geometry-and-trigonometry

[upper blaze]

I am working on something and I am not able to compare anything in sense of inequality

I can use only equations

I know when given 3 points how to check if triangle is orientat clockwise or not

Thats easy

but I need same thing but without comparing anything

Just =

[wintry harbor]

Um

So like, "if a=b, then positive" kind of thing?

Can you take absolute values

I guess that also involves comparing doesn't it

[upper blaze]

no

I mean yes that type of thing

but no for absolute values

The thing is I am in coordinate geometry and I have some polinomial ideals and I cant have equations with absolute value (I have them now) and I ened to get rid of it

Like Area of triangle ABC, A(0,0) B(1,0) C(u,v) is half the absolute value of v

and I need to get rid of that absolute value and it only depends on orientation

[dark sparrow]

sounds like there's some context we're all missing here.

can you show the exact problem you're facing?

spare us no details.

[upper blaze]

Yes. I have some ideal $I$ that is generated by finite number of polinomials. And I am trying to take geometric statement about areas of triangles and find equation or equations that match the statement. For example $AB=BC$, where points are given or arbitrary, can be expresed by distance formula and it forms a polinomial P(something)=0. And in my statement about areas. After I determine the equation I try to see does that polinomal belongs to my ideal $I$.

[somber coyote]

MotionMath

[upper blaze]

And I cant check that if it has absolute values or any type of roots it has to be polinomial

[dark sparrow]

well unfortunately the area of your triangle isn't a polynomial function of the coordinates of its vertices.

[upper blaze]

yea I know. But If I can find a way to convert statement "Triangle ABC is positivly orientet or any equivalent statement" I can make it work for me

everything I can find on orientations

works with inequatilies and one thing that doesnt uses angles which I cant realy use

[dark sparrow]

you cannot avoid a sign check when looking at the orientation

[upper blaze]

😦 Okay I will have to find another way around it thanks

[dark sparrow]

this is still something that strikes me as fishy tbh

[young inlet]

i'm having trouble solving this, i've gotten the equation of line 2 but im not sure what i should do next

[dark sparrow]

find the coords of P and Q

[young inlet]

i feel like there's something wrong in my calculation (rewrote it for over an hour) and i'm still very lost

[silent plank]

show your work and attempt (you should include a diagram)

[young inlet]

sorry its rly confusing

[silent plank]

do you have a better diagram?

because that looks very wrong

wait

[young inlet]

nope :( i tried graphing from the answers i got but i think some of them are wrong so

[silent plank]

why do you have two different equations for line 2

[young inlet]

ahh i havent updated that yet

i did a miscalc before

so now im just really lost

[silent plank]

argh 1 sec

misread some stuff a pain to read sideways

lets start from the beginning

[young inlet]

ahh sorry ill send another pic

[silent plank]

you messed up your sign there

[young inlet]

thanks for pointing it out!!

should it look like this?

[silent plank]

you don't really care about the y-intercepts

you stuffed up the signs for the x-intercept of line 2

same sign mishap you had before

[prime egret]

Does anyone here have any interesting problems involving Viviani's theorem

[young inlet]

ah forhot to change that on the paper

ah wait

i got it now thank you sm!!

[mossy jackal]

any geometry worksheets for a stupid yr 7 recommendations? or ways to get better at geometry

[gloomy vessel]

search up year 7 geometry worksheets

[onyx cloud]

@lapis mist have you tried drawing a picture?

[lapis mist]

i did

oops sorry i deleted it

i keep getting 11/7 as my x

ive done it 3 times

[onyx cloud]

your picture is mislabeled

[lapis mist]

how?

[onyx cloud]

read through the side lengths again

and compare them to your picture

[lapis mist]

omg

thank u sm im so sorry

[onyx cloud]

no need to be sorry

here to help 🙂

[north cosmos]

I need help with quadrilateral proofs

[kind shoal]

[north cosmos]

hey can anyone help with quadrilateral proofs

[nocturne remnant]

[onyx cloud]

don't ask to ask, just ask whatever question you have and we'll see if we can help you

[upper karma]

your pfp >:O

(sorry unrelated to your question)

[proper totem]

pls help guys i am stuck on this question all i have is alpha + beta + gamma = 3lmd/lmd+1

[tropic oracle]

Try substituting alpha beta Gamma in Euler from

[sage flume]

https://dontasktoask.com/

[tranquil birch]

A colleague from work showed me this, and I was looking it for weeks and couldn't solve it smartly. So I wrote an integral, and the result is 7. Now, it would be nice to know the catch. Anyone?

[graceful spear]

https://www.youtube.com/watch?v=mOxnEn9qWwo

[tranquil birch]

Fk ty

[upper karma]

"You should be able to solve this" videos are not for everyone. The titles are misleading.

[onyx cloud]

more accurately, the titles are clickbait

[upper karma]

https://cdn.discordapp.com/attachments/871212849802932324/878808032606318682/unknown.png

why cant this also be c

[onyx cloud]

because C is true

[upper karma]

whoopsmb thats cause of a proof right

[onyx cloud]

that's the definition of a rhombus

[upper karma]

oops

wb b

[onyx cloud]

if you take any rectangle and do what they said, the diagonals will always be perpendicular

i'm not sure about a proof but it intuitively makes sense if you draw it out

[upper karma]

errrr

i dont get it lmao

[onyx cloud]

which part

[upper karma]

answer choice b

[onyx cloud]

yes, but what about it don't you get

[upper karma]

like how would u construct a diagram

[onyx cloud]

if the midpoints of the sides of a rectangle are joined together

do you understand that?

[upper karma]

no

[onyx cloud]

which part

[upper karma]

like how would u construct that

[onyx cloud]

can you draw a rectangle?

[upper karma]

yes

[onyx cloud]

can you find the midpoints of each side?

[upper karma]

yes

[onyx cloud]

now connect them going clockwise

[upper karma]

OHH

yeah

that works

so only D wouldnt make sense rite

[onyx cloud]

hm

i think D is true

[upper karma]

yeah D is right answer ig

[onyx cloud]

nono

[upper karma]

huhh

[onyx cloud]

D is true, so it's not the right answer

[upper karma]

wut but thats the answer

cause

for d to be true wouldnt each side of the rhombus be equal to the diameter

that wouldnt work

[onyx cloud]

hm?

