geometry-and-trigonometry

[stuck sleet]

Help

the answer should be: 31,111 cubic yards

[pseudo junco]

Help
The answer should be 3

[woeful pike]

im sure there was an easier way to solve this lmfao

this is why i need paper

[stuck sleet]

Help

[stuck sleet]

please

[vernal pilot]

I’m not sure exactly how your curriculum is set up, but that doesn’t seem that bad. Most topics usually don’t require a lot of discussion, and you probably end up focusing too much on some topics. If you are given homework sheets they probably have the sort of questions you need to study. Also any assigned textbooks are probably more useful than others for the purpose of doing well.

[queen flax]

hello mathematicians! i have a question related to proving trigonometric identities. can i ask it here

[grave pond]

Most likely you can.

[vernal pilot]

This is the #geometry-and-trigonometry channel

[vernal pilot]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[fleet shore]

im extremely confused
my thought process is that i can prove PT is congruent to QU using CPCTC and then use SSS to prove that the triangles are congruent
but i don't know how to prove that triangle PRT is congruent to triangle USQ

nvm i should prob post this in the help channel

[gentle haven]

Pretty sure that you can use ssa congruence for psq and rtu because the angle used in 90 degrees

That should be a good starting point

[vernal pilot]

Try proving TR=QS? That was my first thought.

[floral cairn]

can someone help

i cant prove all sides are congruent

to prove that it is a rhombus

and i cant think of any other way

nvm

[somber glade]

definition of square

lol

[spring panther]

i swear that angle P is 69

how is it 61

what i did is 2x-26 = x+16

i got x = 42

so now we know that the top side is 42

triangles have a total measure of 180

so i did 180-42 and got 138

since its an isosceles that means LN or PR have the same angles

so i did 138 / 2 and got 69

how the hell is it 61

[brittle crest]

ok

so

I suck at trig

ill give it a shot

[spring panther]

i need someone that dosent

[brittle crest]

Ill give it a shot and tell me if it makes sense but ask someone else for confermation sorry mate

[unborn rain]

#❓how-to-get-help

[brittle crest]

I got 61

[spring panther]

how

tell me

[brittle crest]

I think I know how

[spring panther]

well first you did 2x-26 = x+16 right

what u get

[brittle crest]

x = 42 is right

[spring panther]

what next

[brittle crest]

tell me if I js did sme completly random but

[spring panther]

i mean the correct answer is 61

😐

[brittle crest]

cuase I suck at trig but basciclly, if 2x-26 = x + 16, then x has to be 42, 2(42) - 26 = 58, and 42 + 16 = 58 so that checks out. Next the angles of a triangle should add up to 180, and its an isocolese triangle so the bottem two angles would be the same. Knowing the first angle, 58 degress, the other two angles combined would have to be 122 degrees. Theres two angles so you div by two and get 61 degress

tell me if thats js random shish

your pfp is kewl btw

[spring panther]

wth

[spring panther]

where did you get...

why did you do 42(2) - 26

what

[spring panther]

who is kewl

[brittle crest]

you

[brittle crest]

becuase the expression is 2x-26

[spring panther]

OHHHOOHOHOHOHHOHO

OHOHOHHOHOHOHOHOH

OHOHOHOHHOOHOHOHOH

OHOHOHOHOHOHH

[spring panther]

I AM SO STUPID

[brittle crest]

what

wait

so was I doing the right thing

[spring panther]

YES

[brittle crest]

I've never done trig

[spring panther]

YOU SUBSTITUED

[spring panther]

I FORGOT ABOUT SUBSTITUTION

[brittle crest]

Im in middle school

im

I just guessed

It made sense to me

ij

uh

ok

[spring panther]

im in highscool

i mean if your in 8th we are pretty close

[brittle crest]

dats kewl

nah 7th

[spring panther]

and its not that hard, i just forgot to substitute

[brittle crest]

lol

dude I make stupid mistakes like that all the time

which grade are you in?

[spring panther]

9th

im cooler

@brittle crest

uh

i need more help

[brittle crest]

Ok

Sorry

I left for a min

Im tryn to major in robotics from MIT and I need a really good grade so im tryn to learn calc. I got like a 98 in math the 1st semester but if you could could you help me w calc time to time

[brittle crest]

noit fair

@spring panther

[spring panther]

oh i could help

im so stupid

but

i am a very fast visual learner

[brittle crest]

Twinsy

[spring panther]

i learn fast

[brittle crest]

Same

SMAR

OMGAH

ARE YOU OLDER ME

[spring panther]

so if you give me VIDEOS to watch, ill memorize

and learn

i am NOT able to read texts, AT ALL

[upper karma]

what is the value of a three inch by three inch square

[brittle crest]

Huh

9 inch?

[spring panther]

27

[spring panther]

he said squared

[brittle crest]

Oh

[spring panther]

im pretty sure thats like

3^3

right

3 to the power of three in other words

[brittle crest]

I thiught he ment 3 by 3 in a square shape

[spring panther]

ah

[spring panther]

add me rn

we gonna help eachother

your smarter but ill catch up

[brittle crest]

YAY

Nuh uh

I aint smart

Im

Stupid

[spring panther]

yes you are

98%

wth

ADD ME

[brittle crest]

98 percent in 7th grade math

I DID

[tidal canyon]

is there a formula for proofs&reasons or an easier way to solve them

[vernal pilot]

no, proofs are dumb and dont have formulas. there is no easier method, and it is actively trying to kill you.

jkjk

but there is no easier method other than just pattern recognition and approaching the proof in specific ways

for instance if im doing a trig proof i will focus on the more complex side

[tidal canyon]

aight ty, that sucks tho 😭

proofs are gonna fuck me over this year i swear

[vernal pilot]

is it geometric or trigonometric proofs?

[tidal canyon]

geometric

[vernal pilot]

rip

[tidal canyon]

🥲

[vernal pilot]

learn congruency.

[tidal canyon]

bet

[vernal pilot]

so thats your SSS, SAS, ASA.

[tidal canyon]

got that all down but its still kinda confusing

[vernal pilot]

yeah ig

[tidal canyon]

had to learn sss sas asa ll cpcte and some other stuff before doing proofs

[vernal pilot]

also complementary angles, opposing angles, alternate angles, corresponding angles, ... ect

[brittle crest]

I hate proofin

[tidal canyon]

me three

[tidal canyon]

alr

[brittle crest]

I

suck at trig

[vernal pilot]

i only ever had to do 2(?) geometric proofs in exams

[tidal canyon]

my next test has like 4 figures with p&r and its like 35% of the test grade

[vernal pilot]

ig more like 12 if you count circle theorems, but they were beta questions meant for beta males.

[brittle crest]

lol

[maiden brook]

proofs are better

[vernal pilot]

you are objectively wrong

[maiden brook]

def niot

its math

[untold relic]

can anyone help me with this?

[dark sparrow]

@untold relic what's the goal? simplification?

[untold relic]

yeah

[dark sparrow]

any progress thus far?

[untold relic]

im getting cos theta+sin theta

not sure if its correct

[dark sparrow]

,w is cos(x)/(1 - tan(x)) + sin(x)/(1 - cot(x)) = cos(x) + sin(x) true for all x

[somber coyote]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by ann_dec

[dark sparrow]

damn fuck

it used to understand queries like this

hold on

yeah ok

checks out

your answer is correct

[untold relic]

thank god

i got it on the 4th try

[woeful pike]

tried it for myself and thats a nice problem with pretty annoying factorizing lol

[pseudo junco]

Need Help with this one
Can't move forward

[wild fossil]

Could anyone help me with trigonometry,

the help i need is with the equations and how to plot them

[tame urchin]

This question confused me

On the previous examples it showed when triangles were separated and had 6 vars but now I only have 5 so $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$ WONT work

[somber coyote]

Steve7108

[tame urchin]

Would prefer how to get to the solution rather than the solution itself

[grave pond]

Hmm, I don't think that can b solved with additional assumptions (such as AB being parallel to DE).

