geometry-and-trigonometry

[upper karma]

|| B_A ||

| B_A|

[zealous egret]

B_A means what?

proj b onto a?

[upper karma]

projection of B onto the line A

yes

[zealous egret]

well it would be b.a divided by the magnitude of a?

if you're looking for the scalar value

[upper karma]

yes

so B_A = (b.a)/|a|

[zealous egret]

so | b |cos(theta) / | a |

yea

[upper karma]

now do the same for C_A and (B+C)_A

and use that B_A + C_A = (B+C)_A

what do you get if you subtitute all

[zealous egret]

so I don't look at the vector proj

[upper karma]

you do

you look at B_A

[zealous egret]

I look at the magnitude of the projection

[upper karma]

oh

sure

[zealous egret]

ahh

I was thinking of the projections in terms of

let's say b onto a

(b.a^)a^

sorry for the wrong notation

[upper karma]

this is a bit problematic with negative values tho

[zealous egret]

but you know what I mean

[upper karma]

sure

[zealous egret]

and what law do I have to state that a.b = b.a

even though it's like common sense

is there a formal term for it

[dark sparrow]

symmetry of the dot product

[zealous egret]

it feels like I've just repeated what it's asked me to show though

ah thanks @dark sparrow

and thanks a lot @upper karma

[upper karma]

well sure, it requires you to know about projections

[zealous egret]

is that answering the question though?

idk just feels wrong since it's like rewriting what they just asked me to prove

But I'm just a noob, what would I know 😄

[upper karma]

pff... I agree it's a bit wishy washy

but I think that's what they want

[zealous egret]

it's not really presented as a formal question

just an optional exercise

like "discuss the results alpha = whatever" type of things listed at the bottom

[upper karma]

oh ok

should be fine then

[zealous egret]

Btw another question

How is the dot symbol introduced in the second line

[upper karma]

the first line is wrong

between those huge brackets

should be a dot

then the first to second line is just distributivity

[zealous egret]

The first -> second line is just the way you can manipulate summations, right?

[upper karma]

yes, but it relies on knowing the dot product is distributive

write it out in full if you don't see it

you just pull the sums out of the dot

[zealous egret]

Hmm I'm actually a little confused about how the summation works

if you had something like

∑n∑m

n=1 to 3

and m = 1 to 4

can that also be simplified as above?

[upper karma]

is that a product? You should write your brakcets better

[zealous egret]

yup, sorry

[gilded ore]

b) Use the triangle above to show that (sinv)^2 + (cosv)^2 = 1 for v = <0,90>

[astral hornet]

hint: think of the pythagorean thoreom

[gilded ore]

@astral hornet Im aware that i need 90 degrees to use the pythagorean theorem but i dont understand how this helps us since we are doing the v angle

well a^2+b^2=c^2

[astral hornet]

in (sinv)^2 + (cosv)^2 = 1 replace sinv and cosv with the actual ratio based on the triangle

[gilded ore]

and sin^2 + cos^2 = (a^2+b^2)/c^2=1

how do i make this nice to look at`^^

[astral hornet]

move c^2 over

[gilded ore]

There is a bot that can show it nicely

i dont know the command

[hard gale]

$aaaaaaaaaaaa$

[somber coyote]

emeric75:

[hard gale]

summoned

$\sin^2(u)+\cos^2(u) = \frac{a^2+b^2}{c^2} = 1$

[somber coyote]

emeric75:

[gilded ore]

$sin^2 + cos^2 = \frac{a^2+b^2}{c^2} = 1$

[somber coyote]

Phil:

[gilded ore]

yeet

[astral hornet]

after you move c^2 over you get a^2+b^2=c^2, so substitute the value of c^2 into (a^2+b^2)/c^2=1 and you end up with 1=1

[gilded ore]

what about for v = <0,90>

I understand

this can only be applied from values between 1 and 89 due to the pythagoras theorem only being usable on triangles with 90 degrees

[upper karma]

Hey I'm confused about the dumbest question ever in trigonometry

why does this happen

my textbook here says "prove the following identities" and it says this 1/(1 + tan(x)) + 1/(1 - tan(x)) is congruent to tan(2x)/tan(x)

why would they use congruent instead of equal to?

please @ my when you're here. Thanks

[slim goblet]

hi

[upper karma]

hello!

errr, are you here to answer my question?

[slim goblet]

no lol, im here to ask my own

sorry

this is gonna look very basic but i dont get how to solve this problem, i found the answer by guessing (x=28 not 35, but i have to find the length of that entire line)

[upper karma]

@slim goblet

can you send the entire question?

[slim goblet]

yea

sorry, was on my bus

@upper karma

[upper karma]

thanks you

thank*

[dire rampart]

@upper karma combine the 2 fractions and simplify

then compare it to tan2x/tanx

[upper karma]

umm that's not my question

why did they use "congruent to" instead of "equal to"

lmao I can solve it but I'm confuesed by that

confused*

also @slim goblet

you see that tangent line

[slim goblet]

yea

[dire rampart]

oh my bad I misread

[upper karma]

what angle does that make?

[slim goblet]

it doesnt say

otherwise i could just use sine cosine or tangent

[upper karma]

ummm the rule is

it's 90 degress

[dire rampart]

I guess it's just another way of saying "it's the same thing but written differently:

[slim goblet]

oh thats what you mean

yea, its 90 degrees

i thought you meant what angle does it create when it connects to the darker line

[upper karma]

so what else can you find?

