geometry-and-trigonometry

[trail tendon]

i always thought if you turn something into a series its called series decomposition

maybe thats something idfferent 💀

[faint pasture]

Nah i was talking about repeteadly using sin(a+b) 💀 it was the only real way to approximate before calc came sooo i mean yea

It is tedious but uea

[trail tendon]

ohhh 💀

what do you know?

[faint pasture]

Is there a better way though? Consider that calc and graphs dont exist

The other 3 are just reciprocals of these

[trail tendon]

sin, cos, tan?

you know that sin(theta) is adjacent/hypotenuse from theta?

[faint pasture]

Look into the unit circle definition

[trail tendon]

yeah

learn the other three trig functions

then look at unit circle

and memorize the values

[faint pasture]

Btw viper this

[trail tendon]

cosecant

(csc)

[faint pasture]

Cosecant or simply csc

[trail tendon]

i don't know of any 💀

i mean

[faint pasture]

XD

[trail tendon]

how do you know sin(1 deg)?

and cos(1 deg)?

if you're solving for a side, and you know theta and the length of one side, and the unknown side is in the denominator if you use sine, you can use cosecant instead

[faint pasture]

For all purposes except being rigorous i just let it = 0 lmaooo

If not using series then i guess there isnf any real way too calculate it except solving the 45th degree polynomial

Oh wait

We know sin(18)

[trail tendon]

well then ur not gonna get anywhere

yea

[faint pasture]

Boom we just have to solve a 18th degree polynomial now

[faint pasture]

We got 2k years before newtons arrival thats plenty of time xD

Extremely

[trail tendon]

i mean theres trig identities

[trail tendon]

really?

[faint pasture]

Citruz. Lmao

[trail tendon]

💀

[faint pasture]

Nvm since we know sin18 we know sin 6

Since we know sin6 we know sin 2

[trail tendon]

so wait

[faint pasture]

Since we know sin 2 we knos sin1

Aha

[trail tendon]

ur assuming sin(1 deg) = 0?

[faint pasture]

Ez pz lemon squeazy

[faint pasture]

Nah i was joking about that

[trail tendon]

💀

[sharp ruin]

Heyy what's going on here?
Couldn't sleep so thought of checking here

[trail tendon]

we're sleeping

as you should be

[sharp ruin]

Lol man sleep talking math
You're wild

[faint pasture]

Just thought of a way to calculate sin(1) give me my nobel already

[trail tendon]

sin(1) = 0, heres ur nobel prize

💀

[faint pasture]

Lmaoooo it is close enough for most purposes tho xD

[sharp ruin]

Can't you just call it zero like we physics people do

[faint pasture]

💀

We can

But you see if we do

Then thatd mean sin3 =0

Then thatd mean sin 30 =0

Soo shit just breaks

XD

[sharp ruin]

No it doesn't
Trust me
Small angle => Take the shit to radians

[faint pasture]

Calculus hasnt been invented yet

[sharp ruin]

Or just say zero and move on

[cunning lion]

sin(1 degree) = pi/180

[faint pasture]

We live in 500 BC

[trail tendon]

🤯

[faint pasture]

Again calc hasnt been invented dont cheat using the taylors series

[sharp ruin]

Ohh then how are we here?
The internet?
🤓

[trail tendon]

did pi need taylor series?

[faint pasture]

I- oky-

[sharp ruin]

It doesn't I guess

[faint pasture]

Sin(theta) = theta

Needs taylor series

I just told you-

[sharp ruin]

Lemme tell you something
Pythagoras wasn't a real person 💀

Believe me he really wasn't

[trail tendon]

how come?

[faint pasture]

We know sin 18, then we know sin 6, then we know sin2, then we know sin1 and since we know sin 1 we know all integer values

[sharp ruin]

It was a cult that called themselves Pythagoreans and had an imaginary idol
They worshipped math

I'm not even lying

[faint pasture]

Lmao wtf

[sharp ruin]

There are literal articles about this

[trail tendon]

the weird thing is we're taught math backwards ._.

[faint pasture]

Articles arent a trusted source tho

[faint pasture]

Are we?

[sharp ruin]

Maan ask any historian

[trail tendon]

you think you're joking...

[sharp ruin]

Or history enthusiast

[trail tendon]

but honestly not a bad idea, to an extent 🤔

[faint pasture]

Or conspiracy theorist

[sharp ruin]

You can teach them graphs though
Really not a bad idea

[trail tendon]

yes

[faint pasture]

Tbf yea itd simplify a lot of shit

[sharp ruin]

Oh is that so?
Show me a record that a person named Pythagoras existed

[faint pasture]

Like no you dont need to teach the kids integration by parts or shit, but telling them that they can integrate a function to find area under the graph and teaching them the basic ones could be nice

[sharp ruin]

It really does and that is wild

[trail tendon]

why not teach them IBP?

its not hard 💀

[faint pasture]

Show me a record that the "village" or "cult" exists

[sharp ruin]

Yeah don't they teach graphs and all here in 8th or 9th grade?

[faint pasture]

Its not, but its not needed at the basic level

[faint pasture]

They do?

[trail tendon]

who said anything about basic level 💀

[faint pasture]

I have been doing graphs since lile 2nd grade

[trail tendon]

real

[sharp ruin]

But they were like simple bar graphs and shit

[faint pasture]

💀 why th would yoy want a 7 year old kid acting like an integral freak

[sharp ruin]

That's whataboutery man

[faint pasture]

They are the foundation of all of calculus though

[trail tendon]

literally why not

[faint pasture]

Lmaooo okay dude go sleep

[faint pasture]

Id get an inferiority complex thats why xD

[sharp ruin]

Like I'm serious about this lol

[trail tendon]

😂

dude i feel like it doesnt even matter how old u are lol

unless ur like 3

but at a certain point your brain is good enough to be able to understand

so liike

why not

[faint pasture]

Hmmm function takes in a value and returns another value

That is all it is

Nice

[faint pasture]

Lmaooo you are wrong though

[faint pasture]

But like you need to know multiplication of polynomials before integrating the said result

[sharp ruin]

Okay then just a sec

[faint pasture]

Lmao

[trail tendon]

yeah

i dont mean skip to calculus

i just mean teach it faster 💀

[faint pasture]

I mean there wont be any proof

[trail tendon]

exactly

💀

[faint pasture]

Kids struggle with the current slow curriculum wdym teach it faster- 💀 math is one of the most difficult subject for half the kids in any gjven room

(Your ed system sucks then- we were taught algebra in 6th grade so like 10?)