What does the answer on page 134 say?

[tame urchin]

BC = 2 and ⅖ and CE = 2½

[grave pond]

Hmm, that's not particularly enlightening.

[tame urchin]

I can show prior pages if that could help?

[grave pond]

If BC is supposed to be 2, then ABC is isosceles, but CDE definitely isn't, so AB is not parallel to DE ...

[tame urchin]

Here is the previous example problem

Just realized that image is bad

[grave pond]

Whoa, bad book.
It doesn't state explicitly that ABC and DEF are similar, but just starts acting like it knows they are similar in the solution.

I don't have any better guess than: You were intended (for no good reason) to assume that ABC and ECD are similar, but there must have been a typo in the givens so they meant "DE=5 and CD=5".

[tame urchin]

Oh

Well that makes more sense

Ot tool me a week to find any book about geometry and it could be inaccurate, finan go to Khan academy

[steady merlin]

brain ticklers ☠️☠️

[somber glade]

are they similar triangles

[woeful pike]

Lazy creators lol only needed to add 2 right angle symbols

[grave pond]

But then their stated solution would be completely wrong.

[woeful pike]

Wait yeah

They can’t be right angle triangles

[red flax]

They all intersect or no

[woeful pike]

Well they meet at a point

Idk if I’d call that intersecting but I can see how a someone else might

Just go with what you think

[red flax]

idk its confusing cause they do all meet at a point which makes then intersecting right

[jagged wyvern]

Can someone clarify ? https://discord.com/channels/268882317391429632/1186517077218033664

[red flax]

does anyone know what the answer is too this 😭🙏

[grave pond]

My answer would be yes too, but it's really mostly a question of whether you define "intersect" to include line segments with a common endpoints or not. The only real doubt is about that word, not about the geometry.

[red flax]

yeah idrk I would select all as they all intersect but idk what they want but logically speaking I think they all should intersect since they all meet at a point

[grave pond]

Um, you don't want to have those screenshots with a realname in them up, I think.

[red flax]

idm

should I delete them

?

[grave pond]

I think yes.

[red flax]

okayt

😔

idk how to get my name out of them

[grave pond]

Do you have some image editing software avaialable? Paint? Gimp? Otherwise crop them to just the question and right answer and type "I answered such-and-such because ..."
Inclusing your reasoning will also enable people to give more targeted help.

[red flax]

yes I am doing it rn

And then there’s this one that had like 4 questions of it and I’ve gone to the class everyday but she’s never taught us this

[red flax]

and theres a couple others like this idk if they are right or wrong though I feel like I did it right

and basically I shouldve gotten a 49/54 which is a A but I got a 42/54 instead

Here’s this one too idk if I’m right or wrong I followed the way they taught us and the formula idk I would appreciate some help if I did them wrong

[woeful pike]

You take the reciprocal and multiply it by -1 to get slope of perpendicular line

You are only multiplying by -1

[woeful pike]

Here if you can’t use your calculator you can form an equilateral triangle with any side length you want (I’d use 2 for an equilateral triangle), cut it in half to get a right triangle with angles 30 60 90 and side lengths (in form h, o, a in regards to the 60 degree angle) h = 2, a = 1 and o = sqrt(3). Then you can use this to solve for sin cos and tan.

[woeful pike]

And here you can use the ratios from the last question and since h = 2(sqrt(3) but the ratio from our last question = 2 multiply everything by sqrt(3) to get h = 2(sqrt(3)), a = sqrt(3) and o = 3
Since we are looking for the side opposite the 60 degree angle we will use o which = 3

[red flax]

I know which got me the opposite right cause it should be y=(-9/2)x-21.5 and the other is y=(9/2)x-21.5 or am I tripping cause if u put the -2/9 slope in and reverse it, it literally cannot touch the points at all and the points are (7,10) and (5,1)

and the one with the (1,12) and (9,3) is y=(9/8)x+13.125 and the perpendicular line is y=(-9/8)x+13.125

[red flax]

yes but we were shown with a special right triangle formula where the hypotenuse is sqrt(3x) or x(sqrt(3) and in that image it shows 2(sqrt(3) with would make 2=x and g=2(x) so wouldnt it be 4 since g=2(2) . and I've never learned H, O, and A

[red flax]

and I havent learned that either its just finding sides of special right triangles

and I used the formula from the special right triangles

which my teacher taught us, it wouldnt make sense to give a final exam where there are things we have not learned and have not been taught

idk I might just be dumb and not understand it

[pseudo junco]

Can't prove it

Trigonometry gawds
Come to rescue please

[velvet relic]

It's not even a puzzle there's just a theorem you're supposed to use which is kinda boring

[pseudo junco]

Yeah theorem I know
But directly using it is not what I want

Isn't a way to derive it from this

[woeful pike]

in a 30 60 90 triangle h is always 2x, the side touching the 60 degree angle is x and the side opposite the 60 degree angle is xsqrt(3)

[gritty mulch]

The hypotenuse of 30-60-90 triangle is 2. The leg opposite to 30° is 1 and the leg opposite to 60° is √3. Here, all sides are multiplied by factor of √3. G is opposite to 60° so √3(√3) = 3

[celest timber]

I need help to find the elegant eclipse funtion but i need the function so i can put in into desmos (i need it for code but if its in desmos its easy)

[red flax]

Ahh okay I see I looked at it wrong

[raw kraken]

An easy problem

[woeful pike]

i like how x isnt a weird number like sqrt(3/2) or something

[maiden brook]

assuming its a square

[maiden brook]

i got it with coord geo but how do u do it without that

[woeful pike]

you cant solve it if its not a square lol

[maiden brook]

well eys

yes

but how did u do it without coord geo

[woeful pike]

i will quickly do a sketch of how i did it

[maiden brook]

u dont wanna see how i did it

[woeful pike]

u can show

im just drawing in microsoft paint how i did it

[maiden brook]

one sec my phone is dead

[woeful pike]

my method

[woeful pike]

i dont get how u did it with co ordinates lol

[maiden brook]

was having dinner

[hard coral]

does anyone know can I get a vector to start somewhere else other than the origin?

[pseudo junco]

They always start from origin

You can rotate them though

[raw kraken]

Here's another puzzle:
Given a perpendicular to the diameter of the largest circle (CF), obtain the sum of the areas A4 and A1.

[stuck sleet]

why the 1/2 have vanished???

<@&286206848099549185>

[stable void]

what is that 4u?

[raw kraken]

The length of the CF segment, where u represents any unit, as it is not utmost importance to the problem.

[pseudo junco]

Haven't you taken theta to be at wrong vertex on the left side?

[stuck sleet]

can you explain more clearly... thanks

[pseudo junco]

You took angle at vertex A to be theta, why?
Take angle E to be theta,

[stuck sleet]

that is not my answer

[stuck sleet]

that answer is from the internet

my answer is 6sqrt(2)

[stuck sleet]

what I am pointing out is the algebra wrong here?

@pseudo junco

[pseudo junco]

This is obviously wrong

[stuck sleet]

yeah thanks

[median vapor]

hello. I've been going over some HS geometry (bc I had a really awful HS geometry teacher) and ended up coming across the law of cosines/cosines formula again and I wanted to ask how much further the pythagorean has been extended beyond just non-right triangles? I think it has been extended to all normal polygons, but has it been extended beyond this?

[nocturne remnant]

What do you mean “[the Pythagorean theorem] has been extended to all normal polygons?”

[unreal geode]

Dam this is hard

[vernal pilot]

Wdym?

[vernal pilot]

Pythagorean theorem and by extension the consine law only applies to triangles. Ofc for any right angle triangle you may use Pythagorean theorem, and for non-right angle you must use cosine law, which works for any value of theta.

[potent yacht]

is this Cosine rule or sine rule?

[stable void]

cosine rule

[potent yacht]

this one?

sine or cosine?