@slim goblet

[slim goblet]

idk, the radius equals x but idk how i would solve it from; 21^2+x^2=(x+7)^2

[upper karma]

have you learned quadratics yet?

[slim goblet]

$21^2+x^2=(x+7)^2

yea

a long time ago

barely remember it

[upper karma]

$21^2+x^2=(x+7)^2$

[somber coyote]

🅱мерть - единственная константа:

[upper karma]

so here

you can expand the left side

do you know how to distribute?

[slim goblet]

to yea

[upper karma]

what would you get?

[somber coyote]

Smig has 2 dads:

[upper karma]

ummm

expand that

[slim goblet]

more?

[upper karma]

yes

[slim goblet]

a girl in my class solved it with the pythag theorem

[upper karma]

$x^2 + 14x + 48$

[somber coyote]

🅱мерть - единственная константа:

[upper karma]

is the expanded form

[slim goblet]

oh

[upper karma]

49*

[slim goblet]

yea

[upper karma]

sorry

now you can cancel the two $x^2$

[somber coyote]

🅱мерть - единственная константа:

[slim goblet]

okay

yea

i think i can solve from here

[upper karma]

you probably can lmao

[slim goblet]

yea

but that gets me 28.07 something, would you know how to solve it with the pythag theorem?

because that got them a clean number (28)

[upper karma]

I don't knwo sorry

know*

[slim goblet]

damn, alright

[thick magnet]

How do I do this?

[upper karma]

@thick magnet

use trig ratios

0.52359877559

is pi/6

[thick magnet]

What do you mean?

By trig ratios?

[upper karma]

have you learned trig yet?

and also is it pi/6 radians?

[thick magnet]

I’m in algebra II right now

[upper karma]

ummm

I'm guessing that means no

[thick magnet]

I think that’s next year

[upper karma]

errr

I don't know how to solve without trig

sorry

[thick magnet]

In my school I think it’s combined with precalc

Could you briefly explain

If not, it’s fine

[upper karma]

sin = opp/hyp cos = adj/hyp, tan = opp/adj

those are all the things you need for the question

wait

@slim goblet

come back!

[slim goblet]

yuh

[upper karma]

I got a better solution than pythagorean theorem

do you know any circle theorems?

[slim goblet]

for example?

[upper karma]

this one is the most useful for this question

[slim goblet]

oh lol, im in geometry one

[upper karma]

so basically

the theorem states that AB^2 = AC * AD

$AB^2 = AC * AD$

[somber coyote]

🅱мерть - единственная константа:

[upper karma]

I love latex

[slim goblet]

in this case what is a,b,c and d

[upper karma]

ummm

can't spoon feed you

just look at the image I sent you

AC is the secant

[slim goblet]

okay, is AB the 7 ft one, bc the dark long one, cd the dotted long one and AD the dotted short one?

[upper karma]

no AB is not 7 ft

[slim goblet]

this?

[upper karma]

?

[vagrant elk]

questionable

[slim goblet]

our teacher taught us this,

$AC=X AB=Y and B*C=Z$

oops

wait a sec

[upper karma]

yeah

I guess

[slim goblet]

$AC=X^2, AB=Y^2, and B*C=Z^2$

[somber coyote]

Smig has 2 dads:

[upper karma]

but just look at the image

wait what

[slim goblet]

this

oh

[upper karma]

just ignore that for now I don't knwo what you're talking about

yeah that image

[slim goblet]

okay

[upper karma]

you see how AB is tangent

[slim goblet]

yea

[vagrant elk]

oh boy circle theorems

[upper karma]

$AB^2 = AC * CD$

[somber coyote]

🅱мерть - единственная константа:

[upper karma]

I hate them too @vagrant elk

https://cdn.discordapp.com/attachments/326138757474680852/570726372662837309/Photo_on_4-24-19_at_3.png

going back to that

[vagrant elk]

t b h

[slim goblet]

so 21^2=7*x

[upper karma]

7x = AB^2

HOWEVER

[vagrant elk]

ye

[upper karma]

AB = 21

21^2 = 441

441/7 =

wait what

[slim goblet]

63

[upper karma]

I got 63

[slim goblet]

thats very wrong

[upper karma]

are you sure the answer is 28?

[slim goblet]

yes

you can check it

[upper karma]

I trust in my circle theorems

[slim goblet]

i dont

[vagrant elk]

I mean, what dost thou mean "you can check it"

[upper karma]

lmao

I can't check it

Darkrifts you believe in my theorems right?

[slim goblet]

try $21^2+28^2$

[somber coyote]

Smig has 2 dads:

[upper karma]

where tf did you get 28

[vagrant elk]

same

[slim goblet]

thats the radius

[upper karma]

no pythagorean theorem

yeah but

we're trying to solve for the radius

[slim goblet]

yea

but 28 works for the radius

[vagrant elk]

You don't have the radius, so you shouldn't have 28 I think

[upper karma]

why don't you just say the answer is 28

I'm 99% sure the answer is 63

[slim goblet]

its not lol

[upper karma]

says your etacher

[vagrant elk]

I mean, do you have something which says the correct answer?

