geometry-and-trigonometry

[grave pond]

Yes.

[upper karma]

But I just am not getting why I was given sides ratio then?

[grave pond]

That's just a red herring.

[civic rock]

sometimes the problem can give extraneous info to mess you up

[upper karma]

Ahh yeah okay

[civic rock]

the segment ratios dont interfere with the angles in this case

[upper karma]

Just one more thing, If I know that those angles are equal, I can say that they are parallel?

[civic rock]

yes

[upper karma]

Okay

Okay thank you so much

[civic rock]

np

[upper karma]

Here I know that Ab is parallel to cd and pq. They're parallel to each other, now how do I take the ratio using similarity so as to find cd :pq?

[dark sparrow]

24?

[civic rock]

use intercept theorem in both parallel pairs and see what you get

[upper karma]

Yeah

[upper karma]

Alright

[civic rock]

if we have AB/CD, where should you apply it first

[dark sparrow]

yeah ok i was gonna say the same thing

AB : PQ and CD : PQ are both things you can get a handle on

[upper karma]

Okk

[upper karma]

Umm I'm thinking

[civic rock]

you have AB//CD

you got 2 choices: P and Q

[upper karma]

Umm I really don't know what to do here

[civic rock]

so ignore the point Q for now

[upper karma]

Okay

[civic rock]

you have AB//CD and you have AD intersect BC at P

what does that tell you

[upper karma]

Ad is transversal?

[civic rock]

so is BC

[upper karma]

Yes but how does that help?

[civic rock]

use the intercept theorem

what does that tell you

to give you more hint, the ratio PB/PC?

[upper karma]

It's 3:1

[civic rock]

yep

now we need to find CD/PQ

[upper karma]

So 3/4?

[civic rock]

CD is clearly longer than PQ

[upper karma]

4/3?

[civic rock]

why is it 4/3 then

[upper karma]

Bcoz cb is 4x and pb 3x

Then using similarity

[civic rock]

alr that's enough to tell me you solved the problem correctly

[upper karma]

Great

Thankss

[upper karma]

How do I use pq +qr= ab to solve this?

[upper echo]

Continue PQ to a point D such that QD=QR. Then BPDA is a rectangle and AC is a bisector of ∠RAD. So, ∠BCA=∠CAD=(90°-20°)/2=35°.

[upper karma]

How do we know that bp = da

Da must be perpendicular so how do we prove that

[upper echo]

BA | | PD and BA=PD, so, BPDA is a parallelogram with one angle 90°. So it is a rectangle

[upper karma]

Oh okayy

[upper karma]

Can you please also explain why ac is bisector and 90-20/2?

[upper echo]

Because the triangles ARQ and ADQ are equal

[upper karma]

Okay yes through sas

[upper echo]

tbh through SSS

[upper karma]

Why
Angle adq= Angle arq
Side aq=aq common
And rq = dq

[upper echo]

Then it would be ASS 🙂 but it has other options than equality of triangles.

[upper karma]

Oh I forgot that there has to be an order sorry

[upper echo]

It is in fact SSS because it has two sides equal in the right triangles. So, by Pythagoras the third sides are equal as well.

[upper karma]

Okayy. I find geometry interesting now. Thanks

[sly hazel]

Hi everyone

Can you give some tips to study

The geometry in the space

[dark sparrow]

#study-discussion

[hexed wraith]

Yes break it down what you need more specific to know

[empty yew]

Can anyone figure my level of math with a few questions?

[hexed wraith]

What chapter are you at right now

[maiden brook]

why?

on what topic?

[hexed wraith]

How much are you typing fam

[empty yew]

I am at a whole different chapters simultaneously, like limits & continuity(in one front), complex numbers(on other front) and lastly matrices(methods of solving linear equations which I haven't completed in my school in my fair note) lmao

[empty yew]

Just wanna know

[obsidian hornet]

If u mean geometry level, then i can suggest you a few amount of problems

[maiden brook]

but not on geometry?

cause it’s in this channel

[empty yew]

I am not strong enough in geometry as much I am in other fields of my level

[empty yew]

sorry for causing commotion

cause this is the only channel that is active at this time

atleast a moment before now

[maiden brook]

so u want problems on limits/continuity, complex numbers, and matrices only? Ok I don’t rlly know matrices but I can give some for the other two

[empty yew]

Nope i don't want that lol

I just wanna figure my total level of math

anyways it is pretty late here i will be back tmrw

sorry

[maiden brook]

with problems on what?

otherwise it’s too vague

[zealous pike]

The question asks to check if the function is odd or even.

90 cos(sin(x)) is even, right

[exotic yarrow]

You can check this yourself in #bots

,w is cos(sin x) even

[somber coyote]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[zealous pike]

Ohh

Thx

[dull hill]

wheres D

[civic rock]

extend PQ itself

take a point D along the ray PQ so that QR=QD

[dull hill]

So D is created

[civic rock]

yes

[dull hill]

thought i was blind

[zealous pike]

The question asks to see why the reduction formulas sin(π + t) = -sin(t), cos(π + t) = -cos(t) and tan(π + t) = tan(t) are valid.

The answer:

We know that a full revolution on the unit circle equals 2π. Half that revolution, π. It is easy to see that cos(π + 2π) = -1, sin(π + 2π) = 0, tan(π + 2π) = 0.

We can conclude that the sum π + t as input for a trigonometric function outputs the reflection of t through the origin

[strong summit]

Hello! I am learning geometry soon, and i want to be prepared for my classes in the future so I have to ask, what are the main and most valuable things you need to understand about geometry and some key things to not forget?

[dark sparrow]

• definitions of every concept (parallel lines, perpendicular lines, alternate angles, interior/exterior angles for polygons, types of quadrilateral etc.)
• triangle congruence rules
• area formulas, principally those of the triangle and the rectangle, as well as how to (re)derive the ones for more complicated shapes
• circle theorems (regarding inscribed angles and the like)
• trigonometry basics as well as sine law & cosine law

[civic rock]

you should also analyze every segments' property and extract as much info from them as you can (even the most obvious things)

[dark sparrow]

also one big thing is don't be afraid of making additional constructions on diagrams

sometimes dropping a perpendicular somewhere can help a lot

[civic rock]

you also should know in what case you should make additional constructions (drop perpendiculars, creating parallel lines and such)

for simpler problems remember the methodology of proving stuff (eg. to prove perpendicular lines you can use angles, pythagoras theorem or drop perpendiculars from two sides)

[strong summit]

And these are the basics?

[left mauve]

<@&268886789983436800>

[civic rock]

[lunar hull]

Hi, please review the #rules , academic dishonesty is not allowed

[slim violet]

Translated: In the drawing on the right, which is not to scale, it is given that 𝑃𝑇=𝑄𝑇=𝑇𝑆, 𝑄𝑆=𝑆𝑅, and angle ∠𝑃𝑄𝑇=20∘ .What is the value of 𝑥?

A) 20 B) 25 C) 30 D) 35 E) 40
Any idea? I just can't figure out how to tackle this problem, although it's supossedly an easy one

[civic rock]

mark the equal angles

[zealous pike]

How do I solve this one
I have no clue how to approach it

[civic rock]

group sin pi/100 and sin 199pi/100

[viscid nimbus]

probably gcf them

[civic rock]

sin 2pi/100 and sin 198pi/100 and so on

[zealous pike]

oke

[zealous pike]

oh. Every value after sin(pi) seems to be a reflection through the origin (a reflection through both the x and y axis), does that means the final result is 0?

