geometry-and-trigonometry

[plucky cape]

how

[main fox]

by memorising it 💀

[dark sparrow]

memorize only a few basic ones, and learn how to DERIVE the others from them

[main fox]

okay im being real, if you practive enough (diffrent for diff ppl) then you will automatically see the pattern, you will build muscle memory, hence youll not need to memorise it

[plucky cape]

I know a few but sometimes I get stumped

[main fox]

so, PRACTICE is the key

[main fox]

in which std are you?

[plucky cape]

?

[main fox]

class/standard

[plucky cape]

calculus

[whole imp]

I don't understand trigonometry

I have a hard time with $sin\theta := \frac{\text{opposite}}{\text{hypotenuse}}$

[somber coyote]

Andy

[plucky cape]

what do you mean

[whole imp]

With the quotient rule...

I am having trouble understanding ratios and things that come from geometry

[plucky cape]

what concepts of trig do you need to know

[dark sparrow]

this is not usually called the "quotient rule"

anyway maybe review the very very basics then. working with identities will take more firepower!

[whole imp]

Yes. I don't really understand trigonometry

I want to

[dark sparrow]

if you're going to self-study, have you tried Khan Academy?

[whole imp]

No

[plucky cape]

hey! im doing khan academy rn

[dark sparrow]

aight then go learn yourself some trig on khanacademy

[deft basin]

https://youtu.be/vRgJ1QbOXz4?feature=shared

[peak prairie]

take pi as 180

holy it was 6 hours ago mb

[somber coyote]

sophia

[sterile kayak]

hola hola coca cola

[rigid sinew]

a=20301125

[queen hemlock]

https://tenor.com/view/gunna-fire-writing-heat-gif-25524005

[latent finch]

circle theorems are soooo hard

[whole flame]

Easiest

There is more to it in higher grades

[azure rivet]

exceptionally when we use them in complex plane and coordinate geo

it becomes manageable

[whole flame]

Frr

[long zenith]

Guys help

[worthy eagle]

draw the diagram first

[kind lagoon]

correct answer is (2) 110

[visual flume]

help pls

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[visual flume]

2

[sterile kayak]

i feel like the triangle could be important...

[visual flume]

well im just angle chasing rn

[sterile kayak]

if you had the angle Z or Y you could def do the problem easily

but idk any other way

[exotic yarrow]

Draw a circle

Specifically note that ||if you define the top right and bottom right vertices of the rectangle to be B, C (respectively), then B, C, X, Y, Z are all concyclic||

[opal dawn]

hi

can someone quiz me i have a test tmr

adding angles

[opal dawn]

Can someone check them and let me know what I got erong

[latent finch]

what the sigma

[golden needle]

can someone please tutor me?

[sick wraith]

what grade 😭 5?

[opal dawn]

10

[sick wraith]

bro

i leared all that shit in 6th

[opal dawn]

can u help then?

[sick wraith]

im srufyin

studying

i have exam

today

[visual flume]

woahhh

ty

[visual flume]

tysm

[pearl hinge]

1st page:all good

[main fox]

bro 10th grade, rly!?

[pearl hinge]

Here though the answer is right,the marled equation is wrong

[main fox]

ok , but lets be honest, gravity isnt a force, Sir Newton

[sick wraith]

ik it isnt

general relativity says so

[main fox]

yep

[sick wraith]

u a nerd too?

[main fox]

like i am intrested in nature of matter

[main fox]

what is 'nerd' for you?

[sick wraith]

quantum feild

[sick wraith]

like studying and reserching

[main fox]

QFT?

[sick wraith]

yea

scary for u?

[main fox]

ok then i might be inclined towards that

[sick wraith]

i am

today i had phy exam

[main fox]

nah, intresting, i have learned about feynman's path integral and bohr atomic model , that would embarked the intrest even more

[main fox]

damn, how it went?

[sick wraith]

hmm

[sick wraith]

good

i have to keep my name

slept at 12 woke at 3

studied

[main fox]

u r in which standard?

[sick wraith]

lets not go into ages

💔

[main fox]

broo

[sick wraith]

yeah?

[main fox]

ok, (if that is personal)

[main fox]

is a range good?

[sick wraith]

wdym

[main fox]

not age but grade,std,class

[sick wraith]

yeah ig

[pearl hinge]

2nd page all good.@opal dawn

[main fox]

you could say like im in high schoool, university etc

[sick wraith]

,uni?

[main fox]

k

[sick wraith]

wha

nvm

[main fox]

do you have more intrest in mathematics or in physics?

they are highly interdependent but then also

[sick wraith]

yeah

a lot

im up to some work ima catch u later

[main fox]

got it, cya

[sick wraith]

cya

if u need me dm me

[main fox]

accept Friend req

[sick wraith]

doen

done

[pearl hinge]

Why the percent sign

( ur name is hard to ping)

[simple vigil]

Can I dm ? For physics help 😅

[sick wraith]

any time

(was talking to someone else)

[summer cradle]

np

[wise badger]

He hath been summoned

[sick wraith]

wdym

he hath?

[wise badger]

Eh was tryna joke idk should have done it on #chill

[dire goblet]

i thought i could understand what this means

[fierce karma]

helpp

[grave pond]

What is the question?

And/or: What are you trying to do?

[fierce karma]

you have to find the length of the red line

given these are squares and theres a right angle triange

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[grave pond]

Hopefully you remember that the altitude of a right triangle splits it into two smaller triangles that are each similar to the one you start with.

[pearl hinge]

Determine yhe height of the right angles triangle using similarity of triangles and since the part of the red line is perpendicular on on noth sodes of the square ,u can js say that it is equal to its length i.e. 5

[whole flame]

4√2+10/2

??

Just a rough calculation

Not sure

[wary bluff]

the part of the line crossing the square is not parallel to the length of the square

wait hold up

i think he drew it crooked

but that doesn't mean the line ends at the center of the square

[opaque ember]

Ignore the solutions in pink

I got that 4x+5y=90 due to it being shown

But idk about k and c

[obsidian harness]

180 - (4x + 8y) = a and 90 - a = k

c you should be able to deduce from 4x + 8y as well

so you should actually find c first to make it easier for yourself

[opaque ember]

Bro I'm lost

How do you solve for x

[obsidian harness]

(4x + 8y) and c are what types of angles?

[opaque ember]

and y

Like what equations would you set up

[opaque ember]

Oh I see it now

[obsidian harness]

I'm prompting you right now

[opaque ember]

What about solving for x and y

[obsidian harness]

yes, so you get y = something * x if you do that

but now you also have the equation 4x + 5y = 90

so that leads to a linear equation in x

[opaque ember]

Yes

But I need a second one

to eliminate the x or y

[obsidian harness]

you already know that c = 7x + y/2

[opaque ember]

Yes

[obsidian harness]

and you know that c = 4x + 8y

so 7x + y/2 = 4x + 8y

[opaque ember]

Ok

So what would that be

[obsidian harness]

you try it yourself first

[opaque ember]

3x-7 1/2y ?

[obsidian harness]

where did you get 3x - 7?

[opaque ember]

7x+y/2=4x+8y

[obsidian harness]

ah, there's no equal sign here, so I got confused

you meant 3x = (7 + 1/2)y

yes, that's correct

[opaque ember]

Yes

If you added them to the same side

Wouldn't it be equal to 0

[obsidian harness]

yes

or you meant 3x - (7 + 1/2)y = 0, yes that's also correct

[opaque ember]

so then do I just solve for x

by elimination

[obsidian harness]

sure

[opaque ember]

Yeah I got it wrong

y=18

nevermind

[obsidian harness]

did you get x = (5/2) y?

