geometry-and-trigonometry

[covert hazel]

you need to understand the middle

[late glen]

...

[covert hazel]

let me guide your logic for you.

[late glen]

found it

[covert hazel]

since a ratio like that has to be in the same units, you need to either convert feet to inches or inches to feet

aaaaand it doesn't matter, I guess

[late glen]

im pretty sure that this is geometry since this is in the geometry book apparently

[prisma spruce]

Its an algebra concept tho

[late glen]

and this "homework" is all geometry

im in 9th grade

...

and im already having this

[prisma spruce]

You repeat stuff from alg in geometry

[late glen]

ok then

well that was a year ago

im not gonna remember that

[covert hazel]

did you keep the notes you took in algebra 1?

[late glen]

anyways @covert hazel even tho i never understood one bit about what you told me i thank you for helping me

no i skipped algebra

[covert hazel]

bingo

[late glen]

i only did pre algebra

...

[covert hazel]

khan academy is your new best friend

[prisma spruce]

I did that in pre alg

[late glen]

@covert hazel no im not doing that

i despise it

and in what grade are you? @prisma spruce

[covert hazel]

welp, then you're not going to make it out of 9th grade math

[late glen]

1 semester has already passed

pretty sure i can get out of this one

: )

[covert hazel]

high school algebra is the foundation of the rest of the math you'll need. it's essential that you learn it

[prisma spruce]

How do you just skip alg 1

[late glen]

||||

¯_(ツ)_/¯

[prisma spruce]

Thats like the most important foundation

[late glen]

idk

guess that they thought that it wasnt important

[prisma spruce]

Im in 10th and im dubling with honors geometry and algebra 2 online

[late glen]

wait u did geometry when

???

last year?

[covert hazel]

I've taught a lot of math, and I've taught a lot of high schoolers math. I've also taught a lot of college students math. and the #1 cause of people failing and dropping out of their calculus classes in college is that they do not understand high school algebra

[late glen]

...

well guess that im screwed and im not getting out of college then

[covert hazel]

I strongly recommend that you brush up on the stuff you skipped, because you have knowledge gaps that you don't know you have

[late glen]

well i guess that ill have it maybe next year

idk

[prisma spruce]

No im doing geometry now, im new to the country and they put me in lower classes when I got here so I wasnt able to do it earlier @late glen

[late glen]

all ik is that my geometry book is originally for college

and we're using it

[prisma spruce]

Its also different in every state

[late glen]

ok then

[covert hazel]

what geometry book is it? can you take a picture of the cover?

[prisma spruce]

@covert hazel also thanks, do you have an idea of #16

[covert hazel]

also I highly recommend this site: http://tutorial.math.lamar.edu/Classes/Alg/Alg.aspx

[late glen]

this one

[covert hazel]

@prisma spruce you've got the right idea. for this one, you can't assume it's isosceles, so you have to run through the pythagorean equation for all three sides

[late glen]

thats the book

@covert hazel

[covert hazel]

oh, I actually have a copy of that

I've never flipped through it though

lemme see if I can find it on my shelf

[prisma spruce]

Wait they still do geometry at colleges?

Elementary geometry

[covert hazel]

at my uni, we teach elementary geometry to education majors. these are people who are studying to become k-12 teachers.

k-12 teachers are famously petrified of math, a ton of them couldn't pass their own classes if we didn't teach them their own content in college

[prisma spruce]

Omg you‘re so right

My hs teachers are good but in ms one of my teacher had no idea what she was doing

She was using calculators for stuff like 17+7

[covert hazel]

the actual undergrad geometry class for math majors is axiomatic and proof-based. we go from basic principles and assume nothing (including things like "the paper is actually flat")

[prisma spruce]

I hated logic so much

Logic test was my first math test ever I scored below A

[nocturne musk]

hi by any chance can some one help me with a geometry question?

[prisma spruce]

Ye

[nocturne musk]

its this

I know z is 70

because the straight line is 180 and the angle is 110 so the interior angle would be 70

but I dont know what to do after

like how to get y

[prisma spruce]

You add all the interior angles up and get 262

Then 360-262=98

[nocturne musk]

thank you!

[nocturne musk]

hi

can anyone help me again?

[covert hazel]

[prisma spruce]

Shore

[nocturne musk]

its this

I genuinley dont know this

[prisma spruce]

Its a heptagon

All angles add up to 900 iirc

[nocturne musk]

ohhhh

[covert hazel]

wait wait wait

don't give answers

give methods

[nocturne musk]

yeah

[prisma spruce]

Okkkk

[nocturne musk]

I dont need one now

Im good

thanks guys!

[covert hazel]

ugh, give a man a fish

[prisma spruce]

[covert hazel]

not ugh at you

just ugh at people who are looking for people to do their homework for them rather than actually help them with it

[nocturne musk]

its 141

I get what you mean

I want to understand it before I do it on a test

I always just want an explanation on how to do it so I can do it myself

[covert hazel]

if I have a 12-sided polygon, what do the internal angles all add up to?

[nocturne musk]

900

[covert hazel]

12, not 7

[nocturne musk]

oof

[covert hazel]

well, let's look at it like this.

a triangle's internal angles add up to 180°

a square, 360°

what about a pentagon?

[nocturne musk]

the one in the picture I sent is 7 though right?

540 btw

[covert hazel]

good. now one more, what about a hexagon?

[nocturne musk]

its 720 right?

