geometry-and-trigonometry

[tall fractal]

so

i think x = 5 ?

[cerulean socket]

x is 5

yes

so AC and BD are the diagonals right?

so AC and BD should be 18

did you get that?

[tall fractal]

Oh it's not ac and bd = 32?

[cerulean socket]

i thought it looked something like this

AC and BD are just two times 9

since AE is 9

[tall fractal]

Ohhh

OKAY GET IT

[cerulean socket]

since it is a rectangle

the diagonals are the same

[fallen sleet]

AGD, FGC, and BGE don't look colinear so I assume not

especially easy to see with BGE

aight ill just get it wrong and ask the teacher

[snow kite]

When will this formula ever be useful?

don't you want to have a common angle?

[brazen citrus]

Point A, which lies inside the dihedral angle, is at a distance of 2 and 3 from the faces and at a distance of 6 from the edge of the dihedral angle. Find the dihedral angle.
Round your answer to the nearest whole degree. The use of a calculator is allowed.
i want to check my answer

[boreal pollen]

why is r put to the 2 or 3 power in geometry

[humble pulsar]

cause that's how the formulas work

assuming you mean for stuff like A=pi*r^2, V=(4pi/3)r^3, etc

[boreal pollen]

no l mean like formula in a cylinder

[humble pulsar]

Same thing

[boreal pollen]

R=r squared times pi time H?

[humble pulsar]

V=, but yes

[boreal pollen]

okok

how though

[humble pulsar]

wdym

Like any prism, $V=A_{\text{base}}h$

[somber coyote]

Mosh

[boreal pollen]

do l have to calculate r sqaured

[humble pulsar]

take r... and square it

clearly

r=10, r^2=100

r=7, r^2=49

etc

[boreal pollen]

okok

ok

[signal spade]

Were you helped??

[craggy river]

Well that's equivalent to the tan version

Uhhhh wait no not directly

It's related tho

[craggy spear]

use cos(A+B)/sin(A+B)

[storm portal]

@eternal sable ^^

[upper merlin]

I have no clue about vectors but extending MN to meet XY at P and using melenaus' in triangle XOY finishes the question. There might be a vector equivalent of doing something like this 🤔

[serene iron]

Can someone explain how they got 49

<@&286206848099549185>

[quartz bough]

The ratio of similarity is equal to the side of one triangle divided by the side of the other (assuming they are similar sides)
So the ratio here is ML/QF = 42/6 = 7
The ratio between the areas is the ratio of similarity squared, so here that's 7^2=49

[serene iron]

Thanks

Finally Some1 useful in this darn server

[signal swallow]

is that a quiz

oh its the result

nevermind

[quartz bough]

I didnt even realize it said quiz

[serene iron]

Dw guys I would never cheat

[quartz bough]

oh im not worried

im just surprised that i cant read

[serene iron]

It's all good

Thanks tho

[serene iron]

Why did you pick ML and QF

Why not another side

[quartz bough]

what other sides could i have chosen?

[serene iron]

Oh I just wanted or know the reason

Did u pick it bcz it's divisible?

I think I get it

Nvm I don't

[grave pond]

QF is the only datum you have about QDF.

[quartz bough]

the only two sides given that are similar to one another are the sides i chose

[serene iron]

OMG I GET IT

silly me

Like this

[hollow granite]

whats a circle

[grave pond]

A circle is a curve consisting of all the points that have a certain fixed distance (called the circle's "radius") from a fixed point (called the circle's "center"), and no other points.

So, continuing from #proofs-and-logic message ...

Is it possible that a circle whose center is (2,4) can contain both the points (-1,5) and (5,1)?

[bold yew]

can someone explain bearing on law of cosines to me?

[grave pond]

wdym "bearing"?

It lets you solve SSS and SAS triangles.

[humble pulsar]

Bearing is just how to measure angles in a specific way, google resources

Has nothing to do w/ law of cosines directly

[ionic bluff]

ASS

lol

[dusky grove]

How can I build a linear function if I have following information:
b = -1.5
f(2.5) = 16

or
f(3) = 4.5
f(-9) = 8.5

[dark sparrow]

start with the general formula for a linear function, f(x) = mx + b

then translate your information into equations in terms of m and b

and solve said equations

[dusky grove]

Thanks

[rich parcel]

(cosec A + cot A)^2

can i expand that like how I would expand a quadratic, like (a+b)^2

[upper merlin]

yes

[paper grotto]

can someone help on some grade 7 geometry 💀

[upper karma]

I have a proof (that might sound dumb to some of u) that surface area of a sphere is actually
pi^2* r^2
And not
4pi(r)^2
I might be wrong somewhere tho lol…

[grave pond]

If you post a sketch or writeup of your proof, people here might be able to help figure out what goes wrong in it.

Immediately my guess would be that that you've inadvertently come up with something like a 3D variant of the "Problem Archimedes?" meme (https://i.redd.it/1j5iaaeh7pe61.jpg), but we'd need details to be sure.

[upper karma]

Lmao not like this, ill snd it in some time…

[real gull]

Can the sine and cosine rule be applied to an isoceles triangle

?

[grave pond]

Sure.

[humble pulsar]

they can be applied in any triangle

[real gull]

tysm

[humble pulsar]

proof by peel an orange

[upper karma]

If u dint understand why X = pi*d^2/2
If u arrange all the radii in a row starting from 0 to d back to 0,
Ull get a semi circle with radius d
Is this a valid proof tho?