[opaque ember]

I'm still confused

about getting 18 for k

[opaque ember]

No

I got x=15

https://cdn.discordapp.com/attachments/326138757474680852/1420990470925848606/Screenshot_2025-09-26_002051.png?ex=68d76800&is=68d61680&hm=77f7cbce4ec4d7106b7de0e294962db2cefb228790db8460a9cf05114c8f9af7&

so 4x+8y = 108

the top angle of the triangle is 90

so 90+108+a=180

[obsidian harness]

how did you get 108?

[opaque ember]

wouldn't that make a a negative?

[opaque ember]

108 = c

[obsidian harness]

yes but that's the teacher's solution right

[opaque ember]

Yes

I just got that as well

[obsidian harness]

from the diagram you can't know that

[opaque ember]

I just got it

[obsidian harness]

ok fair enough

[opaque ember]

But the math doesnt work out

90+108+a=180

[obsidian harness]

4x + 8y = 108 and 4x + 5y = 90

[opaque ember]

Yes

[obsidian harness]

then you can figure out y

[opaque ember]

I have y

Y is 6

[obsidian harness]

yes, then sub that into 4x + 5y = 90

so you know x

[opaque ember]

What I want to figure out though is how I can get it from the triangle without sub

[obsidian harness]

yeah x = 15, y = 6

[obsidian harness]

oh jeez

[obsidian harness]

it's so much easier if you sub though

[opaque ember]

Ik but shouldn't k come out to 18

If that were true

108+90+k=180

[obsidian harness]

oh, so you want to know what the problem with this step is

one sec

[opaque ember]

I got k=-18

[obsidian harness]

well by angles on a straight line, a + 108 = 180

[opaque ember]

Isn't a triangle 180 though

[opaque ember]

Oh I see it now

Ok thank you

I need to wake up in 6 hours

Thank you for the help

[obsidian harness]

no worries!

[azure yacht]

In my idea of partial dimensions we take finite cartisan plane 1 D and make it so only 28% of 1D can extend into 2D
For fractal dimension 1.28
There are many to ways to do this but if you distribute it randomly you and draw any shape
The shape will become fractal
Just like how line like fractals are generated in computer
Lets say 28% of space can access 2D in interval [x1,X2] on a line
28% of region can access 2D for all distance function D(x1,X2)>4

[upper karma]

why trig isn't anymore in here

[azure yacht]

I mean they are talking about angles I just wanted people's opinion on my idea since other channels are busy I posted it here also since I am yet to go to college it fits

[high vault]

\begin{gather*}
\text{Inverse trigonometric identity chart.} \
\text{Have fun!} \
\
\color{red} \sin(\arcsin(x))=x \
\color{red} \cos(\arcsin(x))=\sqrt{1-x^{2}} \
\color{red} \tan(\arcsin(x))=\frac{x}{\sqrt{1-x^{2}}} \
\color{red} \csc(\arcsin(x))=\frac{1}{x} \
\color{red} \sec(\arcsin(x))=\frac{1}{\sqrt{1-x^{2}}} \
\color{red} \cot(\arcsin(x))=\frac{\sqrt{1-x^{2}}}{x} \
\
\color{orange} \sin(\arccos(x))=\sqrt{1-x^{2}} \
\color{orange} \cos(\arccos(x))=x \
\color{orange} \tan(\arccos(x))=\frac{\sqrt{1-x^{2}}}{x} \
\color{orange} \csc(\arccos(x))=\frac{1}{\sqrt{1-x^{2}}} \
\color{orange} \sec(\arccos(x))=\frac{1}{x} \
\color{orange} \cot(\arccos(x))=\frac{x}{\sqrt{1-x^{2}}} \
\
\color{yellow} \sin(\arctan(x))=\frac{x}{\sqrt{1+x^{2}}} \
\color{yellow} \cos(\arctan(x))=\frac{1}{\sqrt{1+x^{2}}} \
\color{yellow} \tan(\arctan(x))=x \
\color{yellow} \csc(\arctan(x))=\frac{\sqrt{1+x^{2}}}{x} \
\color{yellow} \sec(\arctan(x))=\sqrt{1+x^{2}} \
\color{yellow} \cot(\arctan(x))=\frac{1}{x} \
\
\color{green} \sin(\arccsc(x))=\frac{1}{x} \
\color{green} \cos(\arccsc(x))=\frac{\sqrt{x^{2}-1}}{x} \
\color{green} \tan(\arccsc(x))=\frac{1}{\sqrt{x^{2}-1}} \
\color{green} \csc(\arccsc(x))=x \
\color{green} \sec(\arccsc(x))=\frac{x}{\sqrt{x^{2}-1}} \
\color{green} \cot(\arccsc(x))=\sqrt{x^{2}-1} \
\
\color{blue}\sin(\arcsec(x))=\frac{\sqrt{x^{2}-1}}{x} \
\color{blue}\cos(\arcsec(x))=\frac{1}{x} \
\color{blue}\tan(\arcsec(x))=\sqrt{x^{2}-1} \
\color{blue}\csc(\arcsec(x))=\frac{x}{\sqrt{x^{2}-1}} \
\color{blue}\sec(\arcsec(x))=x \
\color{blue}\cot(\arcsec(x))=\frac{1}{\sqrt{x^{2}-1}} \
\
\color{magenta}\sin(\arccot(x))=\frac{1}{\sqrt{1+x^{2}}} \
\color{magenta}\cos(\arccot(x))=\frac{x}{\sqrt{1+x^{2}}} \
\color{magenta}\tan(\arccot(x))=\frac{1}{x} \
\color{magenta}\csc(\arccot(x))=\sqrt{1+x^{2}} \
\color{magenta}\sec(\arccot(x))=\frac{\sqrt{1+x^{2}}}{x} \
\color{magenta}\cot(\arccot(x))=x \
\end{gather*}

[somber coyote]

₊‧°𐐪♡𐑂°‧₊ Usagi ⋆౨ৎ˚⟡˖ ࣪

[fierce karma]

thats not right because the drawing is not accurate, the red line is not the same length as the part of the hypotenuse u picked

[worthy eagle]

All you need is to make a triangle

[fresh birch]

Yes, my mistake, the answer is 35/3

[obtuse citrus]

Hey can anyone tell me how to solve this?

[whole imp]

It's not a big deal if you understand them...

in fact, there are many, many more than that

[whole imp]

just calculate it.

[exotic yarrow]

https://www.youtube.com/watch?v=SnA059gbpig&msockid=8fdd29f19b0e11f099477db4d718abe1

this video does a good job of explaining it

[obsidian harness]

@split lance it's definitely Pythagoras from my previous experience, probably something like this

wait I'm blind

the radius is 2, totally forgot

so that means triangles ORQ, OQP are equilateral

yes, then if you can find the 'height' of trapezoid TSRP, you're good (there's a nice way again with Pythagoras)

[rocky thicket]

$\sin(\frac{5\pi}{3}) = \sin(\pi + \frac{2\pi}{3})$

[dark sparrow]

equals, not implies

[rocky thicket]

oh

wait ye

[rocky thicket]

and is it better if i rewrite it as $2\pi-\frac{\pi}{3}$?

[somber coyote]

Kaladin.

[dark sparrow]

if it is, then not by much.

i wouldn't worry about it.

[somber coyote]

Kaladin.

