geometry-and-trigonometry

[spark yacht]

no everyone uses it

like these stuff

[granite marsh]

r ur exams done

[spark yacht]

no its on 17th

and 19th

then over

[granite marsh]

ohk

[spark yacht]

what grade are u

[granite marsh]

same

from where did u get that question?

[sudden solstice]

Guys pls help me I need help 😭 on this stupid stuff I am cooked

[granite marsh]

?

[spark yacht]

ask

[sudden solstice]

I am amazing in algebra tho

[spark yacht]

which

[spark yacht]

try that fe

[granite marsh]

this

[sudden solstice]

I got a 109 in algebra I think in 9th

Overall grade

[spark yacht]

mods bot

[sudden solstice]

But in geometry I am cooked

[spark yacht]

where are u from

[sudden solstice]

Wdym

[spark yacht]

country

[sudden solstice]

USA

[little oriole]

bruh!!!!

[spark yacht]

nah wait ive done this

nvm

[spark yacht]

oh nice

[granite marsh]

send the solution

becoz I didn't understand the question

[sudden solstice]

Guys is there a way I can create more of these problems to study?

[granite marsh]

just search online

[little oriole]

ye there is actually

[sudden solstice]

How

[little oriole]

what do u wan it for ?

[spark yacht]

search class 9th ncert

open triangles chapter

100s of these qn

[little oriole]

wait

what graders r u all ?

[sudden solstice]

Huh? Wdym @spark yacht

[little oriole]

is any1 of u a master level person ? ........or r u all high schoolers ?

[spark yacht]

all hs

[spark yacht]

ignore it

[little oriole]

i m in the wrong channel for some reason then

[sudden solstice]

Does this look appetizing?

[spark yacht]

haha

[livid plaza]

Not everyone here is in high school

[solemn lagoon]

there is a bit of it but its not insanely important I would still make sure you understand the material well though because itll help with some of it

[solid needle]

its abstractly important in the sense that you should have a good intuition of what these graphs look like

actually graphing them in and of itself is not super important

what is important is understanding the properties of these functions, and many of their basic properties can easily be understood if you can visualize it in your mind in some way

[solid needle]

thats good

again, more importantly than merely what it looks like is their properties

for instance, where the asymptotes are, what are their periods, are they increasing or decreasing

so if you know those youre good

[little oriole]

ohk

[spark plover]

yeah in college theres math 127 which has trig

im in highscool tho soooo

anyways im sped and dont understand why i cant use tangets values

i did the whole method and put in the values and got it wrong and didnt understand why

so i checked examples and it just says that its not defined in this equation

the equation btw

[dark sparrow]

cot(0) and cot(pi) don't exist.

https://tenor.com/view/shrimp-as-that-clash-royale-hee-hee-hee-haw-gif-25054781

[spark yacht]

Do u have math 55

[spark yacht]

cotx + √3 = tan^-¹(0)?

[spark plover]

AH

i see

[spark yacht]

Wtf did they do in example 2

Does inscidbdd angle also works like this

[spark yacht]

I got it dont mind

[granite marsh]

?

bro its NCERT

the easiest book

do RD or RS or any other book

[spark yacht]

ikr

I know everything but geometry

or everything but circles

[stoic sun]

Hi what’s your take on this problem? For question 2 the official answer key for length of segment BD is 3 (6:2) and segment AD is the square root of 74

[spark yacht]

What info you have on the triangle

Just sides or any angles too?

[stoic sun]

Just sides

[obsidian harness]

AD is the angle bisector right?

[stoic sun]

Yeah and on the answer key since it’s the bisector the segments created is equal that’s why they just divided it by 2 and I don’t think thats correct as it must follow the proportionality rule

[solid needle]

correct

very obvious that angle bisector doesn't split the opposite side in half if you check an extreme case

[upper karma]

i tried applying the sine rule as well and i dont get 3 for bd as well

[obsidian harness]

yeah the angle bisector theorem states that BD : DC = 8 : 10 actually

[stoic sun]

Yeah lol our team leader is arguing that the segments should just be divided by 2. And that doesn’t follow the angle bisector theorem

It’s the official answer from the book publisher

[obsidian harness]

bruv. when people don't know maths

there was a really nice exercise from a textbook in the help channels

that showed exactly WHY this logic was flawed

[stoic sun]

Yeah I should check that out

[obsidian harness]

can't find the exact q sorry

[stoic sun]

Can you help me confirm the length of AD also. So that I can support my answer

[obsidian harness]

,w sqrt(ab(a+b+c)(a+b-c)) / (a + b) where a=8,b=10,c=6

[somber coyote]

Results provided by WolframAlpha

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

Requested by sooouth.

[solid needle]

@stoic sun draw this triangle for your team leader

a very short but very long right triangle

ask him to draw the angle bisector and demonstrate that the hypotenuse is split in half

very obvious counterexample that disproves his nonsense

with a 90 degree angle here, its very easy to even rigorously disprove his claim even without much geometry at all

set the right angle corner to be (0,0)

the angle bisector is y=x

pick two points like (1,0) and (0,100)

the hypotenuse is y=-100x+100

the intersection of the angle bisector and hypotenuse is (c,c) where c = 100/101

if his claim was correct, the distance from (1,0) to (c,c) is equal to the distance from (0,100) to (c,c)

which is clearly absurd and not true, since the first is way less than 10 and the other is way bigger than 90

[mellow quest]

how to slow this $$x^8+98\cos(4x+3)=98\cos(x^2)+(4x+3)^4$$

[somber coyote]

olivka

[mellow quest]

I found x = -1, x=-3, x=2-sqrt7, x=2+sqrt7

[obsidian harness]

cursed question

how did you get those answers btw

[obsidian harness]

wow

[mellow quest]

its from the Russian exam

[obsidian harness]

,calc (-3)^8 - (4 * -3 + 3)^4

[somber coyote]

Result:

0

[obsidian harness]

ah ofc

[mellow quest]

does that give you any information?

[ashen lotus]

Can anybody solve this question for me? As in which quantity is greater or smaller or equal or relationship cannot be determined

[maiden brook]

The larger the difference between each angle and 90 degrees, the longer the respective side is

[narrow widget]

Bro its basic.If he don't underrstand then give him some time

[silk jacinth]

In an assignment, I should construct a simple sinusoidally oscillating model for atmospheric CO2 concentrations over 12 months.

The oscillation's magnitude is 5 ppm, the average concentration is 420 ppm and the highest point is in May and this is my solution (t is time in months). I think it's right, but in the assignment solution, the c parameter not -2 but 10. Am I missing something or is the solution stated by assignment wrong?

[rapid valley]

30-60-90?

And 45-45-90?

Or just like

Right, acute..

Because if the bisector is just splitting it in half both triangles should be right triangles

and if so, the base of both should be equal, and they both share a bisector, so the base^2+ bisector^2= H

And if they are equal, the sides are equal

Therefore it should be an equilateral triangle if everything is equal

But in the image it doesnt show that

I just need more details

[upper karma]

geo

[stoic sun]

This is the answer from the book

[spark yacht]

Yeah this is correct

The figure is deceiving tho

[stoic sun]

AD as bisector. BD =3

[silent plank]

work is incorrect

the diagram indicates that AD is the angle bisector of angle A,
(not the same as the median from A to BC)

D won't bisect BC here. it's not as simple as just dividing by 2

[trail tendon]

what part indicates that it is the angle bisector?

[silent plank]

the arcs at angle A

[solid needle]

didnt we just go over how this is wrong if this is an angle bisector?

[trail tendon]

what about them?

