geometry-and-trigonometry

[hybrid shoal]

can anyone help

[dark sparrow]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[dark sparrow]

@hybrid shoal ?

[limber lantern]

Can any relation be drawn between orthocenter and centroid apart from the fact that they lie on the same line

[dark sparrow]

did you post the right image

<@&268886789983436800> troll

i'm a she.

let the record state that @dark herald posted an image that is just obama's face projected onto a pyramid.
the image is not a geometry problem by any description.

[past geyser]

just ask lil dawg

[solid crypt]

Tan^2theta=1 could anyone help me solve for theta?

Tan^2theta=1 is just tantheta=1

[dark sparrow]

no, it's tan(theta) = ±1

[solid crypt]

OH

Then thats why I wouldn’t get the right answers

Why is it that so I don’t make the same mistake in another context?

[dark sparrow]

the equation x^2 = 1 has two solutions

1 and -1

[solid crypt]

It’s like that for everything? I must have missed class that day

Thanks

[ember grove]

yo is there a way having sin of an angle to derive its angle? note that i'm not saying having the ratio of the sin but directly the sin itself like 0.45225829372. i just wanna if there's a way to avoid memorization

without

calculator

[sturdy stone]

There isn't a way without calculator or memorization

Especially if it's in decimal form

[ember grove]

damn all right so i guess i will memorize

[sturdy stone]

Just use a calculator

[next phoenix]

it's always like that if the power of your vairable is even

meaning the solutions for x^4 = 1 is 1 and -1 either

obviously not the same on odd numbers since you are essentially multiplying (-) and (+)'es

are you struggling with anything specific

[bleak tangle]

maybe you could use the taylor series for arcsin x

[ember grove]

Is basically that i need it for a test because i can't use a calculator

But im probably better off memorising this tims

[twin crag]

how do i approach this problem

maybe it involves rotation?

[nimble dagger]

I need some help with this

[hybrid shoal]

prove that

[upper karma]

The Question is incomplete

the text says "find the value of x"

[thorny compass]

need some help getting 30, and 90

@upper karma this crap is due in 5 MINUTES, help me sirr

[upper karma]

the rules of the server though

x= 2root2

y=root6

eh

cos(60)= root(2) / x

sin(60)=y / x

dude

what are you talking about

the answer is right

you have to simplify it

2 root 2 means 2*1.414 = 2.828

root 6 = root 3 times root 2 = 1.414*1.732

calculate it

thats right!

you can write root 3 times root 2

try that

what

[cunning lion]

<@&268886789983436800>

[thin sentinel]

,prune 100 --from 871944516079329290 --forace

[somber coyote]

This may only be done by a moderator!

Feb 7, 2024, 6:00 AM

[thin sentinel]

,prune 100 --from 871944516079329290 --force

[somber coyote]

This may only be done by a moderator!

Feb 7, 2024, 6:00 AM

[thin sentinel]

bruh

[upper karma]

hehe

thank you

[dark sparrow]

was this a troll or a test cheater

[cunning lion]

troll

[mystic charm]

yea

[keen vigil]

I'm trying to find that unknown red dot's location using 2 known points and which way they are both facing, i've watched countless videos that either aren't what im looking for or just don't have the same parameters as i do. please help guide me in the write direction! thank you!

[dark sparrow]

585°???

[keen vigil]

58.5

sorry

i guess that wasn't apparent

[dark sparrow]

ok, and those are what, bearings?

[dark sparrow]

it was indeed not

[keen vigil]

yes

[dark sparrow]

do you know your way around equations of straight lines

point slope form etc

[keen vigil]

yes

[dark sparrow]

here is a not to scale diagram of one possible idea

[keen vigil]

ok thank you

[ivory sphinx]

how?

[dark sparrow]

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

Y/N

[ivory sphinx]

Yes

[dark sparrow]

ok

the shaded area equals the big sector minus the small sector.

they have the same angle 80°, but different radii: 14 inches for the big sector, 4 inches for the small

[ivory sphinx]

ohhh

thank you lmao

[solid crypt]

Sintheta=-0.525 I need to solve for theta in radians

I got arcsin(-0.525)=-32 degrees

(Rounded up)

There’s supposed to be 2 angles since sin is negative in quadrants 4 and 3

According to Mathway the 2 thetas are 212, and 328 degrees.

[dark sparrow]

and how did you get arcsin(-0.525) = 32 degrees? did you use a calculator?

[solid crypt]

Yes

[dark sparrow]

how come you did not set your calculator to radian mode?

given that you want to find θ in radians...

also what range do you want to solve over? 0 ≤ θ < 2π?

[solid crypt]

I used Mathway

[solid crypt]

Yeah rhat range

[dark sparrow]

ok,

then you will want arcsin(-0.525) + 2π and π - arcsin(-0.525)

these are your two solutions

in quadrants 4 and 3 respectively

does mathway have a radian mode or equivalent setting?

it probably should

[solid crypt]

I’m not sure

[dark sparrow]

then perhaps you could use something other than mathway.

you could maybe use a regular scientific calculator,

or http://desmos.com/ if you want something online

[solid crypt]

Yeah it gave me -0.55 rad

[dark sparrow]

sounds about right.

[dark sparrow]

what i said here still stands.

[solid crypt]

I’m just confused because Mathway said the 2 angles for Sintheta=-0.525 are 212 and 328

[dark sparrow]

because mathway thinks you want the answer in degrees.

and apparently there is no option that you know of to force it to return an answer in radians instead.

[solid crypt]

There was but it gave me a gibberish answer

[dark sparrow]

can you show exactly what kind of gibberish it was?

i want a screenshot which shows exactly what you put into mathway and exactly what it gave you.

[solid crypt]

Oh its because I put it as degrees and to convert to radians

[dark sparrow]

"No, I cannot show you the screenshot you ask for. Or maybe I can, but I don't want to."

[dark sparrow]

hm. ok, that is strange indeed.

can you have it ONLY do arcsin(-0.525)?

i want to see what it gives you.

