geometry-and-trigonometry

[upper karma]

yup

circle

👍

[inland canyon]

a radian would be like

man

whatinthefuck

my teacher hasnt shit about a radian

we've learned about pi and shit tho

[upper karma]

ah ok

i should probably get a different formula then sry

give me a few minutes - and i will ping you

[inland canyon]

ight

wait a fucking secind

is a radian like

this little bit

[upper karma]

let's see

[upper karma]

not necessarily

[upper karma]

🤔

yes

[inland canyon]

so 180 is 1 radian

[upper karma]

180˚ is π radians

so

[inland canyon]

how the fuck is pie

pi*

[upper karma]

?

[inland canyon]

a radian

like

[upper karma]

it is by definition?

[inland canyon]

would u do the radian times pi

or like

[upper karma]

ok

so

[inland canyon]

so

[upper karma]

180˚, halfway around the circle

is π radians

so

[inland canyon]

imma get a drink \

[upper karma]

the blue angle is 1 radian, or about 57˚

[inland canyon]

back

[upper karma]

hi back

i'm sir topham hatt

[inland canyon]

lol

[upper karma]

anyway

specifically 1 radian is 180/π degrees

t!math 180/π

rip that isn't in there

[inland canyon]

wait so like

180 divided by 3.14?

[upper karma]

that is an approximation

3.14 ≠ π

3.14 is about π

[inland canyon]

so a radian is basically just like

the piece cut out basically

is the radian?

[upper karma]

from the picture i can't necessarily tell what the exact angle measure of the blue region is

lemme see if i can draw a picture

[inland canyon]

we can back to my problem

hold on

[upper karma]

r is radius btw

[inland canyon]

mhm mhm

ok so

okso

is the little

0 witha line through it

the thingy

[upper karma]

$\theta is a variable that typically represents an angle$

[somber coyote]

Sir Topham Hatt

[inland canyon]

thats just a varible isnt it

[upper karma]

fuck

$\theta$ is a variable that typically represents an angle

[somber coyote]

Sir Topham Hatt

[upper karma]

yes exactly

[inland canyon]

fuck it bruh

[upper karma]

called "theta"

[inland canyon]

could have just said x

😔

[upper karma]

welp - that's how it's defined

[inland canyon]

alright

so

[upper karma]

¯_(ツ)_/¯

[inland canyon]

for me

[upper karma]

sry

[inland canyon]

it would be like

[upper karma]

for you, $\theta = 147˚$

[somber coyote]

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

[upper karma]

bruv

[inland canyon]

147/360 * (2πr)?

[upper karma]

exactly

arc length = $\frac{(\theta)(2πr)}{360}$

[somber coyote]

Sir Topham Hatt

[inland canyon]

so do i like

[upper karma]

arc length = $\frac{(147)(2πr)}{360}$

[somber coyote]

Sir Topham Hatt

[upper karma]

you would also need radius to solve for arc length

[inland canyon]

so do i times 147 x 2pie

or like

i keep saying pie

breh

[upper karma]

¯_(ツ)_/¯

all good

arc length = $\frac{(147)(2πr)}{360}$

[somber coyote]

Sir Topham Hatt

[inland canyon]

so do i do 147 times 2pie times radius

[upper karma]

yes

divided by 360

[inland canyon]

mhmmhm

lemme try this out rq

gotta go get calculator

okso

it doesnt give me the radius

can u find radius with centeral angle?

u cant right?

it only gives me the centeral angle

[silent plank]

what's the original question?

[inland canyon]

whats the measure of arc ab?

[silent plank]

the measure of the arc will be the same as the measure of the central angle it subtends

in this case: 147°

[inland canyon]

wait wait wait

u mean to tell me

the fucking arc

is the same measure of the central angle

@upper karma

[silent plank]

(it subtends, yes)

[inland canyon]

thank u for like

dumbing that down

alot

thanks

thank u to top hatt but man u made that shit 10x more complicated then it needed to be

[silent plank]

tbf your intial question wasn't clear

[upper karma]

i thought you meant arc length bro

[low scaffold]

@inland canyon For these kinds of questions I find working in radians from the beginning is easier

[upper karma]

yea - had to explain what a radian was, though

[low scaffold]

My bad I meant Viper

[upper karma]

oh ok

[main lintel]

Wait, the arc has a measure of length, it is not the same as angle.

The respective angle and the arc length are not the same.

[silent plank]

arc measure is different from arc length
they were asking for arc measure

[main lintel]

I'm not native english, but I'm not sure that is correct.

"An arc measure is an angle the arc makes at the center of a circle, whereas the arc length is the span along the arc" - indeed you are correct

I think in my language you would say something like "an angle measure of the arc"

[inland canyon]

it was correct

also

ineed more help lmao

@silent plank think u can help me with this one too?

[main lintel]

Just post the question, you shouldnt ping specific users with new questions.

[inland canyon]

fair enough

What is the measure of minor arc KM?

ik the formula for like

inscribed arcs that are on the outside

m1 = 1/2 (x-y)

[main lintel]

What do you mean by KM, there are no K nor M on the image

[inland canyon]

i mean ab

i wrote ab insted of km

would u do like

(x-y)/(2)=75

x-y divided by two = 75?

