geometry-and-trigonometry

[trail tendon]

" who can solve this"
theres just a picture of a shape with a few line segments

what we solvin for bruh 😂

[upper karma]

Lol

[sly crown]

ik this is probably a stupid question but theres b level problem in my book that says: draw a segment that has the length of 1, use the segment to construct a segment of length √15. i was thinking of extending the segment by adding a segment length 15 then using geometric mean to find a segment of √15, but idk if im just allowed to extend it like that

[trail tendon]

thats a dumb problem

uhh

[trail tendon]

not what you said i mean the problem XD

i should say the problem

💀

uh

i would use the fact that a 45-45-90 triangle has side lengths 1, 1, sqrt(2)

actually

that doesnt even help

idk 💀

[sly crown]

its ok idk what to do without adding more info but ik im not allowed to do that

[maiden brook]

no it’s not

[trail tendon]

"use the segment" is kind of vague

[trail tendon]

maybe i don't understand ?

[sly crown]

then help hockeydude85

[maiden brook]

use the Pythagorean theorem

[sly crown]

idk if i have enough info for that

[maiden brook]

?

what?

[trail tendon]

tHeY dOnT haVe AnglEs

[maiden brook]

huh?

[sly crown]

i just have a segment

[trail tendon]

you need a 90 degree angle to use the pythagorean theorem

and multiple segments

[maiden brook]

bruh I don’t know what u r saying

[trail tendon]

it says "use the segment"

[maiden brook]

if u have a segment of length 1

extend the segment

[trail tendon]

by sqrt(15)- 1

boom ur done

[maiden brook]

and create another segment of length 1

[trail tendon]

😂

[maiden brook]

u can’t

[trail tendon]

you only have 1 br

[maiden brook]

wtf are u saying

[trail tendon]

"use the segment"

[maiden brook]

have u not done a construction before

[trail tendon]

not use multiple of the same segment

[sly crown]

so im allowed to extend it?🧟‍♀️

[maiden brook]

u can copy the segment

[trail tendon]

it should say that but ok

[maiden brook]

no all u can do is keep the segment where it is and make the segment of length square root of 15 appear

it’s not hard

[trail tendon]

thats how i read the problem 😂

[maiden brook]

that’s stupid af

I was joking

[trail tendon]

and thats why i said the problem is dumb

because thats literally what it tells you to do

[maiden brook]

that’s definitely not what it means

[maiden brook]

no

[trail tendon]

thats not what it MEANS but thats what it said 😂

well

its not specific enough

on what it means by

"use"

"use the segment"

[maiden brook]

It is what it means

[trail tendon]

ok what else can you use

can you use angles?

[maiden brook]

the segment and a straightedge and a compass

duh

[trail tendon]

thats not obvious to me

how is that duh

can i use a ruler?

[maiden brook]

what do u mean can u use angles?

u have to be given a segment or u can’t do it

[maiden brook]

a straightedge

it’s a construction

u can’t try to measure the lengths of segments with a ruler

[trail tendon]

i guess you're just used to this type of problem so you automatically know what its talking about

i haven't seriously used a compass in like 5+ years 💀, kind of forgot they existed

💀

[sly crown]

guys does my solution work or not 💔

[maiden brook]

what did u do?

[sly crown]

backread

[trail tendon]

p

[maiden brook]

I don’t see u saying what u did

[sly crown]

i said it after i said the problem

its in that message

[maiden brook]

this is your solution?

[sly crown]

yep

[maiden brook]

oh oops

[sly crown]

HELP

does it work

😭

[maiden brook]

just use the Pythagorean theorem that’s simpler

[sly crown]

but it still works right?

[maiden brook]

yeah

[sly crown]

bc ik in construction u can pick an radius as long as you stick to it i just didnt know if that logic applied

ok thanks guys !

[maiden brook]

what does this mean?

[sly crown]

like any value

as long as the compass is locked

[maiden brook]

ok?

[sly crown]

okay

[maiden brook]

your construction is pretty good because 15 + 1 is a power of two so if u construct like a segment of length 2 (easily) u can just copy that and copy the length 4, etc

[north kindle]

what tangent line is it referring to exactly?

[north kindle]

thanks

[vernal pilot]

Someone is standing in a lighthouse and sees two boats, they are exactly 480m apart along the west-east line. Boat A is directly north with an angle of depression of 22', Boat B is directly west of Boat A with an angle of depression of 15'. What is the bearing of Boat B?

[obsidian harness]

angle of depression

have you tried drawing it out?

if you make a triangle at sea level, then it will be right angled cause of the west-east line and directly north bits

and the way to do that is to use right-angled trig, for instance if you label the flat distance from the lighthouse to boat A d

then the height of the lighthouse is d tan 22

so you can use that for the other triangle for boat B to get the flat distance there

and then Pythagoras

[warm shuttle]

is this supposed to be spherical geometry?

probably won't change the answer very much?

[obsidian harness]

nah in TheLord26's question you of course assume the Earth is flat

480m isn't enough for any kind of curvature to matter

it's just like some funky textbook trig problem

[formal geyser]

Hello

How do i deal with this kind of equations: sin(2x)=sin(pi/4 - x) without subtracting sin(pi/4-x) from both sides and applying sum to product formula? I tried writing smth like 2x=pi/4 - x and solving for x, but i messed it up when it came to adding 2pik radians

[silent plank]

you'd introduce that 2pi*k in that initial step

[undone quest]

@trail tendon all wrong

[trail tendon]

?

[undone quest]

I asked for Radius

How do u say 12

:+

It is closer to 21 but not 21

Hint: diagonals of a rectangle

[trail tendon]

you didn't ask for anything

i gave you a random number 😂

so ur tryna find the radius?

you said "lets see who can solve this" and you didnt tell us what we solvin for 💀

[undone quest]

I told rsidus bro

Try to move ur eyes little downwardd

[trail tendon]

how much downward

you just said radius != 21, which is obvious but u didnt tell us to solve for it 💀

[maiden brook]

shouldn’t power of a point work?

