geometry-and-trigonometry

[regal cave]

so teach me

plzzzz

those are kisi kisi

[dark sparrow]

https://www.youtube.com/watch?v=PUB0TaZ7bhA

[regal cave]

what im confused about are the akar

what is akart in english bro

[dark sparrow]

don't call me "bro", please.

also i don't speak that language.

[regal cave]

dang itt, please gal help me

i am 100% i will able to answerother similar question

[dark sparrow]

can you maybe show where those "akar" are in the question?

or where you see them

[regal cave]

ok ok, the ab is 2 akar 3 and bc is 4 akar 3

4 akar 2* not 3

[dark sparrow]

oh.

these are roots.

square roots, to be more precise.

[regal cave]

oh yeah yeah

roots

please help

[dark sparrow]

so your issue is that you don't know what a square root is?

[regal cave]

it's 0:06 am and i am so sleepy.

no what

i dont know

bro please just teach me

;-;

[dark sparrow]

i said not to call me bro.

i am trying to find out where to start.

[regal cave]

oh sorry

im used to call all people that,e ven classmates

[dark sparrow]

well, not everybody likes it.

anyway, i already linked you a video that explains the basics of trigonometry.

[regal cave]

pleaseee

i need this specifically

[dark sparrow]

sending a thousand messages containing "PLEASSEEEEEE HELP ME PLEASEEEE I AM BEGGING YOU" (or something to that effect) won't speed up the process.

in fact it'll only slow it down.

[regal cave]

ok ok, ill shut up

[dark sparrow]

also you probably should have come here earlier than midnight for help, just saying.

sleepy is not a good state for doing math.

ok

let me ask a question bluntly

do you know what sin, cos, sec, etc. mean?

yes/no

[regal cave]

uh

sinus cosami secan

[obtuse field]

If you want help, don't spam random letters in the chat

[regal cave]

sin= front and miring/ ac cos= side and miring/ac and tan= front/side

[regal cave]

sorry but its urgent, sorry if it bothers you

idk whats miring in english

but its mostly on ac

that most farthest line

[dark sparrow]

the line opposite to the right angle?

that is called the hypotenuse.

okay, so you know that the trig functions are ratios of sides.

[regal cave]

yes that

[dark sparrow]

great.

so then all that remains is to find the hypotenuse, and you will have all the numbers you need to compute all these ratios.

(and it becomes an exercise in keeping straight what is where.)

[regal cave]

what?

id understand

and thats exactly wht im having a hard time on, like root front+side= hypotenusa right?

[dark sparrow]

root front+side= hypotenusa
this is misremembered or just poorly written.

given that you admitted earlier that you are "used to cheating", it's kind of expected.

[regal cave]

the person behind me is chinese, so he is smart

[dark sparrow]

behind?

[regal cave]

really glad and regret at thes time

[dark sparrow]

how would you cheat off of somebody behind you?

[regal cave]

when in class, i look behind

[dark sparrow]

it seems inconvenient unless you have a 3rd eye on your back.

[regal cave]

now im at home

[dark sparrow]

anyway

[regal cave]

yes yes, what is the value of the sin cos tan

[dark sparrow]

!noans

[lime crown]

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

[regal cave]

yes ok

i will recieve step by step

[dark sparrow]

no.

[regal cave]

cuz i gotta wake up at 5 am anyway, ima just push it

[dark sparrow]

you'll receive guidance.

but you WILL NOT receive a full step by step solution

[regal cave]

ok

[dark sparrow]

and frankly i kind of dont want to guide you anymore bc this attitud is just not welcome here

[regal cave]

oh ok then

sorry

[dark sparrow]

the PYTHAGOREAN THEOREM, which is what you tried to state here,

says that in a right triangle, with legs a and b and hypotenuse c, we have: a^2 + b^2 = c^2.

in particular, when the other two sides are known, and you want to find the hypotenuse, you may write that same equality as: c = sqrt(a^2 + b^2).

look up "pythagorean theorem examples" on the internet if you need to.

find the hypotenuse in your triangle.

then calculate all the ratios.

[regal cave]

i know all that, what im having problem is calcuatliong the hypontesuie

[dark sparrow]

what is your problem with that?

can you write down the expression you need to calculate, without doing any calculations?

[regal cave]

what?

that 2 root 3 and 4 root 2

equals what hypotenusa

[dark sparrow]

this "and"

ok

you just are not reading what i say

[mortal crow]

So you’ve been given “a” and “b”, right?

[dark sparrow]

even though i gave you the theorem directly and you do not even have to google it yourself

@mortal crow this is his triangle

[mortal crow]

Oh yeah, so what’s his issue?

What’s he trying to do?

[regal cave]

the value of sin cos tan alpha and cosec sec and cotan beta

i am completly not know this

[dark sparrow]

he's trying to find the hypotenuse

[mortal crow]

Just that?

[dark sparrow]

he said that he has trouble with it but so far has not told me where exactly.

[dark sparrow]

no, also a bunch of trig ratios.

[regal cave]

yes, idk the value of sin cos tan alpha and cosec sec and cotan beta

[mortal crow]

You know what the trigs are defined as right?

Like sine is opp/adj, you know the definitions?

[stiff glen]

I solve

#1207734428965666916

[upper karma]

Anyone good at inscribed angles?

[dark sparrow]

!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]

What is the use of cyclometric functions, besides being a formal way to convert values into angles?

[dark sparrow]

"cyclometric functions"?

do you mean like inverse trig?

arcsin, arccos, arctan? these? @upper karma

and what do you mean by a "formal" way? formal as opposed to what?

[upper karma]

so instead of just saying that cosx=1/2 then x x = 60 deg, we actually can express it as x = arccos(1/2)

[dark sparrow]

that doesn't answer my question of "what do you mean when you say 'formal'"...

[upper karma]

I cant find any way to express it better

[cunning lion]

if we have an equation [\cos(x) = \frac{1}{2}] then one solution (x) (out of infinitely many possible solutions) is given by [x = \arccos\ab(\frac{1}{2})]

[somber coyote]

cloud

[lone panther]

they both use x
use AEB to find x
then use x to find DE

[dull anvil]

how do I calculate drop factors

[silent plank]

wdym by drop factors>

[ocean lava]

Someone pelase help

...

why bully

im just asking for help

[fickle rose]

whats the problem here

it seems you filled out the proof

oh the last one is wrong since theres an x

there is no place in the proof where you stated 2 triangles as congruent

[upper karma]

How do I get the perimeter of this sector

[dark sparrow]

have you found the length of that arc yet?

[rose raven]

guys whats the best method for remembering the inverse function values on the unit circle

[dark sparrow]

they are almost the exact same angle-ratio pairs that you (presumably) have already committed to memory.

except backwards.

