geometry-and-trigonometry

[unborn holly]

I saw a thing that represented midline as v for a sinusoidal equation. Does anyone know why?

[pastel anvil]

wat

ok so I think u do it like this

so if I make s as the side

[tidal river]

I attemped it

wanna hear it

[pastel anvil]

ya go ahead

[tidal river]

so we know the total periemeter has to be 90

and its a rectangle

and we a given 18

so

[pastel anvil]

90 ft2 is area

[tidal river]

I did 90 -18 -18

[pastel anvil]

is not the perimiter

if its units^2 that means area

[tidal river]

umm then I dont know

I was doing it wrong then

how did you do it

[pastel anvil]

the P=18+18+2s

also 18s=90

f*uck

lol

so s=90/18

I censored it

[tidal river]

thats better

looking out for you

man or woman

[pastel anvil]

anyways then the P=18+18+2(90/18)

thanks bro

[tidal river]

so the final answer is

[pastel anvil]

wat

[tidal river]

whats the answer

[pastel anvil]

it's 18+18+2(90/18)

[tidal river]

oh

k

how do you do the server calc

I wanna try it

[pastel anvil]

I dunno but there's a robot to do it

,calc 20+18

[tidal river]

, calc 18+18+2(90/18)

[somber coyote]

Result:

38

Result:

46

[tidal river]

nvm

[pastel anvil]

woah

[tidal river]

I figured it out

[pastel anvil]

technology is amazing

[tidal river]

I know right

[pastel anvil]

u have more problems

[tidal river]

yeah hold up

a sec

[pastel anvil]

so u want to find the area of the entire rectangle then minus the area of the 6 circles

I forgot how to minus

oops

u know what the formula for a circle is

[tidal river]

I cant find the area of 1 cirle

[pastel anvil]

well what's the area forumual

[tidal river]

I know that

But we dont have the radius or the daimeter

nvm

I can do 36/2

to find the diameter

[pastel anvil]

yeah

that's it

[tidal river]

then radius

[pastel anvil]

yep

[tidal river]

, calc 6^6^3.142

[somber coyote]

Result:

5.9982085016251e+216

[tidal river]

technology is disapointing some times

[pastel anvil]

lel

[tidal river]

, calc 113.112^6

[somber coyote]

Result:

2.0943636365788e+12

[tidal river]

bru

Im not getting the correct answer

I did 36/2

18

/2

to get 9

as the radius

[pastel anvil]

yea

[tidal river]

then 9 times 9 times 3.142

and got

254.502

times 6 becuase therer are 6 cirles

I got 1527.012

is that how much wood will not be used

[pastel anvil]

that's how much wood is used

u have to minus this from the area of the rectangle

[tidal river]

so I subtract it from 54 times 36

[pastel anvil]

yea

[tidal river]

I got 417.19

well actually I got 416.988

but that was the closeset multiple choice answer

so it prob correct

This is the last one

I have no clue how to do this one

I know theres a formula but

I dont know it

[pastel anvil]

bruv this is like the easiest one out of all of them lol

ok

so the formula for arc length is this:

for a circle

n/360° × 2πr

ah damn nevermind wtf

which one is ST the shaded one or the not shaded one

[tidal river]

We need radius or diameter

for circumference

[pastel anvil]

when it says arc ST which one is it referring to

[tidal river]

the shaded area

[pastel anvil]

ok so the small one

right

[tidal river]

yeah

[pastel anvil]

ok

so use the formula for circle arc length:

n/360 × 2πr = L

solve for the radius then u can circumference ez

[earnest echo]

2pi times r is already circumference

[pastel anvil]

o yea that's right solve for 2πr

[tidal river]

is 9 pi

the arch length or the sector area

[pastel anvil]

length

[earnest echo]

It clearly says the length of arc ST

[tidal river]

my

bad

So the final answer for the circumference I got was 23.312

correct me if Im wrong

[austere dragon]

what's the symbol k for if we already have symbol AB to denote it?

[sand cairn]

I would say Line segment k.

[silent plank]

to denote the actual line, you'd need to write $\overset{\leftrightarrow}{AB}$

[somber coyote]

ramonov:

[silent plank]

using a single letter without arrows can make it look simpler

[sand cairn]

I unequivocally concur with Ramonov.

[austere dragon]

where does they get 7 from and how did they get 8ps/7 (ps is 9)

[dire sand]

huh

8*9 = 72

72/7

Areyou able to show the parent problem

[dark sparrow]

$PT - \frac{1}{8}PT = \paren{1 - \frac18} PT$

[somber coyote]

Ann:

[dark sparrow]

does that make it clear @austere dragon

[austere dragon]

1 is 8/8 correct?

[sand cairn]

Yes, @austere dragon.

(8)/(8) simplifies to 1.

[austere dragon]

🙏 thank

[sand cairn]

You are welcome.

[austere dragon]

wait but why is pt = 8ps/7?

[dark sparrow]

PS = 7PT/8

multiply both sides by 8/7 what do you get

also, uppercase letters

[austere dragon]

oooOOOO

🙏 thank

[austere dragon]

"what is the maximum possible number of points at which at least two of the five figures intersect?"
what does the final question mean?? i thought they meant if i choose 2 of the 5 figures mentioned what would be the maximum number of points they intersect which is 2, but apparently they asked to try to intersect all of the figures mentioned? what?

[earnest echo]

For maxima, all the figures need to intersect each other

2 circles can intersect each other at most of 2 points
2 lines can intersect each other at most of 1 point
And a line and circle can intersect each other at most of 2 points

So we have
Intersection of circles, intersection of lines and intersection of lines and cirlces

[livid moss]

I think I need to draw a Venn diagram to make sense of all these intersections

[earnest echo]

Not really

We can just count

[livid moss]

I was kidding, also not the one who asked the q

@austere dragon

[austere dragon]

oh sorry i went to eat porridge lemme read it

[earnest echo]

If you can figure out, then all well and good otherwise ping me

[austere dragon]

oooooooo ok make sense, thank 🙏

[earnest echo]

Np

[supple wedge]

can i use discriminant to solve part 2

centre is (1,0.5) and radius=sqrt(11.25)

i used it and the answer was nice so i assume its correct??

