid shuttle]

it gives me x = -2 and y = 2

so, these are P coordinates

is it ok now?

[dark sparrow]

okay, sorry for the delay on my part

(-2, 2) lies on both S and S'

so yes @lucid shuttle so far so good

[forest raft]

if i have 3 vertices that form a triangle, is there a simpler formula to find the area of the triangle other than this

[crisp oxide]

If the vertices are (A,B),(C,D),(E,F)
, then the area is

12det∣∣∣∣ACEBDF111∣∣∣∣

let me put that out of laTeX

note the answer is determinant form

[rotund arrow]

how to calculate all angle using directional cosine

[short echo]

Oh my god

I’ve realized

I’ve made graphs for trig of all kinds

But never have I done so in a 3D space

[obtuse patio]

Dot product

[whole night]

But this proof is not complete (alpha, beta assumed acute, and so is their sum; it relies on geometry)

For any alpha, beta, and alpha + beta < pi/2 there is this proof

This is for the addition case

[golden plume]

OHH

[steep lake]

hi how do u do this lol

[bitter siren]

i’m on the same task as you

i’m trying to graph the complex results of the arcsin function but i can’t find a 3D grapher

[upper karma]

Is anyone ok with dming me and like helping me with really basic trigonometry

Im just getting started

[timber cargo]

Sin of what is root 2 over 3

[steep lake]

thanks

[timber cargo]

Then you find sec(pi/3)

oh wait

I read that as root 2 over 2

[hasty karma]

arcsin meant

i prefier arcs to -1

[steep lake]

Oh wait its okay i got it

[hasty karma]

it is confusing for me

because 1/cosx != arccosx

dont know why it is even used

[timber cargo]

No it ain’t

[hasty karma]

[timber cargo]

Also is that supposed to be a factorial

[obtuse patio]

it means not equal

its programming terminology

[timber cargo]

Oh, that’s interesting

[obtuse patio]

you might see these too in math: & for AND, | for OR

[hasty karma]

like, sin^2 x is just (sinx)^2

but sin^-1 is not 1/sinx but arcsinx

[timber cargo]

Yeah, there should be a more clearer interpretation of the two

[hasty karma]

wikipedia says it came from english math school

and also that they havent became very popular

arcfunctions win

[hasty karma]

yeah, but i sure it is possible to make something similar for any angle

also interesting

textbook has no derivatives

oh, the whole thing isnt like that

looks like some lighter proof/explanation used

[past jolt]

Why aren't vectors and matrices generally taught at this level? The students might have sufficient geometric and algebraic knowledge and students can intuitively grasp the basic concepts through time. I believe it would vastly help future linear algebra courses.

[hasty karma]

funny thing is when you learn calculus after it becomes useful a lot

like, you first learn how things accelerate etc

without propper math base

sad people have to memmorize all formulas and suffer from not understanding why things work that way

while i can just

meow, distance travelled with given v(t)? (sounds of area under curve being calculated)

[outer junco]

Help

[timber cargo]

The inscribed triangles are similar to the circumscribed triangles

[floral shale]

Geometric mean theorem

[smoky vector]

Is this right

[near harness]

Don't you need the perimeter or side length to calculate the area?

[smoky vector]

Yes I did that

[near harness]

What is the side length then

Cuz it's not given in the question

[smoky vector]

Let me check rq

14.6

[near harness]

How did you find it?

[smoky vector]

Using the apothem and tan

To find the base of half of the pentagon’s base

Then 2x

[near harness]

Ohk.

Let me check

[smoky vector]

Alright

[near harness]

If the side length is really 14.6 then yes the area is 365units

[hybrid mural]

Anyone able to help. This is exam prep

[near harness]

I'm trying to solve so gimme a out 5-7min

[smoky vector]

I think it is

[hybrid mural]

np

I haven't found a good way to solve it after almost an hour

[near harness]

Well it is quite a difficult problem. However after producing KM, ABKC is a parallelogram cuz diagonals are bisected

[hybrid mural]

I got that

[near harness]

I got a relation between ac and ab

AC/sin(alpha) = AB/sin(beta)

= 2m/sin(angle ACK)

And angle ACK = angle ABK

We have to find AC and AB in terms of alpha, beta and m right?

[short echo]

Geogebra can do 3D graphs

Not sure how good it is tho as it’s the only one I’ve used

[bitter siren]

can i have the link in my dm?

[hasty karma]

[solemn perch]

anyone able to study trig?

[upper karma]

1357 anyone?

[torn gazelle]

Can someone help me find x

[raven parrot]

whats the sum of 2 directed angles

(OI;OM)+(OI;ON)

OI OM and ON are vectors

illustration:

[outer sinew]

Now that I think of it.... how do we define discs (ovals)? Was pondering this since definition of a circle was rather straightforward, but how would we define discs such that we can discern it from any other geometric shape?

[ruby osprey]

Hey guys. I'm entering geometry next year. Would anyone like to help me get started with learning ahead tomorrow?

[short echo]

If you are talking about ellipses there are two points called foci, where if you add their distances from any point on the ellipse together it will remain constant.

[outer sinew]

Ahhh right

Foci

Thanks

[short echo]

Why did you respond so fast

[outer sinew]

Trade secret

[short echo]

It has been 3 hours since you posted that

How long were you pondering this

It’s like me tryna sleep

My brains just like oh I wonder

[outer sinew]

Minutes, because my brain doesn't really care much about geometry and would rather ponder over analysis

But the question keeps popping back at times

[short echo]

Well, there you go

[outer sinew]

thanks

[wild zodiac]

Place midpoints D on AC, E on CB and F on BA. This way you'll be able to trigger Ceva's theorem

[torn gazelle]

X=5.6

I already found it

[wild zodiac]

Great

You too used Ceva's theorem?

