geometry-and-trigonometry

[upper karma]

Yeah, that makes sense

Okay

[deep trail]

https://gyazo.com/5d9ebf83d14aafffa8e4295c208edd1f how would i do this proof i forget all the proofs help pls

[upper karma]

$aπ=πr²-π(\frac{l}{2})²\implies a=r²-\frac{l²}{4}$

[somber coyote]

Al𝟛dium:

[upper karma]

@deep trail occupied channel, move to #help-0

Following? I divided both sides by π

It gets cancelled

And then the square part

Yeah, and then u powered by 2 r and l/2

Still making total sense to me

I did nothing with the r though, it was already squared

y

Are you asking why?

No, it was mb

Oh lol

I explained it badly but i got it

So far, everything good then

Okay

Now the "big" move comes in

Let me get a pic

Kk

They are lowercase L's btw, just in case.

Okay my method starts with:
Let r be the radius of the big circle
Let l be the length of the square
Remember the callings

And now the next move would be do pythag

Yeah so u would get the value of r in function of the lenght right?

Yes.

Try to do pythag

And isolate r in terms of l as you said

Post what you get

@upper karma whoops, typo

It should be 2r not r

But yeah tag me when you've tried pythag and posted the result

Kk

Mb its

Yeah

Let me check

Its this one

Yeah totally correct

You got it!

Thanks for the help buddy, I appreciate the kindness u had while explaining to me what to do

Yw!

[deep trail]

https://gyazo.com/dc7a6d67a5a90ee6892edf0795cd28e7 help I don't know any proofs to find angles that are not congruent to the angle i am comparing it too

[upper karma]

@deep trail if you know how $\angle{9}\cong\angle{13}$ then consider that 13 and 14 are supplementary angles, to then substitute for 9 and prove it.

[somber coyote]

Al𝟛dium:

[deep trail]

ok

[lament shell]

https://gyazo.com/fbd8717d7f1097a97e5f2880b7e2c321

welp

[rose tulip]

How did he get 30 square root of 3 and 3 sqroot 3 as 10?

https://youtu.be/7bO0TmJ6Ba4

5:18

Or is it just dividing 30 by 3?

[acoustic jungle]

(30sqrt3)/(3sqrt3)=10

[supple wedge]

can i say angle YWU= angle UXY bc angle in same segment

can u use that for triangles

[supple wedge]

<@&286206848099549185>

[silent plank]

no

[final bronze]

So, I accidentally got myself in a shit position, and I need help, can anyone hep me with basic geometry?

[silent plank]

since you don't know whether WXYU is a cyclic quadrliateral

apply properties of parallelograms and parallel lines

[full zephyr]

this question is gonna be the death of me

[upper karma]

You want to find the perimeter of that?

@full zephyr what’s giving you trouble

[full zephyr]

honestly I’m so lost, I just want to try and figure out how I can solve it.

if it’s to find the perimeter then yeah : )

[upper karma]

Well yes it looks like it is asking for the perimeter

So you need the lengths of the 3 sides

2 of them are easy to find

[full zephyr]

Wait rlly? Maybe I’m looking too deep into it, I’m looking at the tip and the two endpoints

I got 19, 13, but I think I got that wrong lol.

[upper karma]

That’s too big

Well

Wait no those are still not the numbers you want

Mind the scale

[full zephyr]

ah, I think I’m thinking too deep about it

[upper karma]

19 13 are not the side lengths but they are almost two numbers you need

One is off by 1

[full zephyr]

so is it by looking at where the endpoints are? Or do I need an equation to find the last one

I thought the top was 15

[upper karma]

Well what does the “top” being 15 mean?

[full zephyr]

could it mean the distance or the...

[upper karma]

I may know what will help

One moment

[full zephyr]

okay!!

[upper karma]

Can you tell me what this length is

[full zephyr]

(15,19)?

[upper karma]

That’s a point, not a length

And I think you mean (19,15)

[full zephyr]

oh..

hmmm..

[upper karma]

15,19 is not a relevant coordinate here

[full zephyr]

okay, well

it says to the nearest meter

And it asks for the length

[upper karma]

Try tunnel visioning on finding the length of the segment I outlined in orange

[full zephyr]

ok

[upper karma]

You could even remove all other parts of the triangle

They don’t matter right now

[full zephyr]

ok, and you don’t subtract in this case

[upper karma]

Not exactly sure if I know what you mean, but I will say no, you don’t

[full zephyr]

I meant like, on my other problems I subtracted (x1,y1) from (x2,y2)

some kind of equation for segments

doesn’t look like I’ll be doing that here

[upper karma]

You are probably overthinking this

[full zephyr]

yeah 😔

[upper karma]

It’s a straight vertical line, from 0 height to... what?

[full zephyr]

from 0 to 15?

[upper karma]

Yes

The length is 15

[full zephyr]

hmmm okay !

wait

that’s it?

[upper karma]

That’s one of the side lengths of the triangle

[full zephyr]

ohh.

[upper karma]

You need two more, then adding them up will give the perimeter

[full zephyr]

omg oh

and the straight one on the bottom is at

13 and 19, would I subtract and get the length 6?

[upper karma]

Not 19

[full zephyr]

oh

20?

[upper karma]

Yes

[full zephyr]

okay so I got 7 and 15

[upper karma]

This is correct

[full zephyr]

the last one is curved

[upper karma]

I wouldn’t say that

Slanted

[full zephyr]

but it looks like 13 and 15

oh yeah

which gives me 2

[upper karma]

Well, do you think it has length 2?