[upper karma]

yeah

[onyx cloud]

i'm assuming you meant diagonal not diameter

in which case no

that wouldn't be necessary

but i think you're confusing what the question is asking

it wants us to find the statement that is false

[marble thistle]

how can a rhombus be inscribed as a circle

[silk patio]

You squash it

[upper karma]

You make it a square

[quiet bramble]

is the answer for this really c?

[keen nova]

AC is a line not a ray?

this seems like a mistake on the test makers part

[quiet bramble]

ok thanks

[silent plank]

well the ray AC does exist within the diagram

[keen nova]

oh yeah right

[keen nova]

yeah no guess you were wrong lol

[twin plinth]

can someone help me with the last part of the question (C)? Im confused

I have no clue how to start to attempt it

I just need a general Idea on what exactly to do

[silent plank]

Are you referring to part C)

[twin plinth]

yes

[silent plank]

well they're asking for cost/(cm^2) right

so first determine the total surface area of the solid

and then do the division

[twin plinth]

The thing thats confusing me is the flat part of the hemisphere. Do I count its area? or do i ignore it?

[silent plank]

what flat part?

you mean the face where the cone and hemisphere join together?

you ignore it. It's not part of the external surface

[twin plinth]

yes

[twin plinth]

ok ty

for the help

[silent plank]

it's deep within the solid

How are you supposed to paint that

[twin plinth]

im dumb and tired lol.

but thank you for the help

[wintry nymph]

Is there enough information to prove that lines p and q are parallel? If so , what postulate or theorem would I use?

[ionic bluff]

uh sure

use vertical angle

then alternate interior

@wintry nymph

[dark sparrow]

@wintry nymph if all you know is that the marked angles are equal then no

p and q will intersect if the angles are anything but right

[ionic bluff]

doesn't alternate interior go both ways tho

[dark sparrow]

the angles are marked as equal not as adding up to 180 degrees

[ionic bluff]

if a pair of alternate interior angles are congruent then the lines must be parallel

[dark sparrow]

youre thinking of angles positioned like this i think

[ionic bluff]

oh right i see nvm

[crimson dust]

@dark sparrow any idea

I already got f a

[dark sparrow]

???

why are you pinging me

[crimson dust]

Sorry

Do i not ping

[silk patio]

What did you try

[crimson dust]

Well first I started off going for f to a

Which would be 10^2 - 6^2

8

[crimson dust]

After that i am not really sure how to approach this

Ok got it

[icy leaf]

u've got any idea bout this..?

[crimson dust]

I can try it

[crimson dust]

Is this the full question.

?

I think i got the answer

Do I just send it to him

Or lead him through it

Ok i think he left

@icy leaf are you still here

What have you tried so far

Explain your thought process

My man out sources his math hw to nerds on discord

I just rearrange it into ax^2 +bx + c

And just used the quadratic formula

Holy shit typing math on phone is hard

Sorry i have a shit memory and dont remeber the names

Anyways walkin corpse are following?

Honestly i gave up a long time ago in getting the names

Of common methods

Because there are deviations in names for all of them

Anyways think we lost walkin corpse

Are you still in hs?

Yup just started my gsce

Gcse

Seems relatively easy

So far

For context my parents have forced me to do math olympiads all my life soo

I enjoy it but i kind of feel i was conditioned too

In asia pretty much everyone is good at math so the only way to be good is to be uber well

For example my school only offers extended and additional extended math

Some parents conditioned there kids to extreme levels eg if you dont score top 10% in ukmt globally they break dowm

Down

That actually seems relatively calming

I generally don't like the academic maths because I just can't see the practical applications

But I really enjoy cryptography and programming related math because there is an end product

And its alot easier to see a end result and feel accomplished

My school offers a cs course but its incredibly basic concerning programming and mainly geared towards hardware, I think the furthest you go is nested if statements

Ithere was one video which sparked a love for geometry as stupid as it was, someone rendered a 3d donut in c

Which really gave me a different look into geometry

Here is the code I was talking about

Renders a 3d donut

[woeful frigate]

:O

the code itself is a donut

[crimson dust]

Which really demonstrate the power of geometry and how powerful someone good at it can be

Its all the way from 2011

https://www.a1k0n.net/2011/07/20/donut-math.html

Here is a detailed explanation

Yeah I think its just a general good example of how math and programming can go hand in hand

Dont really have much more to add other than I think its an excellent must read as the author explains the concepts at a rather low level and throughly

Ok looks interesting, also i forgot to ask are you in uni or already graduated?

If you don't mind me asking where do you go?

Ok

[icy leaf]

Hey..

[icy leaf]

Ya

Ye but i had just learnt ax2 + 2hxy + yx2 = 0

[upper karma]

a cos x + a sin x = a(sin x + cos x) , how to derive?

wdym how to derive?

you mean how to differentiate?

You derive it from distributivity and commutativity of addition if that’s what you’re asking

[sleek panther]

Ohh that's an accurate answer

🔥

[zinc egret]

is this some kind of joke or did you mean how to differentiate?

[onyx cloud]

from their replies above it seems they wanted to know how they factored that out

[sturdy umbra]

Michael from VSauce says triangles actually have four sides, is that true

[dark sparrow]

no

[ionic bluff]

a triangle is just a quadrilateral with a straight angle

[dire obsidian]

Can I get help w this

[crimson dust]

yeah

I can help in around an hour

@dire obsidian

[dire obsidian]

Oki

[alpine dove]

did you watched the video actually? its not hard to understand that it is just a meaningless trolling

[crimson dust]

is it 6.25?

do you have an answer key

[sturdy umbra]

Okay

I can't believe he deceived me

[crimson dust]

Are you trolling?

[sturdy umbra]

Trolling about being deceived or about the triangle being a square thing

[crimson dust]

Both

[sturdy umbra]

If brother_frost is correct that it was trolling, then I am not trolling about being deceived. The square thing, there is an video about it from VSauce's Michael.