[slim goblet]

i trust pythag theorem more than something i havent learned

no i dont

[vagrant elk]

hmm

[slim goblet]

this is tonights homework

[upper karma]

you can't use pythagorean theorem

[vagrant elk]

brb gonna actually look at the problem

[upper karma]

you can't even use trig

[slim goblet]

plus multiple people . got 28

[upper karma]

this is most likely a circle theorem problem

they might be wrong lmao

[slim goblet]

why cant i use pythag? i have a right triangle with 2 sides

[vagrant elk]

Yeah no you can't use 28 @slim goblet
28 gives you the distance from that point connected to the 7 and the point connected to the 28

[slim goblet]

yea

[vagrant elk]

Which isn't the radius

[slim goblet]

yea it is

what do you mean

[vagrant elk]

No, that's a line near the circumference

You're finding the wrong hypotenuse

[slim goblet]

what?

[slim goblet]

yea

[vagrant elk]

You can't find the radius by pythagorean theorem

[slim goblet]

my teacher said that the point in the middle is the center

yea but thats how im checking it

not how i solved it

[vagrant elk]

How'd you solve it?

[slim goblet]

quadratics

[vagrant elk]

How so

[slim goblet]

@upper karma actually did that with me

[upper karma]

I was just going along with ya

[vagrant elk]

What you should solve for is $x^2 + 21^2 = (x+7)^2$

[upper karma]

how did you even get that quadratic equation

[somber coyote]

Darkrifts:

[vagrant elk]

maybe if I can read the paper right

[upper karma]

?

[slim goblet]

yea

thats what .i did

i think

[upper karma]

how did you even get that quadratic equation

[vagrant elk]

no fucking clue I just pulled it out of literally nowhere

he sure went with it tho

[slim goblet]

using pythag

[upper karma]

oh shit

[slim goblet]

thats what pythag would be in this situation

[upper karma]

yeah you can do that but I prefer not doing quadratics

[vagrant elk]

Anyhow, what would solving that give you anyway?

[upper karma]

the circle theorem method makes mroe sense

the radius

@vagrant elk

[slim goblet]

28.07 i think

[vagrant elk]

I know, I'm gonna actually solve for it myself

for

a u t h e n t i c a t i o n

[upper karma]

ask one of the <@&286206848099549185>

[slim goblet]

$x^2+21^2=x^2+14x+49$

[somber coyote]

Smig has 2 dads:

[upper karma]

21^2 = 441

49 -441

[vagrant elk]

Funnily enough it looks to me like 28

[slim goblet]

hah

[vagrant elk]

h m m

[slim goblet]

your circle thing was wrong

[upper karma]

my circle theorems are not wrong

[slim goblet]

[vagrant elk]

Either that or misapplied

*or misremembered

I don't memorize circle theorems because lmfao circles

[slim goblet]

probably a good thing i didnt trust something i didnt know

just remember to fear the unknown

[upper karma]

you trusted Darkrifts though

[vagrant elk]

^

honestly not the best call

because i have like 3 brain cells

[slim goblet]

yea but he doesnt use stupid font affects so he is more trustworthy

[upper karma]

using font effects are irrelevant

[vagrant elk]

they are in fact not irrelevant, just not very relevant

[upper karma]

let's ask @worthy igloo

he's I N T E L L E C T U A L

[slim goblet]

yea but intellectuals dont need font affects to communicate their thoughts

[vagrant elk]

I'm busy looking up the theorems for circles because I've never needed them

and to see if that one shows up

does it have a fancy name

[upper karma]

https://www.mathplanet.com/education/geometry/circles/advanced-information-about-circles

it's here

[vagrant elk]

I see the problem

You misapplied it Vmert

it's to an exterior point

we need the circle

This doesn't look like the center to me

Theorem's right, wrong application

q e d

[slim goblet]

YOURE SMART

[upper karma]

oh shit the 90 degrees only applies when tangent touches the radius

[slim goblet]

YEA

[upper karma]

what do you mean yeah?

you didn't even know that rule until I told you

[vagrant elk]

ke

"YOURE SMART"
i had to look this basic ass theorem up lmfao

and I had to be told what it was

[slim goblet]

im slow lol

compared to you guys, im retarded

[vagrant elk]

It took me hours to prove that if I can break apart any two objects, I can break apart 3

you aren't that bad

[thick magnet]

Oh wait I know that @upper karma

I just don’t know what to do with it

[upper karma]

I need to know

is is pi/6 raidans or just pi/6

[keen aspen]

what question

[minor arch]

Can someone help me understand the logic for the range of Trigonometric Functions? It's easy enough to memorize, but I think it'll serve me better if I have a full understanding of the subject matter.

how did they get it so that:

-1 <= y/r <= 1

???

r = distance. and the distance, I thought, was always a positive.

[vagrant elk]

@minor arch in this case, r is the distance found by the pythagorean theorem and, while it is always positive, y is not

[minor arch]

Ah. So the -1 comes from divided the pythagorem theorem by the distance? @vagrant elk

[vagrant elk]

Nope

the -1 comes from the fact that the y coordinate may be negative

[minor arch]

Ahhhh. Okay, makes sense now.

Perfect sense actually. since y can't be greater than r.

ty.

[soft condor]

hey can i post pics in here

[vagrant elk]

scrolls up literally like 8 messages to picture

yes

[soft condor]

https://cdn.discordapp.com/attachments/568257939002097675/570773627125891085/unknown.png

im sorry i have no common sense

[vagrant elk]

Well, I'd start by finding the radius of that circle

and good ol pythagoras shows up to steal every other theorem's thunder

[soft condor]

im sorry i dont wanna sound mean or harsh but can you please give me the answer

[molten kayak]

there is no better way to guarantee that someone doesn't just give you the answer than saying what you just said

[vagrant elk]

wow ok

[soft condor]

its just ive had a very eventful day

kinda stressted

[minor arch]

Giving the answer just ensures you won't learn. Find the radius, first, and check here to see if you actually got the correct radius.