[full zephyr]

well, I thought you would add it to

[upper karma]

Looks longer than the other ones to me

[full zephyr]

oh really

[upper karma]

So it shouldn’t be less than 15

Let alone 2

[full zephyr]

omg you’re right...

so it would uh

[upper karma]

Right triangle with 2 known leg lengths and a hypotenuse to solve for

Does this ring any bells?

[full zephyr]

hmmm

theorem?

[upper karma]

Yes

Does that word ring more bells?

[full zephyr]

Pyathagrem theorem?

[upper karma]

Yes that’s it

[full zephyr]

I did not spell that right at all

omg

okay

[upper karma]

I like the spelling

[full zephyr]

so 17 and 15

And adding the theorem

[upper karma]

Uh wait, 7 and 15

[full zephyr]

oops sorry I wrote that wrong

7*

7|~ 3?

[upper karma]

What does that notation mean

[full zephyr]

I think it butchered it horrible

[upper karma]

This is not the same as your problem

If it was an attempt at a solution to the one you posted, the person who wrote it did not do it correctly

[full zephyr]

I think this one looks a little more accurate, also I put the variables myself and It calculated it for me which ended up weird lol

[upper karma]

These numbers look very off

[full zephyr]

oh no I meant that’s how it’s supposed to go right

You can disregard the numbers, that was just an example

[upper karma]

Then yes

[full zephyr]

I got 274

and then

[upper karma]

Does it want a decimal?

[full zephyr]

Oh it says to the nearest meter

so I would round it to 17?

[upper karma]

Oh then yea

Round up

[full zephyr]

16.5 goes to 17 which then would be the missing variable?

omg

[upper karma]

17 you can consider the length of the last side, yes

[full zephyr]

and since I got the length, I would add 17 and 7 and 15?

[upper karma]

Yes

[full zephyr]

omg

39!

I think that it’s 39

[upper karma]

Me too

[full zephyr]

omg!!!

Thank you!!

[upper karma]

No problem

[scarlet grove]

hii

i have trigonometry question

how do i find the functions of a negative angle

[undone dragon]

nvmmm

[sly steeple]

So I'm given the area A in red which is 81/4*pi and the radius 9. Considering the area of a circle is pi r^2. The small sections edge should be pi/4. Since pi/4 * 9^2 would give me the area 81/4pi

So if that small A sections circumference is pi/4 then the remaining circumference is 7/4 pi

Therefore the angle of the major arc is 7/4 pi

But that's wrong, apparently the answer is 3/2 pi

Where did I slip

[earnest echo]

Area of sector is given by\
$A=πr^2\frac{\theta}{2π}$

[somber coyote]

The Godfather:

[earnest echo]

Theta is the angle subtended by arc at the centre

[sly steeple]

Aaah of course... thanks

So I'll just solve for theta and then subtract it from the total 2pi

[earnest echo]

Yep

[dark sparrow]

@earnest echo did you really need to include those pi's in there and turn a blind eye to them canceling out

[earnest echo]

I just included them to make the formula more familiar

It's easier to remember that way

[dark sparrow]

is it really

[sly steeple]

Godfather is right actually, I might have not noticed how pi's disappear

[dark sparrow]

$S = \frac12\theta r^2$ is much shorter and also makes it much more clear that radians are the superior unit

[somber coyote]

Ann:

[earnest echo]

Is There a hierarchy in Units
\j

[sly steeple]

for a noobie like me it's nice to see everything that is happening in the formula

[dark sparrow]

well godfather

nobody is stopping you from measuring length in pico-lightyears if you so insist

[earnest echo]

My professor might get annoyed though

[dark sparrow]

point is

degrees suck

[earnest echo]

Once you reach Calculus,
They certainly do

[paper vale]

lol degrees are actually defined in terms of radians because of how inferior they are

[sly steeple]

I got the right answer now Thanks

[sly steeple]

Here I:

1. Found the angle of the large red area by solving for theta, using the area and radius.
2. Subtracted the large angle from total to get the small angle for l.
3. Plugged the small angle into the circumference formula and found the arc of the l.

Are these the correct steps or is there a quicker and better way to find the answer?

[acoustic jungle]

yes

[upper karma]

@upper karma you still need help?

[prime jasper]

If i would of done this in the actual exam i would of only gotten around 3/5 or 4/5

I feel like i am really close to the answer

And sorry for messy handwriting .3.

[paper vale]

you can say x ∈ R/1,2

or in the way the book wrote it, {x:x≠1,2}

[upper karma]

Okay ugh

[paper vale]

well if u want to be more picky you need to define what number system x is in, so maybe reals here

[upper karma]

Yeah

That's what i was gonna say

Yeah do so

[pallid edge]

How many sides would there be in a convex polygon if the sum of all but one of its interior angles is $1070^{\circ}$?

[somber coyote]

Josiah Mo:

[pallid edge]

help just help

wait

is texit saying

1070 to the circleth power lmaooo

[junior light]

The remaining angle certainly can't exceed 180 degrees

[pallid edge]

well yes...

[junior light]

So it seems to be an Octagon(with the sum of interior angles being 1080 degrees), with 8 sides

[pallid edge]

wait wait wait

how do u know that

[junior light]

You can use the formula for sum of interior angles of a polygon

[pallid edge]

we dont know what the other angles are

yes ik

(n-2)(180)

[junior light]

But we know their sum, right?

[pallid edge]

well

n-1 = 1070

[junior light]

Yes, correct

[pallid edge]

but we dont know what n sides adds up to

also i like the pfp i should change mine to doge

[junior light]

The remaining angle can't exceed 180 degrees

[pallid edge]

yes...

so it could be

1071 - 1150

[junior light]

An Octagon has sum of its interior angles as 1080, while the next polygon has sum 1260. Notice that 1260-1070=190 degrees. Getting my point now?

