geometry-and-trigonometry

[vapid kettle]

local rings are ones with

unique max ideal?

[upper karma]

Yes

[cedar prawn]

All these terms and ideas... so head-spinning... no matter how much I understand of math, I will always feel as if a lot of it is way over my head, I think

[vapid kettle]

so local fields?

total guess from nothing: fraction field of some local domain?

[upper karma]

Yeah

[vapid kettle]

;0

[upper karma]

Basicially if your local ring is generated by Pi you add Pi^-1

[vapid kettle]

hmm

some of this sounds vaguely familiar xD

the word uniformizer is coming to mind but I could be way off 0.o

[upper karma]

Yes

This element Ould be called a uniformizer

[vapid kettle]

algebra is fun

[upper karma]

Yeah i like when theres a bit of both

Elliptic curves are great for that

[vapid kettle]

theory is so nice

[upper karma]

Still need to grasp more algebraic geometry

Though

[vapid kettle]

oh I missed all the top matter conversation about things with MSR

[upper karma]

Those sheaf and schemes stuff

[vapid kettle]

yeah the elliptic curves class I "did"

was computational focused

and only assumed a semester of algebra

of course the first day they're talking about Galois actions but mm

"One semeseter of algebra"

[upper karma]

That seems like

Infinitely low

For all there is with localisation, extensions order divisors

[vapid kettle]

didn't even go over divisor groups

[upper karma]

Oh 😦

[vapid kettle]

interesting class though

well definitely had a style

like

first few classes were all about

how do computers do integer stuff

[upper karma]

Its the kind of thing that amazes me

[vapid kettle]

and how do computers do finite field stuff

[upper karma]

Weddfine a formal group over the points of the curve and get so much out of it

We can associates so much info into them

[vapid kettle]

mm

well I must be off to have lunch/do boring writing :(

^one of these things is much more fun than the other

I'm guessing Moon Shadow Rayne poofed off but I'll leave a couple comments

Yesterday you had some question about Graham-Schmidt

but I read through and just saw "My code works perfectly and I can't figure out the problem."

when I see something preface things by "it works perfectly" I automatically go

I don't think I can help this person :p

[waxen sorrel]

How would I approach this

[dark sparrow]

try finding the area of a pentagon in terms of its side

[waxen sorrel]

ik that it forms 5 triangles, each with angles 75, 52.5, and 52.5

ok

[dire bronze]

anyone have any stuff that I should definetly remember before i do my geometry final today?

(i have the formula for shapes)

[minor quest]

So you can't use Pythagorean to get the height?

[dark sparrow]

here you can use the pythagorean theorem to get the leftmost side of that trapezoid

and figure out the big triangle bit's area

but the small one remains unknown

[minor quest]

oh i tried to cut the big one to another one

with one of the side as the height

[dark sparrow]

yeah nop

[minor quest]

wat 😦

why not?

[dark sparrow]

you don't know how the side marked with a gets split

[minor quest]

oh so i was assuming?

[dark sparrow]

you wanted this triangle, didn't you?

[minor quest]

yes yes i did

[dark sparrow]

yeah no you don't know its lower side

[minor quest]

i thought a-c = missing side

[dark sparrow]

yeah no

[foggy oxide]

✧Gamercrew999✧ - Hoje às 15:16
anyone have any stuff that I should definetly remember before i do my geometry final today?
(i have the formula for shapes)

trigonometric relations

like sin law cossine law and right triangle relations

also some trigonometric identies like sin(arcsin(x)) and the other respectives also arc subtraction identities

cos(a +/- b), sin(a +/- b), and tg (a +/- b) is useful too

@dire bronze

also some properties of the notable points of a triangle (barycenter incenter etc.. ) u should google about them too

[dire bronze]

@foggy oxide i just came back from the final

LOL

[foggy oxide]

lol i'm late

[dire bronze]

Thx doe, i got my state test tommorow

[foggy oxide]

well u should definitely know alot about triangles

[dire bronze]

Kk

[foggy oxide]

.0.

[dire bronze]

oh

[foggy oxide]

i never saw this emoji before

is it a guy point his finger up ?

[dire bronze]

Wut

[dire bronze]

Lol

Can u help me with a question

[foggy oxide]

maybe

[dire bronze]

I got it on my final

And idk if im right

[foggy oxide]

send it

is it about geometry

?

[dire bronze]

Yes

It was like which trig equation can help you find X

And yes, A was there

[foggy oxide]

wait

38 is the angle ?

[dire bronze]

And the options were like tengent 52

38 angle

[foggy oxide]

what this question wants to know ?

[dire bronze]

What trig equation can find x

[foggy oxide]

oh

[dire bronze]

And it then listed options

Then said none of the above as a option

I choose none of the above but im not so sure

[foggy oxide]

well, it's the tg (38)

tg(38) = 14/x

[dire bronze]

oh

[foggy oxide]

but well

you can use other ones too lol

since u have the hypotenuse

[dire bronze]

Why is it that

I thought u would do that to find a whole side

[foggy oxide]

wait, what the heck is the point A ?

[dire bronze]

EXACTLY

A was there on the test

I was so confused

[foggy oxide]

this diagram is confusing

[dire bronze]

ye

thats why i choose none of the above

but idk if its right

[foggy oxide]

and you have so many options to find x that i have no idea

x = __BA __ ?

[dire bronze]

probably but it was trig stuff

[foggy oxide]

x is the whole side

or only BA segment ?