[lime dune]

pythagoras

[worn stag]

Fix your attitude a bit

[opal salmon]

Could someone explain this question, I do not understand how the andswer is x+2/5x

nevermind i figured out

[gilded bronze]

Made this triangle calculator in excel so I never have to do trig again, thought yall would appreciate it

[vernal pilot]

Trig isn’t that difficult.

[lost wyvern]

You sure?

[vernal pilot]

I’m assuming by trig calculator you mean you plug in some sides and some angles?

[lost wyvern]

Well

Harder trig

[gilded bronze]

Well that's kinda calculus territory

The idea of the calculator is it instantly graphs the triangle

You can play around and see what happens, throw in negatives, use whatever values

[lost wyvern]

Hmm

[gilded bronze]

Does like 5 steps instantly

[lost wyvern]

Any of yall know how I can do this?

[vernal pilot]

If you have access to a calculator this wouldn’t be that difficult.

[lost wyvern]

My final is comin up I have 28 hours left 💀

[vernal pilot]

No need for a specific excel spreadsheet.

[gilded bronze]

You're very negative, don't underestimate the visual aspect

[vernal pilot]

Then study, don’t waste time making calculators.

[vernal pilot]

:shrug:

[lost wyvern]

I'm not the one who made it

Lol

[vernal pilot]

.

[gilded bronze]

I'm an engineering student

[vernal pilot]

You said you did.

[gilded bronze]

That's me lmfao

[vernal pilot]

Omg why do people have the same damn basic pfp

[gilded bronze]

Nice pocket on the job no wifi insta analysis tool

[lost wyvern]

So anyone can help me out here?

[gilded bronze]

I mean it's a factorial problem

[lost wyvern]

So why am I expected to solve it 😭

[gilded bronze]

What class is this?

[lost wyvern]

Pre calc

[gilded bronze]

Yeah that's differentiating or integrating

Either of those sound familiar?

[lost wyvern]

No

But why do you need to do that

[lost wyvern]

Someone earlier told me to use this formula

I tried this but got stuck lol

[gilded bronze]

That's kinda bs for pre-calc, I honestly don't even know. I'll try to figure it out a bit longer, I'm intrigued

[hoary prism]

try to ask a question lil bro

[lost wyvern]

Alr..

Fr bs

[golden copper]

what is lil bro yappin about

[gilded bronze]

@lost wyvern https://youtu.be/-HwwY4czx_E?si=9J7-DYYZF-wexGUr

[lost wyvern]

Let me take look

[gilded bronze]

Skip to like 7:40

[timid cobalt]

Answer is B I think

I don't have paper on me cause I'm about to sleep

Use the formula sin(90-theta)=-sin(theta)

And similarly with cos too

So for the top write sin(90-89) in place of sin(1) because then you can cancel it out with sin(89) you see?

If you do this then sin45 will be the only one left

Sin45 times 2 is rad 2

And the cos on the bottom all cancels our to 0 so denominator is just 1

So x= rad 2

So the answer is 2 so B

[undone quest]

@golden copper i clearly wrote there Radius !=21

What can you concludr from this ?

Yes u need to find Radius

[golden copper]

?

Who are you

[undone quest]

I am me

[golden copper]

no one cares + didn't asked

[undone quest]

.

[golden copper]

I asked who are you and said "I'm me"

Like no one knows your you

[undone quest]

Me means I

[golden copper]

Greatest invention of all time

[undone quest]

C'mon dont be too serious It was just a question

Just to test patience level

[golden copper]

Ok

[undone quest]

United by maths differentiated by country

[golden copper]

Your Indian?

[undone quest]

Yup

[golden copper]

Me too

[undone quest]

Are bhau

United by country also

[golden copper]

aap nam kya hai?

[undone quest]

Ye angrez logon ko kuch bolo sale bhadak jate

@golden copper myself deb

[golden copper]

Aacha

[undone quest]

@worn stag I have my attitude fixed

[golden copper]

What you want to find in this one

[undone quest]

Radius

[golden copper]

You only stated the small information

[undone quest]

As i said earlier

[golden copper]

Alr

[undone quest]

Must have been a reason for me to state Radius is not 21

[golden copper]

I was confused when you said "!=" And I realised it was not equal to, because I usually use cross in equal sign to denote it

[undone quest]

I should have given Radius = ? But it was too straight forward

In computer ?

[golden copper]

Can you give me the picture again

I think it was just easy

Because you can derive them

Alr

9th

Maybe the radius is 12? Just asking because recantangle in bottom is 12 so another side also have to be 12

I think the bottom doesn't mean radius

Oh ok

[mellow dawn]

I have seen this question somewhere

Is it from mind your decision

?

Is it 21.25 cm

As you can do it from intersecting chord theorem

[tame bobcat]

Hey

Indian

Wahh

kya question he?

be

Yeh kahi to dekha

21

Wo he?

[mellow dawn]

You can do it by Pythagoras also

[tame bobcat]

Approx nikala

[mellow dawn]

But intersecting chord theorem Is an alternative solution

[tame bobcat]

nhi meh approx bola

[mellow dawn]

English please

[worn stag]

Doesn't matter

Speak English here

I know who can

[mellow dawn]

This server is international

[worn stag]

Irrelevant

[tame bobcat]

9 ^ 2 + (x + 16) ^ 2 = (9 + 12) ^ 2 + x ^ 2 x = 3.25 r ^ 2 = (9 + 12) ^ 2 + 3.25 ^ 2 r = 21.25

[mellow dawn]

Check #rules

[worn stag]

What are you yapping about

[tame bobcat]

You should use English yeah

I'm sorry too

[mellow dawn]

Check #rules

[tame bobcat]

🙏

[golden copper]

@undone quest Let me solve this problem later, Im currently studying factoring

[tame bobcat]

We got

21.25

Using intersection chords theorem

[worn stag]

?