[covert hazel]

indeed!

so let's make a chart, and see if we can find a pattern here

Sides | Angles
--------------
3     | 180
4     | 2*180
5     | 3*180
6     | 4*180

notice anything?

[nocturne musk]

woah that was quick

[covert hazel]

I already had it typed out

[nocturne musk]

everytime we go up by one more side we multiple by one more number to 180

[covert hazel]

precisely

[nocturne musk]

so 7 sides would be 5*180

[covert hazel]

I put it to you, that the internal angles add up to (n-2)*180°, where n is the number of sides to your polygon.

[nocturne musk]

which the sum of the interior angles would be 900

alright

[covert hazel]

and so for a 12-gon, what would the internal angles sum up to?

[nocturne musk]

1800?

sorry Im just not very good at math

[upper karma]

(N-2)*180

[covert hazel]

1800 is correct

[upper karma]

(12-2)*180

10 * 180

1800

[nocturne musk]

that was my thought process

[covert hazel]

and there's no such thing as not being good at math. you are unpracticed. keep at it, and you'll begin to see these patterns for yourself

[upper karma]

^^

[nocturne musk]

okay, thank you!

also if I see you, Ill have a fish ready for your previous request

[covert hazel]

excellent, good luck 🙂

[upper karma]

Seems like an interesting convo...

I had a question, does anyone have any website recommendations where there are trig problems? Specifically for graphing sinusoids. I am in need of practice.

[covert hazel]

khan academy is your friend

[upper karma]

Indeed

[late glen]

is this correct?

[thorn talon]

Ehh

Are you missing an equals on line 1?

And no

Couple of errors

Well, 2 errors

[late glen]

this is for 196 rght?

that's my doubt

its 4 sqrt 49?

[thorn talon]

No

sqrt(4) * sqrt(49)

[late glen]

separate?

[thorn talon]

What do you mean?

[late glen]

isnt it 4 sqrt 49?

[thorn talon]

No

[late glen]

cause ik that 4 * 49 is 196

[thorn talon]

Yes

[late glen]

but sqrt of 196 is 4 sqrt 49

[thorn talon]

But why would you square root the 49 and not the 4?

[late glen]

¯_(ツ)_/¯

[thorn talon]

sqrt(ab) = sqrt(a) * sqrt(b)

For at least one of a or b non-negative

We can let a = 49, b = 4

And that's the first error

A second error

[late glen]

this is what it gives me

i need to solve that

4/x = x/49

x^2 = 196

then i find sqrt

x = 4sqrt49

no?

[weak shoal]

[late glen]

and it isnt 14

because i did it rn and its wrong

[weak shoal]

So, the sqrt(196) = 4*sqrt(49)?

[late glen]

mhm

[weak shoal]

Yes, because there are two solutions

Let x = -14. That still satisfies what you want

[late glen]

how about 2nd one?

in that one i still have 3 tries

[weak shoal]

What have you tried for that one

Work through it slowly

[late glen]

in the 1st one i have 1 left

and i dont want to risk it

[weak shoal]

Ffs

You have the equation

x^2 = 196

Correct?

It’s quadratic

So there are two solutions

[late glen]

mhm

[weak shoal]

You know that x = 14 satisfies it. x = -14 also satisfies it

[late glen]

im sorry but this is literally my first time doing this

[weak shoal]

So those are your two solutions

Ah

Fair enough

[late glen]

so i type both in?

[weak shoal]

What do you think

[late glen]

¯_(ツ)_/¯

idk

[weak shoal]

I just said that they are solutions

So what do you think

[late glen]

YAY

i got it correct thanks

[weak shoal]

Lmao

[thorn talon]

sqrt(x^2) = abs(x)

[late glen]

second one

is that ok or no?

[weak shoal]

No

[late glen]

this is bs

lmao

i just cant

[weak shoal]

Your writing is crap lol

[late glen]

i skipped algebra

[weak shoal]

Form an equation properly

[late glen]

how am i supposed to know this shit

???

[weak shoal]

$\frac{12}{x} = \frac{x}{3}$

[somber coyote]

Abhijeet Vats:

[weak shoal]

Multiply both sides by 3x

[late glen]

i only did pre algebra and my school literally said algebra doesnt matter so throw it into the trash and lets go straight into college geometry

ok

both sides so 12/3 and 3/3?

no

that's wrong

right?

[weak shoal]

Huh what

Wtf

[late glen]

idfk

dont judge me lmao

[weak shoal]

Bruh, what is $3x \cdot \frac{12}{x}$

[somber coyote]

Abhijeet Vats:

[late glen]

12 * 3x

and 3x * x?

[thorn talon]

Huh?

[late glen]

still wrong?

@weak shoal what's the answer?

[weak shoal]

We don’t give out answers here

Where are you stuck, exactly?

[silent plank]

why are you giving 2 separate values?

@late glen

[late glen]

@weak shoal everywhere

i dont understand shit

my brain cant process it

: (

[silent plank]

what's $2 \times \frac12$

[somber coyote]

ramonov:

[silent plank]

@mint sierra not meant for you

[mint sierra]

Sorry

[silent plank]

@late glen

[late glen]

2/4? @silent plank

simplified 1/2?

[silent plank]

are you saying that 2 * 1/2 = 1/2
implying that 1/2 = 1?

do you know what multiplication is?