[earnest echo]

Yeah, you can use discrimination

[supple wedge]

aight thx

[dark sparrow]

@earnest echo wrong word

[earnest echo]

Lmao, My fucking autocorrect
I'm not even gonna edit it

[paper vale]

[mortal mesa]

Hi. I was solving the following trig equation without squaring both sides: $1-\sin 2x = \cos x -\sin x\Rightarrow \sin^2 x +\cos^2 x - \sin 2x = \cos x -\sin x\Rightarrow (\sin x - \cos x)^2 + \sin x - \cos x = 0\Rightarrow (\sin x - \cos x)(\sin x - \cos x + 1)$ and I got $x=\pi/4 + \pi k, -\pi/4 + 2\pi k$.
However, $x=0$ also solves it, but I didn't get it. Can you please tell me why?

[somber coyote]

idan:

[final aspen]

quick question, i always used (x,y,z)=(x,y,x) + k(x,y,z) to define the plan in space

[silent plank]

@mortal mesa show your work for the last part

[final aspen]

oh didnt realize there was a quation already on going sorry

[mortal mesa]

@silent plank Sorry for the mess..

$\sin x - \cos x = 0 \Rightarrow \sin x = \cos x \Rightarrow \sin x = sin (\pi/2 - x) \Rightarrow x = \pi/2 - x + 2\pi k \Rightarrow x=\pi/4 + \pi k$

For the second part:

$\sin x - \cos x + 1 = 0 \Rightarrow \sin x - \cos x = -1 \Rightarrow \sin x - \tan \pi/4 \cos x = -1 \Rightarrow \sin x \cos \pi/4 - \sin \pi/4 \cos x = -1 \Rightarrow \sin (x - \pi/4) = -1 \Rightarrow x - \pi/4 = -\pi/2 + 2\pi k \Rightarrow x = -\pi/4 + 2\pi k$

[somber coyote]

idan:

[silent plank]

um use \\ to start new lines

the arrows make it very hard to read

[livid moss]

But, one comment, sin(α) and sin(β) are not only equal when the angles are a multiple of 2π apart. You could have α = π-β + 2πk, too

[silent plank]

fk that was hard to follow

i see the mistake

you changed the value of the LHS

i.e. you effectively multiplied the left side of the equation by (sqrt(2)/2 by introducing cos(pi/4) and sin(pi/4) but left the right side untouched

instead you should have: $\sin(x - \frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$

[somber coyote]

ramonov:

[mortal mesa]

@silent plank
I didn't multiply LHS, I just replaced the coefficient 1 of the cosine with tan(pi/2)

[silent plank]

you mean pi/4?

that step was still legitimate, its what happened after that

[mortal mesa]

@livid moss
I'm not sure which part you're referring

@silent plank yeah sorry

@silent plank oh great, you're right

[silent plank]

i also prefer using the cosine form of that identity since its more convenient to write its general solution

[mortal mesa]

thanks very much! again, sorry for the mess

[silent plank]

using tan like that was unnecessary

the identity pretty much abuses the fact that sin(pi/4) = cos(pi/4) = sqrt(2)/2

[mortal mesa]

Just trying new things.. I don't agree about the abuse, tho.
It's a valid way

[rotund cypress]

Been working on this for a personal project, I don’t know really know if it’s solvable 😅

After playing with it for a while I think it’s equivalent to these

[upper karma]

Maybe this will help

Dividing 360-phi into 3 parts

alpha_1 and alpha_2 are easy to find

alpha_3 will need some calculations

@rotund cypress

[rotund cypress]

hmmm

im not sure what we know about alpha3 triangle

we could calculate the length of the bottom side

but i dont think that's enough to find the thoer angles

[upper karma]

If we get all 3 sides (we can get 2 easily) then you use cosine law and you can get alpha3

[rotund cypress]

yeah i should have made it clear in the orignal picture that i think the common tangent between the two cricles lenght can be calculated

so we can find 2 sides of the alpha3 triangle but i dont see a way to find the 3rd side

[upper karma]

You can in terms of phi2

So at the end you'd get phi1 in terms of phi2 and phi2 in terms of phi1

[rotund cypress]

hmm i see so there is a family of solutions

[upper karma]

System of equations

I think so

But I think the solution has a horrible form

[rotund cypress]

is there some sort of wolfram alpha for plane geomtry?

[upper karma]

Wait

This is cleaner

This should be doable

[lean monolith]

Hello

[rotund cypress]

hi

[lean monolith]

Do u teach calculus?

Hehe

[prisma cape]

Go to #calculus

[lean monolith]

Will they teach me there?

[prisma cape]

idk

[rotund cypress]

they might lol

[upper karma]

@rotund cypress can you continue from there?

[rotund cypress]

@upper karma im trying, i dont see it yet but im slow

[upper karma]

I labeled

DEAC is a rectangle

So CA=DE

[earnest echo]

What is that
A question from JEE advanced

[upper karma]

By Pythagoras you can easily get AB and CB

then cosine law/rule to get angle <BCA

Angle <DCF is easy to get

Same thing for phi2

[rotund cypress]

oh okay ill try that and get back to you, thank you so much

[upper karma]

Np

[rotund cypress]

@earnest echo no, this for a physics lab, I'm an asistant and was told to "make sure these things fit right" this geometry problem is a very simplified picture of 2 toruidal rings that shot jets of vapor around in a ring

which is why didnt even know if the problem was well posed/ solvable

[upper karma]

I think it is

[rotund cypress]

being a stupid physics person i just loaded it in cad and came up with a numerical solution but then i thought i should try to understand it better

[upper karma]

lol

[upper karma]

@rotund cypress it should look like this

[somber coyote]

HoboSas:

[rotund cypress]

Oh nice!, I just finshed redownloading mathematica

is the 270 in degrees?

also is L3 the distance between circle centers?

