geometry-and-trigonometry

[keen osprey]

$cos(x) = \frac{1-\sqrt3}{2}$

[twin crag]

so i just take arccos of that?

[keen osprey]

yes

[somber coyote]

Aman

[twin crag]

wait i get the same thing though

[keen osprey]

wdym

[twin crag]

oops sorry i didnt read the question

yeah 1-sqrt3/2 is correct

it was asking for all values of cosx 😭

wait how do i get a correct answer for cos(theta)

i got cos(x)=0

[keen osprey]

double angle formula for sin

[twin crag]

2sinxcosx=tanx

and then i put tangent in terms of sine and cosine

so 2sinxcosx=2sinx/cosx

then multiplied by cosx to get 2sinxcos^2x=2sinx

[keen osprey]

can divide 2sinx on both sides

[twin crag]

yep i did that

wait

no i idivided by 2sinx

yeah

[keen osprey]

yes thats g

[twin crag]

i got cos^2x=1

sqrt

so cosx=1

then took arccos

but 1 is wrong it says (its asking for values of cosx)

or its not the only value

[keen osprey]

cos(x) could be -1 as well right?

[twin crag]

for cos(0)?

OH

i see

[keen osprey]

cool

[twin crag]

ty

[nocturne remnant]

E must be the midpoint of the chord FH because EO is perpendicular to FH

[hot token]

I'm trying to help my niece with Geometry but I'm not sure what these questions are asking us to do? Can somebody help clarify?

[twin crag]

im pretty sure for the first one it means reflect pentagon QRSTU over line B to form pentagon Q'R'S'T'U'

[twin crag]

and for the second one its reflect ABCD over line K to form A'B'C'D

[hot token]

that's what i thought too but i'm not sure how to do that accurately without a coordinate system

[autumn gorge]

I need help with this

i need to solve in interval 0 < theta < 2pie

[weary torrent]

what's the problem?

[vernal pilot]

hey man, is this supposed to be $\cot(\theta+1)\cdot\csc(\theta-\frac{1}{2}=0$

or $(\cot(\theta)+1)\cdot(\csc(\theta)-\frac{1}{2})=0$

[somber coyote]

TheLord26

[weary torrent]

seems like it

[vernal pilot]

i hate LaTeX

im going under the assumption that it is the second one

$\cot(\theta+1)\cdot\csc(\theta-\frac{1}{2}=0$

[somber coyote]

TheLord26
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vernal pilot]

WHY DOESNT THIS WORK

[thick fable]

$\cot(\theta+1)\cdot\csc(\theta-\frac{1}{2})=0$

[somber coyote]

🐱!Yajat! 【Catfan1398】🐱

[weary torrent]

what are you trying to do

[vernal pilot]

help this dude with a math problem, and then getting severly side tracked with LaTeX and then forgetting to help the dude. dammit

[vernal pilot]

Okay, so I'm assuming that the question is asking for (cot(theta)+1)(csc(theta)-1/2)=0

[weary torrent]

i feel that this is the question

we cant't do anything about until the one who asked it comes online

[vernal pilot]

immajust do both

[weary torrent]

what is there to do in this

it's already solved

[vernal pilot]

what do you mean?

[weary torrent]

the csc of any real angle can't be less than 1 so we can just neglect it

[vernal pilot]

the dude obviously doesnt knwo that bro

so yes, there is a way to help him

[weary torrent]

ohh

sorry my bad

[vernal pilot]

So to start, what you want to do is see that either $(\cot(\theta+1))$ or $(\csc(\theta)-\frac{1}{2})$ must equal 0. From there, you can test to see whether $\cot(\theta)=1$ and $\csc(\theta)=\frac{1}{2}$ turn out to be.If you start with $\csc(\theta)=\frac{1}{2}$ you will find that it has no solutions $\in{\rm I!R}$.

[somber coyote]

TheLord26

[vernal pilot]

i could see the LaTeX gods typing in the chat, oh god

This makes logical sense, if you see how $\csc$ is defined, that being $\csc(\theta)=\frac{1}{\sin(\theta)}$, and as there is no way $\frac{1}{n}$ can ever equal 0, it must mean that $\csc(\theta)$ can never equal $0$.

[somber coyote]

TheLord26

[vernal pilot]

So, you should test $\cot(\theta)=-1$, using some simple algrebra, you can rearrange to make the equation $\arctan(-1)=\theta$, and if you know your exact value triangles, or you use a calculator, you will find $\theta=\frac{3\pi}{4}$.

[somber coyote]

TheLord26

[vernal pilot]

You can double check the answer by plugging theta into the original equation if you want. But if that was the incorrect equation to being with, and you needed to the other equation, (cot(theta+1))*(csc(theta-1/2))=0, well, dam ig?. @autumn gorge

[haughty lichen]

Thankyou guys @smoky jetty @heady scaffold @nocturne remnant

[smoky jetty]

does that mean the intersection of any chord perpendicular to a radius is always the midpoint the chord?

[haughty lichen]

Yes , a perpendicular through centre to chord bisect the chord hence the intersection point is the midpoint.

If you meant the radius is perpendicular to the chord (as i couldn't understand the sentence in the message) , it is not

[nocturne remnant]

Yes and can you not ping reply me for everything

[smoky jetty]

ahh mb

[smoky jetty]

i see, but if it were the radius that's perpendicular to the chord, would it also bisect it?

rusty on quadratics stuff, but I got 0 = (x+16)(x-4), how do u get the desired value of x for it?

[haughty lichen]

Radius goes through the chord so if radius is perpendicular to chord it also bisect the chord.

[smoky jetty]

alrighty, thanks for the clarification!

[haughty lichen]

X has to be 4 as it can't be -16 so .

[smoky jetty]

so u just make the -4 positive?

[haughty lichen]

Um its not negetive 4,

[smoky jetty]

in the RHS?

[haughty lichen]

Yes the expression contain -4 but for equation to become zero x has to be either -16 or 4. Since x can't be -16 so it has to be 4

[smoky jetty]

but for equation to become zero x has to be either -16 or 4. Since x can't be -16 so it has to be 4
sorry but I didnt quite get that. Could u elaborate it further?