[solid crypt]

It forces me to chose an option

That one worked

[dark sparrow]

ok, so it does output arcsin in radians by default.

can you show your exact input for the trig equation?

maybe that was messed up

[solid crypt]

Yeah okay

I did both

[dark sparrow]

ah,

well it looks like it understands "solve for theta in radians" just fine

[solid crypt]

Yeah I just don’t k or how it got 3.69

[dark sparrow]

.

[solid crypt]

Oop I missed

[rose raven]

why is the unit circle so congested

theres so much on it

i understand everything but the pi fractions

[solid crypt]

It’s because half the circle is pi, the whole circle is 2pi

[solid crypt]

The pi fractions are just degrees

1pi=180 degrees

2pi=360 degrees

Then pi/2 would be 90 degreee and so on

[rose raven]

i understand that pi fractions are degrees

just not why they are like that

one thing about them is i know xpi/6 are the ones closest to 180 and 0 but not why they are over 6

oh hold on

its xpi/3 for 60 degrees and xpi/6 for 30 degrees, i guess thats one thing i noticed

[dark sparrow]

pi/6 itself is 30 degrees.

30 degrees is one sixth of 180 degrees.

pi/6 is one sixth of pi.

[languid tusk]

anyone can help? ignore the blue writing

[mystic charm]

The triangle ABE is isosceles (why?)

Concluding that the angles E and A are equal

🔔

Hint: try to find B first

[languid tusk]

angles E and A cant be equal

A is 150

[mystic charm]

Ok?

Oh I meant in the picture

Yeah wrong letters were used

[languid tusk]

oh my bad

so theyre both 15 is what youre saying

[mystic charm]

Hmm, 15 is not a good number thi

Tho

Gimme a second to think

[languid tusk]

it must be

because B(in your drawing) is 150

so if they are equal they must be 15

yk

[mystic charm]

Let's approach this question in a different way

Let me send a picture real quick

[languid tusk]

how old are you btw

[mystic charm]

Bruh why do you ask, I'm 17

[languid tusk]

nah nah sorry didnt mean it like that

[mystic charm]

Start off by drawing this line

Then consider this triangle

What do you think we can do from now on?

To find the shaded area

[mystic charm]

How old are u tho

[languid tusk]

14

[mystic charm]

Hmm, hard question for that level

[languid tusk]

well

its a right triangle

so easy trig

but we dont have the angles

[mystic charm]

We have

Look at your question

[languid tusk]

we dont

[mystic charm]

Let me send a picture

Look carefully what happens here

It’s the same thing happening in the previous part

[languid tusk]

i dont think so

oh shit im a dickhead

youre right

but the top one is 60

and the bottom is 30

i believe

[mystic charm]

yes I changed their position so you do it yourself

[languid tusk]

what find the area of that triangle?

[mystic charm]

Now try to find the area of the blue part

🔔Hint: you can find the value of base and height of the big triangle using the angles

[languid tusk]

i dont get it man

i can find the area of the big triangle

I can find anything of the big triangle

but i dont get how you get to the shaded one

[mystic charm]

Do me a favor and use sine and cosine rules to find b and c

[languid tusk]

dont need them in right triangle trig

[mystic charm]

What do you mean, you shouldn’t use sine and cosine?

[languid tusk]

b is 4

[mystic charm]

And c

?

[languid tusk]

6.9(not even trolling)

[languid tusk]

yes but not the cosine or sine rules as it is a right triangle

[mystic charm]

4sqrt(2), don’t use numbers for square roots unless the question asks, yes

Now we are left with this

[languid tusk]

how is it 4root2

[mystic charm]

What is the sine of 60 degrees

Oh

Bruh

[languid tusk]

0.8666666

[mystic charm]

Yeah sorry

4sqrt(3)

Mb

[languid tusk]

yes

but why would u do it like that

just make it a decimal

tf

[mystic charm]

No

Don’t do that

keep it in sqrt

[languid tusk]

i dont get why

[dark sparrow]

well these are the wrong decimals, for a start

but also working with decimals like this is both hellish and imprecise

[mystic charm]

I messed up, can u solve the question?

[dark sparrow]

,calc sqrt(3)/2

[somber coyote]

Result:

0.86602540378444

[dark sparrow]

just for the record

$\sin(60\dg)\neq 0.86666\dots$ at all

[somber coyote]

アンナ

[languid tusk]

i was joking with all the 6's

[dark sparrow]

i can take a look at the question if you remind me of the original problem

[dark sparrow]

and we were meant to know that how?

[languid tusk]

but u get it its 0.86...

[mystic charm]

Still stuck cuz we can’t find the area of red part

[mystic charm]

Yeah sure

[mystic charm]

.

[languid tusk]

the blue part is what were trying to find

question b

try to igore the blue writing

[dark sparrow]

uh huh

let's see...

[dark sparrow]

so the way this diagram fits in with the problem is that the part labeled "8" is AE and the hypotenuse of the yellow region is AB?

[languid tusk]

yep

[mystic charm]

The way I was trying to solve seems wrong

Since we can’t find AP

[dark sparrow]

or can you

triangles APE and QBE are similar

[languid tusk]

it cant be this hard im only 14 doing 9th grade maths 😂

[dark sparrow]

does 9th grade math include similarity of triangles?

[mystic charm]

Oh damn it

yea

[dark sparrow]

you're gonna have to fuck with roots a little bit

[languid tusk]

i dont think so

[dark sparrow]

the blue area is 1/2 * AP * AQ by the way.

[dark sparrow]

?!

[languid tusk]

what

[dark sparrow]

i am surprised

that you would study geometry and not know of the concept of "similar triangles"

[languid tusk]

dunno

[languid tusk]

how do u know

[dark sparrow]

half base times height

you know that one, right?

[languid tusk]

yes ofvc

[dark sparrow]

thank god

[languid tusk]

but its irregular

[dark sparrow]

... yes so what?

[languid tusk]

i just need to find AQ

[dark sparrow]

AQ is known

AP is the problem

[languid tusk]

AP**

yes

[dark sparrow]

ok wait

this is gonna be hacky

do you know basic trigonometry

like soh cah toa shit

[languid tusk]

yup

[dark sparrow]

wow, you know that but not similar triangles.