[late glen]

yes

ik what multiplications is

[mint sierra]

What's 1 times 1 then

[silent plank]

describe it pls

because 2 * 1/2 is supposed to be trivial
and if you don't realise that you megafked something up when you said its 1/2 then...it seems you don't

or how about this:

do you think there's a difference between: \ $2 \times \frac 12$ and $\frac22 \times \frac12$

[somber coyote]

ramonov:

[silent plank]

or if put it in words

what happens when you put two halves together

[upper karma]

oof

[upper karma]

hi

If I were to write a similarity statement

For DCE

Wouldn’t it be DCE is similar to FGE

[dark sparrow]

would it?

[upper karma]

?

I think it would

But am not sure if it’s right

[dark sparrow]

what's making you unsure

[upper karma]

My friend said otherwise

But I’m pretty sure I’m right

[dark sparrow]

that's good bc you are

[upper karma]

Ok thank you

[nocturne remnant]

angle D = angle F

angle C = angle G

('cause of the parallel lines)

so the triangles are similar

[steel eagle]

SOMEONE HELP PLS

[dark sparrow]

WHY ARE YOU YELLING

WHICH OF THESE DO YOU NEED HELP WITH

[steel eagle]

im scared sorry

i need help with questions 3,4,5,6

[dark sparrow]

alright

well

let's start with Q3, i suppose

just to go through these in order

[steel eagle]

okay

[dark sparrow]

so i see you've alrady found angle 1

as 92°

[steel eagle]

ya

[dark sparrow]

and you're having trouble finding all the rest?

[steel eagle]

uh

thing is

im not sure if its right

[dark sparrow]

what's making you unsure if you're right about angle 1 being 92°?

[steel eagle]

dont know

180-88

Gives 92

[dark sparrow]

,calc 180-88

[somber coyote]

Result:

92

[dark sparrow]

indeed

[steel eagle]

So uh how do I find the other ones

[dark sparrow]

well you've got that pair of parallel lines there

maybe put that to use?

[steel eagle]

huh

Confusion

[dark sparrow]

do you know of any theorems involving parallel lines and their angles with a secant line

[upper karma]

Can somebody help me solve the questions in the middle page?

Plz somebody ;-;

[upper karma]

Plz somebody help

[dark sparrow]

do not post your question across multiple channels

[upper karma]

Can I get help tho

;-;

[dark sparrow]

what exactly are you having trouble with

[upper karma]

15-18

[dark sparrow]

i understood that. you said "middle page".

[upper karma]

I don’t know how to solve them

[dark sparrow]

what about these questions is giving you trouble? what have you tried, and where are you stuck?

[upper karma]

I have no idea where to start

[dark sparrow]

okay let's focus on Q15 at the moment

does your answer of having no idea where to start apply to this question too

[upper karma]

Yes

[dark sparrow]

how familiar are you with the basics of trigonometry

[upper karma]

Trigonometric ratios

[dark sparrow]

how familiar are you with the basics of trigonometry?

please answer my question

[upper karma]

I learned trigonometric ratios

That’s it

[dark sparrow]

ok what did you learn about trigonometric ratios

[upper karma]

Calculating tan cos and sin

[dark sparrow]

ok, so am i correct in assuming that you know which ratio involves which sides of a given right triangle?

[upper karma]

Yes

[dark sparrow]

okay so now look at your triangle in question 15

there's an angle here marked as 55°. the side _____ to it is marked as having length 3, and the _____ is marked as having length x.

fill in the blanks.

[upper karma]

Opposite side and hypotenuse

[dark sparrow]

okay, great

so what trig ratio can you write down now, given this information?

[upper karma]

Sine

3/x

[dark sparrow]

let me clarify

i'm expecting you to write an equation.

involving the angle, a trig function, and your two marked sides.

[upper karma]

Sine(55)=3/x

Hello?

Rip the person helping me left, can somebody else help me?

[upper karma]

Anybody?

[upper karma]

;-;

[thorn talon]

That's fine

[upper karma]

Please somebody

I’ve been waiting hours for help

[upper karma]

Please someone help

[remote heart]

you can use sin(55) which is 3/x then multiply by x, and divide by sin(55) giving you x=3/(sin(55)) for 15

and for #18 you can use law of sines

for number 17, in the latter part of the question, you can use Pythagorean Theorem, and for the former part, you can use inverse cos

[potent steppe]

Big brain to help me pls

[upper karma]

√8 @potent steppe

[late glen]

sqrt 4?

or no

pls help

[steel eagle]

Is it plus or minus 4?

[upper karma]

Which kind of expansion is this?

[potent steppe]

:0

Thankz @upper karma

[upper karma]

@upper karma Taylor expansion I guess

@potent steppe ^^

@upper karma Thanks for the quick help

@upper karma^^

[lavish solar]

Hi guys, I was wondering if someone could help me understand some trig identities?
I don't understand how 1 + cot^2θ is equal to cosec^2θ

cot is just 1/tanθ, and cosec is 1/sinθ
where does the 1 go? and how on earth are they equal?

[weak shoal]

Are you aware that $\sin^2(\theta) + \cos^2(\theta) = 1$?