[silent plank]

there are two arcs both coloured yellow implying congruence

[sly urchin]

Given ∆ABC. M is a point inside ∆ABC. AM, BM, CM intersect BC, CA, AB in order at I, J, K. The line through M parallel to BC intersects IJ, KI at E, F. Prove that ME=MF.

I think i'll definitely need to draw extra lines here

I've been trying to do so, idk which lines to draw

Anyone knows? My intuition isn't great

[covert tinsel]

<@&286206848099549185>

I need help with these two questions I wanna know the answers

[restive terrace]

we aint gonna give you answers but can guide you so you know how to do it

[sage radish]

i dont even know what angle a is measuring here lol

[zealous pike]

I understand it for those angles. How would the trigonometric ratios be calculated for triangles of different angles, like a 55°, 90°, 35° triangle?

[sage radish]

calc which is short for calculator btw

[zealous pike]

but what if I don't have a calculator?

[zealous pike]

here for example, it is easy to know how sin 30° = 1/2, as well as for the other functions or angles shown.

[sage radish]

🤷‍♂️

[zealous pike]

but I have no idea how to evaluate the trigonometric functions for other angles

[solid needle]

in general, you would need the trig functions themselves, sin/cos/tan/etc

there is no general method otherwise

only very specific angles are even expressible with the 4 basic operations + radicals and such

there is, for example, a half angle formula

so if you have like 45-45-90 ratios

you can also do 22.5 degrees

and repeat, so you can also do 12.75, etc

there are some other very specific ones, like 72 degrees is also doable, but not nearly as easily

[zealous pike]

I want to learn those methods, seems interesting

[solid needle]

if you want to know what values are even doable at all, i recommend going here: https://en.wikipedia.org/wiki/Exact_trigonometric_values

look at the section on constructible values

not sure what the values are in general, that you'll have to research yourself

[zealous pike]

Tyvm

[covert tinsel]

Bruh I already solved it and just wanna check besides that’s no even a hw im js testing myself

[solid needle]

im not sure we know that

you could just say what your answer is and roughly how you got it and we would be happy to confirm i think

[upper echo]

Draw a line through A, parallel to BC. And let continuations of rays IK and IJ intersect it at D and G respectively. Then you have similar triangles and find AD/AG=1 by similarity and Cheva. So IA is a median of the triangle IDG, so ME=MF.

[strange solar]

could anyone realistically have the time to simpliy this

p not equal to 0 and ±90

[chrome token]

yes

[strange solar]

well i made it to be like

the

identities thing

pythagorean identity

but extra steps

[chrome token]

one important thing is you dont need to convert 1-sin^x into cos^2x

if you already cancel it out

[upper karma]

@foggy galleon here's some cursed math for you.

[foggy galleon]

WOW

heres this https://www.desmos.com/calculator/aih4o6kisg

[strange solar]

wtf

[foggy galleon]

bro went fom integs suck to dont suck

[strange solar]

my one is harder

chatgpt couldn't simplify mine

[foggy galleon]

same

[strange solar]

indee

o

mine is just the pythagorean identity tho

[foggy galleon]

mine 2+2 deluxe edition

[strange solar]

lol

wtf is yours

i expanded it more and

i need to up my game

[foggy galleon]

i made it for fun

[strange solar]

bruh

this is all you do

ok bet

im adding contour integrals

[foggy galleon]

idk how to add that

[strange solar]

but desmos doesn't understand that

[foggy galleon]

stupid desmos

they need to add it

[strange solar]

i know

for complex analysis

[foggy galleon]

yeah (even tho i didnt study that 💀)

[strange solar]

https://www.desmos.com/calculator/vtsma8i8js

[strange solar]

it's university stuff i think

[foggy galleon]

possible to combine them?

bro my torture thingy turned into text is so long discord wont let me send it

[strange solar]

hm

combine what

[foggy galleon]

all 3 things inside something

[strange solar]

wait

i combined ours

wait

desmos doens't let me save it

https://www.desmos.com/calculator/u2r8oopw0i

heh

[foggy galleon]

i see rn

nice

[strange solar]

go to the bottom

it's not long enough

[foggy galleon]

longer, more complex

[chrome token]

take a snipping tool

and then send

[strange solar]

i must

[foggy galleon]

then give it to ur kid if he wants mcdonalds

[strange solar]

add complex numbers

nvm too lazy to remove all the i stuff

[chrome token]

desmos cant do that

[strange solar]

from the sums

[foggy galleon]

ohk

[strange solar]

ou can

you can

[chrome token]

how?

[strange solar]

in the settings

turn on complex numbers

and i will be turned into root negative 1

[foggy galleon]

ima turn it on rn

HELL YEAH NOW I CAN PUT sqrt-1 LESS GO

imaginar unit

[strange solar]

it takes 2 minutes to save

https://www.desmos.com/calculator/orfhybcbaz

new and improved

[foggy galleon]

i go see

[foggy galleon]

id give it to my son if he crying to go to mcd, solve to eat

[strange solar]

lol

modern discipline

realistically every one of the ones in my desmos thing is just 1

the answer is 1

[foggy galleon]

yes, if no solve id give him mine + yours

[foggy galleon]

lol

[strange solar]

and yours is 4

[foggy galleon]

yeah i saw

[foggy galleon]

yep

[strange solar]

your one is mad

[foggy galleon]

i can make it worse

but id do it tommorow

[strange solar]

ok

same

[foggy galleon]

either way ima js make little cursed things to combine with my big one

[strange solar]

ye

[foggy galleon]

yes this looks good

desmos needs to struggle

[foggy galleon]

ima borrow ur equation a lil

[strange solar]

ah

ok

[strange solar]

wtf

with respec tto h with respect to g

with respect to a

with respect to b

wiath respect to d

[foggy galleon]

lol

[foggy galleon]

https://www.desmos.com/calculator/p2fmxerqgj

[strange solar]

send this to

um

norman levinson

[foggy galleon]

where he at

[strange solar]

20 hours work

[strange solar]

heaven

[foggy galleon]

oop nvm let him miss out

[strange solar]

hahah

jk

[foggy galleon]

dam

[strange solar]

this should be in museum

[foggy galleon]

yes

i sell for 3 million dollar

imagine giving it to an mit professor

[livid plaza]

what the heck!??

[foggy galleon]

i want to sell it for 3 million

[latent sable]

[foggy galleon]

cmon it aint that bad take a look

[livid plaza]

USD?

[foggy galleon]

yes

its so worth it

trust me

it wont knock your socks off

[livid plaza]

I already copied the equation and pasted it in a LATEX compiler. i.e I'm selling it for 1 million USD. if anyone's interested 💀

[granite marsh]

?

[foggy galleon]

no your stealing my hard work

[misty elk]

https://archive.org/details/100GreatProblemsOfElementaryMathematicsDoverHeinrichDrrie/page/17/mode/1up

How to derive that formula? I can't make sense of it

I know cos a-cos b=-2sin[(a+b)/2]sin[(a-b)/2]
and cos a + cos b=2cos [(a+b)/2]cos[(a-b)/2]

[tame oriole]

Yo how do i prove this

[foggy galleon]

calculator

[solid needle]

do you know your double angle formulas

[tame oriole]

Yes

But i get stuck after doing it

[solid needle]

stuck how

show work?

