geometry-and-trigonometry

[upper karma]

you can post those questions that you are stuck with and we can help you.

[dense marsh]

perfect then .........

i quit got a lot of them tbh

[upper karma]

you mean integrating those functions?

[dense marsh]

so alright

yeah exactly

[dense marsh]

exactly

but

[upper karma]

well first let's start with your question on complex numbers unless you wanna dive into calculus first

[dense marsh]

i dont wanna just keep saving things like sublime text do ....

not really

i'll start it with complexe numbers .........

so

[upper karma]

well do you know the form for how complex numbers are written?

[dense marsh]

writing any complex number under the trigonometrical form ..........

yeah

i've studied that 2 weeks ago

[upper karma]

so you know z = a + bi

perfect

[dense marsh]

right at the night of my big brother's birthday ...........

yeah

i know why did they made it

and well

i can calculate and multiply

things like that

usual ...........

and when i figured out abot the exponontial form i

i knew that m gonna need first to see how it works for the trigonometrical form first

that's why i got a bit stuck ........

and youtube didnt help thta much

[upper karma]

well here's the general "formula" for how the trig form should look like (using polar coordinates, smth I don't have that much intuition in unfortunately)

$z = r(cos\theta + isin\theta)$

[somber coyote]

Venti

[upper karma]

where r is the modulus of z

[upper karma]

you get r from this form

the modulus is represented as

$|z| = \sqrt{a^2 + b^2}$

[somber coyote]

Venti

[upper karma]

so that's your r

and theta

is a bit more complicated

$\theta = \arctan(\frac{b}{a})$

[somber coyote]

Venti

[dense marsh]

yeah i can see

so first

[upper karma]

or you could just write it as tan(theta) = b/a to make it look simplier

[dense marsh]

i gotta find [r]

[upper karma]

yup

[dense marsh]

does the formula of the arctan always works !?!?.............

[upper karma]

it should, arctan is just inverse tan

[upper karma]

in the arctan one i just made theta the subject

when you move a function to the other side of the equation you have to make it, its inverse

[dense marsh]

but hey

where is the relation i mean

[upper karma]

I wish I knew, but I don't really have the intuition for that

something I'll also be looking into so thank you

[dense marsh]

what interpretation that would help me to get why did we wrote that theta equal to arctan(a/b)

how old are you mate btw

[upper karma]

16

sweet sixteen

[dense marsh]

i hate when it comes to work with someone i have no idea about ........

16 years and you know about that

[upper karma]

still learning

[dense marsh]

16 years old kid here still face problems with polynomial divisions ...........

tbh

m 12

or 13

yeah who care time is relative anyway

[upper karma]

I wish I started when I was your age but all well that ends well I guess

[dense marsh]

right i

[upper karma]

so good job

[dense marsh]

i can see that .........

[dense marsh]

mom is hell of a mom

tbh

[upper karma]

uhh well I gtg eat dinner, try and solve this I guess -4 + 4i

[dense marsh]

well

solve like that

that's a complex num so

what should i

oh yeah

i get it

so it'll be i guess

[upper karma]

and you can ask your calculus question in #calculus and ping a helper if no one responds cuz I may not be around

[dense marsh]

basing on what you said ........

[upper karma]

[dense marsh]

r would be sqr(32)

and ..........

alright i gotta need a paper

[upper karma]

what are you asking here?

the derivation of the formula that was shown?

[upper karma]

also this isn't true, tan itself doesn't have an inverse, if we restrict the domain to just (-π/2, π/2) then it does have it's inverse, arctan. this also explains why if you "arctan" both sides or vaguely "do the inverse" with tan and arctan, you will restrict the domain to (-π/2, π/2), so you won't receive any solution outside this range.

[upper karma]

ah I see, thank you

[placid gate]

help?

[upper karma]

What have you tried

[placid gate]

i don't even understand it

[oblique grove]

do u still need help

well if u see this u basically need to find the unknown angle in the top triangle first

then that angle's opposite angle is also the unknown of the bottom triangle

then use the fact that all the angles in the angles have to add up to 180 and write an expression for all the angles in the bottom triangle

which needs to be equal to 180

then find x from that

[placid gate]

not anymore

it was D

[oblique grove]

ye

[near lichen]

how can I find the volume with just the width and height

[dark sparrow]

this doesn't seem like enough information.

is there anything here you're cropping out?

[onyx cloud]

if those ends are circular there's enough information, otherwise ann is right

in the case that they are circular:
you need to find the area of the shape from the top view, and find the volume by doing area*height

as for finding the area from the top, you need to somehow find the radii of those half circles at the end

[olive cove]

Also there is no side view so who knows what the back looks like. There might be a hole or a slope. No way to know.

