geometry-and-trigonometry

[pastel verge]

no but that was before she found out DHE was 90

90 for chg was the first thing she wrote

is it because its a perpindicular bisector?

[whole venture]

Which angle are you confused about?

<@&268886789983436800>

[pastel verge]

🤓

[pastel verge]

CHG

[whole venture]

OK so do you believe that GHE is 90 degrees by the vertical angle theorem?

[pastel verge]

no you dont understand what im saying

what either CHG was nor DHE was

so she couldn't use it

[whole venture]

I'm trying to show you how to figure out that CHG and DHE both have to be 90 degrees given that CHD is 90 degrees

[pastel verge]

The only thing she was givin was CHD

wouldn't it only prove that GHE is 90

degrees

[whole venture]

Yes but if GHE is 90 degrees then CHG and DHE have to be 90 degrees because they are a linear pair with GHE meaning that the measure of GHE + CHG = 180

[pastel verge]

ah

Linear pairs have to add up to 180

okay makes sense

[pastel verge]

does it only work

with the diameter

[whole venture]

Any two intersecting lines form a bunch of linear pairs. For instance angles 1 and 2 must sum to 180 degrees. There's nothing special about the line being a diameter.

[unreal tinsel]

Hello everyone, I am looking for a student who knows English well (preferably native). I have a high level of mathematics and I want to learn how to teach lessons in English. I offer you free math lessons if you help me develop my English and teaching skills.

[pastel verge]

whats ur native language

just curious

[upper karma]

Can I get help?

The two questions above

I have solved the first one but my solution has no relation to the second (I think that the solution of the first ought to, in some way, get you the solution of the second)

Here's the first

Um, why can't I send an image here?

[dark sparrow]

you should be able to

something fucked up on your end

,rccw

[upper karma]

oh

'ere if anything is intelligible

[unreal tinsel]

Rus

[maiden brook]

can you make a regular hexagon from any three points? what about four?

[strange quarry]

thats the most cursive writing ever ive seen

[proper sierra]

Anyone know of any good resources that has a good amount of practice problems for pre calc,trig and calc 1?

[upper karma]

@strange quarry Newtonian!

@strange quarry What about this then?

[signal night]

Does anything in trigonometry derive from algbera 2?

I.e. can I have studied 0 alg 2 & proceed from alg 1 + geometry -> trig

[upper karma]

yes and no

in American high schools

"Algebra 2" is essentially an intro to compex numbers and is combined with the same class as trigonometry

or you could say no because they are treated as two completely different subjects

but you could also say complex numbers and trigonometry are very similar and closely related

take note of the identity eiθ=cosθ+isinθ

do you understand what I am getting at @signal night?

[maiden brook]

can someone please explain why AEDB is a rectangle?

(the figure shown is a cube)

[grave pond]

Because it lies in a plane and each corner makes a right angle.
For example, ED is perpendicular to everything in the BCDG plane, so in particular to DB.

[maiden brook]

oh yeah i see

[maiden brook]

more accurately ED is perpendicular to everything in the BCDG plane that runs through D, am i correct?

[grave pond]

Well, its direction is perpendicular to the direction of any line in that plane -- but they only intersect if the line passes through D of course.

[maiden brook]

ah i sorta see

[pastel verge]

Would it just be 180?

I remember someone briefly saying that the diameter would be 180 but i dont know why

[timber cargo]

Circle is 360 degrees, since the diameter splits it in halve, halve of 360 is 180

[crude inlet]

factual

[timber cargo]

^ @pastel verge. If you want a more technical explanation. Use s=rθ. S being the sector length of the circle, r the radius, and θ in radians.

Let r be 1

set s = pi times r

r cancel out

you're left with pi=θ

pi (from radians to degrees) is 18-

180*

[upper karma]

i just wanna thank y’all for helping me lol, i definitely would not have a B+ in this class without you

[runic flume]

What class are you in rn?

Or grade u might call it

[signal night]

What has complex numbers have to do anything with trigonometry?

I guess they both involve some sort of rotation but other than that I feel like they're independant.

[signal night]

isn't euler's formula buried somewhere in calculus anyways?

[dusty sun]

Hello everybody!

I am quite new here.

Can anyone here recommend me an easy book to understand the basic and integral concepts of trig

[lethal moss]

#book-recommendations You'll find some help here

[upper karma]

they are, but as i said i was talking about american high school curriculum 😭

[trim meadow]

Yes and no (it’s complicated lol)

[copper flare]

why does subtracting X from pi not change the graph?

[somber coyote]

Dissrupt

[merry parrot]

or you could transform the graph to confirm

[restive fern]

help pls

[lost jungle]

So I’ve been surfing through khan academy and now I’ve stumbled on to this equation. y= 1/(2(b - k)) * (x - alpha) ^ 2 + 1 2 (b+k)

I need help understanding what it means

[dense tulip]

Ok so for the first triangle it’s 45 45 90

I don’t remember the formula for that but what I do know is that you can find the sides by using the trig ratios

In case you forgot

Sin:opposite/hypotenuse cos:adjacent/hypotenuse tan:opposite/adjacent csc:hypotenuse/opposite sec:hypotenuse/adjacent cot:adjacent/hypotenuse

[teal bison]

@thin sentinel hallo

[mental pelican]

Guys how do you find the area of that triangle ik we can use herons by finding the diagonals but the diagonals are in roots so herons is too much work and definitely not intended

[lapis torrent]

Try using similarity

And after that heron's formula

[lapis torrent]

U dont need to find the diagonals

[upper karma]

how do i do these

nvm

[whole marlin]

correct option is (b)180

because circle has 360 degree ,diameter is half of it that is 360/2

=180

[vernal pilot]

For the left triangle a very simple solution would be to use Scale Factor. Assuming you know your exact value triangles 8/sqrt(2) should be the answer for both a.

This is because sqrt(2) is side length when a=1, so when you divide 8 by sqrt(2) you are basically seeing how much larger this triangle is to the exact value triangle. For the right triangle you would then have to multiply the result by the usual value.

Alternatively, use trigonometry.

[vernal pilot]

You need to know the specific rules for each of them. Google Circle Theorems or tangent lines to a circle rules. An extensive list should show up.

[upper karma]

yeah i figured it out

[dense tulip]

How do you find the period

I know how to find the amplitude and I know that b is 2 pi divided by the period

But I always confused on how I’m supposed to find the period

Especially with this small graph

I’m on question a btw

[lethal moss]

Can you use your imagination to visualise how this graph would look like if we extend the x axis?

[dense tulip]

I came up with this

[lethal moss]

Not ideal but for our purposes it's ok (sorry for not replying, I had to go somewhere)

Now, please tell me what a period is

[lethal moss]

I came up with this

[dense tulip]

W period is a full cycle

[lethal moss]

Now, look at the image, after what x value does the graph repeat itself?

