geometry-and-trigonometry

[grave pond]

Oh duh, power of a point strikes again.

Yeah, makes sense. 👍

[smoky jetty]

thats pretty clever

didnt realize the tangent&secant line till u mentioned it (not as complicated unless u do it the other ways ig)

[fallow marsh]

Is it ICSE?

[river abyss]

What do I have to know about circles. My teacher ran out of time to teach it this year and she told us to teach ourselves. But I don’t know what I need to learn. For reference I was taking Honors Geometry. Thank you!

[grave marten]

help

my midterm is tomorrow

[silent plank]

!help

[lime crown]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[fallow palm]

Use power of point and win

[compact tangle]

Rhombic dodecahedron

[mortal thistle]

how was this solved?

[lucid rapids]

that actually looks vile

[mortal thistle]

nah its kinda easy

somwhat

[rocky widget]

helppppp

[mortal thistle]

im to tird

do tis

[mortal thistle]

first part here

[vernal pilot]

<@&268886789983436800> this person has opened 3 help threads and posted the question in nearly every channel.

[neon prairie]

Albungus?

[vernal pilot]

Yeah

[neon prairie]

Please keep your questions to a single channel at a time.

I usually give it an hour or so if I post a question and get no responses.

Sometimes people don't understand your question or don't find it interesting or whatever. It can be a little hit or miss at times.

It helps to try your best to make your question as easy to read/understand as possible too.

[lime dune]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[polar belfry]

can anyone help me with melaneus theorem of triangles

because i am confused with the sign of =-1 and =+1

https://en.wikipedia.org/wiki/Menelaus's_theorem

[obsidian hornet]

Just count in directional segments

[polar belfry]

But how

[obsidian hornet]

Try to use this on theorem

[undone quest]

All subjects r sh*t but maths is their father

[upper karma]

Silly question but the way they wrote "triangle ABC is similar to triangle ADB" makes me think that AB in the first triangle is exactly the same as the AB in the same triangle, right?

no i mean

i'd literally write out AB/AD = AC/AB right?

[undone quest]

What's this

?

Btw

[upper karma]

but shouldn't the bolded AB be different from the regular AB cuz well they're 2 different triangles, no?

[upper karma]

are green and red supposed to be the same thing?

cuz they used the same letters otherwise they could've changed it?

hmm i never seen it like that

which is why i was curious if that means to

like consider ADB inside ABC

then there's no "conflict" well unless you physically separate out the triangles

wot

they're definitely not congruent lol

AB is like 6

if they were congruent triangles then AC = AB = 4

u need all three sides to be congruent for SSS congruence condition to hold

anyway

with only one side? surely not

unless u have two other angles

okay anyway so like anyway i'm just wondering why they reused letters

why not just introduce new ones?

right sure

no i know how to solve it

it's super easy

that isn't my problem

[upper karma]

?

well okay sure i guess

[gloomy eagle]

wha..

u can use angle size angle

-> congruent

[upper karma]

read the context

besides i did say that

[upper karma]

check this

[gloomy eagle]

i dont understand what u guys were saying 💀

[upper karma]

the message right after that lol

[gloomy eagle]

oh sry my very bad 🤟

[upper karma]

well we were just saying that if you have that two triangles are similar

and one corresponding side on both triangles are the same -> the ratio is 1

then those triangles are congruent

since all congruent triangles are similar by virtue

in which case you only need one side ratio between the two figures

[gloomy eagle]

[gloomy eagle]

It does

The way to write similar triangles

Is to write by the corresponding sides

Which was exactly what the question did

[plucky abyss]

[plucky abyss]

Good geometry question

[kind rapids]

Why do we need to prove a triangle is a triangle?

[sturdy stone]

We don't

[undone quest]

Looking for solution

[grave pond]

There's not information for that.

You can pick any value of x between 0 and 180-43, and then there will be a matching y,

[solar hare]

is 43 here the whole angle or a divided angle?

[plucky abyss]

Left to the reader

[solar hare]

hmm?

[kind rapids]

I guess it is a divided angle

Looking at the diagram it is obvious that's definitely 43

[solar hare]

hm

not enogh info

ig

[viral path]

i solve it

give me a second

x = 94; y =43

[grave pond]

Why?

We could just as well have, for example, x = 90, y = 47.

[grave pond]

Those angles don't give a parallelogram. The angles of the quadrilateral according to your numbers are 86°, 86°, 94°, 94° with each 86° being opposite from a 94° -- but in a parallelogram opposite angles are equal.

[viral path]

but isnt it?

[grave pond]

I'm talking about the angles in the outer parallelogram.
With the lettering in your later post, you're claiming than angle DAB is 43°+43° but angle BCD is 51°+43°.

[mellow quest]

maybe
x < 137
0<y<137 ?

[grave pond]

But angle DAB and BCD should be equal in a parallelogram.

[viral path]

ohhhh ok i get it, i just gave more attention to the triangles than the parallelogram

but it was worth the try

[jade flax]

hooo

[plain scarab]

Is this the first 2 images?

I know 2nd is right idk about the first

it’s either between the first or last

sorry

the question says which two pictures show a circumcised circle

[rough crater]

Am i misunderstanding this question? My answer is 338

The book says 192m

[undone quest]

This question has rot my brian

Trigo and calculus better!!

[untold relic]

how to proceed

[sour sand]

Find the maximum heights of 5 distinct fireworks online and record them. You need to determine how far away the audience has to sit from the point where the fireworks will be lit. To help you do that, you will make a detailed list of the angles of elevation the audience will have to use to see each firework if they are sitting 50 feet away. Assume that the fireworks are shooting straight up so they form a 90° angle with the ground. If they need to look up at an angle that is more than 75°, increase the distance from the launching site to the audience until they can see the firework using a 75° angle or less. Record your list and the distance from the launching site to the audience in your plan. Remember that the audience should be sitting at the farthest distance you found while trying to get them to see the fireworks by looking up at a 75° angle or less.