Maybe its a trick question

[soft arrow]

Hey
I'm trying to convert 3ds max camera fov to lens.. but I'm failing... can any1 help ? >

def camLens():
    #a = 0.5 * (rt.GetRendApertureWidth() / math.tan(rt.selection[0].fov / 2.0))
    # Lens = 0.5 * (GetRendApertureWidth() / tan(FOV / 2.0))
    # and backwards
    # FOV = 2.0 * atan(GetRendApertureWidth() / Lens * 0.5)
    print("\n\n\n")
    print("Aperture : ",rt.GetRendApertureWidth())
    print(" Cur Fov : ",rt.selection[0].curFov) # the fov can be specified as horizontal/vertical/diagonal 
    print("     Fov : ",rt.selection[0].fov) # horizontal fov val
    print("    Lens : ",0.5 * (rt.GetRendApertureWidth() / math.tan(rt.selection[0].fov/ 2.0)))
    print("     Fov : ", 2.0*math.atan(rt.GetRendApertureWidth()/50*0.5))
camLens()

Aperture :  35.970001220703125
 Cur Fov :  94.07093048095703
     Fov :  39.56732177734375
    Lens :  13.298355324706613
     Fov :  0.6905799705312679

[soft arrow]

😄

[upper blaze]

Guys, does anyone remember is Center of triangle ABC only point that satisfy TA+TB+TC=0 where those are vectors?

[slender wasp]

there are so many different 'centres' of triangles that the question isn't specific enough

[upper blaze]

Haha i found it and i though in english when you say center you mean intersection of medians xd

[acoustic jungle]

How do you prove the larger the length of a chord on a circle, the larger the segment without using x-sin(x) (radians) is always increasing

[pure cape]

dont you just need the chord length and arc length formula?

as the chord increases, i.e 2rsin(x/2) increases, x is increasing since r is staying constant

hence the segment, or rx will increase

hmm nvm, maybe proving that the arc length increase might not be enough

[upper karma]

dis would be 2.87 right

[dark sparrow]

that doesn't sound right at all

not only is it a decimal (which you weren't asked for), it's even bigger than 2.

how can the sidelength of this octagon be longer than that of the square enclosing it?!

@upper karma

[upper karma]

but the area is 5

*4

sorry

oh wait :skil

💀

i meant the area of the hexagon would be 2.87 right

[dark sparrow]

what hexagon?

[upper karma]

I meant octagon

sorry its 12am in the us

[dark sparrow]

the area of the octagon is definitely not 287/100.

how did you get that value? do you have any work to show?

[upper karma]

ahaha my work looks super funny

I um divided it into two squares and 6 trianles

prob not the best method

ahahha- I have no idea how to do dis besides using the 45,45,90 method

but than I hit a stagnant and I have no idea what to do afterwards

[upper karma]

ahdkeed wait

[dark sparrow]

you may start by doing this

i am using the letter s for the side length

[upper karma]

I also think thats another method

to appraoch this problem

[dark sparrow]

this does not really get you any closer to finding the sidelength of the octagon.

oh wait, nevermind.

[upper karma]

sorry im really new to advanced geo

[dark sparrow]

this is doable, sure

it'll just take more know-how

[upper karma]

ahhaa

[dark sparrow]

do you understand what i did? (yes/no/i need time to digest it/i want to try something on my own)

[upper karma]

errr I would prefer an explanation

actually wait

ahah I have no idea why u did that

[dark sparrow]

did what?

[upper karma]

why s/sqrt 2 works

[dark sparrow]

and did you mean why or how? sounds like you meant how.

[upper karma]

sorry

I meant how

yeah

[dark sparrow]

be careful to say exactly what you mean and mean exactly what you say in the future. not just with me but every time you speak to someone else.

makes things easier on both you and the other party.

anyway...

you see those little white triangles in the corners?

[upper karma]

sorry sorry

[upper karma]

yes

[dark sparrow]

you see how they are 45-45-90 triangles with hypotenuses equal to s?

[upper karma]

yes yes

[dark sparrow]

and you know that in a 45-45-90 triangle, the hypotenuse is sqrt(2) times either leg, yes?

[upper karma]

yes

[dark sparrow]

do you understand why i marked the legs of those little triangles as s/sqrt(2) now?

[upper karma]

wait how can we determine that this is a definite hypotenuse

[dark sparrow]

what else could it be? the vertex opposite to it is the corner of a literal square.

[upper karma]

sorry im a bit foggy rn

I assumed that the side opposite of s/sqrt 2 is the hypotenuse

[dark sparrow]

sides are not opposite to sides in a triangle...

[upper karma]

sorry im thinking of soh cah toa

[dark sparrow]

also why did you assume that

when

[dark sparrow]

i literally said this

[upper karma]

sorry

[dark sparrow]

do you understand where the s/sqrt(2) marks come from now, Y/N

[upper karma]

yes

[dark sparrow]

okay

do you understand how to get the value of s now?

