geometry-and-trigonometry

[haughty prairie]

Sounds like it fails to avoid confusion

it’s like having a fork in the road and then asking you which one goes straight

bad example

oh well, not my homework

[dusky locust]

wow i legitimately was not aware that this server had a thisnt emote

[dark sparrow]

!original

[lime crown]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[dark sparrow]

they might be implying the adjacent leg

[haughty prairie]

It was on that mans homework

[dark sparrow]

right hm

not good on their part

[trail aurora]

Any tips on how to solve this one? the question doesn't provide any sides

[rare siren]

posted in help but they might be a bit too hard. they're from a Romanian ninth grade olympiad

[fringe smelt]

Easy example but i just want to check if i got the right answer. So please if some one got time: Calculate the area of this object.
I am preparing a pub quiz and i am not sure if this is to hard for people which do not rly use maths since they got out of school 😄

[dark sparrow]

are the curvy bits on the left half-disks?

[fringe smelt]

i dont know what disks means but those are 3 half-circles 🙂

[dark sparrow]

a disk is the shape enclosed by a circle

i mean you know what a "disk" is in its everyday sense right

but ok yeah that answers my question

id say for a non mathematician it would take a bit of time to do this

even i would prob need pen and paper to work it out

i think this is over caliber for a pub quiz

also the answer would involve pi, which is kinda unpleasant

[smoky jetty]

the slight imperfections of those semi circles trigger me XD

[fringe smelt]

Yeah i worked with paint to get it done fast 😄

[bronze wind]

I am just entering my trig unit. Are these answers correct?

[trail tendon]

yes, but #10 isn't done

[bronze wind]

So they are all correct so far? (I know its not done but thanks)

[trail tendon]

yes 👍

[bronze wind]

thanks man

[dark sparrow]

(currently being helped in #prealg-and-algebra )

[raw lotus]

can someone please help me understand trigonometric addition and subtraction identities

[trail tendon]

you mean like sin(A+B)?

or...

[raw lotus]

yes

like sin(A+B)

or cos(A-B)

[trail tendon]

you mean you want a proof or sumthing?

cuz if so i'd just look it up lol

[raw lotus]

i need help understanding it

[trail tendon]

do you know what sin(A+B) = ?

[raw lotus]

let me think

i think it is sin(A)*cos(B)+cos(A)*sin(B)

but i might be mixing it up

they're so confusing

[trail tendon]

yes thats correct

[raw lotus]

oh i got it?

[trail tendon]

yeah

honestly

its better just to memorize XD

like you can look at the proof once but i wouldn't memorize the proof

just like the quadratic formula

yk?

[raw lotus]

i've memorized the quadratic formula

but ik what you mean

[trail tendon]

yeah

i know it can be hard to remember like

which one is which

since they kinda similar

[raw lotus]

do you have any tips for remembering which trig addition or subtraction identity is which?

[trail tendon]

the sign of sine is always the same. like sin(A - B) = sin(A)cos(B) - sin(A)cos(B)
thus, the cosine is the opposite sign.

um

[raw lotus]

that makes sense

[trail tendon]

cosine for some reason likes itself. like cos(A-B) = cosAcosB + sinAsinB

[raw lotus]

that is a pneumonic ic an use

[trail tendon]

the cosines stick together seperately from the sines that stick together when its cos(A+B) or cos(A-B)

whereas with sin(A+B) it uses both sine and cos for each term

i'm just saying the similarities and differences here ngl 🤣

[raw lotus]

i just need a catchy phrase for "the cosines stick together seperately from the sines that stick together when its cos(A+B) or cos(A-B)"

[trail tendon]

🤣 🤣

[raw lotus]

maybe that the sines are abbreviated "sin" and therefore they sin and the cosines don't want to be near sinners so they self seperate

[trail tendon]

i'm not going to lie i just repeated the trig identities for a couple days over and over until i got them, and then kept doing it 💀

[trail tendon]

LOL

i mean if it works...

XD

[raw lotus]

i guess i could just write them down 20 timers every day for a year

times*

[trail tendon]

🤣

actually though like

i had a sheet

with trig identites

it was double sided

i feel like i memorized them in those two days

and i haven't forgotten because i kept going over them

and now i actually use them sometimes ☠️

(very rare)

[raw lotus]

nice

[trail tendon]

just gotta get like a learning strat

idk

xD

[lament pendant]

anybody know any real life objects that take a similar shape to this

[trail tendon]

a snow man 🤣

with a very tiny head 💀

[lament pendant]

okay maybe something else lol

[smoky jetty]

a legless geometric ant XD

[lament pendant]

BRO

[sonic warren]

some Chinese lanterns resemble to the shape.

[lament pendant]

OO

WHATS THE NAME OF THE LANTERN

[dark sparrow]

resemble this shape*

[upper karma]

My hb with a hat

[river solstice]

Water bottle

Those weird ones that they take to the desert

[silent plank]

^ gourd water bottle

[icy zinc]

Trigonometry is the bane of my existence

[ionic adder]

There's a Japanese pot that looks like this

[ionic adder]

Maybe not pot, but it's Japanese gourd

[dark sparrow]

is it a vessel of some sort?

for water, maybe?

[solar yoke]

fancy way to store water

[upbeat light]

ignore the background

how how do i do C

[lament pendant]

Hum, anybody else?

[trail tendon]

why do you ask?

[lament pendant]

Well I have to create a real word problem from that shape and I would have to identify that shape as something first

Tbh that japense water bottle thing seems the closes but one circle is bigger and such

I was trying to find a water bottles brand name or something that is in this shape. I'm sure there are some but I am unsure of them

[trail tendon]

oh 🤣

you could just use something that can be any shape

yk

[south plume]

Would a snowman with a hat be a suitable answer?

