geometry-and-trigonometry

[gritty topaz]

that's ceva problem

[faint pasture]

op isnt responding so there isnt any point anyways

[faint pasture]

its relatively doable angle chasing 😭

[gritty topaz]

really?

haven't looked into yet

[gritty topaz]

can you give a solution

[faint pasture]

oh

[gritty topaz]

yeah

[faint pasture]

the xs cancelled out- nvm

[gritty topaz]

also answer is 10 not 20

[faint pasture]

doesnt this kinda stuff usually indicate multiple solutions?

[gritty topaz]

yeah

quick solution

@formal geyser

x=10

to the scale if anyone else wants to try

[plucky abyss]

Wow nice you good at maths

[stuck sleet]

My answer is approx 444.4862 ft^3 am I right?

[hard moon]

alternatively, you can use law of sines

Because $\tan^{-1}\left(\dfrac{\sin(50^\circ)}{\dfrac{\sin(20^\circ)}{\sin(10^\circ)}\dfrac{\sin(70^\circ)}{\sin(30)^\circ}+\cos(50^\circ)}\right)=10^\circ$

[somber coyote]

allan

[plucky abyss]

Oh you smart in maths

How to calculate 10 ,20 and 50 degree value

[faint pasture]

calculator is the best option at that point

or you could use sin(3x) identity as you know sin(30 degrees)

tho solving the cubic will be tough

[plucky abyss]

I see thanks

[plain scarab]

What am I missing

[obsidian harness]

tan F = 4/3

[modest marsh]

how should i prepare myself as i go in to geometry

[pliant kettle]

Just start

[plucky abyss]

Practice questions

and their answers

[plucky abyss]

It is two tangents to circle from external point

They have same length according to theorem

[sturdy stone]

"according to theorem"

[maiden brook]

the ice cream cone theorem yes

but that’s right

[frozen night]

can u guys help me solve this please

[sturdy stone]

Is this name standard? I've never heard that before

[maiden brook]

nah it’s a joke it’s called the two tangents theorem I think

[sturdy stone]

Lol I've never heard that name either

Ngl i think there is a tendency to give obvious geometric facts names when they really don't need it

[maiden brook]

what should it be called

[sturdy stone]

It doesn't need a name is my point

[maiden brook]

this theorem is often just stated quickly without the name

[sturdy stone]

Yea I think that's fine

[faint pasture]

just say obv and move on

[trail tendon]

each one has paid 9 dollars for a total of 27 dollars, and of those 27 dollars, and the gatekeeper took 2 of those 27 dollars so in total they paid 27 dollars and kept the extra 3 :l

like the 2 the gatekeeper took is part of the 27 dollars they paid, its not extra

[frozen night]

im sorry me not understanding

[ornate vigil]

Who can do my virtual work fir geometry geometry I'm on unit 10 trying to get to unit 12 who can do it for me?

[trail tendon]

they pay 10 dollars each, so they paid 30 dollars. the guy takes 2 of these 30 dollars, and they have still paid 30 dollars. he gives them back 3 dollars, so now the guy has 27 dollars. he also takes 2 of these 27 dollars. they still paid 27 dollars. they paid 27 dollars, and the gatekeepr has kept 2 of these 27 dollars, you don't add 2 to 27 because 2 is included in the 27.

he took 2 dollars of what they paid

[plucky abyss]

Bryson, people can help you but no one can do it for you

[trail tendon]

so they paid 27 dollars, he took 2 dollars of that and gave them back 3 dollars

[frozen night]

can we vc please the math is not mathing 😭

[trail tendon]

br

[trail tendon]

A, B, C, give D 30 dollars. D takes 2 of these 30 dollars, so now the cash register has 28 dollars, D has 2 dollars. he takes 3 dollars from the cash register and gives them to A,B,C.
now the cash register has 25 dollars, D has 2 dollars, and each of them have 3 dollars back (and they have paid 27 dollars).

[frozen night]

i get it now

tyyyyy

[plucky abyss]

[copper terrace]

how the hell did you get 20 I got 6.91751116597

[gritty topaz]

answer is 10

[copper terrace]

how

@gritty topaz how

[gritty topaz]

this

bunch of other solutions possible but i won't go through them

[copper terrace]

i always never liked geometry

[gritty topaz]

that's like olympiad entrance level

[trail tendon]

i never always liked geometry 😔

[copper terrace]

oh

[copper terrace]

wait actually

[trail tendon]

i'd believe it

[copper terrace]

my geometry teacher sucked

she always made mistake

s

[gritty topaz]

i didn't have someone to teach me geometry

[trail tendon]

how do you know that the point E creates an equliateral triangle and two isosceles triangles?

[gritty topaz]

consider yourself lucky

[copper terrace]

you dont even use geometry in higher level maths

you learn trig in precalculus

[trail tendon]

oh i guess it depends on what you consider "higher level maths

[sturdy stone]

Orz

[copper terrace]

everyone forgets the geometry proofs

[trail tendon]

well ok thats fair

[copper terrace]

in Linear algebra you just learn them again usualyl

[trail tendon]

really?

[copper terrace]

or you take a proofs and logics class

[trail tendon]

i mean they do actual proofs so why even bother 💀

[trail tendon]

they actually do the geometry proofs again?

[copper terrace]

nooo

[trail tendon]

maybe ur class was wack or smthn 💀

[sturdy stone]

calling two-column geometry student-torture "proofs" is a misrepresentation of reality

[copper terrace]

not geometry proofs

[trail tendon]

oh like full on proofs

yea

but the geometry proofs are barely proofs imo lol

[copper terrace]

they are baby proofs

I just do paragraph proofs these days

for most proofs

[trail tendon]

they like hand you half the stuff that you just have to memorize instead of actually using your own brain to do the proofs lol

[copper terrace]

I didnt memorize them at all

[sturdy stone]

that's cause high school geometry never goes beyond the inscribed angle theorem

[trail tendon]

then how did you do it 😂

[copper terrace]

I barely did proofs

[sturdy stone]

like cyclic quadrilaterals aren't even mentioned

[copper terrace]

in geometry

[copper terrace]

????????????????

[sturdy stone]

?????????????????????

