geometry-and-trigonometry

[runic jackal]

== 8/cosd(28)

[charred spear]

9.06056041

[runic jackal]

there

[restive void]

oh calculator in wrong thing

@runic jackal do i have to add 3.6, subtract 2.5?

[runic jackal]

yeah

to get the hyp of second triangle

keep doing that

[restive void]

then cos(11) = x/10.2

[broken sierra]

Yo everyone I have a question. What's the difference between cosine sine and tagent

Tangent***

[waxen gorge]

The difference is

[pure agate]

SOH CAH TOA

[waxen gorge]

On the unit circle

Cos is x

Sin is y

Tan is y/x or slope

soh cah toa is for plebs

[broken sierra]

Thx

[waxen gorge]

It's also racist

[broken sierra]

Agreed

Can u define unit circle

[waxen gorge]

It is

[pure agate]

a circle radius 1

centered about 0,0

[waxen gorge]

A circle with radius 1 :p ^

[thorn talon]

Yeah

[broken sierra]

Thx

[waxen gorge]

and all points on it's circumference r the right triangle with angles of some value

[pure agate]

also how is SOH CAH TOA racist 😂

[waxen gorge]

Search up "why is trig racist"

You will see

[thorn talon]

Its because of the way it is

[waxen gorge]

... it's horrors

and true meanings

(plz don't hate me after seeing the truth)

[pure agate]

makes no sense

[waxen gorge]

It's makes a lot of sense

[pure agate]

how can pure logic be racist

it's just a good way of remembering the ratios

[waxen gorge]

No.

I'd like you to think about it, please.

[pure agate]

yeah i have, just explain if it's so obvious

i can't find any articles explaining how it's racist

[waxen gorge]

http://www.washingtonexaminer.com/its-not-just-a-joke-anymore-theyre-actually-claiming-math-is-racist/article/2638468

That's the more broad one

There are some on trig

[pure agate]

that's the worst article i've read in some time

i bet it's from the US

absolute garbage

[waxen gorge]

I bet your absolute garbage

Btw

Terrence tao is in the us

[pure agate]

he's australian

[waxen gorge]

But he's in the us

[pure agate]

that makes no sense

he could move anywhere he wanted

who cares

[waxen gorge]

Lies

Trig is plain racist

Absolutely racist

and everything on that website is tru

[pure agate]

are you trolling?

also as someone from a mixed racial background coming from Brazil to the UK and spending a large amount of my time in mathematics education I can safely say mathematics is possibly the least biased subject

[waxen gorge]

I'm not trolling

[pure agate]

i feel sorry for you

do you believe in the illuminati too

[waxen gorge]

No.

[pure agate]

what about lizard people

[waxen gorge]

No

[pure agate]

you realise the article is tongue in cheek

[waxen gorge]

Nope

[pure agate]

well i guess you wouldn't be able to tell, you're from the US

feelsbadman

when your comedy requires laugh tracks to indicate when something is funny 😂

[waxen gorge]

...

Your ignorance is quite bothersome

[pure agate]

LOL coming from you

you actually believe maths is racist?

boggles my mind

[waxen gorge]

It is

You cannot deny it

[pure agate]

well i can, and i do

absolutely

give me one point you think can convince me maths is racist

some evidence

[copper valve]

i am math.
i am not racist.
therefore math is not racist.

[waxen gorge]

Jkjkjkjkjk

U fell for it xd

[pure agate]

trolled again

feelsbadman

[waxen gorge]

Lol

[upper karma]

Let RS denote the line segment that is an external tangent to both circle M and circle N which are tangent at only one point. Given that radius of circle M is 5 and N is 3, what is the length of segment RS?

I think the answer is 8, but how do I prove it

try doing this on a coordinate plane then just use the distance formula

that's quite time consuming but I am not sure how otherwise

i am pretty sure there is a simpler way to prove it without using coordinate grid

and i am not even sure how to do this in a coordinate grid

nvm i probly do know

yeah it's not hard

[runic jackal]

is not 8

radius of N is 3 right? in the question u said its 8 lol

[upper karma]

oh yeah i messed up wording

you are right

[upper karma]

it was actually a multiple choice question

sqrt 68 is not in the options

[runic jackal]

heh

[upper karma]

the options were

A) -2sqrt(15) B) 4 C) 2sqrt(15) D) 8

i wonder if anyone picked A

[runic jackal]

ohhhh

im sorry

its C

[upper karma]

but how

8^2 + 2 ^2 = 70

[runic jackal]

drawing

[upper karma]

sqrt 70 = sqrt 2 * 5 * 7

[copper valve]

nice drawings

[upper karma]

i agree

[chrome fiber]

that's what i was wondering lmao.

[upper karma]

how do you draw your circles

[runic jackal]

idk

[chrome fiber]

how could there be two right angles in a triangle lol.

[upper karma]

where do you see two right angles in one triangle 🙄

[copper valve]

u can have 3 if it's spherical 🤔

[runic jackal]

first drawing lol

[chrome fiber]

i'm referring to @runic jackal's first image.

yesyes.