[lyric quest]

can someone help me with this kangaroo contest question

[lime dune]

area ratios

if two triangle have the same base length then the ratio of their areas is the ratio of their heights

similarly if two triangles have the same height then the ratio of their areas is the ratio of their base lengths

[lyric quest]

thank you

[teal badge]

Do you know the formula for circumference? Use that. Keep in mind that 45 degrees means that particular sector is 1/8th of a circle edit: shoot I didn’t realise that was sent almost 11hrs ago 💀

[placid fiber]

how can i prove the first one, been trying with cubic sume but i get stuck with cosines

[dark sparrow]

!nosols

[lime crown]

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

[placid fiber]

bruh

[solid crypt]

Does anyone know if there’s an alternative to arcsec because my dumb calculator only does cos sin and tan as well as their inverses

And I will need arcsec arccot and arccsc

I know I can use the reciprocals

But not the reciprocal’s inverse

[sturdy stone]

arccos(1/x)=arcsec(x)

similarly for the other two

[solid crypt]

Thank you

[dark sparrow]

for what will you need them, i wonder

[solid crypt]

Instead of writing 1/tan84 or smth like that (also didn’t know what Daniel said)

Cot84 for efficiency

[dark sparrow]

that's for forward trig functions

and why is efficiency such a concern exactly

what problem(s) are you solving that make a difference of 2 keystrokes significant

[prime monolith]

How would i prove that b3=b1

[lucid rapids]

bruh 💀 😐

[lime dune]

we are not a “do your homework for you” server

fundamentally speaking

[lethal needle]

using your eyes

[loud night]

Holy trigonometry

If I do pre calc outside of school am I screwed for calc ab

[jagged wyvern]

#1208904196603256842

[keen talon]

If I'm learning trigonometry, should I use a calculator to solve stuff like "sin(35degrees) = ac/5" or Is it better if manually solve it to learn?

[cunning lion]

as far as i'm aware there isn't a commonly-known value for sin(35°). you should definitely know the "common values" for angles like 30°, 45°, and 60°, and you should know how to solve for ac to know what to plug into the calculator, but you're not expected to be able to find values of trig functions for more uncommon angles

e.g. in this problem you'd be expected to go from
sin(35°) = ac/5
to
ac = 5 sin(35°)
then just put the right side in a calculator

[keen talon]

alright thanks

[cerulean turtle]

Do all circle have the same curve?

[lime dune]

no, a circle of radius r will have curvature 1/r

intuitively, consider a very very large circle

say, of radius 100000000

it basically behaves linearly unless you zoom super far out

now how about a very small circle?

radius 0.00000000000001

you have to zoom very far in to get that same linear behavior, otherwise it will much more noticeably “curve” than the big circle

[limber basalt]

Any advice on finding the area of this trapezoid?

[dark sparrow]

half sum of bases times height?

[limber basalt]

It doesn't matter that it's diagonal? Can I use distance formula to find the length of the bases and height?

[dark sparrow]

hardest thing will be finding the distance between the green and blue lines, and even that's easy bc gray is their common perpendicular

no, it does not matter which way the trapezoid is facing.

[dark sparrow]

yes, the bases are P1P2 and P3P4

the grey is perpendicular to green and blue?
yes
because green and blue are parallel?
but not for this reason

[limber basalt]

so base1 = P1P2 = sqrt[ (2-4)^2 + (6-4)^2 ] = sqrt(8)
so base2 = P3P4 = sqrt[ (1-2)^2 + (3-2)^2 ] = sqrt(2)

Maybe not. I am thinking of the flip and switch

[dark sparrow]

the what...

[limber basalt]

oh flip and switch is flip the slope, multiply by -1. so 1/(-1) * -1 = 1

[dark sparrow]

here what's known are the two slopes, and the perpendicularity is what needs to be found out.

[limber basalt]

Then
base1 = P1P2 = sqrt[ (2-4)^2 + (6-4)^2 ] = sqrt(8),
base2 = P3P4 = sqrt[ (1-2)^2 + (3-2)^2 ] = sqrt(2),
height = P2P3 = sqrt[ (4-2)^2 + (4-2)^2 ] = sqrt(8).

Which means

area of trapezoid
= 1/2 (sqrt(8) + sqrt(2)) * sqrt(8)
= 1/2 (8 + 4)
= 6

very nice.

[dark sparrow]

you may need to use the point-line distance formula

and then that will be your height

[cerulean turtle]

Is my markings correct?

[dark sparrow]

can we see the full problem?

as-is: the fact there's two apparently different points both named A, and the same for C, looks sus.
the markings appear to be adequate based on the perceived sizes of the angles and sides.

@cerulean turtle

[clever socket]

We know that alpha + beta = 90°, what if we have -sin10° is that cos 80°?

[wanton tinsel]

Hello ! Is there any book that contains proof for geometric formulas such as areas and volumes? I find it hard to, for example , to understand why the area of a rectangular is different from a parallelogram or why the area of a circle is radius* hald the circumference.

I know how to find some of the areas with integral calculus but i also dont know how the volume integral works. In my book the volume integral is given by a formula that also contains basic volumes knowledge , which i dont have.

[lime dune]

you absolutely do not need calculus for the elementary area formulas

a lot of the basic ones for quadrilaterals just boil down to

"chop this triangle off and glue it back on somewhere else to turn the parallelogram/trapezoid into a rectangle"

some of the nontrivial volume formulas (pyramids/cones/spheres) do require integration potentially

[cerulean turtle]

congruency

BAC = DCA

[dark sparrow]

you didn't show the full problem.

[dark sparrow]

...

this looks like your solution to something

and makes nothing clearer. at all.

[cerulean turtle]

I mean is the markings correct? I'm having trust issues on myself

[dark sparrow]

i can only repeat myself

if the diagram is to scale then yes they're correct

BUT there are two pairs of same-name points. which is STILL suspicious.

you have decided not to comply with my simple request to show the original problem (or give me some reason why you can't, like "i made this up myself" or "i lost the original".)

[cerulean turtle]

Do we include the C and A?

[dark sparrow]

what do you mean by "include"?

and was there anything besides this diagram? like some words about some shit being equal to some other shit?

[dark sparrow]

...

i have no clue what you are talking about.

and you are not answering my questions with any measure of clarity.

so i'm out.

[austere widget]

It looks like you're dealing with similar/congruent triangles, right? Idk if this is intentional, but you also have another pair of points named A and C. I would change them to E and F to reduce confusion. Or, you can instead name the points on the right-hand trinagle A', B', and C' so the correspondences are more obvious. But your proof is correct. BAC ≈ DCA because you showed that all the other angles (A and C, B and D) are the same.

