geometry-and-trigonometry

[sick veldt]

nobody likes them

[hard gale]

i like proofs

[sick veldt]

oh

[hard gale]

but 2cols are shit

[sick veldt]

So are these right?

I have other options to switch in as well

first one might be corresponding

but idk

I usually use the corresponding and interior postulates in transversals

[hard gale]

seems right

[sick veldt]

oh really

shit i never get these right

alrighty well

you know the rest of the drill,

3 more 😛

[hard gale]

i'm not native so i actually suck at understanding geo in english

[sick veldt]

shit

hmm well I'll take the risk

m<d + <dab = 180

consecutiveeeeeeeee

oh shit

96 + dab = 180

=84

reeeeeeeee

cmon octoniion

only 2 more

this and one more

[hard gale]

84 yas

gtg to the grocery store

[sick veldt]

SHIT

I got one more

pls

For the last box

AC

would it be AC or AE?

[umbral snow]

Real proofs are a thing of beauty. 2 column proofs are one of the worst things that will ever happen to anyone ever. Just putting my opinion out there

[sick veldt]

yeah.....

[umbral snow]

AE = CE

As E is the midpoint between A and C

How does she know the coordinates of the midpoint of BD are the same as the midpoint of AC?

Sounds like she should be trying to prove that, not assume it

[sick veldt]

jesus christ im dumb

yeah

it'd be AE

its right there too haha

could you check one that the other did wasn't good at?

this one

its fine

I got 100

[umbral snow]

Nice nice

[woeful flame]

hi
" If: sin (a) = 0.6 , So: sin (360° - a) = -0.6 "
that is true or wrong?
I answered "true" but its seems to be wrong
and i dont understand why...

[dire rampart]

that is true

sin(-a)=-sin(a)

since sine is an odd function

and since the period of sine is 360

sin(360+a)=sin(a)

[woeful flame]

sir, That was not my question @dire rampart

my question was different

[dire rampart]

it is

im saying what you stated is correct

[woeful flame]

"sin(360+a)=sin(a)" That is completlly understoood

[dire rampart]

yes

[woeful flame]

but that is not what I asked

i talked about 360-a

[dire rampart]

yes

sin(360-a)=sin(-a)

do you agree?

[woeful flame]

i dont know

i think

[dire rampart]

it is

the period of sine

is 360

in degrees

[woeful flame]

sir wait please

there is a problem in the book? or I am wrong?

[dire rampart]

youre not wrong

[woeful flame]

I asnwered "True"

[dire rampart]

im saying youre correct

it is true

[woeful flame]

because

360-a != a

[dire rampart]

what

[woeful flame]

a-360 == a

[dire rampart]

no

sin(360-a)=sin(-a)

[woeful flame]

that isn't my question sir

[dire rampart]

ik

im getting there

bear with me

[woeful flame]

that is NOT prove that if sin(a) = 0.6 so sin(360-a) = -0.6

that is different cases

[dire rampart]

im gonna get there

dont worry

first

do you agree that sin(360-a)=sin(-a)

just answer this

[woeful flame]

yes

agree

[dire rampart]

ok

now

sin(-a)=-sin(a)

since sine is an odd function

[woeful flame]

😭

its not my question sir

please

[dire rampart]

meaning that sin(360-a)=-sin(a)=-0.6

yes? no?

agree?

[woeful flame]

but why do you do it in this way its very complicated

[dire rampart]

how is it complicated?

[woeful flame]

its very easy to understand that if sin (a) = 0.6

so sin (360-a) should be -0.6

[dire rampart]

yes but why

[hard gale]

well you're asking the question ?

[dire rampart]

is the question

im answering

[woeful flame]

cause 360 - a = -a

and sin(-a) = -0.6

[dire rampart]

not true

sin(360-a)=sin(360-a)

[woeful flame]

so why do you have to say the -sin(a) = sin(-a)

?

that is true but not answering the qestion

[dire rampart]

ok

how would you

answer it?

prove to me

[woeful flame]

if sin(a) = 0.6

[dire rampart]

yes

[woeful flame]

so i know that sin(-a) = -0.6

and i know that

-a = 360-a

[dire rampart]

how do you know that tho

im asking you to prove that to me

not simpy tell me

[woeful flame]

so sin(360-a) = -0.6

its a mathematics prove sir

step by step

[dire rampart]

where are the steps

[woeful flame]

and i didnt used: sin(-a) = -sin(a)

[dire rampart]

this is probably a language barrier, do you speak french?

perhaps oct

[woeful flame]

sin(a) = 0.6 ----- Data
sin(-a) = -0.6
sin(-a) = sin(360-a)
SO sin(360-a) = -0.6

[dire rampart]

perhaps

[woeful flame]

😼

so are you sure that the book is wrong and not me sir?