[rocky thicket]

okay

[main fox]

here is my way of doing it,
sin(5 pi /3)
sin(300 deg)
sin(90 x 3 + 30)
since quotient is odd, we convert the parent func into its co func and since the ray lies in 4th quad, add -ve sign.

so

-cos30
= -root 3/2

[main fox]

btw how did you used teXit? what was the command

[grave pond]

There's no command to use -- it acts automatically on any post that looks like it has TeX formulas in it.

I believe it triggers on the presence of either dollar signs or \( or \[ (and their right-end counterparts).

[main fox]

sin x

it didnt worked

[grave pond]

You didn't enclose your formula in dollar signs.

[main fox]

$sin(x)$

[somber coyote]

Aeronix

[main fox]

wait, it works!!!

[main fox]

tysm!

how to type pi?

$pi$

[grave pond]

There are links to many tutorials in the pinned message in #latex-help; otherwise use #latex-testing for experimentation.

[real sentinel]

My way of doing it would be the classic standard unit circle method . Can be written as 2π - π/3 i. e. 4th quadrant where sinx is negative. So it would make it -(sin{π/3})= - √3/2.

[formal pollen]

I need a playlist for trigonometry which is like not too long

[split lance]

ok

[zenith lance]

Hello

is anyone here good at geometry tutoring?

I need to pass my Geometry EOC retake because I got 395 on the first one

[hasty juniper]

twin we rocking on the same boat im taking geometry and statistics honors

[glad barn]

tryna make a 4D game was a horrible idea

4D cubes are managable, but beyond that?

gulp

[dark sparrow]

i would imagine you need a good grasp of 3D geometry and/or 3D modeling software before advancing to 4D.

[glad barn]

its a grid based system

mostly just works by shuffling arrays in order to display 3 of 4 axses at once

[slender sparrow]

Problem VI:
Given acute angled triangle ABC (AB < AC).
Let K be a point on line AD (K != A, K != D), and M be a point on line DC (M != D, M != C).
Ray BK intersects AC at point E, and ray CK intersects AB at point F.
Perpendicular bisector of EF intersects AM, MK, AD at G,I,H(respectively).
Perpendicular bisector of HD intersects EF at point T.
Prove that (AEF), (AIG) and (THD) meets at a common point.

My idea is that we are going to prove ANFE and ANIG are cyclic quadrilaterals, then the proof is complete

I'll now let N be the intersection of TA and (THD)
so we are going to proof TN.TA = TI.TG = TF.TE

but Im kinda stuck here. We can also let S be the intersection of EF and BC, and we have angle HDS = 90 and TH=TD then TH=TS

but still stuck. anyone has an idea? thanks

[zenith lance]

Dang

Hopefully we pass the EOC

I almost passed by 9 points 😭

[dire goblet]

hi

anyone give me a question and ill solve it

[silent tangle]

Hi I got stuck on this :
cos^2(x/2)+sin(3x)=1/2
=> cos(x)+2sin(3x)=0 and I don't see how can i simplify more...

[obsidian harness]

this is a nasty problem

probably cos^2 x = 36 sin^2 x - 94 sin^4 x + 64 sin^6 x and then it's a cubic

oh wait let me fix the equation

you need to use the triple angle identity to expand sin(3x)

then square both sides

[silent tangle]

I saw this and then I said to myself sin^6 there is no way

thanks I will try with that

[obsidian harness]

no, sub in sin^2 x = t

[obsidian harness]

actually, easier is $6-\frac{8}{1+\cot^{2}x}=-\cot x$

[somber coyote]

south

[obsidian harness]

yeah then it's a quadratic, there you go

[raw frigate]

We can't even imagine 4D so how is it possible to jump on 4D or create 4D games

[obsidian harness]

also, screens are 2-dimensional

so they use perspective drawing and other visual tricks to make 3D objects look read on a 2D screen

[raw frigate]

True

[obsidian harness]

yeah so you'd need to have separate 3D views and that would be as far as you could go

individual 3D cross-sections of the 4D space, projected onto 2D

[dark sparrow]

i mean, look up 4D golf

its creator made devlogs about it

[raw frigate]

Is their a 4D game also

[dark sparrow]

i don't understand your question.

4D Golf is a video game, and yes it is 4D. the genuine kind.

[main fox]

by that logic, we would need a holographic screen to display a 4D obj, isnt it?

[visual flume]

how do you even draw this without a=c😭

[exotic yarrow]

<@&268886789983436800> spammed across many channels ^

[cobalt arrow]

I do Not want to write two column proofs mr teacher guy

[fallen pilot]

Yeah ong

I think a reflection by definition means that if A goes to B then B goes to C

[kindred elk]

guys help

im trying to figure out the anme of two angles

when two parrel lines intersect one

[worthy eagle]

anme?

[kindred elk]

name

are they vertical angles or

tranversal angles?

tranversal lines is what im talking about

whats the name, im on the geometry aops book rn

[worthy eagle]

this?

[kindred elk]

oh, so vertically opposite angles

yes

this is in chapter 3 of the aops geometry book btw

thx

[simple vigil]

Who is studying this ??

[worthy eagle]

@silverrolt

[simple vigil]

Class 7th students😂

[worthy eagle]

[obsidian harness]

probably

[robust dagger]

Hi

[civic rock]

||Lấy J là hình chiếu của K trên AT. Lấy giao điểm của BC và EF là L.
Các hàng điểm điều hoà cần biết đối với bài này: (CBDL) = (EFHL) = (GIHL) = (AKHD) = -1
Dễ thấy TJKN nội tiếp -> AJ.AT = AN.AK = AH.AD (hệ thức Maclaurin)
-> TJHD nội tiếp.
Biến đổi góc rồi chứng minh ∆TJH ~ ∆THA
-> TJ.TA = TH^2 = TF.TE = TI.TG (hệ thức Newton)
Từ đó suy ra được 3 đường tròn ngoại tiếp đã cho ở đề bài đều phải đi qua điểm J||

@slender sparrow ping me if you have any other questions

[fresh birch]

Lol

[south quartz]

Trillions of points

[sick wraith]

oh my what is this gibberish

[simple vigil]

Which concept this this. 💀💀💀💀💀💀

[river mulch]

I had a headache plus it’s not in English fuck I’m dumb AF lol

[vapid portal]

From what I’m looking at it might be harmonic bundles and power of a point but idk

[civic rock]

exactly

[silver basalt]

Is this correct?
False
False
True
False
True

[kindred elk]

IM BACK. can someone help me with how to prove that the 2 opposite side angles of a isosceles triangles are equla

[exotic yarrow]

Reposted here: #1422429158188908595 message

[real sentinel]

Yep

Start by drawing a median to a side

This will divide the triangle into 2 triangles

These triangles have common sides and 90° angles , they will be identical by hypotenuse side test.

So , by this, corresponding angles will be equal too

[alpine raven]

Should I create a separate user account on my PC for gaming, and why would that be helpful?

Is it necessary to use a game controller for PC games, or is a keyboard and mouse enough to start?

[rain scaffold]

Wrong server.

[rain scaffold]

Nope, use your school account.

[real sentinel]

Wrong question dude

[whole flame]

Someone needs a ban

[slender sparrow]

cảm ơn nhé

[wooden pendant]

Can anyone proof why sin 0 is 0

[whole flame]

U can do that using a right angle triangle or a standard unit circle

[mystic folio]

how do i prove that l1 l2 l3 are concurrent with an equation connecting a b and c?