Can someone tell whats wrong in this proof

[humble pulsar]

Sounds like you tried to do integration w/o any calculus knowledge tbh

[upper karma]

I dint do any calculus in this lol

[humble pulsar]

Yeah, that's the problem

[upper karma]

Why

[humble pulsar]

your argument sounds like the one from calc

[upper karma]

Idk

[humble pulsar]

then... didn't follow through with it

[upper karma]

I have not taken any calc courses yet

[humble pulsar]

and spat out a random equation for s

You defined variables then gave a random equation

[upper karma]

Am only in 9th grade…ik integration but not complicated integration

[upper karma]

Thats why I added that sentence above it

[humble pulsar]

Yeah that sentence doesn't make too much sense imo

[upper karma]

Is it not visible or is it wrong

[humble pulsar]

Cause you can easily double count or triple count (or I think infinitely over count) the actual SA

[upper karma]

Dint understand

Whats wrong with that sentence

[humble pulsar]

If I take my circles horizontally, I get the SA sure
But then I also take the vertical circles and get the SA again

I now have 2*SA, but still have infinitely many circles to consider

[upper karma]

Infinitely many yes…infinite radii…together they form a semicircle

[humble pulsar]

(We're also ignoring the fact you'd get 0 from a bunch of circles)

[upper karma]

There are infinite lines in the semi circle

[upper karma]

0 is the radius of the uppermost circle I mean…
Ok not exactly 0, but the next real number after 0 is what I meant

[humble pulsar]

No, your construction actually gives SA=0

Since a circle has no area from the circumference

[upper karma]

Bruh how

[humble pulsar]

They need to actually have width to get SA

[upper karma]

So is that sentence wrong?

[humble pulsar]

The whole thing is just nonsense imo

[upper karma]

Pain

Well i dint get how the csa can be 4pir^2 so i tried to derive something

But ok yeah

[upper karma]

is 2 times the hypotenuse always greater than the sum of its legs?

[grave pond]

Since each leg is shorter than one hypotenuse...

[upper karma]

it depends?

[grave pond]

a+b < c+b < c+c = 2c

[upper karma]

what does that mean exactly?

[grave pond]

a, b, c are the side lengths of a right triangle with c the hypotenuse.

[upper karma]

why is this true?

actually, it makes sense

[near sand]

Hi, does anyone know where to go from here? I'm solving using Property C, Property F and Property Z

,rotate

[cerulean socket]

if they are both rectangles than angle 1 is also 10

and isnt angle two just 90 degrees

[near sand]

angle 2 is basically angle SDC

my picture isn't accurate

[cerulean socket]

ooh

[near sand]

mhm

[cerulean socket]

i thought it was the right angle

let me see

since its a rectangle so ADC is 90 and RDA is 10 so SDC is 180 - 90 - 10

which is 80

am i correct?

[near sand]

i don't think i could solve it like that

i was suppose to use those properties and other theorems

[cerulean socket]

its kinda like property c, angle 1 plus angle 2 plus angle ADC equals 180

[near sand]

but i put my work away already, will just work on it tomorrow morning

[cerulean socket]

ok

[near sand]

thank you tho

[cerulean socket]

ill keep looking to see if i can use a property

[near sand]

I appreciate it

[cerulean socket]

your welcome

[near sand]

thank you

[upper karma]

Question 13. Is tan positive?

[warped ginkgo]

You can check quadrent

[fallen sleet]

Is quadrilateral, pair of sides, other pair congruent, and opposite angles acute a sufficient condition for parallelogram?

[upper karma]

Isn’t that the same way how the volume of a cuboid is formed? Multiply the height by the area of the base…? That’s turning it from 2d to3d, and in my proof i had gone from 1d to 2d…. Sry if I’m wrong

[wintry wraith]

Q1) If asin2(x)+bcos2(y)=casin2(x)+bcos2(y)=c, bsin2(y)+acos2(y)=dbsin2(y)+acos2(y)=d and

atan(x)=btan(y)

Then prove that : a2/b2=(d−a)(c−a)/(b−c)(b−d)

Pls

[upper karma]

Not a pro but, wont bcos2(y) cancel from both sides of the first equation?

[wintry wraith]

Wait

I think

I pasted wrong

[upper karma]

Yeah its ok

[tall glacier]

geometry is for betas

[nocturne remnant]

And alphas…and thetas…and whichever other symbols you would use for angles…

[humble pulsar]

Both use similar ideas, however you need some amount of thickness to make the argument hold water

(ie integration)

[upper karma]

Ok but can it be done without calculus? We can take the example of a cylinder now? If u multiply the area of the base by height u get volume…but there is no vertical height gor a single circle…how does it work

[humble pulsar]

The formula V=Ah is a subsequent of integrating

[upper karma]

Ok how would u restate my proof with calculus?

[humble pulsar]

Do the integration for SA

[upper karma]

Ah idk integration…sry lol…still learning it

[tall glacier]

https://media.discordapp.net/attachments/773116361307062286/773857639226277938/Hmm48.gif

[serene dagger]

bruh

[serene dagger]

see in this diagram

[wintry wraith]

Yuo

[serene dagger]

they made two lines parallel

u constructed them ryt?

so

[wintry wraith]

Yup

[serene dagger]

ad=12

lets say

[wintry wraith]

Ok

[serene dagger]

i think u should use

Pythagoras theorem since its a right angle triangle

ok?

and then

see how it goes

ok?

[wintry wraith]

Bro I was bit sleepy

Sry

[serene dagger]

let us consider the two circles of radii 12cm and 4cm with centers P and Q respectively.

now since the length

[wintry wraith]

Yup

[serene dagger]

of the common tangent of both the circles is 15 cm

cd=15

ok?

now

cd=be

[wintry wraith]

Ok

[serene dagger]

be=15

since

ae=

ac-ce

[wintry wraith]

Ic

[serene dagger]

ae=12-4

=8

ab=?

now

use

pythagoras theorem

[wintry wraith]

[serene dagger]

ok?

[wintry wraith]

I m feeling very sleepy

[serene dagger]

r u ok bro..

[wintry wraith]

I will come later

[serene dagger]

bro x is 17cm

[wintry wraith]

Its just night time

[serene dagger]

okay

[wintry wraith]

Ok

[serene dagger]

for me its day so

xd

[wintry wraith]

O

[serene dagger]

xd

okie

see u later bro

[wintry wraith]

Bye

[upper merlin]

this scared me ngl

[upper merlin]

@wintry wraith the main idea is the find the area of those 2 sectors and subtract it from the area of the trapezium [A B O1 O2]

[viral hamlet]

I need to find a here, the solution that I got to is " a = 1 + 2k ", Anything wrong here ?

