geometry-and-trigonometry

[charred spear]

Support the bot on Patreon: https://www.patreon.com/dxsmiley

[upper karma]

you see that if beta=1

and $$\alpha\to +\infty$$

[upper sedge]

That's really cool.

[upper karma]

then you get $$1=r\dfrac{\pi}{2}$$

[upper karma]

so this means as alpha groes

so this means as alpha groes

r approaches

$$\dfrac{2}{\pi}\approx0.6366...$$

[ocean quartz]

@upper karma it seems like it actually does.

[upper karma]

oh cooool!

that's really cool

[ocean quartz]

It's a bit higher, because it is not ideally 1 here.

[upper karma]

but probably better use the formula with arctan as it has a unique limit

but yeah seems works fine anyway

[ocean quartz]

I will. Thanks!

[upper karma]

nice to see this 😃

how did you make these animations?

like I know C language but how did u do the animations?

a library? or output values for a different software?

[ocean quartz]

I used Houdini.

Probably something much better exist specifically for visualization.

[upper sedge]

There are some nice vector animation libraries out there, but I can't remember their names.

[upper karma]

oh thank you 👍

[ocean quartz]

For Python there is Bokeh and Matplotlib.

[upper sedge]

Are those python-specific, or are they ports of C-libraries?

[ocean quartz]

They are python-specific, as far as I know.

[upper sedge]

Then I'd need to include them in the shed skin, I guess. Sad day.

[ocean quartz]

I guess there still many options exist, maybe even better. I'm not really into visialization, and used only a few tools so far.

[upper sedge]

I'd love to make some animations of different things, but I haven't had the time to sit down and learn the libraries.

I idolize the work of 3blue1brown when it comes to math communication.

I'd like to apply it to logic.

[ocean quartz]

@upper sedge 3blue1brown uses Blender, afaik.

[upper sedge]

Really? I didn't know you could use Blender for 2d animations.

I suppose that it's an animation platform at all should allow that to be set up.

[ocean quartz]

If you render with simple flat colors, it will appear like it is 2d.

[upper sedge]

I'll have to play with that, then.

[ocean quartz]

There is different workflows in graphics package and visualization library like matplotlib. Library will give perfect tools and most straightworward way to visualize data. Graphics software, while much more cumbersome, has better control over picture, resulting in nicer graphs that can be rendered more original and sexy.

I would still make use of both though.

[upper sedge]

I see, I see.

[ocean quartz]

Correction: "I create the animations programmatically, using a python library named "manim" that I've been building up." 3blue1brown actually uses a library. I saw something very similar in terms of appearance not so long ago made in blender as author stated in comments.

[upper sedge]

Excellent. I wonder if "manim" is open source. I'll have to look into it.

[brittle kraken]

Can someone help me with this question? I literally have no idea as to how to solve this.
http://prntscr.com/j9g2n9

[thorn talon]

@brittle kraken still need help with it?

[brittle kraken]

Yes if possible.

[thorn talon]

do you see two triangles?

[brittle kraken]

Ye, one on the left and one on the right

[thorn talon]

what's the relation between the two?

[brittle kraken]

Relation?

I'm sorry I do not know

[thorn talon]

they're similar triangles aren't they?

[brittle kraken]

Oh

Ye

[thorn talon]

so

you can setup a ratio

to solve for AD

[brittle kraken]

I see
So for example, like 482/720 = 480/723 = 459/AD?

[thorn talon]

close

[brittle kraken]

Something like that?

[thorn talon]

you need to match the sides properly

[brittle kraken]

Ah

How so? By finding opposites?

[thorn talon]

480 matches with 720

cause they're opposite equal angles

[brittle kraken]

Ah

Okay

Then 482/723?

And then 459/AD

[thorn talon]

yeah

[brittle kraken]

Alright what do I do afterwards?

Do I cross multiply a specific ratio?

[thorn talon]

480/720 = 459/AD

you can just flip

720/480 = AD/459

then multiply through

by 459

[brittle kraken]

Okay so I cross multiply those then whatever I get for AD, I multiply it by 459?

[thorn talon]

hmm?

[brittle kraken]

Nvm

So I can choose either?

Since they're the same ratio?

[thorn talon]

yes

[brittle kraken]

Okay I got it. Thanks

[remote kiln]

hi all, im working on a problem that involves the regular polytopes in 4-dimensions (hypercube, 120-cell, 600-cell). The wikipedia entry for each of these polytopes gives formulas for determining all of the vertex positions (usually involving permutations of 4 coordinates). However, what i really need to know is, which of these vertices belong to the same cell. For example, a 120-cell is composed of 120 dodecahedral cells and 600 vertices total. So i need to group these 600 vertices by cell. Does that make sense? Wondering if anyone has any thoughts here...

[neon fossil]

depends on the shape I guess

not clear what you want to do exactly

might be more useful to state your problem

[remote kiln]

@neon fossil ok sure, i have a huge list of all of the vertices that make up a 120-cell (there are 600 unique vertices). A 120-cell is made up of 120 dodecahedrons. For each of the Dodecahedral cells, i want to find which of the 600 aforementioned vertices are part of that cell

there should be 20 vertices per dodecaheral cell. I just don't know which of the 600 they are

[neon fossil]

is this a computational question

[remote kiln]

i suppose its a mix of computational and geometry.. you think there's a better place to post?