[dire bronze]

i am assuming just BA

because no point of A to be

way over there

[foggy oxide]

that's so confusing because there is no point on the vertex of the big triangle

[dire bronze]

yeah

ima ask more people

[foggy oxide]

i would say this is a bad elaborated question

[dire bronze]

but it didnt say given:X=BA

but no point of A to be there

but i cant assume stuff

idk

it was so confusing

[foggy oxide]

🤔 how do i face this problem

[dire bronze]

ye

[foggy oxide]

btw

can u show me another question from that exam ?

that u remember

so i can evaluate the questions difficulty level and stipulate if this one is really bad elaborated

[dire bronze]

To be honest

I dont really remember others

there were some find the radius if surface area was blah blah

some chord and radius questions

*tangent

*chord and tangent

my brother said i was right since u cant use trig equations for that

[minor quest]

Any ideas how they make 2 triangles at the bottom left of the rectangle?

One I supposed to have sidelengths 1 and 3 , 2 and 1

[surreal bolt]

yes. connect (-1, -2) to (-1, -3) ... it gives you a trapezoid and a triangle.

[minor quest]

Well they claim them all to be triangles

Even the one on line AT they did a triangle calculation

[dark sparrow]

why not (-1,-2)-(-2,-3)

that gives you two triangles

[minor quest]

Yea but it's not in b X h

[dark sparrow]

both of which have a side aligned with an axis

[minor quest]

They calculated all of these in b X H

[dark sparrow]

what do you mean

using the formula A = bh/2?

[minor quest]

Yes

[dark sparrow]

okay so

let's consider the triangle (-2,0)-(-1,-2)-(-2,-3)

what is the length of its vertical side?

[minor quest]

3

[dark sparrow]

and how far away is (-1, -2) from that side?

[minor quest]

1?

[dark sparrow]

yes

so now you have a base and height for this triangle

what is its area?

[minor quest]

1.5

[dark sparrow]

good

can you do a similar thing for the other triangle in that corner, (-1, -2)-(-2, -3)-(2,-3)?

[minor quest]

Yes

Aight thanks!

[minor quest]

The word said 2 to 3 but why is it 3:2?

[dark sparrow]

idk, may be a typo

:p

[winged flicker]

http://prntscr.com/g3z6h3

so I know that is a dilation of 1/3, but only from the side lengths

how do they locations of the vertices allow that to be 1/3

[dark sparrow]

hm?

[winged flicker]

should S' be at y = -2

[dark sparrow]

it's dilated and translated

[winged flicker]

oh my

god

thats what was drivingf me crazy

I misread it

thank you

[foggy oxide]

@minor quest some cases calculating triangles/whatever area would be harder so i recommend you to know the formula for the area of a quadrilateral when u have their vertex points

[minor quest]

How would you solve this?

[dark sparrow]

okay so

let's call the altitude from C to AB h

1. what is the slope of the line AB?
2. since AB is perpendicular to h, what must the slope of h be?
3. what are the slopes of the lines given in your answer options?

[minor quest]

Uh both are 0

[dark sparrow]

what?

[minor quest]

I mean Slope of ab is 0

[dark sparrow]

how did you get that?

[minor quest]

So slope of h is also 0

[dark sparrow]

no

no

no

how did you arrive at the slope of AB being 0?

[minor quest]

I use y2-y1 /x2-x1

[dark sparrow]

...

you're missing a crucial sign there

=tex \mathrm{slope} = \frac{y_2 - y_1}{x_2 - x_1}

[dark sparrow]

=tex = \frac{-6-2}{2 - (-6)}

[minor quest]

Uh

[dark sparrow]

=tex = 0?!

[minor quest]

I think I use the wrong x and y lol

...

Alright thanks anyway

[dark sparrow]

so...

shall we continue working on the problem now?

[minor quest]

I think I'll try it myself first

[dark sparrow]

ok

i will have to go to sleep somewhere between now and 50 minutes from now though

[bronze shoal]

how do I solve this?

[pseudo smelt]

They have given you two right angles what dose that tell you.

[bronze shoal]

that they're similar

[pseudo smelt]

unless stated one is not an enlargement off the other so no.

Something else that you do with right angles.

[bronze shoal]

hmm

[golden crater]

They look similar to me....

[bronze shoal]

yea i thought they're similar

x and 7 and 13 and 8+x

[golden crater]

Precisely

[bronze shoal]

any clue how to solve it?

[golden crater]

x/7 = (x+8)/13 maybe?

[pseudo smelt]

how do you know that their similar.

[golden crater]

They're drawn overlapping and they both have right angles marked

So they share 2 angles

[pseudo smelt]

yes but mathematically have you proven that ABC is a result of an enlargement to ADE

First off if your given right angle and they ask for X

[bronze shoal]

so the answer would be 28/3?

hm that doesnt seem right

[pseudo smelt]

so what does that mean?

[golden crater]

I'm still getting 9.33333 which is 28/3

[pseudo smelt]

answer is 56/13 btw or 4.3 to 1 dp.

well x = 56/13

enlargement for smaller to bigger is 13 /7

being tired proves one thing you do bizarre things used trig and rules for work out that..

[bronze shoal]

the answer is 28/3

[golden crater]

Cool

[trail vigil]

Wasn't cool

[golden crater]

It wasn't?

[trail vigil]

No

[golden crater]

Awww that's a bummer

[trail vigil]

?

