geometry-and-trigonometry

[minor quest]

@past mantle I tried this :0

So the total internal angles equals 720 because it's a hexagon we have 2 right angles so that's 180 now u just combine all the terms and set them up equaling to 720 to solve for x

[minor quest]

How many integral triangles are there with perimeter that is less than 100?

I need help with this :c

[tropic stirrup]

integral triangles?

[minor quest]

@tropic stirrup triangles with side lengths that are integer

[tropic stirrup]

Oh

[minor quest]

@tropic stirrup do you know how?

[tropic stirrup]

Well

[minor quest]

@tropic stirrup well? xD

[copper valve]

that's a cool question :o

a≤b≤c
a+b>c
3≤a+b+c<100

that's all i know how to start xd

i guess can search for how many triangles have side length a=33, then 32 and so on xd

[tropic stirrup]

let a + b + c = 100 and that c is the largest side so we get c > 33.
Then we get that a + b > c, so 100 = a + b + c > 2c.
Then 33 < c < 50.
Now if c = k, then a + b = 100 - k and a, b < c = k.
From this we get a = 100 - b - k > 100 - 2k, so 100-2k < a < k would do.
The number of a that satisfies this inequality is trivially k - (100-2k) - 1 = 3k - 101.
Therefore, the number of triples would be the sum of 3k - 101 from k = 34 to 49.
The sum of 3k - 101 from k = 34 to 49 can be calculated easily via the transformation k = t + 33,
so that it equals to the sum of 3t - 2 from t = 1 to 16,
which is equivalent to 3 * 16 * 15 / 2 - 16 * 2 = 328.

[copper valve]

😮

[tropic stirrup]

Pls correct me if I'm wrong x'P

[copper valve]

u are good 😮

idk if it's right or wrong but it looks so tidy xd

[tropic stirrup]

@minor quest

[minor quest]

How do you determine that c>33? I’m already lost there xd and then I don’t get the 100= a+b+c>2c ..... @tropic stirrup

[tropic stirrup]

a + b + c = 100, and c is the largest side
Which means c is larger than or equal to a and b.

=tex a \le c,~ b\le c \
a + b \le c + c \
a + b + c \le c + c + c \
100 \le 3c \
c \ge 100/3 = 33.333... \
\text {So, } c > 33\text{ or } c\ge 34

[tropic stirrup]

=tex \text {Also, from } a + b > c, \
a + b + c > c + c \
a + b + c > 2c \
100 > 2c \
\therefore ~ c< 50

[tropic stirrup]

@minor quest

[minor quest]

Sorry but once again could u go in depth on how you get these steps 😭🤔

@tropic stirrup

[shy shale]

can someone help ne with mathia

this online math

homework

[minor quest]

Is it just me or those labels on the picture look misleading?

[upper karma]

it's very clear lol

not misleading at all

[minor quest]

@tropic stirrup hey

[tropic stirrup]

Question) ABCD is a rectangle. Find the angle ? precisely without using Pythagorean theorem or trigonometry.

Hav fun uwu

[minor quest]

@tropic stirrup mind dming me the explanation for yesterday’s problem?

[tropic stirrup]

I gotta go soon so I'll do that later

Meanwhile try and solve that >w>

Anyways, tooodles

[unique prism]

would anyone mind drawing/constructing this figure?

---

a figure ABCD is AB 9 cm, BC 5 cm, angle ABC 90, angle CAD = angle ACD 45.

---

[minor quest]

@alive-_#7540 like this?

[unique prism]

ooh

oh

thank you

i really appreciate it

sorry

[foggy shell]

@minor quest It wouldn't be a rectangle, then.

[unique prism]

?

[vocal cipher]

ye he's right

[unique prism]

hm

[vocal cipher]

angle cab= angle acd

for angles dac and dca to be 45, the side ab and bc have to be the same lenght

[unique prism]

i just saw this in the section that i never look at (because i don't want to cheat)

this should be right, then?

[foggy shell]

Yes.

Two different triangles sharing AC.

[vocal cipher]

yep

[unique prism]

would anyone mind explaining the instructions to me?

sorry, i'm quite confused

[vocal cipher]

ok so you know what he drew

ok so the thing was

only a square

would have the angle acd and acb be 45°

[white swift]

wats this very complicated problem

whats happening

[vocal cipher]

gah damn other language idk words for it

[unique prism]

i'm just being stupid

[vocal cipher]

sec i'll draw it

[unique prism]

ah ok

[white swift]

holy shit

[vocal cipher]

http://prntscr.com/hipjlv

[white swift]

this drawing is so bad

[vocal cipher]

ok so what is an angle of 90°

it's when 2 lines are perpandicular to each other

so that means a square has 4 of those inside

so the sum would be 360

but what also means then

since the sides are eaqual lenght

that means if you connect a line between opposite sides

the angle gets cut in half

as you can see from the picture i terribly drew

so here's the problem then

[foggy shell]

I don't think that alive needs that part explained.