[somber coyote]

Abhijeet Vats:

[late glen]

@steel eagle plus

4

its positive

[weak shoal]

You can derive an expression relating cot(theta$ and csc (theta)

[late glen]

also pls @ me so that I get notified

the bot didnt work

[steel eagle]

oh okay

[lavish solar]

@weak shoal yeah I'm aware of $\sin^2(\theta) + \cos^2(\theta) = 1$

[somber coyote]

Coach Leyton:

[quiet mason]

whats do you get when you divide both sides by sin^2(theta)

[late glen]

pls gimme answer

because ive tried god knows how many times and it is still wrong

ive checked book

and i still dont know how to do it

pls help

[silent plank]

show your work so we can see what mistakes you need to fix

[signal perch]

HI there. How many solutions does (tan(x)^4)-(tan(x)^2)=1 have. My calculator shows four but someone younger me has 7. I've rearranged it to be (tan(x)^4)=(sec(x)^2) and lots of others but I can't get it to be 7. Which me is getting stuff wrong?

I mean I am the someone, previous me got 7 solutions.

[dire rampart]

$\tan^4(x)-\tan^2(x)=1?$

[somber coyote]

the one n only:

[signal perch]

Yes that the one. Thank you for tidying that up.

[dire rampart]

uh

what's your range

[signal perch]

0-2pi

[dire rampart]

4 sounds about right

,w solve tan^4(x)-tan^2(x)-1=0 from 0 to 2pi

[somber coyote]

Display results online and refine query

Requested by the one n only#3070

[signal perch]

So previous me is not to be believed then.

That guy! I'll get even with him some day.

Thanks for the help, and for some insight into how to use Wolfram!

[dire rampart]

np

[signal perch]

I was worried I'd missed some fancy footwork designed to defeart a calculatored answer.

This may be a bit philosophical, but is there any requirement to simplify both sides of a proof when doing trigonometry (assuming there is no parameter like prove both equal something)? For example if it reads uvwx=yz and I just manipulate the uvwx so it becomes yz does that count?

[dire rampart]

yea

[upper karma]

@dire rampart thanks

[dire rampart]

uh

what for?

[upper karma]

Solution

Also , hi

[upper karma]

May I ask how do you use tan cos and sin to find triangle side lengths I can't seem to understand how no matter how much I try

[thorn talon]

Give an example question

[upper karma]

Ok

You have right triangle

Lemme draw brb

(ignore my terrible drawing skills)

Finding xy

From the adjacent angle

*adjacent side

And an angle

Using cos tan or sin

[thorn talon]

Ok

You have the adjacent side, and want the hypotenuse right?

[upper karma]

Yea

[thorn talon]

Which ratio should you use?

[upper karma]

Hm

Sin?

Wait

[thorn talon]

You have the adjacent side and the hypotenuse

[upper karma]

Cos

Sin is opposite and hypitouse right

[thorn talon]

Yeah

So we want cos(70) = something

What is that something

(70 is in degrees)

[upper karma]

And then you'd get it to cos(70) = 12/h h for hypotenuse

[thorn talon]

Yes

Exactly

[upper karma]

Then h*cos(70) =12?

[thorn talon]

Yeah sure

[upper karma]

What would you do after this

[thorn talon]

Well

You want h by itself

[upper karma]

How would you simplify?

[thorn talon]

It's multiplied by cos(70)

You want to undo this multiplication

[upper karma]

So H = 12/cos(70)?

[thorn talon]

Yes

[upper karma]

Then you would get calculator and simplify the cos 70

[thorn talon]

Depends

But sure

I would just leave it like that

[upper karma]

Yes but the things want simplified number to nearest 100

[thorn talon]

Yeah, then do that

[upper karma]

So

12/0.633

≈18.96

?

[thorn talon]

Your calculator may be in the wrong mode

[upper karma]

What mode would I put it in

[thorn talon]

Degrees

[upper karma]

Yea mine was in rad

[thorn talon]

Yeah

[upper karma]

So 12/0.34

[thorn talon]

If you round, you should generally round at the end

[upper karma]

Ohhh

Aight

≈12.77

Is what I got

[thorn talon]

Hmm?

[upper karma]

Hm as I got wrong?

[silent plank]

what did you put into the calc?

12/0.34 approx 36

[upper karma]

I did cos(70)

[silent plank]

w/o even looking at a calc

[upper karma]

Wait

I messed up again

Sorry

35.08 as I got the 0.34.... from cos 70 put it in as 12/cos(70)

Lemme double check

Yea it's prolly 35.08

[silent plank]

did you round properly?

[upper karma]

Yea

Wait

35.09 as 2 numbers after is a 6

And that would all build up to that

Thanks for help

:D

[silent plank]

wdym by 2 numbers after?

[upper karma]

It's 35.0856

Because of the 6 the 5 rounds to 6

And the the 6 makes the 8 a 9

[silent plank]

you don't even need to look at that

generally, 5 already rounds up

[upper karma]

Alright thanks so much

I now have a decent grasp non how to do dis

[wind heart]

Would it be correct if I said <ATP and <ETP are right angles?

[jaunty sail]

<ATP and <ETP?

[wind heart]

Si

[jaunty sail]

Do you mean <APT and <EPT

[wind heart]

no

why would it be that

[jaunty sail]

Well <ATP and <ETP are referencing the acute angles at the top of the triangle

[wind heart]

So wait like

In a proof

It’s be

It’d be

[jaunty sail]

The center letter of the angle is the angle that is referencing

[wind heart]

<APT and <EPT are right angles
<APT = <EPT?