[astral shadow]

How is AD equal to BE?

💀

<@&286206848099549185>

oh wait

I think I know '

[sturdy stone]

It isn't unless there's more information given

[dark sparrow]

that can't be concluded from this diagram

[austere widget]

If triangle AEF is equilateral or isoceles, then AD = BE.

[astral shadow]

sorry sorry i forgto to aactually give details💀

DC=27, AF=46

[sturdy stone]

they aren't equal then

[astral shadow]

BE is 27 though right

[sturdy stone]

yea

[astral shadow]

lit. hey also is kahn academy a good supplemental site for geometry or alg 2

[sturdy stone]

yes

khan academy is good for everything high school math

[astral shadow]

do you include calc to be high school math too

calc AB/BC

[sturdy stone]

yes

[astral shadow]

ok thanks

[mortal crow]

Those are specifically high school

[astral shadow]

yeah

[mortal crow]

They are 💀.

[astral shadow]

ik I was agreeing 😭

[mortal crow]

You said you guess so 😭!

[astral shadow]

nah bro youre just tripping

trust I never said that

[mortal crow]

………you edited it!

[astral shadow]

nah nah you are just crazy I never said that

[mortal crow]

No!

[astral shadow]

Mwuhahahaha

[mortal crow]

No!!

[astral shadow]

what point of concurrency on a triangle is in the center of the circumscribed circle?

<@&286206848099549185>

nvm i got it

[astral shadow]

@upper karma sorry but I just stopped having access to that help channel for some reason

Where we left off though, slope of BC is 1/-3

[limber basalt]

How? I tried using area of equilateral triangle

[dark sparrow]

how what, exactly?

[limber basalt]

Area of the hexagon

[dark sparrow]

can i have a complete sentence please

[feral steeple]

hi

I am so lost

[jagged wyvern]

a.) they're already giving you the point vertices

Also for b.)

[pulsar idol]

Can someone help me find mind x y and z

[plain talon]

Because this is an equilateral triangle and we know one side:24. We can set the side with x to 24 and solve. Similarly, we can set the side with y to 24 and solve

For z, each angle is 60 degrees since it’s an equilateral triangle. So simply set the expression to 60

[pine mango]

Anyone able to help me with some law of cosines

[vocal bison]

hey guys any important theorems i need to remember for tangent lines, arcs & chords, or inscribed angles? cuz i hava quiz tmrw (ping)

[limber basalt]

I don’t know how they computed the area of the hexagon. I tried applying the area of equilateral triangles and multiplied that by 6, but it doesn’t match their answer.

[dark sparrow]

show your work?

i don't think they did any computations per se.

they just broke both shapes into equilateral triangles of side length 2, and made it apparent from the picture that the triangle is made of 4 of those, and the hexagon is made of 6.

from which the ratio of their areas is obvious unless you overthink it (like you seem to have done).

[limber basalt]

Oh yes. I just read this poorly. Thanks.

This isn’t the area

Explicitly

[dark sparrow]

yeah i think you just overthought it imo when it was right there. real "a picture's worth a thousand words" moment etc.

[limber basalt]

Do you have any recommended resources to learn geometry “rigorously?” Do you think it’s worth studying geometry in isolation? Or it’s better to pick up on things as you need it? @dark sparrow

[lime dune]

the traditional model of high school geometry

is absolutely horrible

[dark sparrow]

¯_(ツ)_/¯

in america they put you through the wringer of 2-col proofs

or so i hear

[limber basalt]

Is there a textbook people recommend?

[lime dune]

it focuses way too much on pedantic shit like “all rIGhT anGleS arE cOngrUenT”

[dark sparrow]

idk of any textbook

[lime dune]

and not enough on the big ideas

[limber basalt]

That’s correct

[lime dune]

the best teacher is

do lots of problems

[limber basalt]

Euclid’s? 😂

[lime dune]

no

😭

it’s written in incredibly dated prose

and is not necessarily the best introduction

[limber basalt]

Yeah just do like 500 geometry problems? Proof based or computation or both

[dark sparrow]

khanacademy has some geometry problems

idr if it has proof stuff

[lime dune]

same reason I roll my eyes at people trying to learn big subjects “from first principles”

that’s better for a second pass maybe

the first time

[limber basalt]

What reason? Because they’re reading old text?

[lime dune]

you risk getting so bogged down in the formalisms

that you lose sight of the big important ideas

[dark sparrow]

yeah

[limber basalt]

Sounds like you’re advocating HS approach

[dark sparrow]

like 2 col proofs are just bad

[lime dune]

^^

[dark sparrow]

and in america you are forced to distinguish an angle and the measure of an angle

[lime dune]

and yet the majority of American high schools still use it

[dark sparrow]

which like

[lime dune]

which is dumb

[dark sparrow]

who cares??

[limber basalt]

Okay then what? Khan academy? Just run through a ton of geometry questions?

[dark sparrow]

yeah ig

for a start

[lime dune]

I staunchly refuse to put “m” before my angles LMAO

[limber basalt]

Oh I think that’s so dumb.

[dark sparrow]

yeah like

you can just take it for granted at that point that angles with the same measure are congruent

devoting maybe a minute or two to saying it in class

[lime dune]

my class had us do “flowchart” proofs which were just as bad

no, I don’t fucking care about “definition of congruent segments”

[limber basalt]

I was helping someone with HS geometry, the little pieces I know. They started talking “measure of angle” I was like it’s just an angle lol -_-

[dark sparrow]

and then just write shit like $\angle A = 40\dg$ and not worry about that shit

[somber coyote]

Ann

[lime dune]

please stop making me rewrite that over and over 😭

[dark sparrow]

and also $AB = 5~\mathrm{cm}$ or whatever

[somber coyote]

Ann

[limber basalt]

You missed the bar at the top. Is this AB a segment?

[lime dune]

the only reason teachers “care” is to produce more meaningless busywork for students

[dark sparrow]

serious or joking?

[limber basalt]

I don’t joke, except when I am

[dark sparrow]

that doesn't answer my question

[limber basalt]

My geometry is weak. I’m not afraid to admit it

[lime dune]

the only time I put the overlines is if im referring to parallel/perpendicular or defining a point as an intersection of two segments

dont waste your energy on this pointless notational pedantry

[dark sparrow]

so like congratulations for finding the one answer to my either-or question that's not only uninformative but infuriating in doing so

[limber basalt]

I’m confused lol

[dark sparrow]

i asked you "[are you] serious or joking?"

[limber basalt]

I wasn’t joking. Your answer you seek.