[trail tendon]

yea no i never really went farther than high school geometry lol

idk how important stuff is after that

[sturdy stone]

it's not important for higher maths but it's interesting

[trail tendon]

yea...

its not as practical lol

[sturdy stone]

geometry can be a fun subject in its own right

[trail tendon]

for most things

[copper terrace]

even for topology or for differential geometry its useless

[trail tendon]

no yea

[limpid onyx]

number theory was once considered useless and now look. geometry will have its redemption

i believe

[copper terrace]

geometry has had like 2000 years to do its shit but never shined

its like the first maths

[limpid onyx]

patience is key

[gritty topaz]

created an equilateral triangle on this side and connected it to the corner

[limpid onyx]

i refuse to believe geoemtry is useless, its just too rich

[gritty topaz]

since the corner has 30 and equilateral has 60 angle

[copper terrace]

whats good about it

[limpid onyx]

like this problem is insane

[gritty topaz]

i'll give letters wait a second please

[limpid onyx]

too rich

too beautiful

too nice

[copper terrace]

thats just competive geometry

for competitions

calculus is way more useful

[limpid onyx]

lmao this way too hard for competitions

thats what you think, just wait

[gritty topaz]

ABD is equilateral i constructed
aADB=60 and aACB=30 which implies DC=AB=AD=DB

[trail tendon]

well its not THAT hard but like cmon this is hardly practical 💀

[copper terrace]

calculus will always be more useful than geometry

[limpid onyx]

it is THAT hard

try it

playing with primes for the last 1000 years was hardly practical

[gritty topaz]

you can't build a rocket without geometry and trigonometry

[trail tendon]

i'm not a geometric competitor

[limpid onyx]

even for a geometric competitor

[trail tendon]

you probably don't need this ^ geometry to build a rocket 💀

[gritty topaz]

that's true

[sturdy stone]

yeah you need more

[copper terrace]

prime number has a use in number theory unlike geometry which has no use in real life or pure math

[gritty topaz]

but what i do doesn't take importance from geometry

[trail tendon]

you need different

[sturdy stone]

rockets aren't made out of triangles

[limpid onyx]

number thoery was famously considered useless until many many years later, thats what im saying

geometry will become important

[trail tendon]

i didn't mean you don't just need this geometry, i meant I don't think you would even encounter a problem similar to this lol

weird angle chasing by creating more triangles

[sturdy stone]

geometry is pure math though

[copper terrace]

if i told you to build a rocket you wouldnt be able to do it because you need physics and more math like that

[sturdy stone]

most pure math is also useless outside pure math

[limpid onyx]

reminds me of adventitious quadrangles

[copper terrace]

but it doesnt have a use for higher maths than it

[gritty topaz]

what i do and don't doesnt take anything from geometry

[gritty topaz]

pretty closely related

[trail tendon]

he didn't say you only need geometry and trigonometry, he said you can't do it without it

but i'd argue you don't need this kind of geometry, you need different more practical kinds of geometry

[limpid onyx]

yeah most of geometry is currently absolutely useless

[trail tendon]

you dont need competition-math geometry

[limpid onyx]

but theres just way too many coincidences and too many properties to be useless

[gritty topaz]

it's fun seeing weird angle problems and solutions for me

i can't call something useless if it's fun

[sturdy stone]

yea exactly

[trail tendon]

i wouldn't call it useless, its just impractical

[sturdy stone]

i don't get the obsession with something needing a real-world use or application to some higher math

[limpid onyx]

well its practically useless

[sturdy stone]

it's fun and interesting; that's enough

[gritty topaz]

however topology is useful

[limpid onyx]

which is barely related to this kind of geometr

[trail tendon]

math is just defined in a certain way, whether it is useful or not. its literally like you can make an imaginary world, but at the end of the day the imaginary world isn't practical. however, with applicational math, you can actually do cool things in real life like building a rocket ship and going to the moon. So for the people that want application they just prefer going to the moon instead of pretending you're theoretically going to the moon ☠️

[sturdy stone]

in my humble opinion, inventing your own world is far more interesting than going to the moon

Euclidean geometry is far richer than the moon

[trail tendon]

yep, just depends on what you want

[gritty topaz]

such simple problems

require such weird solutions

it's like chess

[trail tendon]

i don't like geometry

💀

[limpid onyx]

most normal adventitious quadrilateral problem:

[gritty topaz]

it's insane that the answer is an integer

[trail tendon]

i don't think i was really taught it, just picked up bits and pieces lol

[trail tendon]

thats the real mystery

💀

[gritty topaz]

and it's more insane that there's people solving those purely geometrically

[limpid onyx]

exactly

[fossil shard]

is there anywhere i can get help at learning geomatry from scratch.
i just finished algebra 1 a few months ago

I am checking the help forum but it's not loading up

[gritty topaz]

youtube has infinitely many geometry teachers

[fossil shard]

oh yeah. I forgot about youtube completely

[upper karma]

pick up a book bro

i was just about to ask which one to pick up

GUYS WHAT GEO BOOK SHOULD I BUY

[soft roost]

khan academy might prove some help

[formal geyser]

Hello how do i find the rectangle's area

[grave pond]

The three right triangles are similar, so their side lengths must be in the same proportions as the diameters of the inscribed circles.

[gritty topaz]

how can you show that

[dusky locust]

"The three right triangles are similar"

[gritty topaz]

i'm asking how to show that

is it given

how do you even know there's 3 right triangles

[obsidian harness]

the one with the red circle is right-angled cause it's assumed to be a rectangle

lazy assumption I know but this is Catriona Shearer on fricking twitter, not your textbook or a contest question

[gritty topaz]

it could be shared with context then

[obsidian harness]

oh lol that makes so much sense now

[faint pasture]

Why though

How can i prove that the perpendicular bisectors are concurrent? Its like yes ik theres only 1 circumcentre but yea why

[undone quest]

If AD, BE, cf are medians of TriABC
prove that 3(AB²+BC²+CA²)=4(AD²+BE²+CF²)

One of the interesting proofs

As well as lengthy

[grave pond]

I suppose I'll have to assume that all the angles that look right in the diagram actually are right.

[faint pasture]

Yea its given that its a rectangle

[undone quest]

yoplz solve my question

[faint pasture]

Id guess repeated application of appolonius

[undone quest]

What??

What's appolonius

[grave pond]

In that case the triangle that makes up half of the rectangle is similar to the entire other half, and I consider it "well known" that an altitude divides a right triangle into two smaller right triangles that are each similar to the larger one.

[undone quest]

Bro don't forget Pythagoras

[faint pasture]

Its a theorem google it

[faint pasture]

Yea

[grave pond]

That's why the perpendicular bisectors are concurrent: the point where two of them intersect is the center of a circke that contains all three vertices, and therefore it is also on the third bisector.

[faint pasture]

😭

Wait which que were you doing

The rectangle one

Or mine

[grave pond]

Here I was just answering your question about how to know the perpendicular bisectors in a triangle meet at a point.

[faint pasture]

But thats just assuming that there is only one such circle

The proof that three points can only lie on one circle depends on the circumcentre thing too so seems kinda circular (no pun intended)

[grave pond]

I make no such assumption.