[soft condor]

welp figured it out its 15

right?

[molten kayak]

looks good

[vagrant elk]

ye

i can see them but not read them

9, 14.7, 20.5

[thorn sorrel]

not 100% sure if this is the right channel, but if anyone could help me that would be great.
how do i rotate an image 90 degrees clockwise about a point that isnt the origin?

for example, how do i rotate the point (4,7) 90 degrees clockwise about the point (4,1)

[vagrant elk]

Well, for this specific case, you can use graph paper if that's your preferred method and visually do it, but uh
They have the same x coordinate, so you know they end on the same y coordinate
The point is 6 units away from (4,1), so it stays that far away
and you should end up with (10, 1) i believe if I'm doing it right in my head

You know you could just use translations and move the points you're rotating around to the origin, find where your important points land, then move it back

[thorn sorrel]

i know i can but my teacher forces us to rotate them :/

what if they werent on the same x coordinate?

like what if im rotating (7,6) by the point (4,1) 90 degrees clockwise

[vagrant elk]

then you translate to the origin, rotate it, and then move it back before anybody notices

or, since this clearly isn't super advanced geometry, use graph paper

[thorn sorrel]

im kind of dumb, i know how to rotate 180 degrees but not 90 degrees

what are the steps to rotating 90 degrees by the origin then

[vagrant elk]

smh
so, rotating 90 clockwise you can do $(x, y) \mapsto (y, -x)$ if I remember correctly

[somber coyote]

Darkrifts:

[thorn sorrel]

ok i got it, thanks

[vagrant elk]

If you want, you can include the translation in that definition by doing something like

including the translations, then applying that rotation and bundling it up at once

[harsh urchin]

Is this correct? https://i.imgur.com/2v7ecrD.png

7.5 length of flag pole

[upper karma]

Yes mam

[soft condor]

halp meeee

[devout shell]

use this theroem

Let me get the image really quickly

this one is the relevant formula:

[upper karma]

$36=16+4x$

[somber coyote]

Meliodas:

[upper karma]

https://i.imgur.com/CkgjP7G.png

[devout shell]

confused or something about it?

[upper karma]

Yes, how are they distributing here, and canceling the cos(theta).

If that makes sense.

[devout shell]

can we see original expression

[upper karma]

The original expression: tan^2(theta)cos(2theta)

[devout shell]

$\tan^{2}(\theta)\cos(2\theta)$

[somber coyote]

⚡Amphy⚡:

[upper karma]

Correct.

[devout shell]

have to look up identities, ew lol

reduce it how though?

a certain expression they wanted?

[dark sparrow]

rewrite cos(2θ) as 2cos^2(θ) - 1

[upper karma]

They wanted it reduced to the answer above. I attempted to use the 3rd ford of cos(2theta) = 2cos^2(theta)-1, but when multiplying it out, I'm only left with, sin^2(theta), not tan^2 trailing behind.

form*

[dark sparrow]

you would get tan^2(θ)(2cos^2(θ) - 1)

= 2tan^2(θ)cos^2(θ) - tan^2(θ)

@upper karma

[upper karma]

Thank you.

[oak minnow]

how do I do these type of questions?

[devout shell]

you know any equations for this?

[oak minnow]

I haven't done these questions yet

I dont know how to ;/

all i know is that it has something to do with pi

[devout shell]

you can do this by converting degrees to radians so then arc length is equal to r * angle

or if you want to use degrees, you will need to do a proportion to solve it

[ember condor]

Oooh an active person

Sorry for off topic but can i get help in #prealg-and-algebra

[devout shell]

you have to post it first lol

[dark sparrow]

@oak minnow do you know what an arc is?

if you know what an arc is, then it should be quite obvious for you how to do the question

[oak minnow]

a circle

[dark sparrow]

no, an arc is not a circle.

[oak minnow]

erm

it sorta is

[dark sparrow]

no

[idle portal]

is it? :o

[devout shell]

arc a rounded line segment

[oak minnow]

its the edge of a circle itsn't?

[ember condor]

Yeah sorta

[dark sparrow]

an arc is a part of a circle.

but it is not itself a circle.

[ember condor]

An arc is part of the CIRCUMFERENCE

[oak minnow]

yeah

the border of the arc?

so... how do I do the question, can i get some help

[dark sparrow]

okay so like

let me ask you another, hopefully simpler question

what is the length of a 180° arc of a circle with radius 10 cm?

(remember that a full circle is 360°)

[oak minnow]

i actually dont know since i havent done arc problems before

90?

[dark sparrow]

how did you arrive at that

...please don't tell me you just guessed

[oak minnow]

yeah i guessed it sorry

[dark sparrow]

also, why don't you think instead of going "oh i haven't done this before"

a 180° arc is exactly half of a circle.

[oak minnow]

yeah

[dark sparrow]

are you able to find the circumference of the whole circle?

[oak minnow]

because you said a full circle is 360

[dark sparrow]

are you able to find the circumference of the whole circle?

[oak minnow]

thats what i dont know how to find...

im not smart sorry

[dark sparrow]

what do you think pi is?

