geometry-and-trigonometry

[hybrid kestrel]

I just knew the useless thing wouldn't have a preview

[cinder dagger]

What am i looking at bro

[chilly wasp]

the question says "Consider points P(-2, 3, 5) and Q(1, -4, 3). Find the point where PQ meets xy, yz, and zx plane". im lowk confused how one line can meet all three planes

also how do you solve this

[exotic yarrow]

As a lazy example,

[buoyant socket]

How can I tell which side is hypotenuse and which side is adjacent

[lone panther]

longest side = hypotenuse

[chilly wasp]

oh.

thanks 😭

[buoyant socket]

Ah okay thank you

The one opposite from it is adjacent and the one left is the opposite one

[chilly wasp]

i think im supposed to solve it using the section formula

[lone panther]

adjacent = the side next to the angle

adjacent is different from the hypotenuse

[exotic yarrow]

[lone panther]

opposite = the side not next to the angle

also different from the hypotenuse too

[exotic yarrow]

yeah you could do that since one of the coordinates is zero, thereby giving you the ratio

[lone panther]

this also means if you choose a different angle, you get a different adjacent and opposite

[buoyant socket]

It depends on the angle right

[lone panther]

yea

[buoyant socket]

Okay!

[lone panther]

if you pick the other angle in the triangle, they swap

notice that they do that

[chilly wasp]

hmm do you think there could be another way to solve it

[buoyant socket]

Ohhh…

Right.

[exotic yarrow]

Maybe I wouldn't do it with the section formula explicitly, but I'd definitely use some kind of ratio logic

basically testing what multiple of a given direction vector is needed to "travel" along to reach a plane

I suppose you could also find the equation of the line and set each of the coordinates equal to zero

but even that is kinda still doing ratio stuff, albeit less explicitly

[chilly wasp]

whaaaaattt

hold on i gotta comprehend this

[lone panther]

CSP takes 30 whole seconds to find the Normal Way to Solve the Problem

[chilly wasp]

whats csp

[lone panther]

Im gonna give you 30 seconds to look for a three-word phrase in this area

that starts with C, S, then P

[chilly wasp]

OHHH

😭 😭

[lone panther]

bro that was 40 seconds

[chilly wasp]

whatever im slow

why do you think im asking questions in the first place

ANYWAY

[lone panther]

the normal way you usually do this is first find the line through P and Q

[exotic yarrow]

https://giphy.com/gifs/sitewwwreactiongifsus-lcySndwSDLxC4eOU86

tbf the first time someone abbreviated me like that, it also took me a second

[lone panther]

its a 30-second question after all

not a 10-second one

[chilly wasp]

and then

[lone panther]

you at least gotta find the line

[exotic yarrow]

meh maybe I'm a bit slow with numbers but conceptually yeah

[lone panther]

now you have some choices on what answer you ultimately get for the line

these kinds of choices

thats not a concern for now, get a line first then we can try out stuff with that

[chilly wasp]

i got the line

(x+2)/-3 = (y-3)/7 = (z-5)/2

[lone panther]

thats interesting

you think youre missing something there or

[chilly wasp]

eh

i think that should be the line equation

let me check

[lone panther]

youre missing (s

for your )s

usually theyre written in pairs

( ... )

[chilly wasp]

ah right

i got lazy

[lone panther]

yea

[chilly wasp]

there

[lone panther]

alr with this, heres how the Normal Way does it

you remember how you find how 2D lines intersect with the x- and y-axis?

how do those go?

[chilly wasp]

yeah you for x intercept you put y = o and opposite for y intercept

[lone panther]

now this time, we arent exactly dealing with axes

were using planes this time

planes are similar to axes in that you set something to 0 though

use the xy plane for example

what do you think you set to 0 to get the xy plane

[chilly wasp]

z

duh

[lone panther]

yep

[chilly wasp]

im so smart

😎

thanks

[lone panther]

np

[chilly wasp]

OH MY GOD IT WORKS

THIS IS SO COOL

🤯

[lone panther]

you deal with the same ratio-related business either way, but this shows the fastest way to use them

usually textbook ways show the most direct or easy way you can find something

[chilly wasp]

yeah that was so much easier

[lone panther]

theyre rather specific though, Ill warn you that

[chilly wasp]

than the textbook solution

[lone panther]

yea we used the textbook solution from the 2D case

you still gotta pick and choose what you want best

its really ideal you went for a line equation btw

if you went for parametric this wouldve been messy (and Id have to convince you to use the other version)

[chilly wasp]

parametric?

[lone panther]

theres another way you can draw lines

in a way its more direct

say you start at P

then you add (Q - P) t

[chilly wasp]

uh huh

[lone panther]

where t is any real number, its a variable

[chilly wasp]

oh right

i know this

[lone panther]

for example t=0 gets you P, t=1 gets you Q

and since its a line, real number t gets you the line between P and Q

between them, and outside of them

now seeing when that crosses the plane is messier

[chilly wasp]

i kinda forgot tho

let me check

the ohter way of expressig lines

i remembered

[lone panther]

cool

now checking when this intersects the plane isnt as easy

[chilly wasp]

it was (P)t + q i think

where p is a vector along line and q is point on line

is that what ur talking about

[lone panther]

we have two points though, not a point and a vector

we need a vector along that line there

this vector is to tell us what direction the line's gotta be in

"direction vector"

easy way is to do P - Q

or Q - P

[chilly wasp]

yeah

[lone panther]

both of these are vectors that start/end at P and Q

so thatll always work

P + (Q - P) t

Q + (Q - P) t

lol

Q + (P - Q) t
Q + (Q - P) t

these all work as lines, the same ones as that equation you have

[lone panther]

to get a point on this line, you put in a value of t

[chilly wasp]

oh well if you already know the vector and a point its better it use the other form

ur right

[lone panther]

you saw how the equation was more straightforward

once you have the vector and the point, you can still just write the equation version anyway

[chilly wasp]

yk yesterday i was doing some physics and in it there was a question about finding the magnetic inside a circular loop at a distance r where r<R (R is radius of loop). now in the solution it says to use ampere's law and it treats the current being spread over the whole area of the loop but tbh that seems wrong

so i was thinking

more accurate would be the small circle of radius r on the perimeter of the bigger one

and then find the arc length which is enclosed by the small circle

[lone panther]

dont know physics unfortunately

maybe the Pigeon does

[chilly wasp]

no its more of a math question atp

what will be the arc length enclosed by a small circle of radius r situated on the perimeter of a bigger circle of radius R

yk

[lone panther]

youll have to draw this

[chilly wasp]

hold on

this

and you gotta find the arc length of red circle inside the blue

take radii R and r

R>r

[lone panther]

theres certainly one messy way you could solve this

there should be something convenient though

[chilly wasp]

what is it

[vernal notch]

Can someone help me in topology?