[copper valve]

oh

[runic jackal]

sryyy

[upper karma]

oh yeah he oriented it the unusual way but its allright

thanks thinking caps

btw its 12:30 here

[tropic stirrup]

Lmfao

[upper karma]

what time is it in your place

[runic jackal]

oh yeah gurl

u rember the 7, 14, 21 geometry question

@upper karma no prob lol

[tropic stirrup]

=tex t_e=\sqrt{d^2-(R-r)^2} \
t_i = \sqrt{d^2-(R+r)^2}

[tropic stirrup]

Yes

[runic jackal]

is the ans 186

[tropic stirrup]

Yup

=theme dark

[charred spear]

Your theme has been set to dark.

[runic jackal]

ok

solved it yay

[tropic stirrup]

😒

[runic jackal]

@tropic stirrup more

[tropic stirrup]

ABC, ADE are equilateral triangles with side lengths 6 and 8 respectively, where E is on line segment AB, and D is not on line segment AC. Let the intersection of line CE and BD be F. Find the value of FE * CE, with explanations on the steps to achieve such value.

[runic jackal]

hmm wait

BD and CE extended right

just making sure lol

@tropic stirrup

[tropic stirrup]

Good job!

[runic jackal]

@tropic stirrup more

im bored

[tropic stirrup]

Lol ok

[runic jackal]

ugh im on the bus ill try solving more

without paper lol

[dark sparrow]

multiplication cross

[runic jackal]

yesh

[dark sparrow]

\cdot or bust

[tropic stirrup]

(diagram not drawn to scale)
AD is the bisector of angle BAC, where BDA = 90°.
E is a point on BC such that DE // AC.
Given AB = 10, DE = 3, find the value of BF / CF.

[dark sparrow]

\times only for cross product

[runic jackal]

what is cdot or bust

[dark sparrow]

$$\cdot$$ or bust

[runic jackal]

3 bust 5 is 15

[dark sparrow]

no that's not what i meant

[runic jackal]

2 bust 2 is 4

so what do u mean

[dark sparrow]

i meant like

use \cdot or get out

lol

[runic jackal]

@tropic stirrup 😄

[tropic stirrup]

How'd ya know BC = 2x?

Oh nvm I see what you did

[runic jackal]

let DE extended cuts AB at G, i know that GBD = GDB

so they are isosceles, but also since GDA = GAD, also isosceles lol so BG = GD = GA

[tropic stirrup]

Mhm

[runic jackal]

and each of them is 5

and since DE is 3, GE is 2 and yeah

BE : EC = 1 : 1

[tropic stirrup]

Ye ye I see that

[runic jackal]

yeah i got 5 : 2

is it correcc

[tropic stirrup]

it is

There are two ways of solving this

[runic jackal]

ooohh

[tropic stirrup]

one is extending DE, one is extending BD and AC

[runic jackal]

is there a simpler method

ohh extend them until they cut?

moreee

[tropic stirrup]

Wait, 5:2?

[runic jackal]

yes

[tropic stirrup]

BF/CF

[runic jackal]

what did u get owo

[tropic stirrup]

Oh yes nvm x'D

[runic jackal]

more geometryy

omg i love geometry

[keen aspen]

Until you have to do two column proofs

[runic jackal]

noh

[eager pendant]

nice

[tropic stirrup]

I have more challenging problems lel

[runic jackal]

giveee

[tropic stirrup]

There's a triangle with A = 70°. Pick a point P in the interior and Q in the exterior such that ABQ and ACP are similar, and line segments PQ and AB intersect. Given that AP * BC = BP * CA = CP * AB, find the size of angle BPC.

[runic jackal]

woaaa i might need paper

if i draw on phone i cant get it right

[tropic stirrup]

Hav fun~

[runic jackal]

yayy

is it more difficult than the other two

[keen aspen]

Very

I'm trying to do it myself xD

[tropic stirrup]

It is lol

[keen aspen]

How did you come up with such an elaborate question?

[tropic stirrup]

Mmm.... thinking? x'D

[runic jackal]

i cannot

easier pls

or give hints

[runic jackal]

@tropic stirrup

[tropic stirrup]

Ok I'll give you easier one

For an equilateral triangle ABC, there are points D, E and F on AB, BC and CA respectively such that FC = 5, CE = 8, and DF perpendicularly bisects AE. Find the length of BD.

(Try your best not to use trigonometric ratio except the 30° and 60° ones)

[runic jackal]

ohh hai

later

[runic jackal]

@tropic stirrup

[tropic stirrup]

Good job uwu b

[runic jackal]

moreee

@tropic stirruplllllllllll

[tropic stirrup]

Ok

[runic jackal]

that level pls

dont give me questions like the crazy one

[tropic stirrup]

In a triangle ABC, there exists a point D such that AD is a bisector of the exterior angle of A. It is true that AD // EB, AB // EF, AD // FG, where E and G are on AC and F is on CD. Also, BD = 11, CB = 7, AB = 9, CD = 18. Find the length of CG.