Might I try?

[vocal cipher]

sure

[unique prism]

i don't mind

but maybe that's true

but i don't want to be mean or anything

i appreciate it

[foggy shell]

Let's start with the half of the figure that we already know.

We have a right triangle with bases 5 and 9.

We are disregarding the hypotenuse for now.

Let's add congruent 45° angles.

[minor quest]

Oh I forgot about triangles ratio didn't I...

[unique prism]

ohh

i think i understand a bit

thank you for explaining

thanks everyone who has helped or tried to help

[minor quest]

Lul 😭

[unique prism]

a

?

sorry

[foggy shell]

You are most welcome.

[white swift]

did you sort it out?

[unique prism]

yes, i think

[unique prism]

so uh idk if any of you have the time and patience but i have another figure that i need to make, but unlike the other one i kinda did it but there isn't any example in the back of the booklet (idk if it's called that) so i'm unsure if it's correct and i don't trust myself.

[foggy shell]

Okay.

[unique prism]

--

construct the figure ABCD where AB = 5 cm and AD = 4 cm. Angle A = Angle B = 90 ° and Angle D = 120

--

[vocal cipher]

well draw it and let's see

[foggy shell]

Show us what you have and we'll verify it.

[unique prism]

aah but what if it's wrong

[foggy shell]

Then, we'll say that it's wrong and help you from there.

[vocal cipher]

if it's wrong we'll see where you did it wrong and help you not make that mistake again

[unique prism]

aa i'm scared but okay

also my handwriting is very ugly

sorry

and it's terrible probably

sorry

i've been pretty braindead recently

[vocal cipher]

ye that's good

[unique prism]

ah

[foggy shell]

That looks correct.

[unique prism]

sorry for talking like this

thank you

[foggy shell]

You're doing well. 😄

[unique prism]

ah thanks

[west dune]

Hey guys,

It's a trapeze which his "legs" (Don't know how to translate to en) equals

I need to calculate the bases using the expressions```

How to?

=tex [x + 3 + 2(2x - 1)] - 2(2x - 1) = x + 3

Is it possible here?

[west dune]

@dark sparrow What did you come to write? 😅

[dark sparrow]

well i mean that eq you posted is trivial

anyway see that triangle in the right? it's right isosceles

you have that 45° angle there

[west dune]

Ok thanks I got it

[devout zephyr]

I'm trying to find the relation between angle O and angle t when folding a cone.
http://cdn-4.analyzemath.com/Geometry/area_cone_1.gif

I found a formula, t = 2pi*sin(0.5*O), but I'd like to know how they got to that formula since I can't do it myself

[slow pendant]

Sin(O/2) is just the radius.

Because of this, all you are doing is finding the circumference.

Since when cut, the circumference is equal to the central angle in degrees, the circumference, in degrees, gets you t!

[bleak rivet]

Having some trouble with this one

Can anyone shed some light on this for me?

I know I can add 41 and 30 to get 71

then do 180 - 71 = 109

to find the third angle of the left triangle

but I really dont know after that

[copper valve]

there's a triangle with angles 30°, 36°, and (41+x)°

adding them all up we must get 180°

cause all angles of a triangle add to 180°

(in euclidean space)

so we have: 30+36+41+x=180
107+x=180
x=180-107=73°

[bleak rivet]

Thanks man

[copper valve]

np

[forest dove]

Does anyone know how to prove that the rotation group of the cube is S_4 using as little geometry as possible?

I've already been convinced that it has 24 elements

[vapid kettle]

well like

[forest dove]

But I can't visualize the rest of the argument about opposite corners because that just short-circuits my mind

[vapid kettle]

your first comment made me think

just prescribe an explicit isomorphism

it wouldn't use that much geometry

(tis not a great argument/is very tedious though)

[forest dove]

Yeah, also this is for, if this comes up on a test, how to do it?

If push comes to shove

I will just write the words "the action on opposite corners of vertices is faithful, and since the groups are the same size, this is an isomorphism" without really being convinced of it

But I will feel bad doing it

[vapid kettle]

hmm

ok here

you have an argument where

you convinced yourself that the group has 24 elements

lets black box that

now

given two pairs of opposite corners

give an argument that you can swap them

with some rotation (fixing everything else)

[forest dove]

See opposite pairs of corners is exactly what I can't process

[vapid kettle]

this should be doable by a visualizable type of argument which you can then formalize better than "action is faithful"

well if you can get that down

processed

[forest dove]

My visual processing skills are basically negative

[vapid kettle]

it gives you an arbitrary transposition so you're good!

hmm

you have magnets?