[main fox]

take any point on unit circle, now it makes an angle x from the initial ray, hence its coordinates are cosx,sinx
meaning the height of the point from the initial ray is sinx, now watch what happens when x approaches to 0, then the height of that point decreases, hence sinx also decreases(i.e it approaches to 0) so when x = 0, sinx = sin 0 = 0

[real sentinel]

Exactly, it's the most basic thing taught in trigonometry

[whole flame]

Since all three lines pass through the same point (say O), the angles around point must sum to π on one side of a straight line.

If you start with three arbitrary lines, their pairwise angles won’t necessarily satisfy this relation.
But if they are concurrent, then necessarily:

$a+b+c=π or 180°$

[somber coyote]

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

[real sentinel]

Using vertically opposite angles, a+b+c = 180, which is condition for concurrence of lines passing through a point

[exotic yarrow]

Reposted here: #precalculus message

[north spruce]

stupid question but what is this called in English

hard time fidning anything because my translations are off

[primal radish]

unit circle?'

[real sentinel]

It's a trick to remember signs of trigo functions in all 4 quadrants

Eg. Sign is positive in 1st and 2nd quadrant

The trick is as "Add Sugar To Coffee"

A = all positive in 1st qd
S = sin (& cosec)positive in 2nd quadrant
T = tan(& cot) positive in 3rd quadrant
C = cos (& sec) positive in 4th quadrant

[wooden pendant]

All sin tan cos rule

[real sentinel]

i see, mine's was a short trick to remember it

[north spruce]

Thanks loki

🙏

[whole flame]

Other trick is

After School Than College

[tawny loom]

I haven't heard this mnemonic in my trigonometry class before that's neat

[gloomy spear]

can some1 explain how to do this?

i could only get to BAD = 102 and BCD = 78

[frozen pilot]

is there any way to complete the square in desmos?

i know how to do it manuallyfor SAT but just curious if possible with regression or something

[desert torrent]

Graph it and find the vertex, and then use another equation to solve for slope?

[vapid portal]

you don't have to idt

js use the equation of the circle as (x-x_0)^2 + (y-y_0)^2 = r^2

and then expand that

oh mb

didn't see the manually part

[south quartz]

You can find bcd = 180- edc and then find cbd and finally cad

[teal torrent]

Do not know where to start

[gloomy spear]

why is bcd 180-edc

[south quartz]

Oops my bad i meant bdc

[gloomy spear]

ok so 33

but why can you get cbd from that

is there some theorem im missing?

[south quartz]

Triangle cbd => sum of 3 angles=180

[gloomy spear]

oh

ok

ok i get it now thx

[wooden pendant]

Sat is so easy dudee,

[valid pond]

@royal grail hello there.

I haven't seen your work, but I'll give you some clues.

Try adding the common angles like angle ABD and ABC share a common vertex and line.

In simple terms, if angle ABC is divided into two parts, 60 and 20, adding them both would give you back the whole.

It would be 80.

It would apply to most given, including angle BAC.

Would be 80.

If you recall, the sum of the three interior angles in any Euclidean triangle is always 180 degrees.

[whole flame]

Yo supe

I guess this guy just wanted someone to solve it

Didn't need help

[exotic yarrow]

https://math.stackexchange.com/questions/3823143/finding-an-angle-in-a-80-circ-80-circ-20-circ-triangle-variant-involvi

(pure angle chasing doesn't suffice without some kind of construction)

[sick wraith]

.

[remote estuary]

i believe it is x = 60degress

[fallen narwhal]

why angle odb is 12 degrees

pls explain every step of solution !!

[south quartz]

Obf=90, dbf=102=>obd=12
Ob=od=>isosceles triangle=>obd=odb=12

[frozen pilot]

ok so i dont recall asking

like what could possibly be the point of responding to this if you have nothing to add

[royal grail]

whoa you actually did it

now that's actually impressive

[inner lark]

the massacre

[summer cradle]

ODB = OBD = 102° - 90° = 12°

[high wadi]

Okay this isn't for any help, but does anyone else just forget basic math during a test?

[fallen narwhal]

bro i have no words to express the feeling

[south quartz]

I mean depends on what you mean by basic

[real sentinel]

never, that's js in veins now'

[obsidian hornet]

build up to polygon and its done

[old marsh]

hello?

[soft lance]

langleys adventious angles problem

[formal jewel]

Hello, Could someone explain Thales' theorem? I'm studying and looking for other ways to explain the theorem

[south quartz]

The circle one?

Or the proportion one

[south quartz]

Those are certainly writings

[night cape]

hi, is geometry or triganometry the subject that teaches about angles

[south quartz]

Yes

[night cape]

great :D

[high wadi]

I had a question that said “if the width of a rectangle is 5 more than the length and the perimeter is 106 what are the dimensions of the rectangle?”

I completely forgot how to do it lol but then I figured it out later in the day

[formal jewel]

This

[night cape]

perchance were the length 8.33 and width 44.16

[tidal basalt]

Through:(4,0)and(2,1)

[zealous pike]

The perimeter of the shaded area is 2π + 8 and the lines PQ and PR are tangent on Q and R, respectively.

Because the lines sre tangent on the circle, then the radius is perpendicular to the lines. Call the point of the center of the circle C.

If RC is perpendicular to PC and RQ perpendicular to PQ, then we have that PRCQ is a square, and PQ = Radius of the circle. Call PQ = x

Help im stuck here

[tidal basalt]

Its geometry for me

[zealous pike]

It's to find the radius

If the arc length is xπ/2

Then I have that

2x + xπ/2 = 2π + 8

4x + xπ = 4π + 16

x = (4π+16)/(4+π)

Is this the answer

[shy grotto]

There's a little more simplifying you can do, but yeah

[zealous pike]

How

[shy grotto]

4pi+16=4(pi+4)

[high wadi]

Sorry I took forever to answer but the length was 24 and the width was 29

[zealous pike]

So it becomes 4(4pi)?

[shy grotto]

4(pi+4)/(pi+4)=4

[jagged agate]

area

[regal kraken]

can any1 explain to me why the image of orthoconter of a triangle about any of it's sides lies on the circumcircle of the triangle

[whole flame]

Is it a 3d figure

??

I guess it is

Divide the area into 3 parts

[real sentinel]

I'm getting the same

So yep

[night cape]

nuuuuu im wrong 😭 what formula did you come up with

also no worries :D

[pearl mantle]

Can anyone help me revise on triangles and circles?

[winter gazelle]

You still haven’t post the work you’d like to be assisted with.

[pearl mantle]

like helping me see the connections in general

i just get lost once I see the shape

[winter gazelle]

Do you have a sample question?

It’d take a lot of time for us to guide you from the very beginning

[pearl mantle]

Alr

Lemme get the question book

[round beacon]

‼️

there was a question book

[pearl mantle]

Like a book that's just questions

[round beacon]

well yes but if you had a question to pull up I could’ve helped in #help-48

[pearl mantle]

like in general, i jus tcan't see which ones connect w each other

[spiral lodge]

Still, too open as a question. It's almost impossible to do this via Discord

That's why we're asking you if you are finding difficulty with some specific exercise

Of course, precise theory questions are much appreciated too

[pearl mantle]

For example, we have a triangle ABC, a circle (O) with diameter BC intersecting AB, AC at D, E. Call the point that which BE and CD intersect H and the point where AH and BC intersect F. Through D draw a line that is perpendicular with BC that intersects (O) at D1. Prove that E, F, D1 are collinear.

like there is info, i just can't connect them

[jagged agate]

I tried that but it didn't work

[whole flame]

Show me what u did

[jagged agate]

It would be easier if it went through the center
I divided it into these three areas

[whole flame]

I think it needs to be divided into 4 parts... Other part would be the blue part extended inside... But it would be of different lengths so u need to use some geometry or integrate using limit.

