geometry-and-trigonometry

[chrome kelp]

Is all trig covered in Precalc?

[vernal pilot]

No.

[median niche]

26.3?

[formal moth]

there is no way that that is a square

[median niche]

ah then its 26

[median niche]

yeah I think so aswell

now that i think ab it

I'm getting 18

[pliant sky]

Can anyone do part (ii)?

And tell me what ans they’re getting

I wanna verify my ans

Nvm it’s correct(im 99.9% sure)

[woeful pike]

well what sur answer

i think its 1/6 but im probably wrong

[cerulean timber]

why is geometry congruence so hard

[maiden brook]

wdym

[stuck sleet]

Can someone draw this:

A spherical ball of radius R is dropped into a vessel in the form of an
inverted right circular cone. Find the radius and altitude of the cone, if when
three more balls each of radius R are dropped into it they form a layer on top
of the first ball such that all four balls are tangent to each other and in addition
each ball of the upper layer is tangent both to the side and to the top of the
vessel.

Thanks!

[cerulean timber]

Why is geometry congruence so hard is what I mean

[vernal pilot]

Congruency is very easy. It’s just when two shapes are the same.

[cerulean timber]

yes congruency itself is easy but like

proofs

rn im trying to do prove parallelogram properties

that's what I find the hard part about it

[vernal pilot]

That’s pretty easy.

[pliant sky]

Also 1/6

I’m now 99.9% sure it’s correct

[woeful pike]

Idk how you did it but the way I did it was by just using a square. It gave me 2 nice right angled triangles. It works cus a square is technically a parallelogram and if it’s meant to remain constant it should be the same for a square.

[pliant sky]

Yup

Thanks

[twin crag]

isnt the answer for this going to be (sqrt (2), pi/4)

it marks it as wrong

[onyx cloud]

what is cos(pi/4)

[twin crag]

sqrt2 / 2

[onyx cloud]

what is sqrt(2) * sqrt(2)/2

[twin crag]

1

[onyx cloud]

what were you supposed to get

[twin crag]

-1-i?

[onyx cloud]

well yes, but this should have given us -1 to match the original real part

so evidently (sqrt2, pi/4) is wrong

[twin crag]

wait can i tell you how i got it

[onyx cloud]

sure

[twin crag]

alr

[twin crag]

wait i think i did something wrong in my steps

what is the complex conjugate of (-1-i)

is it (1+i)

[onyx cloud]

no

you only flip the imaginary part

[twin crag]

ok so i did that part right

hmm

let me show my steps

okay so to find theta, you can do arctan (1/1)

which is how i got theta = pi/4

and next, to find r, you can find the magnitude of -1-i

which is equal to sqrt( (-1-i)(-1+i)) =sqrt(2)

Now we have (sqrt2, pi/4)

[onyx cloud]

arctan only gives you an angle between -pi/2 and pi/2

you need to adjust it to the correct quadrant

which quadrant is pi/4 in

[twin crag]

1st

[onyx cloud]

how about the point (-1,-1)

[twin crag]

3

[onyx cloud]

so you need to adjust

[twin crag]

do you just do pi - pi/4

[onyx cloud]

add pi

[twin crag]

so its 5pi/4?

[onyx cloud]

test it out

[twin crag]

yep its correct since it equates to -1

ty

[potent igloo]

helo, im having some trouble with geometry and idk if anyone shares my experiences, so I just wated to ask if someone does. Basically whenever I hear a geometric construction or so, I have absolutely no intuition for it, with algebra or so, I can see a series or an equation and my brain immediately goes "ok, I think I should square this, and... no actually, let me write that as an integral, and then integrate the sum, I think that will help", like an intuition for what to do. Whereas with geometry, solving problems just feels like drawing the problem, and then just staring at it for 4h, until you notice the one thing, and go "oh, yeah missed that, problem solved i guess", like it isnt a process and its just frustrating, (ironic as geometry is supposed to be so intuitive). And then there are the endless theorems and formulas to memorise, and bc I have no intuition, its just mind numbing busy work.
Ive been looking at byrnes elements, and its helping a bit, but still

How do I develop an intuition for geometry, I think that would help a lot, and if you do have good intuition, how long did it take to develop?

[onyx cloud]

what is geometry here

what are you referring to

[potent igloo]

Like this stuff:

not the "triangle has these side lengths, calculate....." type

[onyx cloud]

i can’t speak to how to get intuition for this too much as i don’t have it

but why do you feel the need to gain this intuition

[potent igloo]

Because solving these problems without intuition is just applying theorems for 4h until you just happen to find the correct one, I have no guiding light as to what I should do

[onyx cloud]

but why do you want to solve them in the first place

[potent igloo]

Because I need to know geometry for math competitions

[onyx cloud]

those questions are inherently trick based

so the best you can do is just expose yourself to tons of problems and practice

i’m not sure if there’s intuition to get rather than pattern recognition for problem types/theorem applications

or at least not a substantial amount

[potent igloo]

Well, Im much more familiar with real analysis and algebra and that stuff, and there I do have an intuition for what I should do, what theorems will be important, what structure the proof will have, and without that, it would be incredibly time consuming and frustrating to do RA/Alg

[potent igloo]

I mean, being able to imagine the construction in your head, having a feel for what quadrilateral might be cyclic, maybe you should extend this line with a length of x, and that allows you to construct a triangle that is helpful, I am sure that some people have that, and it would make proofs a lot more enjoyable and easy

[onyx cloud]

a large part of that is seeing more examples and constructions

[potent igloo]

yeah ig thats always the answer

it just sucks and is really testing my determination to learn this stuff

but anyway, thanks for taking the time

[twin crag]

for de moivres theorem to work, should there be no "r" in front of (cos theta + isin theta)^n

or can i just use de moivres theorem, then multiply the value by "r" afterwards

because i got -sqrt(3) for this

[onyx cloud]

-sqrt3 for r or for the real part of the whole thing

[vernal pilot]

Competition math is dumb

Anyways, if you want to start getting good at then you should create a process you can follow to solve them.

[twin crag]

real part

[vernal pilot]

For instance step 1 might be lost everything you know about the proof

Step 2 might be seeing what you can easily prove

Ect

[vernal pilot]

DMT: z^a=r^a(cos(ax)+isin(ax))

[twin crag]

oh shoot wait i accidentally wrote the trig form of (sqrt3 -i) wrong

[twin crag]

wait thats not de moivres theorem though

isnt it this

and since this is the trig form of (sqrt3-i)

canw we just do cos (2011 * pi/2)

since cosine is the real part

[vernal pilot]

That is DMT

[vernal pilot]

And so is this

It’s just that that version is better

[twin crag]

but how do i compute the real part using the other version of it?

[vernal pilot]

Idk, I couldn’t see the problem cuz image didn’t load.

