geometry-and-trigonometry

[upper karma]

@pure cape
How can I reach the answer tho?

The correct procedure.

Its thales theorem tho

[pure cape]

Yes

Thales theorem is involved

But you can find a and b

Currently occupied and wrong topic

[upper karma]

hey

[upper karma]

Okay better question.

Can you guys please show their procedures?

[obtuse tapir]

what procedure?

[ivory wraith]

Im so lost i multiplied 200 rpm by 2pi then multiplied that by the radius 60 and now im stuck

[pallid dome]

You're given that side TW is congruent to side GD, so that's one statement. The next could be that angle DGH is congruent to WTP, then for a finish-off, side PT is congruent to side HG!

[upper karma]

ty

[pallid dome]

Of course!

[upper karma]

It wasn't that I didn't know that I just forgot how to word it but still thankyou

[pallid dome]

I got that the linear speed is 5184 MPS, however I think this is DEFINITELY wrong lmao

But maybe not

[ivory wraith]

oh its ok i got the answer

and ya thats very far off

[pallid dome]

LMAO

Anyway that's good

[upper karma]

@quasi sleet please don't multipost, please read #❓how-to-get-help .

[empty cairn]

Can someone help me with this page please?

[upper karma]

show your work.

[wintry tundra]

@empty cairn what exactly do u need help with

[empty cairn]

2-5

also if u can double check 1 for me

[slow elbow]

HOw do i solve this <@&286206848099549185>

[twin prawn]

Hi Ray from Promised Neverland

[slow elbow]

hi

I solved it

I just found the perpendicular slope which was -2 and found b which was 6 so it was -1, 6

[twin prawn]

🎉

[green osprey]

im tryna not to die

so like there is a triangle

QSR

SQ is 33 mm

SR is 65mm

s is a 90 degree angle

find angle Q

[green osprey]

<@&286206848099549185> please 😭

[sleek depot]

Do you know trigonometry

Or the pythagorean theorem

[green osprey]

i know both

i need to use an inverse trigonemtric function

but i triend tan^-1(65/33

[green osprey]

<@&286206848099549185> please

[sleek depot]

Yeah thats correct

[green osprey]

damn ok thanks

I ated the mods twice and still got no responce lmao

[sleek depot]

Yea the helpers never respond

Answer is approx 63 degrees

[green osprey]

tysm

[upper karma]

Tyty in advance

[sullen cipher]

prove triangles afb and afd are congruent

which is pretty easy

so you have AB = AD

and then triangle ACB and CED are similar

and so you have CB/AC = ED/CE

and then just do some switching sides

you get the answer

[hearty hedge]

Need help

[viral orbit]

I need help

[storm portal]

picture picture picture

[upper karma]

I don’t get this ?

<@&286206848099549185>

[vital nest]

is there any particular problem you're confused with

because with an angle and a side of your right triangles, you can just use your trig functions to solve for missing sides

or, better, since they're all special right triangles, you can use those

[upper karma]

Still not understanding wym

And I’m doing #2

[vital nest]

umm

think about it this way

[vital nest]

see the 45-45-90 triangle in the top left corner here?

that triangle is congruent to the one in question 2

so basically the x is just replaced by an 8

[upper karma]

Right

[vital nest]

aand I probably kinda gave away the answer there, but if you're still confused, lemme know

[upper karma]

Thanks xx

[vital nest]

glad I could help

[upper karma]

would that be SSS, SAS, ASA, or AAS? and also how would yu prove ACB~CED similarity?

[sullen cipher]

ACB CED is AA

there is a right triangle

and the big triangle is an isosceles which was determined by the previous proof

and for that

wait

... theres a flaw, bruh, i just need to prove that the afb and afd are congruent

wait there's no flaw

you can use ssa for afb and afd

since angle afb and angle afd is always larger than 90

theres no flaw

[upper karma]

#6

[grizzled lantern]

@upper karma what rules define quadrilaterals and parallelograms

[upper karma]

i just dont know how to do 12b+ 8 = 2a

but i know the rules that define it tho

[grizzled lantern]

could you possibly use substitution to solve?

for example solving in terms of a or b to find what one angle is

[upper karma]

tytytytytyty

[sullen cipher]

np

[rugged marsh]

Or also that complementary angles

With both angles containing b

[river brook]

can somenoe explain how to solve this hahaha

actually i just got it! 😊

[green osprey]

what would x be?

<@&286206848099549185>

pls

please

[obsidian moon]

use the intersecting chord theorem

[green osprey]

ok

also this

this hw assignment is fucking killing me

[obsidian moon]

look above at the image i provided with all the circle geo theorems

which one fits best

theres one where the diagram is literally identical

[green osprey]

ok

ill look through

its just all the lines are confusing lmao

would it be the equal cord one?

[vocal lodge]

help?

<@&286206848099549185>

[patent plume]

Please do not ping immediately

wait 15 min

[vocal lodge]

oh sry

[minor sail]

Could someone please help me in some geometry?

[patent plume]

uhhhh try one of the questions channels

[patent plume]

Alright, so what cna you tell about the triangles?

[minor sail]

Sure

[vocal lodge]

i think 2x-8 and x+6 are equal

[patent plume]

Mm no

[vocal lodge]

wait

i understand

[patent plume]

oh?

[vocal lodge]

2x-8>x+6

[patent plume]

mm not quite

[vocal lodge]

wat

dang

[patent plume]

close, but not quite

Alright, let's go about this a little more systematically

What sort of triangle is the triangle to the right?