[torn gazelle]

Yeah

Thank you!

[whole bane]

hi guys may i ask for help?

what is trigonometric angles?

[wild zodiac]

"Angles given by the ratios of the trigonometric functions"

You should memorise 0°, 30°, 45°, 60°, 90°, 180°, 270° and 360°

[pure vigil]

I have a great resource for memorising these values

[wild zodiac]

to be honest, one of the best ways of doing so is the unit circle

like this one

[wild zodiac]

oh that's nice

[pure vigil]

For values >90, you can use the graph and the values you have already

[wild zodiac]

actually, very nice

[wild zodiac]

the function keeps repeating after one "slide" so to call it

[last quiver]

This is how I remember it, when the point moves the trig functions values changes accordingly, and it also tells when a particular trig function is not defined for example tan(x) is not defined when x = 90° more generally, (2n-1)×90°; n is integer. I infer a lot more info than this from this particular image.

[wild zodiac]

Well, this image shows how these functions change when the point travels the unit circle accordingly

Moreover, it also includes the secans and cosecans functions

[rare elk]

Let ABC be an acute-angled triangle
with A = 60◦
. Let E, F be the feet of altitudes through B, C respectively. Prove
that CE − BF =3/2*(AC-AB)

Can anyone hel;p

havin a real hard time gettin good

at eometry just never have any decent ideas

[vast geyser]

Would someone mind helping me understand the first step of this verification?

I'm not sure how they reach this point in a single step, or more importantly how I am to know that I want to get here without a lot more experience and experimentation than can be afforded in the time i have

[sick hemlock]

You could reach that conclusion without the natural log, then simply apply ln on both sides

Nvm you changed your question

[vast geyser]

Yeah sorry, i realized i worded it incorrectly lol

its really two questions, I don't know how to get cscx-cotx to end up in that form, then the second question is even if I did, how can I know i need too?

theres a bunch of options to transform the expression, but only one leads me to this point

[sick hemlock]

Alright so you know csc²x - cot²x = 1 right

[vast geyser]

yes

[sick hemlock]

You could simplify this as (cscx + cotx) (cscx - cotx) = 1

[vast geyser]

i think i may know where you're going

[sick hemlock]

by using the a² - b² identity

Do you see why cscx + cotx = 1/(cscx - cotx) ?

[vast geyser]

yeah it's the reciprocal?

[sick hemlock]

Yeah, or you could reach to this conclusion by dividing both sides by any one of the expressions here

Sry my phone died

So that's one part of the problem dealt with

[vast geyser]

no problem

[sick hemlock]

Now, the question is asking you to prove ln(cscx - cotx) = -ln (cscx + cotx)

As you can see, the expression you need to reach includes cscx + cotx

But you're starting with cscx - cotx

What way do you know of expressing cscx - cotx as csc x + cotx?

[vast geyser]

the negative reciprocal of csc x - cot x,

so -1/csc x - cot x = csc x + cot x

[sick hemlock]

Well that's one way, or you could simply start with cscx - cotx = 1/(cscx + cotx) and proceed from there

But no, it's not a negative reciprocal of csc x - cot x

csc²x - cot²x = 1
(csc x + cot x)×(csc x - cot x) = 1
csc x - cot x = 1/(csc x + cotx)

There's no negative sign involved here

Once you get to this point, the going gets easy

[vast geyser]

yeah you're correct, I forgot that something raised to -1 is the same as it's reciprocal.

So when we get to ln[1/csc x + cot x] we can rewrite it to:

ln[csc x + cot x]^-1

which can be rewritten as:

-ln[csc x + cot x]

[sick hemlock]

There you go

[vast geyser]

I can feel my brain warming up lol No idea why this one stumped me for so long its really simple in hindsight.

I think its because this is my third week of learning trig concepts and this is the first question they've asked that included thinking about regular algebra again

[sick hemlock]

One thing to keep in mind is that if you can express a in terms of b, and the proof is asking for the result in terms of b, it's advisable to express a in terms of b

If that makes sense

[vast geyser]

im not sure it does lol

[sick hemlock]

Say, A = B + 4
and the RHS in the question/proof includes B instead of A, it's better to express A as B + 4

This is pretty obvious, but you'll see what I mean when you do similar questions like this one

You started with csc x - cot x, but the proof demanded a result in csc x + cot x, so it was better to express csc x - cot x as 1/(csc x + cot x)

[vast geyser]

yeah, I think the issue was that I just straight up didn't even think of taking the reciprocal. Also, now that I get it

I don't really understand why they chose to take such a roundabout way in the solution

They would not have been able to show the first step unless they had already showed ln(1/csc x + cot x)

then they spend several steps demonstrating a phytagrious identity to get back to 1

Anyway, thanks for the help

[stuck lark]

Join AO, BO and CO

Since arcAB = arcAC, angleAOB = angleAOC, thus AB = AC (equal cords subtend equal angles at the centre)

@upper karma

O is the centre

What's that RO in the centre

Ah, my bad, I thought it's the centre

[normal tangle]

can (1+tan)/(1-tan) be simplified as -1, or am i just dumb

[outer sinew]

It can't

What made u think so?

[normal tangle]

I don't know to be honest. I haven't really been focusing recently and i was just told by somebody else that I told them it was correct

I was pretty sure it couldn't have been simplified fruther, much less to -1, I just wasn't sure why I would tell somebody else that or if i was just misunderstood

[outer sinew]

You're right

[sick hemlock]

Simply do it your way, you reached the answer in 2 steps, that's enough

It also shows good thinking on your part

[weak vault]

hey can anyone help me estimate the function of a graph: i need to know the function of the red graph (the green one is just e^x) technically i know every point which is one this graph

[timber cargo]

$x/(x-1/2)$

[somber coyote]

Daniel S.