[haughty lichen]

Could you once revise the ' zero product rule ' from the book.

On the rhs there are two expressions. One of them has to be zero so their product will become zero.

So if x is 4 there is only possible situation for the Rhs to become zero.

[smoky jetty]

oh i see now

def didnt learn that in alg before, lol

So I looked up the rule u said. And wrote my understanding upon it

since the chord lengtb camnot be negative, we choose 4 right

so essentailly in the LHS, we must make either factors =0, and since the factors are in the form of (x+a), we get the inverse/negative of a to cancel it out and thus becomes (0) as a factor

then that inverse of a, will be a value of x, then same concept applies to the other factor, giving us the 2nd value of x

so yeah, thx for the help!

[heady scaffold]

idk what i did but yw

oh ya this is that problem

[smoky jetty]

welp I did a simple proof to see it better, and found out that if the radius is perpendicular to chord and we make two triangles out of the chord segments and the radii, then the two triangles are coungrent by SSA (for right triangle)

[upper karma]

help pls

<@&286206848099549185>

[keen osprey]

draw a diagram

label the relevant sides in terms of the information given

[upper karma]

thats what I have trouble with

[keen osprey]

maybe send a photo of your diagram if possible?

[smoky jetty]

it says that it's side AB that is bisected perpendicularly

[upper karma]

but

it intersecs BC

[smoky jetty]

yeah

[upper karma]

so how would it look?

im confused

[smoky jetty]

hmm this is what I got apparently

smth like this?

[upper karma]

ok

[smoky jetty]

did u get the answer? just curious

[smoky jetty]

Hoping this is correct (dont click if u havent solved it yet) ||By connecting A and D, we can infer that AD=BD (SAS congruence), making triangle ABD an isosceles. Hence, the Perimeter of triangle ADC = AC+CD+AD or AC+CD+BD = x. Now, since AD=BD, and BD is part of triangle ABC's perimeter, we can say that the Perimeter of ADC + AB = P_ABC = x + 12. Therefore, AB= 12.||

[upper karma]

yea I was thinkin the same thing

i wanted to clairify

[smoky jetty]

sure

[hazy moat]

lemme get this straight: when the pronumeral is the denominator, you just rearrange the equation to divide the numerator by the trig function, but when the pronumeral is the numerator you multiply it by the trig function?

[grave pond]

What is a "pronumeral"??

[native jetty]

an unknown such as a letter

you rearrange to make A the subject by multiplying or dividing

[hazy moat]

so i am correct right

[hasty kelp]

If you are asking question c and want to know the length of A, then it is correct.

[north wave]

There is an SSA congruency??

I have only known AAS SAS SSS ASA

And RHS ofc

Yep

[north wave]

Me talking abt this

Everything else is correct except idk there is an ssa congruency

[nocturne remnant]

Didn’t you use SAS?

[hazy moat]

right ok

thx

[autumn gorge]

why is this possible to do and why do you do this?

[smoky jetty]

oh right, yeah it's SAS btw

[smoky jetty]

afaik SSA isnt a congruency criterion for triangles except for right triangles

[smoky jetty]

seems right. Just remind yourself that you'd have to transform that equation into smth that would isolate that unknown value either in RHS or LHS

[hazy moat]

ye ok

[north wave]

Lol

[jagged venture]

in the trapezoid ABCD, with the bases AB > CD, we note M,N the means of AD, respectively BC. The bisectors of angles A and B intersect MN in E and F, respectively. Prove that EF= AB+CD/2 - AD+BC/2

AB+CD/2 is equal to MN

and AD+BC/2 is equal to AM+BN

So we basically have to prove that AM = MF

and that BN=NE

[wheat sinew]

what is a polynomial

[heady scaffold]

an equation with multiple terms

[wheat sinew]

wow

[heady scaffold]

what?

[nocturne remnant]

That is a very bad answer if I’ve ever seen one

[heady scaffold]

😭 im sorry

[real widget]

An equation in the form ax^n + bx^(n-1) + ... + k

Like for example quadratics like x^2 + 2x + 1 are polynomials but so are cubics (like x^3 + 6x^2+x-4) and you can really have any power of x in there

[heady scaffold]

ya

and n is called the degree of the polynomial

[real widget]

ax + b is also a polynomial and so is just a constant "a" but I'm not sure if that's relevant to your specific needs or not

[compact thistle]

What is 1 + cos(x)?

an exo saying it is cos(x/2) which is kinda weird, no?

[thick fable]

no 1+cos(x) can not be equal to cos(x/2)😭

[compact thistle]

EXACTLY, I have been solving an exercise in physics and when I looks at the solution if I find it...

@thick fable Look at that:

[thick fable]

is this somehow related to circular motion

@compact thistle

[compact thistle]

Yep.

[thick fable]

what concept is this i dont get it, also what is $\vec{V}_{\theta}$?

[somber coyote]

🐱!Yajat! 【Catfan1398】🐱

[compact thistle]

angulaire velocity.

cuz:

[thick fable]

isnt angluar velocity $\omega$

[somber coyote]

🐱!Yajat! 【Catfan1398】🐱

[thick fable]

and also how can the time derivative of the velocity vector be equal to the velocity vector

its just too confusing

[compact thistle]

he made a writing mistake then fam.

[thick fable]

it should be acceleration tangential

[compact thistle]

$\ aren't Vc and omega difference?$ Because: $ Vc = frac{ds}/{dt}$ while $omega = frac{dteta}/{dt}$ ?

[thick fable]

yea they are different

[compact thistle]

$\ aren't Vc and \omega different?$ Because: $ Vc = frac{ds}/{dt}$ while $\omega = frac{dteta}/{dt}$ ?

[thick fable]

one is angular velocity and another is linear velocity

[somber coyote]

Анатолив.

[compact thistle]

Oh my lord.