[storm portal]

There’s no k anywhere in the problem, so you shouldn’t get that in your answer

[viral hamlet]

the general solution

I used this thing

[storm portal]

I’m assuming your problem is given that g(pi) = 0, and g(x) = 1 + cos(ax), find a

Oh yeah right - cosine is cyclic

[viral hamlet]

so it's right ?

[storm portal]

Looks good to me

k is any integer 👍

[viral hamlet]

👍

[silent plank]

trig ratios and segment subtraction

[lament fjord]

I need help pls

[fallen sleet]

is quadrilateral, pair of sides congruent, other pair is parallel, and two opposite angles are acute a sufficient condition for parallelogram? wHY oR WHY NTO?

[upper karma]

Guys do You have an idea of how I should start?

[craggy river]

Can't understand it, is it deriving the bottom from the top?

[vagrant flume]

what is the answer? i want to check to see if i got it right firsy before i explain it if i did get it right

and is it deriving bottom from top like someone else said, that’s what i did

[upper karma]

The anwer is D

Answer*

I got b)

I got like 1-cos0/sin0

I dont know why The answer is D

[dapper otter]

I have two linear functions, F1 & F2. I want to draw F3 exactly in the center between them, but don't want to use trigenometry. So if I know k1 k2 , how can I get k3 without using tan?

This is not a homework, just a random question

The way I did it here, was using angles. I just used the average angle to draw the third function

[dark sparrow]

well you can express tan(x+y) in terms of tan(x) and tan(y)

though doing tan( (x+y)/2 ) may be a bit painful

[dapper otter]

Yeah but how would I do that without trigenometry?

if it is even possible

[solid bridge]

pls help

[uneven bloom]

Is there a name for a polygon where all the interior angles are the same except for one?

[dark sparrow]

"a polygon in which all the interior angles are the same, except for one"

[feral gate]

why does the number of (congruence classes of) regular polytopes in real euclidean space differ so much by dimension

is that question too advanced for this channel

[willow anvil]

anyone down to voice call to help me understand what coplanary vectors are?

tag me if yes!

5 - 10 mins max

[feral gate]

please don't post the same thing in multiple channels

@willow anvil

[willow anvil]

oh i apologize!

[dusky grove]

Let's say I have a equation x + 4y - 6 = 0, y = -0.25x and a point P(-6|1).
How can I prove that they are parallel to each other? I assume I have to take the x and y from P and insert them somehow in the equations, but not sure how exactly.

[craggy river]

I have no idea what the point is doing there lol. There are multiple things you could do, but imo the easiest is to isolate y in the first equation and show that the slopes are equal

[blissful monolith]

this is a bonus our teacher gave us, he said its quite complicated and took him a while himself. I know i have to apply Thales theorem but im not getting sure of my answer, i think its 0.6

<@&286206848099549185>

please @ me once you have tried solving this ex, thankyou have a good night

[upper karma]

Is there really a difference in using dilation vectors to map a dilation around a center of dilation that isn't 0, 0 or just straight up counting the distance?

Also unrelated question can a scaling factor be considered a scalar seeing as how a R^2 plane is still considered a vector-space and we are vividly multiplying values with a scaling factor?

[formal bolt]

Call $O'$ the point under $O$, between $B$ and $D$.\
Let $H_1 = AB, H_2 = CD$ and $l_1 = BO', l_2 = O'D$.\
Being the triangle $ABD$ similar to $OO'D$ and the triangle $CDB$ similar to $OO'B$ you get
$$\frac{H_1}{h}=\frac{l_1+l_2}{l_2}, ; ; ; \frac{H_2}{h}=\frac{l_1+l_2}{l_1}$$
Hence
$$\frac{H_1}{H_2}=\frac{l_1}{l_2}$$
But then
$$\frac{H_1}{h}=\frac{l_1+l_2}{l_2}=\frac{l_1}{l_2}+1=\frac{H_1}{H_2}$$
And so
$$h=\frac{H_1 H_2}{H_1 + H_2}$$

[somber coyote]

Mat

[formal bolt]

@blissful monolith

[quasi galleon]

can someone help me with this?

(for the select ones) by the (1. distance formula , 2.converse of the triangle proportionally theorem 3. triangle proportionally theorem). A is similar A by the (1. symmetric , 2. reflexive , 3. transitive ). So ADE is similar ABC by (1. SAS 2. AAA 3. SSS)

[signal swallow]

try turning tan into sin and cos

[simple rain]

guys

pls am poor

how to find concluding statement

[storm portal]

Try to match hypotheses (conditions) and conclusions (results)

Using contrapositives

[craggy river]

$W \to C,
G \to H,
C \to G,
B \to W$

[somber coyote]

hiiistrex

[solid bridge]

somebody please help this assignment's been due since last week and i just wanna pass it

[dark sparrow]

what is the assignment?

are you instructed to check/grade the work presented here?

from what i personally can see, there is a problem there together with its solution

[solid bridge]

i've been tasked with finding the other interior angles of questions i have to make

i've been so stressed because of this damn thing

let me make some first

[dark sparrow]

you've been tasked with... what exactly?

hold on

[solid bridge]

i have to create my own questions

and i have to solve them too

[dark sparrow]

you have to create your own questions and solve them.

[solid bridge]

yes

[dark sparrow]

and all of them have to center around interior angles of polygons

[solid bridge]

sides and angles too

[dark sparrow]

wym sides

some of your questions need to involve calculations of side lengths?

sounds like trig hell if you ask me

[solid bridge]

wait wait

here's the guidelines

"Create and solve three THREE (3) real-life problems that you and your family members encounter and apply your knowledge of polygons, sides, and angles."

[dark sparrow]

boy does that sound vague...