[oak minnow]

erm

it is the ratio of a circle

[dark sparrow]

the ratio of what?

a ratio is always between two things.

[oak minnow]

diameter

and circumference

??

[dark sparrow]

so say it once more

what is pi

[oak minnow]

pi is the ratio of the diameter to its circumference of a circle

[dark sparrow]

no, that's the wrong way around

it's the ratio of the circumference to the diameter.

$C = \pi D$

[somber coyote]

Ann:

[dark sparrow]

so now, do you know how the diameter relates to the radius?

it should be obvious if you know what the diameter and radius are and have the appropriate mental image in mind

[oak minnow]

circumference = pi x diameter

[dark sparrow]

that is true, but no, that is not what i am asking you right now.

[oak minnow]

what did u ask

[dark sparrow]

why don't you read it once more?

[oak minnow]

no

i dont know

because you havent told me

[idle portal]

'so now, do you know how the diameter relates to the radius?'

[dark sparrow]

here's a circle, with a radius and a diameter drawn inside

can you tell me which is which?

[oak minnow]

oh

so the line that goes through the circle is the diameter

[dark sparrow]

yes, so now, with this picture in mind, how does the diameter relate to the radius?

[oak minnow]

erm

[dark sparrow]

it doesn't take an expert to recognize the relationship, both because it is crystal clear from the picture and because it is not particularly complicated per se

[oak minnow]

because the radius is the distance from the center of any point of the circle

?

[dark sparrow]

that is true but it is irrelevant to what i am asking you right now

[oak minnow]

what tf

you asked me how do they relate

[dark sparrow]

okay, let me put it this way: how do the lengths of the diameter and the radius relate?

if you know the radius of a circle, is it possible to know its diameter? and conversely, if you know the diameter, is it possible to know the radius?

[oak minnow]

noz

no

because the radius

this isn't helping

i asked how to do these type of questions

[dark sparrow]

patience, my friend.

we'll get to your original question in a few minutes.

i want to help you reason and know why things work the way they do. if all you know is how to apply formulas, then you're little more than a meat calculator.

anyway, what you said is false. it is very much possible to find the radius from the diameter and vice versa.

[oak minnow]

how would you find it

$C = \pi D$

[somber coyote]

Fuhrer1:

[oak minnow]

with this

[dark sparrow]

no, we are not talking about the circumference right now.

we are talking about just the radius and diameter for now.

here's that picture i sent before, cropped for convenience.

notice how the green diameter passes through the center (the black point)

the center cuts the diameter into two segments.

[oak minnow]

yes ok

[dark sparrow]

and each of those two segments is a what?

[oak minnow]

diameter

[dark sparrow]

no

[oak minnow]

radius

[dark sparrow]

yes, each of those is a radius

the center cuts the diameter into two radii.

do you see it now?

[oak minnow]

i think so

[idle portal]

:o

[dark sparrow]

as an equation, this would be D = 2r. but i want you to NOT memorize that.

[oak minnow]

why not

[dark sparrow]

because you now know exactly why the diameter of a circle is double its radius

(and the radius is thus half the diameter)

[oak minnow]

yeah\

[dark sparrow]

and it should make perfect sense to you, thus obviating the need for memorization.

so with that in mind, we can now return to talking about the circumference

$C = \pi D$

[somber coyote]

Ann:

[dark sparrow]

we had this

with what we said above in mind, we can rewrite this as follows:

$C = 2\pi R$

[somber coyote]

Ann:

[oak minnow]

2 pi r

thats a full circle

[dark sparrow]

yes

that's the circumference of a full circle.

[oak minnow]

the border

[dark sparrow]

also, quick note:

"circle" refers specifically to the curve, as shown on the left. the inside of a circle is called a disk, as shown on the right. though this might be less common at your level. but the distinction between the two needs to be made.

anyway

what i want you to do now is to find the length of a 120° arc of a circle with radius 20 ft.

again, remember that a full circle is 360°.

you now know everything that is needed to do this.

[oak minnow]

no not really

whats the formula given

[dark sparrow]

don't use any formulae!

[idle portal]

t!cookie Ann#8192

[loud cedar]

🍪 | Vulcan213, please mention a valid user!

[oak minnow]

i dont know man

[idle portal]

:(

[dark sparrow]

okay

[oak minnow]

wait

120 x 20pi/360

[dark sparrow]

almost.

[ember condor]

divide 120 by 360

[dark sparrow]

the circumference is 2 * 20pi.

[ember condor]

i think?

no wait

[oak minnow]

what

[dark sparrow]

also don't use x for multiplication

[ember condor]

The ratio of the ARC to the CIRCUMFERENCE (2rpi) is the same as the ANGLE to 360.

[oak minnow]

ok

[ember condor]

That is the only thing you need to know

Try turning that into a formula

[dark sparrow]

i said the radius is 20 ft.

[oak minnow]

120 * 20pi/180

[dark sparrow]

sure. that works.

i would have preferred if you wrote $\frac13 \cdot 2 \cdot 20\pi$, but your answer is correct.

[somber coyote]

Ann:

[ember condor]

Woo hoo

Keep practicing you'll get there

[oak minnow]

why 1/3

[ember condor]

Props to you for asking for help

[dark sparrow]

because a 120° arc is exactly 1/3 of a full circle

[ember condor]

120/360=1/3

[dark sparrow]

^

[ember condor]

Put the angle over 360, and multiply that by the circumference

[oak minnow]

we times it by 2 because

[dark sparrow]

because the circumference of a circle is 2πr.