Whups

That may be a... bit harder if you don't get the point

[runic jackal]

is this difficult

okay

[tropic stirrup]

If you get the point, super easy
else, super confusing

[runic jackal]

im looking forward to a super confusion!!!

[copper valve]

yay points

[runic jackal]

omg

its AC extended isnt it

i was like how can E be on AC if AD // EB and AD is a bisector of angle BAC

[tropic stirrup]

WAIT

SHIT

💩

[runic jackal]

can u like fix lol

[tropic stirrup]

IT'S THE BISECTOR OF THE EXTERIOR ANGLE OF BAC

[copper valve]

@tropic stirrup when u do geometry problems, do u always draw a picture

[tropic stirrup]

Of c o u r s e
but for this problem drawing first without reading through everything is a dumb thing to do >w>

[copper valve]

ya im just wondering like

if some people don't

do they imagine it

or just work from the axioms

like, the info ur given but not visualizing it

thinking about it more abstractly i guess. not really ​sure

cause normally when im doing a proof (I don't really do euclidean geometry) I don't need to imagine or visualize things

sometimes i do but it's not like, necessary

[runic jackal]

i dont even know how the diagram looks like

[copper valve]

i was wondering if that kinda thing happens in geometry too

[runic jackal]

do u need to extend AC so u can place the E

[copper valve]

or if it's like, super duper visual

[tender geyser]

If you want a little easy question, i think you can try this one.
P is the circumcentre of acute angled ∆ABC with circumradius R. D is the midpoint of BC. Show that perimeter of ∆ABC is 2R(Sin A + Sin B + Sin C)

[tropic stirrup]

sin law

[celest swan]

I need to build a rubik’s cube solving arduino robot

For that I’ll need what requirements?
Matrices? Symmetry in molecules? Arithmetic?

[copper valve]

google

:P

im not sure ;o

[celest swan]

Ok

[neon fossil]

you need to know how to implement and manipulate graphs

and the fact that you're always a few positions away from a solution basically

or, learn how to solve the cube, and implement the actual procedure

[copper valve]

that works too ;0

[celest swan]

I need to teach a program how to do it

Not sure if it’ll understand graph theories

Plus, it must adapt to different positions.

[outer grove]

I'm a cuber

im here

no matrices

no arithmetic really

just make some kind of cube class

that you can do 6 turns on

that changes the internal state

then there's a thing called Kociemba's algorithm, which is what computers/robots use for solving

[celest swan]

Yes symmetry

[outer grove]

it's two steps

no no symmetry stuff needed

I don't know the slightest about group theory and I can code kociemba's

[celest swan]

Kociemba’s algorithms are based on symmetries

[outer grove]

no

no they're not

[celest swan]

...

[outer grove]

also it's not plural

it's singular

Kociemba's algorithm for solving the cube

steps:

[celest swan]

I’ll do more research

[outer grove]

1. brute force a way to reduce to the <U,D,R2,F2,L2,B2> state

[celest swan]

I know the algorithm

[outer grove]

2. brute force a way to solve the cube from there

[celest swan]

G1

[outer grove]

yes

ok look

let me explain a bit more

here's the only notation you'll need:

[celest swan]

I’m pretty sure it’s based on the M_48 symmetry group of cubes thought

[outer grove]

nope

I don't even know what that means

[celest swan]

...

[outer grove]

yet I can code kociemba's 100%

easily

ok here

[celest swan]

It’s the 48 symmetries possible for a 9x9x9 prism

[outer grove]

what

that has nothing to do with solving the cube lmao

[celest swan]

Yes

[outer grove]

dude

I'm the one

who can solve the cube

with kociembas

you're the one who can't

you don't know what you're talking about

[celest swan]

I’m the second. And i know how too ^^

[outer grove]

well clearly not

anyway here

you know U D R F L B right

[celest swan]

Yes

http://kociemba.org/symmetric2.htm

[outer grove]

ok

no that's herb kociemba's analysis page

[outer grove]

not his page on solving the cube

[celest swan]

I know

[outer grove]

ok

then listen

do you know <*,*,*,*> notation?

[celest swan]

Yes

[outer grove]

ok

what does it mean

[celest swan]

G1 = <...>

[outer grove]

no

tell me what <U,D,L,R> means

[celest swan]

It’s a directive

[outer grove]

no

<U,D,L,R> is the set of all possible sequences of moves that only contain the moves U, D, L, R

[celest swan]

OHHHH

[outer grove]

the scrambled cube is in a <U,D,F,B,L,R> state

[celest swan]

AND G1 IS ALL MOVES

BECAUSE ITS THE SET CONTAINING ALL MOVES

[outer grove]

the cube after the first step of kociemba is in a <U2,D2,F,B,L,R> state OR a <U,D,F2,B2,L,R> state OR a <U,D,F,B,L2,R2> state

the cube after the second step of kociemba is in a <> state

aka solved

[celest swan]

So <...> are the possible steps to do before solving

[outer grove]

the first step of kociemba means getting from a <U,D,F,B,L,R>-solveable state to a state solveable by one of the 3 sets above

yes

[celest swan]

So my bot must reach Kociemba

Then analyse the cube

[outer grove]

no

[celest swan]

So my bot must reach Kociemba

Then do the right one of the three

[outer grove]

kociemba isn't a group

kociemba is a person

[celest swan]

I know dude

[outer grove]

how can your robot get all the way to kociemba

[celest swan]

I abbreviated

[outer grove]

well what do you mean by kociemba

[celest swan]

I meant kociemba’s first step

[outer grove]

this guy trynna make a solver bot

@ebon vapor

@celest swan ok yes

that's it

robot

[celest swan]

Yes

[vast pasture]

coplanar?