[neon fossil]

it looks like hell anyway

no I mean I'm not sure how to help you

your description of the vertices should include the faces

you can compute it too

find all hyperplanes which contain at least 4 points

make sure they're boundary hyperplanes

and the points in them are the faces

[remote kiln]

right so i was thinking about that

but how do i go about computing the boundary hyperplanes?

oh wait, i suppose its a generalization of finding the boundary planes of a dodecahedron, no?

as in the "facet-defining equations" here https://en.wikipedia.org/wiki/Regular_dodecahedron#Cartesian_coordinates

[upper karma]

for many reasons:

when you have something of the kind

$$X^2>a$$

[upper karma]

well

if a is nonpositive then this is verified for all X

but if a is positive

you can't say

THUS $$X>\sqrt{a}$$

[upper karma]

the reason being

the reason being

the initial inequality would still be true if we take

$$X<-\sqrt{a}$$

[upper karma]

to see this

actually

when you have $$X^2>a$$

[upper karma]

we said a positive so

we can write

$$(X-\sqrt{a})(X+\sqrt{a})>0$$

[upper karma]

a positive product

so you know that this is true iff both factors are positive or both factors are negative

same thing in your case

$$(y-2)^2\ge -x^2+1$$

[upper karma]

who told you that -x²+1 is nonegative to begin with? (so that you have the right to take square root)

and even if it's nonegative

then

either $$y-2\ge\sqrt{-x^2+1}$$

[upper karma]

or $$y-2\le-\sqrt{-x^2+1}$$

[upper karma]

ah ok I see

so you paied attention to where is -x²+1 is nonegative 👍

oh what did you not understand?

well if you're not confortable enough with taking square roots with inequalities

then always tackle it through

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

[upper karma]

you have $$y^2<1$$

[upper karma]

Thus $$(y-1)(y+1)<0$$

[upper karma]

yep

and then actually

you gonna find that

how does it what?

didn't understand

oh

$$y^2<1$$

[upper karma]

$$y^2-1<0$$

[upper karma]

$$(y-1)(y+1)<0$$

[upper karma]

well

you've got a product of two things right?

this product is negative

this means that one factor is positive and the other one is negative

$$ab<0\implies (a<0\text{and}b>0)\text{or}(a>0\text{and}b<0)$$

[upper karma]

so

either

y+1>0 and y-1<0

or

y+1<0

and y-1>0

the first case gives you -1<y<1

the second one gives you 1<y<-1 which is a contradiction (1<-1)

this means that only the first case holds

and then actually

one might notice that

y+1>y-1

this means that if one of them is positive

it has to be y+1

hence necessarly y-1<0 and you get the desired result

you're welcome 😃

glad to see that made things clearer

well actually roots can be involved

like

I leave you do y^2<2 😉

goodnight!

I go sleep

[covert osprey]

http://prntscr.com/j9tgge

Would I use Sin (-7/25 + - 7/25)?

[thorn talon]

no

[covert osprey]

oh

I dont know how to do this one lol

[thorn talon]

there should be an identity for this

[covert osprey]

this?

http://prntscr.com/j9thsi

[thorn talon]

yeah

[covert osprey]

im getting it wrong

[thorn talon]

what did you get?

[covert osprey]

-.0097736882

[thorn talon]

how?

[covert osprey]

I did 2sin(-7/25)cos(-7/25)

[thorn talon]

that is not right

[covert osprey]

wait

Do I need the inverse first?

[thorn talon]

no

[covert osprey]

of the SinA

[thorn talon]

you know what sinA is

it's just -7/25

[covert osprey]

yeah

[thorn talon]

so you know that value already

[covert osprey]

wait do I solve for cosine

[thorn talon]

so find cosA

yes

just use pythagorean theorem

[covert osprey]

24/25

Cos A = 24/25

[thorn talon]

yes

so you have all the values you need

now use the identity

ehh

cosA = -24/25

[covert osprey]

wait why is it negative

[thorn talon]

cause it's in quadrant 3

[covert osprey]

2sin(-7/25)cos(-24/25)

[thorn talon]

no

sinA is -7/25

cosA is -24/25

[covert osprey]

yeah

[thorn talon]

so use the values

[covert osprey]

inverse now?

do I solve for A

[thorn talon]

no

2(-7/25)(-24/25)

simplify

[covert osprey]

.5376

[thorn talon]

sure

[brittle kraken]

Anyone help me with this?
http://prntscr.com/j9v5w6
I don't know what sides to choose

[thorn talon]

you know two sides of the large combined triangle

you can probably try filling in more until you got enough sides

[brittle kraken]

How can I fill it in?

[thorn talon]

the right side can be done through pythagorean theorem

and then you can do something with similarity from there

[brittle kraken]

Alright, by using that, I got 29.2, do I have to round that?

[thorn talon]

that does not sound right

wait

[brittle kraken]

I used a^2 + b^2 = c^2

[thorn talon]

the most right side?

[brittle kraken]

After getting the numbers, I square rooted it

[thorn talon]

that doesn't sound right

[brittle kraken]

Do I have to input numbers in a specific way? After all, 18 and 23 are the only numbers given on the question

And well x

That's why I used those numbers

[thorn talon]

a^2 + 18^2 = 23^2?