Anyways good luck

[snow kite]

I'm watching this video https://www.youtube.com/watch?v=WUvTyaaNkzM&list=PLZHQObOWTQDMsr9K-rj53DwVRMYO3t5Yr where he talks about dividing a circle into rings of dr=0.1 thickness. He then says you can imagine straighetning the rings and approximationg their area with dr*2pi*r

fair enough

THen he says that if you reduce dr, the approximation will become more accurate. I have a feeling this is true, but I can't really put it in words. Can anyone help me explain :^)

tell if this is wrnog channel

[woeful pike]

this is what i got (i havent been taught this stuff properly in school yet so i might be wrong(

[woeful pike]

well every approximation leave some room not filled under the curve, as you get a smaller units of approximation you can start filling in those holes

[snow kite]

this is the part am confused about https://youtu.be/WUvTyaaNkzM?si=IYgvETihdwEOJLiI&t=218

am kinda confused what it means to straigten out a ring... like how does that work

[brave spade]

I want to learn periodic function

It is very confusing sometimes

Suppose I need to find period of cos^4x

[cerulean timber]

it probably is easy but khan academy sucks at teaching it

[limpid onyx]

If I have a statement in Euclidean geometry, if I construct everything hyperbolically (per the klein Bertrami model), will the statement hold?

[limpid onyx]

hmm doesnt look like the euler line holds but im not sure if the circumcenter is still the intersection of the perp bisectors

[versed zephyr]

weird way to say $$e^[i[pi/2]]$$

[somber coyote]

puzzleyman

[versed zephyr]

oof it got messy 😭

$$e^(i(pi/2))$$

[somber coyote]

puzzleyman

[versed zephyr]

HOWWWW

$e^[i(pi/2)]$

[somber coyote]

puzzleyman

[versed zephyr]

WHY ISN'T IT WORKINGGGG 😭

$$e^i(pi)$$

[somber coyote]

puzzleyman

[versed zephyr]

oop i forgor squiggly brackets

$$e^{i\pi}$$

[somber coyote]

puzzleyman

[versed zephyr]

YAYYYYYYYY

now

$$e^{i(\frac{\pi}{2})}

grr

$$e^{i(\frac{\pi}{2})}$$

[somber coyote]

puzzleyman

[versed zephyr]

finally

something good enough

bye

[brisk cloud]

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[brisk cloud]

,help help me with basic trigonomitri

[somber coyote]

Command help me with basic trigonomitri not found!
Use the ,list command without arguments to see a list of commands.

Jan 7, 2024, 11:08 PM

[brisk cloud]

,help trigonometry

[somber coyote]

Command trigonometry not found!
Use the ,list command without arguments to see a list of commands.

Jan 7, 2024, 11:08 PM

[brisk cloud]

,list

[somber coyote]

My commands!

Use ,ls to obtain a briefer listing, and use ,help <cmd>to view detailed help for a particular command, or ,help to view general help.

If you still have questions, talk to our friendly support team here.

LaTeX Rendering

Render LaTeX code and configure rendering options.
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​ ​ ​ ​ ​ ​ ​ texdoc: Searches the texdoc
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Page 1/4

[brisk cloud]

this hurts my brain

[stuck sleet]

Can someone draw this:

A spherical ball of radius R is dropped into a vessel in the form of an
inverted right circular cone. Find the radius and altitude of the cone, if when
three more balls each of radius R are dropped into it they form a layer on top
of the first ball such that all four balls are tangent to each other and in addition
each ball of the upper layer is tangent both to the side and to the top of the
vessel.

Thanks!

[twin crag]

didnt learn that until today lol

[rotund thistle]

I need help with homework, about circles

[vernal pilot]

is T supposed to be the origin?

for the second one

[rotund thistle]

second pic?

[vernal pilot]

yes

[rotund thistle]

I think so

[vernal pilot]

For the second image, you can easily discern angle RSO as being 90 degrees, this is due to a thales law

[rotund thistle]

Ohhhh

[vernal pilot]

And for angle RTA, you know that OTR is 130, so 180-130=50 degrees.

[rotund thistle]

so rta is 50 degrees?

[vernal pilot]

dont mind my shit drawing skills

[rotund thistle]

Its ok lol

i see

[vernal pilot]

hmm, id have to think about the others a bit more

[rotund thistle]

alright

[vernal pilot]

idk man, i cant be bothered doing any more geometry.

[rotund thistle]

Its alright, thank you for the help

[rotund thistle]

<@&286206848099549185> hi anyone pls help?

any grade 10 students here

[latent aspen]

i can try

gimme a second tho lol

[rotund thistle]

alright

[latent aspen]

also w geto pfp

[rotund thistle]

Ty

[vernal pilot]

Gojo is superior.

Geto got done so dirty in the manga.

[rotund thistle]

real

[latent aspen]

ART and RTA add to equal 130 right

[rotund thistle]

for rta thats my answer too

i just

I’m not sure

also for the first pic idk if its correct i literally just answered 70 degrees for each question

[vernal pilot]

this gives me an aneurysm, you should really start with the same letter.

[rotund thistle]

so for every question on the second pic

i need to like start measuring the angle

Starting from

T right

bc it at the center

wait

nvm

[latent aspen]

but they're different angles lol

[rotund thistle]

Ohh

[vernal pilot]

yes, but you should generally start with the same letter, so it would be ART and ATR

[nimble anchor]

am in 10th lol

[rotund thistle]

Alr hi

[rotund thistle]

just if u dont mind maybe help me with this homework

[latent aspen]

leave it to him

circle geometry is something i cannot think about this late at night

[rotund thistle]

alright

[nimble anchor]

am too bad with geometry lol

[rotund thistle]

lol same

[nimble anchor]

sorry I dont think I'll be able to solve it

[rotund thistle]

its alright

[nimble anchor]

am in 10th but I can help with calculus and algebra 😭

[latent aspen]

im not bothered to get out the scrap paper

my mental maths is not good enough

[nimble anchor]

wait, in India, we haven't been thought stuff to solve using arc degrees 😿

[vernal pilot]

Ls

[latent aspen]

this is circle geometry

just remember a bunch of properties

u dont need arc degrees or whatever

am i dumb or something

[nimble anchor]

but we'd be able to find the answers only by using the given 130 and 110 arc degrees right?

[rotund thistle]

i think so

[vernal pilot]

my next thought when i look at this diagram is to draw line TS, as that will give us a triangle TSA.

[latent aspen]

and the circle angle properities

[nimble anchor]

yeah

could you just mark the given 110 degrees in the diagram

[rotund thistle]

wdym

[nimble anchor]

nvm, we haven't even been thought this concept

[rotund thistle]

Ohhh

[nimble anchor]

and am not interested in geometry lol

like I've learnt calculus on my own out my interest

[nimble anchor]

but geometry is not something of my taste :-;

[rotund thistle]

btw for the these 3

[latent aspen]

damn same lmao

[rotund thistle]

idk if im correct but

i just answered 70 degrees to all of them

[nimble anchor]

wait what does m stand for?