We had stats and probability and shit before that

Not here

[trail tendon]

they just dont each it well or smthn idk :l

[sharp ruin]

I found somewhere it was written that Pythagoras was a real person and he created that cult 💀

[trail tendon]

you can learn that during/pre-kindergarten

bruh

[faint pasture]

Lmaoooo i mean logical and creative thinkers are seperate

[sharp ruin]

That's an year or 2 later I guess but not too late in my opinion

[trail tendon]

i feel at least for me i could have been taught things a lot faster

💀

[faint pasture]

I mean sure but its set up as it was required? If kids seemed to understand it in like a year they would shorten it, but it is the kids who are dumb not the tchrs-

[sharp ruin]

Seems okay to me
Why are you guys so pissed about being taught slowly lol

[faint pasture]

I mean you are in mathcord arguing about the ed sytem so yea you are a logical thinker-

[sharp ruin]

I mean all of the systems eventually throw students at the same level when it comes to university

[trail tendon]

am i though 💀

human, to be exact

[faint pasture]

90% of the ppl here are-

[faint pasture]

Irish human to be exacter

[sharp ruin]

Check his sect now lol

[faint pasture]

I hacked you🙃 (you said so above lmao)

[faint pasture]

XD

[trail tendon]

guys linear alg or multivariable first

[faint pasture]

Lmao i even know your religion

[sharp ruin]

💀

[faint pasture]

Linear is funner imo, never tried multivariable

[sharp ruin]

I guess linear algebra first is a requirement isn't it?

[trail tendon]

not at my school, thats what worries me 💀

[faint pasture]

#geometry-and-trigonometry message

[trail tendon]

so im hearing linear algebra first...

[faint pasture]

Are you suree? 🤡

Maybe you are and you just dont know

[sharp ruin]

Like it gives you intuition to visualise vectors and all that
It's needed in multivariable I guess

[faint pasture]

I am in here rn

Its fun

[faint pasture]

That too

[sharp ruin]

Man why are you leading him like that
Lol it's so funny xD

[faint pasture]

XD

[sharp ruin]

Then you should probably learn philosophy
You might wanna kill yourself early 💀
Sorry what the hell did I just say

[faint pasture]

You try to go in space, i try to go below earth (earth is flat obv so i can go below it duh)

Nah earth is just a pringle

[sharp ruin]

It might be a snickers bar

[faint pasture]

New theory just dropped

Ni-

[sharp ruin]

Nickers 🙂

We will be right back (Music plays)

[faint pasture]

this user has been arrested by the fbi for using the n word without a pass

[sharp ruin]

Man it literally means shorts doesn't it?

[faint pasture]

Are those rsndomly chosen too?

[sharp ruin]

Naah obviously not
That's dumb

[faint pasture]

💀

[sharp ruin]

Do USA and China first

[faint pasture]

Austin

Sydney

Melbourne

Hiroshima

[sharp ruin]

Aaah no man not Australia

[faint pasture]

Nagasaki

[sharp ruin]

Berlin

[faint pasture]

Osaka

Beijing

Shanghai

Taipei

[sharp ruin]

New York 💀

[faint pasture]

Mumbai

Macau

[sharp ruin]

Man what the hell lol

[trail tendon]

Perth

[sharp ruin]

Canberra?

[trail tendon]

i don't know australian cities

[faint pasture]

I know like 0 australian cities

[sharp ruin]

Which one is remaining then

[trail tendon]

exactly 🤣

google time

[faint pasture]

Lmaooo

[sharp ruin]

What the hell is that?

[faint pasture]

Sounds like spanish

[faint pasture]

Lawl

[sharp ruin]

It's Australian and no cricket matches I've heard being played there

Doesn't even qualify as Australian

[trail tendon]

😂

[faint pasture]

Lmaooo the criteria is crazt

Ulanbatar was mongolia right?

And which country was lagos in?

[sharp ruin]

My man took out the most underground cities from the whole world just to confuse us 💀

[faint pasture]

Tbf everything except bendigo is kinda famous

Ulaan bataar is the capital iirc

Lagos was in the news a couple years ago

[sharp ruin]

Still not like DC famous or Delhi famous

[faint pasture]

Seattle is seattle

Bangkok is bangkok

Australia perth

Oh lol

[sharp ruin]

I guess I know why he wanna poison that

[faint pasture]

That was quick

[sharp ruin]

China from Asia

[faint pasture]

Dont poison them

[sharp ruin]

I meant Shanghai lol

[faint pasture]

For europe bern

Lol

Frankfurt

[sharp ruin]

Pls now do berlin

[faint pasture]

Berlin

[sharp ruin]

Europe then
Is it done already?

[faint pasture]

I know 2 damn german cities

And its not one of them

So yea

Oh-

Lmao

Africa cape town

Wtf lmao

[sharp ruin]

Nice man

Everyone deserves to die here

So that we can know if it's a simulation or not

One small sacrifice for the bigger fact

[grave pond]

Could we perhaps try to walk the conversation back in the rough vicinity of the channel topic?

[faint pasture]

Oh fuck i forgot that sorry

[wise nest]

chat i searched up 25/35 questions on my geo final yestarday

i think i passed fr

[maiden brook]

hopefully not during it right

[upper karma]

hey just a question about a graph, are these graphs possible(beside circle graph) to get through formulas? the unit circle can be graphed using sin and cos but the following waves used to graph other the shapes I had not been able to find any formulas that represent these shapes. Could they just be demonstration graphs for learning trigonometry? all helps appreciated

I ask this question because like the unit circle I was hoping to draw a square in desmos without many steps

[burnt crescent]

im not very familiar with this but they kinda look similar to square waves and functions from fourier analysis

no idea though i could be wrong in thinking theyre related

[hoary totem]

try and compute the equation for a square

u will see its just a piecewise function involving regular trig in the end

[hoary totem]

they are not very related

[upper karma]

oh yes, I had seen these just recently and they seemed pretty similar, trying to graph them in desmos they even resulted in a graph of the corners of a square so probably the closest so far

[grave pond]

It looks like you're letting a ray from the origin sweep around with constant angular velocity even though that means the point on the polygon moves with a varying speed, right?

[upper karma]

I think that may be the way to go, of all of the graphs ive seen so far this one sorta looks like it could be made with a piecewise function for the different parts

[grave pond]

I'm fairly sure you'll need to define your functions by a case analysis according to which side of the polygon your ray hits at each given angle.

For each of the sides (except horizontal one) the expression in the corresponding case becomes a segment of a scales and moved tangent function.

(But for the hexagon and upwards that will look pretty much like straight lies anyway, so you could just cheat and interpolate linearly between points on a sine curve).

I ask this question because like the unit circle I was hoping to draw a square in desmos without many steps
Oh, I just saw this. I don't think you'll find anything that's shorter and (especially) clearer than drawing the sides of your square one by one.