[main lintel]

What do you mean by y?

[main lintel]

(anyways, I dont think this has an answer without more data)

so y is the longer AB arc?

[inland canyon]

ye

[main lintel]

This is not uniquely defined.

You can have a lot of points from which different arcs are seen at 75 degrees

[inland canyon]

bruh

the fucking

this angle is 75 degrees

i cant like

dumb it down anymore

this is all the info im given

i need to find x

the angle inscribed is 75 degrees

[main lintel]

What do you mean by the angle inscribed?

Is 75 the central angle?

[inland canyon]

the formula is m1 = 1/2 (x-y)

dawg

thats not a centeral angle

centeral angles are like this

from the center

its inscribed

[main lintel]

So you are drawing it incorrectly

[inland canyon]

no

dawg

those are two diffrent things in 1 picture

a is a inscribed angle

b is a centeral angle

[main lintel]

You cannot draw the angle from the point outside of the circle, you draw it from a point on the circle that is not within the arc.

[inland canyon]

inscribed angles can be on the outside or inside

man

look i may be retarded

but ur more retarded

[main lintel]

Lol, you need to chill a bit.

You are pretty rude.

[inland canyon]

inscribed angles

all on the outside

man im just being real

and im not being like un-chill

[main lintel]

You are being real by telling me that I'm retarded?

and I'm trying to help you

[inland canyon]

im not being rude homie im just tsyaing

ur not understand

ur understanding less then me

therfor u cant exactly help me

[main lintel]

ok

[indigo chasm]

i just need help on how to find the frequency and period on these two and also the amplitude

[round isle]

have you been taught about frequency, period, and amplitude?

[indigo chasm]

yes

[round isle]

good

so you should know how to find those

[indigo chasm]

i am confsued tho

@midnight socket dont be a dick

[round isle]

huh?

don't be rude

we are just trying to help

[indigo chasm]

your being rude

[round isle]

go and complain about me to the mods in #discussion

leave this channel open for others to ask a question

[main lintel]

Lol, half an hour ago a guy called me retarded, now you are called dick 😄

what's going on 😄

[round isle]

sometimes ungrateful people enter this server

they think we will give them answers straight away

that is not the purpose of this server

we start by gauging their understanding of the problem

and then guiding them appropriately

this person believes they are entitled to an answer

[main lintel]

It's ok. Combos I think you should appologize and then state what you know to do with the problem, and then what is giving you troubles. 🙂

[indigo chasm]

im not looking for answer im looking for help and you just say well then you should know it

@round isle

[main lintel]

I think he meant for you to continue with what is troubling you.

[round isle]

yes I said that because I wanted to see how you are trying to solve the problem

[main lintel]

Missunderstanding 🙂. Appologize for the rude words and go on 🙂

[round isle]

don't take out your frustration on us

[indigo chasm]

@round isle well im sorry for the inconvinence i hope u can forgive me

[main lintel]

@indigo chasm Do you know to find the amplitude?

@inland canyon For an angle to denote the inscribed angle its vertex must be exactly on the circle.

[late breach]

Can someone help me with a math problem? I am trying to convert this to rectangular

                  1
r =   ____________________________
          5sin(theta)-cos(theta)

[pine brook]

guys, what is the sine, cosine, and tangent for again? i'm new to trig and i don't really remember

[slate blade]

sin cos and tan are special trig "ratios" that kinda display a relationship between the sides of a triangle and its angles.

[pine brook]

and what are those ratios?

[slate blade]

It helps to remember this Mnemonic:
SOH-CAH-TOA

bruh wait hold on let me use bot properly

[pine brook]

ok

[slate blade]

[\sin{\theta} = \frac{opposite}{hypotenuse} ]

[somber coyote]

Doodaide

[slate blade]

[\cos{\theta} = \frac{Adjacent}{hypotenuse} ]

[somber coyote]

Doodaide

[slate blade]

[\tan{\theta} = \frac{opposite}{adjacent} ]

[somber coyote]

Doodaide

[slate blade]

and this is the triangle that u wanna use

[pine brook]

so sine of theta is the opposite divided by the hypotenuse, the cosine is the adjacent over the hypotenuse, and the tangent is the opposite over the adjacent?

[slate blade]

well kind of.

[pine brook]

so basically I just divide the two and BAM i get the ratio?

[slate blade]

yeah. It's the relationship between the side lengths and the magnitude of the angle

[pine brook]

and the θ is the angle, or the "variable"?

[slate blade]

Here's a more detailed diagram

well any of the three can be the variable, you just need to rearrange for whatever you need

if you are given the sides, you can solve for the angle

by using an inverse function

[pine brook]

so basically, it's like algebra

[slate blade]

yeah It's cooler algebruh

[main lintel]

It is usually the angle, but you can also consider it as an abstract variable unrelated to angle.

[pine brook]

and what are the ones i see on a calculator with the sine raised to -1?

[slate blade]

that's the inverse function. Called inverse (sin/cos/tan) and that allows you to find the angle if given the sides

[pine brook]

so if you know the angle you can find the sides?

[slate blade]

[sin^-1 (\frac{opp}{hyp}) = \theta ]

[somber coyote]

Doodaide

[main lintel]

If you know the angle, you can use regular sin/cos/tan to get the "ratio of the sides", not the sides themselves without more information.

