geometry-and-trigonometry

[quiet mason]

lol

[cinder portal]

if you care a LOT about geometry, then make sure to pay attention a LOT, it's probably going to be like the only geometry math class you'll take throughout all of high school (well assuming your school doesn't offer any more geom courses)

[fossil ledge]

Only geometry course but you will end up doing more geometry in your other courses so it’s important to pay good attention

[small raptor]

I think it's a good idea to study as a whole as everything in life can be abstracted into a shape of some sort

Specifically circles and squares/triangles

[quiet mason]

...

[fringe dirge]

What

[small raptor]

I'm speaking probably from an artist background. Everything eventually starts as a shape.

You can literally construct a head with a circle and triangles

[fringe dirge]

What

[quiet mason]

answer my question then whats a shape

as im going towards higher leveks of sciences

i realize theres nothing known as an explanation

its an infinite deal of nothing

and even after knowing everything we dont know anything

[small raptor]

I rather eat ice cream and brownies. 😦 Just look up any tutorial of figure/portrait drawing. The usual start is a circle, lines, then you mark the nose from the mouth/chin with a triangle. The eyes another circle, and you can even do an upside down triangle for the pupil so you can line up the eyelids of the eye.

At least, this is how I do things when I'm proportioning my drawings

Not perfect, but circles, triangles, and rectangles. Looks like Cell or someone wearing a helmet due to the rigid nature, but you see the general construction.

If you were to go very strict, you could probably use the properties of geometry, specifically Triangles and Circles, to get exact measures/proportions

post

So, you can still see the shapes position and how it relates with more 'concrete' aspects of life

There are other mathematical properties attributed in art itself like the Golden Ratio which has Geometry foundations

[upper karma]

u guys are gonna be my help for geometry thank god i joined this

i had a rlly bad time in algebra

i got a B in the second quarter and C's in the rest

[stuck torrent]

Oof

[covert rune]

@cinder portal Wouldn’t it be easy to review geometry later on?

Maybe an online course or something

[cinder portal]

Mmmmm maybe? I wouldnt recommend it tho, stick with your hs curriculum, and if you really love it, pursue it in undergrad

[glad snow]

Given a 2D image A. The content of the image is the position and orientation of objects in the whole image. The style of the image is the color and smoothness of shapes in a small area. The problem is that the function f(image) is sensitive to the angles of style. How to create a fake image B such that it has both the content and style of A but the style is generalized/normalized with respect to angle

[lusty quest]

@small raptor thats pretty cool

[half saffron]

Are rhombus's parallelograms?

[dark sparrow]

rhombi

the plural is rhombi

[half saffron]

oh

Thank you but are all rhombi technically parallelograms?

[dark sparrow]

idk are they

why don't you attempt to prove it

[half saffron]

Hmm my only response is I'm unsure of the official definition of a parallelogram. Given a parallelogram ABCD, is it required AB≠CD in order for it to be parallelogram?

[fossil ledge]

Well is a square a parallelogram

[worthy root]

For a quadrilateral to be parallelogram it's opposite sides mist be parallel .

[quiet mason]

no

[half saffron]

I think a square is just a rhombus..? Which is a parallelogram?

[sly marlin]

yes

a square is a special rhombus, which is a special parallelogram

It would be annoying to say "parallelogram or rhombus or rectangle or square" when you are talking about all parallelograms

[cloud merlin]

Cos of 51 times 58

[devout shell]

Yes, that’s the answer

[summer plume]

I don't understand, how did they get the 0.9?

[stuck torrent]

You see this point?

That is the point if you rotated it at 0.4 radian

Do you agree? @summer plume

[summer plume]

I do

[stuck torrent]

Ok

In the unit circle, what is the cos of an angle

[summer plume]

Adjacent/Hypotenuse?

[stuck torrent]

That is a definition of cos, but not the one I was asking for

What is cos of an angle in an unit circle

[summer plume]

Depends on the degrees doesn't it? If it's 30, sqrt3/2. 45, sqrt2/2. 60, 1/2. 90, 0.

[stuck torrent]

Well, that the cos of specific angles

I'm talking about general definition

The cos of an angle in an unit circle

See if you have notes on that

[summer plume]

The only thing I could find in refrence to The Cos of an angle in an unit circle from my note and my book was the section talking about Circular Functions. Where it says Cos s = x and that later leads into specific angles and the connection of x^2 + y^2 = 1 to cos^2 s + sin^2 s = 1

If you're meaning to tell me that I take that function and plug things in, then why is it 0.9 and not just 9? Because the radians are spaced between two-tenths?

[stuck torrent]

Um

The cos of an angle in an unit circle

You take a point, and rotate it for theta radian

Then the y value of the final point is the sin(theta)

The x value of the final point is cos(theta)

[summer plume]

O

k

[lusty quest]

in this question

the formula contains "sin2pi"

wouldnt that turn everything 0?

[tawdry pivot]

no, it's $i=I_{max}\sin(2\pi ft)$

[somber coyote]

EpicGuy4227:

[lusty quest]

so how would i figure out t then since t is in the sin func

hmm interesting

[cloud merlin]

Look at the bottom the cos of an angle is the bottom of the triangle like the x axis

You see how it’s at 9

That’s why it’s 9

So just look at the .4 mark and look at the x axis

[dark sparrow]

by the looks of it

it's not 9

but .9

aka 0.9

but written in that nasty no-leading-zero format

[cloud merlin]

Oh yea

It wouldn’t make much sense to be 9 since it’s a ratio or fraction

[dark sparrow]

no, it wouldn't make much sense to be 9 because it's a unit circle and so the thing is definitely less than 1

"fraction" and "less than 1" aren't synonyms!!!!!!!!