[outer grove]

rubik's cube solver robot

btw jsyk sotto here is like the master at cube theory compared to me kek

[celest swan]

I needed help with logic behind the algorithms

...

[outer grove]

he thought you needed something to do with M49 9x9 prism symmetry or something

just read the backchat

[celest swan]

*m48

[outer grove]

it's not too much

@ebon vapor

yeahthat

[celest swan]

Mmh

But symmetry would have worked no?

Yes

Like with a molecule

H_2O

[outer grove]

^ what I've been dealing with for the past 10 mins

please have sympathy

[celest swan]

Is symmetrical on an axis cutting O

Well in certain (every) situation the cube is symmetrical on at least two axised

*axises

And when solved, it’s symmetrical on every axises

(Its pattern)

Well

[copper valve]

[celest swan]

Say you get a symmetry on two faces — say they are similar in pattern but not in color

From there, you can solve the whole cube

If all of your movements are symmetrical too

You build the right pattern then you apply that

I gtg

[main sluice]

prove the minkowski sum of ->the boundary of a convex set<- with itself is a convex set

[main sluice]

idk what to do

[slender gyro]

I know nothing about this but here's what I'd do.
Step1: The sum of two convex sets is convex, therefore: A convex => 2A is convex
Step2: If D =bound of convex set A. Then 2A=A+D

Step1 I assume maybe you proved it already, have no clue.
Step 2: Obviously A+D is in 2A
Step2,2. But if b is in 2A then it's also in A+D. This is because if b=a1+a2, you can move begin moving a1,a2 further apart until either one of them is on the boundary. And now by definition b is also in A+D.
QED

My proof of step2,2, has an issue. What if they both hit the boundary at the same time? Then you get a1,a2 in the boundary, but if the set is open that means neither a1 nor a2 are in A, which means they're not in A+D. Not sure what to do about that but yeah

What class is this for btw?

[agile blaze]

I don't understand!!!! 😭

[gritty flare]

oh

dont worry

😦

they triangles look the same

do you see that

[agile blaze]

i will never get anywhere in life 😭

yes they are similiar angles

but not congruent!!!!!!!

[gritty flare]

it scales

what do you mean

[agile blaze]

the corse is simialraties

similiar angles in that task

[gritty flare]

ok

if there is a case that:

[gritty flare]

then we can say that the triangle scales

by SSS

if they have A scale factor

they are similar triangle

so if they have 3 x's that have same number

[coarse hazel]

Sarah, if make a triangle 5 times bigger, each of the 3 sides becomes 5 times bigger. And (this part is important) the shape of the triangle does not change. So now you have 2 triangles and you're asked if they are similar triangles. By looking at the ratio between each of the sides, so if you look at the ratios of the side in the small triangle and the side of the big triangle PQ/LM (right side), PR/LN (left side) QR/MN (bottom side), and if you find out that each side is "made bigger" (scaled) by the same amount, you will have proven that the big triangle is a big version of the small triangle. AKA they are similar. And the amount by which the sides are multiplied is the scale factor.

[gritty flare]

ye better explanation

[coarse hazel]

Oh and if every side is scaled by the same factor, that automically means the shape of the triangle is not changed, in other words, they are similar.

[gritty flare]

ye but she lost hope i think

o.o

@agile blaze hello

[coarse hazel]

no way, my explanation was masterful 😎 reading it is the logical equivalent of grasping the concept perfectly

actually that makes no sense even if it really was perfect

[gritty flare]

eh i just cant explain

p.p

[barren moat]

help me with this

Three problems A, B, and C were given on a math competition. All 25 students solved at least one of these problems.
The number of students who solved B and not A is twice the number of students who solved C and not A.
The number of students who solved only A is greater by 1 than the number of students who along with A solved at least one other problem.
Among the students who solved only one problem, half solved A.
How many students solved only B?

[celest swan]

Yes?

[coarse hazel]

you can probably use symbols from set theory and propositional logic to clarify this a lot

[celest swan]

Something around 2.25 or 6% of the students? Im prolly wrong

[coarse hazel]

but im currently eating chocolate and have no paper with me ...

[barren moat]

write with chocolate on the table

[coarse hazel]

but its such a waste 😱

[barren moat]

lol

jk

it can't be 2.25 since we're talking about people

assuming the problem is not racist or sexist ofc

[coarse hazel]

lol

[celest swan]

2.25*25/100

[coarse hazel]

uhm.. do you at least know how to solve it?

