geometry-and-trigonometry

[upper echo]

tan^-1(3/4) has some measure too, but you won't express it in nice terms and you need not do that

[golden knot]

that is so cool wow

wowwww omg

that's SO COOL

thank you!!

[quaint chasm]

i might have missed up a term but i did the taylor series for sin(1) by hand

oh i did 1/8 instead of 3/8 oops

tiny error

[quaint chasm]

skme other errors too oops

its close tho

[opal dawn]

Hi guys

geometry test tmr wish me luck 🤞

[wanton hawk]

Taylor series is genuinely gonna make me commit why-did-I-take-this-class-ide

[wanton yacht]

😳

[wanton hawk]

https://tenor.com/view/taylor-series-expansion-trigonometry-sin-67-67-meme-gif-13604115539856438070

[wanton yacht]

LOL WHAT

Where did you find that

[west narwhal]

lol

[winged oyster]

rawr

[quaint chasm]

real talk why dont we just take the modulus of the input for the taylor series for sine and cosine
its a periodic function so its not like it changes as you go out further

oh i tried it and its weird

theres a gap

this loops fine tho

[winged oyster]

ig bro

[rare agate]

idk, periodicity isn't a part of a polynomial so actually eventually will eventually fly off to +inf or - inf. But what you said is how computers calculate sine and cosine, its range reduction.

[pulsar hedge]

yeah cause the approximation at x=pi isn't perfect

[mossy jacinth]

Very easy

:opencry

[ionic stream]

I’ve literally never been so confused

[exotic yarrow]

In the future, please use the ,rotate command.

,rotate

[exotic yarrow]

Looks good so far. I'd use $\cos 2\theta=\cos^2 \theta-=\sin^2 \theta$ for the denominator.

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

Then consider what you can divide the numerator and denominator by

[raven pollen]

I am looking for a Gina wilson worksheet?

[silent plank]

whos that?

why do you need worksheets from her specifically

[cinder spear]

Searched her up, she’s the writer behind something called “All Things Algebra”

I don’t know that textbook/workbook but Mrt.231 is asking about that

[rich nexus]

Anyone got state testing next week

[slim plinth]

What state are u from

[rich nexus]

Nj

[slim plinth]

Omg me too

[slim plinth]

I assuming u are a freshmen right

[rich nexus]

8th grade

[slim plinth]

What’s it called is it NJSLA

Or it starts with P

[minor heron]

thankfuly no my state is doing testing rn but high schoolers onlu have to do it in 11th, i do have to trek rlly far to another classroom since my precal class has a lot of juniors testing and the classroom being used

[slim plinth]

Are u from NJ

[minor heron]

no

but igs all states r doing state testing rn

[rich nexus]

Njsla

I lwk hvnt practiced for the test

[minor heron]

ehhh i didn't rlly study for 8th grade state testing either, im sure you'll be fine. glgl!

[rich nexus]

They giving us 40 questiobs this time and the practice tests were hella hard

Questions

But gl

[minor heron]

oof icic

for math english or science?

[rich nexus]

Yo

Guys can I say something

Math

[minor heron]

what is it

[rich nexus]

Srry that was my friend

[slim plinth]

Ohhh friend me

[modest mortar]

I hate vectors without coordinates

[spiral lodge]

Why? They are easy

[tiny acorn]

is there even a general method to working backwards my geometry problems? i feel like i can not work backwards due to 1. requiring theorems i learnt last year and even b4 which i forgot 2. these proofs being extremely nontrivial and requires a combination of the above theorems. My teacher recommended me to learn how to work backwards and analyze these problems but i cant seem to do so

[tiny acorn]

for example this
Translation
Let ABC be an acute triangle (AB < AC) circumscribed by the circle (O;R). The altitudes AD,BE intersect at H. Let M,N denote the intersection of the @jovial sinew BE , AD respectively (M != B , N != A)
a) prove ABDE is cyclic, find the centre I of this circle, then use that do prove DE is parallel to NM
b) Draw the diameter CK of (O), prove ABKH is a parallelogram and H,I,K are collinear
c) Let BCA = 60 deg. Prove |DE| = 1/2 |AB| and calculate the lengths of the chord DE and minor arc DE of (I) wrt to R

i am stuck at the second part of A

cant figure out how to prove is parallel to NM

like not even able to work backwards

[low ocean]

Working backwards gives you something to start with instead of looking for smth unclear . Say, showing parralel, you would naturally ask which theorem you have learnt related to parralel segments, here first thing to come up in my mind is BPT, and what's required? Same angles or ratio ( basically similar triangles ) if so, how would we show 2 angles share the same measurement? Here we have circles so maybe something with circles. We are actively working backward here

[tiny acorn]

1 what is BPT exactly, English isnt my native language
2 yes but i cant find the relavent theorems to the problem at hand most of the time

[low ocean]

Tho geometry is just like any topic in math, the ( in my opition ) most important thing you need is the intuition built up from solving problems

[low ocean]

Yeah right, we call it Thales Theorem

[tiny acorn]

... Thats the intercept theorem?

Idk man screw these names

[low ocean]

You just lack the intuition, dw,I suggest write down every single theorems you have learnt on a paper and look at them whenever you solve a problem. If you can remember them right away then that's great. The goal here is to build the intuition

[low ocean]

Định lý Thales xD

[tiny acorn]

... Twin i didnt know your vietnamese too

:3

wait did you use fricking google translate

[low ocean]

ofc not

I used to suffer from these euclidean geo

I feel you mate

[winged oyster]

what

[violet sluice]

oh ok

[hybrid trench]

Can you guys use p1p2= √[(x2-x1) ²+(y2-y1) ²+(z2-z1) ²] to find the center of a circle?
or do u have to use the (x-h) ² one

[iron violet]

for 2d use (x-h)^2 for 3d use euclidian

[hybrid trench]

😔 what if we don't know the dimension

[iron violet]

circle exists on 2d only 🥀

[hybrid trench]

...

I forgot about that

[vocal sentinel]

why do you use z here

[hybrid trench]

... That's what was in the review sheet

[vocal sentinel]

Why not use $r^2=(x-a)^2+(y-b)^2$ for the circle centre formula?

[hybrid trench]

It just said distance between two points

[somber coyote]

Minλ

[vocal sentinel]

Interesting, but isn't the third variable redundant, at least in the 2D space?

[hybrid trench]

Is it different from the x-h one? Isn't it js the variable name

[iron violet]

its for sphere

or just 3d distance for points

[hybrid trench]

Is this for 2d?

[hybrid trench]

yeah

[iron violet]

Yes

[vocal sentinel]

Yes, unless you ask for a "sphere"

[hybrid trench]

How is the center 0,0,0

For the top one

cause √(x2-0) ²+(y2+0) ²+(z2+0) ² =4?

