geometry-and-trigonometry

[floral shale]

It's still a unit cube

As long as each side is 1 unit in magnitude, you'll have a unit cube

[young lintel]

I think this is a fine way to use chat gpt actually

As long as you verify after the fact (which is what I'm doing now)

This kind of definition-to-name is exactly the kind of thing that Google is bad at

[young lintel]

Each side is length 2 in this cube though

[floral shale]

Well if every vertice was (+- 1, +-1)?

Well you'd have a cube whose centroid is the origin

In simple terms

[young lintel]

Right

[floral shale]

Now if you google "alternating cube" a bunch of Rubik's cube ads pop up

So idk where it's even getting that info

[young lintel]

It probably hallucinated it

[floral shale]

Lmao yeah

Someone gaslit ChatGPT again

[young lintel]

There's probably something else called the alternating something

That's \pm 1 over and over, idk what it would be

[lethal jackal]

idk what this is but anyone know this one?

it has something to do with combination and permutation

[upper karma]

Find the minimum value of 2^sin^2theta + 2^cos^theta.

please help me on this

[dark sparrow]

your question belongs in #discrete-math

[late tangle]

find critical points and use first derivative test ?

nvm that would be calc way uh i think best way might be to graph it if thats allowed

[lethal jackal]

ok thanks

[upper karma]

8sin A - 6cos A=?
The value of A should be such that the answer of equation is highest

Also y^2 is indirectly proportional to x^3 graph

[upper karma]

Alr

[reef glade]

geometry is kinda hard

[slim goblet]

can someone write a respose for me

its so hard

[young tiger]

Draw the figures

A chord is a segment that connects two endpoints that are on the circle.

[young tiger]

Let O and G be the centers of each circle

Notice how OA, and OB are the radii of circle O because they connect the center to a point on the circle.

[slim goblet]

yeah

i see now

[young tiger]

The radii of the same circle are congruent. This means:

[slim goblet]

but

[young tiger]

👍

?

[slim goblet]

wait but

ahhhh im so confuseed

so we have an isocoles triangle right

[young tiger]

Yup for △AOB

[slim goblet]

why not for GCD tho

[young tiger]

Now if you notice that GC and GD are also radii of circle G, you get:

△CGD is also an isosceles triangle

[slim goblet]

ohhhh

but how does that help us find out

if the arcs are also congruent

cuz we dont know the values

[young tiger]

So arc AB is intersected by chord AB, right?

[slim goblet]

yeah

[young tiger]

And arc CD is intersected by chord CD

[slim goblet]

mhm

[young tiger]

The measure of the arc is equal to the measure of the angle is subtends or "bounds"

So this means that m∠AOB = measure of arc AB

And m∠CGD = measure of arc CD

[slim goblet]

but

i thought that

[young tiger]

So we just need to find out if m∠AOB = m∠CGD

[slim goblet]

i thought that the angle

is half of the arc

[young tiger]

Nope that's an inscribed angle

∠AOB and ∠CGD are central angles

You see here

∠ABC is half the measure of arc AC

∠ABC is an inscribed angle

An inscribed angle is an angle formed by two chords and has its vertex on the circle

And you can see here

m∠AOB = measure of arc AB

∠AOB is a central angle

A central angle is an angle formed by two radii and has its vertex on the center of the circle

Do you understand now?

[slim goblet]

ohhhhh

ty

[slim goblet]

so that arc/2=angle thing doesnt apply for central angles

[young tiger]

Right. Only for inscribed angles

[slim goblet]

okkk

so anyways

so arc AB= Angle AOB and Arch CD = Angle CGD

but is arc AB = arch CD??

[young tiger]

We will find out if you can prove that m∠AOB = m∠CGD

since m∠AOB = measure of arc AB

and since m∠CGD = measure of arc CD

You can do it if you can first prove that △AOB ≅ CGD

There are 5 ways you can prove triangles to be congruent:

SAS (Side-Angle-Side) ≅

SSS (Side-Side-Side) ≅

HL (Hypotenuse-Leg) ≅ - only use for right triangles

AAS (Angle-Angle-Side) ≅

ASA (Angle-Side-Angle) ≅

So are you familiar with proving triangles to be congruent?

[slim goblet]

i think so yeah

[young tiger]

Alright

[slim goblet]

but we have no way of knowing

to provve

[young tiger]

So is there any way we can prove that △AOB ≅ CGD?

Exactly

[slim goblet]

so that means

they arent congruent

[young tiger]

So this means that m∠AOB ≠ m∠CGD

Right

[slim goblet]

ohhhhhh

W persons bro tyy

[young tiger]

ofc

[slim goblet]

i dont want to disturb you but I also had this question on my mind im not sure how to do it

[slim goblet]

🙏

[thorn hedge]

Using Tangent Secant Theorem
12^2=7(7+x)
144=49+7x
95=7x
x=13.57

[young tiger]

And round it to the nearest tenth

So it'd be 13.6

Here's a picture of the tangent secant theorem xrayne referred to:

[smoky jetty]

how come the arc measures are not congruent if the radii of both circles are congruent, so is their chords?

cant it be proven using SSS criterion?

[snow crystal]

Can someone help pls

[young tiger]

The question says that the two circles have different radii 🙂

[smoky jetty]

yeah, but i didnt get why you illustrated the two circles as having congruent radii

[young tiger]

No I didn't

You see here

The radii for circle O are OA and OB. OA = OB.

The radii for circle G are GC and GD. GC = GD.

OA ≠ GC, OA ≠ GD, OB ≠ GC, OB≠ GD

[smoky jetty]

oh wait gosh

lmao, I thought the 3 lines on CG and DG were only two

my bad

[slim goblet]

help 😦

[acoustic knoll]

first find the areas of the circle and rectangle
which would be
area of circle = 28.27(28 approx.)m^2
area of rectangle = 216m^2
probabilty = 7/54
that would be option 2

[slim goblet]

how did u find the area of the circle

ohw ait nvm

but once i got the area what do i do from there

[mystic charm]

the area of the rectangle is 216 and the area of the circle is 28, as concoran_edmund said. the point P is gonna be certainly placed in the rectangle and the circle only covers 28/216 m^2 of the rectangle, so there is 7/54 chance that the point P will be placed within the circle

[slim goblet]

ohhhhh

ty

[upper karma]

isn't 1-sin^2x equal to cos2x already?