[fickle rose]

@potent yacht a rule of thumb is that if ur given SAS and SSS u use law of cosines. when given AAS or ASA or SSA then use law of sines

[misty tusk]

you can extend the pythagorean theorem into 3d using De Gua's theorem but thats all i know

[rich crow]

#calculus message

can sm1 exlpain this?

[rich crow]

can sm1 guide me to find $csc^2(\theta)=?$

not proving but solving it, using $r^2=x^2+y^2$ etc...

[somber coyote]

milkshake

[misty tusk]

wdym how to find it

like how do you want it

like $\csc^2(\theta) = \frac{1}{\sin^2(\theta)} = 1 + \cot^2(\theta)$

[somber coyote]

architecture2

architecture2

[misty tusk]

which is fairly simple if we start with $sin^2(\theta) + cos^2(\theta) =1$

we can say that if we divide both sides by $\sin^2(\theta)$ we get this

$\frac{\sin^2(\theta)}{\sin^2(\theta)} + \frac{\cos^2(\theta)}{\sin^2(\theta)} = \frac{1}{\sin^2(\theta)}$

which simplifies to

$1 + \cot^2(\theta) = \csc^2(\theta)$

[somber coyote]

architecture2

[floral cairn]

Help

[median vapor]

I meant to say "regular polygons" so that if you have a regular polygon on each side of some right triangle instead of a square you can still do similar stuff (find the areas of the regular polygons via some reworking of the Pythagorean theorem).

[gentle haven]

Because MN and JN are congruent, what does that tell you about triangle JMN?

[floral cairn]

it is isosceles

angles NJM is cong to angle NMJ

converse base angle thereom

[gentle haven]

Yes

[floral cairn]

but idk hwo to prove

tho

[gentle haven]

Wdym

[floral cairn]

idk hwo to prove triangle NOP is equilateral

[gentle haven]

Oh

So what can you say about angles PON and OPN

[gentle haven]

Based on this

Remember OP is parallel to JM

[floral cairn]

ok

[floral cairn]

ok

wait

are

angle MJN and angle NOP congruent corresponding angles?

[gentle haven]

Yes

[floral cairn]

OHHH

THAN KYOUU

[gentle haven]

Np

It took me a while to find it too

[nocturne remnant]

You mean, if I erected regular pentagons from the sides of a right triangle then the sum of areas of the two small pentagons is equal to the area of the big one?

That just follows from the fact that the area of a regular pentagon is proportional to the square is its side length

And you could have replaced the pentagons with any three similar figures

[woeful pike]

I’m pretty sure that a triangle like that is a impossible as the givens can’t happen. OP can never be = to ON.
The smaller triangle will always be similar to the bigger triangle so if that assumption is true the big triangle is also equilateral

So that means x^2 = x^2 + x^2/4

Can’t happen unless x is 0

[merry pecan]

I tried proving angle ONP as 60° by getting angles KNJ=LNM= 60° but it turns out they are tan^-1 (2) so yeah not equilateral

[gentle haven]

Lol I was about to say something similar

For angle JNM I got (180-2arccos(1/sqrt(5)))

[merry pecan]

[gentle haven]

Idk how to put in texit

[gentle haven]

This is abt 53 degrees

[floral cairn]

Talk are overthinking it

Y’all

[merry pecan]

pretty much the easiest method to prove it is equilateral proved that it actually isnt, how is that 'overthinking it'

and theres no way to prove that it is equilateral

because

it isnt

[floral cairn]

There is

U could prove triangles are congruent

[merry pecan]

[floral cairn]

Then prove the diagnosis are congruent

And since the diagnosis are congruent, the base angles are congruent

And then op parallel to Jim

[merry pecan]

youre just trippin

[floral cairn]

JM

[merry pecan]

how can there be two different results theN? either mine, or your method is incorrect

[floral cairn]

I think yours because we didn’t even touched trig yet

We are still on proofs

[merry pecan]

so the question is designed specifically to be solved by what u were taught recently

[floral cairn]

Yes cuz it’s quadrilateral proofs

And it’s a two column proof

[woeful pike]

Makes sense if u haven’t touched trig

So they don’t really need the diagrams to be accurate

Or even possible yet

Lol

The proof works perfectly based on the givens so that’s all you need

[merry pecan]

you should try to disprove it, it would be a nice exercise

[rich crow]

Where’d I go wrong?

I should be able to find that it equals to = $1-2cos^2(\theta)$ right?

[somber coyote]

milkshake

[vernal pilot]

What did you do? If this is a trigonometric proof you can easily use the trig identities. Whatever you have done just overcomplicates things.

Google “Pythagorean identities”

The tldr is cos^2(x)+sin^2(x)=1

But there is another one with cot and csc

[woeful pike]

are you trying to find sin^2(x) - cos^2(x) = 1 -2cos^2(x)

[woeful pike]

use the identities and then look for a way to make a 1

thats what i did and it was easy

(remember the pythagorean theorum)

[gentle haven]

The only reason you can prove the triangles are equilateral in the problem is because OP is congruent to ON

[rich crow]

So I just memorize the Pythagorean theorem identities?

I’m just tryna like understand the concept yk anywhere y’all can point me for that?

[midnight knot]

Can anyone help with these questions please

[woeful pike]

Well you know pythag right?

And I use sohcahtoa to remember identities

S is O / H, C is A / H, T is O / A

From there the question becomes (O^2-A^2)/H^2

Pythag states H^2 = A^2 + O^2

So O^2 - A^2 = O^2 + A^2 - 2A^2

Plug in to get H^2/H^2 - 2(A^2/H^2)

Plug in cos^2 = A^2/H^2

To get 1 - 2cos^2(x)

[woeful pike]

The X^2 in the denominator top right should be an R^2 based on the letters u use for the trig identities

That’s what’s wrong

[median vapor]

oh. thanks. sorry if i abused the word 'extend' btw.

[gentle haven]

For the second one, i'd say you should start by splitting the figure into two triangles and using the law of cosines to fill in the missing information for each triangle.

[floral cairn]

FUCKKK

[trim trellis]

can someone help

nah nvm idek how to translate that shit

[floral cairn]

I messed up on my geo test

Welp

90/100

[timber cargo]

https://tenor.com/view/jump-slapped-slap-lunch-slapped-gif-15374199

[woeful pike]

Do you know corresponding angles?

[stuck sleet]

what is the one that is finding here??? I cannot understand the question

[limber kraken]

https://www.desmos.com/calculator/ha3ltzvcz1 have a cool graph

[limber kraken]

its hould 11, 7+4, they are similar triangles

[silent ermine]

I have an brain error

If i have a sphere

and i cut it into flat surfaces

the middle one is pir² right

/the biggest one

now

i take this biggest one

and i rotate it pi times around the axis

what part is missing

cause it aint pi * pir²

(area)

oh got it

the further areas travel faster than the areas closer to the center

which creates holes

i think

[stuck sleet]

I solved #18 but #19 confuses me, what does it mean by "circle of contact"

[dark sparrow]

the sphere touches the cone in a single point at the bottom and in a circle further up

[stuck sleet]

can you draw it?

Thank you very much!

[dark sparrow]

like this

[stuck sleet]

If a plane cuts there, would'nt be the question is wrong because the cone becomes a frustrum, so the ration will now be the volume of frustrum and volume of the sphere?

[dark sparrow]

they're talking about the cone ABOVE the plane.

[stuck sleet]

Thank you!

Though I still cannot get the right answer. 😦

[tranquil sparrow]

Hey everyone, I was just doing some work on khan academy and I ran across this problem that seemed a bit odd to me, In the solution the measure of angle 1 is 90deg but in the diagram it is clearly more like 94-95deg, But it does not state anywhere that the diagram isn't to scale with the solution, Shouldn't it? or is that unnecessary?

[floral cairn]

u dont find angles in two column proofs.