[forest dove]

Unfortunately, no

[vapid kettle]

get magnets and make shapes :0

[devout zephyr]

@The Dark_Speed_Ninja/ShadikkuX#5846 I don't really understand what you mean with that

I don't see how Sin(O/2) is the radius, and I really don't see why the circumference wuold equal the angle

[upper karma]

Why are SSA or ASS not a postulate for congruent triangles?

[upper karma]

thank u

[wintry ether]

ok so I want to punch holes in a circled bottom, and have those holes align with the piece I am going to wrap around the circle.

Is there any cool mathematic formula for this, atm ive tried to punch holes, count the holes then divide the holes with the length of the piece that is going to be wrapped

Did this make any sense?

Do I post this in questions, I thouight maybe it would fit in geometry

[brave barn]

Yo

@wintry ether dm me the questions

[wintry ether]

well its

^

right there

[copper valve]

what is circled bottom :o

[brave barn]

O

[wintry ether]

Its

just a circle

O

[brave barn]

Fuck english terms

[wintry ether]

and then u have [=====]

long ass square thingy that u wrap around

circle

I know that to get the length its thingy x 3.14

and u get the radius

thingy

math term

[brave barn]

Uh huh

[wintry ether]

Me no do well

But then Iw ant to make holes

[brave barn]

Holes?

[wintry ether]

in the circle and the thingy I wrap around

yah

wait

[brave barn]

To equal what

[wintry ether]

But I cant get the holes

to align

[brave barn]

So u want same space wholes?

[wintry ether]

I want each hole to align with the hole int he circle

[brave barn]

Like same distance

[wintry ether]

yah

pretty much

[brave barn]

How many wholes

[wintry ether]

Depends I guess on how big the circle is

[brave barn]

Well you can never achieve that xD

[wintry ether]

🤔

[brave barn]

Cause π

[wintry ether]

well thats nice

to know

lol

cause ive gotten

pretty close

at times

but I always have like 1 or 4 holes

[brave barn]

BUT π

[wintry ether]

2 much

is there a formula to get close to it tho?

[copper valve]

u can lay the circle bottom on a protractor

[wintry ether]

Or is that the, holes divided on the length, then u get the distance each hole should have?

[brave barn]

Exacrly

[copper valve]

then think how many holes u want

then mark it using pencil or marker

[brave barn]

Divide the diameter of tje holes and the lengjt of the circle to the hole you wanna make and bam

[copper valve]

oh i think i know what your asking

[wintry ether]

Ic

[copper valve]

maybe. lol

u can do the holes on the circle first

[wintry ether]

Yah thats what I did

then counted them

[copper valve]

then measure diameter of circle

[wintry ether]

and took the amount of holes, divided on legth

oh, that I didnt do

or wait I did

[copper valve]

multiply by 3.14 and have your strip roughly that long

[wintry ether]

Yah I did

THen u get length

of what the piece that should be wrapped

[brave barn]

Yeah but you can never be correct

[copper valve]

then divide that length by # of holes

[wintry ether]

THen u divide that length with holes

[brave barn]

Because the thingy tjst youre counting the length

Might not be accurate

[wintry ether]

ic, but then my next question, should I take the diameter

of

[copper valve]

stelio that doesn't matter

[wintry ether]

the entire circle, or where I punch th eholes?

[brave barn]

IT DOES

I'm a perfectionist

[copper valve]

shouldn't it be the same?

[wintry ether]

dont think it will ever be

[copper valve]

punching holes shouldn't change the diameter :o

[wintry ether]

🤔

Thats why im here 😂

[brave barn]

In which circumstances

[copper valve]

idk if im getting what u mean lol

it's the same circle, just with little holes on the edge

[brave barn]

Yeah but

[shy shale]

can someone help me with this

[copper valve]

not now

[shy shale]

I need to finish this section

its due tomorrow

[wintry ether]

well cheers guys, do u mind if I ask one

last question

or maybe a picture

[copper valve]

sure

[wintry ether]

will make more sense

[brave barn]

Ok

Go.for it

[wintry ether]

does

the distance

here

matter

?

[brave barn]

WAIT

[wintry ether]

:waits

[copper valve]

kind of

[brave barn]

You can do stuff with the radius

[copper valve]

lemme think

[brave barn]

And the distance of the radiuses

Like @copper valve do u get me?