[thick fable]

also what is $\vec{e}_{\rho}$ doing there in the equation and wtf is even this

[somber coyote]

🐱!Yajat! 【Catfan1398】🐱

[compact thistle]

$\vec{e}{\rho} $ = $\vec{u}{\rho}$ it is just a notation.

[somber coyote]

Анатолив.

[thick fable]

what is u now😭😭

yall use very wierd notations

[compact thistle]

yeah because u studied english math and I studied french.

Same but different notaions.

it is just like i and j in carteesian coordinates.

ngl you are giving me heart attack, I feel like all my notes are wrong...

like is $\vec{a}_{\t}$ = R . TETA ?

[somber coyote]

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

[thick fable]

like is $\vec{a}_{t}$ = R \cdot \theta$ ?

[somber coyote]

🐱!Yajat! 【Catfan1398】🐱
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[thick fable]

you mean this?

[compact thistle]

yes.

[thick fable]

no

it is $S=R \cdot \theta$

[shut oxide]

Help me out with proofs please I almost finished it im just stuck

[somber coyote]

🐱!Yajat! 【Catfan1398】🐱

[thick fable]

its better if u just post 2 pics of it

[shut oxide]

one sec

[shut oxide]

helpppp

[thick fable]

wait 15 min of my lecture is left

[compact thistle]

then if I'm not wrong, then $\vec{a}_{t}$ = R \cdot the derivative of \omega$ ?

[somber coyote]

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

[compact thistle]

the derivative of omega.

[thick fable]

okay so what have u tried so far

yes $a_t=R \cdot \frac{d\omega}{dt}$

[somber coyote]

🐱!Yajat! 【Catfan1398】🐱

[thick fable]

as $\frac{d \omega}{dt}=\alpha$

[compact thistle]

what is the notation of the derivative of omega?

[somber coyote]

🐱!Yajat! 【Catfan1398】🐱

[compact thistle]

BRAH

Please, is there a pdf or a book pdf that shows all laws only?

[thick fable]

u mean formulaes?

[compact thistle]

y

[thick fable]

oh

i dont know about that sorry

i means its simple, the derivations to these formulaes are hard but if u just want the formulaes u can just search them on google, ur sstudying undergraduate circular motion so i dont have much idea about what topics r in ur syllabus

[shut oxide]

I tried everything that makes sense

You can look at my steps

[thick fable]

i really dont wanna look at all those im lazy what what u can do is, u can prove the triangles BAC and DCE congruent, with the help of RHS congruence rule, from that u can conclude that angle BCA and DCE are equal

from this can u tell me the relation between

angle BCF and DCF?

[compact thistle]

I appreciate your efforts, and I indeed already downloaded a pdf and I'm learning currentyl.

[lapis belfry]

I don't know if this is the right place to ask this, but I don't know if {4, 4/2} is a regular polyhedron or not.
Like, it's face-, edge- and vertex-transitive, but according to wikipedia, the vertex figure also has to be a regular polygon (which it is not)
So it meets one definition, but not the (supposedly equvalent) other definition. So either it isn't f,e,v-transitive (and thus not a regular polygon), or the definitions aren't equivalent, and it is a regular polyhedra, but I cannot find it on any list.

[shut oxide]

@thick fable still have no clue

[heavy oar]

16?! Most of my proofs were only 2-6 long

[shut oxide]

Yes lol I managed to figure it out

[red flax]

Yo can someone help me figure out how too use the distance formula on this I forgot how too on this equation 😭

Would be appreciated

[warped badger]

u would just plug the coords for the endpoints of each segment

for example for BD it would be sqrt((-5-(-2))^2 + (-2-1)^2)

[red flax]

ohhhh bro

I am so like

stupid I swear

idk why my brain didnt calculate that

[warped badger]

lol no worries

i hate distance formula so much lol

so many bad geometry memories

[red flax]

man I hate geometry 😔

I accel at math but not graphing or geometry there my weaknesses

[warped badger]

same

i just dont like proofs in geometry

[red flax]

😭 I dont like anything about it

well I like a little

but not much

[obsidian hemlock]

Can anyone figure out (xiv)??

[hazy moat]

anyone know how to help me with this? im an advanced 7th grader doing year 8-9 work

i nearly cried doing this today

[vagrant onyx]

You can use the pythagoras theorem

The part of the straw inside the glass, the height of the cylinder and the base diameter of the cylinder form a right angled triangle

[woven estuary]

can someone help i barely understand triangle proofs

idek how to start

[vernal pilot]

$c=\sqrt{a^{2}+b^{2}}\newline c=\sqrt{2.5^{2}+(4\cdot\sqrt{6})^{2}}\newline c\approx10.11\newline10.11+5=15.11$

[somber coyote]

TheLord26

[thick fable]

okay do u know how do we prove any 2 triangles congruent

[woven estuary]

u mean like the SAS ASA AAS stuff?

[thick fable]

yes

[woven estuary]

yes

i just cant find the connections sometimes

to prove theyre congruent

[thick fable]

well u know that VW and UX are parallel

and VZ WY are perpendicular on it

[woven estuary]

yea

[thick fable]

what can u say about thier lenthgs then

[woven estuary]

its a rectangle

and the opposite sides r congruent

[thick fable]

yes

so you know that VZ is congruent to WY and UV to WX and UZ to YX

then can u prove this triangle congruent?

[woven estuary]

SSS

[thick fable]

yes

and then from CPCT u can prove those angles equal too

[woven estuary]

ooh

okay

tysm

can u help me w another problem?

i know ur supposed to solve the big trangles first

i think

oh wait

all the given stuff

[thick fable]

one angle in common if u see carefully

[woven estuary]

and angle w is congruent to angle w because of reflexive

so how do i go from there?

[thick fable]

well u see angle W is common in both the triangles

[woven estuary]

yes

[thick fable]

then what the part troubling you

you have 2 information given in the question itself

[woven estuary]

hmm

well we have one angle of the triangles given already

and theres vertical angles

then what

oh

[thick fable]

well you see youve given a side congruent

[woven estuary]

yea

so vz is congruent to ux

?