[dark sparrow]

your answer of "i don't joke, except when i am" did not tell me that.

anyway, now that you've confirmed you are serious

what could "AB" possibly refer to other than a segment, tell me

[lime dune]

in the us they typically exclude the overline if you’re talking about a length

[limber basalt]

If we were in person, you’d have your answer, even with that sentence.

[dark sparrow]

given that what i wrote is a statement about AB equalling some length

[dark sparrow]

no, i wouldn't actually

i would be confused still.

[limber basalt]

A lot is communicated through body language, but anyways, the over line

[dark sparrow]

what about the overline

[limber basalt]

Oh

I read your last sentence.

Yeah the over line is stupid.

[limber basalt]

The product of two matrices is a scalar 5 cm? 😂

[dark sparrow]

matrices
in fucking geometry

[limber basalt]

Just a lesson in context

Okay so khan academy. Maybe Olympiad geometry has nice problems? Or is that torture

[lime dune]

if you’re lacking foundational skills in geo

Olympiad geometry is absolutely not the play

[limber basalt]

If you took a poll of all the people who are frequent helpers on this server, do you think the distribution of time spent studying, practicing, thinking about math, would have low variance for that group?

[lime dune]

just hop on KA and drill your foundational skills first

AoPS alcumus also has some nice problems that are a bit more challenging but still computational in nature

[limber basalt]

Art of problem solve? Okay. Computation is nice. Proofs are nice when you’re concerned with congruence of shapes

[lime dune]

AoPS has a textbook on basic geo as well

that you can find online

it has its issues but all things considered it’s a decent introduction I think

[limber basalt]

Thanks

[hollow barn]

If I’m proving that the LHS = RHS, and the RHS = tan(A/2), and I have 1 within the left hand side. Can I say 1 = sin^2(A/2) + cos^2(A/2)?

Instead of 1 = sin^2(A) + cos^2(A)?

[limber basalt]

How do you find the radius of this blue circle?

[dark sparrow]

which one

oh the blue one mb

pythagoras

the segment from (-a,-a) to (a,a) is its diameter

[solid crypt]

I’m supposed to write the equation Cotθ=x/y for the 5 other trig ratios

Having trouble after finding tan

[signal venture]

uhh... does anyone want to help with this?

[mortal crow]

Then no.

[dark sparrow]

<@&268886789983436800> test cheater

@silent radish this is a bannable offense jsyk.

[devout prism]

cant you compare those two?

[devout prism]

yeah compare them and find intersection points

[plain veldt]

cud also use distance formula right?

[dark sparrow]

same difference lmfao

[obtuse vector]

hey i can help

so firstly what I would do is realize that 2*sqrt(2) + 1 and 4xcosine(x/2) + 1 share some similarities

firstly, you could rewrite 2*sqrt(2) as 4xsqrt(2)/2

and then you see that the only difference is really instead of cos(x/2) it is sqrt(2)

so you want to find values of x such that the cosine of (x/2) > sqrt(2)

@signal venture

[limber basalt]

Okay so draw a big triangle there. You have

2a^2 + 2a^2 = 2a^2
4a^2 = 2a^2
…

Yeah this is wrong. If the diameter is -a to a, doesn’t that make it length 2a?

[dark sparrow]

2a^2 + 2a^2 = 2a^2
???

where, pray tell, do you get the right-hand side from as "2a^2"??

[limber basalt]

Prayers aren’t needed here. I’m lost lol

[dark sparrow]

"pray tell" is a fixed expression to which i ascribe no religious meaning.

it is like a more polite version of "where the fuck"

(2a)^2 + (2a)^2 = diam^2

4a^2 + 4a^2 = diam^2

diam^2 = 8a^2

diam = sqrt(8) a = 2 sqrt(2) a

does this answer your question

[limber basalt]

Thanks for being polite

[limber basalt]

Doesn’t the diameter go from -a to a? Oh no of course not. It’s not a “90 degree” line. It’s the hypotenuse. This gets back to our previous discussion…

[dark sparrow]

.....

[limber basalt]

If you don’t like the way I phrased that don’t worry. I understand now

[limber basalt]

Yes thanks. I made the mistake of thinking the hypotenuse was 2a. Very wrong.

[primal jasper]

Can’t remember the reason for step 10 of the proof, and can’t find it in my notes either.

[broken sand]

Can some1 please help me with this one

The chinese characters at the top mean "as shown in the figure"

[subtle fiber]

Can someone help me with this problem? I keep getting large negatives and idk why

[dark sparrow]

are you using a calculator?

if you are: check whether you're in degree mode or radian mode.

[subtle fiber]

Ah thx

[dark sparrow]

,calc sin(35)

[somber coyote]

Result:

-0.42818266949615

[dark sparrow]

yeah, sin(35 rad) is negative huh

[shut bronze]

,calc sin(35 degrees)

[somber coyote]

Result:

0.57357643635105

[shut bronze]

🤯

[buoyant cypress]

,calc 5+5

[somber coyote]

Result:

10

[buoyant cypress]

,calc sin(pi)

[somber coyote]

Result:

1.2246467991474e-16

[buoyant cypress]

,calc 180

[somber coyote]

Result:

180

[buoyant cypress]

,calc sin(180)

[somber coyote]

Result:

-0.80115263573383

[buoyant cypress]

,calc (90 radians)

[somber coyote]

Result:

90 radians

[somber coyote]

Result:

-0.87926487577869

[fossil delta]

@primal jasper i don't fully get what you are asking but here a step wise solution anyway: the 10 th step is just taking the ratio of AB/CB=2CE/2CD(given/trival) which is equal to CE/CD which gives similar by SAS

[gentle lotus]

,calc 60+9

[somber coyote]

Result:

69

[gentle lotus]

cool

[lean rapids]

🙂

[gentle lotus]

I'm sorry okay 😔

[lean rapids]

Did you know

69 = (6×9) + (6+9)

[gentle lotus]

no, thank u for that

[jagged wyvern]

Use your notes

[tacit gale]

?

[lime dune]

this is just definition pushing

it's not that deep

[ivory cypress]

anyone able to help me?

[mortal crow]

But you didn’t ask a question….

[ivory cypress]

peek

the question

im lost in the sauce

on this whole quiz

[mortal crow]

This is an actual quiz for school?

[ivory cypress]

yes

its a check

but

same thing

[mortal crow]

But is it graded?

[ivory cypress]

no

[mortal crow]

I was about to say…

[ivory cypress]

but the test in 3 days is lol

and idk what im doing

[jagged wyvern]

Study

[ivory cypress]

yk i would if ik waht to study lol

or how to do it

or anything

idk im sick rn and the teacher dont post nothing

so

[jagged wyvern]

Find worksheets online

That's what im doing in my spring break right now

Gotta use that time wisely

[unique seal]

@jagged wyvern dont be an ass and help him out thats why we and him are here in the first place

in reposnse to OP's question here are the answers:
a) None of the above
b) 1 and 4 I think
c) 3
@tacit gale

[lofty skiff]

can anyone help me with this

[ivory cypress]

anyone know pythagorean identities?