[upper karma]

Yes and yes

Do you also need the other one?

[rotund cypress]

That would be amazing 😅

[upper karma]

Brb

[somber coyote]

HoboSas:

[upper karma]

Here you go

Nothing too hard, just cosine law and definition of tan

You should try to get those results yourself

[rotund cypress]

Thank you so much! Is it bad form to use mathematica to solve systems like this?

[upper karma]

It's not a system

Just plug in the numbers

[rotund cypress]

Oh I see

I guess the hard part was picking the right triangle to focus on

[upper karma]

Yes

Constructing the rectangle was the one

[rotund cypress]

Yeah it took me a while to realise that it had to be a rectangle

[versed river]

pythagoras theorem ig

on one of the smaller right triangles that make it up

[tidal river]

@brisk palm could you help me with this question

[umbral snow]

Can you form a few equations? What have you seen so far?

[boreal adder]

Y is between X and Z. If XY = x + 5 ,
YZ = x , and XZ = 14 , find XY and YZ.

this is from my math warm up and cant figure it out

[tidal river]

how do I find the area for figure C

I ignore all the other figures around it

[versed river]

XY+YZ=XZ

this is for @boreal adder

[brisk palm]

Quantum, the figure is made entirely out of right triangles, and since you know how to find the area of a right triangle, try to find a way to add/subtract right triangles to get the area of the shape

[tidal river]

Could someone help me with this question

[upper karma]

There's a chord theorem that says: BRxRD=ARxRC

From which you get the diameter

[tidal river]

what does b r d a and c stand for

[upper karma]

@tidal river BR is the segment from B to R, same thing with the rest

[tidal river]

oh thanks @upper karma

so the equation would be 3 times x = 4 times 4.5 I would get x and add br or 3 to it

and that would get me the daimeter

from which I could use to find the radius

Correct me if im wrong

[upper karma]

so the equation would be 3 times x = 4 times 4.5 I would get x and add br or 3 to it
$$\overline{BR}\cdot\overline{RD}=\overline{AR}\cdot \overline{RC}$$ $$3\cdot \overline{RD}=4\cdot 4.5$$ leave it like RD, no need to be called x

[somber coyote]

Al𝟛dium:

[upper karma]

But yes it is 3* RD=4*4.5 as you said

@tidal river

[tidal river]

btw how did you make the bot do that

Whenever i use it

It acts preety stupid

[finite sky]

https://cdn.discordapp.com/attachments/576559291952136194/743391027037077544/unknown.png

[upper karma]

There is document at #resources that deals with TeXit

@finite sky it is not free, move to another channel

@tidal river are you good with the problem? now that you have RD, simply add it with 3 as you said and divide the result by 2 to get the radius

[tidal river]

yeah

I also have another problem I did it all but could you like check it to make sure I did it correctly

thats the pproblem

@upper karma so 9is the diameterr

becuase algebra

[upper karma]

,w [[(4*4.5)/3]+3]/(2)

[somber coyote]

Results provided by WolframAlpha

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

Requested by Al𝟛dium#4084

[upper karma]

Yes.

[tidal river]

nice thx

also could you check a problem that I did

its above this message

I already did all of it

just make sure its correct

[upper karma]

Sure

Let me see

Okay a is correct but on b:

You have to put parens in plain text so that they know you mean $\frac{y_2-y_1}{x_2-x_1}$. So you shouldn't be writing it as \verb|y2-y1/x2-x1| which can be understood as $\frac{y_2}{x_2}-\frac{y_1}{x_1}$ and not $\frac{y_2-y_1}{x_2-x_1}$ which is what you actually want.

And also at b you used the point (-12,0) instead of (-12,6)

[tidal river]

@upper karma at b I did use (-12,0)

[upper karma]

Yeah so use (-12, 6) not (-12, 0)

[tidal river]

why

Alo btw

[upper karma]

Bc the stop point is (-12,6) not (-12,0)

As you said correctly in a

[tidal river]

on ur message you wrote " use the point (-12,0) instead of (-12,6)"

[upper karma]

I meant you used

Let me fix it

[somber coyote]

Al𝟛dium:

[tidal river]

my bad just got a litle confused

[upper karma]

Okok dw

Replace the point and use the appropiate parens

(y2-y1)/(x2-x1) and NOT y2-y1/x2-x1

[tidal river]

ok

also

how tf do I solve for part b in this quesion from earlier

[upper karma]

Sure so it's using another circle theorem

[tidal river]

go on

[upper karma]

Okay wait a sec

The first one

(i gtg go rn) hope you find what you want, remember the alternate angles for this one too

[solar hollow]

Anyone here?

[upper karma]

[cosmic pebble]

From

#❓how-to-get-help

Rule 1

The help channels are solely for help with math, so feel free to post your question. Asking whether you can ask a question or if anyone knows about some specific topic is unnecessary, so please try to avoid questions of that nature.

[calm shoal]

does anyone know how to solve this?

[earnest echo]

First of all fuck this imperial measurement system

Now coming to your question

Speed=radius x angular velocity

@calm shoal

[quiet mason]

[upper karma]

I only use length of 3 American football fields per average time to cook a medium-rare 1 inch steak

[little osprey]

I use 1/8 of shrek time to cook a steak

[rose tulip]

In geo will there be a case where they dont give you any numbers.

Just x and y

and no protractor needed

[upper karma]

Yes

[rose tulip]

How would you solve those?

[upper karma]

What kind of question is that

It depends

[rose tulip]

Im not at those questions yet. Just wondered how to solve no numbered questions.

[upper karma]

It's mostly proofs

So you go step by step

Using theorems and postulates

[rose tulip]

Pythagorem theroms?