[pallid edge]

ahhh i see

[junior light]

So it can only be an octagon

[pallid edge]

very smart

cuz if it was 9

itd be 1260

if it was 7

itd be too small

so it has to be 8

soooo

[junior light]

Correct!

[pallid edge]

answer is 10??

cuz

8-2 = 6

*180 = 1080

-1070 = 10

oof

i got it wrong

i get second chance tho

[junior light]

You've been asked about the number of sides, right?

[pallid edge]

i is huge stupids

[junior light]

I hope you got it now?

[pallid edge]

yes th

x

[junior light]

Np :)

[fathom root]

https://gyazo.com/2e51d35fccef17b6dfefe9d525d59765

where am i going wrong im confused

[upper karma]

Well

$\left(\frac{a}{b}\right)²=\left(\frac{a²}{b²}\right)$

[somber coyote]

Al𝟛dium:

[upper karma]

@fathom root

[fathom root]

Thanks!

[distant tartan]

for geometry, is anyone able to explain how to list the angles in a rotational symmetry cause it was never explained to me

[acoustic jungle]

this? https://www.google.com/search?q=rotational+symmetry&rlz=1CARWXF_enCA890&oq=rotational+&aqs=chrome.1.69i57j0l7.2694j0j7&sourceid=chrome&ie=UTF-8

[distant tartan]

yeah

how do you find the amount of degrees it’s rotated by?

its easy to find in a triangle but in weird figured

figures its kinda dumb

all i know is that it has to be lower than 360 and greater than 0

[acoustic jungle]

you connect all the vertices to the orthocenter of the polygon or something

for ex equilateral triangle would be 120

square would be 90

hexagon would be 60

pentagon would be 72

yeah.

[distant tartan]

i guess that helps a bit, hopefully she doesnt give weird shapes

[acoustic jungle]

if a weird shape has rotational symmtry it probably contains one of those regular polygons inside it (I think)

[distant tartan]

so do you just count the amount of lines and divide that by 360?

[acoustic jungle]

no I connect it to the circumcenter of that shape.

[distant tartan]

what would it be for this figure?

cause this is one of the examples

she said its 120 and 240 but

idk how she got there

[acoustic jungle]

E is the circumcenter

wtf

that should be 120

it's basically an equalateral triangle

proof by intuition honestly

[distant tartan]

oh i kinda see it now lul

[acoustic jungle]

another way to see it is

how many rotations can it do before it goes back to it's starting point

and you do 360/that

so for the triangle it's 3

360/3=120

[distant tartan]

how about this figure?

would you just see that it forms a square outside

so would it just be 90

[acoustic jungle]

yes you can rotate it 4 times so 360/4=90

[distant tartan]

im confused cus she put 90, 180, and 270

[acoustic jungle]

oh those are the degrees of rotations that keep the shape the same

[distant tartan]

ah ok

another one was a figure of a snowflake which is basically a hexagon right so it would 360/6

[acoustic jungle]

and the degrees would be the multiples of 60

[distant tartan]

ok i think i understand it a bit better, thanks lol

[acoustic jungle]

basically what your teacher wants is listing 60,120,180... etc probably

[upper karma]

anyone know any good calculators for geometry. trying to find some good calculators online for my lil bro. ping me dm me.

or any good youtubers or vidyas or whatever?

[upper karma]

anything in general for umm geometry, not really sure what hes doing in it rn

thing hes doing in equalities

im guessing hes in advanced algebra and geometry?

[ruby bramble]

Hey guys, I am in calc II rn and we are going over finding the domain and range of functions and I have forgotten how to find them for trig functions...

for instance h(x) = 1/(1-sin(x))

[earnest echo]

Do you what domain and range is?

[ruby bramble]

I want to understand how to get the domain and range for trig functions essentially

(thats an example question)

[austere mica]

i would note that -1<sinx<1

and from there think about 1/1-x for -1<x<1 for the range

bc its slightly easier to deal with

for domain, think about what it cant be

sinx cant be 1 or you divide by 0

@ruby bramble

[upper karma]

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[supple wedge]

hi

Circle $C_2$ is a reflection of circle $C_1$ along the line $y=x+1$, find the equation of $C_2$.

[somber coyote]

Hmm:

[supple wedge]

eqn of $C_1$ is $(x-5)²-(y-0)²=9$

[somber coyote]

Hmm:

[supple wedge]

yeah in general how do i do these kinds of problems

[upper karma]

try drawing the circle and the line

[supple wedge]

aight

,w graph (x-5)²-y²=9

[somber coyote]

Results provided by WolframAlpha

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

Requested by Hmm#2554

[supple wedge]

the fuck

let me go to desmos brb

[upper karma]

it's +

$(x-5)²+(y-0)²=9$

[somber coyote]

HoboSas:

[upper karma]

otherwise it's a hyperbola

[supple wedge]

its + oops

ik

how do i "reflect"

@upper karma

.

zzz

any1

<@&286206848099549185>

reeee

[past geyser]

huh

what even is your question

give me more context

[supple wedge]

sure

The lines $x=2$ and $y=3$ are tangents to a circle $C_1$. Given the centre of the circle $C_1$ lies on the positive x-axis, find the equation of $C_1$. Circle $C_2$ is a reflection of Circle $C_1$ along the line $y=x+1$, find the equation of $C_2$.