[snow crystal]

it is

but what should i do next

I thought it was easier tu substitute 1-2sin^2x with cos^2x-sin^2x

[upper karma]

I'll try to do it but it probably doesn't have a easy solution sadly

so its gonna be a bit painful

[snow crystal]

ok ty lol

[upper karma]

maybe we can do it from the square difference formula?

will be equal to cos2x itself

[remote wharf]

ill do us a favour and turn that image sideways

[upper karma]

already did dw dw

its sending wait

[remote wharf]

okk

[remote wharf]

plus it wouldnt let me download it

[upper karma]

heree

[snow crystal]

i don't see difference of squares here tho

[upper karma]

oh damn wait

ok I have seen something wrong

that doesn't work

I thought of the 2's like they do not affect the square

[snow crystal]

I'll get back to it in 2 mins

maybe ill figure it out

[upper karma]

this makes (1-cos4x/2)*(1+cos4x/2) ig

[snow crystal]

that's funny, I've never seen those formulas before lol

[upper karma]

and thats equal to cos2x

[snow crystal]

I was trying to solve the equation with these

but I guess they dont work

[upper karma]

same but I think its not gonna give anything I believe since sin2a*sin2a is sin^2 4a and not sin^2 2a

the own formulas itself still are painful tho

[snow crystal]

the funny part is that the textbook where I found the equation didnt mention those formulas

idk how they expect me to solve it lol

[upper karma]

maybe there is another way that we don't see?

[snow crystal]

probably

[upper karma]

but I can't see any other way doing it without the formulas

[snow crystal]

what can we do now?

@upper karma any ideas?

[upper karma]

hmm if its -cos4x = cos2x

maybe we can write -cos2(2x) = cos2x

so does that make -2cos^2(2x)-1=cos2x

-2cos^2(2x)-1-cos2x=0

-(2cos^2(2x)+1+cos2x)=0

I was trying to find a smiliar question and found the equation for cos4x=cos2x

https://byjus.com/question-answer/find-the-general-solution-of-the-equation-cos4x-cos2x/

can it lead to somewhere?

[snow crystal]

lmao

[upper karma]

ohh

[snow crystal]

I can ask my teacher tomorrow, thanks for trying to help tho

[upper karma]

yw but still sending these just in case it can help

by the way if you guys solve it tomorrow, can you also send it here I am really curious lol

[snow crystal]

yeah sure

[upper karma]

thanks a lot

[snow crystal]

post it somewhere else bc it doesnt belong in this channel

and ill try to help

[snow crystal]

<@&268886789983436800>

[upper karma]

that looks hard

[snow crystal]

try finding the coordinates of the vertex

[upper karma]

i don't even know what that is

i think it's trigonametry

[sturdy jackal]

what is g(x)=?

[snow crystal]

idk why they posted it here but it's not trigonometry

[sturdy jackal]

this? @snow crystal ?

[snow crystal]

yeah

[upper karma]

oh damn I really did not notice that I sent it here Im sorry

but I already sent it to a friend so I guess there is no need for help

still ty ❤️

[snow crystal]

yw

[upper karma]

ew

[snow crystal]

@upper karma turns out you can substitute sin^22x with 1-cos^22x and you get a quadratic formula

[vagrant isle]

I have finally finished my trigonometry assessment

I can finally finish up for the night and sleep

[fickle void]

Does anyone know how I can determine the area of this figure?

i really forgot how to do it and my friend is asking me for help with it

[snow crystal]

Try dividing it into smaller figures

[fickle void]

couldn't manage to

[dark sparrow]

is this the only pic you have of it

also should we assume everything that looks like a right angle is one

[snow crystal]

also is that an actual textbook exercise or did you think it up

not all side lenghts are given too

<@&268886789983436800>

[exotic yarrow]

<@&268886789983436800> spammed advertisement ^

[snow crystal]

?

[hollow badger]

?

[wind coral]

<@&268886789983436800>

[upper karma]

yo just wondering what we be learning in geometry

anyone wanna hit me up?

alr ty

[reef glade]

Geometry is pretty easy if you compare with algebra 2 and other algebra classes

But I still prefer algebra than geometry

[floral shale]

Pain

You wanna know what you learn in geometry?

Pain

[reef glade]

“Proof me why is this a triangle”

[floral shale]

3 sides motherfucker you can count them. One, two, fucking three

Gonna prove my teacher why his wife divorced him

[reef glade]

You learn geometry to build a house

You can build your own house

[pure ether]

bro...

geo so hard

[acoustic knoll]

eh both are equally easy

[solar skiff]

hello guys

im in class 8

is there any good book

i can practice trigonometry from

[reef glade]

i'm not required to take trig so i never took it so i'm not sure

[acoustic knoll]

You can just google worksheets . There's plenty of challenging as well as easy ones in those online worksheets .

[proper sierra]

What exactly about geometry do i need to know or would i be fine if i skipped geometry and went straight into algebra 2? The goal is to be ready for pre-calc/calc

[calm dawn]

i wanna know too

[proper sierra]

Hopefully we find out🫡

[calm dawn]

🫡

[reef glade]

i took algebra 2 first and then geometry

i prefer taking algebra 2 first than geometry

because algebra 1 skills can be used in algebra 2 even though i think algebra 2 is basically the same as algebra 1 but i think if you took algebra 1 ( honors ) then algebra 2 would just be a easy class for you to take.

[proper sierra]

Ive graduated not too long ago but i had no idea what i wanted to do for my future, now im going to college for comp sci and im currently trying to catch up on math but idk how badly i need to relearn/know geometry for pre-calc/calc 1

[reef glade]

dam

that's tuff man

computer science is pretty good

i'm taking business class so i have to learn about math in business

Yeah

Especially when they want you to proof it in a paragraph

Is like another English class

Use your eyes

We did flow chart proofs and paragraph proofs and 2 Column proofs

Oh okay I need a magnifying lens then

[proper sierra]

All my homies hate proofs

[reef glade]

You need to write soo much is crazy

Is not hard to understand the thing is different subject have different proofs

And you want us to remember all that and they be like you can build a house with geometry

Cya

[upper karma]

uh

this might be stupid but

if pi is an infinite number

wouldn’t it be true that there is a 100% chance that there is an infinitely long chain of 0s

which means it’s a constant

[lucid condor]

If that were the case, pi would be rational, at some ratio with an arbitrarily long numerator and denominator, but it is possible to prove pi as being irrational

[lucid condor]

Pi is nonetheless still a constant

Pi never changes according to some variable

It is simply a number

Even if it is irrational

[upper karma]

I meant finite

Not constant

[lucid condor]

I see

[lapis vault]

can someone teach me the aspects of geometry

cuz i am preparing myself

for the finals

[manic hollow]

Guys, In triangles is there a relation between the orthocentre and and its vertices which shows the type of triangle(equilateral, isosceles, scalene )?.