[civic rock]

ye

lol

[obsidian harness]

indeed, common comp maths trick

[zealous pike]

I would cry tears of fear if that question was on my exam

[civic rock]

students when they see 1+2+3+...+100 before 1777:

[zealous pike]

Gauss was the true man of culture

[vague marlin]

um guys can someone help me understand what does this question wants?
im still new to international math

[upper echo]

They want you to find the area of a cross section, which is a right isosceles triangle for each t with legs sqrt(r^2-t^2). So, their areas are (r^2-t^2)/2

[swift phoenix]

if you cannot visualize things, draw a picture. It may help

[vague marlin]

i see

[vague marlin]

thanks for the tips

but wheres the C tho? im a bit confused with the C

or what do i dowith the c

[swift phoenix]

C is a part of the cylinder

in 2.

[upper echo]

sorry for the rough sketch. C is that volume under the slanted plane y=z and above XY plane

[vague marlin]

t is the radius of the cylinder ?

[upper echo]

nope, it's a parameter for the cross section

[vague marlin]

Ohhh

[vague marlin]

in which the max of the t is less than equal to r?

[upper echo]

not the max. Just -r<=t<=r

t is between -r and r

[vague marlin]

oh ok

i think im starting to get it

btw where can i learn this type of question

i want to know more about this type of geomtry

[upper echo]

maybe practicing some problems on 3d geometry

bc it's easy if you can visualize 3d objects

[swift phoenix]

if you want ot get better, use some 3D apps to help you visualize the objects

[swift phoenix]

I could not agree more

[vague marlin]

I see thanks for the tips

and for helpimg me understand it

[proud jacinth]

Hi people

Some one speaks spanish?

This is chemistry

[heavy axle]

im spanish, what do u want me to translate?

@proud jacinth ?

[proud jacinth]

Yes pls

my english is not the best

[heavy axle]

1 - An aquiarium has to keep the salt concentration similat to the sea's, this is 1.8 g of salt dissolved into 50 g of water. ¿What is the percentage in mass of the salt in the solution?

[west berry]

Yoo what apps do u use in visualizing 3ds

Yoo anyone wanna discuss trig and exponential functions have something in mind that's bothering me

[native pier]

geogebra is pretty nice, but it's sometimes slow. you can click the dropdown menu to change it from graphing to 3D calculator

[native pier]

sure, but I might need a refresher

[vague marlin]

is desmos 3d graph good?

[native pier]

I haven't tried honestly, so I'm not the one to answer.

[vague marlin]

ohh ok

[native pier]

@vague marlin I just tried it, and it's better than geogebra 🙂

[native pier]

so yeah, desmos 3D

[vague marlin]

damn

yesterday i tried the translucent mode

but it doesnt work

[native pier]

huh. how do you get that?

[west berry]

Thanks Marianne..

[vague marlin]

on the top right

theres a tools settings

wrench logo

[native pier]

it does work for me

[vague marlin]

whaat

wait

[vague marlin]

OMG IT DOES WORKS

after testing it

[west berry]

Lol the name reminds me of Marianne beaulieu

[vague marlin]

but yesterday doesnt seem to work

this is yesterdays graph with translucent mode

well atleast it works now

[native pier]

-# 🙂 it's also the statue of liberty for France

[west berry]

Ouuh wow learnt something new
Alr

[civic rock]

desmos does all of its calculations client-side while geogebra does them server-side

[native pier]

ooh, that's nice to know

[civic rock]

it's way faster and also doesnt have dumb bugs

[native pier]

-# geogebra has better ui tho just saying, but that's not the important part

[civic rock]

like if I rename too many points the keyboard on geogebra just softlocks

[civic rock]

a bit sucky imo

[deft sail]

Some people told me that you can approximate trig functions to 3dp. But I just can’t find it 😔. So does anyone know how??

[maiden brook]

you can approximate trig functions to any degree of accuracy

do u see a good approximation for cos x around x = 0

[civic rock]

wait how do you prove lim x->0 ((sin x)/x) = 1

[whole marsh]

its 0/0 so you can do l'Hopital

[civic rock]

no you cant

it'll be circle reasoning

[whole marsh]

it works tho

[maiden brook]

why?

[civic rock]

lemme ask you why is (sin x)' = cos x

[whole marsh]

idk that

[maiden brook]

u can prove that using angle addition

[civic rock]

derivative I mean, how does angle addition work here

[maiden brook]

how?

[civic rock]

^

[maiden brook]

oh alright

so

[maiden brook]

at that point its best to prove this through geometry, its a very good exercise. people say l'hopitals cause once u know that it gets u the result immediately if u dont recall it

[civic rock]

Like I see the geometic proof

idk if Taylor series also leads to circular reasoning

[maiden brook]

that works yeah

[civic rock]

the idea of Taylor series also stems from derivatives iirc

lemme refind the geometric proof

[maiden brook]

whatever. u can look up other proofs of sin x / x if this is such an issue

[civic rock]

ye ik

but why should the length of arc BC smaller than BD

oh wait

that makes sense actually

[civic rock]

ye cuz I see one of the proofs rely on this and thought it's not convincing

[tawny rapids]

cuz when you plug in f(x)=sinx into the definition of a derivative you get cosx

[civic rock]

if you use the definition of derivative you have to use this to get cos x

so it's circular reasoning

[flat lily]

Is it me or this channel's system has been altered? I am lost and needed help with a geometry question that should be easy for true mathematicians like the ones we find here.

[obsidian harness]

you can post here

[flat lily]

And nope, it aint an underhanded tactic to goad into getting help... I am a layman... So here is my issue... I have a rectangle 62779214.6497993 micrometers wide (x-axis) and 88783216.7928779 micrometers long (y-axis), it is surrounded by an obround which may have half ovals rather than half circles at each end, the furthest the obround gets from the rectangle’s far side in the x-axis is 9268577.57727846 micrometers away from the midpoint in the rectangle’s x-axis, if I instead measure from the rectangle’s centre in a straight line to the corner I get a measure Z and if I extend that line to the arc that forms the outer limit of the obround the final line would measure “1.2 times Z”, what is the distance in micrometers, orthogonal to the 9268577.57727846 micrometers, from the perimeter of the inner rectangle and the outer obround? Alternatively the outer figure is not an obround, but a figure composed of four quarter ovals taken from a yet indeterminate oval whose orthogonal radii now meet in the corners of the rectangle above and whose corners between arc and radii are linked by lines parallel to the length of the rectangle. I am looking for the shortest distance from either of the longest sides of the rectangle and the closest parallel straight line in the outer figure, I intuit that 20% of half the hypotenuse of the two straight angle triangles into which the triangle can be split is the average radius of a oval whose smalle axis' radius is now the distance I am looking for... But I am at a loss on how to turn this geometry challenge into pure algebra I can solve on excel. I also need the square root (in micrometers) of the area of the larger figure... I am quite confused about how to solve it myself so I can't proceed without asking the mathematicians.

[left mauve]

This is pre-uni, I don't think most people here know what an obround is

[obsidian harness]

A shape consisting of two semicircles connected by parallel lines tangent to their endpoints.