[crimson dust]

Ok

[sturdy umbra]

What is this
https://youtube.com/shorts/cFd2c5txIvo?feature=share

Did I get click baited

[upper karma]

lmao

[dark sparrow]

did someone watch too much vsauce

[stuck turtle]

Can someone watch too much vsauce?

[nocturne remnant]

*vsauce music starts

[silk patio]

4th side… seems fun

I get it, nice

[whole shard]

Hi all, I have a right triangle. I know the coordinates of the two points A(x1,y1) and B(x2,y2). These points form the hypotenuse. I know the length of each of the cathetus (actually, they have the same length CLen). How can I calculate the coordinates(x3,y3) of the last point?

[dark sparrow]

there are two possible positions for the last point.

@whole shard are you ok with a little bit of complex numbers?

the positions admit a somewhat elegant description in terms of those

[whole shard]

@dark sparrow I'm a programmer and I don't remember the school days 🙂 So I just want to formula to solve my issue)

[dark sparrow]

welp ok

wait, hold on

you're given that AC = BC, and you even know their length?

as in, you have their length as an input or something?

cause it seems a bit redundant. if you have a right triangle, that length is just AB/sqrt(2).

[whole shard]

yeah. I know:

1. C = 90 degrees
2. AC=BC and I know the length of it
3. positions of A(x1, y1) and B(x2, y2)

I want to calculate the coordinates of C

[dark sparrow]

what happens if the length you know from (2) contradicts the positions of A and B from (3)?

[whole shard]

it's impossible I have a rectangle picture. I know the coordinates of the center and I know the coordinates of the top left angle of this rectangle.
one of the sides of the rectangle - my face. The middle of the face - the point C that I want to calculate

[dark sparrow]

also i have a high chance of being wrong here but do you happen to speak russian

[whole shard]

yes)

[dark sparrow]

а можешь показать ту картинку? может быть, все сразу станет ясно

[abstract forge]

I don't if this will help you or not, but your point C will be a point on the circle with AB as diameter

[whole shard]

саму картинку не могу, но попробую объяснить на похожей. Например, у нас есть картинка машинки https://img1.freepng.ru/20180421/ieq/kisspng-3d-car-simulator-driving-simulator-simulation-2d-furniture-top-view-5adacf2824eba9.1320720615242893201512.jpg
очевидно, у нее есть лицо - морда машины. Конкретно - в середине наверху если картинка повернута как показано сейчас.
картинка у меня всегда квадратная.
Я знаю координаты центра картинки
Я знаю координаты верхнего левого угла картинки
картинка может поворачиваться под разными углами
я хочу знать координаты лица в любой момент времени

[abstract forge]

Because angle subtended from diameter on segment is 90 degrees always

[whole shard]

I don't if this will help you or not, but your point C will be a point on the circle with AB as diameter

no, diameter AC

[dark sparrow]

а

то есть картинка поворачивается на произвольный угол, и нам требуется найти, куда при этом повороте отобразится середина верхней стороны

я правильно поняла?

[whole shard]

верно!

[abstract forge]

If C is the 90 degree vertex, wouldn't the diameter be AB?

[dark sparrow]

ага, ок, сек

(x1,y1) и (x2,y2) - левый верхний угол и центр, да?

[whole shard]

If C is the 90 degree vertex, wouldn't the diameter be AB?
if C - is 90 degree, AB - hypotenuse in this case

(x1,y1) - center
(x2,y2) - top left

[dark sparrow]

а, наоборот

угол поворота по часовой стрелке или против?

[whole shard]

в любую сторону может быть

[dark sparrow]

ну, понятно, что угол - величина со знаком

если угол положительный, в какую сторону должен быть поворот?

[whole shard]

вправо

[dark sparrow]

то есть по часовой стрелке

[whole shard]

да

[dark sparrow]

ок

$\begin{cases} x = x_1 + (y_2-y_1) \sin(\varphi) \ y = y_1 + (y_2 - y_1) \cos(\varphi) \end{cases}$

[somber coyote]

Ann

[dark sparrow]

тогда формула будет выглядеть вот так

фи - угол поворота

[whole shard]

спасибо большое! сейчас попробую

[abstract forge]

Wait, I think I got it:

1. Make circle equation with AB as diameter (x-x1)(x-x2)+(y-y1)(y-y2)=0
2. Get equation of perpendicular to AB line through midpoint of AB (centre of circle) (y-(y1+y2)/2)= -((x1-x2)/(y1-y2))(x-(x1+x2)/2)
3. These two equations will intersect at two points, which are the two possible positions of C

[whole shard]

@dark sparrow не подходит формула 😦

 var newX = car.x + (car.topLeftPosition.y - car.y) * sin(angleToRotate);
 var newY = car.y + (car.topLeftPosition.y - car.y) * cos(angleToRotate);

[dark sparrow]

из видео как-то не очень понятно, почему не подходит

[whole shard]

я нажимаю кнопку, туда поворачивается машинка
когда поворот окончен, она прыгает на новое место
в этот момент я знаю координаты верх лево и центр
так же знаю длину катетов
и в этот момент формула выше начинает считать новые координаты

[dark sparrow]

насколько я вижу, машинка прыгает на новое место в нужной ориентации

[whole shard]

да, лицо - это зеленый квадратик
и после прыжка желеный квадратик уходит в непонятную позицию

[dark sparrow]

так

а может в прыжке дело?

может быть имеет смысл хранить вместо абсолютной позиции "лица" его положение относительно центра?

то есть $\big( (y_2-y_1) \sin(\varphi), (y_2 - y_1) \cos(\varphi) \big)$, если использовать те же обозначения, что и раньше

[somber coyote]

Ann

[dark sparrow]

и прибавлять это дело к новому положению машинки после прыжка

может быть даже лучше это расписать через длину изображения машинки

[whole shard]

@dark sparrow спасибо за помощь
я другим способом, но решил свою проблему

[dark sparrow]

рада была помочь

[feral musk]

can someone help me to find what "x" is?

[upper karma]

have you tried anything?

@feral musk

[feral musk]

yes, but i don't really find the correct trigonometric identity

my teacher said i should take this into account

[upper karma]

i mean you can consider the trig addition and substraction formulas

the identities are pinned at this channel too.