[fringe basin]

the equation is (x+8)/13=x/7

[trail vigil]

x =

13x = 7x + 56
6x = 56
x = 9.33

[minor quest]

I don't understand how they got the values of -15 and -29 to plug into the equation

[naive scarab]

I think it might be an error

https://www.desmos.com/calculator/elbv9ku4id

if you plot out the three points and their answer it doesn't work out, unless I did something wrong

[minor quest]

Feelsbadman

[naive scarab]

what you would want to do is get the slope of the line perpendicular to AB, and then plug in the the point at C to point slope form

[minor quest]

I need to understand all of these in one go to finish the course fast enough 😦

[wise sand]

yeah they just like

[naive scarab]

they get the slope right, but instead of -15 and -29 it should be 33 and 19 respectively

[minor quest]

I see

[naive scarab]

here's what you get when you do that: https://www.desmos.com/calculator/untgiz7wcp

[minor quest]

I'm trying to learn how to do it to solve this

A similar problem

[naive scarab]

if you want the altitude from AB to C, you need two things

first the slope of the line perpendicular to AB

then the line with that slope which passes through C

does that make sense?

[minor quest]

Well the slope for Ab is -1 idk the reciprocal for that

Is it also -1 or 1?

[naive scarab]

it's -(1/-1) = 1

so now try to find the line with that slope through the point C

[minor quest]

Do I have to do point slope or can I jump right into y=mx+b

[naive scarab]

you can always go back and forth between the two, what do you think is easier to start with?

[minor quest]

Y=my+b I'm more comfortable with

[naive scarab]

so um, I don't see the correct answer among the choices again I think

[minor quest]

Yea

I was confused

Thought it's x-y=14

[naive scarab]

that's right

[minor quest]

That's why I'm asking for help

[naive scarab]

😦

[minor quest]

Lul online program be like

[naive scarab]

so while you're here, I might as well point out something about point slope vs the regular form

the regular form wants you to know the slope and the y-intercept

point slope wants you to know the slope and a point on the line

point slope is a lot easier to work with in that way, and you can immediately rearrange it to be the form you know

cause if you have

=tex y- y_1 = m(x-x_1)

[naive scarab]

this is the same as

=tex y = mx + (y_1 - mx_1)

[naive scarab]

just by some simple algebra

so even if the regular form is easier to graph, point slope is easier to write an equation for, and then you can rewrite it immediately in the normal form

[minor quest]

But isn't it faster substituting a point to find the y int rather than doing the slope form and then convert?

[naive scarab]

I mean in practice you just have to do simple arithmetic to convert

[minor quest]

Don't you also do simple arithmetic for filling in a point to find the y int?

[naive scarab]

I guess, idk it depends on perspective

maybe I'm just as comfortable with point slope now that I've seen both for a while

I had similar feelings at first just cause it felt different

even though they're really the same

[minor quest]

I'll have to test around em more

Wait so do I yolo answer the closest one? X-Y=2?

[naive scarab]

probably, the slope was right for the first problem which was also wrong

so that's the only sensible choice

[minor quest]

This is why I hate online course

[naive scarab]

this is pretty bad

[brazen roost]

=tex \pi

[brazen roost]

is pi

It's also the ratio of a circle's diameter to its radius

[maiden trench]

It's as Aozora said. It's a number that happens to be the ratio of a circles circumference to its diameter. If I had a circle of diametre 1, it's circumference is pi.

If I increase its diameter to 2, it's circumference is 2*pi

Is it? Circles are the same shape so it makes sense that circumference should increase proportionally to diametre.

I used the wrong word. Sorry. I meant circumference should increase proportionally to diameter.
I don't get what you mean by a "squared circle"

That's for area.
For circumference and diameter it's C = pi*d

Huh?

Yes, I see.

What do you mean by there aren't fixed points? Also if they square has side length 1, the radius of that circle is 1/2.
And we can measure circumference as well as we can measure the side length of the square. If we had perfect instruments we could measure it exactly.

Pi was first approximated by fitting other shapes over the circle. I'll see if I can find a link.

Because that's how they drew it.
http://personal.bgsu.edu/~carother/pi/Pi3a.html
Why should the circle jump to either square?
That link shows a picture.

Basically you can put one regular shape inside the circle and another outside it and then you can approximate pi from the perimeters of those shapes. The more sides, the better the approximation.

Anyway, I'm off. Good luck understanding circled.

[golden crater]

If anyone has any questions about pi, I may be able to help

[dark sparrow]

it's the ratio of the circumference of a circle to its diameter

...??

1. i just did

2. i don't see the connection you're trying to make

i mean, surely you know that numerically π is approximately 3.14159?

???

?????

define sophisticated?

[maiden trench]

Those are both nonsense.

[dark sparrow]

π is a number

so?

...

i do understand the meaning of sqrt(2)

it's the positive number that when multiplied by itself gives you 2

=tex \sqrt{2} \cdot \sqrt{2} = 2

[dark sparrow]

that's what

a number x multiplied by itself is x * x

i don't know what you're trying to get out of me, and i'm starting to suspect you're a troll

a real life example of what?

finding the area of a square from its side

see what?

[maiden trench]

She just said ...

[dark sparrow]

yes

i did

[maiden trench]

To square a number is to multiply it by itself.

[dark sparrow]

=tex x^2 = x \cdot x

[golden crater]

Pi is my favorite number... so I mean maybe it's the singularity of my personal universe... is that more what you're looking for?

[dark sparrow]

you're talking nonsense

so what does it mean, to multiply something with itself?
we've told you.