[vocal lodge]

equilateral

triangle

wait

i flipped it

2x-8<x+6

ez

cuz 60>52

ez right?

am i right

[patent plume]

uhhhhhhh

[wintry tundra]

2x-8 = x+6 i thought

[patent plume]

Why do people think that?

[vocal lodge]

im pretty sure its 2x-8<x+6

[wintry tundra]

nah nvm im wrong

[patent plume]

and that's wrong too rip

I mean it people, pls let's go about it systematically

[patent plume]

Right, it's equilateral. So all sides are equal right?

[vocal lodge]

yes

so all the angles r equal too?

[patent plume]

So that means that one of the sides of the triangle to the left is x + 6

[vocal lodge]

so each angle is 60 degrees

[patent plume]

dw about the angles, although that is true

Jesus

[vocal lodge]

ya ok continue

sry bout that

[patent plume]

so what sort of triangle is the triangle to the left?

[vocal lodge]

acute isoceles

[patent plume]

Alright. So if one of the sides is x + 6, the other is x + 6 as well? (by the diagram)

[vocal lodge]

yeah

[patent plume]

So the triangle on the left is x + 6, x + 6, 2x - 8 right?

Now do you know about the triangle inequality?

[vocal lodge]

2x+12>2x-8

right

cuz they have to be bigger3

than that last side

[patent plume]

so like, that's not helpful. It means that there's no limit to how large x can be(or is there?)

so we go to inequality 2

[vocal lodge]

ok

[patent plume]

Since the triangle on the left is isosceles, what can you tell me about it's angles?

[vocal lodge]

its two base angles r equal?

[patent plume]

yes.

So you know that one angle is 52, and that the sum of the angles is 180?

[vocal lodge]

yes

180-52

then divide taht by two

[patent plume]

Yes

What do you get?

[vocal lodge]

64

[patent plume]

Right. So within a triangle, the largest angle will also have a side opposite to angle that is larger than the others

So 64 > 52 right?

[vocal lodge]

yes

[patent plume]

So x + 6 > 2x-8

[upper karma]

no

[patent plume]

Rip hi Hobo

[upper karma]

go on

[vocal lodge]

yes i see

[patent plume]

x + 6 > 2x - 8
-x > -14
x < 14

So we're done

[vocal lodge]

smart

[upper karma]

wait

[vocal lodge]

but i thought you need two inequlaityies

[upper karma]

why not triangular

[vocal lodge]

im tryna find a range

[wintry tundra]

4<x<14

[patent plume]

Now, to explain a couple of things,

1. you did say that x + 6 > 2x-8, but the reasoning isn't right
2. I used the triangle inequality cause it's usually a better inequality, but it doesn't work here rip

Oh rip I forgot about the reverse

[vocal lodge]

wat

[patent plume]

Man I should sleep

[upper karma]

because you dont know which one is the bigger one

[vocal lodge]

k im confused

[wintry tundra]

if x is less or equal to 0 it wont work since sides have positive length

so any x input that makes 2x-8 negative or 0 wont work

[patent plume]

^

[patent plume]

you need to read up lol

[vocal lodge]

oh okkk

makes sense

ty

[upper karma]

nah im good thanks :yaw:

[patent plume]

fair enough

[vocal lodge]

so the range is

0<x<14

?

[wintry tundra]

no

[vocal lodge]

why?

[wintry tundra]

4<x<14

2x-8 cannot be less or equal to 0

[cold siren]

can anyone explain this?

[wintry tundra]

we just did

look in chat history

[cold siren]

oh alr ty

[wintry tundra]

do u see how 2(4)-8=0?

[cold siren]

jesus are you like in my class

[vocal lodge]

lmfaooooo

[cold siren]

i-

[vocal lodge]

thats so funny

we got the same problem and i dont even know you

[cold siren]

im curious are you tho?

[vocal lodge]

nope

idk who u are

[cold siren]

makes sense

this isnyt my real name tbf

[patent plume]

2x - 8 > 0, otherwise that side length doesn't exist

2x > 8

x > 4

[cold siren]

the two inqualities would be 2x>8, x>4, can you explain?

[wintry tundra]

where did u get 2x>8

[cold siren]

uhh i was just looking above

[wintry tundra]

oh ok

[cold siren]

im extremely confused lmao

[wintry tundra]

2x>8 is the same as x>4

they are the same thing

[cold siren]

oh lol ur right i wasnt paying attention

[wintry tundra]

two inequalities is 2x-8<x+6 if i recall

and x>4

[patent plume]

yes

[wintry tundra]

so just isolate x in the first inequality and then put together both pieces to have a range

[cold siren]

ohhh tysm

[wintry tundra]

no prob

have fun with geometry lol

[vocal lodge]

geometry worst math

[cold siren]

fr

[patent plume]

lol

[wintry tundra]

im abt to start calc later today since im teaching myself and its just geometry but more algebraic and also harder apparently so

[cold siren]

yikes

[wintry tundra]

but its cool to be able to find how many units are in a certain part of a graph

[patent plume]

Mm no, from my experience calc isn't so bad

and not particularly related to geo

at least, not the basics

[wintry tundra]

i have no experience with calc so im just infering from first impression and such

[patent plume]

ig. For the beginning, the only geo is probably the definition of the derivative, and that's debatable

Integration I'm not so sure about, but I don't think it's going to be very different

[wintry tundra]

def of derivative is df/dx = nx^(n-1)?