[dense tulip]

Does it repeat on -2 and 2?

[lethal moss]

No, it has to include the "upper slope" too

But you were on the right track.

[dense tulip]

Ohhhh now I get it

[lethal moss]

So...?

[dense tulip]

So it’s -2 and 6

[lethal moss]

Yes

[dense tulip]

Which 6-(-2) equals 8

[lethal moss]

Exactly

Problem solved i guess🙂

[dense tulip]

Yep thank you

But I do have a couple more question if you have the time?

[gentle haven]

is there any way i can practice proofs in geometry

im struggling a lot with them

[vernal pilot]

A whole lot of practice. It’s good to just list off facts and see which parts seem like they could fit in a rule.

[pine mango]

no idea lol

[solid coral]

Can someone help me with graphing sinusoidal functions with phase shift?

[pseudo nebula]

God that is hard

[pseudo nebula]

I think i'm going crazy

[pseudo nebula]

Assuming z is minimized when obtuse angle = 180
x^2 + y^2 + 2xy = (z+11)^2
x^2 + y^2 = (4z-6)^2
assuming z is minimized when x = y
3x^2 = (z+11)^2
2z^2 = (4z-6)^2
2(z+11)^2/3 = (4z-6)^2
2(z^2 + 22z + 121)/3 = 16z^2 -48z + 36
z^2 + 22z + 121 = 24z^2 - 72z + 54
23z^2 - 94z - 67 = 0
47/23 + 25sqrt(6)/23 < z < 17/3
(Yea don't trust me idk wtf im doing)

[worn hollow]

Does anyone have access to the Eurographics 87 archive? I need an academic article from the proceedings and can't find it anywhere online. I'm willing to buy it.

[crude vault]

Is this still unsolved?

[pliant onyx]

gravitional pull of the sun

[somber coyote]

zouiro

[tepid turtle]

Can one help me with this one

[pulsar nebula]

Is the helium molecule at the center most point of any star

Any different that the helium molecules at the outer perimeter of the star ?

anyone know ? i saw someone talking about the sun / gravity

geo-spatially they would be different but in how we express the helium in a thermodynamics model of the solar system the star sits at the center of ...

is it okay to just treat all helium the same in the math ?

[upper karma]

I have two coins in my desk. The one in the bottom is slightly larger than the other. I draw a cone between the borders of the two circles.
As I go up and lose the first coin of sight, the cone seems to get straighter. This is obviously an illusion because both coins have the same size and are only moving in the y direction.
When will the cone ever seem to draw a straight line / cone instead of a "triangular shape"? And will it ever seem bigger than the original coin (on the bottom)?

[humble peak]

I'm a bit confused what you're talking about. I tried it in real life, but I couldn't replicate it (but i'm pretty short)

So I remade it in blender

are you talking about the second ring got smaller as we got farther away?

[upper karma]

You have two coins. The little one is on top of the other. You bring the little one on the top up 5 cm in the air and draw a cone with the one on the bottom.

Does that make sense?

let me do it on blender

[upper karma]

straighter lines. to infinity?

[south summit]

AKBP is blocked in a circle, how to prove that E is part of the circumference of the circle?
and its also given that E split MN in half

[hardy bay]

Is MK=NK supposed?

[strange quarry]

its really pretty

[undone escarp]

do you still need an answer?

[south summit]

no

[lucid moth]

anyone know what theorem this is

i think its alternate segment theorem but im not sure since its outside the triangle

[bleak blade]

EDF is tangent, so we know from tangent lines that angle CDF is equal to half the arc CD

using the parallel lines, we get angle ACB is equal to angle CAD

finish from there

[upper karma]

conider a rectagular lamina is inserted inside this cube which is obliquely restin over the surface ABCX(the corner without any name, at the back).

[upper glen]

To solve this geometry question, you will need only 3 things, segment angle, internal alternating angles and inscribed angle, the inscribed angle is always half of the arc, the segment angle will also be half of the arc and internal alternating angles are equal, see my solution

[spring mist]

interesting

[upper glen]

the CDF angle will be half of the CD arc, the same thing goes for the DAC inscribed angle and then we conclude that they are equal, and because they are parallel straight (AD and BC) we use the internal alternating angles and find that DAC is equal to ACB

[lucid moth]

I see thanks for your explanation

[upper karma]

i need help

is ANYONE online

[lyric sonnet]

yes

[upper karma]

the radius of this is 5/2

right?

[merry parrot]

how?

[lucid moth]

radius is 3 no

6/2=3

[lyric sonnet]

and...

[upper karma]

yeah

[upper karma]

i was js checking

wait lmmie send the question

[lyric sonnet]

ok

[upper karma]

how do i do this

[lucid moth]

V=1/3Bh

where B is the area of the base

so 8x8=64 64=B

h= 3 in.

64x3=192

1/3(192)=64

V=64

[upper karma]

ohhh

can u help with this

and this

@lucid moth

[lucid moth]

erm

[lucid moth]

whats the question to this?

[upper karma]

find x

[lucid moth]

okay

so from what I can see

a 128 degree rotation is being performed

[upper karma]

yeah

[lucid moth]

A rotation is a rigid motion meaning size, shape, angle is perseved

so you just set the corresponding sides together

6x-19=2x+1

[upper karma]

ohh

[lucid moth]

6x-2x=4x

1+19=20

20/4=5

x=5

and if you want to find the side lengths just subtitute x

[upper karma]

ahhhh

[upper karma]

is this B

[lucid moth]

yes I believe

[upper karma]

oh okay

thank you sm

[lucid moth]

yeah np

[upper karma]

isn't this js 47?

[lucid moth]

measure of arc YZ would be arc YX+ arc XZ

47+129=176

[lyric sonnet]

yes

[upper karma]

yeah i fugured it out

[dense tulip]

Your welcome

[upper karma]

sorry this is my last question

[upper karma]

Thanks lol

[lucid moth]

ok so

V=πr^2h

is the equation for the volume of a cylinder

The question states that it consists of half so

[upper karma]

yeah

[lucid moth]

V of half cylinder=1/2πr^2h

[dense tulip]

Thanks ai you’re the 🐐 of geometry/trigonometry

[upper karma]

ahhh

[upper karma]

Fr tho

[lucid moth]

Do you want me to go on or you got it

[lyric sonnet]

h is length of 28 r is 14/2

[upper karma]

go on

cuz im still kinda confused

[lucid moth]

ok

[upper karma]

ahhh so

okay wait lemmie solve it rq

[lucid moth]

sure

[upper karma]

yeah wait

nvm

i'm losr

lost*

[lucid moth]

What are you lost on

finding the answer of the half cylinder?