[mortal crow]

Yes, sin^2 + cos^2 = 1 | sec^2 - tan^2 = 1 | csc^2 - cot^2 = 1

But ya can just look them up you know

[dark sparrow]

what's troubling you?

[vernal pilot]

Take the inverse sec function of 1.348 on your calculator.

[weak torrent]

Can someone pls help me solve this? It’s trigonometry and you are meant to solve for X and when you are looking for the answer you have to round to the nearest hundredth :’) I’ve tried everything and I still keep getting it wrong

[sonic warren]

tan

tangent

tangent 36 = x/45

[sonic warren]

use a calculator

[blazing oak]

45/x = cot36

[orchid scaffold]

Can anyone help with this I've been stuck on this for an hour and a half

[shut bronze]

what does m mean, and do the numbers represent the lenghts of the arcs?

[orchid scaffold]

I think m is the measure and the numbers are arcs

[limber basalt]

Any tips on this question? I am sure a triangle is involved

[shut bronze]

use cos and sin to find the distance travelled east and south seperately

add them together

and then use pythagoras to convert back into angle and distance

[dreamy blade]

you'll want to have multiple triangles I believe, and utilise the cosine law

the diagram shows the angles and distances given by the problem

[fossil delta]

@orchid scaffold if 31x+7 is the circumference for the circle and TU and UV are tangents then : let O be the center of the circle and so Angle TOV=$\frac{11x+17}{31x+7}\cdot 360$ now angle VTU=(1/2)angle TOV so Angle TUV=180-($\frac{11x+17}{31x+7}\cdot 180=180(1-\frac{11x+17}{31x+7})$
That's what I got

[somber coyote]

ideal_37

[upper karma]

x=2pi/7

no calculator ofc

[dark sparrow]

what is the symbol at the very left of the image?

[upper karma]

that is capital Q @dark sparrow saying question and then :

[dark sparrow]

ok

[indigo karma]

can someone help me with this: (sinx)^1979 + (cosx)^1991 + sin2x + cos2x = 1 + sqrt2

[dark sparrow]

looks ugly

[dusky locust]

consider what is the maximum possible value of (sinx)^1979 + (cosx)^1991

consider what is the maximum possible value of sin2x + cos2x

[upper karma]

did anybody made progress on my question ?

[tacit gale]

oh ty. very helpful

[dark sparrow]

did you?

[upper karma]

i tried using double angle formulas

i tried to apply it to every single term but that aint possible

cant think of anything other

[frank hound]

Vector question: given a vector $\overrightarrow{AB}$ in component form $\langle a,b \rangle$ find a vector that is perpendicular to it

[somber coyote]

Cewkins

[frank hound]

Is it correct to use the slope * perpendicular slope = -1 to find the line that is perpendicular to it?

And then any vector that is equivalent to that line can be considered perpendicular to AB

Which means that there is an infinite number of answers as long as they are on that line

Also assuming that the vectors start from the origin point (0,0) you can easily obtain the slope of the first vector using y/x

I did it using this method but i am not sure if that counts or of there is a more ideal solution

Also using the dot product it ends up being $(a)(x)+(b)(f(x))=0$ where $f(x)$ is the slope-inercept line with a slope of $\frac{-1}{m_1}$

[somber coyote]

Cewkins

[sick kraken]

2 x Angle U = (31x-7)-(11x+17)

Formula is 2xAngle=outer arc-inner src

[sick kraken]

^^^

Am I right

[median ledge]

25/24 is not the same as 7/x

enjoy my paint skills, blue triangle and red triangle are congruent

so $\frac{25}{24}=\frac{7}{y}$

[somber coyote]

647TheSheep

[median ledge]

express y in terms of x in this equation and solve for x

or just use the result you already got and solve for x using pythagoras

[sick kraken]

Oh i just noticed

Ty

[orchid scaffold]

yeah but every time I try to solve it I get 0.5 and Im not sure if im correct or not

[frank hound]

I can discuss vectors here right?

[dark sparrow]

yeah that's fine

the worst thing that can happen is you get redirected

if your stuff is sufficiently high caliber

[frank hound]

Oh okay i thought i did something wrong posting about them above

[dusk ingot]

I need help I dokt understand how to do this

[silent plank]

which one?

[winter gazelle]

Do you know sine law?

[sick kraken]

No their hypotenuse are different?? Or am I wrong

Also would it be

so $\frac{25}{24}=\frac{7}{25-x}$

[somber coyote]

holi

[sick kraken]

?

[solemn hollow]

you're going to use SOHCAHTOA to solve for the missing side. For instance on number 12, you're trying to find the angle but you already have the opposite and adjacent so in order to find the that angle you're going to set it up like tan(x)= 8/20, then you have to use inverse trig function to find the angle so on like calculator you're going to type in Tan^-1 8/20 in order to find the angle.

[cerulean turtle]

Hallo

Need help

[orchid iron]

does someone have a copy of trig identities

[silent plank]

its pinnned

@orchid iron (pinned in the channel)

[frank hound]

The traingles are similar right?

[sick kraken]

Yes

[frank hound]

Should the ratio be the sides that are the same to each other? i thought 24 lines up with 7 not 25

Or maybe i am missing something

[sick kraken]

Ratio should be the given sides of that triangle. Big over small

But one side needs to contain the x value in order to find it

[real dome]

guys do you know how to do no,(10) and (12)??
I can't use intersecting chord theorem

[frank hound]

Makes sense

[frank hound]

What i did was split the two triangles, the big parts for big triangles and small one are 24 and 7 respectively and small parts 25-x and x respectively

What is the exact answer btw?

[frank hound]

Btw how is the smaller side 25-x? Shouldn't it be just x?

The answer i am getting for x is 1.96, and for the last side its 6.72

[frank hound]

Which lines up with this

[vocal bison]

help pls ping

[sturdy stone]

Power of a point

[rugged shuttle]

Help

[dark sparrow]

!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

[dark sparrow]

oh wait this was 6 hours ago

@rugged shuttle do you still need help w/ this

[rugged shuttle]

yes step 1

i dont really understand

[dark sparrow]

#8 or #9?

if both, which first?

[upper karma]

hello peeps

just a small question

Did i word/do this correctly

[dark sparrow]

looks overcomplicated

and clunky

and the variables are not really clearly defined ...