5 times already.

[golden crater]

Well I do share my universe with everyone else

[dark sparrow]

you do realize that numbers don't need to represent an amount of apples, right

what?

what singularity?

why is geometry flawed?

[maiden trench]

Also why is it obvious ... I'm sure we'd be in a lot of trouble if it was obviously flawed.

[dark sparrow]

what?

how?

what?

you're talking gibberish

what pixels?

space is continuous

there's no "smallest length"

...ok

i've had enough

👢

[minor quest]

what is the equation for a line that have the slope of 0 and passes through (-4,-1)

is it just y=0-1?

[dark sparrow]

i mean

yes

yes it is

a line with a slope of 0 is a horizontal line

so yes it's just y = -1

[minor quest]

oh right

[brazen roost]

So

Dox got kicked?

[dark sparrow]

yes

[brazen roost]

Either he's troll talking bullshit

Or genuine idiot talking bullshit

[dark sparrow]

and then you went on and on and started rambling on nonsensical topics

mate

you questioned SQUARING

and you asked the same question AT LEAST FIVE TIMES over and over and have received an adequate answer every time you asked it

[dense crater]

But you're not questioning math.

You're questioning already defined and well known topics.

Which are certain

And defined

This is the geometry channel, so this is off-topic, but back when I was around 15 years old, I went through this "anti-logic" phase.

And essentially tried to convince a group of people to think "anti-logically"

It didn't make any sense

And my answers fell flat quickly

That's what I feel I'm watching back now.

But what does that even mean?

...What?

But it's two separate statements..

I'm so confused

But this is off-topic- we can move to general if talking more.

[dark sparrow]

what assumptions do you think should not be made, and about what?

[surreal bolt]

Hmm just adding empathy. I had a RL student argue with me about what the definition of the median in a triangle was. Man ...

[fringe basin]

as in the geometric center?

[surreal bolt]

no given triangle ABC, it is the segment connecting the midpoint of one side, say AB, to the other vertex, in this case C.

[minor quest]

What was his definition? @surreal bolt

[surreal bolt]

I don't remember exactly. I think he was confusing angle bisector with median -- but being very adamant that he was right.

[surreal bolt]

well I was tutoring at the time. Just said thanks but I'm not going back :/

[minor quest]

Am I on the right track?

Normally it just something like 8/5 * 5 and then you get 8

[dark sparrow]

what was the problem @minor quest

[minor quest]

uh finding b

but i think i got it

u just keep going

[dark sparrow]

...no, what was the problem statement?

were you asked to find the eq of a line perpendicular to one with slope 3.2 and passing through (-3, 4)?

[minor quest]

yes

[dark sparrow]

right

that's what i thought

then you're all good

[brazen roost]

Anto

pls

[dark sparrow]

although i'd probs rewrite -1/3.2 as -5/16 :p

[brazen roost]

staring at a computer screen right after waking up is unhealthy

[dark sparrow]

???

got any sources for that?

[brazen roost]

Uhh

general experience

[dark sparrow]

@upper karma you still here? let's discuss your problem here rather than #help-1

[upper karma]

Okay

[dark sparrow]

reposting your picture

[upper karma]

So sin 58degrees = AC/AB

[dark sparrow]

yes

but that's not helpful

you don't know AB

[brazen roost]

You don't need to know AB

[upper karma]

Couldn't I get it with pythagoreon theorem

[dark sparrow]

that'd require you to know AC as well

you don't know AC either

@brazen roost i'd prefer it if you didn't interfere while i'm helping someone.

anyway

[brazen roost]

Ok

[upper karma]

back to square 1

[dark sparrow]

=tex \frac{AC}{BC} = \textbf{?}(58^\circ)

[upper karma]

I'm trying to find the trig function

But what do I find it relative to

58degrees?

[dark sparrow]

i mean, 58° is your angle

really, you only have six possible responses to my question

sin, cos, tan, cot, sec, csc

we've ruled out sin already

[upper karma]

Tan

[dark sparrow]

no

[upper karma]

whattt

[dark sparrow]

what two sides' ratio does tan give you?

[upper karma]

Opposite over adjacent

[dark sparrow]

what is the opposite side here?

[upper karma]

Opposite to 58degrees is BC

[dark sparrow]

yes

and the adjacent?

[upper karma]

wait a minute

I don't think there is one

[dark sparrow]

yes there is

cot

[upper karma]

If I had to guess it'd be X

AD

[dark sparrow]

what?

[upper karma]

So i'm looking at AC, and the closest angle or length I can find is X

[dark sparrow]

=tex \frac{AC}{BC} = \cot(58^\circ)

[dark sparrow]

@upper karma

[upper karma]

I forgot cotangent existed

[dark sparrow]

okay

so, AC = BC cot(58°)

does that need explaining?

[upper karma]

Yeah pls

[dark sparrow]

=tex \frac{AC}{BC} = \cot(58^\circ)

[dark sparrow]

is this clear?

@upper karma

[upper karma]

Sorry one moment

Okay

https://cdn.discordapp.com/attachments/326138757474680852/343382411377246209/latex.png

[dark sparrow]

so you understand this, right?

[upper karma]

if BC is adjacent to AC

Then what would be opposite to AC

[dark sparrow]

BC is adjacent to THE ANGLE

sides are called adjacent and opposite TO ANGLES

[brazen roost]

Anto

Remember how dox was like

[dark sparrow]

what?