You should always be prepared for possible problems. One risk in the firework show is that the fireworks might not shoot straight up. Make a list of the distances between the audience and each type of firework if it didn't shoot straight up and instead formed a 68° angle with the ground.

[solar hare]

hmm

lets take 5 fireworks

whose max heights are

A = 600 ft.
B = 800 ft.
C = 1000 ft.
D = 1200 ft.
E = 1500 ft.

Distance for 75 degree elevation for each firework is:

tan(θ) = height/d

so,

d = height/tan (θ)

calculate each height for this elevation where,
d= distance and θ= 75 and height = max heights for each firework

now

to calculate for 68 degree angle with ground

use the same formula as above :

d = height / tan(θ)

where θ = 68 in this case

comparing the distances for 75 degree elevation

and hoping that you calculate the distances and estimates correctly

and in case fireworks are not shot straight up

they can sit on the corresponding distances

[undone quest]

How's the angle beside 43 is 43

[silent plank]

its not, illegal math was done

[undone quest]

Yes

That's why

From 2 hours I am figuring it out

The 43 beside black colored 43

[obsidian harness]

isn't this question unsolvable tho

[silent plank]

as mentioned, it's not solvable

[obsidian harness]

oh thanks

[undone quest]

Oh

Still my math teacher

Figured it out. -.

And fooled us

[silent plank]

well they must've done some invalid math

[undone quest]

Any tips on how to rock geometry?

[grave pond]

It's easy to find some solution -- such as, as I proposed, x=90°, y=47°. It just won't be the only possible solution.

[undone quest]

My question is

How angle beside 43 is 43

[obsidian harness]

it's not

that person was wrong

[undone quest]

Prove it

[obsidian harness]

like this

that's also possible

[undone quest]

If it is 43

Then x ,y has solution 90 and 43

[grave pond]

Then x=90° alright, but y becomes 47°. So it's possible for that angle to be 43°, but it is not necessarily 43°.

[undone quest]

Sorry 47

Type mistake

[grave pond]

We could also have, for example, x=47°, y≈61.8°.

[undone quest]

And x is also not necessarily 90

There is a no of solution

[grave pond]

Angle BDC will be 43° if and only if x is 90°.

[undone quest]

Yes

[grave pond]

(x determines y but not the other way around. Note that y=47° can go with either x=90° or x=94°, corresponding to the special cases of the parallelogram being either a rhombus or a rectangle).

[solar hare]

yep

[undone quest]

Ramunajan summation s=1-1+1-1+1-1+1.....infinity

What will be the answer for this

Solution saythat 1/2 is invalid

[plucky abyss]

Wdym .... infinity

[undone quest]

Extends to infinity

[plucky abyss]

I see

[rough crater]

Bump

[jade sphinx]

How do we know that F has the same angle with regards to the elevated plane, as the elevated plane has to the ground?

[grave pond]

Is F defined in the problem to be horizontal?

[undone quest]

You troposphere

[jade sphinx]

Yes

[undone quest]

Is this related to force?

[jade sphinx]

I mean this is the actual problem I am having;

I have no idea, why I can assume the angles are as described in the teacher's notes (second picture).

[jade sphinx]

Yirh

[undone quest]

And friction

[jade sphinx]

No. Just the geometrical stuff.

[undone quest]

Vector?

[obsidian harness]

cause of the normal vector to the plane, that's the component of weight perpendicular to the slope

and by Pythagoras the component of weight parallel to the slope is mg sin theta

[undone quest]

U have to study invarience in vector

[jade sphinx]

Ehhh, you got a link or something?

Like to the rule.

[undone quest]

See this u will understand

[obsidian harness]

https://www.youtube.com/watch?v=ufgY237M5KQ

[jade sphinx]

Alright cool, thanks! 🙂

[obsidian harness]

nwnw hope it helped!

[rough crater]

Anybody mind checking this?

[plucky abyss]

It's lengthy

[frosty swift]

@jade sphinx

[frosty swift]

Calculate the angles which I have marked

[hollow edge]

helpp

[acoustic ravine]

Ok

So outline the planes as shapes right

Wait nvm im mis understanding

If you outline the planes it’ll become transparent

[trail tendon]

if you extend EAB and DCG into two dimensional planes, the planes are parallel so theres no intersection, they put in #prealg-and-algebra

[acoustic ravine]

That’s what I was trying to say

I’m just exhausted and couldn’t articulate the words😭

[trail tendon]

yea 😂

[acoustic ravine]

😭😭

[plucky abyss]

Everything good

[upper karma]

Can someone explain

Why no solution

[faint pasture]

well sin(x) = 1/sqrt(2) implies that x = 2npi + pi/4 or 2npi + 3pi/4
however this implies that 2x = 4npi + pi/2 or 4npi+ 3pi/2
butt for both these values tan is not defined

hence there cant be any solutions

[undone quest]

The height of triangle is 4.3
(equilateral)
And if we add the length of perpendiculars drawn from it we get the same length as height

Can anyone tell me how's is this happening ----

Wow I made a new theorem. -.