[ember condor]

The diameter is 2 times the radius

[dark sparrow]

or that.

[ember condor]

Pi is the ratio of the DIAMETER to the circumference, not the radius.

[idle portal]

wot, no

[dark sparrow]

other way around.

[ember condor]

yeah sorry

Which is funny because we use radius-es a lot more

[dark sparrow]

circumference to diameter

also, radii

[ember condor]

I know

[dark sparrow]

one radius, two radii

[ember condor]

That's why i said radius-es and not radiuses

[oak minnow]

2 radius

[dark sparrow]

the plural of radius is radii

pronounced "rady-eye"

Latin

[ember condor]

Per my last message i know that

[ember condor]

hmm

[oak minnow]

so back to this

[ember condor]

Yep

[oak minnow]

would it be

[ember condor]

the Angle, you have

[oak minnow]

105/360

[ember condor]

Yes

And then 2pi(r). We have r

[oak minnow]

2/24

[dark sparrow]

huh?

[oak minnow]

2/24 *2 * 8pi

[dark sparrow]

wait wait wait what

where are all these numbers coming from lmao

[idle portal]

magic

[ember condor]

that's my line

[dark sparrow]

$\frac{105}{360} \cdot 2 \cdot 25\pi$

[somber coyote]

Ann:

[oak minnow]

yeah

i put 8

was a mistake

[ember condor]

I think that was just arith

[soft condor]

Anyone up for a constructed response

[devout shell]

There is Theorem you can use

Let me get it quickly

[frigid wasp]

anyone know how to find the radius of circle centre q?

[jaunty horizon]

https://i.imgur.com/Tozi0Oq.png

use trigonometry

@frigid wasp

[frigid wasp]

ye but then what do u do

i got that much

non calc btw

[jaunty horizon]

ik

you want to get the length of the opposite side and you already have the angle and the adjacent side

What do you use? sin cos or tan?

[frigid wasp]

how do u have the adj

14 is the radius of the big circle

but tan

@jaunty horizon

[stoic wraith]

Hey guys, practicing for my final exam, I'm having a bit of trouble grasping trig. I'm trying to find the length of a side of a right triangle given the length of the adjacent side. In this case, the hypotenuse. I assumed that I need to set the cosine equal to the adjacent side's length over some variable to represent the hypotenuse's length, but i can't figure out how to get that variable alone now. Any ideas? I think I might have just fried my brain lol.

[upper karma]

@stoic wraith

You have hypotenuse.

You have a degree.

$ \cos{x} = \frac{adjacent}{hypotenuse}$

[somber coyote]

rudy:

[upper karma]

where x is a degree.

Now

This is a fraction = to a fraction

Just cross multiply

cos x * hypotenuse = adjacent

[stoic wraith]

Am I able to send images in this channel? I am reading your messages by the way.

[upper karma]

Yes, send an image

[stoic wraith]

This is what I'm working with, for exam practice. I'm being asked to find the length of x. https://www.sebz.xyz/i/firefox_April_25_2019_EminentIggypops.png

[upper karma]

Perfect.

So you have the adjacent side, finding for x

what function has that?

[stoic wraith]

I thought i was cosine, right? Adjacent over hypotenuse?

[upper karma]

Yeah😎

$\cos{38} = 20/hypotenuse$

[somber coyote]

rudy:

[stoic wraith]

Precisely, that's what I've got on my notes here.

[upper karma]

,w cos(38) in degrees

[somber coyote]

Display results online and refine query

Requested by rudy#0446

[upper karma]

gusssing ur using degrees

[stoic wraith]

Yes, it wants my answer in degrees, good guess. 😃

[upper karma]

You have, $54.72h = 20$

[somber coyote]

rudy:

[stoic wraith]

But in practice, to find that, would I divide both sides by twenty first?

[upper karma]

,w 20/54.72

[somber coyote]

Display results online and refine query

Requested by rudy#0446

[upper karma]

What do you mean?

[stoic wraith]

https://discordapp.com/channels/268882317391429632/326138757474680852/571125325279395841

I thought it would copy the message, my bad.

[upper karma]

Oh, no. You want to solve for the hypotenuse

[stoic wraith]

But anyway, from that point, without the help of wolfram or anything of that sort, just using a TI-84.

[upper karma]

So, cos(38)h=20

[stoic wraith]

OH.

Thank you!

[upper karma]

h = 25.380 or so

[stoic wraith]

I've got it now, you're the best.

Hoping I do well on this final 😓

[upper karma]

gl, trig is one of the most beautiful of maths

[stoic wraith]

thank you

[stoic seal]

hello

can someone help me with an equation trigonometry

cos^2 x+1=sin⁡x+2,x ϵ [0,2π]

im stuck

someone please

[stoic seal]

hello

[slow charm]

cos²x-1=-sin²x

[willow shoal]

can someone help me with math I have this assignment I need to get done ASAP so I can get a class next year its geometry and I dont understand the lesson, but if you can break it down and tell me the equation i'll be really thankful

[devout shell]

Post it

[willow shoal]

theres 4

[devout shell]

post it in a question channel down below then

[willow shoal]

any?

[devout shell]

yea

[willow shoal]

Alright I posted in a or whatever

[slow charm]

Alpha

[oak minnow]

how do I find the chord length of this

[oak minnow]

i thought it was 2rsin(0/2)

[devout shell]

you can do it using radian measure as well

but is said degrees so I posted this formula

[oak minnow]

yea

ok

arc measure is my final answer for a right

what would the circumference be

[devout shell]

arc measure is what they want

you are given a radius...