[upper karma]

there are many theorems in geometry

And in maths in general

[rose tulip]

Ok. I was confused.

[rose tulip]

https://youtu.be/Ld7Vxb5XV6A
Where did he get 180 from? At the end of the equation.

[upper karma]

They are supplementary angles

[rose tulip]

Yeah

[upper karma]

Which means both angles add up to 180°

That's why

[rose tulip]

Yeah

you gotta write the 180 at the end?

[upper karma]

As it is true that both angles add up to 180. This equation must be true:

[first angle]+[second angle]=180

What he did basically

[rose tulip]

So if we are solving for a complemantary angle we gotta add that =90 at the end

[rose tulip]

And what did he do to get -90?

[rose tulip]

180-270 right?

[tidal river]

Could someone help me with this

and yes I know this a probabilty

but it showed up on my geometry course

[upper karma]

“X out of Y” rings a bell for me

But I haven’t taken a geo course yet so idk

[hard forge]

how can you derive the sum/difference trig identities

[lost osprey]

Geometrically

[hard forge]

oh

what is the process

[lost osprey]

Set up a unit circle diagram where you partition an angle into two arbitrary segments

[hard forge]

ok

[lost osprey]

And find the sine of their sum in terms of their individual sines

[hard forge]

huh

that doesn't seem so bad

[lost osprey]

No not bad at all

Very clean derivation

[hard forge]

yeah

thanks!

[versed river]

https://themathpage.com/aTrig/sum-proof.htm

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:trig/x9e81a4f98389efdf:angle-addition/v/proof-angle-addition-sine

both of those go over it

oh it was 3 hours ago

kek

[austere dragon]

how did they get "x + y = 60"?

[upper karma]

3x+3y=180

From first equation

[austere dragon]

wait

can you continue from 3x+3y = 180 to x + y = 60

my algebra rusty

[upper karma]

Bruuuh

3(x+y)=3(60)

[versed river]

diwide both swides by thwee

[austere dragon]

oooo so u factor it out

ok thank

[ashen axle]

I am generally looking to pay someone to do my geometry class. 200$. Its a decent amount of work, not sure if this is allowed in this discord if not I apologize. Im just not smart in math and i dont have a teacher so id rather just pay someone than have a huge stress on my back. Dm me if intrested! Thanks.

[dark sparrow]

uhh yeah this is not allowed

[upper karma]

Rule 2: Requesting or offering the exchange of money for completing homework assignments is a bannable offense.

On #rules

[ashen axle]

ok ill head out ✌️

[livid moss]

Lmao, not even helping with an assignment, just doing the whole class

[little osprey]

ikr

[supple wedge]

bruh

[brisk palm]

$200 for a whole class?

[paper vale]

[covert rune]

I’m more worried as to how a grade 8 student has 200$ to spend

[surreal bolt]

Eh Good Will Hunting ... quote about learning and a few bucks in library fines

[upper karma]

Hello everyone, I have a trigonometry-related question.
While trying to find the arctan of -1, I looked at the unit circle trying to find angles whose sin and cos values are the same as far as the number part goes, but with opposite signs. And I noticed that this occurs in both the 2nd and the 4th quadrants. I would like to understand the reasoning behind choosing the correct angle.
Thank you in advance. :)

[upper karma]

Arctan has a limited range

(-pi/2,pi/2)

Ah, so that limits it to the 1st and 2nd quadrants?

Because tan is bijective between (-pi/2,pi/2)

Ah, so that limits it to the 1st and 2nd quadrants?
Yes

That's a convention

Tan is also bijective like between (pi/2, pi)

My knowledge in this department has gotten a bit rusty, but I think I now understand which direction I should go. Will revisit bijective functions to fully grasp this. Thank you a ton!

As a convention you use this

No problem

My knowledge in this department has gotten a bit rusty, but I think I now understand which direction I should go. Will revisit bijective functions to fully grasp this. Thank you a ton!
Look up also inverse functions

How they relate to the original

I absolutely will!

Actually at least injective, (which includes bijective)

For a function to be invertible

I should perhaps go back to surjective functions too, since I remember bijective functions are the ones that are both injective and surjective

[upper karma]

Correct

[hard forge]

how do you derive the power reducing identities?

[paper vale]

just work backwords from the double angle formula

[hard forge]

uh

ok

so for sine

$\sin^2(\theta)=\frac{1-\cos(2\theta)}{2}$

[somber coyote]

Rubidinium:

[hard forge]

and

$\cos(2\theta)=\cos(\theta+\theta)=\cos(\theta)\cos(\theta)-\sin(\theta)\sin(\theta)$

[somber coyote]

Rubidinium:

[hard forge]

$\cos(\theta)\cos(\theta)-\sin(\theta)\sin(\theta)=\cos^2(\theta)-\sin^2(\theta)$

[somber coyote]

Rubidinium:

[hard forge]

so I should just plug that in to the power reducing identity?

$\sin^2(\theta)=\frac{1-\cos^2(\theta)-\sin^2(\theta)}{2}$

[somber coyote]

Rubidinium:

[hard forge]

oh

[paper vale]

nah u cant do that

just add 1 on both sides then divide by 2 in the double angle identity

then use 1 = sin^2 + cos^2 on RHS

then rearrange

[hard forge]

uh

$\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)$

[somber coyote]

Rubidinium:

[hard forge]

so I add 1 to both sides here?

[paper vale]

yea im talking about that

[hard forge]

oh ok

$\frac{\cos(2\theta)+1}{2}=\frac{\cos^2(\theta)-\sin^2(\theta)+1}{2}$

[somber coyote]

Rubidinium:

[paper vale]

yea use the identity that i said for the 1 in RHS

[hard forge]

RHS?

[paper vale]

right hand side

[hard forge]

oh ok

wait

how can you use that Pythagorean identity?