[timber cargo]

That’s what I think it is

Vertical asmyptote at x=.5

and a horizontal asymptote at y=1

[weak vault]

its close but not the same

i tried changing some of the parameters and still doesnt match up

i played around with trig functions in the exponent. If you have a constant times x^sin(ax) (a->∞) then you get some graph looking like a surface. This surface can when using lnx as your constant enclosed by lnx and -lnx. In the end i try to find a patern for the function enclosing this "surface"

https://www.desmos.com/calculator/w202wn0cyp

[upper karma]

Can anyone help with this?

[near harness]

Sorry man. Idk matlab. If it were some other language I would have been able to help

[upper karma]

It's probably doesn't matter if its matlab or not, its just finding angle beta shown in the code and where it is on the diagram

[near harness]

Ig. But still a bit too complicated for me.

[upper karma]

So far, heroes formula and the geometric proof

[dark sparrow]

heron's formula*

[upper karma]

Never heard of it

[fervent palm]

HEY

HELO ME LEARN COORDS

[loud pike]

Goood Morning Everyone!! 😀

[outer sinew]

it states that the area of a triangle can be given by:

where a and b and c are different sides of a triangle

[upper karma]

Is that how the area formula is derived? A= 1/2 bh

[outer sinew]

not quite, this formula was derived after that formula came to existence

[hallow ermine]

Ooo, I remember using this during math history class years ago. Used to approximate the area of a triangle? Please correct me if I'm wrong.

[outer sinew]

Not only approximate, it gives the exact area

[ebon patio]

help-22

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[vagrant ore]

For this assignment could I use desmos and put y = cot x and y = csc x and then draw on these graphs what shows on desmos?

[upper karma]

prove that the angle is the same

[trail birch]

A chord is 22 cm long. It is 60 cm from the center of the circle. What is the radius of the circle? Could I have help for this?

[timber cargo]

Well, draw it out

[trail birch]

Thank you :)

[forest wave]

Who can help me in pythagorean theorem

Its just 1 question tho

Oh thank you

What should i do if theres a square root

@everyone

[timber cargo]

a^2 +b^2 =c^2

The square of 5* root 3 is 75

[forest wave]

Ty

Then it will be 75²-60²

[timber cargo]

no

c is your hypothenuse, a and b are your other side lengths

[forest wave]

But im getting the length

[timber cargo]

I don’t understand where you’re getting 60 from

and why you’re subtracting things

[forest wave]

Oh my bad

But the answer will be 45 right?

[worn skiff]

have you learnt the special triangles

[forest wave]

Not yet

Im at 9th grade

[outer sinew]

Special triangles?

[worn skiff]

like the 30-60-90 triangles, right isoceles, etc

[worn skiff]

ohhh ok

[forest wave]

Oh i forgot we learned special triangles

[outer sinew]

Ah

[forest wave]

Sorry bout that folks

[worn skiff]

np

[worn skiff]

You can connect one of the chord ends to the centre

and use pythagorean

[covert estuary]

hi, i am new to the trigonometrie and i want to learn more about that stuff so can one of you recommend a free online book for this purpose

[upper karma]

how do i start trigonometrie

[sick hemlock]

how do i trigonometrie

[timber cargo]

How do I spelly trigonmetrieei collectrty

[gilded cipher]

why use this formula instead of A = 1/2 bh

nvm figured it out

[carmine plaza]

Guys I have a question but I don't think it's worth taking a math help for

Photomath said u can combine two solutions into one

Does anyone know how I do this

Or am I not supposed to ask this in here

I mean it is geometry

[floral shale]

Well, they're within the same set like

They'll just overlap

It's more by analysis than raw algebra

[carmine plaza]

Yeah i just kinda dont understand how it got to that final x=

I mean the way i see it if i count both those answers up and divide them by their most common factor (4) i can get to the answer down there

[vagrant ore]

I got this assignment and I know how to find the other trig functions but can someone explain to me if/why the quadrants they’re in matter?

[sick hemlock]

because the quadrants determine the sign of trig functions, in simple terms

We were taught it as the CAST rule, which means certain trigonometric functions are positive/negative in certain quadrants

[unique mirage]

I have some clue on how to find far south and far east for group a but cannot find out how to solve for group b

[vagrant ore]

oh ok if I find the other trig functions they would be negative or positive depending on the quadrant?

[sick hemlock]

yes

[vagrant ore]

ok thank you!!

[potent sedge]

in this simplification, where does the3sqrt2^2 and -sqrt2^2 come from on the bottom?

fricking rotating conics man

[hybrid mural]

Let ABC be a triangle with incenter O. Given that the area of triangle AOC is 2, the area of triangle COB is 3, and the area of triangle AOB 4, find AB, BC, and AC.

Anyone have an idea?

[upper karma]

/imagine Pell's equation

[glass obsidian]

Guys ı Need fast help

How many times √15 is the perimeter of a triangle with heights 15, 20, and 30

[uneven ermine]

Any idea to solve it?

[dark sparrow]

this is not an AI art server.

[lost dove]

so the grid.. is it detailed...i mean is it just the blocks or does it have more division inside the cube?

[uneven ermine]

As u can see it cuts the squares and generate different lengths

[lost dove]

can u send a pic of just the diagram?

@uneven ermine ?