[obsidian harness]

congrats, yuo rediscovered an existing theorem

https://en.wikipedia.org/wiki/Viviani's_theorem

[plucky abyss]

No problem

[undone quest]

Wtf

I thought it was mine -.-

[obsidian harness]

there are 8 billion people alive yk that right

and 2500 years since Euclid

[faint pasture]

We all have that moment one day or the other lmao

I thought the fact that sum of consecutive odd numbers is a perfect square was something i discovered

[lime dune]

LMAO same

😭😭

tbf i was in like

fifth grade at the time

[faint pasture]

I was in 8th but okay 💀

The fact that my textbook had me prove it like the next week

[onyx tiger]

we all laugh till they proved riemann

[trail tendon]

ah yes, i loved that day when they proved Bernhard Riemann

😭

[onyx tiger]

[trail tendon]

guys remember when they proved euler

fire bro

[shy moss]

which one

💀

there are quite a few theorems

[trail tendon]

exactly 💀

[onyx tiger]

the 1 million one

[shy moss]

the hypothesis?

its not proven yet

[onyx tiger]

yeh exactly that's why I said we all laugh till they proved that one(it's actually their own discovery and not found by some ancient babylonian 2000 BC)

[trail tendon]

you mean we all laugh until they prove that one

proved implies they already did 💀

XD

[plucky abyss]

[green current]

can someone teach me how to do this i literalyl did it like 5 minutes ago and then forgot

[south nymph]

find ratios of the sides

if they're equal they're similar

[green current]

so like

[south nymph]

15/10 and 10/8 are not the same

therefore these polygons aren't similar

[green current]

OHHH THGANBKJ YOU

THANK U SM

I KNEW THAT WQAS IT BUT I WAS DOING IT ONT HE WROGN SIDES

[south nymph]

no worries :D

[frozen hatch]

need help with 13

[mortal crow]

Welp, geometry trauma is coming back to me 💀

Oh, I don’t need help by the way

[flint shale]

For these, as they are congruent you can make two equation with the corresponding sides

[grave pond]

What is "HL", though?

[flint shale]

i'm not sure

i thought at least you could just make the equation

[grave pond]

Yeah, that would be my best guess too.

[flint shale]

mhm

[trail tendon]

hypotenuse-leg theorem apparently

[flint shale]

Oh

ok

is that another way to proove that two triangles are congruent?

[frozen hatch]

Wdym?

I don't get it

[flint shale]

Do you know what the HL means?

[frozen hatch]

Yeah

[grave pond]

I think that'd just "if the hypotenuses are equal and a leg in one triangle equals a leg in the other, then the right triangles are congruent".

[flint shale]

That sounds about right

ohhh

i just realised in austrlia we use RHS

[frozen hatch]

How do I setup the algebraic equation though

[grave pond]

There'd be another form of congruence where the two known legs are different legs -- but with the expressions here that doesn't seem to lead to positive lengths.

[flint shale]

you would just want to get the corresponding sides

[frozen hatch]

Equal them to each other?

[flint shale]

so y-x and x+5 are an example

[flint shale]

yes

[frozen hatch]

You can't slove a equation with 2 variables tho

[grave pond]

You have two equations with 2 variables.

[flint shale]

yeah

[frozen hatch]

Then I equal them and slove for x and y correct?

[grave pond]

Probably yes, but you're expressing it in a funny way.

Can you show the equations you're going to solve?

[frozen hatch]

This

Sorry for me acting dumb

My teacher didn't teach this to me

[torn yoke]

x+3=3y and y+1=x

make that into a system of equations

basically in HL the hypotnuse and one leg should be equal to each other

[rough crater]

"an angle whose radian measure is theta is subtended by an arc that is the fraction theta/2pi of the circumference of a circle. Thus, in a circle of radius r, the length s of an arc that subtends the angle theta is s = theta/2pi * circumference of the circle"
What is theta / 2pi?

Idk why Im having trouble with this wording

theta/2pi is the arc? and s is the length of the arc?

[torn yoke]

@winged cradle do you get it?

opps wrong adam 💀

[torn yoke]

ngl i never learned this

at least ive never seen the word theta

[rough crater]

or is it saying the arc is theta/2pi of the circumference of the circle

theta is just an angle

I'll wait for someone

so its kinda like saying the fraction angle/2pi of the entire circumference is our arc. And if we do 2pi/2pi of 2pi*r, we just get the circumference

I think it just made sense to me, maybe I'll wait for confirmation

[frozen hatch]

Yeah tsym

[winged cradle]

bruh

[torn yoke]

mb 😭

[radiant spruce]

is there a algebric way to prove pythagorean theorem?

[trail tendon]

is there an algebraic way to do a geometric proof?

uh 💀

[radiant spruce]

that's what im asking

[trail tendon]

a^2 + b^2 = c^2 works for right triangles

if you remove the geometry part (the right triangle part), then the algebra part is just a^2 + b^2 = c^2

which is kinda meaningless without the geometry 💀

[radiant spruce]

ok i see it now

i need to find a way to prove pythagorean theorem for a school project 💀

[trail tendon]

can you use law of cosines 😂

[radiant spruce]

uhhh

idk

[woeful pagoda]

you can use the big square (a+b)²

and the little inside c²

then (a+b)²- 4*(ab/2) = c²

[radiant spruce]

idk which one is the hypotenuse bro

in this case is c the hypotenuse?

[frozen hatch]

Does sas apply to right triangles?

[upper karma]

It applies to all triangles

[frozen hatch]

Alright thanks

[plucky abyss]

No problem

[faint comet]

Can someone explain question 4 for me please

[upper karma]

let matrix be

A B

C D

A(1)+B(2)=4

C(1)+D(2)=4

A(-1)+B(3)=1

C(-1)+D(3)=6

[runic turtle]

Is my diagram right for this trig question

And if yes how do I find h

[obsidian harness]

that's correct

so if the foot of the perpendicular from C to line AB is called D

then let AD = x, and hence B = 50 - x

[runic turtle]

Wait

[obsidian harness]

ok

[runic turtle]

Is that right

The do I make both equal to h

Then make those equations equal to find x?

[obsidian harness]

yes that's correct

[undone quest]

(x+1)(x-2)>0

Find possible value of x

[grave pond]

Divide according to the possible signs of x+1 and x-2?