[languid tusk]

yeah im good at trig

[dark sparrow]

[languid tusk]

like sine rul

cos rule

area of non right triangles

[dark sparrow]

AP/AE = BQ/QE = tan(AEP)

[languid tusk]

etc etc

[dark sparrow]

from this you get AP = BQ * AE/QE

[languid tusk]

that symbol means multiplication right

[dark sparrow]

yes

do you understand how what i wrote came to be?

[languid tusk]

i have to keep scrolling

can u put the diagram again

[dark sparrow]

angle EAP is not marked as right here, but it is right.

[languid tusk]

yo im gonna go to sleep man

it cant be this complex

[dark sparrow]

it is not that complex

you claim to be good at trig

[languid tusk]

i am

[dark sparrow]

AP/AE = BQ/QE = tan(AEP)
this should be clear then

is it?

[languid tusk]

weve never done similar triangles

[dark sparrow]

i am talking trig

literal soh-cah-toa shit

[languid tusk]

but youre saying theyre similar triangles right

[dark sparrow]

i was

but i have abandoned that idea

[languid tusk]

but how do u know they are the same ratio

[dark sparrow]

they are tan of the same angle

[languid tusk]

oh yeah im actually a drone sorry

ive done this for too long

i have to go to sleep

fuck this question

[next phoenix]

yeah

[next phoenix]

how tf does one know trigonometry but not similiar triangles

[strange spear]

how do i learn trigonometry and geometry

like i wanna relearn

Where do i do that

[blazing veldt]

https://www.khanacademy.org/math/geometry

[lime dune]

KA has long since been outpaced by other sites imo

[cunning lion]

such as...?

[steady socket]

does anyone know where i can find sources to help me with Proving a Quadrilateral Is a Parallelogram

[fickle rose]

probably khan academy but it mostly stems down to understanding the theorems and the properties of a parallelogram

[steady socket]

okay thank you

[opaque frigate]

Would anyone be able to help me write this equation in a way edfinity likes?

$\cos\left(\frac{\pi}{2}\left(x-3\right)\right)+13$

[somber coyote]

Jimmy'

[buoyant cypress]

can anyone help further of this

[dark sparrow]

i can see sin^2(x) - cos^2(x) factorizes as (sin(x) - cos(x))(sin(x) + cos(x))

[humble flint]

I’m so confused

I need to find angles from SSS

Sides are already given

[vocal bison]

help pls this was my warmup in math and i still dont knwo how to do it

[vocal bison]

Uh i got 108 pi minus 18 radical 3
i dont tihnk thats correct tho

[silent plank]

how are you getting the 18sqrt(3)

[vocal bison]

30 60 90 triangle

when i split 120

[silent plank]

give the full explanation of how you reached that value

[vocal bison]

i make the altitude so it splits 120 into 60 adn 60 on both sides so the other angles have to be 30 and 90
to get the sl i did 18 divided by 2 = 9 and then sl to ll i multiplied by radical 3 so it gives me 9 radical 3 on both sides, that added up is 18 radical 3

[upper karma]

i am here to ask some non hw related qstns

caus like i got nothing else 2 do

[silent plank]

adding up sides doesn't give you area

[upper karma]

i thought area was just l(w) or somthing

wll

thats 4 lik

i think just rectangles lol

[vocal bison]

uh idk thats what i was taught so
if i did area of triangle it would be 1/2 bh right so like 162 but aint that wrong too

wait

[silent plank]

well those b and h would need to be perpendicular

[vocal bison]

so then what would i do bc i wouldnt do pythagorean theorem

[silent plank]

why not

instead of pythag, you could do trig again to get the other side (which will in this case be the altitude)

there is a more convenient method of applying the trig formula for area of a triangle

[vocal bison]

but i cant use a calculator

[silent plank]

who said anything about using a calculator

[vocal bison]

whattt aunt the trig for area ofa tirangle 1/2 sin something

[silent plank]

yes

[vocal bison]

im lost

[silent plank]

look up trig area triangle

[vocal bison]

it still invovles me putting it in teh calcaulator unless its calc ready eq

[silent plank]

no calculator needed

[vocal bison]

bro waht

[silent plank]

120° is a nice angle

[vocal bison]

ok guys but i havent even learned that circle memorizing thing

[silent plank]

you can get the sine of that from unit circle/reference angles

/ supplementary identify

[vocal bison]

so like how cna I do it without using that convenient trig formula

[silent plank]

you're not really supposed to memorise the whole circle, but rather just the first quadrant
and you can apply the relevant identities/properties to get everything else as needed

well as you initially set it up

[vocal bison]

so was i right or wrong

[silent plank]

the start of the calculations were fine

you just needed to do more

instead of pythag (what's wrong with using pythag), you could do trig again to get the other side (which will in this case be the altitude)

[vocal bison]

isnt that what i did

i give up

[silent plank]

no

you found the length of the chord

and then not much else

if you want to use A = bh/2
you'd need the length of the altitude you drew (which you didn't find yet)

please read the convo

[vocal bison]

i did tho wasnt it 9

[silent plank]

did you make a pic to go with your work?

[vocal bison]

|uiy\

chol on

[silent plank]

and now apply the area formula

[vocal bison]

um

81 radical 3

[silent plank]

yes

[vocal bison]

wait si ti suppose to be minus

[silent plank]

yes, that's what you should be subtracting

[vocal bison]

ok

thx

[ivory sphinx]

How do I do this> Given sen x = - 3/2 times square root of 3 and pi<x<3pi/2 what is the value of y=(1+cos x) times ( 1-cos x)

[glass needle]

can someone help me with some circumferences problems?

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[upper karma]

Stupid bot lmao

[silent plank]

stupid how?

[dark sparrow]

consider, if you haven't already: (1+cos(x))(1-cos(x)) = 1 - cos^2(x)

however! $-\frac{3}{2}\times \sqrt{3}$ doesn't lie in $[-1, 1]$. are you sure you typed it correctly?

[somber coyote]

Ann

[glass needle]

Let a, b and c be three positive real numbers such that a^{2}+b^{2}=c^{2}.
Sketch the circumference of equation: x^{2}+y^{2}-2ax+2by=0.
Also indicate their points of intersection with the coordinate axes.