@past geyser

[somber coyote]

Hmm:

[supple wedge]

the eqn of c1 is above

[austere mica]

reflect the centre in the line @supple wedge

the radius is obviously unchanged

[supple wedge]

how does that work

@austere mica

[austere mica]

consider the perpendicular distance from ur centre to the mirror line...

[supple wedge]

oh

[supple wedge]

so i use the same x coordinate and sub into y=x+1

i find the distance btwn the centre and that coordinate

[austere mica]

not exactly

[supple wedge]

like

x coordinate of centre

plug into y=x+1

bc u said perpendicular

so the x coordinate doesnt change

[austere mica]

no perpendicular to x+1

see my diagram

[supple wedge]

oh

[austere mica]

brb

my phone is on low battery

[supple wedge]

shiet

tesr

rip

[supple wedge]

ima head to bed

[ruby bramble]

@austere mica Oh I know pi/2 would make it divide by 0 which means that it is not in the domain

However its not only pi/2 , how do I find the other values?

like the formula

[upper karma]

What’s poppin

Oh

Well

Set the denom to 0 for domain, solve for solutions

[heady lynx]

Okay, so I have a problem on a homework that I did about a week ago, I got the answers both right, but the second part, I did not understand HOW I got the answer(honeslty I looked at the example question and copied the basic answer from there). I am looking more for what exactly was done to get the answer, more than the answer itself, it was a bit confusing for me

"Choose an expression in terms of R and (theta) for the distance between M and N"

how?

[pliant osprey]

they're taking CM-CN

[acoustic jungle]

CN = Rcos(theta/2)

shouldn't d be 972

[pliant osprey]

it doesn't seem greater than the radius though

[acoustic jungle]

lmao I was in radians

google calculator sucks

search bar calculator I mean

yeah it's 284

never trust search bar calculator

[heady lynx]

Lol

[craggy wedge]

pls help

[simple leaf]

Tell me what you’ve done so far

[craggy wedge]

idk how to do it

lol

[pliant osprey]

well okay so firstly your answer is going to be something like (y+3, x) right? (just to be clear about what single coordinate notation is)

[craggy wedge]

ye

[pliant osprey]

okay so take the first step. what are the coordinates of (x, y) after reflection across y=x+2?

[craggy wedge]

um

(-x,-y) right

?

[pliant osprey]

okay, i didn't get that in one step either. is there a line like y=x+2 that you know how to reflect (x, y) across?

(-x, -y) would be y=-x i think

[craggy wedge]

ehhhh im bad at math so idrk

[pliant osprey]

i lied it's not y=-x. it's more like rotation by 180 degrees

[craggy wedge]

yeah

[pliant osprey]

no worries, let's think

so what line is like y=x+2 but simpler?

[craggy wedge]

y=x?

[pliant osprey]

yeah

what is (x, y) reflected across that?

[craggy wedge]

(-x,-y)

?

[pliant osprey]

hm okay let's think. what is (1, 2) reflected across y=x?

[craggy wedge]

ehhhhhh

[pliant osprey]

okay, let's look at (1, 1). (1, 1) is on y=x, right? so its coordinates can't change

after reflection

[craggy wedge]

yes

[pliant osprey]

so we know it can't be (-x, -y)

[craggy wedge]

yyes

[pliant osprey]

so we see that (1, 2) becomes (2, 1) after reflection

[craggy wedge]

ohhhhhhhhhhhhhhhhhhhh

so it is (x,y)---->(y,x)

[pliant osprey]

right, yeah!

so we have that transformation over y=x. but we're given y=x+2 instead

[craggy wedge]

yeah..

[pliant osprey]

note that y=x+2 is just a translated version of y=x

[craggy wedge]

yes ik

[pliant osprey]

so if we just push our whole coordinate system down two units

then we really won't know the difference between them

[craggy wedge]

yeah

[pliant osprey]

like y=x before and y=x+2 after the push

so let's push the coordinate system down, reflect our new (x, y), and then push it back up again

what does (x, y) become after the push?

[craggy wedge]

(x,y+2)

maybe

[pliant osprey]

hm, close. so note that we're pushing y=x+2 down to y=x

so all of the y-values decrease by 2

[craggy wedge]

so its

more like (x,y-2)?

[pliant osprey]

yeah, basically. do you see that?

[craggy wedge]

yeah

[pliant osprey]

okay, great. so now flip it across y=x, what do you get?

[craggy wedge]

wait flip waht across

[pliant osprey]

(x, y-2)

[craggy wedge]

oh

[pliant osprey]

that's our new (x, y) after the push

[craggy wedge]

(y-2,x)

[pliant osprey]

okay, awesome. now we're pushing the coordinate system back up, so what happens to that point?

[craggy wedge]

it moves up 2 units?

[pliant osprey]

yeah, exactly

actually i get that my terminology is very ambiguous, when i say "coordinate system" i really mean the plane, oops

so like the coordinate system stays in place

[craggy wedge]

yeah

[pliant osprey]

and we're moving the plane with its lines and points around

okay, great. so what's our new point after reflection across y=x+2?

[craggy wedge]

ummm

(y,x)

oh wait no

thats wrong

[pliant osprey]

^ that would be across y=x.

[craggy wedge]

(y+2,x)

[pliant osprey]

okay, wait

remember how we said reflection across y=x+2 is the same thing as pushing the plane down, reflecting the new (x, y), and pushing the plane back up?

so after pushing the plane down and reflecting across x=y we got:

(y-2,x)
as you said

and then when pushing the plane back up you'd said it meant that the point:

it moves up 2 units?

oops sorry for the ping

[craggy wedge]

yea

[pliant osprey]

so what are the coordinates now?

