geometry-and-trigonometry

[hidden fern]

the root of both sides?

[wintry tundra]

there we go

see

to get the root of the other side just put it in a calculator

so u dont have any complications id simplify the right side into one term

[hidden fern]

thank you

[wintry tundra]

ur welcome man!

trig can be weird bc of how much memorization there is

but remember

[hidden fern]

didnt have class for an entire year cause pandemic

[wintry tundra]

its algebra with a calculator

[hidden fern]

and didnt have math for online classes cause it got cancelled

[wintry tundra]

theres a lot of calculator using in trig since ratios of non integers are a bit strange sometimes

[hidden fern]

The teacher aint too good tbh

[wintry tundra]

hey it happens

[hidden fern]

just sends us videos

[wintry tundra]

hmm

it sucks for people who have to be online all year

[hidden fern]

sometimes its not the complete unit

[wintry tundra]

my school is in person again so im pretty lucky

[hidden fern]

we still in lockdown for another 20 days

and even then, only online classes after

[wintry tundra]

hm

[hidden fern]

the rest of the school year is online cause everyone is getting sick

[wintry tundra]

sorry i was helping someone in #prealg-and-algebra

it is flu season so

[hidden fern]

thank you for the help

[wintry tundra]

a lot of false positives happening bc of it

also yeah no problem man

[hidden fern]

I am now dealing with trig ratios now, which thankfully I remember

[wintry tundra]

soh cah toa

[hidden fern]

yup

[wintry tundra]

and when u have $sin^1$ or something just flip the variables

[somber coyote]

hiidostuff

[wintry tundra]

$sin^{-1} i mean$

[somber coyote]

hiidostuff

[wintry tundra]

$sin^-1$

finally

[somber coyote]

hiidostuff

[wintry tundra]

AUGH

[hidden fern]

I remember something about when something equals to anything below 1

[wintry tundra]

ok u get the gist

[hidden fern]

use sin-^1

on both sides

[wintry tundra]

well $sin^{-1}$ is h/o

[somber coyote]

hiidostuff

[upper karma]

that is NOT TRUE

be VERY CAREFUL with that notation

[versed river]

ho

[wintry tundra]

hohoho

how is that not true

[eager plover]

HELP

[sleek depot]

Q = S as angle of rectangle is 90

basic property of rectangle

[silent plank]

you aren't told that it's a rectangle

apply properties of perpendicular lines

[sleek depot]

oh right

[steep temple]

h/o is $$ \frac{1}{\sin\theta} = \csc\theta $$

[somber coyote]

dlp

[steep temple]

@wintry tundra

it's not the case that $$ \sin^{-1}\theta \equiv \frac{1}{\sin\theta} $$

[somber coyote]

dlp

[hollow blade]

Hi guys

I don't understand

[upper karma]

Please don't multipost. Read #❓how-to-get-help

[verbal holly]

#geometry-and-trigonometry i just need help kinda i am a beginner to this math olympiad but dont have even knowledge of the letter g of geometry after seeing the amc 10 past papers pls suggest me some #resources for it 🙂

[covert rune]

https://artofproblemsolving.com/store/item/aops-vol1

It's a fantastic starting point for people interested in math olympiads. It goes over most topics you'd need in a great amount of detail.

[silk patio]

@verbal holly which question in particular in geometry scared you? I’m curious

[verbal holly]

@silk patio bruh me is a beginner so u can probably understand

[verbal holly]

dont u feel being an indian it becomes too much expensive for me ahh!!! i need something in pdf format and cures away the need of aops books

[grand yacht]

Hi all

[noble lion]

So uh, I believe b = (180 - 80) - 40 so 60

And for a, I have no clue how to work it out, I just know it's an angle < 80

[grand yacht]

M new here

Plz guide

[noble lion]

Maybe a = 40 too?

[polar summit]

@noble lion a=b

they subtend same arc

[noble lion]

its 60 haha yeah

[verbal holly]

can u pls tell the reason for it

[alpine cliff]

theres different chats for different topics, go to a general chat for help

read the introduction

[steep temple]

How the heck do I work out the exact value of $$ \tan(\frac{-5\pi}{4}) $$

[somber coyote]

dlp

[steep temple]

If it was positive, I would do something like:

[upper karma]

I'd just recommend spitting tan into sin and cos

So that if it's the negative what's disturbing you, you can notice if sine or cosine is odd/even/etc. And get rid of the inside negative

[steep temple]

$$ \tan(\frac{5\pi}{4}) = \tan(\frac{4\pi}{4} - \frac{5\pi}{4}) = \tan(\frac{-\pi}{4}) = -1 $$

[somber coyote]

dlp

[earnest echo]

$\tan(-x)=-\tan(x)$

[somber coyote]

The Godfather

[steep temple]

I can't use my positive approach here because then I'd arrive at $$ \frac{-11\pi}{4} $$ and I don't have that in my table

[somber coyote]

dlp

[upper karma]

Are you ignoring our suggestions?