[feral musk]

oh yes, i didn't think of that

thank you

[north cosmos]

can someone help me solve this 👀

[trim breach]

A psychiatrist can.

[weary drift]

@north cosmos don't shitpost

[dire obsidian]

@crimson dust yes it is 6.25

If you don’t mind walking me through it, that would be great

[crimson dust]

ok so just use ratios

3:4:5

for the traingle

[dire obsidian]

heh i havent taken geometry in 4 years

so i kinda need someone to hold my hand

LOL

[formal tartan]

Did you solve it?

[vestal lily]

@feral musk

[silent plank]

what have you tried?

[edgy juniper]

the question is “describe the shaded region”

can someone help with this, i got (b) correct, A’ intersection B
and when i got to (d) i thought it was the same, so i put X intersection Y’
but the book tells me im wrong on (d) and that the correct answer is X’ intersection Y
i thought (b) and (d) were the exact same..? any explanation?

[keen nova]

I'm pretty sure you're right

looks like a typo

[molten delta]

On the bottom example

Why isn’t sin just 2/29?

Because sin is O/H and the O from tan is 2 and the H from cos is 29

[dark sparrow]

no

[molten delta]

So shouldn’t it be 2/29?

[dark sparrow]

by that logic you could manipulate sin(x) into being almost any value you want by changing the way 2/5 is written

what if it was 10/25 instead, or 60/150, or 420,690/1,051,725?

if all you have is the trig ratios, even if you imagine the angle as part of a right triangle it is impossible to get the length of any one side

[molten delta]

Oh

I think I get it now

You need the length of at least one side

[dark sparrow]

yes exactly

like, you can fix the length of one side in your triangle if that makes it easier for you

just be explicit about it and stick with your choice

[molten delta]

Ok ty

[placid gate]

help?

[molten delta]

Length of the the base is 3, and the length of the height is 4

And area of triangle is bh/2

So (3*4)/2

[outer ember]

Yep

[placid gate]

how?

[molten delta]

Not the best visualization but I hope it helps

[placid gate]

B and C is 4

how is that 3

[rare sierra]

because the actual triangle itself starts at 1

[silk patio]

It’s obvious

Consider areas

What does GD/AD represent

Yeah, that’s what it is

Nice

Simplicity is an artform

[dire obsidian]

@upper karma yesh I know those fundamentals

I’ve seen people solve this problem multiple ways

And I just wanna see how others do it to see what works for me or clicks the best w me

[outer mural]

Anyone got a clue if this is trig? I feel like it's a scalene triangle and I got no clue how to solve for one of those with only this information

[pure cape]

seems like its missing information, does it mention anywhere else in the question at which height the person is standing at?

[outer mural]

nope, that's an entire screenshot of all the informatin

I think my professor forgot to proof read this question because there's absolutely no way to do trig on this problem without atleast one side of information (more than what there is already). Guess I'm SOL until the lecture tomorrow

[pure cape]

welp its missing information then

[outer mural]

Gotta love college, paying to be confused XD

[pure cape]

or you can call an arbitrary variable for the height at which the person is standing at and find the distance with respect to that variable

[outer mural]

wdym arbitrary variable? is there a way to find the height of the 15 degrees and the 4 degrees in some sort of ratio that adds up to 170

^^only thing i can think of

[pure cape]

arbitrary variable as in unknown variable

and there is no way to find the height with only these informations

however you can find an equation in terms of the distance between the building and monument and the height at which the person is standing

[outer mural]

I'm guessing that's not part of trig tho, this is a trig class so if that's the case then the question just doesn't have enough information to complete and that's okay.

[pure cape]

yeah, im just saying

[outer mural]

no worries man. thankyou for your time 🙂 '

[sleek cargo]

anyone know how to do simple geometry proofs?

[nocturne remnant]

What problem are you encountering in particular

[sleek cargo]

Everything

[keen nova]

you're gonna have to be more specific or at least give an example if you want us to be able to help

what so unrigorous about HS geometry?

*unrigorous

[trim breach]

What part of the area are you stuck on?

[upper karma]

math is painful

[trim breach]

Properties of a kite.

[upper karma]

um idk the properties of a kite!

[trim breach]

It is easy to visualize if you extend ZY and RS.

[upper karma]

um im not allowed to do that!

[silk patio]

Congruent triangles

[upper karma]

uhh so like

tso and xoz?

umm

are any two distances from the center to the chord equidistant

[upper karma]

<@&286206848099549185>

[proper totem]

@upper karma perpendicular from centre of circle bisects a chord so YZ=RT and as OT and OY are 5 so using pythgrs theorm u get ro=oz , u can also use SAS congruency (yz,90,oz = tr,90,or)

[upper karma]

thank u

uhh can u help with this too

nvm

[ocean snow]

<@&286206848099549185>

[proper totem]

here find the area of the sector subtract the area of triangle and add the area of trapezium .. here the non parallel sides are tangent to circle so the radius at the point where tangent meets curved part is perpendicular to tangent

[upper karma]

<@&286206848099549185>

[silent plank]

what have you tried

[proper totem]

base perimeter = the perimeter of the sector (240).. u can find radius of base and using pytgs theorem .. u can find height

[north cosmos]

🤓

[upper karma]

💀 I er ended up only figuring out that once the radius coincided that the sector could be used to fill 2/3s of the bottom part of the cone so I realized I needed to use s=rtheta and that I needed to convert the 240 into radians

[upper karma]

can you explain in more detail if possible

[silent plank]

did you find the arc length of your sector?

[upper karma]

ye

I got 4pi/3

oh wait no

i got 12pi

4pi/3 was what I got from converting the degree measure to radians

and than I just used the formula S= r *theta to find arc length

[silent plank]

So you see that this will be the circumference of the base of your cone

[upper karma]

oo so I just find the radius than right

[coarse raptor]

Hello

I have this inequation to solve in $[0;2pi]$

[somber coyote]

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

[coarse raptor]

$\cos(x)+pi/3<-1/2$

[somber coyote]

σρhιδιοη

[coarse raptor]

I found this , it’s right ?

[keen nova]

just say it in text

[coarse raptor]

Ah te « ,ask cosx ….. » ?