[trail tendon]

bro thats like what i said

except

different

💀

[south plume]

[trail tendon]

😔

[empty yew]

What are the conditions necessary to determine that it's a square?

Anyone!

[trim shoal]

In triangle (ABC), (AB = 8), (BC = 15), and (AC = 17). Determine whether triangle (ABC) is a right triangle. If it is, find the measure of the right angle. If not, explain why.

anyone try

[empty yew]

It's a right triangle

But what is the measure of the right triangle! 😁

Is it perimeter?

[trail tendon]

it is a right triangle, because it is applicable to pythagorean theorem (8^2 + 15^2 = 17^2) and pythagorean theorem only works for right triangles. the measure of any right angle is 90 degrees 💀

[trail tendon]

all sides must be equal, all angles must be 90 degrees

[empty yew]

Ohh thank you

[lament pendant]

well i dont think a snoman has that small of a hat and i dont think it has 2 circles

[trail tendon]

it COULD tho

🤔

whats the area of the snowman from the perspective of the onlooker yk yk 🤔

[maiden brook]

is this product-to-sum/sum-to-product?

havent gone over it yet

[south plume]

Write it as a summation then express it using the double angle identity

@maiden brook

[maiden brook]

like this?

[south plume]

Yep

[faint comet]

Hey guys does anyone here have 1 min for me to explain to them about the unit circle I want to make sure I got it correct

[trail tendon]

sure

[south plume]

Go ahead

Oh lol

[trail tendon]

but you can't use the double angle identity... its k*5

unless im missing smthn

i mean i probably am

just yea

[south plume]

It still works

[trail tendon]

it does?

[south plume]

Because the double angle is (1+cos(2*whatever’s inside))/2

So we’d get sum from n= 1 to 17 of (1+cos(2*5n))/2

(Idk how to use latex bot lol)

[trail tendon]

wait

what

the double angle identity is cos(2k) = cos^2(k) - sin^2(k)

[faint comet]

So like this photo here, I understand the unit circle fully but I didn’t know how he got the triangle number, I’m pretty sure now that since he knew tan=OPP/ADJ he drew those numbers in and he knew that with thoes numbers that triangle looked similar to the 30,60 degree triangle so he added in the two and then he knew for 1/ square 3 works for the 30 degree so we knew it must be pie/6

[maiden brook]

so this but then what?

[south plume]

Which is this one then?
I’m a fraudddd

[south plume]

Apply the summation to each term
i.e sum (cos(10k)/2) + sum(1/2)

[maiden brook]

whats the cos sum?

[south plume]

Do you know the values of cos0 & cos180?

[maiden brook]

cos 0 = 1

cos 180 = -1

[south plume]

👍

You can piece it together from there

Instead of 0 and 180 we’re dealing with 10 & 170

[trail tendon]

the half angle identity is cos(k/2) = +- sqrt((1+cosk)/2)

[maiden brook]

yeah i have to learn sum-to-product

[faint comet]

Am I correct ?

[trail tendon]

yea thats basically the idea

the dumb part

is that

[south plume]

Checks out👍

[trail tendon]

their tangent in the picture and the tangent in their equation just happened to be the same lol

[maiden brook]

the cos part cancels out

its 17/2

[faint comet]

Ok and how do I know what reference angle to use when dealing with angles like 240 like why is it 60 not 30?

[south plume]

Subtract 180 from the result

[faint comet]

Wouldn’t it be 270 becuse it’s in Q3 ?

[south plume]

Yep. There you go

[south plume]

You said it was 240 in the question above

But it wouldn’t matter

Depending on the question/application, sometimes we only care about the angle if its between 0 and 180

[faint comet]

I think I understand now

[south plume]

Therefore if we get an answer greater than 180, we get it’s reference angle by subtracting 180 from our result

[faint comet]

Yeah yeah

Thanks bro

[south plume]

👍

[trail tendon]

do you need help with that or u just givin us a problem?

[north kindle]

Is the answer 318? My reasoning is that you can pass a plane through any set of 3 vertices, no sets of 5 vertices, and only the faces for 4 vertices. It also seems impossible to pass a plane through 3 of the vertices connected to a face without passing through the 4th, so I got 876 - 24 + 6 = 318

Wait fuck

Wrong questions

[trail tendon]

what kind of math is this

[north kindle]

I couldn’t find an elementary combinatorics channel and this seemed like the next best thing

[lime dune]

#discrete-math #competition-math

[north kindle]

Ah, I didn’t know this was discrete math

[stray tangle]

I need help

[trail tendon]

do you know how to find the slope between two points?

[trail tendon]

ohhhh

[dark sparrow]

@stray tangle do you still need help with this?

[upper karma]

i dont like geometry :(

[maiden brook]

why

[upper karma]

because i dont want to memorize hundreds of formulae in conic sections

and circles

[north kindle]

Just rederive them when you need them

[upper karma]

google time pressure during examinations

[ionic adder]

imagine rederiving equation of tangents and the point of intersection of tangents in ellipse and hyperbolas

[north kindle]

Skill issue /s

[upper karma]

dont forget length of chords, they are the worst

what the fuck is $L = a|t_2 - t_1|\sqrt{(t_1 + t_2)^2 + 4}$

[somber coyote]

deltaG

[ionic adder]

that's easy

[upper karma]

there's more formulae in polar form, slope form etc

[ionic adder]

skill issue if you don't remember it

[upper karma]

agreed

atleast 3d geometry formulae are easy to memorize

[solar yoke]

how can i make this 3d

my geometry isnt enough for this

[dark sparrow]

wdym by "make this 3d"?

what do you want to accomplish, exactly?

[upper karma]

i suppose they want it to have height

[dark sparrow]

let them explain what they want.

don't try to guess, it just muddles things.

it is boz's responsibility to clarify their own request.