[keen aspen]

:Anngery:

I really want this to be an emote

[autumn tapir]

How am I supposed to do this?

I tried expanding cos and sin but it ain't working

Cant get rid of the weird thetas and 4√2

And oh btw

You're supposed to find general solutions for the given equation

[narrow sleet]

Hint: Let x = 5(theta)+2

[autumn tapir]

Ooooooh

That is brilliant!

Lemme try that out

Will 4√2 get sorted too?

[narrow sleet]

You will see

[autumn tapir]

Got it!

Thank you so much wolf!

A quadratic was formed

And that gave me the solution

[narrow sleet]

👍

[tribal sun]

J’ai une question ?

[hard gale]

non

enfin vas-y pose la par contre je suis assez busy là

[tribal sun]

J’arrive pas à trouver l’angle ω

J’ai

-> cos ω=−√5
->sinω= (√5)/2

Vu qu’il n’est pas dans la table trigo je rame un peu .. merci de votre réponses à l’avance

[hard gale]

cosinus = -sqrt(5) < -1

tu vas avoir du mal à trouver un angle avec ce cosinus

[tribal sun]

Oui c’est pas normal y’a une erreur quelque part

C’est défini sur -1 et 1

[hard gale]

[tribal sun]

Voici ma forme algébrique

Du coup je me demande est ce que je réécris pas la forme algébrique sous la forme 1-(6/4)i je pense que c’est pas ça qui faut faire

J’ai un doute

[hard gale]

c'est comme dire que $\frac{1+2}{3+4}=\frac 13 + \frac24$

[somber coyote]

emeric75:

[hard gale]

enfin bon le but c'est d'essayer de virer ce i au dénominateur

[tribal sun]

Oh non c’est totalement faux ce que je dit

Merci

[hard gale]

multiplier par le conjugué du dénominateur ça marche plutôt bien

[tribal sun]

C’est ce que j’ai fait

Ça donne -0,6+1,2i

[hard gale]

ça a l'air ok

[tribal sun]

Sauf que pour trouver l’équation cartésienne

Il faut calculer calculer le module

[hard gale]

équation cartésienne de quoi?

[tribal sun]

Mettre sous forme cartésienne la forme algébrique

J’ai fait ça concrètement

Mais il faut déterminer l’angle

[hard gale]

ah j'étais en mode wtf

cartésien/algébrique c'est la même chose pour moi lel

ok ok....

[tribal sun]

Du coup faut déterminer l’angle

Pour avoir x= r cos (a) et y = r sin (a)

Sauf que √5 apparement n’as pas d’angle 🤔🤔

[hard gale]

ok je revérifiais les modules et tout...

nan c'est juste ta division à la fin qui est foireuse

[tribal sun]

Et où v

[hard gale]

$$\frac{-3}{5} = \frac{3\sqrt{5}}{5}\cos(\theta)$$ $$\frac{-3}{5}\frac{5}{3\sqrt{5}} = \frac{3\sqrt{5}}{5}\frac{5}{3\sqrt{5}}\cos(\theta)$$ $$\cos(\theta)=\frac{-1}{\sqrt 5}$$

[somber coyote]

emeric75:

[hard gale]

jsp comment t'as fait, mais les fractions dans les fractions c'est pas trop ça j'ai l'impression

et pareil pour le sinus

[tribal sun]

Ah oui je suis debile ...

Bah je l’ai pas fait à la main je l’ai fait de tête

Merci

[worthy root]

[narrow storm]

yo

@mild cargo

am i allowed to @he1pers?

help me out here ples

<@&286206848099549185>

[stuck torrent]

Do you know euler's formula

[trail minnow]

plis

whoever no

power reduction

[stuck torrent]

oh

[trail minnow]

look at the channel

[stuck torrent]

I just thought of that

[narrow storm]

Don't think Euler is relevant to this

[trail minnow]

you can use it

[stuck torrent]

Yes, but using it you can get the result really easily

[trail minnow]

if you know it then do that

it's much easier than the alternative method

$\sin^2 x = \frac{1 - \cos 2x}{2}$

[somber coyote]

hegel:

[dark sparrow]

[trail minnow]

hint

$\sin{3x} = \sin{(x + 2x)}$

[somber coyote]

hegel:

[trail minnow]

rip question asker

2019-2019

[narrow storm]

helo

@trail minnow

went for some soup

[trail minnow]

kay

[narrow storm]

so i separate into s^2(x) * sin(x)

right

[trail minnow]

ye

[narrow storm]

then aplly formula

apply

then put the sin(x) in numerator?

[trail minnow]

well

actually

there's an easier way to do this

but it's kinda gamey

[narrow storm]

because its i cant do it that way

i gotta do it using these formulas

¯_(ツ)_/¯

[trail minnow]

do you know the sum of angles identity

$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

[somber coyote]

hegel:

[narrow storm]

yes

[trail minnow]

so try plugging in x for alpha and 2x for beta

if u knew demoivre's theorem we could do more but you probably do not if this is an alg II class

[narrow storm]

why am i plugging that in

and how is this addition

[trail minnow]

sin(3x) = sin(x + 2x)

i don't really know how to explain this intuitively

[narrow storm]

sin^3(x)

not sin(3x)

[trail minnow]

yes i know

but the thing is that sin(2x) expands to 2 sin x cos x

and cos 2x expands to 1 - 2sin^2 x

if you follow this through you'll see : p

[plucky marlin]

eulers formula plis

[trail minnow]

$\sin(x + 2x) = \sin x \cos 2x + \cos x \sin 2x$

they can't use euler's formula

[plucky marlin]