[dusk edge]

Can someone help with this Geometry problem? It is under the section Other angles where 2 chords are 1/2 the sum of the intercepted arcs

[tulip wasp]

I have an intriguing problem that I wonder if y'all would know how to solve.
I want to place 2 circles inside of a triangle using only a compass and straightedge such that they both fall on the midline of the triangle, one touches 2 sides of the triangle, the other touches the remaining side, and both circles are the same size. I attached an image in case my description doesn't make sense

[wise pawn]

is the triangle equilateral or just isosceles

[tulip wasp]

Equilateral

[wise pawn]

that makes it easier at least, idk I'm starting by writing some equations for things just to find points of intersections since ultimately we just need to construct the centers of these circles and we can easily construct the midline

x^2+(y-a)^2=a^2 and x^2+(y-3a)^2=a^2 for instance, make a line to represent a triangle with unit side lengths to find the tangent point etc

there may be some super elegant solution that doesn't require grinding through this grunt work lol

but I think it's probably best to just roll up our sleeves and get it over with

[tulip wasp]

I did some experimenting in geogebra and apparently if I divide the midline into 5ths, the first 2 segments make up one circle while the next 2 segments make up the second and the 5th segment is outside the circles

this is possibly the oddest pattern I've ever seen and I have no idea how to prove it

[floral shale]

Is this true for all triangles?

[tulip wasp]

I have no idea. not likley, probably just equilateral

[upper karma]

need helppppp

<@&286206848099549185> PLSSS

[crisp sand]

use one of the available help channels

[worn arch]

!help

[lime crown]

Please read #❓how-to-get-help

[radiant wren]

yes
because if you draw a line here

the centre of that circle is now the what's it called

the

CENTROID

yes

and that's 2/3 the way from one vertex to the midpoint of the opposing side

so it's 2 + 1 + 2 from the other circle

= 5

[tulip wasp]

isnt the centroid where all 3 midlines intersect?

[radiant wren]

yes

the 2/3 thing is a theorem

[tulip wasp]

that, annoyingly, is not the centroid

[radiant wren]

it is if it's an equilateral triangle

[tulip wasp]

its not

if it was this problem wouldve been easy to solve

[radiant wren]

no
C1 is the centroid of the partial triangle I indicated with my line

[tulip wasp]

oh wait I see

intriguing

could this be used to draw this thing on paper?

[radiant wren]

you can divide the midline into 5ths I t h i n k ?

[tulip wasp]

I'm pretty sure I could, but that is one heck of a grueling process

[radiant wren]

not really

[tulip wasp]

its not 👀

[radiant wren]

first step

(CB is the midline)

so as you can see I constructed a segment parallel to CB

I will now be using that

now use that segment to make 4 more equal copies of itself on the line

[tulip wasp]

ohhh I see why my way is so difficult. I started with the triangle and put the circles inside, requiring me to use thames theorem to break the midline into 5ths

your way makes much more sense

[radiant wren]

and finally this

so as you can see it's clearly not a hassle

in any way shape or form

/sarcasm

I am not good at geometry but construction 💯

[tulip wasp]

All of that for one alchemy circle

[remote pebble]

:0

[faint agate]

Can anyone help me make sense of these? Im unable to get the conditions imposed on x and y. Is there a shortcut ?

[elfin sandal]

Can someone help me understand this?

[upper karma]

$tan^{-1}(\frac{4}{3})$

[somber coyote]

Bandera

[grave pond]

Looks like you did 4/5 (which would give sin V) instead of 4/3

[elfin sandal]

Alright thank you 🙏

[civic bay]

i don't know how

[upper karma]

I need help

[obtuse patio]

!show

[lime crown]

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

[obtuse patio]

@upper karma!help

[upper karma]

yes?

i need help tho

[obtuse patio]

!help

[lime crown]

Please read #❓how-to-get-help

[upper karma]

I NEED HELP

[obtuse patio]

Did you read #❓how-to-get-help

[upper karma]

no

[obtuse patio]

then do that

@upper karma

[upper karma]

k

[bitter nymph]

@faint agate looks like a bunch of u-substitution, properties of even and odd functions and piecewise functions

[frozen ocean]

there's a video on the right corner

[shut venture]

hey where can i ask for a hand with some geometry in space?

3d geometry

[echo karma]

im completely lost

its the last questioon of my hqw

pls help

[austere meadow]

hi, I have this one problem from my homework that I don't understand. Even if I check the right answer, I don't understand where I failed.
" 2tan2(x-π/4) ≤ 2√3 "

[dark sparrow]

@echo karma do you know how to find the area of a parallelogram?

[bitter nymph]

A parallelogram is just a skewed rectangle

[lucid trellis]

i have a question what year level or grade level you start geometry?

[rotund grove]

7th grade

[dark sparrow]

probably depends on where you're from

[chrome furnace]

For me it’s 10 grade bruh

But yea it depends

[lucid trellis]

rlly i am doing geometry at year 5 bruh

[wintry wigeon]

<@&268886789983436800>

[faint agate]

Yea I think I kinda got it yesterday one guy spent a lot of time explaning it to me, Im gonna solve questions related to it today

[lime pond]

1. What are the given?
2. What condition/s is appropriate to solve the problem?
   3 why are you using that condition? Explain and justify.
3. Explain your solution.
4. Give your conclusion.
5. What type of person os Alice? What values does she have?
6. Do you see yourself as the reflection of alice?

Can somebody help me with this to how to explain please

[pure vigil]

<@&268886789983436800> year 5 is ages 9 to 10, therefore this person is under 13

[dark sparrow]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dark sparrow]

@meager sundial

and yet.

i saw that.

i can put in a word for you.

<@&268886789983436800> ?

[calm hazel]

trigonometry

to find AC, VX

[reef flume]

@calm hazel try using the Pythagoras theorem to find the following

or try using the proportionality equivalence

[upper karma]

@calm hazel Use the Pythagorean Theorem to find AC and VX. Then find the scale factor, using AC to DE, to find the remaining sides.

[violet kiln]

Use similarity rules

[violet kiln]

Any 2 SIMILAR shapes have sides that change with the SAME RATIO

[wintry wigeon]

I randomly came up with a construction problem which actually felt pretty inetersting.