[dire rampart]

thats exactly what i did but in different order

[woeful flame]

ok

so are you sure that the book is wrong and not me sir?

cause i tried to understand for 4 hours why i wrong

i was sure that I understood everything after learning that for alot of hours and then this question made me think that all this was spent of time

and i dont understand anything

[dire rampart]

what does the book say

[woeful flame]

@dire rampart

the book say that the answer is wrong

that " If: sin (a) = 0.6 , So: sin (360° - a) = -0.6 " is Wrong.

I think there is another mistake sir

but im not sure

the book say also

"If A - B = 180" so "sin A = sin B " is Wrong.

but I say its true

who is wrong me or the book??

[upper karma]

A = 270 B=90

sin(A)=1 sin(B)=-1

unless 1=-1 you're wrong

😬

actually I wrote that backwards sin(270)=-1 and sin(90)=1

[thin wasp]

I'm really bad at geometry, can someone help

In a triangle ABC with <ABC=90, point K lies on AB. Point L is chosen on BC such that CL=AK. Point M is chosen on AC such that KM=ML. If <KML=90, prove that AK=KM=ML=LC

I've tried drawing all types of constructions, I know that MLCK is cyclic but I can't really abuse that, and I think we might be able to do this through some form of Thales'.

<@&286206848099549185>

[dire rampart]

@droreh#9503 book is wrong for the first one but right for the second

[hard gale]

[dire rampart]

@woeful flame

[thin wasp]

Can anyone help?

[upper karma]

State your problem and we shall see

[thin wasp]

@upper karma I stated it above

CyriusToday at 7:23 AM
I'm really bad at geometry, can someone help

In a triangle ABC with <ABC=90, point K lies on AB. Point L is chosen on BC such that CL=AK. Point M is chosen on AC such that KM=ML. If <KML=90, prove that AK=KM=ML=LC

CyriusToday at 7:36 AM
I've tried drawing all types of constructions, I know that MLCK is cyclic but I can't really abuse that, and I think we might be able to do this through some form of Thales'.

CyriusToday at 7:45 AM
<@&286206848099549185>

[little osprey]

Have you drawn it/

Can I have a look?

[thin wasp]

here is the picture

https://gyazo.com/b8afa1d4f9f033e4b342c75c6bcf0a85

[little osprey]

Ok lemme try

Have you recognised the fact taht AKM and MLC are right angled?

[stable tusk]

hey guys i have this
tan(2x) - tanx = 0

[little osprey]

Go somewhere else Rambo

This room is currently being used

[stable tusk]

i end up getting this
(tanx+tan^2x) / (1- tan^2x)

[little osprey]

:/

[stable tusk]

wait

wat

[little osprey]

@thin wasp Dude are you here?

Ig not.

@stable tusk If you'd like you can use #help-4

Or beta

Basically any room that is not currently being used.

[thin wasp]

@little osprey Hi

AKM is not right angled, nor is MLC, I'm not sure what you mean

[little osprey]

Use flipping Deduction

BKML is a square

[thin wasp]

No it isn't

It's cyclic, but it isn't necessarily a square

[little osprey]

Ok, tell me

what is the proof of a square?

How do you deduce that a quadrilateral is a square/

[thin wasp]

If all angles are 90 degrees and all sidelengths are the same

[little osprey]

Yeah so basically it's a rectangle with 2 adjacent sides equal

So since KMLB has that property

It is a square

[thin wasp]

By your definition, this would be a square

https://gyazo.com/7c6234560cc3c0293152dfff9ba18fab

[little osprey]

It might look werid

[thin wasp]

Just because opposite angles are both 90 doesn't mean its a square

[little osprey]

Lmao

Dude

I said

It has all the properties

of a

rectangle

[thin wasp]

BKML does NOT have the properties of a rectangle

1. <KBL = <KML = 90

[little osprey]

Yeha

[thin wasp]

2. MK =ML

That does NOT imply that:

[little osprey]

Dude I'm soo sorry but I really gtg

You came at areally badd time

[thin wasp]

Can I ping someone else

But I'm really curious what you learnt at school about geometry

If you think that's a square

@little osprey The problem is, you're assuming KMLB is a rectangle when it isn't

https://gyazo.com/b8afa1d4f9f033e4b342c75c6bcf0a85

[atomic geyser]

You can’t assume that it is a rectangle. It looks like a right triangle but it very well could be at 91 degrees or 89 degrees. There is no right angle marking that would prove that it is a right angle

[thin wasp]

@atomic geyser Can you help with the problem?

And it is a right triangle by the problem statement

But BLMK is simply cyclic with <LBK=<LMK=90, not a rectangle

[atomic geyser]

Is this trig? I’m not in it until January-February

[thin wasp]

No?