[undone quest]

Umm

Is it so

[plucky abyss]

There are many possible values but since you want just one 3 is the answer

[obsidian harness]

draw a U shape as your quadratic is concave up (you have x^2 so that is positive, not negative)

and then greater than 0 means it's not in between the roots

so just x < -1 and 2 < x

[plain scarab]

What are these graphs called?

[sonic bramble]

3d vector graphs or SVGs or just 3d graphs

[noble knot]

Whats the difference between (angle of ABC) and (angle of BAC) when identifying the adjacent and opposite side?

[silent plank]

adjacent and opposite sides are relative to the position of the angle you're using

[noble knot]

Let's say you have a triangle which has the following sides: AB=13, BC = 5 and AC = 12, the question is what's sin(angle of BAC)

[silent plank]

it would be helpful to first draw a diagram

[silent plank]

can you identify which angle <BAC is refering to?

[noble knot]

Not sure, but this is the problem I'm refering to

I just don't get the angle notation here, because if it was (angle of ABC) the solution would be different

[silent plank]

<BAC is the acute angle formed between segements BA and AC

the angle is at the letter in the middle

[noble knot]

That means that the vertex is A?

[silent plank]

yes

[noble knot]

Oh, so the angle would be referring to A, meaning that sine is sin(5/13)?

[silent plank]

no

bad notation

trigfunction(angle) = ratio

sin(<BAC) = 5/13

[noble knot]

Alright, so it would be arcsin(5/13)?

[silent plank]

wdym by it

[noble knot]

To calculate the angle

[silent plank]

to calculate <BAC, yes

but you aren't asked for that

[noble knot]

Just trying to find a connection between the two.

Anyway, thanks for the answer.

[kindred lily]

can anyone help me

[kindred lily]

please

@silent plank

[pliant kettle]

You can subtract DF from both the sides

The other angles will the same coz linear pair

And BD = DF coz isosceles triangle

Now you've got 2 sides and a angle to be equal

Your proof will be completed

[pliant kettle]

AF = DF
Subtract DF on both sides
AF - DF = DH - DF
AD = FH

[kindred lily]

oh thanks

i figure that one out

i need help on this one now

is this right?

so far

do i subtract jc from sc?

then subtract jm-km

?

[pliant kettle]

Wait

You subtract SC from JC

[kindred lily]

ok

[pliant kettle]

And BC from CD

[kindred lily]

CD-BC

i word it wrong mb

[pliant kettle]

So thats one side equal

Another is adding KM

To the base sides

[kindred lily]

i write km and km is reflexive

[pliant kettle]

So now we've got 2 sides and an angle as equal

[kindred lily]

wait

why do i add km

[pliant kettle]

Why not just write addition to get the answer

[kindred lily]

ok

[pliant kettle]

JK = MD
JK + KM = MD + KM
JM = KD

Btw angleCJD and angleCDJ are equal

Isosceles

[kindred lily]

how do i prove that

oh nvm

[pliant kettle]

good

[kindred lily]

does everything have to be in order

except the prove

[pliant kettle]

Side, angle, side

That's the usual order

i.e 2 adjacent sides and the angle between them

[kindred lily]

ok

[pliant kettle]

Alright

Goodluck

[kindred lily]

thx

[plucky abyss]

[pliant kettle]

[kindred lily]

😐

[pliant kettle]

Ok how do I explain

Basically the long triangles are congruent

[kindred lily]

what should my frist step be

[pliant kettle]

Addition of angles

Prove that BDM is congruent to ACM

You got it?

[kindred lily]

ok

[pliant kettle]

Ok

[pliant kettle]

Hope you got what I meant

[kindred lily]

bro wtf do i do

i just added

1 and 3

2 and 3

[pliant kettle]

Lol

What's 3

[snow karma]

what is

3

lmao

[pliant kettle]

1 = 2
1 + AMB = 2 + AMB
BMD = AMC

[pliant kettle]

After adding the angles as mentioned you'll get that the angleA and angleB is equal

So all 3 angles equal

Congruent

Then then correspondence with parts

You get the answer @kindred lily

[snow karma]

or js DM = MC

cause of bisector

[pliant kettle]

Oh yes

I didn't see that lol

[snow karma]

js hope mac understood lol

thats ur goal

[pliant kettle]

Ayoo don't pin it on me 💀

[snow karma]

all u

[pliant kettle]

No problem

But lol

[pliant kettle]

Cat pfps are as wild as ever

[snow karma]

i was forced

[plucky abyss]

[sacred oar]

Is this 5?

[pliant kettle]

You've been on roblox for 16hours!?

[snow karma]

no, the game menu is js kept open

for 16 hrs

[pliant kettle]

Sed

[plucky abyss]

[sacred oar]

<@&286206848099549185>

[obtuse field]

yeah (you can check your answer on a graphing calculator like Desmos)

[sacred oar]

Yeah I just got confused with what a root means

I figured it just meant solutions

Got one more trig

[obtuse field]

ah yeah that's correct, it just means solutions

[pliant kettle]

How do you find no of solns then?

[sacred oar]

You get a total of 5 solutions

[pliant kettle]

How'd ya get sinx = 0?

[sacred oar]

See the fourth line

I form a quadratic (kind of) equation

[pliant kettle]

Oooh

I just cancelled out the sin at the start lol

Well yes I do get it now ig

[plucky abyss]

[pliant kettle]

[plucky abyss]

Sin cos tan I am trigonometry #1 fan

[pliant kettle]

LMAO just wait

[snow karma]

E = D * I and H = D * B and DB = 2DC

If R is the rotation by angle pi/2 around the center H, how do i prove R(I) = E?

[obsidian hornet]

If i understood correctly, then <DCH=<DIH=45, and EH//BI=>IHE=90, so, problem solved.