[left mauve]

O

[flat lily]

I had to google for the word obround... I am not even sure I want an obround but I cant find the name of what could be a more fitting shape.

In Uni I study philosophy, not math, so I would be amazed if I stumbled with a question that involves post-degree uni stuff.

[left mauve]

What is your purpose with taking it though, it seems like you're just taking it and measuring distances from the rectangle to it...

!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.

[flat lily]

Well, I calculated an average football field from wikipedia's article on it... And the answer was ambigous due to the end zones... (more than the fact that more than one sport has that name) I was using it for a standard jousting field or a standard coliseum in my worldbuilding, I know this may not seem anything but frivolous to people dedicated to mathematics but I cant stand not being able to solve it.

And please, someone tell the bot that no one stated this problem to me, I came up with it in the process of developing stuff for my sci fantasy worldbuilding.

[civic rock]

[flat lily]

I really suspect its pretty basic level stuff if you really do hard math daily... If its a bother 'cause it aint that basic I gonna understand.

[left mauve]

Do you have a diagram drawn

[flat lily]

yep but not a scanner, I can draw one to show you if you are willing to help me

[left mauve]

Idk if i will be able to, but posting it here would just make it easier for others to as well- anything's better than reading that textwall

[flat lily]

ok

will work on it right now then

The image is a bit wonky

[grave pond]

Long wall of text. Lengths measured to a precision much smaller than the size of a single atom. Definitely troll.

[flat lily]

I am not a troll. What is your problem you are insulting me?

I use precise numbers because I will round up the solutions, but I can't tell how they would cause problem in the process.

Ok, someone that takes this seriously, can anyone help? Axe Cutter? You seemed willing to aid... But I wont ping ya in case you had to leave, I dont want to be rude...

Baricentre, that black circle represents the baricentre or centroid.

never mind, I am on a help channel now,

[flat lily]

@feral cloud I am really thankful with your help, really, you cleared the difficulties that got me stuck.

[mystic umbra]

you could've easily found it out by realising the fact that its just midpoint of a hypotenuse so just divide the hypotenuse by half

[deft sail]

1?

[mystic umbra]

engineer grindset (or rather any applied maths grindset)

sinx ~ x, tanx ~ x and cosx ~ 1

[flat lily]

it was a tad more than that... I wasn't seeing other issues involved.

[mystic umbra]

erm alright

[obsidian harness]

correct but it won't work

you need one more polynomial term

the problem is that y = 1 is a constant function which is ass

I mean from sin x ~ x, and sin^2 x + cos^2 x = 1, you can figure it out methinks

let cos x be approximated as 1 + ax + bx^2, and you know that a = 0 because cos is an even function

[zealous pike]

How do I make it show me the point of intersection ;-

[left mauve]

Click the point

[civic rock]

click at the intersection

[zealous pike]

it doesn't wokr

work

[civic rock]

reload the page

[left mauve]

It does I just checked

[zealous pike]

I've zoomed about 100x and it isn't showing me ;-

[left mauve]

[civic rock]

^

it might save your work

[zealous pike]

it didnt

and it didnt work either, but it doesnt matter

It is close enough to the point I want it to intersect so I'll consider it is alright

[left mauve]

Huh

Just use desmos

[zealous pike]

desmos worked

thx

[zealous pike]

I checked them up in the graph but I'm still uncertain. My answer for 51-54 is:

51: 4 sin(x) ; 52: 2 cos(2x)

53: (2/3) cos(3x) ; 54: 3 sin (x/2)

are they alright?

[native pier]

You go to tools --> point--> and then click the point of intersection. then you go back to the algebra panel and it'll show you the x and y coordinates in decimal form

[zealous pike]

Thxx

[zealous pike]

I can't seem to solve 55. Can I consider that -π/3 and 2/3 returns the minimal?

[native pier]

I don't see why not

-# but you should also probably get other people's opinions because I'm not confident in my answer lol

[civic rock]

maybe yeah

it's better if they shifted the graph a little so we can see a bit clearer

[karmic steeple]

small angle approx

in rad

fucking love thats shit

[mystic umbra]

yum yum sinusoidal waves

[spiral crescent]

how to do rational number

[left mauve]

Elaborate.

[spiral crescent]

what does rational numbers mean

[mystic umbra]

q = a/b where a,b are integers

that's it :3

[spiral crescent]

?

[mystic umbra]

that's it

[spiral crescent]

I am studying year 9

[mystic umbra]

yeah that's what it means

literally

set of all numbers that can be expressed as a ratio of integers

[spiral crescent]

integer=3
Irational number =1.4444444444444444444

?

[mystic umbra]

no

if it repeats indefinitely, its rational

if it ends, its rational

if it doesn't end and doens't repeat, its irrational

oops

[spiral crescent]

thanks

thats what I was asking for

[civic rock]

this is definitely geometry

[mystic umbra]

shhhh

[spiral crescent]

then which chat do I ask this question

[mystic umbra]

prealg or precalc

maybe

[civic rock]

#prealg-and-algebra

[spiral crescent]

tqsm

how old are you guys

?

[civic rock]

sqrt(-1)

[austere lichen]

hi guys, i have a question if somenone is willing to help

[civic rock]

what's your question

[austere lichen]

it's a geometry problem, one second

the task is to find the area of this piece

and i keep missing for about 5-10m^2

[civic rock]

can you show your work

[austere lichen]

sure

[civic rock]

two things

1. You should not approximate along the way, only do it at the very end
2. It's not recommended to use sine theorem to find angles as the angle can either be acute or obtuse

[austere lichen]

the approximation i get, but how do i solve this without the sine theorem?

[upper karma]

@mystic umbra trying to understand this

[haughty sequoia]

Imo the easiest way is to decompose it into rectangle triangles.

I see there are a few.

Then from there you can calculate a few sides that can help you get the needed areas.

[mystic umbra]

what it means is that squaring is irreversible in the sense that taking a square root gives two roots

since we started from one root we cant possibly have two roots

or something like that

[upper karma]

i see i see..

[mystic umbra]

sinx = sqrt ( 1 - cos²x) isnt an "identity" is what it means... i think so

[haughty sequoia]

Pretty much, I suppose. sin's and cos's domain are R, whereas the sqrt's domain is R+. If you were to calculate sin(-pi/2) you'd be getting from the root a necessarily positive value, whereas given the sin's periodicity you can have sin(-pi/2) = -1, when outside the root. Yes, with the sqrt it's not a proper identity anymore.

[civic rock]

you should use cosine theorem instead

find the side lengths

and don't abuse Heron's formula as it's time consuming to put into calc

you should use Area=ab.sin(alpha)/2

[austere lichen]

okay, that formula does make sense

but i don't have the side that i labeled as z in my sketch

[civic rock]

^

[austere lichen]

and i can't get it with the cosine theorem because i don't have the beta

[civic rock]

the angle here is 63°25'

[austere lichen]

yes, and i have x (got it with the cosine theorem) and the 125, but it uses the angle between the sides no?

[civic rock]

doesnt have to

[austere lichen]

really? it works with any angle?

that would make it so much easier

[civic rock]

oh yeah it uses the angles between the sides

you have to plug in and solve a quadratic equation in your calc

but it doesnt have the same problems as using sine theorem

[upper echo]

when computing alpha, it is 63,.. not 62...