[lone panther]

you can ask it as a question #❓how-to-get-help then wait for someone to help you out

@chilly wasp it turns out the "messy" way isnt that bad

Ill show you what it is then

[lone panther]

you know the equations of the circles you used?

[chilly wasp]

hold n

[exotic yarrow]

https://giphy.com/gifs/viralhog-viral-hog-pigeon-catches-the-sydney-train-kS1rILGKhd0JGsKuNZ

[lone panther]

no way, its the Civil Service Pigeon

z^2 what

[chilly wasp]

the radii are kinda random here tho

[lone panther]

why not capital and lowercase r

[chilly wasp]

cuz z is cool

[lone panther]

bro

this is for your own good, you gotta name them reasonably

[chilly wasp]

😭

[lone panther]

go expand the second one, then substitute the first one
solve for x

[chilly wasp]

okai

[exotic yarrow]

me being a celebrity

eh ig I write mini physics text walls in hwh every now and then but that's rlly it

ok I leave y'all to it

[lone panther]

cya

[lone panther]

I mean granted you did do a face reveal by riding a civil service as a pigeon

its hard not to point that out

[chilly wasp]

x = R - (r^2/2R)

[exotic yarrow]

[lone panther]

alr with this

I want you to think about how you can get the angle you need

to figure out what that arc inside the big circle is

[chilly wasp]

alright

[lone panther]

you should try drawing some lines first to see what angle you want

then you need a way to get angles from lengths of lines

you gotta see how you can get an angle first before looking for ways to get it from your r and R

remember, the bottom half is identical to the top half, so youre getting half of the angle you want
thats alr we just * 2 at the end

[chilly wasp]

lemme see

omg bruh

i need to find the length of that blue verticle line

🤬

[lone panther]

do you have to?

alr what are you thinking of to get an angle

[lone panther]

what tool is gonna tell you it

[chilly wasp]

cos inverse

[exotic yarrow]

wolfram alpha

[chilly wasp]

wait no

[lone panther]

thats the good choice

arccos will do it

[lone panther]

next up, theres a reason I chose this triangle

[chilly wasp]

yeah

[lone panther]

its to prevent you from seeing right angles where there are none

the only right angle is between that red line and the x-axis

no other angles are right

[chilly wasp]

yep

thats the prblem agghhhh

[lone panther]

so if we need a right triangle,

how about you draw one here

one that includes the angle we want

theres a grand total of 2 possible right triangles

[chilly wasp]

if i can find that small length on the axis between the verticle and the red perimeter

[lone panther]

(the bottom half is just the same 2 right triangles but flipped)

[lone panther]

alr thats a good move

surely theres an easy way to do that

what do you know about the red radius? and about where the blue line's at?

also

there's a good reason we dont just copy-paste decimal numbres

[chilly wasp]

huh

[lone panther]

think about this

is "3.03714" really gonna tell you the method to calculating it beyond having the calculator do it for you

[chilly wasp]

oh

ur right

[lone panther]

also it looks ugly as all hell

[chilly wasp]

i should be using the radii

right

wait

IM CONFUSE

how did i even get the slope

no

i do know

[lone panther]

you shouldnt need the slope btw

[chilly wasp]

why not?

[lone panther]

the black and green lines are entirely unnecessary

erase them

[chilly wasp]

🤯

[lone panther]

you had a different right triangle in mind

why bring in these other lines outside the right triangle?

[chilly wasp]

?

[lone panther]

so all you really need now is this

surely theres an easy way to figure this distance out

its on the x-axis

[chilly wasp]

hmmmmm

[lone panther]

and you know where the blue and red are at

so whats the big deal

(youd know where the blue is after turning that 3.07 whatever to an actual formula)

(so that if you were to drag your R and r sliders around, you can see the blue line move along with the circles, so that you know what it is)

[chilly wasp]

the values made it harder

maybe i should re write it

ima cry yo

what is this

😭 😭

[lone panther]

ok tell me

what was the formula again

[chilly wasp]

for?

[lone panther]

the formula you told me earlier

the only one

lol

[chilly wasp]

x = R - (r^2/2R)

this?

[lone panther]

yea

[chilly wasp]

alright

[lone panther]

so change your r and z to R and r

then go put that in

[chilly wasp]

.

[lone panther]

lol go type it in with your actual hands bro

you just copy-pasted it

[chilly wasp]

😭

[lone panther]

press the / key on your keyboard

[chilly wasp]

that maeks a difference?

okay

[lone panther]

desmos reads the latex of the stuff you type
this aint latex, so it doesnt work

desmos thinks this is R - r^2 / 2 * R

since by default / only does the closest two things it sees

thats what the {}s are for in latex

,,R-\frac{r^2}{2R}

[somber coyote]

mtt

[lone panther]

if you write it like this

[chilly wasp]

okey here

[lone panther]

didnt you say r < R

or did you swap them

[chilly wasp]

i think i mightve

😭 😭

OH RIGHT

i did

oops

whyd it overlap

[lone panther]

wdym

we solved for it

do you know what that R - r^2/(2R) was for

[chilly wasp]

line through points common on both circles?