[paper vale]

what do u mean how

just use it

[hard forge]

but it's cos^2x-sin^2x

I thought u could only use the identity for sin^2x+cos^2x.... which equals 1

[paper vale]

yea so use it

i dont know what u are confused about

[hard forge]

oh wait

do u want me to substitute the +1 on the end with sin^2 x + cos^2 x

[paper vale]

yea

[hard forge]

ohhh that makes sense

$\frac{\cos(2\theta)+1}{2}=\frac{\cos^2(\theta)-\sin^2(\theta)+\cos^2(\theta)+\sin^2(\theta)}{2}$

[somber coyote]

Rubidinium:

[hard forge]

cool

$\frac{\cos(2\theta)+1}{2}=\frac{2\cos^2(\theta)}{2}$

[paper vale]

no

[somber coyote]

Rubidinium:

[paper vale]

yea now that is it

[hard forge]

yeah I made a mistake with the latex lol

[paper vale]

for sine one, just use this identity again now

[hard forge]

ok

[paper vale]

cos^2=1-sin^2 and rearrange

[hard forge]

sure

$\frac{\cos(2\theta)+1}{2}=1-\sin^2(\theta)$

[somber coyote]

Rubidinium:

[hard forge]

makes sense

thanks

[bronze onyx]

Hi @hard forge

[hard forge]

hello

[rose tulip]

Prolly a bad time but could this go both ways?

https://youtu.be/6s1CI3uuhko

With Z and Y being the Sup angle.

Oh shit it is

Fuck sorry

[supple wedge]

pt 1

how do i approach this

@ me if ur helping

[surreal bolt]

Do you know what theorems might be helpful here?

[supple wedge]

nope

my geometry is dog tier

[surreal bolt]

This is for class right?

[supple wedge]

nah just a practice exam paper

[surreal bolt]

How soon do you need this?

[supple wedge]

like anytime but i wanna do it now

[surreal bolt]

sure. If you want to try it yourself. I have two hints.

[supple wedge]

go ahead pls

[surreal bolt]

1. separate the two triangles (draw them on different pieces of paper even) and 2) google "similar triangles" or "similar triangles proportions"

[supple wedge]

aight

[surreal bolt]

come back in 20 minutes 🙂

[supple wedge]

ok so i gussed

DCE and ABD are similar

and AC:CE is also sqrt(5):3

@surreal bolt

[surreal bolt]

er AC: CE isn't important 🙂

[supple wedge]

@surreal bolt they asked me to find CE tho

idk im just guessing xd

[surreal bolt]

AC isn't a line in the picture 🙂 and you don't need it 🙂

[supple wedge]

oh fuck

like EC:AB

i type and saw wrong

[surreal bolt]

it's fine

[supple wedge]

so EC:AB is also sqrt(5):3

amirite

[surreal bolt]

I dunno what other ratio are you comparing it to?

[supple wedge]

the one in the qn

[surreal bolt]

what is the ratio in the question?

[supple wedge]

sqrt(5):3

[surreal bolt]

What sides does that ratio describe?

[supple wedge]

nothing else

so am i right or not

[surreal bolt]

(read the problem). What sides does sqrt(5):3 describe?

[supple wedge]

AE:ED

[surreal bolt]

(I know you don't know me, but i wouldn't be asking you so many annoying questions if you were right.)

[supple wedge]

guess im wrong

[surreal bolt]

at the moment maybe!

work at it 🙂

[supple wedge]

bruh

ok

[surreal bolt]

AE ED is nice.

but is that really what you want?

[supple wedge]

no how do i get the other ratio

[surreal bolt]

Did you draw two triangles?

[supple wedge]

no

[surreal bolt]

lol

[supple wedge]

can see it tho

[surreal bolt]

pls follow directions.

just humor me 🙂

[supple wedge]

then what now

[surreal bolt]

okay is AE part of one of the two triangles?

like is it a FULL side>

[supple wedge]

no i drew them seperately

ill add it in i guess

[surreal bolt]

My point is find AD.

[supple wedge]

the fuck whats ur plan now

[surreal bolt]

find AD:ED ...

[supple wedge]

aight

[surreal bolt]

Hey I gotta go.

[supple wedge]

bye

[surreal bolt]

Two fucks a night is enough 🙂

[supple wedge]

sure

ill do other questions and if anyone else wants to help me ping thx

[earnest echo]

@supple wedge you here?

[supple wedge]

ye

@earnest echo

[earnest echo]

Do you know about similarity?

[supple wedge]

??

sort of

[earnest echo]

Do you know AA similarity condition?

[supple wedge]

heard of it before

[earnest echo]

Do you know the concept or not?

[supple wedge]

aka teacher taught it but idk what it is

nope

[earnest echo]

First you need to learn what similarity is........
Go on YouTube, there are many videos on similarity
Try to understand it and then try the question again and if you are still stuck, then post your question again

[supple wedge]

link one if u dont mind?

[earnest echo]

https://youtu.be/reGN77SiESA

This might help

[supple wedge]

thx

ima watch

holy shit 1 hour

[supple wedge]

Find, in radians, the two principal values of $y$ for which $tan^2y+tany-6=0$.

[somber coyote]

Hmm:

[supple wedge]

whats a principal value

do i just solve normally

[dark sparrow]

uh

is... that exactly what the thing says

[supple wedge]

yes

[dark sparrow]

does your book ever define what it means by principal values

[supple wedge]

let me find my book which i havent touched for like 5 years

be right back

[dark sparrow]

i have a hunch it might be asking you to solve this thing for y ∈ (-π/2, π/2)?