[floral shale]

The location of the orthocenter can help you determine whether a triangle is obtuse, acute, or right

Uh but I do I know what in equilateral triangles, the orthocenter is also it's centroid

@manic hollow

As far as isosceles and scalene triangles go, not too certain

[vivid plinth]

irrational*

[vivid plinth]

It's a constant anyway since its value never changes

[vivid plinth]

Oh I didn't read this sry

[vivid plinth]

I don't think u can prove a number with this property is finite

[slow hazel]

How would you find a tangent line at a sharp corner of a curve that only intersects the point at the corner? If not the tangent line, then a line that behaves at described

[vagrant jacinth]

It's not always possible:

[slow hazel]

fair, I guess im looking at something more like this, blue representing the tangent line I'm getting at the moment

[vagrant jacinth]

ahhh

if you know the curve is smooth to the left of the point (or the right) you could calculate the left-derivative of the curve

or the right depending on the side

like if you look at f(x) = |x|

the left derivative at x = 0 would be -1 and the right would be 1

[slow hazel]

would the output of that be something similar to the blue line depicted:

[vagrant jacinth]

no, it would be like your first picture, basically the direction the curve is headed if you approach the sharp point from the left or right, the problem is that if the curve is sharp there's no real way to get a "tangent line" at that point, precisely because there's no derivative at that point

you could do some things like take the average of the left and right derivatives, but that won't always give you what you want, since curves with a point don't have tangents at that point

[bitter nymph]

Just put a normal over everything and hope for the best

[pure ether]

not possible most of the time

theres a physics phenomenon which also

explains how u cant find the direction of normal force from an edge

which is kinda similar that ur looking for tangent lines

thats why an absolute value graph is not differentiable for all values of x

bc of the (0,0)

[slow hazel]

Understood thanks for the help!

[floral shale]

Can't. You'd have an infinite number of tangent lines. And we hate infinity

That's the easiest way to describe it

The moment a mathematician rotates 8 90°, they start shitting themselves and screaming

[pure ether]

wat

does that mean

[floral shale]

Good question

[manic prawn]

Can you divide a square into 3 equal area triangles?

[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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[slow hazel]

Why do they hate infinity?

[floral shale]

Infiniteez nuts

Nah but when it comes to tangent lines

By definition there only should be one locally

At corners you'll have an infinite number of tangent lines

[storm grotto]

Anyone able to help?

[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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[storm grotto]

1

I know the given but the rest is lost on me

😭

[reef yarrow]

@storm grotto What do you need in order for a triangle to be isosceles?

[storm grotto]

2 equal sides and 2 equal angles

[pure ether]

no just one

its an if and only statement

so u only need one to prove the other'

[reef yarrow]

Yeah, those both have to be true but if you show one then the other has to be true

Sorry I misread it anyway. They've already given you the steps of the proof, you just need to label each step

I'm guessing you have a textbook that gives you names for the different rules you can use in a proof

[vast saffron]

can I ask about vectors here 👀

[gaunt nexus]

that goes in algebra, ig

[vast saffron]

ah

vectors are geometry right 🤔

[gaunt nexus]

no they are linear algebra

[vast saffron]

hmm

[graceful vale]

Is there some Chinese in this group,i want help

With a math video that is in Mandarin possibly
It's the only video on that topic that i can find.....

https://youtu.be/AP8Wvn9X6v8
This is the video

The time duration where i need clarification is from 15:00,from 25:45 and from 27:20...

Here he discusses
1)Some Rhombus sort of structure,i want to know with what arguments did he arrive at that conclusion

2)He talks about Cauchy Crofton Formula,what context is he talking about it.....(like why is using that there),i have rough idea of what Crofton Formula is...

3)third in 27:20 one,He talks about evaluating integral(it comes up in his slides) integral(1/(1+x^3)) how does it come up there

Help with these,even if you can just translate and type what he said,i would be very thankful

[wind coral]

<@&268886789983436800>

[tawdry condor]

Depends on what you do with them

Many geometric theorems with vector in R^2, but if you have a question that involves vectors in an algebraic way then algebra channel

[snow crystal]

guys I'm lost, what am I doing wrong here??

[nocturne remnant]

The sum to product thing doesn’t work because there’s a sqrt(3) attached to the left term

[snow crystal]

so what can I use instead?

[nocturne remnant]

You need some other observations to solve it

Consider writing sqrt(3)cosx + sinx as 2sin(pi/3)cosx + 2sinxcos(pi/3)

Then use compound angle formula in reverse

[snow crystal]

wow that's pretty smart

thank you

[upper karma]

i need help

[wooden patrol]

Send me math problems to solve so I can get better and better!!! :)

[old geyser]

@wooden patrol are you still there?

I'm having trouble with analytic geometry

look

Do any of you get it?

[floral shale]

Not sure why you need to use vectors and linear algebra to prove triangle OAB is equilateral

if e1 and e2 are unit vectors with an angle of 60°, it should be easy to prove that AOB will be a unit equilateral triangle

[upper karma]

someone help me with geometry

is the answer of this D?

[dark sparrow]

how did you get D?

[upper karma]

so

I did 12x+8=10x+15

and the -10x so it became 2x+8=15

and then i subracted 8

so it was 2x=7

then divided by 2

so it was x=7/2

then i plugged that into the equation

and got 50

wait so is the answer C?

[upper karma]

someone pls help

[dark sparrow]

what equation

sorry for the delay on my part -- was busy

[upper karma]

it's okay

[upper karma]

12+8

12x*

i plugged it in 10x+15 too

and got the same answer

[dark sparrow]

yes, so that way you know your x value is correct

note that the two sides of the equation 12x+8=10x+15 are AM and MB

you are asked for AB

[upper karma]

yes

so c is correct or no?

[upper karma]

okay what do i do next?

@dark sparrow C was wrong, so is the answer D?

the answer is A nvm

[dark sparrow]

no, it is not.

in fact you had the right answer the first time round

you just could not justify it properly

[upper karma]

so D is the answer?

but I'm still confused

how did we get 100?

[dark sparrow]

we got x = 7/2

we got AM = 12*7/2 + 8 = 42+8 = 50
thus also MB = 50

AB = AM+MB = 100

[upper karma]

OHHHH

omg thank you sm

[upper karma]

how do i find the answer to this

someone please help me

i have 5 mins before the assinment is due

[viral merlin]

pythagoras

[upper karma]

is that a^2+b^2=c^2

[viral merlin]

ye

[upper karma]

wait so

what numbers do i put

[viral merlin]

that's for you to figure out

[upper karma]

okay thanks

[fresh wedge]

help

[pseudo viper]

the laws of sine / cosine are to be used when you're not working with a right angled triangle, right? otherwise you work with the whole sohcahtoa thing right?