[snow karma]

how did u find that those two angles are equal to 45 degrees

[obsidian hornet]

DB=2DC

[snow karma]

uh ty and what else

[obsidian hornet]

DC=DH

[snow karma]

so DCH are DIH are equal angles

?

[obsidian hornet]

DCHI lie on one circle

With diameter CH

[snow karma]

ahh two right triangles sharing same hypotenuse

ty

[obsidian hornet]

Need to keep explaining?

[snow karma]

no tyy

i believe u wanna say EHI is right isoceles in H so its a rotation

[obsidian hornet]

By the way, id like to share some notice with u: if u see a quadrilateral with two same angles like this, so, It is worth reacting instinctively that 4 points lie on one circle.

[snow karma]

oh so in our case its angles HDC and HIB

alright thank youu

nvm its another angle outside the quadrilateral without a name

[maiden brook]

or more commonly that the opposite angles are supplementary

[abstract apex]

I'm on unit circle rn and this isn't making any sense

Like I get it that it has a radius of 1 unit but how does that help when all of it is in raidans?

[grave pond]

What is not making any sense?

[opal ibex]

I think the unit and the radians are separate things.

Hello everyone, I am new here and having trouble working out this formula here. Particularly the arithmetic within the parentheses. I can swear that I am doing it all correctly with PEMDAS in mind each time I determine theta, but each time I compute the radian, I get the wrong answer despite following the exact given steps. If anyone might be able to help me resolve the issue, it would be much appreciated. Thank you.

[opal ibex]

I think I figured it out. Typo in the system :P

[plucky abyss]

Congratulations

[opal ibex]

Thank you!

[remote creek]

I sound goofy for asking this but is the period for secant and cosecant both 2pi?

[snow karma]

yep

[remote creek]

ok thank you

[plucky abyss]

Sin cos tan I am trigonometry #1 fan

[faint comet]

Can somone please Tell me why this function has no VA?

[plucky abyss]

Because it's invisible

[upper karma]

What is pemdas

[undone quest]

(Parentheses exponents multiplication division addition subtraction ) just like bodmas

[plucky abyss]

[upper karma]

how much of what is drawn on a geometric diagram is to be taken at face value?

For example, if i see two lines that kinda look like they're pointing in the same direction then i should not assume that is the case instantly right?

https://cdn.discordapp.com/attachments/238956921581862913/1263148677090902026/image.png?ex=66992e60&is=6697dce0&hm=91a8ffa99503e034f4c6737e304a4817df145fb4711506835d9115927999911f&

In specific, how to know that RZ is an extension of RQ?

[jade flax]

yo need help?

[jagged wyvern]

How do you memorize the unit circle

[wicked egret]

with 30, 45, and 60?

[jagged wyvern]

That part that gets me is the radians

I cant use a calculator

@trail tendonIm desperate

[trail tendon]

yo?

[jagged wyvern]

I have a quiz on the filling out the unit circle

[trail tendon]

pi/6, pi/4, pi/3

[jagged wyvern]

Without the use of a calculator

[jagged wyvern]

What about the other quadrants?

How do you determine that

[trail tendon]

multiples of these

[jagged wyvern]

Multiples?

[trail tendon]

2pi/3

5pi/3

ect.

and you just gotta find out which ones are bigger kinda

[jagged wyvern]

Its really difficult

[trail tendon]

its really not

[jagged wyvern]

Is there like a special way

[trail tendon]

you're probably just unused to it

[trail tendon]

i think theres like a finger trick or smthn 💀

[jagged wyvern]

A finger trick?

What is it

[jagged wyvern]

What do you mean man

[trail tendon]

https://www.youtube.com/watch?v=LE6dmczMc68&ab_channel=ItsMsPruitt i think this video is on it

[jagged wyvern]

Thank you

[jagged wyvern]

I mean all in all

It shows a process of memorizing the coordinates

But she never says anything about memorizing the radian pattern across the circle

[trail tendon]

oh

[jagged wyvern]

Yes I did watch the video

If there really is a correlation between the coordinates and the radian pattern pls let me know

[brittle hatch]

just draw the unit circle many times and practice

the denominators ending in 6 tend to be closest to the x axis i think

[wicked egret]

@jagged wyvern i gotchu

lmk when youre on

[opal ibex]

Oh, my current instructor actually taught me this at the beginning of the semester. You have your hand open with all 5 fingers, and counting from the thumb down to the pinky are the cosines from 1 to 0. And then the reverse but in the same order of coordinates are sines from 1 to 0 from the pinky to the thumb.

[jagged wyvern]

Sorry I’m back

I thought you were gonna leave a long message

[trail tendon]

yea that was what was in the video

but he means the radians not the coordinates

like the angles

[dim void]

Anyone can help me

[snow karma]

at what

[dim void]

Math

[snow karma]

js send the problem then

[trail tendon]

no

[wicked egret]

so with the unit circle being added up to 2pi

at each 45 degree incriment you add pi/4 each time you go around

Same sort of increments for the 30 degree increments. Except you add pi/6

However you can simplify a lot of these

Also for the cos/sin lengths, you create a triangle with one of these angles, and the longer length will be root3/2, shorter 1/2

That’s my quick mental exercise to remember all of them

Root3/2 is arround .866, and the other length is 1/2 = .5

Helps to remember the unit circle is a circle with radius = 1 too

[plucky abyss]

[snow karma]

hello trig fan #1

[plucky abyss]

Hello

[wicked egret]

after tan(pi/4) is shifted to 1

why is it tan(x) + 1, not tan(x) * 1

oh god nvm

[upper karma]

you're supposed to know that RZ is an extension of RQ from the problem though because they use that as a hint for one of their problems, so i thought i was missing something

[wicked egret]

its addition to begin with

[upper karma]

no i'm sure we are

Lol

because it uses it as a hint :V

wait i'll show you

Look at b)

[snow karma]

u can prove it btw

[upper karma]

hmm how so?