[upper karma]

fair enough

[solar yoke]

uhh

let me show an example

[dark sparrow]

brb

[solar yoke]

not only this figure

we can adjust the side count

using the sliders on top left

[dark sparrow]

... that's a truncated octahedron

[lucid valve]

hey can anyone help me with this

suppose there are three vectors of equal magnitude which subtends equal angles with each other then the resultant of any two of them will subtend what angle with the third one??? will that be half of the angle between them?

[dark sparrow]

idk how you plan on "adjusting the side count" on that

what is this all for? knowing this ought to make it possible to figure out what you really want.

[solar yoke]

i maxed the side count now it resembles circle a bit more.

[upper karma]

i think it'll be the same, if im wrong please correct me
i took an example of i, j and k (unit vectors representing x y and z axis respectively). resultant of i and j is i+j which is still perpendicular to k

[solar yoke]

rewrite this in a simpler english

[upper karma]

i think it may differ at different angles

[ionic adder]

they mean if you tell us what you are trying to achieve, then it will be easier for us to help

[lucid valve]

i think thats a special case for right angles

i am reffering to acute angle

[solar yoke]

okay i'll be back

my brains boutta explode

[echo beacon]

how do i do question 26

[dark sparrow]

so then, will you tell us what you're trying to achieve?

[upper karma]

i considered 3 vectors in space:
A = ai + bj + ck
B = αi + βj + γk
C = pi + qj + rk
such that their magnitude is equal to λ and angle between each one of them is θ.
resultant of A and B = A+B
the dot product of (A+B) and C = A.C + B.C
since they subtend equal angles and have equal magnitude, A.C = B.C
therefore (A+B).C = 2AC
|A+B|.|C|cosδ = 2|A||C|cosθ (let δ be the angle between A+B and C)
|A+B|cosδ = 2λcosθ

rest is trivial

like, $|A+B| = \sqrt{2\lambda^2 + 2\lambda^2\cos\theta}$ where $\lambda$ gets cancelled out and you're left with $\cos\delta = \frac{\cos\theta}{shit goes here}$ and its independent of magnitude

[solar yoke]

i'll do that eventually. give me some time to think about this problem omo

[somber coyote]

deltaG

[dark sparrow]

so when you came here with the problem, you didn't have any idea why you were doing it.

[lucid valve]

so it varies with theta right?

[upper karma]

im too lazy to solve further

[solar yoke]

it?

[upper karma]

idk, simplify and check man

[lucid valve]

i was thinking more of a geometrical approach

[upper karma]

i dont like geometry

[lucid valve]

like the resultant of two will have magnitude of 2acos^2 theta/2

[solar yoke]

i had no idea how i was supposed to do it. i still dont and i also have no idea how describe this in english even in my native language. so yes i guess

[lucid valve]

it is an identity for equal magnitude

[dark sparrow]

what is your native language?

[solar yoke]

you dont know it

[upper karma]

thats why Ann is asking

because they dont know

[solar yoke]

[upper karma]

are you doing this for fun/timepass?

[dark sparrow]

how can you be so sure?

[solar yoke]

i just know this ok

[solar yoke]

yes

im doing math for fun

[dark sparrow]

can you tell me what it is anyway

bc now im just curious

[solar yoke]

dm

i need to learn 3d geometry

[upper karma]

@lucid valve i got curious and solved it further

i got $\delta = \arccos{(2\cos{\frac{\theta}{2}} - \sec{\frac{\theta}{2}})}$ where $\delta$ is the angle between resultant of 2 and the 3rd vector

[somber coyote]

deltaG

[upper karma]

when $\theta = \frac{\pi}{2}$\
$\delta = \frac{\pi}{2}$ which is obviously true

[somber coyote]

deltaG

[upper karma]

yes.

[frozen oriole]

https://tenor.com/view/putin-sigma-mewing-gif-13750899674712754521

[dark sparrow]

!redir

[lime crown]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[dark sparrow]

<@&268886789983436800> scam

[upper karma]

okay i've seen people doing this emote quite a lot now, what does this even mean?

[ionic adder]

mewing, looksmaxxing

[little anvil]

i solved the other ones, i just need help on 3.3.2

[ionic adder]

but again, this is not the right channel

[ionic adder]

do you still need help?

[little anvil]

yup!

[upper karma]

isnt that like a 2014 4chan trend

why is it still alive

[ionic adder]

wait ill send the sol

[ionic adder]

im not familiar with internet trends

but i still have a banana cat pfp and name

[little anvil]

ok tyyy

also if u could guide me thru it too cuz im tryna learn how 2 do it aswell

[upper karma]

amazing

[ionic adder]

it basically uses the idea of congurency

like you have ab = ac

and angles same

so prove that tri BAM is congurent to tri CAM

[little anvil]

ohh ty ill try that!!

bro. how ru so smart thats crazy

[ionic adder]

np lol

also is that AoPS?