[lone panther]

yep

[chilly wasp]

yeah

oh right

[lone panther]

alr

now you know where that red line is

you also know where the red circle ends

[chilly wasp]

OH

OH

🤯

[lone panther]

bro couldve subtracted them 20 minutes ago

and all you had to do was use the actual formula for it (and not swap R and r)

[chilly wasp]

its r^2/2R

that small length

💔

omg

im stupid

[lone panther]

its really clean how that distance turns out though

usually its ugly

thats partly why I didnt want you to try this out at first

r^2/(2R) though is small and neat

alr now go take it home with the arccos

[chilly wasp]

yeah

[lone panther]

you have all you need

[chilly wasp]

NICE

[lone panther]

remember to 2 * the angle at the end

this is just the upper half of it

[chilly wasp]

wait

okay nvm

i thought about using arcsin for a second

[lone panther]

arcsin would need that vertical distance which then requires you to deal with the y-coordinate

the y-coordinate is not nice, not when the x-coordinate and hypotenuse already are clean

[chilly wasp]

yeah

the angle is arccos(1-(R^2/2r^2))

for one side

[lone panther]

1 - shoudlnt be necessary

[chilly wasp]

arc lenght isss

[lone panther]

neither is the ^2

[chilly wasp]

eh

why not

[lone panther]

as a reminder

arccos needs you to get a cosine for it

whats the cosine of that angle

[chilly wasp]

1 - (R^2/2r^2)

yeah

[lone panther]

whats the adjacent of that angle

[lone panther]

and remember, your right triangle would be here

[chilly wasp]

oh i took the other one

[lone panther]

also your r and R are swapped again

[lone panther]

this is the old image

[chilly wasp]

angle between the black line x axis as theta

[lone panther]

terrible idea

[chilly wasp]

why tho

[lone panther]

that finds you the angle of a completely different circle

[chilly wasp]

i can use r*theta for arc length

[lone panther]

unless you want to say you need to find the arc length of the red one, and you never clarified the entire time

[chilly wasp]

yeah 😭

[lone panther]

given you swapped r and R...

yea thats what happened

tell me again, which arc length do you need

blue or red

[chilly wasp]

red

[lone panther]

jesus christ

[chilly wasp]

[lone panther]

then you still swapped r and R in the image anyway

thats the old one from when r and R were swapped

you have the vertical line correctly graphed rn, what formula does it use

[chilly wasp]

this

[lone panther]

you swapped r and R again

[chilly wasp]

did i

[lone panther]

you said r < R right

and R referred to the centered circle

[chilly wasp]

oh

oops

[lone panther]

and r referred to the smaller one, yes?

[chilly wasp]

i guess so

[lone panther]

didnt we fix that last time, why is it still here?

[chilly wasp]

😭

cuz i changed the formula instead

[lone panther]

bro what

youre like that lawmaker who tried to rule π as 3.2

[chilly wasp]

😭

[plucky canyon]

Yall im a trig beginner can yall teach me

[lone panther]

maybe someone else can but rn I gtg soon

[plucky canyon]

Ok bro

I don’t get sine cosine and tangent like why is it like 1/2 and allat

[chilly wasp]

thanks mtt

[lone panther]

gtg soon means Im still here for a bit

[chilly wasp]

i mean i got the answerrrr

yay

[lone panther]

which is?

[chilly wasp]

2R*arccos(1-R^2/2r)

[lone panther]

you forgot the * 2

[chilly wasp]

RIGHT

thanks

[lone panther]

np

cya

[chilly wasp]

goobdye

[chilly wasp]

how would you prove two points are on the same side of a plane

i was think like before

define vector between the two

find equation of line

and then see if a point on the line satisfies the equation of plane

is that it

[winged oyster]

in yalls opinion is it easier to find a right triangle through pythagorean theorem or perpendicular line formula

[winged oyster]

would assume so

[chilly wasp]

wdym perpendicular line formula?

[winged oyster]

if they have the same coordinates itd also be obvious i think

[winged oyster]

like

2 lines with perpendicular slopes

[chilly wasp]

oh

i would trust perpendicular line formula more

whats formula in it tho?

if one is m other is -1/m

just gotta check that right

[winged oyster]

yeah

[chilly wasp]

nice

yeah i like that more

what about you

[winged oyster]

i like perpendicular slopes more cause its faster

[chilly wasp]

nice

[winged oyster]

so same as you

[chilly wasp]

im dying to get a proof in a test

so i can write some dumb stuff to end it off

but sadly

school is over

no more proofs

[chrome glacier]

formula to determine what?

[tiny acorn]

Heres a problem that im stuck on right now.
"to make pencil shells, they uses rectangular wooden sticks with bases that are 8mm by 8mm squares and 185 mm long. How many pencils can be made with 10m^3 of wood, knowing that 12% of the wood is lost in processing?"

[dawn pond]

So we assume x such that it represents the wood cost if a single pencil, we get that x = 8² × 185
And then we assume y to be the amount of pencil we get, wouldn't that be
y = 10¹²-x(1.12)

[tiny acorn]

Wouldnt y be 10^12 × 88% / x?

[dawn pond]

That was gonna be my second idea

But the thing is

[tiny acorn]

?

[dawn pond]

Is that 12% relative to the whole quantity or to the x

[tiny acorn]

Thats ambiguous

I think its like to the 10^12

So only 88% of the wood is actually used

[dawn pond]

So if you use x amount, would that require 112% of x or will it leave you with -(x+12)%

[tiny acorn]

Truly no idea

[dawn pond]

Because my idea is that you're using 112% since that's more realistic

One pencil should cost 11840, which is nowhere near 1% of that whole quantity let alone 12%

Sounds more realistic

But since you pointed out, my approach seemed terrible

Something like y = 10¹² /(1.12)x might be the better answer

Thoughts?

[tiny acorn]

So with this you mean each pencil shell uses up an extra 12% of x as waste?