[tranquil sparrow]

Yeah i know but at the end you prove angle 1 is 90deg its just not included in the screenshot

[tranquil sparrow]

Also does not really seem to be the point of the question?

[blazing veldt]

"As long as the answer is right, who cares if the question is wrong?" -Dodecahedron

[vernal pilot]

Unless it is given a value, then do not think of an image as accurate.

[tacit jasper]

Hi, I'm trying to figure out if a line segment is intersecting a circle, I don't really need the points of intersection just whether or not there is intersection. Also it must be a line segment and not an infinite line. I've been trying to figure this out for a bit and feel like I haven't made like any progress at all, and I just need a pointer in the right direction

I think the main thing is that I'm trying to figure out if a line segment intersects a circle and not a line in general, since everything I found basically has to do with an infinite line

Like I think I could just find the projection of AP onto AB where A is one end point and B is the other end point and P is the center of the circle and then finding the distance of that subtracted from AP... but I feel like there has to be an easier/faster way

[grave pond]

There are two ways an intersection can happen:
a) One end of your line segment is inside the circle and the other end is outside.
b) Both ends are outside, but the perpendicular line from the circle center intersects your line segment at a point that lies inside the circle.

[tacit jasper]

are you almost done typing

[grave pond]

Alternatively: Parameterize your line segment as A+t(B-A) with t in [0,1], and then write the distance from a point on the line to P as a function of t. In fact, the square of that distance will be a nice quadratic function of t, so you just need to be able to compute out whether this function takes the value r² anywhere on [0,1]. That's just a quadratic equation -- or, just to find if a solution exists, check if you can find two points with values on different sides of r² among 0, 1, and the vertex of the parabola if that falls between 0 and 1.

[tacit jasper]

lol sorry I didn't know if you were actually typing or if discord was just bugging

I guess what I'm going for is the most performant one since I would be calculating this whenever a circle and a point could potentially be intersecting in a world (just writing a program to do this)

[stuck sleet]

Thanks I got the answer and the right solution!!!

[stuck sleet]

Each of the twelve edges of a cube of edge a is tangent to a sphere. Find the volume of that portion of the cube which lies outside the sphere.

Thanks for the help!

[brittle crest]

I just found this peoblem on youtube, if anyone wants to give it a shot feel free

Answer: ||20||

[upper karma]

by far the easiest method to solving this problem is to click on the spoiler

[brittle crest]

Smart

[grave pond]

I don't think the answer in the spoiler is right, though. Yes it was; I mistakenly made 10 the radius of the circle instead of the diameter.

My procedure would be to start with the blue 1:2 rectangle and find the circle that goes through the three points we see the circle intersect the rectangle in. Clearly, then, some diameter of the circle will pass through the last corner. Finally scale everything so the diameter becomes 10.

[woeful pike]

link or method pls

[brittle crest]

i dont know how to do this saw it on a youtube short

sorry

[viral mauve]

https://www.youtube.com/watch?v=VMiS7G8xqQs

[grave pond]

Interesting, that's completely different from what I did.

[brittle crest]

thats the vid nice

[viral mauve]

yeah just google reverse image search

[keen turtle]

Without tracing paper how to find point of rotation

[undone oar]

fjghn

[maiden brook]

what did u do?

[grave pond]

This.

[upper karma]

shit uh

Perpendicular bisectors of points

U get the 2 shapes and connect their corresponding points

And draw the perpendicular bisectors of the lines u made

Then they should all touch at some point

and that point is ur centre of rotation

[keen turtle]

Can i inbox you plz

[upbeat light]

How would I solve this exactly?

[dark sparrow]

@upbeat light do you still need help with this?

[upbeat light]

Yeah

[dark sparrow]

any progress?

[lyric sonnet]

is this from a test

[exotic flame]

how can this be solved? sine rule?

[grave pond]

Yeah, for example decide that c=1 and use the law of sines to derive a and b -- you'll find it is a pretty strange triangle.

[exotic flame]

like that $\frac{1}{3SinA} = \frac{1}{2SinB} = \frac{1}{6sinC}$

[exotic flame]

like that ?

[grave pond]

Hmm, not quite -- since the equation you already have says that the three denominators there are equal, your equation would lead to a=b=c=1, which is definititely not the case.

[somber coyote]

Adam

[exotic flame]

like that right since i let c =1 then therfore the rest should be equal right?

[grave pond]

What you'd do is something like
$$\frac{a}{\sin A} = \frac{c}{\sin C}$$ and then, say, divide both sides by 3 to get $$\frac{a}{3\sin A} = \frac{c}{3 \sin C}.$$
Now you can use that you know $3\sin A = 6\sin C$, and therefore $$\frac{a}{6\sin C} = \frac{c}{3\sin C}.$$
Then just multiply by $6\sin C$ to get $a$ in terms of $c$.

[somber coyote]

Troposphere

[exotic flame]

oh so u used the fact that 3sin A = 6sin C after diving by 3 to get a common denominator (sinC)

but still how to get m(<A)

[grave pond]

My plan was to get the side lengths first and then use the law of cosines.
But something strange happens when you have those side lengths.

[exotic flame]

i will have to ask my teacher about this, (hes the one who wrote it)
maybe theres something wrong in this problem

anyways man thanks for the help

[deft drum]

A very naive question, Why the triangle OAP is right angled that too angle OAP?

I know it is something related to ⟂ projection, but do not know how to fit in argument

[grave pond]

OA is perpendicular to he entire plane of the front face of the box, isn't it?

[ember stream]

If you have a line, it will just keep going forever

But if you put a bend in that line, it will become a circle and just keep on looping on itself

Is there a way to calculate “bendiness” and the resulting radius of the circle from this number?

It would have to be the same for every point

[grave pond]

You may be looking for "radius of curvature".

[ember stream]

Oh on thank you

Ok

[grave pond]

It's common to define "curvature" to mean the reciprocal of the radius of curvature. Then you get a number that is 0 for straight lines, changes smoothly to a small number for slightly bent curves, and becomes larger the more, um, curved the curve iss.

[upbeat light]

Yeah I figured it out, just converted it to Cartesian and drew it on an argand diagram to visualise it

[upbeat light]

A past paper I think, I can send you the link if you like

[ember stream]

I dont understand how he knew those two sides were equal

What do those dots on the rectangle denote?

Why can we assume the circle cuts the top side in half?

Does the dot denote the midpoint between the corner and where the rectangle intersects the circle?

Im not familiar with this notation

[gentle haven]

Do you mean the things on the sides of the rectangle?

[ember stream]

Yah

[inner escarp]

is this correct? I'm using these variables later for cosine rule to cancel out a and c

[lyric sonnet]

oh ok dw

[upper karma]

law of cosines as in a^2 = b^2 + c^2 -2bcCosA ?

u can just rearrange

and get the angle

(b^2 + c^2 - a^2) / (2bc) = cosA

[deft drum]

Oh because AP lies on the front face, OAP is right angle

[grave pond]

Yeah.

[celest wave]

evsxp;=[l

[keen turtle]

I have difficulties identifying corresponding points while doing transformation question in igcse 0580 extended math, can someone tell esay way

[heavy anvil]

I would appreciate it if you could solve it simply on paper, I don't understand the keyboard sometimes.

[subtle tartan]

2

[modern yoke]

for this do u have to do 12sin(65)

[grave pond]

What the heck is the text outside the hypotenuse supposed to say?

[modern yoke]

cm

im bad at writing

hypotenuse is 12cm

[grave pond]

Then 12sin(65°) will indeed find the horizontal leg.

[modern yoke]

alr ty

[timid thistle]

so i have four points like this and i'm tryna find an ellipse that runs through them

and i have no idea how to do it

[ember stream]

Tada

[grave pond]

Four points are not enough to define a unique ellipse -- there will be many (or, in some degenerate cases, none) ellipses passing through those four points.