[trail minnow]

this is why we must alg II

bad class

[plucky marlin]

ban em

[trail minnow]

big sad

de moivre's theorem kek

then equate imaginary parts

[narrow storm]

im actually in college but my credits didnt cary over so i gotta redo all the shit i sorta forgot

[somber coyote]

hegel:

[trail minnow]

RIP

again euler's theorem is the correct way to do this

but i dun think you know it

[narrow storm]

never learned that

[trail minnow]

anyway $\sin 2x = 2\sin x \cos x$

[somber coyote]

hegel:

[trail minnow]

so $\sin 2x \cos x = 2\sin x \cos x \cos x = 2\sin x \cos^2 x = 2\sin x(1 - \sin^2 x) = 2\sin x - 2\sin^3 x$

[somber coyote]

hegel:

[trail minnow]

u see where the sin^3 comes in now

and then for $\sin x \cos 2x = \sin x (1 - 2\sin^2 x) = \sin x - 2\sin^3 x$

[somber coyote]

hegel:

[trail minnow]

sorry this explanation is basically me throwing shit at the wall but im really tired and there's no good intuitive explanation for this property T . T

[narrow storm]

its just hard to follow

[trail minnow]

ye

[narrow storm]

its fine though

thanks though

[upper karma]

friends

im having to calculate the rectangular coordinates when given polar coordinates. easy enough right?

but my question is, if R = 25 m/sec squared, do i use 5

[dark sparrow]

uh

[upper karma]

i know its probably obvious.. i think its a trick question

[dark sparrow]

what

can you post the entire question

exactly as stated

[upper karma]

yeah

question 4 d

a was easy enough

[small raptor]

at 210 degrees. That's an important bit of information.

[upper karma]

why

[dark sparrow]

if the question didn't have the "m/s^2" bit

would you be able to do it

[upper karma]

yes

[dark sparrow]

then do that

the m/s^2 is just the unit

[upper karma]

ah ok

wasnt sure if it was a trick q

i guess the answer will remain in that unit then?

for vectors x and y

im super new to this shit lol.. still studying algebra 2 but this other class is having us do basic trig

[dark sparrow]

yes

[upper karma]

thanks ann

😃

[dire rampart]

yw

[dark sparrow]

:| @dire rampart

[dire rampart]

u snooze u lose

[dark sparrow]

heck u

[upper karma]

@dark sparrow whats the best way to convert a negative value to a degree if the degrees are going counter clockwise

[dark sparrow]

??

[upper karma]

So like, I got -74 for the angle , because it had a negative x coordinate

But the angle was really 106 degrees

Know what I’m sayin?

[formal bolt]

Mm they're not the same angle, probably you didn't need to "convert" it but to find its supplementary: 106+74=180

[upper karma]

Ahhh

@formal bolt in this class we are taking the angle from counter clockwise

Dunno if that matters

I’m pretty confused tho 🤔😂

The angle I was looking for was definitely 106 but my calculator told me -74

180-74 eh

[formal bolt]

Did you evaluate it through arctan ?

[upper karma]

Yeah

That’s what my textbook said to do

But maybe the example problem didn’t involve a Resultant vector ending in the negative x plane

I’m thinking that’s why I got a weird answer

[formal bolt]

Positive counterclockwise angles is common

Also arctan has (conventionally) codomain ]-pi/2,pi/2[, if it is the inverse function of tan(x) restricted to ]-pi/2,pi/2[
But tan(x) is pi-periodic so you can add pi to your solution and still find an angle with the same tangent. You have to decide which one to choose from the situation

[upper karma]

I can’t say I understand but thx for trying to explain hehe

I’m brand new to trig. Actually still doing the beginnings of algebra 2, but my other class has us doing some trig stuff

[queen hawk]

X is 10 degrees right?

Someone posted this on another Discord and asked for help

Was wondering if x is 10

he assumed it was 20 and it worked too

Wondering which is actually correct

If someone answers please ping me

[keen aspen]

Oof its this meme question again

I swear its the 3rd time now

[tawdry pivot]

I remember this, forgot the name

Langley’s Adventitious Angles

[worthy root]

#shitpost

[tawdry pivot]

I found the problem pretty interesting tho

the one posted by olnir is a bit different to LAA tho

[tawdry pivot]

welp I'm not good enough, anyone got hints for this one? (btw a geometry program gives us evidence for the value of x)

oh ye looking back

https://www.mathpages.com/home/kmath734/kmath734.htm

$f$ is the solution to the equation $\sin(e+f)=2\cos\left(f+\frac{a-e}2\right)\sin\left(\frac{a+e}2\right)$ isn't the actual general solution

[somber coyote]

EpicGuy4227:

[tawdry pivot]

using law of sines, we get $\sin h\sin f\sin(c+d)=\sin c\sin(g+h)\sin e$

[somber coyote]

EpicGuy4227:

[tawdry pivot]

then substituting the values from @queen hawk 's problem, guessing x=20, some manipulations later (using double angle formula for sin, cos(90-x)=sin(x)), I got that x=20 is indeed the solution.

[tawdry pivot]

we can also see that a,b,c,d,f determine the quadrangle, and that the solution to the generalized problem is the solution to sin(a+b+c)sin(f)sin(c+d)=sin(c)sin(a+b+f)sin(b+c+d)

[queen hawk]

I don't understand anything

[tawdry pivot]

well, you need to be familiar with the sine rule to get started. You'll also need to do a bit of algebra to follow along with me

[cinder portal]

No trig solution plz

Lol

[trail minnow]

kek

[tawdry pivot]

ye a geometry solution would be interesting, anyone got one for olnir's problem?