[dawn pond]

Yes

So ur wasting 12% of x for each craft

[tiny acorn]

Maybe

[dawn pond]

Hence why you need 100% and another 12%

Which gives 1.12

[tiny acorn]

Ah

,calc 1.12*11840

[somber coyote]

Result:

13260.8

[tiny acorn]

,calc floor((10^12)/13260.8)

[somber coyote]

Result:

7.5410231e+7

[dawn pond]

Sounds about right

[topaz basalt]

i want tricks for squares an cubes

can anyone help ?

[foggy heron]

wdym?

[fluid elm]

Can someone help me with vectors and the law of cosines?

[zealous pike]

It says that it's not possible to find x because you can't affirm that the middle line is parallel to the base. Is that really the case?

[fluid elm]

Yes, I can't explain it properly, but I've seen this question before and some mathematicians have discussed it, and the answer was that it's impossible to find x.

[grave pond]

If you don't know the lines are parallel you can move the point marked "z" in pencil up and down without changing anything about angle y -- so in that case, there is indeed not enough information for a solution.

[zealous pike]

Isnt the sum of the angles of any trapezium 360°?

[grave pond]

Yes, every quadrilateral even.

[zealous pike]

Hmm

[grave pond]

But that doesn't help you here, because all that tells you is that z+x = 130° (which you already knew from the angle sum of the outer triangle anyway), and you need x alone.

[zealous pike]

It's unsolvable because although you can find a value in case the middle line is parallel, you can't find anything in case it isn't parallel, so it's uncertain?

[grave pond]

I'd say that adding an assumption that the middle line is parallel to the base will select a single one of the many solutions the problem would otherwise have.

(If they're parallel, then x=y, and we've presumably already figured out that y=60°, so that settles the sum too).

[zealous pike]

And it fits the criteria for the trapezium, 110+120+60+70 = 360

But

interesting question

If I ever become a math teacher Ill give some of those for my students

[short swift]

anyone have a good explanation of the Law of Sines and Law of Cosines

[upbeat lava]

you mean sine rule and cosine rule?

[grave pond]

There are things called specifically the law of sines and law of cosines.

[upbeat lava]

Oh, mb

I don't know anything about that 😅

[grave pond]

In an arbitrary triangle you have a/sinA = b/sinB = c/sinC, and also a² + b² = c² + 2ab·cosC.

[upbeat lava]

How does that differ from the sine rule and cosine rule?

[grave pond]

If you've learned the same theorems under those names, then it doesn't.

I've seen others answer the same question, assuming the asker meant how sine and cosine give ratios in a right triangle.

[upbeat lava]

Kk

[upbeat lava]

Isn't cos rule A squared = b squared + c squared -2bcosA?

[grave pond]

You can switch around the names of the sides to get three instances of the rule.

[upbeat lava]

Oh, okay 👍

[grave pond]

The one I quoted is common because it reduces to a² + b² = c² when C=90°, thus generalizing the Pythagorean theorem.

[upbeat lava]

Mb then, that's just the one I've learnt 😅

[vocal sentinel]

<@&268886789983436800>

[ionic stream]

how do i memorize the unit fast circle no borax no glue

[dusk pecan]

Are there ways to visualise the hyperbolic angle? For example, for Euclidean angle, if one says 0 degree, 45, 90 degree or really any degree, I have an idea about it, but I have no way of visualising the hyperbolic angle immediately until I compute the area manually?

[rocky sedge]

just learn the first quadrant pattern, then copy it around the circle and change the signs.

[rare ether]

is someone here able to help me with the brief explanation on radians? possibly examples? formulas?

[sharp stag]

Hi guys can someone help?
Let ABC be an isosceles triangle with base AB.
Consider a point P inside ABC such that the angle PAC is congruent to the angle PBC.

Prove, in this order, that:
a. ABP is isosceles with base AB
b. P lies on the bisector of angle ACB.

[dawn pond]

It's generally a great first step to draw a diagram

Alright so we have some useful information about our triangle(s)

So you're given a description of your triangle, ABC is isosceles with base AB

What information can we conclude from this?

[sharp stag]

That AC=BC

and angBAC=angABC

[dawn pond]

Very well

[dawn pond]

Now what properties do we have of the point P

[sharp stag]

Only that angPAC=angPBC, isnt that it?

[dawn pond]

And what does that imply

[sharp stag]

I mean, logically i'd say that that implies that PA=PB but I wouldnt know how to prove that

Or that it lies on the median (which is the bisector and height)

[dawn pond]

Let's take a more angle based approach

So we know that PBC = PAC
And that CBA = CAB

[sharp stag]

Ohh so PBA=PAB

[dawn pond]

If you have a diagram, you can notice that there's a correlation between those 4 angles

Which is that CAB = PAC + PAB
And CBA = PBC + PBA

[sharp stag]

Yes

[dawn pond]

Which essentially gives this

[sharp stag]

So using the 4th theorem we can say that ABP is isosceles right?

[dawn pond]

And that's enough to prove that the triangle is isosceles

[sharp stag]

And it also proves that its isosceles with base AB

[dawn pond]

PAB is isosceles with base AB

[sharp stag]

Now onto part b?

[dawn pond]

And that's task 1 completed

[dawn pond]

Let's take a look back at the properties of an isosceles triangle

[sharp stag]

Well, the height, median and bisector are the same

[dawn pond]

There's an interesting property about its angle bisector

[dawn pond]

Correct

So if we can prove any of those, we prove all of them

[sharp stag]

(median of base, height of base and bisector of top angle)

[dawn pond]

Which one seems most obvious to prove?

[sharp stag]

Id say the median

[dawn pond]

And I would agree

[sharp stag]

Since if you can prove that the point M where the medians end is the same you prove it right?

[dawn pond]

You can use cos/sin

Are u familiar with those?