[tame oriole]

Hold up

I think i was supposed to solve it not prove it

[zealous pike]

Do you have more information or is it just that

[sudden solstice]

Guys am I cooked I don’t even know how to do congruent angles theorem and also ASA and putting the given for them

[frigid comet]

The given for ASA would be, for example, angle J is congruent to angle K, and a side has a tick. To determine that a triangle is congruent by ASA, their must be a congruent angle, a tick-marked side, and a congruent angle, in that order. I’ll send a pic to show you because im bad with words

I hope this helps

Wait mb AAS does prove congruency

Here’s the more detailed explanation if it helps

[primal flare]

Plss hiw do u do this

Ik it's ljke 400+x^2=(x+12)^2

Omg nvm I forgot how to do algebra

[narrow sail]

hi

is there any way to prove the Pythagorean theorem using trigonometry?

[trail tendon]

the law of cosines

that probably wasn't the answer you were looking for but 💀

[misty elk]

Yes
https://www.cut-the-knot.org/pythagoras/TrigProof.shtml

[primal flare]

omg please help

[narrow sail]

thanks

[strange solar]

@foggy galleon

[livid plaza]

that's how the world works

[strange solar]

lol

[foggy galleon]

i been awakened

[strange solar]

did you work on it

[foggy galleon]

with patent i ruin you

[foggy galleon]

yeah im working on it

[strange solar]

ok

[foggy galleon]

making a different torture equation to put on the original one

[strange solar]

ok

[somber coyote]

integrals don't suck

[strange solar]

texit can't compile this masterpiece

[paper spear]

what

wwhy

what does it mean

its MATHS so its NUMBERS why is it all LETTERS

😿😿😿

[foggy galleon]

yes, very nice

@strange solar

[strange solar]

\left( \frac{\sin^2(p)}{1 - \sin^2(p)} \right)^{ - \left( \int_0^1 \frac{d}{dx} x , dx \right)} \left( \frac{ \left( \frac{d}{dx}(x) - \cos^2(p) \tan^2(p) \right) \left( \frac{\cos^2(p) \left( \frac{\sin^2(p)}{1 - \sin^2(p)} \right)}{\cos^2(p)} \right) }{ \frac{ \left( \frac{d}{dx} x \right) - \frac{1 - \cos^2(p)}{\frac{1 - \cos^2(p)}{ \left( \int_{\frac{d}{df} 0}^{\frac{d}{dt} t} \left( 2 \left( \frac{d}{dx} \left( \frac{ \frac{d}{dx} \left( \frac{x^3}{3} \right)}{ \left( \frac{d}{dx} 2 \left( \frac{d}{dx} \left( \frac{x^2}{ \frac{d}{dx} 2 \left( \frac{d}{dx} \left( \frac{ \frac{d}{dx} \left( \frac{x^{ \int_{-2}^{1} dt}}{ \int_{-2}^{1} dt} \right) }{\int_{-1}^{1} dt} \right) \right) \right) \right)} \right)} \right) \right) \right) dx \right) - \left( \left( \frac{\sin^2(p)}{1 - \sin^2(p)} \left( \cos^2(p) \left( \frac{\left( \frac{\sin^2(p)}{\tan^2(p)} \right) \left( \frac{\sin^2(p)}{\cos^2(p)} \right)}{1 - \sin^2(p)} \right) + \left( \frac{\sin^2(p)}{\frac{ \left( \cos^2(p) \tan^2(p) \right)}{\cos^2(p)}} \right) \right) \right) \left( \left( \frac{d}{dx} x \right) - \sin^2(p) \right) \right) \right) \right) } \right) \frac{\sin^2(p)}{1 - \sin^2(p)} \right) \right) \frac{d}{dx} \left( x \left( \frac{ \left( \int_{-1}^{1} dx \right)}{ \frac{d}{dx} 2 \left( \frac{d}{dx} \left( \frac{ \frac{d}{dx} \left( \frac{x^3}{3} \right)}{2} \right) \right) } \right) \right) \right) \left( \left( \frac{d}{dx} x \right) - \sin^2(p) \right) \left( \frac{\sin^2(p)}{1 - \sin^2(p)} \right) + \left( \frac{d}{dx} x \right) - \sin^2(p)

[strange solar]

make trig functions and stuff

[foggy galleon]

(trust me i didnt study triginometry)

[strange solar]

oh

$$
\left( \frac{\sin^2(p)}{1 - \sin^2(p)} \right)^{ - \left( \int_0^1 \frac{d}{dx} x , dx \right)} \left( \frac{ \left( \frac{d}{dx}(x) - \cos^2(p) \tan^2(p) \right) \left( \frac{\cos^2(p) \left( \frac{\sin^2(p)}{1 - \sin^2(p)} \right)}{\cos^2(p)} \right) }{ \frac{ \left( \frac{d}{dx} x \right) - \frac{1 - \cos^2(p)}{\frac{1 - \cos^2(p)}{ \left( \int_{\frac{d}{df} 0}^{\frac{d}{dt} t} \left( 2 \left( \frac{d}{dx} \left( \frac{ \frac{d}{dx} \left( \frac{x^3}{3} \right)}{ \left( \frac{d}{dx} 2 \left( \frac{d}{dx} \left( \frac{x^2}{ \frac{d}{dx} 2 \left( \frac{d}{dx} \left( \frac{ \frac{d}{dx} \left( \frac{x^{ \int_{-2}^{1} dt}}{ \int_{-2}^{1} dt} \right) }{\int_{-1}^{1} dt} \right) \right) \right) \right)} \right)} \right) \right) \right) dx \right) - \left( \left( \frac{\sin^2(p)}{1 - \sin^2(p)} \left( \cos^2(p) \left( \frac{\left( \frac{\sin^2(p)}{\tan^2(p)} \right) \left( \frac{\sin^2(p)}{\cos^2(p)} \right)}{1 - \sin^2(p)} \right) + \left( \frac{\sin^2(p)}{\frac{ \left( \cos^2(p) \tan^2(p) \right)}{\cos^2(p)}} \right) \right) \right) \left( \left( \frac{d}{dx} x \right) - \sin^2(p) \right) \right) \right) \right) } \right) \frac{\sin^2(p)}{1 - \sin^2(p)} \right) \right) \frac{d}{dx} \left( x \left( \frac{ \left( \int_{-1}^{1} dx \right)}{ \frac{d}{dx} 2 \left( \frac{d}{dx} \left( \frac{ \frac{d}{dx} \left( \frac{x^3}{3} \right)}{2} \right) \right) } \right) \right) \right) \left( \left( \frac{d}{dx} x \right) - \sin^2(p) \right) \left( \frac{\sin^2(p)}{1 - \sin^2(p)} \right) + \left( \frac{d}{dx} x \right) - \sin^2(p)
$$

[somber coyote]

integrals don't suck

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[strange solar]

watch

bruh

[foggy galleon]

HAHAHAH LOL
stupid bot load its not hard

[strange solar]