[oak minnow]

30cm

[devout shell]

circumference is 2pi * r...

[oak minnow]

oh

i forgot about pi for a second

im not getting it still

[dark sparrow]

DON'T USE ZERO FOR THETA WOW

[oak minnow]

THEN WHAT DO I USE?

[vague pagoda]

Theta ofc

[dark sparrow]

if you can't type θ then at least write theta

[oak minnow]

@devout shell

[supple abyss]

Ø

[oak minnow]

95.5°/360° = 50/188.5

do I cross multiply

[devout shell]

explain how you got that proportion

[oak minnow]

which proportion

[devout shell]

the one you are asking me about

[oak minnow]

oh to cross multiply?

because the answer is meant to be in ° form

[devout shell]

the right hand side should be a ratio of the arc length to circumference to circle

the units are cm

[oak minnow]

idk

[devout shell]

the units are given to you in the problem and they are cm, then you gave me a proportion with no unknowns and you want to solve for something? Solve what though

[oak minnow]

i dont know how to solve for chord length

in simpler terms

[devout shell]

it's x/360 = 50/188.5, so solve that for x

[oak minnow]

wait so were not meant to imply the answer for a in x

[dark sparrow]

what problem are we doing right now

are we doing 2a or 2b?

[oak minnow]

https://cdn.discordapp.com/attachments/326138757474680852/571186636428279808/Screenshot_2019-04-26_at_13.50.32.png

2b

yeah that gives me 95.5

i want the chord length not the arc length

[dark sparrow]

what

you're given the length of the chord.

[oak minnow]

radius of the chord length

the radius is 30cm

[dark sparrow]

"radius of the chord length" makes no sense

you're asked for the angle made by that chord

[oak minnow]

yea thats what i dont know how to get

https://cdn.discordapp.com/attachments/326138757474680852/571187824880254987/unknown.png

[dark sparrow]

well do you know how to get the chord length from the angle and radius?

i.e. if you have a chord subtending an angle of θ in a circle of radius r, are you able to find its length?

[oak minnow]

no

i dont have the angle

[dark sparrow]

no! no! no! no! no!

forget about the problem for a moment!

i asked you a question!

if you have a chord subtending an angle of θ in a circle of radius r, then are you able to find its length?

this has nothing to do directly with the problem you posted

but i want to make sure you have that down before attempting the reverse

[oak minnow]

yes, i think i know how to find it

[dark sparrow]

well

how do you find it, then

[oak minnow]

centeral angle/360 = arc length/2pi*r

[dark sparrow]

that's... not...

DO YOU EVEN READ THE MESSAGES I SEND

BECAUSE IT SURE AS HELL SEEMS TO ME THAT YOU DON'T

[oak minnow]

jesus no need to yell

[keen aspen]

:AngeryAnn:

[dark sparrow]

you do realize it's really fucking frustrating when this happens, right

when i'm trying to help and you do your best to NOT COOPERATE

[devout shell]

I've given up for the night as well

[flint compass]

I used to think that a period of a sine function was a ratio between 2pi and the number of revolutions it makes. But I realized that it couldn’t be divided by the number of revolutions because a normal sine function’s period is 2pi/2pi with only one revolution (not 2pi revolutions) so what does x in 2pi/x represent?

[dark sparrow]

revelations?

[flint compass]

What do you mean by that?

[dark sparrow]

are you sure you didn't mean "revolutions" or something

i am having trouble parsing your question

[flint compass]

Oh, yeah, sorry I made a typo

I fixed it

[dark sparrow]

But I realized that this didn’t make sense to what does x in 2pi/x represent?

this does not parse

[flint compass]

What do you mean by parse?

What part does not parse?

[steady sleet]

Parse = does not make sense, it is an incoherent sentence

And they quoted the entire part that does not make sense

[flint compass]

I edited the question.

[steady sleet]

It's still not really clear what you mean imo

Perhaps draw a picture?

[flint compass]

I’ll just rephrase my question. Is the period a ratio?

[past mantle]

This picture is triggering.

[flint compass]

So is the period just number of oscillations in one revolution?

[dark sparrow]

no, the period is the amount of time it takes your sinusoid to complete one cycle

[winged saddle]

yo i dont rly understand why there is 2 solutions

isnt there just 20x=0+2pi n

why is there also 20x= pi+2pin

ok nvm

i need to think more

because sin a = sin(180-a)

k IM DUMB

[wet moth]

Hi guys

Anyone wanna help me with this?

[olive solar]

so you want to invert the equation $A = \frac12s^2$

[somber coyote]

collouarfloride-a brown stain:

[olive solar]

so that s is a function of A

what do you think would be the first step here

[wet moth]

switch sides?

then multiply both sides 2 i think

2 x 1/2 s = s^2 and A becomes 2A

@olive solar

[olive solar]

yes

okay

[wet moth]

so the answer is C?

[olive solar]

no

so you have $2A = s^2$

[somber coyote]

collouarfloride-a brown stain:

[olive solar]

now what would you do from here to get rid of the exponent on s^2

[wet moth]

s = sqrt of 2A?