[hazy moat]

did that

[hazy moat]

sorry waht?

[thick fable]

no and why would u need that

[woven estuary]

idk

cus sides

[thick fable]

well u dont need that

[woven estuary]

oh

then what

[thick fable]

u have to prove these 2 triangles equal

congruent*

ASA?

what do u think about it

[woven estuary]

oh wait

i thought we were trying to prove the 2 tiny triangles congruent

oops

nvm

tyy

[thick fable]

no

np

[woven estuary]

yeah i read it wrong

[thick fable]

what is part troubling you

[hazy moat]

help again pls-

[hazy moat]

doesnt matter now i figured it out

im doing a different one now

[thick fable]

you are writing the angles wrong

[woven estuary]

is this right

[thick fable]

,rccw

[thick fable]

yes

[woven estuary]

okk tysm

[thick fable]

wait i dont know what reflexive are but they common angles

[woven estuary]

my teacher said its just used to prove something is equal to itself

he didn't really explain much other then that

[thick fable]

oh ok

then its right

[thick fable]

you need to write angle 1 congruent to angle 2 thru CPCT

[hazy moat]

then what is it cos i dont understand it

[thick fable]

why are u doing that question then

[woven estuary]

dont you still have to prove that the small triangles are congruent? or no

[thick fable]

if your school havent taught u yet

[thick fable]

no

[woven estuary]

oh

[woven estuary]

omg r u doing proofs too

[woven estuary]

so how would i continue

[thick fable]

your already done

youve proved those triangles congruent

[woven estuary]

i only proved the big ones

[thick fable]

now u know what is meant by CPCT?

[woven estuary]

yes

[thick fable]

yea that what is needed

[thick fable]

then u can just use that to prove the angles equal

[woven estuary]

so i would just say theyre equal due to cpctc

[thick fable]

ye

[woven estuary]

thats all right

[hazy moat]

its an entrance ticket, meanign taht it gives u a preview of what ur gonna learn

[vernal pilot]

Use Pythagorean thereom, and then add 5 to it to account for the additional length of the straw.

[thick fable]

.

[unreal sinew]

shouldnt this equal 81 degrees?

finding the sum of the interior angles?

nvm i did smth wrong

[hazy moat]

i did that

[vernal pilot]

Not a triangle

[spice escarp]

i need help

it says to prove that triangle BEC is congruent to triangle DEA

[thick fable]

!status

[lime crown]

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

[thick fable]

nice pfp

[spice escarp]

2

thanks

[thick fable]

well could you show me the question?

like is there anything given congruent

[spice escarp]

alright

[thick fable]

okay so you need need to write all those steps u wrote after the step 1

they r of no use actually

well

u know what vertically opposite angles are?

[spice escarp]

no

[thick fable]

something like this

not sure if that's visible

better one

can u do ur question now

[spice escarp]

ohh ok

[sudden glacier]

yo that happens to be what im doing in geometry atm

im very bad at it though

[thick fable]

i see a lot of ppl here facing problems in these provings, i means its so simple if you have the knowledge about the relations between the angles on a transversal

and the rules ofc

[sudden glacier]

Can anyone help me with this(explain it like a baby please)

[thick fable]

not the best place to ask such physics question but okay what have you tried so far

[sudden glacier]

idk

i dont know how to solve it

also how do I apply for another channel like physics or engineering

I was thinking cos30-10 for the left one

[thick fable]

you wont be able to solve this if u dont know about vector projections

[sudden glacier]

so what do I do

[thick fable]

hmm

other than that one way I could think of is by using lami's theorem

[sudden glacier]

hold on brb

[thick fable]

you can check #old-network and find the physics server there

[sudden glacier]

is there an engineering server

[thick fable]

there are specified engineering servers

[solemn hollow]

can someone help me with geometry please

#geometry-and-trigonometry

[thick fable]

@solemn hollow what have u tried so far

[solemn hollow]

i tried drawing a triangle and labeling the sides and the angle measure

[thick fable]

yea

show that please

[solemn hollow]

this is what i did

its messy- sorry

[thick fable]

ok do u know that the angle oppsite to the longest side is the largest?

[solemn hollow]

yes, i know that

[thick fable]

so you that angle ACB should be greater than greater than 70 degrees

so

it can just be a lil bit greater than 70 too

so for angles A only 40 degrees are left

less than 40 precisely

[solemn hollow]

is it bc 70+70=140 and 180-140= 40?

[thick fable]

yes

[solemn hollow]

so the highest it can be is 40?

[thick fable]

and that angle C can just be lil greater than 70 degrees too

less than 40 actually

A being 40 means

C and B ar both 70 degrees

[solemn hollow]

wait i have a question.. how do i figure out the lowest it can be

[thick fable]

which cant be

ok

like u know that B is 70

[solemn hollow]

oh that makes sense

[thick fable]

yea

so A cant be 40

because increasing C will decrease the angle of A

[solemn hollow]

so it has to be less than 40

[thick fable]

yea

[solemn hollow]

then what does it have to be greater than?

[thick fable]

sorry i dont get what u r asking

[solemn hollow]

like a range you know "< angle < " in that form it would be <A<40 but what would be on. the other side

[thick fable]

well u know it can obviously cannot be 0

[solemn hollow]

oh-

thank you

[thick fable]

np!

[solemn hollow]

can you help with the other one please

[thick fable]

well its the same concept

u know that Z has to be the smallest

that is less than 30 degrees

so u know that sum of the angle Z and Y is less than 60

so similarly u can say that

X can be greater than 120

[solemn hollow]

ohh that makes a lot more sense

i understand it

thank you so much!:)

[thick fable]

oh i just realised you love messi

i shouldnt have helped you

[solemn hollow]

oh damn-

[thick fable]

[solemn hollow]

my baddddd

[thick fable]

change it to ronaldo🔫

[solemn hollow]

noooooo

[verbal sluice]

can anyone help with trig ?

its this one

[frozen ocean]

,tex .recip trig

[somber coyote]

Akira E-Girl

[frozen ocean]

you need to use these^

but please don't dm me for your question

[sick kraken]

Hmm

When do you learn that

Algebra 3?