[barren moat]

not really...

[celest swan]

I used logic. But i don’t have a sheet of paper near me so im prolly wrong

[coarse hazel]

first you put everything into formulas and then idk (maybe i should understand problems better before deciding to pitch in)

[barren moat]

I think it should be like

N_a and stuff like that

[coarse hazel]

so if a=the number of students who solved A, etc.. you can translate the story to formulas, like (b and not a)=2x(c and not a)

okay screw chocolate

wow this is hard

[barren moat]

ikr

[coarse hazel]

i hope nobody finds it before me, because this one is fun :D

[barren moat]

just don't look in discord until you give up or solve it. So you don't spoil the answer to yourself

[coarse hazel]

true. okay from now on i ignore everybody until i find the answer

[barren moat]

or give up

[copper valve]

someone already asked this question before 🤔

[barren moat]

who?

[copper valve]

i didn't look at their name. it was like a month ago

[coarse hazel]

im still solving it, and this time it doesnt feel like a dead end

as opposed to the other roads xD

[barren moat]

[coarse hazel]

omg this is going to be beautiful

[barren moat]

you're still doing it?

[coarse hazel]

yeah :D

i dont know if ive ever had this much fun doing math :P

[barren moat]

in b4 i spoil the solution

[coarse hazel]

wow you found it already?

and please dont

[barren moat]

ok

[coarse hazel]

well congratz if you figured it out

[barren moat]

🇹 🇾

[coarse hazel]

group theory?

[barren moat]

no it's like I solved a problem

:/

[coarse hazel]

i dont get it

[barren moat]

it's like solving a task

solving problem or a Rubik's cube

[coarse hazel]

og

i see

time for a break.. im tired :(

[barren moat]

lol and you should be

[runic jackal]

@tropic stirrup

[tropic stirrup]

o_o

[upper karma]

gurl give me a geometry problem

without proofs

[tropic stirrup]

Jesus that's a long way around

[upper karma]

please?

[tropic stirrup]

What do you mean by "geometry problem without proofs"

[upper karma]

geometry problem that doesn't require you to prove something

[tropic stirrup]

Then that's just computation, not geometry

[upper karma]

just give me a geo problem

[tropic stirrup]

Find the area of a cyclic quadrilateral with side lengths 4, 5, 6, 7

[upper karma]

nvm

how do u solve the triangle one

[tropic stirrup]

Heron

:D

Second one is brahmagupta? can't remember

[upper karma]

nvm

[tropic stirrup]

Heron's formula

[upper karma]

yeah I see

now

9.92157

is the area

[charred spear]

7.5

9.92156742

[upper karma]

wtf

I rounded

isn't that obvious

no buli

you mean engineering discord

baddum tsssssssss

wait when u say area of a cyclic quadrilateral is it the area of the circle it is inscribed in?

[waxen gorge]

What

No

Just it's area @upper karma

[upper karma]

ok

what's the point of it being cyclic then

[waxen gorge]

It might help with the problem

Yep

[upper karma]

all I found was that opposite angles add up to 180

how do I use that

tbh I could maybe draw it

but my notebook is far away

[runic jackal]

nice propertiesss

@tropic stirrup any hints on how to do the cyclic quad problem

[tropic stirrup]

hmm

cosine rule

😛

[runic jackal]

ok

tf didnt expect that

[tropic stirrup]

and what ali said

;p

[runic jackal]

done lolol

is it not an integer

its sqrt(840)

[tropic stirrup]

Yep

[runic jackal]

hmm

yay

give more

[tropic stirrup]

Brahmagupta's theorem: if a cyclic quadrilateral has side lengths a, b, c, d, where s = (a+b+c+d)/2, then the area of the quadrilateral is sqrt((s-a)(s-b)(s-c)(s-d)).

[runic jackal]

aw

[tropic stirrup]

Setting d = 0 lets you know that it's a generalization of Heron's formula

;p

[runic jackal]

omg

mind blown

more problemsss

[tropic stirrup]

You

[runic jackal]

😒

more geom problems pls

im bored

af

[tropic stirrup]

😛

[coarse hazel]

@runic jackal for which n does the following equation have integer solutions? a^n + b^n = c^n

im sorry, thats only funny if youve never heard of it >.>

[runic jackal]

need hints lolol

@tropic stirrup HINTS

I CANNOT

[coarse hazel]

do the numbers mean that the wave at 8 is twice as long as the wave at 4?

[tropic stirrup]

Power theorem

:p

[keen aspen]

The round the three decimal places throws me off

I was thinking 14 but nvm

[runic jackal]

ugh

[keen aspen]

Ok I give up for today

Ill try later its 5am here

my brain needs sleep

[upper karma]

Did you know that a^2+b^2=c^2?

[keen aspen]

@tropic stirrup You should send me the answer, i'd love to show that to some classmates

It was a very interesting problem

[runic jackal]

@keen aspen what problem

[keen aspen]

His geometry problem

with the power theorems

[runic jackal]

nah dude thats not fun

[keen aspen]

I was working with it for about 30 mins

[runic jackal]

which one

[tropic stirrup]

Lel

It's, well...