[upper karma]

yeah

[lucid moth]

π·7^2·28≈4310.26512

4310.26512/2=2155.1325... cm

[lyric sonnet]

total volume= volume of 1/2 cylinder plus volume of rectangular block $ig$

[upper karma]

oh i forgot to multioply with pie

[somber coyote]

David

[lucid moth]

pi but yes

[upper karma]

yeah mb lol

thank you sm

[lucid moth]

np

[dense tulip]

The 🐐 once again helping people

Is it an honor to be in your presence

[upper karma]

this is wrong btw

[lucid moth]

thats not the total volume for lunch box

the volume for the rectangle prism is V=LWH

14x28x17=6664 cm

6664+2155=8819 cm^3

[upper karma]

ahhh

[spring mist]

confia q ele é o cabritum

[lucid moth]

fr he is a gold medalist according to his about me

[spring mist]

yeah he is the goat here on my country and his

[bleak blade]

which user

also gold medalist for what?

[lucid moth]

@upper glen

For a math competition prolly

[heavy shale]

I tried simplifying and putting values for m but still not working

how do you solve this

[upper karma]

B
area is xy=500 divide by x, y=500/x
the length around the area is 2y+2x minus the side against the barn, 2y+x=2*500/x+x=1000/x+x
set f’=0
0=(1000/x+x)’=-1000/x^2+1 subtract one, -1=-1000/x^2 multiply by x^2 and cancel the negatives each side, x^2=1000, x=sqrt(1000) and y is 500/sqrt(1000) which is close to b

But that should probably be in #calculus

[heavy shale]

someone plz help

[warm falcon]

'm a bit baffeled as of now.
I like checking out graphs of euqations(simple one in general). It is generally easy to understand why a graph looks the way it does for linear and a simple 2nd degree equations, even trigo functions makes sense.
But ones graph im not able to understand is graphs of equation of form of ax^2 + by^2 + cxy + dx + ey + f =0.
I've started a chapter of circles and parabola so i wanted to experiment. I dont understand why the graph changes the way it does when i change the values of the constants. another things i dont understand is why this equation in specific has such a specific maxima even though it has such a simple equation.

[tame breach]

Give me some difficult geometric question please

[uneven totem]

I think this maximum point can be found using the quadratic formula. Just like x^2+y^2= 2, it looks irrational. Anyways, what you have discovered are the famous conic sections. They have three forms: ellipses, parabolas and hyperbolas. Perhaps you can search up more about them now that you know what they are called.

Oh, its not irrational

[uneven totem]

Like, euclidean olympiad style geometry?

[tame breach]

I have absolutely no clue, i have to give this dude something to get busy with

[uneven totem]

Who is he

Difficalt is vwry relative

Here: How large portion of the rectangle is shaded?

[dark sparrow]

oh this is actually an interesting one

[frozen ocean]

Show work

[pseudo nebula]

are there any givens?

[warm falcon]

Thats pretty cool. Never could have thought it to be part of cone. very cool. thanks!

[warm falcon]

417/2000?

i didnt google it up. i just used calculator a bit

i used coordinate geometry and tried. bottom left as 0,0 and top right as 4a,b

[uneven totem]

Coordinate geometry is not a bad idea. However, it seems like you made a mistake somewhere.

I drew the picture so you cannot google it up.

Hint: ||dilation||

[smoky jetty]

https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem I just learned the very basic of this theorem. feels great to know it exists

[upper karma]

how do i do 3 and 4

[silent plank]

identify the corresponding parts based on the congruent angle makings
and set up the ratios accordingly

drawing rotated/reflect triangles may help

[solar garden]

Can I get help with this please?

[lucid moth]

yeah so as given B is the midpoint of AC and AB=DE

a midpoint divides a segment (AC) into 2 equal parts

[solar garden]

Right

[lucid moth]

so if we know AB is congruent to BC (def. of a midpoint)

AB=DE, AB=BC => BC=DE

The transitive property states that “if two quantities are equal to the third quantity, then we can say that all the quantities are equal to each other”.

[solar garden]

I shoulda specified that I don’t know what to do with the answers themselves. Where it says reasons and lists them then wants me to match them to the letters

[solar garden]

So 3 would be answer B?

[solar garden]

Alright cool thanks

[lucid moth]

the last statement should be what you are proving

yeah np

i edited it so it should be easier to comprehend

[solar garden]

There’s one more I need help with

I don’t really know how to attach statements to reasons but I got #6 because of what u told me last question

Oh I used the wrong second picture

[lucid moth]

yeah

[solar garden]

One sec

There we go

[lucid moth]

ok one sec

[solar garden]

I appreciate you bro, saving my skin

[lucid moth]

np

[short crag]

hi everybody, i'm new here. I will do my best to help as much as I can.

I'm facing a problem right now with this:
Two perpendiculat lines pass through the point A(3,3). What is the area of the triangle whose third side is the abscissa (base of triangle is on x)?

i would really appreciate some help

[lucid moth]

Is that all the information they gave you?

[short crag]

yes i have picture also

[lucid moth]

ok could you send

[short crag]

i've been trying all day to do this
i googled, i asked on forums and nothing helpful yet

[lucid moth]

Ok let me see if I can help

[short crag]

i can tell that h = 3

but i can find a for the formula of area P = (a*ha)/2

thaaaank you sooo much for trying to help me

[lucid moth]

ok so I think to find the are you first find the area of the box around it

which from what i got was 19.5

and then you subtract the area of the smaller triangles

A= 1/2bh

[short crag]

how did you get 6.5 if you dont mind me asking

[lucid moth]

sorry im using pad to write

im assuming that is 6.5 since its in the middle of 6 and 7

[lyric sonnet]

why can't you do (3/2)*6.5
If i may ask?

[short crag]

unfortunally its just a sketch it can have many options

[lucid moth]

I'm using the formula A=bh to calculate the area of the square around the triangle, b=~6.5. H=3

I'm not sure where you got 3/2

[lyric sonnet]

1/2 * 3

*6.5

[short crag]

i'm gonna buy a book just because of this

not enough informations to slove this

professor said its in the book so...

thanks for trying

🥺 ❤️

[lucid moth]

yeah np..sorry I couldnt help

[lyric sonnet]

wait

I'll solve it

[short crag]

i'm waiting

[lyric sonnet]

Area = 12.37 sq. units

[short crag]

do you have steps

[lyric sonnet]

yes

[short crag]

can you send them please

[lyric sonnet]

ok

I am coming brb

[lyric sonnet]

according to this the two lines that pass through a are perpendicular
thus the triangle is right angled at A

the x-axis is the hypotenuse

use distance formula to find the length of the two perpendiculars which are your base and height

[short crag]

i will try

[lyric sonnet]

ok

[short crag]

i have only h = 3

you used pitagoras th?