[vocal sentinel]

That is a sphere

The first one is because $(x-0)^2+(y-0)^2+(z-0)^2=2^2$

[somber coyote]

Minλ

[vocal sentinel]

Then centre is 0,0,0

Second one is to complete the square

[hybrid trench]

So the r for the second one is √11?

[vocal sentinel]

No

You haven't complete the squares for other terms x y z into the identitiés

[hybrid trench]

Isn't it just x(x-4)+y(y-4) +z(z-4)?

[vocal sentinel]

No....

You want into this form $r^2=(x-a)^2+(y-b)^2 +(z-c)^2$

[somber coyote]

Minλ

[tiny acorn]

for the thales theorem/intercept theorem/BPT/whatever this is, it says that

$$AC \parallel DB \implies \frac{| SA |}{| AB |} =\frac{| SC |}{ | CD |}, \frac{| SB |}{| AB |} =\frac{| SD |}{| CD |} , \frac{| SA |}{| SB |} =\frac{| SC |}{| SD |}$$

however is the converse true?

$$AC \parallel DB \Longleftarrow \frac{| SA |}{| AB |} =\frac{| SC |}{ | CD |}, \frac{| SB |}{| AB |} =\frac{| SD |}{| CD |} , \frac{| SA |}{| SB |} =\frac{| SC |}{| SD |}$$

[hybrid trench]

... What is this

[hybrid trench]

Okay so a is 2

[somber coyote]

multiplexer

[tiny acorn]

meth

[vocal sentinel]

correct

now complete the squares for b and c

[hybrid trench]

both are 2 too

[vocal sentinel]

correct

So you have to add 3 times of 4 and minus 3 times of 4

move -12 to other side of equation, we have -11+12 = 1, that is radius

[vocal sentinel]

Unecessarily true, only holds iff the points laying down in the correct orders on the same sides of the vertex S

[tiny acorn]

wdym in the correct orders

[vocal sentinel]

Suppose you have points D C on the line DS, if you reverse their orders, sure, the ratios still correct, but the line are not gonna parallel anymore

[tiny acorn]

ok sure

thx

[tiny acorn]

wait but hold up

wont they just intersect?

if they dont intersect thats fine no?

[vocal sentinel]

You want AC and BD to parallel, not crossing each other, and for thales to work, they must parallel

[tiny acorn]

hmmm okay but i wanna prove the converse

like in some problems

""if ratio then parallel"

not "if parallel then ratio"

if you know what i mean

[vocal sentinel]

isn't it already a statement, no?

[tiny acorn]

hm?

[vocal sentinel]

ah yes, you need to prove it

maybe use contradiction?

[tiny acorn]

nvm i was trying to ask about the application of the converse thales theorem to this problem, specifically a)

[vocal sentinel]

You don't really need talet for a?

[tiny acorn]

how do i solve it then?

[vocal sentinel]

chứng minh đồng vị

[vocal sentinel]

DAB= DEB do nội tiếp

mà D nằm trên MA

[tiny acorn]

then what

[vocal sentinel]

vậy DAB=MAB, mà MAB= MNB do cùng chắn cung MB, nên DAB=MNB

=> DEB=MNB

bên còn lại y chang với góc EDA và NMA

[tiny acorn]

SO ITS FRICKING ANGLE CHASING ALL ALONG?

[vocal sentinel]

là qed

[tiny acorn]

SCREW ME

[vocal sentinel]

thường câu a chỉ là tìm góc nên khá dễ

[tiny acorn]

blud

[vocal sentinel]

còn tâm đường tròn chắc ngon r

[tiny acorn]

its trivial bruh

midpoint of AB

[hybrid trench]

Ooo okay

[final comet]

Good afternoon. For a project I am researching billiards in different objects. While focusing on billiards in an ellipse, I found out about Poncelet Porism, which I understand up to some point. However, the papers I've read then jump to a different concept called the 'Poncelet grid', which takes a poncelet $n$-gon inscribed by an outer ellipse $\Gamma$ which is inscribed by inner ellipse $\gamma$. Then taking two edges of the $n$-gon that are a fixed index $k$ away and taking the intersection of their extentions, you get a set of points $P_k$ that all lie on a confocal ellipse. The papers I read just claim this without much explanation, as if the result is very logical from what was discussed before, however I cannot find the logic behind them.

I can individually prove that given two tangents on an ellipse, you get that their intersection lies on an ellipse depending on the difference in the parametrization of $\gamma$. However, this implies that they are notconfocal, since their foci are scaled by parameter $\sec(\delta/2)$.

I also know Graves' theorem, which states that a constant-length string wrapped around an ellipse pulled taut traces a confocal ellipse. From this angle of proof we would need to show that every point of $P_k$ preserves the length of this string, which I struggle to find at this moment.

Could someone perhaps help me understand how this works?

[somber coyote]

AwaknThundr

[past quarry]

guys what book do you recommend for me to master geometry and trig before i move onto precalc

[edgy nest]

cengage by g tewani

[strong wind]

Realest

Ever

[edgy nest]

I worship cengage 🛐

[strong wind]

Hahaha

Ur from india too right Ankur?

[edgy nest]

yessir

[strong wind]

Nicee

[edgy nest]

and a jee student

[strong wind]

^^

Oh me too hehe

[edgy nest]

Lol, 27?

[strong wind]

I just came to 11th

😅

So ill give 2028

[edgy nest]

Oh nice man

[strong wind]

^^ im supposing ur older lol

[edgy nest]

good luck, and be disciplined and consistent otherwise you will regret it lol

[strong wind]

Yess ^^

[edgy nest]

i gave jee this year, ill be giving advanced in 3 weeks lol

[strong wind]

Nicee

How did it go? ^^

If u dont mind me asking :D

[edgy nest]

i had 97.3, 42.5 rank

[strong wind]

Thats rlly awesome ^^

[edgy nest]

general male but i have delhi HS so ill prolly get DTU mech

[edgy nest]

nah bro, if youre general male, even 99.5 doesnt guarantee best thing

[strong wind]

True that lmao

Im also general male 😓

[edgy nest]

exactly man
our situation is fked, we gotta work hard
or you can manage 98+ in mains and work your ass off for under 2k in advanced

[strong wind]

Fr

Good thing im talented enough for my age

🥹🥹

I just need to build discipline

I know it will be tuff but ill stay strong

😌

[edgy nest]

ayy lets go

[strong wind]

Hehehe

[edgy nest]

keep working hard man, even i though im too talented ill surely get under 10k but that did not happen for whatsover reason

[strong wind]

🥹

Best of luck for advanced 🫂🫂🫂

I hope you get a good rank 😄

[edgy nest]

Thanks man, appreciate it, and likewise!