[brittle kraken]

Wait

I don't touch a?

I mean

I don't add a number on a?

[thorn talon]

what do you mean?

[brittle kraken]

a^2 + 18^2 = 23^2?
You inputted the number on the c position instead of the a position

[thorn talon]

we know what the hypotenuse is

[brittle kraken]

I see

so I get 14.3

Am I on the right track?

[thorn talon]

something like that

[brittle kraken]

Alright, do I implement 14.3? Or do I have to round that up?

[thorn talon]

i would leave it exact

but it doesn't really matter

[brittle kraken]

http://prntscr.com/j9vk79

so is it going to be like that?

[thorn talon]

yeah

[upper karma]

correct but maybe better leave exact expression rather than replace it with an approximation

[thorn talon]

^

[upper karma]

so that you get the exact value of x at the end 😉

[brittle kraken]

I'm confused now lol

[thorn talon]

a^2 + 18^2 = 23^2

a^2 = 23^2 - 18^2

a = sqrt( 23^2 - 18^2 )

a = sqrt(205)

i assume you got something like that anyway

[brittle kraken]

I did

Now I don't know what to do next

Scientifica - Today at 8:14 PM
correct but maybe better leave exact expression rather than replace it with an approximation```
Sorry but I got thrown off here

[upper karma]

I'm sorry. What I meant is that

don't continue the exercise with the value 14.3

keep the sqrt(205)

[brittle kraken]

Alright then

So what do I do next?

[thorn talon]

similar triangles

and you have enough sides to setup a ratio

this question doesn't ask it, but it can be useful to know how to prove similarity here

[brittle kraken]

so x/18 = 23/sqrt(205)?

[thorn talon]

ehh

so you see how the big triangle is made up of two smaller triangles?

[brittle kraken]

Yeah

[thorn talon]

this is quite a common problem

but basically the left triangle is similar to the right

[upper karma]

@thorn talon Oh that's a nice remark! (I thought of using the Pythagoras theorem to get to the 1 degree equation, but your remark is much better 👍 )

[thorn talon]

ty

[brittle kraken]

Well I was taught differently the other day. One of my classmates just told me to do x/18 = 18/23, with that giving me 14.1 by cross multiplying. Is that right?

[thorn talon]

hmm

[upper karma]

oooh yes

actually

[brittle kraken]

The example used was this:
http://prntscr.com/j9vskm

[upper karma]

there are some sorts of relations like that

but pay attention to how you use them

like in your exercise you don' know the value of the the height of your big triangle

[brittle kraken]

Yes

[upper karma]

ooooh wait

[brittle kraken]

I don't

[upper karma]

cosine

the cosine

indeed x/18=18/23 you're right

because

[brittle kraken]

mb, here's the continuation
http://prntscr.com/j9vtuc

[upper karma]

you know the angle between x and 18

[brittle kraken]

yes

[upper karma]

call it A

you have cos A=x/18

but if you see it in the big triangle

you also get

cos A=18/23

[brittle kraken]

I don't really see how cosines are supposed to be used in these types of problems though

[upper karma]

so you're right

this means that x/18=cosA=18/23

[brittle kraken]

Oh

It's to prove that the answer is that exactly?

[upper karma]

yep

[brittle kraken]

Since our geo teacher said that we're not going to learn law of sines/cosines yet

I see

Thanks

And how about this? What should I choose?
http://prntscr.com/j9vuvr

[upper karma]

Similarly 😃

you can always go tryhard using the Pythagoras theorem

otherwise if you memorize the quotient relations in cases like these

keep playing with them

I don't memorize them

but as I konw the cosine, sine and tangent functions

I can find them again xD

if you don't know them then it's alright

look again at your course

actually

look again at the example pic you sent us

https://prnt.sc/j9vtuc

it contains the answer 😉

to your new question

so review the relations you have in your course

Why NC/NA=NA/NB?

[brittle kraken]

It's what we're learning, we're not as advanced yet

That's why we weren't introduced to using the pythagorean theorem in those types of problems yet

https://images-ext-2.discordapp.net/external/4_zoRJLQccyHrC7-UmiDDnJp_QNTrnYSYBp4o96f714/https/image.prntscr.com/image/FuVQNNowRvW5ihXHahBBew.png
I really don't get it now since I just found out the bottom side adds up to 20

And x/20 = 20/5 or x/5 = 5/20 gets me the wrong answer

[upper karma]

@upper karma let's call

$$\alpha=\angle{ABC}$$

[upper karma]

you see that

$$\tan\alpha=\dfrac{NA}{NB}$$

[upper karma]

and that

$$\dfrac{1}{\tan\alpha}=\tan\left(\dfrac{\pi}{2}-\alpha\right)=\dfrac{NA}{NC}$$

[upper karma]

that's because $$\dfrac{\pi}{2}-\alpha=\angle{ACB}$$

[upper karma]

and that's how you get your relation

[upper karma]

Ok thx

[upper karma]

You're welcome 😃

[upper karma]

@everyone who can help I’ll PayPal u money if u can help me

[brittle kraken]

Lol

Btw thanks scientifica

And kangaroux

[upper karma]

@brittle krakenwelcome 😃

[upper karma]

send help

i need to learn geometry in 2 days

FINAL EXAM

books videos?

help

[upper karma]

@upper karma what geometry?