[ionic adder]

oh i remember, its something like lightskin stare thing

[little anvil]

yup its for my tutoringgg

so

hard

[ionic adder]

nice

it def is hard considering you are an 8th or 9th grader

but keep at it

[little anvil]

tyt

uhh i still cant figure it out

im a lil sped

[hybrid lodge]

y=1/2sinx+([sqrt3]/2)cosx

i cant figurre it out

[ionic adder]

could you send a figure of what youve tried?

it will be easier to help that way

[little anvil]

I SOLVD IT

WOO

because of sas, bam and cam r equal

and so thas how m is the midpoint

for b its like cuz its a straight line and the 2 triangles add up to like 180 and just sum yap stuff i dont fully get it but im fine w that being incorrect atp

[ionic adder]

yes correct

[ionic adder]

what

[little anvil]

Idek

[upper karma]

you mean $\frac{1}{2} \sin{x}+ \frac{\sqrt3}{2} \cos{x}?$

[somber coyote]

deltaG

[hybrid lodge]

ya

convert to sine fuction

[upper karma]

i hope you know that $\sin{30} = \frac{1}{2}$ and $\cos{30} = \frac{\sqrt3}{2}$

[hybrid lodge]

ya

[somber coyote]

deltaG

[upper karma]

also, you can write it as $\cos{60}\sin{x} + sin{60}cos{x}$

[somber coyote]

deltaG

[hybrid lodge]

ok

Oh i see now

thanks

[lament tree]

I found a simple synthetic proof that a tangent to a circunference forms 90º with the radius
basically a proof by contradiction and it's cus you can construct from any acute angle and a side lenght an isosceles triangle with this length and 2 of this angle
the details are given as an exercise for the reader cus I'm hungry and too lazy too write out the rest of the proof bye

someone asked something like that here like last year and I couldn't find one so I finally could invent one yey

[maiden brook]

aops nice

[little anvil]

yess

aops is fun but also my death

[raw lotus]

i don't knnow what this is called

i need to look it up so i can watch videos

these specific types of problems

17 to 22

[lime dune]

"angle sum identities" and "angle difference identities"

[raw lotus]

can i have some help with these problems?

[ionic adder]

The guy above you just told you something

[trail tendon]

use the trig identity sin(a+b) = ... or cos(a-b) = .... (called the angle sum identities and angle difference identities that elrichardo mentioned)

[dark sparrow]

@raw lotus do you still need help with this

[devout gale]

can anybody help

[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

[devout gale]

2

[empty yew]

A cone, a hemisphere and a cylinder have equal bases. The heights of the cone and the cylinder are equal and are same as the common radius. Are they equal in volume?

If the volume and surface area of a sphere are numerically equal, then the radius of the sphere is ____ .

What is the ratio of the volume to surface area of a sphere?

The volume of a sphere is the product of its surface area and ____ .

Anyone!!!

[dark sparrow]

@empty yew what exactly do you wish to accomplish by rapidfiring these questions here? do you expect to be handed answers to all of them?

[dark sparrow]

ok, show your current progress?

[empty yew]

🫤

Help with that questions please!!

[dark sparrow]

do you have any progress yourself? do you know the formulas for the SA and volume of these solids?

[empty yew]

What's SA?

[dark sparrow]

surface area

[empty yew]

Then yes

[dark sparrow]

ok then these problems can all be solved by writing these formulas out

and making sketches where necessary

such as for that first one

[empty yew]

Ok

[empty yew]

They're not equal.

Is it right?

[dark sparrow]

show your work

[empty yew]

One min

[dark sparrow]

ok yeah this checks out

what is up with that r in the third line though

[empty yew]

Can't understand!

[empty yew]

Yes

[silent plank]

that was what

what is up with that r in the third line though
was addressing as that looks much more like another pi than r

[empty yew]

Ohh sry for the misunderstanding

👍

[empty yew]

Is the radius is equal here?

[dark sparrow]

equal to what?

[empty yew]

Radius of volume and radius of SA.

[nocturne remnant]

The sphere in question is the same right?

Then yes

[empty yew]

That was the total question!

So only I am asking

[nocturne remnant]

Wtf are you saying bro

[snow plinth]

Here's a proposition in my lecture notes and an extract from the proof:

[somber coyote]

Philip

[snow plinth]

Why is it that, when we iterate this process arbitrarily many times, we obtain a countable collection of sets?

[empty yew]

I am confused

[dark sparrow]

@snow plinth you might wanna move that to #real-complex-analysis

[dark sparrow]

there is only one sphere

[snow plinth]

ok 😢 will try

[empty yew]

Ohh ok

Can a ratio contain a variable?

[dark sparrow]

yes

in fact it has to, in your case

[empty yew]

3r:1

[dark sparrow]

show your work

[empty yew]

Sorry it's r:3

[dark sparrow]

show your work!

[dark sparrow]

ok yeah this is correct

[devout gale]

I got through the first part by doin 12* 2 pi and then I got stuck on the second

[dark sparrow]

if the curve was y = sin(x), where would it have a minimum point?

[devout gale]

At y = -1?

[dark sparrow]

both coordinates

[safe wyvern]

I don't know if this is the right place to be asking this 🤔 but can some one help me out and explain this to me

[dark sparrow]

for everybody else's benefit, i've cropped off the 90% of the image that didn't contain the problem...

@safe wyvern judging by your "wtf is this" caption, you have zero idea how to begin with this problem. am i correct?

[fluid charm]

Hello can anyone help me with this. I can only get the apparent power but i dont how to get the true power when only the voltage and current are given in a waveform. The apparent power is 2080 by multiplying the voltage and current by .707 to get the effective value but i dont know how to get the true power from.there. please help.

[solar yoke]

what i wanted to say yesterday was, i want to create a sphere like one in this guys profile picture. i want to be able to adjust the tile count, radius of the sphere and the angle which the sphere is created accordingly(the angle at we see the sphere).

using plain js on browser

i think hardest part is angle part. i dont know how i shall rotate the sphere by a specific angle. if im not wrong there are two angles in 3d plane(even though this sphere will be created in 2d plane. thats what makes it even harder)

ann doesnt reply its not looking good for me

[opal carbon]

were you planning on like writing the 3d rendering from scratch

[solar yoke]

i just want to render a basic sphere

[opal carbon]

using what though?