[solid bridge]

man if i could i'd turn this stupid thing with nothing in it but it's gonna screw my grades badly

[dark sparrow]

and i take it you are stuck not knowing what kinds of problems the teacher expects of you with such wishy washy directions

can't blame you there really

[solid bridge]

i don't understand the lesson also

it's literally 10 pm in my time and i just wanna get this thing over with and sleep

can you help me make questions and help solve them pls

[dark sparrow]

i have no idea what kinds of questions are expected either

[solid bridge]

it says "real-life problems that you and your family members encounter"

[dark sparrow]

best i can suggest is the same question you sent but with different numbers and/or different shapes

[solid bridge]

okay sure i'll try that

[dark sparrow]

dressed up in some sort of narrative involving you or your family

[solid bridge]

how do i solve it though

[dark sparrow]

i thought the solution written up there was made by you??

[solid bridge]

no no

it was an example made by my teacher

[dark sparrow]

you should have been told a formula by the teacher

that in any simple n-sided polygon the sum of all interior angles is (n-2)*180°

[solid bridge]

hm okay thanks

i'll try

well i finished it

thank fucking god

thank you btw you just saved me from hell

[finite saddle]

not sure if this is the appropriate place to ask, but can you suggest a video/article/whatever, which explains how sin, cos, tg and ctg work ( for example how to find the value of cos(-45), tg (30) and so on )
i can solve problems with them using formulas, but i lack some basic knowledge

[graceful pulsar]

hey can anbody help me?

pls

BRo

can anyone help:?

[silent plank]

depends

[graceful pulsar]

wdym

[silent plank]

we have no idea what you need help with so we won't know if we can actually help

#❓how-to-get-help

[graceful pulsar]

oh my bad

I need the area

so can u help me?

@silent plank

[silent plank]

split your shapes into smaller/simpler shapes

[graceful pulsar]

I did

but I still dont get it

[young rampart]

for the first shape

there are two rectangles

can u spot them?

for the second shape, there is a rectangle and a triangle

[graceful pulsar]

yea

I see

it

[young rampart]

so do u get it?

[graceful pulsar]

but finding the numbers are kinda hard

[young rampart]

hol up

lemme draw a diagram for u

now do u see it?

[graceful pulsar]

uh yea

i think so

[normal iron]

Given that 12 sin X = 12, what will be the value of sec X and cot X ? (in orders)

[hollow raft]

Can someone help me with angle sum and addition

[inland solstice]

I can

[hollow raft]

thanks

@inland solstice here

Wrote out my formulas

No idea where to go from here

[inland solstice]

I would probably use the unit circle to help find those exact values. Since sin is the y value and cos is the x value on the unit circle

[hollow raft]

for problem one which part of the formula do I sue?

Use

Should I do 2(-3/5) - 1

would that get me my cos2a?

getting me -1.72

[inland solstice]

I believe so, you should probably double check with a "helper" though

[upper karma]

Is this right solution.i mean did i solve it right or not

I need some help did i solve it right? Right.

[upper karma]

Guys any one there

[upper karma]

Maybe you should try khan academy or YouTube .for me khan academy helped me lot understanding trigonometry

[digital talon]

hi, can you fit this curve to a sin fucntion in the form y = asin(bx+c)+d

[upper karma]

Is my solution right or wrong someone please tell me

[young abyss]

i think this is right

[craggy river]

5beta + 30 = 180

Double check your work

[quasi galleon]

i need help with finding the volume of the pentagonal prism

[cerulean socket]

just search up the formula and memorize it

[craggy river]

Idk if that's a particularly good way to do math

It'll get the right number but you won't understand much

Although imma be honest I forget how to find the area of a regular pentagon

BUT once you find that, since it's a prism, you can multiply by the 'height', which in this case is 15cm

[wise pawn]

I'd do it by making a point in the center and then breaking it up into 5 triangles of the same area, then use some trig to work out what the area of one of those triangles is

[dark sparrow]

bad attitude.

[rustic thistle]

quadratic formula vs quadratic function

[dark sparrow]

those two refer to different things lol

[young rampart]

quadratic function is ax^2+bx+c

quadratic formula is the formula to calculate x

[rustic thistle]

this might sound dumb

is the function used to find the roots?

[young rampart]

you use the formula to find roots

or factor the function

[jolly plume]

Can someone please explain to me what the mesurement unit Minute is?? denoted as x' where x is a real number

[silent plank]

1 (arc) minute is 1/60 th of a degree

[craggy river]

Unfortunate

[craggy river]

^ then (arc) seconds are 1/60 th of that, denoted x''

They don't get used much besides global coordinates

In other words, it's another unit to measure angles

But small

[limber ice]

Hello, if I have the value of sin(theta)=z/2

how can I find sin(2theta)=?

[young rampart]

theta=arcsin (z/2)

i think

so

2theta=(arcsin z/2)/2

[compact jasper]

Have a sum of two lines (A+B) of a triagle and all angles . How to find 3rd line? Stuck a bit... heron formula?

[frank condor]

Hi, can anyone help me, it’s a simple calculation, but I just don’t know why this is

Expand (2sinAcosA)sinA

How does sinA times 2cosA equal to 2cosA, what happened to the sinA?

[silent plank]

Have a sum of two lines (A+B) of a triagle and all angles . How to find 3rd line? Stuck a bit... heron formula?
you can apply sine rule and solve a system of equations to determine the lengths of A and B
and then cosine law to determine the final line

[upper karma]

Isn't sin(A)*2cos(A)=sin(2A)? Because of the double angle formula

[frozen garnet]

Probably quite a simple question, but

Two circles of radii 10cm and 15cm respectively are placed inside a square. Find the perimeter of the square to the nearest centimetre.

I've figured that the diagonal is 50cm

But I've got no idea how to find the sides of the square

[slow spruce]

How are they placed? If you know the diagonal Pythagorean theorem will tell you the side

[frozen garnet]

So if c=square root of a^2 + b^2

Well, it's pretty much c = a^2 + a^2 given they're even

But a is what I'm struggling to find out

[slow spruce]

a^2+a^2 are liked terms, so you can combine them

See if you can see how to solve it after

[frozen garnet]

So c=square root of a^4?