[ruby swallow]

^

[upper karma]

@upper karma just 10th grade regular geo

[keen aspen]

Post it @upper karma

[upper karma]

Wdym

@keen aspen

[keen aspen]

Post what you need help with

[upper karma]

Everything

[keen aspen]

That's very broad

[upper karma]

@keen aspen

[keen aspen]

Give me a standard to start

[upper karma]

I have to learn geometry alg 2 and trig in 2 days or else I’m going to kill my self

Like congruency

Similarity

I have a book full of standards and stuff

[keen aspen]

Why do you need to learn all ?

[upper karma]

Dude

you can't learn all of that in two days

lol

[upper karma]

helloi

[upper karma]

@keen aspen placement exam

@ nick have some faith

@upper karma

Pls ping me

From now ont

Thnx

I would love to have hope

but those topics are too big

Lol

I know

I can try tho

Alg 2 can’t be that hard right

Like

Trig alone

can't be mastered in two days

[keen aspen]

Yeah

in a week you could

[upper karma]

Maybe

Trig equations can get annoying

but I mean

practice enough and you'll be fine.

Two days for trig, not enough.

Most certainly.

[brittle kraken]

http://prntscr.com/jb3vb9
Will someone help me solve this? I have to solve 6 of these problems as my teacher requires me to show step-by-step procedures to solve all these problems.

[lunar raven]

Alright, do you need to show why A and B are wrong as well?

[brittle kraken]

Ye

[tropic stirrup]

[A]

1. Prove that the perpendicular bisector of AD is also the perpendicular bisector of BC.
2. Demonstrate that (A, D) and (B, C) interchange pairwise.

[B]

1. Prove that triangle ADC and ABC are of different areas.
2. Demonstrate that D cannot reflect onto B.

[C]

1. Demonstrate that the lines that pass through A, B, C, D respectively while being perpendicular to BC cannot intersect with each other.
2. Prove that D reflects onto neither A, B, nor C.

[this assumes that the given trapezoid is not a rectangle.]

[dusk quartz]

hello guys, i need a little help, how do you calculate the volume of a pyramid given the sides lenght, and the base. I found this on how to calculate the area of a triangle : https://www.mathopenref.com/heronsformula.html, but that's a triangle, not a pyramid, is there a way? thanks

[zenith ember]

https://www.mathopenref.com/pyramidvolume.html

[dusk quartz]

alright, but what if i don't know the height? is there a way to solve then?

[zenith ember]

It depends on what you have.

[dusk quartz]

well, lets say i have all of the slope heights and the base, do i need anything more?

[zenith ember]

You will likely need to use the pythagorean theorem.

Possibly more than once.

[dusk quartz]

:(

[zenith ember]

Is the tip of the pyramid directly above the center of the base?

[dusk quartz]

no, sadly it isn't

[zenith ember]

Ah.

Hmm.

Can you show me exactly what you know?

[dusk quartz]

well @zenith ember , basically, here i took a screneshot of the link you posted, i basically know all of the blue lines' lenght, and don't know the red lines's lenght

hopefully you get the picture

i could even maybe get the angles, if it's neccesary

[zenith ember]

And the sloping lines are all different?

[dusk quartz]

well, Im pretty sure they are, since the tip of the pyrmaid wont be above the center of the base

maybe it would be better if i could just calculate the height of the pyramid using the slant height

[zenith ember]

The base isn't square?

Or rectangular?

[dusk quartz]

well, sadly it wont be in all cases

[zenith ember]

Oh. This is a general thing.

What are you trying to do?

[dusk quartz]

what do you mean? do you mean i should ask in #math-discussion ?

[zenith ember]

Usually, there is a specific homework problem people are looking to solve.

So I sort of assumed that you had specific values for all of these.

[dusk quartz]

ah, okay, well, i am using this for programming purposes, that's why i say "i could even maybe get the angles", since this is in a programming enviornment

i mean, i can plug some values if you want

[zenith ember]

Ok.

What is it you are trying to do, ultimately?

[dusk quartz]

like what am i going to use this for in programmming?

well, i am basically going to use this for mesh deformation, it's a long story, and it's not really that important for this problem i don't think, but basically I need to be able to calculate the volume of the mesh, and then after removing one vertex calculate the one that is more closer to the original value

[zenith ember]

Hm. Ok.

So you have 5 points?

You have the 5 vertices of the pyramid.

Are 4 of them guaranteed to be in a plane?

[dusk quartz]

yes to the "You have 5 points?", but the problem is ,that it can differ with each mesh, so i need a calculation that is compatible with 4 sides on the base, 5 sides on the base, 3 sides on the base, 10 sides on the base ect, im not sure if this is possible, but i think it may be possible if i could calculate the height

[zenith ember]

Are the points on the base guaranteed to be in a plane?

If so, you could find the normal vector of the plane.

And use that to find the height.

[dusk quartz]

hmm, i see what you mean, that may work, i can't try it out just yet, as im going to be off to bed soon, but i think that will work. Thank you very much!

[zenith ember]

You're welcome.