[high wadi]

We’re given the perimeter, which is 106 and we know that width is 5 more than the length. So we can setup an equation based off the formula to find the perimeter of a rectangle which is P = 2L + 2W, so we can plug our numbers in, 106 = 2L + 2(5+L). Solving for L gives L = 24 and we know that the width is 5 more than the length so then just do L + 5 which gives 29 for our width.

[night cape]

ahhhhh awesome! I see what my issue was, i added all sides and forgot that rectangles have their own perimeter calculation

:D thanks!

my second mistake was multiplying by 5 and not adding 5

[gloomy spear]

can anyone help me with this question pls

[kindred ocean]

if the radius of the spheres is r, then the height is 4r, the length and width are both 2r, because the spheres exactly fit

from there, you should be able to find expressions for the area of the cuboid and the spheres

[whole flame]

Find the volume of the box then find the total volume of the two spheres and subtract it from the volume of box... Divide it by volume of box and multiply by 100

So basically u use percentage formula

Final-initial/initial X 100

[summer cradle]

Red trapezium:
tan(27°) = 0.0625/x
x = ...
Base = b = 0.25 + 2x
Area = A2 = ½(0.25 + b)(0.0625) = ...

Thin orange isosceles triangle:
2 same sides = radius = 0.241
Base = b = 0.25 + 2x
Height = h
h² + (b/2)² = 0.241²
h = ...
Area = A4 = ½bh = ...
(or use Heron's Formula directly)

Light blue reflex circle sector:
Peak angle of thin orange isosceles triangle = a
½(0.241²)sin(a) = A4
a = ...
(or use Cosine Rule)
Reflex angle of circle sector = t = 360° - a
Area = A3 = (t/360°)(pi)(0.241²) = ...

Green part:
AD = 0.25/2 = 0.125
tan(27°) = DE/AD = DE/0.125
DE = p = 0.125 tan(27°)
EA = q = sqrt(p² + 0.125²) = ...
Triangle ABC:
AB = m
BC = n
ABC ~ CFE
AB/BC = CF/FE
m/n = (0.125+m)/(p+n) = tan(27°)
m = ...
n = ...
AC = sqrt(m² + n²) = ...
Radius = R = EC = EA + AC = q + AC
Area = A1 = (78°/360°)(pi)(R²) - ½q²sin(78°) = ...

Total area = A1 + A2 + A3 + A4 = ...

[summer cradle]

Small triangle:

70° + 30° + y = 180°
100° + y = 180°
y = 180° - 100° = 80°

[keen barn]

thxs

[summer cradle]

np

[worthy eagle]

use angle sum property to find the unknown angle in the triangle then use corresponding angles property to find y

[civic rock]

<@&268886789983436800>

[nimble fulcrum]

,rccw

[nimble fulcrum]

,rccw

[nimble fulcrum]

so we know that the volume of the box is LWH

@gloomy spear

Since the spheres are identical

And they touch the top and bottom

the diameter is x/2

so we have

,,LWH = \left(\frac x2\right)(W)(x)

[somber coyote]

calvin

[nimble fulcrum]

the width is the same too

so we have

[somber coyote]

calvin

calvin

[nimble fulcrum]

so thats the volume of the box

For the spheres

since the diameter is x/2 the radius is x/4, half of the diameter

[somber coyote]

calvin

[nimble fulcrum]

so we have

[somber coyote]

calvin

calvin

calvin

[nimble fulcrum]

Since there are 2 spheres in the box

We multiply this by 2

Giving us

,,V = \frac {\pi x^3}{24}

[somber coyote]

calvin

[nimble fulcrum]

we then divide this by the original amount

The amount of the box

[somber coyote]

calvin

calvin

calvin

[nimble fulcrum]

@gloomy spear

Oh yeah then times that by 100

[somber coyote]

calvin

[gloomy spear]

Ohhhhhh ok I get it

Thx

[high wadi]

Oh my god sorry for the super late response but, you are very welcome!

Gotta love math

[storm ore]

I know this

[night cape]

:D thank you! and right? im learning to love it!

[high wadi]

I used to dread any kind of math from elementary to the beginning of high school, but when I got into my sophomore year, we were learning basic trig, just how to find the angles or sides of a right triangle. All of a sudden, everything clicked. I went from not understanding any math to it being my favorite thing to learn. Now I'm in Honors Precalculus; sadly, my school doesn't offer any higher math classes.

[minor heron]

awww thats sad u gonna do dual?

im doing ap precalc right now c: so i guess we're about the same

[high wadi]

I was gonna do dual for math, but I decided I would take a comp sci and mandarin class for dual enrollment

The only complaint I have for my precalc class is that we get into actual precalc next semester, so the first half year of my school year is gonna be review

[empty remnant]

What is trigonometry even about? Why do we need it for linear algebra en calc1

[minor heron]

lol

ours is ap so igs we js follow the general curicullum yb college board

its so ez rn i have a 99%....

[high wadi]

LOL yeah my precalc H is actually so ez its not even funny

I wish my school offered more AP classes we literally have THREE APS

THAT IS IT

is there a general chat in this server?

[minor heron]

damn

[minor heron]

try #discussion

we have like

4 math aps alone

precalc ab bc stats

i feel bad for you...

[high wadi]

hm I have no access to that cahnnel for some reason

[minor heron]

oh

maybe its bcs of roles

[minor heron]

i have "i like math and i'd like to chat about it" selected that might be it

[high wadi]

I have the same things

[minor heron]

hmmm

[cunning lion]

you have the studying! role due to selecting "hide the discussion channels" in id:customize

[minor heron]

ohhh

[high wadi]

it says I have it off?

[minor heron]

hmm

try it again

[cunning lion]

you do now

[minor heron]

#discussion

[high wadi]

I turned it on and off

IT WORKS

[minor heron]

yayy

[high wadi]

As I was saying before, the only APs we have are AP Lit, APAAS, and AP Gov

[minor heron]

damn

[high wadi]

I go to a ghetto school, so I'm not surprised tbh

I knew I should've gone to that tech school

[night cape]

thats so cool! its absolutley awesome when once the idea clicks you start wanting to do it with earnest

and congrats aswell :D

[tropic cairn]

Yall i did this trigonometry question like this, i asked chapt gpt if this was another correct way of doing the question but i could not like exactly comprehend my answer so like idk if this way of doing this question is correct or not, can any1 tell? (I did it in another way than shown on google etc)

[dark sparrow]

!nogpt

[lime crown]

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

[civic rock]

[simple vigil]

Did any of you studied
Sin(a+b) and cos(a+b) in class 10 ?? 😭

[summer cradle]

I do

[simple vigil]

Seriously 😳

[summer cradle]

either grade 10 or 11

[simple vigil]

It

Must be 11th

[upper karma]

idk i did this

in 9th grade

[simple vigil]

🥲🥲🥺🥺

I can't do this in 10th

[upper karma]

what are you finding hard about it

@simple vigil

remembering it?

[simple vigil]

Yeah

Cos(a+b)

[upper karma]

sin(a+b) remember it like sinacosb + sinbcosa, so put sin and cos in the first and you give them the respective angles

for the next one

switch them up

thats how i remember it

[simple vigil]

This one is easy

But cos ones

[upper karma]

cos(a+b) = cosacosb - sinasinb lmao

thats the easiest

just cos with both angles

and sin with both angles

[simple vigil]

Maybe when I will study in unit circle then is might be easy to understand

[upper karma]

just remember cos in the first and sin in the second

and for each u give two angle to both of them

ngl i still remember it this way lmao

it just never went away

[simple vigil]

Habit 😅

[upper karma]

but i don't have ot remember them anyways i got a sheet full of trig functions

im gonna neet

just keep on using them made them remember lmao

thats the trick

don't even learn it just keep using them you will learn them

[simple vigil]

Bro but one thing I find hard is identify proving question

Like how would I know we have to divide by cos²A 😭

Can I share pictures in this channel ?