Have fun, poppy x'D

[keen aspen]

That's all I can do this late

[tropic stirrup]

Lol x'D

[keen aspen]

Give me answer

I want to learn off of it

[tropic stirrup]

hmmmmmmmmmm

Ok I'll send you by pm

[keen aspen]

Ok

[runic jackal]

xd

[keen aspen]

Hehe good luck

[celest swan]

How can i find the area of the darkened section?

<@&286206848099549185>

[neon fossil]

it's a well known problem

[hazy field]

area of the circular section minus area of the triangle

[neon fossil]

^

[celest swan]

Could you show me the path to follow? Say r=3, n= 30 degrees

[chrome fiber]

do you know how to find the area of a sector?

[celest swan]

Yes

n/360 =As/At

[chrome fiber]

and the area of a triangle, when you know a side and the angle it subtends?

[celest swan]

No, that’s where i fail

I mean wait

I can only find the angles

[chrome fiber]

for a triangle having two sides, a and b, with the angle between between a and b being x, the area will be:

$$\frac{1}{2}ab\sin(x)$$

[celest swan]

Ok

[chrome fiber]

use that.

[celest swan]

Yeah i see now

[chrome fiber]

you know the area of a sector, and the trianlge having angle n and the two sides being r.

[celest swan]

So the difference is what im searching

[chrome fiber]

yup!

remember, since the segment (that's what the shaded region is called btw; minor segment to be exact) is a part of your sector, you'd have to subtract the area of the triangle from the area of the sector.

[celest swan]

What about the same situation but in 3d and with a sphere and the section being a differenciable function of x and y?

To find the volume, can i simply apply disk method on the area obtained earlier?

??

@chrome fiber

Yes?

[chrome fiber]

i don't know a lot about 3d shapes, i'm sorry.

<@&286206848099549185> maybe?

[violet nest]

wheres the problem

[celest swan]

It should prolly go in calculus anyways

[violet nest]

can you post the problem again

[celest swan]

Ok im gonna post the whole shit. That was just the first step.

[neon fossil]

it should work

[celest swan]

But it’s long, i tell you guys

[neon fossil]

if you set it up properly, anyway

[upper sedge]

big volume - small volume.

[waxen gorge]

Hai

[celest swan]

Wait

Is there somewhere where it wont get flooded?

Like

Where the channl wont get other messages

[charred spear]

Query made by @ebon vapor
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=steradian
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[celest swan]

Ok look

Im gonna create a discord

[violet nest]

._.

[upper sedge]

o.o...

[celest swan]

It’s a hella long problem

[upper sedge]

I want to be involved, but I need to step away.

[celest swan]

Sure

[upper sedge]

Before I go, tripple integral with spherical coordinates.

[celest swan]

?

[fallen ivy]

@agile blaze electrical engineering? No

also the law of cosines is this

=tex c^2 = a^2 + b^2 - 2ab\cos(C)

[fallen ivy]

where a, b, and c are the lengths in a triangle and C is the angle across from c

[final prairie]

(i.e., between the a and the b)

[agile blaze]

oh

[restive void]

@spice pewter

wait

wrong tag

sorry

lol

@unreal needle

yes

you

[unreal needle]

look for me?

[restive void]

ill brb, give me a min

[unreal needle]

ok! ill grab my laptop

[restive void]

so

these are the first three

i got mn to be 16.6

when rounded anyway

[unreal needle]

Yeo!

Yep!

thats what i got

[restive void]

i got m<M = 65.0

i think

[thorn talon]

Yeah

[unreal needle]

Law of Cosines right?

[thorn talon]

Just inverse tan

[unreal needle]

oh yeah

[thorn talon]

Its a right triangle

[unreal needle]

But yeah

its 65.0

[restive void]

sorry back hi

n = 25 lol

so when u solve for right triangle, you have to find every angle and every side

[unreal needle]

yeo

[restive void]

so

so inverse of one of the functions is for angles right

[unreal needle]

Yeah

[restive void]

and then tan, sin, cos are sides

is PO sin(37) = x/22?

[unreal needle]

Yeah...

[restive void]

i got for the missing legs 13.2 and 25.66

or 25.7

is that what you got

[unreal needle]

yeah thats what i got

[restive void]

k good

so next triangle is this

np i got with pythag so 21.1

[unreal needle]

yup

[restive void]

is <p tan-1 18/11

[unreal needle]

Yea

[restive void]

okay so 58.6

[unreal needle]

it should come out to 58.6

[restive void]

then <p = 31.4

<n**

[unreal needle]

yep!