[austere lichen]

omg

thank you so much

i'mma head off to jump off a local bridge

[upper echo]

and you don't need to compute z. Just use formula for the area of a triangle ab*sin(t)/2

[zealous pike]

Hi

[zealous pike]

I wonder if the correct identity wouldn't be sin x = ± sqrt(1 - cos² x)?

[cunning lion]

i think the best way to express it would be as
[ \sin^2 x = 1 - \cos^2 x ]
or as
[ \abs{\sin x} = \sqrt{1 - \cos^2 x} ]

[somber coyote]

cloud

[haughty sequoia]

Indeed it would be that, but as Cloud said the best way is without the root to avoid complications.

Cool one with the absolute value.

[cunning lion]

the second one is closer to what you were going for, but i think the absolute value is a more precise way of expressing it than plus-minus

[deft sail]

So then what’s b?

[grave pond]

Ideally something so the derivative of the approximation looks like -x, since cos'(x) = -sin(x) and sin(x) looks like x.

[maiden brook]

graph cos(x). It should be kinda clear

[grave pond]

Honestly, just looking at a graph wouldn't get me much closer than "it must be negative".

[patent siren]

Hi guys, so I’m currently in geometry and right now I’m covering angles of triangles I covered triangle sum theorems but I’m trying to study for every theorem in my chapter 5 right now I’m doing exterior angle theorems does anyone have any tips? I know that an exterior angle of a triangle is equal to the sum of the two opposite interior angles

But any tips are appreciated I have a huge test tomorrow.

🫶

[zealous pike]

[patent siren]

me when math

[zealous pike]

Gobble down any theorem you see ahead of you. Congruence theorems, similarity of triangles, 90 - 30 - 60 triangles, 90 - 45 - 45 triangles, some trigonometry too, like law of sines, law of cosines

[timid blade]

Hello could anyone help me with my geometry work? I’m taking in the summer so I don’t have to have any math classes my senior year but it’s kinda kicking me in the butt rn lol

[loud sky]

Shd we remember each and everyone of those theorems stewart,meneleau and appolonius theorem...

[zealous pike]

Y not

[civic rock]

you don't need to remember stewart

The other one you have to remember, especially Melenaus

[obsidian harness]

that's your job to find using algebra

[deft sail]

Oh what I wasn’t thinking 😭😭🙏

[obsidian harness]

the key is you want x^2 + (1 + bx^2)^2 to have 0 as the x^2 coefficient

you can never get exactly 1 but that's how you make it the closest

[obsidian harness]

from sin^2 x + cos^2 x = 1

[deft sail]

Yeah that makes sense

[obsidian harness]

it's public knowledge what the approximation for cos x is around x = 0

but not many people know how to get there without Taylor series

so sin^2 x + cos^2 x = 1

up to and including the x^2 term, x^2 + (1 - x^2 / 2)^2 ~ 1

[maiden brook]

yeah sure it doesn’t tell u the coefficient, but u know it’s negative (but more than -1) but that the linear term is likely 0. I mean justifying the approximation through calculus is probably beyond what he wanted

anyway

[inland yarrow]

i love china tst

just wanted to share a question which I thought was very very nice

if you have solution dm me

[civic rock]

translation?

@inland yarrow

[inland yarrow]

my bad

P is a point on 9 point circle of triangle ABC, connect AP and draw the perpendicular line to AP at P which meets BC extended at Q. Let X be on PQ such that XA is perpendicular to AQ. If H is the orthocenter of ABC, and D and M are the midpoints of BC and AQ respectively provs that's HX perpendicular to DM.

apologies

[civic rock]

shit 9 point circle

I haven't read about it yet

[inland yarrow]

oh

[civic rock]

ik what it is but

never solved a geometry problem related to it

[inland yarrow]

essentially midpoint of sides, height, midpoints of AH BH CH concyclic

[civic rock]

ye

had to go get my papers to prepare for gaokao tomorrow 😭

[inland yarrow]

Tomorrow??

Isnt it over

[civic rock]

vietnamese gaokao I mean

[inland yarrow]

oh okay

Good luck!

[civic rock]

I'm genuinely scared of literature

[civic rock]

thx

[inland yarrow]

bump if anyone wants try

[maiden brook]

probably try comp math channel if u haven’t already. I don’t think many people could get it (including me)

[inland yarrow]

its okay i just wanted to share a really nice problem

Thanks tho

[maiden brook]

do u have the solution?

[inland yarrow]

Yes

[civic rock]

did you solve it or?

[inland yarrow]

i did

but i missed some stuff on my first try

[civic rock]

alr I just came home

will try later

[inland yarrow]

after i looked at the first few lines of solution i did it again

and i completed it yay

[civic rock]

getting pretty close to solving this

just need to prove one more ratio

nvm I just solved it I am an idiot

@inland yarrow

will send you the solution later once I finish dinner

not as hard as expected

[mystic umbra]

#competition-math could be a bit more of a help... maybe?

[civic rock]

nah they did solve it

[inland yarrow]

Nah just wanted to share a nice problem

[chrome shore]

this might sound cringe to ask but, what do you think about "if the world around us is like a perfectly built machine, then geometry is one of its major guidelines".

[mystic umbra]

if

[chrome shore]

still an interesting if though i guess

in fact, even if the world may not be/ seem like a perfectly built machine , wouldn't geometry still be one of its major guidelines?

[tawny blade]

This was the exam i did a few days ago. Is this a hard or difficult exam?

[zealous pike]

Given a triangle of two equal sides, the opposite angles to the included angle between the equal sides are always equal?

[left mauve]

Decent, but that's subjective- depends on your country

[mystic umbra]

yeah

[zealous pike]

Thx

[mystic umbra]

you can use sine law to prove it for iff

i think

[half bone]

havent gone through the horrors of final exams but still chúc anh thi tốt ạ!!!

[civic rock]

cảm ơn chị/em

[half bone]

ur welcome (i'm younger than you for sure)

[tawny blade]

Its from south africa lol

I had a look at the final matric paper for maths (P2 i think) and there was a cubic graph

[zealous pike]

Given the triangles ||obviously not to the scale|| in the image, where AB = PB = BC and PC = CD and given angle APB = 20°, find angle DPC

Given that AB = PB, then triangle ABP is an isosceles triangle. If triangle ABP is isosceles, then the angles opposite to angle ABP are both 20°.

Angle ABP = - 20*2 + 180 = 140°.

Lines from sides of a triangle are straight lines, therefore, angle PBC is a supplementary anglr of angle ABP. We have that

PBC = 180 - ABC = 180 - 140 = 40°.

Triangle BPC is an isosceles triangle. Opposite angles to angle PBC are equal, therefore:

BPC = BCP = (180 - 40)/2 = 70°.

We find that angle PCD = 180° - 70° = 110°

Angle DPC = PDC = (180 - 110)/2 = 35°. Therefore, angle DPC = 35°.

Summing the angles of the triangle APD we have 35+70+20+35+20 = 180

Is it right?

[tawny blade]

is that like a tan function?

[tawny blade]

Where did you find this question?

[zealous pike]

Here

[tawny blade]

💀 I meant the exact question

[zealous pike]

Well I come up with it in my head using a question I found here as a reference

Lemme scroll up a lil bit

[tawny blade]

Oh ok thanks

[zealous pike]

Basically it but with different points

[civic rock]

literally the same problem?