[lyric sonnet]

h is the length any one of the perpendiculars obtained with distance formula
base is the length the other of the perpendiculars obtained with distance formula

[short crag]

this is h

not that you draw

[lyric sonnet]

different diagram from this #geometry-and-trigonometry message

[short crag]

😦

[lyric sonnet]

oh sry

[short crag]

thank you soooo much for what you've done ❤️ don't waste any more time on this i will ask professor

[lyric sonnet]

[lyric sonnet]

you can't that as height use that because the length of the x-axis base is not given assuming the length of the x-axis base 6.5 and height is three will give a wrong value thus the one I drew is more accurate ig

[lyric sonnet]

hope you get the answer to your question.

[lyric sonnet]

or use a help channel like #help-9

anyways bye 👋

[vale shuttle]

Anyone who have doubts in following topics
1 calculus
2 algebra
3 Cordinate geometry
4 vectors
5 trigonometry
6 stats and probability
Wheather related to assignment or tutoring can msg me

[short crag]

hi guys i solved it. i'm so happy so i wanted to share that with you. it's 9sqrt(2)

[frozen ocean]

What about complex analysis 👀

Also you seem like a scammer

[wind coral]

yeah, he is banned lol

[lucid moth]

Nice!!! Could you maybe show me the work so I could understand

[short crag]

y-y1 = k(x-x1)

y-3 = k1(x-3)
y-3=(-1/k1)(x-3)

when you solve this, you will have that both equations are equal to 3 so you can equal them to eachother
-xk1+3k1+y = yk1-3k1+x
if y = 0 you can get k and put it to the equation above so you get x1 and x2
i will post a pic from geogebra

so you get this

x2-x1 is the base and now you have base and height so there is area

what do you say

[upper karma]

Hi my book says that for f(x) = cos x the roots are x = pi/2 + n * pi. The variable n also shows up for the maxima and minima formulas. Can someone explain to me what n represents?

[upper karma]

n is an integer

so it can be any integer

yes

does it have to be an integer within the period of the graph

if the graph is horzontally stretched

no literally any integer

so how do you find the roots

[upper karma]

it still is but ur just gonna have to modify the equation

oh ok

so it asks like find the roots between 0, 2pi

would it be integers between that period

oh if the domain is limited like that it has to be specific integers

so the integers would be from 0-6 right?

[upper karma]

No

wait lemme see

yeah if its in [0,2pi] its less than that

1 and 0

i think it would be 3

isnt it 0 and 2 actly

roots of cos is (2n+1)pi/2

oh ok

ty

actually its 1 and 0

oh yeah

that makes sense

[lucid moth]

Hmmm interesting ty for sharing

[dense tulip]

0,pi is the inverse of cos

[upper karma]

?

[lyric sonnet]

¯_(ツ)_/¯

[smoky jetty]

late response but another approach could be by finding the coordinates of the lines' x&y intercepts with the slope formula, then using distance formula u can find the lengths and since the lines are perpendicular, A=(bxh)/2

[smoky jetty]

would u mind elaborating how u formed that diagram below?

and is "k" the slope? sorry im kinda confused but curious to know

[vivid plinth]

yup, k is the slope

[ionic summit]

hey guys anyone can help?

[mystic charm]

What do you think about the similarities among the triangle ABH and AHC?

Wait I did something wrong

Nvm

[ionic summit]

right angled triangle?

[ruby elm]

prove (cos x)/(1 - tan x) + (sin x)/(1 - cot x) = sin x + cos x

[exotic yarrow]

Reposted here: #help-19 message

[hollow echo]

Make use of trigonometry.

[upper karma]

can some one pls post the solution?

[lyric sonnet]

Hint:||sine rule||

[lament tree]

anyone knows how to calculate the volume of a cone without integration?

I tried doing it with cavilieri's principle but didn't work out

I know calculus tho I'm just really curious of how this could be done

[heavy shale]

there is a formula

[lament tree]

ok but how do you derive it without integration

that's what I want to know

[heavy shale]

oh i dont know that

[lament tree]

I have an idea

[lament tree]

I do not

I do

[vast quiver]

sup

[lethal moss]

hi

[pseudo nebula]

https://math.stackexchange.com/questions/623/why-is-the-volume-of-a-cone-one-third-of-the-volume-of-a-cylinder apparently make it a square pyramid

[versed peak]

hey

can someone please help me out with calculus

[warm falcon]

unfortunate

[warm falcon]

tf is dilation

[lyric sonnet]

yes

[lyric sonnet]

yes

[lament tree]

:0 nice thank you

love cavilieri's principle

[frozen ocean]

#calculus

[hushed summit]

Can anyone confirm answer is b?

It asks for R

I guess the rest of the info is understandable although in different language

[upper karma]

could you translate it?

[crystal glen]

Yes i can

[oblique imp]

Hi everyone im new here and need help with a pythagorean theorem question !

[snow cedar]

@copper comet

[copper comet]

?

hello, why did you ping me

[snow cedar]

Sorry i made a mistake

@oblique imp

[copper comet]

ah ok

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[snow cedar]

Okay

[fervent yacht]

Anyone from India having chat gpt 4 pls let me know

[south summit]

$2\cdot \sqrt{10-6\cos \left(b\right)}=\sqrt{13+12\cos \left(b\right)}$

[somber coyote]

Roto

[south summit]

how do i solve this

[upper karma]

square both sides and its pretty straightforward

[thick storm]

given equilateral trapezoid ABCD (AD=BC) and that sides AB and DC are a and b respectivly how do I find AD and BC?

and angle ADC is 60

[upper karma]

draw altitudes from A and B to the side DC, name the altitude from A as "AE" and the altitude from B as "BF"
AB=EF (ABFE is a rectangle)
so the length of side EF is "a"
the rest involves some congruency and trig and you should be able to do it

[hushed summit]

In the graph, point A is a tangent line and the angle formed by lines AB and BM is 90°. If AM has length 5 and MC length 4, find the radius, R.

[frozen ocean]

don't use gpt for math, it's stupid

[lucid moth]

,rotate

[lucid moth]

nvm i got it

[tidal canyon]

this dosent make since to me, "divide a 5 inch line into two parts so that one part is (a) 2 1/4 inches short than the other (b) 3 times the other" is the last part a printing error??

the rest of it makes since but (b) is ??

[wind coral]

what is confusing you about (b)?

[lament tree]

what's wrong?

[tidal canyon]

3 times the other it dosent seem finished? first year of geometry btw so im kinda dumb, prof never said anything abt it either

[lament tree]

if it's in two parts, the second part is 3 times the first one

[tidal canyon]

ty

[tall coral]

What does that arc over the variables mean in geometry
(I can't think of a better way to explain it)

[wind coral]

represents the arc ABC

[tall coral]

oh ok

thanks

[wind coral]

np

[maiden brook]

I asked this question earlier in another channel but didn’t get a complete answer: Given are parallel lines m and n and point P between the lines. Construct an equilateral triangle such that one vertex is P, one vertex is on m, and one vertex is on n.