[supple wedge]

thats what i thought too

not sure tho

ok found my book

it says

wait let me test something

$\deg$

[somber coyote]

Hmm:

[supple wedge]

fuck

how do you do degree in latex

[dark sparrow]

^\circ

[supple wedge]

oh

For the function $y=sin(x), -90^\circ\leqx\leq90^\circ (or -\frac{\pi}{2}\leq{x}\leq{\frac{\pi}{2}}$ are chosen to be the principal values of $x$ or $sin^{-1}y)$.

fuck

[somber coyote]

Hmm:
Compile Error! Click the reaction for details. (You may edit your message)

[supple wedge]

whatever

@dark sparrow

lets see

For the function $y=sin(x)$, when ${-1}\leq{y}\leq{1}$, the principal values of $x$, or $sin^{-1}y$ are -90^\circ\leq{sin^{-1}y}\leq90^\circ$

[somber coyote]

Hmm:
Compile Error! Click the reaction for details. (You may edit your message)

[supple wedge]

For the function $y=cos(x)$, the principal values of $x$, or $cos^{-1}y$ are $0^\circ{\leq}cos^{-1}y{\leq}180^\circ$

[somber coyote]

Hmm:

[supple wedge]

For the function $y=tan(x)$, the principal values of $x$, or $tan^{-1}y$, are : $-90^\circ<tan^{-1}y<90^\circ$

[somber coyote]

Hmm:

[supple wedge]

@dark sparrow

that is what my book says

[dark sparrow]

bad latex

[supple wedge]

i know

xdd

@dark sparrow so how do i do that qn

[dark sparrow]

solve $\tan^2(y) + \tan(y) - 6 = 0$ for $y \in (-\pi/2, \pi/2)$

[somber coyote]

Ann:

[dark sparrow]

y in radians obviously

[supple wedge]

aight

[upper karma]

what are some ways i can calculate the area of an irregular polygon?

[upper karma]

calculate

[dire sand]

tf is your username lol @upper karma

𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫

[austere dragon]

how am i supposed to figure out the sum of the interior and exterior angles of this hexagon can someone give me a hint this is nearly impossible

its real quite here

i dont wanna cheat by just looking up the sum of the interior angles first because i think the exercise assumes that i do not know the sum of the interior angles of a hexagon

[lean plaza]

What is the question

[versed river]

you have 6 triangles, so every angle in each of those triangles should add to 180

[austere dragon]

thanks

[versed river]

that gives you 6*180=1080 in total, but thats including some angles you don't actually carte about

those angles are all in the centre in a circle, so they add to 360

so the sum of the interior angles is 1080-360=720

[runic beacon]

Or you can just consider one vertex of the polygon and then draw diagonals to the rest of the vertices of the polygon from that vertex.

This will consider every angle of every triangle that you will make.

So then it would just be (number of triangles )* 180

By number of triangles I mean when you will be dividing the polygon the triangles formed then

[dark sparrow]

@austere dragon in euclidean geometry the exterior angles of a convex polygon add up to 360° always

[austere dragon]

🤯 wow what

thats weird but awesome

[rich wolf]

Makes sense if you think about it

Exterior angles of a convex n-gon form supplementary angles with the interior angles of the polygon

So you can subtract the value of the sum of the interior angles from 180n

So you have

,w 180n - 180(n-2)

[somber coyote]

Results provided by WolframAlpha

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

Requested by AMD#4434

[austere dragon]

wuh

what if it was an octagon wouldnt it be over 360 then

checkmate mathist

[dark sparrow]

i said exterior angles

there is nothing special about the case n=8 here lol

[austere dragon]

too good to be true

[dark sparrow]

ok yknow what

think of it this way

imagine you're driving a bike along the polygon

exterior angles are the angles by which you need to turn on the corners

but by the time you made one full lap around the polygon you will end up facing the same way you started

so all the turns you've made at every corner, combined, give you 360 degrees

[austere dragon]

then why doesnt the interior angles sums up to 360 too

[dark sparrow]

why would they

[austere dragon]

cuz they go around too

[rich wolf]

pretend those are circles

[dark sparrow]

@austere dragon no??

the angles by which you turn are the exterior angles, not the interior angles

[austere dragon]

ok im understand now thank

[austere dragon]

why is d + dab = 180 and b + bad = 180??

[upper karma]

Notice that it is a parallelogram

[austere dragon]

yea?

what u trynna prove dude

[upper karma]

...?

Properties of a parallelogram...

@austere dragon

[austere dragon]

uh im missing the point sry

plz explain

[upper karma]

https://images.app.goo.gl/SVy2QdcCSgWSfNPh7

Click on it

[austere dragon]

wow what i didnt know that

thats pretty cool

[austere dragon]

is it bad to guess in geometry

like if theres an obvious right angle in a diagram you shouldnt guess that its a right angle and do it the hard way instead

cause ive been doing that method

[earnest echo]

If it's not clearly mentioned that a given angle is right angle, then don't assume it to be right angled

[upper karma]

Yes it is in general bad to assume with no argument

[austere dragon]

🙏 thank guys

[solid sequoia]

For #11, I have 1/6th of the circle is the arc, but that would mean 1/6 x 36pi would be 6pi and that isn’t an answer

This is confusing

[upper karma]

Yes

Ik

[silent plank]

how are you getting 1/6 th of the circumference for the arc?

or more specifically why are you doing 60°/360° ?

[solid sequoia]

I got it by doing the 2 angle thing, 30 + 120 = 150 which would be the outer angle of the circle x 2 = 300, leaving the other arc as 60 so 60/360 of the circle

Looking at it now, the graphic doesn’t really match up, so how would you do this the proper way?

[silent plank]

@solid sequoia
what "2 angle thing"?
you seem to be just doing random things.

do you know which angle you're supposed to be using to get the arc length?

[solid sequoia]

When the 2 angles in the triangle can find the outside angle @silent plank

I'm not sure since I haven't taken Trigonometry thoroughly so this is probably wrong

[silent plank]

when calculating the arc length AB, the angle you want to use would be angleAOB

[solid sequoia]

So 120?

[silent plank]

yes

[solid sequoia]

omg

im so stupid i know this and i didnt even see it sorry

so it's 12pi...