[cinder portal]

Would systems of equations help? Naming two different, maybe three different angles

Just an idea

Actually nop

Another idea i had was inscribing it in a circle. Dont havr pencil or paper around me

[tawdry pivot]

you'll find that only using the identities that only involve adding angles won't get you anywhere

[cinder portal]

Ah i figured as much

My last thought would probably be the ratio between 20/60 and 10/70

[worthy root]

Why did they substutrd x for the expression except 2 .

1/(2+x)

[granite mortar]

what do you mean?

this is a continued fraction, it "appears in itself"

hence the x=1/(2+x)

[quiet mason]

how can you do this level of trig without doing these type of suns

sums*

[worthy root]

Kyu ki alag majab tere bhai ka

They didn't teach this to us

[dark sparrow]

what language is that

[worthy root]

Hindi written in English letters

[gritty siren]

[worthy root]

Imagine Chinese language being written in English letters .

[gritty siren]

wo shi fa guo ren
dui bu qi, wo de zhong wen bu hao

[worthy root]

Chen pen kimung ta

toung eeees ae neard

[gritty siren]

[dark sparrow]

tu parles chinois tuong ?! o.O

[gritty siren]

a tiny little bit

[worthy root]

Illegal Google translate

[quiet mason]

@worthy root whats majab

i understood youre sentence

[hard gale]

Imagine french written with Chinese letters tho

[quiet mason]

baal ka laora

boka choda

lmao

[worthy root]

Don't abuse man.

Should I ping moderators for this

[quiet mason]

idc

why shouldnt i

[worthy root]

Or just ping @upper karma because only he understand

[quiet mason]

its just slangs in a diff lang

[olive solar]

@worthy root chinese language can be written in "english" language

[upper karma]

hmm

[worthy root]

Letters*

[upper karma]

it's not really a slang lionel

[olive solar]

it's called han yu ping ying

yeah

[quiet mason]

i literally know these slangs

[olive solar]

romanized characters

[quiet mason]

nd words like bhai

chutiye

[worthy root]

It's a bad word

[upper karma]

lol

hahahahahah

[worthy root]

Bhai means brother

[olive solar]

han yu ping yin is basically what you're thinking of

it just includes the tones for each word

[quiet mason]

isnt chutiye a slang

[upper karma]

no it's an insult

and if you say it to someone on the street you will get lynched

[quiet mason]

and i tried to create my own sentence too

mera bara bara hai

barra*

[upper karma]

hahahahahaha

[olive solar]

wait

oxide

you're telling me there's a speedrun strat to getting lynched?

[upper karma]

@olive solar l y n c h e d

[quiet mason]

lmao how bad is chutiye

lmao stain

youre sick

[upper karma]

colo the fastest speedrun strat is just

[olive solar]

o right

if u want something done right

[upper karma]

wear muslim attire in public and say you eat beef

[olive solar]

i'll stop i don't want to get obliterated by mods

[quiet mason]

LMAO

like thats getting killed

easy

[worthy root]

But the words u said were quiet inappropriate.

[quiet mason]

idc

[worthy root]

You are immature

[quiet mason]

do what you want

idc

[olive solar]

i'm glad they're not loud inappropriate

[upper karma]

lol

[olive solar]

anyways, this isn't really #geometry-and-trigonometry

[worthy root]

I used to say these words when I didn't knew what they meant. You are immature

[olive solar]

if u want to be degenerate, do so in chill pls

that's where it belongs

[worthy root]

#bot

[quiet mason]

krishna nobody cares bruv

people can use whatever they want

[worthy root]

Ok

Bye

[hard gale]

Life Speedrun - lynch%

[olive solar]

[upper karma]

lol emeric

[worthy root]

Rreee

[queen hawk]

It was a simple

Math question

Why you had to make it 100% more complicated

[raw willow]

Is there any possiblity to get hexagonal and squared 2d grid both on the same grid? (it's so diffuclt to form a proper question about that in english for me)

[old ferry]

It would look bad, wouldn't it

[raw willow]

I don't know, I try to imagine that

[old ferry]

I think I might have misunderstood the question

You need a grid, and that grid should be both hexagonal and squared at the same time?

[raw willow]

yes, exactly

I got only something like that

which is... ugly

[old ferry]

Why would you need that?

[raw willow]

just for a game idea

I'm just curious is it possible

it's nicer

I'm looking for other solutions

We can deform a hexagonal shape, but squares must be the squares

I killed the conversation ;_;

[devout shell]

[rustic temple]

;3;

[digital stirrup]

Can the squares be diagonal?

For example, I just made this:

into this:

So the hexagonal grid is oriented from left to right, but the square grid is turned 90 degrees or so

The squares can be 1/4 of the size to make it seem denser

This is another option, if you want the squares aligned: Edit - Should have made the squares another color lol

[hushed nexus]

could anyone help me with understanding how they got the lengths of the line segments? i.e. how they got to AF = s - b

[old ferry]

Call the line segments from |AF| to |EA| counterclockwise d,e,f,g,h,i.
d+e+f=g+h+i=f+g+h=d+e+i=d+i+h=e+f+g=s.
Remember b was h+i and s=d+i+h. this means d = s - b.

[summer plume]

Hi, so I'm working on a word problem where I use a pleasure wheel (Like a Ferris Wheel, but made out of wood and turned by hand) to answer the questions below. One of them asks, "The path that each passenger takes is circular. Give the equation for the path using x and y - coordinates." Now, I thought it would be the equation y = A * cos(Bx + C) +D, but later I'm asked, "Determine an equation containing trigonometric functions that models the path taken by a rider." Making me think I had it the other way around. How do I answer the first question? If anybody could help......

[tawdry pivot]

a^2

[analog sluice]

Is there a way to calculate the area of a polygon given it's 4 points?