[verbal sluice]

this is trig i managed to figure it out ! im first year in college

[sick kraken]

Oh ok

I haven’t gone through it that’s why im asking

[verbal sluice]

all good

[woeful python]

anyone know how to do this?

[silent birch]

Can someone tell me how parametric points on parabola come from?

like what's (at^2, 2at)

[steady mason]

where do i ask about kinematics

i have a question about somethign

[fickle rose]

don’t ask to ask just ask

[woven estuary]

how sid he get this answer

wait

am i dumb

[upper karma]

remember that the sum of all angles in a triagle equals 180

for this we construct a formula

solving for the first triangle as an example

$x=180+(8y-7)+(\frac{6y}{5})$

[somber coyote]

Mr. Macro

[upper karma]

this might help

[woven estuary]

yes tyy

i didnt see that the big one was isosceles

[fickle rose]

can someone explain to me both the geometric mean theorem for altitudes and the geometric mean theorem for legs?
didnt get it while i was in class and the theorem sheet confuses me

[exotic yarrow]

<@&268886789983436800> a question from the current AMC (which is happening until the end of today)

[obtuse field]

Wait, how do you know?

[exotic yarrow]

The testing portal has a very distinct format

And this question doesn’t show up on any of the practice tests on said portal

Btw they also posted this in a help channel

[obtuse field]

ah

banned

[quiet sandal]

does that mean it was confirmed question?

[upper karma]

Hello! So we had a test today, but i cant seem to move after it. So basically, there was a problem where you need to calculate the area of the triangle. V and B were unknown at the start of it. Then, i found out if you make it into 2 smaller triangles, you get the area of 84cm2. But if you calculate it as a whole triangle it is 85cm2. All my classmates are saying it's 84, who's in the right?

Apologizing for bad image

Since b is 10 and a is 17, so the area should be (10x17) : 2

Which is 85. But if you put it into two triangles

so it would be 8x15 : 2 + 6x8:2

Which would be 60 + 24

Which is 84cm2

how??

[urban ivy]

Hi

I can't read the image, could you write it more clearly?

[upper karma]

Sure 1 second

So v2 is 289-225 which is 64

so it is 8 centimeters

[urban ivy]

Yes

[upper karma]

Then we calculate b and that is 8x8 + 6x6

Which is 36cm + 64

so 100

And then we calculate the area of the triangle

Which is ab : 2

So 10 x 17 : 2

which is 170 : 2

So 85cm

Right?

[urban ivy]

Is AB =21?

[upper karma]

a is the one on the top right

The side on the top right

[urban ivy]

Yes

[upper karma]

But if we take it as 2 triangles

which are

So 8 x 15 is 120 : 2 is 60 right?

• 48:2

which is 24

is 84cm2?

It does it was like that

it split the side to 6cm on the left

and 15cm on the right

How could it be wrong

its just calculating an area

[urban ivy]

My bad, I copied the values incorrectly

[upper karma]

its nothingg

Everyone got 84cm2 but i was the only one to go calculate it without splitting it into 2 triangles

i have no idea what is happening

even google

[urban ivy]

My friend

That isn't a right angle

[upper karma]

what.?

[urban ivy]

17,10 and 21 can't be the sides of a right angled triangle

17^2+10^2≠21^2

That was what i was saying

[upper karma]

Ohhhhh i didnt understand you

[urban ivy]

The perpendicular doesn't split the side as 6:15

[upper karma]

so 10 must be incorrect?

[urban ivy]

The question is wrong

[upper karma]

i swear it was like that

[upper karma]

thank you for explaining

i swear it is something wrong with it

[urban ivy]

You either copied it incorrectly or you're getting a bonus

On your test

[upper karma]

yup i hope so

i think i didnt copy it incorreclty

[upper karma]

Ironic

because they all got 84cm2

So it must be like that

should*

[urban ivy]

Yeah, but if two sides of your right angle triangle is 17 and 21, the third side is √152 not 10

[upper karma]

Exactly

[urban ivy]

There is no right angled triangle with one side being 17, hypotenuse being 21 AND the perpendicular splits the hypotenuse as 6:15

[upper karma]

so if it is like that it is a win win basically?

Even if i calculated it wrong?🤣

[urban ivy]

If you copied the question correctly your teacher should give you free marks

[upper karma]

Niceee

[upper karma]

Im 10000%

[urban ivy]

Turns out everyone was wrong

[upper karma]

lol

[upper karma]

Lmao

[upper karma]

How can i prove it to the teacher if he says im speaking nonsense?

like can i draw it somehow or something?

[urban ivy]

show him that b=10 by your method, then apply Pythagoras on the whole triangle ABC and show him that b=√152.

So either √152=10 or the perpendicular doesn't divide the hypotenuse as 6:15

Ask him to choose

[smoky jetty]

is CA and the perpendicular line to AB the unknowns?

yeah the given values arent accurate

[smoky jetty]

u can argue that the ratios of the given side lengths of the triangles arent in proportion, when they must be since the 3 right triangles must be similar

[hard coral]

I could use some help with this here. This is calculating light ray reflections using vector fields to simulate light rays. The issue I am having is solving for the function regarding reflection angle. However I have such defined though inside it, is another function that measures the angle from the reflection vector to 2pi. I have drawn out a table and a graph of what the graph looks like according to the table. As to how I got these values, I just used my intuition as to what the reflection angle would actually be when the surface is angled at certain angles. The function is taking in a input of a angle, which goes from the normal vector to the X-axis. I need help finding this function F.