[craggy wedge]

wait

i actually i think i solved the problem

it is (-x-2, y-2)

[pliant osprey]

yup! awesome

[craggy wedge]

yay thanks for the help

[pliant osprey]

np! hf with that

[prime jasper]

I really need help

I can't see

[paper vale]

@prime jasper i have a proof

OCB is isosceles so OCB=OBC=46, so COB=88

therefore CAB=44

now u can use ACB being isosceles to work out ABC=ACB=68

therefore ACO=22, and also from opposite angles, CFO=107

now we know 2 angles in OFC, so we can discern FOC

now we know FOC and OCD(90degrees) therefore we can get CDO

[prime jasper]

Ahh i see

[paper vale]

basically what motivated this solution was trying to work out FOC from the start

[prime jasper]

Thanks man

[paper vale]

np

[sharp narwhal]

Say you are given a solid in three dimensions. Can one obtain the rotation along the xy plane using some combination of rotations along the xz and yz plane?
Like for cube for instance
Can you get roatation by 90 degrees in xy as some combination of rotation in 90 degrees in yz and rotation by 90 degrees in zx

[paper vale]

yea lol doing 2 out of the 3 is the same as doing just the 3rd

[stone salmon]

omg im having so much trouble with this

whats the angle? i thought it was like 14.... but it has to be a fraction

[pliant osprey]

how did you get 14?

[stone salmon]

sin^-1 (1/4) @pliant osprey

but it has to be like a fraction

[pliant osprey]

that's probably in degrees then. it's asking for an answer in radians

[stone salmon]

i know

i cant figure it out

im looking at the unit circle

[pliant osprey]

yeah then multiply by pi/180, right?

[stone salmon]

yea i know that rule

but how would i apply it here

[pliant osprey]

so we have 14 * pi/180 radians, right?

[stone salmon]

hmm i dont think so

like if it was 1/2 then i would answer pi over 6

[pliant osprey]

right, yeah

i don't get the discrepancy though

[stone salmon]

so what would the answer be for sin (angle) = 1/4

in radians

im so confused

[pliant osprey]

well if we simplified it would be 7pi/90, right? or we could just plug it into the calculator with it set to radians

[stone salmon]

ok but where did you get 14 from

its not 14

its like 14.475......

[pliant osprey]

okay i took the 14 from your 14, but it's fine. you can plug in 14.475 if you like

[stone salmon]

thats wrong

i already know it

[pliant osprey]

lol okay, i meant plug it into the expression like (14.475*pi/180). but either way i did it on the calculator

and it comes out to 0.25268 ish

yeah the other expression comes out to that too

[stone salmon]

dude

the answer is a fraction

that obv cant be a fraction

ahhh i just need to find it whats sin angle = 1/4 in radians

why is this so hard

[silent plank]

wdym by it has to be a fraction

[stone salmon]

ok let me post the full thing here then

the answer to this has to be a fraction

[silent plank]

you're not asked for theta

you're asked for cos(theta)

[stone salmon]

ik but it would make my life way easier for the other questions

there is like 4 more involving it after this

[silent plank]

which can be calculated without evaluating theta

[stone salmon]

how?

[silent plank]

by using famous pythagorean trig identitites

[stone salmon]

we did do that in class

[silent plank]

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

[stone salmon]

but i dont remember it

rip

[somber coyote]

ramonov:

[stone salmon]

ok so now what

[silent plank]

use theta instead of x,
solve for cos(theta) and apply the appropriate sign based on the location of theta

[stone salmon]

ok yea i didint understand one bit of that

😭

can you demonstrate how i would do the first one and then i can try doing the rest alone

@silent plank

[mellow ravine]

Can someone explain how this top identity turns into (cosx-1)^2

[north epoch]

a^2 - 2a + 1 = (a-1)^2

it’s the same here, except a = cos(theta)

[mellow ravine]

Does that identity have a name or is that something I'm just supposed to know

[north epoch]

uhhh not really

it’s just something you’re supposed to know

[pliant osprey]

^ you can think of it as factoring the quadrilateral

and @stone salmon okay so say sin(theta) was 1/2. then i could say (1/2)^2 + cos^2(theta) = 1, so cos^2(theta) = 3/4. say theta is in the first quadrant, so i know cos is positive, and cos(theta) = rt(3)/2.

[stone salmon]

@pliant osprey wdym rt

[pliant osprey]

square root

[stone salmon]

ok i understand that

but i still dont understand how to solve my question

[pliant osprey]

instead of 1/2, you have 1/4

[stone salmon]

ok i will give it a shot

LETS GOOO

whats the thank function again in this discord?

Thanks @pliant osprey

Thank @pliant osprey

thank @pliant osprey

looks like there is no thank function

[pliant osprey]

haha your gratitude is appreciated anyway

[stone salmon]

the lim x -> -1 -

its equal to

-2 right?

im kinda confused because there are like 2 dots

[pliant osprey]

if you get arbitrarily close to -1, the function gets close to what?

[stone salmon]

whats the point of the solid dot tho? @pliant osprey

im not sure if its -1 or -2

[pliant osprey]

i mean who knows?

if you get arbitrarily close to -1, the function gets close to what?

[stone salmon]

-2 i guess

[supple wedge]

.

did no one really talk here for 10 hours lmao

anyw i still need help so

[stone salmon]

i talked like 30 sec ago

ah man the subsequent prompts are so much harder

[supple wedge]

oh shitcord broke

lmao

[stone salmon]

@pliant osprey how is the second one not 1/4

oh cuz its only angle nvm

but still how would i go about solving that

AAAAHHH why is this so hard

[pliant osprey]

you can transform sin to cos using a horizontal translation @stone salmon. maybe try searching up their relationship?