How do you solve this with reflections?

[lucid moth]

gl to those taking the regents tmr

[maiden brook]

this is not for geo regents

but gl

to those

[maiden brook]

the geo regents has much simpler constructions than this (even though this is still a relatively basic one) — I took it myself in June

[crisp edge]

um whos in geometry

like currently right now

[pseudo nebula]

you should state your intention

[maiden brook]

can someone answer this pls?

[lucid moth]

I am

[upper gyro]

No way someone1010????

[upper karma]

i am

[upper gyro]

Anyway can some person give me a few circle theorem problems?

I am trying to brush up on it

[pseudo nebula]

this discord is scary good at math lol

im terrified

[upper gyro]

It is

[upper karma]

https://i.imgur.com/2RSqLgw.png

how can i figure out what those 2 are?

shit does not make sense

[pseudo nebula]

trig

Or just take a guess bc it's fairly easy to guess

[half viper]

Triangle angles add up to 180°

[upper karma]

i alr got the angles

i need the line lengths

[upper karma]

ik i could guess but im tryna figure out how to do it

[upper gyro]

Just use sin and cos

$\sin{25^{\circ}} =\frac{8.9cm}{xcm}$

[half viper]

SOHCAHTOA

[pseudo nebula]

Brute force with Pythagorean?

[somber coyote]

Professional Procrastinator

[fresh lava]

my brain isnt grasping how this looks can someone draw this out for me

[lyric sonnet]

coming

[fresh lava]

what

[lyric sonnet]

to draw it out

[fresh lava]

thank you

[lyric sonnet]

I hope its clear

[fresh lava]

and daniels at the top left

?

[fresh lava]

it is thank you very much

[lyric sonnet]

yes

[lyric sonnet]

np

[tall coral]

what does it mean when the points of a line are formatted like this?

[dark sparrow]

that... looks like a fraction, not formatting.

@tall coral can you show where you saw such notation?

[crude lark]

<@&268886789983436800>

[smoky jetty]

hey uhh, i know it's been days already but if u wont mind, can u elaborate "y-3=(-1/k1)(x-3)
when you solve this, you will have that both equations are equal to 3 so you can equal them to eachother
-xk1+3k1+y = yk1-3k1+x
if y = 0 you can get k and put it to the equation above so you get x1 and x2"? my initial understanding of k1 = (-1/k1) is that 3,3 is the intersection of those perpendicular lines, so its slope would be -1 * k1 = -1. But i couldnt really think of anything beyond that. 😅

i tried isolating k1 in the given equation, but it didnt seem correct

i instead tried to set slope to -1 for red line, and 1 for blue line and got this. the hypotenuse is still on x axis and Area = 9?

the y intercepts were found with equation y=mx+b, although this def need correction?

[tall coral]

That's what I was thinking
I can't show where it was annotated though 😓

[dark sparrow]

why not

[solid coral]

can someone help me with this question?

[solid coral]

nvm solved

[tall coral]

because it was on a pre-exam I can't access anymore

I think there was a similar question though earlier in this chat hold on

[tall coral]

_ _

[dark sparrow]

those look like just fractions to me

[lethal moss]

And mine is this

I solved it in only 3 hours of trial and error

And the craziest thing is that I came up with this monstrocity myself

I'm on the dark side now

Oh, Maybe you know a formula for these? (I came up with this myself too.) $\begin{cases}a_1x^2+b_1y+c_1&\ a_2y^2+b_2x+c_2&\end{cases}$

[somber coyote]

themathboi #2137

[lucid kestrel]

Hello

Can someone help me with trigonometry?

[wind coral]

Send your question

[lethal moss]

What is the problem we can help you with?

and btw welcome to the server

[lucid kestrel]

Prove that secA(1-sinA) (secA+TanA)=1

[lucid kestrel]

Thank you 😊

[wind coral]

Have you done anything yet?

[lucid kestrel]

Nope, I don't know how to even start the problem

[wind coral]

rewriting sec(A) in to 1/cos(A) would be a good place to start

[lucid kestrel]

I don't get it

[wind coral]

You know that sec(A)=1/cos(A) right?

[lucid kestrel]

Yes

[wind coral]

So that means you can interchange sec(A) with 1/cos(A)

For example if I have 3sec(A), I can replace sec(A) with 1/cos(A) giving me 3*(1/cos(A))

[upper karma]

I need help with a certain problem of analytic geometry
(a) Write the equation of the parabola P whose vertex is V(0, -3) and directrix is 𝑥 = −1, and determine the equations of the tangent lines to parabola P through the point P(-4, 0).
My problem is when I need to determine the equations of the tangent line
I don´t know what I need to do
I am thought use dy/dx but I think that I am doing it wrong
I already have the parabola

[pseudo nebula]

Can you show your attempt?

[upper karma]

Sorry, I am already tired

[proven schooner]

hey i need to solve cos(x)=0 then sin(x)=0 on each a,b,c,d,e

what should i do exactly

[upper karma]

do you know the periods of sin and cos?

[pseudo nebula]

I'm don't quite know what the question is asking for, the point is not on the parabola? But I can take the derivative $\ \frac{d}{dx}(y^2+6y+9=4x)\ 2y\frac{dy}{dx}+6\frac{dy}{dx}=4 \ \frac{dy}{dx}(2y+6)=4\ \frac{dy}{dx}=\frac{4}{2y+6}$

[somber coyote]

someone1010

[proven schooner]

i know cos(x)=0 can be if x= pi/2 for example

idk if thats what u mean

[pseudo nebula]

Both are 2 π

[proven schooner]

oh yeah

[upper karma]

I need the lines are tangent to that parabola by that point

[pseudo nebula]

,w graph y^2 + 6y + 9 = 4x

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by someone1010

[pseudo nebula]

(-4,0) isn't on the graph

?

Ohhh nvm

Being dumb xd

[upper karma]

You find it?

[pseudo nebula]

I can but it's messy as hell

[upper karma]

Yeah

I have the same problem

[pseudo nebula]

Incredibly messy

[upper karma]

Okay thanks You

I Will try it

[proven schooner]

whats the best way to write down the answer ?