Stop pinging me

[golden copper]

yappin the familia

LOL

[worn stag]

Or atleast say something coherent

[tame bobcat]

16(16+2y) = 12(30)

[worn stag]

I said stop pinging me

[tame bobcat]

Bro let me finish

TF you guys yapping

[worn stag]

or atleast say something relevant

[tame bobcat]

It's a doubt server

[golden copper]

It's nice to see catgod crying

[worn stag]

You're annoying so stop it

[golden copper]

LOL

[tame bobcat]

Please

[worn stag]

[golden copper]

lol

[worn stag]

<@&268886789983436800>

[golden copper]

Do it again

[tame bobcat]

16(16+2y) = 12(30) 16(16+2y) = 360
16 + 2y = 22.5.
y = 3.25 therefore, radius = √(9^2 + (16+3.25)^2) = 21.25

@undone quest

Read

[golden copper]

Your doing it rn

Bruh

[worn stag]

It's a good thing I have them blocked

[golden copper]

Don't ping him

[tame bobcat]

Using intersection chords theorem we got 21.25

[golden copper]

LOL

He got purged

Oh he aint

[tame bobcat]

What's a purge?

[worn stag]

Mass delete of messages

[golden copper]

Deleting things

[tame bobcat]

I see..

[golden copper]

Nope

[spark stag]

please get back on topic

[tame bobcat]

Yesssir

I gave him the theorem and solution

Yet he continued yapping

[golden copper]

Purge means deleting something massive not only messages, example humans

[tame bobcat]

[golden copper]

[worn stag]

Mind your decisions have good problems sometimes

[tame bobcat]

Alright no fighting

But where do I find more help channels

I couldn't see em

[lost wyvern]

How will rewriting them all as 90-theta cancel anything?

[timid cobalt]

Here

So if we replace sin(90-89) for sin(1) it will be written as -sin(89)

That cancels with the sin(89) in the end

[lost wyvern]

What

Why negative?

[timid cobalt]

There is a proof on that

I don't remember it on the top of my mind

[lost wyvern]

If sin(-x) then -sin(x)

But you can't have sin(x) be -sin(x)

[timid cobalt]

Oh shoot

[mellow dawn]

You can use Sina +sinb formulas

@lost wyvern

[lost wyvern]

But what's the point of rewriting it all as Sin(90-89) + Sin (90-88) ..etc

[timid cobalt]

Nah nah forget that

I mixed up my formulas

[lost wyvern]

You still got the right answer

😭

[mellow dawn]

Use Sina +sin b formula

[vagrant snow]

The best answer is use calculator 🔥

[lost wyvern]

I tried but it didn't help

[mellow dawn]

Can you show your workings

[lost wyvern]

I factored out 2sin45 then got stuck

[timid cobalt]

My formula is right Google is dumb

I wish I had pencil to show you

[lost wyvern]

Rip

[mellow dawn]

How you took 2 as a whole common?

[timid cobalt]

@lost wyvern do you have paper

[lost wyvern]

It's already in the problem

[lost wyvern]

Ye

[timid cobalt]

Ok here

[mellow dawn]

Send one more question

[timid cobalt]

So write this out for sin1

Sin(90-89)

And then that will simplify to -sin(89)

sin(2)=sin(90-88)=-sin(88)

[lost wyvern]

Oh

[timid cobalt]

Sin(88) is the second to last term in the sum

[lost wyvern]

OMG

[timid cobalt]

It all cancels out except sin45

And all the cos cancel out to zero

[lost wyvern]

So they all cancel out with a Sin45 left

[lost wyvern]

Same rule?

[timid cobalt]

Yes

[lost wyvern]

Cos 1 is Cos (44-1) which is -Cos(44)??

That sounds wrong 💀

Ok ignore that for a second

[timid cobalt]

No

I am wrong sorry

[lost wyvern]

How would you write cos like that

[timid cobalt]

I think cos(90+x)=-cos(x)

[lost wyvern]

Yes

[timid cobalt]

Let me find my trig book

[mellow dawn]

It should be (2cos44 + 2cos43 ......2cos1 ) + 1 in numerator

While I have solved

[lost wyvern]

Why the 2's in front?

[mellow dawn]

2s?

[lost wyvern]

The formula is 2Sin(etc)*cos(etc)

[mellow dawn]

Yes

Then

?

[timid cobalt]

@lost wyvern I'm so stupid

[lost wyvern]

Why is there a 2 in front of all your cos values

[timid cobalt]

I keep on messing stuff up

[lost wyvern]

What did you discover

[timid cobalt]

sin(90+x)=-sin(x) right

[lost wyvern]

Yes

[timid cobalt]

I'm so lost

Google be tweaking

[lost wyvern]

How did you get the right answer lol

What did you do

[timid cobalt]

I did my method what I told you

But apparently I did it wrong

So I have to dig out to find the right formula

[lost wyvern]

Cos(90-x) = Sin(x)

Cos(90 + x) = -sin(x)

[timid cobalt]

Let's think of another way

[lost wyvern]

Can we do it from the formula of SinA + SinB

Ahhh

[mellow dawn]

Because of the formula

[timid cobalt]

I did it

[timid cobalt]

Yes use that

[lost wyvern]

There we go

[mellow dawn]

Wait why did you do that

[lost wyvern]

Everything cancels out so then you just have 2sin45

Which is sqrt2

Then you can do logbase sqrt2 of 2 which is equal to 2 and that's the answer

[mellow dawn]

How 1/2 came?

[lost wyvern]

Factor out 2

See how I have 2/2 *

[mellow dawn]

What?

[timid cobalt]

Cancel out numerator and denominator

They are the same thing

[mellow dawn]

But how 1/2 came?

[lost wyvern]

Because when you do the formula everything has a pair

But sin45 doesnt

So when you factor out 2sin45 your doing sin45/ 2sin45

Which is 1/2

[mellow dawn]

But there is already 2 in common

It is in question

[lost wyvern]

2 isn't in common for Sin45

[mellow dawn]

So

It can be 1 right?

[timid cobalt]

I think it is 1

[lost wyvern]

2 ( 2sin45 * cos44 + 2sin45 * cos 43.. )

But when you get to 45 degrees, there's no 2 angles to add up to 90. 45+45 is 90 but you only have one 90.