[snow karma]

now assume that q indeed belongs to RZ

Q*

[upper karma]

sure

[snow karma]

then by pythagorean theorem in triangle RXZ

I MEAN THALES THREOM IM SORRY

[upper karma]

first of all, if we assume that q indeed belongs to RZ

then it's obvious that RQ = RZ - QZ

i don't see why we have to do anything

Lol

big line = sum of two small line segments that is part of the big line

[snow karma]

i mean yes

but whats the point

[upper karma]

my question was whether RQ and QZ are two different lines

or is RZ an extension of RQ

and how would one know precisely

[snow karma]

well tbh in this question not only can u prove it

but also its the 'use ur eyes and know urself' kind of thing

[upper karma]

i do that shit before and i got trolled

in one of their own questions lol

"you can't assume two lines are collinear" literally was the same context too

[snow karma]

dang two lines are colinear is a big assumption

[upper karma]

no i mean it was literally

the same scenario as RQ and QZ shit

it literally looks like it's one line

[snow karma]

well the RQ and RZ crap is js intersection proving thats it

[upper karma]

but if you break it into two lines then you can assume they're "collinear" , can you?

no i meant RQ and QZ 😭

are they two lines or one line

notice how they talked about extension of PQ

[snow karma]

ohhhhhhhh

[upper karma]

why they didn't do the same thing for RQ

now we don't know if it's an extension or two different lines

or do we? we probawbly do but i don't see why off the top of my head

[snow karma]

their knowledge is too powerful thats why

[upper karma]

💀

[snow karma]

u cant understand them

[snow karma]

yeah u dont see it

im telling u, their knowledge is way powerful than urs

[upper karma]

so their question is just flawed

[upper karma]

tbh they made me like this

i just look at the diagram

but they did the whole bs of "it can be two diff lines and u can't assume collinearity"

and also it's sus bc they talked about the extension of PQ

why not also say RQ is extended too

i'm thinking surely there's some way you can infer

[snow karma]

this collinearity crap reminds of how stupid the 'figure drawn not to scale' thing

[upper karma]

yeah i'm not sure about that either

so what's the "line"

am i not supposed to trust the diagram at all?

[snow karma]

yeah

ur eyes are lynig

lying*

[upper karma]

💀 but they say the diagrams are drawn as accurately as possible

if we're talking about the topology of the diagram

bruh

[snow karma]

ur retina is lying to u

[upper karma]

anyway if someone knows how do you know that RZ is an extension of RQ only based on what the question provides you then i gotta know it too

[snow karma]

tbh

yeah no idk

but u can prove it

[upper karma]

surely u can

unless they're just wrong

i doubt they're wrong though considering you got IMO peeps making this shit

[polar belfry]

What will be the ratio of in radius and exradius of right angle triangle?

[polar belfry]

√2+1:1:?

[snow karma]

why not

[obsidian hornet]

its different for each right triangle, but equals, for example, p-a/p, p=a+b+c/2; a-side in front of right angle.

[spiral lodge]

<@&268886789983436800>

[warm shuttle]

I wanna figure out geometry (because i'm bad at it), but the problem i have is that i try to prove something, but then i rely on some other theorem and then another one for that and so on and i don't know if there is circular reasoning going on so i am not satisfied with that. I took a look at Euclid's Elements, but i don't really like that as either. For example the very first construction he starts with a line segments AB and circles centered on each of A, B, with radius |AB|, and then just assumes that they intersect. Is there a more modern axiomatic system for geometry (that doesn't involve starting with set theory)?

[upper karma]

it doesn’t say that

[snow karma]

i never understood what projection u were talking abt precisely. but yeah now i see it, alr

[jagged wyvern]

Tysm

[spark stag]

Is there a more modern axiomatic system for geometry (that doesn't involve starting with set theory)?

Yes https://en.wikipedia.org/wiki/Tarski's_axioms#The_axioms but I don't really think it'll help with your core issue here

Or you can take a look at an annotated edition of Euclid, such as https://farside.ph.utexas.edu/Books/Euclid/Elements.pdf. They'll usually point out issues like the one you noticed — e.g. if you look at page 8 there's a footnote commenting that the assumption that the two circles intersect should indeed be a separate postulate

that said, I'm unconvinced that a more axiomatic approach will make geometry more clear to you — that isnt really the goal

[wicked egret]

Do not forget though that each 45 degree angle is root2/2 for both sin and cos but depending on the quadrant it’ll be neg/pos

[haughty arrow]

Am I doing this wrong?

I keep putting the answer as pi/6+2Kpi , 5pi/6+2Kpi and it keeps saying it’s wrong.

[warm shuttle]

why /6?

[warm shuttle]

what do you think i should do instead?

[remote creek]

in trig, when considering the notation "arcsinx" to represent the inverse function to sine, does the lowercase letter denote a relation or does it still denote a function regardless of the capitalization?

[trail tendon]

uh

oh i see

i didnt read properly 💀

the problem is sin(theta) = sqrt(2)/2 is not pi/6 or 5pi/6, and also u didnt put plus or minus when you took sqrt but yeah

[nocturne remnant]

I don’t understand this, what do you mean by “the lowercase letter”

[remote creek]

me too, I finished a quiz and apparently I was WRONG for saying that arcsinx represented the inverse to sine. Apparently the system said the reasoning was that the lowercase, which im assuming the notation, only denotes a relation. I plan on contacting my teacher to recover that point but I want to get a fact check first.

[trail tendon]

maybe more context is needed...

could have wanted arcsin(x) instead of arcsinx

or like

yea idk i could speculate all day 💀

[remote creek]

I said what was on the q word for word. Also, arcsin(x) is still lowercase

[trail tendon]

did it ask what the inverse of sin(X) was?