[keen nova]

to confirm, you wanna find the solution to $\cos(x)+\frac{\pi}{3} < -\frac{1}{2}$

[somber coyote]

Ninja

[keen nova]

where x is in [0,2pi]?

[coarse raptor]

,ask$\cos(x)+\frac{\pi}{3} < -\frac{1}{2}$

[somber coyote]

Results provided by WolframAlpha

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

Requested by σρhιδιοη#0111

[coarse raptor]

X IS IN $[\frac{\pi}{3};0]$ !

[somber coyote]

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

[coarse raptor]

I will prove it lol

[dire obsidian]

How do you find the radius

It’s either C or D

[silent plank]

distance formula

[dire obsidian]

😆 I keep thinking it’s

Sqrt x_2+x_1

But it’s a

[dire obsidian]

Can you show how it’s done for distance formula

I keep@messing up

[upper karma]

Is it d

[dire obsidian]

Ya I figured it out

[umbral viper]

How do I do this question?

"A is the top of a vertical radio mast AB standing on level ground. Two points C and D are on ground level such that C is due east of B and D is 500m due north of C. The angles of elevation of A from C and D respectively are 11° 13' and 8° 14'. Calculate the height of the tower to the nearest metre?"*

[nocturne remnant]

Set height = h and BC = c, BD = d
You have:

• the elevation angle relating x and c
• c^2+500^2=d^2
• the elevation angle relating x and d

3 unknowns, 3 equations baby

[upper karma]

Are there anyone here interested in geometry???: ) If yes, dm me.

[keen nova]

dm you why?

[upper karma]

actually i grow a geo server , so ....

[keen nova]

that's server advertisement 👀

[upper karma]

yes i can't post it here

so ,...

[keen nova]

well like, idk if talking about it is much better honestly

[molten delta]

Can someone explain to me why the csc(-30) = -2?

When ever I do it I end up with $\frac{2\sqrt{3}}{3}$

[ocean seal]

hi, we speak spanihs?

[somber coyote]

Crimson Blitz

[dark sparrow]

are you sure you are not calculating sec(-30°) by mistake @molten delta

csc(-30°) = 1/sin(-30°) after all.

[molten delta]

Oh

[ocean seal]

I need help, but I don't speak English

[molten delta]

Yeah I mix those up all the time

Ty

[ocean seal]

Determine the trigonometric ratios of the angle in normal position that intersects the circumference at the point
indicated

How is it? or how is it done?

[quiet blade]

When you draw the unit circle, you will see that those points lie on that circle

Now, starting from the coordinate beginning, and taking the positive x-axis, you draw the angle between the x-axis and the point

in general, you can see that all points that lie on this circle can be written down as ( cos(x), sin(x) )

where x is the angle in question

[jovial axle]

i have i think a dumb question

but im struggling with graphing $h(x) = 2 \sin (x) - \sin(2x)$

[somber coyote]

jan Niku

[jovial axle]

is there a nice way to do this through identities or is it just a find the points you can know and fudge the curve between kind of thing

actually thinking now thats always what you do

but maybe if i could get this looking like something a little more normal

[rugged trail]

sin(2x) = 2sin(x)cos(x)
=> h(x) = 2sin(x) - 2sin(x)cos(x) = 2sin(x)(1 - cos(x))
and from that you can find zeroes more easily

where sin(x) = 0 or cos(x) = 1
x = pi * n or x = pi * 2n
so just integer multiples of pi

[jovial axle]

i guess im more looking for characteristic points

i think ive found the 0s

just need to remind myself of some identities to sort out the max

oh, its just a zero factor thing again

[rugged trail]

You can use the first/second derivative

[jovial axle]

$x=\frac{2k\pi}{3}$ are the candidates

[somber coyote]

jan Niku

[jovial axle]

but in my interval is just 2pi/3 and its negative

nice

thanks 🙇

[jovial axle]

fortunately my interval is only -pi to pi

[rugged trail]

One of the latter two is the maxima, and the other the minima, I'll leave finding that to you

[keen wind]

So I was reading this out of interest and now I'm curious

When multiplying matrices, why would be used dot product instead of something else?

I literally just learned that when adding matrices, they have to be the same size, and I think I understand why. However for multiplication, why is it done like that? From what I know dot product is another way of saying the f(g(x))

[dark sparrow]

how much do you know about vectors?

if you know your way around vectors you may want to watch 3b1b's essence of linear algebra, he explains the motivation behind matrix multiplication

[keen wind]

So right now, I'm in college calculus 1 and calc based physics 1, so I'd say probably very little to nothing

[dark sparrow]

right

matrix multiplication is kinda hard to motivate at your level then

[keen wind]

I feel like I kind of get the gist, because someone recently explained to me how matrices are used in certain computer science subjects like ML and the spinning donut created in 2006

[keen nova]

3b1b's essence of linear algebra is good for understanding this

[upper karma]

<@&286206848099549185>

[keen wind]

Alright thank you!

[severe python]

What is the surface area of this ? I got 3990Pi but The closest choices available are 4090pi and 3890pi ?!

<@&286206848099549185>

[dark sparrow]

how did you get 3990pi?

[severe python]

2.25.pi(40+25)+2.10.pi(32+10)-10.10.pi

sorry it's a bit messy without math keyboard

[dark sparrow]

...

are those periods meant to represent multiplication

[severe python]

yeah

[dark sparrow]

,calc 2 * 25 * 65 + 2 * 10 * 42 - 100

[somber coyote]

Result:

3990

[dark sparrow]

well your arithmetic is okay at least

i see the issue though

you should have subtracted pi*10^2 twice

because that area where the cylinders touch was counted in both the lower cylinder and the upper cylinder's surface areas, but it's not part of the whole thing's surface at all

[severe python]

hmm, why ? isn't only the bottom surface that should be ommited ?

[dark sparrow]

no see the thing is

imagine gluing this shape out of two paper cylinders

[severe python]

ahh I understand

[dark sparrow]

and you need to make it so that there's no paper walls inside

[severe python]

the same area cut of the big and the small cylinders

[dark sparrow]

you need to remove the bottom lid of the small cylinder and cut out a circle from the top lid of the large cylinder

thus subtract 2 * 100pi

[severe python]

I see, thx a lot mate !