[vocal lodge]

can someone explain this?

[patent plume]

uhhhhh no

that's the power rule

[wintry tundra]

cool

yeah im telling u i only know a small bit

naptime is that a test

[cold siren]

nah deltamath uses assignments

[vocal lodge]

no its a practice sheet my teacher uses

[wintry tundra]

hmmm

[vocal lodge]

yes its deltamath

[cold siren]

p sure we have the same assignment

[wintry tundra]

ok then

[cold siren]

cuz i did that one too

[wintry tundra]

well i mean draw the triangle

[cold siren]

yeah i drew it out and it made sense

[wintry tundra]

and look at the lengths

[upper karma]

Yes math

[wintry tundra]

see which line would be connected to which vertex to be a midline

yes blownbck math indeed

[vocal lodge]

yeah nevermind i dont need help

[upper karma]

Ah yes finally a server with 300iq members

[cold siren]

man

[wintry tundra]

u would be surprised

i legit said 2x-8 = x+6 earlier which was pretty clearly wrong so i mean my brain die

[upper karma]

299iq members

[wintry tundra]

lol

[cold siren]

literally i need to study more my last assignment i got a 67% im glad it wasnt a test

[upper karma]

Pft

Lucky

[wintry tundra]

what was the topic about

[cold siren]

it was about triangle inequality

my teacher really doesnt teach so i have to study myself but im bad at knowing where to look lmao

[vocal lodge]

yes me too

i dislike math very much

i dont rly pay attention in class so i just teach myself

[wintry tundra]

hm

math is cool when u get to the right parts

[vocal lodge]

whats ur favorite parts of math

[wintry tundra]

like trying to find the cosine of i is pretty cool

[cold siren]

literally my last quiz grade

[vocal lodge]

sheeesh

good luck

[wintry tundra]

the main reason math is disliked tho is because itll 50% be too easy and 50% make no sense

but its more fun to learn something u like

so u either 1. havent found a part of math u actually like or 2. just dont like math at all which is ok

[vocal lodge]

hopefully 1 for me

[wintry tundra]

well what do u like right now

cuz there might be a kind of math that pertains to it

[vocal lodge]

algebra

algebra was so ez

like algebra 1

not algebra 2

[wintry tundra]

yeh i like algebra a lot

[vocal lodge]

idk algebra 2 yet

[wintry tundra]

im in algebra 2

[vocal lodge]

is it ez?

[wintry tundra]

i mean im a math buff so for me yes

but idk about other people

u learn about inverses

so like lets say u have y=x^2

an inverse would be changing it to x=y^2 and isolating y

so the inverse of y=x^2 is y=sqrt(x)

[vocal lodge]

hmm seems complicated cool

[wintry tundra]

yeah

because of inverses there is a new function that will be learned next year

lets say u have y=e^x

do u know what e is?

[vocal lodge]

no

is inverse switching x and y

[wintry tundra]

before we go on ill just say what e is algebraically

$e=(1+\frac{1}{\infty})^\infty$

[somber coyote]

hiidostuff

[wintry tundra]

e is continuous exponential development

its specific value is 2.7182818284590....

but back to the original problem @vocal lodge , inverse of y=e^x

[humble pulsar]

[wintry tundra]

lol we just having a conversation in here now

i can move it if needed

[humble pulsar]

$\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n = e$

[somber coyote]

moshill1

[wintry tundra]

yeah i know i know

[humble pulsar]

1/inf

[wintry tundra]

but im just saying how it would be seen to like geometry level

[vocal lodge]

ok man this is tuff

[wintry tundra]

yeah see for now u dont know how to invert y=e^x

but we need to isolate y in x=e^y

so theres a new function called a logarithm: the inverse of y=e^x is y=ln(x)

ln being natural log having base e

if u wanted to invert y=10^x u would have y=log(x)

if u wanted to invert $y=2^x$ u would have $y = log_2(x)$

[somber coyote]

hiidostuff

[wintry tundra]

and logarithms have a lot of properties

well not a lot i should just say "unique" properties

but if u wanna know more google is ur friend, especially khan academy

[upper karma]

can someone help me with this?

[olive cove]

Do you know how to calculate the area of a parallelogram?

[upper karma]

i figured that one out but can you help me with this one?

[olive cove]

It's not right angled right?

Well it doesn't really matrer

Let's say the hight of the triangle is h and the base is 8

You can use sin to calculate h

And then apply the formula for the area of a triangle

[upper karma]

how do i do that?

8sin(40)

wouldn't that just be the unknown side tho

[olive cove]

No that's not correct

Do you know soh cah toa?

[upper karma]

yea

opposite = x (unknown) and hyp = 8?

[olive cove]

No hypotenuse = 4

And we want to know h

So you know sin(40) = h/4

[upper karma]

so 4sin(40)

[olive cove]

Sorry I wrote it wrong

No wait

What you say is correct

So h = 4sin(40)
And now you do 0,5x8x4sin(40)

[upper karma]

i got 10.2846

[olive cove]

So the answer is 10

[upper karma]

the area is only 10?

[olive cove]

Apparently

Well a bit more but you have to round it

[smoky sleet]

what formulas do i use in order to get the answer, im a bit slow lol

[silent plank]

trig formula for area of a triangle

[smoky sleet]

Thanks, I got 30

[rugged parrot]

anyone able to help?