[upper karma]

ah well darn, I suppose now getting into more complex math it would probably be the easiest way. I appreciate your insights on this, Im not too familiar with some of the terms mentioned, mathwise im still a bit new but ill look into what you said, I think I understand what you are saying though by using segments to then use interpolation to find a function that matches the goal. I will probably stay with drawing the sides then for now, mission continues...

[lime dune]

for what troposphere described you primarily need trig stuff

[hoary totem]

i did it

[upper karma]

oh my gosh

[hoary totem]

i drew a square using c(x) (square-cosine) and s(x) (square-sine)

c(x) and s(x) are graphed too

[upper karma]

you have really done!

thank you!

[upper karma]

Im unfamiliar with the math or maybe need more time to understand how its been done but it is really impressive. I really appreciate your time working out the problem , I am assuming they are the piecewise functions you had been talking about earlier, I suppose now the mission is complete

now that the problem is solved I suppose it is just a matter of learning how, but yeah, thank you for showing me the way

[hoary totem]

i implore u to figure it out urself

[upper karma]

count on it, I am really interested in learning to do it, maybe even achieving more shapes too

[hoary totem]

it should be easy to see the parts where its just 1 and -1

the parts where its not 1 and -1, draw a small triangle

work out the geometry

[upper karma]

I think Im beginning to understand, ill keep in mind drawing out the triangle, I will experiment in desmos once im able and see

[hoary totem]

the hexagon was harder

i only made half of it

because... the rest seemed like a waste of time

the hexagon isnt rotationally symmetric by 90 degrees

which is why the "cosine" and "sine" functions dont look similar

[twin crag]

is k = 1/60? that feels wrong ngl

i used the formula for the period of a sine wave

2pi/A (where A is part of sin(Ax))

then did 2pi/A=120pi

A=1/60

oh wait

this isnt a sine wave

uh

[hoary totem]

well

the particle moves in a circle

[twin crag]

yeah 💀

[hoary totem]

u should recognize that this is the parametric equation of a circle

and uh

well it moves a distance of 120pi every second

which means it made... 120pi/2pi = 60 revolutions in 1 second

[twin crag]

i was just looking at the equation for y my bad

[hoary totem]

so the period is 1/60 hertz

[twin crag]

oh so my logic still applies?

[hoary totem]

the answer is not 1/60

u need to convert from hertz (1 rev/s) to rad/s

[twin crag]

oh alr

thanks

is there a way to convert hertz to radians 💀

oh

wha

[hoary totem]

yes

tbh

wait i dod this wrong

the radius of the circle is 6!

so the circumference is 12pi

[twin crag]

oh so 10 hertz

[hoary totem]

which means the distance of 120pi is only 10 revolutions

[twin crag]

wait would i show this as 1/10
or just 10

[hoary totem]

its 10 revolutions per second

[twin crag]

ok so just 10

[hoary totem]

so the period is 1/10

[twin crag]

oh

[hoary totem]

so wait yes

its 10 Hz

[twin crag]

ok ty

[hoary totem]

convert to radians

[upper karma]

I understand now I believe, I suppose lots of trial and error was required to get the hexagon to appear, half of it though may even be enough though maybe changing one of the points to negative could just flip the graph over and save the extra time. Even on the square its a bit time consuming to sort of sculpt a result out of these graphs

[lime dune]

you don’t need trial and error for this, look up parametric curves to see how they did it

[upper karma]

I think ill check into this for the moment

I will see what I can do but I do think they could be pretty helpful for these graphs

[hoary totem]

yeah it did not involve trial and error for me

[silver fable]

Hi I have two intersecting lines (p0, p1) and (p2, p3) and a radius (R) I need to find the center of the circle tangent to both lines, can anyone help me?

[faint pasture]

P0, p1 refers to points on the line?

[shut bronze]

are there not infinitely many circles tangent to any two given lines

[faint pasture]

Radius is given

[shut bronze]

even so isn't there like 4

and infinite or 0 if they're parallel

[faint pasture]

Yea thats true

[faint pasture]

Lmao but its given that they are intersecting so yea

[shut bronze]

oh true

[faint pasture]

Hmmm

Well the problem is rather doable.if yk the formula for distance between point and line

[shut bronze]

perpendicular distance formula

[faint pasture]

Hmm

[shut bronze]

|ax + by + c|/sqrt(a^2+b^2)

[faint pasture]

Yea

[silver fable]

Yes

[silver fable]

Yeah they are finite lines but it's ok to consider it infinite as I just want to fit a circle between it

[faint pasture]

Well get the equatiom of line from there

[silver fable]

Is there like a direct formula or should I solve the line equation? It's for programming purposes

[shut bronze]

if its finite wouldn't you have to check all 4 and see which ones have the tangent points actually on the line

surely there's an easier way

[faint pasture]

Well try it the normal way and see if you can derive the formula

[silver fable]

Yeah that's also an issue but I was thinking if I solve it for an infinite line first I can get an idea on how to make corrections to have it on the side I want

[shut bronze]

yeah try that

[silver fable]

Also there is one case, one if the lines is always perpendicular to X or Y axis

[faint pasture]

No?

[shut bronze]

how else do you do it

[faint pasture]

Also there should logically be 4 but the method only gave 2 so idk whats the problem

[shut bronze]

what method

[agile slate]

I was messing around with topology, trying to understand the high level overview (i'm a math newb).

After looking at sheaf theory, ring theory, e.t.c, is it safe to say that we are basically trying to model the behaviours of certain shapes and movements/processes that have physical existence using numbers? (for ex. the drum vibrating problem)

[faint pasture]

The circles you get both have the point on line

[shut bronze]

absolute value is on both sides so shouldn't it give 4 results

[faint pasture]

But a=b and -a=-b is the same thing

[faint pasture]

This definitely doesnt belong in this channel

[agile slate]

Sorry again, where should i post it?

[faint pasture]

The finding equation of line then calculating distance and letting it equal to R for both the lines

Then solving the linear equations

[faint pasture]

Iam not aware of that tbh

[agile slate]

alright, thank you anyways

[shut bronze]

oh i figured it out

if you use radius you get 4 solutions

because instead of |...| = |...| you have |...| = |...| = r

[faint pasture]

Hmmm

Isee

[echo beacon]

i got 50 but comment section is all over the place

wait

i’m lost

beauty

[vernal sapphire]

Should be 40

I think

[shadow depot]

Yeah it’s 40

You end up with an equilateral triangle when you draw a line on the right side. The 50-80 line and one of the sides of the triangle produces an isosceles triangle, so you end up getting x + 30 = y, and x + y + 70 = 180

So x is just 40

I feel like I cheated though because I used sin(30) = 1/2

[heady kiln]

Add all the different angles and add to get 360

[trail tendon]

you mean the four angles of the entire quadrilateral?

cuz thats not possible because you only know 2 of the angles

[heady kiln]

wdym? there's four angles visible

[trail tendon]

you mean 50, 70, 80, 60, and ? ?