[sharp stag]

Yes but not in this context

[dawn pond]

It's alright

We can use the property that PA = PB

[sharp stag]

Wait

if I proved that they're both isosceles triangles with base AB

Then the median point is the same for both

[dawn pond]

And similarly CA = CB

[dawn pond]

That works too I believe

[sharp stag]

What was the method you were gonna use?

[dawn pond]

Since AC = BC, this implies that C is a point on the median of AB

Similarly PB = PA

So P is a point of the median of ABC, which is also the bisector of ACB

[sharp stag]

Ohhh yeah that works

thanks so much anyway couldnt have figured out by myself

[dawn pond]

Give yourself some credit, u practically solved everything

I just pointed out some basic properties ^^

Again, I suggest using a diagram in case you're not. Geometry in particular can give very important insights through visual observation

[sharp stag]

Oh yeah I mean Im drawing the given information

[loud sphinx]

Guys can u help me i just wanna make sure that i understand this well is Like is angle theta how we view things and its mesuewred and determined by how big the thing is divided by our distance and we can see these stuff in a bigger way using lenses that show enlargements using dilation that will apply ln the other angle

[grave pond]

It is common to use the letter theta to refer to any which angle you need to give a name in whatever you're doing.

There is no single meaning of "theta" that is always the one that is meant when speaking about telescopes.

If you continue to refuse to show the diagram you're asking about, we cannot tell you whether you're right about what theta means in the diagram you're keeping secret from us.

[dawn pond]

It can also refer to temperature but that's irrelevant

[strange narwhal]

I've converted it into this, so the blue region equals the circular segment of the smaller circle minus the circular segment of the bigger one (which is in green)

finding any angle on vertices A or C will solve the problem

[grave pond]

I think that's the reasonable approach.

[strange narwhal]

thx, but I'm still struggling

[grave pond]

I think I'd just coordinate bash the intersection points.

[strange narwhal]

coordinates 😂 😭

yeah, I've heard of such a thing (coordinate bashing, I mean 😂 )

but I neither dunno much about it nor like coordinates much

but I'll struggle some more

don't spoil it haha

[grave pond]

Alternatively, you know all the side lengths in the triangle between one intersection point and the two circle centers, so you can also use the law of cosines.

[strange narwhal]

yeah, I just used it

lemme see where I can get with it

[spring hazel]

i remember something like this from geometry last year

good times

[grave pond]

Pretty sure it is not the first time the area of those particular lunes have been discussed here, yes.

[spring hazel]

no it was an exam question

i wasnt on mathcord last year lol

ah they're called lunes! i remember now

[strange narwhal]

yeah, a year ago when I was studying geometry I intended to derive the general formula for the intersection of circles, but I stopped studying euclidean geometry in general to focus on trigonometry for calculus since I was going to uni

but I think I'll attempt that later

[spring hazel]

is the triangle equilateral?

no

[strange narwhal]

probably almost

EFB might be

[spring hazel]

i think maybe you extend ac? is that helpful

[strange narwhal]

what I had done

[spring hazel]

is that 2 (alpha+ beta)?

[strange narwhal]

yeah

[spring hazel]

are the trangles similar?

try extending middle line

[strange narwhal]

then I can express the blue area as $$\frac{5^2}{2}\left(2(\alpha+\beta)-\sin2(\alpha+\beta)\right)-\frac{10^2}{2}\left(2\alpha-\sin2\alpha\right)$$

[somber coyote]

Vanellope von Schmugz

[queen juniper]

i js went ahead and ||integrated it since the blue area is just the area between 2 circles||

[spring hazel]

was that easier?

no like i havent done calc i only know power rule from 3blue1brown

[strange narwhal]

I had drawn it with extended first but I can't see how it could help

[queen juniper]

not rly, but it works ig

[strange narwhal]

if you already know calculus then it won't be that hard, probably

you won't have to figure much out, just compute

[strange narwhal]

but like, will I eventually need to compute a trigonometric function for a non-notable angle?

[grave pond]

I would expect so, but I don't remember the solution.

[strange narwhal]

because I'm trying to avoid that

[grave pond]

I'm not optimistic that it is possible to avoid it.

[strange narwhal]

yeah

[strange narwhal]

maybe it's possible to develop using identities and find some substitutions, so one doesn't need to find the angles

but it seems at least one function will have to be computed

[spring hazel]

what about inscribed angles

is there a way to use those?

no its not

[strange narwhal]

thought about them, haven't really tried

just used, might have helped

wait a bit

[spring hazel]

,tex

[somber coyote]

Please give me something to compile, for example latex ,tex The solutions to \(x^2 = 1\) are \(x = \pm 1\).See ,help and ,help tex for detailed usage and further examples!

Apr 5, 2026, 10:23 PM

[spring hazel]

,tex $ 100 \pi \cdot \frac {2 \alpha}{360}$\

[grave pond]

(My gut feeling is that is feels unlikely that the area is even algebraic over Q[pi], in which case it will be impossible to compute the solution without evaluating at least one trig function.)

[spring hazel]

how do i do pi and alpha

[grave pond]

\pi and \alpha, but in #latex-testing.

[spring hazel]

thanks

[strange narwhal]

'tis what I said, man

[somber coyote]

ragav

[spring hazel]

that's the sector so you'd still need ef

[strange narwhal]

you should write 2pi instead of 360 in this case

you could use the degree mark, but you'd be mixing units and radians are better to operate

[strange narwhal]

luna = moon 👍

[spring hazel]

yeah

[spring hazel]

can i turn this into a kite?

[strange narwhal]

you already have a kite, that's why I concluded the angles are such

[spring hazel]

oh i'm slow

it's a been so long since geometry

[strange narwhal]

no problem haha

for me too

[strange narwhal]

that was 1 and a half year ago

and I had never been specially good at it

[spring hazel]

ah pretty much the same for me

[strange narwhal]

but after dedicating some time with it I improved considerably

very fun to deduce the theorems from the axioms

[strange narwhal]

it's not possible indeed, it seems

there are some cool cancellations I could have done, but still, there would be some arc functions to be evaluated

@spring hazel @queen juniper

[queen juniper]

hmmm

[spring hazel]

thanks

[strange narwhal]

my naming of the angles could have been better

[dawn pond]

Oooo fast

Thanks 🙏

[sullen palm]

Hello yall if yall remember the other LazyGenius it was me

I had problems with trig

[sullen palm]

Area of the shaded region?