[upper karma]

err no

[dark sparrow]

you see the side of the square is broken up into pieces of length s/sqrt(2), s and s/sqrt(2), yes?

[upper karma]

yes

[dark sparrow]

can you write that as an equation?

[upper karma]

o yes

idk how tho-

[dark sparrow]

you have just sent mixed signals

you said "yes" and then immediately followed up with "no"

please do not do that

[upper karma]

sorry

im a freshman in hs

[dark sparrow]

unless you are deliberately trying to confuse me, in which case you should admit it explicitly

[upper karma]

so i am actually very confused

[dark sparrow]

if you mean you can't write the equation then say you can't write the equation, there's no shame in that

[upper karma]

yeah thats what I mean

sorry for using the freshman in hs excuse

[dark sparrow]

if a line segment is broken into several non-overlapping pieces, and you know the length of each piece, how do you find the length of the whole thing?

(i'm looking for a description in words here)

[upper karma]

do u add all the pieces

[dark sparrow]

i don't know, do you?

[upper karma]

um yes

idk

[dark sparrow]

so if i gave you a segment made of a 9ft piece and a 3ft piece you would be uncertain whether or not the whole thing has length 12ft, did i get that right?

[upper karma]

no

u would def add

all of them

to find the whole length of a line segment

[dark sparrow]

great, so why feign uncertainty like you just did?

[upper karma]

sorry

i just realized what u meant

[dark sparrow]

and now let's take that 'add all the parts to get the whole' idea and apply it to your problem

[upper karma]

so s/sqrt 2 + s/sqrt 2 + s

right

[dark sparrow]

that's the sidelength of the square in terms of s, yes.

now make that into an equation.

use the fact that you know how long the side of the square is.

[upper karma]

ummm so (s/sqrt 2 + s/sqrt 2 + s)^2=4

?

[dark sparrow]

why did you square it

what stopped you from writing $\frac{s}{\sqrt{2}} + \frac{s}{\sqrt{2}} + s = 2$?

[upper karma]

becuz wouldnt u square one side to get the area of a squre

[somber coyote]

Ann

[upper karma]

OHH WAIT

ye

mb

[dark sparrow]

WHO SAID ANYTHIGN ABOUT AREA LMAO

[upper karma]

that is the faster way yeah

[dark sparrow]

at no point

in our discussion

did area

EVER come up

in any way

[upper karma]

im kinda dumb pls bear with me xd

[dark sparrow]

nor did the problem ask for the area of anything, not even indirectly

[upper karma]

sorry

xd

[dark sparrow]

i don't like how you keep apologizing. it strikes me as the wrong thing to do if i were you.

though i can't say i can suggest a good alternative.

anyway, now we have the equation.

are you able to solve it for s?

[upper karma]

ye

so x = -2 + 2 sqrt 2

I mean s

[dark sparrow]

that was fast.

[upper karma]

yeah

I figured it out

[dark sparrow]

did you do the work for it as we were talking?

[upper karma]

💀 er i tried figuring it out on my own

as well

xd

[dark sparrow]

but yes, s = 2(sqrt(2) - 1).

[upper karma]

i used my calc and got 0.82

0.83*

so thanks

[dark sparrow]

who said anything about decimal approximations?!?!

[upper karma]

um isnt that what ur supposed to do

[dark sparrow]

no

"simplest form" does not mean "approximate to two decimal places"

you are asked for an exact answer

[upper karma]

"cru

oh sorry

@dark sparrow wait sorry for the ping but is it alright to assume that all values of s are the same length

[dark sparrow]

what?

what do you mean "all values of s"

s doesn't have multiple different values

[upper karma]

sorry im a bit lost because like idk if we can assume all values of an octagon are equal

[dark sparrow]

if several different sides are marked with the same letter it means they all have the same length

[upper karma]

but thats the thing

[dark sparrow]

idk if we can assume all values of an octagon are equal

[upper karma]

the question didn't mark them all

[dark sparrow]

you meant sidelengths

[upper karma]

ye

sorry

side lengths of the octagon

[dark sparrow]

you don't need to assume anything

you are told

the octagon is regular

[upper karma]

oo right

sorry

[dark sparrow]

it helps to read the problem

[upper karma]

err are u from the us

its like almost 1am

in the U.S

[dark sparrow]

no, i'm not from the US

[upper karma]

ooo where are u from

(if u dont mind me asking)

[dark sparrow]

russia

[upper karma]

o.O u live in russia?

[dark sparrow]

yes.

i was born here.

[upper karma]

10/9

use soh cah toa

[dark sparrow]

please don't present obviously incorrect answers as fact

they are being asked for sin(θ), while the calculation you made gives tan(θ).