If you know the ratio, you can use sin/cos/tan ^-1 to get the angle. So the inverse.

[slate blade]

Well not you need 2 variables to find the other one. Else you need to simultaneous eqns

[pine brook]

ohhhhh ok i get it now

well that was a little complicated
so the regular sin, cos, and tan are for finding the ratio
and the inverses are for finding the angle

[slate blade]

yeah pretty much

sometimes instead of seeing inverse sin, you see arcsin or arctan or arccos

[pine brook]

wait a minute, i found this on the pinned messages

[umbral snow]

Fastest lesson in trig I've ever seen haha. I understand if you have any questions

[pine brook]

there's a whole list of trigonometry stuff and the unit circle too i guess

[slate blade]

lol I have no clue where this guy is level wise It took me way longer to understand this stuff

which one the unit circle or the dark latex text

[pine brook]

you mean me?

[slate blade]

yeah

[pine brook]

i'm only a freshman in highschool actually
i got used to complicated maths growing up so it seems kinda easy

[slate blade]

well that's good to hear

so yeah you should start doing intro to trig and stuff. It's actually really fun once you get to the cooler stuff. Master the basics first tho

but back to your queries which picture the unit circle or the black text

[pine brook]

what do you mean? i don't have a question about the pics

[slate blade]

oh? that's good I think I'm having a stroke lol

[umbral snow]

It's natural to have questions on this stuff, haha. It's just a bit more advanced trig.

[pine brook]

i go to a special school where everything is much harder than the regular curriculum, so i am ded

ok bye guys

[slate blade]

CYA

@late breach
[ r = \frac{1}{\sin{\theta}-\cos{\theta}}]

[somber coyote]

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

[slate blade]

lol close enough

You then do this

\begin{align}
r\times (\sin{\theta} - \cos{\theta}) = 1 \
\therefore r \sin{\theta} - r \cos{\theta} = 1 \
\because r\cos{\theta} = x, and, r\sin{\theta} = y\
\therefore y-x = 1\
\end{align}

[somber coyote]

Doodaide

[slate blade]

this is what you wanted right? I learnt a different terminology from polar to coordinate form. Hope this helps lol

[upper karma]

What angle, measured in radians, does point travel through when the wheel makes of a complete counterclockwise revolution?

[main lintel]

2pi. It is irrelevant if the wheel is running on the surface or not, it just makes a complete revolution (that is rotation)

A complete rotation is 360 degrees of rotation or 2pi rad.

[inland canyon]

What is the measure of minor arc AC

can someone just tell me the answer

i have 6/10 i need 7/10 to move on to my next assigment

would be greatly appreciated

<@&286206848099549185>

[slate blade]

bruh u need to draw it a bit neater and where is the 206 supposed to be

[inland canyon]

DA

[slate blade]

so is line BA tangent to the circle at point A?

[inland canyon]

ye

[slate blade]

and what is C supposed to be?

[inland canyon]

BD

idk bruh im just copying the grapgh i was given

man its all good

i figured it out

and got 7/10

so i can move on

[upper karma]

Hello I need help is anyone there

<@&286206848099549185>

[main lintel]

You should post your question, and if noone answers in 15 min you can ping helpers. Aren't these the rules?

[upper karma]

Anyone able to help me out for these

bad image

?

[main lintel]

Yes, please if you can delete the above images, and past first image correctly rotated, and second image a bit more readable.

@upper karma

[upper karma]

Just these 2

<@&286206848099549185>

[slate blade]

for 5:

[
\frac{(x^3)^{-2}}{x^2\times x^5}
]

[somber coyote]

Doodaide

[slate blade]

there we go

You do a bit of a combine right.

[
\frac{x^{-6}}{x^7}
]

[somber coyote]

Doodaide

[slate blade]

which simplifies to

[
\frac{1}{x^{13}}
]

[somber coyote]

Doodaide

[slate blade]

therefore k=13

and for 6) it's 4

I don't wanna type it all up cuz I'm a lazy bastard but if you need the working just holler

@upper karma

also this ain't geometry or trig

[kindred rapids]

How would I find the stationary points from this equation y = x^3 - 12x + 5 ?

My teacher said we should get two values for x and two values for y ... but how?

[silent plank]

what do you know about stationary points?

[fast tide]

@kindred rapids that is a strange way of describing the problem but I think you are looking for the values of x where the derivative of the function = 0

stationary applies in physics where the position is y time is x

this is either a physics problem or a calc 1 problem

all you need to do is find the derivative of the function. set the function = 0. then solve for x.

the derivative function is $g = 3x^2-12$

[somber coyote]

galois

[fast tide]

solve for x $3x^2-12=0$

[somber coyote]

galois

[fast tide]

x=2,x=-2

use that to solve this y = x^3 - 12x + 5

[kindred rapids]

Yes thank you
I worked it out

[kindred rapids]

Sorry...I only now started the topic, I haven't gotten used to the language yet that's used when talking about this...