[torpid horizon]

wtf did i just see someone try to call that a square

all sides need to be equal and it doesnt even remotely

[grim sorrel]

Well you can see:

<MKA + <CLM = 180°

and that

<AMK + <LMC = 90°

Does that help anything

Then you could flip those triangles around

and see that all those sides are same

because those are always true:
<MKA + <CLM = 180°
<AMK + <LMC = 90°

[torpid horizon]

um

[grim sorrel]

From the problem statement

BLMK angles are 360° and <KML = 90° aswell as <LBK = 90°

So they will be supplementary

in all cases.

Helped @thin wasp

?

[thin wasp]

I don't follow

[grim sorrel]

BLMK has four angles so their sum is 360°

KML and LBK are given 90°s

so only ones left are the supplement angles MKA and CLM

well

the adjacent ones

but you can flip around and they still equal 180° in total

[thin wasp]

Uh

What have you actually proved

I follow all of your statements about the angles

But i don’t see how you’re using that to prove the question @grim sorrel

[torpid horizon]

cyrius

question

is this a geometry question your teacher gave

or what

[thin wasp]

No, why?

[grim sorrel]

Just by the definition of those angles you can see (by moving the triangles together) that AKM and LMC always form a right triangle if you move the asked sides together ML to AK or MK to LC (Maybe also mirroring triangles)

[torpid horizon]

just curious

also what are you trying to prove and what are you given?

um

module

question

[grim sorrel]

Look up a bit he posted those @duck

?

[torpid horizon]

why do you keep switching the angle letters to the point where the actual angle is different?

like cla and alc are the same but you're switching the middle letter too...

[grim sorrel]

Not angles those are triangles

[torpid horizon]

oh

that

[thin wasp]

In a triangle ABC with <ABC=90, point K lies on AB. Point L is chosen on BC such that CL=AK. Point M is chosen on AC such that KM=ML. If <KML=90, prove that AK=KM=ML=LC

[torpid horizon]

how does that prove its 90 degrees

[thin wasp]

@torpid horizon

[grim sorrel]

You can get all the angles just from the Problem statement really

after you know those two are 90

I mean of course not exact values but that which ones are supplementary or complementary

[thin wasp]

Ok so once you know those angles add up to some number

You can make a new triangle by joining them

How does that prove that they are isosceles?

[torpid horizon]

ok

so what i see

bm is perpendicular to

ac

[grim sorrel]

<KAM + <MCL = 90°
<LMC + <AMK = 90°
<MKA + <CLM = 180°
<BKM + <MLB = 180°

[torpid horizon]

never

mind

it is not perpendicular

module

do you know how euclidean geometry works

?

[grim sorrel]

Tell me

[torpid horizon]

<KAM + <MCL = 90°

tell me how this can logically equal 90

[grim sorrel]

Well triangle has angles sum to 180 and <CBA = 90

[torpid horizon]

youre basically telling me

[grim sorrel]

sooo

[torpid horizon]

angle A and angle C of the two triangles is forming a irght angle

[grim sorrel]

Yes yes

[torpid horizon]

Well triangle has angles sum to 180 and <CBA = 90

???

[grim sorrel]

So the rest equal 90

that is those angles

[thin wasp]

@grim sorrel I agree with all your angle identities, I don’t know how you use it to prove the question

[grim sorrel]

I didnt

I havent randomly assumed anything

Yes it was

It is in the problem statement that <ABC = 90°

well

Ah that I didnt know

[thin wasp]

@duck#7999 Not necessarily

Also I think I’ve proved it, thank you all

[grim sorrel]

Tell me

The problem is interesting

Yes I know that but triangles

[thin wasp]

@duck#7999 Olympiad geometry probably differs American curriculum geometry

[torpid horizon]

do you use proofs in it

[thin wasp]

@grim sorrel you reflect m over l

[torpid horizon]

oh

so this picture did not need proofs?

[thin wasp]

What

[torpid horizon]

like

do you have to use steps

[thin wasp]

You’re given that certain angles are 90 degrees

[torpid horizon]

and justify each step

or no

[thin wasp]

Give me a moment

[torpid horizon]

thats why i asked you if this was a question your geometry teacher gave you

[torpid horizon]

i see

[thin wasp]

Stuff like q9

[torpid horizon]

ok

dool

cool*

[thin wasp]

This isn’t like the American system (I think?) where you mindlessly bash out “proofs” about angles being equal by some arbitrary postulate

[torpid horizon]

euclidean geometry is pretty much accepted everywhere tho

i believe it was the basis for all geometry

[thin wasp]

Yes, but all those derivations here are assumed

In olympiads, that’s the easy bir

[torpid horizon]

ok

how are the postulates arbitrary though?