[jaunty sail]

Do you know trigonometric functions

[wind heart]

Nope

[jaunty sail]

You can prove this to yourself

[wind heart]

No idea

[jaunty sail]

Oh

[wind heart]

I’m a sophmore

If I learn that, it’ll probably be near the end of the year or in college because that sounds complex

[jaunty sail]

If you’re in geometry

You’ll learn them at the end of the year

[wind heart]

Ok

[jaunty sail]

I’m referring to sine, cosine, and tangent

[wind heart]

Oh Ok

[jaunty sail]

How did you come to the conclusion that they are right angles

[thorn talon]

Is there any more information?

[upper karma]

Does anyone know how to derive v² = v₀² + 2aΔx purely from geometry with the constraint ȧ = 0 ? I know how to do it with algebra but I can’t figure out how to do it with geometry

[median crown]

acceleration =0?

[upper karma]

time derivative of acceleration = 0, so acceleration is constant

[median crown]

ok

we'll take a similar approach to the pythagorean theorem

lets say

we have some non zero initial velocity and and a positive acceleration

so

we have this

now in the formula, the first clue is v^2

aka lets make a square

this square represents v final^2

side g and h are the same length

now lets include v0^2 inside of that bigger square

sorry that took so long

my computer was going sped

this picture represents v0^2 as a part of v^2

now, since you proved it algebraically, you should know that delta v is a separate component from v0, but is also squared

delta v is the vertical distance between point f and line segment bd

so lets get that in there

now, we have asserted that the bottom left corner and the top right corner squares exist separately but also make up v^2

now we just have to get the side rectangles

now, the dimensions for each side rectangle is v0* delta v

and there are two of them

so...

v0^2+ 2v0(delV) + (delV)^2 =v^2

but going back to the proof, what is 2v0(delV)?

2v0(delV)=2v0(at)

remember the 2v0(at)

2a(delX)= 2a(1/2at^2+v0t)

2a (delX)= (at)^2+ 2av0t

the a and t in 2av0t multiply to = delV

so now you can see that 2v0(at) as a component of 2a(delX), which is from the original equation

once you add v0^2 to (at)^2+ 2av0t, you can factor down to (v0+delV)^2= v^2, which is a true statement

@upper karma you get that?

[upper karma]

@median crown Thanks for the geometric interpretation, unfortunately I was unable to fully follow, the main thing I got was you started with a graph of v = v₀ + at and then constructed a square of that, but I don’t fully understand how you condensed it from there and got rid of time

[upper karma]

How do I solve #12 and #14???

[upper karma]

Hello?

[grim cairn]

Can anyone good with conversion please help me? PM me please. Thank you

[shadow wraith]

@upper karma
#12 you can use the secant-tangent theorem, that is the length of tangent (external point to point of intersection) squared equals the external segment multiplied by the whole secant. In your case x^2 = 5 * (5+7)
x = sqrt(60) = 2sqrt(15)

#14 you can use the cyclic quadrilateral angle theorem, that is opposite angles of a cyclic quadrilateral are supplementary.
<b = 180 - <d
<b = 180 - 25 = 155

[grim cairn]

anybody know what to do for this?

[cinder portal]

K lemme help u out a bit

What is a radian defined in mathematical terms? Hint: It's a ratio between 2 things of a circle

@grim cairn

Ping me back

[grim cairn]

ok hold on

@cinder portal sector legnth?

[cinder portal]

Mmmm be a bit more specific

What is a "radian"?

[grim cairn]

a radian is a unit for measuring circles

[cinder portal]

What is it mathematically, as in, give me the definitio of a radian

It does SOMETHING

[grim cairn]

it measures an angle

[cinder portal]

Like for example, Celsius has some purpose, 0 is freezing, 100 is boiling for water

Yes we know it measures an angle, but how does it do it?

[grim cairn]

through the arch length?

[cinder portal]

Ur getting close

The radian is the ratio of 2 things

Or is defined as the ratio of 2 things of a circle

[grim cairn]

radius and diameter

?

[cinder portal]

Mmm not quite, if it was just the radius and diameter, it wouldnt be too meanigful

I mean the diameter is ALWAYS twice the size of the radius

Nothing too interesting about it

[grim cairn]

im stumbled

[cinder portal]

Do a lil bit of searching and then tell me what the radian measures

This is incredibly important

[grim cairn]

ok im going to eat breakfest real quick so it may be a while

[cinder portal]

Cuz if u dont even know what a radian is, ur goinf to be doing math by memory ehich isnt good

[grim cairn]

i will ping u

[cinder portal]

Yea sure, ping me back when youve got the answer

[hollow harness]

ahem nothing happened

[cinder portal]

Oh bruh

[hollow harness]

totally didnt just misread an entire conversation

[grim cairn]

@cinder portal a radian is used to measure a circle

[long sapphire]

Outstanding

It takes one Google search to answer goon's question that he posed to you

[grim cairn]

thank you boomer

[grim cairn]

so now wot

[cinder portal]

Uhh not just measure

Like i said, a radian is the ratio of two things, what two things is it

I know it measures circles, but what part of the cirvle does it measure? How do we get that number/value?

@grim cairn

Be more specific, radians have a very well defined definition

[dark sparrow]

i wouldn't say "a radian measures circles" is a good description in the first place tbh

[grim cairn]

@cinder portal a unit of angle, equal to an angle at the center of a circle whose arc is equal in length to the radius.

[cinder portal]

Yes

YES YES YES

GOod job

now, using that, can you do your homework now?