[strong wind]

Yessss

^^

Accha ill go now-

I gotta help mom

Byeeee

[edgy nest]

See ya bro

[alpine bramble]

Where to find trig Questions. Word problems, Any resources

[winged oyster]

khan academy

or youtube

or ask google

[marsh steppe]

any ai can give you problems to solve

but yeah

youtube content creators also explain trig

[winged oyster]

or this

[marsh steppe]

yes

[tiny acorn]

!noai

[lime crown]

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

[wanton hawk]

Is this bot talking about not trusting bots

[marsh steppe]

ok thanks sorry im new here

[fervent stump]

You dont "have to" give the JEE

It has very to toxic culture around it, especially when it comes to the subject and the way you are asked to rote memorise.

[edgy nest]

Real

[edgy nest]

its too much hyped up by coachings to make tons of money and pressured by parents for respect in society

[upper karma]

CENGAGE MENTIONED 🗣️ 🗣️ 🔥 🔥 🛐

[balmy star]

θ

[winged oyster]

how do you get that

[tiny acorn]

either charmap ,copying from some unicode site or greek keyboard

[strong plume]

Honestly speaking, Out of 14 lakhs students who write it, Only 2-3 lakhs are actually serious about it

[edgy nest]

2-3 lakh are serious and working, next few lakhs prepare but get a reality check
last 2-4 lakhs just filled the form for fun

[strong plume]

Facts

Did u write jee?

[edgy nest]

Yeah

[edgy nest]

wbu?

[strong plume]

Na I started my 11th grade this year

[strong plume]

Did u write it this year or are you already in college?

also, Any tips?

[edgy nest]

i wrote last month

[edgy nest]

if you are serious about it - discipline and consistency.

[strong plume]

Its so hard to be consistent honestly

Ever since boards ended, i have kind of been distracted

school only starts in june

[edgy nest]

i few days break is fine, but when you are back at it, dont stop. leave no fking stone unturned

[strong plume]

Results of mains came out right? Are you writing advanced then?

[strong plume]

True

[edgy nest]

yeah im writing advanced

result hasnt come out yet bro 💀 im writing
no idea what will happen

97.3

lmao what was your rank?
mine was 425xx

damn no way

2s2

mere do dost hai, ek ki 98.2 and 98.54
their ranks are 27k and 23k respectively
so hows your 21k 😭

sir what?

you cant be serious dawg

btw shifts are normalised and age is no longer used in tie breaker

[uncut skiff]

Anyone have a good geometry question?

[upper heron]

the answer key says that the height of the COM is r(1-cos(θ)), why is this?
as i remember, cos(θ) is adjacent/hypotenuse

and i tried experimenting in desmos, and it does hold true
but i still dont get why

[cinder spear]

The change in h is the initial depth - final depth
The final depth is rcosθ, and since the center of mass always stays at r from the pivot point, the initial depth is r
From that, find Δh

[upper heron]

oh I see, its actually just r-rcos

[cinder spear]

Mhm. Which is the same as r(1 - cosθ)

[zealous oyster]

question regarding sines

what stops someone from making a formula for sin(ϴ/3) using the formula for sin(3ϴ) and the depressed cubic formula?

[exotic yarrow]

You could, but it would be quite useless

The cubic formula requires you to compute a cube root of a complex number

To find the cube root, you need to use De Moivre’s theorem, which in turn requires calculating the sine or cosine of an angle divided by three

Which gets you right back where you started

[zealous oyster]

not the cube root but the roots of a third order polynomial

isn't there a general formula for that?

[exotic yarrow]

Yeah

The cubic formula

… which involves cube roots.

[zealous oyster]

Ah like that. yeah

thanks for the elaboration

[exotic yarrow]

[rich nexus]

can someone help me with this

[torn gulch]

they um say the steps

the point P refers to the center of the circle

from there any point on the circumference of the circle connected in a straight line segment from point P is the radius, PQ

then use the compass to go from there

and the next few steps would explain it

[rich nexus]

ight ty

[tranquil ocean]

would vectors be considered geometry

[grave pond]

Depends on how abstractly you're working with them, but if you're asking here rather than in #linear-algebra, then the answer is most likely "yes".

[wanton hawk]

I mean they’re also a precalc topic

[unique wren]

Can somebody explain to me the relationship b/w imaginary numbers and trig functions
Like i heard in some class that i(iota) is just 90° rotation
And trig works with degrees

[iron violet]

How do people understand geometry without loosing hair? (Seriously, I spent alot of time but still i cant seem to understand it)

[trim stump]

Once you get the hang of it, it'll be easier

One of the major factor of being good at geometry is just having good memory and familiarizing other formulas that might be a solution to the problem

[iron violet]

Oh thanks for the advice

Ill work onto it

[trim stump]

This is how I do it idk
Idk how to explain better but think of a circle with the center being the origin ad its radius is 1
Look at (1,0), if you multiply it by -1, it would be (-1,0)
Therefore multiplying by −1 corresponds to a 180° rotation.
Now look at i=√-1
i²=-1
So:
doing i twice gives 180° total rotation
That means:
each i step = half of 180° = 90°
Soooo (1,0)(i)(i) should be (-1,0)
Qnd (1,0)(i) should be (0,1)
Am I making sense or am I hella confusing

[iron violet]

1, -1, i, -i make a circle in the graph

if you connect them

[unique wren]

Thanks i remember a video which explains this, you've helped my revise it

[balmy star]

sin(θ)

I'm gay

[somber pendant]

woah

[unique wren]

Why is this guy under so many supervisions

[tiny acorn]

<@&268886789983436800> really suspicious person
No actually this is a textwall

[sharp monolith]

Even here

[finite widget]

Does anyone know how to find trigonometric angles? I'm fairly new to trigonometry, and I haven't gotten the baseline of triangles and such (ironically enough, I take calculus, which has and goes over trigonometry). Can someone give me a step by step on how to find an angle like this?:

(I know the Pythagorean theorem, but I cannot execute it.)

[grave pond]

What do you mean by "find an angle" when it's given in the diagram as 43° explicitly?

Do you mean the unmarked angle? That must be 180° - 43° - 90°, due to the angle sum of triangles.

[tiny acorn]

which angle?

if you mean by the missing one then do it as tropo said

[finite widget]

stuff like that. i mean, i don't know specifically what i want to solve, but i want a clear way on how to since most of the time trigonometric angles have sides as well, and you have to execute both to get your answer.

i didn't clarify this before, but that's what I meant.

[finite widget]

Yea, pretty much but theres still something I kinda wanted to be answered, in which I already asked Tropo

[tiny acorn]

this?

[finite widget]

Yea something like that

[grave pond]

I'm afraid I don't understand what your question is.

[tiny acorn]

just take the sin/cos/tan of the angle as needed and use the chart above to solve duh.