[restive void]

i think i can do these next few, ill tag you in the next section which looks really confusing

[unreal needle]

alright! it's really refreshing to do straight trig haha

[restive void]

@unreal needle not done with the sections, but how do i find the angles here

if im trying to find u

would i just do sin(u) = 19/28

i mean inverse sin

[unreal needle]

you could use pytha of the last side and just use tan-1

or that

[restive void]

yeah i got um

20.6

what would i do

[unreal needle]

tan-1 (19/20.6)

i got 42.7

[restive void]

same

is that <U

[unreal needle]

lemme check with the sine way real quick

yes

i got the same thing with sine, so yeah

thats <U

[restive void]

@unreal needle been awhile but how to do this

[unreal needle]

im not quite sure how to do this

[thorn talon]

I would just inverse everything tbh

[restive void]

so 13

sin-1(.36)

[thorn talon]

Yes

[unreal needle]

yah, inversing makes sense

[restive void]

i forgot terms

help

[thorn talon]

Ok

[restive void]

obv draw a picture

but idk where to start

[thorn talon]

What's the difference of height between the two buildings?

[restive void]

120

[thorn talon]

Yes

So you can form a triangle with height 120 right?

[restive void]

yeah, so the hypoteneuse is going from top left to bottom right cuz angle of depression right

[thorn talon]

Yes

Well

To the left tip of the building on the right

And then you can figure out the angles

One of them is complementary with 55

[restive void]

i dont know how to draw this..

[thorn talon]

From the top left corner of the right building

[restive void]

holy shit the next 7 are all questions like these

[thorn talon]

Just draw a straight horizontal line to the left building

You can see a triangle form right?

[restive void]

like this?

[thorn talon]

No

As in like

[restive void]

or this

or niether

[thorn talon]

That yes

So obviously we already have a 90 degree angle

Now what about the top left angle?

[restive void]

its 55 or 35

[thorn talon]

35

[restive void]

okay

[thorn talon]

And the length of the left side is?

[restive void]

120

[thorn talon]

Yep

[restive void]

and just solve from here?

[thorn talon]

You now have all the necessary information

Yep

Solve the bottom length

[restive void]

okay i can do

tan(35) = x/120

[thorn talon]

Yep

[restive void]

84

[thorn talon]

Yeah

[restive void]

hypot is 146

[thorn talon]

Yeah, though not necessary for the question

[restive void]

oh

[fast stratus]

hi

acq78ver h9w 9od are y9u

woops

[waxen gorge]

"kangaroux how old are you"

[thorn talon]

[past mantle]

I need help with something

Why $$\cos(60)=\frac{\sqrt{3}}{2}$$

[robust socket]

you should post this on #precalculus

[runic jackal]

@lethal silo

[lethal silo]

Hm

I don't understand that :|

[runic jackal]

hmm

is similarities

which part did u not understand

[tropic stirrup]

everything

[lethal silo]

yes

@tropic stirrup do you have a better solution?

[tropic stirrup]

That's literally how you solve it

[lethal silo]

there must be another way too.

[tropic stirrup]

Well there is

involving trigonometry shit

[slender gyro]

Can you categorize all the angles whose sines we humans can calculate through ordinary means? I'm being intentionally vague

[neon fossil]

you can probably calculate it for all elements of Q*pi

it looks trascendental in general

[slender gyro]

The first technique only gets you to sin(n/2^m) right? Which is sender but yeah

Oops i forgot a pi.

[neon fossil]

google sines and cosines of fractions of pi

then google sin(pi/7)

[slender gyro]

Nice, thanks

[upper karma]

how to demonstrate the relationship of chasles?

[bronze silo]

What’s a good way to learn Unit Circle?

[gritty flare]

oh

i just memorized 3 values

[gritty flare]

and I just altered it around

1/2 is smallest (30 deg)

sqrt(2)/2 is mid

(45 deg)

and sqrt(3)/2 is highest

60 deg

[upper karma]

Can someone please give me a challenging problem in which I prove proportionallity or similarity in a triangle? I'd like it to be more focused on the theorems, etc, though.

[gritty flare]

uh

Theorems?

[upper karma]

y'know

wait nvm

just anything where you prove similarity or proportionality in a triangle

@gritty flare

[gritty flare]

ye

what

[upper karma]

is that ok now?

[gritty flare]

im thinking

o.o

[upper karma]

ook

[mossy vine]

http://puu.sh/zd4Vw/62d4e165af.png

http://puu.sh/zd4YP/ca372dece2.png

[gritty flare]

the same problem?

srsly?

where do you find that

[upper karma]

o.o

ill try it soon

[mossy vine]

?

[gritty flare]

Im sure it was posted before and no one could solve it

[mossy vine]

idr posting them

they're not too bad though

[upper karma]

o.o

if nobody on here can solve it
then rip me

[mossy vine]

I can :^)

[upper karma]

but you are
legendary

[mossy vine]

I would expect gurl can too

[upper karma]

also
legendary

anything more two-columny? D:

[mossy vine]

:c

[upper karma]

ik ik

[mossy vine]

honestly they're not that bad

[upper karma]

b-b-ut solarkoid couldnt solve it
and soldarkoid is smart

o does this require knowledge of cyclic quadrilaterals or inscribed shapes?