[echo beacon]

WAIT YEAH IM SO DUMB I DREW THE LINE AS WELL 😭😭

[trail tendon]

bruh

[echo beacon]

it just cuz i was doing it during my lunch break

i was half eating half doing it

on my phone mentally

wait

i’m so confused

[heady kiln]

👍

[trail tendon]

50 + 70 + 80 + 60 = 260
360 - 260 = ? = 100
but the answer is not 100, why do you think this method ^ would work?

[echo beacon]

nvm i got it

with my dads help

so real

bro saw that it could be on a circle

[chilly pasture]

Anyone have any advice on this problem? I think this has to do with trig identities but im unsure what im doing wrong.

[gentle haven]

First thing I can think of is use the quadratic formula

[obsidian harness]

Correct so far, now if tan x - 8 = 0 then tan x = 8

not -8

Oh also there are infinitely many solutions if your domain is the real numbers

Yeah so you need to be adding +n pi to your 2 principal solutions

just arctan them and ye

[olive jetty]

is this a construction of a median?

[olive jetty]

@obsidian harness

[obsidian harness]

this is a median, so no

[olive jetty]

oh so it has to be B then, correct?

[obsidian harness]

nope

those two lines are definitely not parallel

[olive jetty]

is it a perpendicular line?

[obsidian harness]

yes

[olive jetty]

ty🙏

[digital pebble]

yo @everyone, anybody willing to help me for my hw

[gentle haven]

Perpendicular bisector I think

[obsidian harness]

actually no

basically you have one point A in the diagram, and not two points say C and D

ofc you construct the perpendicular to point A by using your compass to mark out equidistant points C and D

[gentle haven]

Oh wait I just realized it's not a line segment

No bisector then

[obsidian harness]

mhm

[tall coral]

I have literally found everything BUT what I'm supposed to

[shut bronze]

did you use the fact that those two specific segments are equal in length to find any of these

i feel like if you remove that restriction all of this still holds

[tall coral]

nope! I just used a bunch of angle equivelencies

[shut bronze]

well yeah im pretty sure if you allow moving the point at the bottom there than all of these can hold with varying values for the missing angles

you need to somehow use that fact

sounds like trigonometry

[trail tendon]

not everything

[tall coral]

idek what I am doing anymore iguohfdszaOL

[grizzled hornet]

is the ans 30?

[tall coral]

not from what I got

I just got it

[grizzled hornet]

do you know the ans?

[tall coral]

Here's what I got (quite a convoluted model, but it's apart of the process)

[trail tendon]

how do you get 30?

[grizzled hornet]

i dont know

[faint pasture]

Who said the quadrilateral is cyclic

[tall coral]

one of the triangles has two equal angles meaning it's iscosoles

[grizzled hornet]

@tall coral how did you get 70 on the upper triangle?

[faint pasture]

Which one?

Oh nvm

I saw it

[trail tendon]

i dont understand the blue 70 degrees

i think its gotten from the blue 120

but idk how they got the blue 120

[faint pasture]

Me neither

[trail tendon]

unless its the arc degrees

in which case idk what that has to do with the angle

[faint pasture]

I just saw the isoceless thing

[grizzled hornet]

yes i also dont know

[tall coral]

I thought opposite angles of a cyclic quadrilateral add to 180

[faint pasture]

But the thing is its not neccessarily a cyclic quad

[trail tendon]

whats a cyclic quadrilateral

[grizzled hornet]

aah yess

[trail tendon]

oh

[faint pasture]

Since the green point can move to anywhere out of circle

[grizzled hornet]

opp angles make upto 180

[faint pasture]

And shit still satisfies

[faint pasture]

Vertices on a circle

[trail tendon]

a quadrilateral inscribed in a circle, where all the vertices are touching basically?

ah

[faint pasture]

Hmm

[trail tendon]

yeah the opposites would add to 90

but yes we dont know

[tall coral]

it can't, or the given information would be wrong

[faint pasture]

How?

[tall coral]

the orange 80 degrees would change if the green point moved

[faint pasture]

It wouldnt?

[tall coral]

this green point?

[faint pasture]

Green point as in the top right ome

[faint pasture]

No-

[tall coral]

oh

moving that would change the red 70 degrees then

[faint pasture]

So how do you know thr current configuratiom leads to it being cyclic?

[tall coral]

let me clean the model up a little with geogebra

[faint pasture]

Ill be back in a bit

[tall coral]

oh wait I see where I went wrong

[faint pasture]

Hmm got any other leads yet?

Angle chasing is half the reason i despise geometry

[faint pasture]

Also idk if you know about that but i think i saw a theorem for this exact kind of situation where the bisector of a side is equal to the bisected lengths it was related to angles too i dont really remember its name or what it stated tho if im being honest

[tall coral]

just more lines

if I can prove these two are equal, then the whole thing comes together nicely

no wrong angle

if I show all this is true, we're gold

[trail tendon]

you mean if you can prove that those are simlar triangles or smthn?

no

thats not what u ment

wait yes maybe

[faint pasture]

Guyz isnt geometry so easy

✨

[tall coral]

wait that wouldn't even work

[trail tendon]

i would say geometry is harder than algebra or precalc, imo 💀

i know like

[faint pasture]

I was being sarcastic 💀

[trail tendon]

you can pick "hard" problems from any one of them

[trail tendon]

i know 💀

[trail tendon]

but the geometry ones are most often the hard hardest

[faint pasture]

Cuz you can literally pick an IMO problem and it still is just geometry xD

[trail tendon]

💀

[tall coral]

let me start over

[tall coral]

I'm ginving up tonight, I got nothing.

[faint pasture]

or just hear me out lets just call that problem unsolvable and move on

[hoary totem]

ezzzz

took about 5 mins with a friend

[faint pasture]

who said its a cyclid quad though

[hoary totem]

its gotta be cyclic

cuz of the equal lengths and also the right triangle

it must be cyclic

[faint pasture]

argh iseee

[echo beacon]

yeah but how did u guys realize it was a cyclic quad

i would never think of it

[echo beacon]

the two equal bases form the diameter

[hoary totem]

...because of the equal lengths?