[finite widget]

So, I was asking just how to solve stuff like that in general. Like a step by step guide on how to solve an entire trigonometric question

[finite widget]

Okay

[tiny acorn]

trig is an incredibly diverse subject

so that question cant really be answered

[grave pond]

I don't even understand what "stuff like that" means? The only stuff you have shown was a diagram where the angle was given explicitly, so there's not even anything to solve there.

[finite widget]

Sorry, I didn't know it was a diagram. However, you could answer that diagram yea, but the thing is I wanted a step by step guide on how to solve them. Like how to use the Pythagorean theorem and such and execute them to solve actual trigonometric problems.

[grave pond]

Sorry, I didn't know it was a diagram.
Um, wut?

[tiny acorn]

for these kinds of questions then

notice how this can be directly derived from sohcahtoa

well c is just the pythagorean theorem

[finite widget]

Okay! Thanks for helping me, and sorry for the confusing explanation D:

[true compass]

Can someone explain why my first approach isn’t valid?

The 2nd approach leads me to the correct answer as I’ll just do sqrt(17^2 - 15^2)

Then do 12 - ^

[tiny acorn]

First approach is the one on the left right?

You cant just assume BC = 8

[tawdry silo]

Didn't he do the pythagoras theorem to find it out?

[tiny acorn]

What you do is smh computing AD-BC (which you did in the second one) and then use that to find out BC

[true compass]

Yeah I did

[true compass]

Yeah I didn’t assume, I did Pythagoras but I’m struggling why it isn’t valid

[true compass]

That doesn’t tell me where I went wrong tho, that’s obviously the 2nd approach I’ve done

[tawdry silo]

How did you get 17 in the first one?

[true compass]

Yk how DC = 17 I assumed that line across would also be 17

[tiny acorn]

Define C' (projection of C). you assumed C'B = 17

[tiny acorn]

There is nothing saying that DC'BC is a parallelogram lmfao

[tawdry silo]

But i don't think you can assume it like that

[true compass]

Yeah that’s fair enough

Idk why i approached it like that but fair I get it now

[tiny acorn]

The exercise didnt give DC' and BC are parallel and equal, u just assumed that

[true compass]

Yeah true

Cheers guys

[tiny acorn]

Congrats

[tawdry silo]

Could i pls know the answer?

[true compass]

4

I can share working out if you like

[tawdry silo]

Please do

I want to see

[tawdry silo]

yeahh, nice

i did it this way too

[true compass]

Nice one

[tiny acorn]

Yea thats what im thinking

[wintry stone]

Hey guys, I wanted to ask whether it's possible to represent 3d rotations in tilted planes centered at the origin as rotations in the independent planes that make them. For example, if we want to rotate a point (any point) in a plane that is tilted from the positive z axis in the YZ plane by 20 degrees by 45 degrees (in that plane), can we represent this rotation as first a rotation in that YZ plane and then a rotation in a base plane like the XY plane in this example by 45 degrees. Or switch the order of rotations as 3d rotations are non commutative.My question is, if this is possible how do we find the necessary planes to rotate in (XY, YZ or/and XZ), and the angles to rotate by?

[wintry stone]

oh and the order in which to rotate in those planes by?

[tiny acorn]

Bruh

Dont ping us b4 15 minutes

[wintry stone]

oh sorry

[tranquil ocean]

I wasnt sure to be honest... I've been working with a lot of the sin cos and tan ratios, as well as the tip to tail and tail to tail addition/subtraction rules I thought I would ask here

[tranquil ocean]

I feel as if I also shouldve thought of this since im literally in the vectors part of my calc+vectors course rn... 😭😭

[wanton hawk]

https://tenor.com/view/malicious-monkey-thinking-monkey-freaky-monkey-gif-17111090804698831003

[tranquil ocean]

my excuse is i'm sick and my brain is mush ✌️🥹

[quaint chasm]

does anyone know the equation for how many unique faces you can form using n vertices? including intersecting ones

from what i can tell the relation of vertices to faces goes something like this

1 vertex : 0 faces
2 : 0
3 : 1
4 : 5
5 : 26
6 : 54

6 could be wrong but im pretty sure im right about the previous ones
i think there's a factorial term to it somewhere but idk

quick sketch of what i mean for 4 vertices

[grave pond]

The upper left and lower left and lower right are the same in different rotations, so you shouldn't multiply them all by 5.

I would count it as "all the ways to select some of the vertices, except the ones that select only 1 or 2 vertices, or none".
That yields 2^n - 1 - n - n(n-1)/2.

[quaint chasm]

true my bad

i was drawing that in a rush before class

[quaint chasm]

awesome

surprising that that’s not a factorial since i all combinations of n values is usually factorial

i also see the formula for edges in there

[grave pond]

So I get this result:

$ perl -l -e 'for $n (1..7) { print "$n ",(1<<$n)-1-$n-$n*($n-1)/2 }'
1 0
2 0
3 1
4 5
5 16
6 42
7 99

[quaint chasm]

ok i was initially right with 16 faces for 5 verts then

[grave pond]

Presumably we don't want a "face" whose sides cross each other, so once we've decide which vertices we're using, there's only one order to connect them in. So we're interested in subsets without ordering. Factorials are for when the order of picking matters (or else we need to correct for overcounting later).

[quaint chasm]

ic

[wintry stone]

<@&286206848099549185>

[restive crest]

i got pinged twice in here

in this same exact channel

[wintry stone]

It's been like 50 minutes i swear, i can ping now right?

[restive crest]

usually people ping in help channels

but im not sure in channels like these

[wintry stone]

this is an open ended question i think which is against the rules to post in help channels right?

cuz there is like not really one answer, idk?

anyways, can anyone help me on this?

plzzzzzz

pretty plzzzzzzzzzz

[tiny acorn]

[wintry stone]

hey, could u like provide a direction to pursue in this question, like is representing 3d rotations in this manner possible and how might I try to do it without mixing up the order of rotations?

[tiny acorn]

I have no idea how, but asking random ppl and posting textwalls aint gonna help you

[wintry stone]

but you just helped someone on their trig question. : (

[tiny acorn]

As they said, helpers are volunteers, we choose whether or not to help

[wintry stone]

IG I will have to open a help channel now, even though I think it's against the rules for open-ended questions but okay

[wintry stone]

ig you have a point

okay

well, here is the link :https://discord.com/channels/268882317391429632/1499084623739748452 if anyone wants to help, they can do so here, regards.