[upper karma]

What do u mean sides are called adjacent and opposite

[brazen roost]

The viewpoint and singularity thing

[upper karma]

They can be both?

[dark sparrow]

no

"opposite" and "adjacent" are words that refer to the positions of a side relative to an angle

[brazen roost]

You know

You can't solve that problem without being familiar with trigonometry

[upper karma]

I'm lost

[brazen roost]

Go learn that first

[upper karma]

I learned trigonometry but not well

[brazen roost]

Try to find similar, but simpler problems

[upper karma]

See I can solve those

I just have no idea how to do this problem

[dark sparrow]

okay

do you want me to walk you through this one?

[upper karma]

Yes please

[dark sparrow]

https://cdn.discordapp.com/attachments/326138757474680852/343382411377246209/latex.png

is this clear?

[upper karma]

Not really

Be right back literally

[dark sparrow]

ok

[brazen roost]

So anto

[dark sparrow]

pm

@upper karma is it the cot that was throwing you off?

...i have to disappear for some time, ping me when you're back

in case you were confused by the cot:

BC/AC = tan(58°)

[dense crater]

Is Cotan = tan^-1(x)?

Or is that something else?

As in, it can also be written like that

[dark sparrow]

no

cot(x) = 1/tan(x)

=tex \tan^{-1}(x) \text{ denotes something different, because bad conventions}

[dense crater]

Right

[dark sparrow]

specifically, tan^-1(x) is a notation for arctan

arctan(x) = y iff -π/2 < y < π/2 and tan(y) = x

[dense crater]

Ahhh Arctan

[brazen roost]

Somehow I felt deja vu about this conversation

Well, I can see where they're coming from

[dark sparrow]

yes

f^-1(x)

[brazen roost]

=tex f^{-1}(x)

[brazen roost]

Why -1 tho

[dark sparrow]

because f^2(x) is used in the same context to denote f(f(x))

=tex f^1(x) = f(x) \ f^{n+1}(x) = f(f^n(x))

[dark sparrow]

f^n is f composed with itself n times

[brazen roost]

16th century logic

[dense crater]

@dark sparrow When you take pictures on your phone, do you copy them here via Dropbox, or?

Upload to Imgur and then link?

Or?

[dark sparrow]

upload directly

[azure storm]

drag'n'drop

[dense crater]

Ah that's cool

Okay

[foggy oxide]

LOL this is funny

well i don't think this is a level 3 question actually (max is 5)

this would be a little bit harder if they didn't say the circles radius

so ppl would have to notice it

[final prairie]

Interesting tbh

[upper karma]

are these angles equal?

[dark sparrow]

which ones?

[upper karma]

the pointy ends

[dark sparrow]

can you mark the angles you're referring to?

[upper karma]

the topmost angle and the angle in the bottom left corner

the topmost acute angle

[upper karma]

fuck that was fast thanks, yes, CAB and CDE

it took you like 5 seconds to do that how did you do it

[dark sparrow]

ms paint

lol

anyway, you're saying AB is perpendicular to BD and AE is perpendicular to DE?

[upper karma]

yes

and then there's the fact that CAB and CDE are acute

i mean if you treat DB as a rectangle's diagonal you see that CDE is equal to the angle formed at the opposite end of the diagonal, on the opposite side, and that angle is actually CAB after you use (sum of angles in triangle is 180 deg) rule

[upper karma]

nice, thanks, your solution is much simpler

i was totally ignoring that opposite angle

[dark sparrow]

DCE and ACB are vertical angles

:p

[upper karma]

so there's a name for that?

i was just thinking about the opposite angle thing in lines and stuff like that

[dark sparrow]

the angles made by two straight lines like the ones i marked α in the pic are called vertical angles

...as a pair, that is

[upper karma]

i was doing it like this but i knew there was a simpler solution, thanks for showing me

i mean, i knew there had to be one

ah right, i see

in my country we just say they're opposite so they're equal

if they're formed by 2 lines

[dark sparrow]

diagrams without names for points 😓

...also, where are you from, if you don't mind sharing?

[upper karma]

romania

[dark sparrow]

ah

[upper karma]

sorry i felt that points would clutter the drawing

[dark sparrow]

labeling in a different color helps

[upper karma]

too poor / lazy to buy colored pencils / pens lol

where are you from btw? american?

[dark sparrow]

nope

lol

i'm russian

[vapid kettle]

Definitely not :p

[foggy oxide]

that's one way to show heron's formula

[upper karma]

Hi,
Well, I know more about how we do but basically,
I need to know the position of the point H which is the projection of the point A on the axis Bv. So, blabla the calculations is obviously BA. V, but here is how I calculate the unity of the axis v on the classical orthogonal axes X and Y. I have to make vx = Cos (Alpha) and vy = Sin (Alpha)? Or I have to calculate the slope of the style vx = DeltaY / DeltaX of the red line ?

In the example of the video below, the 4 points of the rectangle of the mouse must be projected on the other rectangle along the red axes (only the min and max points are retained), but when the rectangle rotates, exceeding one Certain value.

https://www.youtube.com/watch?v=vs4GFdhJYKw

for information
 I do

xv = Cos(Alpha) yv = Sin(Alpha)

And the calculations are

`BH = ( ( xA - xB )xv + ( yA - yB )yv ) / ( Sqr ( ( xvxv) + (yvyv) ) )

xH = xB + ( ( BH * xv ) / ( Sqr ( ( xvxv) + (yvyv) ) ) )
yH = yB + ( ( BH * yv ) / ( Sqr ( ( xvxv) + (yvyv) ) ) )`

[minor quest]

How to convert? To that form

[naive scarab]

expand the (x-h)^2 part

[minor quest]

ok?