[feral geode]

is euclidean geometry really much harder than the geometry for A level maths

[hollow wadi]

i took that topic over the summer

is basically honors geometry i'm not sure if is hard but it was kinda confusing

[upper karma]

is there an efficient method to memorize the xy coordinates of a unit circle? especially regarding the irrational values

[rugged hearth]

well for 1 u only gotta memorize one corner

easiest is top right

because then u can just do some flipping to get other corners

[rugged hearth]

and for the angles just know that 2pi is a circle so therefore pi is a circle and all that

and for the 30 degree and the 60 degree angles

what i do to remember them is only remember the 30 degree one

because i can just flip the x and y coords to get the 60 degree one

[upper karma]

I was confused on how the measurements become square roots

[rugged hearth]

i cant explain why its exactly sqrt3 and sqrt2

but you know that

the points represent the x line of the triangle and y line of the triangle

where the hypotenuse is 1 (length of unit circle)

and pythag theorem is a^2+b^2=c^2

so if u do some shifting and stuff u end up with square roots

[upper karma]

got it, thank you!

[upper karma]

np

[lyric hedge]

Has anyone read Elementary Geometry from an Advanced standpoint by Edwin Moise. I was wondering if this book would give me a good grasp on Geometry and Trigonometry or if I should buy a sepereate Trignoemetry book

[vernal pilot]

Google exact value triangles. Then to find any point just do (cos(x),sin(x)) using those triangles.

Also if you are new to trig (which I get the feeling you are), learning exact value triangles is more important than knowing each point on the unit circle.

[sweet shell]

Im struggling to understand why I cant give answers to trigonometry questions in decimal form.
Example.
Calculate the area of a triangle where each sides length is sqr(3) and the angles are all 60.

My answer ends up at 1.2990381
Which is correct
BUT
They want me to basically answer
(3*sqr(3)) / 4

Why is this? I don't know how to convert the decimal 1.2990381 to 3 times square root of 3 divided by 4

[nocturne remnant]

you are probably supposed to get the exact answer by not using a calculator and just manipulating the surds

the answer is 1/2 sqrt(3) * sqrt(3) sin 60

and sin 60 = sqrt(3) / 2

so you can simplify

[sweet shell]

alright

thanks

[wooden furnace]

Can we define a 3d vector with just the direction cosines?

[upper karma]

Update: Yup, the teacher accidentally put the square on the top, so it should not be a right triangle. He said he will take 84 and 85 as a correct answer. No bonus for figuring it out though. Im very happy anyways!

[upper karma]

How do u do tolerance

[upper karma]

say you had a diagram like below. How would you take moments from bottom to top and vice versa for the wall one. Is there like a formula.

[upper karma]

so solve the theta?

[placid fulcrum]

can a kite have parallel sides

[thick fable]

kite does have parallel sides i believe

[placid fulcrum]

would you say that’s a always sometimes or never statement

[thick fable]

ahhh

kite

cant have

parallel sides

im sorry lol

[placid fulcrum]

ok because I put never on my quiz lol

[smoky jetty]

I've posted and discussed it here before, but now I did get a response from the teacher. So, you're supposed to compare the volume of the box (assuming width=height) and the volume of ellipsoid... Now, I've learned that the statement by my teacher, "if volume of shape_A < shape_B, then shape_A fits" is not valid for all problem, now we've got an ellipsoid. How do u even make an ellipsoid hole on the ground

[placid fulcrum]

a kite never has parallel opposite sides

[thick fable]

not this question again gawd😭😭😭

[smoky jetty]

fr

[placid fulcrum]

whag math class are u talking

[thick fable]

yea they dont

[placid fulcrum]

taking

[placid fulcrum]

ok good thanks

[smoky jetty]

I either completely misunderstood this or the process we were instructed to see if box fits is just utterly ridiculous

[placid fulcrum]

what math class is that problem from

[smoky jetty]

conic sections

[placid fulcrum]

ummm what’s that

[smoky jetty]

A conic section, conic or a quadratic curve is a curve obtained from a cone's surface intersecting a plane. The three types of conic section are the hyperbola, the parabola, and the ellipse; the circle is a special case of the ellipse, though it was sometimes called as a fourth type.
Wiki

[placid fulcrum]

oh gotcha

does anyone wanna take a stab at this proof??? my teacher said that the point of doing it was to do it without using isosceles trapezoid conjectures for example the isosceles trapezoid diagonals conjecture

it’s for extra credit and I need the answer

[rugged hearth]

also another way to help is through using this combo move where you relate the points on the unit circle to 45 45 90 triangles and 30 60 90 triangles

[upper karma]

I think I bombed my trig test today because I switched up sin = y and cos = x

[rugged hearth]

F but next time instead of memorizing (cos, sin) try thinking of it like making a right triangle with the hypotenuse being your radius, and theta being the angle closest to the center of the circle and the right angle being at the bottom right, and finding the side lengths of each of the legs will give you cos for the x leg, and sin for the y leg

like instead of straight up memorizing everything try to find where they come from first

[torn basin]

Hey can anyone help me with proofs?

[thick fable]

just ask it

[torn basin]

What do you ean by that?

[thick fable]

ask ur question

i mean

never

ask if i cant get help

jusr directly ask ur question

and if u dont do that ur banned🔫

[smoky jetty]

I finally get troposphere's solution to it weeks ago

[thick fable]

damn

hard work

[smoky jetty]

the dimensions can be strecthed to transform the shapes to circle and a square

@grave pond thanks for the solution a few weeks ago (sorry for the ping, just a relief from the problem)

[smoky jetty]

but if a rectangle or a square is placed at the origin, would the origin always be the midpoint of the diagonal?

I guess so, if we use the midpoint formula for coordinates

[rugged hearth]

wtf

[grave pond]

Not necessarily, but if the midpoint of the rectangle is not at the center of the ellipse, you can slide the rectangle closer to the center without hitting the opposite side of the ellipse.

So the rectangle fits at all if and only it fits in the middle.

[stiff plaza]

I need a refresher in the basics of geometry, does anyone know any good books or sites?

[quaint sinew]

any idea how i'd go about solving this

[teal imp]

is that sinC + cosC =1 or sinC + CcosC = 1

[teal imp]

evan chen's Euclidean Geometry

[thick fable]

@quaint sinew Hint: ||divide whole equation by sqrt(2)||

[rare shuttle]

Anyone know if there's anywhere to get Algebra and Trigonometry with Analytic Geometry 13th addition free?