[stone salmon]

nvm i got all of them @pliant osprey

fantastic i completed the entire thing but the only thing stopping me from submitting it is an impossible question

[austere breach]

so the exercise goes: Find the symetrical dot to dot M1(1,2,8) in relation to the line $$ p: \frac{x-1}{2} = \frac{y}{-1} = \frac{z}{1}

[somber coyote]

Lonac:
Compile Error! Click the reaction for details. (You may edit your message)

[dark sparrow]

point

not dot

[austere breach]

Yes, my bad

now from here i know that

$ \overrightarrow{Sp} = (2,-1,1)$ and that because the line from point M1 to line p is orthogonal that $ \overrightarrow{Sq} \perp \overrightarrow{Sp}$ meaning $ \overrightarrow{Sq} * $ \overrightarrow{Sp} = 0

[somber coyote]

Lonac:
Compile Error! Click the reaction for details. (You may edit your message)

[austere breach]

and then if we mark $ \overrightarrow{Sq} $ as (x,y,z) we get 2x-y+z=0, so can i say (2,-1,1) is $ \overrightarrow{Sq}$ ?

[somber coyote]

Lonac:

[austere breach]

And then from there i could get the line q, and since M1 is an element of that line we get $$ q: \frac{x-1}{2} = \frac{y-2}{-1} = \frac{z-8}{1}

[somber coyote]

Lonac:
Compile Error! Click the reaction for details. (You may edit your message)

[austere breach]

But i think im making a mistake when going from $ \overrightarrow{Sq} \perp \overrightarrow{Sp}$ to $ \overrightarrow{Sq} * $ \overrightarrow{Sp} = 0

[somber coyote]

Lonac:
Compile Error! Click the reaction for details. (You may edit your message)

[austere breach]

So to rephrase the question, if i know two lines are perpendicular and i have $ \overrightarrow{Sp}$ how do i find $ \overrightarrow{Sq} $

[somber coyote]

Lonac:

[regal meadow]

what would the diagram be for this question?

[upper karma]

Something like this

AB is the common chord

[upper karma]

calculate: .....

[acoustic jungle]

probably: largest distance between the intersections.

which can be calculated using pythagorean

[glass pawn]

trigo distances and heights

can someone help

[heady lynx]

The best thing to do is draw a diagram

I.e two triangles one with base x and one with base x +10

[peak mesa]

Sorry guys I forgot the theoretical name of this and I don't remember how to calculate someone can help me?

[umbral sparrow]

hypotenuse

lol

[peak mesa]

xDDDDD

yes

[umbral sparrow]

a^2 + b^2 = c^2

[peak mesa]

that

[umbral sparrow]

pythagoras

[peak mesa]

sure

thks

[paper vale]

[upper karma]

pi

[lament shell]

need help pls

[pale cargo]

@lament shell First find X it will be easier

the square next to it means you have a 90 degree angle so the function $(10x-24)$ is equal to 90 degrees

[somber coyote]

Albot1288:

[pale cargo]

and a straight line is always 180 degrees

@lament shell ?

[lament shell]

im back sry

um

[pale cargo]

alright do you understand so far?

[lament shell]

y3ah

[pale cargo]

ok so you have x now right

what did you get for each answer?

[wild storm]

if I wanted to find 20 % of the area of a circle, do I multiply that area of that circle by .20 to get it ?

[acoustic jungle]

yes

[upper karma]

<@&286206848099549185>

New to this stuff, need some help on what to do

[past geyser]

why is this in trigonometry channel

also start off by labeling the parts that you know on the diagram

[upper karma]

alright!

sorry, i didnt know where to put it

[past geyser]

also what have you tried so far

also don't ping helpers until you waited 15 minutes and no one responded

[upper karma]

oh sorry thanks for telling me

i didnt do anythin yet bc i have no idea how to start

im a beginner so thats why

i labelled everything i know of the diagram

[acoustic jungle]

use pythagoras

cuphosie cook. I can't tag you.

[upper karma]

Thats okay

[upper karma]

what do i do first

whats the first step

im confused, im new to this

well its been over 15 min ;/

<@&286206848099549185>

i rlly need help on this question, ive been waiting for more then an hour

[north glade]

Hint: RO is the radius of circle R

And as new identity said, you can use Pythagoras

[proven vine]

I don’t get this. I put 1/6pi but it was wrong l, so I tried -1/6pi and it was right

[upper karma]

@upper karma just start filling in numbers, eventually something will click. And if it doesn’t, start with your goal and try to figure out how to get there, and if you can’t do that we will provide guidance

[proven vine]

How is it negative if the cosine is going to the right

[upper karma]

Angle is in 4th quadrant including boundaries.
Cos(x) = (sqrt(3))/2
if x = -pi/6 then it is in fact in quadrant 4.

I could draw a diagram if I am not incorrect

[proven vine]

I mean

So cosine has to positive in quad 1 and 4

Does it not?

[upper karma]

yes

[proven vine]

I am confused as to why it’s negative if it’s in quad 4

[upper karma]

The angle is negative

The cosine of the angle is positive

cos(-pi/6) = (sqrt(3))/2

-pi/2 < x < 0

[proven vine]

So if the angle is negative

Then the exact value is automatically negative?