[wooden quiver]

is considered the regular quadrilateral pyramid VABCD with the peak in V knowing that m = 2V6 cm and l = 4cm calculated Ap

[fast tide]

@kindred rapids I don't think you did anything wrong. I think your teacher is trying to give you physics but leaving out a lot of important details.

maybe the next physics teacher you get will be better

[kindred rapids]

Oh this is Add math

[fast tide]

attention deficit disorder?

[kindred rapids]

Additional Mathematics

[fast tide]

like AP math HS level

advanced placement

[kindred rapids]

I'm not sure what that is...

[fast tide]

in america we call it AP classes. this gives college credits while in high school

[kindred rapids]

Oh well I'm in the Caribbean...they just call it Additional Mathematics class. I'm in the science class and it's just one of the eight subjects that I do

[fast tide]

this is what it means https://www.calculushowto.com/types-of-functions/position-function/

stationary means the speed of the object is 0

x = time

when the speed of an object is 0 the position function will not change at all. the position is constant.

[visual trail]

how do i solve tan(a)*sin(a)?

[round isle]

tan = sin / cos

[visual trail]

sin^2/cos

yes

[upper karma]

how do you check an ambiguous case for law of sines for 0 1 or 2 solution?

[tender prawn]

sin x + cos x = -1/5, find cos 2x

i tried squaring both sides and got sin (2x) = -24/25

yep im extremely stupid nvm

[waxen vale]

could anyone help me with these, I always seem to mess up which quadrant the rotation lies in.

[humble pulsar]

Rotation matrices take the form cos sin -sin cos, so what quadrant is cos and sin both negative @waxen vale

[waxen vale]

wait am I wrong, or is cosine positive and sin negative

meaning that it lies within the second quadrant, correct?

[humble pulsar]

the (1,1) entry is cosine, so cosine is negative

the (1,2) entry is sine, so sine is negative

[inland canyon]

<@&286206848099549185> bruh i get this is super easy i just forgot how to do it, but like how does one find an arc given this information? its not like the arc is twice the size of the inscribed angle is it?

[barren meadow]

How does -cosx/sinx-1 become cosx/1-sinx ?

[woeful sleet]

multiply -1 to both numerator & denominator

[barren meadow]

ohh ok

[grizzled lantern]

@astral merlin u still need help?

ok well i usually figure out what i use based on the angle that im given

for the first one you can see that the side x is adjacent to the angle 63 and we also have a hypotenuse

So i think "hey, i have an adjacent and i have a hypotenuse. oh i can use cos because of cah"

yeah basically

np

then u'd use an inverse tan, cos, or sin

it would be the same process

look at 7

you have an opposite and a hypotenuse so i'd use sin

but you need to use inverse sin

yep

[wild tundra]

https://media.discordapp.net/attachments/763611244694994964/783845526327656448/POTD-66-0.png

someone might know where I got this from..

I got

tan x = (1+-sqrt(5))/2

[upper karma]

May someone help me with this?

[oak citrus]

you kind of doxxed ur principal lol

[edgy slate]

Anyone here?

Can someone explain to me what 't' or the parameter means in a parabola?

[wild tundra]

t is parameter used to represent the parametric form of coordinates in a parabola

like how in circle you use rcos theta and rsin theta

@edgy slate

[edgy slate]

Yeah but in circles I knew r meant the radius

So what does t mean? @wild tundra

[wild tundra]

t doesn't mean anything

It's a general parameter

[edgy slate]

Then?

[wild tundra]

if someone asks you in what form the coordinates of a points of a parabola be ?

[edgy slate]

at^2, 2at

[wild tundra]

yeah

[edgy slate]

What is t here??

How do I find t?

[wild tundra]

it's a parameter ! you can find it only if some condition and the equation of the parabola is given !

for example, take y^2 = 4x

[edgy slate]

where a=1?

[wild tundra]

if a point (1,4) is given, you have to compare it with ( at^2, 2at) and find t.

[wild tundra]

yes

value of t changes with each and every point on the parabola

hope u understood

[edgy slate]

Kinda

[upper karma]

dont cheat, this is from MODS

[wild tundra]

Kekw

@upper karma solved that question ?

[upper karma]

yes

[wild tundra]

Nice

[upper karma]

domain is very large

[wild tundra]

Wont it be enough to find how many values of x occur in 2π and then we can just find for 50π ?

[wild tundra]

I got stuck here

Just help me out after today

[upper karma]

hint: tan is periodical

[hallow bay]

Are there any more values of theta I haven’t spotted? I know I’m yet to subtract 30 from the final values, but I’m unsure as to whether I can subtract the initial value (36.9) from 360 as well as 180.

[supple onyx]

No this is fine

[upper karma]

,w sin(theta + 30)=3/5

[somber coyote]

Results provided by WolframAlpha

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

Requested by Sol Invictus#3760

[upper karma]

@hallow bay

...

[hallow bay]

That means very little to me

This, however, is cocking me up no end.

[humble pulsar]

@hallow bay which question in particular?

[hallow bay]

@humble pulsar Managed (a). Can't do (b); got -60 which for p which it clearly isn't. Haven't attempted (c).