[covert pond]

Greetings everyone! Since there doesn't seem to be an introductory channel, I am posting this where it may be relevant. I am a mathematician/scientist and I am seeking anyone who may be interested in a serious examination of the Goldfinch Conjecture. You may respond to me here or message me directly. Hope everyone is having a wonderful day! While I realize this channel is for pre-university subjects, I am not yet able to post in the higher math channels. And, as some of what I have to discuss has to do with geometry, I am posting this here.

Also, the following geometry problem is from a college level course. I am issuing a challenge to provide a satisfactory solution to 3.8 in the following:

[old fable]

How do I do 1.5

[humble pulsar]

what have you tried?

[old fable]

I tried out writing everything out and trying to see some pattern

But without much luck

[humble pulsar]

$\norm{A-B}=\norm{A+B}\iff \norm{A-B}^2=\norm{A+B}^2$

[somber coyote]

Mosh

[humble pulsar]

since norms/lengths are non-negative

[old fable]

Ok so what do I do next

[humble pulsar]

well what do you know about the norm of a vector squared?

[old fable]

It’s the dot product with itself?

[humble pulsar]

$\norm{u}^2=u\cdot u$ yes

[somber coyote]

Mosh

[humble pulsar]

so you have $(A-B)\cdot(A-B) =(A+B)\cdot(A+B)$

[somber coyote]

Mosh

[humble pulsar]

both sides you can expand w/ properties about the dot product

[old fable]

Which property do I exactly use then

[humble pulsar]

"you can expand"

dot products distribute as you'd expect with FOIL

[old fable]

So I calculate te magnitude of A - B right

[humble pulsar]

no

there's nothing to calculate

A and B are generic vectors

[old fable]

Oh I just foil

[humble pulsar]

Yes

[old fable]

Thanks

[delicate lintel]

im going into geometry, can anyone make practice problems

[analog owl]

Uhhhh

[river stream]

there are books

[shell fulcrum]

why is there a timer on your screenshot?

[silk patio]

They’re speedrunning their homework

Trying to find new strats to beat their pb

[delicate lintel]

what are proofs pls help

kinda confused

[silk patio]

Arguments for why something is true

[glad bluff]

Hola! quería preguntarles cómo hallarían una formula general para la relación entre el área de dos polígonos irregulares? Es decir, una fórmula que con sólo los datos del primer polígono, pueda calcular el área del proporcional...

[silent plank]

English?

[glad bluff]

Hi there! ask them how they would find a general formula for the relationship between the area of ​​two irregular polygons? That is, a formula that with only the data of the first polygon, can calculate the area of ​​the proportional ...

[silent plank]

are you told anything at all about the second polygon?

like is it an unrelated shape and has nothing to do with the first?

oh, the sentence is cut off at then end

the polygons are similar

whether the polygons are irregular doest matter

if the proportion of their sides is
1:a
the proportion for the area will be
1:a^2

[glad bluff]

look...

My task is to find an equation that calculates the area of ​​the blue polygon, knowing the are of the red one

[glad bluff]

Yes, I had thought something like that

I thought about finding the constant of proportionality (k), and doing:
Blue area = (Red area / k^2) -Red area

is okey?

the task is to find the area of ​​the blue polygon that surrounds the city (red polygon), which is where you cannot spray pesticide

[dark sparrow]

is the blue polygon a constant width around the red?

@glad bluff

[glad bluff]

yes

has to be a thousand meters away from the red sides

if it's difficult with meters, let's use another scale

[dark sparrow]

lmao no metric best

[nocturne remnant]

“Let’s use feet”

[dark sparrow]

i think the amount of added area ought to be approximately 1km times the perimeter of the red

[glad bluff]

okey

thanks!

[kind rain]

Can someone help me with this I do t know what equation to use?

[dark sparrow]

you do not use any equations

do you know what it means when an angle is bisected?

[kind rain]

Nah

[dark sparrow]

to bisect something means to split it into two parts of equal size.

[kind rain]

So PQR is 126?

[dark sparrow]

indeed.

[kind rain]

Alright I need help with another problem though.

I get to dividing then it doesn’t make sense

[dark sparrow]

"get to dividing"?

so you have some work to show?

[kind rain]

Yeah first you subtract 13

Then it leaves you with -16x=177

[dark sparrow]

...can you show what equation you started with?

before you subtracted 13 as you said

[kind rain]

(-16x+13)°=180

[dark sparrow]

this is wrong.

[kind rain]

How so?

[dark sparrow]

you basically just said that angle LMP takes up the entire 180 degrees

which is not the case, as can be seen from both the diagram and the prompt (angle LMN is a straight angle)

[kind rain]

Yeah which led me to think it was 180 so is the angle 90

[dark sparrow]

no it's not 90 degrees either.

[kind rain]

Because I’m trying to find LMP

[dark sparrow]

you're trying to find x first and foremost.

once you have the value of x, the angles themselves are not hard to calculate.

[kind rain]

So what should I do?

[dark sparrow]

consider that angles LMP (-16x+13 degrees) and PMN (-20x+23 degrees) add up to 180 degrees.

[kind rain]

Yeah so I need to solve the equation to find the numbers but I need it to = something to find it

I don’t know what it equals

[dark sparrow]

read what i said again.

[kind rain]

Yeah so wouldn’t the equation be (-16x+13)°=180? I don’t understand what the number should be in place of 180

[dark sparrow]

why are you insistent on throwing angle PMN in the trash?

what's stopping you from writing down (-16x+13) + (-20x+23) = 180?

[kind rain]

Ok

[dark sparrow]

no seriously

has it really not occurred to you even after i said it almost explicitly?

i even bolded the words add up.

[kind rain]

I got 36

And 144

[dark sparrow]

i'm getting the feeling that my questions are getting ignored.

[kind rain]

What

[dark sparrow]

what's stopping you from writing down <equation>?
you could have at least said "i have no idea" here

[kind rain]

I wrote it down and I got 34 and 144

Is that right

[dark sparrow]

did you mean to say "i got LMP = 36 and PMN = 144"?