[naive scarab]

try to do it. I don't want to be mean but it's not that difficult. It's best if you at least mess around with it before asking for more help.

I'm assuming that you're trying to get from the second to the first right?

[minor quest]

first to second

[naive scarab]

oh

it's already in that form

a = 3, h = 0, k = 5

[minor quest]

oh ..

[naive scarab]

lol it's ok, it happens

[grave echo]

expanding (x-h)^2 part?

like uh

x^2 + 2xh - h^2?

[dark sparrow]

no

x^2 - 2hx + h^2

(x-h)(x-h)

[grave echo]

well wrong placement it seems

[minor quest]

What would be the closest answer for this???

none of them seems to be correct or am i missing something?

[azure storm]

I also find an answer different to all of those 😮

though one is "closest"

[minor quest]

alright thanks! yea u can see the review for the activity is 1.9 stars lmao

[surreal bolt]

hehe. even + even = even, I'd go with 3x + 2y = 20

[azure storm]

i'd go with "there is no such thing as "closest answer", every answer is equally wrong or right" 😄

along with a "let's put it to 1.8stars"

[surreal bolt]

Just saying. The ones with odd constants are totally wrong ... soo 😃

[minor quest]

@surreal bolt wait what?

oh

[surreal bolt]

(the other one is wrong also)

[foggy oxide]

@minor quest what exactly did u get

[surreal bolt]

um?

I said b, c, and d are wrong. But it is a bad question. All of the choices are wrong so process of elimination doesn't apply. 😦

[foggy oxide]

?

[foggy oxide]

this question is from ITA (Brazilian institute of aeronautical technology)

well

let me translate it

fuck can't find it~

A right triangle with hypotenuse "c" = 2(1+ sqrt6) is circunscribed on a circle with unit radii. Find the surface area from the conical shape formed by the rotation of the right triangle around it's biggest leg.

very nice question btw

Obs : ITA is probably one of the best universities from my country and this is a dissertation question soo ...

[eager saddle]

With a triangle of sides
a = 24.6
b = 15.9
c = 39.4
I use the law of cosines and end up with angle A as 16.7 degrees...

But that doesn't sound right, because when I try to solve the rest with law of sines, my angles do not add up to 180.

I calculated
A = 16.7
B = 10.7
C = 27.4

Can I have some help, please?

Okay, my C was miscalculated, and should have been about 153.6 degrees.

[minor quest]

What does the hint means?

Not necessarily congruent to the sides of the pyramid?

[dark sparrow]

the base and a side face need not be the same triangle

@minor quest do you know how to calculate the area of one side face?

[minor quest]

Yes

[dark sparrow]

do you know how to calculate the area of the base?

[minor quest]

Yes

[dark sparrow]

do you know how to calculate the surface area of the pyramid?

[minor quest]

I thought since its equilateral then all sides of the base are the same?

[dark sparrow]

all sides of the base are the same yes

[minor quest]

Nope they showed me some Pythagorean theorem

[minor quest]

They did b*h /2

[dark sparrow]

AB = BC = AC, but ABC and PAC need not be congruent. that's what the hint said

[minor quest]

I did heron formula

[dark sparrow]

the area of an equilateral triangle is a special case worth remembering

=tex a^2 \frac{\sqrt{3}}{4}

[dark sparrow]

where a is its side

[minor quest]

Well I did the find s and then square root of s(s-a)(s-b)(s-c)

Where abc are the side lengths

[dark sparrow]

needlessly complicated

:p

@minor quest

[surreal bolt]

needlessly complicated .. 😛

You should be given the "stereotypical math instructor" award 😃

[dark sparrow]

huh?

in what way am i behaving stereotypically?

[foggy oxide]

@minor quest Heron formula should be used when u have all 3 lenghts soo ... or when you have the area and want a lenght etc...

and s = a+b+c/2 (semi perimeter)

[dark sparrow]

(a+b+c)/2

[foggy oxide]

lol

just by the fact of me typing semi perimeter it can be understandable that is meant to have parentheses lol

now i have school

goodbye

[minor quest]

Alright thanks!

[minor quest]

Can I get some clarification for this tip?

[dark sparrow]

what exactly do you want clarified?

[minor quest]

the corss sectional polygon will never have more sides than the polyhedron has faces @dark sparrow

[dark sparrow]

what's unclear about that?

@minor quest

[minor quest]

it said the cross sectional polygon will never have more sides than the polyhedron has faces? so the polyhedron only have 3 faces so the cross sectional polygon sides can only be under 3?

[dark sparrow]

i mean, a polyhedron can't have only three faces :p

but yes, for example if your polyhedron has say 8 faces you can't have a cross sectional polygon with more than 8 sides

[minor quest]

alright so a tetrahedron have 4 faces so can a pentagon be in there?

[dark sparrow]

well

how many sides does a pentagon have?

[minor quest]

5\

I see

While you're at it do they want me to write (10/12)^3 or 125/216?

[dark sparrow]

the latter

i assume it's implied that the boxes are cubes

[minor quest]

Alright

[foggy oxide]

i hate when i'm doing geometry and my intuition see something but i can't prove

😭

or when i see the solution using analityc geometry / calc

;((((((((((

this is even worse

;(((((((((((((((((((((((((((((((((((

[fringe basin]

I really like it when I my sketches are good so that I can see what an angle is then I check and I'm right

[foggy oxide]

what about the angles that are very unusual ?

like the arctg(1/2)

you can guess that's the ratio but well,

the angle is very strange when working with degrees

[minor quest]

@dark sparrow how did they get 1/3 for the P(A and B)?