[timid thistle]

how can i find an ellipse passing through the 4 points tho

[grave pond]

I'm not sure this works in all cases, but one simple-minded approach could be to decide that your ellipse will have an equation of the form ax^2 + bx + cy^2 + dy = 1.
Plugging your four points into that will give you four linear equations in four unknowns (a,b,c,d) that can be solved with standard techniques (unless you're unlucky and hit something degenerate). Afterwards you can then complete the squares to get it into the form a(x-p)^2 + c(y-q)^2 = h.

The resulting ellipse will always have its major/minor parallel to the coordinate system axes.

[lament tree]

fun little proof I made some time ago that an isosceles trapezoid with the average of the parallel sides being the height is 45º, or maybe the converse I'm not sure of a detail but well
from a square, extend the diagonals until C and D such that CD is parallel to AB, from ABCD construct an isosceles trapezoid, it's pretty obvious this quadrilateral is indeed an isosceles trapezoid and that the diagonal forms a 45º angle with each base, so now I will prove the average of AB and CD is the height
specifically by proving that the parallel in the middle, GH, is equal to CE, which will be proven to be the height, which since CBE is a right angle with 45º angles, would mean would be equal to BE, which is the height by definition
from H draw EF parallel to AC, where F lies on a straight line extended from AB, since H is the midpoint of BD, EH and HF will be equal, which means the quadrilateral BFDE is a rectangle, since it would fit in a circle with center H, implying BEF is a right angle (because of thale's theorem)
that means BE is a height, and since AC is parallel to FE and GH is parallel to CE, GH is equal to CE, which by what I said above means the average of the parallel sides is the height
now my hand-wavey argument is that by extending AD I could make any possible angle of isosceles trapezoid greater than 45º, and that any trapezoid with such properties would fit in one constructed that way

spent some time trying to decipher what that proof drawing I made in the middle of the year was about lmao

[frozen ocean]

geometry dash theme

[stuck sleet]

Show that the areas of zones on the same sphere
or equal spheres are to each other as their altitudes.

[drifting laurel]

Now that im actually taking the time to patch up my trig, im realizing how beautiful it is

[light shoal]

everyone is doing that while i just wanna know how to calculate the area of a triangle 🫤

and idk if im even in the right place 😭

[ripe bison]

Pythagorean's theorem, and yes, you're in the right place, all good!

[light shoal]

phew

[ripe bison]

You can also open a chat at the top of the server for more specialized help, but I don't know how that stuff works, as I haven't done it yet

[light shoal]

ok

i thought it was bxW

base times width*

[ripe bison]

Oh wait, I'm a moron, my bad, yeah, that's what it is

[light shoal]

oh

[ripe bison]

I wasn't even thinking there, my head's in the clouds today

[light shoal]

lol its ok

[ripe bison]

What type of triangle is it?

[light shoal]

right

[ripe bison]

It'll be the base times height, then divide by 2

[light shoal]

ok thx

[ripe bison]

For reference, base times height without dividing by 2, only finds the area of a rectangle

[light shoal]

because the triangle is half ofthe rectangle?

[ripe bison]

Yep

[light shoal]

thx

[ripe bison]

Np!

Trig is unfortunately one of my weaker points, so I'll definitely be asking questions here a little bit. I've already taken pre-calculus, so I'm just self-learning calculus at the moment, but the book I'm going through has kind of like a pre-calculus chapter 1 section (which I hear is very normal), so I'm just going through that at the moment

As for why I'm using this chat and not pre-calc, well....to be realistic, it's more active than the other chat, so...yeah!

[maiden brook]

it depends on what ur given but yeah thats one way

[vernal pilot]

Keep it in the correct channel.

[vernal pilot]

That is true for every triangle, A=bh/2.

[ripe bison]

This technically is the right channel though. It's a trig section of a pre-calc section

[vernal pilot]

Fair enough

But for anything like derivatives or integrals or anything similar about trig functions belongs in #precalculus

[ripe bison]

No sweat; Pre-calc where I live doesn't even go over derivatives and integrals, lmao

[vernal pilot]

if im not mistaken, precalc should go over derivatives, and the basics of integration/antiderivatives.

[silent plank]

once you're doing derivatives, integrals
you're already well in calculus territory

[rose raven]

So far in my precalc we haven't done any derivatives or Integrals

I asked my teacher if we would and he said we wouldn't, but we will do limits

[pseudo nebula]

I feel like you maybe would depending on your teacher, but usually not considered precalc

[slim sundial]

I do the main three calculus topics at the end of the year

Limits
Derivatives
And
Integrals

[rose raven]

Nice

[final moss]

What is precalc

why can't we just go to calc without precalc

[orchid current]

it's like skipping the prequel and tutorial together

[final moss]

so we don't need it

I actually always skip prequels and tutorials

[silent plank]

its like trying to learn multiplication before addition

[final moss]

Algebraic Modeling
Functions Defined and Notation
Linear and Quadratic Functions
Power Functions and Variation
Polynomial Functions of Higher Degree
Real Zeros of Polynomials
Complex Zeros
Graphing Rational Functions
Solving Rational Equations
Solving Rational Inequalities
Exponential and Logistic Functions
Exponential and Logistic Modeling
Properties of Logarithmic Functions
Solving Exponential and Logarithmic Equations
Vectors in the Plane
Dot Product of Vectors
Polar Coordinates
Complex Numbers in Trigonometric Form
Solving Systems of Two Equations
Matrix Algebra
Matrix Row Operations
Systems of Inequalities in Two Variables
Geometry of a Parabola
Geometry of an Ellipse
Geometry of a Hyperbola
Translation and Rotation of Axis
Polar Equations of Conic Sections
3-D Cartesian Coordinate System
The Binomial Theorem
Sequences
Series
Limits, Motion, and the Tangent Line
Limits, Motion, and Areas
Limits
Graphs of Trigonometric Functions

No way

Is this precalculus

I know only half of it maybe

But somehow understanding calc 1&2

[dark sparrow]

what exactly goes into "precalc" varies a lot from place to place

generally it includes:

• equations beefier than quadratic
• trig
• exponentials & logs
• maybe basics of complex numbers
• maybe basics of vectors
• maybe some combinatorics

[orchid current]

we dont have any proper course such as precalc. everything is mixed up together in sequence

yeah topicwise the required sequence is followed but a seperate course is not followed in my education system

[final moss]

We had quadratic and cubic equations, inequalities. Trig, log, ln, exponential equations, inequalities. So we can call it precalc

[pseudo nebula]

I've heard of basic proofs being in precalc, overall there seems to lack a standard and just depends on what your teacher wants to include

[limber kraken]

from what I can see u just created random parralel lines with similar lengths and angles, labeling the intersection points , hmm its a bit facinating though that you used a ruuler, which is very rare

[lament tree]

wdym?

well I used a ruler just cus my very objective was to make it pretty

my entire objective was to make a pretier proof coming from a square

[vocal yarrow]

can someone help me with this:
On a fine circle with centerpoint M. A, B, and C are chosen so that in radians, CMB = BMA = 1.13. Determine the size of angle ABC in radians and degrees, rounding to 2 decimal after comma

[limber kraken]

You don't get it do you, when creating a sketch / sketch diagram, basiclaly nobody uses a ruler, cause ou wanna save time, here I guess its acceptable cause ur sharing it with other people but seeing someone actually use a rular when doing no formal proofs is crazy

[lament tree]

the actual reason is that when I do purely geometric reasoning I like making everything very pretty

and I only labeled the points to share it in here lol

by wdym I wanted to know what you meant by

u just created random parralel lines with similar lengths and angles

like is the proof wrong

are the lines not actuall equal cus my argument seems solid

that stuff are equal

at least to prove something constructed that way has the properties I wanted to

[limber kraken]

That's not what I mean, I am trying to say, your proof is a bit abstract and compact such that you wouldn't usually use it during a normal session working with trapezoids, which is fine and all , but its too abstract in a sense such that a very specific scenario is need for it to work, I like it cause it doesn't have to use a lot of algebra, basically none at all but maybe you can generalise this a bit more, such that the angle is theta, the lengths are a bit messed up,etc