[steep temple]

I am not sure what this means

[upper karma]

tan(x)=sin(x)/cos(x)

This is just preference

[steep temple]

and I am familiar with this rule, but I would have liked to try a modified version of my approach first

sure, but what the hell is cos(4)

[upper karma]

I don't memorise tan values, i just split them into cos and sin

[steep temple]

are you saying $$ \tan(\frac{5\pi}{4}) \equiv \frac{\sin(5\pi)}{\cos(4)} $$

[upper karma]

No

[somber coyote]

dlp

[steep temple]

ok then can you elaborate

[earnest echo]

You could always write
tan(-5π/4) as tan(-(π+π/4))=tan((-π)+(-π/4)) and then use your identity

[upper karma]

$\tan\left(\frac{5π}{4}\right)=\frac{\sin\left(\frac{5π}{4}\right)}{\cos\left(\frac{5π}{4}\right)}$

[somber coyote]

Al𝟛dium

[steep temple]

ah right yes of course

[upper karma]

There are a lot of ways of doing this

I'm just suggesting the one i'd do

[steep temple]

but then i'd have to solve for sin(5pi/4) and cos(5pi/4)

[deep trail]

how i got oa=1 i think and then dont know the formula or whatever to find the blue line

[upper karma]

Yeah, as i said

[steep temple]

i think it would take equal to or more steps than https://cdn.discordapp.com/attachments/326138757474680852/800811848550055946/180738241824030720.png

[upper karma]

sin and cos values are more known than tan for me, i don't like memorising stuff and i don't encourage it whenever it's possible

[steep temple]

i dont memorise tan pi/6 i just know 1/sqrt3 is 1/sqrt3

[upper karma]

Surely, there are a lots of ways of doing this.

The one best suits for you is the one you want to take.

[steep temple]

yes this is what i was thinking of thanks

[upper karma]

As i said, you can apply this logic if you get the negative out by knowing tan(-x)=-tan(x)

[steep temple]

yes yes of course

thanks

[glossy vigil]

how do i do this

[zenith garnet]

how do i prove perpendicular ac and bd here?

[upper karma]

have you done (a) ?

consider triangle ABC, what is BE ?

@zenith garnet

[zenith garnet]

i found it

how bout this one

@upper karma

[upper karma]

look at the hint

perpendiculars from A and D

[zenith garnet]

i did that

not sure where to go from here

i know that the right angles formed from the perpendiculars are equal

[fair plover]

Let ABCDEF GH be an octagon inscribed in a circle. Four of the sides of the rectangle AB=BC=CD=DE=1, and the remaining four sides DE=EF=FG=GH=√2, find the area of the octagon.
I know that the central angle between the chords length 1 and sqrt2 are 90 degrees, but i dont know what to do with that
how do i use the chord ratios to find the angles of the triangles the octagon is divided into to find their heights?

hey, this question is the only one giving me trouble right now. some help would be really really appreciated.
ping me if you know

[upper karma]

you also know that the heights are congruent

[zenith garnet]

and i know angles B and C are equal

yeah

thats a-a-s

[zenith garnet]

doesnt angle angle side not prove congruence

[upper karma]

yes it does

[zenith garnet]

oh shit

mb

[silver slate]

anyone active in here?

[wintry tundra]

@silver slate this whole server is pretty active, if you have a question someone (including me) is bound to help u if they know how to

i cant help with everything but i will try my best

[silver slate]

Okay thank you

I just need help on trig specifically

I’m absolutely terrible at it

[upper karma]

what's giving you issues?

[wintry tundra]

that is not trig

not at all

[silver slate]

Ive missed a bit of class so I dont know whats the first step to take to solve the equation

[wintry tundra]

well what do u think the solution is

can u even guess or no

[silver slate]

and its in my trig class so i kinnda just assumed

[wintry tundra]

nah thats basic algebra

[silver slate]

oh im sorry

[wintry tundra]

nah ur fine

u can still ask here

what do u think the solution is tho

[silver slate]

maybe...a2 and b2

[wintry tundra]

this is actually something u should know by heart if u wanna get these kinds of squares done fast

a2 and b2?

what is the and part

• or -

[silver slate]

I dont have a button to make then expoents

[wintry tundra]

u type a^2 and b^2

or $a^2 + b^2$

[somber coyote]

hiidostuff

[silver slate]

ohhh

yeah thats what i was thinking

[wintry tundra]

i just put a plus so itd work tho

do u think its a + or - between the a^2 and b^2

[silver slate]

positive

because the -b^2 would make a positive?

[wintry tundra]

i suppose in the solution to $(a-b)^2$ you would add a^2 and b^2

[somber coyote]

hiidostuff
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wintry tundra]

but theres another thing that you subtract from a^2 + b^2

do u know what FOILing is?

or just FOIL

[upper karma]

for math mode use $ on a^2 and b^2 too so that it doesn't give error

[wintry tundra]

FOIL is usually taught for algebra 1

thx

FOIL is what to do when u have to distribute a binomial

like $(a-b)^2$

[somber coyote]

hiidostuff

[silver slate]

sorry i dont

[wintry tundra]

First
Outside
Inside
Last

[silver slate]

so what would the first step look like

[wintry tundra]

$(a-b)^2$ can also be written like $(a-b)(a-b)$

[somber coyote]

hiidostuff

[wintry tundra]

because when u square a number u multiply it by itself

[silver slate]

so (a-b)(a-b) be the answer?

[wintry tundra]

no

it can still be simplified

[silver slate]

aww man

[patent plume]

have they taught you distribution?

[wintry tundra]

do u know what distribution is?

[silver slate]

Yeah

[patent plume]

Well, you distribute here

[wintry tundra]

so FOIL is how u distribute binomials

[silver slate]

so is the answer a^2 -b^2

[wintry tundra]

no

u multiply the first numbers

then the outside numbers

then the inside numbers

then the last numbers

[silver slate]

but a and b are letters

how do I distribute that?