[dusk quartz]

thanks again

alright then, goodbye

[zenith ember]

Night!

[signal bobcat]

I don't have a specific problem I'm doing, but I need a little help understanding something...

Let's say I have a pyramid, but the base of the pyramid is an octagon

How would I go about finding the volume of it?

[zenith ember]

I believe it's still base area times height over three

[upper karma]

The cubic polynomial f(x) is given by f(x) = 2x3 + ax2 + bx + c, where a, b, c are constants. The graph of f(x) intersects the x-axis at the points with coordinates
(– 3, 0), (2∙5, 0) and (4, 0). Find the coordinates of the point where the graph of f(x) intersects the y-axis.

[upper sedge]

I want to plug in all of the points to set up a system of equations, and then solve the system of equations for a, b, and c.

That will probably be the easiest way to do it.

[sudden stream]

Hey guys. I have a take home quiz for geometry and I need someone to hop on a call with me and help me

It would be very very helpful

@zenith ember Think you can? It would be most appreciated.

@neon fossil How about you?

[neon fossil]

probably not

just ask your question here

someone will probably help

also helping on take home quizzes is icky

[sudden stream]

How come?

We are allowed to use the internet & resources

++ its timed

thats why

i'm stressed people wont be here in time to help me

[neon fossil]

won't be any faster unless you ask your question already

[sudden stream]

@neon fossil

[keen aspen]

Are both triangles congruent?

[sudden stream]

I assume so

Yes

They are

[keen aspen]

Yes by SSA

Now since both sides are congruent, what can you do to both expressions

[sudden stream]

answer is 60 right

i think

set them equal to each other

@keen aspen Do you know how to do this one?

[keen aspen]

What are your options?

[sudden stream]

Equal, Greater Than, Less Than

[keen aspen]

Oh okay

Use trigonometry

[sudden stream]

@keen aspen What would we use?

I'm new to Triginometry

[keen aspen]

Law of Cosines

to find the angle measure

a^2=b^2+c^2-2bccosA

[sudden stream]

So what would we input

[keen aspen]

Well look at triangle RWT

We are trying to find RTW first

so 34^2=14^2+35^2-2(14)(35)cosA

Find A

[sudden stream]

Um

How?

[keen aspen]

Solve for A

[sudden stream]

23.4

rounded to the tenth

[keen aspen]

==34^2-(14^2+35^2)

[charred spear]

-265

[keen aspen]

==-265/(21435)

[charred spear]

-0.27040816

[keen aspen]

=pup evaluate arccos(-.2704) in degrees

[charred spear]

Query made by @keen aspen
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=evaluate+arccos(-.2704)+in+degrees
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[sudden stream]

105 degrees?

[keen aspen]

Yes

[sudden stream]

thats RTW?

[keen aspen]

Yes

[sudden stream]

Okay. Now how do we get TWR

[keen aspen]

Now TWR

35^2=34^2+14^2-2(34)(14)cosB

[sudden stream]

How did you use the bot again?

[keen aspen]

==

[sudden stream]

For this kind of thing

Oh

[keen aspen]

or =pup evaluate

==35^2-(34^2+14^2)

[charred spear]

-127

[sudden stream]

Then?

[keen aspen]

==-127/(23414)

[charred spear]

-0.13340336

[keen aspen]

=pup evaluate arccos(-.1334) in degrees

[sudden stream]

=pup evaluate arccos(-0.13340336) in degrees

[charred spear]

The bot is already processing a Wolfram|Alpha query for this server.
Try again in a moment.

Query made by @keen aspen
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=evaluate+arccos(-.1334)+in+degrees
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[sudden stream]

oh u beat me

[keen aspen]

Lol there

[sudden stream]

RTW is bigger

[keen aspen]

Yep

[sudden stream]

Correct?

[keen aspen]

Yes

[sudden stream]

Okay. Last problem, thanks so mcuh by the way

[keen aspen]

np

[sudden stream]

This is definitely helping

[keen aspen]

Well I think you can evaluate without doing math

[sudden stream]

BC is bigger?

[keen aspen]

Yes

[sudden stream]

I just didn't want to

[keen aspen]

Because the BD is slanted

[sudden stream]

You know

Assume

Could we do math just to check it?

If you don't mind

@keen aspen

Sorry for tagging

It's a timed homework thing so

[keen aspen]

hmm im trying to think

Yeah idk how to implement math into this

[sudden stream]

but you're positive

[keen aspen]

I can see that BXC is a larger angle

[sudden stream]

it's BC is bigger

corrrct?

[keen aspen]

and with law of sines, larger angles have larger sides

[sudden stream]

So we're 10000000% positive

[keen aspen]

If angleDXC is smaller than BXC, then Line BC is bigger than DC

Yes

LOL

[sudden stream]

Okay

I'm submitting

Lets see

[keen aspen]

If I get it wrong i'm sorry

[sudden stream]

my

teacher

sucks

BRUH

[neon fossil]

destroyed

[sudden stream]

@neon fossil can you check this one? me and my friend are skeptical. it was answered above but we're not sure if it was correct or not, we just need a second look on it. Thanks so much for being helpful btw

https://cdn.discordapp.com/attachments/326138757474680852/440671376383279105/Screen_Shot_2018-04-30_at_6.25.07_PM.png

[keen aspen]

That one i'm confident with

it's the other one that I can't prove mathematically

[neon fossil]

it's like sin(a)/34 = sin(b)/35 or something

[keen aspen]

can't use that with two unknowns

[hoary burrow]

Show the proof again?