[wintry tundra]

what exactly is the isue

issue*

[fair ermine]

Im still up lol

Anyways

2 is pretty easy

All you gotta do is split it up into a rectangular prism and a irregular pryamid

[vapid stag]

need help on #65

50/x=100/120?

[sullen cipher]

no

100/ 100+120

100/220

wait

1 sec

i think i made a mistake

i think ur correct

wait

bruh whats wrong with me xd

wait no i think im correct

[vapid stag]

i think u're correct lmao

50/100=x/(100+120)

x=110

[pure bronze]

if you're using Thales' theorem, then it should be
50 / x = 100 / 100 + 120

since you're using the first expansion of the theorem itself (because you had to include the inner segment)

[obtuse tapir]

Good geometry questions pleaaaseee

[crystal quail]

Any ideas on how to explain an axis of symmetry for 3-dimensional shapes? 2 is easy enough. Fold object along line and each point matches to another. But in 3-D it's a bit more complicated and I'm not sure how to word it

[vapid stag]

my fx-115ES Plus calculator does not seem to have cot

how do i do this question if so?

[crystal quail]

tan(x) = 1/cot(x)

[vapid stag]

so tan(x)=1/-2.5?

that doesn't sound right

still confused on how to apply it on the calculator

without the cot button

[crystal quail]

I'm fairly certain there's a way to do it, but if not then you have to use that identity

well more of a definition than identity, but still

[vapid stag]

this is what the manual says.

[crystal quail]

tan^-1 is cot

[vapid stag]

oh thanks lol

i just started trig class so i have no idea what's going on

[crystal quail]

Yeah so sec is cos^-1 or 1/cos

and csc is sin^-1 or 1/sin

we gave names to these trig functions because they come up often enough to not like writing the inverses in fraction form

[wintry tundra]

sec is $1-\frac{2!}{x^2}+\frac{4!}{x^4}...$

[vapid stag]

hmm

[somber coyote]

hiidostuff

[crystal quail]

oh I would not bring Taylor Series into this dude's life rn

[wintry tundra]

Lmao

[vapid stag]

u guys sure tan^-1 is cot?

i must be doing something wrong

[wintry tundra]

Tbh I don't even know how Taylor series ends up being able to solve this

[crystal quail]

why are you doing something wrong?

[vapid stag]

this is the problem i have the do. the answer seems to be -.4

i did tan^-1(-2.5)

and it came up weird O_O

[wintry tundra]

cot is tan^-1

[crystal quail]

nonono

dont do tan^-1(-2.5)

that's saying x = -2.5 when it's cot(x) that's = -2.5

[wintry tundra]

Or 1/tan

[crystal quail]

so like

@vapid stag

tan(x) = 1/cot(x)

and if cot(x) = -2.5 you just plug in

tan(x) = -1/2.5

which is -0.4

[vapid stag]

tan(x) just means i press the tan key right

[crystal quail]

yeah and you should be able to input a value into a parentheses

[vapid stag]

for some reason tan(-1/2.5) becomes -6.981.......

[crystal quail]

again that is not x

you're not solving for x

or theta in this case but im saying "x" because i dont wanna type theta

your job is to find tan(x) yes?

[vapid stag]

yeah

[crystal quail]

tan(x) = 1/cot(x) by definition

and we know cot(x) = -2.5

plug in

tan(x) = 1/cot(x) = 1/-2.5 = -0.4

no calculator needed

[vapid stag]

oh wait

it's legit just 1/-2.5

[crystal quail]

yessir

[vapid stag]

ah ok i see

i was confused because the reciprocals of the problems had fractions all the time

and had a hard time wrapping my head around is basically 1/something or something/1

[crystal quail]

cot is a fancy way of saying tan^-1 or 1/tan and vice versa

yeah

[vapid stag]

i see. thanks for the help!

[crystal quail]

it gets ugly and dumb, which is why we just gave it that name

[vapid stag]

so tan(x), given cot(x)=-2.5

becomes 1/-2.5

now that looks like ezclap

[crystal quail]

easiest of claps

if this helps at all i'll say it, if it confuses you forget i said it
csc = 1/sin
sec = 1/cos
cot = 1/tan
can also think of it as
sin*csc = 1
cos*sec = 1
tan*cot = 1
I always got confused at which csc and sec were inverses to, so just think that s's and c's need to both be there or something 😂. Sin starts with an s so it's inverse must be the one that starts with a c, and blah blah blah

[vapid stag]

yup

i already got that in the book chief

[crystal quail]

🤙

[vapid stag]

i don't know how quickly i might forget about this as soon as i'm out of math courses in the real world however.

i feel like these things are ought to be used or forgotten rather quickly

[crystal quail]

Depends on your field. I went into physics and I've used these for 8 years. Not much in the last 1.5 years since i finished with school but after that long it's hard to forget

Many many fields use tons of trig

[vapid stag]

that's a rather long time.

I just want to be a hypetrain software engineer lol

[crystal quail]

Is that a train that's solely powered off the hype of the passengers?

[vapid stag]

i thought it's just one of the most popular occupation thus far

with a higher chance of seemingly "limitless" scalability

it's not so much like jobs as a nurse or a doctor, where the amount of clients are limited by interacting with one user at a time.

[crystal quail]

me after all these years of doing triangle

[tiny snow]

The angle was unnecessarily constrained. This is better. Ping me if you attempt it.