[craggy river]

No that's not how exponents work

Try again

[slow spruce]

a^2+a^2=a×a+a×a, which isn't the same thing as a^4=a×a×a×a, but it is the same thing as 2×a^2

[tribal sentinel]

if a point was a center of gravity of a triangle

how can i discribe that a straight is the center of h

come on help me on my advanced middle school geometry 😅

[nocturne remnant]

What’s a culmun 💀

[tribal sentinel]

u know

the thing

/

[nocturne remnant]

:/

[tribal sentinel]

i mean the straight

or the Height to be precice

[nocturne remnant]

You can make use of similar triangles

Excuse the bad quality

[tribal sentinel]

i don't care about quality

i just wanna finish my homework

btw i don't see how can i make use of those triangles :/

[nocturne remnant]

Uh

Can you see that they’re similar

[tribal sentinel]

they are both 90 ° ?

[nocturne remnant]

Similar triangles dude

Triangles with the same shape

[tribal sentinel]

ik they are and ... ?

[nocturne remnant]

Because they are similar, the ratios of corresponding sides are the same

Can you recall some geometry

[tribal sentinel]

recalls

...

squeasing his brain

[nocturne remnant]

In this case, the ratios of the two “bottom sides” are the same as the ratios of the two “right sides” of the triangles

Which we can use to solve the problem

[tribal sentinel]

i think we didn't learn this law yet :/

we are studing about tales theory

[nocturne remnant]

whaT

Tales?

[tribal sentinel]

no no we passed that chapiter

we r studing about tryingles and the center of gravity

[nocturne remnant]

Did you mean Thales’ Theorem

💀💀

[tribal sentinel]

yes

[nocturne remnant]

LMAO
guy really said “tales theory”

[tribal sentinel]

sooo funny

[nocturne remnant]

Well you should’ve learnt about similar triangles

[tribal sentinel]

i can't stop laughing ha ha ha

come on i have this shape on my note book rn

how can i prove that i is the center of [bg] ?

[nocturne remnant]

Awwwwww

Funny joke

But anyways

Added some letters

[tribal sentinel]

i don't mind

[nocturne remnant]

Well first thing is to prove that triangle FHK is similar to triangle FGI

FGI is easier I just realised

[tribal sentinel]

oh sorry man but this thing doesn't exist

i just got a raindom photo from google :/

[nocturne remnant]

Um

What is the problem that you’re given

A photo would be cool

[tribal sentinel]

how can i prove that i is the center of [bg]

[ea] is 1 cm ||a is the center of the triangle||

[eg] is 3 cm

[nocturne remnant]

Do you have a picture?

[tribal sentinel]

i don't have a webcam

[nocturne remnant]

Welp
Can you use a phone or sth

[tribal sentinel]

i only got some bad quality pictures

[nocturne remnant]

Hmm
So this is the diagram you’re given (except for the b and h stuff)?

[tribal sentinel]

exactly

[nocturne remnant]

Cool

I’m trying to think of an easy way to do this

[tribal sentinel]

the center of the triangle is called a

[tribal sentinel]

cool

[nocturne remnant]

Uh
Do u remember sth called “intercept theorem”

[tribal sentinel]

that math that i am studying is arabian math

so let me read about this on google

is this is what u r refering to so yes i did

[nocturne remnant]

Hmm still thinking of easy way

[tribal sentinel]

according to this what i am given is not logic

[nocturne remnant]

Why not?

[tribal sentinel]

it's 1/3 not 1/2

[nocturne remnant]

Yeh but EG = EA+AG

so the math does check out

[nocturne remnant]

Well
Idk, can we use this fact directly?

The 2:1 thing

Is that allowed

[tribal sentinel]

this fact can't be used it's just a prove that none of this is logic

it just make us struggling

[nocturne remnant]

It is logic tho
EG is 3 not AG is 3

[tribal sentinel]

yes

that's what i said

[nocturne remnant]

Ok I think I have sth

Since EA:AG = 1:2,
area of EAF:area of AGF is 1:2 aswell

[tribal sentinel]

ea:ag is 1/3

[nocturne remnant]

It’s not

Read the q carefully

[nocturne remnant]

But since EF:BF is 1:2,
area of EAF:area of BAF is 1:2 aswell

So combining the two we have area of BAF = area of AGF

[tribal sentinel]

what is the q ?

[nocturne remnant]

The question

I meant the diagram

[tribal sentinel]

i litterly used egx1/3=ea to prove that a is the center of the triangle

oh wait forget it

[nocturne remnant]

Lol?

[tribal sentinel]

i just had to read ur idea again

or u know what don't forget it

ea is 1 cm and eg is 3 cm

how ea:ag=1:2

[nocturne remnant]

ea:eg not same as ea:ag

:/

[tribal sentinel]

i used a ruler

ag is 2 cm

yes what u said is right

but how can that help ?

[nocturne remnant]

I’m drawing sth

Wait

I constructed some area argument

Wait I made mistake

Square brackets mean “area of”

[tribal sentinel]

ok

[tribal sentinel]

u made no mistake

[naive timber]

How do I simplify this

[somber coyote]

KaiML

$$\tan{\alpha}=\frac{f'(\theta)2\tan{\theta}-f(\theta)+f(\theta)\tan^2{\theta}}{{f'(\theta)+f'(\theta)\tan^2{\theta}}}.$$

[naive timber]

I want to show that

[somber coyote]

KaiML

$$\tan{\alpha}=\frac{f(\theta)}{f'(\theta)}$$

[nocturne remnant]

Bruh

[tribal sentinel]

i went to a bean break and i find out that my brother typed things :/

[naive timber]

o

[nocturne remnant]

Rip

[tribal sentinel]

i hate sibilings

[naive timber]

Any ideas on how i should simplify this disgusting fraction?