[humble pulsar]

b is just find when the curve = 0

0=2sin(x+60)

sin(u) = 0 when u is an integer multiple of 180

$x+60=180n, n \in \mathbb{Z}$

[somber coyote]

moshill1

[humble pulsar]

so yeah, when n = 0, you get x = -60, but n = 1 yields x = 120

[hallow bay]

@humble pulsar As I didn't really understand that, I just settled for the logical method of showing that the graph had been translated 60 degrees to the left. For (c), what do I do after I have subbed in -1.6 for -1.6 = 2sin(x + 60) ?

[humble pulsar]

you solve for x again

[hallow bay]

@humble pulsar I am, but I get x = -113.1

[upper karma]

Yo

I need help boys

[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.

[upper karma]

Anyone know this

Is this a test

[oak citrus]

what are the different types of triangles

[upper karma]

Naw

[oak citrus]

ie. scalene , etc etc

[upper karma]

Its hw

Idk the triangle

That’s all it gives

[oak citrus]

ok what do all angles need to add up to in a triangle

[upper karma]

180

[oak citrus]

ok

so now you have 3 angles

and you know one is 60

[upper karma]

Yeah

Would they all be 60

[oak citrus]

they are all equal

so that means x = ?

[upper karma]

60

So would the whole thing just be 180

JKL

What is even going on here

[oak citrus]

idk

aldium take over

thanks

[upper karma]

Yeh

Uh ok

So you know all the angles in a triangle add up to 180° @upper karma

Yeh

Yeah but we don't have the 3 of the triangle yet

Yeah

We have 2 of them, the 3rd one can be found by something you can try to notice

Well wouldn’t all the angles be 60

[oak citrus]

ok yeah i see why im wrong aldium

[upper karma]

Cause the triangle looks equal

Focus here

[upper karma]

No that's bad, don't assume things if you don't have a reason to believe them

Ok

So how would I use that to solve

[oak citrus]

im so dumb i helped someone with a question like this yesterday and helped them correctly

[upper karma]

If the question doesn't tell you it is an equilateral triangle, even though it looks like one, don't assume anything without a proof

Yeah so how would I solve it using the 3x-70

X is -67?

Can you stop filling this channel with nonsense @viscid dagger

@upper karma no

Oh Ight

That wouldn’t equal x tho

I'm gonna assume you are trolling due to your behaviour on other channels, please stop

If you really need help, don't interrupt this conversation

There's literally 0 drama

Uhhh I’m just tryna solve a question lol

Anyone know the answer

I'm pointing out to stop interrupting us

Ayo?

<@&268886789983436800>

[deft blaze]

@viscid dagger troll.

[spark stag]

killed

sorry about that folks

[upper karma]

Thanks

[deft blaze]

good game

[upper karma]

Ight can we help me again lo

Lol

Ok so back, @upper karma i'm not giving away the answer, this server isn't to do so, we promote understanding and critical thinking, so if you want i may continue helping you

Ok

So what do I do now

With the 3x-70

[upper karma]

Here, notice we want the inner angle

Ok

Ok, tell me, how many degrees are there on a straight line?

180

Great

So would you agree then that y and (3x-70)° add up to 180°

Yeah

Hence, $y+(3x-70)°=180°\implies y=180-(3x+70)°$

[somber coyote]

Al𝟛dium

[upper karma]

Do you agree with this so far?

Yeah i do

Don't tell me "yes i do understand" without understanding it, if you have any doubts let me know

Okay

No I get it

Are you able to simplify for a little y=180-(3x+70)

I don’t think so

I think it wants the exact answer

we'll get to the exact answer, but i want you to show some effort.

Yeah I have it written down

this is algebra, 180-(3x+70) can be simplified if we consider the parenthesis

what can you do with it

notice there's a - in front

Yeah

Would u have to distribute

uh yes

can you try and see what you get

if it's wrong don't worry, i'm here to guide you.

So it’s 180 x 3 and 180 x 70?

ok no, notice that this isn't 180(3x+70), but instead 180***-*** (3x+70)

Yeah so how do I distribute that

you'd distribute the -1, remember that 180-(3x+70) is the same as 180-1(3x+70)

or simply notice that when a minus is in front, we change the signs of the parenthesis, if you insist

as such 180-(3x+70)=180-3x-70

Ok

So 110=-3x

this was purely algebra, i suggest you to go to review it after

Yeah I will

[upper karma]

why equal sign though, remember we are simplifying the right side of y=180-(3x+70)

it should be simply y=110-3x

Ohhh ok

do you agree? recap of what we'd done so far:

• We called the inner angle of the triangle y, and noticed that on a straight line there are 180 degrees, hence both angles that "form" the straight line must add up to 180 hence we did
  y+(3x+70)=180
  y=180-(3x+70)
  y=180-3x-70
  y=110-3x

Yeah

What should I do now

if you notice, we have now all the angles of the triangle

would you be able to name them all?

as in for example, the angles are: 90, x and 3x

[upper karma]

@rugged shuttle ??

[queen quest]

Hi folks, I'm trying to figure out how, given a point in 3d, along with a pitch, yaw, and speed, to determine the new x y z after a given period t.

[sleek dove]

I'm so confused

[jovial axle]

bro its a test lol

we cant help w tests

[west basin]

i like the part where it says tests

[jovial axle]

obviously

[west basin]

yeah we cant do test help

[deft blaze]

you should delete that

[sleek dove]

My guy its not that serious but ok

Don't u ever wonder that u just help people with these problems but they don't tell u its part of a test

[upper karma]

how do i write the polynomial equation for the least degree in each set given: 3,3,1,1-2

[queen quest]

I think I figured mine out.