[faint pasture]

Yea so why cant there be multiple circumcircles

[grave pond]

I'm not claiming there can't (even if that is actually true too).

I'm assuming it known that the perpendicular bisector of AB is exactly the locus of centers of circles that pass through A and B.

So let the intersection between the p.b. of AB and the p.b. of AC be P, and draw the unique circle with center P that passes through A.

Because P is on the p.b. of AB, there is a circle centered on P that passed through A and B. But we have just drawn the only circle that passes trough A and has center P. So that must be the one that also passes through B.

Similarly, the circle also passes through C.

[faint pasture]

Yea so how can i prove that 😭

[grave pond]

We now have some circle with center P that passes through B and C -- which shows that P must be in the locus of points that are centers of such circles. But that is just the perpendicular bisector of BC.

[grave pond]

If you have two circumcircles, each of their centers must be on the perpendicular bisector of AB, and also on the perpendicular bisector of AC. However two lines can only have one point in common (unless they coincide completely, but that only happens if B=C, in which case there's no triangle), so the centers must both be that one point. So your circumcircles both have the same center and have the same radius. That is, they're the same circle.

[faint pasture]

Yea so how can i prove there may exist such a circle

[grave pond]

That's an axiom. Euclid, book I, postulate 3: To describe a circle with any center and radius.

[faint pasture]

😭 that was unsatisfying to say the least
How about like angle bisectors or the altitudes or the medians?

[grave pond]

They meet at different points, for different reasons.

[faint pasture]

Any hint on how to prove concurrency?

[undone quest]

Which one

[grave pond]

For which of them?

[undone quest]

I can help

[grave pond]

The goalposts seem to be moving around a lot here.

[faint pasture]

All of them seperately

[undone quest]

@faint pasture ?

[faint pasture]

All of them seperately

[undone quest]

See

[faint pasture]

sorry ig 😭

[undone quest]

If u change the figure without changing the dimensions

Something can de done

Because sides r not given

In these type of problem we generally change positions of dimension

Wait

[faint pasture]

😭 how do you plan on doing that this is like the most general case

[sturdy stone]

angle bisector concurrency is basically the same as perpendicular bisector concurrency

[faint pasture]

How? Incircle is also an axiom?

[sturdy stone]

what?

[grave pond]

I don't think I can actually prove off the top of my head that the altitudes are concurrent ...

[sturdy stone]

the incircle wasn't even mentioned

[faint pasture]

The perpendicular bisector conncurrency was proved that way thpugh

[grave pond]

The axiom I referred to is: Once you have the point P and the point A, there exists a circle with center P that passes through A.

I then proved that circle also happens to pass through B and C.

[sturdy stone]

No
The perpendicular bisector proof is:
The perpendicular bisector of AB is the line consisting of all points equidistant to A,B.
The perpendicular bisectors of AB and BC intersect in a point, O. This point has the same distance to A and C. Therefore it is on the perpendicular bisector of AC, hence the perpendicular bisectors concur at O.

[undone quest]

https://youtube.com/shorts/sEwcNAX91UM?si=ZzkME82nt_zl45Jp

I Am a ninth grader so I cant help u much hope it helps

[sturdy stone]

I say angle bisector concurrency is basically the same as this because the angle bisector of ABC is the line through B consisting of all points equidistant to AB and BC. So the argument can be repeated with just a few minor tweaks.

[undone quest]

@faint pasture

[faint pasture]

Oooooh i got it now

Had to reread i read it wrong the first time for some reason

[undone quest]

Bro I can't even understand this question

0_.

[faint pasture]

Hmmm that makes sense

Now how about the altitudes or medians

[grave pond]

Not exactly, because the perpendicular to the angle bisector (which bisects the supplementary angles) also have some of the points that are equidistant to the two infinite lines ...

So we need some pesky footwork, though I'm confident it can be done with enough care.

[faint pasture]

Hmmm

[grave pond]

For medians what I think of intuitively is something like

There exists an affine transformation that makes our triangle into an equilateral triangle. Affine transformations preserve the medians, and in an equilateral triangle they have to meet in a point, by symmetry. So they must also have met in a point before the transformation.
But "affine transformation" is not really in the classical Euclidean vocabulary, so it doesn't really make a proof in that setting, without a huge detour of defining them first ...

[faint pasture]

Hmmm idek what that means so ill be back in a bit after googling

Yea no i clearly dont know enough in order to understand what means

[sturdy stone]

You can do medians by showing that any two medians divide each other in the ratio 1:2 (using similar triangles). There is only one point on a line segment that divides it into ratio 1:2 so they must concur.

Also it can be coordinate bashed kinda easily i think

[faint pasture]

Oh that was smart

[faint pasture]

Coordinate bashing could be done for all the problems i just gave though i wanted a euclidean solution

[sturdy stone]

Yea but it gets messy for a lot of them

Like angle bisector would be a nightmare i think

Unless you allow trig

Then it's probably not that bad

[faint pasture]

💀 tbfh my tchr did them by coordinate in the classes and yea it got shitty but it was a proof nonetheless

I wanted a pure euclidean based solution tho

[sturdy stone]

You can also use ceva's theorem

[faint pasture]

For like altitudes i guess the solution is rather simple too if you allow coordinate

[faint pasture]

I kinda can i just really really dont want to

[faint pasture]

Yea its allowed

Anything except coordinate bashing is allowed

[sturdy stone]

Draw lines through A,B,C parallel with the sides opposite of A,B,C and let A_1, B_1, C_1 be intersection points as in the figure. Then ABC and any one of A_1,B_1,C_1 is a parallelogram. So this gives C_1B=AC=BA_1 and similarly for the other sides of A_1B_1C_1. Hence an altitude in ABC is a perpendicular bisector in A_1B_1C_1 and since the perpendicular bisectors concur, so do the altitudes.

[faint pasture]

That was beautiful, thanks for your time!

[sturdy stone]

No problem

[grave pond]

Ah, nice!

[plush mirage]

need help guys

[grave pond]

Split each of them into two rectangular boxes.

[gritty topaz]

volume = surface area times depth

[nova raptor]

hey

[trail tendon]

is for horses

[lone panther]

@north kindle here

[blissful timber]

so... @lone panther wanna do it?