[thin wasp]

The harder path is trying to find what constructions/modifications you can make

[torpid horizon]

they are assumed but they havent been disproven

and most of them are theorems btw which are proven

[thin wasp]

You’re getting into a different area

Their only ‘proven’ because of assumption

I have to go for now

[teal dragon]

it isn't "American" to use widely recognized postulates

nor is it mindless

[upper karma]

help

im bad a proofs

at

hmm

@ me

plz

[quaint ruin]

@upper karma

What is a right angle congruence theme

Theorm

[torpid horizon]

@thin wasp

yes

but geometry is all based off assumption...

and i mean the beginnings

which are the planes

lines

i forgot the rest

but after those assumptions everything else can be proven
if you want to take the perspective with geometry as that everything is based off assumption, so is science.
everything is based off something that was assumed and then was expanded

@quaint ruin

right angle congruence theorem is if two angles are right angles then they are congruent

[quaint ruin]

Oh

Thanks

[lament bay]

Can I use AC=AC and BD=BD via reflexive property ?

[quaint ruin]

I dont think so

it looks like a parallelogram

[lament bay]

But I'm trying to prove the 2 triangles are congruent

[quaint ruin]

Oh

Hmm

[lament bay]

So what should I do instead if it's wrong?

[quaint ruin]

i think u should put

Something about the midpoint

Like definition of midpoint

[lament bay]

Its given already

[quaint ruin]

so like de equals eb

oh okay

Fml

umm

U should go with vertical angkes

im no help sorry 😦

i failed geometry

[lament bay]

Lol

It's cool

[split vessel]

please help

on this question i am about to post

https://media.discordapp.net/attachments/502677275536261121/517905014228779038/unknown.png?width=495&height=309

[serene field]

Use whatever tool you like best to graph the function.

Look at what the function does as x approaches the 2 points from either direction.

[split vessel]

i dont understand what you just saied

can you please elaborate?

[serene field]

It says you can use a graphing utility to graph the function.

I assume you're working on 56, with secant.

Use your favorite graphing tool, Desmos, Geogebra, TI-83/84/89/Inspire, whatever you'd like, to graph sec(x).

Look at what the y-value does as x approaches the two values in the question from either direction.

[split vessel]

i still dont get it

i graphed it on my calculator

but i dunno what do i do next?

[serene field]

It's asking you where the graph goes as x goes towards the 2 given values from the left or right.

[split vessel]

so

[serene field]

For example

[split vessel]

what should my answer look like?

[serene field]

As x approaches 0 from the left

It goes to - infinity

As x approaches 0 from the right

It goes to +inf

Just as an example.

[spare garden]

why is tan(30) a negative number? Im trying to make Nicolas Cage Pong and the camera should be at an angle of variable angle and at a distance of variable Distance but the sphere should be at the centre of the screen, meaning that I have to set my y axis high enough to do so. So I made Distance 10 and angle 30, which should mean the y axis should be 5.77, but it was -6.57.

[serene field]

*note that this is in radians.

[spare garden]

does that mean the a my answer shouldn't be tan(angle) * distance?

[serene field]

I think you may want to find out where the camera is initially pointing

[spare garden]

I used Debug.Log(tan(angle)) and it returned -6.57

[serene field]

A sneaking suspicion it's initially pointing straight up or straight down.

[spare garden]

its rotation starts at 0,0,0,0

[upper karma]

@spare garden do you still need help?

[spare garden]

yes

Btw your rocketship is bae

[upper karma]

Awww

Okay

You want to know whyTan(30) is negative right?

[spare garden]

yes

[upper karma]

Wait

[spare garden]

I also want to know how I can position my camera properly

[quaint ruin]

Uhh

[upper karma]

I don’t know much about camera positions, but tan(30) shouldn’t be negative

But it is

Which is weird

[quaint ruin]

tan 30 isnt negative?

[upper karma]

Are you in degrees or radians?

[quaint ruin]

uhh

How u tell?

[upper karma]

What calculator do you have

[quaint ruin]

Dgrees

im in degrees

[upper karma]

Yeah, that makes sense

[quaint ruin]

switch to radian?

[upper karma]

Yes

[quaint ruin]

oh

-6

[upper karma]

Mrcuddles wants to know why

But I’m confused myself

Tan(30) shouldn’t be negative

It’s in the first quadrant

Ohhhhhhhh

I’m so dumb

[quaint ruin]

...