[lament tree]

oh sure

wdym "session"

the whole point was making a geometric proof with no algebra just for the fun of it, I know a way easier proof

and constructing the trapezoid from a square for specific reasons

tbh in this context I think proving one proves the converse as happens often in geometry so I don't really care if my proof works for what I wanted to directly or just for the converse

[limber kraken]

People do math for fun, for jobs or to discover stuff, a session of math will be when ur playing around with math, maybe for a purpose maybe not

[lament tree]

oh sure

[lament tree]

and if this doesn't work as a proof for anything I'm just happing with making a cool construction

also I think at the time part of the reason I used a ruler was to explain the proof to people without writing anything

I just wanted to point out each part and use my fingers to explain stuff

[limber kraken]

Yea sure, but limiting yourself to a small plane might give you temporary happiness but the world of math is bigger, stuff we don't know exists, math hasen't been solved yet, and it will never be, exploring will gain you two things, knowledge and fun of mystery/discovery, ur proff its great and all but its a bit specific, just explore other stuff, break down stuff to create something else

[lament tree]

I know lmao

[limber kraken]

Yea I guess I can agree on that part, its easy to explain, explaining is a part of math after all

[lament tree]

tho I once in a while like to make proofs like I was an ancient greek I'm also studying stuff like topology and abstract algebra I know how limited this is

[limber kraken]

It dosen't realluy matter if it is useful in a normal session or not, its more about the knowledge you gain, the more open it is the more knowledge you will gain

[lament tree]

I mean generally when I actually want to know something about geometry and not reprove something I already know how to prove in a prettier way I either do a lot of algebra with lengths and angles or use analytic geometry

and there's also trigonometry

[limber kraken]

Yea, I guess, but you can't have infinite knowledge anyway, at a certain point you will probably encounter geometry with incomprehendible proofs, take the grazing goat problem for example (Numberphile) it has a very complex proof, and your not gonna spend time checking every step, at that point all you can do is say yes, nothing more to it just accept it

[lament tree]

indeed

lots of proofs about tiling are incomprehensible to me for example

[limber kraken]

Looking at where ur heading have you looked at 3 blue 1 brown and the pi^2/6 lighthouse problem, it has geometry trigonometry and it as open as you would expecct it to be

[lament tree]

oh yeah I've seen that one

it's cool

[limber kraken]

what aout the highly more complex pi digit, box collision one

[lament tree]

I think I've watched literally all 3b1b videos

yeah

that one actually was one of the reasons why I started liking math

it had so much of what math has to offer:
abstraction
wonder (why tf is pi there)
reframing a question

[limber kraken]

ok, lets see what I can reccomend you, uhh.. have you seen mind ur decisions, he might not be the best explainer but gives out a decently hard question most of the time, he does logic, geometry and calculus in some parts

[lament tree]

I know about that one too

[limber kraken]

well, now ur at a astandstill, either you go to simple but harrd question such as ones Numberphle offers or go to the deep debts of reading books

[lament tree]

ym depths?

or debts

cus both fit if you don't just download pdfs as I do

[limber kraken]

forget my spelling but yeah reading books is wild, the explainations are a different area of math by themselves

[lament tree]

indeed

understanding how to understand math is a whole skill

[limber kraken]

do you know a lot of calculus by anychance, tahts a area im not familier with but makes up basically 50% of all math, if ur done with calculus its time for you to create ur own math, if ur not learn calculus, unless if ur like me and go all the way down into 1 singular subject before learning calculus

[lament tree]

yes Ik calculus

at least in 1d

and real analysis too

which is like calculus but buffed

[limber kraken]

well I guess look at IJMO papers/ IMO if ur a bit older

[lament tree]

I'm not that interested in international math olympiads

[limber kraken]

but the math is quite unique

[lament tree]

ik

but I like what I'm currently doing

aka seeking advanced topics and studying them

the cool math olympiad I have in my country doesn't require basically any content, if you know what a quadratic is you basically know all there is to know

[limber kraken]

have you looked at the millienium questions, they seem to be the perfect bit, but they are very distant from actual math

[lament tree]

wdym distant from actual math

but yeah I know about those

[lament tree]

yet I never even passed the first phase (tho one of the times was actually because the test answers were leaked and I decided not to cheat for some reason)

(and in my school only the first and second best scores would pass)

(and they cheated obviously)

[limber kraken]

wdym wdym from actual math, look at soemthing like the reimaan hypothesis and the trivial zeros or P vs NP, they doont fit in any category, they could be consider calculus, or complex analysis

[lament tree]

math in general doesn't really fit in categories I think

[nocturne remnant]

The Riemann hypothesis, at least, is not “distant” from actual math in any sense

[lament tree]

categories are made to structure study

but I'm pretty sure no researcher can be in one "category" alone
and also higher math start to become everything all at once

like I was learning linear algebra and a lot of stuff was about calculus

[limber kraken]

well it is, non trivial zeros are just weird, i know it is a variation f calculus but the main problem isnt

[lament tree]

since derivatives are linear operators and integrals can be used as inner products for functions

[nocturne remnant]

Just because it doesn’t fit in a typical college textbook doesn’t mean it’s not “actual math”

The hypothesis itself is a very fundamental problem with many implications if proved

[limber kraken]

actual math and categorical math are not the same really

[nocturne remnant]

Yeah so why are you arguing it’s “not math”?

“not actual math”

[limber kraken]

I never quoted that

[nocturne remnant]

I mean uh

Why would it be distant from actual math

[limber kraken]

This is weird to explain ngl but it provides a question with a answer that is not constructed through math but from alternate logic of the universe, math is kind of the observation of us humans of the universe, but the reimaan hypothesis is distant from the universe in a sense, there are other factors that prove stuff, that can also be prooved by the reimaan hypothesis, but the other proofs are constant and have no trivials

[nocturne remnant]

Ok I have no idea what you are trying to say, I’m just gonna stop here

[limber kraken]

english just sucks ngl

Elon mask, wait that was my friend's nickname😂

[vocal yarrow]

can someone help me with this please

[limber kraken]

can you make a graph/diagram, its really weird with the text and stuff

[heady walrus]

Supposing that A, B, C are points on the peripheral, then |MC|=|MA|=|MB|=r, since CMB=AMB (the angles), then triangles BCM and AMB are both isosceles and congruent, the rest should be easy to figure out.

[lament tree]

are you just saying that the riemaan hypothesis is abstract?

that seemed like saying that with a lot of words

[limber kraken]

kind of is though

[limber kraken]

yes/no

[lament tree]

yeah like a lot of math questions

[limber kraken]

is it something like this

[lament tree]

but tbh the RH is pretty simple by itself, it's just a question about the 0s of a function which have a lot of implications for reasons that are tbh understandable

most people don't know them but they are understandable

like

[limber kraken]

yeahh.. say that to a mathematician and you will be very happy

[lament tree]

yk how a polynomial can be written in terms of its roots?

[lament tree]

simple to understand the question =/= simple to find the answer

[limber kraken]

ahh yes like the grazing goat problem

[lament tree]

how do you double a cube with ruler and compass is very simple to understand but took more then a thousand years for people to prove it's impossible

[lament tree]

like, as a product of x- its roots

[limber kraken]

hat do you mean double a cube?

[lament tree]

from a cube, construct another with double the volume

basically, from 1 conscruct the cube root of 2 with ruler and compass

that was proven to be impossible

with a lot of algebra

[limber kraken]

oh yeah cause pythag dosent apply

[lament tree]

that isn't the reason

[limber kraken]

hmm?