[kind rain]

Yeah

[dark sparrow]

no, that's not right.

if you could show your work, i could point out where you went wrong.

i must emphasize that i want you to show, not tell.

so a picture would be nice.

[dark sparrow]

missed minus sign in **-**20x + 23

||and even disregarding that, "4x = 144" does not mean "angle PMN = 144"||

[kind rain]

Well now I have -36x=144

[dark sparrow]

that's correct, now solve for x.

you're one step away from it

[kind rain]

So -36 divided by 144?

[dark sparrow]

think about that for a moment

[kind rain]

Multiplication?

[dark sparrow]

i too can say words.

[kind rain]

Well I don’t know

[dark sparrow]

no, the solution to -36x=144 is not x = -36/144.

if you had the equation 2z = 10, how would you solve it?

[kind rain]

Divide 10 by 2

[dark sparrow]

great

so why not do the same here

[kind rain]

Ok

[dark sparrow]

you have the unknown multiplied by a number

[kind rain]

So I just did -36x/144 and got -0.25 that’s where I’m stumped

[dark sparrow]

........

okay so you haven't been reading the conversation that happened between us right here right now

like at all

the kind of mistakes you keep making cannot be explained in any other way, except maybe severe absentmindedness

sorry but i can't continue like this

[kind rain]

I mean just tell me what to do

No need to try to lead me on

[dark sparrow]

if all i do is tell you what to do, you will learn nothing

and we can't have that

[kind rain]

No cause then I’ll know what i did wrong

[dark sparrow]

"do something wrong" does not mean "stray from a strict list of steps somebody gave you"

in any case i need to sleep right now so

[kind rain]

Well can you tell me to divide or multiply because I don’t know what to do

[dark sparrow]

okay, so taking into account your inability (or unwillingness) to draw analogies...

to solve the equation -36x = 144, divide both sides by -36 to get x = 144/-36.

[kind rain]

And then I get x=-4 I’m so confused

[fallen girder]

Now plug x into your original angle equation to solve for LMP and PMN

You found x, now use it

[kind rain]

So would the equation be (-16+-4+13)

(-20+13)?

Or (20+13)=180?

[fallen girder]

What is your starting equation?

Which angle will you do first?

[kind rain]

The starting equation is (-16x+13) and (-20x+23) and it says LMN is a straight angel.

https://cdn.discordapp.com/attachments/326138757474680852/881973665103773767/image0.jpg

[fallen girder]

Yes. Since the first one is LMP then let’s do that first

What is LMP’s equation?

[kind rain]

So (20x+13) because you add -4 which makes it a positive?

[fallen girder]

Look at your picture

What angle corresponds to LMP?

Left side or right side?

[kind rain]

Left

[fallen girder]

Yes

What is the equation for the left side?

[kind rain]

(-16x+13)

[fallen girder]

Yes

Now plug in x

[kind rain]

So (-20+13)?

[fallen girder]

Write it out

[kind rain]

Or should I do (-16-4+13)

[fallen girder]

(-16*x +13)

(-16*-4 +13)

[kind rain]

So -16 times -4?

[fallen girder]

Do you know order of operations?

PEMDAS or whatever it’s call now

[kind rain]

Oh ok

So

-16+9?

[fallen girder]

No.

Parenthesis, exponents, multiply/divide, add/subtract

You definitely multiply first

[kind rain]

So 4+9

[fallen girder]

That’s right

[kind rain]

So 15

[fallen girder]

Yep

[kind rain]

That makes no since though because it’s out of 180 and no way 15 is one of those angles

Am I missing a step?

[fallen girder]

#1: I said multiply and you add

#2: 4+9 is not 15

Not sure if you’re trolling or what

[kind rain]

So 45?

[fallen girder]

Yes

@kind rain did you get it right?

[kind rain]

I don’t know

I still have 8 questions left

[fallen girder]

Okay good luck

[old fable]

Can I get some help with the question above 1.16

[old fable]

<@&286206848099549185>

[rare sierra]

e

[upper karma]

help

"Calculate the x and y values ​​on the EFGH parallelogram"

[livid moss]

Don't multipost and I already told you to use the fact that opposite angles are equal.

If you are still stuck, explain what the problem is

[upper karma]

i'm not good at this kind of math please try to help me

[onyx cloud]

do you know what the sum of all of the angles in a parallelogram is?

[upper karma]

no

[onyx cloud]

the sum of all the angles in a parallelogram is 360 degrees

can you write an equation for me now that you know that?

[humble pulsar]

I'd say what Luna had (already) said is more helpful imo

[onyx cloud]

oh i see

[humble pulsar]

rather than directly using angle sum

[onyx cloud]

i thought opposite angles were in terms of 2 variables

[humble pulsar]

nope opposite angles gives 2 1-var eqn's

[onyx cloud]

yup i see that now, i thought it'd end up being a system of equations

@upper karma

do you understand what they said

when they told you opposite angles are equal?

[kind rain]

I need help with this

[rare sierra]

hm

[livid moss]

#❓how-to-get-help @kind rain

[rare sierra]

Well I'd say go and add up all the parts

and then solve with 90 Degrees s the answer

as*

@kind rain m<RSQ= 32 Degrees I think

[kind rain]

Thanks

[rare sierra]

np

[kind rain]

How did you get that though?

[rare sierra]

well you need to first add up everything there

and since you know that the total sum of degrees has to be 90

you can sort of mix and match the answer

You should get 23x-43+18=90

then you can simplify it 23x+25=90

or wait

I think I messed up

hold on

wait no I didnt

so 23x-25=90

then you can add 25 to both sides to cancel it out

23x=115

115/23

and that gets you how much x is worth

then you apply that to the angle

and solve

[kind rain]

Ok

[rare sierra]

tell me what you get for TSQ

and ill compare

[kind rain]

58

Can you help with this one @rare sierra

[rare sierra]

Sure

alright so this one isnt a right angle

but we know both sides must be equal by definition of an angle bisector

you follow?

[kind rain]

Ya

[rare sierra]

ok

so we need to find the degree of one of them first

lets start with ABD

well its pretty obvious we can't really simplify that one

same as the other

hm

i need think

ok got it

we can set it up as 6x+14=3x+29

got it?