[dark oasis]

Hint: try taking common factors and using trigo identities

[naive timber]

would it be a good idea to multiply everything by cos theta?

[naive timber]

I dont see any common factors tho.

[dark oasis]

Observe nicely 👁️👁️

[naive timber]

uhhh

[naive timber]

?

[dark oasis]

It depends whether it will make your question simpler or more complicated

[naive timber]

well all I want to do is get rid of all the trig functions

kinda

[dark oasis]

Tho I don't think that would be a great idea

[naive timber]

ok

tho note that f is different from f'

[dark oasis]

Ikr

[naive timber]

i dont see any common factors

Im probably just dumb

[dark oasis]

I can see f' common in denominator.
And f common from some terms of num.

[naive timber]

okay...

give me a sec

[naive timber]

oh wait I see something

@dark oasis I get this

ok... I dont see how I can proceed yet.

[dark oasis]

^

[naive timber]

ok trig identities huh

pulls out document i made a long time ago because I can never remember trig identites

[dark oasis]

Lol

Well understanding>> memorizing

[naive timber]

true.

But remembering the identities is importent cuz you cant use them if you dont remember them

ok

so

tan^2(theta)+1

is

sec^2(theta)

um

[dark oasis]

Ye

[naive timber]

2 tan theta

is tan 2 theta * 1-tan^2 theta

umm

that seems like it complicates things

[dark oasis]

Yea yea

[naive timber]

umm

[dark oasis]

Open tan θ

[naive timber]

?

[dark oasis]

Tan theta=sin/cos

[naive timber]

riiight

didnt mean to reply dont get confused

[dark oasis]

[naive timber]

[dark oasis]

I see

[warm arrow]

I got a sin 2(thetha) lol

[naive timber]

im not supposed to get that AHHH

[warm arrow]

and a -cos2(theta)

I got the pattern but can't get the final

[naive timber]

hmmm

[dark oasis]

I m getting sin 2θ - f(θ)/f'(θ)

[naive timber]

My problem is the sin 2\theta

[dark oasis]

[naive timber]

DELETE

[tribal sentinel]

wait wut

👀

[warm arrow]

I'm getting -f(θ)/f'(θ) × (cos²θ) + sin2θ

[naive timber]

what.

[dark oasis]

Lmao

Can you recheck question once again

[naive timber]

okay.

Its a calculus question though

so

[dark oasis]

?

[naive timber]

its just the last step

[warm arrow]

I'm kinda wanna know your working @dark oasis

[naive timber]

I can post it here or in the calculus channel

[dark oasis]

Sure

Wait a sec

[naive timber]

@dark oasis do you want me to show you the question? I checked all of my work and it seems fine

@warm arrow could you tell me the steps you took?

[warm arrow]

I change the tan²θ+1 to sec²θ

[naive timber]

ye

[warm arrow]

I'm trying to text in LaTeX but I don't know how to type on discord haha

[naive timber]

you just uhh

text the way you would using latex

like $\LaTeX$

[somber coyote]

KaiML

[dark oasis]

$$\frac{f'(\theta)2tan\theta - f(\theta) sec^2\theta}{f' (\theta)sec^2\theta}$$

[somber coyote]

hoi

[dark oasis]

$$\frac{f'(\theta)2tan\theta}{f'(\theta)sec^2\theta} - \frac{f(\theta)}{f'(\theta)}$$

On simplyfing this we get ^

[somber coyote]

hoi

[warm arrow]

$$\frac{-f(\theta)(cos2(\theta))}{f'(\theta)}+sin2\theta$$

[dark oasis]

Yeah if u can confirm once

[naive timber]

huh

[naive timber]

wdym

[dark oasis]

Question recheck

[naive timber]

yea I checked it

[dark oasis]

Oh

[somber coyote]

skypirate

[naive timber]

oh WAIt

i think I found a mistake

shoot

[warm arrow]

Finally get the right one 😅

[naive timber]

oh i messed

up

AHHH

[warm arrow]

Ohhh boy

[naive timber]

Ok. If you two would give me a few minutes I think I might be able to sort this out

Shiiittt

[warm arrow]

I knew the question had something wrong 🤣

[naive timber]

[warm arrow]

Well I already spent so much time on this, better be worth it

[naive timber]

IM SORRY

[dark oasis]

No need for that, but tell us what sin you committed just now

[naive timber]

uhhh

I may or may not

have

done

[dark oasis]

So you did or didn't?

[naive timber]

$\frac{\frac{dx}{d\theta}}{\frac{dy}{d\theta}}$

uh

Yea

[somber coyote]

KaiML

[naive timber]

sooooooo

we dont talk about that

for now

[dark oasis]

[naive timber]

i screwed up big time

[warm arrow]

Well

Can you kindly send the whole question?

[naive timber]

ok

[warm arrow]

So that I can 'die' peacefully

[naive timber]

here

[warm arrow]

[naive timber]

fml

i got it

im so sorry

[warm arrow]

Well I don't get it, teach me

@naive timber

[naive timber]

basically

by making a mistake on the calculus step, i asked for help with the wrong fraction.

It is basically the same steps

but I just screwed up in the beginning

$$\tan{\alpha}=\frac{\frac{f:'\left(\theta \right)\tan \theta +f\left(\theta \right)}{f:'\left(\theta \right)-f\left(\theta \right)\tan \theta }-\tan \theta }{1+\frac{f:'\left(\theta \right)\tan \theta +f\left(\theta \right)}{f:'\left(\theta \right)-f\left(\theta \right)\tan \theta }\cdot \tan \theta }$$

[somber coyote]

KaiML

[naive timber]

^THis is the right fraction

@warm arrow

Then you multiply in the denominator

and then you use common denominator

simplify

again

then simplify more

and you get the answer

[warm arrow]

Okkkk

[naive timber]

i have so many things to rewrite

[upper karma]

idk is this relevant but does anyone know what this kind of shape is called?

[craggy river]

circles

Looks like an epicycloid if you sretch the definition pretty far, other than that I don't have anything

[upper karma]

okay thank you so much!