[upper karma]

Uhh wdym

[proven vine]

Why is the answer

Not Pi/6

Why the negative sign

It’s in quad 4

[upper karma]

Because pi/6 is not in quad 4

-pi/6 is however

[silent plank]

because that's what the question stated/wants

[upper karma]

^^^

It only wants the angle in quad 4

[proven vine]

Oh I see. Retard moments

[upper karma]

Lol

You’re fine

Anyways good luck brööther

[proven vine]

Give me a break ;-; I reviewing this shit for calc 1

[upper karma]

dude ur doin good

We all have those mega brain moments

I once spent 45 minutes factoring and then unfactoring a poly, turns out I wrote the question down wrong too

[proven vine]

Yeah fuck polynomials

Does calc 1 go past quadratics?

I don’t want to review that too ;-;

[upper karma]

Uhh yes lol

Functions are a big part of it

If you need help they’re fairly simple to understand

I’ve spent some time studying them

[proven vine]

Alright I’ll have to study that too 😛

[upper karma]

Kewl 😄

[unborn karma]

Does anyone here know the game euclidia?

[dark sparrow]

have heard of it, have briefly played it, didn't get into it much

[unborn karma]

Well im stuck on a level that i honestly dont have the slightest idea of how to pass

And i feel like this is the best channel for me to go to because its basically all geometry in the game

[dark sparrow]

what are the L's and E's again

[unborn karma]

Thats how they score it, L is lines, E is energy

Energy is used for tools like the perpendicular line tool uses like 2 energy

[dark sparrow]

ok

how much energy does it take to make a circle

1?

[unborn karma]

Pretty sure just one

Yup

[dark sparrow]

does the perpendicular tool also cost 1 line

[unborn karma]

Yup and 2 energy

[dark sparrow]

anyway ok the idea here would be to make a right triangle with legs both equal to 1, and the right angle at B

[unborn karma]

The midpoint tool uses 1 line and 3 energy

[dark sparrow]

its hypotenuse will have length sqrt(2)

[unborn karma]

Like that?

[dark sparrow]

yeah

wait did that already use up the lines and energy

[unborn karma]

Lol yeag

But im not worried about that rn

[dark sparrow]

maybe you don't even need to trace that hypotenuse specifically

[unborn karma]

Okay

[dark sparrow]

just make a circle with A as the center passing through that other point

[unborn karma]

Alr

Wait dont ya mean b as center?

[dark sparrow]

did you start over

[unborn karma]

Yeah

[dark sparrow]

ok start over AGAIN

[unborn karma]

Lol sorry, did you mean to just add that circle?

[dark sparrow]

yes i did

[unborn karma]

Ok

[dark sparrow]

passing through that other point

not through B

through the point at the top

[unborn karma]

Oh ok

Okay i solved it

But... why does that equeal radical 2?

But thank you for the help, i got stuck on that problem and ended up quitting the game because i couldnt find anyone to help me for a few months lol

[dark sparrow]

pythagorean theorem lol

[unborn karma]

I think i did just work it out in my head lol

Thank you for the help and onto new levels

[night snow]

J

[upper karma]

can someome help me with this?how did this thing do it?

Btw thats a trigonometric identity

[silent plank]

which step don't you understand

[upper karma]

Everything

Idk pi/2<t<pi means

Tan in terms of sin t

[silent plank]

pi/2<t<pi
t is between pi/2 and pi, effectively telling you t is in quadrant 2

[upper karma]

i dont understand a bit 🤔

[silent plank]

Tan in terms of sin t
expressed using sin(t) and constants (no other trig functions, different arguments of sine etc)

are you familiar with the unit cirlce

[upper karma]

Yeah

[silent plank]

then you should be able to understand what i said

[upper karma]

ye i get it now

thx

[upper karma]

someone could help me @!@

Wtf are those problems

from ig @upper karma

How would you even start

thats why i posted

Wtf are those problems
@upper karma go ig and go to gercekboss page

[upper karma]

idk probably a lot of trig involved

@upper karma yah

[covert tendon]

which identity im supposed to use to prove this 1? im confused

[upper karma]

@covert tendon maybe this tan(x+y) = (tan x + tan y)/ (1−tan x •tan y) / and this tan(x−y) = (tan x–tan y)/ (1+tan x • tan y)

[covert tendon]

@upper karma oh i made a mistaken when i used this 1 .... mb , i will try again thanks !

[upper karma]

@covert tendon np budd 👍

[wind gale]

@upper karma
Dude sorry for tagging but I only see u

https://cdn.discordapp.com/attachments/724154447974105173/754011643628879952/unknown.png

Anyone know this

[upper karma]

bro there's 7k people here

Also no-name aren't allowed

I'm pretty sure

yes

[wind gale]

ma name is .

ill change it lol

[upper karma]

well

@wind gale use this

[wind gale]

@upper karma THANK YOU !!

[upper karma]

@wind gale np 😋

[covert tendon]

@upper karma bruh i literally wrote 3 pages trying to solve it but couldn't... can u help me with it ?

[upper karma]

xdddddddddddd

@upper karma bruh i literally wrote 3 pages trying to solve it but couldn't... can u help me with it ?
@covert tendon yah sure

ah w8 a sec

[covert tendon]

so how im supposed to solve this 1? i feel horrible lmao

[upper karma]

which prob u mean

[covert tendon]

this 1

how can i prove it

wait imma just go 5 mins drink some water and do something , this 1 literally consumed my energy ..

[upper karma]

k

[covert tendon]

@upper karma im back , so ye how can i solve it ? , sorry for annoying u tho , but i literally tried my best XD

[upper karma]

@covert tendon im tryin @!@

[covert tendon]

oh u solving it on ur own ? LOL

[upper karma]

huh ?

oh u solving it on ur own ? LOL
@covert tendon yah ?