[copper valve]

no

not at all

[brave barn]

Draw as many radisues as the holes he wants with the same distance in between each other

SoI he will draw a polygon

Inscripted polygon

[copper valve]

why would u do that

[brave barn]

So you can be precise

[copper valve]

I'd do that if i didn't know what π was lol

[wintry ether]

ok this might be rly stupid

but what is this sign

oh

thats 3.14

pi

[copper valve]

lol

yea

[brave barn]

Π

[wintry ether]

dont even know how to make that on a keyboard

[brave barn]

π

[copper valve]

need greek ^^

[brave barn]

I'm greek so I have it xD

[wintry ether]

Ic

Fotia

Kalimera

[brave barn]

:O

[wintry ether]

😛

ALl the greek I know

[brave barn]

Καλημέρα

Φωτιά

[wintry ether]

Anyways, so does the distance

[copper valve]

the difference is not too important @wintry ether

[wintry ether]

from the edge matter

[brave barn]

it is

[wintry ether]

Cause the thing is

[copper valve]

just measure the biggest one

[wintry ether]

if I have a piec eof leather

that is

4mm thick

[brave barn]

You can't undo it

[wintry ether]

I tend to want to have the stitching line 4mm from the edge

[brave barn]

T H I C C

[copper valve]

hmm

[wintry ether]

so I was wonderin gif

lets say the leather is 6mm thick and I end up making the line 6mm inside

of the circled thingy

wouldnt that screw up

the whole formula

or do I have to figure out the diameter

AFTER

ive done the stitching line

and take the diameter of the circle within the circle

if u get me.

[copper valve]

u can measure from the middle

like

take the width of leather strip

here it's 4mm

then whatever diameter u get of the circle

minus 4mm from that

[wintry ether]

Cause I did like this

prly cant see it, but measured from the circle within

the circle

[copper valve]

measure the outside

[wintry ether]

Ahh aight

[copper valve]

like from the edge of circle to other edge

[wintry ether]

cause I was thinking the further inn the holes go

the fewer holes

[copper valve]

then just subtract 4mm from that

[wintry ether]

u are gonna get

[copper valve]

ya

[wintry ether]

ohhhh

wait

so when u get the distance

u just subtract

whatever stitching length

I have?

so lets say the holes are 6mm inside

[copper valve]

idk anything about stitching lol

[wintry ether]

oh the stitching line is just the ... lets say

like u said 4 mm

is where I make the holes

so I get 20 holes

[copper valve]

is it the same as thickness of leather?

[wintry ether]

divide it with the length, then subtract 4mm

well thats what u usually do, but u can do 10mm

or 3, its just a thumb of rule

[copper valve]

u subtract 4mm first then u divide for the holes

[wintry ether]

now the diameter is 8.2

so u subtract 4mm, before u multiply by pi?

[copper valve]

ya

[wintry ether]

so 7.8

[copper valve]

then multiply by pi

and that's the length of strip to cut

[wintry ether]

24.492

so 24.5

[copper valve]

ya

once u have that u can divide by number of holes and evenly mark them

[wintry ether]

ic ic

oh, and when I mark the holes

I take half the distance it gives on the first hole

right?

cause the 2 ends are gonna meet on the middle

[copper valve]

the what o.o

[wintry ether]

wait ill do a picture

im horrible at explaining

lets just say the thingy said the distance between each hole

should be

1.4

the first mark on the straight thingy

would be halfed no?

[copper valve]

oh ya, also the last one

should be 0.7 away

[wintry ether]

yah and the last one, and then I start marking

inbetween those 2

marks

[copper valve]

ya

[wintry ether]

Ayyy

so I kind of

did get to the right answer

but instead of just subtracting

the distance

I was measurring

from within the circle

😛

Like the pic I posted earlier

[brave barn]

Still here?

[wintry ether]

Yah

I think it all makes sense

I spent 2 long

[copper valve]

stitching is wizardry

[wintry ether]

trying to get to this, but rly lit a light

for me, how useful math is

😂

[copper valve]

^^

[wintry ether]

Cause before, I just meassured distance between the holes

and did the same on the piece that was gonna be wrapped

and ended up with

[brave barn]

NO

[wintry ether]

100 more holes

than I needed

[brave barn]

Never

[wintry ether]

I KNOW

STELIOS

😛

[brave barn]

XD

[wintry ether]

I lived to learn

When I was wrapping it I instantly saw

that I goofed

[copper valve]

ya it's cause on a circle it's hard to measure the distance between holes

[wintry ether]

but yeah

[copper valve]

cause the curve on the circle is always gonna be a bit longer than whatever u measure with a ruler

[wintry ether]

I made some things with this formula I just did some mistakes along the way I think

like forgot to do half of the length

on each side

so I ended up with some more holes

but it will never be perfect

correct?