So then at the end there's a ... + sin45

Then when you take 2sin45 common

It becomes 2(2sin45) [cos44 + cos43.. + 1/2]

[nimble sonnet]

I have school starting next month and my math is geometry. I want to be as prepared as possible for it. Can anyone give me ideas to get prepared?

[timid cobalt]

I see what you did

But that means you cannot cancel with denominator because 1/2 and 1

It's a 1 at the end

I have to go sleep it's 1 am here lol

[mellow dawn]

Study

[nimble sonnet]

What else? Get a notebook to take notes, don't almost fail like Algebra? (itz cuz i didnt do my homework)

[mellow dawn]

What grade are you in?

[nimble sonnet]

going into 8th

[mellow dawn]

Don't waste your time

[nimble sonnet]

wdym? its a highschool credit. I need to pass cuz i dont want to retake

[nimble sonnet]

I'm still gonna study and take notes

[golden copper]

Do homework

It make you practice maths

[nimble sonnet]

lol

I will

[golden copper]

@nimble sonnet your 8th and your undergraduate?

;-;

[nimble sonnet]

I FIXED MY ROLE

sorry

[worn stag]

That’s normal for some people ig

[golden copper]

Also changed your name because it's hard to spell but it's your choice just saying

[golden copper]

I see

[worn stag]

Kids these days are smart sometimes

[lime dune]

HS geometry is such a joke of a class that you don't need to do much in the way of preparation

[queen cosmos]

guys, why is the principle value of tangent is only related to sin(x)

[lime dune]

they slow walk the shit out of it anyway

[queen cosmos]

like why only -p/2 p/2

[golden copper]

Like me when I was kid I thought moon has infinity gravity so it not haven fall

LOL

[queen cosmos]

but not p, 1

[lime dune]

just be able to do basic algebra and you'll be fine

[nimble sonnet]

Yeah but I got a C in Algebra 1, C- is failing in my school

[queen cosmos]

i think cos should also be considered because tan is sin/cos

[lime dune]

then brush up on your algebra

[nimble sonnet]

ok

[queen cosmos]

can someone explain this question pls

[golden copper]

You should know prealgebra before diving into algebra 1

[queen cosmos]

nibbas answer the question pls

[nimble sonnet]

I knew x + 8 = 12 solve for x when I went into algebra 1

[worn stag]

It’s defined that way

[queen cosmos]

yeah but why

[golden copper]

Nah bruh that is not only covered in prealgebra

[queen cosmos]

like i want to understand it

[golden copper]

It like ton topics

[queen cosmos]

why only -p/2 p/2

[lime dune]

if it weren't that way i could give you infinitely many angles with a given trig ratio

[queen cosmos]

why?

[nimble sonnet]

sin, cos, and tan, are part of triangles right? or division

[lime dune]

the circular trig functions by themselves are not invertible

[worn stag]

Try plotting its graph

[lime dune]

we restrict their domains so that we can invert them on those restricted domains

[worn stag]

It’s defined to make it continuous for all values in its range

Because those ugly asymptotes

[obsidian harness]

,w arctan(-21)

[somber coyote]

Results provided by WolframAlpha

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

Requested by sooouth.

[obsidian harness]

by definition, if you take the principal angle, you are just adding or subtracting multiples of 2pi

so the principle angle has the exact same position on the xy-plane

*principal

and hence tan, sin, cos, csc, cot, sec or whatever will also be the same

for this question, you just needed to find arctan(-21)

and all possible angles that give the same value when tanned (the inverse of arctan, to get back -21) are $-\arctan 21 \pm 2k \pi$, where $k$ is any integer

[queen cosmos]

thank you @obsidian harness

god bless you

[obsidian harness]

no worries!

[somber coyote]

south

[nova kraken]

How do you learn math effectively?

[nimble sonnet]

Book Approach: https://www.youtube.com/watch?v=_-GQ_T_0dgU
https://www.youtube.com/watch?v=fb_v5Bc8PSk
https://www.youtube.com/watch?v=J2OT5HAMYyw

[lyric epoch]

GUYSS!! i have an observed formula for

sin values

can anyone tell me a method to refine it??

[lost wyvern]

It's a 1/2 because you factored out 2sin45 while it's just a single sin45. Sin45/2Sin45 becomes a 1/2.

Every angle had a pair adding to 90 so you could apply the 2sin45cos(etc) formula but sin45 didn't which is why it's alone

@anirudh.4.real

[dapper geode]

help

what is the area of a 25 sided polygon with a radius of 8

[faint pasture]

Well no

Cuz 90

This is like saying

Sin(alpha) = 0.34 is a formula sure it works for infinitely many values but it doesnt for way more

Tho this is kinda close to the first term of taylors series of sine so maybe

(Pi/180 ≈ 1/60)

[obsidian harness]

Google exists

[warm shuttle]

subdivide into triangle pieces and you have A = br/2
where b is the perimeter of the polygon and r is the radius of the inscribed circle

(τr)r/2 for the area of the circle

as for the perimeter we have
tan = opposite/adjacent
tan(τ/50) = opposite/r
b = 50 r tan(τ/50)

A ≈ 202.127

A (circle) ≈ 201.062

[inner lark]

find k so that y +x=3k is tangent to y^2+x^2=k

[inner lark]

googles explanations r kinda lacking ngl

[obsidian harness]

sub in y = 3k - x into y^2 + x^2 = k and then you have a quadratic in x

thus the discriminant b^2 - 4ac must equal 0 to be tangent

[warm shuttle]

if a + b = 90°, then
tan(a)tan(b) = 1
is there some way to generalize this to triplets of angles that sum to 180°?

assymetrically is fine

[obsidian harness]

https://math.stackexchange.com/questions/477364/tan-a-tan-b-tan-c-tan-a-tan-b-tan-c-in-a-triangle?noredirect=1

black magic

[warm shuttle]

this breaks if you have a 90° angle tho

[obsidian harness]

I mean this also breaks if a or b is 90

[warm shuttle]

dividing both sides by tan(C) sort of fixes it

if you are willing to say that finite/tan(90°) = 0

[outer wadi]

why is tanc=0,tanc=-1 rejected

[hoary totem]

tanC = 0 implies C = 0
tanC = -1 implies C = -45°

both do not make sense in a triangle

tan C = -1 can also imply C = 135°
which could make sense or not depending on the triangle

[twin spindle]

Did I get this question correct? I calculated C to see if its the seconds highest side, which it was, but still not 100% sure

If you anyone can't read the shitty handwriting btw, x=80.49 and c=92.54

[outer wadi]

Got it thanks

[fluid jackal]

Need help

[obsidian harness]

PR and SQ intersect at right angles, cause it's a rhombus

so what must angle x be?