[covert tendon]

rip thanks dude

[upper karma]

lmao @!@ xddd

[covert tendon]

dont try so hard tho, fuck it idc anymore im really so god damn tired lmao

[upper karma]

dont try so hard tho, fuck it idc anymore im really so god damn tired lmao
@covert tendon nahhh i'll sove it

[silent plank]

express the tans in terms of sine and cosine and then apply product to sum formulae

[covert tendon]

@upper karma i hope so tho LOL

@silent plank i tried

i literally tried

but

im doin something wrong

i cant get it at all idk why , i solved a lot of questions but this 1 ... is a nightmare

[silent plank]

$\frac{\tan(\alpha + \beta)}{\tan(\alpha - \beta)} = \frac{\sin(\alpha + \beta)\cos(\alpha - \beta)}{\cos(\alpha + \beta)\sin(\alpha - \beta)}$

[somber coyote]

ramonov:

[silent plank]

did you do this?

[covert tendon]

wait lemme make sure

lemme look at my notebook

[upper karma]

@covert tendon use this identity

[silent plank]

that typesetting 😦

[upper karma]

so.

[covert tendon]

i think no i didnt ... i mean how did u change it to that 1 ? i think i need a break LOL im getting stupid @silent plank

@upper karma did u solve it using that 1 ?

[silent plank]

definition of tan

[upper karma]

nah but i found it

[silent plank]

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

[covert tendon]

but u wrote it wrong then didnt u ? @silent plank

[silent plank]

i simplified it

[covert tendon]

it should be like :

(sin(a+b)/cos(a+b))/(sin(a-b)/cos(a-b))

?

wait what ?!

why ?

[silent plank]

oh edited now

[covert tendon]

oh mb

XD

[silent plank]

you had insufficient parentheses before

[covert tendon]

ye ye i forgot them,mb

[silent plank]

yeh. and getting rid of the skyscrapers, you'd get what i typed

[covert tendon]

oh ye i tried this 1 too

but didnt work

idk if i made a mistake or something

i was so god damn tired XD

[silent plank]

maybe it'd be more helpful if you showed your work

[covert tendon]

that's the problem i cant , cuz i canceled it with a big "X" lmao and nothing is clear in the pic

[silent plank]

if you applied the product to sum/difference formulae properly,
you'd end up with double angles
and then reach the result in the step after

[covert tendon]

i wish i can send a pic tho

i think i will take a break

and try to solve it again later

but ye thanks everyone , i really appreciate it

[pale cargo]

I have to be doing something wrong.

[silent plank]

is there a better image

[pale cargo]

I'm using law of Cos to get they hypotenuse by $(\pi/3) * 6$

[somber coyote]

Albot1288:

[silent plank]

that's not how cos works

[pale cargo]

wait hold on I'm explaining it wrong and skipping steps

it would be $(\pi/3) = 6 / C$ so I move the C and then divide by 6 right?

[silent plank]

no

[somber coyote]

Albot1288:

[pale cargo]

what would be my first step then?

[silent plank]

the cosine function isn't even in your equation

the angle by itself doesn't give you the ratio of two sides

[pale cargo]

well we have a theta and the adjacent side so I thought we could use CAH

[silent plank]

"CAH", can be used, but you did not apply that properly

$\red{\cos}\br{\frac{\pi}{3}} = \frac 6c$

[somber coyote]

ramonov:

[pale cargo]

This is what I have set up

[silent plank]

you NEED the trig functions

by your logic,
pi/3 would also be a/c as well as a/3

because soh and toa

[pale cargo]

so X will equal 12

[silent plank]

what's X

[pale cargo]

12

C

[silent plank]

yes

[pale cargo]

then you para theorem to get a

[silent plank]

would be one way to do it

[pale cargo]

It seemed the easiest

thx

[mental vessel]

Okay I’m confused as hell

[upper karma]

?

[mental vessel]

What am I supposed to put on those boxes?

[upper karma]

bruuuh im now solving trig problem so........

when i finish i'll try it

[mental vessel]

I’m guessing is it’s asking me how to determine the opposite side of the angle and the hypotenuse from the angle

But I don’t know how to do that

Okay thanks

[mental vessel]

Is this right

(I made up the number 7 for the adjacent and the angle just to see if I could solve it)

[proud gust]

Can you guys help me with this

hello

this is very urgent

i have a quiz on monday

[upper karma]

show proof that it is not a test

[proud gust]

How

[upper karma]

this one

a pic of a place where it says "homework" or "assigment" works

[proud gust]

Same picture hold up

There

Hello

[upper karma]

uh okay fair enough

have you tried anything out

if no, go ahead and draw a diagram to have a visual understanding of it.

[proud gust]

yes but i keep getting -73.4666667

[upper karma]

can i see what you did

[proud gust]

my handwritng sucks can i just explain

so i did: 7/15 (94-15) -94 = -73.4666667

i subracted 50 from 94

then mutiplied by 7

then divied by 15

then subtracted 94

[bitter jetty]

can someone explain this identity
cos4θ − sin4θ = 1 − 2sin2θ;

[mental vessel]

did i do this right

[mental vessel]

i can make it brighter if you want

[oak citrus]

No its ok

I had brightness low

What made you choose tangent

[mental vessel]

okay

and because the only numbers that were there was the angle and 7

and tan = opposite over adjacent

and i already had the adjacent number

right?

[oak citrus]

Yes

[mental vessel]

okay

cool