[patent plume]

they stand for numbers

[wintry tundra]

a and b are letters

so to foil $(a-b)^@$

[somber coyote]

hiidostuff

[wintry tundra]

oops

replace @ with 2

u would multiply a and a, a and -b, -b and a, -b and -b

then add

[silver slate]

so what would be the answer

[wintry tundra]

well add after u multiply

here we dont give answers we teach u how to do it

intuitive learning is more beneficial than just spewing answers

[silver slate]

I understand

[wintry tundra]

oh u understand the problem now?

or are u just saying "oh ok"

[silver slate]

but if i mutiply i would still get letters

[wintry tundra]

yeah

[silver slate]

And i can't really add letters?

[wintry tundra]

the answer is just variables

well when i say add i mean put them together

[silver slate]

so like ab?

[wintry tundra]

yeah

u can have ab

[silver slate]

so ab^2

?

[dense vine]

add then multiply

[wintry tundra]

yeah

[dense vine]

I call it mathematical

[wintry tundra]

wait no

multiply then add

[silver slate]

but im not adding

[wintry tundra]

so a times a (first), a times -b (outside), -b times a (inside), -b times -b (last)

tell me what each one is

[silver slate]

so a^2+b^2= ab^4

[wintry tundra]

no

go thru what the product of each one i told u are

[silver slate]

it would a^2

and then +b^2

right?

[wintry tundra]

ok so a^2 is the product of the firsts

now tell me the product of the outsides

[silver slate]

when you say outside do you mean the b

wouldn't be b^2

[patent plume]

What about a * b?

@silver slate

[silver slate]

i never thought about that

so ab^2

[silver slate]

?

[patent plume]

(a-b)(a-b). So we do a*a + a *(-b) for the a stuff. For the b stuff, we do (-b) * a + (-b) * (-b)

[silver slate]

Like this?

[patent plume]

ypu

exactly like that

[silver slate]

So now that I have this]]

Now what?

[patent plume]

Well, you know that 2*3 = 3 * 2 yeah?

@silver slate

[silver slate]

yeah 6

[patent plume]

Well, would a*b = b * a then?

@silver slate

[silver slate]

Yeah

[patent plume]

So - ab - ba = - ab - ab?

[silver slate]

yeah

[patent plume]

which is........?

[silver slate]

-2ab = -2ab?

[patent plume]

Yeah

So it ends as a^2 - 2ab + b^2

that's it

[wintry tundra]

so theres many... i think they are identities that u should know

[patent plume]

identities yes

[silver slate]

so this is the answer?

[wintry tundra]

$(a+b)^2 = a^2 + 2ab + b^2$

[patent plume]

yup

[somber coyote]

hiidostuff

[wintry tundra]

$(a-b)^2 = a^2 - 2ab + b^2$

[somber coyote]

hiidostuff

[silver slate]

okay thank you guys for bearing with me

Im not very smart\

[patent plume]

$(a-b)(a+b) = a^2 - b^2$

[somber coyote]

MasterTernimus

[wintry tundra]

^

no problem man

[patent plume]

np

[wintry tundra]

come back if u need any help

[patent plume]

^

[wintry tundra]

and i advise memorizing these identities

[patent plume]

^

[wintry tundra]

its not as hard as u would think to know these

[silver slate]

so this is correct right?

[patent plume]

Uh no

(a-b)^2 = (a-b)(a-b)

[silver slate]

aww man]

[patent plume]

not (a-b)(a+b)

[silver slate]

ohhhh I get it now

[patent plume]

yup

[silver slate]

Now that I see the equation

can I get some help on a different question?

[patent plume]

sure

[silver slate]

thx

[patent plume]

This is from $(a-b)(a+b) = a^2 - b^2$

[somber coyote]

MasterTernimus

[silver slate]

But starting off how would i do a

Like questio9n a

[wintry tundra]

so what is the first term

in a)

[patent plume]

Same thing as b btw

[silver slate]

k^2

[wintry tundra]

yep

what is the 2nd term

[silver slate]

-49

[wintry tundra]

hmm

lets go with that for now

[silver slate]

ok

[wintry tundra]

so what is the square root of $k^2$

[somber coyote]

hiidostuff

[silver slate]

k?

[wintry tundra]

yep

i wont ask u the square root of -49 yet

but you know the identity we showed u where its

$(a+b)(a-b) = a^2 - b^2$

[somber coyote]

hiidostuff

[silver slate]

Yeah

[wintry tundra]

a represents the first term

b respresents the second

so k is the first term right?

[silver slate]

So do I use this formular for everything?

[wintry tundra]

or the sqrt of the first term

yeah if u find 2 terms and the second one is being subtracted

for the others u gotta see if the middle term is adding or subtracting

but anyways k is the sqrt of the first term right?

[silver slate]

yes

[wintry tundra]

so where would u put k

[silver slate]

Where a is on the left side

[silver slate]

of this

[wintry tundra]

yes

sorry internet started acting up

[silver slate]

its okay

[wintry tundra]

so does k substitute into a or b

[silver slate]

a

[wintry tundra]

yep

so now we have $(k + b)(k - b) = k^2 - b^2$

[somber coyote]

hiidostuff

[wintry tundra]

a is solved

but now for b

do u see how $b^2$ is subtracted

[somber coyote]

hiidostuff

[silver slate]

yes

[wintry tundra]

what was the second term u told me?