\documentclass{article}
\usepackage{amsmath}

\begin{document}

[
\left( \frac{\sin^2(p)}{1 - \sin^2(p)} \right)^{- \left( \int_0^1 \frac{d}{dx} x , dx \right)}
\left( \frac{ \left( \frac{d}{dx}(x) - \cos^2(p) \tan^2(p) \right)
\left( \frac{\cos^2(p) \left( \frac{\sin^2(p)}{1 - \sin^2(p)} \right)}{\cos^2(p)} \right) }
{\frac{ \left( \frac{d}{dx} x \right) - \frac{1 - \cos^2(p)}{\frac{1 - \cos^2(p)}{
\left( \int_{\frac{d}{df} 0}^{\frac{d}{dt} t} \left( 2 \left( \frac{d}{dx} \left(
\frac{ \frac{d}{dx} \left( \frac{x^3}{3} \right)}{ \left( \frac{d}{dx} 2 \left(
\frac{d}{dx} \left( \frac{x^2}{ \frac{d}{dx} 2 \left( \frac{d}{dx} \left(
\frac{ \frac{d}{dx} \left( \frac{x^{ \int_{-2}^{1} dt}}{ \int_{-2}^{1} dt} \right) }{
\int_{-1}^{1} dt} \right) \right) \right) \right)} \right)} \right) \right) \right) dx
\right) - \left( \left( \frac{\sin^2(p)}{1 - \sin^2(p)} \left( \cos^2(p) \left(
\frac{\left( \frac{\sin^2(p)}{\tan^2(p)} \right) \left( \frac{\sin^2(p)}{\cos^2(p)}
\right)}{1 - \sin^2(p)} \right) + \left( \frac{\sin^2(p)}{\frac{ \left( \cos^2(p)
\tan^2(p) \right)}{\cos^2(p)}} \right) \right) \right) \left( \left( \frac{d}{dx} x \right) - \sin^2(p) \right) \right) \right) \right) }
\right) \frac{\sin^2(p)}{1 - \sin^2(p)} \right)
\frac{d}{dx} \left( x \left( \frac{ \left( \int_{-1}^{1} dx \right)}{
\frac{d}{dx} 2 \left( \frac{d}{dx} \left( \frac{ \frac{d}{dx} \left( \frac{x^3}{3}
\right)}{2} \right) \right) } \right) \right) \right)
\left( \left( \frac{d}{dx} x \right) - \sin^2(p) \right)
\left( \frac{\sin^2(p)}{1 - \sin^2(p)} \right) + \left( \frac{d}{dx} x \right) - \sin^2(p)
]

\end{document}

[somber coyote]

integrals don't suck

LaTeX source sent via direct message.
```Compilation error:```! LaTeX Error: Can be used only in preamble.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
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l.49 \documentclass
                   {article}
Your command was ignored.
Type  I <command> <return>  to replace it with another command,
or  <return>  to continue without it.```

[foggy galleon]

lol

bot gave up

[strange solar]

ik

[foggy galleon]

imagine the bot tryna render the torture equation

[last parrot]

in set notation, is there a way to say A and B is mutually exclusive>

i js dont feel like writing allat

i want sumthing short like the A union B

[somber coyote]

integrals don't suck

LaTeX source sent via direct message.

[strange solar]

@foggy galleon it didn't even do the whole thing

[foggy galleon]

ik bc it was so long discord wanted me to buy nitro to continue

[strange solar]

ye the bot also didnt're nder the second part

[foggy galleon]

oh btw here is some of the work: https://www.desmos.com/calculator/wzmxkruim2

[foggy galleon]

ik

[strange solar]

i also provided some of it in latex

[foggy galleon]

oh nice

the math equation on 33% zoom

@strange solar

discord automaticly compressed it into a file XD

[dark sparrow]

what is this

[foggy galleon]

the equation on top

[dark sparrow]

i mean what's the whole point of this burj khalifa-sized monstrosity of a formula

are you in a "who can make the most complicated formula in existence" competition

[foggy galleon]

no im here so if any of you kids want mcdonalds, give them that and tell them to solve it for the mcdonal

[dark sparrow]

and this, according to you, is peak comedy?

[foggy galleon]

yes

[dark sparrow]

how old are you again?

[foggy galleon]

14

[dark sparrow]

oh. expected, i guess...

not really in a position to tell others on here "you kids" then, are you, kid?

[foggy galleon]

im 1 nanoseconds old (nah jk)

[livid plaza]

$\left( \frac{\sin^2(p)}{1 - \sin^2(p)} \right)^{ - \left( \int0^1 \frac{d}{dx} x , dx \right)} \left( \frac{ \left( \frac{d}{dx}(x) - \cos^2(p) \tan^2(p) \right) \left( \frac{\cos^2(p) \left( \frac{\sin^2(p)}{1 - \sin^2(p)} \right)}{\cos^2(p)} \right) }{ \frac{ \left( \frac{d}{dx} x \right) - \frac{1 - \cos^2(p)}{\frac{1 - \cos^2(p)}{ \left( \int{\frac{d}{df} 0}^{\frac{d}{dt} t} \left( 2 \left( \frac{d}{dx} \left( \frac{ \frac{d}{dx} \left( \frac{x^3}{3} \right)}{ \left( \frac{d}{dx} 2 \left( \frac{d}{dx} \left( \frac{x^2}{ \frac{d}{dx} 2 \left( \frac{d}{dx} \left( \frac{ \frac{d}{dx} \left( \frac{x^{ \int{-2}^{1} dt}}{ \int{-2}^{1} dt} \right) }{\int{-1}^{1} dt} \right) \right) \right) \right)} \right)} \right) \right) \right) dx \right) - \left( \left( \frac{\sin^2(p)}{1 - \sin^2(p)} \left( \cos^2(p) \left( \frac{\left( \frac{\sin^2(p)}{\tan^2(p)} \right) \left( \frac{\sin^2(p)}{\cos^2(p)} \right)}{1 - \sin^2(p)} \right) + \left( \frac{\sin^2(p)}{\frac{ \left( \cos^2(p) \tan^2(p) \right)}{\cos^2(p)}} \right) \right) \right) \left( \left( \frac{d}{dx} x \right) - \sin^2(p) \right) \right) \right) \right) } \right) \frac{\sin^2(p)}{1 - \sin^2(p)} \right) \right) \frac{d}{dx} \left( x \left( \frac{ \left( \int{-1}^{1} dx \right)}{ \frac{d}{dx} 2 \left( \frac{d}{dx} \left( \frac{ \frac{d}{dx} \left( \frac{x^3}{3} \right)}{2} \right) \right) } \right) \right) \right) \left( \left( \frac{d}{dx} x \right) - \sin^2(p) \right) \left( \frac{\sin^2(p)}{1 - \sin^2(p)} \right) + \left( \frac{d}{dx} x \right) - \sin^2(p)$

[somber coyote]

RedCode

LaTeX source sent via direct message.
```Compilation error:```! File ended while scanning use of \frac .
<inserted text> 
                \par 
<*> 1084689263817396244.tex
                           
I suspect you have forgotten a `}', causing me
to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.```

[livid plaza]

anyways, thanks for another masterpiece @strange solar

[harsh viper]

what IS this? wheretf can i learn about this

[brave gulch]

How do I prove this like the problem for me is that I don't see any like powers or anything so I have no idea where to start

[silent plank]

many of these trig identity proofs involve stuff like
difference of two squares / pythagorean trig identities
here consider multiply by (1+sin(theta))/(1+sin(theta))

[foggy galleon]

in my brain

jk i made it with the help of @strange solar

[solid needle]

its meaningless nonsensical garbage, there is nothing to learn

dont waste your time

[foggy galleon]

yes, by the mile, i only made this for my fun

thx for telling him

[plain dirge]

I hate radicals😭🙏

[foggy galleon]

which one better, degrees or radians

[foggy galleon]

yo yk that equation you gave me where no matter the input of "p" its always 1

i made the same thing but different

the simplest one yet

[uncut crest]

Can anyone help me with this?