Okay lemme try

[thorny cargo]

can anyone teach me trigonometry? i couldnt understand a single thing which my teacher said if you want to then dm me

[muted mango]

whats the information given about the arc thingy inside?

is it allowed for me to use calculus ?

@strange narwhal

[sullen palm]

Yall teach me too

[strange narwhal]

quarter circle

[strange narwhal]

nope, but it's a way to solve it, so try that either way

[sly urchin]

(I) inscribed in ∆ABC and touches AC at E. G is the intersection of the A-median and BE. Prove that GI, C-intouch-chord and C-midline concur

[obtuse walrus]

Is it B-bisector or just BE?

[sly urchin]

Fixed it

[celest path]

How can I know how many degrees I need to rotate around a point in a rotation transformation problem and in which direction (clockwise or counterclockwise)?

[fallen violet]

Hi

[strange narwhal]

crazy construction

[lusty igloo]

Hi my name is Shawn Mpuuga
I am in grade 11 in South Africa I will be learning about trigonometry and geometry at school and want to help
I also want to do some self study particularly the Taylor series, Frioer Series and Chebysiv polynomials
Anyone here learning about them please tell me and we can help each other
Please send me a dm
And tell me when you are available
Will be busy when I go back to school and will try to look at your messages on the weekend

[upper karma]

Hi! I am Anusha, I am a junior in HS, i have geometry, trigo, calculus in my syllabus... ive been studying math as a hobby since the past few years and I'd def want to contribute here..

[junior solar]

HII guys

[stray copper]

Soh-Cah-Toa

[forest dew]

bro literally explained trigonometry

even dog understand it

[stray copper]

Exactly

😂

[silent spear]

how much's covered in your syllabus ?

just basic identities and all?

[forest dew]

but i forgot a gain

what was soh cah toa mean

oh

i remembered

[silent spear]

9th-10th?

[forest dew]

sin = op/hyp

[silent spear]

op?

[forest dew]

opposite

[stray copper]

cosine equals adjacent/hypotenuse

[silent spear]

ah u guys call perpindicular opposite right ?

[stray copper]

Tan equals opp/adj

[forest dew]

me ❤️ trigo, calculus, functions and geometry

[upper karma]

yes (im currently in 11th)

[forest dew]

but me hates percents

me hates problems

[upper karma]

yes

[silent spear]

ah so pcm ,i kind of wish i took pcm instead of pcb

[forest dew]

like if this headphone is 2000 dollars

they do 20% off

2000 x 20/100

[upper karma]

get it changed

[forest dew]

ez

this is easy but

i hate

the hardest ones

[silent spear]

finished my 12th this year , gave jee too , maths would have sure helped bcz now i am thinking to get into some automotive field and take up some courses

[stray copper]

Its 1600

[upper karma]

oh

[silent spear]

havent studies maths in 2 years , got cooked by maths 😂

[forest dew]

right

eliminate zeros

then multiple

[silent spear]

didnt even attempt most of it tbh

[upper karma]

if u dont mind what were ur grades like in 11th

[silent spear]

got kind of decent in chem , physics went well

[forest dew]

2000x20= 16.000/100 zeros gone

[silent spear]

97%

[forest dew]

1600

[upper karma]

wht abt 10th?

[forest dew]

my overall score is 96,4/100

[stray copper]

I just did 2000 times 80%

[upper karma]

im sorry if this is rude to ask

[silent spear]

93%

[forest dew]

from all studies

[silent spear]

bad at langs

[forest dew]

i can help on langs

english or

[upper karma]

fair score yea

[forest dew]

Anusha

you love same things with me

it always same

trigonometry

calculus

geometry

[silent spear]

english is good , i gave my sst and reigonal language exam without studying

not interested in them

[upper karma]

huh

[upper karma]

yea i get it

[silent spear]

my percentage was only good because i got full in maths and science lol

[upper karma]

maths was weird this yr i hate geometry lwk

ill get like a 96/95

or sum

[silent spear]

u gave ur 10th boads this year?\

[upper karma]

yes

[silent spear]

ur reigon?

[upper karma]

up

[silent spear]

35 and 34 guys got real sht paper

[upper karma]

i think

idk

i had 5.3 i think

[silent spear]

95 is extraordinary for ur reigon ngl

[upper karma]

i hope so

[silent spear]

that question where u had to proof the

parallel lines

that was a good one

[upper karma]

i dont think i had it

cause all of the proofs in my exam were fairly easy

[silent spear]

ah i see

9th-10th was good for me olympiad wise

gave lots of them for maths and science

after i took pcb i only gave of physics

[upper karma]

9th/10th for me was cbse alligned w ecs

loved the subjects i studied so went into depth for a few

pcm specifically

[silent spear]

ah i see

[upper karma]

never had the jee dream

[silent spear]

i just gave for fun

[upper karma]

i mean obv

pcb right?

[silent spear]

i wanted to see how good my physics was jee wise lol

[silent spear]

yes

[upper karma]

well good luck for neet (if yet to take)

[silent spear]

not gonna give it , not interested

[upper karma]

what will u give then?

[silent spear]

i got like 92 /100 in physics in jee

[upper karma]

i think thats amazing?

[silent spear]

nothing , i am looking for some automotive courses

[upper karma]

oh

[silent spear]

like 99.98+percentile in physics

[upper karma]

wow man

[silent spear]

gave olympiads and stuff so it helped tbh

[upper karma]

makes sense

[silent spear]

same stuff helped in 10th ngl

u ever give olympiads ?