[upper karma]

30 radians

Not 30 degrees

[quaint ruin]

No ur not dumb

[upper karma]

@spare garden

[quaint ruin]

oh

lol

[upper karma]

Okay so tan(30) is 30 radians

I was wondering why

[quaint ruin]

uh can u help me on 3 packets

Its about to be a all nighter for me

😞

[upper karma]

So the reason for this is is 30 radians is about 1718 degrees

That amount is in the fourth qurasrant

[quaint ruin]

wth

oh

[upper karma]

That means that any tangent in the fourth quadrant is going to be negative

What else do you need help with

[quaint ruin]

I need help with 16

Im just guess and checking

my eocs are next week im panicking so much

[upper karma]

Eoc?

[quaint ruin]

End of course

like finals

but

we have 2 finals

One is state required

And the other one is also state required but the teacher makes it

stupid education

[serene field]

From SOHCAHTOA, we know that the sine of an angle will yield the ratio of opposite/hypotenuse.

Furthermore, we know the hypotenuse is the longest side of a right triangle.

[upper karma]

Can you show me your process Mr cuddles

[spare garden]

uhh

[upper karma]

Wake

Wait

I thought I was helping you but I’m helping @quaint ruin

Do you have any work to show for how you got the answer Gary

[quaint ruin]

Hmm

nop

I think i have to use laws of sines

but i pretty much frogot that

My schedule atm

:/

[upper karma]

Rip in biology

[quaint ruin]

I mean

Its only one unit

In ecology

the only promblem

My teacher loves to put trick questions

[upper karma]

That frigginis sucks

Back to the problem though

I forgot the law of sines

[quaint ruin]

hmm

Same

[upper karma]

Okay just looked it up

sinA/a=sinB/b=sinC/c

[quaint ruin]

Oh yea

well fuck this shit

Im searching it up

screw studying

[upper karma]

Noooo

Study

[quaint ruin]

i mean

I will

[upper karma]

@quaint ruin okay I figured out the problem

[quaint ruin]

I honestly gave up on life 😦

[upper karma]

No no

[quaint ruin]

im about to go mental

[upper karma]

It says that the sine of an angle is .28

What is sine? It’s just opp over hypotenuse

[quaint ruin]

yea

[upper karma]

All the opp sides are the first numbers and the hypotenuses are just the last numbers

[quaint ruin]

Oh

[upper karma]

Just kept dividing the first by the last until I got .28

It’s C

[quaint ruin]

ok

THANK YOU SO MUCH

U PROBABLY JUST SAVED ME FROM SUICIDE

[upper karma]

Lol you’re welcome

You can use a calculator on the exam right?

[quaint ruin]

yes

[bright nebula]

then use it lol

[quaint ruin]

Yea

but

when u dont know how to sue it

use

[bright nebula]

ok nvm

[quaint ruin]

and also

like geometry

is stuipd

SO much fucking ruless

[bright nebula]

you don't need to remember them like in organic chem tho

you just need to be familiar with them

[upper karma]

Ohhhh fuck

Don’t say that my dude

If you ever plan on taking calculus at least

When you take calc, you use everything

[quaint ruin]

Im taking calc next semester

[bright nebula]

HAHAHAHAHAHAHA calc will fuck anyonw with that attitude

[quaint ruin]

Algebra2/Pre Calc

[bright nebula]

it will make u your bitch

[quaint ruin]

im only in 9th grade dude

[upper karma]

Not precalc

[quaint ruin]

HELP ME

[bright nebula]

precalc is easy

[upper karma]

Don’t take precalc

It’s pointless

[quaint ruin]

i have to

[upper karma]

Take trig

[bright nebula]

jsut limits and continuity

[quaint ruin]

Im doing that rn

[upper karma]

Wait

[quaint ruin]

its with geometry

[upper karma]

Really??

[quaint ruin]

geometry/trig

yep

[bright nebula]

you guys had choices lul

[upper karma]

You don’t need a whole course for that

[quaint ruin]

I learned about triangles

[bright nebula]

it was shoved down my throat

[quaint ruin]

trig functions

yea

all of those shits

[upper karma]

Limits is like 2 weeks and continuity is like a frickin day

Lmao

[quaint ruin]

fml

[vale inlet]

Calc isn't that bad, just make sure you understand the proofs in class/books, and do lots of examples o:

[upper karma]

Who takes an entire course with just that

[quaint ruin]

im in a special program

for spsecial people

[upper karma]

Yeah, take trig

[bright nebula]

hehehehehe

[quaint ruin]

already took it

[upper karma]

It’s extremely important

[quaint ruin]

its with geometry

[upper karma]

What?

[quaint ruin]

Im taking it rn

[upper karma]

As a separate class?

[quaint ruin]

no

[upper karma]

In conjunction with geometry?

[bright nebula]

someone saying calc is easy is either doing masters or he hasn't faced off against real calc yet

[quaint ruin]

yes

[upper karma]

As like a separate semester?

[quaint ruin]

no

same calss

class

same period

same time

[upper karma]

Are you sure?