[rotund thistle]

measure

[latent aspen]

measurement of

[nimble anchor]

lol dude, even I love the topic entropy

am into this space stuff, and entropy is my most fav topic

[nimble anchor]

how are we supposed to use the arc degree of KS?

[vernal pilot]

TYR is 90 degrees cuz of thales law again lol

[nimble anchor]

yeah

[latent aspen]

WAIT

NO

OMG

THE CORNER PROTUDES FROM THE CIRCLE

[vernal pilot]

bro just had a realisation

[latent aspen]

MY OCD GAHHHHH

[nimble anchor]

waitttttt

[vernal pilot]

HAHAHA

[nimble anchor]

SAME

HERE

[latent aspen]

i don't actually have ocd but

WHY

[nimble anchor]

WTHHH

[nimble anchor]

I Kinda have OCD

[latent aspen]

invalid question

next

[nimble anchor]

like I check my door for 1000 times if its locked

surely not a symptom of OCD right?

[latent aspen]

everyone does that lol

[nimble anchor]

bro literally made me toh sit and learn a new concept when my prebaords are just round the corner

[latent aspen]

also

my confusion during this question

is it just me or are all the answers wrong

[rotund thistle]

idk

[nimble anchor]

rushes to search for notes

[rotund thistle]

btw am i correct here on the measure of angle KES and KUS

both are 70 degrees

?

[latent aspen]

i have no idea

i havent done it

lol

[rotund thistle]

alr

[nimble anchor]

wait, it should be x^3/3 right?

[rotund thistle]

hmm

[nimble anchor]

😭

[nimble anchor]

+C*

[latent aspen]

exactly

[nimble anchor]

wait lemme learn the concept and try to solve it ;--;

[rotund thistle]

alright

[nimble anchor]

crying with happiness after finally solving an integral qn

[latent aspen]

how long have u been learning for

[nimble anchor]

calculus?

I just watched a 4 hr long yt lecture 3-4 months ago while preparing for an international phy competition

in which I'd to solve qn from Bcs physics ;---;

[latent aspen]

interesting

[nimble anchor]

it was actually a grp competition'

and guess my rank

[latent aspen]

and how old ru

15 oh

[vernal pilot]

all wrong.

[latent aspen]

LMAO

[nimble anchor]

206 out of 430 groups 🙂 👍

[latent aspen]

nice damn

[vernal pilot]

i dont think so

[rotund thistle]

Oh

[nimble anchor]

yeah unfortunately

[vernal pilot]

they dont pass through origin, so i dunno if its actually possible.

[nimble anchor]

i wish if I was older enough to solve calculus on full time basis

[latent aspen]

im pretty sure i've covered the context for calc 1 and 2

after like

[rotund thistle]

what do u think the answer is

[latent aspen]

4 months maybe

[vernal pilot]

i think its impossible

[nimble anchor]

its like I've covered everything, but not deep enough

like am not good with logarithm qns

[vernal pilot]

because the 70 degrees refers to KOS, where O is origin. And that is quite clearly different to KES.

[latent aspen]

surprise surprise im actually an english main lmao

[vernal pilot]

cool ig

[nimble anchor]

whats that thing, which is the opposite of chain rule which is used in multiple integration?

[latent aspen]

u sub?

[nimble anchor]

💀 ☠️

[vernal pilot]

english is a mid subject.

[latent aspen]

i agree

[nimble anchor]

idk lol, I haven'y done that too

[latent aspen]

just

u= something

ez

[nimble anchor]

lol in india

there's a board named CBSE

and like 80% of students study in it

like we literally have literature for our english exams 😭

like how the hell will literature help me to get into MIT?

[vernal pilot]

i once read half a book and got a B+ on an exam about that book.

[nimble anchor]

that too childish STORIES

[latent aspen]

i once searched up the wikipedia summary of a book and got full marks on the quiz 😭😭

[nimble anchor]

and a few robert frost poems bruh

[vernal pilot]

bro in all of the quizes about the book that we did in class i would always end up in the top 3 cuz we did some weird quiz website that added powerups LOL

[nimble anchor]

@latent aspen u love pace stuff too?

[latent aspen]

ive played that too

blooklet right

[latent aspen]

hm?

[vernal pilot]

no it wasnt blooket, but it was similar

[nimble anchor]

nostalgia lol

[nimble anchor]

space*

[latent aspen]

i used to

more of a quantum physics guy now

[nimble anchor]

USED TO? 💀

[latent aspen]

😭

old interests

[nimble anchor]

☠️

[latent aspen]

did u just L me

[nimble anchor]

a bit of us moment ;--;

[vernal pilot]

bruh physics is so annoying, you gotta learn like 50 equations and processes about 1 topic then only use 5 of them, but hope you got the correct 5.

[nimble anchor]

well am still into space stuff, but finding quantum phy more interesting 👀

[latent aspen]

i like how its turned into everyone flexing about their strengths and trying to cover it up with modesty

[nimble anchor]

wait wth, why english main?

[rotund thistle]

real

[latent aspen]

i happened to be good at it

[vernal pilot]

i feel like we should stop. cuz i dont want lobotomy kaisen infiltrating the sacred #geometry-and-trigonometry channel

[latent aspen]

yes

[rotund thistle]

i think earth science and biology

is far more easy

[nimble anchor]

+1

[rotund thistle]

than

physics

[latent aspen]

i published two books 😭

they suck

[nimble anchor]

what's ur age bro?

[latent aspen]

ah

14

(lmao)

[nimble anchor]

☠️

I thought I was the youngest guy of my school to learn calculus ;--;

[rotund thistle]

i want this school quarter of mine here

to end quick

physics and polynomial stuff and circles

aaaaaaahgahgsgsh

[latent aspen]

i am currently in beijing, china

[rotund thistle]

ohh

[latent aspen]

i left my laptop at home

[nimble anchor]

neighbor 🫂

[rotund thistle]

teach me math, sensei.

[rotund thistle]

Jk

[latent aspen]

call me yaga

[rotund thistle]

yaga

[latent aspen]

getos sensei

[rotund thistle]

O

[latent aspen]

i thought u were indian-

[rotund thistle]

Ok u got me there

[latent aspen]

what did u think it was 😭😭

[nimble anchor]

yeah I am

[latent aspen]

u also on holidays?

[nimble anchor]

like india is a neighbor of china right?