[lone panther]

you can begin, yea

[blissful timber]

okkk

trignometry 101

[zenith minnow]

will you also teach sim sum (not sure what do you call it) etc :d

[blissful timber]

so, trig is all about right triangles.
whenever you have a right triangle all the side lenghts are related to the angles in specific ways: these are the trig functions

let me make a drawing rlquick

[lone panther]

is this it

[north kindle]

why is tan(x) so weird compared to sin and cos when you graph it

[lone panther]

you can see from the fractions above that tan(x) = sin(x)/cos(x)

once you see that, then notice that cos(x) has zeroes

every zero cos(x) has is an asymptote of tan(x)

now why sin(x) and cos(x) have the shapes they are is ahead of the lesson

let max-cat continue

[blissful timber]

you can remember it with SOH CAH TOA

its a simple way of memorizing what sides go in which fractions

Sine = Opposite / Hypotenuse (S O H)
Cosine = Adjacent / Hypotenuse (C A H)
Tangent = Opposite / Adjacent (T O A)

[wintry wigeon]

Because tangent has cosine in the denominator, which causes it to diverge to infinity.

[blissful timber]

these functions relate the angle with the ratio between to angles. So, like how with the pythagorean theorem you can find a missing side with the other two, with Trignometry you can find a missing side with an angle and one of the other sides.

you can find all the angles only given the side lenghts aswell

[north kindle]

that seems extremely powerful

[blissful timber]

it is! But takes practice

the last thing you need to know is that the functions have relations between them selves

Sine / Cosine = Tangent (this one is just playing with fractions)

[wintry wigeon]

Hence why high schools spend a whole year on it

[blissful timber]

Sine^2 + Cosine^2 = 1 (this one is provable with the PT)

trig is a powerful tool in geometry and physics because we are always drawing triangles and with little info we can know all about the angles and side lenghts

all it takes is patience and a bit of practice

Here's proof for one

[gritty topaz]

at first you think trig is all about right triangles

but you know you understood trig once you realize it's all about circles

[plush mirage]

oh ok thanks

[mellow quest]

<@&268886789983436800>

[thick violet]

how do i improve in manipulating trigonometric equations to get the desired ans

which all qs?

[worldly forum]

wdym

[thick violet]

like all questions?

or specific ones

[worldly forum]

yeah its suppose to be like SSS theorums and postulates

[thick violet]

yeah but did u get some one ur own

like the 7th one is sss

[worldly forum]

thats whsat i got im js tryna make sure it correct

2nd is ASA

[thick violet]

ahh peace

mhm

[worldly forum]

right?

[thick violet]

yeah

not sure abt 9th

[worldly forum]

yeah idk either

[thick violet]

yeah

10th is sas

and last is aas

[worldly forum]

alr

Thank you bro

[thick violet]

np bro

[mellow quest]

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

[wise robin]

how did you write a as $sin^2\theta$?

[somber coyote]

菜月

[wise robin]

same with b

[teal lintel]

can anyone tell me what is the golden ratio used for?

[ornate vigil]

What's the answer to this

[faint pasture]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[gloomy spear]

think there is less information

like about the 10.2 yards

[faint pasture]

I dont think so

[sterile horizon]

any tips how to be an expert an math?😅

every time when were doing a recitation, im scared raising my hand so thats why im not learning that well

[plucky abyss]

😆

[sterile horizon]

😓

[plucky abyss]

But before becoming expert you have to go step by step

[sterile horizon]

math is rlly hard but it u have to go through it

step by step

[plucky abyss]

Yes it's a sad subject

[sterile horizon]

can anyone tutor me

im dumb ngl

cant even do fractions

im turning grade 8

☠️

[plucky abyss]

Watch basic videos

[sterile horizon]

i forgot how to do fractions

[sterile horizon]

copy that

[plucky abyss]

You can ask here doubt

[sterile horizon]

im always shy asking help for here

cus i look like a dumb person who doesnt get the topic

[plucky abyss]

No need to feel that

It takes time

[sterile horizon]

+ive been struggling in math

[floral siren]

area = (b1+b2)/2 * height

[sterile horizon]

dyk like

[sterile horizon]

when u get the topic but tmrw or the other day you'll forget it

even tho u read notes

i have this like, short term memory lost

i forget things easily

[floral siren]

u can try asking for extra work after class for practice

[sterile horizon]

thx for the advice!

and im also planning this s.y ill advance read, and do some notes

[floral siren]

that's what im doing rn TT

[sterile horizon]

cs last s.y was trash

[floral siren]

gosh same

[sterile horizon]

i failed half of my test

[plucky abyss]

Hey don't worry too much

[sterile horizon]

parents got dissapointed, and i hate it

they taking away my phone soon if i fail this s.y again

[plucky abyss]

Just start doing simple problems by watching khan academy or something similar

[sterile horizon]

i didnt actually fail but I js maintain my grades

im js an average student

[floral siren]

average is fine tho

better than failed :/

[plucky abyss]

Most of the people in the world are average only 😊

[sterile horizon]

half of my cm failed so im lucky

cm's*

[floral siren]

js dont be shy to raise ur hand during class

[sturdy stone]

grades aren't distributed based on luck

[sterile horizon]

cant do that lmfao

i remember one time i raised my hand and decided to do the math solution, i got the solution wrong but i got the answer right so my classmates laughed at me

called me stupid for answering the solution wrong

[floral siren]

nah making mistakes is fine

[plucky abyss]

True

[floral siren]

at least u volunteered and tried to solve it

[plucky abyss]

Don't take too much stress enjoy the precious life

[sterile horizon]

too bad

my mom pressures me

i cant enjoy my teenage life

im always locked up in the house

cant even go outside 😓🙏🏻

[mellow quest]

😢

[upper karma]

what country?

fractions should be friendly to you bro.. in grade 9 you will struggle more

How to find the domain of a trig function ? is there a step by step way to do it ?

[floral siren]

i find gr9 math easy tho

[kind rapids]

Most of the time just remember it. But if you want a way to visualize then you can check unit circle

[upper karma]

well not for him..

I'm alr in 9th grade maths..

[floral siren]

this is 9th grade math right?

[mellow quest]

7-8 grade

i think

[upper karma]

that's 6th

[sturdy stone]

I learnt it in 3rd grade

[floral siren]

yoo what

[sturdy stone]

Yea it's just addition, division and multiplication

[upper karma]

I was thinking more like these

[kind rapids]

natural domain of arcsin is [-1; 1]

So you can solve inequality: $-1 \leq \frac{x+1}{2x^2} \leq 1$

[somber coyote]

Closer

[upper karma]

thanks

[upper karma]

Lmao ur better than them u at least volunteered

If it were the other way around they would have done it

[sterile horizon]

Philippines

[upper karma]

bro you should study

don't be like other kids spending alot of time doing unnecessary things, well if you're that bored you can just go out but you need discipline

[safe raptor]

You can't guarantee that in the Philly

[vestal kelp]

Can someone help in on the help channel

[upper karma]

You can guarantee that if you're disciplined

[north kindle]

im confused on this problem, im getting that the segment from the intersection point of the two diagonals of the kite to C is a negative value

wait nvm

[pliant kettle]

Ok

[upper karma]

How much time does learning 11th grade co ordinate geometry takes?