[fluid jackal]

Can Angle sum property of a triangle be used?

[obsidian harness]

ofc

[fluid jackal]

So x 50°

[obsidian harness]

yep!

[fluid jackal]

So y?

[obsidian harness]

also SR and PQ are parallel

[obsidian harness]

honestly if you guess you'll probably be correct

[fluid jackal]

Is it 40°?

[obsidian harness]

yeah

[fluid jackal]

Alternate interior angle?

[obsidian harness]

a rhombus is just a special parallelogram so opposite sides are parallel

so by alternate angles, angle QPR = SRP = 40

[obsidian harness]

ye

[fluid jackal]

K thanks for the help!

[obsidian harness]

no worries!

[flint marlin]

Joining bp

Or assuming o to be the centre and joining ob and op

Nothing worked

I guess

[pearl plaza]

Yo whatup

[exotic yarrow]

[multiposted here: #help-33 message]

[thorn aspen]

[fringe steppe]

thank god you corrected yourself

[trail tendon]

bro replying to something that happened four years ago:

[rare egret]

this is hard for me

it look easy but i cant get it

can somebody help

[grave pond]

Cut triangle CED off, turn it and reattach it in an appropriate place ...

[warm shuttle]

How do i know AE?

oh

[upbeat pond]

Hey can anyone suggest a good video for trigonometry... I am in junior year

[pearl osprey]

guys I know this is kinda physics but related anyways

isn't the depth the z axis and the height the y axis?

[obsidian harness]

I'd advise you to stare at this picture until it makes sense

btw this is maths, it has applications in physics with the magntiude of vectors of course

[fringe steppe]

iik

[warm shuttle]

there's the whole

(i am not Freya Holmer)

[obsidian harness]

LOL

[proper wren]

Can anyone give me a complete roadmap to trigonometry ?

[obsidian harness]

https://archive.org/details/algebratrigonome0000stew_w1b2

Stewart's algebra and trigonometry

chapters 5, 6, 7 are on trig, including double-angle, half-angle, and product to sum equations

[proper wren]

W man (Wow I am talking like a cool boys)
I always would love to solve some tricky questions

[obsidian harness]

btw Google can make pie charts for you

of course if you want to do it properly you need to calculate the right angles and then draw them with your protractor

[fluid jackal]

Didn't know that thanks

[obsidian harness]

yeah Google Sheets

[fluid jackal]

K

[obsidian harness]

https://support.google.com/docs/answer/9143036?hl=en#zippy=%2Cmonthly-expenses

check otu the examples

np

[exotic laurel]

I can't believe I'm asking this, but this is unsolvable, right?
You can imagine sliding the point x on its line and it wouldn't affect the given angles. This means there simply isn't enough information.

[ripe loom]

Point x is fixed though since the length above it is equal to the base

[mellow quest]

https://www.youtube.com/watch?v=5vhklRWogzo

classical problem

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

[exotic laurel]

Do those small lines on each of those segments denote that they are equal?

[nocturne remnant]

yes

[proper wren]

20 degree is the answer i think

[obsidian harness]

40

[nocturne remnant]

this is an impossibly bad guess

[obsidian harness]

wait

[nocturne remnant]

its a different problem

btw

😭

[obsidian harness]

ah yeah no wonder

[exotic laurel]

Fuck my life. Had I know that, I would have already solved it. I spent 30 minutes feeling like an idiot just because I didn't know about it.

[proper wren]

Sorry

It's not your fault

You just have to manipulate your brain to do this

[exotic laurel]

I hate this. Imma be back.

[proper wren]

Good luck

[exotic laurel]

My geometry is a bit rusty NGL.

Never done a lot of it since secondary school.

Actually, did some in high school but not much.

I gotta practice some before I even attempt the easiest of problems or I will end up spending a lot of time feeling like an idiot.

[glass grail]

Hiii

I need a refresher on geometry (:

I’m decent at the basics

[deep kettle]

does anybody have tips for determining how many degrees an 1 object has rotated from another object?

^^ "determining rotations

[upper karma]

guys I just started trigo today

i mean I have done it before a bit (cuz IMO) but proper start was today

[obsidian harness]

in how many dimensions?

you should just post the question that made you ask this question

the original one

[urban axle]

I just started reading Euclid's elements, I'm wondering why does he reference circles using three points on the circumference when he earlier writes that a circle can be described by a center and a rdaius? Why not reference the circles like A, AB or something?

is that how circles are referenced in modern math, with 3 points on the circumference?

[grave pond]

There's no deeper point to it -- it's just a practical way to explain to the reader which of several possible circles in the diagram he's talking about.

[brazen epoch]

How can i reliabily find the angle of rotation of a conic with a xy term?

Just for curiosity

[urban axle]

Thank u

[sand patrol]

How to find AOC

[grave pond]

There are seveal ways. One of them starts by noticing that triangle AOB is isosceles.

[echo canyon]

Oh yeahhhh right

i didn't think like that. I just took AOC and ABC as 2 triangles

[twin spindle]

Im confused on how to set up the equation

Cause their are 2 angle of elevations, from the observers POV

And by the way I drew the picture, their can only be one angle

Akso "BL" stands for balloon.