Please help me with that, you can't use calculus

[placid latch]

How will you make $x^2 + y^2 - 2ax + 2by = 0$ an equation of a circle?

[somber coyote]

YoAsassin!

[placid latch]

What can you tell from the fact that AC = BC?

[ivory sphinx]

It's 3 over both two and the square root of three multiplying the 2

[manic charm]

isosceles

idk how to say it

[placid latch]

Can you say anything about the angles?

[dark sparrow]

ok then you wrote it improperly!

-3/(2 sqrt(2))

[manic charm]

<D=<E

[placid latch]

About the isosceles triangle

In a $\triangle ABC$, if AC = BC, then $\angle CAB = \angle CBA$

[somber coyote]

YoAsassin!

[placid latch]

Do you know this property?

[ivory sphinx]

So given that sen x = - 3/2.sqrt of 3 and that it lies on the third quadrant find the value of y = (1+cos x) (1-cos x)

[manic charm]

no

[placid latch]

It’s pretty basic, you should keep it in mind

[manic charm]

what is it

[obsidian harness]

In an isosceles triangle, the base angles are the same

[manic charm]

so

<CAB=<CBA

[obsidian harness]

Yes

[manic charm]

what is the reason

[obsidian harness]

RHS congruence

Let M be the midpoint of AB

Then triangles ACM and MCB are congruent by RHS

@ivory sphinx Have you tried using (1+cos(x))(1-cos(x)) = 1 - cos^2(x) as Ann said?

As in multiply top and bottom by 1 + cos x

[lapis hornet]

hey guys I'm applying to a 2 yr junior and senior boarding school next school year for math and science and since covid happened I am trying to catch up on all my math in order to be ready for advanced classes. Over the summer I need to learn geometry so I can double classes. Any resources or advice?

[severe quail]

Khanacademy has a nice geometry course. check it out and see if it works with your learning style.

[glass needle]

I will send the message again because I got asleep yesterday and I don’t know what happened next.
Let a, b and c be three positive real numbers such that a^{2}+b^{2}=c^{2}.
Sketch the circumference of equation: x^{2}+y^{2}-2ax+2by=0.
Also indicate their points of intersection with the coordinate axes.

Please help

[silent plank]

approch with completing the square to get the equation into standard form

[glass needle]

What’s is the standard form? They haven’t teach that to us in my class

[silent plank]

centre radius form
if you don't understand certain terminology, do a quick search

[ivory sphinx]

what do I do with that? Sorry if it's basic I need to know how to do that but I haven't reached this part yet and it'll probably take some time

[obsidian harness]

Do this @ivory sphinx

[ivory sphinx]

Ok

Now what after that?

[upper karma]

Someone help plis

[upper karma]

a levels?

[obsidian harness]

The key part is cos^2 x + sin^2 x = 1

And both sin and cos x are negative in the 3rd quadrant

Solve for cos x

There's actually no need to multiply by 1 - cos x actually

Just substitute the value of cos x in

[lapis hornet]

Thanks

[ivory sphinx]

Thank you!

[obsidian harness]

Npnp

[buoyant cypress]

can u share a complete fig?

i cant see the rhs

and is there a certain point u are stuck on or u cant find how to start?

[robust estuary]

arnav mujhe mat mar

[shadow fable]

Basic Mathematics by Serge Lang will cover that and more. It's a pre-calc textbook. Then use something like Khan Academy for videos and actual practice.

[lapis hornet]

Thanksss

[indigo needle]

deleted : i found out the problem for the site not accepting my answer was that they rounded DOWN in example answer despite it ending in 7

[hard trout]

Hi all, finding the equations for hyperbolas; all's fine and understood; but these questions; I'm missing out on the visualisation of what this questions asking? It just hasn't clicked in my brain;

Would the two points be the foci?

I clearly struggling with the description; I understand hyperbolas; but not the question lol 😛

[dark sparrow]

indeed they would.

[hard trout]

So okay, you can then easily work out centre, and C, but the 6? 2b?

Sorry; 2a?

[astral mural]

Does anyone know how to do this 😭😭

[vernal pilot]

!status

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[vernal pilot]

Also which part specifically?

[astral mural]

I’m stuck on number 2 because we’ve been learning how to find area with similar triangles but I’ve never done it with a pyramid before with cross sections and stuff so I’m just like completely lost 😭😭

[vernal pilot]

It’s a very similar concept. The ratio between the height and area of base will stay consistent.

,,\frac{h_{1}}{b_{1}}=\frac{h_{2}}{b_{2}}

[somber coyote]

TheLord26

[upper karma]

hello

[urban hemlock]

whats up

[deft drum]

I want to learn more about the origin of word affine and how it is related to the origin. Coming from the topic, shifting the origin is also called affine transformation from one of my discussions on the machine learning lecture.

Hermann Weyl used affine geometry to introduce vector addition and subtraction.
-- Wikipedia https://en.wikipedia.org/wiki/Affine_geometry#History
Wasn't vector operations already defined before 1918?

[zealous pike]

Can I ask about coordinates geometry in here?

It wants the distance between the point B and the point D. Is "c" the answer? given:

$\sqrt{(-120 - 3)^2 + (47 - 80)^2 + (32 - 55)^2}$

[somber coyote]

Pi

[shut granite]

Bro can someone check me?? Im literally helping a friend with their geometry homework and the system keeps saying their answers are wrong

[hard trout]

Can someone explain here what might be an approach to this (question 23) -Hyperbolas. I have established it's a rotated hyperbola; (tan-1(3/2)) rotation simplifies it to be on the X axis, with foci at +/- sqrt(13). I can then workout 2a=4, but all this gives me a b that's a negative; and it all falls over. 😜 What am I missing? The answer given is in the standard form. Anyone willing to point me in the right direction? Half familiar with rotating functions in the standard form, but working backwards from these points is a non starter for me

Have been trying the approach of rotating and translating to make it easy in the general form so I can expand it then figure out how to rotate it back, but clearly I'm not at that point

Sorry foci at +/- sqrt(13)/2

[hard trout]

That looks right to me, assuming z and y are parallel; which there's nothing to suggest they are not. Didn't check all the basic addition, just looked through. Seems legit.