[upper karma]

i dont

i js learn for fun

[forest dew]

YOOO

nice tag

F1

dude

[silent spear]

well

thanks

ig

[silent spear]

what type of problems do u like in trigonometry

[forest dew]

PYTHAGOEREAN THEOREM

idk how to spell

ill just say PİSAGOR

in turkish

[silent spear]

ah

u guys get like those proof problems ?

[queen folio]

little maths challenge for you

[forest dew]

next year

we get pythagoeran

[silent spear]

ah alr

[tulip umbra]

can anybody help me with this?

[tulip umbra]

<@&286206848099549185>

[tulip umbra]

mb

[winged oyster]

what are the aaa, asa, sas, sss stuff

i know a stands for angle and s stands for side

[strange narwhal]

similarity of triangles

criteria for telling they're similar, more precisely

and similar meaning "of same proportions"

so 2 triangles are similar if you can transform one into the other by just rotating, reflecting and/or scaling

[winged oyster]

hmm ok

[minor heron]

yeah its where u know like. AAA - you are give that two triangles share all three angles

or such

[rapid valley]

You mean congruent

By rigid motions

Asa,sss, sas, etc.. are triangle congruence theorms no?

[strange narwhal]

"scaling"

[rapid valley]

You wouldn't be able to do SSS, or ASA by scaling lmao

If you have an included side that is 7, and another that is 12, you cant just scale them to be similar

[strange narwhal]

kind of, but look at the message you mentioned

since I said scaling, I was talking about triangle similarity

nevertheless, the criterion SSS need not to mean equal sides, but proportional ones

[rapid valley]

And he was talking about ASA, SSS, etc.. no??

Are those not triangle congruence theorms?

[strange narwhal]

so if you have sides a, b and c for a triangle and x, y and z for another, having a/x = b/y = c/z or any permutation would be a SSS criterion for similar triangles but not necessarily congruent ones

[rapid valley]

I was taught similar and congruent are different

[strange narwhal]

congruent triangles are simply a particular case of similar triangles

[rapid valley]

Erm, so youre saying that if I can dilate a triangle to match another using SSS they are congruent?

No?

Because if they are equal, you wouldn't scale lol

[strange narwhal]

what I'm saying is that all congruent triangles are similar but not all similar triangles are congruent

that's what "particular case" means

[rapid valley]

Thats what I was saying lmao

[strange narwhal]

not really. You said "dilate"

[rapid valley]

SSS, ASA, etc.. are "congruence" theorms

NOT similarity

[rapid valley]

Hence my input on this

It confused me when u said scaling and similar

[strange narwhal]

Those letters compress what is actually being said, but what is being said depends on the context and who's saying it. The criteria refers to sides and/or angles, but what does it say about it? That they're equal or that they're proportional? Yeah, at times they're called "congruence" criteria for triangles, so they would mean that the sides are equal, but you do not need to use the criteria like that. At times they're called criteria of "similarity", so what they mean is that the correspondent sides are proportional (and possibly, but not necessarily, equal)

[rapid valley]

Buddy, if ASA is proven, and ASA is a theorm that is used to prove is triangles are "Congruent", would that not mean everything is equal??

ASA, Angle-side-Angle, is used to prove if 2 triangles are congruent to each other

You can understand that with a quick Google search

Ive never seen someone use ASA or SSS to see if triangles are similar instead of congruent

[strange narwhal]

if 2 triangles aren't congruent, you wouldn't use congruence criteria to tell they're congruent, would you?

But suppose they're similar but not congruent, how would you tell?

Isn't the idea of "similarity criteria" reasonable to you?

[strange narwhal]

if those letters mean congruence criteria, then yeah, everything would be equal

[strange narwhal]

I already understand well enough what congruence and similarity mean

[strange narwhal]

I did not pay attention to the specific criteria you wrote, since my argument works in general, but yeah, if you're proving similarity and you have 2 angles, then the side is not important for two reasons:

the third angle is already determined;

you'd only have one ratio for sides.

[rapid valley]

Wow

[strange narwhal]

so yeah, for this specific criterion, with the side included, it's more relevant for congruence rather than similarity, but I maintain that those criteria can and are used for similarity

[strange narwhal]

what?

[rapid valley]

I know

[strange narwhal]

know what?

[rapid valley]

Broad definitions aren't very good

[strange narwhal]

what definitions of mine were broad?

actually, they aren't even "mine" lol

general, maybe, but not broad

I talked about similarity instead of congruence because of that

because the ratio need not be 1

[rapid valley]

I just needed to make sure you knew similar =/ congruent

[strange narwhal]

You did terribly

You misread what I said multiple times and interrupted me

[rapid valley]

Glad I dont accept reviews

I dont think I interrupted you

My apologies

[strange narwhal]

you insisted even after I demonstrated that I knew the difference

[strange narwhal]

sending messages while one is still typing is interrupting

[rapid valley]

Glad you know the difference

[strange narwhal]

Likewise.

[rapid valley]

Geometry isnt hard

Unit circle is where it gets complicated

Trig, wow

[strange narwhal]

What do you find complicated about it?

[rapid valley]

The memorization ofc

[strange narwhal]

c'mon, man

you don't need to memorize anything

plus, you can translate it all into ratios and Pythagoras' theorem so you don't have to use the notation

[rapid valley]

I never understood how to do that

[strange narwhal]

not that this is always better, it's just an alternative way to do it

[rapid valley]

I barley got through the unit circle section

Polar coordinates were good though

[rapid valley]

I took Calc BC junior year and I barley got a 4 on the AP exam

I completely forgot the formula for half angles and doubles angles

[strange narwhal]

about one and a half year ago I didn't know much geometry, let alone trigonometry

it's not that hard, really

trigonometry is just a generalization of rectangle triangles

a generalization of their ratio

[rapid valley]

Keep typing, I was just preparing something

(If you thought I was gonna send a message)

[strange narwhal]

I was kinda done lol

[rapid valley]

Oh lmao

[strange narwhal]

but you can just see sine and cosine as projections

[rapid valley]

Trig sub?