[craggy river]

Keep in mind that's not an epicycloid, it's just the closest thing I can think of

Glad to help regardless

[upper karma]

alright

[upper karma]

I need quite a bit of help with spatial geometry

since I am not that good at understanding some of its rules to say the least

[craggy river]

Like what specifically

[upper karma]

ok back
so how do I calculate the distance of a point from a line in space

[bright whale]

can I see the question

[upper karma]

alr brb

it's in french so I'll translate "we pose that A(1,2,-1) and D the line in which its parametric equation is the following :"
"a)determine the coordinates of point H the orthographic projection of A on D
b)calculate the distance of point A from D"

question b's answer is easy as it ties to a but then I don't understand how I can solve a so yeah

woops forgot to post it without the weird tilt

[upper karma]

here

[bright whale]

sorry cant help you

[upper karma]

fairs

[jade ember]

guys

is cos4 theta = 1 - 2sin^2theta

[delicate dove]

Sorry but this pic was taken from a CHROMEBOOK smh

[slow spruce]

The triangles are similar (why?)

recall what that means about the sides

[jovial pendant]

What do you need help with

[delicate dove]

I tried getting help already and just decided to ask a teacher on Tuesday

Dont worry bout it

[jovial pendant]

Pretty simple problem I can help rn, do you just not understand where to start?

[delicate dove]

Heres what I tried

Quadratic formula

And we have to answer in radical form

[jade ember]

use quadratic formula

[indigo forge]

hello

can someone help me with calculus hw?

[dark sparrow]

-> #calculus
-> post your problem(s) there

[indigo forge]

my b I meant trig

I got 135/2 for all of these but they are incorrect and I need help

[dark sparrow]

you entered 135/2 into each of these answer boxes?

[indigo forge]

For the first one I put sin (135/2) and so forth

[dark sparrow]

so you did not actually calculate the sin, cos and tan of 135°/2 as you were asked to do

[indigo forge]

That’s what I came out with

[dark sparrow]

so you did not actually calculate the sin, cos and tan of 135°/2 as you were asked to do

you were asked, "what is sin(135°/2)?" and you answered "it's sin(135°/2)"

[indigo forge]

No I was asked what is sin(67deg 30deg) and I came out with 135/2

[dark sparrow]

first off, 67°30' is not "67deg 30deg"

second, as i keep trying to tell you, you did not do what was asked of you

[indigo forge]

Even when trying to do it like that I cannot figure it out

I get a decimal and this does not take a decimal as an answer

[dark sparrow]

i mean yeah when you're asked for an exact value of course it's not gonna take a decimal as an answer

that's not something that should surprise you

@indigo forge are you able to at least state the half-angle formula for sine?

[indigo forge]

No I am confused with with the entire problem

[dark sparrow]

$\sin^2(\theta/2) = \frac{1-\cos(\theta)}{2}$

so this is foreign to you?

[indigo forge]

Yes

[somber coyote]

Ann

[dark sparrow]

my bad, i screwed it up

but if you don't know even this formula then you cannot do the problem

[indigo forge]

Yeah no I do not

[dark sparrow]

then you do not have the requisite knowledge to do this problem

[royal citrus]

Hey

what is another way to prove parallel property of two lines

without showing that they have alternate angles?

[nocturne remnant]

For angle properties you have

• alternate angles
• interior angles
• corresponding angles

[upper karma]

If two lines have a transversal which forms alternative interior angles that are congruent, then they're parallel, could try that

[nocturne remnant]

If you’re looking for non-angle properties the only easy thing I can think of rn is parallelogram

[royal citrus]

oo

[royal citrus]

well not exclusively looking for non angles just something beside alternate in case i get stuck while trying to show two angles alternate

[upper karma]

hallo didnt go to a class once we learnt this, can anyone explain what to do?

[silent plank]

for the first one, start by drawing a diagram

the fact that B is 90° makes the problem considerably easier (and you should be able to see that a famous theorem would be applicable here(

and for the second one, apply cosine rule

[silk jacinth]

isn't this just plain wrong?

I keep on getting 5.23... when calculating this with SpeedCrunch

oh nvm

I kept squaring and not cubing that 1.5

[dark abyss]

isnt cd 3.26496554346 becuase my math program says that isnt right

nvm

[upper karma]

Side CD?

[upper karma]

yeah it is bc pythagoras theorem

[upper karma]

yeah 3.2649... is right bro

[gleaming nova]

no need to include all the decimals

just write it as a square root

[abstract fog]

there is something in my math class I love to call

the unexplainable go fuck yourself bullshit

this is the part of the math where the notes don't do shit to help you and it exists for the sole purpose of making you stress out

for example this :)

fuck you

you know why?

because we could easily solve the 45 degree by doing 2.4 minus 0.9

but nooooooooo

lets throw this bull shit of square 2 even though 2.1 is supposed to take care of that

because you know why?

fuck you thats why

[upper karma]

I mean simpler maths questions are written in a certain way to include things that may throw you off, the simpler questions test your interpretation and application more than your actual knowledge, igy tho

[abstract fog]

getting stuck

mentally hurts me sometimes

and I have a hard time copping

because sometimes asking for help does not work

and sometimes looking online for help does not work

and if it does work sometimes it can take too long and waste tons of time

while everyone else in the class can do it in one shot

[upper karma]

everyone's different, those people who can do one thing in one shot probably get stuck on something else for a while - I have asperger's and LLI so I'm basically wired to be obsessed with numbers and sequences and I still get stuck sometimes, it's normal bro don't stress about it

[dry python]

yo

what shapes make the faces of the 3d shapes of a triangulaor pyramid

help pls

[craggy river]

It's a pyrimid, so the "sides" (everything but the base) will be triangles

And it's a triangular pyramid, so the base will be a triangle as well

[dry python]

ok

i need to need the shape

like how much triangles

and how much shapes are there

[soft marten]

i dont understand trig

[grim plank]

Figure out the angle between the diagonals.