(my geography sucks)

[latent aspen]

oh

i am australian

speaking chinese

[rotund thistle]

o

[nimble anchor]

wait, I was learning about neurons when I've my maths exam day after tommorow, and now am learning about degree measure of an arc which will be taught to be in 11th 😭

[latent aspen]

interesting

[nimble anchor]

ching chang choo cing

;--;

(no offence)

[latent aspen]

interesting

[nimble anchor]

well I visited china like 5 years ago

my first foreign trip ;--;

I visited hongkong. macau and then thailand ig

[latent aspen]

i visited china 11 years ago and boy am i not glad to be back

[nimble anchor]

i dont even remember lol

[nimble anchor]

lol u must 3 years old back then

[latent aspen]

yeah

[nimble anchor]

wait lemme learn this thing and get back to you

;-;

[latent aspen]

isnt circle geo a 9th grade topic

[nimble anchor]

it is, but we weren't taught degree measure of an ARC

its a part of our 11th grade

wait wth

[latent aspen]

just divide the arc length by the circumference then multiply by 360 degrees?

[nimble anchor]

this video is in chinese ;---;

[latent aspen]

rip

[nimble anchor]

wait she speaks both english and chinese

ok it was her intro

which was in chinese

[nimble anchor]

well, we dont have circumference

checkout the qn posted earlier

[latent aspen]

then do u have the radius

[nimble anchor]

in the same channel

[latent aspen]

wiich one

[nimble anchor]

.

[vernal pilot]

As the mathematician looked at the diagram he asked: "Is it Thales' Theorem because the center of the circumcircle is on the triangle, or is the center of the circumcircle on the triangle because of Thale's Theorem". The geometric responded: "Stand proud, for you have learnt circle theorems". Unfortunately the mathematician wasn't aware of cyclic quadrilaterals.

[nimble anchor]

@latent aspen

[latent aspen]

interesting

geto doing the floss is forever sacred

[nimble anchor]

what punett?

[nimble anchor]

she speaks chinese or korean ;--;

[nimble anchor]

itadori ;-;

[vernal pilot]

anyways, im going.

good luck on this "geometry"

[latent aspen]

mk byeee ^^

[bronze wind]

how do I find the measure of this?

[vernal pilot]

O is origin btw

[bronze wind]

what does that yellow text say?

[vernal pilot]

2x

[bronze wind]

where did you get that from

[vernal pilot]

circle theorem.

[upper karma]

angle POM = 2 angle PQM ig

[vernal pilot]

Angle Subtended by an Arc at the Center is Twice the Angle Subtended by it at the Circumference

[upper karma]

I derived this formula to find the side of a triangle with the other 2 sides and the opposite angle know, and I really thought that I did something. But then I realized that the formula looks kinda familiar 😭

[maiden brook]

bruh

u didn’t know law of cosines was a thing beforehand?

[upper karma]

No 😭

But the derivation was funn 🤷

[upper karma]

Can i say an triangle in an equilateral if it has all equal angles?

[vernal pilot]

yes.

[upper karma]

Kay

[upper karma]

taking geometry hn over the summer, is there anything I should do to prepare beforehand (im in algebra 1 hn rn?)

[grizzled star]

Oh there's a geometry and trig section here

[grizzled star]

learn how to draw rigid transformations.

[sturdy stone]

learn geogebra

[deft jolt]

does anyone know which side of angle i use if its cut by a bisector?

like if i were to find measure GRA, which side of the angle do i use?

[robust plover]

I’m pretty sure it depends on what points you reference, so if you say angle GRA then it’s the entire thing, but if it’s GRO then it’s the left side, or if it’s ARO then it’s the right side

[vernal pilot]

That’s angle GRA

[fickle tangle]

hello

can someone help me with quadratic expressions

preferably explain in vc if could

[woeful pike]

Wdym like the quadratic equation?

[regal phoenix]

i guess you could choose depending on how much information you have

[exotic moss]

Does this channel include Plane Geometry?

[sturdy stone]

Yes

[regal phoenix]

is the triangle (a=0, b=1, c=1) technically considered a right triangle?

because of circle theorem

[vocal light]

anyone can help?

[rose raven]

X=0 is the line of symmetry

Just look at where it reflects on the y axis

[vivid token]

Help please

[upper karma]

What does the variable adjacent to sin, cos and tan mean?

I see sin A, or cos B
is this referring to an angle in a triangle or just the variable that equals the result?

[grim niche]

the second thing

[ocean kernel]

anyone need help while i’m waiting for my question to be answered

[ocean kernel]

you would just change each point to: (-y,x) instead of x,y. Ex. (2,8) becomes (-8,2)

[vernal pilot]

thats whats called a line.

<@&268886789983436800> spam

[woeful pike]

isnt it a thing called a depressed triangle?

something liek that

[woeful pike]

degenerate triangles

[vernal pilot]

yes, but no one is going to call it that for a very obvious reason; its a line.

[woeful pike]

hes asking if its technically a triangle. the answer is yes.

even if no one calls it that

it technically is

[maiden brook]

depressed triangle LOL

[pine mango]

Not sure how to complete 45-45-90 triangles, my work and problem is posted. Any tips?

[vernal pilot]

There are 3 ways you can approach this question. First way is using the pythagorean theorem.

$c^{2}=a^{2}+b^{2} \newline \newline \text{Of course, a=b, so: } \newline \newline c^{2}=2a^{2}$

[somber coyote]

TheLord26

[vernal pilot]

Plug in a=26 and you can easily find the answer: $\newline \newline 26^{2}=2a^{2}$

[somber coyote]

TheLord26

[vernal pilot]

Generally you should try to simplify equations as much as you can before solving them, as it will make it much easier in the solving process.

That is how I would have solved it, however I think you solved it using using ratios. Which is a very nice solution, and there isn't much simplifying during it, until the very last step.

[pine mango]

Okay, I tried doing the square root of 2 and had that divided by my hypotenuse

[vernal pilot]

yes, if you were doing an exam or something, leaving it as 13*sqrt(2) would be your answer

[pine mango]

As if I'm correct the hypotenuse is square root of 2 times the leg but the problem is the legs are unknown

[vernal pilot]

if you recall the 45-45-90 triangle then you can see that the side ratios are 1-1-sqrt(2)

[pine mango]

Right

[vernal pilot]

I'm wondering, are you currently learning about trigonometric functions? like sin(x), cos(x) and tan(x)?

[pine mango]

Yeah

Basic trig as of now

[vernal pilot]

understandable, it's hard to say what you are expected to do to solve this question, whether you were meant to use trig functions or any trig/geometry technique. overall i would say good job.

[pine mango]

Thank you

I feel like it's a bit confusing but I just used the notes we got and it made a decent amount of sense

[vernal pilot]

i obviously dont know what your notes are, but if youa re confused about anything feel free to ask questions

[pine mango]

Yeah the unit is basic/inverse trig, elevation/depression, and special right triangles

[pine mango]

Will do thank you

[vernal pilot]

well in that case there is a way to solve this question using trigonometric functions.