[upper karma]

Hey guys i have another arc question

[abstract narwhal]

whats the question

i'll go after you lol

[upper karma]

lol tysm

this is the question 🙂

$s=r\cdot \theta$ where $\theta$ is on radians and s is the length of the arc

[somber coyote]

Al𝟛dium:

[upper karma]

@upper karma

How do we determine the radian length though?

I mean the radians

What?

[abstract narwhal]

it tells you the radian measure is between 3pi/4 and 5pi/4

[upper karma]

Oh oops im so dumb

is the answer B

[abstract narwhal]

👍

[upper karma]

yay thx guys

Yep

i love this discord server already 😄 the helpers dont give you the answers, but help you find the answer on your own which is awesomeee

[abstract narwhal]

yeah thats the best way to learn

:)

[upper karma]

So true :)))) do people here actually like teach classes?

they seem so professional

[abstract narwhal]

probably some do

i'm only 18

[upper karma]

I believe some do, but not much of the ones you'll see active helping

But u guys r sooo helpfull u shud become part-time educators

lol

welp here i am as a 13 yr old and im learning from some of the best teachers ever 😄

[abstract narwhal]

lol

alright am i good to ask now?

[upper karma]

Yeah

Depends on the time needed to help you, i will help or not bc i gtg sleep

If not anyone else will

[abstract narwhal]

alright np. thanks

[rotund hull]

I can help

[upper karma]

I'm gonna see if i can

is this a bad time for me to ask another question

[rotund hull]

I got you kewl

[upper karma]

I got you kewl
@rotund hull thx bro

Yes, but on another channel you are free to do it @upper karma

[rotund hull]

@upper karma lets go see if a #questions is open

[upper karma]

Oh kk

[abstract narwhal]

let say you have an irregular tetrahedron with faces a, b, c (these are 3 triangles meeting at one vertex). i have found out that you can caluculate the dihedral angle between each triangle using the formula below, where A is the angle of face a at the vertex, etc. θbc means the dihedral angle between faces b and c

is there such a formula to calculate dihedral angles between the triangles when there are more than 3 meeting at a vertex (e.g. irregular octahedron)?

i cant find anything similar online so i'm beginning to think it's not possible

[abstract narwhal]

<@&286206848099549185>

[onyx cloud]

@abstract narwhal i couldn't find a formula for an irregular octahedron or other polyhedrons but how did you go about solving it for the tetrahedron. maybe you can apply a similar method to other polyhedrons

[abstract narwhal]

@onyx cloud well i didn't come up with the formula myself, but it's derived using the normal vectors to each plane.

and it doesnt specifically have to be octahedron; just any 4+ angles meeting at a point

[onyx cloud]

it might be a good idea to try the same approach (using the normal vectors to the planes) to find the dihedral angle for an octahedron, or any other situation you were mentioning

[abstract narwhal]

yeah okay

i will try it

[glad veldt]

I was really surprised when I could not figure out this question that my younger sister asked me to solve for her because she didn't know how. I thought that their was not enough information given but i would like to see if any of you can solve it; If an isosceles triangle has an area of 4 units squared, find its exact perimeter.

I don't know if i am somehow missing something

[surreal bolt]

um, pretty sure that depends.

Sorry. As in you are right. Are there restrictions on the sides otherwise?

[glad veldt]

nope thats all it gives

[wary wind]

Hmm....I don't recall any relation w.r.t area and and perimeter of isosceles triangle.

I think there's some algebra involved here to figure this one out

[paper vale]

umm maybe it assumes integer side length for no reason

[surreal bolt]

Draw a 4 x 4 square. connect the bottom two corners to the midpoint of the top segment.

oops 🙂

nvm it's late and my brain is mushy

[paper vale]

bruh

[wary wind]

Like, let's say -

Area = 1/2 bh = 4

So, b = 8/h

Then you would have to apply Pythagorean theorem to find out what's the hypotenuse side (let's say a).

After which you have both a and b - so you can find perimeter as 2a + b

[paper vale]

it is pretty clear that there are infinite solutions

[wary wind]

I think that's what the problem is going for though?

I might have to solve this get a better idea of what's going on

[glad veldt]

yeah i got to 2a + b before and i didnt know where to go from their

it asks for exact perimiter so i was expecting some surds or somthing like that

[surreal bolt]

Draw a 4 x 2 rectangle and connect the two bottom corners to the midpoint of the top side.

Ditto for an 8 x 1 rectangle. Sorry for the wrong answer earlier.

[paper vale]

it doesnt have to be integer sides

[glad veldt]

ive tried drawing boxes around the triangle but i still get stuck

[surreal bolt]

I know I didn't make it as such. hmm one sec.

[wary wind]

There's one slight issue with that approach, @surreal bolt. You assume the triangle to be right isosceles in nature - Which may not be true

[glad veldt]

yeah well, thanks for the help, i guess there are many solutions then

[surreal bolt]

Think this is right.

[paper vale]

there are infinte bruh

[surreal bolt]

I know. But you only have to show two to poke holes in the problem's logic 🙂

[paper vale]

an example, just make the base bigger, and the height smaller

this is clearly a monotonic increase in the permiter

but constant area

[surreal bolt]

The original question was ... the poster thought there was more than one solution ...

[paper vale]

[surreal bolt]

ok ok

[lost granite]

what is the number of faces of an n-cube?

i meant like sides

so for the square it's 4

for the cube it's 6

is it just 2n?

[wary wind]

It doesn't work that way.

For square, you are asking for many edges are there.

While for a cube you are asking for number of faces.

And Faces != Edges.

[lost granite]

yeah i don't know how to express that

is there a term

[sour jacinth]

is there a term
@lost granite What do you mean?

[lost granite]

the sides or faces idk of a hypercube

[sour jacinth]

the sides or faces idk of a hypercube
@lost granite Depends on the dimension of the hypercube

[upper karma]

Can someone tell me what Trig identities and the Unit Circle are applicable to in the outside world? Thanks.