[thick fable]

we dont help ppl with pirating things

[rare shuttle]

oh mb

[upper karma]

I need help with my geometry

[thick fable]

!da2a

[lime crown]

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

[upper karma]

This

This question is confusing me

[thick fable]

i dont think this is geometry

[upper karma]

It is

I’m just confused on it

[thick fable]

just count the total number of fork, knifes and spoon

and make it in a ratio form

[upper karma]

How do I do that?

[upper karma]

Nvm I found out

[lunar folio]

What's the best way to learn the entire geometry unit? Tutor, online (youtube/khan academy/online textbook), or zmth else

Cause I'm skipping from alg to alg 2 and need to self teach myself geo

[rugged hearth]

get a textbook if u can and use youtube and khan academy if u stuck on something and if u really dont get something then you might needa tutor

[smoky jetty]

Ah, I see. But, will the condition, "if the diagonal of a rectangle or square is less than the diameter of the circle, then it fits," be valid?

so, like regardless of how it would be placed in the circle, whether centered or not so centered

[grave pond]

Yes, that's exactly it.

[smoky jetty]

ohh omg, thx so much!!

[lunar folio]

Thanks

[cedar wren]

Help

<@&286206848099549185>

[silver prism]

how do i calculate h in the pyramid on the left?

[foggy parcel]

if a line, A(x1,y1)B(x2,y2) is divided in a proportion of m1:m2 at the point P

Coordinate of P
= [ {(x1m2 + x2m1) / (m1+m2)},
{(y1m2 + y2m1) / (m1+m2)} ]

[graceful talon]

would i be right here

[upper karma]

yea

[smoky jetty]

so this was my full solution regarding it

However, when I compare the ratio of the scaled shapes' volume (assuming that before scaling, width = height) to the ratio of scaled shapes' area, I couldnt get the same answer

So idk if this one is correct

[fallen sleet]

what is my teacher saying

the thing she wants us to do for SSA is check how many cases there are (0,1,2) BEFORE we actually solve

fine

but the handout she gave us says

that the only two solution case is a<b and a>h

aka h<a<b

(we are given A,b,a and h=bsinA)

???

b is less than a here

[obsidian harness]

yeah then there is no ambiguous case for this triangle

[fallen sleet]

this is when A is acute?

the diagram you posted?

[obsidian harness]

yeah

[fallen sleet]

ok the conditions she gave for A is right or obtuse are:
a leq b -> no solution
a > b -> one solution

are these correct

[obsidian harness]

a <= b no solution is correct

as that means A must be acute

[fallen sleet]

ok

[obsidian harness]

this is correct

the other solutions for this will give A is acute

also you can never have more than two possible triangles for SSA if you've noticed

it has to do with the fact that sin(x) = c only has two or fewer solutions (if 0 <= x <= 180 deg)

[fallen sleet]

yes a circle and a line can't have more than two intersections

[obsidian harness]

interesting

[fallen sleet]

?

[obsidian harness]

yeah I presume your geometry class doesn't think of this using trig

as in sine rule, cosine rule

[fallen sleet]

ok thanks

[obsidian harness]

no worries

[wise vale]

Greetings. I'm 13 years old, I'm in 7th grade and I started studying geometry, but I don't really understand what should I do?

[thick fable]

study

[lunar folio]

khan acadmey

[slender gulch]

Help pls

[maiden brook]

x = (180-50)/2 = 65

y + y + 50 + 65 = 180 -> y = 65/2

[frozen ocean]

!nosols

[lime crown]

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

[maiden brook]

Triangle ABC has AB = 15, BC = 13, and AC = 14. Let O be the orthocenter of this triangle, and let the reflections from the orthocenter across sides AB, BC, and AC be points D, E, and F, respectively. Find the area of ADBECF.

[regal wedge]

have you made a diagram?

[maiden brook]

with geogebra

[regal wedge]

i mean it’s labeled wrong but sure we can use it

[maiden brook]

yeah

the labeling was annoying

i didnt want to construct all of this with a compass

[regal wedge]

in the car rn so a bit hard to think about it

let’s see

if you can find the diameter, you should be able to do anything you want

[fathom geyser]

can somebody explain to me why exactly it holds for all numbers of the form n = 3 x 2^m? i tried it for example with 3 x 2^2 = 12, which gives pi/6

and can't really see the connection

[maiden brook]

oh it should just be 2(84) = 168

[fathom geyser]

by similarity of triangles do they mean these two triangles?

[heady scaffold]

if AB = 1/3BC then AB = 1/4AC
Now we just solve like normal

6-2 = 4
4/4 = 1
1 + 2 = 3
x is 3
1-(-7) = 8
8/4 = 2
y is -5

if im not wrong

[upper karma]

I’m really confuse

[thick fable]

use similarity on the last one

[graceful talon]

how would i solve #2

[thick fable]

have u tried drawing any diagram

[graceful talon]

wdym

[thick fable]

a diagram?

[upper karma]

What about the others

[thick fable]

same concept for the second one

equating ratios of sides equal

[upper karma]

Bro tbh I really don’t understand what that even means

Like I have 0 knowledge on this topic

@thick fable

[thick fable]

u should watch any tutorial on it then

[upper karma]

Can you send one please

@thick fable

[thick fable]

lmao chill im here only

[upper karma]

My fault

[thick fable]

u can just watch any on yt sorry im not on my laptop rn, so im not gonna do it, its very hectic yk

but there r

a lot of them on yt

just search similarity in triangles

[upper karma]

Oh shi

It’s due in a hour

Alr bet

[thick fable]

nice

[upper karma]

Ty

@thick fable yo

[thick fable]

?

[upper karma]

I don’t feel this is right

X as in TR

[thick fable]

it should be 9+ alpha

[upper karma]

It should

Idk what I did wrong

[thick fable]

okay did u watch any video?

[upper karma]

Yea

Wait hold up

6/0.9= 6.66

Ah nvm

Dosent make sense

[thick fable]

well okay

so

yk that triangle HDM is similar to triangle HTN

right?