[grim cairn]

no

thats why i asked here

anybody know how to solve this?

[dire rampart]

how is the arc related to the radius and angle of a sector of a circle?

[umbral snow]

@grim cairn
Just a very simple formula:
rθ = s

[grim cairn]

do i have to convert anything @umbral snow

[umbral snow]

@grim cairn
θ is in radians. The formula there is a big reason radians are important

[vagrant steeple]

Did I do something wrong here? I need help with this proof

[thorn talon]

i don't believe that's how fractions work

[median crown]

True

[vagrant steeple]

alright

[idle bloom]

Yea your fraction rules are all over the place

[grim cairn]

so for this I should use rθ = s, but how would i plug 3 and 61 in if they are both radians

anybody know?

[frigid obsidian]

61 is length, not radians. And the formula works only when the angle in radians

[grim cairn]

so im using the wrong formula? @frigid obsidian

[frigid obsidian]

no, right one

but you'll need to convert 120 deg to radians in the next question

[grim cairn]

ok so what about the first one tho, do I put 3 in for θ and 61 is s, so it would look like r3 = 61?

@frigid obsidian

[frigid obsidian]

nah

61 is said to be the radius

r = 61

[grim cairn]

oh shoot im stupid

so 61 * 3 = s?

[frigid obsidian]

yes

[grim cairn]

so the answer would be 189?

[frigid obsidian]

why 9?

[grim cairn]

huh

oh wait

183

right?

[frigid obsidian]

yes

[grim cairn]

ok

[frigid obsidian]

the formula L = 2pi r is a special case of your formula, by the way

[grim cairn]

ok thank you

so for the second one I should convert 120 to radians then multiply by 14

[frigid obsidian]

that's right

[grim cairn]

so 120 converted is 2.1 and 2.1 * 14 = 29.4?

[frigid obsidian]

how did you convert?

[grim cairn]

@frigid obsidian

[frigid obsidian]

hmm

I'd just leave it as 2pi/3

[grim cairn]

so 2pi/3 * 14

[frigid obsidian]

yes

[grim cairn]

how do you multiply that

[frigid obsidian]

28pi/3

[grim cairn]

thats the answer?

[frigid obsidian]

Yes, I'm not sure you can write it in that text area, though

[grim cairn]

i just put it into a calculator and it calculated as 29.32153143

so if i rounded the answer should be 29.3

@frigid obsidian

[frigid obsidian]

yes, if they need a rounded answer

[grim cairn]

welp we will see

i got it right lets goo, thank you @frigid obsidian

what would the formala for this be?

[median crown]

Arc length formula

L= theta*r

[grim cairn]

so 8=5r?

[median crown]

In radians

[grim cairn]

oh

so 8=0.0872665r? @median crown

[median crown]

Sure

,w calc 5*pi/180

[somber coyote]

Display results online and refine query

Requested by maleb1964#6739

[grim cairn]

so i should do 8/pi36? @median crown

[median crown]

Sure

[grim cairn]

so its 91.7

that doesnt sound right

@median crown

[median crown]

,w calc 360/5

[somber coyote]

Display results online and refine query

Requested by maleb1964#6739

[median crown]

,w calc 72*8

,w calc 72*8

[somber coyote]

Display results online and refine query

Requested by maleb1964#6739

[grim cairn]

huh

why did you divide 5/360

[median crown]

,w calc 576/2/pi

[somber coyote]

Display results online and refine query

Requested by maleb1964#6739

[median crown]

Looks fine to me

[grim cairn]

ok

so it says . (Note: You can enter π as 'pi' in your answer.)

so i should put 288/ pi

[median crown]

Yeah

[grim cairn]

sweet thank you @median crown

,w calc -360*180/pi

[somber coyote]

Display results online and refine query

Requested by motxch#9236

[grim cairn]

,w calc -180*180/pi

[somber coyote]

Display results online and refine query

Requested by motxch#9236

[grim cairn]

,w calc 660*180/pi

[somber coyote]

Display results online and refine query

Requested by motxch#9236

[novel loom]

I can't understand one thing. r = sin25°r +13,3sin25°
(1−sin25°)r =13,3sin25°

Why does the 1-sin25 come there?

And where does the other r go in the second last step?

[gaunt raft]

subtracted sin25'r from both side in 3rd last line and took out common r in 2nd last

[novel loom]

can i use just r here instead of 1?

[silent plank]

use just r where?

[novel loom]

If I understood correctly, they put 1 in instead of r? Im I just this lost?

[silent plank]

where?

@novel loom

(which line?)

[novel loom]

third last to second lsat

[silent plank]

r = r sin(25°) + 13.3sin(25°)
subtracting r sin(25°) from both sides:
r - r sin(25°) = 13.3sin(25°)
factoring out the r:
r( 1 - sin(25°)) = 13.3sin(25°)

[novel loom]

ohh, thank you, didn't understand the factoring

[vernal cobalt]

hello, can someone help me with my hw

problem is asking that given sample space U = {(x,y): |x| + |y| <= 2}, and C = {(x,y): x^2+ x^2 < 2}, what would the compliment of C be?

[minor dawn]

So the complement is everything not in C , but inside the sample space, so C' = {(x,y): 2<= x^2+ y^2 and |x| + |y| <=2}? @vernal cobalt

I'm assuming you meant x^2 + y^2 and just made a typo.

[vernal cobalt]

oh, so there was no need to think of a function for the complement

?