[tall coral]

[frosty steeple]

Did you cry on ur paper

[proven schooner]

ahahahahah

[frosty steeple]

😭 its understandable if u did

[proven schooner]

nahh

its prolly oil or smthng i ate

🤣

[frosty steeple]

BRUH u be wiping ur oily hands on the paper

thats so funny

[proven schooner]

AHAGHAHAHA

brooo

never

but its only trainning exercize after all

[frosty steeple]

okay okay good, imagine ur turning the paper in like that

[proven schooner]

☠️

[frosty steeple]

sadlly i dont even understand what is on ur paper, i jus know its trig

[proven schooner]

its really easy to understand

thats a basic exercise

[frosty steeple]

let me go look again

[frosty steeple]

excercise 28

u must be rlly good

[proven schooner]

😭

nah

its only trainning exercises from high school

[frosty steeple]

one thing abt me is im no inconsistant, and for math, practice is key, and i dont have the key

[proven schooner]

before getting in real things in 10 days

[frosty steeple]

ur not from the US are you

[proven schooner]

im not

im french

why

[frosty steeple]

yup that makes so much sense

in america, they dontmake u do this (which is rlly a bad thing)

i wish they rlly did make us do what u guys are doing

itd be more effective

[proven schooner]

what do you guys do instead

[frosty steeple]

we just learn something in class, then our school dont even give us hw

they just make use practuve in class

[proven schooner]

hw ?

[frosty steeple]

homework

[proven schooner]

ohohm

my bad

[proven schooner]

its not like i did a lot my homeworks during highschool tbh

[frosty steeple]

ur good

[frosty steeple]

really 😭 most ppl do have alot of things other than school, like sports and other things that prevents them from doing hw

(and is it okay to be having a conversation here like this )

[proven schooner]

yeah its sad

i wont dodge anything this year tho

lmao

[frosty steeple]

good, do your very best, and for everthing

[proven schooner]

what u wanna do after highschool

[frosty steeple]

university, major in CS

as for right now

how about you

[proven schooner]

i'm going for preparatory school

43h/week of maths/physics/chemistry science of engineering

2 years

then competition to join engineering school

based on ur ranking u get the better school

[frosty steeple]

wow this is amazing

[frosty steeple]

we also have this test we have to take called SAT and ACT, and u have to excel a certian score to be accepted into good schools

[proven schooner]

i just looked

but thats a test you do after highschool ? @frosty steeple

[frosty steeple]

nope you do it while ur in highschool

[proven schooner]

ah yeah

cant compare then

[frosty steeple]

you do yours after?

[proven schooner]

yeah ofc

prepa is after highschool

you do 43h/week during 2years then u do the competition

[frosty steeple]

ohhh okay i understand now

[proven schooner]

you get accepted to prepa on ur file

and ur results in highschool

[frosty steeple]

that is pretty intense but that is amazing, competitions help you (if u do well)

[proven schooner]

https://www.concours-centrale-supelec.fr/CentraleSupelec/2023/Multi/M048.pdf

thats how it looks

[frosty steeple]

oh wow thats alot, i wish you luck on your exams

[proven schooner]

thats in 2 years ahaah i have time

i dont know anything yet

ngkl

ngl

i start in 10 days

[frosty steeple]

im sure you will do great

[frosty steeple]

have confidence in urself

[proven schooner]

yeah ill see

how much years left in hs ?

[frosty steeple]

for me?

[proven schooner]

yes

[frosty steeple]

i have 2, there are 9,10,11, and 12 im in 11th grade

how about you

[proven schooner]

at what age do you finish highschool in the us

[frosty steeple]

18

[proven schooner]

im done

told u i join prepa

[frosty steeple]

yes you did, i have such bad memory :,(

[proven schooner]

😭

dx

dw

its fine

[frosty steeple]

ahhhhhhhhhhhhhhhhhhhhhhh

but ill be 18 then at 18 ill start collage then collage is 4 years as well

then ill be 21

what age do you have to be to join prepa @proven schooner

[proven schooner]

theres no age

its after the highschool

[frosty steeple]

ahh i see

[proven schooner]

if u jump classes u can be there at 16

[frosty steeple]

ill come and take it

jk i would not do good

[proven schooner]

some people join there at 16-17 and do the competion after only 1 year and join the greatest engineering school at 18

even before

lmao

[frosty steeple]

really? truly talented people, youll become amazing and join a great school

[proven schooner]

they are just childs of engineers or elites

social reproduction

[frosty steeple]

ahhh they have it easier

[proven schooner]

they just get into it since they are born

they are taught maths in the bed at 5

lmaooo

[frosty steeple]

for real, they dream in numbers

would u say you would want to be born in a family like that?

[proven schooner]

i dont know

lol

[frosty steeple]

honestly, both worlds have their own problems and good things

[proven schooner]

yeah

[frosty steeple]

YOULL DO AMAZING

[proven schooner]

stop saying that 😭

[frosty steeple]

its because i relly belive u can

[hard trout]

I don't suppose anyone's got some pointers on this one? Solving trig equations and this has me stumped as to the next step

I'm guessing I just haven't thought of some identity to rearrange it? Or....mental blank happening here 😝

Obviously the three black lines were just rearranging it to see if i spot something.

Stumbled down to this point. 😮‍💨

Should be "+ πn" on the last line. Not 2. Spotted that

[toxic mural]

hi, anyone know how the substitution integration method can get 6-u^2? I didnt get that

[gentle haven]

In a quadrilateral, if just one pair of opposite sides are parallel and equal, does that prove that the quadrilateral is a parallelogram?
Or are there any examples where it doesn't work?

[hallow crypt]

Actually, you may be right I was going to say rectangle and trapezium, but I'm pretty sure those count as parallelograms

[haughty canyon]

how do i solve Q15 and Q17

i’m only 14

[timid merlin]

can anyone tell how to solve this?

my friend send me and asked me

idk how to

[fluid stream]

#chill

[wind coral]

well your firend is trolling you this is the collatz conjecture

[upper karma]

[timid merlin]

wow m gonna kill em

m just 13 tho idk what was this

[wind coral]

good luck with that lol

[upper karma]

for 15, let angle BAE=EAD=x and CDE=EDA=y
given that BA and CD are parallel to each other, this means angle BAD and CDA must be supplementary (co interior angles), i.e. their sum is 180 degrees
clearly BAD=BAE+EAD=x+x=2x
and similarly CDA=2y
so 2x+2y=180, and hence x+y=90 degrees
now in the triangle AED, the sum of all three angles must be 180 degrees
so x+y+AED=180, put x+y=90 here and you get 90+AED=180, which means angle AED=90 degrees
and hence triangle AED is a right angled triangle

[old cradle]

I am having a doubt with this
A point is such that (abscissa of the point, other than zero) that it equals to the ordinate of the point. In which quadrants can the point lie?

[old cradle]

anyone pls help

@old cradle

@old cradle

[woeful pewter]

Anybody?