If you want to fiddle with it: https://www.geogebra.org/geometry/j5b2jgum

[crystal quail]

@tiny snow Do you do a lot of algebra in this geometry course or is it mostly axioms and normal geometry stuff? I'm giving it a shot but don't wanna get too crazy complex if it's avoidable

[crystal quail]

I'm almost certain im really close to solving it but it's well past my bed time. are you good if i get it done tomorrow (probably within the next 12 hours or so)

[tiny snow]

@crystal quail Sorry for the late reply, something came up here. The students are well-versed in both algebra and geometry.

[pseudo wing]

oh you're the teacher

[tiny snow]

@crystal quail Absolutely, I will go about finding a solution of my own.

@pseudo wing I am 😄 Setting up some problems for an assignment.

[pseudo wing]

ahh

go easy on them

[tiny snow]

@pseudo wing Just need to make sure they're not too hard to solve (or even impossible).

[crystal quail]

are you sure it's possible within 90 degrees?

i got 2/3*pi radians which is a bit over

[tiny snow]

@crystal quail Note that I changed it to 0 - 180.

[crystal quail]

oh

so i may have got it then

[tiny snow]

I swapped the image, may have gone unnoticed. Sorry for that.

[crystal quail]

teacher i lock in my answer at whatever 2/3*pi radians is

[tiny snow]

@crystal quail I'll let you know what I get later today. Check back tomorrow and get some sleep now.

[crystal quail]

@tiny snow I was thinking about it more in bed and I think i goofed. new answer is 8/27*pi radians. will try to confirm tomorrow. zzz time now

[tiny snow]

Ended up with this equation. Wolfram gives the numerical solution of x ≈ 0.255096785031812... Since <CAB = 2x, converting to degrees results in an angle of 29.23 degrees. Checking this in GeoGebra it looks correct, but is it possible to get an exact value for x?

Here's the link if anyone wants to fiddle: https://www.geogebra.org/geometry/npevnekj

Radians works too, but Wolfram is normally rather good at figuring out any exact solutions, so it's looking like this won't be a problem for the students after all.

[upper karma]

hi help

[neon hamlet]

How would you solve for this?

[haughty ore]

@neon hamlet you gotten helped with this one yet or solved it?

[neon hamlet]

nope

didn't solve it yet

[haughty ore]

okay is there anything in specific you get stuck with?

[neon hamlet]

just making the picture

I can solve it, im just not sure how I would make a picture and get the ratios

[haughty ore]

so i just googled the term angle of depression (i'm from Sweden so never heard that one before) but if it's a picture you need i think i would look something like this:

since you have the angle from the horizon and downwards rather than from the building

[neon hamlet]

right so how would we go about solving this

[haughty ore]

with trigonometry ofc 😄

[neon hamlet]

is there an equation?

[haughty ore]

@neon hamlet okay so

you have sin, cos and tan to choose from

᲼
sin v= far Cathetus/Hypotenuse
cos v= close Cathetus/Hypotenuse
tan v= far Cathetus/close Cathetus

᲼
and we know the "far Cathetus" that's our building (if you look at the dotted line triangles)

and we know the angels. what we want to know is the "close Cathetus" so we can use tan formula for that:

tan (32)= 600/a
and tan(20)=600/b

you following me? i'm not using the correct terms since i don't know them i just translating

after this you you need to do algebra to make x alone and solve the fraction on the other side with a calculator

hard to know what if you don't tell us ^^'

1.a) ? or the both sub questions?

they are asking what the size of the angle centred at the corner B is

[vapid stag]

so tan is given

this means cot = 1/squareroot of 3

and it becomes squareroot of 3 /3

right?

so now we have values of x and y which x= squareroot of 3 and y=3

in the III quadrant this means x and y are both negative

now we find the value of r

please tell me i was on the right track

[hearty hedge]

Can anyone help me with this?

[haughty ore]

you can probably use "Law of cosinues" to solve this. just plug in you values then and the only non given number would be cos of the angle.

[placid path]

https://gyazo.com/ccdf37765fe7fd01f1348fd5132d90a4

im stuck on the last part

any suggestions on what to do

[humble pulsar]

evaluating Rsin(x+t) at x=pi/4 gives sqrt(30)

so likely using that you show it

[placid path]

oh thanks

[upper karma]

What is the formula for the volume of an irregular anti-prism (vertically stretched or squashed)?

[placid path]

@upper karma i think its this

[torpid hearth]

your pfp scares me lmao

[split talon]

This is for Math History

What the heck is "trisection method"??? I have never heard of this and I cannot find it online anywhere

[stark scroll]

How

[upper karma]

I forgot what to do

[wintry tundra]

pretty sure two diagonals bisect each other

so set each half equal to each other

ill get u started with x=y+3

@upper karma

[upper karma]

ok

[slender sonnet]

Could some one please help me

[wintry tundra]

@slender sonnet sure go ahead

[slender sonnet]

Stuck on the stupids one

I got 15 times 34 but what is half circle and trying to find the area of all of it

[sullen cipher]

do u know how to find the area of a circle @slender sonnet

[vocal lodge]

can someone help me with 12

[wintry tundra]

all sides are similar in number 12

do u know why?

[vocal lodge]

no

[wintry tundra]

well firstly it shows that two lines are parallel and K is touching JL and P is touching JM so KP is scaled from LM

and KJ and PJ are parts of LJ and ML so

[rocky tulip]

Guys, can I ask how to study chebyshev distance? Have you learned it in college?