[pine mango]

I don't wanna over complicate it

[vernal pilot]

ok

fair enough

[pine mango]

As long as I can get the question right I don't mind using whatever methods

[vernal pilot]

idk how your exams are structured, but you should try doing things in different methods to get accustomed to solving questions using whatever method you might be asked to use

[pine mango]

Yeah, usually 50 questions at the end just on the whole course

Headed to bed, thanks for all the help @vernal pilot very much appreciated 🫡

[vernal pilot]

np, although i dont think i helped that much lol

[woeful pike]

degenerate as i said after

[upper karma]

i think u will graph it down

[vocal bison]

Help plz ping ik u set them equal but I’m still confused

[mystic charm]

What are you confused on? like why we set them equal?

[woeful pike]

3c/9 = 4-c

U can simplify the left first

[rose raven]

and 3c/9 can be simplified first into simplifying it using an LCM of 3

which would make the fraction c/3

[tidal zephyr]

Hello everyone in the group, could anyone recommend a book on geometry?

[regal phoenix]

oh i'm also curious about this one

[frozen ocean]

#book-recommendations

[proud adder]

Can someone please help me

[proud adder]

,rcw

[upper karma]

yeah one sec i havent done geo in a bit but ill find something

[woeful pike]

Draw lines OF and OH and it might be easier

[somber coyote]

EEE

EEE

EEE

EEE

[upper karma]

Looks simple but...

Find the value of x?

[subtle inlet]

GQ is 6cm or GD?

[upper karma]

GQ

[stuck sleet]

What does this mean?

"Angles may be congruent of course, without being the same angle"

If they have the same measure the are the same right???

[subtle inlet]

maybe like for example \angle BAC = \angle BCA = 60. They are congruent but not the same.

[stuck sleet]

why they are not the same?

[subtle inlet]

because the vertex is not the same i think

[wet mural]

anyone got any idea?
given:
DEFG is a rectangle trapped in ABC
DE = 3EF
AH is the height to BC (AHB=90)
AH = 6cm, S(ABC) = 27 cm^2

[woeful pike]

Whats the question

And is S(ABC) like area or perimeter

[mystic charm]

Did you find the answer? If no, I can explain you what to do

Some information are missing

[woeful pike]

Yeah can u show us what the text is on the side

[sour mortar]

hello

can someone help me with this

[unkempt kestrel]

st/tr

cpst (corresponding parts of similar triangles)

idk if thats standard notation tho

[restive island]

is that khan?

[sour mortar]

yeah i figured

thanks

[sour mortar]

no delta math

they assigned us it cus of snow day

[maiden brook]

ew

[rose raven]

i love deltamath

[maiden brook]

why

[maiden brook]

@rose raven

[rose raven]

Because the math on their is easy

[maiden brook]

its easy yet still tedious

the worst type of math

[regal phoenix]

preferences

[maiden brook]

how could it be considered good

[upper osprey]

I have did the calculations on a 2V dome but I haven't found info on the formulas for a 3v or even grater. Can anyone point me to where I can get this info. I have found wonderful sites but I don't get the formulas on how they did it.

[knotty sinew]

find the area of the orange triangle and divide it by area of rectangle

if u cut the triange from the top, u can see the parts of the triangle are half of the corresponding blue part

so i suspect its 50 percent

ye its 50 percent

wut do u guys think?

so essecntially if u cut the triangle u can fit it into the blue parts

so its 1/2

or 50 percent

[knotty sinew]

means angles are measured the same regardless of size cuz its a consant

constant

i thinkt thats radians

[mystic current]

so i need help with this assignment and im have to submit it by 12 and it's 11: 33 can someone help me

[weak torrent]

I really need help over these 2 math problems I’ve been crying over them pls :’)

[woeful pike]

Ur meant to show the problem lol

[weak torrent]

Sorry I forgot to send them but I figured them out finally😭

[north moth]

Hi, I need help with these two problems, I don't know how to reach the solutions provided

[silent plank]

do you knwo how to calculate the area of a sector?

[north moth]

I got the answers, and aced my quiz. I wasn’t converting degrees to radians properly.

Ty

[blazing walrus]

yooo

I may need help with these

[vernal pilot]

Do you know the law of cosines?

[blazing walrus]

NVM

dw

youtube helped me

this

https://www.youtube.com/watch?v=5tp74g4N8EY

[wide field]

Is it possible to describe polyhedron with spheres of different radii (at least 2 different spheres)?

[tawdry patrol]

is arccos different than cos^-1? im using the microsoft math solver on the web to do some trig problems but gave me an answer that looked wrong and then I used the app version which is more basic but has trig functions which then gave me a better answer. (Sorry in advanced if I used this channel wrongly)

[sturdy stone]

they are the same

[digital nexus]

can someone help me start this problem? I can probably get it after two steps but I dont know how to start

[mystic current]

can someone help me with this test

[frozen ocean]

we don't help in tests/quizzes

[knotty sinew]

cheating

[knotty sinew]

chatgpt thinks im a child prodigy

lol

[willow bolt]

How do I prove this?

[royal cedar]

Try cross multiplying

[maiden brook]

yep

[dark sparrow]

!nogpt

[lime crown]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[latent aspen]

first off if you're 11 thats against discord TOS and u could get the entire server shut down

[latent aspen]

and this ^^

[dark sparrow]

oh yeah good point

[latent aspen]

do we ban

[dark sparrow]

<@&268886789983436800> we got an underage over here!

[latent aspen]

idk

also calculus is actually not that hard

its not a flex

[spark stag]

let the kid think he's smart

[latent aspen]

mbmb

[dark sparrow]

the kid admitted to being 11 years old did he not

[latent aspen]

lmao yeah

[spark stag]

yeah i banned

but telling an 11 year old theyre not a genius wont get you anywhere

theyll learn that naturally once they grow up

[latent aspen]

interesting

stuff like calculus and quantum mechanics tho u can learn that anytime as long as you have an interest

and thats good

[vernal pilot]

!nogpt

[lime crown]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[vernal pilot]

lol

[vernal pilot]

Don’t try cross multiplying, that’s not how you do proofs

[upper karma]

not sure but I think you have to solve both LHS And RHS
$$ \frac{sin \theta cos \theta}{sin^2 \theta + cos^2 \theta + cos \theta} \boxed{[\therefore 1= sin^2 \theta + cos^2 \theta]} $$
$$ \frac{1 - cos \theta}{ \frac{sin \theta}{\cos \theta }} , \boxed{\therefore tan \theta = \frac{sin \theta}{ cos \theta}}$$

[somber coyote]

Snöwdinger

[upper karma]

or multiply and divide LHS by 2 that will form sin(2x) ¯_(ツ)_/¯

[dusky dirge]

Y'all ... I tried this like a million times ... I think that I don't understand the word problem:

[solid nebula]

its a triangle

if u look at a compass

and look at where east and south is

you'd find yourself a triangle when you create a line between them if that makes sense just look at a compass

[junior meteor]

Anyone know how to do this?