[remote creek]

IG the best option now is to js contact my teacher

[remote creek]

nah it was a truth or false

[trail tendon]

do you have a picture or smthn 💀

[remote creek]

Ill get that in a bit but it basically said Said "T or F, does the notation arcsinx represent the inverse function to sine

and naturally I put t which was wrong

[trail tendon]

very strange

i mean its not a perfect inverse but still it is considered the inverse 🤷‍♂️

[nocturne remnant]

I have no idea why the lower/uppercase should matter

[remote creek]

same, which was why I was very confused

[grave pond]

Did it just want T instead of t?

[remote creek]

There, and the explanation was that a lowercase denotes a relation not a function but honestly no clue as to what lowercase was being mentioned

[trail tendon]

no yeah thats weird

technically it's not a perfect inverse function but it is THE inverse function that's accepted

i'd talk with ur teacher about that yeah

[remote creek]

alr thanks for the response

[grave pond]

I'm now fairly curious about seeing the system explanation that talks about lowercase letters.

[chilly whale]

That's interesting, ive heard something similar about say Arg and arg for the argument of a complex number

[river abyss]

I’m reviewing Geometry and I realized that my teacher never taught us circles, surface area, volume, circular functions, and radiance measures. Can someone show me what I should learn about theses topics?

[warm shuttle]

i don't think radiance is a concept from geometry

[silent plank]

it's directly related to circles

khan course should be sufficient

[warm shuttle]

oh you mean radians

[maiden brook]

3D geometry is generally not so important although how could u not learn about circles?

[plucky abyss]

Got stuck in loops

[stuck sleet]

My answer is Circumcenter is at the point (0, 2b/3)

Am I right or wrong? Thanks!

[plucky abyss]

Take a,b=1 and draw and check

[bitter notch]

Sorry i forgot wat is isosceles

Just recheck until

That

Gg

[stuck sleet]

My answer is this:

[faint pasture]

looks correct

[frank heart]

@formal apex

[coarse bison]

how can I find the perimeter of the red circle centered at the red dot?

let's say the radius of the red circle is R and the distance from the vertex of the circular sector is \tau

[hoary prism]

youre asking for the circumference of the red circle?

whats the significance of the black thing

circumference just depends on the radius

so in this case the circumference would be $2 \pi R$ because R is the radius of the red circle

[somber coyote]

chipotle

[hoary prism]

or in general $\pi$ times diameter (which is twice the radius)

[somber coyote]

chipotle

[coarse bison]

the significance is that you are not counting that part

so, the perimeter of the red circle should depend on \tau and on R

and on \beta, the angle of the circular sector

[hoary prism]

is the black one supposed to be overlain on the red one

like

is the black one just supposed to be the red circle with that sector cut out

[coarse bison]

I want the perimeter of this green circle

which has the center at the black dot

notice that this is different from the arc length of the circular sector because it is not centred

in fact, this becomes a circualr sector when \tau = 0. That is, the distance from the center of the green circle and the vertex is zero.

[acoustic ravine]

how do i graph sine cosine and tangent functions

[obsidian harness]

1. review your rules of transformations: there's translation, compression/stretch, reflection (which is just a stretch with a negative scale factor), in both the x- and y-axes

2. you can use Desmos or GeoGebra to graph them

[plain scarab]

Could anyone confirm this?

Nvm im good

[jaunty mountain]

I believe it’s 4/3PiR^3

[trail tendon]

yep, its right

[fathom meteor]

Step 1: Graph the function with no transformations Ex: Just sin(x)
Step 2: Add the amplitude modification if applicable Ex: 2sin(x)
Step 3: Add frequency modification if applicable Ex: 2sin(2x)
Step 4: Add horizontal translate if applicable Ex: 2sin(2x+pi/3)
Step 5: Add vertical translate if applicable Ex: 2sin(2x+pi/3)-4

[plucky abyss]

But how to do step 1

[lime dune]

look it up

[plucky abyss]

Okay

[snow karma]

B and B' and B'' are aligned.
ABCD and AB'C'D' and AB''C''D'' are all squares.
How do I prove that D, D' and D'' are aligned?

[plucky abyss]

Can we assume A at (0,0) and some random lengths for squares

[autumn olive]

Yo

[snow karma]

this was on an old exam in my highschool

and we don't use proofs by considering graphs

[plucky abyss]

Okay

[snow karma]

Also A(0,0) what about the other axises

[lime dune]

it’s some sort of spiral similarity configuration

dunno what specifically to do with it tho

[snow karma]

we didn't study that too

sadly

[lime dune]

[plucky abyss]

Then we can take x,y,z as lengths of squares and get coordinates for other points

[snow karma]

okk but the x axis and y axis

u didn't define them

u only defined the origin of the repair A

[plucky abyss]

By distance formula

[snow karma]

no

like

it ain't no graph without an x axis and a y axis

[plucky abyss]

Okay I not solved it's just initial thought

[snow karma]

I should give u a trig question to put u out of ur misery

ur a fan of those

[plucky abyss]

Lol it was just a joke

[snow karma]

yea ik

[plucky abyss]

I don't know any math

[snow karma]

o

[plucky abyss]

[autumn olive]

Distance formula

What?

[snow karma]

idek lol

[plucky abyss]

[warm shuttle]

is DD' perpendicular to BB'?

[snow karma]

well not mentioned

[formal geyser]

Hello, guys

There is one semi circle and two rectangles

What is the area of shaded green rectangle

[warm shuttle]

there's a couple similar triangles

width green : width red = height green : 3

[obsidian hornet]

Rotation homothetics and cyclists lemma I consider.

[snow karma]

tf did u js say

[warm shuttle]

what's a rotation homothetic?