[silver slate]

-49

[wintry tundra]

what do u think the second term is actually now that u see that b squared is subtracted

[silver slate]

-7

or +7

cause thats the square root

[wintry tundra]

yeah the square root of 49 is 7

so when u see a negative on the right side

its actually a positive being subtracted

so take the root of the positive and dont use i

[silver slate]

so what would the equation look like

[wintry tundra]

cuz $\sqrt{-49} - 7i$

[somber coyote]

hiidostuff

[wintry tundra]

i meant equals on that above one

so u got 7, and since a is used now it has to go on b

so if we are using (a+b)(a-b), what do u think it is?

[silver slate]

so (k+7)

[wintry tundra]

yep

(k+7) for the first

and whats the second

[silver slate]

(k+7)^2

idk im kinda getting lost

[wintry tundra]

no cuz its + and -

$(a+b)(a-b)

$(a+b)(a-b)$

[somber coyote]

hiidostuff

[wintry tundra]

this

so the first one is k + 7

so the 2nd has to be k - 7

[silver slate]

(k+7)(k-7)

[wintry tundra]

theres ur answer

[silver slate]

so what do with it

[wintry tundra]

i gotta get ready for school tomorrow so excuse me for a sec

[silver slate]

oh

[wintry tundra]

just put it down

[silver slate]

okay]

[wintry tundra]

thats the answer

[silver slate]

thanks so much for you hlep

[wintry tundra]

welcome

[hearty pond]

can anyone help me do my triangle homework

[wintry tundra]

show me and ill see what i can do @hearty pond

[wintry tundra]

ah the proofs.... its all coming back to me

[hearty pond]

we are doing proofs and i know that if i find that either triange pqv and rqs are congruent or triange qvt and qst are congruent then i could use CPCTC to say that VQ is congruent to SQ

[wintry tundra]

first u could say that angle VQT = angle SQT

[hearty pond]

ok what is the statement for that

[wintry tundra]

what do u think

[hearty pond]

definition of bisector

[wintry tundra]

oh shoot

i forgot about statements

i thought u were asking me how u write it

but yeah definition of bisector

[hearty pond]

oh haha

[wintry tundra]

from there u can go angle PQT = angle RQT

[hearty pond]

isnt that also definition of bisector

[wintry tundra]

yeah

so id guess u would want to get ASA to prove the equivalence

[hearty pond]

yeah

[wintry tundra]

so u should try to prove either that triangles PQT and RQT are equiv or VQT and SQT are

[hearty pond]

it has to be VQT and SQT because i need to prove that VQ is congruent to SQ

[wintry tundra]

oh my bad

i meant triangles PQV and RQS

[hearty pond]

yeah

but i dont know what else i can prove

[wintry tundra]

its been like a yearish since i was in geometry

[hearty pond]

oh i see

[wintry tundra]

well i guess 1.5

[hearty pond]

well i know that PQ is congruent to RQ

[wintry tundra]

thats true

u should also say QT = QT

bc geometry is redundant

[hearty pond]

yeah

[wintry tundra]

and u want to prove those large triangles as equal to get some sides to be equal

and now u can prove the big triangles equivalent with SAS

[hearty pond]

i guess i could prove those large triangles to be equal to get angle VTQ and STQ and then prove the small triangles

[wintry tundra]

yeah

u gotta prove that PV = QS tho

and idk if u can

to get VTQ and STQ

[hearty pond]

I got it i can explain

So prove big triangles which gives you both the angles by T

and then once you have that you can prove triangle VQT is congruent to SQT with ASA

ans then with CPCTC you can prove that VQ is congruent to SQ

[wintry tundra]

oh yeah

i see

i missed that

[hearty pond]

its all good thanks so much for the help

[wintry tundra]

alright welp looks like u got it handled

yeah no problem

[hearty pond]

have a good one

[wintry tundra]

u too

[hearty pond]

thx

[spark star]

<@&286206848099549185>

Can someone explain this to me?

[upper merlin]

so do u know about the property that the largest side is opposite the largest angle and the smallest side is opposite the smallest angle in a triangle

[spark star]

No?

How does that work?

[upper merlin]

this isnt a rigorous proof but u can think of it like this, if the angle is large then as the two rays extend they get farther away from each other quickly

and the length of the 3rd side will be long

if the angle is small then they dont diverge as fast and the side is small

if u want a rigorous proof u can use the law of sines

do u know what that is

[spark star]

Does this property work for this?

[upper merlin]

yes

[spark star]

It's only shortest to longest.

[upper merlin]

sorry i should say this property is for triangles

[spark star]

I just don't get what they mean by that.

[upper merlin]

they want u to determine the shortest, middle, and longest sides and order them in that way

[spark star]

Yeah

But how does it work?

I get the shortest with the second shortest?

[upper merlin]

no u get the shortest side length with the smallest angle

so the smallest angle is b which is 48 degrees right

[spark star]

Yeah

[upper merlin]

so the smallest side will be the one opposite to that

which is AC

[spark star]

Oh I see

Then what will determine the middle?

[upper merlin]

the middle angle determines the middle side

so the middle angle is 52 degrees

[spark star]

Yeah

[upper merlin]

and the side opposite to that is AB so thats the middle side

[spark star]

So that goes with the shortest angle?