[solar yoke]

html, css and plain js

[opal carbon]

rendering 3d graphics isn't exactly a trivial problem

[solar yoke]

on browser

[opal carbon]

https://threejs.org/

maybe look up some three.js tutorial or something

[solar yoke]

i could render a circle

[solar yoke]

okay

[opal carbon]

ya but like youre planning on writing a whole renderer from scratch? that's a pretty involved project...just for this one application? it's like blacksmithing your own axe to cut one tree then never using it again

just a bit insane

[solar yoke]

lmao

well

[solar yoke]

if i could create this in like 30 or less ines of code. how many lines of code would it take for the sphere?

im not planning on making a fully functional 3d renderer

i just want to create a sphere

[cyan latch]

whats that one website coders use to practice?

[solar yoke]

code pen?

[cyan latch]

and holds enough leverage to be hired by companies

i think it was code pen

some asian dude holds 94th rank on it

and is also a world class champion in ping pong

[opal carbon]

well when rendering 3d points you can think of the screen as basically a section of a plane and you are projecting the vertices of your shape onto that section of plane

if you are extremely fluent in vector math like projection type stuff you can probably reasonably code a thing that can render simple wireframes as a small project. doing anything that will like actually color faces or lighting is a whole other level that would take a lot of knowledge of the theory of computer graphics

[cyan latch]

leetcode

[solar yoke]

u said website coders brah

[cyan latch]

:p

[vocal bison]

Help pls im confused

Ping

[drifting cipher]

Does anyone know how to draw this

[obtuse quiver]

anyone know the angles I'll get from this one? double angles and any other range than 0 to 360 deg make my mind shut down

[nocturne remnant]

Take it one step at a time

First find the possible range of 2x-60

Then find the values of 2x-60 in the range that make the equation true (use a graph if you feel the need to)

Then solve for the corresponding x

[lament tree]

I know how to prove geometrically that if two angles a and b such that a + b < 180º and a < b, then 90º - a/2 < b
but how can I prove it algebraically?

[nocturne remnant]

It’s not true

[dark sparrow]

what is your geometric proof?

[faint comet]

How do I get better at trig proofs ?

[dark sparrow]

practice

[plush badge]

Which one is correct?

[silent plank]

don't use chat ai for math

and they're both wrong

[faint comet]

If I put that cos on the right over cos to make the denominator the same does that cos become cos squared over cos?

[next phoenix]

yup

[dark gyro]

it becomes sec

[next phoenix]

to get them on the same denominator essentially what you are doing is multiplying cos by cos/cos

hence cos square divided by cos

which gives you 1/cos after summing it with 1 - cos^2 + cos^2

so Sec

[thin sentinel]

$\cos\theta = \frac{\cos^2\theta}{\cos\theta}$ yes

[somber coyote]

new_nick: haprzTypo

[dark gyro]

adversary orz

[next phoenix]

orz?

[faint comet]

Ok thank you guys

[next phoenix]

wha

[dark gyro]

look at orz closely

[next phoenix]

LMAO

i'm not an orz tho-

[dark gyro]

its alright

we use orz quite a lot when someone provides some Aha! solution

[next phoenix]

I have never heard orz ever in my life and I'm chronically online

(sometimes)

(never not)

[dark gyro]

well now you have
now you can use it

[vernal pilot]

It’s like unlocking the national Pokédex in Pokémon, you’ve caught all the normal internet lingo, now you gotta catch the orz.

[lament tree]

whay

*what

you sure

[nocturne remnant]

bruh

if a = 1, b = 2

then it doesn't work

[lament tree]

aaaaaa

[dark sparrow]

!xy

[lime crown]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[dark sparrow]

<@&268886789983436800> scam ad

[brave widget]

hi

anyone knows a playlist on yt where i can learn trigonometry?

i struggle with it

ping me if you answer

[brisk ore]

does anyone knnow why for part b its multiplied by -4/5

[exotic yarrow]

law of cosines: $$BC=\sqrt{AB^2+AC^2-2 \cdot AB \cdot BC \cdot \cos \theta}$$

[somber coyote]

Civil Service Pigeon

[brisk ore]

wouldnt it be +-?

[exotic yarrow]

It says that BC is the longest side, so chances are, you want the larger angle

which gives the negative soln

[brisk ore]

ohh

so if they want the longest side we solve to find the largest angle

[brisk ore]

if i cant tell this part straight away, would it be fine to solve for both then use the larger one as my answer or would i lose method marks

[exotic yarrow]

idk how you're marked

[brisk ore]

nvm but thanks for the help i was stuck on this

@exotic yarrow for part a) i got 51.1 but wouldnt 128.9 degrees be the angle DCB? so it would be irrelevant for the question

[exotic yarrow]

it's obtuse in the diagram

so you take the obtuse case from the sine rule

[brisk ore]

is there a reason why we do that?

or is it one of the rules

[exotic yarrow]

edited

it should make more sense now

[brisk ore]

oh ok

so if the triangle given was only ACB

the answer would just be 51.1

[exotic yarrow]

you wouldn't have sufficient info then

you would need something to indicate if it's obtuse or acute

since you can't reject the obtuse case off the angle sum of the triangle

[brisk ore]

alr

thanks

[lament tree]

found out what was wrong

[brisk ore]

for part b when finding the minimum length would that just be the sum of the lengths?

[exotic yarrow]

If you wanted to do AC+CD sure

or you could do AD all at once

up to you

[brisk ore]

i was thinking

if i did AD all at once

would i still take 27 to be the angle for DAB

and 7 to be the length of DB

[exotic yarrow]

yeah you could do cosine rule on angle DAB

[lament tree]

so what if a + b + c = 180º and a and c are less than b, now that should be true

[brisk ore]

lemme try that one second

[lament tree]

that 90º - a/2 < b

[brisk ore]

when i use cosine dont i need the angle of B?