@upper karma

[upper karma]

Yea

I figure that part out

[thick fable]

ale

alr*

now

what can u say about thier ratios of thier sides?

[upper karma]

Mhm

[thick fable]

HM/HM+MR=DM/TR

right?

[upper karma]

Oh yea

Hm/hm?

Huh

[thick fable]

now just plug in the values

what?

[upper karma]

Alr

[thick fable]

also that is in the denominator

HM/(HM+MR)

[upper karma]

What do I do next

[thick fable]

what is that

[upper karma]

Wait what

Am I trippin?

[thick fable]

yea

https://youtu.be/YiFwvAFk-xs?feature=shared

watch this man

[upper karma]

thats the one i watch

[thick fable]

lmao

[upper karma]

well only the frst clip

first*

bruh i hve no time to watch that long video

lterally do in 30 min

oh nah my grade bout to drop

damn

[thick fable]

if it is so, then just plug ur values in here and get the answer

[upper karma]

BRO

i did that and u said its wrong

like what is going on

[thick fable]

wtf is this man

[upper karma]

YOU TOLD ME TO DO THAT

[thick fable]

they r just 2 expressions written

LMAOOOO

[upper karma]

WHY CANT U JUST SAY IT

[thick fable]

SAY WHAT

[upper karma]

like bro im obviously sturggling

NAH

THE DAMN ANSWER

IVE BEEN SITTING HERE FOR 1 HOUR

ON ONE QUESTION

[thick fable]

I AINT TELLING NO ANSWER

[upper karma]

BRHRUIHF

bro

[thick fable]

LMAOOOOOO

[upper karma]

ok lets make a deal

nah chill

we cool fr

[thick fable]

ALRIGHT what

[upper karma]

look bro like u gotta understand my sitation

i have 0 tme

time

to watch a video

im trying to do my best

[thick fable]

🤣🤣

ok ok

[upper karma]

i dont want my grade to go back to a 60

it litteraly just got to a 70

[thick fable]

i can make u an equation

thru latex

u can solve thay

that

in 1 min

[upper karma]

uh alr

...

[thick fable]

alr dawg

just for you

im here on my pc now

okay wait

i wanna know how did u got that

[upper karma]

oh thank you bro i apperiacte it

[thick fable]

alpha sign

in the equation

first just tell me what question r we dealing with

[upper karma]

So far the first one

With the stick figure

[thick fable]

the one with the racket in his hand right

[upper karma]

yea

[thick fable]

,, \frac{6}{9+6}=\frac{0.9}{x}

[somber coyote]

यजतलमाओ

[thick fable]

now u just simplify for x

[upper karma]

do i add 9+6

[thick fable]

bruh r u fkin serious

lol

[upper karma]

so i add it?

[thick fable]

that's just normal algebra now

yeaa obv

[upper karma]

alr

bro

im done

i got 2.5 💀

[thick fable]

2.5?

out off?

[upper karma]

i do cross multipaction right?

damn i got 20 mor emins

more mins*

[thick fable]

yea

u cross multiply

[upper karma]

i did 6 * x. which equals 6x

then i did 9+6

which is 15

then i did 15*0.9

which is 13.5

then

[thick fable]

,w 15*0.9/6

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by yajatjamal

[upper karma]

6x = 13.5

Huh

oh wait

is that the answer?

nvm

thank you

[thick fable]

yes

that is the answer

np

[upper karma]

im going to attempt the 2nd problem i'll tell u if i need help

[thick fable]

use help channels for it

[upper karma]

Wait bro

Ah damn

Anytime I post on the bell channels no one even says anything

Help*

[thick fable]

no

they do help

youll get help faster there

[upper karma]

Hmm Alr

[wintry aurora]

umm guys I need help right now

Diameter of cylinder A is 7 cm and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area

[maiden brook]

What’s the formula for volume? Given this, which would affect the volume more

[timber cargo]

pi(r)^2 *h

[wintry aurora]

@wintry aurora Greater the Volume also has greater Surface area

[proven sphinx]

I guess cylinder B will have a greater volume because in the formula for finding the volume of a cylinder the radius is squared and thus the increase in radius would be more effective than the increase or decrease in height

[forest ocean]

@upper karma which grade r u in

[proven sphinx]

Guys! I always wonder how the angle between the line of vision of the viewer and the top of a tall object is measured is there a technique or a particular device for it? anybody has any idea about it?

[thick fable]

that is just trigonometry

[proven sphinx]

Ik but how can you measure the angle between a tall building and your eye without knowing the height of that building.

Can you pls explain it to me?

[silent plank]

protractor

[grave pond]

I'd recommend a sextant rather than a protractor.

[silent plank]

(in)clinometer

[pastel brook]

sup bois

why is 0! = 1

[blazing veldt]

https://www.youtube.com/watch?v=X32dce7_D48

[untold valve]

this might be better suited for #linear-algebra but how would I go about finding the difference in coordinates of the two ends of an arc of a circle, assuming I have the length and the radius of the circle, and the coordinates of the first point
I mention linear algebra because technically I'm using vectors for all this - I'm writing a program where I have an object at a coordinate (which c# stores coordinates through x and y components of vectors), and need the object to move around a circle at a given speed.

[hardy solar]

help me pls

im on trigonometry

and stuck on this one question

[slow moon]

What is it

[hardy solar]

i need help w number 8

[candid peak]

literally my geometry teacher doesnt even teach so idk how to help

[slow moon]

Use inverse trig

[hardy solar]

what's inverse trig

[slow moon]

It's how you solve for angles instead of side lengths

[hardy solar]

huh

[heavy oar]

My geometry teacher is getting to the point that she can't even teach properly anymore. She's allegedly retiring at the end of this year. The entire class is stuck on translating shapes even though we're in quarter two.

[slow moon]

ur using law of sine right

or cosine

[hardy solar]

i mean i've learnt it it just idk what to use

i've used both and didn't get the answer

[slow moon]

Try this video

https://youtu.be/BQIownZd16Y?si=eo4VqP8AIKntZ09Y