[north crescent]

no

[jolly nymph]

how do you use sin-1?

or cos or tan-1

when is it applicable

[little remnant]

I need some1 to help me with this

[mint nexus]

how do i do this??? im so confused

[jolly nymph]

HK is 2/3 the length of HT

[mint nexus]

right, but how do i apply that?

[jolly nymph]

it gives u HT, and it gives you a way to find HK

6y-10 = 2/3(27)

[mint nexus]

OHH

[jolly nymph]

memorize this

[mint nexus]

i did it the other way☹️

per question with that similar question do i apply that logic?

[jolly nymph]

im sorry huh

[mint nexus]

do i apply 2/3 logic to those questions

[jolly nymph]

i used this theroem

[mint nexus]

ohhh

the thing is its 6y=28 which gives me a decimal

[jolly nymph]

shouldnt be too hard to do it still

decimals can b the right answer still

[mint nexus]

but idk if its supposed to be a decimal

4.6666666667 doesnt sound too true

4.7

idk

[maiden brook]

14/3

[jolly nymph]

since you know the 2/3 rule, you could even say (6y+10) + 0.5(6y + 10) = length of HT which is 27

[jolly nymph]

yea for irrationals do a fraction form

[jolly nymph]

the other option is better tho cuz it uses the theorem

[mint nexus]

i dont think it wants a fraction according to the format

i feel like in missing something

[maiden brook]

irrationals?

[mint nexus]

what about like

(2/3)6y-10=27(2/3)

4y-10=18 -> 4y=28 = y=7

it was 9 wtf

[jolly nymph]

something like 14/3 where decimal form is 14.33 repeating

[maiden brook]

thats not irrational

[jolly nymph]

is it not?

irrational numbers are numbers that never terminate

right

[maiden brook]

smh

irrational numbers are numbers that cant be written as the quotient of two integers

[mint nexus]

it was 14/3

except it asked for like

another number

making the answer 9

[maiden brook]

very few rational numbers terminate, anyways

[mint nexus]

ohhh wait

kt= 3(14/3) -5=9

[jolly nymph]

oh oops

[junior meteor]

Pls help

[maiden brook]

why is stuff crossed out on the screen

whats the reason for a)

[junior meteor]

That’s what I had written down

A is given and B is def bisect

I just don’t know the rest

[lost drift]

First of all we are going to use Basic Propotionality Theorem[BPT] to prove the above question.

BPT explains that If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

To prove AE/EC = BD/DC, we need to prove that ED is parallel to AB.

In order to prove that we follow the aboe given steps

So, a) AD bisects ∠BAC = It is aldready give in the question.

b) Since AD bisects ∠BAC, it automatically means that ∠BAC= ∠BAD+∠DAC, hence ∠BAD=∠DAC

c) AE=ED is also given in the question.

d) Now since AE=ED inside the triangle AED, by isosceles triangle property[The two angles opposite to the equal sides are congruent to each other] , ∠EDA=∠DAC.

e) Since ∠BAD=∠DAC, we can say that ∠EDA=∠BAD

f) Finally, since the results of (d) and (e) are only possible if ED and AB are parallel line[Alternate angles are equal], we can say that ED||AB.

g) AE/EC=BD/DC by Basic Proportionality theorem by Thales which i have explained above.

Hope this helps

👍

[junior meteor]

Thanks

[lost drift]

Formula= θ/360 x πr^2

[bleak agate]

Anyone know the answer for this?

So is it cos80

Or

[maiden brook]

use law of sines

[keen latch]

do you know the cosine and sine rule?

[narrow bear]

just take 2/3 away from both sides by dividing by 2/3 on both
6y-10=27
answer would be 37/6 or 6.1666667

[dusky dirge]

HALP PLZ

[proper echo]

Alright, I have a geometry test tommorow and have a review paper.

[vocal bison]

Help plz ion get this

[dusky dirge]

PLZZZZZ HALP (my homework is due like 2 days)

[vernal pilot]

!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.

[vernal pilot]

We ain’t doin’ your homework

[vernal pilot]

consider the triangle KMN

It has two sides that are equal and one side which is different, this means its a iscoceles triangle. this means that angle KMN and KNM are equal.

As we know that angle KMN = 50, then angle 1 is 180-2*(50), or 80.

and angle 2 and 16 is 50

idk if you know this specific proof, but the quadrilateral formed by the midpoints of a kite is a rectangle, so you can just state that angles 3, 15, 7 and 11 are 90 degrees.

you now know angles 2 and 3, and the sum of angles 2, 3 and 4 is 180, do some math and you can find that angle 4 is 40 degrees.

with this info it should be relatively easy to get the other angles

[vocal bison]

alr thanks didnt know abt that proof
it helps

[vernal pilot]

its definitely a major shortcut, i think it would be possible without know about it, but definitely harder.

[tender bridge]

What are the main identities that one should know for trig? I know like, there are a lot, but most of them are rearranged versions, so what are the main ones I should know?

[dark sparrow]

Pythagoras, angle sum/diff, sum<->product

[tender bridge]

angle sum is alpha+beta, right??

[dark sparrow]

f(alpha+beta), where f is sin, cos or tan

[tender bridge]

do the half and double identities matter?

[dark sparrow]

yes, but double is rederivable from angle sum, and half from double

[tender bridge]

ok

[waxen moss]

How do I do this please help

Finding area of AFB

[dark sparrow]

,rccw

[dark sparrow]

feels like missing info

[dusky locust]

Are there any right angles in this diagram?

[cobalt ingot]

i would like some help with the first problem

[snow plinth]

How many zeros does sin(2(n+1)x) have in the interval 0<x<=pi/2?

Well, whenever 2(n+1)x= k pi there is a zero, but I'm dumbstruck on how to go further.

I need to find in what set k belongs to basically.

I think I figured it out actually, so nevermind my question!