[hard trout]

The centre point in that equation should be 80-43 I think.

**80-47 sorry

[hard trout]

No, disregard, your right. Looks correct to me.

[mellow zodiac]

Hi

[junior sluice]

this geo?

[mellow zodiac]

Yeah, high-school geo

[junior sluice]

why is the wording so bad 💀💀

[mellow zodiac]

Idk it's a headache

[junior sluice]

which one we looking at

[mellow zodiac]

Basically angle X would be 2C, but how do we figure out C

[mellow zodiac]

In the figure O is the centre

[junior sluice]

lemme see

ima draw it out rq

[mellow zodiac]

Alright

These are the formulas I tried using

[junior sluice]

i haven’t done geo in a long time i don’t remember every single formula tho

[mellow zodiac]

Same

It's for a friend of mine

[junior sluice]

oh

we know that o is the middle so from OA and OB is the same distance

bc their both the radius

[mellow zodiac]

Mb go on

[junior sluice]

oh

it’s 4 sides so it’s PROB that all angles added together =360

[mellow zodiac]

What about the angle on the inside of x?

[junior sluice]

obtuse

wait

over 180

WAIT

IM SMART

WHAT IF WE DRAW A LINE FROM A TO B

[mellow zodiac]

Wouldn't that be useless though

[junior sluice]

no watch

gimme a sec

oh yeah wait ur right

[mellow zodiac]

Cause we still wouldn't know the full <a and <b there is the separate 20 and 30°

If we're to draw two chords from a point outside the circle we would get 90° on both ends where the radius touches

[junior sluice]

i’m smart

[mellow zodiac]

Nice, but what about the centre two angles

[junior sluice]

we’ll figure it out

ok wait this would be a square right?

wait no i’m bugging

suppose we draw a line through the middle of the square

[mellow zodiac]

360- (70 +60)'

50?

[junior sluice]

no

[mellow zodiac]

I think the answer is 50 here?

[mellow zodiac]

Well yeah actually it doesn't seem like 50 here tbh

[junior sluice]

🧠

the right side works the left doesn’t

[mellow zodiac]

💀

[junior sluice]

💀

lets pretend the left side is a 30 60 90

my drawing could be inaccurate so that’s why it don’t look like it

[junior sluice]

i got 60 💀💀💀

not even an answer choice

[mellow zodiac]

Perfect 💀

[junior sluice]

lets see what went wrong

[mellow zodiac]

It looks like it is more than 80°

[junior sluice]

x?

[mellow zodiac]

Yeah

[junior sluice]

lets think outside the box

[mellow zodiac]

Circle*

[junior sluice]

sure

[mellow zodiac]

360 - 70+60

230/2

115

It's closer to 125

So 125 it is

[junior sluice]

idk man

[mellow zodiac]

Maff 👍

[junior sluice]

oh forgot all my formulas so ima try and use another way

[dark sparrow]

what kind of guesswork is this

can i see the original problem

cause ngl i am curious now

[junior sluice]

up here

it’s solvable i’m just dumb

me using triangles to solve this proves every theory wrong

[junior sluice]

WAIT

I DISCOVERED A BREAK

[dark sparrow]

i've recreated the diagram bc the one in the photo was tiny and badly angled and horrible to look at.

i have a feeling y'all have both overcomplicated it.

[mellow zodiac]

That feeling is correct lol.

[junior sluice]

bc we forgot the formula 💀💀😭🙏

[mellow zodiac]

WOW what did you use for that?

[dark sparrow]

look at the interior angles of quadrilateral ACBO...

• angle A = 20° (given)
• angle C = x/2 (inscribed angle thm)
• angle B = 30° (given)
• angle O = 360° - x (full circle minus angle x)

[dark sparrow]

MS Paint.

[dark sparrow]

what formula did y'all forget

[junior sluice]

💀💀

idk what’s going on

[junior sluice]

how am i supposed to know what formula i forgot?

[mellow zodiac]

Ok so x is 2c so 360-2c isn't it?

[dark sparrow]

why the skull reacts on this btw?

slippery

[junior sluice]

we forgot it exists

[junior sluice]

all this has to add to 360

WAIT IM COOKIN

[mellow zodiac]

Seemed like a mathematics modeling software

[junior sluice]

(i think)

[dark sparrow]

y'all are still overcomplicating

you specifically @

sfgjgladf

@junior sluice

you dont need any extra constructions here

[junior sluice]

💀

[dark sparrow]

i have listed all four interior angles of ABCO

and it's THOSE that need to add to 360°

once you write that down you get a simple linear equation in x

[junior sluice]

that’s what i said at the beginning

[junior sluice]

wait wah?

[dark sparrow]

.

can you write down, as an equation in x, the statement "The sum of all four angles in that list equals 360°" ?

[junior sluice]

A+B+C+O equals 360

wait wrong letters

[dark sparrow]

as an equation in x

please read the things i send fully.

also the equals sign is a thing.

[junior sluice]

oops

ik

💀

[junior sluice]

i’m going to bed bruh

ain’t no way i spent 5 hours studying on a weekend night

it’s 3 in the morning

see y’all tmrw 🙏🙏🙏💯💯🔥🔥🔥

[mellow zodiac]

Goodnight!

[mellow zodiac]

@dark sparrow what am I doing wrong

[dark sparrow]

uhh

third step?

how did 70 happen

also your handwriting (or is it fingerwriting?) is bad

from $410 + \frac{x}{2} - x = 360$, how did you get $70 + \frac{x}{2} - x = 0$? @mellow zodiac

[somber coyote]

Ann

[mellow zodiac]

Sorry about it.

It is 100

[mellow zodiac]

410-360

[dark sparrow]

410 - 360 is not 70.

[mellow zodiac]

😀

[junior sluice]

💀

[mellow zodiac]

It is 50 💀

[dark sparrow]

by the way you could have also just cancelled out the 360's right from the get-go.

in doing so, you would have not even set yourself up for such a silly arithmetic misstep.