Given a point A and two parallel lines b and c, such that all three do not coincide, construct an equilateral triangle ABC such that B is on line b and C is on line c

[runic pawn]

how did you construct it?

like

with euclids axioms

[wintry wigeon]

yeah just regular compass-straightedge rules

[runic pawn]

that follows euclids axioms

[wintry wigeon]

But I took shortcuts using geogebra

[wintry wigeon]

yes

[runic pawn]

im assuming you used circles?

[wintry wigeon]

yes

[wintry wigeon]

[runic pawn]

i just wanted to see the constructions you used

is what i meant

[wintry wigeon]

Oh

I'll have to do it again

I wrote it down somewhere, hold on

[runic pawn]

bet

[wintry wigeon]

I take shortcuts for orthogonal lines and copying lengths

Just to reduce tedium

[runic pawn]

ill accept that

wait, copy a length?

[wintry wigeon]

This option

[runic pawn]

oh i see, could you do it without that?

[wintry wigeon]

yeah

[runic pawn]

bet

[wintry wigeon]

I forget the rule, but it can be done

[runic pawn]

you can chain circles

[wintry wigeon]

yeah it's been a while

[runic pawn]

but they have to be in specific spots and there are positions you cant necessarily reach all the time

@ me when you send it

[wintry wigeon]

@runic pawn

[runic pawn]

thanks

[runic pawn]

very interesting

[wintry wigeon]

[indigo shell]

why did Pythagoras come and make up crap like this

[wintry wigeon]

I found a much faster and easier construction

[runic pawn]

lets see it

[wintry wigeon]

Initial problem

hmm I may have made a mistake somewhere

I marked the wrong side of the blue circle

One of the purple triangles will be equilateral. Which one exactly depends on initial placements

[wintry wigeon]

The solution generalizes nicely to non-parallel lines

[frozen ocean]

thats pretty cool

[runic pawn]

It is

[hidden cobalt]

does this look more like a triangle or a rectangle?

@everyone

[toxic crow]

I would say find the area of the closest rectangle and triangle, then compare it to the actual area

[dark sparrow]

you realize everyone pings are turned off on this 100k+ strong server for a reason right

[toxic crow]

Would have to say triangle

[nocturne remnant]

Yeah this isn’t really a math question

[silver prawn]

hello

[near harness]

Hi

[modern egret]

okay so i missed a day of class and they had gone over a completely new subject

can anybody explain to me how to do this?

i really only need an explanation for the first one—i think i could figure out the others

(please ping me, too! I'm working on other things as well)

[wintry wigeon]

To figure out the first one, create an equilateral triangle of side length 1, then draw a bisector through one of its sides.

[modern egret]

thank you!!

I didn't realize that these also had to do with the unit circle😭 that's my bad

[upper karma]

Can someone help me with the rest I already did the first steps

[pastel vector]

@upper karma Yea bro I gotcha basically u just gotta plug in the y and x coordinates into ur new equation. Like for the first one u gotta do y = -(1/2)x + b, so it would be 3 = (-1/2)(-5) +b. Then solve for b. Tada new equation. Then u graph both

[faint sun]

The perpendicular slope would be the negative reciprocal meaning what do you need to multiply m by to get -1
Then find the perpendicular slope and use the point-slope formula: y - y1 = m(x - x1)
For the first question you found the perpendicular slope: -1/2
Use the perpendicular slope in the point slope formula and use the point (-5, 3) in the point-slope formula
It would be y - 3 = -1/2 (x + 5)
Simplify from there

[manic mango]

Hi everyone! Kind of a multipost but... does anyone have any good resources about circles? A resource that is from high school geometry or aligned with the AMC8 is really what I am looking for.

I just want to make this topic about circles really intuitive for me since I am learning it right now and I am taking the AMC8!! Thanks in advance ❤️

[upper karma]

Ty all

[thin locust]

idk

isnt amc 8 just stuff which is covered in hs

[manic mango]

i mean for the most part in my experience but there are some things u have to learn on ur own that isnt taught in hs

[thin locust]

yea idk how bad us schools are

if they are worse than i thought then u prolly have to learn some stuff by urseld

[manic mango]

lmfao yea true

[outer sinew]

Post in chill not here smh

[upper karma]

its geometry

[outer sinew]

this is not the place to troll

[zinc belfry]

Would anyone mind giving me the equation on how to find the central angle?

[near harness]

Area of sector = πr^2 x ∅/360°

4π = πr^2 • x/360

If you know the radius you can use this equation

Radius is 6inches?

[dark sparrow]

don't use the empty set symbol as a substitute for theta.

[near harness]

Oh right.

My bad

I'm on my phone and that somewhat resembled theta

I didn't realise I was using the wrong symbol

Accidentally used phi 😂

[dark sparrow]

this isn't phi either

[sick hemlock]

I'm pretty sure the empty set symbol is phi (pronounced fai)

[grave pond]

You're wrong then.

[sick hemlock]

You know, you could correct me rather than simply telling me I'm wrong

we were taught it as phi in school

[grave pond]

The empty set symbol was introduced in Bourbaki's set theory volume, and the responsible author, André Weil, is on record saying it is taken from letter Ø used in Norwegian (and Danish).

Ø is a vowel completely unrelated to the Greek consonant phi.

[sick hemlock]

well, we're being taught it as phi in school so ig the schools don't really care

[grave pond]

What can I say other than your school teaches you falsehoods?

https://jeff560.tripod.com/set.html gives the relevant quote from Weil.

[sick hemlock]

fair

[grave pond]

Actually the typographical shape of the empty-set symbol is closer to what is used to represent "diameter" in engineering than it is to an actual letter Ø, but the origin appears to be the Ø.

[near harness]

Seems like phi

[grave pond]

Neither of those look like an empty-set symbol.