[slim plinth]

Check my work

[static mason]

i need help on this question

[spare wharf]

I came up with 105 as well

And I did it in a different way so you're good 👍

Triangle is 180°

55 + 50 = 105
180 (triangle) - 105 (2 of three angles) = 75
180 (flat line) - 75 = 105

[glacial garnet]

PLLLLSSSS

[regal berry]

what do you have to find

oh x I see it now

ok so is ABCD a rectangle

so angle ABK is 36 degrees

wdym

ok

so EC is an angle bisector of course

[pulsar fractal]

is this correct?

[exotic yarrow]

This makes no sense

The tangent-secant theorem applies when the tangent and secant meet outside of the circle

Also you don’t even have a tangent with what you used

Consider this triangle

[pulsar fractal]

i thought AB is a tangent?

[exotic yarrow]

Tangents meet the circle at exactly one point

AB meets the circle at two points: A and B

[pulsar fractal]

oh shoot i meant secant

oh wait i see

[grave pond]

Looks more like a diameter on those pictures ...

[winged oyster]

it is

[wanton hawk]

https://tenor.com/view/jerry-meme-tom-and-jerry-meme-jerry-points-get-a-load-of-this-guy-this-guy-gif-11384860625931393733

[opal dawn]

guys is Abe 63

angle

[solid pivot]

guys

can someone help pls

its a kahoot assignment, its not a test or anything

idk how to get x

[exotic yarrow]

straight recall:

[opal dawn]

can someone help: i got 63 for Abe idk if its right

[solid pivot]

ohhhh is it ratios?

[solid pivot]

lemme look

[exotic yarrow]

You could write it as
$$\frac{PA}{PC}=\frac{PD}{PB},$$
which is an equality of ratios.

[somber coyote]

Civil Service Pigeon

[solid pivot]

yea, i got 63

[opal dawn]

is this right

i dont get the second image

[solid pivot]

because the 86 is inscribed with abc and then you half it and then you add it with 20 which equals 63

[opal dawn]

can u check the two

plz

[solid pivot]

idk man

lemme finish my hw

im struggling with the same topic

[opal dawn]

can u check the second one

it is multiple choice

[solid pivot]

i looked at it but couldnt figure it out

sorry

[exotic yarrow]

Hello?

[opal dawn]

someone help PLEASE

i dont understanddd

[solid pivot]

hey pigeon, i did this thing and i got it wrong

because i got 7.5 and the answer was 4

[exotic yarrow]

!show

[lime crown]

Show your work, and if possible, explain where you are stuck.

[solid pivot]

ok

[exotic yarrow]

... actually do either of those make sense

[solid pivot]

i did

6/8 = x/10

and it gave me 7.5

but the answer was 4

i did that because of the ratio thing u sent

[exotic yarrow]

because if the numeric lengths correspond to these highlighted things

then you should use the intersecting chords theorem

[exotic yarrow]

well this is also a different configuration tbf

you can't apply theorems in places where they don't apply

[solid pivot]

welp

u gave me the wrong theorem

sooo

is it my fault?

[exotic yarrow]

`the secant secant theorem applies when you have two secants

you don't have two secants here

you have two chords

[solid pivot]

yea, i used that for a dif one

the secant thing

i used the secant one for this one

and i got it wrong

[exotic yarrow]

oh

did you just send that chord thing by accident

ah

[solid pivot]

no, that was the next problem

and i still didnt know how to do it

bc circles r hard

how would this one work?

[exotic yarrow]

Read carefully - PA and PC are the parts of the secants outside the circle and PB, PD are the entire secants

so it should be 6/8 = (6+10)/(x+8)

[solid pivot]

oh

so i just did part and part

not part and whole

thanks

how would this one work tho?

[exotic yarrow]

It takes time to get images

https://www.onlinemathlearning.com/tangent-secant.html

[solid pivot]

oh sorry

[opal dawn]

help meeeee

pleaseeee

im so confused

are these right

@exotic yarrow

[exotic yarrow]

You good?

[opal dawn]

pigeon can u help me

ive been on this for hours D:

[opal dawn]

?

[exotic yarrow]

[opal dawn]

:(

[exotic yarrow]

straight recall, though the proof is also quite short tbf

[slim plinth]

Also bearing depends on that

Like shape “c” and “z”

That’s what I found

What’s this now

[exotic yarrow]

Are you trying to draw a parallel line and do something with the alternate interior angles theorem or smt ... ?

if so, you should have these pairs of angles - your diagram has incorrect labeling.

[slim plinth]

It’s all alternate angles are congruent

Bearing has that

[exotic yarrow]

yeah see what I said above then

[slim plinth]

Okay

[slim plinth]

What are some advices of this bc I’m having a quiz on Friday

[feral canyon]

basics of vector algebra?

[slim plinth]

Never did

[feral canyon]

but i see some diagrams involve directions and may be magnitude and angles

so

ok dude my bad homie

[slim plinth]

It’s okay

[slim plinth]

Bearing is an angle

Btw

[stone lark]

help pls

[civic urchin]

You found how many turns each gear will do ?

[stone lark]

yes

The angle for the gear with radius 3 is 2π and the angle for the gear with radius 2 is 3π/2, right?

[queen glade]

How can i create laggy stuff in desmos

[low ocean]

Use sum, prod and floors func

Also trig xD

[queen glade]

I know sum and prod but whats floors

[spiral lodge]

Wdym laggy?

[queen glade]

Like very intricate stuff

[spiral lodge]

Oh nice I didn't know that word

[low ocean]

Floor(x) return the greatest integer less than or equal x ( fixed :p )

[queen glade]

Elaborate please

[spiral lodge]

Take a number x

Whichever you like

[queen glade]

Ok i choose 3

[spiral lodge]

Mmh ok

What is the greatest integer which is less than or equal to 3?

[queen glade]

Uhh

2??

[spiral lodge]

And why not 3?

[queen glade]

Cause im dumb idk what an integer is im in 6th grade

[spiral lodge]

Integer means natural numbers but their negative version as well

[queen glade]

Oh okay

[spiral lodge]

So 0, 1, 2, 3, ... and also -1, -2, -3, -4, ...

[queen glade]

I get it now

So i still pick 2

?

[spiral lodge]

@queen glade

[queen glade]

Oh

3 then

[spiral lodge]

Read this one more time

[spiral lodge]

Correct. Do you see why 2 wasn't correct?

[queen glade]

Because it wasnt equal to 3

[spiral lodge]

Yup awesome 👍

[queen glade]

Yay

[civic urchin]

I think that :
The gear with a radius of 3m does a 60° rotation (counter-clockwise)
The gear with a radius of 2m does a 90° rotation (clockwise)

[spiral lodge]

So the floor of 3 is 3 itself

What about the floor of, let's say, 7.3?

[queen glade]

7.3?

[civic urchin]

Since g3 did 1/6 of its rotation and g2 did 1/4 of its rotation

[spiral lodge]

Nope, 7.3 is not an integer

[queen glade]

Oh 7

[spiral lodge]

Yup perfect 💪

[queen glade]

Okay i get it thanks

[spiral lodge]

Yw

[queen glade]

So the floor of pi is 3?