@keen aspen

[keen aspen]

For which one?

This one he just send?

[sudden stream]

2

[keen aspen]

sent*

[sudden stream]

yes

[hoary burrow]

Yeah.

[sudden stream]

tysm man

sorry for wasting time

if i am

[keen aspen]

I basically just used the law of cosines

Given SSS

You can use it to find the angle value

[hoary burrow]

Show me, I don't know it :/

I'm nothin' but a fool.

[keen aspen]

35^2=34^2+14^2-2(34)(14)cosB

Where B is angle RWT

[keen aspen]

yes

[hoary burrow]

RT^2 = RW^2 + WT^2 - 2(RW)(WT)*cosW?

[keen aspen]

yes

[hoary burrow]

Okay and how are you certain of B here?

[keen aspen]

What do you mean "certain of B"

[hoary burrow]

B's value.

[keen aspen]

==35^2-(34^2+14^2)

[charred spear]

Invalid syntax: Imbalanced braces

[keen aspen]

==35^2-(34^2+14^2)

[charred spear]

-127

[keen aspen]

-127/(-23414)

==-127/(-23414)

[charred spear]

0.00542411

[sudden stream]

Where are you getting 23414

[keen aspen]

wtf discord

no it does that when you put x

==-127/(-23414)

[charred spear]

0.13340336

[keen aspen]

=pup evaluate arccos(.1334) in degrees

[charred spear]

Query made by @keen aspen
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=evaluate+arccos(.1334)+in+degrees
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[keen aspen]

there

[hoary burrow]

Okay, and for T...

[sudden stream]

@keen aspen But

Earlier you didnt get 82.33

You got 97 and 105

[hoary burrow]

Is this right?

That's < RTW

[keen aspen]

sin(82.33)/35=sin(t)/34

=pup evaluate (sin(82.33)*34/35

[charred spear]

Query made by @keen aspen
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=evaluate+(sin(82.33)*34%2F35
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[sudden stream]

wat

[keen aspen]

=pup evaluate arcsin(.586805)

[charred spear]

Query made by @keen aspen
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=evaluate+arcsin(.586805)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[hoary burrow]

Dude

https://media.discordapp.net/attachments/326138757474680852/440682641071800321/unknown.png

[keen aspen]

=pup evaluate arcsin(.586805) in degrees

[hoary burrow]

This is RTW

[charred spear]

Query made by @keen aspen
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=evaluate+arcsin(.586805)+in+degrees
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[sudden stream]

@keen aspen are u ok

[keen aspen]

xD

Hold on

[sudden stream]

ur numbers

are going wack

everywhere

[keen aspen]

it's hard to do math without paper

[hoary burrow]

Dude

I think that RTW < TWR

Not only by the Hinge Theorem

But through Law of Cosines

[keen aspen]

TWR is greater than angle RTW

[hoary burrow]

As RTW is approx. 82.33

Right

[keen aspen]

Because of law of sines

[sudden stream]

@keen aspen so u gave me the wrong answer

[keen aspen]

Bigger side length; bigger angle

[hoary burrow]

Right

[keen aspen]

Shit did I actually?

[sudden stream]

yes

[hoary burrow]

I said that RTW < TWR

[sudden stream]

u said put this

[keen aspen]

Thats right

[sudden stream]

Nope

u didnt tell me to put that

[keen aspen]

Oh shit

[hoary burrow]

Idk what you did PJS

[keen aspen]

Less than sign should've been correct

[sudden stream]

damn it

[hoary burrow]

Why didn't you just use hinge theorem, @keen aspen

[keen aspen]

Where did you get the first screenshot from

[sudden stream]

thats what i dont know

its sitting on my desktop

[keen aspen]

oh my bad dude

The rest I am certain is correct

and idk what the hinge theorem is @hoary burrow

[hoary burrow]

Larger side = opposite larger angle

[keen aspen]

Oh, idk why

[sudden stream]

@keen aspen its alright man that just kinda sucks

[keen aspen]

Is this a class where you can redo

[sudden stream]

we all make mistakes

no

[keen aspen]

What college is this for?

[sudden stream]

lol

it's for high school

freshman

[keen aspen]

My dual enrollment website layout uses that layout

[sudden stream]

canvas?

[keen aspen]

==-127/(23414)

[charred spear]

-0.13340336

[keen aspen]

hm

[sudden stream]

r u checking it over again

[keen aspen]

yeah

[sudden stream]

okay

i jusyt really hope

its right

[keen aspen]

No it's not

[sudden stream]

oh

rip

this is what we put btw

[keen aspen]

Yeah RWT has the larger side meaning it has the larger angle

[sudden stream]

damn it man

[keen aspen]

I don't know why Itold you otherwise

I guess my calculations were wrong in some way

sorry dude

[sudden stream]

meh

it's alright

im on the edge in my grade now thio

tho

[keen aspen]

On the rest i'm certain is right

[sudden stream]

hey wanna check the other questions for me tho

number 4 and 5

[keen aspen]

Sure

[sudden stream]

do 4 first then 5

i forgot what i answered but

once you tell me what u got

im sure i got it

[keen aspen]

=pup evaluate 33/tan(49)

[charred spear]

Query made by @keen aspen
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=evaluate+33%2Ftan(49)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[sudden stream]

er

lol

=pup evaluate 33/tand(49)

[keen aspen]

=pup evaluate 33/tan(49) in degrees

[charred spear]

The bot is already processing a Wolfram|Alpha query for this server.
Try again in a moment.