Ohh

Then across that is

[upper merlin]

no the shortest angle goes with shortest side, middle angle with middle side, and largest angle with largest side

[spark star]

BC

[upper merlin]

yeah so largest angle is angle A so largest side is BC

[spark star]

Yeah

Wow thanks

[nocturne thicket]

for #23 ONLY, I got 9, however, my teacher says it's 3?

[upper merlin]

i got 9 too

[alpine cliff]

hmm

its 9

cannot be 3 by triangle side inequalities

[subtle nest]

its 69

[alpine cliff]

great thanks

I didn't think of that lmao

[verbal holly]

ac ab bc is the increasing order of sides btw this is based on triangle inequality concept

[verbal holly]

which pdf btw?

[nocturne junco]

Why is DC and PC equal? What property could explain this?

[versed river]

because the angles opposite them(beta) are equal

oh unless that much isnt given

[nocturne junco]

thats why im confused hahahaha

[silent plank]

PDC is an eq triangle

[olive jacinth]

how do u do this

[patent plume]

@olive jacinth the 28' is like decimal point. But instead of it going from 0.99 to 1.00, it goes from 59' to 1 degree

[alpine cliff]

' represents minutes " represents seconds. there are 60 minutes, 60', in one degree and 60 seconds, 60", in one minute.

[frozen island]

how did they get rid of the 1

which identity is that

nvm got it

[steep temple]

https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

[storm portal]

This should be pinned.

[upper karma]

I am trying to solve for all angles from the following equation

[somber coyote]

Elfen

Elfen

[upper karma]

can someone please help me on two problems?

[heady bronze]

Uh so my friend needed some help with work and heres the assignment, could someone explain to me how to do it?

[wintry tundra]

OCB is 2 degrees it seems

If OA and CB are perpendicular

[oak frigate]

B

From inscribed angle theorem

[inland condor]

Can I get some help with this? Geometric probability question

[silent plank]

where did you get stuck?

[wanton storm]

find the areas of the circle and triangle and then subtract them to get the white area

Then use bayes theorem to find the probability

[sleek depot]

do u mean P(A) = n(A)/n(S)?

i didnt know its called the bayes theorem

[upper karma]

So I failed at trisecting an equilateral today

How do I prove this?
What do I need to search/study to be able to prove & calculate this?

[upper karma]

Nvm jus realize basic trig and cosine ryle can do the job

[hasty wyvern]

how to solve for x

ping if you respond

[sleek depot]

Question seems incomplete

[hasty wyvern]

thats it

it just says solve for x

[sleek depot]

x can be 1,2,3,4,5 and so on

[hasty wyvern]

it then just says solve for X no decimals

exactly

im so confused

on that

[clear lodge]

this is the full question

plz

[storm portal]

It's still incomplete

[sleek depot]

Even the sentence sounds incomplete

[hasty wyvern]

ihave no clue hwo to dit

yeah

[storm portal]

^^

[clear lodge]

^

[sleek depot]

U can write values as a set

A = {x| x€Z, x<0}

[hasty wyvern]

I just put x is negative infinity to 0

[sleek depot]

To -1

[clear lodge]

[hasty wyvern]

but then you have to find tangent of the triangle formed based on that soooo

[sleek depot]

Not 0

[hasty wyvern]

ok

[slim saddle]

hello

i had a question

if there is something like

(10y)/(293+y)

could you cancel out the y?

[clear lodge]

@slim saddle could you answer my question first

@hasty wyvern needs it

[slim saddle]

ill try

try refreshing

it seems like an imcomplete question

[sleek depot]

No

U can't split the denominator

Splitting is only applicable for numerator

[slim saddle]

okay thanks

[storm portal]

$\frac{10y}{293 + y}$ cannot be simplified, reduced, factored or have any terms cancelled

[somber coyote]

Shen

[storm portal]

$\frac{10y}{293 + y} \ne \frac{10}{293}$

[somber coyote]

Shen

[upper karma]

I have a super basic vector problem I have a question about

[humble pulsar]

I might have an answer, but you never posted the question like #❓how-to-get-help says

[upper karma]

I'm curious to see what your answer is

[humble pulsar]

ok well what do you think the answer is?

[upper karma]

I said 126 miles , not because you add them. But because east would be considered negative in a coordinate plane right ? So shouldn't it be 63 - - 63 = 126

I mean west

My bad lol

[humble pulsar]

126 is the total distance

0 is the total displacement, since 63 + (-63) = 0

[upper karma]

But isnt it delta r?

Final - initial

[humble pulsar]

No this is just a vector sum

[upper karma]

In phyiscs?