[spiral lodge]

Exactly! Very well 🤗

[queen glade]

Yay

But how would i write a floor equation

[spiral lodge]

$\floor{x} = 3$

[somber coyote]

Alberto Z.

[civic urchin]

[3.2] = 3

[queen glade]

Okay

[low ocean]

Btw floor(-2.3) is -3

[queen glade]

Oke

[spiral lodge]

Yeah good observation

[low ocean]

Which is kinda funny

[queen glade]

Negative numbers get less the further you go

[low ocean]

exactly, good one

[spiral lodge]

Especially due to the fact that sometimes floor is compared to integer part. But this is true only for non negative numbers

[queen glade]

I have another question now

[grave pond]

Okay, so you take the largest of the two numbers on each side of the dot and preserve the sign, right?

[queen glade]

Whats ceiling

[low ocean]

You can probably guess it

[queen glade]

The integer above or equal?

[spiral lodge]

ceil(x) return the smallest integer greater than or equal to x

[queen glade]

Okay

[grave pond]

Ceiling is where floor is .

[queen glade]

Okay

[spiral lodge]

The lowest integer above or equal, yeah

[grave pond]

(As shown by Alberto's reaction).

[queen glade]

Okay

[grave pond]

You can also describe it as "if it's not already an integer, round up (for ceiling) or down (for floor) until you reach one".

[queen glade]

[tepid charm]

Fun puzzle: Solve for all solutions of $\tan x = x.$

[somber coyote]

LemmaLover

[grave pond]

I doubt the solutions of that have any nicer description than "the numbers for which tan x = x".

There's exactly one within pi/2 of every integer multiple of pi, and the distance between two successive solutions is always more than pi, but beyond that?

[vital bluff]

how do you even begin to approach this

solve for all solutions of x=x ahh

[civic urchin]

@stone lark do you still want to know how I found these values or no ?

[dense dawn]

who can help pls

[vestal basin]

sen is sine?

[manic grove]

yes

[vestal basin]

do you know trig identities

[stone lark]

I got the result that g3 rotates 1/6 if gears g4 and g3 are connected by a pulley.

But g4 is connected to g1 by a pulley, not to g3.

[civic urchin]

I didn't even notice that there was a g1

[stone lark]

Yes, sorry for the delay.

[stone lark]

That's why this problem has me so hooked.

[civic urchin]

What is the kind of connection g1 has with g3

[stone lark]

a common axis

[civic urchin]

G4-G1 is a pulley
G3-G2 is friction

[stone lark]

yes

[civic urchin]

Both will have the same angular rotation

Knowing that you can find the arc of the rotation

[stone lark]

well yes

but i still cant resolve it

😭 ✌️

[civic urchin]

You are complicating your life a bit too much

So 1- find the distance of rotation of G4

Or you can directly skip that
G4 does 1/8 of its rotation
Since G1 is 4x smaller it will do 1/2 of its rotation so 180° or pi rad

With that you can continue the problem

(I think that) I solved it if you want to see how I did it

[stone lark]

I would really appreciate it, bro

[civic urchin]

I'm sorry I had to do it on paper because my laptop is dead

Hope its readable

And understandable

[stone lark]

Oh, so the movement of P2 and P3 originates from point P

I thought they started from where their radios were initially marked.

[civic urchin]

That's what I've understood

[civic urchin]

That certainly make the problem a bit harder

[stone lark]

Well, they may not have framed the problem correctly; doing it the way you did makes a lot of sense.

[civic urchin]

It could've been explained a bit better I won't lie

G1 being on the rotation axis as G3 was mentioned nowhere

[stone lark]

yeah thats so weird haha

I'll stick with how you did it, now I'll continue practicing with other exercises, thanks a lot bro.

[civic urchin]

No problem and good luck

[dry root]

does anyone have a good practice book for trig problems

[velvet coral]

very good one

[willow kiln]

Most people don't know who created the hardest chapter Trigonometric Ratio

[trim stump]

who created the hardest chapter Trigonometric Ratio

[willow kiln]

Aryabhatta

Most prebably

[trim stump]

Most people also don't know who solved the general formula for cubes

Say who

@willow kiln Say who!!?

[hasty stump]

its not even that hard wdym 😂

its a piece of cake

[trim stump]

Tell that to Aryabhatta

[scenic mason]

Hi

[unique wren]

For finding point of concurrency of three lines can i just solve the lines the traditional way or should i use the matrix method

[scenic wyvern]

i am super stupid ive been slacking all year just using ai and getting good grades on my math stuff but now i have a huge geometry test in 4 days and i dont know a single thing on it and im gonna fail like no one has ever failed before im trying to learn the entirety of geometry in the next coumple of days can anyone help me or gimme like some pointers or some yt vid links on how to learn geometry cause im watching like the organic chemistry tutor and stuff and its just not clicking to me can smn plz help

[civic urchin]

Matrix with Gauss-Jodran would be faster

[civic urchin]

I think I can help you to go over what you don't understand
Also try to get the notes of your classmate if possible

[true compass]

Struggling with this question

Surface area I get is 22,000cm

It wants 5 shapes so 22,000 x 5 = 110,000

6m^2 to cm = 600 x 600 = 360,000

So it would be 1 but it's incorrect. I'm going somewhere wrong in the 6m^2 calculation but can't see it

[spiral lodge]

1 m² = 1 • (100 cm)² = 10000 cm²

6 m² is not (6 m)², rather 6 • 1 m²

[left depot]

use total surface area

u will get 2.2 msquare for one

so for 5 it is 11 msquare

then 11 divide 6 is 1.83

which is roughly 2 pots

[spiral lodge]

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.

[left depot]

sry my bad boss

[spiral lodge]

Don't worry, it's fine.
This isn't the case, but sometimes there are ultra lazy people who come here just to have their answers given, and we don't like to do that

[true compass]

my brain is not braining but if you had 2^2 it's 2 * 2 = 4

so surely it's 600 x 600 (600cm in 6 metres)

what am i doing wrong here

[left depot]

aa think more clearly

its 6 metre square

not 6 square metre square

or 6 metre whole square

[spiral lodge]

The square is only written in the m of meters, not on the 6

[true compass]

shhiiit

cheers guys

[tiny acorn]

What is this math slop

[fallow forum]

prob 114

[grave pond]

Opposite angles in a cyclic quadrilateral add to 180° due to the central angle theorem; from threre it's just iscosceles triangles and a bit of algebra.

[tiny acorn]

I could do it, but im too lazy

[grave pond]

Ah, then I'll be lazy too.