[obsidian hornet]

Similarity Transformation

[warm shuttle]

so are you saying squares are in fact similar?

[snow karma]

oh affine geometry

[warm shuttle]

what

oh

so they don't actually have to be squares?

[obsidian hornet]

Yes, in this picture you rotate and shrink squares with A center

[snow karma]

so like is there a specific transformation function that aid in this question

[warm shuttle]

i can't find mentions of cyclists lemma

[snow karma]

actually nvm I solved the question

it's easier than it looks

[snow karma]

ty for kinda showing the way, I thought this was plain Euclidean geometry

[obsidian hornet]

This is just a transformation in Euclidian geometry

[warm shuttle]

i mean i know how to solve it by throwing coordinates and trig functions at everything

but i still have no clue what the nice solution is

[snow karma]

well we consider this a whole other domain in math

affine geometry

(in my country)

[obsidian hornet]

Understood

[snow karma]

dang

[warm shuttle]

affine geometry is just euclidean, but you don't use all axioms

i guess so is hyperbolic... and finite

[obsidian hornet]

Two circles are given, intersecting at points A and B. Two cyclists ride along these circles (each in his own way) at constant speeds and in the same direction (either both clockwise or both counterclockwise). They leave point B at the same time, make one turn and return to B. Then there will be a fixed point that is equidistant from cyclists all the time.

[warm shuttle]

crazy shit

but not necessarily at the same speed?

[warm shuttle]

where is the fixed point?

[warm shuttle]

oh but they have the same velocity:radius of the circle

this isn't affine geometry, right?

clearly you gotta look at the angles ZOA and ZO'A'

[obsidian hornet]

Homotehetics is a special case of affine transformation. So, its just a composition of rotate and homothetics in Euclidian geometry.

[warm shuttle]

but rotating isn't

[obsidian hornet]

Yes

[warm shuttle]

wait

is the proof literally that it's a composition

so you decompose and find center of homotheityty?

however you write that word

[obsidian hornet]

Just consider that u rotate and shrink squares like on this picture

So, A is a center

And 3 points are collinear

[warm shuttle]

So A is Z?

[obsidian hornet]

Yes

[warm shuttle]

why are they collinear?

[obsidian hornet]

Cyclists lemma

[warm shuttle]

i don't understand too much of a logical jump for me

[obsidian hornet]

By virtue of the transformation, images and prototypes of points will be colliniar

[warm shuttle]

oh

but

why

i know this isn't true of pure rotation

like (1, 0), (√3/2, 1/2), (√2/2, √2/2) are not collinear

wait

yeah

because you aren't comparing points in the same transformation

like if you said points A, B, C are collinear

here i apply an affine transformation

A', B', C' are still collinear

but you are saying A, A', A'' are collinear

(names are examples, don't relate to the diagram)

[warm shuttle]

What's the solution?

[warm shuttle]

the problem is that i see 0 points that remain collinear in this diagram

(that are moving in a line that is)

oh but AB->AD is a transformation and in fact is just a rotation of the whole plane by 90° counterclockwise about A

And if you belive that you can get a side of square by rotating it's adjacent side like this then clearly if we are given AB', then this transformation gives AD'

But it also preserves lines

being linear

[snow karma]

oh mb

so lets say that r is the direct quarter of a turn around the center A

we can see that r(B) = D and r(B') = D'

[warm shuttle]

yeah

and r(BB') = DD'

yay

[snow karma]

yep and they are perpendicular

do same thing to two other points

[warm shuttle]

because it's a 90° rotation

[snow karma]

theyre colinear

yeah or a direct quarter turn lmao i call it that

[warm shuttle]

because you are just taking the set of possible values of B

and then passing it through r

but since it's a line

[snow karma]

ya

[warm shuttle]

it just stays a line under rotation

And so do all the C's

[snow karma]

The Cs are actually not aligned

[warm shuttle]

but they are

[snow karma]

u cant prove it tho

[warm shuttle]

same argument with 45° rotation

(and scaling by sqrt(2))

[snow karma]

oh wait huh

[warm shuttle]

not only that

[snow karma]

yeah this drawing is shit then

from 2012

[warm shuttle]

yeah it is

[snow karma]

was fking 2 at that time damn

[warm shuttle]

it's actually true for any shape, not just square

[snow karma]

hm

terrific honestly

affine geometry amazes me

[warm shuttle]

it's not affine geometry tho?

[snow karma]

are rotations transformations

[warm shuttle]

yes

[snow karma]

arent transformations in affine geometry

[warm shuttle]

no, affine transformations are in affine geometry

(which rotations happen to be, but that's...)

[snow karma]

ahem

[warm shuttle]

but like

affine geometry is actually opposite

it's discarding all the stuff you can transform between and just looking at them as equal

like all circles are similar

all triangles are "affine"

[snow karma]

idk why lmao but what ur saying to me sounds like thkse excuses masteroogway used on twitter to not get cancelled

[snow karma]

well lets be fair

u admitted rotations are afine geometry related

affine*

[warm shuttle]

i mean

for example

if you have a theorem of affine geometry about squares, then it also is true about parallelograms

[snow karma]

sure

[warm shuttle]

so angles are not preserved

and in this cases they strictly are, which is why i'm unsure about calling it affine geometry

[snow karma]

lets say we have euclidean geometry ok?

[warm shuttle]

planar* euclidean geometry

[snow karma]

where u have to solve for x for a mathematical expression that play the role of a segment's length

And jt requires algebraic manipulations

so many of it

is it geometry

or algebra

(or algebraic geometry like what my country calls ig)

[warm shuttle]

(btw i don't think euclid really separated segments and their lengths)

[snow karma]

No?

[warm shuttle]

i didn't read the whole thing yet tho

[snow karma]

i never did

the only thing ik abt euclid's laws is probably euclids algorithm