[echo beacon]

and there is a 90 degree angle

[hoary totem]

and the right angle?

i said so

[echo beacon]

yeah

[hoary totem]

yeah

[echo beacon]

and then

the equal length at 50 80

radius

[hoary totem]

and then it was obvious

[echo beacon]

bro I WOULD NEVER THINK OF THAT

[hoary totem]

that its cyclic

[echo beacon]

HOW DO U GUYS

THINK OF THAT

😭😭😭

[hoary totem]

it was the obvious next step

lol

[echo beacon]

[hoary totem]

idk what to say

[echo beacon]

[violet vale]

hi

[sturdy stone]

are you the real doctor house

[nocturne remnant]

It kinda looks like concyclic + there’s not really another easy way to do the problem if it’s not concyclic

[deft drum]

I am unable to find another answer from trigonometric and algebraic manipulations. Please help me find another option (not the answer directly, but an approach)

Currently the expansion of $(\cos x + \iota \sin x)^3$ looks like $$(\cos^3 x - 3 \cos x \sin^2 ) + \iota (3 \cos^2 x \sin x - \sin^3 x )$$.

[somber coyote]

ĐARK々MÁTTER

[faint pasture]

well compare the coefficient of the imaginary parts

[faint pasture]

.

[deft drum]

Yes but there is one more correct answer

[faint pasture]

oh replace cos^2 (x) with 1-sin^2(x)

[deft drum]

Ah using cos^x + sin^x = 1 identity

These are too hard for me to remember, I will check

[faint pasture]

hmm

[faint pasture]

give me a challenge for trig (not directly related to triangles or other geometric figures)

[chilly pasture]

Thank you that really helped!

Now im still stuck on the solutions. arctan (1) is just pi/4, but arctan(8) Im not sure. My graphing calculator gives pi/2 -arctan (1/8). The question requests it be in radians so the answer will be something like (?) +pi(n), pi/4 + pi(n)

[faint pasture]

arctan 8 is not simplifiable

,w calc arctan(8)

[somber coyote]

Results provided by WolframAlpha

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

Requested by enderdoesntmind

[faint pasture]

unless you wanna give this as an answer that is

[mellow quest]

∠BAC = 60 ∠AHD = 51
H - orthocenter
D - incenter
∠ABC = ???

[formal geyser]

I got no idea how to solve it

I mean, i calculated the 120 angle at the top of bottom triangle

And 60° angle also

But it didnt help

[faint pasture]

sine rule

too find the length of the bisector

then cosine rule in upper triangle

[tall furnace]

Hi , I think u can find the solution easily with this figure

[faint pasture]

elaborate please

[tall furnace]

all the angles O=60

so X//Y

[faint pasture]

what- why

[tall furnace]

um... if X//Y, all Os=60

[faint pasture]

not neccessarily?

[tall furnace]

so right-bottom tri's angles are 40,60,80

[mellow quest]

idk

[grave pond]

How do you get the angles to the left of the vertical line there?

[mellow quest]

https://en.wikipedia.org/wiki/Orthodiagonal_quadrilateral

[grave pond]

Can you explain a bit more?

I see there's a kite in your diagram, but not how you conclude anything about the angles to the left of its vertical diagonal.

[mellow quest]

guess I was wrong

[grave pond]

Crunching trigonometry in a coordinate system says the top angle of the triangle is 87.88° rather than 90°, so your numbers would look almost correct on a scale drawing.

[plain nexus]

please help me

If you can answer in Turkish, translate it into Turkish

[subtle birch]

is english fine?

[subtle birch]

first use law of consines to find DB/EB

then find <ACB

then use law of cosines again to find AB

and at last use law of sines to find x

[worthy orchid]

X=26

[tall coral]

how did you prove that that was an equilateral triangle?

OHH I FORGOT ABOUT MY SPECIAL TRIANGLES

UGHHHH

[humble lily]

@heady token

[verbal moth]

hi im new to discord, can anyone help with taking pictures and sending it to the group, im also on computer

[bronze owl]

Are the angles the same?

[humble lily]

aaaaaa

yes

[bronze owl]

You can use the angle bisector theorem to find this

4/3 = x/9

Same for the other triangle

[obsidian harness]

Download the Discord app on your phone so you can upload pics directly

that's the easiest way

[humble lily]

Yay

[bronze owl]

Not exactly

You can have an angle bisector that divides it into two triangles that arent similar

But the a proof of the theorem does involve similar triangles

So pretty close

[cold stream]

why is it b²+c²=a² in the Pythagorean identity sin²+cos²=1 and not a²+b²=c²?

or its just how my teacher put it

[maiden brook]

what?

b^2 + c^2 = a^2 in a right triangle just means the side with length a is the hypotenuse

still valid

[cold stream]

aight

[faint pasture]

A,b,c are just variablrs

They can mean anything

[cold stream]

aight

i just was a little bit confused

[severe temple]

Hey uh sorry but can you explain something to me? I have to prove that if a×tan(A)+b×tan(B)=(a+b)×tan((A+B)/2) in a triangle, then a=b and for some reason I get a/b=-1. If the first 2 pluses would be replaced by minuses I think it would work...? I can show you what I wrote but handwriting is really bad and I use romanian notation for trig functions. (Like tg instead of tan)

[heavy lava]

I have test tommorow and still confused on how to solve trig equations, could someone help explain a few of these review questions?

[wicked jetty]

Stuck on this

[velvet pendant]

I HAD MY MATH FINAL EARLIER TODAY! IT WAS EASY, I THINK I’M GONNA END THE SEMESTER WITH AN A!!!

[velvet pendant]

Oh hi, what identity are you attempting to verify?

[trail tendon]

ayyy nice 😂

[velvet pendant]

Frrrr

[trail tendon]

what was it on?

[velvet pendant]

Probability and stats and basically all of trigonometry

[trail tendon]

ayy XD

[velvet pendant]

Well I calculated the score I need on it to end the year with an A

I basically only need to get a 30/45 or a D

And I prolly got an A on it

So yay

I ended up completing the entire AP stats khanacademy course to study for it 😭😭😭😭 even tho it was just a mini unit and I’m not even in ap stats

Can’t wait for ap calc A and pch next year

[bronze owl]

What da question

[fleet compass]

Has anyone ever done the flvs geometry 1 and 2 I am struggling hard with it if anyone could help out with it.

[cold sun]

how do i use cosine rule to find angles?

whats the thing for that again

[bronze owl]

If you already know the lengths of all the sides, isolate the cosine function and then inverse cosine both sides

[velvet pendant]

On its m<c = acos((a^2+b^2-c^2)/(2ab))

You can get that by rearranging the law of cosines, which is c^2=a^2+b^2-2abcosC

[cold sun]

ended up rearranging it yea

[velvet pendant]

Oh ok

[cold sun]

thanks

[velvet pendant]

Someone pls verify that
5sin38/sin17 = (-5tan55tan38)/(sin55(tan38-tan55))

(degrees not rads)

[weary remnant]

Just plug it into a calculator

?