[upper karma]

Yea for right angled triangles you can just use the ratios of the sides to the angles (sohcahtoa)

[night idol]

Someone help please, I’m not sure how to do the second and third one

[upbeat helm]

For 2 you should first find the length of the bottom side in the top picture

The “ends at the same point on the ground” part and then just use Pythagorean thm

3 is similar

[night idol]

Alright, thank you!

[pine river]

did i do this right?

[obsidian goblet]

yea

[neat python]

In the isosceles triangle ABC, B=120 CD is the height of the triangle,

Find AD if the length of the altitude through the vertex B is 10.

<@&286206848099549185>

[zealous stag]

ok

let's see

[neat python]

Another programmer nice XD

[zealous stag]

lol

To find AD, Let's break down the given information:

[neat python]

Sure

[zealous stag]

In isosceles triangle ABC:

1. B = 120 degrees (angle at vertex B)
2. CD is the height of the triangle

right?

[neat python]

Yes

[zealous stag]

We're also given that the length of the altitude through the vertex B is 10 units.

[neat python]

Yes

[zealous stag]

Let's denote this altitude as BE, where E is the point where the altitude intersects side AC.

[neat python]

Sure

[zealous stag]

Now, since the triangle is isosceles, we know that angle A = angle C. Thus, angles A and C are equal.

[neat python]

Yes

[zealous stag]

Let's denote AD as x (the length we need to find). To find AD we need to determine the relationship between x and the known lengths in the triangle

AND

Since CD is perpendicular to AB, angle CDB is a right angle. Angle CDB = 90 degrees.

right?

[neat python]

Yes

[zealous stag]

Since angle CDB is a right angle and angle B is 120 degrees we can find angle C using the following equation:

angle C + angle B + angle CDB = 180 degrees

angle C + 120 + 90 = 180 degrees

then solving we get angle C = 30 degrees

as angle C = 180 - 120 - 90

right?

[neat python]

It's -30

😀

[zealous stag]

ops

tf

wait

a min

[neat python]

Lol

[zealous stag]

it seems there was an error in the problem statement

or i am wrong

😦

[neat python]

Nah

We should get 30

[zealous stag]

yea but how

[neat python]

I can't get too

😂

[zealous stag]

error in the statement?

[neat python]

I don't think so

[zealous stag]

shet

<@&286206848099549185>

[neat python]

😂

[zealous stag]

am i dumb?

i think i lost my brain cells

[neat python]

I am dumb too

[zealous stag]

lemme open a help channel

[mortal raft]

I am dumb three

[neat python]

Lol i found AD

[mortal raft]

W

[drowsy hatch]

can someone explain me this

in both cases the 2 planes are cutting the same quadric if im not wrong

[mortal raft]

no idea but u should help me #help-2

[pine river]

pls help idk how to do this

[dark sparrow]

inscribed angle theorem for all of these

do you know what theorem that is

[smoky jetty]

have a read on inscribed angles and central angles theorem

[cedar ether]

I dont understand how the top equation can be turned into the lower equation. The video im watching attempts to explain the concept called a "Perfect square" I dont see how that applies here.

[nocturne remnant]

First of all, that's an expression, not an equation.

if you had to turn the lower expression to the upper one, could you do it?

[cedar ether]

I can understand the X squared and the lone four; but I dunno where the 4X came from. sorry for the wrong vocabulary

[nocturne remnant]

Do you know how to expand something like (a+b)^2?

(It is not the same as a^2 + b^2)

[cedar ether]

guess not?

how

[nocturne remnant]

well- what about (a+b)(c+d)? How would you expand that

[cedar ether]

wouldn't you multiply cd by a and b then like add or smt?

[nocturne remnant]

can you try it? Like write out the expanded form of (a+b)(c+d)

if you cant do that then there is quite a lot of catching up to do

[cedar ether]

I dont remember being taught this bud

[nocturne remnant]

also this belongs in #prealg-and-algebra

[cedar ether]

its in a geometry problem nvm

[neat python]

In the triangle ABC, the line AD is perpendicular to the median BM and bisects it.

Find the length of side AB if MC =10.

<@&286206848099549185>

[compact tapir]

excuse the bad drawing

[neat python]

Np

Ty

[neat python]

10 is good but it's wrong XD

Answers are
A)2
B)3
C) 5
D)8

[silk dove]

i tried and got 10 as well , but i might just be bad at geometry, ill draw it one sec

[exotic yarrow]

I also got 10

[neat python]

I got 10 too so idk

😂

[silk dove]

since BM is the median, it dissects AC into 2 equal parts AM and MC, so AM = MC = 10
BO = OM, because AD is perpendicular to BM and bisects it at O
Triangles BOA and MOA share the side AO, and BO = OM and the angle between them is equal (90 deg)

so they have 2 equal length sides and the angle between them is also equal, therefore the remaining sides AB and AM are equal, so AB = AM = 10

sorry if this was worded badly, first time actually trying to explain anything in math in english

tried ,y best to make the drawing fit too

[neat python]

Idk i will ask my teacher in 1 hour

If something is wrong

Wait

[neat python]

Am will be 5

AC=10

[silk dove]

you said MC = 10

[neat python]

Oops

💀

[silk dove]

if AC = 10 then AB = 5

👍

[neat python]

Done XD ty

[silk dove]

no worries, happens

[near mica]

(x+2)^2= (x+2)*(x+2) , you can write like that: (x+2)*x + (x+2)*2 = (x^2 +2x) + (2x + 4) and here we come to x^2 + (2x+2x) +4 = x^2 + 4x + 4

[tepid inlet]

How would you go about finding DEG in this photo

[zenith escarp]

🙂 DEG = 31*

[tepid inlet]

Thank you I see now

[wind knoll]

can someone help my friend with this question

[silk dove]

v: Angles on a straight line always add up to 180º, so 30º + v = 180º, v = 150º

y: Angles 30º and 2y are called alternate interior angles, because they lie on the opposite sides of a line that is passing through two parallel lines, they are equal. 2y = 30º, y = 15º

z: Angles on a straight line always add up to 180º, so z + 130º = 180º, z = 50º

w: Angles 130º and w are called co-interior angles, because they lie on the same side of a line passing through two parallel lines, they add up to 180º. 130º + w = 180º, w = 50º

x: Angles on a straight line always add up to 180º, so x + w + 2y = 180º, x + 30º + 50º = 180º, x = 100º

[wind knoll]

tysm!

[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]

tell your friend to come here and ask for help herself

[silk dove]

my bad my bad i thought it was better to explain stuff but that makes sense

[dark sparrow]

well you basically did the entire thing for op

which is no bueno

[silk dove]

Would it have been better if i just said the answers? should i also give the rules used to find the answers, but not tell them in which steps exactly the rules are used?