[restive jungle]

4.5 no? for the chord length

[dense carbon]

Apothem is distance from midpoint to midpoint of one of the sides. This thus splits the pentagon into a 5 congruent kites which can be split again into a total of 10 congruent right triangles with a base of 5.
I’m assuming you can use trigonometry so
The angle of angle adjacent to the base is 360/10 because 10 triangles
Depending on whether you are using degrees or radians I will just assume you are using degrees and if you aren’t multiply the degrees by pi/180 to get radians
sin(360/10) =something idk
5/(cos(360/10)) =hypotenuse
(5sin(360/10))/(cos(360/10))=opposite side aka half of side length
(25sin(360/10))/(2cos(360/10)) =area of one of the right triangle slice
25/2 x tan(360/10) simplification
25tan(360/10)=full triangle slice
125tan(360/10) full pentagon area

I could be wrong

But just to make sure

I’ll do some tests

Let me just try again

Something seems wrong

cos(180/5)=base

cah
cos=a/h
a=5
cos=5/h
1/cos=h/5
5/cos=h
5/cos(180/5) x sin(180/5) x 5 x 5
125/cos(180/5) x sin(180/5)
125tan(180/5)

Yeha it should work

Why do I feel like it’s now

Not

I guess it’s correct

Since the work is correct

[strange solar]

hm

[strange solar]

p can be any value for the trig one

except 0, or 90 or -90

[foggy galleon]

alr

ima test mine brb

mine can be any input besides 0

https://www.desmos.com/calculator/xuzldlacgq

[obsidian harness]

,calc sqrt(3.9 * 1.3)

[somber coyote]

Result:

2.2516660498395

[obsidian harness]

yes that's correct

you need to set $\lambda (a + b)$ equal to $b + mu (-b + a)$

[somber coyote]

south

[obsidian harness]

from which you should find lambda = mu = 1/2

[dense hollow]

any1 help

[granite marsh]

@dense hollow

bro

send the translation of question

some parts are written in the language which I can't understand, it's maybe chinese or japanese or some other language

?

@dense hollow

[obsidian harness]

just pay attention to the English part

the Chinese part says exactly the same thing

probably just angle-chase till you find all the angles in a triangle I bet

exterior angle theorem is a nice shortcut

[granite marsh]

I think I have done this question

wait lemme again do

[dense hollow]

i have no idea, but the topic is isosceles triangles

[granite marsh]

is it 180/7?

@dense hollow

@dense hollow

@dense hollow

@dense hollow

found it

?

[granite marsh]

180/7 degree

[dense hollow]

lemme try

@granite marsh yes, but how u get it

[obsidian harness]

lambda and mu are just variables

okay so I've let the origin be in the bottom left corner

the red line is lambda(a + b)

the black line is b + mu(-b + a)

you need to travel vector b for the start of the black line, then the direction vector of the black line is -b + a

you're missing the fact that you can describe any point on a line with the direction vector of the line * scalar multiple

two vectors are parallel if they are scalar multiples of one another

say, 3i + 5j and 6i + 10j are parallel

okay then maybe it was just the variable names $\lambda, \mu$?

[somber coyote]

south

[obsidian harness]

those are real numbers

compare coeffs of a, b yep

[somber coyote]

Vanellope von Schmugz

[obsidian harness]

yeah I see okay okay

cool okay I think I don't need to explain anything more

that black vector doesn't start at the origin

you need to do position vector + direction vector

it's like y = 2x versus y = 2x + 3 or something

otherwise you're stuck with the vector through the origin all the time, see

[granite marsh]

DM me

can you teach me vector

@strange narwhal

[foggy galleon]

kids will remember what will math do to them when they're older

[exotic sun]

What is trigonometry

[pastel stump]

Guys you know which topic is this? (I know it´s analytic geometry, but idk which topic specifically) It´s the only excercise I can´t solve about an.geom.

That´s from J. Stewart btw

[exotic sun]

That looks like algebra

[pastel stump]

Nope, mate

That excercise is in the analytic geometry part

[normal plover]

https://youtu.be/LTtNzt7ZDxA?si=384XG_OAWoFYQq_U

[uncut crest]

Thanks

[dense carbon]

Np

[fresh edge]

can anyone help pls 😥

<@&286206848099549185>

[pastel stump]

<@&286206848099549185>

[kind lake]

Hellooo

I need helpp

[woven bronze]

Hello, does anybody know what will happen on a Trig Function Graph if the period is negitave?

[solid needle]

the period cant be negative

so if you have something in mind youre gonna have to ask more descriptively

[graceful talon]

the period as in the coefficient of x?

,,\cos(-x)=\cos(x)\\sin(-x)=-\sin(x)

[somber coyote]

0_א

[graceful talon]

sine flips

[rotund snow]

what grade are yall in just wondering

[graceful talon]

,w plot y=sin(-5x)

[graceful talon]

beyond grades

[somber coyote]

Results provided by WolframAlpha

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

Requested by lkjsfkjsiu

[rotund snow]

idk what am I doing here in grade 9th

[graceful talon]

to learn?

you just joined. welcome

[rotund snow]

ty ty

I was just doing matrix its lowkey fun tho kinda hard for me

[graceful talon]

do u see the other channels?

u can take a look at #linear-algebra

[lone solstice]

how would i type this into the calc

[median iron]

33/tan 19

[rotund snow]

yeah

[graceful talon]

make sure you're in degree mode to have the same results

[lone solstice]

k thx

[obsidian harness]

,calc 33/(19 * pi/180)

[somber coyote]

Result:

99.513722312196

[obsidian harness]

tan x = x confirmed /j

[ocean galleon]

This is actually pretty easy

[lone solstice]

so dtm

[ocean galleon]

want a tut

[sage radish]

is ts true?

[dark sparrow]

well alpha = 90° and sin(90°) = 1, so...

[sage radish]

yeah but i mean is it true for all angles

[dark sparrow]

like if you have an angle inscribed into a unit circle like this and are looking at the chord that subtends it?

[sage radish]

yeah

[dark sparrow]

well

ok the central angle will be 2alpha

and. yeah

it will be true still

[sage radish]

okay thanks

[half pike]

Anyone wanna help? I have nothing ive been trying to do it for 2 whole days

[tame oriole]

Yo anybody knows how to solve this : P

[royal fulcrum]

Are you sure you have to prove it?

Bcuz if you put x=45° it doesn't satisfy the condition

[tame oriole]

Uuuuuh

Not prove it

I need to solve it i think

[royal fulcrum]

Yh that would make sense

[upper echo]

You can notice that MBA1 and CBB1 are similar. So you express BB1 in terms of a which solves it all.

[brave elm]

Try converting in the terms of tanx

[burnt lagoon]

please help

[uncut crest]

How do I find the measure of arc DE? The given part is the arc length of DE and the radius

[royal fulcrum]

You can use the formula
Arc length= 2pir (angle DQE/360)

[uncut crest]

That’s what I did and it turns out I typed it in the calculator wrong

And I got 490 degrees or something

Got it now though

Thanks

[sage radish]

490 kinda sounds weird to me lol

[trail tendon]

it should sound weird cuz the arc doesn't even complete a 360 degree revolution xD

[dense carbon]

Maybe it completes 360 and then another 130 for the 490 degrees

[dense carbon]

I have no idea how to solve it so here is my blind attempt
10 inch radius
uhhhhhh
Degrees to radians is degree * pi/180
so radians to degree is radians * 180/pi
Makes sense
90*pi/180=pi/2
pi/2 * 180/pi =90

8.73 * (180/3.1415926535) =500.192

Oh dear

That isn’t right at all

Nope still the same

My autocorrect calculator works

Oh wait no I have to divide it by the radius

Taht why

I forgot teh radius existed

So

8.73/10 * 180/3.1415926535897932384 =50.0192

See makes much more sense

[dense carbon]

Arc = pid (diameter) and then multiply by angle
So divide it by pid
Arc/pid = angle

8.73/pi * 20 =55.577

But not since it’s over 360

So multiply it by 360???