Anyone has the idea?

For like super tough examination

[pliant kettle]

It depends on the person

[upper karma]

I guessed so-

[pliant kettle]

Ofc

[plucky abyss]

It takes 4 days 10 hrs 22 mins 15 sec

[faint pasture]

It took me 4 days 10 hrs 22 mins 18 secs so idk man

[faint pasture]

Just say jee

[pliant kettle]

that's assuming a lot of things ig

[faint pasture]

When have you ever seen an indian considering something other than the olympiads and jee hard for math- 💀
And olympiads dont test coordinate so yea jee it is

[sturdy stone]

Olympiads don't explicitly test coordinate stuff but it is useful to know

[faint pasture]

🤡yall are just trying to prove my arguement wrong when we all know it isnt with a 90% certainty

[pliant kettle]

Why are assuming he lives in india?

[faint pasture]

His name

[pliant kettle]

I don't get it?

[faint pasture]

Its a classic indian name

[pliant kettle]

Ohh

[faint pasture]

And theres a chance he is an immigrant

So i said 90% certainty

[pliant kettle]

Yes

Alright cool bud

[plucky abyss]

[pliant kettle]

[upper karma]

HELP IM USING BETA

what r these acronimsbro

this is why i study math

https://tenor.com/view/tom-jerry-tree-hit-funny-gif-18830046

[upper karma]

...yeah

[blissful timber]

sorry, didn't saw it.

$(sin(\theta))^2 = sin^2\theta$ they mean the same

[somber coyote]

Max-Cat

[flint shale]

Hello,
I am having a bit of trouble with this question:
Use the formulas $\sin(\theta)=\frac{O}{H}$ and $\cos(\theta)=\frac{A}{H}$ to prove that $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

Thanks in advance!

[somber coyote]

coffee maker

[flint shale]

Nevermind

I got it!

[wise robin]

thanks, this wasn't what I meant but I got my answer

[wise robin]

you can prove it using both ASA and SAS

but I don't think there's a way to prove BA = BD though

If you were given that triangle ABD is an equilateral triangle, then you would have been able to say that BA = BD, maybe that's what your teacher wants you to use

[obsidian harness]

AC = CD and BC = BC and angles ABC and DCB are both right angles hence equal

[vestal kelp]

ok thank you

[upper karma]

it takes a lot of time

last year when i went through jee coordinate geometry, we took like 4 months to complete the whole unit

3-4 months

[smoky jetty]

we have a sh!tty math system

[obsidian harness]

I remember helping someone from PH and they were doing mean median mode in 10th grade or something.....

[smoky jetty]

basic educ. system here commonly teaches those at 10th grade

[obsidian harness]

yikes

[wise robin]

is it late?

[obsidian harness]

yeah you could easily learn this stuff in middle school

or even earlier

different education systems have different priorities

[wise robin]

ours is pretty bad

[gentle haven]

Bro what

[obsidian harness]

yes

[faint pasture]

💀 it might be just them tho

[light grail]

could someone tell me what parts are factored?

[faint pasture]

$\frac{-(sin^2(x)cos^2(x) + cos^4(x)) - sin^2(x)}{cos^2(x)}$

[somber coyote]

Ender Doesn't Mind

[faint pasture]

now you factor a cos^2(x) from the bracket part

[light grail]

and then factor out cos^2?

[faint pasture]

hmm

[light grail]

great minds think alike

so i would guess the best first step is to rewrite it

[viscid osprey]

i am so cooked for my entrance exams

[obsidian harness]

it was their homework and they didn't understand it

[faint pasture]

thats sad

[faint pasture]

aint we all

[pliant kettle]

Sed

[outer wadi]

how do i find which quadrant this lies in

[silent plank]

add/subtract integer multiples of 2pi

[outer wadi]

will it be in 2nd quadrant?

[pliant kettle]

Yes

[outer wadi]

ok thanks @silent plank @pliant kettle

[pliant kettle]

Alright

[outer wadi]

what does the line above 30 represnt

[abstract cave]

22.5 degrees

[outer wadi]

whats the conversion like

[abstract cave]

60 minutes= 1 degree

[outer wadi]

got it thanks

[gentle haven]

In Euclid's 5th postulate, does it mean that if any pair of angles on the same side sum to less than 180, the two lines intersect?

Meaning it includes consecutive interior/exterior angles?

Not corresponding angles too.

[upper karma]

i think its the first one

since you know it will make a 90* angle on both portions and that the bases will be equal also the bisector side will be equal

[lilac peak]

Can anyone help me with trig identities?

[trail tendon]

do you have specific problem?

[lilac peak]

Yes, if p=sin40, how do I write an expression for cos140 in terms of p?

Those are degrees btw ^

[trail tendon]

is it just me or is that kind of a vague question 🤔

[lilac peak]

That's what it says

[trail tendon]

like what counts as "in terms of p"

ok i guess uh

[lilac peak]

Uhh

[trail tendon]

do you know the cos(a+b) identity?

[lilac peak]

For a previous question the answer was -p

[trail tendon]

just if it has p in it? 💀

[trail tendon]

hmm

maybe it wants more p lol

[lilac peak]

Like the cos(90+theta) etc..?

It might be vague but that's the type of questions I'm doing

[trail tendon]

nah i was thinking like cos(a+b) = cos(a)cos(b)-sin(a)sin(b)

maybe this isn't what they want you to do

[lilac peak]

Im from Asia so our questions are a little different

[trail tendon]

oh

wait

LOL

[lilac peak]

Idk that but maybe I can try?

[trail tendon]

alr

have you seen it before?

[lilac peak]

Nope

[trail tendon]

it puts it in terms of p so it satisfies what they wanted you to do but it may not be exactly what they had in mind 💀

[lilac peak]

Okay I'll try rq 😭 ty

[trail tendon]

you know

they probably wanted you to do cos(140 deg) = -cos(40 deg) 😭

then -cos(40 deg) = -sqrt(1-sin^2(40 deg))

probably simpler 💀💀

[lilac peak]

OHHHH

TYSM

[twin spindle]

Im confused why the answers -4/3

Ignore the 3 I got, I kinda jus gave up

But I thought it would be -1/3

Bcz sec= 1/cos and cos=3

But since its in quad 3, I said -1/3

But im not sure how I was wrong

[grave pond]

A cosine is never 3.