I wouldn't say identities are directly useful in real life but trigonometry in general is very used

As in, will I be using it in engineering? Will it be useful to know if I am building a drone?

Very much

If angles are involved, then trig comes in

Much appreciated

Trig is also very useful for calculus, analysis and many more subjects

Thanks very much!

Np

[sour jacinth]

As in, will I be using it in engineering? Will it be useful to know if I am building a drone?
@upper karma Of course. For example, if your drone is travelling in a certain direction, you can know the components of that force in the x, y, z axes by using trigonometry and algebra

[split escarp]

Help me

I got ED

It's 4

[sour jacinth]

How did you get it?

[split escarp]

Isosceles triangle

BDC IS ISOCSCELES

9=5+4

[sour jacinth]

5 is lenth BE?

Or ED

[split escarp]

Yh

BE

Y u delete?

[sour jacinth]

I realized I don't have enough data to state what I wrote

It's easy, it's just similitary and congruence

[split escarp]

Ik

That's the test topic

I didn't know how to reason and f one out ans

[sour jacinth]

I didn't know how to reason and f one out ans
@split escarp Don't worry, we'll do it

Final question: Is angle BCD equal to angle CED, right?

[split escarp]

Yh

[sour jacinth]

OK, give some mins

Thinking where we can start from

[split escarp]

k

[sour jacinth]

@split escarp Is A collinear to E?

[split escarp]

yes

[sour jacinth]

OK... That will help a lot

[split escarp]

on the question it didnt directly state that they're collinear but they can be

you can construct a line

passsing throuhg a and e

[sour jacinth]

Found EC

Trying to see where to go next

[finite sky]

bruh

How

what is ec

[split escarp]

wait

til he finishes

[sour jacinth]

Found DC

[split escarp]

Kool

[sour jacinth]

Trying to see where to go next

Still working on it

Found ED

Only AB is missing

Found AB

@split escarp Done!

Do you have the answers? Just to confirm...

I have AB= 11.25, DC= 5, ED= 4

[split escarp]

I don't haveanswers

Thx

Can u show me if working out

I will try and make sense of it

[sour jacinth]

@split escarp Do you want to do it via voice or written?

[pallid edge]

ok so ik that like

cos sin tan are used in right triangles

(i learned them this year along with solving for side and angle and stuff liek that law of sines cosines)

[vale nimbus]

any triangle works

[pallid edge]

BUT WHY DOES IT APPEAR OUT OF RIGHT TRIANGLES AND IN RANDOM EQUATIONS

[umbral snow]

Have an example?

[pallid edge]

ummm ig?

lemme find one

well like

arent the ratios only for right triangles??

like howcome if u have a triangle

area = .5 a * b * sin(c)

if the tri is not right

[silent plank]

be careful with capitalisation

[pallid edge]

huhh

why does that matter

ohh

right

wait i think i just realized

sin is hte ratio

not the measurement

Ohhh

i get it now

[umbral snow]

These objects are defined originally on right triangles, but if you're clever they can be used on more than that

Worth checking is the cosine law

[vale nimbus]

https://mathbitsnotebook.com/Geometry/TrigApps/TAarea.html#:~:text=AreaΔ %3D ½ ab sin,the area of the triangle.

[pallid edge]

oh ik cosines

c^2 = a^2 + b^2 -2absinC

another question

why is the surface area of a sphere

four times its area

of the middle section

this is a geometry question right

@umbral snow

@vale nimbus

i hope the proof is simple

[umbral snow]

Oh I have no idea why rofl

[vale nimbus]

https://youtu.be/GNcFjFmqEc8

[umbral snow]

I'm sure there's some 3000 year old proof of the surface area of a sphere

[pallid edge]

WAIT IS THAT CALC I SEE ON A GEO PROBLM

[vale nimbus]

idk

[pallid edge]

bruh

[umbral snow]

A lot of geometry ends up being very calculus

You can prove this result with calc, but you don't have to

[split escarp]

@sour jacinth

@sour jacinth I wanted written

soz for the long reply

[sour jacinth]

@split escarp It's a somewhat long answer

Ready?

[finite sky]

Yes

Im his friend sure

[split escarp]

yh

im ready

[sour jacinth]

@finite sky OK

[split escarp]

@sour jacinth

kool

[sour jacinth]

But you'll have to answer my questions, because I'm not gonna hand you the answers in a silver dish

[split escarp]

ok...

[sour jacinth]

Let's start

[finite sky]

Ok

Btw this was from our maths test yesterday

[split escarp]

i told him

[finite sky]

And its suppose to be an 8th grade test

So

[split escarp]

tru

[sour jacinth]

Let's say I have a triangle

[split escarp]

y

yh*

[sour jacinth]

And that triangle has 2 congruent angles inside

[split escarp]

ok

[sour jacinth]

Do you know what happens with the sides that are not common to those angles?

[split escarp]

no

[sour jacinth]

OK

When you have a triangle with 2 congruent angles, non-common sides are equal

[split escarp]

ok

[sour jacinth]

Remember that. It's pretty important

Do you guys understand that?

[finite sky]

Oh o

Yes

[split escarp]

yh

[sour jacinth]

Graphically stated:

Red angles are equal, so the sides with the double mark must be equal

[split escarp]

ok

[sour jacinth]

Why? Fold the triangle in half using its height as an axis and check why

[split escarp]

nice

[finite sky]

Yeah

[sour jacinth]

Any questions so far?

[split escarp]

no

[sour jacinth]

So I'm gonna focus on different parts of the triangle

Right now we have these

[split escarp]

yes

[sour jacinth]

Let me do some graphs in powerpoint

[split escarp]

also what year would you say this question is for

we encountered it in year 8

is it at our standards?

[sour jacinth]

Yeah, but you need to think carefully about it

Finishing the graph

[split escarp]

ok

thx a lot btw

[sour jacinth]

No problem, you're welcome

[split escarp]

also may I ask what level of maths are you studying
university or high school or like what?