Arc = pid(ang/360)
arc/pid = ang/360
360arc/pid = ang

(360 * 8.73)/pi * 20 =20,007.686

You know I feel like that’s slight wrong

Nevermind everything I said was wrong ignore it

[obsidian harness]

hints for you on how to solve this yourself:

1. what must the circumference be, for the full 360 degrees?
2. great, so how many degrees would 1 in. of the circumference be, dividing both sides (in and degrees) by the same thing
3. now multiply both sides by 8.73 and you're done!

search up 'unitary method' on YT if you don't understand

[granite marsh]

I didn't understand the question, like we have to find the measure of arc DE which is aldready given as 8.73 in.

[obsidian harness]

no the measure of the arc is the angle

so that's angle DQE

[granite marsh]

oh

oh yeah

sry

I forgot

[spark yacht]

radius times angle

[lapis moon]

!showwork

[lime crown]

Show your work, and if possible, explain where you are stuck.

[dense carbon]

Uhhhh no

You need to find the fraction of the angle over 360 then multiply by radius

Probably

Ah well I’ll probably forget about this later

[spark yacht]

length of an arc right?

[hoary prism]

they’re asking for the angle DQE

[earnest cave]

Hey could someone just give me a hint on the C task?

(For some reason it changed T to train on the translated one)

[dense carbon]

Yes which means you can probably reverse it for angle

[spark yacht]

yea thats what im saying

[obsidian harness]

given you know what $M_2 T$ is from question b

[somber coyote]

south

[obsidian harness]

you just need $PU = PM_2 + M_2U = \left(a + \frac{b}{2} \right)+ 2 \cdot M_2 T$

then to show that two vector lines are parallel, you need to show that they are scalar multiples of each other

e.g 2a + 6b = 2(a + 3b), so 2a + 6b and a + 3b are parallel vectors

the scalar multiple is 2

[somber coyote]

south

[stable kettle]

A completely geometric equation for the area between two circles. No calculus or anything fancy, just pure algebra, geometry, and trigonometry. Check it out! https://www.desmos.com/calculator/aejqywu3l4

[undone jay]

Let ABC be a triangle with ∠BAC = 70°, ∠ABC = 50°, and M be the midpoint of side AC. Let P and Q be points on lines BC and BA, respectively, such that the circle passing through P, M, B is tangent to BA and the circle passing through Q, M, B is tangent to BC. What is the measure of the angle ∠PQB?

(A) 60°
(B) 70°
(C) 80°
(D) 90°
(E) 100°

[stable kettle]

I think I got it.

In △ABC we are given that ∠A=70, ∠B=50, and ∠C=180-70-50=60

Here is one way to see it:

With M the midpoint of AC, the rays BM, BA and BC form certain angles. Denote ∠AMB = α, ∠CBM = β, with α+β=50°

The circle through B, M, and P is tangent to BA at B. By the tangent–chord theorem, the angle between the tangent and chord BM equals the inscribed angle in the alternate segment: ∠ABM=α=∠BPM.

Similarly, for the circle through B, M, and Q tangent to BC: ∠CBM=β=∠BQM

Coordinate Computation reveals that the position of P and Q are uniquely determined, hence ∠PQB=60.

• Place A = (0,0) and B = (sin60,0)
• With ∠A=70° and using the Law of Sines, C is approximately (0.262, 0.719).
• Then, the midpoint M of AC is approximately (0.131, 0.360).

Next, using the tangency condition (forces the center of the circle through B, M, and P to lie on the line perpendicular to BA at B), the circle meets line BC at a point P approximately equal to (-0.051,1.091).

Similarly, for the circle through B, M, and Q, tangent to BC (whose center lies on the line perpendicular to BC at B), the second intersection (Q) is approximately (-0.680,0).

Finally, computing the angle at Q in △PQB (using the dot–product formula) shows that ∠PQB=arccos(0.5)=60°

Therefore, the answer is A. (This may be wrong)

Also, I learned this in Mechanical Engineering: https://www.desmos.com/calculator/ma0m9iyzat

[undone jay]

Interesting, even though your arguments appear to be right, maybe there is an easier way to do this. I would think of that because this problem is from a quite easy olympiad, for younger people you know, i don't think they would expect anyone to understand law of sines, arccos...

Thank you for helping!

[undone jay]

Yes, I checked the answer here and it is correct, the answer is indeed 60 degrees. Unfortunately this first phase doesn't show any explanations for the solutions

[lavish lotus]

Someone help me

#help-30

[stuck anchor]

∂f/∂x

[obsidian harness]

@autumn heron but basically sine rule gives $\frac{RY}{\sin(180 - 2 \theta)} = \frac{m}{\sin \theta}$

[somber coyote]

south

[obsidian harness]

rearrange this to get RY = ...

and then compare this with what 4.3 tells you to see what's happening

[azure helm]

guys is the angle of (2,1) makes with y axis the same angle that (2,-1) makes with y axis?

like one angle is the reflection of the other wrt x axis

dunno why it is so confusing

[autumn heron]

@obsidian harness Thank you 🙏

[obsidian harness]

no worries!

[ocean spear]

yeah thats confusing

[royal fulcrum]

Yes, they make same angles. You can verify that by finding slopes and equating it to tan(alpha)

[muted sail]

Can someone do this step by step. I’ve spend since this morning at 9 studying trig except this seems difficult to me

[silent plank]

apply pythag

[muted sail]

I’ll try that

[light thorn]

how do i do this question

[muted sail]

What do I do now

[distant monolith]

Expand out all the squared terms and see if anything cancels nicely

@muted sail

[muted sail]

Grand

Is there anything I can do here

[maiden brook]

it should be (x^2 + 8x + 16) = (x^2 - 8x + 16) + (x^2 - 10x + 25)

and also u can/should be able to do each of those expansions in one step since they are simply squares but it’s fine.

do u have any ideas how to continue?

[muted sail]

I’ll change up what you said with this first

[muted sail]

I don’t know how to continue adter

After*

[silent plank]

simplify and solve the quadratic

[muted sail]

What do you mean by that

[silent plank]

which part of the instructions don't you understand

[muted sail]

How do you simplify it

[silent plank]

combine like terms

rearrange to general form

[muted sail]

Can I take away the brackets then

[silent plank]

stuff = 0

here, after squaring, yes

[muted sail]

Squaring?

[silent plank]

what you did

with expanding your (x+4)**^2 **etc

that ^2 is called squaring

[muted sail]

Where do I square it than

[silent plank]

you already did

assuming you've corrected the above errors,
you can safely remove those () and combine / simplify as i've instructed

[muted sail]

Okay I’ll try that

I’ll work with this than

[spiral lodge]

Now bring everything to the left-hand side, making it = 0

And then simplify like terms

[upper karma]

Can someone help me

[upper karma]

<@&286206848099549185>

[dark sparrow]

does heron's formula help at all here

[upper karma]

No clue what that is

😔

[dark sparrow]

formula for the area of a triangle in terms of its sides

[upper karma]

Oh

What is it

[dark sparrow]

https://en.wikipedia.org/wiki/Heron's_formula?wprov=sfla1

[upper karma]

Thanks

[sage radish]

this system seems underdefined🤔

[uncut crest]

Someone help

[stable kettle]

#❓how-to-get-help

[raven wraith]

evaluate each value as it is and then rationalise the denominator, see if that works

[sinful schooner]

good luck soldier

[uncut crest]

😭

[maiden brook]

any ideas?