[zealous pike]

Ye

[civic rock]

you see TP=TQ=TS

[zealous pike]

Yes

Is my solution wrong?

[tawny blade]

🔥

[civic rock]

you got the correct answer

[zealous pike]

Math is always full of "it is obvious but let's state it either way" stuff right

[tawny blade]

What grade level in high school is this usuallu taught?

[zealous pike]

Dunno

I'm not in high school anymore

I'm self learning

[tawny blade]

Yeah i want to teach myself calculus this holiday

I just finished my midyear exams

[civic rock]

1+1=2, is that an obvious thing?

[zealous pike]

Ye, and that one mathematician wrote a whole book proving it

[civic rock]

what book is this from btw

[tawny blade]

Rn i just asked chatgpt to solve 1+1 = 2 in the most complicated way possible. Result is funny asf

[zealous pike]

Geometry: Seeing, doing and understanding
Harold R Jacobs

[civic rock]

fundamental theorem of engineering, take the equation on a very big scale, 1 and 2 are pretty much 0 so 0+0=0

[civic rock]

seems like a cool name for a cool book

might take a look at it later

(ages better than Euclid's Elements anyway)

[zealous pike]

I like it because it explains well. I don't like it because some exercises, mainly the initial ones, are like "use a compass and draw this triangle"

Like, I ain't using a compass in my entrance exams. I can't even use a calculator 😭

[tawny blade]

Gpt just defined it using Peano axioms, and used set theoretic definitions for 1 and 2 ( i have no idea what i am saying 😭 )

[tawny blade]

They wont allow calculators?

[civic rock]

ridiculous

[zealous pike]

No

You are expected to know how to handle arithmetics

So good luck

[tawny blade]

gulps

My school allows scientific calculators in exams

But they have to be non - programmable or graphical

[civic rock]

Q1. Approximate 1+2 using Newton's binomial theorem

[tawny blade]

I suppose that means you cant have a calculator that can plot graphs or factorize equations

[zealous pike]

This happens here in Brazil mostly because of military schools that makes the kiddos do crazy arithmetic manipulations. So they are like "if they can, why u can't too"

The hardest exams here are military too smh

[tawny blade]

Seems like they love torturing brains

[zealous pike]

The course given in Brazil for the merchant marine role expects you to know basic calculus stuff before 23 years old.

High School in Brazil barely prepares u for precalculus at all

[civic rock]

assume the path of the rocket is semicircle and the curvature of the earth is negligible, calculate the travel length of the rocket from Brazil to the White House

[tawny blade]

Just watched this story of how this nutter with a samurai sword killed these people in England

How disturbing is this?

[tawny blade]

I guess they place a lot of pressure on science and maths in Brazil?

[zealous pike]

No. Most people here can't even sum fractions

And most people who gets into university is due to their performance on non stem subjects, or because they got to memorize a lot of formulas

[tawny blade]

That system is rigged

[zealous pike]

Indeed

[maiden brook]

This isn’t

[zealous pike]

Isn't right?

[maiden brook]

I don’t think it’s very obvious that the longer side is always opposite the larger angle

[zealous pike]

But the fact that angles of different sides are different is obvious imo

[maiden brook]

Well yeah but it’s saying more than that

[maiden brook]

And that u can prove just as simply as you can state

[proper basin]

a^2 + b^2 = c^2

[zealous pike]

Ic

[zealous pike]

The question is basically: Small circle has radius of 1, bigger circle has radius of 3 and they are tangent at point p. Line r is tangent at points A and B. Find the distance between A and B.

I'm still trying to solve it but I'm organizing my ideas. The point at which the tangent line intersect a circle is always perpendicular to the radius?

[civic rock]

yes

[left mauve]

Not the point, but the line at that point, yes

[civic rock]

you can drop a perpendicular line from the center of the smaller circle to the radius of the big circle

you'll get a rectangle and a right triangle

[zealous pike]

That angle with a question mark is not a right angle, is it?

[civic rock]

def not

[zealous pike]

figured

[wet pivot]

How is art of prblm solving in geometry

[obsidian harness]

it's good

[chrome shore]

found it yet? tried my luck and it seems 2*square root of 3. did it on a hurry so not sure if i'm correct XD

[zealous pike]

No, I'm still clueless in how to solve it

OH I GOT IT

[zealous pike]

Something like this right?

[civic rock]

correct

[zealous pike]

THX BRO

U ARE A FRIEND, FRIEND

[civic rock]

de nada

[chrome shore]

If you want more geometry and trigo problems I recommend the YouTube channel MindYourDecisions. You can find a playlist there for this stuff that gathers problems from all sorts of places. It's quite the way to have fun while learning / testing yourself

Wish that channel was posting them consistently but oh well

[civic rock]

mmm i tried the problems on the channel

just some teasers that I'm not really a fan of

[chrome shore]

it really has quite the diversity, from basic school stuff to famous tests and all. had a fun time with it for real but once all problems were done i started searching again and yeah...nothing that organised came up

[zealous pike]

U right xD

[chrome shore]

Yesssssss

[lone cedar]

anyone want help?

[zealous pike]

Mee

[lone cedar]

ok aske me in my dm

[zealous pike]

Y dm

[lone cedar]

i don't answer publicly

it's my policy

[zealous pike]

Ohhh I understand

Indeed there may be some sensitive data right

[deft sail]

So remember how I said if there way that I can approximate a trig function

Well I think I found a way how

And it doesnt work great now but I think I can update a few things to make the approximation better.

[zealous pike]

"If sen x = 3/5, then what is the value of cos(2x)"

I got at the fact that cos(x) = 4/5, but I can't figure out about cos(2x). Any hints?

[summer cradle]

cos(2x) = 1 - 2 sin²(x)

[zealous pike]

cos(ax) = 1 - a sin²(x)?

How do I demonstrate this identity?

[grave pond]

No.

Unless you fix a=2.

[zealous pike]

Y

How this works

[grave pond]

Depends on what you have to work with.

The angle-addition formula will work (combined with sin²+cos²=1 to make it look neat).

Or De Moivre's formula and multiply out. (And then again sin²+cos²=1).

[grim scarab]

cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
cos(2x) = cos(x + x) = cos²(x) - sin²(x) = cos²(x) - sin²(x) + sin²(x) - sin²(x) = 1 - 2sin²(x)

[zealous pike]

Ohhh

Neat

Thx

[deft sail]

So do you guys like it or

[violet badge]

Cos(pi/7).cos(4pi/7).cos(5pi/7)=?

How to solve this guys (i'm kinda stupid)

🥶

[mystic umbra]

sinx ~ x for x<<1 as always

[maiden brook]

Sorry, it’s terrible

Try 75 °

[signal vapor]

This can simplify to -cos(π/7)cos(2π/7)cos(4π/7)

[lunar hull]

Hi, I'm going to have to remove it from here as well

@wooden valve

Sorry, pirated materials are against Discord TOS 😦

[wooden valve]

ok

[wooden valve]

Hi, any one interested to discuss on the book "The four pillars of Geometry"?

[flat galleon]

How do we go from lhs to rhs? Arcin(1-x) does not equal arcsin x in general

[sleek epoch]

if there are any indians here ive got a great way for u to remember sin cos tan

tag me for details

btw

[dark sparrow]

why Indians specifically

[left mauve]

@sleek epoch go ahead.