[lime dune]

presumably one balloon is right above the other?

otherwise we don’t have enough info to answer the question as you probably noticed

[undone summit]

can someone help me study for geometry cbe please

[pearl osprey]

now idk if to use unity or unreal

guys I think there's a typo on my book:
lim x -> 5 (g o f) (x)
if f(x) = { (x-1)² + 1 if x < 4,
10 if x = 4,
3x - 2 si x >4
}
and g(x) = ((x-1)² - 4/(x-2))

the answer says 14 but I get 140/11

I've asked to gemini it says 140/11 also

desmos says

so you are my last check

🥸

[warm shuttle]

i also say 140/11

[vagrant snow]

Z as up is diabolical

Idk it just feels wrong😭

[sand patrol]

Can Anyone help with this

[obsidian harness]

i) Pythagoras
ii) expand (x - 1)^2 + (x - 2)^2 = 5^2 and rearrange
iii) quadratic formula

then once you find x it's just x - 2 and x - 1 ofc

[sand patrol]

ii one is quadratic formula right?

[warm shuttle]

no?

[zealous pike]

I think I understood how it arrived at the point for pi/4, but I can't find a way to arrive at the points for the other ones in the table. Could someone give me a light?

[vestal hound]

u can just find the side using pythagorus and quadratic and just prove the solution of the equation and the solution of the given equation are equal

right?

[vestal hound]

I also want to learn that kind of maths but I got no clue

[obsidian harness]

the other values use the 30-60-90 triangle

sin 45 and cos 45 use the 45-90-45 triangle

[zealous pike]

I want to understand it so I don't have to memorize the table

[obsidian harness]

so SOHCAHTOA gives sin 30 = 1/2, cos 30 = sqrt(3)/2

[warm shuttle]

https://www.youtube.com/live/yBw67Fb31Cs?t=2238

(timestamped)

[vestal hound]

I want to learn everything but I don't have the teacher..

and don't know where to start

[zealous pike]

I am learning by myself

[vestal hound]

same

I'm kinda lost

stuck at basics

basic trig basic calculus and everything

grade 9 is kinda trash

[warm shuttle]

read some books really

[vestal hound]

no interesting chapters in my book

[vestal hound]

Ok I'll try my best to find some

can anyone tell me what is sin inverse

and are sin inverse and arcsin the same?

[warm shuttle]

probably yeah?

[vestal hound]

take angle ABC as x and angle AOC as 2x and see that triangle AOC and ABC both are isosceles

[vestal hound]

ohh is it used to find ratio

of two sides at a certain angle

[warm shuttle]

it's used to undo sin

[vestal hound]

ohh

how to find the angle if I know two sides

[warm shuttle]

arcsin(sin(x)) = x if 0 <= x < 2π

[warm shuttle]

depends

which angle?

[vestal hound]

like any

[warm shuttle]

and are we talking about a right angled triangle?

[vestal hound]

yeah

I have a formula idk if it already exists but the angle thing could help me

[warm shuttle]

soh cah toa
sin = opposite/hypotenuse
cos = adjacent/hypotenuse
tan = tg = opposite/adjacent

[vestal hound]

i know that but if i know the sides can i find the angles?

[warm shuttle]

yes

[vestal hound]

how?

I don't know alot so please help me gain knowledge

[warm shuttle]

well one entirely valid options that ancient astronomers did is to measure them

[vestal hound]

are there like formulas?

[warm shuttle]

but if you can't for some reason then use that

[vestal hound]

ohh

if sin p/h = 3/5 can we find angle?

[warm shuttle]

huh?

[vestal hound]

idk how to explain

p/h=3/5

[warm shuttle]

no it's like
h = 5, p = 3

[vestal hound]

the angle between them

[warm shuttle]

sin(angle) = p/h

[vestal hound]

ohh

[vestal hound]

yeah that angle

[warm shuttle]

and if want to know the angle you use arcsin

[vestal hound]

yeah i want to learn that superpower

[warm shuttle]

arcsin(sin(angle)) = angle = arcsin(p/h)

actually calculating it is something you would usually leave to a calculator

[vestal hound]

ohh

[warm shuttle]

it's like sqrt

[vestal hound]

i thought we could find it on our own

[warm shuttle]

you can calculate it

but it's pain

[vestal hound]

ohhh

would u tell me how to approx it

[warm shuttle]

https://dsp.stackexchange.com/questions/25770/looking-for-an-arcsin-algorithm

[vestal hound]

thank you so much for your precious time 🙂

I'm determined to learn mathematical superpowers

[obsidian harness]

a mathematician turns coffee into theorems

they can do alchemy

[vestal hound]

I'll try to learn it.

[obsidian harness]

I was joking with you.....

[vestal hound]

ik but I'll try to learn everything possible to know

like I'm addicted to maths

[autumn olive]

Ok

@vestal hound

[lunar jackal]

When writing transformations in order, which order do I go in? My school teaches me to go from H shifts, Reflections or Stretches / Shrinks, and then V shifts. However, two of my tutors tell me to go from left to right: Reflections, Stretches / Shrinks, H shifts, and then V shifts

[trail tendon]

i agree with your tutors but honestly it doesn't really matter, you can kind of do it any order

[sterile beacon]

I thought that perpendicular means when 2 lines like this + has four 90 degrees angles.
I don't see how PC T (upside down) AB

shouldnt it be PP' T (upside down) AB

[trail tendon]

the lines don't have to extend past the point of intersection, they just have to intersect at a right angle

[shadow falcon]

hi, are ther any tips or tricks to easily graph quadratics, at the moment i know how to graph these with relative ease, but was wondering if there are any visual rules of graphs that may help me graph quicker

ik this may be a bit of a broad question

[cunning lion]

usually it suffices to find the vertex and direction, possibly plot a few points near the vertex, then sketch a parabola

[green swallow]

yo anyone got any tips for entering geometry?

[trail tendon]

learn ahead

[green swallow]

okay thank you

[sterile beacon]

oh, thanks!

[fluid jackal]

Can we find the no. of sides of a polygon without finding the exterior angle?