[twin spindle]

But isn't it 3 in this case? cause tan=sin/cos?

[grave pond]

Consider that 0.6/0.8 = 6/8= 3/4 and so forth. Just because you know one division that gives the same result as sin/cos doesn't mean it was that particular division.

And the value of a cosine is always between -1 and 1, so it cannot be 3.

Instead what you'll need is to draw a right triangle with the unknown angle theta in one corner.
You can take the legs to be 1 and sqrt(7)/3; that makes the hypotenuse sqrt(1+7/9) = sqrt(16/9) = 4/3.

[twin spindle]

I see I see, thank you

Wait for hyptenuse why 7/9?

[grave pond]

the square of sqrt(7)/3 is 7/3³.

[twin spindle]

I get why you add the 1, but not aqrt(7/9)

Ohhhhh I see I see

Thank you

[formal geyser]

Hello guys

90% of people under this problem say it is 15cm^2

But i got 25cm^2

[obsidian harness]

the blue area probably doesn't include the red semicircle

[formal geyser]

Oh, i didn't consider it

Thanks

[obsidian harness]

no worries

[sand patrol]

can anyone help

[somber yacht]

PXQ and SRY are congruent.

so PX = RY

[mellow quest]

hmm

i got 15,00004

[obsidian harness]

how did you get that lol, like what kind of software

[mellow quest]

paint

[sand patrol]

how do u say its congurent

[obsidian harness]

oh bruh you rounded early

it's $\frac{\pi (5 \cdot 20 / \pi)}{4}$

[somber coyote]

southy

[obsidian harness]

after you square R and remove the square roots

[sand patrol]

can u say how its congruent

[somber yacht]

ye gimme a min

[sand patrol]

k

[somber yacht]

In parallelogram opposite sides are equal yes, so PQ=SR. then, angle PQX is equal to angle SRY since they are corresponding angles. Since PQ=SR and angles PQX and SRY are equal, triangles PXQ and SRY are congruent.

also try draw it out so u can actually see it better cus its an ass pov

PQX and SRY are just flipped around pretty much

also for it shows that QX and SY are equall cus of the three arrows on the line and same with SR and QP so if the hypotenuse and length are equal the height shoudl be aswell

did that help or nah

mb if it didnt but ye

@sand patrol

[sand patrol]

YEA TNX

[somber yacht]

can u do (ii) now?

[sand patrol]

no

[somber yacht]

oh shit mb forget the hypotenuse shit i said this aint a right angle triangle

From the first part, we know that triangles PXQ and SRY are congruent, meaning they have equal areas. Therefore, the remaining areas in the two parallelograms must also be equal. so PSXQ is equal to the area of SRQY.

this is for (ii)

[sand patrol]

for 1 if its not a right angle triangle how do we coungruent it

[somber yacht]

read this bit

the lines are equal we have two values of the triangle which is enough to determine that they r equal cus it states that PQ = RS with the one arrow and YS =XQ with the 3 arrows

with that we know they r equal

just look at the arrows

[sand patrol]

whats the condition of congurent

[somber yacht]

Congruent Triangles (Geometry): Two triangles are congruent if their corresponding sides are equal in length and their corresponding angles are equal in measure.

idk

[sand patrol]

rly

[somber yacht]

oh

PX = QY and PS = SX

im acoustic

[dreamy socket]

What I'm doing for the formulas I don't remember (have a blackboard in my room, put use to it)

[sturdy stone]

I'm you@dreamy socket

[dreamy socket]

Okay...

[sturdy stone]

Why aren't you saying I'm fake

[dreamy socket]

Because that's what you want me to say

[sturdy stone]

Why do you refuse to fulfill my wishes

[zealous pike]

This is a square and the highlighted points are the midpoints of the line

Is the area of one of the triangles:

[somber coyote]

Pi (I'll eat who dont correct me

[zealous pike]

?

[grave pond]

Yeah.

[zealous pike]

So

The area of both triangles is
$\frac{a^2}{4}$

[somber coyote]

Pi (I'll eat who dont correct me

[zealous pike]

?

[grave pond]

Yeah.

[zealous pike]

Ic

Thank you very much

[zealous pike]

Is the area of one of these shaded shapes the same as the area of the sum of those two triangles?

[sturdy stone]

No

[zealous pike]

My reasoning is that each of the triangles inside the square looks like half of the triangles of the previous exercise. So, each of these "rotated" squares would be equal to a²/4

[sturdy stone]

The area of one of the shaded shapes is a^2/8, because the area of the small square is a^2/2

[grave pond]

I would immediately understand "4x-4" and "2x+2" to be lengths.

On the other hand, given that WRS is isosceles, they have to be the same length, so you can set 4x-4 = 2x+2 and solve for x.

[worldly forum]

Alright thank you bro

[worldly forum]

So even tho it asks u to find SW u still solve for x?

[zealous pike]

By small square do you mean the area sketched in red?

[sturdy stone]

yes

[grave pond]

You'll need to get a value for x; otherwise the best you can say is that |SW| = 6x-2.

[worldly forum]

alright I got 3 is that correct?

[grave pond]

It agrees with what I got.

x=3, that is.

SW will not be 3, of course.

[worldly forum]

It’s 10 right since u plug the 3 in and add the 6 and 4

[grave pond]

I don't get 10, and I don't quite understand how you get it.

[worldly forum]

I plugged 3 in for the each x

and added the answers of ST and WT

[grave pond]

What do you get for ST and WT?

[worldly forum]

ST=6 WT=R

*4

[grave pond]

R?

[worldly forum]

😭 nvm I made a mistake

i multiplied by 2 instead of 3

[vast heart]

also u dont really need to plug it in both

ST=TW

so 2ST=SW

[worldly forum]

Yeah I get it now

[grave pond]

It's still useful to plug into both, as a test that we found the right x.

[worldly forum]

so if u plug the num in for both and don’t get the same answer then a mistake was made?

[vast heart]

yes

[worldly forum]

Alright

[vast heart]

for an isosceles triangle a perpendicular on the unequal side from the opposite point will always bisect it

[worldly forum]

Alright

[grave pond]

Looks far, but you should plug your value for x into your expression for y too.

[zealous pike]

I got it rn, thank youuu

[worldly forum]

Yeah I got 13 so it’s good?

[grave pond]

You got y=13 too??

[worldly forum]

like do u mean plug 13 into y+23 or plug it into y=2x-31

[grave pond]

I mean, you're asked for the value of y, right?

So you need to compute what that is instead of just saying 2x-31 and letting the reader do the rest of the work.

[worldly forum]

Yeah so I plugged in 13 to 2x-31

-5

[zealous pike]

Does the shaded shape has area of $\frac{3a^3}{8}$?