[strange narwhal]

this may look like a weird formula, but it really is just the known bh/2. In this case, b is b and h is asin(C)

that is so because you can draw the height relative to b and it'll be opposite to C and it'll be scaled by a (relatively to the unit circle)

[rapid valley]

Interesting

[strange narwhal]

a better image

the hypothenuse defines the scaling

[rapid valley]

That makes sense

[strange narwhal]

what do you mean?

[rapid valley]

The trig notation to find the area instead of using 1/2bh

[strange narwhal]

oh, substitute, you say

yeah, could be

[rapid valley]

Ok, I gotta go finish burning off my 3D print

It was cool talking to you!

[strange narwhal]

a nice exercise to test this understanding would be to prove (or derive, better) the law of cosines @rapid valley

try to express the side of a triangle in function of the other 2 sides and the angle between them

[rapid valley]

C^2=a^2+b^2+2abcos(c)?

[rapid valley]

Oh ok

[strange narwhal]

almost

[rapid valley]

The other a and b are supposed to be uppercase btw

And thats supposed to be a minus

Im in a rush mb

Gn

[strange narwhal]

'tis the opposite, sides are generally lowercases and vertices uppercase, so it'd be c^2=a^2+b^2-2abcos(C)

[limber dagger]

can somebody help me with this?

[strange narwhal]

inscribed angle therem or something like it

chord intersections

[limber dagger]

i tried that

90+130=220

220/2

110

right

but it says 110 is wrong

[vocal sentinel]

what is total circle degree

[vocal sentinel]

Then if given 2 known arcs, what is the sum of 2 other arcs?

[limber dagger]

im confused

[limber dagger]

well it would just be 360-220

thats 140

[vocal sentinel]

then 2 angles of 5 add together makes 140

[limber dagger]

so the answer would be 70?

yeah it was

how would I do this

[vocal sentinel]

see thats crazy

a chord-tangent angle inscribe and arc will be half of that arc

you are given that arc so?

[vocal sentinel]

you understood it or nah

[strange narwhal]

smth a lil' more interesting

[thorn tartan]

By power of point, DE = 12*25/23.

30 = BCD = BFD = FAB = 180 - BGF -> BGF = 150

[limber dagger]

oh hell nah

[limber dagger]

yes i do

[winged oyster]

qlryaslfxlf

[vocal sentinel]

good job!

[broken crater]

i thought angle kjl would be 52 degrees because it corresponds with the arc but now chatgpt is saying it’s like 76 degrees??

<@&286206848099549185>

and then im confused about when to use the co-sign formula and when to set up a proportion (for the lengths and measurements of chords and arcs)

[alpine garnet]

what is the chord length

[broken crater]

that’s the thing it doesn’t say

i wonder if it’s 52 and it’s just badly drawn because one of the questions is the length of arc kl

[grave pond]

When LLMs say things that sound like nonsense, it's very commonly because it is nonsense.

[broken crater]

lol sounds right

[grave pond]

Do not waste lifetime on trying to figure out ways their semi-random mutterings can be true.

[exotic yarrow]

<@&268886789983436800>

[near vortex]

<@&286206848099549185> I need some help on graphing polar functions. Can someone help?

[lime crown]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[near vortex]

ok

[lunar hull]

Please don't try to work around the restrictions on the factoid commands by forwarding them.

[cyan gust]

Chat

Anyone here?

[serene kernel]

can someone js quickly explain to me the like diagram for this probem
i dont think its like that much of a question to the point where I have to create a post in help-forum
js really need to understand question more
Show that the pairs of bisectors of the angles of a triangle determine on the respectively opposite
sides three segments such that the reciprocal of one is equal to the sum of the reciprocals of the
other two.
this is the problem
it would also be appreciated if someone could js quickly sketch out a diagram for me
bc im quite bad at drawing diagrams

[near vortex]

I dont see an image

[serene kernel]

there is none

thats why i need someone to help me draw a diagram to visualize it

[near vortex]

Oh ok

[cyan gust]

Yo

[alpine garnet]

um shouldnt helper pings restrict to help channels

[lunar hull]

I don't believe this is technically feasible to do. Iirc, the controls over pings is "can the user ping this role in any channel?" And also "can the user ping any role in this channel?"

This doesn't allow for turning off specific roles in specific channels. But please correct me if I am wrong.

[alpine garnet]

true, but you can disable it everywhere and exempt the help channels but that is too much effort

[lunar hull]

Maybe I wasn't clear. I acknowledged this possibility in the message you are responding to, but the only way to exempt the help channels is to allow users to ping any role within that channel. This includes here and everyone.

[alpine garnet]

oh alr

[slim plinth]

How would we do this

[woeful quail]

Calculate the volume of the pool in cm^3 (V=pir^2h) so r=100cm in that case and h=30cm and then divide that by the rate since the product of the rate and a certain x number of seconds will be equal to the volume. Then take that quantity and then use dimensional analysis to convert each second to 1/60 of a minute by dividing by (60seconds/minute) and then round to the nearest 1 minute

[slim plinth]

Would it be first convert seconds to mins

[woeful quail]

It doesnt matter because of propertieeeeews

[slim plinth]

Oh

We can do the work then we convert all the time

I mean at once

[woeful quail]

Yea

[slim plinth]

After that

I got v=300,000 pi cm^3

Do we add

To the 250 cm^3

Then 1 sec = 1/60 mins

[woeful quail]

U could just writr it as 250cm^3/(1 second) and then say 250cm^3/(1/60 * minute) since they are equivalent or 250*60cm^3/minute

[slim plinth]

Oh

[slim plinth]

That’s what I did

[woeful quail]

Yep

Pretty sure 100^2*30 is 300000

[woeful quail]

Wait why is ur cm^3 on the top??

Thry have to cancel out in dimensional analysis

[slim plinth]

I want to keep the cm^3

Is it 1 min / cm^3

But that is where mins cancels too

Can we make it 1 min / 1 cm^3