[tiny snow]

This turned out quite nicely.

You can get a feel for the problem here: https://www.geogebra.org/m/cxayxmka (by moving B or C)

[slim saddle]

I need to find minor arc BD

length

<1 = 30

KM = 8

then KF = 16

as <2 = 60

pls ping me

[tiny snow]

@slim saddle Are BD, AE and HF straight lines?

[slim saddle]

yeah straight line segments

nope @wanton kestrel idk what it is

[somber coyote]

fields!

fields!

[slim saddle]

ok

i thought it was just 8 sqrt(3)

HM = MF right?

i already got the answer

Can you prove arc BD = arc HF

wow i am so stupid

BD = HF was given in the instruction

lol

hehe sorry for wasting ur time

[upper karma]

Nvm I don’t need help anymore

[sleek depot]

Diagonals bisect each other

so x = y+3 and x - 6 = 2y - 15

[upper karma]

Yeah but what do I do after

To solve it

[upper karma]

Thank you. That appears so be for an anti-prism with sides of only length a, but also I guess it would be possible to use trigonometry to derive a "stretching factor" to multiply that by.

How do i do #2

And i got 288 for #1

idk if its right tho

<@&286206848099549185>

[fair ermine]

Hey @slender sonnet ?

[upper karma]

Hi

I am currently studying the subtraction and addition formulas for trig functions

I am wondering how to interprent

$sin^2(v)+cos^2(v)$

[somber coyote]

Fish of the sea

$sin^2(v)+cos^2(v)$

[upper karma]

is it really just

using the formula and square the results?

[somber coyote]

Fish of the sea

i.e for $sin^2(v) = (2sin(v)cos(v))^2$

[upper karma]

woops forgot that it is 2v and not just v

[vestal tartan]

anyway

[upper karma]

Would this be 28/21 considering it moved to Z?

[olive cove]

It's 21/28

[upper karma]

Okay, thanks.

I thought it would be flipped when the angle thing is moved.

[olive cove]

I just use soh cah toa.

Easy to remember.

[upper karma]

Yeah would these be flipped though?

Pls

[low tiger]

i need help in trignometry

[olive cove]

They are two different triangles so you can't really flip anythin

[humble pulsar]

what about it?

[low tiger]

my bad

this

[humble pulsar]

still no idea what you want help with

[upper karma]

can someone help pls

[low tiger]

q1

[upper karma]

@upper karma well a hint

is that you can find the base side of the top triangle using tan

oh it is multiple choice

its to late for a “hint”

[prime linden]

<@&268886789983436800>

[viral jolt]

is this a test?

[upper karma]

Help pls.

Do you do cos-1(0.4593) into the calculator?

[wintry tundra]

yeah when ur trying to find a variable in a sinusoidal function then u take the inverse cosine of the answer

cos^-1 x is the inverse of cos x

[upper karma]

Ok but I got like something over 1 for some reason.

[wintry tundra]

well yeah makes sense

wait nvm no it doesnt

[upper karma]

👌

[wintry tundra]

cos 1 is about 1

hm

[upper karma]

I also got 0.248

[wintry tundra]

nvm ur right

[upper karma]

For which one pls?

[wintry tundra]

did ur answer come out as something like 62?

[upper karma]

No.

[wintry tundra]

welp

its wrong then

[upper karma]

Oh I don’t have it on degree

Nvm.

[wintry tundra]

yeah thatll help

radians > degrees tho

as in i like radians better

[upper karma]

Yeah, thanks!

[wintry tundra]

no prob

[upper karma]

How do I put a inverse on my computer pls.

[wintry tundra]

cos^-1

[upper karma]

Ok ty!

[wintry tundra]

no problem

[upper karma]

What about this one pls?

All 3 sides have a variable so I don’t know which formula I need to use to solve it.

[silent plank]

the question is overspecified. you can pretty much choose whichever trig function/ratio you want

[upper karma]

Wait what?

How do u do that?

[silent plank]

do you know your trig ratios, sin, cos, tan?

[upper karma]

Yeah.

Socahtoa

[silent plank]

they could all be applied there

take your pick

[upper karma]

So just make ratios for example by putting opposite over adjacent?

For each of the variables?

[silent plank]

use any trig function with the appropriate ratio

[upper karma]

Okay, thanks.

[silent plank]

you only need to use 1

[upper karma]

So like something like this pls? @silent plank

[silent plank]

different question?

[upper karma]

No just is this what you mean by solving that last one pls?

[silent plank]

you only need to choose one trig function / ratio

i.e. you could choose to use sine to set up your ratio

and then use inverse trig to get the angle

[upper karma]

Oh!!!!!

[silent plank]

alternatively you could've used cos or tan
(and use inverse trig to get your angle)

[upper karma]

Okay, I see what you mean.

I’m kind of dumb.

So like I use tangent and apply 18/43 to inverse tangent?

Is that basically what I have to do pls?

[silent plank]

wrong wording

[upper karma]

Yeah okay, I got you, thanks!

[silent plank]

and wrong ratio

[upper karma]

Lol, yeah.

[silent plank]

note the location of theta

[upper karma]

That’s not tangent.

So is 18 the adjacent?

[silent plank]

to theta yes

[upper karma]

So I got 65 degrees, was that all I needed to do? @silent plank

[silent plank]

yes

[upper karma]

Okay, thank you!!!