[lament tree]

by symmetry same for 90º - c/2 < b

[exotic yarrow]

set up the cosine rule on angle DAB

you'll see

trust me

[lament tree]

that's equivalent to the "biggest angle opposes biggest side" theorem, that's why I was curious

I am curious, still

[brisk ore]

so a = sqrt 12^2 + 7^2 - 2 x 12 x 7 cos 27

[lament tree]

oh found out why

[exotic yarrow]

$7^2=12^2+AD^2-2 \cdot 12 \cdot AD \cdot \cos 27^{\circ}$

[somber coyote]

Civil Service Pigeon

[brisk ore]

why wouldnt AD be where 7 is

[lament tree]

a + b + c = 180º
b = 180º - a - c > 180º - a - b
2b > 180º - a
b > 90º - a/2

[exotic yarrow]

why would it

AD and AB include the 27 deg angle

[lament tree]

so there's a cool proof for the "biggest angle opposes biggest side" theorem

[brisk ore]

oh yeah

ok wait

[lament tree]

ihu

[brisk ore]

where would you go from here

[exotic yarrow]

it's a quadratic you can solve

you can also angle chase if you want to

and set up the cosine rule on angle ABD

[brisk ore]

what would you get if you did that

i tried but it went wrong

[exotic yarrow]

you should just get what the mark scheme says, no?

[brisk ore]

i meant if we find the angle of ABD separately

[exotic yarrow]

what did you get for it

[brisk ore]

i got 102. i added 27 and 51

im tripping mb

yh i got 102

[brisk ore]

yeah this clears it up

so a = sqrt 12^2 + 7^2 - 2 x 12 x 7 cos 27

over here

it shouldve been so a = sqrt 12^2 + 7^2 - 2 x 12 x 7 cos 102

and that would give me 15.09 and i add the given lengths to get 42 (2sf)

i had a question though. you see how if you have one angle here you subtract 180 from it as its a straight line

is there any relationship here

if we found one angle here

so 102 for the bottom part

[exotic yarrow]

nah you just need to add the two

there's not anything that forces a bisector or anything like that

[brisk ore]

ah alright. that was confusing me earlier on

cause idk why but i took it to be something and i got a weird value

[brisk ore]

thanks for the help though this cleared up a lot

[trail tendon]

what do you need help with?

solving the expression?

[harsh iris]

how to solve the letter c

[trail tendon]

do you know how to simplify 25/12 - 0.25

[harsh iris]

do i subtract is as 25/12 - 1/4?

[trail tendon]

yes

or 25/12 - 3/12 yea

[harsh iris]

so 11/6

[trail tendon]

yes, thats the result inside the parenthesis

so now you need to multiply 11/6 * 4 * pi, right?

to get the inside of the bracket

[harsh iris]

is it 22pi/3? im not sure, also did i insert the right equation for the h(t) that is 25/12? Not just h(12)?

[trail tendon]

i was assuming that your equation is correct. if it is, yes, you plug in h(25/12), not h(12) because it wants to know at 25/12 seconds, not 12 seconds

[upper karma]

Hi

[upper karma]

Can I anyone help me with this one

I'm so dumb

I have one more try

[trail tendon]

can you write out your best definitions for what a quadrilateral is, what a parallelogram is, and what a rectangle is?

[upper karma]

Can you write yours

I need the answers so bad It's gonna be due very soon 😭

[quaint sequoia]

How abt for this. This person told me it was 10.95

I used pythagorean theorem

I did this 💀

[vocal bison]

wut

[quaint sequoia]

I need help

[vocal bison]

dont u just add 6 and the radius

[quaint sequoia]

Huh? 💀

Why would I do that?

[vocal bison]

the radius is given its 7 so gn shoild be 7

un less im dumb

7^2 +b^2=13^2 -> 49+b^2=169 -> 169-49=120 -> sqrt 120=2 radical 30 or 10.95

oop

[quaint sequoia]

Where did you get 13 from?

[vocal bison]

6+7

segment GN should be 7 too if PG is the radius

i think

[quaint sequoia]

But it looks shorter

How would it be 7?

[vocal bison]

i thought is not always up to scale ?

[quaint sequoia]

Um idk but I thought it was 7^2+ b^2= (n+6)^2

[vocal bison]

i prob did it wrong then i did this like a month ago soz

[quaint sequoia]

Wait but what you said does seem right

NO I RHINK UR RIGHT

[vocal bison]

🤷‍♀️

[quaint sequoia]

How about this ?

I already used 360-260 to get 100 but idk what to do next

[magic dust]

hii, any can help me in 2 exercise? plis

[quaint sequoia]

Just send them and someone will help c:

[magic dust]

oooo

ty

can someone help me with the procedure please?

[vocal bison]

ummm like probably 80

[quaint sequoia]

Probably 80 😭?!

That doesn’t remotely look close to 80 😭

[vocal bison]

🤷‍♀️

[quaint sequoia]

HOW DO I DO IT?

[trail tendon]

i'm not 100% sure i am correct

but

if arc BCD is 260, the rest of the triangle is 100 degrees, right?

[quaint sequoia]

Yeah yeah

I did that

But then I get stuck

[trail tendon]

with the circle being 100 degrees, i think the angle BDE would also be 100 degrees.

i actually think it would be 50

but if you do the math thats not an option

so i think its 100

thats why im not sure im correct

💀

[quaint sequoia]

Yeah :c

[trail tendon]

but if it is 100

[quaint sequoia]

Tbh I just put 40 😭

[trail tendon]

then you know ABE is 90 and ADE is 90

nah

look

if ABE is 90 and ADE is 90

and DEB is 100

then you can find the last angle

its a quadrilateral

[quaint sequoia]

But how 😭

[trail tendon]

1 sec lemme draw

[quaint sequoia]

HOW IS IT A QUADRILATERAL?

[trail tendon]

it has four sides lol

[quaint sequoia]

Oh yes yes

[trail tendon]

yeah

[quaint sequoia]

But how do we get angle A?