[minor dawn]

Depends on the setting... In this case I can't think of any single function. Geometrically it is a square with a circle enscribed inside... You would be hard pressed to find such a function.

[vernal cobalt]

Yeh

I would prolly be better of using set operations

Thanks

[silent plank]

wdym works for an isosceles triangle?
if a triangle is within another triangle, it is pretty safe to conclude that they aren't congruent.

[median crown]

use law of sines

big brain move

[upper karma]

anyone understand this

[thorn talon]

Have you tried anything?

[upper karma]

yes ive been repeating on multiple accounts and still failing

[thorn talon]

Well

What's the relationship between the small and large triangle?

[worn wind]

It's asking what the distance between R and S is

[silent plank]

^what have you tried?

[worn wind]

Nothin

[fringe schooner]

It might be useful to look at what there is first: you are given the hypotenuse lengths of two triangles, where the hypotenuse of one is a leg in another

[vagrant steeple]

would i be correct in answering this as 2(sqrt3)

[dark sparrow]

no because $\tan(\pi/3) \neq 2\sqrt{3}$

[somber coyote]

Ann:

[vagrant steeple]

but it says to use double angle identities

and tan pi/3 is just sqrt3

[thorn talon]

🤔

[vagrant steeple]

plug in sqrt3 to the formula?

[thorn talon]

I mean

[dark sparrow]

[thorn talon]

If that's what it equals

Then sure

[silent plank]

how would it know whether you use dbl angle identities or not?

[ember heart]

@upper karma

[plucky basin]

Hey! I wondered how i can find the equation of a sphere when i know the centrum center of the ball where the sphere tangent a plane (x-2y+2z+39=0)

I calculated that the radius or distance from the center of the sphere to the tangent plane is 18, i believe that's correct

Never mind, I'm just stupid 🤦

[upper karma]

Hey you guys know the process of substitution

Like you use it to find the point of intercept

I'm stuck on this

[silent plank]

not trig but ok.

start by choosing either equation and isolate either variable in that equation

use whatever looks most convenient

which seems be be isolating y in the first equation here

@upper karma

[leaden jasper]

if we have sqrt(sinx) then that also means that sinx > 0 but is it the same case with only x? Does x>0 since sin0 = 0?

[dark sparrow]

no

3π/2 > 0 yet sin(3π/2) < 0

[leaden jasper]

yep understandable

thanks

[manic osprey]

Anyone online to help?😅

[thorn talon]

🤔

[dark sparrow]

no, obviously not. the server is barren and desolate, and not a single message has been sent in any channel here for about three years now.

[buoyant forge]

absolutely not the right copy pasta

Ok so, i got this problem

(x+5)(x-1)=0

and i can kinda solve it to x(x+4)=5

and from there i can find x=1

but i couldnt really tell the other x without looking at the answers page

[fringe schooner]

ahh

[buoyant forge]

is there a way to check

[fringe schooner]

there is a different way to solve

[buoyant forge]

ah

[fringe schooner]

so in this case you have two separate things being multiplied, and the product is zero when either of them is equal to zero

does that make any sense?

[buoyant forge]

hmm yeah somewhat

wait yes

that makes complete sense mb

[fringe schooner]

so can you find (x-7)*(x+3)=0?

[buoyant forge]

probably

1sec

🤔

is this solveable

[fringe schooner]

yes

so try thinking about it as a*b = 0

[buoyant forge]

ok so

x = 25

is one of them

-25

wait

-5

is what i meant

thats untrue

[fringe schooner]

ya :/

[buoyant forge]

i just shorted myself

[fringe schooner]

so if this is of the form a*b=0, where a=(x-7) and b=(x+3), can you get any farther with it, without multiplying the two things

just keep them apart

[buoyant forge]

no?

[fringe schooner]

there are going to be 2 solutions, where a=0, and b=0

does that make sense?

[buoyant forge]

yeah makes sense

[fringe schooner]

b/c 0a=0 and 0b=0, and 0*0=0

[buoyant forge]

oooooooooooooh

wait

so is x1=7 and x2=-3?

that feels too easy to be true

[fringe schooner]

yep

[buoyant forge]

wait what

[fringe schooner]

ya

[buoyant forge]

ur a wizard

[fringe schooner]

thanks

[buoyant forge]

well i should be the one thanking

very thanksyou

[fringe schooner]

happy to help

[buoyant forge]

that was legit one of the most magic math ive done in a long time

[fringe schooner]

it gets more fun too

[buoyant forge]

appreciate it

well

im currently speedrunning math so enjoyment comes later for me

no time to savor the moment 😦

[fringe schooner]

ahh

ya same

except I’m procrastinating by helping strangers on the internet

[buoyant forge]

for some reason i didnt pay attention in college so now im speedrunning to be elligable to apply for engineering uni

iq=0

[fringe schooner]

rip

Good luck!