[lone dagger]

presumably it will involve knowing Hindi or some other Indian language

[torpid raptor]

Soh cah toa ?

[worthy eagle]

i feel like directly learning sin=Perpendicular/Hypotenuse, cos, tan.. is much easier than learning anything like soh cah toa or any trick

Same with periodic table, learning directly is easier than "Highly naive kids rub cat's fur"
only VIBGYOR makes sense to me, actually useful

[left mauve]

Nah bro periodic table needs the shortforms

I aint learning a 100 elements without a way to remember them in order

[upper echo]

substitute l-y=t in the second arcsin. Or just note that the function is symmetric to the first arcsin, so their integrals are equal. So, the integral is the same as 2arcsin

[sleek epoch]

cz other ppl wont relate

[lone dagger]

well, tell us :P

[left mauve]

Go ahead then

[zealous pike]

In the image we have a triangle with angles 1, 2, 3, 4.

Angle 1 is equal to the sum of angles3 and4. Already accepted statement: supplementary angles of a same angle are equal.

Proof:

Angle1 is supplementary angle of angle2, hence, angle1 + angle2 = 180°.

Angle3 and angle4 are also supplementary angles of angle2, hence, angle2 + angle3 + angle4 = 180°

We can conclude that

angle1 + angle2 = angle2 + angle3 + angle4

Angle1 = angle2 - angle2 + angle3 + angle4

angle1 = angle3 + angle4

And we have proved our theorem.

is this alright?

[civic rock]

looks alright to me

[deft sail]

I have something better try 0.004(θ-60)+0.886

[signal vapor]

I brute forced the entire periodic table into my brain

no mnemonics

[left mauve]

🥀

[zealous pike]

I also want to memorize the periodic table

[left mauve]

Use mnemonics

Dont do brute force

[zealous pike]

Y

I brute forced the skills of every character in a moba game

[civic rock]

same

[left mauve]

Why are yall like this

[civic rock]

I remember when my literature teacher forced us to memorize 50 essays by heart to prepare for grade 10 exam

[left mauve]

Why dont u want to make ur life easier

[left mauve]

WHAT

thats rigged fs

[civic rock]

I'm not joking

[left mauve]

But why

whats the point

[civic rock]

to secure a better chance to get a decent literature score

[left mauve]

But.. you can just write one yourself

[civic rock]

because literature score is doubled

anyways I finished gaokao today

maths is prob my worst surprisingly

[native pier]

same, never had the willpower to memorize it, but since I'm taking chem next year, I'll probably be forced to 😂

[maiden brook]

This isn’t a good way of going about it

[civic rock]

interesting... but impractical

[formal jetty]

Help!

[worthy eagle]

Do you know the formula for surface area of a sphere?

[formal jetty]

4pi r^2

[worthy eagle]

yeah so substitute r=5 and calculate

[formal jetty]

100 pi

[worthy eagle]

I am confused after seeing the options

[formal jetty]

So I'm not crazy, good.

[worthy eagle]

options are wrong bro

[formal jetty]

Bonus points: find the question for which these options make sense.

This is where I'm stuck.

[worthy eagle]

lol

[zealous pike]

Autism

Maybe

[alpine lintel]

Hi guys

[native pier]

Hello! Welcome to the mathcord

[alpine lintel]

Hi

I am stressing out

To like the extreme rn

I have 45 assignments

Due in 6 days and was wondering if someone can help me do them

[native pier]

Sure! There are many people who are willing to help guide you through math problems here Just check out #❓how-to-get-help and say where you are stuck. However, keep in mind we can't do them for you!

[deft sail]

Hmm ok

[zealous pike]

I have a doubt and maybe it is not what is necessary to solve this question. Could I assume that lines AB and BC are tangent lines to the circle?

[calm owl]

i think they're tangents bc a square has right angles but im not sure 🤷

[zealous pike]

I see. Going from that, can I assume that Q is perpendicular to point m, being MQ the radius, and that QC = 3?

[calm owl]

yeah ig

[zealous pike]

How certain are u about that

[calm owl]

pretty certain

[grave pond]

It seems to be the only understanding of the question that makes sense.

[calm owl]

yeah

[zealous pike]

Thx

Im not at the level to solve this one yet

The radius perpendicular to a chord always bisects it into two equal lines?

[grave pond]

The tangent of angle CQH is 1/3, and since triangle MHQ is isosceles angle DHM is two time angle CQH.

In fact, if you dig up an appropriate double-angle formula for the tangent (or use complex numbers to double the angle), it turns out ||triangle HMN is a 3:4:5 triangle, with the 3 aligning with the QC you already know.||

[zealous pike]

How do you come to the conclusion that ch = 1?

[grave pond]

The diagram is mirror symmetric across the diagonal BD.

[zealous pike]

Then AN = 3?

[grave pond]

Yes. (Sorry for using N for a different purpose in my second diagram!)

[zealous pike]

Np

Im just a newbie trying to understand a problem way out of my current level

If u don't mind me, I'll fill you up with dumb questions now. Tell me to stop if u wish

[zealous pike]

How come triangle MHQ is isosceles?

[grave pond]

MQ and MH are both radii in the circle.

[zealous pike]

OH RADIUS

ic

Thx

[cunning skiff]

asking for a friend are the geometry and algebra 2 courses on khan academy good for prepping for the NY regents?

if someone replies please ping me : )

[zealous pike]

Hi

[fresh birch]

I did this but I'm not so sure actually

[sterile tiger]

can someone help me with that one

and this one

[summer cradle]

focus on right triangle PEO

[fresh birch]

use the antiparallel theorem

[civic rock]

try to calculate the angles

[civic rock]

pythagoras theorem, the 37° is a red herring

[sterile tiger]

thxs

[calm owl]

you could figure out AC with trig but that would be a much longer route

[calm owl]

so i would definitely reccommend pythagorean

[civic rock]

actually because CB and CD are intersecting tangents, angle BAC=DAC=37° but the actual angle of BAC is around 36,8699°

[calm owl]

why?

[civic rock]

triangle BAC and DAC are congruent

[calm owl]

oh

[cobalt notch]

hello

[calm owl]

ohhhhhhhhhhh

[cobalt notch]

Im new in this sv

[civic rock]

giving that DAC = 37° is a pretty horrible clue tbh

which can cause a lot of confusions

[civic rock]

welcome!

[calm owl]

guys how do i convert between radians and degrees

[gloomy badger]

rad -> deg: muliply by 180/pi
deg -> rad: multiply by pi/180

[calm owl]

ohhhh alr thx

[gloomy badger]

alg

[grave pond]

There's a lot of lines on your diagram that probably obscure more than they enlighten, but your basic idea of solving r² = (r-1)² + 3³ is sound.

[zealous pike]

Just to clarify the theorem. It is said that the external angles of an triagle are equal to the sum of the opposite interior angles.

Angles 1 and 3 aren't equal the sum of angles A + C, but the sum of 1 + 3 is equal to the sum of A + C, right?