[mossy vine]

yes

[upper karma]

meh not really used for my test

kek this class is so easy and boring that i never pay attention
but that also means i havent remembered the shit

[mossy vine]

ah rip

[upper karma]

meh
ill just read the list and find my own similar explanation lel

[rare talon]

@mossy vine thanks for the problem

I'll try it

[upper karma]

ill try it one day too
when im not studying for the test 😄

[rare talon]

@mossy vine the only solution I can find

Uses extended law of sines

[upper karma]

what's ext law of sines?

[rare talon]

I don't suggest you bother with the problem first I think

[upper karma]

Lmao

is it law of sines for cylic quadrilaterals?

[rare talon]

Since you're just learning simliarity

Hmmm

a / sin A = 2R

[upper karma]

a?

[rare talon]

Have you ever heard that?

a = length of BC

[upper karma]

I've used law of sines before, not extended tho

[rare talon]

In triangle ABC

Oh I see

Then you can try

😂

[upper karma]

kek

[rare talon]

2 column proof are shit tbh

[upper karma]

ive left this class off, @rare talon

im actually taking honours trig rn as well

[rare talon]

Hmmm I think

You can try that problem

It's fun

At least the first one

[upper karma]

when i have time i might

i need to remember tho

[rare talon]

It's not impossible as well

[upper karma]

lemme copy and paste

[rare talon]

Might be need a stretch of thinking, though for someone who is not familiar with that kind of problem

[upper karma]

ah ok

well looks cool anyways

[rare talon]

I think I can find you some similiarity problem

THe challenging one

Not really 2 column based, though

[upper karma]

Solarkoid suggested one, so I'll do that when I'm done eating

[restive void]

why is <C n either an angle of depression or an angle of elevation?

neither*

[waxen gorge]

Bc

Angle of elevation has a side flat on da ground

And angle of depression has a side on the top

[restive void]

oh

okay i need like help with a serious problem now

[waxen gorge]

Mk

[restive void]

word problem lemme type it out

[waxen gorge]

hopefully it's fun .-.

And make the notation readable k thx

[restive void]

A person starts 17 miles from the base of a tall mountain, and looks up at a 4 degree angle of elevation to the top of the mountain. When they move 12 miles closer to the base of the mountain what will be their angle of elevation when they look to the top? answer to the nearest degree

[waxen gorge]

Ok so

We make the triangles

[restive void]

plural?

[waxen gorge]

And solve

[restive void]

you dont need to use inverse tan?

[waxen gorge]

You fo

Do*

Lol I moved my camera when I took da pic so it's blurry

Too lazy to take another

[restive void]

why do you neeed the first triangle?

[waxen gorge]

One triangle is when u mvoe up

The other is the original

[restive void]

is the angle 13?

[waxen gorge]

Idk

Lol

[gritty apex]

If both diagonals of a quadrilateral are congruent, and so are two opposite sides of it, is it a rectangle?

[neon fossil]

no 😦

[gritty apex]

Can you prove it/tell me where I can find a proof?

[celest swan]

Let L# be a set of triangles, containing L, L_1, L_2,...,L_12. Let L be a triangle of side a, b and c, where a = 30cm, b = 40cm and c = 50cm. At the first turn, we place L on the table. Then, the next turn, we place one of the remaining triangles. The measure of the sides of a triangle are equal to half of the next bigger’s. How much turns are required at maximum before the figure reaches 150 cm3?

[neon fossil]

it can be any parallelogram

like a trapezoid

[rugged moat]

If tanA = a/b, What is the value of sinA/cos^8A +cosA/sin^8A

How would you go about it

[waxen gorge]

So

could you make it more clear

sinA/cos^8A?

Could u send a pic

[tropic stirrup]

FF' = 6, and the shown curve is an ellipse. Given that QF bisects angle PFF', find the length of the major axis of the ellipse.

An easy one tbh

[lethal silo]

😰

[tropic stirrup]

---

OA = 2, OB = 4, where vectors OP and OQ are denoted p and q.

Find the length of the locus of R satisfying this.

[lethal silo]

I can't solve any of those ):

[tropic stirrup]

---

AB = 6, and C, D trisects AB. For a point P on the circle, define theta = PCD. Also, E and F are on AP, BP such that CE and PC are perpendicular and DF and PD are perpendicular. C1 is an incircle of PEC, and C2 is an incircle of PFD. Denote S1(theta) and S2(theta) as the area of C1 and C2.

Find the value of $$\lim_{\theta \to 0^{+}} \frac{S_1(\theta)+S_2(\theta)}{\theta^2}.$$

[copper valve]

[tropic stirrup]

There ya go, all three types of geometry problems

[copper valve]

are u bad at them

[tropic stirrup]

Well I expertise on high school math

[copper valve]

this is highschool

[tropic stirrup]

Ye

[copper valve]

so do u expertise on this

[tropic stirrup]

I kinda do

if it's highschool

😛

[copper valve]

then how come u said earlier that u dont ;o

only expertise on middle school

inconsistent answers! :0

[tropic stirrup]

Eh, I don't in like the competition high school math