[buoyant forge]

thanks u2 :))

[forest dove]

Alright so

[eternal orchid]

good to see the fruits of your research in trigonometry @forest dove

[somber coyote]

Daminark:

[dusk snow]

yes

[eternal orchid]

yeah one of the Chebyshev polys probably

[dusk snow]

the hard part is finding that constant in front

[somber coyote]

Daminark:

Daminark:

[dusk snow]

yes

that should work

[forest dove]

So that does suggest that maybe dealing with complex exponentials is better than inducting like I had in mind lol

[dusk snow]

induction seems particularly useless here since the roots keep changing

unless you are still trying to prove that it's indeed some polynomial in sin(x)

[forest dove]

I just meant inducting to show that it's a polynomial of degree m-1/2

[dusk snow]

then it's fine

[somber coyote]

Daminark:

[eternal orchid]

missing binomial but whatever

[forest dove]

Right

[somber coyote]

Daminark:

[forest dove]

So gg

Ah but the degree

Well that should be clear too

[somber coyote]

Daminark:

Daminark:

[forest dove]

Alright back

So hmm

[somber coyote]

Daminark:

[forest dove]

I'll put this on hold for now and look at the other expression

[somber coyote]

Daminark:

[eternal orchid]

looks like you are going to need to brush up before your trigonometry quals @forest dove

[forest dove]

Apparently yeah

I'm not seeing how to find that the coefficient is (-4)^{(m-1)/2}

Hmm wait so there's something I can do to the OG expression actually

[somber coyote]

Daminark:

[forest dove]

And I guess here the point is you're using linear independence of exponentials

[somber coyote]

Daminark:

[forest dove]

The sin^2(2pij/m) doesn't cause problems because that's a constant and that also gives the linear independence business

@eternal orchid alright ready for the qual now

[carmine sundial]

sir

[carmine sundial]

i need help please. the answer can only be always, someetimes, never

[rich wolf]

@carmine sundial start off by defining all those terms

[dusk wolf]

can I get help on this channel?

Prealg-algebra is being used and got told to pick a new channel

[thorn talon]

Ok

[dusk wolf]

Just trying to figure out what gets plugged into where.

3(x+h)^2-x+h+9/h?

[thorn talon]

You're getting part of right

Sort of on the right track

[dusk wolf]

that is good to know.

[upper karma]

Plug in (x+h) into the function 3x^2 - x + 9

Then subtract the function 3x^2 - x + 9

Simplify, and divide by h

Then let h be 0

[thorn talon]

No

[upper karma]

I misread

nvm

Do everything but don't let h be 0

[dusk wolf]

is the formula I wrote out correct?

[upper karma]

dunno

[thorn talon]

You almost got f(x + h)

[upper karma]

f(x) = 3x^2 - x + 9
f(x+h) = 3(x+h)^2 - (x+h) + 9

[dusk wolf]

alright so I was on the right track with the formula just need to simplify

[thorn talon]

And parenthesis

[upper karma]

Use parenthesis, they make things alot clearer

Heck, before simplifying the whole thing, just start with the square

(x+h)^2 = (x^2 + xh + h^2) and sub that in, THEN distribute

[thorn talon]

Missing a 2

[upper karma]

Rough night 😫

[thorn talon]

It's ok

It's most correct anyway

[upper karma]

(x+h)^2 = (x^2 + 2xh + h^2)

Go back to: 3(x+h)^2 - (x+h) + 9

= 3(x^2 + 2xh + h^2) - (x+h) + 9

distribute and subtract:

f(x+h) = 3x^2 + 6xh + 3h^2 - x - h + 9

• f(x) = 3x^2 - x + 9

---

= 6xh + 3h^2 - h

Division by h yields: 6x + 3h - 1

[quiet mason]

@North Coaster#5611 could you solve that fresh talwhatsoever thing

[upper karma]

No

But you know that dude?

[quiet mason]

cuz i did

and i might help you

[upper karma]

I'm more interested in learning how you got to the point where you could solve it confidently

(also that youtube guy is smug af)

[quiet mason]

i know abit of geometry

it seems brute force

make the base of the 6-6 isosceles

"i tried alot but couldnt solve this problem so i got the solution from a professor"

also opposite angles of a cyclic quad are supplementary

[upper karma]

Really?

That's useful to know

I first extended aline that bisected the 6/6 side and let it go to the opposite angle

[dusk wolf]

still having issue on that (x+h) I am pluggin the numbers in and keep getting a wrong answer

[upper karma]

I gave you the answer

And showed the work

Save yourself the confusion, and divide by h LAST. Don't write it all out over a fraction, it gets messy. Focus on the difference

[dusk wolf]

I need to look over my work again.

[upper karma]

(f(x+h) - f(x))/h is easier to think of as:

(1/h) x (f(x+h) - f(x))

[dusk wolf]

does that -7 get a (x+h)

[upper karma]

What the heck

What are you doing

[dusk wolf]

that is the problem

[upper karma]

You're doing a new problem

[dusk wolf]

4(x+h)-7

[upper karma]

-7 is a constant

You're plugging in (x+h) as your x

[dusk wolf]

the explanation I am getting from my work is confusing more

[upper karma]

Is this precalculus?

[dusk wolf]

algebra

[upper karma]

dm me

@quiet mason meanwhile can you help me work out that geometry puzzle

[lean plaza]

Hey! I'm trying to find the centroid (or mass centre) of a polygon with n vertices that is non-intersecting. I know there's a formula for the 2D, but I want to be able to make it work for a 3D one as well. Is there a general formula, there is for the 2D one?

I might have found a formula if a n-dimensional simplex is what I'm looking for?

Can I simply do a mean of all x, y and z points respectively?

x_center=mean(x), etc?

[wind heart]

So like

[lean plaza]

I think I'm looking for the centroid of a polyhedra

Btw

[wind heart]

Nvm I solved it

It was way easier than I thought