[maiden brook]

set up some variables

[raw forum]

,rotate

[raw forum]

bro pls turn off your light when u take pics

im assuming the angle at the top is P

also what are u supposed to prove

lil zoom out never hurt anybody :)

feel free to ping me when u take the photo again with:

• goal shown
• angle P shown
• no shining light

[sour mortar]

i was trying to crop it sorry

[vapid scaffold]

(1) Draw a suitable Cartesian plane and plot the following points.
A (2, 5), B (4, 3), C (2, 1), D (0, 6), E (3, 6), F (7, 0)
(2) Plot the following points on a Cartesian plane and join them with
straight line segments in the order of the letters and return to the
starting point.
(i) A (1, 7), B (2, 1), C (5, 5), D (8, 1), E (9, 7)
(ii) A (5, 1), B (5, 3), C (0, 5), D (0, 6), E (5, 4), F (5, 5), G (10, 5), H (10, 1)
(iii) A (1, 4), B (0, 4), C (0, 7), D (1, 7), E (1, 6), F (7, 6), G (7, 7), H (10, 7),
I (10, 4), J (7, 4), K (7, 5), L (1, 5)
(3) Shanuka says that “the vertices of a square are positioned at P (2, 2),
Q (2, 7), R (7, 7), S (7, 2)”. Plot these points on a cartesian plane and
verify the validity of the above statement.
(4) Draw a Cartesian plane and plot four points such that the x - coordinate
and y - coordinate values of each point are equal to each other. Write
down the coordinates of the four points.
^5& (i) Plot the points given below on a Cartesian plane and join them in
the order of the letters with straight line segments.
A (4, 1), B (4, 2), C (4, 3), D (4, 4)
(ii) Extend the line that is obtained.
(iii) Write the coordinates of two other points on this line.
^6& (i) Plot the points given below on a Cartesian plane and join them in
the order of the letters with straight line segments.
P (2, 3), Q (4, 3), R (6, 3), S (7, 3)
(ii) Extend the line that is obtained.
(iii) Write the coordinates of two other points on this line.

how to solve thi one guyss

[rough lodge]

i am not explaining in detail but i think you have to give tri pto congruent tri oqr and this will give you <tor =90 degree then its simple subtraction

[dusky locust]

that question solution should be easily accessible online

[onyx fjord]

I’d like to get some advice, I’m confused how PR is a hypo on solution A.?. I’m also confused on letter B. (Solution) if where is the Hypothenus or A or B… Quick question is b Adjacent on the Pythagorean theorem formula pic? ^^

This is Lesson 4: Secant, Tangent and Chord Lengths.

[lost drift]

Well, The longest side in a right angled triangle is the 'Hypotenuse' or easy to remember the side opposite to the right angle [90degrees] is the hypotenuse.

And there is a theorem which states that 'The tangent at any point of a circle is perpendicular to the radius through the point of contact' which makes ∠PQR=90°, so PR is the hypotenuse.

In both the questions(A and B), PR is the hypotenuse

[fluid ridge]

PR is a hypo in solution A due to it being a right angle, as we can see the radius of circle R makes contact with point Q of tangent QP thus making it perpendicular.

Ur 2nd question is a bit confusing however PR is regarded as the hypotenuse (c in the pythagorean theorem) and the QR a side (which can be labelled as a in the pythagorean theorem) is also given. Knowing that it is a right triangle we can apply the pythagorean theorem to find PQ (can be labelled as b). The solution provides the steps in finding b. Lastly we can use the two tangent theorem to state that PQ is congruent to PS therefore solving the question

Abt ur quick question, b is usually regarded as the longer side (opp) however there are some cases wherein a and b are used interchangeably, i dont think it is good practice to do so

[lost drift]

In a triangle, there are three sides and three vertices. Each side is adjacent to the two other sides that share a common vertex with it so the side 'b' is adjacent to both 'a' and 'c'

[lost drift]

Dude thinks that the Hypotenuse is different for the questions 'a' and 'b'

[fluid ridge]

oh so thats their second question? They are clarifying whether the hypotenuse is the same through out the questions?

[lost drift]

yup

[fluid ridge]

ohh I see in that case you already answered their question really well. I'll just leave mine just for extra info to them. Thanks

[vocal bison]

how do u do these

[lost drift]

If P and Q are two points whose coordinates ate (at^2 , 2at) and (a/t^2 , 2a/t) respectively and S is the point (a,0). Show that 1/SP + 1/SQ is independent of t.

Please help me with this question.

[plain cedar]

9th grade
If AE=3
BG = 1
How much is AF?

[coral thistle]

hi does anybody know what “what is the length of the altitude drawn to the longest side” means? the question is “the lengths of the sides of a triangle are 74 ft, 38 ft, and 88 ft. what is the length of the altitude drawn to the longest side?” i don’t understand 😅 are they asking me for the height straight down the middle?

[ember rose]

what math is this?

[upper karma]

Anybody have resources to self-study geometry?

I need to testout within may and im past triangles n' stuff.

Would really appreciate any study forum.

[dire wedge]

Hey I was wondering if someone could help me with a geomatry proof

[sturdy stone]

Not if you don't post it

[lost terrace]

AEF is isosceles triangle in ABC where E and F lies on AB and AC respectively. EF passes through H.Let BF intersect (AEF) at N. Prove $\angle HNF=\angle HAF$

[somber coyote]

lelouchvigeo

[lost terrace]

Can anybody tell this

[regal phoenix]

i tried drawing it but it’s pretty hard to understand

[mint reef]

guys can someone help me with easy as hell geometry hw

idk why but i forgot how to do it

its ez

<OND=60°

and i gotta figure out

<COM

[upper karma]

Just ask

Don't ask to ask

[dusky locust]

@waxen moss which angles are right?

? what do you mean assume

he said there are

[upper karma]

I think they mean 90 degree angles

[dusky locust]

yes i'm asking which angles are 90 degrees

[waxen moss]

Ive solved it its quite easy 😅

[dusky locust]

ah ok that is good to hear

congratulations then

[near glen]

I need help w this problem yo

[regal phoenix]

it can be solved with simple geometry

pythag theorem, square area, right triangle area and semicircle area

[near glen]

bro I need answer bro

[regal phoenix]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[near glen]

God damn it

Damn u good moral demons

screw morals bro that shits fake

What’s not fake is an F

[regal phoenix]

i can help you doing it, if you want

[unkempt kestrel]

!noonion /j

[regal phoenix]

the blue lines are the angle bisectors of the triangle ABD, and the red lines are of the triangle ACD. being AD = 2 and the distance between the interceptions of the angle bisectors equal to sqrt(2), find BC

question brought by @carmine timber

[slim coral]

can I say k is located on the center of the circle whenever I see this?

[vernal pilot]

No

Because the side lengths don’t need to be equal

[slim coral]

got it thanks

[old marsh]

Can anyone help me with the questions please?

[vernal pilot]

Which one? And what do you need help with specifically?

[old marsh]

The ones for which i havent written down an answer.
c (ii) and (iii)
And 7a (ii)

I just want a quick guidance on which rule to apply if possible

[rotund zodiac]

Can you help me with this

AD = DC, S(ABD) = S(DBC)

[bright compass]

Anyone know the answer?

[upper karma]

,ask cos^-1(4.5/6)(180/pi)

[somber coyote]

Results provided by WolframAlpha

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

Requested by rocinantexyz

[upper karma]

sorry, it says the result is in radians but it is in degrees, heh silly computer

[upper karma]

What are SSS, SSA, AAS, etc. ?