[upper karma]

hmmmm

[upper karma]

which ones exactly

[dark sparrow]

l and n

n is apparently the square root of the number of transistors...? but why'd the number of transistors be a perfect square

also why do you need to speak of the transistor count at all

[upper karma]

wait maybe i think its best if i like draw what i mean by n

[dark sparrow]

"doubling the number of transistors while keeping the area unchanged means that each transistor now needs to occupy half of its former area" is not something that needs 17 pages of bureaucracy to justify

[upper karma]

so i guess l really is just reduced by 50%?

or did i misinterpert

[upper karma]

anyways thinking about this more i realise i made a wrong assumption here

by "n^2 squares" i thought like, the amount of columns and rows of squares in the grid are equal

so you have n*n = n^2 total amount of transistors

but thats clearly not a requirement so in reality its just n*m where n represents the amount of rows and m the amount of columns

I understand if i am still overcomplicating this though

ok nvm this is way too dumb i really fucked this up

[dark sparrow]

yes you are overcomplicating

[upper karma]

ok i guess i will start from the beginning

[upper karma]

Im sorry Ann but something is really not clicking about this? Like take for example a 2x2 area where there are 4 transistors, the amount of transistors doubles to 8 but i dont think you can partition the same area with only 8 squares (you'd need to 9, because the square root of 8 is irrational)

Oh wait is my answer correct just to verify?

[fossil mica]

Wassup my guys

I need help with something

[dark sparrow]

yes your answer is correct @upper karma

[dark sparrow]

!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]

ohh okay so im just being unnecessarily verbose

[fossil mica]

Is there a geometric representation of A^3+B^3+C^3=D^3

Like a higher version of pythagorean theorem

[dark sparrow]

that isn't how Pythagoras generalizes

[fossil mica]

Yeah i said "like"

[upper karma]

there is something called fermat's last theorem

[dark sparrow]

you can have a^2+b^2+c^2=d^2 (with exponents 2 rather than 3) be represented geometrically as a box with sides a, b and c, and diagonal d

but with cubes i don't think there is a good visual

[upper karma]

it says a^n + b^n = c^n cannot hold for integers bigger than 2, if thats what u are getting at

[dark sparrow]

he's got three addends on the left

@fossil mica if you were looking for a generalization of Pythagoras, that's the one

if you were looking specifically for a visual to represent your equation with cubes, i don't think there really is one.

[upper karma]

@dark sparrow by the way this is what i had meant in my original attempt. I am going to of course simplify this because its too verbose but just wanted to make sure if my logic here is correct even?

[dark sparrow]

since when do we know they make a rectangular grid tho.

nor do we need to assume that at all.

you are still overcomp.

by a Long shot

[upper karma]

i mean if its a bunch of squares being laid together then surely its going to make up a rectangular shape?

[dark sparrow]

and re: this, that is too small a number. but you could imagine laying them in a 45° rotated grid relative to the original.

[dark sparrow]

there are other things than transistors in your computer surely.

[upper karma]

okay i relent haha how should i word this better and come to the conclusion of 1/sqrt(2) without my mental gymnastics

[dark sparrow]

yes exactly

zero mental gymnastics required.

[upper karma]

yeah but how should i go about it instead? I am misled currently because i only have my mental gymnasticy attempt circulating in my brain right now

[dark sparrow]

.

you don't worry about the shape of the motherboard they tessellate

[frank hound]

Vector question: Given a vector $v$ and the value of the dot product $u\cdot v$ find the component form of $u$.

[dark sparrow]

!xy

[lime crown]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[dark sparrow]

(NEI as-is!)

[somber coyote]

Cewkins

[dark sparrow]

there is not enough info to find u.

[frank hound]

Its not in english originally

[dark sparrow]

If the original problem is not in English, then post it anyway!

[frank hound]

I am aware

Let me explain my thoughts completely first

[dark sparrow]

no

first you post the problem

THEN you explain your thoughts

not the reverse

[frank hound]

Alright

[dark sparrow]

also what language is it in

[frank hound]

Arabic

[dark sparrow]

ok

i don't speak that, but the math notation should still be legible.

so send it still

[frank hound]

Yeah sure give me a moment

[dark sparrow]

so they really do not give you anything, huh.

[frank hound]

So as you can see the value of u * v (dot product) is given along with v in component form

[dark sparrow]

here is the issue with this problem: there are infinitely many values of u in each case.

[frank hound]

And this is what i am going at

[dark sparrow]

is this the expected answer?

[frank hound]

Using the dot product formula and rearranging the terms i get a linear equation, meaning any point on the line should be correct right?

[dark sparrow]

does your book have any kind of answer key at the end?

[dark sparrow]

as-is, yes.

but the problem looks strange to me still... why didn't it say "find all possible values of u" instead of just "find u"

the latter wording to me implies there is only one

[frank hound]

Nope, and the book doesnt provide examples on this problem

It really sounds more like a conceptual thing

[dark sparrow]

wonderful

[frank hound]

I just wanted to ask if this is a valid answer

[dark sparrow]

so your teacher stuck you with an underspecified problem.

[dark sparrow]

if someone demanded an answer from me at gunpoint, i would give them the same answer as you did.

[frank hound]

He asked about it in class, i gave him this explanation but he seemed not sure himself that much about it so i just thought i ask

[frank hound]

Lmao

[dark sparrow]

,rccw

[frank hound]

Yeah the rotate function

[dark sparrow]

can you translate?

[frank hound]

Yes, 38 and 40 "Find a vector perpendicular for each vector"

[dark sparrow]

oh. so there's no implication that there's only one.

ok.

[frank hound]

What i did was find the slope, then find the perpendicular slope and write a y=mx equation

Every point on that line should be valid, and since the vector is assumed to be in standard form then b = 0

[frank hound]

Well theoretically there shouldnt be, even the length of the vector itself is irrelevant as long as it satisfies the conditions

Interesting questions, i have encountered other ones similar to this that are even harder

Thanks for the help

[dark sparrow]

!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

[shut bronze]

that "Thank you." in #6 is so passive aggressive lmao

[dark sparrow]

is it?

it's not intended to be.

6 is kind of a rare status as i've found

[molten gulch]

2

[dark sparrow]

alright

do you have a diagram

[molten gulch]

no

[dark sparrow]

ok then what progress do you have

[shut bronze]

wow - so I read the problem...
5 lines beginning with "Let __ be.." and then "Show that.."

this is why I don't like geometry

[molten gulch]

To prove that MF=MG and that angles ∠PFO and ∠PGO are right angles, we will use properties of the nine-point circle and the characteristics of the given configuration.

Let's denote H as the orthocenter of triangle ABC, N as the midpoint of AC, and K as the intersection of DE and BC.

Observation: Notice that O is the midpoint of segment EF.