A and C are supplementary of angle 2, and 1 and 3 are also supplementary angles of 2. Therefore, 1 + 2 + 3 = 180 = A + C + 2; 1 + 3 = A + C

[civic rock]

ye

[silent plank]

1 + 3 = A + B
wdym by A+B

[zealous pike]

1, 3, A and B are angles

[silent plank]

there are multiple angles around B, which one do you mean

[zealous pike]

The one perpendicular to point B? Angle ABC

[silent plank]

not perpendicular, but you mean "2"?

[zealous pike]

Wait

Oh Ic

Misstype

There

[silent plank]

ok. that makes more sense

[zealous pike]

A + 2 would be wrong anyways

[grave pond]

It is said that the external angles of an triagle are equal to the sum of the opposite interior angles.
I would interpret that as meaning that the green angle is the sum of the two red angles here.

[zealous pike]

It is possible to say that if angle A is equal to D angle B is equal to angle E, but line EF ≠ CB, then triangle ABC ≠ triangle EDF?

[civic rock]

yes

but the property of 2 triangles being different is never that useful

[zealous pike]

Does any triangle have a mid segment?

[grave pond]

What is a "mid segment"?

[zealous pike]

A segment that connects the two midpoints of two sides of the triangle

[grave pond]

In that case, every triangle has three of them.

[zealous pike]

Thx

[fresh birch]

I wanted to verify that the diagonal of the square passed precisely through the intersection point of the radios drawn towards the points of tangency, in addition to being collinear with the center of the circumference xd

[grave pond]

Oh, for that I'd just say: We have two tangents to the circle, so the line from the circle center to their intersection point must bisect the angle between the tangents. On the other hand, the diagonal in a square bisects the corner angles. So those two angle bisectors must be the same line.

[fresh birch]

I hadn't really thought about bisection, I had checked it with a counter-assumption case, if they weren't collinear when applying Pythagoras the result was a negative R, thanks for the idea

🙏

[zealous pike]

I want to solve 2.4, I got at 1 + cot - cosec smh 😭

I dunno how to show my work cus my letter sucks

I'll write it on texxit just a minute

[somber coyote]

Pi, a future fluent jp speaker

[zealous pike]

JUST A SEC

[zealous pike]

wait

I messed up

Im just running in circles ;-

[grave pond]

I'd generally recommend reducing everything to just sine and cosine there.

[tawny rapids]

this identity isn't even correct

where tf did you find this

[zealous pike]

apparently the equality holds

[tawny rapids]

sure doesn't look like it

[grave pond]

The red function should subtract sec x instead of csc x.

[tawny rapids]

oh

[tawny rapids]

I was just looking at this I blame pi

[grave pond]

Fixing that still doesn't make the two graphs equal, though.

[tawny rapids]

I've been fed misinformation wtf

[zealous pike]

$1 +\cot - \csc = \frac{2}{1 + \tan + \sec}$

[somber coyote]

Pi, a future fluent jp speaker

[zealous pike]

It is right now

Is it possible to get at the expression on the right only with what we have on the left?

[grave pond]

In that case we have
$$1+\cot-\csc = \frac{\sin}{\sin}+\frac{\cos}{\sin}-\frac{1}{\sin} = \frac{\sin+\cos-1}{\sin}$$
and similarly in the denominator on the RHS.

[somber coyote]

Troposphere

[grave pond]

Multiply both sides by (1+tan+sec), express in terms of sin and cos, and simplify; you should get 2.

[zealous pike]

Ic

Thx

[grave pond]

The main shortcut during the simplification is probably to notice that (sin+cos-1)(sin+cos+1) = (sin+cos)² - 1²

[tight dock]

Sin is bad
Cos is short for because
Tan is what you get on the beach

[tawny rapids]

I thinking sinning is good actually

[obsidian harness]

It can be proven that if $OD$ is the angle bisector of angle $BOC$, then $OD^2 = OB \cdot OC - BD \cdot DC$, which is a reduction of Stewart's theorem.

However, is the converse true? Given that $OD^2 = OB \cdot OC - BD \cdot DC$, is $OD$ necessarily the angle bisector?

[somber coyote]

south

[obsidian harness]

proof details are spoilered here

[fathom jewel]

An observer is walking along a straight road which is aligned to the north. At one point the observer notices a tall tree on a bearing of 54" with an angle of elevation of 48" (rounded to the nearest degree). After walking 220 metres along the road, the tree is now on a bearing of 160" with an angle of elevation of 25". Let h be the height of the tree.

Need help drawing a diagram

[dark sparrow]

i think you meant ° and not " ??

" is for arc minutes

[narrow sail]

pls prove

(sinA + secA)^2 + (cosA + cosecA)^2 = (1+secAcosecA)^2

[narrow sail]

nothing yet

im too tired to

[dark sparrow]

ok then come back when you are less tired

[narrow sail]

bruh

[dark sparrow]

we are not going to just give you the answer

[narrow sail]

fine.

imma try it first

nvm i got it

[upper echo]

Well, if OB=OC then your relation is true for any D inside BC. If OB is not equal OC, then for D inside BC the only such case is when OD is a bisector. You can prove it by vectors: If OB=a, OC=b then OD =a+(b-a)k with 0<k<1. And your relation means (a+(b-a)k)^2=|a| |b|-|a-b|^2 * k(1-k). So, after simplification you get k=|a|/(|a|+|b|) which means that OD is a bisector. And you get that it is true for any 0<=k<=1 if |a|=|b|.

[zealous pike]

How to solve this?

I know how to solve it, DON'T

I wanted to make a pun after someone asks me "what have u tried?"

Anyways, just checking it out

23, because AD = CE, figure ADEC is an isosceles trapezoid (the book has already introduced stuff about trapezoids and this exercise is on the quadrilaterals chapter, so I think that's what it wants?)

24 Because ACED is an isosceles trapezoid, angle A is equal to angle C, because base angles are equal

25 ABC is at least an isosceles triangle

26 Line DE is 1/2 of the length of AC, if FG is 1/2 of DE then FG is 1/4 of AC

Are those answers sound?

[civic rock]

one more property about FG and AC you missed

[zealous pike]

It is a quarter segment of AC?

[civic rock]

one more property

[zealous pike]

FG is parallel to AC?

[civic rock]

yes

you can't miss that

[zealous pike]

Understood

It is not in the question, but is it possible to prove if angle B is equal or different to angles A and C with what was given?

[civic rock]

if you can't deduce anything about B then it is not part of any special property

[grave pond]

No, B could be anything with the given information.

[civic rock]

yeah

B is just the apex of the isoceles triangle

[zealous pike]

Understood

[grave pond]

You can see that by imagining someone gave you an angle that's supposed to become B and then figure out how you could complete the diagram so all the givens become true, no matter which angle they chose.

[zealous pike]

Ic

Interesting

I'm confused with this one.

In the image there is an infinite series of equilateral triangles. The length of the sides of the largest triangle is 4 each. Each successive triangle is formed by the midsegments of the previous ones. It wants me to guess the sum of the lengths of all sides of the triangles.

For this one should I consider the largest ones or each new one?

[civic rock]

find the ratio of the length of the longer side and the next smaller side

[zealous pike]

Should I consider these ones too?

[civic rock]

no you focus on the perimeter of the biggest triangle

and the perimeter of the smaller one inside

[zealous pike]

Ic

[civic rock]

compare the perimeter of the red triangle and the green one

[zealous pike]

There u go

Yes its wrong

[zealous pike]

Nowhere

I used js to solve this. It seems that the sum approximates 24?