[quaint ruin]

gimme a trig

or something

[bright nebula]

pretty much Indian education system

[upper karma]

What’s the unit circle

[quaint ruin]

uhh

[upper karma]

because that’s literally all of trig

[quaint ruin]

circle...

im learning that rn

Sooo

i havent learned that

i have learned radians arcs and shit

of circl

[vale inlet]

unit = 1

[upper karma]

That’s geometry

[quaint ruin]

more unit left

oh

[vale inlet]

unit circle = circle with radius 1

[upper karma]

Alright, let me give you a crash course in trigonometry

You have a circle with a radius of 1

Divide it into 4 quadrants

Like a graph

Only a circle is surrounding it

[quaint ruin]

t!wiki unit circle

[loud cedar]

📖 | ** https://en.wikipedia.org/wiki/Unit_circle **

[quaint ruin]

@upper karma

[upper karma]

Yeah

[quaint ruin]

lemme show

u

[upper karma]

Does your class say trig/geometry

Because they really should be two separate classes

~~also skip precalc there’s nothing you need to know that’s important in precalc

[quaint ruin]

lemme pulll

up

apparetly i dont take trig

[upper karma]

You’re British?

[quaint ruin]

nope

Im from the states

of georgia

[upper karma]

British literature?

[quaint ruin]

yea

we have to take taht

dunno why

[upper karma]

Me neither

But regardless, if you’re not taking trig, take precalc

[quaint ruin]

yea

i have to

[upper karma]

You should take trig instead of precalc

[quaint ruin]

idunno the school system is so broken

[upper karma]

Are they forcing you to take precalc or can you skip it

[quaint ruin]

where is budget and taxes and mortages

they are forcing me

[upper karma]

You should see if you can take trig instead

Go up to the office or ask a teacher that teaches it

[quaint ruin]

maybe

[upper karma]

If not, precalc is the class you need

Because you do you a lot of trig

And algebra

[quaint ruin]

we have to take 8 ap classes

ugh

and the rest are honors

why did i put myself

in this position

[upper karma]

8 in total or in one year

[quaint ruin]

total

[upper karma]

Oh that’s fine

What grade are you in

[quaint ruin]

thats the minimum

im in 9th

First year

[upper karma]

Just take 3 sophomore 3 junior and 2 senior

[quaint ruin]

Im taking ap human geo

[upper karma]

Human geo?

[quaint ruin]

Yes

like population and shit

Elon are u from USA

[upper karma]

Sounds boring

[quaint ruin]

?

[upper karma]

No I’m from Nigeria

[quaint ruin]

Ohh

yea

Our education is broken

[upper karma]

Nah I’m kidding

[quaint ruin]

flawed

[upper karma]

I’m from South Africa

(Elon is from South Africa)

[quaint ruin]

oh

lol

[upper karma]

Well, best of luck to you

[quaint ruin]

i just hope it gets bettwr for me

fuck

yea

I migjt just drop out

[upper karma]

At least get a high school diploma

If you don’t go to college, fine, but keep the option open for yourself

[quaint ruin]

i already got pre-acceptance into colleges

[upper karma]

How

[quaint ruin]

SAT ig

[upper karma]

SAT freshman year?

[quaint ruin]

yea

and my psat

score

[upper karma]

Interesting

[quaint ruin]

my guessing skills are on fleek

[upper karma]

I have much to learn about the art of guessing

[quaint ruin]

so

whenever they give u

like

solve for AB

and there is an x

just plug each one in

as x

and see

which one u get

and one ususally is x and outcome is the answer

[upper karma]

I need an actual example from a problem

[quaint ruin]

hmm

such as

5(x)+50=

solve for that

[upper karma]

-10

[quaint ruin]

plug each answer in

[upper karma]

Too much time

[quaint ruin]

no

[upper karma]

It’s easier to just solve for it

[quaint ruin]

i mean if they ask for the answer

not x

[upper karma]

If they ask for the answer, it’s stil quicker to solve

[quaint ruin]

hmm

i mean it might be

but when u are doing complicated promblems

and sat is timed

fuck time

it goes too quick

u dont have time to solve all those shits

[upper karma]

Move to mars

I’ll fly you there

[quaint ruin]

lol

[upper karma]

Each day is like 36 hours

[quaint ruin]

yeet

well

i wasted like 50min

time to start my hw

lol

[upper karma]

Bye

[quaint ruin]

wait

so

where are u from

[upper karma]

Cali

[quaint ruin]

(not elon musk)

oh

i see

lucky

its 11 for u

[upper karma]

What time is it for you

[quaint ruin]

,time

[somber coyote]

The current time for Garyxue213 is 1:21 AM (EST(-0500)) on Fri, 30/11/2018

[upper karma]

!