[trail tendon]

thats assuming that the angle is 100

the sum of the angles of a quadrilateral is?

[quaint sequoia]

Um idk

90

[trail tendon]

360 ;-;

[quaint sequoia]

WHAT ARE TOU TALKING ABR 😭

[trail tendon]

a quadrilateral is a four sided shape

like a rectangle

[quaint sequoia]

Yes

[trail tendon]

but not necesarily a rectangle

this one is a kite

[quaint sequoia]

Yes

[trail tendon]

but the four angles still add up to 360 degrees

just like a rectangle

[quaint sequoia]

Si what would be angle A?

[trail tendon]

do you get that ADE is 90 and ABE is 90 degrees?

[quaint sequoia]

Yes yes

[maiden brook]

bruh

[trail tendon]

so angle A is 360 - 90 - 90 - 100

ASSUMING the angle is 100

which im not 100% sure of

[maiden brook]

it is

[trail tendon]

ok

[quaint sequoia]

BED is the center of the cricle soooo

That would mean arc BD is 100 as well

BUT ANGLE A

How do I get that?

[trail tendon]

um

[quaint sequoia]

BC if I multiply 100 by 1/2 it gives me 50

[trail tendon]

we are given that the arc BD is 100

which means that the angle BED is also 100

[quaint sequoia]

And that’s not an option

[trail tendon]

nah

[quaint sequoia]

YES

Ik that 😭

Im asking how to get angle A

[trail tendon]

whats the sum of the angles

[maiden brook]

bruh

[quaint sequoia]

Viper u have me lost

[maiden brook]

just take the quadrilateral lol

[quaint sequoia]

wym?

[maiden brook]

ABED

[trail tendon]

a four sided shape has four angles

[maiden brook]

angles sum to 360

[quaint sequoia]

Yes

[maiden brook]

u have all but one angle

[trail tendon]

so if you have 3 angles

[quaint sequoia]

OOO

[trail tendon]

you subtract 360 by the 3 angles

to get the last angle

lol

[quaint sequoia]

ITS 80

RIGHT?

[trail tendon]

yes

[quaint sequoia]

OMG

[trail tendon]

assuming 100 is the angle

[quaint sequoia]

So smart viper

[trail tendon]

💀

[quaint sequoia]

👏🏼

[trail tendon]

i wish 💀

[quaint sequoia]

How abt this ? This person told me it was 10.95v

Is that correct?

[vocal bison]

why didnt yall just do 260-180

[quaint sequoia]

VIPER

[trail tendon]

um

let me do a calculation to see if its what he said

also i dont know how they got v

but lemme see

they are correct

[quaint sequoia]

OKIE THANK YOU

[trail tendon]

look at it this way

you got a right triangle

you got the hypotenuse length and one of the side lengths

you want the other side length

[vocal bison]

guys i said it already

[trail tendon]

how

[vocal bison]

🤷‍♀️ thats what my teacher taught me

[trail tendon]

that seems very methodical but not logical

like i bet it works for some reason

but we don't know the reason

so

💀

and basically the only reason i bet it works is because it gives the correct answer

[vocal bison]

well as long as i get it tho 🤷‍♀️ now i need help too so

[trail tendon]

but i don't see why

[trail tendon]

help with what

[vocal bison]

i sent this earlier but im jsut confused do i find the lateral and suraface for each figure

like i mean the cube and then the pyramid or like a cone then the cube um

[trail tendon]

um

i mean for the surface area you just add the areas of the 2D surfaces, right?

[vocal bison]

ya

[trail tendon]

so with a square pyramid

you're going to have four triangles and one square

[vocal bison]

ya

[trail tendon]

when it says 10cm cube

does that mean that the lengths are 10cm

[vocal bison]

yes i think so

[trail tendon]

i would assume so but yea

ok

so you know the side length of the square is 10

right?

[vocal bison]

yuh

butlike what do i find

[trail tendon]

does a square pyramid have equilateral triangles?

or isocoles?

cuz if its isocoles this would be much harder lol

[vocal bison]

umm probably equilateral ??

[trail tendon]

i hope 💀

[obsidian harness]

Isosceles

[vocal bison]

OOP

[trail tendon]

dang

ok

rip

well we know the height of the square pyramid is 10

so i guess we gotta find the length of the slant

[obsidian harness]

Even if the square pyramid fits inside a cube (remember a square pyramid is just talking about the base, not the height)

[trail tendon]

just using the pythagorean theorem

[obsidian harness]

Yeah you need the height of the slant

[trail tendon]

right

i guess it doesnt make it much harder actually

[obsidian harness]

Clearly the length of the diagonals of the pyramid is longer than the slant height

[trail tendon]

yeah

the only reason why i asked

was because in the picture

it doesnt look like it goes to the top 💀

so i was wondernig if a square pyramid had to have equilateral triangles or smthn

but nah

um

mimi is this hard to visualize

[vocal bison]

well i tried to draw it

[trail tendon]

show

[vocal bison]

but i did this like yesteday so

ok

[trail tendon]

its hard with 3d lol

[vocal bison]

ignore the work i did taht wrong and the pyramid is suppose to be higher and touch but whatevr

[trail tendon]

ok so

hmm

imagine theres a line in the square pyramid from the tip of the pyramid down to the middle of the square

inside the square pyramid

[vocal bison]

so the height

[trail tendon]

yes

[vocal bison]

ok

[trail tendon]

but if you take that and you look at it from the side in a 2d way

you see a triangle

[vocal bison]

yuh

[trail tendon]

a triangle with height 10 and base 10

you want the side length

[vocal bison]

wiat aint the base 5

wiat nvm

[trail tendon]

you want to cut it in half, yes technically

to solve for the side length

[vocal bison]

ok

wait we solving for hypotenuse sooo like 5 rad 5

wait wut

[trail tendon]

yes

[vocal bison]

oh ok

[trail tendon]

keep in mind this triangle is not the same as the triangle in the square pyramid

all you did was solve for the altitude of the 2d triangle

[vocal bison]

yuh

[trail tendon]

cool

so then you have the altitude of each triangle, and you have the base

you know how to do the rest?