[covert belfry]

do you have work shown or anything yet?

try starting with the triangles area and then the small portions of the circles inside the triangle

[solemn obsidian]

not sure but maybe finding the area of the whole triangle and then finding the area of the arc and then multiply by 3 and then subtract that from the whole triangle

[crystal oar]

can anyone teach me about coordinate geometry of a circle

[obsidian harness]

Why don't you use your textbook and YT?

[crystal oar]

its not effective for me because i cant ask what i dont know

[crystal oar]

how do i know when do i need to find the midpoint, when do i need to find a gradient , when do i need to do completing the square bro

sanity is going insane bro

[dark sparrow]

i mean, the problem will usually tell you what it wants you to do?

if it wants you to find the midpoint of something then it'll say it wants the midpoint of something

[crystal oar]

i dont even know what the question mean

idk what method to use it to find the coordinate

[dark sparrow]

well send the question here.

we can't tell you what to do or how to do it without seeing what needs to be done.

[crystal oar]

The line x+2y = 10 intersections the circle (x-2)62 + (y+1+^2 = 25 at the points a and b.
a) find the coordinates of the points a and b.
b) the perperndicular bisector of ab intersects the circle at the points p and q. find the exact coordinates of p and q.

i know how to do a but idk what b meanas

[dark sparrow]

probably capital P and Q, but ok.

so, have you seen the word "perpendicular bisector" before?

[crystal oar]

no

[dark sparrow]

(also typo? presumably your circle is (x-2)^2 + (y+1)^2 = 25?)

[dark sparrow]

so really that is what you were missing, rather than anything specific to circles.

[crystal oar]

yes

[dark sparrow]

the perpendicular bisector of a line segment is the straight line that is perpendicular to that line segment and goes thru its midpoint.

does that definition make sense to you?

(try not to take it as an instruction)

[crystal oar]

yes

[dark sparrow]

yeah ok

taking into account the whole circle thing you've got here, there's also a shortcut you can take.

a little trick to save yourself some effort

that the perp bisector of a chord in a circle will pass thru the circle's center. (and so you don't actually have to calculate the midpoint of AB here)

[crystal oar]

so what do i have to do

i dont know what do do about finding p and q

[chrome token]

easier would be two assemble these 4 triangles into a square (there is atleast one set of 4 triangles in which you can do this)

[dark sparrow]

how do you know they assemble into a square in the first place

[chrome token]

there is one set which do assemble into a square

not all sets, but one set does

[dark sparrow]

ok but we can't just assume ex nihilo that these particular triangles do

[chrome token]

true

but if the question has a unique answer, then the answer would be correct even in the case of the triangles forming a square

[quaint pecan]

since the radius of each circle is 6 inches, each side of the triangle is 12 inches, and it must be an equilateral triangle, so you can find out the area of the triangle which is 62.352 inches. Now, since its an equilateral triangle each angle of the triangle is 60 degrees, so it makes a 60 degree angle at the centre of each circle, find out the area of that 60 degree sector in the circles which is 3*(3.14 x 36)/6 (multiply it by 3 to find out the area of all 3 sectors together) and the area would be 56.52 inches. Subtract 56.52 from 62.352 and you will get 5.832 inches as your answer

[tribal nest]

Finally did a Hypercube in Desmos that has a full rotation of a₀ and b₀

[crystal oar]

its like everytime i think i already learnt all the stuff, and then i mvoe on to the question, i blurred

[dark sparrow]

sorry, i didn't see this and got busy.

you find the equation of the diameter of your circle that's perpendicular to line AB.

or the same line by any other name.

and then you find its intersections with the circle by the exact same process as in part a, which you said you had no issue with

[pallid roost]

Hey uh

[obsidian harness]

seems like too little information

[pallid roost]

I’ve got an investigation and that’s the on information we’re allowed

[pallid roost]

If that’s like clearer

[obsidian harness]

I don't understand what you need to do for your investigation

[pallid roost]

Basically we loosely take measurements to find the height of the top of a building to the bottom

And we’re only allowed those measurements

[pallid roost]

Is the information there enough?

[silent plank]

from what i'm seeing no

do you have the original problem statement

[pallid roost]

Here

[silent plank]

did they provide a base diagram for you

[pallid roost]

Nope

[uncut crest]

What would the exact answer be?

[quaint pecan]

5.832 is the closest you can get because you have to take the approximate values of Pi and Root 3 which are both irrational numbers

[final birch]

what should I do if I encounter a integral of 1/x³+1

[exotic yarrow]

Factor the denominator using sum of cubes, then partial fractions.

[final birch]

then trig integration?

[exotic yarrow]

Yeah for the $\int \frac{1}{x^2-x+1} \dd{x}$ part

[somber coyote]

Civil Service Pigeon

[final birch]

I see the (x+1) got cancelled

[exotic yarrow]

What?

No I just didn’t talk about that part because it’s not relevant to trig

[final birch]

Oh

👌

[maiden brook]

u shouldn’t give full solutions

[quaint pecan]

oh should i not ? im new out here

i just explained the whole thing to him since no one helped him

[empty yew]

!nosols

[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.

[quaint pecan]

alr thanks for letting me know

[dull socket]

has this been resolved?

i could help!

[severe sandal]

Hi

Is this correct

I can’t do b..

[dark sparrow]

,rccw

[dark sparrow]

@severe sandal you know what a plane of symmetry looks like, right?

[severe sandal]

Kinda

Why isn’t 2b 3

If I cut it diagonal

It kinda works

[dark sparrow]

hmm

what do you mean by "cut diagonal"

actually i don't see where you are getting 3 planes at all

[severe sandal]

OH

NVM

I GET IT

[dark sparrow]

i can only see one: the one parallel to the triangular bases and halfway between them

[severe sandal]

I MEANT LIKE cutting it in the way of an x but I looked closely and it doesn’t work

But only one of the lines from the x

On the rectangle part

Is this correct

The cuboid isn’t right

[dark sparrow]

the cuboid is indeed not right

[severe sandal]

But it u turn it after it could be

[dark sparrow]

what

[severe sandal]

Is the one on top correct

[civic drum]

whats the hypotenus

<@&286206848099549185>

[covert belfry]

hypotenus' are only in right triangles

this is an isoceles triangle

[civic drum]

how would i apply pythagoras theorem

[covert belfry]

you can draw a supplementary line splitting the triangle

and making a perpendicular line from the top to the middle of the line segement below

[civic drum]

ohh

[covert belfry]

i'm not sure if that's clear sorry

but yeah if you have the height of the triangle and the base value you can find the hypotenuse of the split up triangles

[civic drum]

i think i understand it

[covert belfry]

yup yup yup

[civic drum]

thank you

[covert belfry]

your welcome!

[woeful shell]

circle

!!!!!!!

[undone jay]

Let ABC be a right isosceles triangle with ∠CAB = 90°. Let P be a point on line BC, with B between C and P, and Q a point on line AB, with A between B and Q, such that BP = AQ. Let R be a point on line AC, with C between A and R, such that ∠PQR = 45°. If the measure of angle ∠PRQ is m/n degrees, where m and n are positive integers that are coprime, determine the value of m + n.

[dull socket]

the cosine rule is a great way to start in this one!

[steady wyvern]

Hey guys

[steady wyvern]

Why are the bisectors of two intersecting lines perpendicular

[steady wyvern]

?

[graceful talon]

,w define bisector

[somber coyote]

Results provided by WolframAlpha

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

Requested by lkjsfkjsiu

[narrow pasture]

Hii

[twin robin]

i need help guys

[dark sparrow]

do you still need help with this?

[twin robin]

yeaa

[dark sparrow]

have you made a diagram?