[grave pond]

Neither of those look like an empty-set symbol.

[near harness]

What do you say about this? @grave pond

[grave pond]

$\varnothing$ is not $\phi$ or $\varphi$ or $\Phi$, or even {\O}, though it it historically derives from the latter.

[somber coyote]

Troposphere

[grave pond]

I say that the last sentence of that is plainly wrong.

[near harness]

Bro

Nvm

[outer sinew]

💀

yeah

[upper karma]

Cube ABCD.A'B'C'D'. Vecto AB equals vecto a, vecto AD equals vecto b, vecto AA' equals vecto c. Demonstrate vecto AC, vecto B'D', vecto BD', vecto DB', vecto BC' and vecto AD' according to vecto a, vecto b and vecto c.

[humble shell]

Hi there, would anyone perhaps know how to find the area of such a complex shape in terms of r?

[fallen ridge]

null set , no solutions

[dark sparrow]

thanks for mansplaining.

[silent plank]

if you want a cheap substitute for theta, just use the letter t assuming that isn't already used elsewhere

[near harness]

ohk. thx

[thin locust]

hey ash

[mental sapphire]

helpo

[vagrant ore]

Quick question, for this I tried doing cos = 21/29 and I got 0.72 but it said it was wrong. Is there something I missed?

[dark sparrow]

you are not asked for theta itself but just for cos(θ)

you have correctly identified that cos(θ) = 21/29

wait no nevermind

i thought you had taken the arccos or something but thats 0.76

,calc 21/29

[somber coyote]

Result:

0.72413793103448

[dark sparrow]

... this does round to 0.72. the fuck?

@vagrant ore do you have limited attempts on this thing

[vagrant ore]

No not limited but if I get it wrong it docks points

[obtuse patio]

no

#rules

<@&268886789983436800>

[dark sparrow]

maybe it didn't like the format you entered it in? maybe it wanted one of .72 or 0.72 but not the other, or you left a space at the end of it or something.

do you get an error message of any kind

[vagrant ore]

Hm possibly I’ll try it.

No error message

The question before it was to find the sin of this same thing and I got 0.76 for that but that said it was wrong too

So I just skipped that question

[vagrant ore]

It must have just been the format because I logged back in and put in 0.72 and it went through

[upper karma]

what is the area of ​​the circle

[young ermine]

Science rules

[wild rivet]

Can anybody tell me how to solve inverse trig functions(sin inverse, tan inverse, cos inverse) without a calculator but by hand?

[silent plank]

depends

[summer sapphire]

What is the formula for the distance of a point from a line when the information provided is the equation of the line in symmetric form and the coordinates of the point?

[novel cove]

--> Shortest distance between a line and a point is their perpendicular distance (i.e. the line you draw passing through the point and intersecting the original line is perpendicular to the equation)

So:

1. Find the slope of the equation from the symmetric form of the equation, and thus the slope of the intersecting line (their slopes multiplied = -1)
   --> So you have the incomplete symmetric form of the intersecting line (i.e. y=ax+c, where a is known)
2. Substitute the coordinates of the point into the incomplete symmetric form, and you obtain c
3. Do simultaneous equations of the equation obtained with the original equation, and you get the perpendicular distance between the line and the point/required answer
4. Distance formula

'd be something like this if my working is correct. But you get the general method

[low vector]

is it 25

[grave pond]

Before electronic calculators, you looked up the values in printed tables. Computing those tables by hand in the first place would have taken years or sometimes decades of work for someone

[grave pond]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[upper karma]

Please I nead help with geometry problem

I found all angles except EDC and ECD and I need ECD

[magic token]

need help

[frozen ocean]

I got a question

how did you find 70? in E

[upper karma]

It was on the problem

I find the rest of the angles

[upper karma]

except DCE and EDC

[frozen ocean]

okay so I think if you subtract 180 - 70 you'll get 110 for angle C

[frozen ocean]

also where did 50 come from

[upper karma]

All angles on the photo are from the problem

[upper karma]

What?

[frozen ocean]

no wait it should be 45 degrees for C

[frozen ocean]

there's a lot of parts for angles that i remember though

for angle c should be 45 and then after that 180 - 70 - 45 = 65

for angle D

[upper karma]

So it is not a isosceles?

[frozen ocean]

how do you know it's an isoceles?

as for me if it's an isosceles then the angle c should be 30 degrees

[lapis jackal]

Uh hi I have question

How would u solve these

@frozen ocean

[frozen ocean]

basically I have no idea about this because in canada our teacher never taught us like these ones

you could ask help in help channels

#1021175428326633542

[frozen ocean]

@upper karma does this works or still confusing

[lapis jackal]

Ok thx

[frozen ocean]

np and welcome here

[upper karma]

Yeah it ist

[frozen ocean]

oh

then what supposed to be?

a lot of math teachers in Canada told me if you always see an isoceles angle, it could be one of these degrees: 30, 35, and 45, which sometimes confuse people

[upper karma]

I found it

[upper karma]

sorry, I misstouched the the "t"

It was an inscriptable figure

[frozen ocean]

okay

[nocturne remnant]

Hint: ||try to prove that MA=MB=MC=MD||

[hollow moth]

#help-34

[minor mountain]

Could someone please recommend me a really good geometry book(Euclidean geometry)

[wild rivet]

Oh, I know how to do all that with a calculator but thought it would be possible by hand, like there would be some formula

[wild rivet]

yk asians

[grave pond]

Excuse me what?

[outer sinew]

Asian stereotypes:

[dark sparrow]

????

[wild rivet]

What I mean is that Asians aren't allowed to use a calculator. An example would be if someone is from India, for the general CBSE curriculum, you can't use calculators at all.