The solutions I have found online have the limitation where if the polygon is not simple, it is incalculable and values at 0

which is what I'm trying to solve for because the ultimate goal is to see if the lines are collinear

If I use anything with slope involved, it's very easy to get errors because of Nan numbers

[potent mango]

could ya'll help me approach this?

im trying to find a possible solution with law of cosines but im kinda lost

[fringe dirge]

Well can you find any extra information about the diagram

[potent mango]

i know that AD+BC=280

I made an equation with law of cosines and simplified it a ton but it has three variables in it that i cant substitute for anything

[dark sparrow]

ok so let x = AD and y = BC

$\begin{cases} x + y = 280 \ x^2 + 180^2 - 360x \cos(\alpha) = y^2 + 180^2 - 360y \cos(\alpha) \end{cases}$

[somber coyote]

Ann:

[dark sparrow]

are those the equations you made @potent mango

[potent mango]

yeah I just made AD = a and BC = B

[dark sparrow]

ok doesn't matter

watch what happens when you attempt to rearrange the second equation for cos(α)

[dark sparrow]

$x^2 - 360x \cos(\alpha) = y^2 - 360y \cos(\alpha) \ x^2 - y^2 = 360(x-y)\cos(\alpha) \ \cos(\alpha) = \frac{x^2 - y^2}{360(x-y)} = \frac{x+y}{360} = , ?$

[somber coyote]

Ann:

[potent mango]

woah dang i simplified the x^2-y^2 wrong i see that now

bruh so its just 280/360

7/9

😃

[obsidian oxide]

yo

can someone help me with circles

[spark stag]

[obsidian oxide]

dm me

so i can send it

nvm

can u help

[upper karma]

Can u guys help on this?

Im thinking it must involve inverse trig functions

[dark sparrow]

,rotate -90

[dark sparrow]

$x^2 + y^2 - 6x + 8y + 24 = 0$ defines a circle in $\bbR^2$

[somber coyote]

Ann:

[dark sparrow]

from that, you can deduce the possible range of values for $\sqrt{x^2 + y^2}$

[somber coyote]

Ann:

[dark sparrow]

and from that, the possible range of values for $f(\sqrt{x^2 + y^2})$, where $$f(t) = \frac85(2 \cos^2(t) - 3 \sin(t))$$

[somber coyote]

Ann:

[upper karma]

Ive tried that

Maybe its my inability to complete the solution from here on but

This wasnt exactly working..

[dark sparrow]

wdym, "wasn't exactly working"

where did you get stuck

[upper karma]

Well i got at that stage of factoring it

But then i didnt know what to do

Sin can vary from (-1 to1)

So to maximise the stuff under the bracket

It should be one of the extremes

But i dont think thats right

[dark sparrow]

ok before this

have you established that sqrt(x^2 + y^2) varies from 4 to 6

[worthy root]

Ann you BIG bren

[worthy root]

6

How can I prove this

Should I prove that x+y=w+z=180 first .

Then use the axiom

[dark sparrow]

sure

[worthy root]

How can I prove that x+y sum to 180

🤔

[dark sparrow]

well

consider what x+y+z+w might be

[worthy root]

Ok

[dark sparrow]

like, look at the picture

[worthy root]

360

[dark sparrow]

there you go

[worthy root]

I got a feel

How to do it thank you

I told I got a feel about how to do it

Sorry my English is bad

[hot plaza]

hello! I'm minutes new here and idk if this is the right discord or channel to ask for help about some measurements 😅

[dark sparrow]

go ahead

[hot plaza]

alright 😅
I'm spending my vacation time creating some papercraft figures.
right now I had this whim to put a nose to a flat square face. you will get it with the image.
so my issue is that if I put this piece it should look in a side view as a right triangle, but the paper piece has to be folded to fit in that triangle gap i made for the face.
Right now have the measures of the gap and how it would look in the side view, yet I still don't know how to get the other faces of this pyramid.

[small raptor]

What do you mean getting the other faces of the pyramid?

[hot plaza]

well the nose would look like a pyramid.... right?

[small raptor]

I imagine if you created a 3D object to insert, yeah, a pyramid. Alternatively, just fold a square/rect into two triangles to insert.

Although, that wouldn't resolve the bottom nose. 🤔

In any case, if you can measure the cutout of the nose, you have then the base, and sides.

As for construction... 🤷

Probably requires a long rectangle that can form into the pyramid face and a bottom piece to fold into a square

[hot plaza]

which part do you mean by "cutout of the nose"? (sorry, still not as knowledgeful of the English language as I would like to)

[small raptor]

... the area of the square that is cut out for the nose insert

[hot plaza]

👌

[obsidian oxide]

can anyone help with that?

[digital stirrup]

Hey @obsidian oxide , the section of the circular garden is just an 80 degree slice of the whole circle. You can see from the diagram that your circle has a radius of 13 feet. You can use the formula for the area of a circle with a radius of 13 feet to find the total area of the garden. Then, realize that your "slice" of the garden is only 80 degrees. Since we divide a circle into 360 degrees, that means that your slice will cover 80/360, or 2/9 of the circle, which means that the area of the slice you are looking for will be 2/9 of the area of the entire circular garden.

[obsidian oxide]

thanks

[digital stirrup]

no problem

[worthy root]

Just asking if geometry of class 9 and 10 is used in class 11 @upper karma

Anyways how can I do it

I mean there is less information right ?

Any good construction that I can make

Leave I got this

Done

@dark sparrow ann isgeometry used in precalc and calculus .