Claim 1: OM is perpendicular to DE.
    Proof: OM is a perpendicular bisector of EF (as O is the midpoint of EF), and EF is parallel to BC. Therefore, OM is perpendicular to DE.

[dark sparrow]

and you did this all blind?

[molten gulch]

you can considerite like that cause i ve done a bad configuration on paper

[grave pond]

That sounds like you do have a diagram, just one you think is "bad"?

[storm patio]

https://media.discordapp.net/attachments/743817386792058971/1211334832546775040/image.png?ex=65edd279&is=65db5d79&hm=e4408137394981dc652ab7110c6fcc40062fe81ce921aaa10aecc8fb10ba2b01&

how do you prove this? (without descending into madness)

[indigo karma]

i got it thank you (took me 2 days ...)

[dusky locust]

ah congrats 🙂

[dusky locust]

to start,

[carmine vapor]

i need help with the last statements

if anyone knows about this pls help

[hoary lynx]

<D is 90 degrees because that is the definition of perpendicular

<E is also 90 because of the same rule

do you see the solution now?

[nimble spruce]

how to approach this

[dark sparrow]

do you have a diagram?

[deft jolt]

how do i do this

i understand that n = Length C square rooted. Otherwise no idea

[summer ruin]

Wanna tip

I can show you the formula

[deft jolt]

yes pls

i had notes but i fell asleep so theyre all scribbled😭

[dark sparrow]

!nosols

[lime crown]

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

[dark sparrow]

(warning shot)

[dark sparrow]

it looks like there is not enough info as drawn

also not clear what the problem asks you to dod

[deft jolt]

the lesson thing is called pythagoras by proportions, it states to find all the missing values

[dark sparrow]

just that?

just so we're safe can you send the entire page

[deft jolt]

yup and this was literally all I was given😭

[summer ruin]

Everything

[deft jolt]

that is everything

[dark sparrow]

so it does NOT say anywhere that any of these triangles are right triangles, then.

[deft jolt]

my teacher said I just have to ASSUME point C is a right angle

this is a reference photo of the notes we took

[summer ruin]

I think I’ll the first half and any of you can do the other half if someone else joins

[deft jolt]

“Find the missing values for each right triangle including the length of the altitude”

is the question

[summer ruin]

What you think Ann

[deft jolt]

oh i understand what you guys ment now😭 it is “Find the missing values for each right triangle including the length of the altitude

[dark sparrow]

see, this is why we ask to send EVERYTHING, incl. ALL written instructions.

[deft jolt]

completely sorry abt that

[dark sparrow]

ok right

can you show the one you were having trouble with again

[dark sparrow]

ok right

well c should be clear, yes?

[deft jolt]

yes

[dark sparrow]

ok

[deft jolt]

and i understand to find N i would have to square root C

[dark sparrow]

no

that is incorrect

[deft jolt]

o

[dark sparrow]

first off lowercase letters

[summer ruin]

There’s 4 problems in total right

[dark sparrow]

second, n = sqrt(AD * DB) maybe

[deft jolt]

i see

[dark sparrow]

third, sqrt(c) doesn't make sense from a dimensional analysis standpoint (it would not be measured in meters like a length should)

now to actually find n here in a way that doesn't cause suffering: consider that ABC and ACD are similar triangles

[deft jolt]

ok

but how would i find the length of AD and DB to get c?

[dark sparrow]

... did you read my last msg

or did you tunnel-vision on the previous one

now to actually find n here in a way that doesn't cause suffering: consider that ABC and ACD are similar triangles

[deft jolt]

honestly im tunnel visioning

[dark sparrow]

ok well dont

ah shit fuck

ok wait no nevermind you can still cook with that

[deft jolt]

so AC is congruent to CB?

[dark sparrow]

what.

is 8 equal to 15?

no of course they aren't congruent.

[deft jolt]

no but we assume that their similar?

[dark sparrow]

we do not "assume" anything.

[deft jolt]

im sorry if im troubling u im just confused😭

[dark sparrow]

blech

here is the diagram again

i claim that $\triangle ABC \sim \triangle ACD$.

[somber coyote]

Ann

[dark sparrow]

do you know what similarity of triangles means

[deft jolt]

yes

[dark sparrow]

ok right

tell me in your own words then

what does it mean

[deft jolt]

that if they had the same ratio of corresponding sides their angles are congruent?

[summer ruin]

Exactly

[dark sparrow]

that "if" is troubling.

the statement "these two triangles are similar" means two things, each of which implies the other:

• each pair of corresponding sides has the same ratio
• corresponding angles are equal

or congruent if you want to get technical or are forced to draw a distinction between an angle & its measure

anyway

CD/AC = BC/AB.

go nuts.

[summer ruin]

How bout trying a Pythagram theorem on AC and CB to get c

[nimble spruce]

no diagram given

[dark sparrow]

ok, but have you made one yourself?

[upper karma]

guys how can i find the value of trignometric ratios more than 270 degrees for example sin(A) where A is more than 270 degrees.
for eg sin(3700)

[dark sparrow]

what's the significance of 270 degrees as a threshold, as opposed to 360 (a full turn)? are you like, allergic to the 4th quadrant or something?

in any case, sin and cos are periodic and their period is 360°.

as is tan, but it goes through two cycles in a full turn, so its period is 180°.

[past geyser]

for the instance of sin(3700)

you're simply making a 360 rotation 10 times

just like what ann said sin and cos are periodic

[past geyser]

also why you gotta be so rude to this gentleman

[dark sparrow]

what part of what i said was rude?

[past geyser]

"are you like, allergic to the 4th quadrant or something?"

i don't know about you but that comes off a quite condescending

[dark sparrow]

i intended it as a joke, and one that's obviously a joke at that -- since mathematical objects are not physical things and thus can't cause any kind of real allergic reaction

[upper karma]

Chill

[dark sparrow]

@upper karma did i answer your question there btw

[upper karma]

Yes

[dark sparrow]

ok

[upper karma]

Thank

[outer cosmos]

Can someone help me?😔😭

[astral shadow]

<@&286206848099549185>

oh wait

we chillin mb

wait we aint chillin

[subtle prism]

An isosceles trapezoid is a trapezoid like this:

Try giving each length of interest a variable name

[worldly nest]

is there any way

to remember the trig identities

very easily

and the value

[dark sparrow]

do you have any particular classes of identities in mind here or just in general?

also what do you mean when you say "the value"?

@worldly nest

[worldly nest]

like

sin pi/2 = 1

[dark sparrow]

so trig values of simple angles

one moment

https://cdn.discordapp.com/attachments/1064643750493691967/1134924411120648213/20230422_100420.jpg

there is this kind of silly mnemonic which i call the "trigonometric hand"

which also fits into the 1st quadrant of the unit circle the way i've positioned it here in the photo

the others, you derive via symmetries.