[grave pond]

If it seems to be true with the calculator, actually proving it should probably start with sin(17⁰)=sin(55⁰-38⁰) and a sum formula, followed by algebra crunching.

[red isle]

what’s the meaning of the distance of non-coplanar straight lines？

[faint pasture]

Any two straight lines will always be coplanar no?

[red isle]

?

[grave pond]

Not in 3d space.

[faint pasture]

Nvm

[red isle]

in 3d space

not parallel

[grave pond]

If the lines are not parallel, there will be exactly one line that intersects both of them at right angles.

[red isle]

i'm interested in how to prove it.

[grave pond]

The distance between those two intersections is the shortest possible distance from a point on one line to a point on the other.

[grave pond]

I'd use coordinates, with some conveniently chosen coordinate system.

[red isle]

We haven't learned about spatial coordinate systems yet, can we use the synthesis method?

[grave pond]

Undoubtedly, but I can't immediately rattle off a proof. What is the exact claim you want to prove?

[red isle]

“If the lines are not parallel, there will be exactly one line that intersects both of them at right angles. ”this proposition

[fading wing]

Does anyone know where i can study plane geometry and Trigonometry (ratios, functions, equations, identities)? bec i have a uni addmission test in 2 days and i never studied those in school

[lime dune]

if you're just starting to study these two days before the exam

all i can say is

you should've started way earlier

[fading wing]

ik i made a mistake but now i need help where can i study this from?

[devout elm]

demos gone mad'

x^x ?!

[faint pasture]

Its correct though

[trail tendon]

how?

[devout elm]

how ?

so many limites

[faint pasture]

(-0.4)^(-0.4) is defined though

[devout elm]

why

[faint pasture]

I cant elaborate rn, ggl it youll find the way

[devout elm]

(-5/2)^(2/5)

[faint pasture]

Hmm

[faint pasture]

So desmos has just pointed out multiple points like these its all correct

[devout elm]

maby it just cant calculte it

[devout elm]

\sin\left(x\right)\cos\left(y\right)=\tan3

sin(x)cos(y)=tan 3

ANY IDEA WHY

[faint pasture]

,w calc tan(3)

[somber coyote]

Results provided by WolframAlpha

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

Requested by enderdoesntmind

[faint pasture]

Hm since its less than one there are gonna be infinite pairs of x and y

Sooo y not

[devout elm]

yha .. trig funtions are weird

[faint pasture]

I think fun but okay

[devout elm]

im rn in 9th .. so for me .. i dont understand some things

[faint pasture]

Well thats what makes it fun

If you know everything then its boring no?

[devout elm]

very true .. my summer holidays are going on and i have target to finsh calculus

though sometimes its fun to already know \

[lime dune]

that … sounds like a rather lofty goal

don’t rush it

[devout elm]

IK

but if i go slow then ill take me 1 year to do it

[lime dune]

realistically if you’re still trying to wrap your head around trig funcs

im not sure you’ll be able to do it in a single summer

[devout elm]

uhhhhhhh ............

[lime dune]

don’t rush, make sure you actually get what’s going on before you “move on”

[devout elm]

k.. ill do trig first..
any way to learn trig better ?

is there online website ?

i cureently use khan acadmy

[faint pasture]

Idk i did calc 1 and 2 in like a month combined

[lime dune]

it's easier when your foundations are strong

which is not always the case

[faint pasture]

Hmm thats true

I didnt have to look at the precalc stuff during that month

[lime dune]

yea figured

[devout elm]

noice

[tribal cobalt]

I need help with Stereometry. Anyone who understands this? I really need help. tomorrows a exam. :)

[shy cape]

hello

[stray junco]

ive been breezing through my homework till this point i have 0 clue on how to go about this. ive gone through khan academy and everything and i need help with it

🙏

[grave pond]

Do you know how to compute the area of a sector of a circle if you already have the radius and its central angle? Or the radius and the length of the arc?

[stray junco]

i understand how to calculate the area using both but i havent been able to figure out how to use those two to find the radius

i confused myself alot and i tried converting the central angle to degree but still came out wrong

[grave pond]

Can you explain, with either words or formulas, how you would compute the area if given the radius and an angle in degrees?

[shut bronze]

theta/2pi * pi*r^2 = theta*r^2/2

fraction of area * total area

2 pi is 360 degrees

[grave pond]

Great!
So if you insert the known values there, you get the equation
3 = (1/5)r²/2

[stray junco]

thank you guys so much, i finally solved it thank you!

[grave pond]

Whoops, sorry, I confused you for Jess.

[stray junco]

no worries, i already knew how to compute it i just have mad adhd and needed someone to help for a sec

[shut bronze]

oh, he asked the question and you were trying to help him

because you phrased it like a question i was just answering you

my bad

[grave pond]

I must have sounded insufficiently Socratic. :-)

[sacred beacon]

hI

It is a Geometry problem "Draw a plane P and then draw a line and a semi-line on it"
pls help me

[harsh wasp]

So im trying to find how many mm this Colume tappers in relation to the hight so I can get an angle, it's for a wedge that will angle the machine to cut the slope, I was thinking rise over run but idk if I'm using the equation properly.

[steep fossil]

can anyone help me solve these geometry problems 🙏🙏

[brittle bridge]

U gotta be more specific bro

[steep fossil]

sorry ive finished most of that its just these two problems are really confusing me

[azure helm]

How many units apart is any pair of parallel sides of a regular hexagon with side of $6$ units? Express your answer in simplest radical form.

[somber coyote]

studying_calc_real_analysis

[maiden brook]

any ideas?

[azure helm]

you are asking to me?

nah, I dont have enough geometry intuition, can you guys hint me on this?

[maiden brook]

draw a diagram and draw what u want

[azure helm]

mmm I dont have drawing

but each side is 6 units

and we have 3 parallel pairs

[maiden brook]

select two of them it doesn’t matter

draw the segment that marks the distance u r looking for

[azure helm]

its a rectangle

now what

interesting

[maiden brook]

a rectangle?

can u send the diagram u made

[azure helm]

@maiden brook

its a equialateral triangle

but

how do I continue after this?

[maiden brook]

yeah so u want that horizontal distance in the center there that u drew right?

if u take one of those two smaller triangles that make up that equilateral triangle, what can u say about those?

[azure helm]

oh yeah

by pythagoreas we can get the result if we multiply it by 2

basically calculating one of the sides of one of the right triangles

anyways, i got ||6sqrt(3)||

[maiden brook]

yes

[azure helm]

how do I solve this ?

I tried inscribed angle theorem but didnt do the trick I think

[tepid basalt]

Glad I don’t have to learn this 💀

[sand epoch]

OB and OA are both radii

It's an isoceles triangle

180-110-2x=0

From there it's pretty intuitive

[azure helm]

why isosceles tho?

circumcenter propertie?