Cause I tend to ocd over things

But if I know, what I want to achieve can never be perfect

I can let things go

😛

[copper valve]

well depends on how u think of perfect :o

[brave barn]

When it comes to circles I give up

EXCEPT GAUS

[wintry ether]

well perfect

would look like this

[brave barn]

Gaus' spherical geometry Is beautiful

[wintry ether]

every hole

being perfectly ontop

when its wrapped

[copper valve]

i think it's just a matter of practice ^^

[wintry ether]

yes paint skills

are mad

[copper valve]

heh better than me

[wintry ether]

prly, ive only made a handful of round

projects

but ive always liked circles so, I thought I might aswell learn

some formulas for it

I was amazed about the diameter x pi

😂

Before I just made the thing imma wrap way longer

wrapped it around then cut

off anything I didnt need

@copper valve wait, wouldnt it be -8mm

sinc eits 4mm on each side

[copper valve]

no :o

i did -2mm on both sides haha

half the diameter

so it would align

[wintry ether]

wouldnt that be 4 on each side

[copper valve]

i did some math wizardry, lol

I'll try to draw a picture

[wintry ether]

ok I might be over tired from doing 2 much math

cause Im supposed to do, the whole radius first, to find the length of the thingy to wrap

btw just tell me to fok off

if im becoming annoying, or if its 2 late for u

or w/e

[copper valve]

i have too much time on my hands, lol

[wintry ether]

oh but it should be 4, since I'm drawing around the circle

with 4

With one of those pointy thingies

[copper valve]

so if u want the strip centered

[wintry ether]

that u can put a distance on

[copper valve]

then u subtract half the thickness of strip on both sides

[wintry ether]

OH wait, now u get the circle inside the thing u wrap around

right?

[copper valve]

sure

lol

[wintry ether]

Oh

thats cool, imma remember that atm im trying to get it

to be ontop tho

[copper valve]

ontop :o

[wintry ether]

yah, so then u just take

the whole

diameter

right?

[copper valve]

ya

[wintry ether]

Aight so the subtracting thingy before and after

has nothing to do with the holes

aligning?

Correct?

[copper valve]

correct

[wintry ether]

that only decides wheter the circle is getting wrapped inside or not

Wait let me take notes

lol

Imma get a use for this 100%

aight

now is there any trick to get the holes the same distance

around the circle? Cause when Ijust punch them I always end up

with not correct distance on the last 2 holes

where I end and started

[copper valve]

how are u punching the holes?

[wintry ether]

https://i.pinimg.com/originals/ef/ed/89/efed898b6a2fb6e747c57d6bc01a37fe.jpg

with one of these

but I first mark them all up

using this

[copper valve]

so nice :o

[wintry ether]

Like I tend to just

put it into a distance

and then just run it along the line

u know? sometimes it works out

sometiems I get the last 2 holes

2 close

😂

Was just wondering if theres any way to secure it

so it always adds up

like divide radius with amount of holes I want

to get distance between each hole

or something

[copper valve]

maybe but i think at this point it's down to practice xd

[wintry ether]

oh

aight, mark holes, divide holes with leangth of wrapping thingy, get distance, half distance on start and end

Think we got it

Love u man ❤

[copper valve]

^^

[wintry ether]

Here one of the things I made

just so u can see what Im planning

If my phone

could

upload

😅

Well nvm, ill show u the next thing I make using your perfected formula

❤

[copper valve]

Ok ^^

[upper karma]

Anyone @here know if I did numbers 3 & 4 correctly?

I'm sorry, 4a and 4b. Not 3.

[thorn talon]

Looks right

[upper karma]

Great! Thank you.

[thin hound]

lolol

[vast hazel]

School should be giving tests like that

[fluid aurora]

where is the basic math section where you learn before high school

[upper karma]

how many sides does a polygon have if its smallest interior angle is 120 degrees and each successive angle is 5 degrees greater than its predecessor

I need help

6 sides

i think

There are 2 answers

One for concave

One for convex

wait a second

9 sides

But how lol

https://www.algebra.com/algebra/homework/Polygons/Polygons.faq.question.616896.html

probably wrong though

im doing algebra and im in middle school now

geometry aint what im doing

Man

Ok then

i tried

Lol thanks

I need the formula too

try going on khan academy

really helps

Kk

how many sides does a polygon have if its smallest interior angle is 120 degrees and each successive angle is 5 degrees greater than its predecessor

If anyone else comes lol

[dark sparrow]

I need the formula

😒

[upper karma]

Lol

I think it’s 2.5n^2+117.5n

But I can’t figure out the sides from that

I think

[dark sparrow]

naively, you could just

add up 120 + 125 + 130 + 135 + ...

until your sum is a multiple of 180

[upper karma]

Yeah but I won’t get credit

[dark sparrow]

...