[tiny acorn]

First recognize that the double dashed isoceles triangles other angles are 38° and 104

Therefore the opposite angle to it is 76

The angle next to the 76° angle is also 76° due to isoceles

,calc 180 -(76+38)

[somber coyote]

Result:

66

[tiny acorn]

,calc 180-66

[somber coyote]

Result:

114

[tiny acorn]

And this is our result, x = 114°

[tiny acorn]

It is 114

[idle crypt]

hello i need someone teaching me calculus 1 chapter 3 please I HAVE A MID EXSAM

i need to understand it as fast as it gets

[cold elk]

#calculus

[grave pond]

Someone ought to have been paid to teach it to you in digestible chunks spread out over the semester leading up to the exam ...

Also, yes #calculus.

[slim plinth]

Wait

Use desmos geometry calculator

[vocal sentinel]

Logically it would be 114

[stray isle]

i js had a circles test today

ts is something we learned

but something like this would be too easy on the test

[tiny acorn]

Mf i found ts on youtube shorts

[stray isle]

same

but most of yt shorts is under 10 so there's probably like a billion wrong answers

[tiny acorn]

[opal coral]

for negative exponents? what about -2 lol

[low ocean]

~~ 1/arctan ~~

[silent plank]

don't think they ever get used,
people use reciprocal functions or fractions for that

[proper kelp]

Hello I am an IGCSE student 11th grade and im studying trigonometry and im looking for help. Its weird and makes no sense and I am just looking for help in making sense of trigonometry at this stage.

[silent plank]

Post specific questions you have issues with

[unique wren]

Be specific

[proper kelp]

Its the graphs of trigonometric functions chapter.

Can someone hop on voice chat if possible? I would perfer that

[hasty stump]

i can try

[sullen palm]

Umm anyone down to teach me trigonometry

Like dm me and then teach

[winter rapids]

<@&268886789983436800> hacked/scam

💜

[winged oyster]

maybe their teacher is bad

[tiny acorn]

how do i get my line to fit the grid line on desmos

[potent hollow]

whats the equation of your line

[tiny acorn]

sin(5pi)x=y

[tiny acorn]

Whats the equation of the closest grid line anyways

[potent hollow]

i think its 1/sqrt3 *x

[tiny acorn]

Wait no i meant to type sin(pi/5)x

[potent hollow]

tan30

[tiny acorn]

Huh okay

[graceful sable]

guys

i forgot how to differentiate can u guys help me recap rq I'll send basic qns

so it saves me time to do the basic questions myself I can skip to the intermediate ones

can someone help

[tiny acorn]

!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.

[tiny acorn]

Or go to #calculus

[graceful sable]

hi

[graceful sable]

okay thanks

[iron violet]

rate of change of a curve

[tiny acorn]

this

thanks btw

[vocal sentinel]

that is precalc i suppose

[grave pond]

"How to differentiate" is definitely calculus, not pre-.

[tiny acorn]

Uhh "differentiation" without context sounds liek calc

[vocal sentinel]

Differentiate is it supposed to be d/dx?

I believe that term is for year 11-12 maths

[sullen palm]

I’m kinda a desmos newbie so how do I get the circle graph

[tiny acorn]

Click on the wrench your should find the circular grid

[sullen palm]

Ok ty

[stone lark]

help 😭

I get 6.29 m^2

[sullen palm]

https://tenor.com/view/red-paper-coach-meme-gif-11221063440541197155

[stone lark]

oh srry

I forgot

[sleek ridge]

The problem assumes a simplified calculation

So the lengh of the pulleys are not considered

Solving it properly, you should get the masses' coordinates

put them in a matrix and define the area as the module of the determinant divided by 2

but as the problem expect you not to use analytic geometry, you simplify the quadrilateral, getting a trapezoid

[vivid stratus]

is it like find the area formed by n, after the belt move 1 metre

[stone lark]

That's how I did it

I'm sorry for the mess

[vivid stratus]

Oh I thought this

[stone lark]

yes its like that

[sleek ridge]

correct anwser is around 6,6m^2

[stone lark]

How did you solve it?

[sleek ridge]

to get the vertical diplacement, you need to see that each circle has a circular displacement(x in metres) and an angular displacement (theta in radians)

as the 4m circle has a 1m circular displacement, you can calculate the angular displacement

as the 2m circle is inside the 4m one, their angular displacement is the same

so you get the displacement you got

-1/2 and 1/2 for the left ones and -1/3 and 1/3 for the right ones

now you plot them in a cartesian grath and set a reference for y (in my case, the original mass positions)

so the y positions would be (-1/2)(1/2)(1/3)(-1/3)

to calculate the x positions, you use the pythagorean theorem to find the axis distance

it will be equal to sqrt(63)

I'll send an image, just a sec

So, to find the x positions you use the radius of each circle (x1 = 0)

the yellow line is the y and x axis

[void cedar]

GET OUT

[sleek ridge]

x2 = 4
x3 = 1 + sqrt(63)
x4 = 3 + sqrt(63)

so now, you have the four coordinates

use analytic geometry to find the area

are you familiar with analytic geometry? @stone lark

[stone lark]

Not much since I'm just starting to study it, but if you'd like, continue down that path.

I think I've already seen my mistake.

[sleek ridge]

there is a formula for every shape since you know its coordinates.

you use it and get around 6,6

idk why are the alternatives so different

[stone lark]

probably a bad printing in the alternatives

[sleek ridge]

they probably subtracted the diameter of the circles so you get a trapezoid

[wanton hawk]

What was it?

[void cedar]

spam

[wanton hawk]

https://tenor.com/view/batman-i-see-think-thinking-pondering-gif-4994872

[stone lark]

@sleek ridge bro thx so much

[sleek ridge]

anytime

[placid carbon]

https://docs.google.com/presentation/d/1bLAf3nwjLTRVSaQP8CaBvY-f4EdJAOuEnTChdxlMkQE/edit?slide=id.g4dfce81f19_0_45#slide=id.g4dfce81f19_0_45

hi im really stuck on this lesson and i was wondering if anyone could teach me

[minor heron]

wouldn't it be ||29 and 151||

[slim plinth]

When I did transversal rule

[minor heron]

how did u use transversal rule on ts

i dont see parallel lines?

[slim plinth]

I constructed a parallel line

[minor heron]

oh

i just did um

so i got 122 as that angle

[slim plinth]

Did I say it correct

I “constructed”

[wanton yacht]

I got x = 26

Correct me if I'm wrong

[slim plinth]

Then it’s right angle

[minor heron]

like x and the like other angle next to y are equal

cus isoceles triangle

[wanton yacht]

Yes

[minor heron]

58 is like

the exterior angle

[slim plinth]

Like this is this correct

[wanton yacht]

So Jake, have you learned that the sum of a triangle's angles has to be 180 degrees?