[sand epoch]

360-100-110=150

It's a circle

O is the center

[azure helm]

i know coa is 150

but

what about isoceles

why???????????????????

[sand epoch]

If O is the center of the circle, anything line starting at O that touches the circle will be equal to the radius

[azure helm]

wow

[sand epoch]

Wait nvm it's the circumcenter

Lol

[azure helm]

no, I got you

but why isoceles though

[azure helm]

its isoceles because lines starting at circle and ending in circumcenter is radii

anyways I solved, it was 50

but I wanted to ask about this one now, if someone can help me a bit

The first square below is in position ABCD. After rotating the square $90$ degrees clockwise about its center point, the second square is in position DABC, as shown. Next, square DABC is reflected over its vertical line of symmetry, resulting in the third square in position CBAD. If the pattern of alternately rotating $90$ degrees clockwise and reflecting over the vertical line of symmetry continues, in what position will the $2007$th square be? Write your answer starting with the lower left vertex and continuing clockwise with the other three vertices. Do not use spaces or commas when entering your answer.

[somber coyote]

studying_calc_real_analysis

[bronze owl]

You could try analysing the movements of vertex A and B, and finding some sort of pattern in it

[tranquil glade]

in hyperbolic geometry what similarity of triangles hold

[maiden brook]

when do u first get back to the original position?

after how long?

[azure helm]

after 5th is rea-eated

if we 2007mod 5 is 2

cbad

[bronze owl]

It goes back to position 1 at the 5th iteration, but if you take 2007 mod 5, you would be accounting for the original position twice

[obsidian harness]

look up hyperbolic law of sines and cosines

the hyperbolic law of cosines depends on the Gaussian curvature

so SSS doesn't apply

nor does AA

but I think SAS should still work

hyperbolic law of sines is strikingly similar to the regular one btw

[grave pond]

Hyperbolic triangles are similar only if they're congruent.

[obsidian harness]

ah that makes sense

yeah why did I mess up the SAS lol

[faint pasture]

if im given an equation of pair of straight lines

how can i get the two lines from it

(dont suggest quadratic formula- thats too damn slow)

[grave pond]

Just for clarity, you have a single quadratic equation in two variables, and you happen to know from somewhere that its solution set is the union of two lines, and you want to figure out what those lines are?

[meager skiff]

I was wondering if this is SAS?😭

[sturdy stone]

it is

[meager skiff]

Alr thanks broski

[faint pasture]

yea

like the quadratic eqn has been formed by multiplying equations of two lines

and i just want the equations

[sturdy stone]

factor

[faint pasture]

the coefficients arent setup for that to be ez unfortunately

[sturdy stone]

welp

[faint pasture]

;-; is there no decent way then?

quadratic formula's my only hope?

;-;

[sturdy stone]

you could multiply out (ax+b-y)(cx+d-y) and equate coefficients and solve the system

but thats definitely also slow

[faint pasture]

that's worse than the quadratic formula-

[sturdy stone]

yh exactly

[faint pasture]

;-; ah okay thanks anyways

[grave pond]

I don't think you can avoid solving at least one or two single-variable quadratics along the way ...

If your equation is, say, px² + qxy + y² = 0, then the result is two lines crossing at the origin, whose slopes are the two solutions of m²+qm+p=0.

[faint pasture]

those are rather doable but the ones with px^2 + qxy + ry^2 + sx + ty+u =0 are too time consuming

[grave pond]

Okay, if solving single-variable quadratics is acceptable, then we have something to work with.

View your equation ry² + (qx+t)y + (px²+sx+u) = 0 as a quadratic equation in y, and set its discriminant to 0. That gives you a quadratic equation in x that you can solve to find the x-coordinate of the intersection point. (This should actually be quick, since if your original equation is indeed two intersecting lines, the equation in x has a double root!)
Then plug that x in and solve for y. You already know the discriminant is 0 now, so the calculation is again simple.
To find the slopes, now you can simply ignore the terms of degree <2 -- this corresponds to switching variables to translate the intersection to the origin, but you don't need to actually do the calculuations, because they don't change the degree-2 terms, and you know the lower-degree coefficients will end up at 0.

And now the only full application of the quadratic formula will be finding the slopes from px²+qxy+ry²=0 as above.
Then you can write down the equations of the lines in point-slope form.

[faint pasture]

why do we set the discriminant equal to 0? it can just be a perfect square no?

[grave pond]

We want the x-coordinate where there is exactly one matching value for y.

[trail tendon]

complete the square.... naj:kskull:

[faint pasture]

OH i seee! thank you soo much

[grave pond]

(You'll need special cases if the original y² coefficient vanishes, in which case one of the lines is vertical).

[faint pasture]

💀 i- try that lmao

[grave pond]

(Also, if "discriminant(x)=0" ends up having no x² term, then the lines are parallel -- or else your original equation wasn't for two lines at all but for a parabola).

[grave pond]

The algebra actually works out pretty nicely, giving the intersection point
$$ (x_0, y_0) = \left( \frac{2rs-qt}{q^2-4pr}, \frac{2pt-qs}{q^2-4pr} \right) $$
and the common denominator $q^2-4pr$ is also the discriminant in the equation you need to solve for the slopes afterwards.
If it is negative, the conic was an ellipse (or a single point or nothing), and the above expression is its center.

[somber coyote]

Troposphere

[grave pond]

It the equation was a hyperbola, the very same procedure gives its asymptotes!

[devout elm]

hiii ...
how to visualise the trig functions on unit circle
(i dint get tan and cot )

[devout elm]

yha.. but why

[obsidian harness]

similar triangles

so for example, CBD is similar to BED for tangent

[devout elm]

yes .. but.. then what

[obsidian harness]

you'll honestly learn more if you try it out by yourself

[devout elm]

k.. lemme see

[obsidian harness]

e.g does CB/BD = BE/ED work? and so on

[devout elm]

it works as it is similar

[obsidian harness]

I mean if you only know that CD = cos x and BD = sin x, can that tell you that BE = tan x?

(and that CB = 1 ofc)

[devout elm]

yes becouse it is slope

[obsidian harness]

😭

[obsidian harness]

okay yes tan = sin/cos is the slope, so the slope of line BC yes

but that's completely unrelated to BE

in maths there's not always an easy answer or a 1 line reason for everything

[devout elm]

the thing i dint understand is if i make a tangent to the hypotaneus .. how can we say its sin/cos

the tangent if extended can also be cot

[obsidian harness]

so as before, you can show that BE = tan x using similar triangles CBD and BED

assume you only know that CB = 1, BD = sin x, and CD = cos x

[devout elm]

if they are similar .. i can say sin / cos = DE/ sin ?