[obsidian harness]

yes

interior angle of a regular polygon = 180( n - 2)/n

[fluid jackal]

Ohh forgot that

[obsidian harness]

it's actually derived from the exterior angle formula you know

[obsidian harness]

180 - 360/n

= (180n - 360)/n

= 180(n - 2)/n

so honestly just use the exterior angle formula cause it's more natural

[still sun]

Find the function R as n approaches infinity. R raised to the power n are iterations of the function R

[hollow flame]

Bros

[obsidian harness]

search up De Moivre's rule or whatever it's called in German

https://de.wikipedia.org/wiki/Moivrescher_Satz

[lucid panther]

can anyone help me by explainting what the Congruent Diagonals Theorem is?

[mellow silo]

You mean like within certain quadrilaterals?

Basically means that in squares, rectangles, etc there are diagonals (from vertex to opposite vertex) of the same length, so if I draw a line between two opposite vertices in a few types of quadrilaterals it will be the same as any other line drawn.

[ashen garnet]

is it not possible for it to be outside A and B because if it was outside than XAC+BAC wouldn't equal 180 and thats not possible

[obsidian harness]

no, that's not the right reasoning

the key is that ACB = 100

observe that any point outside the circle will make ACB be acute, less than 90 deg

and any point inside the circle will make ACB be obtuse

clearly if X is outside line AB, there is no point on line g that will be inside the circle

if X is between A and B, then AXC + CXB = 180

CXB is a totally different angle to ACB

https://www.geogebra.org/calculator/qejfwhg9

[ashen garnet]

Ohhh ok I think I get it now

Thanks for the help

[obsidian harness]

no worries

[lucid panther]

thanks bro

[mellow silo]

No problem

[fluid jackal]

How? (i)

[silent plank]

do you understand the hint?

[fluid jackal]

Ya

[silent plank]

did you try applying it, where did you get stuck

[faint pasture]

How old are you btw

[fluid jackal]

13

[north kindle]

are all quadrilaterals with perpendicular diagonals kites?

[vagrant snow]

I'm pretty sure one has to bisect the other too

[upper karma]

Tell me it wasn't in class 8th

Ncert

I think that chapter was called

data handling

[fluid jackal]

Correct

[upper karma]

Or statistics

[fluid jackal]

It is

[upper karma]

Total marks 540

540 = 360°

So u can equate ig

[fluid jackal]

Kk I already got it

[upper karma]

ok

Chill

Wait you're 13-

U joined 2021

When u were 11

That's a war crime

[fluid jackal]

What's the prob

[upper karma]

WE GOT CODE RED!!!

it's not allowed

Min age is 13

[fluid jackal]

Bro...

I'm 8th grade

[upper karma]

State

[fluid jackal]

Wut?

[faint pasture]

Not really

[faint pasture]

He 13 now so yea

[vagrant snow]

I mean only one has to bisect the other

They don't need to both bisect each other

[faint pasture]

Yea that is enough

Cuz you can prove by congruence

[grave pond]

No -- it's a simple matter to just pick one that isn't, e.g. the one with vertices at (-1,0), (0,1), (2,0), (0,-2).

[tepid pasture]

does anyone mind if they could give me a brief demo or explaination of trigonometry

[tulip seal]

What is this called

[trail tendon]

two trapezoidal prisms stacked one on top of the other ._.

idk

idevk if it has a name 💀

[urban axle]

was watching this https://www.youtube.com/watch?v=6Lm9EHhbJAY and at 7:00 she shows a cool Euclid proof of how to make a square with twice the area of a given square, and hints that there's a similar way to do so for three times the area. I can't figure it out. Does anybody know what I can search to find it?

[maiden brook]

or two intersecting pyramids

[warm shuttle]

if you call side length of the given square 1, then you just need to construct a line segment that has length sqrt(3). One famous formula in geometry that has sqrt is pythagoras theorem

want: a² + b² = sqrt(3)²
if you have a = 1 and b = sqrt(2), then you should be good

So this gives one way to do it

draw diagonal of the square

draw perpendicular to it

use compas to construct a circle and cut off 1 unit length on that perpendicular

now draw the hypotenuse of that triangle

now construct a square on that hypotenuse

done

This almost certainly isn't the solution that requires least steps, but it's pretty easy to come up with

[urban axle]

Oh snap thank you for this explanation, beautifully put

[upper karma]

pythagorean theorem?

[obsidian harness]

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

[urban axle]

wow

Unrelated but I'm reading Euclid Book 1 Prop 31 - To draw a straight-line parallel to a given straight-line, through a given point. It seems overly complicated? Why can't you just make a perpendicular line through a random point on the given line, and then another perpendicular line through the given point to it?

is it a thing where it's easier to construct that way but more difficult to prove or sm?

[obsidian harness]

https://mathcs.clarku.edu/~djoyce/elements/bookI/propI31.html

what Euclid is saying it can be much more general than that

you can take any angle in fact

as long as the angle DAE equals the angle ADC

if DAE = ADC = 90 then that's a special case

[urban axle]

ah ok thank you

[urban axle]

https://arxiv.org/html/2407.07109v1

[obsidian harness]

looks like Word lmao

but yeah that's cool

[urban axle]

lol

[north kindle]

im having trouble with this problem
i know PQB similar to CQA
CPQ similar to ABQ
CQ/PQ = AC/PB
but im not sure where to go from here, theres no other similar triangles i can find and i cant see how to use what i have here to solve the problem

[obsidian harness]

AQ * QP = BQ * QC by power of a point / intersecting chords theorem as well

[north kindle]

oh true

i just completely forgot that was a thing lol

[obsidian harness]

lol it happens

[north kindle]

still not sure what to actually do with that though

[obsidian harness]

me neither

[north kindle]

wait i might have something?

ive gotten that CQ/PQ + BQ/PQ = s/PC + s/PB which seems close

oh yeah thats it

[obsidian harness]

ah right s/PQ = s/PC + s/PB

nice

[bleak lance]

P+2k?

3n+1=?

[fringe steppe]

hi

How come roq is bisected (into x and x)?
ping me please
This was the first question
this is the second

ANOTHER QUESTION:

How BQ bisects ang B in triangle ABC?