[olive solar]

yes

[wet moth]

thank you

[olive solar]

👍

[wet moth]

brain's not working today sorry for the trouble

[olive solar]

np

[devout shell]

what theorem did you use to get your answer?

perfect, and you followed the procedure and you got the correct answer lol

12.00 if you want to put decimal places

there are probably other problems where you have to round

but this one, was nice to you lol

they try to get you like that lol

[upper karma]

i need help

[devout shell]

post it

[upper karma]

if u have x,y and lets say u have 6 and -2 which sides do they go

[devout shell]

you describing a triangle or something? please show a problem

you mean like: y=mx+b?

and you have a point (6,-2)?

[upper karma]

the last part yea, im trying to help my friend with math but i never remember which side of the coordinates they go in its like f(x) = 2x -6 for example

[devout shell]

well you know what f(x) means?

or are you only familiar with the y=x?

[upper karma]

more familiar with y=x

we didnt have f(x) in school i think

[devout shell]

do you know what f(x) means?

[upper karma]

no

[devout shell]

it just means, where ever you seen an x in this equation, plug in the input value there
for example f(1) means, where ever there is a x in my equation, put in a value of 1 for x
like if f(x)=2x-6, then f(1)=(2 * 1) - 6

[upper karma]

ohhhhhh

[devout shell]

so as a quick check, if f(x)=2x+x^2, what is f(3)?

[upper karma]

can i write the way i calculate it

im used to doing the whole thing

[devout shell]

just write but don't simplify so that I can see if you got it right

[upper karma]

whats the ^

[devout shell]

means exponent, but that's not very important to know about

just show me what f(3) looks like

[upper karma]

f(x)= 2 * 3 + 3^2

[devout shell]

perfect

so just 3 in a three where there is an x

[upper karma]

that makes sense

do you think you could help my friend as well if i make a group

because sometimes i struggle myself

[devout shell]

he not in this discord?

[upper karma]

he is

oh

not in this

but he has discord

[devout shell]

are you familiar with the terminology: variable, input, and output?

[upper karma]

not sure, english isnt my first language and I had it all in finnish

[devout shell]

oh man lel

[upper karma]

yep

[devout shell]

but invite your friend to this server so he can DM without having to send me a friend request

[upper karma]

okay

[devout shell]

ah gosh, you need to be friends to make group DM lol

[upper karma]

he can just dm u

ill invite him

[devout shell]

but you want to know about (x,y) stuff?

[upper karma]

well im mostly just trying to help him

but I like maths

[devout shell]

haha, nice

[upper karma]

yeah

when i had geometry in school I did the graph the opposite way

I got half point and the teacher asked if I was left handed

[devout shell]

just invite him over then and I'll let him read the stuff I told you first then I can explain the original question

[upper karma]

im gonna say tho; hes not very good at math

and im not very good at teaching

[devout shell]

you both learn lol

[upper karma]

yeah

[devout shell]

it's a slow day today anyways, so I have time to sit around for a while and do this

tell him to go to #help-1 and ping me there

[upper karma]

sweet

oh okay

thank you my dude

he has to wait 10 mins to be able to talk btw and Im still not sure if u have a graph which side of the graph the numbers are meant to be, are they always on the same side? @devout shell

[devout shell]

how well can you read English?

[upper karma]

i can understand it pretty well but school subjects are out of my league in english like complicated stuff

[devout shell]

https://www.mathsisfun.com/data/cartesian-coordinates.html

try reading this and I can explain it more if you have trouble

[upper karma]

thanks

for me i just get confused sometimes with like the side

so its why it makes it also harder for me to help

[devout shell]

they explain the standard way to draw it

I'm left handed as well but it never really was a problem to draw stuff oriented to the right lol

[upper karma]

it was a problem for me

i dont know why my brain has hard time analyzing it

[devout shell]

it's just a standard convention you will have to get used to then

[upper karma]

yeah

thank you

[hollow hamlet]

uh

[devout shell]

just scroll up a bit

[hollow hamlet]

What all do I need to read?

[devout shell]

from when your friend first sent a message to now

looks like a lot but should only take like 10 minutes max to read

[hollow hamlet]

Nonsense

It should take 2 minutes max

maybe 3

4 at the very most

Anyway..

what am I looking for

when reading this

[devout shell]

just start here at this message and read because I go over what f(x) means

[hollow hamlet]

oh

ouch

light mode

you must hate your eyes q-q

anyhow time to read

[oak minnow]

3b

is it 1/2*r^2(theta-sin(theta))

[devout shell]

have you referred to your math book at all for help?

[oak minnow]

I dont really understand it

[dark sparrow]

@oak minnow are you still here?

[spiral lintel]

How do you solve the area of two overlapping circles with radius 4 and they go through each other's centers

[dark sparrow]

the what area?

the area of the overlap?

[upper karma]

Help how do i approach this kind of problems

[dark sparrow]

implying this problem admits a clear-cut classification of any kind

honestly play around with it

experiment

[umbral snow]

"irregular hexagon"

Is there such a proper thing?

[dark sparrow]

yes?

irregular = not regular

[umbral snow]

But then is it really a hexagon?

[dark sparrow]

why wouldn't it be?

[vagrant elk]

it has 6 sides

ergo

he e x a go on

[dark sparrow]

would you consider this to not be a hexagon?

[umbral snow]

Well then I'm a Kaynebegon

[oak minnow]

@dark sparrow yes, im here

[dark sparrow]

okay, good.

from the textbook screenshot you posted, what is the first sentence you do not understand?

[oak minnow]

i dont know how they got 2pi/3