[humble pulsar]

in any vector sum

[upper karma]

nvm nvm maybe thats why im tripping

but west is still negative right

So thats why its - 63 + 63 = 0

[humble pulsar]

if you define the x-axis as being parallel with I-88 (with pos x being East) and define 0 to be your initial starting place, then traveling west 63 is equivalent to the vector -63i. travelling east is 63i, so you're adding 63i+(-63i) = 0i = zero vector

[upper karma]

Right right okay

[humble pulsar]

not always cause I can just define going west to be positive

in which case east would be neg x

[upper karma]

Yeah but for a basic physics course I think we follow the standard

lol

[humble pulsar]

there's no "standard"

we say pos x is to the right by convention, but very quickly you realize that you can orientate the axes to fit the situation

[upper karma]

? Im pretty sure in a regular 2D coordinate system going left on the x axis is negative

[humble pulsar]

ie boxes on a ramp / dynamics

we define the axes so one direction is parallel to the ramp and the other perpendicular

[upper karma]

Right, but even though you changed the rotation of the coordinate system it didnt change how it works

[humble pulsar]

right

[upper karma]

unless you just like looking at it backwards for some reason

cause you're crazy

lol

I think my teacher just failed to write -63 and thats what was throwing me off

he wrote 63 + 63 = 0 and I was like............. uhhhh

[humble pulsar]

63[W]+63[E] = 0

[upper karma]

Yeah he didnt write that either

lol

I think he forgot to mention that west changed the value to negative

So I'm not crazy yay!

lol

[spark star]

What class is that for?

[upper karma]

Physics lol

[spark star]

I don't have that LOL

[humble pulsar]

it's not the fact "west is negative", it's the fact "east vectors" and "west vectors" are opposite

[spark star]

Mosh, you know what Earth Science is?

[humble pulsar]

maybe?

[upper karma]

Well however you want to think of it

[spark star]

I don't know.

Coding

[upper karma]

like you said its all just convention anyway

lol

[spark star]

You know what Earth Science is?

[upper karma]

No sir

[spark star]

Ah okay

[upper karma]

Sorry my dude

of girl

or girl

[spark star]

It's alright.

Male, yes.

[upper karma]

Lol

[humble pulsar]

yes the convention is that pos x is to the right of pos y, but you pick the axes that typically makes numbers easier ie positive. so if you're only travelling west you'd just define west as pos

[upper karma]

Like if you wanted to you could called negative and positive " gooble and goooable " its just shit we made up based on obervations

[spark star]

Should I take American Sign language?

I feel like, it would be useless, yet not useless.

[humble pulsar]

letting deaf people communicate isnt useless. .

[spark star]

That's the non useless part.

The Useless part is

I already know 3 different languages and I would forget the sign language easily.

[upper karma]

Its actually quite "usless" since the deaf community is quite small

Thats like learning dutch

thatll go far

lol

[humble pulsar]

that's just wrong but ok

[upper karma]

I'm just saying its not real common unless you're trying to be an interpreter or meet a deaf person once

[humble pulsar]

Cause you're literally saying someone with a disability doesnt deserve the right to communicate

[upper karma]

Thats not what I said but ok

[humble pulsar]

you said a whole ass language is "usless"

[upper karma]

When did I say take away sign language so they cant talk 😂

Its usless to learn if we wants to get the most out of a language in terms of how many people speak it

Or "hand" it

You could reach more people learning sign language than speaking French however

Since French is 200 million people and sign language is 400 million

So really French is usless

LOL

I bet all the people who took French in my HS are sweating rn

[wintry tundra]

I'm taking the khan academy course for trig and I learned about the sines and cosines of special triangle angles on the unit circle, but it only showed a 45 45 90 triangle

What is the cosine of 150 degrees tho? I don't see what it could be

[humble pulsar]

draw a 30 60 90 in quad 2 with angle between the arm and neg x axis being 30

[wintry tundra]

Oh I see

So what would I do then I can imagine it in my head

[humble pulsar]

cos is the length of the side on the x axis

and negative since it's neg x

[wintry tundra]

Oh ok

So I just have to find cosine of 30 then subtract 1 since its negative

Wait no nvm I was thinking something else

So just -cos 30

[humble pulsar]

yes

[wintry tundra]

Ok

[humble pulsar]

-sqrt(0.75)

[wintry tundra]

And cos 30 is 1/2sqrt3 on a unit circle

Wait no

Sqrt3/2

So to get this straight, if I have to find the trig functions of an angle >90, then I should find its supplement and use that with transformations appropriate to the quadrant?

[limpid gust]

i need help pls

[main lintel]

@limpid gust With what? what is your attempt?

[limpid gust]

i don't know how to write an equation, the last sentence is messing me up

[main lintel]

Which problem?

[limpid gust]

4

[main lintel]

Where is your sketch?

https://jspaint.app/

[limpid gust]

wdym?? I am trying to write an equation

i need to find t

which i can do

i think

[main lintel]

How can you solve it without the sketch 😐

[limpid gust]

what sketch?

[main lintel]

sketch of the triangle and the angles

[limpid gust]

just a labled triangle

hows that help

[main lintel]

You need to see where are these angles and how they relate

[limpid gust]

hold on one minute

[inland canyon]

i completely forgot

how todo this shit

someone for the love of god fucking help me i have like an hour to complete this big ass test 😔

[main lintel]

What are these data, and what is x?

29 is the angle, and X and that number on the rigth?

[inland canyon]

number on the right is the arc

@main lintel

x is the smaller arc

is x just 29 bruh

[main lintel]

@inland canyon I dont know if theer is unique solution

You must have more data

Like where is the position of 29° point and so on

Yes, this is a wrong task

[wintry tundra]

What about it

It's epic notes but no context given

[jagged magnet]

Yo do you have any geometry notes

Chapter 6 quadrilateral

[wintry tundra]

<@&286206848099549185> so would sin theta be equal to cos theta only at +/- $\frac{\sqrt{2}}{2}$

[somber coyote]

hiidostuff

[upper karma]

no $\sin(x)=\cos(x)$ whenever $x=4\pi n+\pi/4$, for any integer $n$

[somber coyote]

spinsicle

[upper karma]

@wintry tundra

[wintry tundra]

Yeah I should have specified

They will be root 2 over 2 on the y at the intervals of 4pi * n + pi/4 on the x

Bc of 45 45 90 right triangle theorem

[upper karma]

well you can look at their graphs

[wintry tundra]

yeah i did

[umbral fossil]

@wind vector Could you explain tangent in the unit circle

Like its behaviour

[wind vector]

tangent is tangent to the unit circle

[wintry tundra]

unit circle best circle 😩

[umbral fossil]

??