[twin robin]

Yes, I drew everything. I actually got a lot of progress in the question, but after finding only the quadrilateral of the beams, I was frankly confused

[dark sparrow]

can you show your progress thus far

along with the diagram of course

[twin robin]

one minute

I know it's quite complicated but maybe someone can figure it out

[dark sparrow]

oh god this is worse than i thought

[twin robin]

its okay

[twin robin]

Because I'm working on Olympiad questions

[dark sparrow]

oh that explains it 💀'

[zealous bloom]

Hello, could someone help me calculate area?

The radius is 4 cm

The coloring might be a bit off

But I only need approx

I need the area of the colored area

The area of one circle is 16Pi

[normal sierra]

where tf did a circle come from

[simple sphinx]

I kind of need help with this.

[devout shell]

step 1: draw what we know
did you at least draw the situation of point A yet?

[sinful schooner]

im sorry but wth is that T-T

[devout shell]

@simple sphinx

[timber copper]

hi, I need some help with some math questions. They are in Spanish, I hope it won't be a problem ❤️

[obsidian hornet]

got 46, but I don’t think this is the right answer, due to M should be the incenter of ABC and i don’t believe in this

[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.

[hard hornet]

turkish spotted

[patent knoll]

Does anyone know a website with COMPLEX AND HARD questions for algebra and trig for year 10 maths and above (not university level)

[obsidian harness]

https://madasmaths.com/

if you really want hard however, search up JEE questionbanks

then you'll suffer

[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

[sudden solstice]

Can someone help and explain this to me?

This is a new unit I am doing in math and I don’t understand it

[sudden solstice]

Please help

Me

[willow tundra]

isnt ab here equal to dc?

[burnt lagoon]

please help

[upper echo]

use the fact that 4cos²x-1=sin(3x)/sin(x) and 243*3 mod 360 = 9.

[sudden solstice]

Yes

[dense carbon]

I accidentally made a formula on Desmos to calculate the location of the nth point of any root of unity I think

Except this is not in a unit circle

You need to modify iy slightly to get anything useful

[kind oyster]

should be

[dense carbon]

Yeah probably unless it’s not to scale

[hollow saffron]

can anyone help me lern trignometry

teach*

teach me

[obsidian harness]

bruh, Khan Academy and org chem tutor on YT

[fierce cipher]

yall how do thetas work? i learned about them in class like 5 months ago but didnt pay attention

[trail tendon]

theta is a variable, usually representing an angle

[fierce cipher]

Thanks

[uncut crest]

Help

[dark sparrow]

,rccw

[dark sparrow]

@uncut crest do you still need help with this?

[jovial linden]

This is in my daughter's A Levels book, it seems to me they're using the equality $\sqrt{cos^2{2\theta}}=cos{2\theta}$ (which is incorrect) to get to $\frac{sin{2\theta}}{cos{2\theta}}$.

[somber coyote]

ian

[jovial linden]

And the simplification they give is clearly wrong:

It isn't even the first place the book does this, is the book wrong, or is there some strange difference in notation in the UK? Or am I just missing something daft?

[obsidian harness]

A levels doesn't cover nuances like this

should of course be $\frac{\sin 2 \theta}{| \cos 2 \theta |}$

[somber coyote]

south

[jovial linden]

Ahh, thanks, that would explain it. Seems like they are setting kids up for pain in university though! 😄

[obsidian harness]

yeah the textbook authors should have just put a $0 \le \theta \le \pi/2$ in there

[somber coyote]

south

[obsidian harness]

this laziness only reinforces misconceptions, I agree

[uncut crest]

yeah

[ionic zodiac]

Dawg yk how A level ppl prove a 1 to one function

Legit just horizontal line test

If it only intersects once it is one-to-one💀

[obsidian harness]

fair enough

A level isn't supposed to be rigorous anyways

[obsidian harness]

so they wouldn't actually need to prove it

[ionic zodiac]

I thought it was for like pre university though

[obsidian harness]

they'd just need to observe

[obsidian harness]

pre-university isn't supposed to be rigorous in general

JEE isn't rigorous, it's just spamming a bunch of concepts to crack the question, and so is JEE adv

[ionic zodiac]

U take jee?

[obsidian harness]

no

[ionic zodiac]

Or

Ah

[obsidian harness]

and I'm not Indian

[ionic zodiac]

If you ask an a level student here in Singapore

[obsidian harness]

like in JEE you just observe right?

[obsidian harness]

well very very few people take H3 maths in Singapore

[graceful talon]

spamming is funny

[ionic zodiac]

They probably wouldn’t know what the limit definition is

H2

H3 ppl are ok

[obsidian harness]

yeah that is true, that does tend to get taught more in US calculus classes

[ionic zodiac]

Yeah

I suppose they want us to focus on computation

Rather than intuition

[obsidian harness]

well they want you to focus on computation and intuition

rather than formally justifying everything with respect to some given axioms, which is where rigour comes in

[ionic zodiac]

Sad

But reasonable ig

[real pebble]

please help me

SSOMEONE

PLEASE

my grade depends on it and its already not lookin to good

halp

please

i need halp

😭

[signal vapor]

30 60 45 90 angles have specific values for trigonometric functions

Example sin 30 = 1/2 same goes for the others

[graceful talon]

you're the best to know that

[dark sparrow]

sorry i was asleep. but the idea is you should make a better diagram and put some extra stuff on it like this

[dark sparrow]

A-levels does, this looks like (I)GCSE stuff.

oh no huh it is a-levels

[obsidian harness]

Finding BQ and QC rely on basic right angled trig: you should apply the trig ratio definition, sub in, and solve

The key thing is that BQ = AP

So apply right angled trig again to find DP, where you now know the height and both bases

[lost lotus]

if sin=opp/hyp then what is arcsin

[lapis moon]

'arcsin' is the inverse function of 'sin'

[dark sparrow]

arcsin is the function that answers the question of "what angle gives a sine ratio equal to this number?"

[lost lotus]

oh

ok thx

also how to do hyperbolic secant and cosecant?

[dark sparrow]

define "do"?

[lost lotus]

solve

[dark sparrow]

do you mean calculate

[lost lotus]

ys

yes sry i got my math vocab wrong

[dark sparrow]

sech and csch are defined as 1/cosh(x) and 1/sinh(x)

[lost lotus]

so just same concept?

[dark sparrow]

by analogy with their circular equivalents

[lost lotus]

also how to convert the law of cosines (the one like a^2+b^2-abcos(c) where a and b are sides of a triangle and c is the angle between them) into hyperbolic geo

[dark sparrow]

uhh

[lost lotus]

ive been stuck on that since grade 5

[dark sparrow]

what grade are you now

[lost lotus]

9

[dark sparrow]

why would you even be studying hyperbolic geometry in school

[lost lotus]

no im studying it in my freetime

[dark sparrow]

iirc the hyperbolic law of cosines looks like

cosh(c) = cosh(a)cosh(b) - sinh(a)sinh(b)cos(C)

but don't quote me on that

[lost lotus]

oh ok thx

[dark sparrow]

it is quite strange you would be studying even the Euclidean cosine law back in grade 5

sounds way too early

[lost lotus]

no i did it back in my free time

[dark sparrow]

unless you are like insanely ahead of the curve

[lost lotus]

when i was grade 5 i was a gifted kid

[dark sparrow]

ok so then you are

[lost lotus]

bruh i tried to learn tensors in grade 6 i thought it was just grade 7

was i really that overkill?

[dark sparrow]

i would say kinda yes

[lost lotus]

bruh

[dark sparrow]

tensors though...