[sudden stream]

rip

=pup evaluate 33/tan(49) in degrees

[charred spear]

The bot is already processing a Wolfram|Alpha query for this server.
Try again in a moment.

Query made by @sudden stream
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=evaluate+33%2Ftand(49)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[sudden stream]

=pup evaluate 33/tan(49) in degrees

[charred spear]

Query made by @sudden stream
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=evaluate+33%2Ftan(49)+in+degrees
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[keen aspen]

Wait

[sudden stream]

lol wtf

[keen aspen]

=pup evaluate 33/tan(49) decimal approximation

[charred spear]

Query made by @keen aspen
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=evaluate+33%2Ftan(49)+decimal+approximation
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[sudden stream]

yes

we got 28.7

[keen aspen]

28.7

yea

[sudden stream]

o

k

the other one is 70 something

check 5

[keen aspen]

=pup evaluate arctan(8/12) in degrees

[sudden stream]

not 70

[sudden stream]

40 something i dont remember

[charred spear]

Query made by @keen aspen
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=evaluate+arctan(8%2F12)+in+degrees
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[sudden stream]

33.7

correct

nicee

@keen aspen thanks for ur help

[keen aspen]

np

[sudden stream]

dont worry about getting it wrong

it's great you dedicate your time to helping

[keen aspen]

Come back again if you need anymore help

[sudden stream]

okay

[keen aspen]

Next time I won't be a nonce and answer stupidly

[sudden stream]

we have a take home quiz part 2 SOON

another 5 questions

[keen aspen]

Okay

[sudden stream]

ill tag u next time

[upper karma]

Hello again

[signal bobcat]

So, of a hexagon, if the radius and side length were tripled, how would that effect the area of the hexagon compared to what it was originally?
Ping me with the answer, please.

[upper karma]

the ratio of the areas equal the square of the ratio of the side lengths @signal bobcat

so in this case the area would be 9 times larger

[signal bobcat]

Okay, thank you! ♡

[upper karma]

i’m beginning geometry and just want a clarification: while learning all of the postulates and theorems, there are proofs to go along (they prove why the postulates or theorems work) and it has come to me that it’s a lot of information to process

so is it just memorization?

Not all of it.

But, a good piece.

darn..

okay i’ll deal with it

thank you!

Not a problem.

[lusty sparrow]

How do you do #4

[thorn talon]

hmm

that answer looks correct

not entirely sure how that particular solution works though

[signal bobcat]

If there's a helper who is relatively free, I will need help later today, so if you would dm me, that would be greatly appreciated.

[fallow quail]

not sure if vectors fit here, but please tell me if they dont

so if we have 2 vectors: a and b

and AB vector is : a-b

and BA is b-a

are the 2 vectors the same but with opposing directions?

[zenith ember]

Yes

[fallow quail]

is that their only difference?

[zenith ember]

Yes.

More formally, they have the same magnitude but opposite direction.

[fallow quail]

so if I use vectors in coordinate geometry to find the length of a line, it doesnt matter if I use AB or BA right?

[zenith ember]

If all you care about is the magnitude, that is correct.

[fallow quail]

thank you:)

(still dont have the answer for yesterday question lol)

[upper karma]

Repost it @fallow quail

[dreamy swift]

can anyone solve it?

??

[dreamy swift]

u sure its ratio

?

or i have to do with the Interior Angles

and exterior angle

ohh k i forgot that

so i have to do the ratio

12.5?

@upper karma

for qr

k

and 17 ?

for dc

@upper karma

k k

thank u

[fallow quail]

so there was a chapter in my book about equation of a line, mid points, gradients, etc. with vectors instead of the normal algeabraic way

is there any advantage to use vectors in these questions compared to the normal way?

like using normal vectors for gradient, and vectors to get an equation for the line

[zenith ember]

Yes.

Vector equations will standardize to multiple dimensions.

Normal algebraic equations can also extend to multiple dimensions, but I find the extension less intuitive.

[fallow quail]

ah, I see. Well I find vectors more annoying for one dimension, maybe once I get to more dimensions it will be easier

[normal cloak]

not really

but vectors and matrix makes studying problems and solving them easier

[fallow quail]

thanks

[night lotus]

Where would the UV go?

[zenith ember]

It says "Tangent at point Y"

[night lotus]

So would it be a U and a V on a line?

[zenith ember]

Seems likely

[night lotus]

Is this symbol on RA representing a chord?

[simple coral]

So relating to the past few questions

If a problem asks for the arc length of arc ABC but doesn't specify minor or majot arc, which do you find?

Because it could ask for either

Which leads to different answers.

[charred spear]

Rendering failed. Check your code. You can edit your existing message if needed.