[nocturne remnant]

You already

Sent that

[vagrant ore]

Oh sorry I meant this one

[hollow juniper]

lmao I don't know what formula my prof used for the volume

but like am I just not seeing it XD

[viral hamlet]

is this solvable by squaring both sides? is this even solvable or m just tripping ?

[dark sparrow]

it is very easy to see that this has no solutions

cos(3x) + sin(x) will always be between -2 and 2 no matter the value of x

[viral hamlet]

okay then M fr tripping.

OMG THIS SUBJECT IS CANCER 😭

@dark sparrowty btw

[storm portal]

This is because:

-1 ≤ cos x ≤ 1 for all real x
-1 ≤ sin x ≤ 1 for all real x

[silk iris]

Use sohcahtoa and plug the sides in

[silk iris]

Use sohcahtoa to find the sides

So sin is opposite/hypotenuse, cosine is adjacent/hypotenuse, tangent is opposite/adjacent

[sour void]

or remember the unit circle and remember that sin v = y and cos v = x 😏

[soft marten]

what do the arrows mean

[sour void]

probably that they’re “parallel”

To each other

[finite cove]

Can anyone help me with this

I’m super lost

[quasi galleon]

i need help with number 13

[grave pond]

There are several things you're asked to do. Which of them do you have trouble with?

[upper karma]

Would a scale factor be considered a scalar?

[honest nest]

anyone understand circle theorems?
i have a challenge to make a circle where all 4 main theorems are applicable to solve an unknown angle

[void atlas]

try writing 17pi/12 as (pi+15pi/12)

[void atlas]

use trig identities sin(pi-x) = sin(x) and cos(pi-x) = -cosx

[finite cove]

@void atlas my teacher got 0.268 idk how

[upper karma]

is there a trick for counting the number of interior triangles that share a vertex with a regular octogon

[dark sparrow]

"is there a trick for <problem description that is just specific enough to not be general yet too vague for there to be anything of use to say>"

[upper karma]

??

how can i explain it better

[silent plank]

with the aid of a diagram

[dark sparrow]

^ and/or the exact problem statement

[grave pond]

Are you sure you don't want triangles that share all their vertices with the regular octagon?

But even so, what exactly do you mean by "interior"? Is the triangle allowed to share one or more edges with the octagon?

[clever kiln]

um guys whr do i post my probability doubts?

[grave pond]

#probability-statistics if they are conceptual misgivings.

#❓how-to-get-help if it's "I doubt I can figure out what to do with this homework problem".

[upper karma]

A regular octagon is given. How many triangles are there whose vertices are the vertices of the octagon? [Two congruent triangles at different places are considered distinct; that is, ABC and BCD count as distinct.]

[dark sparrow]

is octogon spelled with an o?

in any case, this is combinatorics, not geometry.

[upper karma]

lol

[upper karma]

oh

i wll ask there?

[soft marten]

how do i get the side lengths when none of them are the same except hypotenuse

[signal swallow]

hypotenuses aren't the same on those two triangles

similar triangles means their ratios are the same

[soft marten]

yeah but which ones do i cross multiple

nvm

[soft marten]

how is this wrong

[craggy river]

That's very not to scale

[grim plank]

cus it isnt 60

[soft marten]

y

[craggy river]

Go to like geogebra or something and make a rhombus and its diagonals. Move stuff around a little and you should see what the right angle is

[grave pond]

sees what you did there

[soft marten]

why is it not 60

[grave pond]

Do you know what a "rhombus" means?

[soft marten]

wdym

CED is triangle

180/3 is 60

60/6 = 10 but apparently not

[craggy river]

Where is 180/3 coming from

[grave pond]

It is a triangle, but nobody said it was an equilateral triangle.

The key word in the problem statement is the claim that ABCD is a rhombus. You can't reach a solution without considering what that word means.

[craggy river]

^ don't rely much on the picture, it's misleading

[soft marten]

idk what to do

[grim plank]

draw your own one

[grave pond]

I will ask once again: do you know what the word "rhombus" means?

[soft marten]

no

not rlly

[grave pond]

Your first step is to look that up, then.

(It's a bit of a douchebaggy problem if you don't have a discussion of rhombuses somewhere in your textbook).

[soft marten]

i dont use textbooks

do diagonals always intersect at 90 degrees

or not

wait so its 15

90/6

i dont understand the diagonals

and length

[grave pond]

That's another property of rhombuses you need.

(Which is, by the way, shared between all parallelograms).

(And yes, the diagonals in a rhombus always cross in a right angle).

[soft marten]

do I multiply the SW to equal RT

[craggy river]

Trust the picture for this one. What does the ratio between RT and SW look like?

[soft marten]

SW is half of RT

[craggy river]

Yep

Use that

[soft marten]

so 2(SW) for rt

how ??

[nocturne remnant]

Kite
Name: doug

[trim gust]

yo I need help with geometry

Im trying to reduce the size of a square plaque from 4.5 inches to 4inches and I was wondering what radius the chamfer would have to be to make it work

I can show yall what Ive started if thatd help

this is the problem essentially

[wintry tundra]

so the distance from the top of the circle to the tip of the triangle is 0.25?

[trim gust]

yes Im looking for the radius

[wintry tundra]

do you know that R value?

[trim gust]

no

[wintry tundra]

is that line supposed to be the diameter

[trim gust]

Can I voice call and share my screen Ill be able to explain it better

[wintry tundra]

hmm

i cant really voice call rn

[trim gust]

Im trying to lower the height by making arcs

well lower length ig

but I need the radius of this arc for the software to understand

[wintry tundra]

im trying to understand

so why would chamfering a square plaque do what u want

[wintry tundra]

because then its not a square

[trim gust]

this here is 4.5

[wintry tundra]

so the diagonal

[wintry tundra]

so now ur making a curved square

[trim gust]

Im trying to add chamfers arcs to make it 4 inches but i need the radius of the arcs

[wintry tundra]

so the area will reduce to 4 inches