[vocal bison]

wait

was my work right

[trail tendon]

your math was correct

[vocal bison]

oop

[trail tendon]

i hope you are understanding conceptually whats gonig on

lol

its kinda hard to visualize

[vocal bison]

no i am conmfused because there is a pyramid inside a cube and i thought u were suppose to find the la and sa of the cube and the pyramid
but i just did the pyramid

wait

ohhhhh im slow

[trail tendon]

uhh i dont know whaat u mean la and sa

[vocal bison]

lateral area nad surface area

[trail tendon]

oh

yeah

we were finding the surface area

[vocal bison]

yeah but why did we jsut do the pyramid inside not the entire cube

[trail tendon]

it doesn't want the surface area of the cube

[vocal bison]

wait im slow

[trail tendon]

just the surface area of "each figure" inscribed in a cub

xD

[vocal bison]

oop my bad

reading iks fundamental

[trail tendon]

XD

[vocal bison]

thanks trho

cleared it up

[trail tendon]

you got the surface area then?

[vocal bison]

yah

[trail tendon]

may i ask what the lateral area is 💀

unless its the real world area then idk

[vocal bison]

100 radical 5

waht

am i suppose to conver thta

[trail tendon]

oh is that the surface area

[vocal bison]

whatt nop

[trail tendon]

the lateral area?

typing be like

[vocal bison]

yeah

[trail tendon]

nah i meant like

what IS the lateral area

like what does that mean

[vocal bison]

its like yk a cylinder but minus the bases so jsut the part where it wraps around for ex

i think

[trail tendon]

really???

[vocal bison]

i think so

[trail tendon]

ohh

i think the lateral area is just the sides

i don't think its a cyllinder

actually idk wait

[vocal bison]

no i said that for exmaple

[magic dust]

hi guys any can helpme plis? i need to resolve 2 exercises of the trigonometry

[trail tendon]

ohh yeah

[trail tendon]

what are the questions

[magic dust]

in the figure, AOB, COD and EOF are circular sectors and the areas of the circular trapezoids CDFE and ABDC
S₂u² and S₃u² respectively. If S₁u² is the area of the circular sector EOF and S₃/5 = S₂/3 =S₁ , find the value of M= L₂/L₁ - L₁/L₃ + L₃/L₂.

please

[trail tendon]

wth

ohhh

[magic dust]

can you help me plis?

[trail tendon]

i can try

[magic dust]

tyyy<3

[trail tendon]

if im smort enough 💀

what is 19/6

the arc length?

oh is it the area of something?

[magic dust]

yesyes

is the valor of the area M= L₂/L₁ - L₁/L₃ + L₃/L₂.

[trail tendon]

is 19/6 the area of the figure inside AOB?

oh

[magic dust]

I need a procedure but I don't know how to do it

I translated what the exercise question says

[trail tendon]

what is L?

[trail tendon]

it's still a weird question

[magic dust]

yes

[trail tendon]

it doesn't define L1, L2, ect.

[magic dust]

I assume it is the length or area of ​​each part

[trail tendon]

i guess it doesnt matter which

thats kinda dumb

ok

[magic dust]

sorry

[trail tendon]

what's all this about the trapezoids though

this isn't possible without the trapezoids

and we don't have any numerical information on the trapezoids

unless the 19/6 and stuff has to do with it

i think this question is too weird and dumb 💀

[magic dust]

does this help? AOB, COD and EOF are circular sectors and the areas of the circular trapezoids CDFE and ABDC
S₂u² and S₃u² respectively. If S₁u² is the area of the circular sector EOF and S₃/5 = S₂/3 =S₁"

[trail tendon]

ohh wait

[magic dust]

L is side

[trail tendon]

i didn't read it carefully

[magic dust]

oooo

[trail tendon]

what is u

S2u^2

why did they square u, and what is u?

[magic dust]

u isss

wait me 1 min please

[trail tendon]

💀

[magic dust]

i dont know

sorry

[trail tendon]

LOL

[magic dust]

[trail tendon]

bro sorry but i think this question is completely stupid

can you move on or something?

it doesn't make sense to either of us

i'm not sure if it even makes sense to anyone 💀

[magic dust]

ooo

u is

[trail tendon]

huh?

[magic dust]

Union of two sets

[trail tendon]

i don't think thats how its being used here 💀

that would be capital U

this is lowercase u, prob a variable

but a variable that represents what?

S1, S2, S3 are variables that represent what?

[magic dust]

hmmm

iss

heyhey

[trail tendon]

yoyo

[magic dust]

Use similarity, so (length 1 / length 2)^2 = (area 1 / area 2)
So the area of EOF is S1
The area of COD is S1 + S2 = S1 + 3S1 = 4S1
The area of ABO is 4S1 + S3 = 9S1
So using the above principle, you can find that value with the ratios of the lengths

that's what someone told me

sorry if I bother you a lot

[rare fossil]

One message removed from a suspended account.

[dark sparrow]

you don't bc it makes no sense

actually 6, 7 and 9 also make no sense

but at least with those the $\cong$ can be replaced with $=$

[somber coyote]

Ann

[dark sparrow]

no such luck here

how can an angle be perpendicular to anything?

where did you get this from? @rare fossil

[rare fossil]

One message removed from a suspended account.

One message removed from a suspended account.

One message removed from a suspended account.

One message removed from a suspended account.