[upper karma]

😅

Well I got it

[copper valve]

The exterior angles will add to 360

[upper karma]

9 and 1

6

9 and 16*

And I have another question

[copper valve]

16 shouldn't work

[upper karma]

Well

I need a concave one too

So I think it would be 16

[copper valve]

cause you'll have an angle of 180

[upper karma]

O

[copper valve]

then 2 edges will actually be 1 edge

if you could have that, then ya 16 sides

[upper karma]

Hm

Eh

I’m going with it

[upper karma]

@copper valve wait would it be 9 and 15?

[copper valve]

no cause then you skip from angle 175 to 185 :0

[upper karma]

AHHHHH

then what is it

Fmlllll

[copper valve]

9

and if they expect concave answer

16

[upper karma]

K

They expect concave

[copper valve]

even tho it doesn't really make sense

just go with it

[upper karma]

K

Thank you

Can you help me with one more

Lol

[copper valve]

maybe!

you'll have to ask the question first

[lost jetty]

protip

paste it into paint and turn it 90 degrees

[upper karma]

Lol

[white swift]

or do it on your phone

[upper karma]

Thank you

[white swift]

dunno I guess you draw ED

maybe thats useless 🤔

you know that AB=CO=OE=OD

[copper valve]

i thought that at first haha. geometry too spooky

=BO=ED

[white swift]

draw everything you know

and see what happens 🤔

equilateral isosceles triangles something something

[copper valve]

this is too crazy o.o

[upper karma]

Ok

[white swift]

this is like 8th grade geometry or something lol

😄

[upper karma]

Yeah it is

[white swift]

were actually bad at geometry

what topic are you on

[upper karma]

Uhh

I’ll chexk

[copper valve]

this is INSANE

what the heck

im bamboozled

[upper karma]

Discovering and proving polygon properties

[copper valve]

im gonna get some paper and draw this out

[upper karma]

Lol

[copper valve]

i wanna say it's 15 but i can't prove it :(

[upper karma]

Hm

can math bot do the calculations?

Lol

I wish it could solve the whole problem

[upper karma]

let angle OAB be x means angle BOA is also x because ABO is isosceles. that makes angle OBE 2x. since triangle OBE is also isosceles, OEB is 2x and that leaves angle EOB (180-4x). angle AOB + angle EOB + EOD = 180. u solve for x and that would give u 20

[copper valve]

@upper karma nice!!

[upper karma]

lol the auxiliary segment is probably the radius OB

[rough juniper]

completely lost here, and my notes are @ school

I think it's simple because the coordinates are the same, but that's probably the wrong answer since it's not found correctly

[upper karma]

lol D and E have the same coordinates

i think the question meant D(-8a/5, 2c/5)

[rough juniper]

yeah most likely

[upper karma]

find the distance of BD and AE and then compare

[rough juniper]

yeah, that's what I need to prove ( not a proof*)

[upper karma]

yeah have u found the distance of both

[rough juniper]

no

it's not solve for a, you just need to prove it

[upper karma]

find it first

yeah find the distance first

[rough juniper]

im not supposed to

My guess is that I can just say it's equal because it's an isoceles

[upper karma]

D(-1.6a, 0.4c) and B(2a, 0)
distance of BD is sqrt(12.96a^2+0.16c^2)

A(-2a, 0) and E(1.6a, 0.4c)
distance of AE is sqrt(12.96a^2+0.16c^2)

[rough juniper]

or that I guess

[upper karma]

literally the same qed

[rough juniper]

I'll put down both and wish for luck,

[upper karma]

u dont go for luck

[rough juniper]

Maybe talking to the teacher to be less vague will be a good idea*

[upper karma]

u go for ur solution

[rough juniper]

Anyways thanks, I'll work on it some more for the confirmed answer

[main sluice]

if the pattern shown above is continued to infinity

in all directions

what is the asymptotic ratio of the green squares to the red squares?

[main sluice]

I think the asymptotic ratio is 1:1

[copper valve]

you can outline a single red square and one of the adjacent green squares; and note that considering this as a single tile, will cover the plane without overlap

[crude kraken]

Not looking for a solution, just figured when I stumbled upon it that you guys might like it.

:p

[tropic stirrup]

Seems like...... ye seems so

[copper valve]

interesting

seems like a complex number question i made

but with geometry :0

[past mantle]

Doesn't look like anything complex number wise

[crude kraken]

In my solution I used complex numbers

[dark sparrow]

thought so

[main sluice]

I've seen that problem

in it's generalised form

the answer uses complex trig bash

[copper valve]

whats the generalised form?

@main sluice

[main sluice]

regular n-gon

pick a vertex

product of all lengths of diagonal from that vertex

[copper valve]

non regular ;o?

[main sluice]

shut up

you saw nothing

[copper valve]

how is that not the same exact question

i think theres a way to do it without trig

or induction

[keen aspen]

I have a question

Does a trapezoid have 1 pair of parallel sides

or at least one pair of parallel sides

[copper valve]

yes

at least one pair

[thin hound]

http://prntscr.com/hlosbp

@upper karma

How do i find out what number i need to multiply after subtracting the two x's to each other?