[somber coyote]

You haven't set your timezone! Set it using ,time --set <timezone>!`

[quaint ruin]

i think ur cst?

or

idk

[soft kraken]

Volume of room ÷ Volume of 1 basketball

[fair lily]

@soft kraken lolwat

That's WRONG

[soft kraken]

That isn't.

Also, don't @ me

[fair lily]

Ok sure by why?

Ping?

Also it's wrong because you need to take the CUBE of one diameter

If you get what I mean

There would be left space

[soft kraken]

It isn't wrong if you can understand what I said.

[fair lily]

Volume of one basketball is different from volume NEEDED to FIT one basketball.

Your statement is not very precise

So it's volume of room / diameter^3

To be exact

[soft kraken]

No, it isn't.

diameter^3 is pretty much wrong.

[dire rampart]

theres nothing wrong with what manis said

assuming ofc the balls are inflated

which i guess is a reasonable assumption to make

[soft kraken]

^

[rugged moat]

There would be leftover space between each ball

So we would have to take a cube

Whose side legnth would be the same as that of the ball's diameter

Unless they do shit like this

[soft kraken]

Oh no

[rugged moat]

Rather than stacking them on top of each other

Technically both of you are wrong if they pull that off

[dire rampart]

its fine, question probably assumes they're tightly packed anyways

[upper karma]

guys

given 2 points on a random window

how should i choose the third so that the circle defined by those 3 points isnt bigger than the window size?

[balmy bison]

put them close to colinear

oh wait

you want it smaller, so make them in like an equilateral triangle or something

[lament bay]

How would I prove DC=FB, using what theorem?

[vagrant bolt]

aint angle dce = angle fbe by the angles in the same segment theorem so that proves all angles are same so it must be congruent

im not sure if u can say they are in the same segment though

actually yea they must be --> coming off the fe chord

[obtuse saddle]

how will two rays, facing opposite directions, intersect

or are they parallel because wouldnt they have the same slope?

[wild hamlet]

yes

they are parallel

[obtuse saddle]

they are?

Well thanks

[raw terrace]

ok so

does anyone have a compiled list of goemetry theorems

cause @lament bay needs help with learning them

uh

wait actually

i might have somehting oof

one minute

https://quizlet.com/314253989/geometry-theorems-flash-cards/

yeah ok thats what my teacher gave at a final review thing

and told us to study the ones we didnt know

and make flashcards

legit math theorems just sometimes have to be memorized @lament bay but I do recommend when learning them to think about the name of the theorem and relate it to what the theorem actually states

[obtuse saddle]

geometric theorems

like proofs?

@raw terrace

[raw terrace]

yeah

its ok

i got dojonny

[upper karma]

hello

I'm trying to solve this equation in [0;2pi]

4 cos^4(x) - 11 cos^2(x) + 6 = 0

if I declare X = cos^2 (x), I can solve it with 0.75

how can I find all the solutions?

<@&286206848099549185> any ideas?

[keen aspen]

Yeah cos^2(x)=X

4X^2-11X+6

What did you get when you factored?

[upper karma]

@keen aspen I nearly finished but I'm stuck with cos x = sqrt(0.75) or sqrt(-0.75)

I can't get the exact values of these (with pi)

[keen aspen]

You can

[upper karma]

Ok cool how?

[keen aspen]

sqrt(3/4) => sqrt(3)/2

arccos(sqrt(3)/2)

Do you know that angle?

[upper karma]

Oh ok

Nope

I never used arccos before

[keen aspen]

Okay well its 30

[upper karma]

Ok

So 30 or -30?

[keen aspen]

Yeah both

Wait hold on

Cant take sqrt of a negative

Ahhh okay

So yeah -30

[upper karma]

Thx for helping+

[keen aspen]

Np

[stark snow]

Easy question:
So, the line ABE is straight and the ratio between AB and BE is 3 : 2
AB is equal to a.

How do I work out the length of BE?

I was thinking that a is 3/5 of the whole length, so... yeah

[teal dragon]

AB/BE = 3/2
a/BE=3/2
BE/a=2/3
BE=2a/3 @stark snow

[sick veldt]

@hard gale hey hey hey

I got 3 questions I need reassurance

[hard gale]

hoihoihoi

[upper karma]

What’s up squad it’s Young don the sauce god

[sick veldt]

hi

getting a tesla soon

[upper karma]

Neat

[sick veldt]

mans 16 and doesn't have a permit

feelsgood

[upper karma]

Don’t worry

Didn’t get my permit till I went to the states

[sick veldt]

lol

are you good at the geometric questions

[hard gale]

there's 2 wrongs in this q

[sick veldt]

B and D ?

ik C has to be right

[hard gale]

d is only true if you have a square yeah

[sick veldt]

so its just C?