[silent plank]

Have the student work through the problem themselves and guide them along the way.

avoid giving answers if possible

[silk dove]

right, so i should've just explained the rules that were used in the solution, but not where to use them

My bad, I'll know next time

[silent plank]

progressively go into more detail as needed

question their understanding of rules, clarify rules if needed
see if they are able to identify where they're applicable

direct them to where to apply them if required,

[silk dove]

alrightt

[upper karma]

can someone help with this?

[patent apex]

just wondering is there is the a way to expand tan(a+b)

similar to expanding sin(a+b)

[last quiver]

yes tan(a+b) is simply sin(a+b)/cos(a+b) now simplify this to get it in tan form

[potent dragon]

[tan(a) + tan(b)]/1-tan(a)tan(b)

[opaque phoenix]

I love maths

Im gonna preach maths

[silk dove]

like as a religion or?

[pure ether]

why do ppl like math so much smh

[timber cargo]

Something that came out of our heads and describes the universe, is deeply mesmerizing

[unreal crag]

bro math is just like every other religion, it explains natural phenomena

[rotund radish]

We are just crazy

[patent apex]

oh yeah completly forgot about that

[foggy parcel]

In the first one, apply the formula s = rθ

[upper karma]

ya I'm struggling w b

[foggy parcel]

Wdym?

[foggy parcel]

Anyways, to get through the 2nd one firstly u have to get the area of AOB and the area of AOC triangle individually
Then summing it all up would give the desired solution u r looking for

For the AOB
Apply (θ/360)* πr^2

For the triangle,
Draw a perpendicular from A to CB and use sine to get the height of the perpendicular
Then Apply the normal formula for triangles.

@upper karma

[upper karma]

ohhh

ok ok lemme go do it

tysmmmm

[foggy parcel]

U r welcome

[smoky jetty]

saw this fun highschool lvl problem online im pretty happy to have solved it and thought to share it to yall

[snow crystal]

I got 36π-56sqrt(3), is that the answer?

[smoky jetty]

close, but nope

[silent plank]

show work

looks off

[smoky jetty]

oh right!

[snow crystal]

,rotate

[silent plank]

6*9 isn't 56

[snow crystal]

oh wait it's 54

yeah

[smoky jetty]

there you go!

[snow crystal]

so it's 36π-54sqrt(3)

[smoky jetty]

btw was this 6(6pi - 3sqrt.3)?

or it's a curly nine

[snow crystal]

it's 9 lol

i cant write

[smoky jetty]

ohh ight. lol my handwriting is no better eitherway

how was the problem?

[snow crystal]

I did something similar a couple of months ago in school

so it wasn't THAT hard

This is what we did

,rotate

[smoky jetty]

thats neat!

what grade was that taught?

[snow crystal]

10th

[smoky jetty]

the 10th grade textbook I have doesnt have those

[snow crystal]

I major mathematics

so that's probably the reason

[smoky jetty]

you're in college?

[snow crystal]

nah I'm in high school

[smoky jetty]

oh lol

[snow crystal]

idk I'm from Poland so it probably looks different from your school

but basically we choose 2 subjects that we're going to major

in high school

[violet kiln]

Help Please #Helpers

[patent apex]

you take the exterior angle

wait no

exterior angle is 18

divide by 360

360/18=20

[violet kiln]

which equation I use

I didn't understand

[mystic charm]

do you know how you can find the interior total angles of a polygon? (for example the angles of a triangle)

[violet kiln]

Yes

I use the equation (n-2) × 180

[mystic charm]

correct

[violet kiln]

Where n is the number of sides

[mystic charm]

so if we divide it by n, it means that each angles is 172, right?

[violet kiln]

Divide what by n??

[mystic charm]

this formula

(n-2)*180/n = 172

now try to find n

[violet kiln]

How to find n??

I use algebra??

[mystic charm]

yes, multiply both sides with n, which will give us, (n-2)*180=172n, therefore 180n - 360 = 172n

now find n with this equation

move 172n to the left side of the equation and 360 to the right side, we will have 180n-172n=360

8n=360, n=45

[wet abyss]

I started with SL loney's plane trig book.
I needed a bit of advice on whether the ones who solved it solved all the questions and how much time it took them to finish the book

[patent apex]

The sum of all the exterior angles of a polygon is 360

So 360 = 8n

[violet kiln]

So n=360/8

[patent apex]

I apologise for being a dumb as

[violet kiln]

n=45

[patent apex]

My brain went 180-172= 18

[patent apex]

Ye

[upper karma]

guys how do I do this

[silent plank]

do you know your parallel line theorems
and/or have access to a list

[violet kiln]

8. x=70 alternate angles are equal
9. x=130 Vertically opposite angles are equal

IDK about 7 and 9

[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.

[floral shale]

Use parallel-transversal laws

[upper karma]

idk what they are

[upper karma]

I know a bit about them

[upper karma]

ohhhh

[upper karma]

x=130 is number 10

[prime musk]

heyy guys

can u please help me with a question?

[snow crystal]

Just post it

[prime musk]

I did in the help-forum.... no one helped

anyways

Prove that sin pi/14 is a root of the equation 8x^3-4x^2-4x+1=0.

This is the question

I would be grateful if u could help....

[snow crystal]

Ok so you have to substitute x with sin pi/14 and see if it's equal to zero

[prime musk]

I did... but the equation is becoming too lengthy... and nor am I able to simplify it

[snow crystal]

Basically when you have such equation and want to check whether certain number is its root you have to substitute x with the given number

[snow crystal]

Hmm

[prime musk]

could u just tell how I can simplify it?

[snow crystal]

Im thinking

,rotate

[snow crystal]

,rotate

[snow crystal]

I hope this helps

@prime musk

[prime musk]

hmm

so these are the roots...

and I basically replace x with sin pi/14 and check right?

[snow crystal]

Yeah

[prime musk]

but I don't know the value of sin pi/14...

So I mean how do I check?

[snow crystal]

I Don't have time rn but try to plug in sin pi/14 and find some trig identities/formulas that will help you to simplify the equation even more

[prime musk]

ohh ok... thanks

[pseudo viper]

the only time you can use the sine rule is when you have two angles + a side opposite one of the angles or two sides + an angle opposite one of the sides right?

[weary hound]

When you're given two angles and one side of a triangle
Or, when you're given two sides and the opposite angle to either one of the sides

Alternatively, when you're given one side of the triangle and the radius of the circumcircle
(Or one angle and the radius of the circumcircle)

[pseudo viper]

" when you're given two angles and one side of a triangle "
the law of sine wouldn't work if you have angle A, angle B and side c bc you need atleast one angle and it's opposite side for the law of sine to work?
side a (unknown) / sin A (known) = side b (unknown) / sin B (known) = side c (known) / sin C (unknown )

unless the resources i've been using haven't explained it properly and i'm wrong

[weary hound]

You can find angle C anyways if you know angle A and B

Because, A+B+C = π
-> C= π-(A+B)

[pseudo viper]

oh true

okay yeah i see it now

forgot about that haha

[dapper dome]

This article has a detailed explanation for this type of question.
https://www.mbacrystalball.com/blog/2015/10/02/lines-and-angles/

[upper karma]

oh thanks

11 is parallel, 12 is neither, and 13 is perpendicular. Did I do these right?