[copper valve]

depends on where they are on the graph

idk what those points ive labelled are called

but they are ¼ of the period apart

so in your case you would multiply the difference by 4

[thin hound]

oo

so look

http://prntscr.com/hlp9gv

this would be 1/2

and i'd multiply by 2?

@copper valve

[copper valve]

yes

[still zodiac]

The bearing of point Y from point X is 075°.
The bearing of point Z from point Y is 125°.
The bearing of point Z from point X is 110°.
Find the bearing of point X from point Z.

please help

[umbral snow]

The opposite of 110

== 110 + 180

[charred spear]

290

[umbral snow]

@still zodiac

[shy shale]

hey

anyone on

Need help

[dark sparrow]

[fierce tide]

can someone help me with this problem?

[crude kraken]

🇯 🇺 🇸 🇹 🇦 🇸 🇰

[fierce tide]

mk

[crude kraken]

Exterior angle being 2a?

[fierce tide]

what does a equal?

[crude kraken]

oh

44 + a + 10 + (180-2a) = 180

54 + a - 2a + 180 = 180

54 - a = 0

[upper karma]

a + 10 + 44 = 2a

[fierce tide]

its not 0

[crude kraken]

-a + 54 = 0

-a = -54

a = 54

[fierce tide]

nope

i tried that too

[upper karma]

2a is 108

[crude kraken]

^

[upper karma]

is the exterior 108?

[fierce tide]

yess

ty

[crude kraken]

😛

[fierce tide]

what is the measure for the exterior angle

[copper valve]

3x+5x-6=90
x=12

[fierce tide]

nope

wrong

i did that 2

[crude kraken]

Which angle is the exterior?

[copper valve]

i told u what x is

not the exterior angle

[crude kraken]

oh I c

== 180 - (5 * 12 - 6)

[charred spear]

126

[crude kraken]

There ye go

[fierce tide]

wt...

ty?

[crude kraken]

Exterior angle = angle outside of triangle

[fierce tide]

ik

[crude kraken]

solving for x isn't enough, you gotta find that 5x-6 angle and then get the angle on the other side by subbing 180

:p

[fierce tide]

60-6 is 54

top 1 is 54

bottom 1 is

36

36+54+90?

[crude kraken]

180 - 54 = 126

[fierce tide]

what about the bottom 1

[crude kraken]

Doesn't matter for the solution

I mean, it's needed to set up the equation

[fierce tide]

to add up to 180 right?

[crude kraken]

mhm

8x - 6 = 90

8x = 96

x = 96/8

exterior = 180 - (5 * 12 - 6)

[fierce tide]

ty

[dark sparrow]

this seems to involve some terminology specific to a course or field

so you'll have to explain what pretty much every term means

[foggy shell]

Which part are you having trouble on...?

@sullen flint Since it's a parallelogram, you can assume that opposite sides are equal in length. Now, you already have the long side. Find the short side AB using the Pythagorean theorem.

You can add a variable x in the process because you can resolve it with the 60deg angle.

[sullen flint]

wait so AD is 8+3 ?

[foggy shell]

AD is congruent with BC.

They're the same length.

[brazen roost]

AD is 8

actually

[foggy shell]

Have you learned about the tangent trig function?

[sullen flint]

nope

wait but

if BE congruent with AD and AD is unknown

ugh

[foggy shell]

AD is congruent with BC.

BC is known.

[sullen flint]

🤔

[foggy shell]

???

[sullen flint]

im sorry but i dont understand how to find AD or BE

is AD = BC ?

oh

🤦

[foggy shell]

Yes, length of AD = length of BC.

[sullen flint]

hmm but it says that BE | AD perpendicular

[foggy shell]

That wouldn't make a difference about the length of AD...

[sullen flint]

yes but i want to find BE

[foggy shell]

I'm thinking of a way to solve this without trig.

[sullen flint]

so ?

[copper valve]

gurl would find a way if it weren't 1am

if possible

[foggy shell]

I've got it.

Sorry, that requires trig as well.

[upper karma]

angle BAE = 60°. cos BAE = 3/BA. BA = 3/cos60° = 6cm so perimeter is 28cm

[foggy shell]

He hasn't learned trig.

That's the only reason why I'm stuck.

@upper karma

[upper karma]

improvise adapt overcome

that triangle BAE can be pictured like half of an equilateral triangle

so AB is simple twice of AE

[sullen flint]

🤔

[dark sparrow]

what have you tried so far, and where did you get stuck?

[fierce tide]

i got stuck with the algebra

[dark sparrow]

can you show me the work you've done so far, then?