[hard gale]

is ABC

[sick veldt]

damn

how is A right though

[hard gale]

B is ofc true cause that's the def of a rectangle

[sick veldt]

cause of it creating a triangle?

cause of the parallelogram forms two triangles from the diagonals its a rectangle

[hard gale]

+the diagonals have equal lengths in a rectangle (+ the diagonals both cut at their middle cause parallelogram)

so if you take any two half-diagonals in a rectangle, their lengths will be equal

[sick veldt]

ah

Oh christ

its a two column

[hard gale]

geez

bha i can try lel

[sick veldt]

not too hard

these are the other options i can drag in

[hard gale]

wtf is cpctc?

[sick veldt]

LOL

corresponding parts of congruent triangles are congruent

I don't understand it fully

but it seems like the only obvious choice out of HL congruence and transitive

not transitive at alla nyways

[hard gale]

and waht's HL?

[sick veldt]

I've been seeing throughout the lesson of them just using the cptc in the last box

dude idk lol

hypotunuse i think

lemme see

ahh hypotenuse leg

The hypotenuse leg theorem states that any two right triangles that have a congruent hypotenuse and a corresponding, congruent leg are congruent triangles

those are right triangles soooooooo out of the question

Idk if the first one is right

cause AD and BC are opposite

but so is AB and DC

hellow

[hard gale]

seems roight

but screw 2cols anyway

at least when you get to do your own from scratch it's ok

but completing a 2col seems like pure bullshittery to me

[sick veldt]

lol

Its gonna be on the state exam that I gotta pass to graduate :PPPP so I gotta love it

6y - 5y = 10

y = 10

[hard gale]

yas

[sick veldt]

cool okay last one

:PPPP

[hard gale]

yas

[sick veldt]

cool 100%

ty m8

See you monday

[hard gale]

[naive anchor]

Hello : ) I am trying to solve a real world problem. I have a round disc measuring 380mm across, but it has a radiused edge, or fillet, applied to the circumference. With the added material and at the widest point, the diameter becomes 386mm. The pic below shows what I am talking about. Is there a way to calculate the actual radius of the fillet applied to the circumference, or r?

note that the fillet is not semi-circular, so I guess this is an arc length problem ... its been years since i did geometry though : (

[hard gale]

do you know the width of your 'plate'?

[naive anchor]

yeah, 18.5mm : )

its one of these 🤣

i'm using it as a temporary base for a robotics experiment and need to 3d print two clamps either side which will allow me to attach shelves for sensors. I could just have the clamps be normal tangent to the fillet, but would prefer a more 'positive' match for the inside of the clamps to the fillet. I could just do it vaguely, but wheres the fun in that

like so

i trundled through some sin and cosine rule, bit of pythagoras, but theres too many unknowns

my other option is to 3d print a bunch of known radius arcs and see which one fits

[hard gale]

yup there's a double pythagoras involved

big mess lel

our unknowns are h and R

[upper karma]

if you literally have the 3D model on your computer, use boolean modifiers to take the difference a shape at the edges there to make your clamp specifically around the shape there

[naive anchor]

sadly that model is one i made, just an approximation

[hard gale]

if you apply pythagoras in CMO you get MO²+CO²=CM², or (9.25)²+(h-3)²=R²

if you apply pythagoras in CMS, $ MS²+CM²=CS²$, or $$\sqrt{3^2+9.25^2} + R^2 = h^2$$

[somber coyote]

emeric75:

[hard gale]

2eq, 2unknowns

[naive anchor]

ahhhh k k : )

I will poke around with some numbers and try it out! Thank you

[hard gale]

[naive anchor]

one question, might be a dumb one ... how can we be sure that M is a right angle?

[hard gale]

shit idk how it's called in english

[naive anchor]

hha ok not to worry : ) Many thanks for your help, its super appreciated!

[glad ocean]

y = x + 1

what is y's value?

[dire rampart]

x+1?

[glad ocean]

in terms of y and z

[rugged moat]

What's z

[azure lion]

y = 1y + 0z

[limpid basin]

Lmfao

[tired stream]

can anyone here help me with some simple trig

triangle that has 2 known sides = 10root3 and 6 with an angle of 60 inbetween

i need to find the length of the third side

I tried using cosine but my answer came back as wrong

[azure lion]

should work

I got approx. 232.08

maybe your calculator was set to radians

?

unless you didn't mean you used the cosine rule

[upper karma]

hey guys

I have this equation : cos(x) = sqrt(75) or -sqrt(75)

and I would like to find its solutions in [0;2pi]

[serene field]

The range of the cosine function is bounded between [-1, 1].

There are no solutions for cos(x) = +- sqrt(75).

[upper karma]

*sqrt(0.75)

my bad

@serene field so there would be several solutions