[dark sparrow]

yes

[upper karma]

okay thanks

[upper karma]

what is the answer to number 4 and 6?

number 1 is alternate interior angles, number 2 is alternate exterior angles, number 3 is corresponding angles, and number 5 is corresponding angles

I can't figure out what 4, and 6 are

[silent plank]

Q5 isn't corresponding

Q6 is similar to one of the ones you'd already completed

there's also a listed option you haven't selected yet,
if you're not sure what they are, its recommended that you look it up

[upper karma]

what exactly do i search up

[silent plank]

a listed option you haven't selected yet,

[upper karma]

ohh okay

[upper karma]

is it consecutive?

[silent plank]

no

[upper karma]

what

[silent plank]

circle/highlight angles 1 and 8 on the pic

and compare that to your parallel line theorems

[upper karma]

I'm talking about question 5 not 6

[silent plank]

you replied to the comment about Q6

Q5 is consecutive, yes

[upper karma]

ohh okay

[upper karma]

oh mb

[upper karma]

it's a vertical angle

[silent plank]

make sure you reply to the correct messages, i'm taking these at face value

1 and 8 aren't vertical angles

[upper karma]

what are they, I'm so confused

[silent plank]

circle/highlight angles 1 and 8 on the pic
and compare that to your parallel line theorems

[upper karma]

wait is question 3 and 5 vertical?

[upper karma]

okay

[silent plank]

note that

Q6 is similar to one of the ones you'd already completed
and answered correctly

[upper karma]

okay will note

Q6 is alternate exterior

[silent plank]

yes

[upper karma]

yea it made sense after

[upper karma]

is Q4 alternate interior

[silent plank]

no

it doesn't sound like you've searched up the option you haven't selected yet

[upper karma]

I did but the pictures aren't like matching

[silent plank]

what did you search up

[upper karma]

Q4 is vertical right?

[upper karma]

vertical angles, and consecutive interior angles

[silent plank]

yes

[upper karma]

ahh okay thank you for your help

[upper karma]

How do I do 19

[grave pond]

The two indicated angles add up to a right angle.

[upper karma]

so is it 75 + (x+8)=180?

[grave pond]

Is a right angle 180°?

In particular, does angle ABC in the diagram look like it's 180°?

[upper karma]

no nvm i got confused with anoter question

[upper karma]

so 90

[grave pond]

Yes.

[upper karma]

75 + (x+8)=90?

like this

[grave pond]

Yes.

[upper karma]

ohh okay that makes sense

then i slove for x and i get x=7

[grave pond]

Yes.

[upper karma]

ahh thanks for the help

[smoky jetty]

always remember that if two lines are perpendicular, then each angle in their intersection measures 90 degrees

[wind coral]

Memes go in #chill

[quaint chasm]

anybody know how to do cross product on a vector of arbitrary dimension?

[grave pond]

The usual concept of "cross product" only applies to dimension 3 (and in a somewhat different guise in dimension 2).

[quaint chasm]

fair enough

[grave pond]

In higher dimension there are different generalizations of the 3D cross product that preserve different of its properties, but none of them give you the whole package.

[quaint chasm]

kk
was just curious cause i have implementations of it in my code for 2 and 3d, but was wondering if it worked in higher dimensions

[solid needle]

not sure if this is the right place for this question but

i have a question that is relatively open-ended and vague

given the radius of this circle in blue

can anyone help me find a "natural" way to construct the line/length in red?

geometric constructions are fine, but i also don't want to limit myself to that, if anyone has any other clever ideas

[winter rover]

can anyone tell me how we use trigonometry in real-life problems? Because I'm not sure how we use it though I know most of the concepts.

[solid needle]

The most direct application is just the geometry

I have a wooden frame that is 3m by 4m, and in order to strengthen it, I want to add diagonal wooden beams connecting the opposite corners. However I need to know the angle of the diagonal beams so I can cut the ends accordingly. How would I calculate the angle?

These types of problems occur all the time in arts and crafts, construction, etc. Anytime you have enough information to determine the exact triangle, but you need the exact measurements of the sides or angles, you need to use trigonometry.

There are slightly more abstract usages, such as vectors, which are important in physics. Im pulling on a string in order to drag a heavy object up a ramp. How much strength is needed to pull this string? We can calculate this using vectors, but we will now need to apply trigonometry in the formula

There are also some much more abstract applications of trigonometry, but these are the kinds of topics that everyone here is excited to talk about so I'll let other people chime in on those

[winter rover]

thank u

[ember stratus]

Hi I'm trying to prove $sin(\theta)= 2sin(\theta/2)cos(\theta/2)$

[somber coyote]

aSome1gussy

[ember stratus]

What I was given is 𝑒𝑖𝜃=cos(𝜃)+𝑖sin(𝜃)
with 𝑒^𝑖𝜃=(𝑒^𝑖𝜃/2)^2
so what am i meant to do next

on second line (this "trick" yields De Moivre's formula).

idu how DM's formula relates to this, apart from maybe the modulus arguemnt form for r cis theta

[dark sparrow]

are you not allowed to simply invoke the double angle identity for sine, of which this is an instance?

also for a lot of it

perhaps, though, you could do this, for example:

let $z = \cos(\theta/2) + i \sin(\theta/2)$; acknowledge that $z^2 = \cos(\theta) + i \sin(\theta)$ by de moivre's, and expand $z^2$ the ``honest'' way and read off the imaginary part that results.

[somber coyote]

Ann

[ember stratus]

Ok, how do you expand z^2

@dark sparrow

I got the rest.

@everyone

[dark sparrow]

don't ping everyone

[dark sparrow]

well, [cos(θ/2) + i sin(θ/2)]^2 = ?

you've got the square of a sum here, do you know how to expand that?

[ember stratus]

bru i'm legit tripping, can you do it for me

i'm literally caffeinated

ok i guess cos^2(theta/2)+2cos(theta/2)(isin(theta/2)-sin(theta/2)^2

r8?

ANQUAN AN NIE!

@dark sparrow

ye?

[dark sparrow]

no

cos^2(theta/2)+2cos(theta/2)(isin(theta/2)-sin(theta/2)^2
yes

[dark sparrow]

what was this about

[ember stratus]

sorry my pleas for help

isn't that contrived solution though?