[queen quest]

i could also use a hand, I think. I have a diagram I'd like to post.

[upper karma]

@molten bronze do you need help

with the drawing part?

Or the whole thing

Prove segment AC is congruent to segment CD

Oh yeah then nvm sorry

then you're good

Like naming the angles?

crap

sorry

I had to go

They are vertical angles yes

[queen quest]

I'm working on this but here's my problem area

my problem set is: given pitch, yaw, and distance travelled, get delta x, delta y, delta z

(imagine objects in a 3d coordinate space, and you're tracking one with a known vector)

I think, that in this scenario, the following equations work:

no, something's wrong.

[upper karma]

Can someone help me with my geometry

[molten bronze]

It’s Eli lemme see I’m not that bad

[queen quest]

i'm so close

[upper karma]

@queen quest can you help me with geo

[queen quest]

honestly I really doubt it. it took me two days to figure out this project I'm working on

[upper karma]

I'm not asking for a help on a project

[molten bronze]

I just helped u :0

Hi can anyone help me

#geometry-and-trigonometry

[barren meadow]

What's the half angle identity for csc(x/2)?

[slate blade]

Correct me if I'm wrong but isn't it just the reciprocal of the normal half angle identity of sine?

[
\pm \sqrt{\frac{2}{1-\cos{x}}}
]

[somber coyote]

Doodaide

[slate blade]

and @molten bronze whattup

[upper karma]

Who can help me with my geometry in the future, I'm also in the 9th grade and I'm not getting taught as well

[steady leaf]

Anyone able to help me with geometry as well in about 16hours from now? Im trying to do this homework assignment and we werent taught all of it yet. Please friend me if possible.

[silent plank]

why 16 hours from now specifically?

[dark sparrow]

find the y-intercept explicitly

[buoyant tapir]

I have like 5 trig problems I need help with of anyone is willing to help

[silent plank]

[turbid shoal]

Yo can someone help me in a call?

I have 5 questions for Similar Polygons/Triangles

[obtuse tapir]

Who can help me with my geometry in the future, I'm also in the 9th grade and I'm not getting taught as well
@upper karma ok ill do it

[turbid shoal]

Karl

Marx

[upper karma]

can someone help please

[earnest leaf]

is it possible to simplify sin(n * pi / 2) for n = 1,2,3,... in terms of -1 and n

[dark sparrow]

yes

$\sin(n\pi/2) = \begin{cases} 0 & n \text{ even} \ (-1)^{(n-1)/2} & n \text{ odd} \end{cases}$

[somber coyote]

Ann

[final parrot]

sin(a) = opposite/hypotenuse = 0.3

Therefore say, 3 = opposite, 10 = hypotenuse

Construct a right angled triangle

Use pythagoras to find the adjacent side

Amd then apply cos(a)

$cos(a) = \frac{\sqrt(91)}{10}$

Yuck

Omg

[somber coyote]

Thirst4Carnage

[final parrot]

Well that's better

[dark sparrow]

and also wrong

$\cos(\theta) = -\frac{\sqrt{91}}{10}$, not $\frac{\sqrt{91}}{10}$ as you wrote.

[somber coyote]

Ann

[final parrot]

Oopsie

[dark sparrow]

and also bad latex.

[final parrot]

Ik

Oh the sqrt goes inside {}

sqrt{2}

And \

$\sqrt{1}$

[somber coyote]

Thirst4Carnage

[final parrot]

Yay

[upper karma]

Thanks man @final parrot

[final parrot]

Thank Ann too

[steady leaf]

At that time it was like 11pm Thursday and I have school Friday, so figured after my school day I’d be free

[upper karma]

help please

i dont think there is a CC" so im very confused

ping me if you respond

[dusky surge]

@upper karma hello there!
CC'' with a line on top means that you can draw a line between C and C''

[upper karma]

@dusky surge thank you so much!

[kindred rapids]

Hi...so I have this question and I'm stuck on part A
So i found dy/dx.....and then i said dy/dx = 0 for stationary points but then I got x = square root of 2/3
And I'm not sure how to go about from there. I'll have to find y coordinate when x = square root of 2/3 ...but when I substitute that into the formula...I'm not sure how to solve for y ....help please?

[kindred rapids]

Sorry I meant when I substitute it into the equation **

[dusky surge]

This is not the right channel, but it's okay I guess ^ω^

[earnest echo]

x=√(2/3) is not the only point

And why are you trying to find the y coordinate??

[kindred rapids]

To state the stationary points....that's what the question is asking

I have to get the x and y coordinate for the stationary point....don't I?

[earnest echo]

I don't think you need to state the y coordinates but there is no harm in stating them

[kindred rapids]

My teacher said that I should

[earnest echo]

Anyway
As I said

x=√(2/3) is not the only point

[kindred rapids]

Negative and positive

[earnest echo]

Then say so

[kindred rapids]

Sorry..

[earnest echo]

Are you stuck in calculation??