[wind vector]

cos becomes smaller when you approach 90

[umbral fossil]

I know cos and sin

Just not tan

[wind vector]

sin becomes bigger

$\tan x = \dfrac{\sin x}{\cos x}$

wait does it not work here

[wintry tundra]

space btwn $ and \tan is what ruined it i think

[somber coyote]

Gamination

[wind vector]

yea lmao

this ^

if u know sin and cos on the unit circle

their ratio is tan

[wintry tundra]

when use x and not theta

[wind vector]

idk hwo to type theta lmfao

[wintry tundra]

$\theta$

[somber coyote]

hiidostuff

[wintry tundra]

look at it

beautiful

[wind vector]

omg

[umbral fossil]

eh

[wind vector]

so curvy

[wintry tundra]

i know its so nice

what a theta

[wind vector]

$\tan \theta = \dfrac{\sin \theta}{\cos \theta}$

[somber coyote]

Gamination

[wind vector]

ahhhhh

[wintry tundra]

i actually only know what sin and cos do on a graph

i still need to do the khan lesson on tan

bc im taking trig and calc early on khan academy

[wind vector]

tan goes to +- infinity when their angle reaches 90 degrees

[wintry tundra]

cuz im impatient and want to start doing physics

[wind vector]

likewise for 270 degrees

[wintry tundra]

hmm

[wind vector]

because tan theta approaches 1/0 format

[wintry tundra]

can you show me how that works with sin and cos

oh ok

[wind vector]

because sin at 90 degrees is 1

and cos at 90 degrees is 0

and their ratio is 1/0

[wintry tundra]

so 1/0 is graphed as infinity

[wind vector]

not infinity

+-

[wintry tundra]

my b

but its just graphed with the range of [-infin,infin]

[wind vector]

it comes out from the negative side when its degree is just higher than 90 degrees

like 91

[wintry tundra]

theoretically its all connected, right?

like some sort of cylindrical graph

[wind vector]

no

the x axis is the degrees

[wintry tundra]

y is tan theta right

[wind vector]

y = tan x

yes

[wintry tundra]

cool

[wind vector]

and tan x = sin x / cos x

[wintry tundra]

i need to see csc theta sec theta and cot theta

yeah

hmm

so if sin^2 theta + cos^2 theta = 1

[wind vector]

https://www.desmos.com/calculator

it can graph functions

[wintry tundra]

then u could say sin theta = 1 +cos theta i

so tan theta = 1+i

weird

wait no

still wrong

binomial division doesnt work like that

so just 1+i all over cos heta

theta

[wind vector]

from where did u drag in i

[wintry tundra]

root -cos theat

theta*

[wind vector]

wait

$\sin^2 x + \cos^2 x = 1$

[somber coyote]

Gamination

[wind vector]

what next

[wintry tundra]

subtract cos^2 x from both sides

$sin^2 x = 1 - cos^2 x$

[somber coyote]

hiidostuff

[wind vector]

and what next

[wintry tundra]

root both sides

[wind vector]

$\sin x = \pm \sqrt{1 - \cos^2 x}$

[somber coyote]

Gamination

[wind vector]

what next

[wintry tundra]

hmm

we would need to find a way to divide cos x over sin x

[wind vector]

what that supposed to mean lmao

what do you have in mind

[wintry tundra]

sin x over cos x

[wind vector]

that's tan x

[wintry tundra]

finding what tan x is

i know

what would it be equal to in regards to this theorem

or identity i mean

[wind vector]

wait

$\cot x = \dfrac{\cos x}{\sin x}$

[somber coyote]

Gamination

[wind vector]

cot x is 1/tan x

@wintry tundra what about it?

[wintry tundra]

i was just wanting to see if you could write tan x using the pythagorean identity

[wind vector]

$\sec^2 \theta - \tan^2 \theta = 1$

[somber coyote]

Gamination

[wind vector]

@wintry tundra

[wintry tundra]

sec = hyp/adj

tan = opp/adj

opp-hyp / adj = 1?

no

hyp - opp / adj = 1

yeah no that makes sense

[wind vector]

the square

[wintry tundra]

and square the values properly

it all comes together

[upper karma]

can any one help me?

[cosmic pebble]

From

#❓how-to-get-help

Rule 1

The help channels are solely for help with math, so feel free to post your question. Asking whether you can ask a question or if anyone knows about some specific topic is unnecessary, so please try to avoid questions of that nature.

[sleek depot]

[sleek depot]

Find A (angles in linear pair)

[sleek depot]

AB = BC
Hence B = C
Let x = B = C
Apply angle sup property of triangle and find C

Apply similar concept in next Q

[upper karma]

i don't understand what yoiu mean

[sleek depot]

What part

[upper karma]

number 1

[sleek depot]

What part of sum 1?