[warm shuttle]

idk the answer, but you will love this one

(drawing is not very accurate tho)

[fringe steppe]

I agree

atlaest for the second one

AC and Ab are tangles.

tangents*

PQ is the diameter

[warm shuttle]

see
https://en.wikipedia.org/wiki/Morley's_trisector_theorem

[stuck kindle]

Could someone give me a hint for 167 here? I circumscribe the triangle then extend MD to get inscribed angles which subtend the same arc, creating equilateral FAD. But, I don't really know where to go or if that's helpful

[warm shuttle]

ADC is similar to AMD by 2 angles. the rest follows directly

[stuck kindle]

where's the second equal angle?

[fringe steppe]

it's given?

one is common and the other is given

[stuck kindle]

oh CAD is common sorry, thank you

[fringe steppe]

np

,w substitute c = root(a^2+b^2 - 2ab*cosA) in root[(a+b+c)/2)(a+b-c)/2)(a+c-b)/2)(b+c-a)/2]

[somber coyote]

Results provided by WolframAlpha

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

Requested by anshulyolo121

[fringe steppe]

tada, 1/2 ab sin(A)

[obsidian hornet]

Q is incenter of ABC

[obsidian hornet]

recommend to see the latest Morley's theorem proof, suggested by Tran Quang Hung

[fringe steppe]

nuh uh

uh huh

yeah yeah

I thought you were referrign to the first one

[fringe steppe]

how?

[lucid panther]

just to make sure the distance formula is x2 -x1 to the power of 2 plus y2-y1 to the power of 2

and then square root right?

[obsidian hornet]

Famous 90+A/2 angle

[fringe steppe]

yeah

[obsidian hornet]

Saving in plenty of problems

[fringe steppe]

that requires bisection of B and C.
But I am literally asking how can we say it bisects B and C?

[lucid panther]

meoove accept my friend request

[fringe steppe]

Okay

I am the type of person who uses the internet in the airplane mode

[obsidian hornet]

No, you are already have that angle in your theorem condition(90-A/2), so, there can be only one angle with measure 90+A/2

[fringe steppe]

okay so something other is provided

another question

in the triangle problem, look at the last line.
How we transition from the last third line to the next one?

[lucid panther]

i think i answered ur problem but i meight be complety wrong

[obsidian hornet]

Sorry, had to go, its simple; (AXC)=(BZC)=(AYB)=2/7=> (XYZ)=1-6/7=1/7

[lucid panther]

help

If x,y are positive reals such that x+y=2
show that (x^3)(y^3)(x^3+y^3)≤2

i dont get it

[obsidian hornet]

Buy the way, in my opinion in this problem its better to calculate (ARX), coz 3(ARX)=XYZ

[grave pond]

One approach would be to write x=1-a, y=1+a in which case x³y³ = (1-a²)³. Then the LHS of your inequality expands to a polynomial in a².

[lucid panther]

omg ur so smart

i did not think of that

thanks brotha

[distant siren]

Can someone help me

[trail tendon]

no

[keen nova]

[distant siren]

what did i do wrong

i used cosine rule they used sine rule but i cant see anything wrong with what i did

but my answer is wrong

[fringe steppe]

oh thanks!

[obsidian hornet]

No problem

[flint marlin]

Yk you could just use routh's theorem

[fringe steppe]

wth

[fringe steppe]

.

[flint marlin]

Prove that √(a+b-c)+√(b+c-a)+√(c+a-b) <= √a+√b+√c where a,b,c are sides of a triangle

[obsidian harness]

good idea

sqrt(2y) + sqrt(2z) + sqrt(2x) <= sqrt(x + y) + sqrt(y + z) + sqrt(z + x)

nearly there, you just need to enforce an ordering on x, y, z since WLOG you can swap a, b, c

probably not

they should just be arbitrary positive numbers at this point

[grave pond]

If they don't satisfy the triangle inequality, one of the LHS square roots will fail to exist.

[obsidian harness]

oh but that's true

[grave pond]

Oh.

[obsidian harness]

idk

it looks easy but it's weird

https://approach0.xyz/search/?q=OR content%3A%24\sqrt{2x}%2B\sqrt{2y}%2B\sqrt{2z}%24&p=1

ah so $\sqrt{\frac{2x + 2y}{2}} \ge \frac{\sqrt{2x} + \sqrt{2y}}{2}$

[somber coyote]

south

[obsidian harness]

sum up and it's done RIP

https://math.stackexchange.com/questions/2113487/triangle-lengths-inequality?noredirect=1

same

[wispy nova]

starting geometry this year, any tips for me?

[high rampart]

yeah, if you're going to be just starting out take it slow and learn at the pace that's fast enough to keep up with your class, but slow enough that you can really take time to study and learn the material well. Remember, if you have any problems you can always come and post stuff in the help chats here

[lucid panther]

pay close attention to details

don't trust ur eyes(if u see a angle that looks like 90 or smth like that) u need to prove it

in geometry u can not trust ur eyes trust the number if yk yk

[warm shuttle]

draw all radii

[fringe steppe]

bruh momemnt

[sonic sphinx]

Have fun with cos tan and sine

[sonic sphinx]

I'm not sure how you study, but I'd recomend Kahn Academy. Im assuming your a freshman in highschool.

[gray osprey]

Can anyone help me with a maths question

[sonic sphinx]

480 what?

nvm doest matter

[sonic sphinx]

Radius

[sonic sphinx]

a is the center

[sonic sphinx]

AE and AF are the same length

[sonic sphinx]

isosceles

[gray osprey]

Yeh that’s what l got up to aswell couldn’t find the length of Ad though?

[sonic sphinx]

tryna work on that I haven't done geometry in a while

[gray osprey]

Ok thanks I would be so grateful if l could get the answer lol

The ratio to the circle radius is 10:7 ratio

[sonic sphinx]

yeah DE is the same as B to the circle

[gray osprey]

Wait really

[sonic sphinx]

I think so... I'm going off old knowledge lol

try it out

[gray osprey]

So 10/17y + 2/5x gives the width of the rectangle