[dark sparrow]

yes

[worthy root]

Ok thank

So should I learn all the chapters of geometry

Or sone specific

Some*

@dark sparrow

[dark sparrow]

if you try to cherry-pick, you'll inevitably miss something crucial

and it'll come back and bite you in the ass later

[worthy root]

Ok

Lol

Geometry is easy will do it in 10days

I thought the same when starting algebra but algebra took 3 months

Let's see wish me luck . That I found it easy

I have learnt a little geometry in my class 10

So I think I will make it

😎

[small raptor]

the rabbit hole is deep.

[quiet mason]

"Geometry is easy"

lets see if you can do it to the intensive level in 3 months

[worthy root]

I hope so

I have done it in class 10 Ok for passing the exam but not intensively

Let's see

,w graph y=e^x

[somber coyote]

Display results online and refine query

Requested by Krishna#4971

[worthy root]

,w graph y=x

[somber coyote]

Display results online and refine query

Requested by Krishna#4971

[worthy root]

,w graph y=e^x and graph y=x

[somber coyote]

Display results online and refine query

Requested by Krishna#4971

[spark stag]

do you somehow forget #bots exists every time you're told

or something

[night wing]

Can someone help me with this?

[spark stag]

label the opposite, adjacent, and hypotenuse sides of the triangle

use that to determine which trig ratio is appropriate to use here

[night wing]

Im just too dumb...

😂

welp I labelled these

just forgot this formula...

aint it..

Tan(31°42' = a/58

so that would equal... 35.8 right?

[idle bloom]

Hmmm

I got something else

Your equation seems right tho

How are you plugging it into the calculator

@night wing

[night wing]

58tan(31°42')

[idle bloom]

Whoops I made a rookie error

Calculator was in radians ;-;

[night wing]

lol

so was I right?

[idle bloom]

Aight yea you're right, my bad

[night wing]

YES!

I hadnt did trig for 3 weeks and got it 😂

ahhh crap...

bearings...

[idle bloom]

My phone calculator has the radians button visible when I'm in degrees so I got confused

[night wing]

oh

could u help me out with this

[idle bloom]

Yea it's dumb lol

Sure

[night wing]

I forogt how to do it

[idle bloom]

So bearings are just more specific directions

You have NESW

Your starting direction

Then you put your amount of degrees

Then you put direction you want to turn that amount of degrees into

So if I was facing pure north and wanted to look 45° to the west

I'd do N45°W

Does that make sense?

[night wing]

Yea

Could u put that in a formula 😂

as a simplified note

[idle bloom]

Now typically your starting direction is always North or South and your ending direction is always East or West

Yea sure 1 sec

[N/S] [number of degrees you want to turn] [E/W]

[night wing]

u dont need a calc for this right?

[idle bloom]

So what's one way you could write the bearing for B

No

[night wing]

lol I have like one last but dont wanna overload u

Hold on

It would be

N35°W is that right?

[idle bloom]

Not quite

Ok so let's start at North

Which way is west and which way is east

[night wing]

west | left, east | right

[idle bloom]

Sí

So which way is the shortest way to turn from North

[night wing]

to b?

[idle bloom]

Yea

[night wing]

East

[idle bloom]

Sweet

So now we gotta establish our amount of degrees east we need to turn

How many degrees so you think we need to turn

[night wing]

125°

to reach B

[idle bloom]

Sweeeeeet

So your first answer is N135°E

That's exactly right

Now do the same thought process but let's start at South

[night wing]

ok

and this is for A?

Because A in the corner is that just saying it is the same as B? or is it 90 degrees?

[idle bloom]

A is the origin

[night wing]

Quick game of hot and cold...

S125W

[idle bloom]

Again not quite

So we're starting south

Which way do we want to turn, east or west

[restive wren]

How did they reach at 4th step

Root(1+t)square×(t+1/t)

[quiet mason]

can you simplify sqrt(1+1/t^2)?

[restive wren]

That part

I am not getting

[quiet mason]

yes now lets do

$(1+1/t^2)$

[somber coyote]

Lionel:

[quiet mason]

cant i multiply and divide by t^2?

[restive wren]

Yes

[quiet mason]

so if i do that ill get
$(t^2(1+1/t^2)/t^2)$

[somber coyote]

Lionel:

[quiet mason]

wont i

[restive wren]

Yes

[quiet mason]

now the above expression is just
$(t^2+t^2/t^2)$

[somber coyote]

Lionel:

[quiet mason]

the numerator thus becomes (t^2+1)

dosent it?

as t^2 and t^2 get cancelled out

agree?

[restive wren]

Yes

[quiet mason]

now after seeing the fraction it is now $(t^2+1)/t^2$

[somber coyote]

Lionel:

[quiet mason]

isnt it?

[restive wren]

Yes

[quiet mason]

so now the expression reduces to
$sqrt(1+t^2) + sqrt((1+t^2)/t^2)$

doesnt it?

[somber coyote]

Lionel:

[quiet mason]

agreed?

@restive wren bro you cant just leave me

come on

[restive wren]

Yes

[quiet mason]

so you agreed?

and understood?

[restive wren]

Yes

[quiet mason]

now suppose we have something like
$a+a/c$

[somber coyote]

Lionel:

[restive wren]

Yes

[quiet mason]

isnt the expression same like $a(1+1/c)$

[somber coyote]

Lionel:

[quiet mason]

agreed?

[restive wren]

Yup

Ok i got

Ir

Thankd man

[quiet mason]

sure?