[silent plank]

technically the triangle looks like a right triangle but isn't actually indicated as one and you should've applied the cosine rule

but it's close enough

[upper karma]

Okay, had me scared there for a second, thanks though!

[silent plank]

,calc 18^2 + 39.05^2

[somber coyote]

Result:

1848.9025

[silent plank]

,calc 43^2

[somber coyote]

Result:

1849

[silent surge]

Can you pls help me find the value of x

[silent plank]

what have you tried?

[sullen cipher]

u can find the rest of the angles

to use law of sines

[dim plaza]

You do 24/2 which is the base of the right triangle then multiply it by the tan of 65 so you find y then use pythagoras' theorem to find the hypotenuse, you should get 21.33
√(tan(65))^2+12^2 ≈ 21.33

tan stands for tangent which is the length of the opposite side divided by the length of the adjacent side

[fickle sleet]

This is confusing to me. Why is the domain (with restrictions of -pi/2 to pi/2) of the sine function not the same as the range of the arcsin function? Aren't they inverses of each other with a restriction? Same for the range.

[pure cape]

because arcsin(3x+1) is not the inverse of sin(3x+1)

try solving for y in x=sin(3y+1)

[fickle sleet]

Not exactly sure how I'm supposed to do that

[pure cape]

take the arcsin on both side?

[fickle sleet]

I completely forgot they cancel out

so the arcsin of sin(3x+1) should be... y = 1/3arcsin(x)+1/3?

[pure cape]

this is kind of the same case with y=1/x, the inverse of y=1/x is y=x, but the inverse of y=1/(x+3) is not y=x+3

no

the arcsine of sin(3x+1) should be 3x +1

[fickle sleet]

I meant inverse my bad

[pure cape]

then yes

actually no

[fickle sleet]

-1/3

[pure cape]

your sign is wrong

yep

-1/3

[fickle sleet]

Gotchu the little steps always get me

Thanks for the help saved me a lot of time

[ivory glacier]

hey, in this triangle the dots are evenly spaced

like it's a 3-4-5 right triangle

how would I find the area of the shaded regio?

[sleek depot]

the shaded is an equilateral right

calculate its area

[tiny snow]

@ivory glacier Introducing that line in red should get you started.

[sleek depot]

oh wait its curved

[tiny snow]

(disregard the circles, they are construction artifacts)

[sleek depot]

ok

[silent plank]

trig formula for area of a triangle will be very useful here

[devout cove]

What is x I’m lost

[vocal lodge]

can someone help me with the proof

[narrow rock]

what do you notice about triangle KLO, that happens to be pretty obvious

[vocal lodge]

it has two congruent angles?

[narrow rock]

yeah what does that mean

@vocal lodge

[vocal lodge]

its isosceles?

[narrow rock]

yeah

so

KL = KO right?

now do you know about the triangle inequality

[lunar sand]

confused

[narrow rock]

wdym

what are you confused abot

which curve most accurately represents the points of data on the graph

use ratios

[sullen cipher]

you should know thale's theorem by now

[narrow rock]

what'

[hearty hedge]

Anyone good at this?

[olive cove]

Yes

[pallid dome]

Hey does anyone have a proof on sec(a+b)? Whether you've seen one or if you can prove it?

[pallid dome]

Specifically a geometric proof

[tiny kiln]

Can someone help me with geometry please

[nocturne vale]

how do i solve for y

i found EB through the trapezoid midsegment theorem

but i cant find 3y-2 or 2y+8 and i dont know where to start

[narrow rock]

did it specify that B and E are midpoints of the trapezoid

or what

because it doesn't say that

also if it did say that then 3y-2 and 2y+8 are the same length

[nocturne vale]

@narrow rock didn't. but i'll assume that it is

thanks

[narrow rock]

i mean if you are doing a class learning about that you can assume that

but in any other case you can't

[nocturne vale]

yeah im doing a class learning ab it

[narrow rock]

just tell your teacher or whoevers giving the homework to specify that

if they wont kill you

[nocturne vale]

they didnt teach us how to solve it if the line isnt at the middle lol

@narrow rock

are TV and SU equal?

seem to be

wish i could have somebody to help me through these problems

[narrow rock]

they are equal if that is an isosceles trapezoid. meaning that if SV and TU are parallel and <STU and <VUT are equal yeah

TV and SU are equal

[narrow rock]

you can assume that ig because no further information is further provided

but you should ask your teacher to be more specific because

[narrow rock]

you should never say this when doing geometry

meaning you already know the problem is messed up, if you have to rely on what it looks like

[upper karma]

help

<@&286206848099549185>

[narrow rock]

go ask a question in a question channel

if you want to ping helpers

[upper karma]

#help-0

thx

[narrow rock]

and stuff

but ping after 15 min

[coarse wraith]

Guys. I need tosa help

[fathom peak]

can someone help me with this like i dont understand how to solve this

[narrow rock]

what dont you understand

did you try drawing out so that θ = 230

that could help

[jaunty sail]

@upper karma do you still need help

See if this illustration helps

[upper karma]

Hi

if one has the following equation

[somber coyote]

Fish of the sea

[upper karma]

and one multiplies both sides by cos^-1

[somber coyote]

Fish of the sea

[slow crest]

What is the half pi symbol?

Or is it a T

[upper karma]

T = 2pi so T/4 = pi/2

[silent plank]

cos^-1 is not something you multiply by

[upper karma]

It is not?

What operation does one do then?

[silent plank]

it is a function

[upper karma]

which accepts cos(x)?