[sick hemlock]

I doubt you're allowed to use calculators in higher level exams in the US either

[wild rivet]

We're not allowed in class, or exams, or anywhere, it's prohibited

[wild rivet]

What exactly do you mean by higher level?

[summer jasper]

wassup anybody here?

[candid schooner]

i am

i havent seen some of these topics yet

sorry

[summer jasper]

np

[thin locust]

has nothing to do with people being asian

its just practice

like multiplying 2 3-digit numbers

or if youre having trouble with that, start with 1-digit then 2-digit

or calcukate powers

like

try to get to 2^20

or 2^25

in your head

or 3^10

oh trig values

idk

you can derive 0, 15, 30, 45, 60 ... degrees

[upper karma]

Bro what 💀

Some asians are stupid to no offense to Asian people, but race doesn’t determine who’s good at what

I’ve seen for god sakes black people getting a 99 in math again, no offense to black people, but if you work hard you’ll be fine

[wild rivet]

ik. You misunderstood what I meant, I didn't mean to say that, I meant many people outside of Asia believe in this stereotype. As an Asian and a human, I don't believe in this stereotype

[wild rivet]

You're right

Actually, It's not related to genetics. I mean it might be influenced by that but by a very minute amount. If you practice a lot, have an interest and are ready to do whatever you can, you can become really good, like really good

I don't believe any of these things to be clear but ik that many people do. It's actually nice to know there're people around here who don't believe in this

Anyways, this is a trig and geometry channel

[tawdry blaze]

I have the solution to this and I have solved for all the angles. Do you need a photo of the working?

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[upper karma]

Hey guys i need help with a trigonometry problem so:

In a right-angled triangle ABG with A=90, BG=10cm and hB=0.8 are given. Find: a) The length of side AG
b) The length of side AB.
c) The other trigonometric numbers of angle B.
d) The area of ​​triangle ABG.
e) The height of AD corresponding to the hypotenuse BG.

[wild rivet]

What does A=90 and hB=0.8 mean?

I'm new to trig so I'm going to try

?

[nocturne remnant]

A also stands for angle at A (angle BAG)

[wild rivet]

oh

[upper karma]

A is the corner of the triangle and hB is a trigonometrical number

[upper karma]

yeah

that

[silent plank]

hB is a trigonometrical number
for what

[nocturne remnant]

Idk what is hB tho lol

[wild rivet]

yeah

[upper karma]

sine

we write it like that in our country

ημΒ

[wild rivet]

Oh

[silent plank]

so sin(B) = 0.8?

[upper karma]

yep

[silent plank]

what have you tried

[upper karma]

nothing cuz its complicating for me 😅 im not good with trigonometry i dont really understand same with Pythagorean theorem

[wild rivet]

B is above point A, right?

[silent plank]

did you draw a diagram

[nocturne remnant]

Have you tried drawing a diagram

[upper karma]

there isnt any diagram in the photocopy

[silent plank]

hence, did "you" draw

[wild rivet]

because then, AG would be 8

[upper karma]

no :/

i didnt

[wild rivet]

which would just form a normal right angled triangle pair, 6,8,and 10

[upper karma]

im so bad at maths

[wild rivet]

height of ad?

[upper karma]

im sorry for the troubles

[upper karma]

we dont know

we have to find it out

thats the thing

[wild rivet]

I'm in grade 7

[upper karma]

there isnt any diagram

[upper karma]

im in grade 8

lol

[wild rivet]

No, I mean the triangle is ABG

[upper karma]

yeah

I think there is another triangle

[nocturne remnant]

That’s why you have to draw it !!!

[upper karma]

But I dont know ! 😦

[silent plank]

have you even drawn a right triangle before

[wild rivet]

doesn't trig start in grade 9 or 10?

[upper karma]

Im only good with equations

[upper karma]

every country has a different curriculum

[wild rivet]

ik

[upper karma]

yes

[silent plank]

can you start by doing that

[wild rivet]

Is it possible for you to send the question, then the community can help

[upper karma]

imma use 3d drawing

[silent plank]

why 3d

this is a 2d problem

[wild rivet]

Yeah

Pls don't tell us you're doing 3d trig

Still that can be done on 2d paper, my apologies

[upper karma]

a right triangle

[silent plank]

label the point at the right angle with A

[wild rivet]

that's just like any other right triangle, making the right triangle and diagram is a skill in itself

[silent plank]

no

[wild rivet]

No

Replace d with a

[silent plank]

do you know what a right angle is

[wild rivet]

and d shouldn't be there

[wild rivet]

we're working with triangle abg

not adg

[upper karma]

its a type of triangle with an angle that has a measure of 90

o

[wild rivet]

Can you share a pic of the question once?

[upper karma]

its in my home languange

greek

[wild rivet]

It's not possible they suddenly mention a random point d out of nowhere

[wild rivet]

oh

[upper karma]

i used google translate

[wild rivet]

google translate isn't that accurate sometimes

[upper karma]

the only time that the problem mentions D is at e

and there isnt any diagram

[silent plank]

label the other two points of your triangle with B,G

[wild rivet]

must be a typo

yeah

[silent plank]

not a typo

[wild rivet]

top one should be b ideally

[silent plank]

D is the point on BG where AD is perpendicular to BG, worry about that later

[wild rivet]

why not?

oh

[upper karma]

thats all we know

[wild rivet]

right

my bad

this is very easy

ag is 8

[silent plank]

recall what sides sine gives the ratio of

[upper karma]

opposite vertical side and hypotenuse

[silent plank]

what's with the vertical

just opposite

[upper karma]

thats how i learnt it in school

[wild rivet]

because opposite(AG)/hypotenuse(BG which is 10) is sin

We can derive that 8/10 is 0.8

[silent plank]

not ideal to use vertical as then you might get mixed up with the line that looks vertical here

[wild rivet]

yeah

[silent plank]

which isn't the side opposite to B

[upper karma]

oh

[silent plank]

the positions are relative to the angle you're using

and how the image is rotated/oriented should make zero impact on your ability to identify them

[wild rivet]

Also, you can use sohcahtoa to remember these ratios

[upper karma]

I will start a tutorial class on Maths this summer i just need to pass the exams

:/

How to find the other trigonometric numbers sine cosine tangent?

actually thats easy

Assuming that B is the right angle, and G is on the hypotenuse, then we can use the sine rule to find the missing sides. We know that sin(90°)=root(1), and sin(90°-x°)=1/2 (x being the angle AG), so the length of AG is AG = 10/2 = 5 right?