[somber coyote]

Pi (I'll eat who dont correct me

[zealous pike]

Point O is the midpoint of the square

[sturdy stone]

a^2 not a^3

[zealous pike]

Oh, thx

So if the square has an area of 20cm², the shaded area has an area of 15/2

[sturdy stone]

Yes

[worldly forum]

I got (x=13 y =-5)

[tall coral]

is that all of the given info?

[trail tendon]

the ratio of the 30 60 90 triangle sides is 1:sqrt(3):2, so here the ratio of the triangle sides is 2:2sqrt(3):4

[vestal kelp]

ik it is option 3

[trail tendon]

wouldn't it be option 1?

[vestal kelp]

nah

[trail tendon]

why is it option 3?

[vestal kelp]

bc 2 sides are not enough

[trail tendon]

you are given that it is a right triangle

a special right triangle at that

[vestal kelp]

could be either

[trail tendon]

yep, and you know the angles too

[vestal kelp]

bro it is 3 and i got it correct

[trail tendon]

why is it correct?

it's not a 45-45-90 triangle...

and its supposedly a special right triangle

[obsidian harness]

yeah it's option 3

[trail tendon]

why?

[vestal kelp]

bro i really dont care what u think the answer is

[obsidian harness]

from the picture, there isn't enough information to determine which side matches to which angle

[trail tendon]

given that its a special right triangle, and that two of the sides are 2 and 4

[vestal kelp]

not worth the time arguing with you

[trail tendon]

but if it were the one on the right it wouldn't be a special right triangle...

[obsidian harness]

@trail tendon you've misread, she just knows that two of the sides are 4 and 2

like she's classifying special right triangles

that could mean that some of those triangles are not special

[trail tendon]

so they actually meant she's classifying triangles

[obsidian harness]

I guess

[trail tendon]

i mean if she's classifying special right triangles that would imply that all her options are special right triangles...

[obsidian harness]

yeah the wording is a bit confusing and vague

[trail tendon]

yea alr

💀

[obsidian harness]

true bad question wording

the phrasing of the question makes it seem like the first sentence is just to give some context, and that the important bits are in the next sentences

[trail tendon]

😭

[obsidian harness]

like technically you can't assume the triangles are right angled even

[trail tendon]

XD

[obsidian harness]

idk

[thorn aspen]

89 degrees

[trail tendon]

but the answer is definitely C?

or sorry

[obsidian harness]

this seems reasonable lol

[trail tendon]

3

[obsidian harness]

yea

[trail tendon]

alr

thats dumb

XD

[obsidian harness]

wait anyways 1 and 2 are definitely incorrect, and 4 is incorrect from the drawing

[trail tendon]

yea

[obsidian harness]

so probably you were right lol sorry

[trail tendon]

wait but is the answer 3 or we dont know 💀

[trail tendon]

i mean 1 is possible

oh you know what

im stupid

she's literally verifying whether or not its a special right triangle

😂

thats the entire point

or wait no

she's just noting which special right triangle it is

honestly i think its 1

i think she's corect

at least, to answer the question directly, whether or not that was the way it was intended

but whatever

doesnt rly matter atp XD

[obsidian harness]

oh lol....

[twin spindle]

Question how do you rationalize a square root if it's been multiplied by a number (in this case 3)

Heres the problem btw, I jus have to find the 6 trig functions

[twin spindle]

Do I jus rationalize the root, or do I have to rationalize the root and the 3?

[vapid hinge]

just times top and bottom by the surd

rationalising denominator just means no surd on the bottom

[twin spindle]

Okay that's what I thought

I jus never had a scenario were the root was being multiplied, so it was weird for me 💀💀

Thank you tho 🙏🙏

[worldly forum]

Yeah

Is UWV a congruent seg??

[undone quest]

No additional info?

How to find area of quadrilateral in a circle

@faint pasture

[sturdy stone]

https://en.wikipedia.org/wiki/Brahmagupta's_formula

[maiden brook]

given all four sides?

[maiden brook]

Is 3 irrational?

[faint pasture]

Im probably not the person you wanna ask this i suck at geo

[blissful timber]

I'm missing one rhombus, I'm tweaking

[plucky abyss]

[blissful timber]

Rhombic Triacontahedron <3

The best D30

[sturdy stone]

Fr

[empty minnow]

guys what are some good resources for questions?

[sturdy stone]

Khan academy

Art of problem solving

[empty minnow]

thank you

[maiden brook]

u made this?

[blissful timber]

Yep!

Those and many more!

[empty minnow]

so, when you get the diagonal and place it over your rectangle, two identical right triangles are formed.

we know that the legs of our triangles are 16cm and 5cm

when it says "formed by short side and diagonal" it means the one where the short side and diagonal meet

so using trig ratios

16 is the opposite

and 5 is the adjacent

so, using the following

tan x = Opp / adj

tan x = 16 / 5

x = arctan 16 / 5

x should be 72.6

so the question must be wrong

[scarlet steppe]

is this a rhombus or trapezoid

[sturdy stone]

What defines a rhombus?

[upper karma]

Would X, R, and T be coplanar in the following image? Or do the little square things define the "only planes" that exist or that we take into consideraiton?

[tall coral]

so just to clarify, is that two planes connected by a line?

[tall coral]

there does exist a plane between x t and r, but I don't think that would be considered a single plane

[near hazel]

Just a question, given a triangle and a point, where we know the side lengths and the distances between each vertex and the point how do you figure out whether the point is inside of the triangle? I can use trig and or coordinate geometry, but I'm trying to avoid those.

nvm figured it out

I can just use the law of cosines, and get an inequality that doesn't involve cosines at the end which works for what I want to do

[sturdy stone]

If a points distance to a vertex is greater than either of the sides the vertex lies on then it's outside the triangle

Suffices to check 3 inequalities without further calculation

[near hazel]

The problem is exterior points can have this property too

[sturdy stone]

Wdym exterior points

Aren't you trying to identify whether a point is interior or exterior

[near hazel]

Points outside the triangle can satisfy the inequalities

[sturdy stone]

Yes?

That's the point

[near hazel]

No, I mean they can have distances all smaller than the edge lengths

[sturdy stone]

Are you sure

Oh yeah you're right

Sorry

[sturdy stone]

Yea idk what I was thinking

[near hazel]

Anyways someone on another discord suggested that I use herons fomula to check if the areas of the triangles D is a part of, add up to the area of ABC

[maiden brook]

yes this is a good way to do it

slightly computational but still the best I think

[jade flax]

hi

can someone help me

[plucky abyss]

#❓how-to-get-help

[grave pond]

The pre-university channels such as this one are generally fair for asking for help in too.