[obsidian harness]

yes!

you're almost there actually

cause you want to find BE which is a hypotenuse

So you need to be using CB from the triangle CBD

[devout elm]

uhhh .. now i confused

ohh got it

now what about cot

[obsidian harness]

sinilar, so if you call the intersection of the pink and red lines F

[devout elm]

ohh yaa got it

[obsidian harness]

triangles FBC and FAB are similar

[devout elm]

thanx a lot 🙂

[obsidian harness]

yeah let me try, so sin/1 = cos/x, x = cos/sin yep

npnp

[elfin cave]

hey uhh, so im kinda stuck on this. tangentX=2/3 then to find the missing angle you would do Tan^-1(2/3) but then the angle would be .588. i have no clue whether im right, and im even more confused how an angle is .588 on a right triangle

[lime dune]

radians!

if you want it to be degrees, multiply by 180/pi

also there's nothing wrong with an angle of 0.588 degrees; certainly it doesn't make sense in the context of the question you're doing

but in general it's pefectly possible for such an angle to exist

[elfin cave]

thank you so much, that clears up everything

[lime dune]

npnp

[stuck sleet]

wouldn't this whole question is wrong because a triangle with side lengths: 3-4-5 is a right triangle?

[silent plank]

why do you think that means its wrong

[upper karma]

there is nothing wrong with it

[fading basin]

what is the best way to do the following
I have a plane
a vector on that plane, v
and integer grid on that plane
I want to find the closets integer vector to v, such that its length is less or equal to length of v

[obsidian harness]

depends on what you mean by 'closest', but one approach is by cosine similarity where you aim to maximise the dot product

out of all the integer vectors that have a similar length

so if your vector v had length 8.7, you would do 8.7^2 = 75.69

so you would look for all integer vectors such that x^2 + y^2 = 76 for example

[fading basin]

I think it is called CVP or something
clostest vector problem
I can see ppl discussing it for arbitraty lattices -- but this is where it gets hard
I only need it in 2d for integer lattice

and this specific one -- people are saying it is easy

[obsidian harness]

ah 2D is nice

if you can find the coordinates of the endpoint of vector v, just look at the four nearest vectors

[fading basin]

yea
(well, I have additional conidion that I only care about the closts vector that is shorter)

[obsidian harness]

oh true hmmmm

[fading basin]

so, what is the algo

where can I read about that

[obsidian harness]

IDK I'm not a CS person

[fading basin]

damn(

[obsidian harness]

it just sounded like an interesting problem

[fading basin]

oh
actually tho
this is just rounds up and down, correct?

and one of them is bound to be shorter?

[wooden river]

why is the answer quadrant 2?

also is it negative beacuse of the < symbol from the question?

[trail tendon]

it's in quadrant four

[wooden river]

why does it say theta is in quadrant 2 though?

[trail tendon]

it says quadrant IV, look closely lol

[trail tendon]

IV -> 4

stil kinda small but yeh

[wooden river]

damn okay

i see that now

[trail tendon]

yeh

[wooden river]

lol thanks....

[trail tendon]

they hab messy writing 💀

[wooden river]

yea

blame my prof

[faint pasture]

thats a W

[trail tendon]

real 😂

[faint pasture]

puns intended

[wooden river]

so is it negative because its less than?

in the question\

negative values

[dull pilot]

how to rotate a rectangle

[trail tendon]

quadrant 4 is under the x axis, so the y values are negative so the sin is negative.
quadrant 4 is under the x axis but to the right of the y axis, so the y is negative and the x is positive, so tangent is y/x is negative/positive -> negative

[wooden river]

or is it beacuse since quadrant 4 is negative

[trail tendon]

for quadrant 4, the y values are negative and the x values are positive

sin is the y
cos is the x
tan is sin/cos is y/x

[wide mason]

hey i nee help in geometrey

Geometry*

if anyone knows pls dm me

i need help

[upper karma]

i have enrolled in geometry

5 minutes ago

[trail tendon]

nice

[upper karma]

i have learnt about

3d space

2d space

1d space

rays

segments

lines

[trail tendon]

oo

[grave pond]

You're a lot more likely to be helped if you actually ask some questions either here or in a help channel (see #❓how-to-get-help).
People are not very likely to promise you they're going to help, when they don't even know if the particular problems you have are ones they can help with.

[harsh crow]

50 degrees because central < BOC is 100 degrees and <BAC and <BOC intercept the same arc and the inscribed angle is half the measure of the central < meaning m<BAC = 100 deg/2 = 50 degrees

[azure helm]

exactly detective L

but that was over a week ago

[harsh crow]

im just scrolling up lmao

[harsh crow]

also i think this one is 2 sqrt 3

because apothem = sqrt 3 so 2 times that is 2 sqrt3

but im not sure

[azure helm]

I got 6sqrt(3)

u sure?

maybe I did mistake

[harsh crow]

maybe i did

holdo n

[harsh crow]

my mistake

yeah its 6 sqrt 3

[azure helm]

[harsh crow]

i got sqrt 30 im not sure tho

[harsh crow]

• sin^2(x) cuz 1-csc^2x is the same as (sin^2(x)-1)/sin^2(x)) and doing division u get -sin^2(x)

[upper karma]

. In the convex quadrilateral ABCD given below, m (∠BCD) = 90°, |AB| = |AC| and AC ∩ BD = K. If the areas of triangles AKD and BCK are 10 cm² and 25 cm² respectively, how many cm² is the area of ​​quadrilateral ABCD?

A

10 cm

D

K

2

2

25 cm

B

C

A) 55

B) 60

C) 70

D) 105

E) 90

[deft drum]

I am learning functions for limits and continuity where the instructor mentions the domain and range of function depends on its nature like what all values it can recieve and return.

Here they mention domain is set of all real numbers for the area of circle, now since they included extra information shouldn't both range and domain be both positive real numbers? Negative radius doesn't make sense to me.

If area of circle wasn't mentioned then we could have concluded domain as all real numbers.

[grave pond]

The expression pi r² makes sense in itself no matter what r is. It's just only when r>0 that it is useful for finding the area of a disk.

[lament shore]

This is a GCSE maths question so it shouldn't be hard for most people here,😅 can someone help me with this?

[faint pasture]

yea soo let the distance between the bases be B
and the height of tower be X
thenn try to draw out a diagram which includes the 19 degrees angle

[meager skiff]

Anyone know how I can survive Geometry in the summer?

[trail tendon]

start working on it rn

[crystal tiger]

are you taking a Mr D course

[tall coral]

what makes undefined terms undefined?
Like what stops me from defining a plane as R^2