[night lotus]

Um I think I'd find ABC

I'd look at the angles of the triangle inside and the arcs that are already given and subtract them from 360

[upper karma]

The notation over RA typically denotes an arc, rather than a chord.

[night lotus]

So how do I find arc RA when RS = 12 cm?

[upper karma]

You'll have to know the angle between SR and SA

[night lotus]

What if I don't?

Here's a picture of the full problem

[upper karma]

Buuut ya do

[night lotus]

Does the arc length change when the chord is 12?

[upper karma]

The radius is 12

[night lotus]

ooooh ok

I keep mixing up the words and it's making the problem harder than it should be

Also did I get 28 a, b, and c correct?

[upper karma]

You didn't take any pictures of these

[night lotus]

How do I find the variables for this problem?

[upper karma]

Have you learned the circle theorem about inscribed angles?

[night lotus]

Idk what that is so probably

it's been a year

[upper karma]

If two inscribed angles share the same arc then they have the same measure

[night lotus]

so that makes angle d 29 degrees?

and arc c?

[upper karma]

Yes angle d is 29 degrees

And so is arc c

[night lotus]

oh okay

thanks

[upper karma]

Np

[sudden stream]

Hi I need some help

http://i.ankatic.ga/3247ff601754a5767e0d064911c59ecd.png

@upper karma Think you can help?

[upper karma]

It’s the centroid so it should be 2/3 of CF

[sudden stream]

Ok

Thanks

[mortal briar]

Question 1. Did I do this right? And will a question like this always be a tan-1

<@&286206848099549185>

[remote edge]

I dont understand how to make an equation

<@&286206848099549185>

[odd mortar]

[upper karma]

Use properties of a centroid

[plain wigeon]

AD is a cevian but if you don't want to use ceva's theorem you can do other things as well

[dusty coral]

!rank

i will pay you 1000 dollars and make you rich

[neon fossil]

dont post this everywhere

[upper karma]

X=60

[dusty coral]

@Smience#3476 nope

[oak fractal]

<@&286206848099549185>

Plz do the helping'

[upper karma]

If you google the pythagorean theorem, this becomes clear enough.

Basically the pythagorean theorem states that in a right-angled triangle.

The square of the longest side is equal to the squares of the two shorter sides.

[oak fractal]

so x=36?

[dusty coral]

Pythagorean theorem: $$\text{Leg 1}^2+\text{Leg 2}^2=\text{Hypotenuse}^2$$

[oak fractal]

46*

[upper karma]

This is a relationship that you can rearrange and exploit to solve all these questions in all kinds of right-angled triangles.

Well no

[oak fractal]

46 squared?

[upper karma]

x^2 = 16^2 + 30^2

Yes

46 squared = x squared

Woah

noooo

[oak fractal]

46 squared is 2116

[dusty coral]

@upper karma any progress on that question from yesterday?

[upper karma]

30^2 + 16^2

== 30^2+16^2

[charred spear]

1156

[upper karma]

Then the square root of this number.

[oak fractal]

im going to fail ugh

Square root of 1156?

[upper karma]

Watch a khan academy video or two on the pythagorean theorem.

==sqrt(1156)

[charred spear]

34

[oak fractal]

I'm literally in the middle of a test

[upper karma]

Lol.

[dusty coral]

?

[oak fractal]

This was my last resort

[upper karma]

Left it until too late then.

[dusty coral]

you're cheating lmao

[upper karma]

Let this be a lesson in preparation for tests.

[copper valve]

oy no cheating 😠

[dusty coral]

you're cheating lmao

[oak fractal]

It's ma

may*

Time to do whatever it takes to pass

[dusty coral]

whats ur grade

[oak fractal]

86.65%

[copper valve]

[dusty coral]

is that good or bad for u

[copper valve]

ur already fine then

[upper karma]

By most standards.

[dusty coral]

im in algebra 2 class rn

[upper karma]

Not to mention if you're still at the level where you're learning pythagorean theorem, you're not by any means screwed.

[copper valve]

is that like group theory, ring theory?

[upper karma]

Lol.

[oak fractal]

It's a final test

I can't do bad

[copper valve]

ok

i cant have u cheating tho

[oak fractal]

:((

I no cheat!

[upper karma]

This is the definition of cheating.

[oak fractal]

I ask for no more help plz don't kick me

[copper valve]

o ur back already

[upper karma]

You kicked him?

[copper valve]

yea

[oak fractal]

I won't ask for more help

[dusty coral]

Solve for $$x$$ in $$\sqrt 3 \cot \left(x+\frac\pi 2\right)-1=0$$ is what we're doing right now in class.

[oak fractal]

sorry

[upper karma]

Y tho

[copper valve]

damn I can't update roles cause i can't type their name on my phone -_-

academic dishonesty is the worst!!

(i think)

[dusty coral]

Solve for $$x$$ in $$\sqrt 3 \cot \left(x+\frac\pi 2\right)-1=0$$ is what we're doing right now in class.

[dusty coral]

Solve for $$x$$ in $$\sqrt 3 \cot \left(x+\frac\pi 2\right)-1=0$$ is what we're doing right now in class.

[copper valve]

murder is also pretty bad

[upper karma]

Lmao. I mean, we're talking about a kid doing pythagorean theorem and flunking a final in middle school 😛