SEEMS CONTRIVED

NAH?

OHHH i get it now ty

[pseudo viper]

law of sines, if im trying to find a side then i should write the law as side over sin(angle) and if im trying to find an angel, i should write the law as sin(angle) over side. is that correct?

[nocturne remnant]

There’s no “correct” way of writing that you “should” use but if you find it easier this way then sure, go ahead

After all, it’s just the same fact written in two different ways

[pseudo viper]

true

ngl trig is way easier than i remember from when i was school
ig i have a different mindsent now and actually wanna learn / understand

[upper slate]

NQ would be an angle bisector right?

[last quiver]

no , it need not be angle bisector

It's a implication of the angle bisector theorem ig

[last quiver]

So in this case NQ is a angle bisector only if NL : NM = 1

[upper slate]

ok thank you

i put angle bisector but i wasnt 100 percent sure lol

[upper karma]

how do i solve this?

[upper karma]

option D is 90

yea

is it 5x+x=90?

ohhh okay

thank you

[upper karma]

the answer is b right

did i do this right?

[fossil burrow]

Correct ans is B

For second one

I dont find any corresponding angles in the 1st question though... there arent even any || lines

[upper karma]

ohhh

[upper karma]

wait how

[fossil burrow]

Corresponding angles need parallel lines right?

[upper karma]

no i dont think so

wait they might

[fossil burrow]

Corresponding angles are the angles which are formed in matching corners or corresponding corners with the transversal when two parallel lines are intersected by any other line

[upper karma]

yea

[upper karma]

what do i put for this tehn

but angle 3 and angle 7 hold the same position meaning they are corresponding

[pseudo viper]

arcsin(0.960672361) returns 74, but it should instead return 106. if you subtract 74 from 180 you get 106, so is there something i don't know about?
am i supposed to subtract the returned angle from 180 ( or 2pi if im working in radians ) if the angle doesn't make sense?

[upper karma]

find "k" (with solution please :'v)

[dark sparrow]

arcsin only ever returns values between -90° and 90°

if you want the obtuse angle that has sine equal to your number, then yeah, subtract 180 - arcsin(...)

[smoky jetty]

https://youtu.be/LOPT0mAsyCM there's this cool video I watched before regarding that

[pseudo viper]

okay so i have to evulate everything and decide whether the angle it gave me is the angle i need, or whether i need to subtract 180, etc
that's a pain

[pseudo viper]

i'll give that a watch on my lunch break! thanks

i see the word calculus and i shiver

[weary hound]

What is (tan 45° -1)

I was excited about this problem then I got extremely disappointed

[pseudo viper]

what does the ... mean?

[last quiver]

||k is 0 obv||

because tan45° -1 = 0

[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.

[weary hound]

To fully express, k=(tan 5°-1)(tan 15°-1)(tan 25°-1)(tan 35°-1)(tan 45°-1)(tan 55°-1)(tan 65°-1)(tan 75°-1)(tan 85°-1)

And that is why we use the abbreviation as ...

Because else my hand hurts to type this entirely

[pseudo viper]

are those standard angles? bc why not (tan10-1) or is that also included

[weary hound]

You assume that the questioner wrote like that believing the reader of this problem notices the pattern of the angles: 5, 15, 25, 35,... and so on until 75, 85

And no ofc we do not know what the heck tan 5° or 25° is for now

This is why we have to think

What is 0× sin 385° × sin 5/8° ×e^(√(2π²-1))

Whatever we multiply anything to 0, 0 is still 0

[pseudo viper]

im not understanding step 3 and 4
if i get angle 60 from my calculations and subtract that from 180 to get a second angle that is 120
if i add 60 and 120 together, of course it's always going to equal 180?? im adding 60 + the difference between 60 and 180 so i don't see how it'll ever be more or less than 180?

unless by original angle it means the angle that i'm starting with and use in my formula?

[pseudo viper]

@rugged wind ^ spam

[smoky jetty]

this is probably about an ambiguous case right (I just read smth about it coz I forgot it already)? I think it means that the original angle and its supplementary angle cannot be in the triangle

does it have any figure u could share? im also confused with the wording

[pseudo viper]

this is the article im looking at
https://www.softschools.com/math/calculus/the_ambiguous_case_of_the_law_of_sines/

just confused bc if i subtract angle 80 from 180 to get an angle 100
adding 80 + 100 obviously equals 180 so i don't get what they're tryna say

[silent plank]

<@&268886789983436800> scamp didn't get purged

[smoky jetty]

do u know the very first step to determine if there are two triangles possible to be formed, one triangle, or none at all?

[pseudo viper]

no

[smoky jetty]

since you're given an SSA case, we wouldnt know for sure if the figure in the article has side b's length, long enough to enclose the triangle or at the shortest length to do the same

for that, we know that sinC/c has to be proportional with sinB/c as the law of sine states. Now, plug the given values and solve for the value of sin B

the value cannot be greater than one (or less than negative one), personally i dont know exactly why rn, but I'd say because 1 indicates sin90 degrees, and any sin of a degree higher (i.e an obtuse angle of a triangle) could be subtracted to 180 deg giving value less than 1

if the value meets this condition, say that the triangle can have 2 possible angles for the 2nd triangle, then the original angle or the given angle, 33deg, added to the possible angles of B (65.19 deg or 114.81 deg), should still give space to the 3rd angle u could easily find by 180deg= 33deg - (65.19deg or 114.81 deg) - angle A

do u get this elaboration @pseudo viper ?

because if the angle is greater than 180, then u basically have a triangle whose two angles are supplementary, and by property of triangle, it's simply false.

imagine a triangle with angles 120,60, and B. Doesn't make any sense right?

however if the second angle had a value when added to the given angle would give 90deg to the third one, then you have a right triangle as the only possible triangle to be formed.

i'd appreaciate anyone's correction on my elaboration

[pseudo viper]

so im guessing by original angle they mean the first angle that we know?

bc that's what im getting confused with

[smoky jetty]

yeah i think so

it's the only way I could say the steps are reasonable, otherwise im also confused by it

https://mathbitsnotebook.com/Geometry/TrigApps/TAUsingLawSines.html u can refer to this aswell

[pseudo viper]

so lets say you end up with two angles
is there a way to actually figure out if one is the correct one
or do you just have to pick one based off what makes the most sense?

ig they're both correct

[smoky jetty]

both triangles formed are correct, again there are two possible triangles, so, u have 2 possible answers

[smoky jetty]

precisely

[pseudo viper]

wouldn't one angle result in one of the lengths being shorter / longer
like if i get 60 and 120 as my angles
the sides with angle 60 would be shorter than the sides with angle 120
so wouldn't that allow you to figure out which is the one you actually wanna find
math is complicated smh

[smoky jetty]

let's say it's angle B that can have 2 possible angles (acute and obtuse), it already has a given side, 10, and the only side length that could have two possible values would be the opposite side to A, since angle A could also have two values

to check if 10 as b' s length is correct according to law of sine:
sin33/6 = sin65.19/10 = sin114.81/10
~0.091 = ~0.091 = ~0.091
(sine of 114.81 is equal the sine of its complementary angle, if you're familiar with reference angle stuff)

also, in the figure, it's side a's length that shortens as angle B becomes obtuse, and vice versa

[pseudo viper]

so to check if your angle matches up with what you already know, use law of sine as if you were trying to figure out side b ( if you just calculated angle B from the law of sine )

[smoky jetty]

considering that u already determined that whatever angle you're checking is valid, then yeah

[pseudo viper]

sweet

thanks a lot for the help, appreciate it!!