[dark sparrow]

how many odd prime numbers are there between 1 and 6?

(also, this belongs in #probability-statistics, as it's a probability question :P)

@minor quest

[minor quest]

uh

3?

[dark sparrow]

are you sure? 😛

what are they?

[minor quest]

oh wait odd and prime

2

[dark sparrow]

yeah

[minor quest]

3 and 5?

[dark sparrow]

yes

[minor quest]

ohhh i get it now thanks! it didn't really make sense when they said odd prime for a moment

[dark sparrow]

why are you posting probability questions here and not in #probability-statistics?

[minor quest]

Right

[foggy oxide]

anyone

have a cool solution

for this one :

A triangle is inscribed into a unit circle, the biggest side of the triangle is 2 and its area is 1/(sqrt(2)), the minor side of the triangle is ?

my solution

needs the use of the quadratic formula so ...

[dark sparrow]

i mean obv the biggest side is the diameter

[foggy oxide]

yes

i think you already know thales theorem

(this one is from a entrance exam too)

[dark sparrow]

i used two facts from trig:

sin(2x) = 2cos(x)sin(x)

[foggy oxide]

i'm back

[dark sparrow]

sin^2(x/2) = (1 - cos(x))/2, cos^2(x/2) = (1 + cos(x))/2

[foggy oxide]

i thought about using trig too, nice solution my man

i ended up solving a bi-quadratic with substitution

in a biggest circle some adjusts would be needed but this method would work too actually...

[foggy oxide]

someone here

if someone have a better idea

ABC is equilateral

P,M are midpoints

PN is the bissector of A ^P C

PM = 1 and i know how to calculate segment PN (by the internal bissector thing formula)

the question asks MN i can use cossine law (because i know PN)

but i would say

i'm probably inattentive and didn't see the easier sol, this is a way more complicated than it need, so if someone see a simpler solution pls tag me

[fluid stream]

is theta given?

[foggy oxide]

no

but

PC is perpendicular

BPM is equilateral too

[fluid stream]

ok wait

[foggy oxide]

PC is perpendicular right ?

am i wrong ?

i slept only 5 hours past night dont judge me 😭

i'm preety sure PC is perpendicular welll

[fluid stream]

what does "PN is the bissector of A ^P C" the caret here mean?

[foggy oxide]

Angle APC

PN is the bissector

[fluid stream]

oh

[foggy oxide]

well the incenter and barycenter of an equilateral triangle is in the same point

brb

[fluid stream]

You could use law of cosines from angle MCN, i think

[foggy oxide]

I was going to use from MPN

if you're saying to think about triangle MCN, i have only MC measure

so it's more logical to use law of sines

i can get the angles

[fluid stream]

You can find AN and NC with angle bisector theorem i think

[foggy oxide]

yes

PN is actually sqrt( 2) (ab) / (a + b) (where a and b are the legs of the right triangle CPA)

one of the solutions i thought was find PN then cossine law on triangle PMN, the second one is sine law on CPN

quite easier

since i know the angles from CPN too

[fluid stream]

Having MC and NC and angle MCN, you could also use law of cosines there

[foggy oxide]

yes, yes, but i think finding NC it's in terms of difficultly equivalent to finding PN or deducing the relation to find PN

the original problem if you liked it :

ABC is a equilateral triangle with sides equal to 2 such that P, M, N lies on the sides AB, BC, AC respectively and P,M are midpoints and PN is the bisector of the angle ApC, find the segment MN

[acoustic meadow]

alright just to be sure i am not the best at math but there is no such thing as a "scalene equilateral triangle"

bcuz a scalene triangle has no sides that are the same but a equilateral triangle has sides that are same

right??!!

[foggy oxide]

yes you're right

scalene means that all sides are different

equilateral means that all sides are equal

it's very easier to explain talking about sides

not angles

but in euclidian geometry the internal angles of a triangle sum up to 180 and in the scalene all angles are different, in the equilateral the angles are 60 (degrees)

[dark sparrow]

ew, degrees

[foggy oxide]

.-.

what do you have with degrees

radians is like "no" ...

[dark sparrow]

degrees are an unnatural and arbitrary unit

what's wrong with radians?

[foggy oxide]

no ??????????????????????????????????????????????

degrees are not unnatural

[dark sparrow]

you're dividing the circle into 360 pieces

the number 360 is arbitrary

yes it has a lot of factors

but it's still arbitrary

[maiden trench]

At least they're not gradians

[dark sparrow]

you accidentally two words

[foggy oxide]

i don't remember why the 360, but i already read "why" in a math history book

it came from the baby... something

babilonians

idk

[dark sparrow]

babylonians

360 is arbitrary no matter how you put it

[foggy oxide]

well, actually radians make sense but no ...

[dark sparrow]

what do you mean by "no"

[maiden trench]

It's only useful because of the number of factors, but pi and circles have a strong link.

[foggy oxide]

idk, when i see radians being used in non trigonometry problems i have a weird feeling

@maiden trench before the "," you were talking about 360 right ?

well radians can be explained a little bit if u think the unit circle has 2pi of circuference

that's why it makes too much sense

(that's how i face radians actually )

[dark sparrow]

makes too much sense

uh

[maiden trench]

Also when would one use radians outside of trig?