[junior sluice]

wait what? there is no way x is 50 x is an obtuse angle 💀

[mellow zodiac]

Oh yeah that would work too though 💀

[junior sluice]

going to sleep fr this time

see yg tmrw

[dark sparrow]

x is 100, but also the diagram's not to scale anyway.

[junior sluice]

wait ur saying c is 50?

i mean o

whoops

[mellow zodiac]

Thanks a lot @dark sparrow
We were trying to complicate it too much here.

[junior sluice]

frfr

ty

[ocean lily]

Nvm

[cloud sedge]

can smbdy help me with my conic sections hw

[dark sparrow]

well you gotta post the problems and all of your progress with them, otherwise nobody can.

[celest urchin]

someone help me i’m begging and i’m desperate

[junior sluice]

wait i’m learning the same thing

[celest urchin]

unfortunately for us

[junior sluice]

real

[lavish patio]

Hello

[weak mauve]

learning how to graph tan waves is te shittiest thing ever

[weak mauve]

If you plot a point midway between C and D, lets call that Z.

You can use SOH CAH TOA to find the length of AZ.

Then you can use Pythagoras to find CZ and ZD then add it up to find the total length of CD

[ocean lily]

what is soh cah toa

[vernal pilot]

It’s a phrase used to remember how trig functions work

,,\sin(\theta)=\frac{opposite}{hypotenuse}

[somber coyote]

TheLord26

[vernal pilot]

Hence the soh part

[ocean lily]

so how do i find the length

[ocean lily]

is it root 124

[vernal pilot]

Idk what the question is

[ocean lily]

he said used sohcahtoa for this

[vernal pilot]

Google law of cosines.

[ocean lily]

what do i do

[vernal pilot]

You know ACD is a scalene triangle.

Solve for some angles and you can get an answer.

[ocean lily]

im a little confused

[fickle rose]

no

sohcahtoa only works for right triangles

[indigo needle]

hello! i would appreciate an explanation as to why a slanted version (parallelogram) of a rectangle with the same side lengths has a different area

[cunning lion]

imagine cutting off a triangle from a parallelogram and moving it over to make a rectangle

[indigo needle]

ohh okay i can see that

nvm! thanks 👍

[astral acorn]

why are these equal?

these as well apparently

[dark sparrow]

sin(u) cos(v) = 1/2 [sin(u+v) + sin(u-v)]

[astral acorn]

doesn't seem like this is why

[dark sparrow]

what do you mean by "doesn't seem"

[astral acorn]

I can't get it from that form to the one I showed

[dark sparrow]

sin( (a+1)x ) cos(ax)

= 1/2 [sin( (a+1)x + ax) + sin( (a+1)x - ax)]

[onyx bobcat]

Guys im lost i dont understand this.

[vernal pilot]

?

[onyx bobcat]

Nah bro,i only understand 3rd Grade Trigonometry,Pythagoras theorem,exponents,tetration(a little) but not this.

[vernal pilot]

Google angle sum identity.

This requires trig identities

[dark sparrow]

3rd grade trigonometry?

[strange spear]

am i right that the angle right after 45 could be 180-45

i think just use sin law

[rose raven]

3rd grade trig, tf?

[strange crater]

waaa i was never any good at geometry… i have a circle S and two points p and q inside S, i need to construct an orthogonal circle passing through p and q and i’m stuck

any hints?

there’s probably something something inversion that would be helpful but i’m not sure of what

[north moth]

so i know i need to multiply by the conjugate to finish this problem, but I'm not arriving at the same solution.

[gray mason]

Ah 3eme?

[vernal pilot]

Which part are you getting wrong?

[solid crypt]

What on earth does it mean by use 80degrees to verify the identity

[north moth]

I worked it out doing this and simplifying. I know thats the longer way but it's what im comfortable with atm

[cunning lion]

plug in theta = 80 degrees to both sides and see if they're equal

[junior sluice]

isn’t this just trig identity?

[junior sluice]

they’re equal we just gotta show how we know

we use theories to make the left equal to secant squared

[junior sluice]

i think i’m lost

"The Pythagorean formula for tangents and secants. There’s also one for cotangents and cosecants, but as cotangents and cosecants are rarely needed, it’s unnecessary." -Clark University 😭

im sped

[lime dune]

write $1=\frac{\cos^2\theta}{\cos^2\theta}$

[somber coyote]

elrichardo1337

[lime dune]

now the LHS becomes $\frac{\sin^2\theta+\cos^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}=\sec^2\theta$ as desired

[somber coyote]

elrichardo1337

[merry mulch]

qno 48 please

[dark sparrow]

!status

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[merry mulch]

2

[dark sparrow]

show current progress?

[merry mulch]

just a sec

this is the solution i found in web

but i have some doubts.

btw is that solution correct?

the name of sides are different but rest is same

[dark sparrow]

this is status 5, not 2

[dark sparrow]

it appears so.

but you should state your doubts.

[merry mulch]

my bad

[merry mulch]

can we write GF as diameter. if yes then how?

[dark sparrow]

what do you mean by "write GF as diameter"?

do you mean "can we claim that GF is a diameter? if so, then how do we justify it?"

[merry mulch]

yes

[dark sparrow]

triangle EFG is an inscribed right triangle, with E as its right angle

[merry mulch]

still confused.

[dark sparrow]

the circumcenter of a right triangle lies on the midpoint of its hypotenuse,

and thus the hypotenuse of a right triangle is a diameter for its circumscribed circle.

[merry mulch]

got it. thanks

[cerulean turtle]

Isn't this supposed to be 180?

[silent plank]

what exactly do you think is supposed to be 180°

[cerulean turtle]

all angles inside a triangle add up to 180

[silent plank]

they aren't quite using that here

[cerulean turtle]

I'm confused

[silent plank]

they're not using angle sum of a triangle directly

they're applying exterior angle theorem

(which is a result of combining angle sum of a triangle and angle sum on a line)

[cerulean turtle]

So is her answer correct?

[silent plank]

yes

the red angle is adjacent to the 115° angle
the interior opposite angles of the triangle refer to the blue and green angles (that aren't adjacent)

look up

exterior angle theorem

[desert sail]

Exterior angle of a triange is the sum of its interior opposite angles.