[upper karma]

🙏🏻 those problems were the end of me in the beginning of school bruh thats a fair issue but onlu thing which got me good was watching one of these youtube videos and solving along eventually i was able to do them MOST of the times

that most of the times was enough to get me good grades

[upper karma]

i think you can idek

u can try

[upper karma]

actually A+

lmao

i just solved them like crazy i eventually got the idea

also cuz some of the less attention student was able to somehow do it better then i was and that shit made me crazy i was like HOW CAN HE DO IT HOW HE GET THE IDEAS

i was locked in

[simple vigil]

See this question there are two problems

1. logic
2. mixture of 2 chapters

[upper karma]

do u mean quadratic as a second

chapter

[real sentinel]

Very basic question

[simple vigil]

Yeah I mean it's basic but still

[simple vigil]

How's that basic if you are doing it for first time 🥺
It took me 5 mins to realise I have to divide by cos²

[tropic cairn]

Can anyone help me figure out if my way of doing this question is correct or not 😭

[tropic cairn]

It is in 10th, im in 10 rn so..

[simple vigil]

Which board?

[tropic cairn]

Cbse

[simple vigil]

Same 🤣

What's your doubt then ? @tropic cairn

[tropic cairn]

Like the question i clicked a pic before of, i js wanted to know if the way in which i did the question was valid or not of trigono

[simple vigil]

Wait let me also solve and send to you 🫡

[tropic cairn]

Alr but i js wanted to know if my solution was currect or not

Correct*

[simple vigil]

I think it's correct but you can do like
Taking value of tan as P & B

Then finding H and using the ratios find the solution

[tropic cairn]

Yes i saw that way on google but I thought this way was easier for me to do yk

[simple vigil]

No this method of P & B is easier we have to just put the values

[tropic cairn]

Tyyyy i understood and tried t9 do it by this method and surprisingly is quite easy!!

[real sentinel]

Dude even a 8th grader can do that with basic trigo knowledge

[simple vigil]

@real sentinel we studied trigo in 10th for the first time

[real sentinel]

that's wierd

[simple vigil]

That's CBSE 🤣

[real sentinel]

stop whining here

doubts only!

Use #chill or #discussion @simple vigil

[simple vigil]

But in exam you will have to use a constant while calculating H

[rotund ice]

I need to learn trigonometrie. Currently I am a high school student. At school we are reviewing trigonometrie but the teacher is not explaining it. I want to learn Trigonometrie from starting, so can someone recommend me any youtube video or help me?

[obsidian harness]

https://www.youtube.com/watch?v=lJ9K4eupKkI

[eternal cloak]

anyone in here?

[zealous pike]

No

[eternal cloak]

anyone else?

[zealous pike]

If u have a doubt just send it

[eternal cloak]

What is the relations between angles measured in radians and degrees

it got sum to do with sinus

like beginner type stuff

[zealous pike]

Radians and degrees are like fahrenheit and celsius, systems to measure angles

180° = π
90° = π/2

And so on

[eternal cloak]

yea I know that Im just trying to understand how do the degrees corrilate with the sinus function

nvm I found that shi

https://tenor.com/view/salsa-dance-salsa-dancing-dance-football-twitter-sports-twitter-dance-sports-twitter-gif-9983991042154241861

[real sentinel]

Dude wrong channel

[tropic cairn]

Can any1 help me w this question

[exotic yarrow]

you don't do this question because it's wrong

one of those plus signs should be a minus sign

[tropic cairn]

Sry wrote the question wrong, this is the right one

[exotic yarrow]

the thing you have to prove has $m^2+n^2$, and you're given $m, n$ explicitly

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

so you might as well calculate m^2 + n^2 directly

[tropic cairn]

But how

[tropic cairn]

Im confused like by using the identities or

[exotic yarrow]

show me what you have so far

[tropic cairn]

I didn’t do anything so far..

[exotic yarrow]

you didn't even substitute in the expressions for m and n?

[fresh birch]

square and add both equations bro

[tropic cairn]

Wait whole sq both eq and add them i did now

[real sentinel]

Just square both sides of the equations and add both of them

[brittle kiln]

quickly i need facotrs of 40

sorry 405*

[spiral lodge]

Google

Or search it on Wolframalpha

https://letmegooglethat.com/?q=factors+of+405

[brittle kiln]

thats rlly funny lol the heck

[spiral lodge]

Also, for the record, when someone writes they want something quickly, helpers are suspicious that it might be an exam and hence they're less prone to help

[brittle kiln]

ah makes sense

wait an exam on saturday

i guess maybe for the home schoolers, but nah im just doin homework

[proper peak]

Seeing sen feels criminal

[cunning lion]

it's just spanish or portugese

[visual flume]

HAHAHHAH

HAHHA

YK WHAT HAPPENED

SO I WAS DOING MY AOPS HW

AND ONE OF THE PROBLEMS WAS RLY WEIRD AND SMTH AND I ASSUMED SOMETHING AND GOT IT RIGHT

RAHHHHHH

[minor heron]

Peak

[sharp tusk]

If I see that on any math exam, I am walking out the door.

[whole flame]

🗿 🗿

[whole flame]

Square the given 2 equations on both sides and add them up... Use the rule sin^2x + cos^2x =1

Easyuyu

[fresh birch]

I don't understand the controversy with the image I sent.

[craggy plover]

Let A be a finite set of points in the plane, no three of which are collinear. Assume there exist two triangles, each with vertices among six distinct points of A, whose intersection forms a hexagon that contains no points of A either in its interior or on its boundary. Prove that A contains a convex hexagon whose interior is free of points from A

help me solve this problem

[whole flame]

Ahhh

I didn't see ur answer

I just scrolled to the question

But the solution is so general

U don't have to think for such kind of proofs

[primal radish]

help can someone psl help me with this

[whole flame]

2 is constant so take it out... Now take cosec^2x as -(- cosec^2x).. reason is bcz derivative of cotx is - cosec^2x... Now assume cotx as t so -cosec^2x dx becomes dt

Change the limits

When cotx= t at m t will become cotm and at π/2 t will become 0 as cot π/2 is 0

Now integrate t and put the limits

So i guess m should be cot inverse √2

[primal radish]

omg thank you so much

[alpine marsh]

can someone help me calcualte the orthocenter for a triangle with the corrdinates A (-2,1) B(0,-2) C(5,-2)

[civic rock]

let the coordinates of the orthocenter be H(x,y) and notice the dot product of orthogonal vectors is 0

[alpine marsh]

what does that mean

[whole flame]

I'll tell u a simple way

Consider orthocentre as H (x, y)

[alpine marsh]

yes...

[whole flame]

Then make equation of AH

BH

CH

[real sentinel]

m should be π/4 or 3π/4 or -π/4

[whole flame]

Or

Consider line AH perpendicular to BC

Bcz lines passing through orthocentre are perpendicular to a side of triangle

[alpine marsh]

yes

[whole flame]

So the product of slope of AH and BC must be -1

[alpine marsh]

they go through the vertex and perpendicular to the other side

[whole flame]

That's property

[real sentinel]

Yep

[whole flame]

Yepp

[alpine marsh]

multiply slope?