[wide rune]

anyone knows where to find derivation and intuition of the condition for general 2 degree equation in x and y to represent pair of straight lines

what does this even mean

[proven blade]

you can represent a polynomial as a matrix like this:

and by looking at the matrix, you can classify what curve/surface the polynomial describes

[remote moth]

hello

[upper karma]

can someone help me with trigonometric identities im supposed to solve cos^2(x)-sin^2(x)=sin(x) and i dont really know how to go about it. Thanks

[thick fable]

i think you should use help channel for this

[smoky jetty]

i think since it says the most no. of lines using "any" two points from the 4 given, it means that line gd = line dg. So there is no permutation i.e a particular order to which they draw the line with any 2 points will be = 1. Then it seems like it asks for how many "combination" of lines created by two points from the 4 points.

[upper karma]

In 17, draw a line through R parallel to both P and S. Try it then.

[hardy bay]

Consider the hyperbola

x²+y²+4xy-8x-4y-2=0.

Calculate the conjugate diameter in the direction of y=4x.

The worked solution says if p, q are the slopes of conjugate diameters, then pq+2(p+q)+1=0. Since p=4, then q=-3/2 so the conjugate diameter is y-2=-3/2x.

My question is where "pq+2(p+q)+1=0" comes from.

I haven't find that formula anywhere outside this solution I looked a lot round the internet. Everything I find is related to equilateral hyperbolas.

The worked solution also uses a similar formula (p-q)/(1+pq)=1 which holds when two conjugate diameters form a 45° angle; and the formula 2m²+0m-2=0 for calculating the slopes of the axes.

Where all these formulas relating slopes come from? I can't find them anywhere I tried everything.

[humble quiver]

hey

[upper kite]

the two circles are internally tangent, AB is a tangent to both circles, CA = 2, DC = 1, CF = 6
need to find DE
gray is what i drew
would appreciate help idk what to do

[tough stream]

Use AB square equals ACCF and ADDE

[smoky jetty]

he's referring to the tangent-secant power theorem where the square of the tangent segment is equal to the product of external secant segment and the whole segment

since the values of the external and whole segment of bigger circle's secant are already given, u can use it to find both triangles' tangent BA. Then with that, u can proceed to the smaller circle

[warm bay]

hi i am doing pretty basic trig stuff and i am trying to find the sin cos and tan of a theta. would anyone be able to help me

[wind coral]

send your question 👍

[lucid moth]

yeah i help with basic trig

[warm bay]

5pi / 3 or 300° for theta

#20

[wind coral]

,rccw

[wind coral]

so have you used the unit circle before?

[warm bay]

yes!

[wind coral]

ok great, im going to send the unit circle so we can evaluate #20

give me a sec

so do you know how to use the unit circle to solve expressions like this?

[warm bay]

sorta. i’m still learning.

i don’t understand how you get the -sqroot of 3/2 from 5pi/3

[wind coral]

ok, well first were gonna try to solve sin(-pi/4)

[warm bay]

so that’s 1/2 correct?

[wind coral]

not quite

[warm bay]

no i’m confused

[wind coral]

were going to need to manipulate the expression a bit

[warm bay]

so the radius is 2 correct cus it’s the denominator

[wind coral]

where did you get that from?

[warm bay]

my teacher told me the denominator is the radius

of my ordered pair

[wind coral]

umm im not so sure what you mean by that ngl

here, i think we might need to get some basics down first

[warm bay]

fr my teacher ain’t even give us the chart u sent me

[wind coral]

mhh, maybe your just learning how to derive the unit circle

but anyways, i have a resource for you that helped me understand trig

[warm bay]

i’ll take anything ur willing to give me

[wind coral]

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

[warm bay]

thank u

[wind coral]

unit 2 goes over the unit circle

[signal lily]

I want lwarn spherical trigonometry soon anyone have a good course?? Like on YouTube

[oblique wadi]

is there a proper mathematical name for this type of 1xn tabular shape?

like in general

[dark sparrow]

array, maybe?

[severe smelt]

as a 11 yr old

[pseudo nebula]

Double precision floating point?

Row vector?

[frozen ocean]

What do you mean?

[lethal moss]

Hi. Can anyone recommend some decent solid geometry tools?
I'm struggling to find some myself.

[lament tree]

how do I prove two triangle horizontal slices are the same if one of the vertices are transported through a parallel line?

basically something caveliery's principle-esk

I've been trying to prove alpha and beta are equal

which I just realised is probably not true actually

shit

anyways

I basically need to prove AB = CD

actually I'm not even sure this is true

[lament tree]

OH ACTUALLY

I got it

I think

pretty sure that if I prove it for paralelograms, I can prove it for triangles

yeah I did it

yey

sometimes a small piece of insight can do wonders

[nocturne remnant]

AB / (base length) depends only on the ratio of distances between the three parallel lines, so it is constant

[lament tree]

really? :0

how so?

is there a general thing about parallel lines and ratios that comes from?

[nocturne remnant]

You can identify a pair of similar triangles in the figure, so the ratio of heights = ratio of base lengths

[lament tree]

I can't ;-;

[nocturne remnant]

;-;

How about this

[lament tree]

the only I see is the intersection triangle and that small one on the top of it

[lament tree]

:0000

OMG FR HOW DIDN'T I SEE THIS

[nocturne remnant]

Basically whenever you have parallel lines and some intersections there are always similar triangles like this

[lament tree]

a lot of geometry is just finding the right triangles 😅

[nocturne remnant]

:p

[lament tree]

makes sense

[nocturne remnant]

It gets easier with practice

[lament tree]

yeahh I need to practice geometry

focused too much in algebra and stuff and I am really bad at geometry

weeeell but I think my proof is very pretty

thank you anyways

good lesson learned: f i n d t h e t r i a n g l e s

the triangles in the triangles inside the triangles

themselves made of triangles

[errant beacon]

question

when an angle is said to be in standard position and this angle is negative, it always begin in quadrant 4?

[frozen sonnet]

how can i know the volume of a cube ???

[wind coral]

<@&268886789983436800>

[gray sentinel]

I have to find the area of triangle ABC but I can only do it by Geometry, I can't use trigonometry, they told me that it comes out by Congruence or Similarity, but no matter how hard I try, it doesn't come out, any idea where to do the lines? BM=MC, BAC = 2 Theta and BMA = 1 Theta

[north kindle]

im having some trouble finding q and r, not sure where to go from here

nvm i got it

[smoky jetty]

observe angles BMA and BAM. Given that BAM=BMA, what do u think are the properties of triangle BAM?

[jovial terrace]

this is just sqrt(101) e^i * 1.47 right

[broken pasture]

Hey guys im kind of a mathematics scrub and i was wondering if you guys could help me understand some concepts applied to programming. Im trying to understand how 3D coordinates could be turned into 2D and then draw a line from one coordinate to another in relation to a viewport in a 3D environment.