[foggy oxide]

well, sqrt2 said to me like i shouldn't use degrees

.-.

i think she hates degrees

[dark sparrow]

i do dislike degrees

personally, i have to think more to visualize 45° than π/4

[foggy oxide]

like Complexlma said the number of factors are a very convincent argument to the use of degrees

well, when i think about 45°

i just imagine a right triangle being distorted

or a right triangle with equal sides

whatever

(thinking about the transformation of 45° )

[maiden trench]

I wasn't saying it as an argument for their use. I just meant that's the only thing they have going for them.

[dark sparrow]

2π/5 makes more sense to me than 72°

[foggy oxide]

well, this is a argument actually

well, you're kinda right in this one

but i hate having to use 2pi/3 to describe 120° and etc..

[dark sparrow]

how come?

[foggy oxide]

what do you think about the use of tau ?

my internet is raging me out

[dark sparrow]

¯_(ツ)_/¯

[foggy oxide]

i have to go ...

cya

~~school ~~

[minor quest]

is there anyway for me to convert e+12 to the actual number?

[dark sparrow]

what?

did your calculator output something like 5.67e+12?

that's short for 5.67 * 10^12

[minor quest]

yea but i want to do the actual 5.67 times whatever 10^12 is but it will just revert back to e+12

[dark sparrow]

...10^12 is just 1,000,000,000,000

you do realize powers of 10 are easy to calculate, right?

[minor quest]

oh right...

[dark sparrow]

and multiplying by a power of 10 is trivial

[prisma abyss]

@foggy oxide tau I guess is only useful in radians

but if you want to do area you have to divide by 2

and if there's anything more ugly then timesing by two is dividing ew

[vale raven]

multiplying is the word

[foggy oxide]

lol

[prisma abyss]

ye I sun goofed

@vale raven is your name a reference to o gochuumon wa usagi desu ka

[vale raven]

No, people ask that a lot xD

I had this name before I found out about the green tea coffee thing

[prisma abyss]

is it meant to be an -autistic- special way of saying coffee

[vale raven]

In the anime?

[prisma abyss]

your name

[vale raven]

My name is a combination of my real name and Dutch for coffee

[prisma abyss]

o

cause I thought you were trying to say coffee in japnese

which is hoohii

コーヒー

koohii*

more like

[vale raven]

うん

[prisma abyss]

あの

[vale raven]

でも、英語を話せない日本人は

[prisma abyss]

そう

私は日本語を話せない 泣く>_< ;

[vale raven]

あっ

[prisma abyss]

でもちょっと

多くないよ

アジア人が数学が好きね。。。

ごめん

[vale raven]

いや、どのアジア人も数学が好きだとは限らない

It's kinda late, so good night o/

[prisma abyss]

日本の友だちはあまり有ります

お休み

night

[fringe basin]

fuck is this weeb shit

[fluid stream]

how does one mass point?

[scarlet quail]

mass point?

[dark sparrow]

@fringe basin uh

can you please not

[foggy oxide]

i always forget the identities for cos(arctg(x)) and sin(arctg(x))

[surreal bolt]

just fill in the right-angled triangle.

[foggy oxide]

.-.

look at that problem

[surreal bolt]

no no ... make the tutors here look good 😃

[foggy oxide]

what ?

[surreal bolt]

nvm

[foggy oxide]

it's too much embarrassing i thought very stupidly

but

is 6.66666666... = 6.67 ?

[surreal bolt]

depends on the class and the test.

and the instructor.

[foggy oxide]

because my area is 6.666... not 6.67

the problem answer is 6.67 (it wanted aproximated to 3 decimal values)

i actually don't remember the rules for aproximation

[surreal bolt]

if it s homework that is usually close enough.

[foggy oxide]

no it's a problem from brilliant

[surreal bolt]

you mean 3 significant figures?

[foggy oxide]

here

you had to type

[surreal bolt]

2 decimal places ... says the problem

[foggy oxide]

the value

yes

idk the rules for aproximation

how does it work ?

[surreal bolt]

that is not what you said earlier ...

[foggy oxide]

what did i said ?

i could just typed wrong

[surreal bolt]

3 places ...

[foggy oxide]

well 3 is very close to (2) on my keyboard, i'm sorry

[surreal bolt]

btw the way you round depends on your instructor.

[foggy oxide]

the answer says it's 6.67

not 6.66

[surreal bolt]

yes I know.

you said that 2 times.

[foggy oxide]

so, are my calculations wrong ?

i think you already know how i did

[surreal bolt]

Are you in the US?

[foggy oxide]

nope

[surreal bolt]

oh. well go ask your instructor.

[foggy oxide]

i don't have a instructor

.-.

[dark sparrow]

@foggy oxide round up at and above 5

[foggy oxide]

i think i got it

[upper karma]

guys

[dark sparrow]

mm?

[upper karma]

ugh.

can I use a link shorter to post an image?

[dark sparrow]

you can upload your image directly

[upper karma]

oh

ok

didn't think I could

[dark sparrow]

okay, those are four lines

[upper karma]

I haven't done anylitic geometry in so long. I am literally learning it now

the y interfect of r3

[dark sparrow]

well, what do you think the slope of r3 is?

[upper karma]

1/5

[dark sparrow]

mhm

[upper karma]

I calculated it

hmm

[dark sparrow]

and you know (4,-1) is a point on r3

[upper karma]

yes