[whole flame]

Yepp

[real sentinel]

Product of slopes of perpendicular lines is -1

[whole flame]

Remember the product of slope of 2 perpendicular lines is -1

So do this for AH BH CH with respective sides of triangle

U will get 3 equations

2 variables

Solve n u will get answer

[copper hill]

can anyone let me know if im missing anything for linear equations and slope of lines ect,? i have the formulas but i struggle with the questions anyhow. I either cant figure out what the question wants, or when i do the question i misunderstood the concept and i try to stick to the formula too hard and end up getting it wrong due to missing something small, any details i need to know

[quasi flume]

From what I see, you have all the notes/formulas for highschool linear equations, the tiny mistakes and rigidity of using the formulas is more of an experience thing, I say just try to figure out what tiny mistakes you keep making/struggling with, and do questions with them until it's memorized

[copper hill]

alright ^^ thank youuu I'll deffo do some practice questions

[quasi flume]

Np!! Gl! Feel free to ping me if u need help with anything like this lol, I can't help with anything university level, but most highschool stuff I can help with

[main fox]

Guys , the question says that I gotta use elementary concepts , but it is way longer, is there any shorter approch, that uses conceptually high topics but get me answer in less steps?

[quasi flume]

Uuuh, for elementary are we counting sine and cosine laws?

Nvm, you need side lengths for those

And honestly there might be, I'll look into it

Yea I think there's just a lot of steps given the questions limitations

[main fox]

nah, just similarity, Congruency, ASP, and basic things

[main fox]

yeah sure, go ahead inform me once you have found it

[main fox]

that is why i was searching for an alternative way to do it shorter doesnt matter if it is conceptually high

[spark yacht]

do some

auxillary constructions

this is a famous problem, idr the name tho

[civic rock]

Langley's adventitious angles

[regal kraken]

substitue cotx as t

[whole flame]

Use the sum of angles of the triangle is 180° and use the exterior angle property...

[obsidian harness]

You can easily let AB = 1 since scaling the figure up or down does not change the angles

[quasi flume]

true

[tropic cairn]

Can any1 help me w this que, i divided both numerator and denominator w cos theta and applied the identity in numerator but im getting tan theta + sec theta at last but i should be getting 1/sec theta - tan theta

[worthy eagle]

you got LHS = tan + sec
now multiply it by tan-sec in numerator and denominator

@tropic cairn

[simple vigil]

It's beyond my level and the hand writing 💀

[spark yacht]

Rationalise

[delicate raft]

is trig is eazy?...

[spark yacht]

kinda

[simple vigil]

Yes but sometimes it is hard
If you have to transform because there are many possibilities of a single question

[delicate raft]

But what does it requires to first jump into trignometry?

[spiral lodge]

#geometry-and-trigonometry message
That's the answer to a very similar question

[delicate raft]

is khan academy is much better than Phsiycs wallah and Icse wallah in youtube?for maths?

[tropic cairn]

Can anyone help me w this que i cant do it after the last steppp

[tropic cairn]

If u get the flow of the identities and like know where to apply each, then it’ll be pretty easy, but practice is needed

[delicate raft]

.

[real sentinel]

Dude just separate the terms first

Write tanx/sinx *cosx separate

And cotx/sinx*cosx separate

it's easy than as hard it seems

[simple vigil]

No it's easy
In this question if you simplify RHS one step you will get the same

[slate thorn]

Anyone around who'd like to walk me through some problems in coms? first time here sorry if this isn't how you're supposed to request aid! I'm more a audio/visual learner

[whole flame]

Post ur question

Here

If anyone knows the answer they will respond

[spiral lodge]

I don't know what they are 😅

[worthy eagle]

Khan Academy is very good for understanding concepts but it doesn't follow your syllabus and also doesn't provide you notes that will help you in your exams so if you have issue in understanding any topic, you can go to khan academy, search that specific topic and clear all doubts. But if you are studying for an exam and want to only learn what's needed PW and others are good.

PW will provide content focused on your exams and khan academy is just like collection of knowledge, if you are interested and have time apart from your normal studies you can look at khan academy to clear concepts.

[fresh birch]

You can do it after using the Monsky transformation.

[fresh birch]

Even using trigonometry it will be complicated, but you could guess the value of the angle if you use the Monsky transformation through a cyclic quadrilateral and then the Langly adventitious angles, but you need to know if they are the general cases and also, if it meets the degrees of freedom to verify if there is only one value that meets the conditions.

[fresh birch]

by congruence it comes out 20

[main fox]

cyclic quad?

but one angle is 40 and other is 30, it should be equal (from same base) to construct a cyclic quadrilal

[main fox]

does it takes long steps?

[fresh birch]

you have to build the cyclical

[fresh birch]

more or less

[fresh birch]

.

[main fox]

?

[main fox]

what is cyclical ?

[fresh birch]

inscribable quadrilateral

[fresh birch]

Here is the solution to the problem

[main fox]

let me take a look

[main fox]

that IS the Long way i was taking earlier

i already know that, but it is very tedious

[fresh birch]

Yes, but it is the most formal way. Another way would be with the Monsky transformation, taking the initial figure to another where you can use Langley's adventitious angles and you will notice that we are talking about an 18-adventitious triangle. You test the values ​​and you get to 20...

[copper hill]

this is my handwriting at its BEST what do u mean 🤬🫵

[simple vigil]

I mean it's my dream to write like that
Sorry 😔

You are elder sorry 😐

[copper hill]

NO

ELDER?!

im in 3rd year im 15😭 that'd be 9th grade for you Americans 💔

[night cape]

are there any rescourses you guys could recommend to learn geomatry as a full course online?

[quasi flume]

KhanAcademy is good for highschool level stuff, and afew of the lower level college stuff #book-recommendations is good for textbook resources as well

[night cape]

thanks :D

[waxen oracle]

I'm in gr6 and I js learned sin cos tan :))

Its rlly hard lmao

I was wondering if anyone here goes to spirit of math?

[whole flame]

Everyone who helps here knows math obvious

[simple vigil]

Nah I m indian and turned 15 2 days ago

[dark sparrow]

you're in grade 6?

... how old are you

[waxen oracle]

11

😭

Ok look

I am currently doing trigonometry

Qnd I was wondering the purpose of sin cos tan

[quasi flume]

Their ratios between the angle and side lengths of right triangles

For example, let's say you have a 45-45-90 triangle

Given one side length, you could find the lengths of all the other side lengths using sine, cosine, and tangent

https://www.mathsisfun.com/sine-cosine-tangent.html

Check this place out, they have an interactive triangle thing (idk what to call it lol), that shows the relation of all sides as you increase/decrease angles

[dark sparrow]

ok yeah im really sorry but you're too young to even be on discord

<@&268886789983436800> underage

[thin sentinel]

they're standard ratios between different side lengths given the same angle (think about triangle similarity). but yes we can't have you here, sorry. good luck. feel free to come back when you're 13+ or the legal age in your country.

[quasi flume]

Wait it isn't 13+ across all of discord and changes depending on country?

Never knew that

[dark sparrow]

iirc the 13+ thing is global to discord

[thin sentinel]

yeah there's a whole breakdown https://support.discord.com/hc/en-us/community/posts/360050817374-Age-restriction but they're all ≥ 13

[quasi flume]

Gotcha

[frank pebble]

If you are asking about what’s useful about them:

In principle you can calculate any angle using them, it’s just the algorithm for doing so is tedious and not taught in school, so most often you’ll see only exact values like sin(30deg)=0,5 etc., and maybe at some point a look up table with values for random angles (like 1 degree, 2 degrees…)

Another useful thing about them you’ll learn is trigonometric identities, useful in geometry problems (like double angle, half angle, etc.)