[grave pond]

Yes.

You're summing a geometric series.

[zealous pike]

Is it possible to use summation notation to describe this sequence?

[civic rock]

yes

[zealous pike]

I thinm I've figured out how, wait a min

[grave pond]

$$\sum_{k=0}^\infty \frac{12}{2^k}$$

[somber coyote]

Troposphere

[zealous pike]

Dont give me the answer bro😭

[grave pond]

That was just notation, not an answer to the problem.

[civic rock]

do you know the geometric series?

[zealous pike]

Given a, n and r both integers and n = 1, a geometric series is a series of the form ar, ar^n, ar^(n+1)…

I don't know much of it though, I have to review it

I just skimmed through it by now

I can figure out the n-th sequence but I can't find out the sum of the nth first terms

[civic rock]

for it to have a converging sum, -1<r<1

[unique lotus]

if anyone would be willing and able to aide in tutoring / helping go through material on the following;

trig angles & measures, trig functions, unit circle, circular functions, odd and even trig function identities, right triangles, trig any angle, graphs of sin/cos, & other trig graphs, and inverse trig functions

was out of town for essentially an entire unit worth of lectures (and I have an exam in two days - I am cooked).

[civic rock]

ar + ar^2 + ar^3 +... + ar^n = a*(r^n -1)/(r -1)

if -1 < r < 1 and n goes to infinity then r^n -> 0

[zealous pike]

In this case, r = 1/2 which is in the interval -1 < r < 1, so r^n = 0?

But how

Wait my brain stopped working

[civic rock]

n goes to infinity

[zealous pike]

That buggies me

[civic rock]

like (1/2)^2 = 1/4

(1/2)^3 = 1/8

[zealous pike]

It is because it approximates 0, right

[civic rock]

it keeps getting smaller and smaller

and it tends to 0

[civic rock]

dont worry it bugged mathematicians and philosophers for thousands of years

zeno's paradox

[zealous pike]

Imagine me who is just an aspiring mathematician

[civic rock]

for falsidical paradoxes like this there will always be some kind of debates remaining

[civic rock]

for now this is generally accepted

[zealous pike]

In that case, the answer for our problem is 0?

Cus if n approaches infinite, then 12*0*(-1 + (1/2)) = 0

[civic rock]

no

[civic rock]

replace r^n=0 into this formula

[zealous pike]

Oh, -1 is not an exponent?

[civic rock]

no

[zealous pike]

Ic

[civic rock]

I should've written it clearer

[civic rock]

if -1 is an exponent I would've written n-1 in brackets

[zealous pike]

It is 24 then

[civic rock]

yep

[zealous pike]

Infinite sums will give me nightmares

[civic rock]

nah if you learn how to deal with them they should be fine

[zealous pike]

I hope so

Thx

[hard knot]

can someone explain me the cofunction theorem? how cos and sin together makes a perfect 90? is it because cos is x and sin is y?

[civic rock]

can you elaborate?

[hard knot]

like these questions

[civic rock]

yeah somewhat

you can use this trig function visualization to see it clearer

the explanation is a bit more complex for angles >90° however

[hard knot]

oh god

thank you 😭 😂

[civic rock]

np lol

[night linden]

Sniped

[fresh birch]

[zealous pike]

hi

[left mauve]

pi

[zealous pike]

If I have a right triangle whose sides are unknown, but the angles are known, is it still possible to find out the ratio of its sides?

[civic rock]

yes

using trig functions

[zealous pike]

But

I'd need to use a calculator

Right

[worthy eagle]

Most of the time, yes.

[zealous pike]

What if I don't have a calculator?

[left mauve]

Try to use standard angles

0, 30, 37, 45, 53, 60, 90

[dark sparrow]

37???

you mention these weird angles of 37 and 53 but not 30????

[civic rock]

biggest sin ever

[left mauve]

forgor

[dark sparrow]

ok but like

,w sin(37°)

[somber coyote]

Results provided by WolframAlpha

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

Requested by ann_dec

[left mauve]

it's 3/5

[dark sparrow]

it's not 3/5

[left mauve]

it is

[dark sparrow]

it's APPROXIMATELY but not EXACTLY 0.6

[frigid gulch]

😭

[civic rock]

it's a bad approximation for sin^-1(3/5)

[left mauve]

bruh i was taught to use them as standard angles as well ok 💀

[frigid gulch]

Cursed teachers

[left mauve]

0.601 is pretty close

[dark sparrow]

ok but like all the others are exact and this one is not

it's ass

[left mauve]

,w sin 45

[somber coyote]

Results provided by WolframAlpha

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

Requested by executor.1011

[left mauve]

oop

eh

does it really matter

like the third decimal place

[dark sparrow]

yes for exactitude it matters

pretending sin(37°) is exactly equal to 3/5 is as wrong as pretending pi is exactly 22/7

[left mauve]

Yes

I agree it's wrong to say it's exactly equal

But im definitely gonna keep using them lol

[dark sparrow]

[insert low-severity insult here] behavior

[left mauve]

Yes, I'm well known for that

[frigid gulch]

,w sin 53

[somber coyote]

Results provided by WolframAlpha

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

Requested by bottle_03

[left mauve]

4/5

like come on

it's 0.799

[dark sparrow]

it's still inexact.

[left mauve]

I dont disagree

[frigid gulch]

I don’t think anyone here is going to change their minds on this matter

[left mauve]

Oh btw

[worthy eagle]

us bro us

[native pier]

what

[left mauve]

EXACTLY RIGHT

it's like a prerequisite for jee

[left mauve]

im searching for a pyq that has 37 and 53 in it hollup

[worthy eagle]

lol fr

[zealous pike]

😭 I'm not equal to no one

[left mauve]

Biased exams need based solutions

[civic rock]

[left mauve]

Like what we gonna do

Being obstinate that they aren't exact is just gonna lead to a -1 or a 0 :)

[worthy eagle]

sin(53°)=3.99/5 I would still use it

[left mauve]

Yuh

[zealous pike]

I don't think that's how it works but sin and cos of 37° seems to be between 0 and 1

[left mauve]

?

Ofc they are

[civic rock]

pi seems to be 31^(1/3)

[left mauve]

No it's sqrt g

||This one is obviously a joke don't cancel me||

[civic rock]

is this some mystical thing god has decided when he created the earth?

[left mauve]

I think so

Circles and gravity

[civic rock]

god why conspiracy theorists didnt think of this before

[left mauve]

They have

Um i wanted to ping a user

his name is pi=sqrt g

but idk how to ping a symbol

[civic rock]

can this lead us to find the true answer to the 1/137 mystery in physics?

and the God equation that Einstein has been chasing for?

[dark sparrow]

-> #chill

[left mauve]

The god equation is
$e^{i\pi} + 1 = 0$

[somber coyote]

Exes are clutter 🌟

[left mauve]

[native pier]

agree

my fav :D

[lilac lynx]

quaderilqaterals are changing the shape of my brain tbh

quadrilaterals i meant

like how am i supposed find 3 different angle sums using only one property when there exterior and interior angles ??????

wiat im supposed to use both properties right ?

oh yeah i get it.

its a question with lots of work and formulas

[thorn narwhal]

what is the angle of emergence

[left mauve]

Yes, that is indeed the question.

[thorn narwhal]

oh

[left mauve]