[silent plank]

that was already in the chat log they posted

[desert sail]

oh mb

[heady cedar]

helloo, what channel should I go to if I wanna ask about permutations and combinations?

[silent plank]

!help

[lime crown]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[heady cedar]

thank youu

[desert sail]

I considered LHS, changed the sec^2 x to to (1 - tan^x) using identity and also the cot x into cos x/sin x

now after simplifying the expression ive gotten, cos x - sinx / sin x/ 2(1 - tan^x)

[dark sparrow]

horrible lack of parentheses. better to write it on paper.

by the way sec^2(x) is 1**+**tan^2(x), but changing it to that would be no good either.

instead, consider writing sec^2(x) as 1/cos^2(x).

[dusky locust]

hm this is a false premise

wonder if should this say “Prove if:”

[dark sparrow]

good point lmao

it doesn't even work if x=pi/4

@desert sail where did you get this from?

[foggy gust]

1. (2 points) On the side AN of triangle AMN, point K is marked such that ∠AKM = β. Find MK, if AN = a, ∠A = α, ∠M = φ

help somebody pls i don't understand

[dark sparrow]

@foggy gust в чем именно затык? не знаешь, как начать? или какой-то прогресс есть, но где-то посередине ты застрял?

и еще такой неудобный вопрос: почему на задаче написано "2 балла"? уж не из контрольной ли она взята?

(у нас с такими вещами строго: если палишься, что пытаешься списать на контрольной (или экзамене, и т.д. и т.п.), которая проводится прямо сейчас, то за это банят)

(но если контрольная уже прошла или это прошлогодний вариант или что-то в этом духе, то все в порядке. поэтому спрашиваю)

[foggy gust]

это после контрольной уже

и не моя даже была

просто таск интересный и я его решить хотел

первое изи

а тут я просто не понимаю что делать

мне чат-гпт че-то предложил

использовать теорему синусов

[foggy gust]

я просто не знаю как к формуле прийти

ваще не сталкивался с такими задачами раньше

[dark sparrow]

эээ

гпт верить нельзя ни в коем случае

[foggy gust]

но вроде как он адекватно написал все

типа

[dark sparrow]

так в этом и проблема

[foggy gust]

mk/sina = an/sin (fi)

а не

забей

не адекватно

мудила

он

[dark sparrow]

ГПТ умеет полный нонсенс написать так, что для незнающего звучит адекватно

[foggy gust]

:blobsob:

[dark sparrow]

так вот

по-моему здесь все-таки кроме теоремы синусов ничего и не надо

применяешь ее к треугольнику AMN, чтобы выразить AM, затем к треугольнику AMK, чтобы выразить MK

[foggy gust]

т.е так?

но ведь

[dark sparrow]

ну и надо осознать, что функция синус обладает свойством sin(180° - x) = sin(x)
то есть N = 180° - alpha - phi => sin(N) = sin(alpha + phi)

[foggy gust]

MK не относиться к треугольнику

[dark sparrow]

??

[foggy gust]

я не понял

:deadge:

[dark sparrow]

ладно, давай распишу поподробнее

применяя теорему синусов к треугольнику $AMN$ (который большой), получаем: $$\frac{AN}{\sin(\varphi)} = \frac{AM}{\sin(\angle N)} \sqb{= \frac{AM}{\sin(\alpha+\varphi)}} \implies AM = \frac{AN \sin(\alpha+\varphi)}{\sin(\varphi)}$$

до сих пор понятно? или что-нибудь из написанного требует объяснения?

если что, это еще не конец

[somber coyote]

Ann

[foggy gust]

всевсевсе

понял

дошло наконец-то

[foggy gust]

я дальше сам могу

спасибо

[dark sparrow]

рада была помочь

[foggy gust]

AM/sinb = MK/sina

AM = MK*sinb/sina

и получается

MK*sinb/sina = a*sin(a + fi)/sin(fi)

и так можно МК выразить да

изи

спасибо большое

[hearty lintel]

What is the Inertia Momentum of the triangle?

Ix.

[desert sail]

Well it was asked on my exam a few days ago, I couldn't solve it and none of friends could, i tried it agai later and still couldn't, so I thought of asking here and since I've concluded it isn't possible, I've notified my teacher she said its wrong question so points will be given to everyone 👍

[sturdy isle]

In code this translates "ÿØÿà"

[sturdy isle]

Nice

Dead chat

[dark sparrow]

a 7 minute silence constitutes "dead chat" by your standards?

[sturdy isle]

Almost

[inner ingot]

используя лемму про теорему синусов, получаем что MK= sina*гипотенуза

[frail frost]

do i have to like geometry to go into an engineering field...

[foggy gust]

какая гипотенуза? не дано же что треугольник прямоугольный

я уже решил кстати, спасибо

[inner ingot]

нужно перпендикуляр провести

ирешать уже с новым треугольником

[dark sparrow]

да какой нафиг перпендикуляр?

усложняешь только

[inner ingot]

мне что первым на ум приходит, так и решаю. Если бы это было для меня усложнением, моя первая идея не заключалась бы в ней

я уже понял что он решил, я просто написал как бы я решил

[vocal bison]

Help plz ping

[weak mauve]

since when this became a russian channel lol

[weak mauve]

we need to know h to answer this

[dark sparrow]

it never was specifically Russian

[junior sluice]

if it’s silent for 3 minutes i assume that chat is dead

[dark sparrow]

[woeful depot]

I'm doing a 3d trig word problem. Is there a way to get AI to check my understanding of the word problem, and check my solution?

[dark sparrow]

no

!nogpt

[lime crown]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[empty sparrow]

Hi can anyone help with the last two??😭🙏

[dark sparrow]

@empty sparrow ok so you know triangles XWZ and ZWY are similar, right

[empty sparrow]

Yeah (I forgot something and was confused sorry thank you very much)

I’m trying to figure out XZ (question e) but I’m not sure how to use the formula in the hint with the problem

[dark sparrow]

you are given WZ and WX

the legs of a right triangle

this is the most direct use case for Pythagoras one could possibly think of

[dark sparrow]

XZW is a right triangle with W as its right angle.

do you know Pythagoras' theorem?