[thin sentinel]

probably soon to be 17+ in Australia

[quasi flume]

Damn really?

[thin sentinel]

depending on how they classify it wrt social media

[quasi flume]

Yikes, I hate all ts going out, but that's not a convo for here lol

[thin sentinel]

i guess i would say sin/cos/tan let you translate angular measurements into rectangular measurements

but i don't think any of these answers really explains why they're useful

[quasi flume]

Yea, it'd be best to just showcase the real world uses of them ig

Not just "oh it works because xyz" and instead the actual applications of them, but ig its too late for that lol

Anyway, I also got questions xd

How the hell do I transform sinusoidal waves- I'm teaching myself this stuff because teachers are all unavailable for the next few weeks to months and have 0 clue what I'm doing (textbook also doesn't really explain it well)

[edgy seal]

So like usually you'll take the equation y=sin(x) and transformations turn it into y=a*sin(k(x-c))+d

so like a transformation is a procedure that eats an equation for a function, and spits out a new function, where you've changed the graph of the function in a certain way

so translation upwards 5 units would eat say sin(x) and spit out sin(x)+5

and if you plot the grpah of y=sin(x) and y=sin(x)+5 you'll see the second graph is the first one shifted up 5 units

so the important thing is to understand what the transformation you're applying does to the functions it eats and then do that to your rule

[quasi flume]

Oh, so it's literally just linear or exponential transformations, what if it's inverted? Same theory as linear or any other graph?

[slate cave]

does anybody know this

if know pls explain

[somber coyote]

Sending query to Wolfram Alpha, please wait.

Something went wrong!

An unexpected error occurred while processing your command!
The error has been reported and should be fixed soon.
If the error persists, please contact our friendly support team at our support guild!

Exception

aiohttp.client_exceptions.ClientConnectorError: Cannot connect to host public5c.wolframalpha.com:443 ssl:default [Name or service not known]

[stone badger]

Do any of you have recommendations for resources to learn these topics, like a good YouTuber or something?

1. Trigonometric formulas for a single angle

2. Sine, cosine, and tangent of the sum and difference of two angles.
   Double-angle formulas for sine, cosine, and tangent

3. Solving trigonometric equations

4. Solving trigonometric inequalities

[simple vigil]

For which class ?

11th ?

[quasi flume]

KhanAcademy for basically anything below university level, they have lessons and quizzes there too, The Organic Chemistry tutor on YouTube also has great videos

[simple vigil]

What is Khan academy? 😅

[quasi flume]

Interactive course thing for basically anything math/science related, they have great videos in there, with practice quizzes/tests and walk throughs for stuff you struggle with, they don't have anything past 1st year university, but it's an amazing free resource

[simple vigil]

Oh 😮
I will also learn from them especially physics and maths

[quasi flume]

Yea they're great, lemme take a ss

[simple vigil]

Ss of what ??

[quasi flume]

Here's their physics modules

There's more as you scroll down

But yea

Great resource

[simple vigil]

Oh thanks 👍😊

I will definitely check it out

[quasi flume]

Np!

Always available if you got any questions on math/science stuff

[simple vigil]

Thanks again hope i could also help
If you need in idk something I can help you in 🥺

[quasi flume]

How good are you with sine waves and precalc?

[simple vigil]

I m in 10th 🥲🥺🥲

[quasi flume]

All good then lol dw

[simple vigil]

Ok 🆗

[timber turret]

Does anyone have tests or like good excersises for derivatives of trigonometric functions

[granite phoenix]

I would do it by compound angle and double and triple arc identities, but I don't know Moivre's theorem.

sen 5x = sen(3x+2x)

[left river]

You know de Moivre's theorem?

Use the theorem then Substitute n=5

Expand the left hand side and separate the real and imaginary parts. Then express everything in terms of sine theta only.

[granite phoenix]

Moivre's theorem is for number complex, sure?

[left river]

Yes but it can be used to derive trigonometric identities for real angles.

Bottom line: it's an ironic theorem

[granite phoenix]

Ty nwn

[left river]

np

[visual flume]

hlep pls ahhh

[slender phoenix]

can anybody help me study trigonometry easily (Im a machining student)

[silver quarry]

If you prefer watching YouTube instead of books https://m.youtube.com/playlist?list=PLhSQQGbiWoTChZSbP-XcXH40IT7vfyyFk

I recommend this one: https://youtu.be/mhd9FXYdf4s?si=MtURf_8IpejY2tzf

[slender phoenix]

Ill take books aswell

[silver quarry]

I'll let someone else recommend as I've not read them

[slender phoenix]

oh alright but thanks for your help

[drowsy bison]

Guys

when proving something of euclidian geo using analytical geometry

how do you do?

like, I need to prove the angle bisector theorem using analytical geometry

and I'm trying to do so using the equation for the angle bisector line

but it has so many things to take into account, like if the eq is of the form Ax+By+C then if C is negative then my eq becomes negative

and so

[summer cradle]

try to set up an equation

[grave pond]

It's often a great help to choose the coordinate system cleverly.
In case of the angle bisector theorem, everything becomes much easier if you pick a coordinates such that the angle bisector is the y-axis. The slopes of the other lines now end up having very simple relations.

[drowsy bison]

yeah?

but the conditions about the formula for the angle bisector line don't affect?

That's what's been puzzling me

Like I can just say "I got this triangle but I don't care about where it's located in the plane, as long as it's a triangle I can choose my coordinate reference"?

and place it as I did in my proof?

[grave pond]

Yeah, that’s what I would do.

[drowsy bison]

Okay, ty

I mean it makes sense but we didn't see the topic with the teacher

and now I got an assignment that's like 100% of my first unit

bc he didn't want to check homeworks

he also said not to cheat but we could use the book

[grave pond]

It's up to you whether you have more purist ideals about what "using anslytic geometry" requires here...
But in this particular case the win from a well-choosen coordinate system is huge -- each of the two ratios in the theorem rurns out to equal |x1|:|x2|.

[drowsy bison]

I don't know what that last thing is but ty

[grave pond]

The ratio of the absolute values of the x-coordinates of the two other corners in the triangle.

[drowsy bison]

oh, ty

ty so much

[drowsy bison]

just got it!

I haven't done that much algebra in a while

[visual flume]

here sin(x)^2 + cos(x)^2 = 1 hehehe thats all you need to know

[fallen monolith]

Hello chat

[summer cradle]

Let A(0,0) and B(0,6)
sqrt(x²+y²) = 3sqrt(x²+(y-6)²)
x²+y² = 9(x²+(y-6)²)
x²+y² = 9(x²+y²-12x+36)
x²+y² = 9x²+9y²-108x+324
8x²+8y²-108x+324 = 0
x²+y²-(27/2)x+(81/2) = 0
This is the equation of a circle with centre (a,b) = (-(27/2)/-2,0) = (27/4,0)
a²+b²-C = r²
(27/4)²+0²-(81/2) = r²
(729/16)-(81/2) = r²
r² = 81/16
Area = pi * r² = pi * (81/16) = (81/16)pi inch²

[mystic zealot]

why into geometry and trig

[real sentinel]

It's easy

Write general term

Solve for power equals to -13 and -15

As you have to multiply (x + 1/x) again

[real sentinel]

2th part of the question can be simplified

Middle terms can be simplified as (x^2 - 1)^7

Split the last bracket into power of 7 and u get [x^4 -1]^7