[restive wren]

Yes

[quiet mason]

np mann

[restive wren]

Really

Thanks

A lot

[quiet mason]

it was comparatively easy to what you were solving tjo

[restive wren]

A lot

[quiet mason]

you became braindesd or what

[restive wren]

Yea

[quiet mason]

lol np man

[restive wren]

Actually when i think about it now
I feel like a stupid i got stuck

It was so obvious

[quiet mason]

lol

[worthy root]

Wait gotcha

I have to extend the TS to some point G

[worthy root]

Reee bro @restive wren your name reminded of my school bully

[restive wren]

Lol really....

Wait is that ncert textbook problem?

[worthy root]

Yup

Yeah it's a ninth grade ncert

Any hint legends

?

Taime has come to poong hulpers

[restive wren]

U got ans?

[worthy root]

Nope

Any hints ?

[restive wren]

Take angle PQR as 2x and PSR as 2y

[worthy root]

That's what I did

[restive wren]

Use exterior angle property on both triangle

One by one and then equate

[worthy root]

Angle QTR=180-(x+y)

Big offf

[restive wren]

U got it

[worthy root]

I forgot the exterior angle theorem

[restive wren]

Yea

[worthy root]

Lo

Lol

[restive wren]

Happens bro....

[worthy root]

@restive wren can you change your name it really makes me angry

[restive wren]

Bro its my name... and i cant change it

[worthy root]

Lol

[restive wren]

Maybe we cam be friend

[worthy root]

Jk

Just kidding

[restive wren]

I kno

Btw

[worthy root]

?

[tepid finch]

@somber nest draw a triangle where tanx = 3

oh jk its asking u to find cos x/2

[somber nest]

I'm gonna post this in math help

[silent plank]

half angle identies

[gritty siren]

You can keep it here @somber nest

please avoid posting the same thing in multiple channels

[somber nest]

okay

anyone know where to start? it's in quadrant 3

[silent plank]

start with half angle identity for cos

ASTC can tell you whether its + or -

[somber nest]

okay one moment

[silent plank]

draw a triangle to find missing info

that should be enough info to solve it

[somber nest]

choice 3 right?

https://gyazo.com/10e6e2be0c2ce04c6530fd091d3c7056

[silent plank]

what's the general solution for tan(theta) = 0?

[somber nest]

0

[silent plank]

what are other solutions?

[somber nest]

solutions as in references angles or?? not sure what you mean

[silent plank]

theta has infinitely many solutions,
eg tan (pi) is also 0

[somber nest]

your talking the range right? all real numbers

anyone got any clues?

[silent plank]

you first need to find an expression that covers all the solutions of tan(theta) = 0

ie. -pi, 0 , pi, 2pi, 3pi ... etc

[somber nest]

no clue

I need something visual

[silent plank]

the choices show can provide some hints,
kpi OR 2kpi where K is an integer

one of those would be the correct choice, the other skips solutions

are you still struggling?

[somber nest]

err, I need steps on where to begin

I can't just go straight to conclusion, I don't understand this

[silent plank]

so to solve for ALL x in the question, you need to find ALL solutions to tan(theta) = 0

[somber nest]

so solve for all quadrants?

[surreal ingot]

Zerpo, how far are you on Trigonometry?

[tepid finch]

its like
tan(0) = 0
tan(pi) = 0
tan(2pi) = 0
...
in general
tan(n * pi) = 0 where n is the subset of all integers

your solution has to solve for EVERY case

[somber nest]

my Professor uses reference angles to solve problems usually

[tepid finch]

but in the specific question you gave it asks you to solve for the general case

[silent plank]

do you see how "tan(n * pi) = 0 where n is the subset of all integers" gives you all the solutions?

[somber nest]

to solve for x, we set 3x + pi/6 = 0? subtract pi/6, divide both sides by 3

[silent plank]

no

[somber nest]

x = -pi/18

[silent plank]

that only gives 1 solution

to find all solutions of x, equate it to all solutions of theta listed above

[tepid finch]

Zerpo do you understand what we're saying regarding there being multiple solutions

[somber nest]

if it's in quadrant 2, pi - reference angle. if it's in quadrant 3, pi + reference angle. if it's in quadrant 4, 2pi - reference angle

like this

???

[tepid finch]

sort of, but you also have to consider that the angle could be beyond just the 4 quadrants

more than 2pi

[silent plank]

eg tan ( 9001 pi) = 0

[somber nest]

it's asking for just tan though, why we need 4 solutions. we need 4 solutions in a question like this https://gyazo.com/9b8d8e6a7eed288fe1f93c9940816e4e

when range is [0,2pi] we find 4 solutions

[silent plank]

that is bounded between 0 and 2pi

the question you posted is for all real numbers

(which is infinitely many)

[somber nest]

the choices only have positive or negative pi/18

[tepid finch]

the question says: " on all real numbers"

[somber nest]

I'm hella confused

[tepid finch]

this means x can be anything

x can be -235823989234

wait is this a different problem

[somber nest]

https://gyazo.com/10e6e2be0c2ce04c6530fd091d3c7056

[silent plank]

the first stage is understanding the general solution of tan(theta) = 0
the next step is pretty simple

[tepid finch]

for instance u got the solution x = pi/18

but I could put in x = pi/9 and that also works

so x can be something other than just pi/18

actually pi/9 doesnt work but you get what i mean

[somber nest]

can you do this on a whiteboard

https://awwapp.com/b/uqjmaqi3j/

everyone have different techiques

[silent plank]

drawing won't really help

[somber nest]

I just don't understand you guys

[tepid finch]

have you learned the unit circle yet?