[lament tree]

and idk the reason exactly either

because it's stuff from abstract algebra

I didn't yet learn

[lament tree]

you can also do that with some functions written as infinite polynomials, the so called analytic functions, for example euler did that with the sin function to answer the basel problem
you can also do that with the zeta function, write it in terms of its roots, but you can also write it using prime numbers and that gives rise to a lot of facts about primes, that's basically the reason why the zeros of the zeta function are important

[limber kraken]

how do you even double a cube with a 2d surafce, like do you ahve like a 3d pen, or is it a imnaginary cube that you construct using a ruler and comapss and tehir rules

[lament tree]

the original idea is that like you have an irl cube and you do something to each length and then connect them

but the actual question, simplified, is that from the side of a cube, you want to conscruct the side of a cube with double its volume

[limber kraken]

ahh ya, so that comes from that the cube root of 2 cant be constructed, i see

[lament tree]

yes

[lament tree]

there's even a weird ass ancient greek tale about that

[limber kraken]

wait

[lament tree]

in which an oracle told people to do that for a cube in a statue of a god, they tried doubling each side and some disaster happened

[limber kraken]

wth, these stories are wild

[lament tree]

the story is like even funnier knowing the problem is impossible to solve

[limber kraken]

wait what if a^3*2=b^3, so your saying that its impossible right

(for rational values)

[lament tree]

what

[limber kraken]

im just doubling the area of a^3 there

[lament tree]

this is impossible because the cube root is irrational

2a³ = b³ for a and b rational
2(c/d)³ = (e/f)³
2 = (de)³/(cf)³
2 = (de/cf)³
basically, if a and b are rational, that would mean the cube root of 2 is irrational

some irrational numbers like the square root of 2 can be constructed from ruler and compass, actually any square root can, but the cube root specifically can't

[limber kraken]

yup, that is true though teh cuberoot of 2 is irratonal anyways its like 11pm here need some sleep

[lament tree]

damn where do you live?

it's 9 am here

[vocal yarrow]

I dont know i only have the text available

[vocal yarrow]

yeah it should be easy but im clearly too dumb to get it

[heady walrus]

An image (not to scale). We can see that the segments AM, BM, CM have the same length. The reason for this is that a circle is defined as the set of all points which have the same distance to the middle point of the circle, for some arbitrary distance (in this case r).

We are also given that the angles AMB and BMC are equally large. Since both triangles ABM and BCM share the same length for two of their sides, and the size of the angle between those, both triangles are thus similar (which means they have the same proportions). As a side note they are furthermore an especial kind of similar, namely congruent (which means they are exact copies of the same triangle).

Let's now take one of the two triangles, say triangle AMB. Since two sides of AMB are equal, this fact gives us that the triangle AMB is isosceles. Isosceles triangles have the property that their base angles, in this case angles BAM and ABM, have the same size. The sum of angles inside a triangle is 180 degrees, but since we are working with radians, the sum of the angles in radians is equal to pi radians. We are given that the angle BMA is 1.13 radians. Since angles BAM=ABM, then BMA+BAM+ABM=1.13 + 2*BAM = pi, which gives us BAM = (pi-1.13)/2.

We now recall that the triangles AMB and BMC are similar, which means that their angles at similar vertices have the same size, the angle MBC is similar to angles ABM and BAM (since these are the same). This means that the angle MBC = ABM = (pi-1.13)/2.

We see that the angle ABC is the sum of angles MBC and ABM, so ABC = pi-1.13. Lastly can you use an approximation of pi, like 3.141592, effectuate the subtraction, which gives 2.011592, which can be rounded down to 2.01 in two decimals.

[maiden brook]

the reason is simple

its that pi is transcendental (i.e. can never be the root of a polynomial with integer coefficients)

and no transcendental number can be constructed

ok why did u just delete that

[lament tree]

I was talking about the cube root of 2

that has no relation to pi

doubling the cube

I think you're messing this up with squaring the circle

[maiden brook]

oh i misread

[lament tree]

happens

and the cube root of 2 is not transcendental

but it also can't be written as it should for it to be constructable

[lament tree]

do you know why tho

(I know why just want to know if you do)

[maiden brook]

i think its cause it has to be the roots of a quadratic to be constructible?

not positive

i think it has to be of deg 2 or less

it has some relation to dimensions of the plane i think

[lament tree]

it's cus the equations for circles and lines are 2 and 1 degree

like

ay + bx + c = 0

and (y-y0)² + (x-x0)² = r²

you basically, when constructing intersect and combine and stuff these equations

[pseudo nebula]

cube roots can't be constructed? well that's fun to know

[stuck sleet]

Help

[vernal pilot]

what do you need to know specifically?

[scenic wigeon]

Why is that
$$\sin y = \sin x$$
is equivalent to
$$y = n\pi + (-1)^n x$$
for all integers $n$?

[somber coyote]

Progameyen

[vernal pilot]

first consider what sin(y)=sin(x) does. for any value of y you get a lot of answers to x, and for any value x you get a lot of answers to y. you can see this visually when you graph in on desmos as it creates a bunch of lines.

Now consider what y=n*pi + x(-1)^n is doing. for every even value of n, you create a line that has a positive slope of 1, and for every value of n that is odd, you get a negative slope of 1.

and the y-intercept is offset by n*pi, which is also the same frequency that sin(y)=sin(x) repeats.

[scenic wigeon]

Okay i tried rewrite:
$$\sin (y + 2m\pi) = \sin (x + 2n\pi)$$
For integers $m$ and $n$. That leads to
$$y + 2m\pi = x + 2n\pi$$
Then, to
$$y = x + 2k\pi$$
for all integers $k$.\
But that only yields positive slope of 1, where is the negative slope of -1?

[somber coyote]

Progameyen

[vernal pilot]

You need $(-1)^{n}\cdot x$

[somber coyote]

TheLord26

[scenic wigeon]

Where?

[vernal pilot]

For gradient

[scenic wigeon]

Huh??

Which part of my working is wrong

[vernal pilot]

You just need to account for it.

[scenic wigeon]

Im still thinking where can i put the negative in my working

Cause if i put at
$$\sin(-y) = \sin(-x)$$
then it will be $-y = -x$ which is same as $y = x$

[somber coyote]

Progameyen

[vernal pilot]

$y=(-1)^{n}x+2n\pi$

[somber coyote]

TheLord26

[vernal pilot]

sin(-x)=-sin(x)

[scenic wigeon]

But then how do i proceed from there (cause i cant cancel sin due to negative at outside)

Lemme revisit my working hmm

[vernal pilot]

Try plugging in negative numbers and you will release you need to account for them.

[scenic wigeon]

Okay...lemme try

OHHHH I GOT IT

Okay it is not 2npi
It is npi

And it makes sense

Thanks a lot

[pallid mortar]

$y=(-1)^{n}x+2n\pi$

[somber coyote]

Mictic

$y=(-1)^{n}x+2n\pi$

[lament tree]

not cube root of 2 at least

[floral shale]

Wdym can't be constructed

Its just x^3 flipped across y = x so you can legit just graph x= y^3

[grave pond]

"Contructed" with compass and straightedge, that is.

[floral shale]

That's what I assumed

But like

I've never used a compass and ruler to graph functions...

Idk if you can for a lot of functions

Can you construct a parabola?

[grave pond]

It's not about graphing the cube root function -- it's about constructing a single function output when the input is given as a length.

[floral shale]

Oh

[grave pond]

I.e. there's no construction that, given two points with a distance declared to be 1, will construct two points with a distance of cbrt(2).

[light hazel]

can someone help me find the value of a

[vernal pilot]

We know that angle ABC=360-2*117

So it equals 126

And we know a+126+DAC+DBC=360

And DAC=DBC=90

So we can do a=360-126-2(90)

a=54

Google circle theorems if you aren’t sure how we get angle ABC.

Also this assumes that CB is tangent to CD and AB is tangent to AD

From how it’s structured it makes sense if they are tangents, but no info on that is given. I still think it’s ace to assume that.

[wanton crane]

is it correct?

[vernal pilot]

no

wait, maybe?

is that an i at the end?