[smoky jetty]

your welcome!

[pseudo viper]

i've come to the conclusion that i hate the law of sine when tryna find an angle
it's a bunch of hassle 😭

[pseudo viper]

okay well that doesn't workl

(5 * sin(74 degrees) ) / sin(36 degrees)
(5 * sin(106 degrees) ) / sin(36 degrees)
both equal 8.17698038688
which i realise why now

so what do you do if you get an ambiguous case and need to know exactly which angle it is 😔

[dense kettle]

hii

[leaden quail]

hello i know this sounds cursed but for my finals they made me find the equation of an inscribed circle

looking at the test answers now, i somehow got it correct, although that subject was a pain

maybe even a big one

i mean all the stuff related to finding equations of tangents from point outside of circle was excruciating

i felt like the universe was fileting my self esteem

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs!
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[leaden quail]

,list

[somber coyote]

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[leaden quail]

wowie zowie

[grave pond]

Please use #bots or DMs for further help interactios.

[sonic mulch]

Does anyone recommend a book for me to start studying geometry for the Olympics?

[pseudo viper]

what do you do if you get an ambiguous case with the law of sine when trying to solve for an angle?
i know you can use the law of cosine to find an angle, but if i only have two sides and an opposite angle, then i can't use law of cosine bc i'm missing a side

i know in a real world scenario, you'll most likely know all the sides of the triangle
but what if you didn't?

[grave pond]

This reflects the fact that SSA triangles generally do have two solutions -- one where the angle at the end of the first side is acute and one where it's obtuse.
Sometimes you can eliminate the "obtuse" case because it would lead to those two angles alone to total more than 180°.
But in the rest of the cases the ambiguity is real and you'll have to hope the problem contains other information that will tell you which of the solutions is the one you want.

[pseudo viper]

gotcha
so it's a case by case basis really

that's annoying but understandable

[upper karma]

How strong does my trigonometry need to be, in order to advance to calculus?

[hollow dust]

tri

angles

[pseudo viper]

apparently you need a very strong understanding of algebra bc that's where everyone falls over

[smoky jetty]

after almost 3 hours of trying to find this only using geometry, I gave up and used law of sines lmao. Was able to find it, fortunately (took some bruteforce coz im dumb)

@pseudo viper wanna give it a try?

there are a few ways the problem can be tackled, im gonna share mine in a few mins

[pseudo viper]

uhhhhhh

yeah idk where i'd even start with that

[smoky jetty]

have u done geometry before?

like in school

[pseudo viper]

i'm like 7 years outta school so anything i learnt back then is forgetten
i did run through the geometry section on khan academay lmao but i quite literally just ran to get some basics down

[smoky jetty]

i see

well i'd say that it's doable with basic knowledge on circles, rectangles, and law of sine

at least on the solution I came up with

[pseudo viper]

hmmmm

[snow crystal]

How did you use law of sines here?

It's not a 30 60 90 triangle

[smoky jetty]

i'll share my solution later

[snow crystal]

Alr

Maybe you just evaluated the angles

But I'm not sure if it's necessary

[smoky jetty]

im trying to make it understandable rn, coz the draft i have involves too much process that wasnt necessary after using law of sine

[smoky jetty]

ooh, welp u could def give it a try and check

[snow crystal]

Ok I will

[pseudo viper]

law of sines isn't only for a 30 60 90 triangle

[snow crystal]

I know

But I want to approach this problem more geometrically first

Without using the law of sines

[pseudo viper]

gotcha

yeah i wouldn't know where to start
p much my only knowledge atm is algebra and trig

should probably spend a lil time on actual geometry

[snow crystal]

Welp idk how to do it with geometry 😸

At least not yet

The answer I got is 4sqrt(3)-4π/3

@smoky jetty is that the answer?

[smoky jetty]

hmm nope

if it's equal to ~ 1.87, it's quite close

ffs i used ballpen and im struggling to make my solution pretty now

[snow crystal]

This is what I did

,rotate

[snow crystal]

@smoky jetty I'll be grateful If you find the mistake (if there is any)

[smoky jetty]

ight im gonna share it

sorry if many parts are simplified to such extents ( i dont have correction tape with me atm, so mistakes were hard to erase, hence dirtier work), so feel free to ask for elaboration

@snow crystal

[snow crystal]

Uhh

I think it would be easier if you look at my solution tbh

[smoky jetty]

was this your final answer?

[snow crystal]

Yeah

That's exactly what I got

[smoky jetty]

welp it's pretty off i'd say

[snow crystal]

But are you able to find any mistake in my solution

[smoky jetty]

im checking rn

[snow crystal]

Cause I've been trying to for past 10-15 minuts and haven't found anything yet

[smoky jetty]

but both of our handwriting looks so uhh lmao

[snow crystal]

Lol true

[snow crystal]

Did u find something?

[smoky jetty]

sorry for the late response, i couldnt find exactly where u got it wrong

hopefully others could try and check

[snow crystal]

So that means the answer I got is correct

Because If not then idfk

[smoky jetty]

no, the correct answer is ~1.252

[snow crystal]

U sure?

Do you have the answer key for that

[smoky jetty]

yeah, the video reference to it has the answer as 1.252 as well

[snow crystal]

That's weird

[smoky jetty]

https://www.youtube.com/watch?v=2Seb863FnfU lol ignore the clickbaity title

had others try it as well (they were pretty much olympiad nerds) and got the same anwer.. lmao they said it wasnt even difficult

[snow crystal]

Ok so tell me what I did wrong

[smoky jetty]

soorry but i just couldnt see the wrong part, and the handwriting also is pretty confusing to me

though certainly something is wrong there

[snow crystal]

Ok Im gonna show this exercise to my teacher and maybe she'll tell me

[smoky jetty]

yeah that'd be better!

nice try tho!

[snow crystal]

Though I'll have to wait for like a week or something