No need to apologise

[kindred rapids]

Yes in the calculation

I'll substitute back into the original equation but when I do...not sure how to solve it with that square root

[earnest echo]

I mean there is not much you can do about it
simplify it as much as you can

[kindred rapids]

Squars root of 2/3 cubed ....how do I simplify that?

Square root of 8/27 ?

[earnest echo]

Yes

8=4×2
27=9×3

[kindred rapids]

Not sure about -4 (square root of 2/3)
How do I simplify that?

[earnest echo]

All you can do this maybe rationalise the denominator but I wouldn't call that simplification

Though you can first and second term

Before rationalising anything

[humble pulsar]

tan = sin/cos

t=12 for part 1, h=30 for part 2

[upper karma]

Who can help me with 2 and 3

[kindred rapids]

Guys does anyone know how to get 2/3 and -2/3 from square root of 2/3 ?

Or from the square root of 4/6 ?

[silent plank]

triangle inequality
apply the relation between the angles and sides of a triangle

---

your question regarding the 2/3 is unclear

[upper karma]

How do I do this?

[silent plank]

what are properties of a scalene triangle?

[upper karma]

all the sides are unequal

it has no line of symmetry

But how do I use coordinate geometry to prove it?

[silent plank]

distance formula

to show that the lengths of all sides are different

[upper karma]

Oh so thats how I can prove it

then my summary sentence will explain if it is or it isnt

depending if all sides are unequal correct.

Would I use the formula for each side?

[silent plank]

yes

[junior ivy]

do I just add x by -2pi and add y by 2pi 3 times?

[upper karma]

Thank you @silent plank

[sullen epoch]

a right?

[orchid perch]

if someone wouldn't mind getting in vc with me to talk trig, I'd really appreciate it

[humble pulsar]

No

[humble pulsar]

just ask your question...?

[upper karma]

++ ask

[feral idol]

can anyone help

[kindred rapids]

..hi I know this question mightn't be clear but
I have the square root of 4/6 ....is there anything I can do to that square root of 4/6 to get 2/3 and - 2/3 ...?

[hollow river]

Hey, I know the answer for PR is 3 but I don't know how to get that, any help?

[odd blade]

Guys did I do this correct ?

[silent surge]

<@&286206848099549185>

Please Idk how to answer

[tender prawn]

we're not allowed to use trig and idrk how to do this

i know there's a 75 75 30 triangle

and i can figure out some side lengths with that

[upper karma]

show diagram

[tender prawn]

oops

[upper karma]

,rccw

[tender prawn]

thx

im thinking it might have something to do with heron's formula?

the (sqrt2)/2 is referring to the horizontal line btw

wait idk if i even got those side lengths right hold on sry

oh the sqrt2/2 is referring to the one perpendicular to bd sorry

A to the midpoint of CD would be sqrt(1+sqrt(3))

[neon hamlet]

I have no clue how to set this up

I just know that angle C is 70 and B being 40

Mind if someone can walk through this?

[desert vortex]

Think son think

[neon hamlet]

That's the thing I can't. I tried to figure out stuff but it isn't reasonable

Just generally confused

[desert vortex]

You must remember

[neon hamlet]

I can't...

[desert vortex]

Gotta remember them rules my boy

[neon hamlet]

well so far x is 70

[desert vortex]

What is z bro

[neon hamlet]

I solved it

but now last hw question

x = 7

How do you find the base?

?

[tender prawn]

err

you have two congruent segments

both of which you know the values of (in terms of x)

[neon hamlet]

Right

[tender prawn]

so what is MK

[neon hamlet]

41

[tender prawn]

mhm

[neon hamlet]

So other side of the triangle is also 41

[tender prawn]

what do u know about the angles of mkl

triangle mkl

[neon hamlet]

60

OH

[tender prawn]

soooo

[neon hamlet]

So equilateral

[tender prawn]

mhm

[neon hamlet]

So 123?

for perimeter

61 * 3

[tender prawn]

i don't think that's 61*3 lol

61*3 is like 183 or smt

i ended up cheating with geogebra

[neon hamlet]

and last question

https://gyazo.com/fe5ef6b60b7f07499a5586b508bde363

https://gyazo.com/0112e6961c1072c71275621d60f3b8ab

That's my proof

I don't feel great about it

[tender prawn]

lmao i was doing the same exacvt thing a month ago

lemme take a look

[neon hamlet]

Alright

[tender prawn]

are you sure it's asa

[neon hamlet]

Might be aas then

[tender prawn]

i think u have aas rn

yea

u can get asa if u use alt interior angles on the pointy angles

[neon hamlet]

what else?

[tender prawn]

i mean i think thats it

[neon hamlet]

it's alt ext angles right?

[tender prawn]

if u do angle 1 and angle 4

u would have asa

[neon hamlet]

ah alright

[upper karma]

Can someone help me please

[upper karma]

Is that a test?

[hollow raven]

highly doubt that this is some ordinary question 👀

[crystal charm]

Unrelated question : what is the tiling displayed on the banner of the server?