[swift frost]

#❓how-to-get-help

[arctic finch]

i made a silly little formula based off the law of cosines

C can be any angle but this way if you're given all the sides it'd be more convienient

cause you can just put this into the calculator

[somber coyote]

Şêro

[smoky jetty]

if you "made" that formula without any prior knowledge that it already existed long ago, then it's pretty impressive!

[cinder badge]

hello, i am new to this server and i was wondering if i can get help with trigometry, again i do not know how this server works, i am in highschool still btw

[arctic finch]

There are many math help channels and a help forum in the server

[cinder badge]

i did do one but nobody is helping lol

[dark sparrow]

help forum is dead

@cinder badge if you have trig question(s) you can post them either here or in help channels

don't forget to share your progress and be as specific as possible with your doubts

[tribal shore]

@near jungle By opposite and adjacent they mean:
You have a rightangled triangle
Then you pick one of the two acute angles
adjacent side is that that touch the peak of that angle.

If in more geometric words.
Angle is 2 rays that have common point that is border of each ray.
If side contains this point of the ray or equivalently If side lies on one of rays of your angle (lies mean that every point of side is in this ray like)

[vapid ermine]

Hello guys I have a bit more unconventional math. I need help loading and saving data for plots in a 3d game. In the game, units are represented by studs on a X,Y,Z coordinate system.

Notes:

1. 'data' on plots refers to the parts that are built onto the plot.
2. The position of parts refers to the very center of the part. For example a part with a size of (5,5,5) would not be influenced by the size and rather be at the point at the very center (0,0,0)
3. This coordinate system can go into the negatives.
4. Each player gets a plot assigned specifically to them.

I am trying to save the parts the player builds on their plot when they leave then reload them when they join again.

I think that in order to do this I would save
PartPosition-PlotPosition = saved position

then when they join the game it would go through this operation:
Check the items that are saved.
Clone this item and put it into the game.
Move it to the saved position + PlotPosition

Would that work?

[balmy otter]

From the top of a cliff, the angle of depression of the top and bottom of a tower are observed to be 45 and 60 respectively. If the height of the tower is 20 m, find : the height of the cliff and the distance between the cliff and the tower

[limpid flint]

Anyone know a good pdf somewhere of geometry identities and theorems that would be useful for a relatively easy high school math team. It requires nothing above like algebra 2 and it’s pretty casual but we do state and stuff.

[upper karma]

imagine if we used tau for this, it would be beautiful

[karmic lagoon]

fyi if that's a roblox game it's not just math, it'll include lua code too. what you can do is have every plot's rotation be the same, allowing for easier saving/placement. therefor you should just have to save the vector3, cframe ect

anyways

this teacher giving hw we aint learn yet

surprised i learned algebra before geometry too

[wind coral]

Not to mention eulers identity would be even more beautiful, e^(i*tau)=1 is just more appealing imo.

[floral shale]

Why tau

Who hurt you

Tau is like the equivalent to the voices in my head

[lyric sonnet]

69 degrees ?

vertically opposite angles?

[karmic lagoon]

yeah

my blind ass didn't see it at first

took me like 3 minutes to register that

😭

[lyric sonnet]

oh

[lyric sonnet]

sry

[karmic lagoon]

not your fault

[gentle haven]

"From a circle of radius 15 cm, a sector with angle 216° is cut out and its bounding radii are bent so as to form a cone. Find its volume."

I dont really understand how i'd get the radius of the base and the height of the cone from the given information

I think the slanting height is 15 cm because that's the radius of the original circle.

[gentle haven]

Nvm i got it

[lyric sonnet]

oh

[crystal shadow]

Can someone help me for Olympiad geometry?

[lyric sable]

hey guys. do you mind helping me out with geometry

can some explain to me, im somewhat lost and like i don't get it

[rich dagger]

can someone please explain to me how this is 0.97 and not 11.5 I'm so lost 😔

[grand shadow]

hey sorry i'm kinda confused here
did you set this up yourself and get 2sqrt(6)/5 and you're wondering why that's the answer

like what happened

[rich dagger]

yeah I got 2sqrt6/5

[grand shadow]

when you type that in a calculator, you get 0.97

[rich dagger]

but whenever I put it in the calculator it's giving me 11.5

[grand shadow]

that's strange

idk how it would be doing that unless you're like typing it in wrong

cosines can't be greater than 1

[rich dagger]

am I putting this in wrong??? like idek

[rich dagger]

yea ik

[grand shadow]

yeah

that outputs the arccosine of the answer

[rich dagger]

how do I put it in?

ohhh

[grand shadow]

so just enter the 2sqrt(6)/5 part and you should be fine

[rich dagger]

omggg bruh I'm so dumb tysm for your help

[grand shadow]

no problem

(also, you aren't dumb)

[rich dagger]

thanks :]]

[grand shadow]

yeah it was definitely outputting the arccosine of the answer in degrees

[tough mesa]

Excuse me

Can someone solve ques no 14

[dark sparrow]

we don't give out answers here

but also the question looks sus as fuck?

[tough mesa]

...

[tough mesa]

It is more like a question of geometry but it's given in combinatorics

[vapid rover]

ABCD.EFGH cube with side length 5 cm. the x and y points are the midpoints of EH and GH respectively. determine the distance from point B to the plane AXYC

Is dis the right way to do it?

[brisk patio]

to calculate the area, you can divide this polygon into 2 different ones and calculate base x height of each and then add the areas. Each small square on the Cartesian plane counts as 1.

after that, to calculate the perimeter, you just need to add up all the sides of the polygon, still with each little square counting as 1

I hope this helps, mate : )

[lyric sable]

thank you so much like honestly, i was lost in it and now i'm not 😃

[gloomy palm]

guys can u guys help me with pythagors and trigonometry

[upper karma]

yeah help me too

[lethal moss]

I know it's a meme, but try to solve it for real.

[lethal moss]

Hi. I am Polish too. The most popular term for both is just circle, but sometimes people call okrąg a ring, however that does not happen often in mathematical language.

||jesteś pierwszą osobą, która to zauważyła 😀||

I forgot to ask, what are you learning right now? I am just curious.

I was asking for more of a general idea of what stage you are on.😆

[lethal moss]

pretty interesting stuff, in my opinion, maybe you would be able to solve this? For now none of the "elite" mathematicians have tried.

[small pebble]

need found the x

the right answer is 3√2

but i always get the 4 (????)?

[lethal moss]

Try describing your method, we will then be able to spot any mistakes together

[lethal moss]

I was thinking more about solving with pythagorean theorem

[small pebble]

damn, i forgot to simplify the 2

[small pebble]

hhm sounds good

Thank y'all guys, helps me a lot

[lethal moss]

no problem 😃

[small pebble]

but the cos45° isn't √2/2?

[lethal moss]

it is

[small pebble]

so why there is a "1" up there?

[lethal moss]

hmm