[empty sparrow]

I thought it would be G squared (I’m using that to represent XZ) = 9 squared + 18 squared

But I got the wrong answer

[dark sparrow]

show work that led to your wrong answer

[empty sparrow]

I did it in the 7th grade but I forgot it now 😭

[dark sparrow]

also use the symbol ^ for exponents

G^2 = 9^2 + 18^2 is correct thus far

show the answer you got and how you got it

we must rule out arithmetic errors

[dark sparrow]

claims to have forgotten it
cites it correctly just now

[empty sparrow]

That’s what I got but the answer is supposed to be 20.1

[empty sparrow]

Isn’t it understanding how to also use a^2 + b^2 = c^2 ? 😭😭

[dark sparrow]

81+324 is not 243. it WAS an arithmetic error on your part.

[dark sparrow]

what, according to you, is the difference between a^2 + b^2 = c^2 and the pythagorean theorem proper?

[empty sparrow]

I subtracted the 81

[dark sparrow]

why would you subtract it??

it's literally being added.

81 + 324 is not the same thing as 324 - 81.

[empty sparrow]

Sorry I got it thank you, idk why I was subtracting it I’ve been doing cross multiplication with the right triangle altitude theorem so I probably just got it confused😭

Thank you

[empty sparrow]

Thank you for your help I got my hw in by 11:59🙏

[steel carbon]

I want to know what I should learn in trig before I move on to linear algebra I currently know, Pythagoras theorem, Pythagorean trig identities, sine law, cosine law, Thales triangle theorem, inequality triangle theorem, intersecting chord theorem and triangle similarity/congruency. is there anything else I need before I move to linear algebra?

[dark sparrow]

trig to linear algebra is a leap

do you know your way around vectors in the pre-linear-algebraic sense?

[steel carbon]

no

im not sure what the next step is so ima try linear

can you tell me though? what i should know in trig before i move on

because i only have surface level understandings of trig and i haven't built an intuition in trig

all i know , is the linear graph equation [ y=mx+c]

[dark sparrow]

this has very little to do with linear algebra

[dark sparrow]

ok then learn vectors.

[steel carbon]

ah ok

[dark sparrow]

and the geometry to do with them

[steel carbon]

thanks

wait... are vectors a partof linear algebra or do you learn vectors in linear algebra?

[dark sparrow]

1. yes, definitely. but the concept of a vector in linalg is much more abstract.
2. depends. but usually, uni students going into linalg have some experience with vectors in two and three dimensions.

[steel carbon]

oh, so... which branch teaches the vectors and their theroms?

wait, would linear algebra go more into physics or is it still pure math?

[dark sparrow]

.....

branch of what

[steel carbon]

math

[dark sparrow]

if you twist my arm i will say (elementary) geometry

[steel carbon]

geometry teaches vectors?

ah wait ight ,i got it, thanks... Ima do some vectors now 😄

[mortal crow]

For me, the end of Precalculus introduces vectors for the first time

[steel carbon]

nevermind, linear algebra from trig is a leap... so yeah, ima prolly do algebra 2 now then 3 n then finaly linear algebra.

[upper karma]

I have a trigonometry problem in #help-23

[fickle pawn]

How in the world do asymptotes work

I actually nuked a test cause I still have no idea how to do them

[sonic bloom]

Geometry #help-32

😔

[vernal pilot]

An asymptote is whenever a graph doesn’t go past a certain x or y value. For instance in the graph log(x), there is an asymptote at y=0

[lime dune]

not quite

a graph can cross its horizontal asymptotes

horizontal asymptotes describe long-term limiting behavior of the graph of the function

[cunning lion]

the main thing about an asymptote is that it's a line that a function gets arbitrarily ("infinitesimally") close to. for example, the function f(x) = sin(x)/x has a horizontal asymptote at y = 0 but crosses it infinitely many times

[mortal crow]

Ya read my mind

[untold tangle]

can anyone help me?

[upper karma]

https://www.cuemath.com/learn/geometry-theorems/ I would recommed reading through all the theorems if you don't know them already.

[untold tangle]

oh, okay

[upper karma]

What are you stuck on?

[untold tangle]

you have to find the angle size of ACB

and im not entirely sure if my answer is right

[mortal crow]

To clarify (I haven't learned this in class yet, but I think I figured it out), the sine function is y = a sin b(x - h) + k (and the cosine function as well), but the tangent function has 'b' in the parentheses | y = a tan (bx - h) + k

Correct?

[sturdy stone]

"The" sine function is just sinx

[mortal crow]

Wait wait wait.....so what is y = a sin b(x - h) + k referred to as? I want to call it the parent function but that doesn't seem right...

[sturdy stone]

Probably as a sine wave or a sinusoid

[mortal crow]

Ohhh yeah, it sinusoid. Thanks!

[bronze wind]

How to solve #20?

[surreal escarp]

@lone panther hope u dont mind the ping because the channel from last night is taken,

thank u for ur help! but for the last one i got the answer for what AEB and DE is, but im confused on how that will help me find DB because AEB is an angle while DE is a line? ty

#help-46 message (og image)

[ebon pier]

alternate angle theorem

the opposite sides of a parallelogram are parallel so we can take the alternate angles to be equal to eachother obtaining 2 equations

hence obtaining the values of x & y

yes

[ebon pier]

in this case it is 4x -8 = 8y - 12 and x/4 = y-8

[surreal escarp]

wait nvm it would be 120 too right because of the bisectors

[regal cave]

p

p

p

anyone here?

pp

please help

no. 5

[dark sparrow]

<@&268886789983436800> spammer

[regal cave]

NO ITS BEACUSE ITS GOING TO BE SOON

I NED HELP

[dark sparrow]

what?

[regal cave]

OR IM GOING TO GET SCOLDED

no.5 pelaseee

[lethal needle]

do you know what sin and cos are?

[dark sparrow]

what's troubling you with problem 5?

[regal cave]

uhhh

everything, I'm so used to cheating so idek what these are

so please bro

teach me or answer the question

i really want to know

[dark sparrow]

well we do not condone cheating.

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[regal cave]

yeah tmrw im gonna have a test