[fierce tide]

i can't its on paper

and im in school rn in study hall

[dark sparrow]

can you not take a picture?

[fierce tide]

nope

[dark sparrow]

...ok

[fierce tide]

let me just write it out

[dark sparrow]

yeah do that

[fierce tide]

x-40+x-38=x

right?

[dark sparrow]

yes

[fierce tide]

ok 2+2x=x

[dark sparrow]

huh?

how'd you end up with a 2 there?

you should have ended up with 2x - 78 = x

[fierce tide]

40-38

[dark sparrow]

no.

x + (-40) + x + (-38)

= x + x + (-40) + (-38)

[fierce tide]

oo

[dark sparrow]

= 2x + (-40) + (-38)

(-40) + (-38) = -78

can you take it from here?

[fierce tide]

yes

ty

[past mantle]

I'm having trouble with this proof ;-;

The last point is B

[opal remnant]

are you not allowed to say it's valid by symmetry or smth?

[past mantle]

Nope

[plush cairn]

a diagnol bisects the angles

so rhx and mhx are the same

hr and hm are the same

hx and hx is the same line

triangles are congruent by sas

therefor the respective angles are the same

@past mantle

[past mantle]

Thx

[copper valve]

for the last question

the triangle with 30°, α+80° and other angle β

β=30° because the triangle​ is isosceles

(cause inscribed in circle or somethin)

so we have 30°+30°+80°+α=180°
140°+α=180°
α=180°-140°=40°

[fleet rivet]

What does cos, sin & tan actually do?

[dark sparrow]

not sure what you mean exactly

[fleet rivet]

E.g. when I type cos30, what does the calculator do

[dark sparrow]

mm well

multiple algorithms / formulae exist for evaluating cosines of arbitrary angles

or any trig functions really

the one i know about the most involve these formulae, which take in an angle (in radians! this is important) and produce the sine and cosine:

=tex \sin(x) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + \frac{1}{9!}x^9 - \cdots \ \cos(x) = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6 + \frac{1}{8!}x^8 - \cdots

[dark sparrow]

these are infinite series and can be continued as far as you need depending on how precise a value you want

[fleet rivet]

ok, but how do you explain it?

What is it doing with my x value

[dark sparrow]

i mean

those are useful for computation rather than explanation

and i'm not sure which one you want

[fleet rivet]

What is useful for explanation then?

[dark sparrow]

well

you know the unit circle definitions of sin and cos, right?

[fleet rivet]

no...

Dya know what GCSE's are?

I am that level

[dark sparrow]

a UK thing

that's as much as i know about that

[upper karma]

i took GCE o levels

is it similar

[dark sparrow]

@Sir Haris The Guy#0141 i assume you know how the trig functions are defined using right triangles

@Sir Haris The Guy#0141

[fleet rivet]

Yep

[dark sparrow]

yeah ok

well

here's a circle of radius 1 centered at the origin of a coordinate plane

[fleet rivet]

mhmm

[upper karma]

why do i see this unit circle everywhere

[dark sparrow]

now we're going to restrict our attention to angles which have one side aligned with the positive x axis, and whose vertex is at (0,0)

[fleet rivet]

ok

So essemtialy on the right side of the y axis right?

[dark sparrow]

something like this

[upper karma]

oh

[dark sparrow]

the side that points exactly towards the right is fixed

while the other one can vary

[fleet rivet]

ahh ok

[dark sparrow]

notice that this other side always intersects the circle at exactly one point

[fleet rivet]

mhmm

[dark sparrow]

and well

we can now redefine cos(θ) to refer to that point's x coordinate

and sin(θ) to refer to its y coordinate

but one thing we want to make sure before we start doing anything with that is make sure this agrees with our old definition which was only valid for angles between 0 and 90°

i drew an acute angle to make this a bit more convenient

[fleet rivet]

mhm

But don't they technically work for angles greater than 90 as well?

[dark sparrow]

well, no, the right-triangle definition sort of fails 😛 since you end up with a triangle whose angles add up to more than 180°

[fleet rivet]

For sin and cos rule though?

[dark sparrow]

our new definitions work for any angles

[fleet rivet]

that being the circle?

[dark sparrow]

yes

and what we're doing now is making sure they give you the same answer as the old ones, wherever the old ones were valid

[fleet rivet]

Ahh right, and since computers cannot accurately calculate a curve, it needs to use the prev equation

[dark sparrow]

no

[fleet rivet]

then?

[upper karma]

what is the significance of the circle having r = 1?

[dark sparrow]

computers have nothing to do with this

if our new definitions gave different answers than the old ones, we would have come up with something completely different, and so we wouldn't be able to call our new functions sin and cos, would we?

they'd be different objects, and thus we'd need to name them differentll

y

[fleet rivet]

mhmm