[worldly nest]

Wow

thanks

but how do i remember the numbers?

[dark sparrow]

they just go 0-4 both ways

if you're talking about the red and blue ones

[worldly nest]

Ok

thank you

👍

[quiet birch]

im having a problem with the last sub question, been stuck on it for almost a day

[inland jay]

Hello people

What do I do with the thirty million trigonometry formulas I have to remember for high school

Am I supposed to memorize them all and be able to actively use them while solving equations, and if I am, how many, roughly?
I am just wondering if I have to eternally solve trigonometry questions or is there an end to it (for high school, at least)

[dark sparrow]

30 million?

where does such a high count come from?

[inland jay]

I was being sarcastic

[dark sparrow]

i made this list a while back that fits on 2 pages

or just over 1 page if you don't count the somewhat pretentious sixth section

(which you can ignore really it's more of a curiosity)

[inland jay]

I see

So learning these will suffice for any trigonometry question I face on the exam (high school)

[dark sparrow]

pythagoras and angle sum/diff are the big ones

think so

i mean maybe triple angle identities too

but you can derive those

from angle sum

[inland jay]

Hmm

I see

[inland jay]

After this, I did an exercise on trigonometry, and it seems they are not that difficult if I am good enough at algebraic manipulation.

It is a relief to hear that the ones I learnt in that chapter alone will suffice for high school.

Well thank you very much.

[solid condor]

Fellas can we ask any trignometry question even if its easy?

[dark sparrow]

(i.e. yes)

[solid condor]

i feel embarrased

actually

easy asf questions for my o levels but still

okay then

[dark sparrow]

we are all at different stages of education

"easy" is wildly subjective

[solid condor]

Agreed

[dark sparrow]

this doesn't look like trig to me.

but ok

progress?

i see some angles marked with something that i read as "8°". is this your doing?

@solid condor

[dark sparrow]

@solid condor are you like, still there or what

[solid condor]

Yeah I saw it but forgot to react
Thanks ;))

I got it now
corresponding angles 🤦

[worldly nest]

Oh wow thanks

[covert quarry]

8 degree 💀

hows 8 degree like an obtuse angle

[lilac narwhal]

I’ve heard of cotangeant, secant and cosecant, but what are they used for?

(currently grade 10 student who learned about sin cos tan recently)

[foggy parcel]

cotangent , secant, cosecant are the opposite trig ratios

like, 1/tangent = cotangent , i.e. 1/tan = cot

[lilac narwhal]

So do they form like another triangle or something lol

[foggy parcel]

u can say so
cuz for example cosec = hypotonous/perpendicular (The opposite of sin)

[lilac narwhal]

Ah, I leaned it as opposite

secant is opposite of cos?

[lime dune]

draw the rest of the height

set up similarity equation

solve

[lusty girder]

Dunno if this channel is appropriate for analytic geometry but gawd dayum these questions were something

The example questions given are definitely not enough to cover this entire thing

[lime dune]

bruh wtf 😭 but yea this is the right channel to put that in for better or for worse lmao

[dusky locust]

loll

[bold verge]

isosceles trapezoid. h is perpendicular. can someone explain? I tried Pythagoras theorem but couldn’t solve it by plugging the values

[hybrid oriole]

hi

any help please?

im struggling so bad lol

!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.

[dark sparrow]

!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

[dusky locust]

pythagorean theorem is right idea

you want to find AB : DC; one way to proceed is let AB = x and DC = y and go from there

[ripe tendon]

Hi

I've been stuck on this question for a few hours

Can anyone help out

I'm pretty sure my answer is wrong

[dark sparrow]

well you miscopied

on your diagram you've got AT marked as x-9

instead of x+9 as it should've been

[plush brook]

It would be really funny i think if that was the only error

[bold verge]

I broke DC like so, then I attempted to do x^2 + (x + y)^2 = (x + 2y)^2. Then I got stuck at x^2 = y(y + 2x). I don’t know how I can find the ratio from this.

[dark sparrow]

x^2 = y^2 + 2xy you say?

add x^2 to both sides of this @bold verge

[neon salmon]

I have an interesting preposition

Why do we not identify triangles as quadrilaterals because they possess 4 sides technically [sherman line]?

[lapis scaffold]

https://tenor.com/view/leonhard-euler-euler-time-to-shred-euler-shred-shredder-gif-24591321

[dark sparrow]

"heehoo i get to proclaim established terminology as wrong on a technicality" is not as interesting a take as you think

[copper comet]

whats a sherman line

[dark sparrow]

some kind of construction that shares some properties with the 3 real sides

[neon salmon]

The Sherman line is a line that shares the same properties as the other three sides

Bisected by the nine point circle tangent to the invircle and a chord of rhe circumcircle

[dark sparrow]

nobody in their right mind would think to call all triangles quadrilaterals based on this. it would be ridiculous.

and if YOU do that, you're guaranteeing that you won't be understood.

[neon salmon]

Although the math is accurate

[dark sparrow]

again, your only motivation for this seems to be to introduce unnecessary confusion into established geometrical language in the name of "technical correctness"

which is just so pointless.

how would you even define a triangle even? if you insist on considering it as a type of quadrilateral?

for that matter what would you need to have as the definition of a quadrilateral which accommodates triangles-with-sherman-lines?

[copper comet]

triangles can be defined by 6 points, does that make them hexagons?

[neon salmon]

Triangles would be renamed to fit the definition of quadrilateral and named on that basis

[dark sparrow]

name me a definition of "quadrilateral" which accommodates both shapes already known as such AND triangles with your sherman line.

clearly you want to redefine that word.

so the stage is yours.

[copper comet]

triangles also have euler lines, why not make them pentagons?

[neon salmon]

I will reply tommorow at this same time

[dark sparrow]

[bold verge]

Thank you!

[opal carbon]

the Sherman line possesses some of the same properties as the 3 sides of a triangle (i.e. tangent to the whatever circles) but it doesn't actually pass through any of the vertexes of the triangle. It's referred to as "fourth side" in quotes, don't take it literally

[dark sparrow]

no you see the fun part is where you take it literally anyway to be self righteous

[ocean oracle]

How to do 6D (diagram on the right)

[lime dune]

what do we know about opposite sides of a parallelogram?

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[ocean oracle]

they are congruent

[lime dune]

and?

there’s one more piece you need for this question

not quite

[lime dune]

they’re also parallel

to get from V to T how do you move in the coordinate plane?

however you move you do the same for getting from R to S

[ocean oracle]

oh yeah

I tried using the distance formula

[lime dune]

you don’t need it for this question

[ocean oracle]

but I got like 50 square root