[abstract flume]

sorry cant help with tests

[lime hawk]

not a test

[hollow raven]

2:3:4 -> scale ratio by a factor of x, which is what we want to find to determine the largest angle
We know that the sum of all the angles is 180 degrees in a triangle, so we will use that to our advantage to solve for x

2x:3x:4x

2x+3x+4x = 180

x = ?

once x is solved, determine the largest angle value. Since its fairly reasonable to say that x will be positive even before we calculate this, the largest angle's value will be equivalent to 4x

the rest you will have to try to figure out

[kindred harness]

Not a test just the last homework problem I need to do

[upper steppe]

Anyone here that can help me please?

Geometry is my hardest one:(

[upper karma]

Help

[abstract flume]

since the two lines bisect eachother DE = BE and CE = AE

@upper karma

[upper karma]

What about the reasoning

[abstract flume]

and then by opposite angles

angle AED = angle BEC

so by SAS triangle BEC is congruent to triangle DEA

[upper karma]

Wow I didn’t understand at all

Lmao

[abstract flume]

if you look up the definition of bisection

[upper karma]

So what do I put for the fist reason in

[abstract flume]

it means split in half

equally

so that means that BD and AC are both split into two equal parts

so thats why DE = BE and CE = AE

does that make sense?

[upper karma]

@abstract flume

So what would the reasoning be

[abstract flume]

are those all of your options?

[abstract flume]

you could just write given

i think its a given

because thats the definition of bisection

is "given" an option?

[upper karma]

No

[abstract flume]

oh i see

its a midpoint divides a segment into two congruent segments

because since the lines are bisecting each other they go through each other's midpoints

or the one highlighted

[upper karma]

Ok about

[abstract flume]

actually more like the one highlighted

yeah

its "a segment bisector divides a segment into two congruent segments"

my bad

thats basically what i said

[upper karma]

Ok

It works

What about the next statement

@abstract flume

[abstract flume]

ok

so now by opposite angle theorem

angle AED = angle BEC

[upper karma]

Wait

Can I call

You

?

[abstract flume]

sorry im not allowed to call ppl i dont know on the internet

[upper karma]

Ah

That’s fine

Ok so keep going

What do I do for the next statement

[abstract flume]

angle AED = angle BEC

by opposite angle theorem

[upper karma]

Ok

[abstract flume]

is that an option?

[abstract flume]

SAS

wait its not the two tirangles are congruent

its that the two angles are congruent

[upper karma]

So?

[abstract flume]

you have to specify the angles being congruent first

because SS is not a congruence

[upper karma]

Ok

So what do I do

@abstract flume

[abstract flume]

you need to say that the angles are congruent

[upper karma]

Ok

So what do I put

[abstract flume]

angle AED = angle BEC

[upper karma]

Ok

And for the reason

I say?

[abstract flume]

opposite angle theorem

do you know that one?

[upper karma]

No

I do not

[upper karma]

Wait

I might

[abstract flume]

angle A = angle B

angle C = angle D

[upper karma]

That’s all I have

[abstract flume]

let me read that

vertical angles are congruent

thats another way to say the opposite angle theorem

it has many names

[upper karma]

Ok

Worked

[abstract flume]

now finally the two triangles are congruent by SAS

[upper karma]

Ok so what do I put

[abstract flume]

since they both share two sides

and an angle

triangle AED is congruent to triangle BEC

[upper karma]

Do I add an angle

Or a triangle

[abstract flume]

$\triangle{AED}\cong triangle{BEC}$

[upper karma]

Ok

Ok

[abstract flume]

uh thats didn't quite work

[upper karma]

What statement

[abstract flume]

$\triangle{AED}\cong {triangle{BEC}}$

no i mean the latex

ok whatever

but yeah the two traingels are congrunet by SAS

[upper karma]

Like this

[abstract flume]

yes

and by SAS

[upper karma]

Ok

[abstract flume]

oh wait

also not only does DE = BE, CE = AE

CE = AE for the same reason that DE = BE

sorry

[upper karma]

So after de

[abstract flume]

yes

[upper karma]

I add ce?

[abstract flume]

de isn't congruent to ce

but ce is congruent to ae

de and ce aren't related

[upper karma]

Whattt??

[abstract flume]

$\side{CE}=\side{AE}$

[somber coyote]

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

[upper karma]

I put that

[abstract flume]

and its the same reasoning

[upper karma]

So I change up

The de

To ce

[abstract flume]

and the be to ae

here let me write this out fully:

[upper karma]

Can you just write it on a piece of paper lmao

Ok thank you

[abstract flume]

sure

[upper karma]

Start from the beginning

[abstract flume]

sorry it's sideways

@upper karma

[upper karma]

It’s fine

[abstract flume]

ok i gtg now

[upper karma]

Ok bye

[abstract flume]

hope i helped

[upper karma]

Thanks

Ye

@abstract flume

Help 😦

[solemn axle]

I wish you luck. Proofs are a very special part of math

As you stop your car at a traffic light, a pebble gets lodged in the treads. When you start off again, the distance y of the pebble to the pavement varies sinusoidally with the distance x that you travel. The period is the circumference of the tire and the diameter is 26 inches.

Anyone know how to get a sin equation for this?

I have totally forgotten how to get my abcd out of this

[abstract flume]

@upper karma oh im so sorry i haven't done this in so long you have to do triangle AED cong CEB

[upper karma]

@abstract flume what