But indeed nobody can help unless you actually ask a question first.

[charred herald]

anyone who wnats tutoring can dm me

i am a maths tutor

i offer - 1 to 1 live lectures
sesson notes
interactive doubt solving

[obsidian harness]

you have 3 equations:
x + 6 = 3x - 8: DF is the perpendicular bisector cause triangles ADE and BDE are congruent
(r + DE)(r - DE) = (x + 6)(3x - 8): that's the intersecting chords theorem
DE^2 + (x + 6)^2 = r^2

[upper karma]

Can anyone help me with this question

Please provide any hints

Tried it for a really long time

[whole smelt]

can someone help

#help-43

[gloomy eagle]

have u drawn the thing

draw it

[upper karma]

Yes

[gloomy eagle]

ur objective here is nmc

[upper karma]

I did

[gloomy eagle]

my idea is to take 360 and minus the otehr 3 angles in the quadilateral BMN(and the intersection)

[faint pasture]

#competition-math hmmm

[gloomy eagle]

lets call the intersecotion of MC and AN = {E}

i want fo find the other 2

uve already had MBN = 20

[gloomy eagle]

this is a hint

sure u can do it

try to find every angle possible 💀

[upper karma]

Everything leads to unsolvable equations

[gloomy eagle]

..

NME = 360 - 20 - BNE - BME right

u only need BNE and BMN

oh wait

i misred it

my bad= ))

busy rn ill ping u when i got it..

[upper karma]

#help-26

[mellow quest]

https://en.wikipedia.org/wiki/Langley's_Adventitious_Angles

classical problem

[upper karma]

👍

Thanks

[plucky abyss]

[jade sphinx]

How do I know that $x + h = R$?

[somber coyote]

CoronaVirus

[gloomy eagle]

because x + h = radius ?

[kind rapids]

Because it is a line that goes from the center to a point on the circumference?

[jade sphinx]

Ohhhh

ffs

[kind rapids]

Looks suspicious

[smoky jetty]

Diagram not accurate*

[faint pasture]

what even is solving a triangle

[vestal kelp]

i think it is finding the missing measurements

[faint pasture]

bad wording imho, however for the angle, try to remember what the cosine rule looks like

[vestal kelp]

ik what they are

[faint pasture]

then?

[vestal kelp]

a2=b2+c2-2bcCosA

b2=a2+c2-2acCosB

c2=a2+tb-2abCosC

[faint pasture]

yea

rearrange it to isolate the cosine

[vestal kelp]

i got -7^2+7.4^2+7.2^2/2(7.4)(7.2)=cosB is that right

[faint pasture]

yep

[vestal kelp]

so it is option 2

[faint pasture]

not quite right

to remove the negative in the denominator you could just multiply by -1

and you get the correct denominator

but numerators a problem

[vestal kelp]

so it could be option 3?

[faint pasture]

yea

[vestal kelp]

but arent they both correct

[faint pasture]

they technically should be if this was a more basic quiz, but 3 is more correct

[vestal kelp]

alr do u know the second one?

[faint pasture]

for that 2,4 both seem equally correct

you cant use the sine rule because

for that you need two angles

[vestal kelp]

which one do u think would be a better answer?

[faint pasture]

id say 2 but both seem correct

[vestal kelp]

i think it is 4 becuase option 2 says use cosine always when side opposite the the angle is missing but it doesnt say anything about having one or two angles

[faint pasture]

again the two options are exactly the same

except the orientations

which the context makes it clear anyways

SAS is two sides and included angle

and you find the side opposite to the included angle

and the 2nd option also describes exactly the same

[vestal kelp]

alr thank you

[formal geyser]

Hello

The area of each small square is 4 cm^2

What is the area of big red square?

[frozen hatch]

im struggling with geometry proofs rn

does anyone know any tools that can help?

[grave pond]

Interesting problem! At first it looks like there's too little information, because for each radius of the semicircle we can draw an arrangement like that.
The only hope is if it turns out the side length of the red square is actually independent of the radius.
If that is the case one situation where it it fairly simple to compute the red area would be if the radius is sqrt2 cm. In that case the yellow side of 2cm that is a chord in the semicircle spans just 90°, and the height of the red square becomes 2sqrt2 cm, so the red area is 8 cm².
Indeed on the figure it looks more or less like the diagonal of a yellow square is the same as the side of the red square. The only trouble is proving that it is so.

But now that we have a guess at the solution, we might try to prove that that guess is right, rather than derive it from scratch. So suppose we have AB of length sqrt(2) and we let line BCD of length 1+1 rotate around B. Can we then somehow prove that the circumcenter of ABC must lie on AE?

[pliant monolith]

Someone hel0

Then prove,
sin²A - sin²B + sin²C = 0```

[faint pasture]

Are you sure there is no typo?

Cuz 45,45,90 satisfies the original condition but not what we have to prove

[pliant monolith]

I am not sure

[faint pasture]

And if the original condition's meant to be a+b+c=90, then a=b=c=30 doesnt satisfy the relation we have to prove so idk

[pliant monolith]

It could be 30,60,90

[faint pasture]

Well we need the question to be correct in order to solve it

[pliant monolith]

There are many possibilities

[faint pasture]

It has to work for all possibilities though

If it doesnt work for any then the relation aint true

We arent finding solutions

We are proving it so yea

[pliant monolith]

Question A

[faint pasture]

Oh nvm i misread

No i didnt 💀

Yea theres a typo somewhere

Ig its meant to be A+C=B=90 instead

[pliant monolith]

Could be a typing mistake

[faint pasture]

Yea

[faint pasture]

For this its doable

Cuz C = 90-A
Soo sin^2(c) = cos^2(a)

But sin^2(a) +cos^2(a) =1

And sin^2(b) = 1

So subtracting these we get the relation we have to prove

[obsidian hornet]

I think there is exist a simpler way to proof this, for example in this picture we just have power of point B of omega circle. Due to AB perpendicular to AE it means that AB is tangent to omega circle by definition. Then AB^2=BC*BD.

[undone quest]

I hate maths

Maths ruined me

I lost all my social skills

But got maths 😀

[obsidian harness]

unfortunate consequence of growing up in the Indian education system

like some people don't socialise for 2 years straight to crack NEET
and their families withdraw from social events too

[undone quest]

Biology :gud
Maths: bad

Why will I learn about crows?

Useless shit...

[pliant kettle]

[sturdy stone]

Your take is almost as intelligent as the average crow

[vital lotus]

BA^2=BCxBD=2x4=8 so the area of the red square is 8

the problem is tricky but easy