Where can they be applied

and where can i learn more about them?

[robust finch]

SSS is side side side, SSA is side side angle, AAS is angle angle side

They can be applied to triangle congeuency

If triangles have those values in those order congruent the entire triangle are congruent

The order matters because some orders do not make it congruent

Such as ASS

I remember it because it spells a bad word

So its useless

Hehehe

You can use this video to learn about them

https://m.youtube.com/watch?v=luSFHh_vcis

[quiet vigil]

an yone know how to do these?

[dire wedge]

SAS

[regal phoenix]

TQ/2, 90, SP/2

oh this is so neat

[exotic laurel]

Is there any explicit formula to calculate the sine of any real number?

[vernal pilot]

Yes, the Taylor series of the sine function.

[exotic laurel]

Oh, no, not the Taylor series. I just began learning about the Taylor series a couple of days back and expect to understand and learn to use it to solve the problems it's meant to solve in a week or two. I want an explicit formula.

[vernal pilot]

Maybe try the complex notation of the sine? Using DeMoivre’s theorem.

Start with $e^{i\theta}=\cos(\theta)+i\sin(\theta)$

[somber coyote]

TheLord26

[vernal pilot]

And with a little rearranging and smart use of algebra+calculus you can make it just about sin

[exotic laurel]

Does an explicit formula not exist that can be used to calculate the value of sin (x), where x belongs to the set of all real numbers?

[vernal pilot]

Computers use Taylor’s series for calculations iirc

But the complex definition works for all real numbers

[exotic laurel]

So, is the answer to my questions a "no"?

[vernal pilot]

Idk

The Taylor series is an “explicit” formula

[exotic laurel]

I'm just asking whether there's an explicit formula that currently exists that can be used to calculate for sine of any real number x?

[vernal pilot]

The complex definition and Taylor series….

[exotic laurel]

Ok, alright. I'll learn about the Taylor series and what you said. Thanks.

[vernal pilot]

If you don’t want to use Taylor serious just use the complex definition

[exotic laurel]

Alright, will try it.

[grizzled acorn]

Hi

[hearty badger]

how do u do this

[frozen ocean]

you don't need to post this twice

but what have you tried so far?

[wary star]

bru why does my calc on the elft show a different answer

they both have the same input

[upper karma]

the one on the left is in radians, the one on the right is in degrees.

but it is the same angle

[wary star]

oh how do i make it into degreea

my first time using this calculator

[upper karma]

I don't know the details for that model, but check the settings for how angles are represented

[quiet birch]

need help with the part in red, missed a class and having trouble

[high drift]

If i was given 4 edges have length a,b,c,d. What are the conditions for those 4 edges to be 4 edges of a trapezoid(Let call that trapezoid is ABCD with AB//CD and AB=a,BC=b,CD=c,DA=d)?

[sturdy stone]

Sum of any three of them has to be greater than the other one

That might not be sufficient but is definitely necessary

[high drift]

yep it is definitely necessary

i am thinking about triangle inequality, i hope it will help me

[vernal pilot]

|a|+|b|>=|c|

That’s the triangle inequality

[sturdy stone]

I am aware

Not sure what your point is

[maiden brook]

no need for absolute value

[vernal pilot]

Someone responded to you talking about triangle inequality, I just thought to put it here. Idk what’s going on tho.

[vernal pilot]

Wrong cuz then people might think it’s funny to put negative numbers in A and B.

And cuz we are talking about a geometry thing, measurements are always positive hence the absolute value.

[maiden brook]

nope it’s redundant

[vernal pilot]

Literally no.

[maiden brook]

no absolute value

<@&286206848099549185> what do u think

[celest gull]

help please

[twin crag]

pretty sure its a rhombus

since the angle information fits the definition of a rhombus, and the diagonals of a rhombus are perpendicular

[celest gull]

you have to choose more than one (all that apply)

[twin crag]

okay then ig its

Rhombus
Quadrilateral
Parallelogram

[keen stirrup]

whats the area of a rectangle im stuck

[covert cape]

are there any youtube videos anyone would recommend for geometry midterm review

[keen stirrup]

wdym

[maiden brook]

whats the definition of area

[keen stirrup]

idk man

[hearty badger]

can some one answer this https://web2.0calc.com/questions/help_85646

[keen stirrup]

any specific topics

[covert cape]

proofs

also what should i look for when i have parallel lines?

[hearty badger]

alt interior angles

[covert cape]

thanks

[worthy stream]

can someone help me please

How do I center it around S 😭

[covert cape]

what are the properties of all parallelograms?

nvm

[covert cape]

is AAA a congruency theorem of triangles?

[hearty badger]

nope

[maiden brook]

think why not

and why its called AA similarity not AAA

[glacial yacht]

i need help w/ this Prove that in any triangle, perimeter is greater than the sum of
its medians

[nocturne remnant]

try to bound the length of each median in terms of the side lengths

[glacial yacht]

like

m1 m2 and m3

in terms of sides?

[nocturne remnant]

ye

[glacial yacht]

ill try

im getting m1 = 1/2 into root of 2b^2+2c^2-a^2

[glacial yacht]

tysm elon

[glacial yacht]

In triangle ABC, ∠C is an acute angle m∠C = 2m∠B. AD ⊥ BC.
Prove that AC = BD − DC can we solve this question other than using trigono

[worthy stream]

tysm

[silent steppe]

does this relation hold?

[manic shoal]

will anyone possibly know how to do this question

😭

[maiden brook]

could u construct a segment through q parallel to rs

[slate trail]

Just extend PQ to join RS, say it meets RS at X

Then c= angle RXQ as they are corresponding angles

Also RXQ form a triangle where a is the exterior angle to the triangle so, the sum of interior opp. angles = exterior angle or simply b+c=a

[regal phoenix]

i'm pretty sure it does

cool variation of the sine rule

[austere cedar]

Can somebody help me on how to do this?

[jagged wyvern]

#1199151309132136548

[jagged wyvern]

@toony8509

[covert cape]

when do u use pythagorean theorem in geometry

[crisp edge]

bruh what

can someone help me please

[woeful pike]

Sin(H)/h = Sin(G)/g

Sin(10)/820 = Sin(102)/g

Or g(sin(10)) = 820(sin(102))

Or g = 820(sin(102))/sin(10)

[crisp edge]

okay thank you!! 🙏

[willow glacier]

I need help with my math homework and Can someone please help me? Consued on what kind of flag to make and hoping someone can give me an example, what they would make or any suggestions. Thank you

[soft trellis]

i can

so basically

you must create

a flag

with different shapes for example triangle rectangle square

at least 3 but you can add more

use 3 colors like green organge red ext

nevermind

i will send you a picture of an idea tommorow

if thats ok