[kind rain]

Ya

[rare sierra]

both are equal

so then we do same as other

cancel out the 6x by subtracting it on both sides

14=-3x+29

-29 on both

-15=-3x

simplify it and cancell out the negative

x=5

got it?

[kind rain]

Ya

[rare sierra]

ok

so now that we know x=5 we can impliment that in

so for abd we can go 6x5+14 which is 44 I think

so abd=44 degrees

DBC is

15+29= 44 too

so both are 88

got it?

[kind rain]

Ya

[rare sierra]

alright and thats all the answers

tell me if ya need me again

[kind rain]

Ok

[upper karma]

Can someone help me with a basic geometry problem

[pure geode]

tell me

[ashen stratus]

How does one draw this

<@&286206848099549185>

[pure geode]

for example

xd

or also like this

with the dot

different examples of rays:

[pure geode]

point A, ray B

[pure geode]

point P and Q, ray r also called PQ (with arrow on top)

[pure geode]

point R and N, ray RN (with arrow on top)

if im not wrong xd

[ashen stratus]

Yes thank you, this was for a friend

We drew the same thing almost

[pure geode]

ok nice

[ashen stratus]

Yesyes

[acoustic wadi]

does complementary have to be 90

[pure geode]

complementary angles add up to 90º and supplementary angles add up to 180º

[acoustic wadi]

okay thank you

[pure geode]

np

[acoustic wadi]

taking class online and cant ask if it has to be 90 or if they can add to anything else

ty

every time my prof says 90 he says real quietly 90 in a high pitch voice

[harsh plaza]

Anyone on

That can help

[nocturne remnant]

[kind shoal]

JUST ASK

!

[untold fossil]

Please help

[nocturne remnant]

Do you happen to know that triangle’s area can be given by
1/2absinC
Where C is the angle included in lines a and b

So our problem simplifies to proving
2BPxBQ=BAxBC
Which should be simple enough using stuff like intercept thm

[fallen girder]

Did you draw a picture?

[storm portal]

@upper karma ^^

[upper karma]

Uhh no

Idk what to do

[fallen girder]

Start with a pic

[wispy fulcrum]

Hello

[heady wave]

I believe the translation vector is how many units right and up it is shifted

Part B is the coordinates of each points after the translation is performed

[wispy fulcrum]

@heady wave

IDK part b

[heady wave]

graph the new triangle after the translation has been performed

[wispy fulcrum]

?

[ocean snow]

On the line joining (4, −5) to (−4, −2), find the point which is
three-seventh the distance from the first to the second.

[dire copper]

AB ┴ BC
|AB| = |BC|
what is the area of the shaded region?
Can anyone help?

[nocturne remnant]

Some cutting and rotating could probably be useful

Nice drawing

[arctic prawn]

need x and e

tried it multiple times just dk how to solve it -.-

[flat tartan]

Is there an intuitive way to remember trig functions ?

The only one that’s intuitive for me is that sin(x)/cos(x) = tan(x) because I know sin(1)/cos(1) = 0, and sin(pi/2) / cos(pi/2) = undefined

[fallen girder]

Soh cah toa

@flat tartan

Or do you mean something else?

[flat tartan]

Maybe it’s not called trig functions

I mean things like tan(x) = sin(x)/cos(x)

These

Maybe this is more of a precalculus question

[umbral scarab]

yea its more precalc

[fallen girder]

I would say you just need to memorize the identities needed for your class. However sin^2 + cos^2 =1 is all you need for that one. Like if you start with it and divide every term by cos^2 then it becomes tan^2 + 1 = sec^2

[tidal marsh]

Helloooo

Anyone here ?

[cloud moth]

ohhh

[tidal marsh]

The hexagon in the figure has 6 angles equal to 120 "and four of its sides measure 5, 8, 4 and 11 like go in the figure Determine the perimeter of the hexagon.

Help

@cloud moth

;-;

[cloud moth]

im not good at geometry

sry

[tidal marsh]

ª

[valid harbor]

@arctic prawn do you still need help

[acoustic wadi]

Every time my trig prof says 90 he says 90 again but in a high pitch quiet voice

Anybody know what I’m talking about or why

[marble thistle]

Yo I got an easy geometry question that I just seem to not answer correctly

Numero 96

[acoustic wadi]

Can someone briefly explain standard position

[brittle wasp]

Can someone help me with my geometry homework please

[valid harbor]

@marble thistle sum up all of the angles and make them equal to 180

[marble thistle]

I did

but the math doesn’t work

4x^2+4x+55=180

4x^2+4x-125=0

U can’t factor this

The answer key says it’s 5,-7

@valid harbor

[valid harbor]

sir

it is $55*2 + 2(x^2 + 2x) = 180$

[somber coyote]

Pealover

[valid harbor]

@marble thistle

[marble thistle]

why are u multiplying

[valid harbor]

$<XYW = 2<XYT$ and $<ZYW = 2<ZYU$

[somber coyote]

Pealover

[marble thistle]

OHHHHH

wait

no

so 2x^2+4x=55

so u substitute that

2x2+4x+125=180

[valid harbor]

that's very false sir

@marble thistle

[marble thistle]

I don’t understand

[marble thistle]

that’s exactly what this is

Wait I added wrong

[valid harbor]

how do you deduce that $2x^2 + 4x = 55$

[somber coyote]

Pealover

[marble thistle]

U can’t

[valid harbor]

what

[marble thistle]

2x(x+2)=55

2x^2+4x-70

2(x^2+2x-35)

(x-5)(x+7)

wow

@valid harbor thanks

[valid harbor]

np

[upper karma]

HI

[umbral snow]

Hi joseph stalin

[whole shard]

Hi all, I have 2 circles. I have coordinates of these circles and I have a radius of each circle. I need to calculate if there are any intersections between them. Is there a formula to calculate it?

nevermind. found the solution

[dark sparrow]

find the distance between the centers and compare it to the sum of the radii lol

circle-circle collisions are the easiest thing to test for

[whole shard]

yeah, I did it exactly this way

[upper karma]

Can someone help me with my geometry homework

[dark sparrow]

@upper karma do you still need help with your geometry homework

if so then you really should have posted it outright

since you know

http://dontasktoask.com/