I think i found all the answers

idk imma ask the teacher

thx guys

ig

[wild rivet]

that's wrong

B is not the right angle

Sorry for the delay, had to go somewhere

[upper karma]

its fine

its just an assumption

since there isnt any diagram

[wild rivet]

you don't need the sine rule for now

[wild rivet]

Basically, yes, this is fine

so in this case

angle BAG = 90

degrees

now you know that sin is opposite/hypotenuse

right?

@upper karma

[upper karma]

yes

[wild rivet]

so we know hypotenuse but we don't know opposite

so far so good?

[upper karma]

yes

[wild rivet]

Now, the hypotenuse is 10, meaning the denominator is 10. now think, what should the numerator be

x/10=0.8
transpose and you get x=8

because 0.8*10=8

So far so good?

[upper karma]

Yeah

I get it

[wild rivet]

x being the opposite side to theta which is AG

so 8 is the length of AG

[upper karma]

why isnt it 5 tho

[wild rivet]

Now you can just use pythagoras even though it's a pair we all know for right angled triangles

[wild rivet]

why should it be?

[upper karma]

i mean if we use the sin rule as i said

[wild rivet]

You don't need the sin rule, and I don't know it yet, you must be making some mistake

[wild rivet]

unless it's written in the question solve with the help of the sin rule

[upper karma]

no its not

[wild rivet]

so, you don't need it

for these kind of simple basic questions, you don't need it

So you found out the length of AG

which is 8 cm

now they're asking you to find the length of ab

[upper karma]

then we need the length of AB

[wild rivet]

So, you can just use pythagoras now

Tell me what answer you get

Found an answer?

10 squared = 100. 8 squared = 64, now 100-64 = ?

[upper karma]

36

sorry for the delay

[wild rivet]

what do you get?

[wild rivet]

ok

[wild rivet]

What do you get?

[upper karma]

36?

right

[wild rivet]

No, like what side length do you get after using the pythagoras theorem

[upper karma]

AB

[wild rivet]

Not the answer to 100-64, the final answer, like the length of ab

[wild rivet]

yeah

[upper karma]

im confused

lol

[wild rivet]

Go back, what's pythagoras theorem?

just tell the equation

no need to explain

[upper karma]

a2+b2=c2

i cant make 2 smaller

[wild rivet]

now plug in the values

so x2+8 squared = 10 squared

[upper karma]

yes

[wild rivet]

which can be said as 10 squared - 8 squared = x squared

so far so good?

[upper karma]

then

100-64=

10x10= 100

8x8= 64

[wild rivet]

yes, you're getting the hang of it

[wild rivet]

solve it

go ahead

you're right

[upper karma]

36

[wild rivet]

so we know that x squared = ?

[upper karma]

36 squared?

[wild rivet]

no

think and type

[upper karma]

do i need to type the equation?

[wild rivet]

no

just answer my question

[wild rivet]

this

[upper karma]

i need to square 36?

[wild rivet]

no, but you're close

if you don't get it in the next try, then I'll tell it

hint

is

that we're working with squares

not square roots

[upper karma]

bru this is so simple and i literally cant find it

i dont understand your question

[wild rivet]

x squared = 36

[upper karma]

36ab?

[upper karma]

wait didnt i say that

[wild rivet]

You said x squared = 36 squared

[wild rivet]

nooooooooooooooooooooooooooooooo

[upper karma]

.

u asked me to solve it

[wild rivet]

That was the answer to 100-64 which is 36

[upper karma]

and i did

[wild rivet]

but then

I asked you

because 100-64=x squared

what is x squared then

[upper karma]

u confused me with your question

[wild rivet]

you said 36 squared

[upper karma]

yeah yeah

[wild rivet]

but it's 36

did you understand it now?

[upper karma]

Yes

[wild rivet]

another example now for Pythagoras theorem to test your knowledge

[upper karma]

I know Pythagorean theorem u just confused me

[upper karma]

sure

[wild rivet]

if hypotenuse is 5 and one side is 4, what's the other side

with explanation

[upper karma]

The other side is square root ( 5^2 - 4^2 ) = square root of 21, which is approximately 3.45

[wild rivet]

no

no

nooooooooooooo

everything is wrong after 5^2-4^2

the square root of 21 is not even related

and 3.45 is not the square root of 21

you clearly need help with pythagoras

tell me what 5^2-4^2 is

@upper karma

[sick hemlock]

5² - 4²

just for clarity

[wild rivet]

yeah

@upper karma

[wild rivet]

tell

[sick hemlock]

what are they facing trouble with? Pythagoras Theorem in general or subtracting squares?

[wild rivet]

First, he asked help for a trig question

He understood till where I explained AG=8

then he had to find AB, which was easy given you have the hypotenuse(10cm) and AG(8cm)

you just need pythagoras

[sick hemlock]

I'd suggest drawing a figure for clarity

[wild rivet]

He has

[wild rivet]

this

[sick hemlock]

Right right

[upper karma]

wdym what is this

[wild rivet]

like solve it

[sick hemlock]

5² - 4² = ?

[upper karma]

9

5 squared = 25

[wild rivet]

Yeah