[wanton crane]

no

';' this

[vernal pilot]

ahh ok

then yes

[ancient furnace]

yes

[coarse willow]

i cannot catch the 3rd line

may i know which formula is used here? pls

[heady walrus]

$cos(u+v)+cos(u-v)=2cos(u)cos(v)$. Let $u=\frac{2 \pi}{7}, v = \frac{6 \pi}{7}$

[somber coyote]

Crystopher

[heady walrus]

Proof :
$cos(u+v)=cos(u)cos(v)-sin(u)sin(v)$
$cos(u-v)=cos(u)cos(v)+sin(u)sin(v)$
Thus
$cos(u-v)+cos(u+v)=2cos(u)cos(v)$

[somber coyote]

Crystopher

[dark sparrow]

@heady walrus \\ for newlines, also \cos(...) and \sin(...) for proper trig function display.

[coarse willow]

thanaks

[tawdry frost]

Hi all, I have a question here: Is there a good way to measure what angles a rotated object (Rx,Ry,Rz) rotates around Z-axis?

[light hazel]

@vernal pilot Thanks for solving my query

[dry junco]

Hello

I have a question

Can anybody help me out

How to solve this one?

[woeful pike]

U know the angles of an equilateral triangle

[vernal pilot]

x=60 because equilateral
y=90-60=30

[robust plover]

Wouldn’t y = 15?

[dark sparrow]

,rccw

[dark sparrow]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[nocturne remnant]

bro is giving out wrong answers 💀

[woeful pike]

How is he wrong

If anything is wrong the problem is wrong

Both of them can be right. One is basing it off the diagram the other is basing it off the text

[nocturne remnant]

oh shit nvm

yeah the problem is wrong

probably a typo

[robust plover]

Oh crap yeah

Mb 💀

[vernal pilot]

No. Where did you get 15 from?

[robust plover]

I’m pretty sure I added angle POQ and angle QOR to get 120 degrees, and then since triangle OPR is an isosceles triangle I just divided the remaining 60 degrees by two and then just subtracted 30 degrees from angle RPS which would be 45-30 = 15 degrees

Idk man I’m running on no sleep and hella coffee rn 😭

[vernal pilot]

you have overcomplicated things

because you know that QOP is an equillateral, that means that all the angles = 60 degrees

and because PQRS is a square, you know that each of the interior angles are equal to 90 degrees

and you can write y as $\angle OPS = \angle QPS - \angle OPQ$

[somber coyote]

TheLord26

[vernal pilot]

in other words $\angle OPS = 90-60 \newline \newline \angle OPS = 30 \newline \newline y=30$

[somber coyote]

TheLord26

[vernal pilot]

so x=60 and y=30

[grave crypt]

Given isosceles triangle ABC ( ∠BAC < 90° )
Construct a circle with its diameter being AB. The circle intersects BC at point D, AC at point E.
Prove that:
a) DBE is an isosceles triangle
b) ∠CBE = ½ ∠BAC

(note: the problem is in another language so i tried my best to translate it)

its already posted in #1192097278714974288 but i thought it would be faster if i posted the problem here as well

and yes use inscribed angles to solve it

[lapis moon]

plz don't cross-post your problem, so that we won't duplicate effort

[grave crypt]

ok thanks

[upper karma]

can i get help please

[rose raven]

!help

[lime crown]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[north wave]

Ur pfp is from one shot..the little girl..neo or somtg

[grave crypt]

niko

[north wave]

Yeah

[grave crypt]

great game

enjoyed every bit of it

wish the game had more recognition

😭😭

[north wave]

True

[grave crypt]

the ending was heart wrenching

[north wave]

I didnt play the complete game

So idk the ending (no spoilers)

[grave crypt]

spoiling time

😎😎

jk

go play solstice

[north wave]

Im just too scared to get a bad ending cz all ur actions are permanent

[north wave]

Never heard of it. Will check

[grave crypt]

you can reset the games files

[north wave]

Also im lazy

[grave crypt]

then idk

bye me sleep

[north wave]

Okee

[somber glade]

anybody know what you usually learn in high school geometry semester 2

?

[slate adder]

Can someone recommend but really good books for geometry? Starting from the very basic and going till graduate level, can be multiple books obviously

[slate adder]

depends on ur curriculum

but mostly graphs and transformations and then conic sections

along with straight lines and circles ofc

[robust plover]

Oh my lord, yes you’re 100% right 🤦‍♂️

[chrome kelp]

What is the hardest trig topic to remember or learn?(in preparation for calculus)

[patent crater]

Nothing is really that hard to learn in trig. Some of the identities can be difficult to remember, but they're easy to derive and you'll remember the ones you use.

[rose raven]

I have a shirt with all the identities

[cerulean timber]

why are sss, sas asa, ssa triangles so confusing

i have no idea what is going on and none of the videos relate to the questions

[chrome kelp]

Nice

[chrome kelp]

What is an easy way to remember the identifies?

[cerulean timber]

why are these so confusing

[gusty forum]

https://cdn.discordapp.com/attachments/899995881636507668/1192239980299419749/wd2d2.png?ex=65a85b02&is=6595e602&hm=4fb0421c3c1daddb675a97aa1bd6ed1db6d934e6187a402c07325679b2595b3b&

[earnest lichen]

that's not a squarre

length is 4r and width is (2+sqrt3)r

[gusty forum]

thats the answer?

i need the steps later its important

i need the answer first lol

[earnest lichen]

why

[woeful pike]

if the square when compared to the circles is rotated it can work but it isnt

id just work on it as if they did that because thats the only possible square but u can just say that

but that would be too complicated as it seems like this is a middle school question

[short turtle]

it is a square

because in the context of the question

22/7 = 3.14

this can be simplified to 1=2 (with some manipulation)

[earnest lichen]

yea makes sense, the existence follows from intermediate value theorem, but idk a nice way to get that exact side length then

[short turtle]

and applying some more logic 4 = 4+2sqrt3

[woeful pike]

...

[earnest lichen]

they ask you to not overthink

[woeful pike]

wjat are u doing

[short turtle]

simple

[short turtle]

meths

[woeful pike]

lol

[keen turtle]

Anyone know, why sometimes in enlargement value of k is taken as negative.

[patent crater]

Writing them down helps solidify it in your brain. Also, it helps to know that most of the identities are fairly easily derived from two identities: sin^2(x) + cos^2(x)=1 and sin(x+y)=sin(x)cos(y)+cos(x)sin(y).

[chrome kelp]

That seems fairly simple. I will try writing them down.

[night violet]

hi

hi

[woeful pike]

sohcahtoa

s = o/h, c = a/h, t = o/a

[vernal pilot]

Don’t ask questions in multiple different channels.

[keen turtle]

Okay

[formal moth]

does anyone know a place where i can find good trig excersises

like proof that this is equal to that sorta excersise

[chrome kelp]

How hard is unit circle?

[silent plank]

depends on how much right triangle trig you already know

[vernal pilot]

Not hard. It’s just a circle with radius 1 and some very useful right angle triangles in it using the radius as the hypotenuse.

[chrome kelp]

All I remember is soh cah toa

[visual pivot]

does anyone have the download of the algebra and trigonometry book 2 solution key?

[limber kraken]

https://www.desmos.com/calculator/1hyw6qfrng, there is a complex part and a normal part tha is hella hard to simplify, sokve for b: assuming all the equasions given are equaland b is a variabble, i used t as a slider so i didnt et bombarded by exclamations while solving for this, yes indeed this is not a challenge but something i need a answer for, i need the answer in form b=...

[vernal pilot]

<@&268886789983436800>

[rose raven]

what did i miss

that was mod ping necessary

[runic beacon]

it was probably deleted by the mods

[vernal pilot]

I called the mods on you. You better run.

[rose raven]

nuh uh

[vernal pilot]

I know what you did November 16th 2011.

[rose raven]

brother i was 4 years old

i probably shit my bed or something