[somber nest]

Professor mainly uses reference angles

no unit circle

[silent plank]

reference angle here would be 0

but you are also going around the unit circle infinitely many times

[somber nest]

tan is positive in quadrant 1 and 3

[silent plank]

in the Q, tan(theta) = 0

which happens at k * pi

[tepid finch]

you might want to understand the unit circle first

its kind of hard to visualize angles larger than 90 degrees without triangles
*with triangles

[silent plank]

i.e
tan(101 pi), tan(504 pi), tan( any integer * pi) all equal 0
and would result in different x values when you equate it to (3x + pi/6)

[somber nest]

hold up I got a similar problems in my notes

I will show you guys

[upper karma]

Is anyone here familiar with directed angles?

[tepid finch]

the same applies to tan(x) = 0

x = 0 + 2kpi OR x = pi + 2kpi

and to solve for tan(3x + pi/6) you just substitute the x above with 3x + pi/6

[silent plank]

(which simplifies to just kpi)

[tepid finch]

yep

[upper karma]

Okay

So basically a quadrilateral ABXY is cyclic iff $\measuredangle XAY = \measuredangle XBY$

[somber coyote]

Big10:

[upper karma]

(Where \measuredangle stands for directed angle)

But I'm not sure how to prove the $\measuredangle XAY = \measuredangle XBY \implies$ ABXY is cyclic

[somber coyote]

Big10:

[upper karma]

Or rather that $\angle XAY + \angle XBY = 180^\circ$ or $\angle XAY = \angle XBY$

[somber coyote]

Big10:

[upper karma]

Directed angle just means positive for counterclockwise between holding XA still and rotating AY and negative for clockwise direction.

So if XAY = XBY that means that in the quadrilaterall ABXY you've got two opposite angles that are in the opposite direction but have the same angle.

So not only is it cyclic - its also a parrallelogram

What?

I'm talking about directed angles mod 180°

Still applies homie

Dude

No

That's wrong

@upper karma

Kk

Look

When <)XAY=<)XBY then we have that <XAY = <XBY+n*180

Then either n is 0 or 1

So suppose n is 0 then <XAY = <XBY

Now suppose that <XAY = <XBY + 180

Then those points lie opposite

Nah homie

Cuz then

<XAY = -<XBY

Cuz thats how directed angles work

What?

Well <XAY = 180-<XBY

Saaame shitskiees

What do you even mean

Do you know how directed angles work?

Yeah fam, just work on it intuitively

<XAY+ 180 = 180-<XAY, so if <XAY = <XBY+180 = 180-<XBY = <XBY - clearly a contradiction imho

How do you even get the first line?

@upper karma

Think

Intuitively

From the heart

The first line is already wrong

@upper karma

How man?

Because that implies that <)XAY=-<)XAY

And that isn't always true

And you're mixing up directed angles with normal ones

Naaah

[summer plume]

Didn't I already answer this?

I did Sin x/1, but that wasn't the answer they were looking for

[fossil ledge]

Yea sin x should be right

[spark stag]

by "unsimplified" i guess it means

rewrite as $\frac{\sin (x) \cos (x)}{\cos (x)}$??

[somber coyote]

Namington:

[spark stag]

poorly phrased, in any case

[worthy root]

How to prove 2)

[silent plank]

after proving congruency in (i), which angles in the 2 triangles are equal?

[upper karma]

Well what do you want to learn for? @upper karma

Like, for which contest

You have already finished other intro/intermediate series books?

Because in that case AoPS Vol 1 would be kind of a waste imo

Well enough for what?

From what I've heard it's like AMC 10 geo

Don't think there is much difference

There is some geo stuff in AoPS Vol 2 but way more in Vol 1

I guess for the basics ¯_(ツ)_/¯

[worthy root]

@upper karma you can learn from ncert if you want and practice problems from aops v1

aops v1 is availabe on the internet

Google it

yes

Serach on google ncert class 9 and clas 10 do the geometry

Ncert is always a good book and is referred by evryone here in INDIA

well\

intutive undesrtanding is important too

And reading aops v1 dont get him, that intututive feeking of concepts

feeling*

I know man

But before solving some problem of high level its always reassonable to solve some basic and mid level

He is knew too the geomtry

new*

[eager pendant]

aops v1 is very basic

[worthy root]

But the theory ?

For example lets say he is studying congruence . AOps v1 will strictly define whats congruence but when he will be asked to solve some real world problems on Conguence he will donnt know how to use the rules he studied

That why intutive feel of concept is very important

EULER TBH aops will be good decision

INTRO TO GEO

I am doing ncert and aops v1

It depends on the person

yea

You needs to start seeing some basic example like this

Anyways using the peoperty of the exterior angle can help you solve the problem

Also use the fact that angles of triangle sum too 180deg

and i have given you enough hint

yEA

THATS what i like about aops wording is awlays clear

the inro book will give you the capacity tooo do so

@eager pendant I better recommend you to read the preface of the aops v1

Its clearly written its for student who have a basic knowledge of the subject and they want to develop problem solving skills

[upper karma]

hi jazza

[worthy root]

Hi jasi

[tacit meteor]

hi

i have invented a way to calculate coordinates of point of intersectino of 2 lines without using parametric equation

now i am attemping to do it on a plane but am having a hard time

using similar method

[idle bloom]

What's the method you use for two lines

[tacit meteor]

https://www.webwhiteboard.com/board/pydr7cjr

i have illustrated it here

please take a peak

yes

there are 3 cases

slope are opposite

slopes are equal

[idle bloom]

Yea I'm very confused

[tacit meteor]

let me break it down

[idle bloom]

Well if the slope are equal they can either intersect 0 times or a infinite amount of times

If it's a linear line that is

[tacit meteor]

i mean both positive not equal

both positive or negative

[idle bloom]

Gotcha

[tacit meteor]

can u go back to the drawing board

[idle bloom]

And are we talking linear equations?

[tacit meteor]

ill start it over for you

no

no equations

[idle bloom]

Ok