[slow crest]

Apply same function to both sides to cancel one side

[upper karma]

oh

so if i understand this correctly

[silent plank]

eg. cos(7) isn't the product of 7 and some mysterious value "cos"

[upper karma]

in this case, it would be 2cos^-1(cos(x))?

it is not?

all I know from the unit circle is that it is the x coord

[sleek depot]

cos is a function, like cos(x) will equal some expression

[sleek depot]

yes

[upper karma]

what is that expression?

[silent plank]

no arccos function isn't linear

[upper karma]

welp

seems like I have a lot left to learn

but thanks for the help!

I'm one step closer to beating this nightmare

[silent plank]

apply the complementary properties of sine and cosine

[upper karma]

I am curious on the second part

How it ended with cos^2 theta -sin^2 theta?

[prime linden]

In the denominator?

[upper karma]

Already got it, should be cos^2 + cos^2 🙂

[prime linden]

I dont see why, in the numerator you just open the paranthesis and 2cos^2(x) gets subtracred leaving you with cos^2 (x) - sin^2(x) in the numerator.

In the denominator use the the difference of two squares.

cos^4(x) - sin^4(x)=
(cos^2(x) +sin^2(x)) (cos^2(x) -sin^2(x)) =
1* (cos^2(x) - sin^2(x))

[steep temple]

wat

shoelace method hmm

[grave depot]

hello! does someone know about fractals?

I have to do a paper in which i do an analysis between fractals and my topic. i'm honestly stuck rn

[upper karma]

Hey

who knows how to calculate centroid of body

[storm portal]

@upper karma #❓how-to-get-help don't ask to ask, just ask

[nocturne vale]

@upper karma thanks m8

[upper karma]

You're welcome

[zinc gale]

What was the difference between an R and an O with a dash going across it?

[forest mica]

What's up with cos(x+y)=cos(x)+cos(y) ?

[silent plank]

wdym by "what's up"?

[patent plume]

@forest mica are you asking why it doesn't work?

The answer to that would be similar to the answer to the question "why is 1 + 1 = 2 and not 1.6"

[upper karma]

Hey! I’m doing a review and I have a lot of difficulty with proofs. Can someone help?

Just to check if my reasons are correct

Can some one help me out?

[slim saddle]

help

i got 83.78

idk if thats right tho

because my smarter friend got a different answer

CO, AO,OD,OF are all radii and have lengths of 8

FO + CO are two radii, so it is Diameter

COD therefore is 90 degrees, and AOF is 120 degrees (AOC is equalateral)

360-60-120 = 150

or u could do 90+60

150/360 = 5/12

64pi * (5/12) approx = 83.78

[humble beacon]

can someone help with this question

i have to find all angles using laws of cos

[patent plume]

Did you guys learn the cosine rule?

[humble beacon]

yes but im still lost

[patent plume]

What is the problem? Like where are you stuck?

[humble beacon]

so i have to find all angles using laws of cos

[patent plume]

Alright

That's not really telling me where you are stuck tho

[humble beacon]

i know i have to find the length of the hypotenuse to use cos

[patent plume]

uhhh

Wait, can you write the cosine rule for me?

[humble beacon]

ok hold on

[patent plume]

Cool. Btw, hypotenuse isn't really used for anything but right angled triangles

So, here you know which side you gotta find first right?

[humble beacon]

just the angles

angle a and b

[patent plume]

Right, but for the angles you need the sides?

Check out the formulas for angle b and angle a

they're all written in the form of the 3 sides

[humble beacon]

yeah

i see that but the equations

have a, b, and c

which i think is referring to the angles

but i have 1 of the angles

[patent plume]

Right. So you have 2 sides and the angle between those sides

Oh btw, the small a,b,cs are the sides

and the capital ones are the angles

[forest mica]

Nah, I'm asking for what x and y would it work for? Like functions maybe. I saw it plugged into desmos, and it was wild looking

[patent plume]

it would be weird yeah

cos(x)cos(y) - sin(x)sin(y) = cos(x)+cos(y) after all

(cos(x)-1)(cos(y)-1) = 1 + sin(x)sin(y)

idk if that can be simplified/if that's the right way to go

[forest mica]

Whats the formula for cos(x+y)?

Oh lol

[jaunty plank]

Someone help I’m stuck

[olive cove]

Let's say the horizontal distance is x

[jaunty plank]

Yea

[olive cove]

Which operator do you have to use if you know the hypotenuse and want to know the adjacent side?

Of the angle

Do you know it?

[open spindle]

Can someone help me?

[olive cove]

@jaunty plank Are you here?

[jaunty plank]

Yea

I was eating my bad

[olive cove]

Ok

Can you answer my question?

[jaunty plank]

No

[open spindle]

Isn’t it o/h

[olive cove]

No you want to know the horizontal distance

[open spindle]

Oh haven’t learned that yet sorry

[humble pulsar]

Yeah that's why the question wasnt for you 🙃

[olive cove]

In this case you want to use cosine

Soh cah toa. Use this if you don't remember.

[humble pulsar]

diagonals bisect each other

[jaunty plank]

Alright

[olive cove]

So what is your next step?

Try to form an equation with x in it

[jaunty plank]

Alright

[open spindle]

I got it, XR=19

[humble pulsar]

,w solve 8x-18=2(2x+5)

yep

[olive cove]

@jaunty plank You get the equation
cos(11,5) = x/40