[obtuse tapir]

some nice intuitive questions please

[upper karma]

no what is the angle of c

@sleek depot

[sleek depot]

C = (180-A)/2

And A = 180-126

[hybrid solstice]

can someone help

pls

[alpine cliff]

What do you know?

[hybrid solstice]

2 apirs of parallel sides

[alpine cliff]

Good

[hybrid solstice]

and 2 sides are eqaul

but theyre right next to each other which hurts my brain

[alpine cliff]

What does 2 sets of parallel lines mean

[hollow raven]

It has 2 pairs of sides that are parallel with each other

[hybrid solstice]

its a parallelogram

[alpine cliff]

Good

What are the properties of a parallelogram

[hollow raven]

Oh nevermind 😳

Carry on

[alpine cliff]

@hybrid solstice igtg soon you gotta be quick

[hybrid solstice]

Opposite angles are equal
Opposite sides are equal and parallel
Diagonals bisect each other
Sum of any two adjacent angles is 180°

[alpine cliff]

Nvm Tyty

Take the second line

Our last bit of information is about sides

So the angle prop won’t matter

The diagonals won’t matter

And so that leaves us with opposite sides are equal and perfect

Mark the diagram

What is UT equal and perfect with

And what is UN equal and perfect with

[hybrid solstice]

te

[alpine cliff]

So

[hybrid solstice]

and

ne

[alpine cliff]

UN = TE

And UT = NE

How could you draw that

Equal and perfect lines

Also called congruent

If I gave you 2 segments, said I know they are congruent, and said show that they are congruent without using words

How would you do it

[hybrid solstice]

UHH

same lengt?

[alpine cliff]

I gotta leave but basically you just tick them the same way

[hybrid solstice]

oh

[alpine cliff]

UN already has one tick

[hybrid solstice]

oh ok

[alpine cliff]

So TE has to have one tick

And same logic for the other pair

[hybrid solstice]

oh ok

[alpine cliff]

So now you have all sides equal and perfect

And 2 sets of parallel lines

Analyze the diagram and see if there’s anymore info to consider

If not find the shape that fits the two statements above

[loud bone]

i feel a bit dumb for not being able to figure this out bit lets say i have a ring and i move that ring at a constant speed (along the x axis), at some point the ring starts to get slid on top of a sheet of paper (its edge is perpendicular to the direction of the rings movement)

how do i calculate the area of the paper thats enclosed by the ring?
its not a linear relation between time and how much is covered right?

here is an illustration and i want to find the marked area as a function of t

[silent plank]

area of sector - area of triangle

[jagged magnet]

can i get some help on this

[alpine cliff]

Do you remember what similarity is or how it works?

[green tree]

@jagged magnet i think its sss

[alpine cliff]

nah

[wide swan]

sas

[upper karma]

I need help creating a scenario that meets these outcomes.

I’ve met all the highlighted outcomes

I just need to meet the two triangle and 3D triangle outcomes

anyone know how to find X and Y?

[vocal agate]

@upper karma to find x you would use tan 76= 29/x and then you can solve for x since there is one unknown

anyone know how to solve in intervals of i

[alpine cliff]

Like in the complex plane?

[wide swan]

are u sure thats not an upside down questoin mark @vocal agate

im p sure thats an upside down q mark

[upper karma]

can someone help please

<@&286206848099549185>

[steep temple]

15 minutes

the wait time is 15, not 9

also consider posting what you have tried

[vocal agate]

@wide swan i think you are right what does an upside down question mark mean

[devout echo]

I don’t understand what the Pythagorean theorem is

My teacher rushed me and I’m having a hard time finding a website or a video that explains it well

[humble pulsar]

if you have a right triangle, you take the legs of the triangle (the sides which make a right angle) and square each of their lengths then add those together. That sum will equal the length of the hypotneuse squared

[devout echo]

Ohhhhh I didn’t know I had to square them

This changes a lot

Thanks

[humble pulsar]

$a^2+b^2 = c^2$

[somber coyote]

moshill1

[wide swan]

it might've printed wrong

[upper minnow]

Can anyone help

<@&286206848099549185>

[grizzled lantern]

@upper minnow what is true about the sides of a square?

[upper minnow]

All equal

[grizzled lantern]

Also what do you know about the angles of a square

[upper minnow]

All 90 degree

[grizzled lantern]

Yep

So 90/2 is?

[upper minnow]

45

[grizzled lantern]

So angle s ant t must be?

[upper minnow]

45?

[grizzled lantern]

Yep

180-90 is?

[upper minnow]

90

[grizzled lantern]

You now have all angles needed

[upper minnow]

So how do I get the perimeter

From the angles

[grizzled lantern]

You can use soh cah toa

Angle a is 90 degrees

[upper minnow]

I never learned that

[grizzled lantern]

Ok cool

You heard of a 45 45 90 triangle?

[upper minnow]

Isosceles?

[grizzled lantern]

Sat is a 45 45 90 triangle

[upper minnow]

Ok yeah

[upper minnow]

Hmm ok I think I got it

[grizzled lantern]

then what do the sides equal?

[upper minnow]

Can u check over the statements

[upper minnow]

I got 24.14 as the perimeter

[grizzled lantern]

what are the sides?

[wide swan]

also try to use exact form not decimals

[upper minnow]

7.07 twice and 10

[upper minnow]

Ok

[grizzled lantern]

square root of 50?