[slim plinth]

Yes

[minor heron]

i got like 58/2=29. um. maybe they meant to divide it by 2 idk

[wanton yacht]

Oh wait I divided incorrectly, my bad

I was thinking 52 in my head

[minor heron]

x + 32 should not add up to 90, as that is not a right angle

[slim plinth]

Also ik the proof is side then angle

[minor heron]

ah

[wanton yacht]

Okay. Because that triangle on the left is isoceles, then you know what?

[civic urchin]

Did you find X = 58 ?

[slim plinth]

Yes

[wanton yacht]

Because I'll tell you, x is not 58 degrees, but 58 degrees is important somewhere

[slim plinth]

It’s this

Then the outer part is “58-x”

[minor heron]

If you think about it... if you have 122 as one of the angles of the ISOCELES triangle

What are the other 2 angles in that ISOCELES triangle... hmmmm

[minor heron]

umm...

that's irrevelant

[slim plinth]

Then it equals 90

[minor heron]

and also incorrect

x isn't 58

[slim plinth]

My idea was make it 32+58+(58-x)=90

[minor heron]

Uhh

well 32+58=90

90+58-x=90

148-x=90

....err... yeah this is....

questionable

just... it isn't necessary

[slim plinth]

x=58

[minor heron]

OKAY lets say x=58

then what would the bottom right (theta) be

if 122+58=180

then theta would be zero

[wanton yacht]

Dawg

[slim plinth]

Angles cannot be 0

[minor heron]

exactly

so x cannot be 58

[slim plinth]

So would it be 58+x

[minor heron]

um...

.....

how about we go back to thinking

if you have 122

in an isoceles triangle

[minor heron]

the other two angles add up to 180-122 and should each be the sum / 2 cus they are equal

[wanton yacht]

If you just consider the one triangle, you end up with this

[slim plinth]

Ohhh

[slim plinth]

Can I make a right triangle then figure out

[wanton yacht]

Do you see how the angles are the same because the triangle is isoceles?

[slim plinth]

Yes

[minor heron]

a right triangle is unnecessary to solve this problem...

[wanton yacht]

Maybe you could, but you're making it harder than it can be

Solving math problems is often about breaking problems into smaller chunks.

[wanton yacht]

Once you figure out that one angle is 122 degrees, and you identify the triangle as isoceles, you should only look at those two facts. The rest of the problem essentially disappears until you figure out what x is. At that point, you're just left with this

I mean, I think what you're saying is clever, and technically accurate, I see what you're saying. Whether you can solve for x, I mean it introduces another unknown because you have

x + y = 58

32 + x + y = 90

Solve for x, but you don't know y

You know what I'm saying?

[slim plinth]

Is it because the proof “if side then angle”

[minor heron]

there aren't any congruence theorems if thats what you're saying?

[slim plinth]

Yea

[slim plinth]

How did u get x+y=58

[minor heron]

well

wait nvm

wrong angle

[wanton yacht]

Just because of how you setup your angles

[slim plinth]

Then the 32+x+y=90

Is it it 90-58

So I will need a right triangle

[slim plinth]

I got y=0

[wanton yacht]

y is not 0 in that case

[slim plinth]

Bc I substitute x=58-y

[wanton yacht]

Let me show you what y I'm talking about

[slim plinth]

32+(58-y)=90

[minor heron]

so the idea is x = 58-y, 32+58-y+y=90. y isn't 0, just canceled out

that woul just lead you to 90=90 btw

which doesn't help

thats why a right triangle is unecessary

cus it doesn't help w/ the problem

[wanton yacht]

Oh wait

I shouldn't use y, it's already being used

x + z + 32 = 90

[slim plinth]

x+z=58

[slim plinth]

^

[wanton yacht]

Without more information, I'm not sure you could say anything about x and z

That's why I'm saying focus on these facts instead
(1) One angle is 122 degrees
(2) Triangle is isoceles, and because of that, the other two angles are equal (theorem)
(3) All angles add up to 180 (theorem)

From that you can make some clear deductions

[slim plinth]

Straight lines are 180 degrees

x=29

[cinder spear]

How did you get that z is 58??

[wanton yacht]

I didn't

That was his labeling

[minor heron]

mmhm

[slim plinth]

Yay

[cinder spear]

Yep, there you go

[slim plinth]

That’s the theorem I used

When I did bearings and angles of depression

[minor heron]

right! yey

[slim plinth]

I used that

[minor heron]

my teacher never taught us whatever bearings are but nice

[slim plinth]

It’s an angle of on a compass

[minor heron]

oh

[slim plinth]

N,E,S,W

[minor heron]

that sounds kinda uh

useless

ngl

[slim plinth]

Example of what I did

(degree)+(min/60)+(sec/3600)

[stray isle]

the parallel lines aren't neccesary tho

[slim plinth]

But this can help

[minor heron]

it vould work but help is sort of well…its faster probably and easier just thinking with isoceles triangle

[slim plinth]

Ok

[torn gulch]

erm

why is there a right triangle

the sum of angles in a triangle is 180

it's an isosceles so the angle X is also the angle there on the bottom so (180-58)+2X=180 => 2X-58=0 => X=29, Y=180-29-151? am I stupid or something

[torn gulch]

watch a youtube video

[wind solar]

try

phi means diameter

[upper echo]

looks like there is not enough info, since the width of the box is unknown

[wind solar]

yeah thats why i assumed that the corner is connected to both centers of the circles

[subtle glacier]

Doesn't seems right.

[wind solar]

idk any other way around it for now

[wind solar]

anyone

[wanton yacht]

There just simply isn't enough information, lol. If you assume the angle between the centers of the circles you can figure it out, but otherwise, no

[slim plinth]

I just want to make sure that if it’s Isoceles is it two angles and two sides are congruent

I had these
• If sides then angles. 
• If angles then sides. 
as part part of my geometry proof

[wind solar]

yeah there's no way to get a concrete solution, you can only assume

[minor heron]

the sides are marked as congruent already, no?

[slim plinth]

Yea

[brisk fog]

Hey guys I am stuck on my math hw and need some help 😓😓
I’m mostly stuck on how to find the angles of the triangle and would appreciate a in-depth explanation if possible ;;;

[minor heron]

Have you learned trig laws? like law of sine/cosine

cus since theres no right angle, im pretty sure u have to use ... law of cosine, i believe

[brisk fog]

Not yet but I tried turning it into a right triangle like this to solve it. Would this help or would I have to search up law of cosine 💔

[minor heron]

Hmm, let me think if that would work.

[tiny acorn]

hint: this might help

[minor heron]

Yes, it should be solvable...

As pointed out, draw an altitude in the middle

You can solve for the height using basic trig ratio - then find the other height using the total (319) minus the value that you find