(plus, 10/9 is bigger than 1, so how can it be sin(anything)?)

[keen nova]

hey I've got this question that I've had for over a year now and still cant prove it, i just came up with this question on my own

"given an arbitrary quadrilateral consisting of 4 distinct points, it can always be translated/rotated around in the co-ordinate plane such that all 4 of its points lie in different quadrants"
intuition is telling me it should be true but i cant seem to prove it.

https://www.desmos.com/calculator/3va2tnek7h

I made a desmos thing to help test it

and thus far it works

[dark sparrow]

this is equivalent to saying there exists a pair of perpendicular lines which crosses each of its sides once, is it not?

[keen nova]

no not really? do you mean each pair of opposite sides?

[dark sparrow]

i mean if you treat the pair of perpendicular lines as a single shape

a plus-shape for lack of a better word

what i mean is there exists a plus-shape whose four arms each intersect a side of your quadrilateral

[keen nova]

well in the desmos graph there is an example that doesn't cross all the sides lol

[dark sparrow]

??

[dark sparrow]

well this one is not aligned as we need

[keen nova]

OH IM DUMB

nvm

[dark sparrow]

theres no vertex in Q4

if you were to move it up tho

hm

[keen nova]

yeah it would ok yeah

[dark sparrow]

this feels like the polygon's convexity may play a role but im not 100% sure

[upper karma]

What about rotating the plane

We could just inter change the positions of the points and prove the congruency of the new quadrilateral with the older.

[keen nova]

https://www.desmos.com/calculator/2jarxzmurf

improved the desmos thing

[upper karma]

@keen nova probably this

[keen nova]

not entirely sure what that means

[upper karma]

And as Ann said we could use the plus as the diaconals

[keen nova]

also don't know what you mean by that

[onyx cloud]

this should be true only for convex quadrilaterals. I'm pretty sure there are concave ones that won't satisfy this for their solutions

[dark sparrow]

can you present an example of one

[onyx cloud]

something like this should do it https://www.desmos.com/calculator/qofdhum9io

unless i'm missing something obvious

[dark sparrow]

hmm

ah yes i think you might be right

[onyx cloud]

your idea should work for convex and some types of concave quadrilaterals, though, so it could still be worth exploring with an amended question

[plush cliff]

how do i do the 8th one

[onyx cloud]

have you heard of the law of sines, and the law of cosines?

[white jasper]

Help pls

I feel embarrased not being able to solve this.

[onyx cloud]

uhhhh

are you given anything else?

those side lengths could be anything

[white jasper]

Nope, Im 20 years old a physics/engineering student. I tutor a kid who asked me this question from middleschool.

[onyx cloud]

well it could be any of those answers

they all make sense

[white jasper]

Can you explain ?

[onyx cloud]

yes. you aren't given any other side lengths, so these triangles could be of any size

[white jasper]

Understood.

[onyx cloud]

i'm baffled by this. there's gotta be more to it

[white jasper]

I guess so, why would a 16 years old kid get this -_-

[onyx cloud]

the better question is, why is a 16 year old in middle school

[white jasper]

Idk how the system works in america bro. I might be telling it wrong. I live in Belgium.

[onyx cloud]

ah nvm then

but yeah this question is odd

[white jasper]

He is not in highschool but the year before it. ( do you understand now 😀 )

He should be in highschool but he didnt pass one year.

[onyx cloud]

ah makes sense

[still mica]

if i pass sin(36.87) into a calculator which gives .6 and Cos(36.87) which gives .8 i know the ratio of lengths of a right triangle. what is the calculator actually doing to get the value of .6 and .8 ?

[wintry tundra]

idk but heres something cool, square the results of both and add them

[fast gate]

There's probably a few ways to do it, and I assume calculators would use one I don't know for better precision and/or speed and/or how much stuff it has to store. One I do know is using Taylor series, which is a way of representing a function (not just sin and cos) as an (in)finite sum. I doubt it is this exactly, because it doesn't work well for x far away from 0 (maybe it adds or subtracts to get closer to 0 because the functions repeat?)

Oh this also isn't in degrees, but calculators can easily convert away from that.

[subtle palm]

i might be able to help

find alpha and just go with the-OHHHHHHHHHHH I SEE THE PROBLEM

damn

uhm

i could assume; key word assume

that alpha must -mmm nah that aint gonna do

[dawn temple]

anyone here familiar with the zBuffer concept for 3D graphics?

[golden holly]

hi

[fading galleon]

<@&286206848099549185>

[upper karma]

so i get the part of trigonometry for the sin(x) cos(x) and tan(x) and how you can use them and the reverse functions to find missing angles and sides etc, stuff like that. i'm familiar with. the thing i dont get is what the calculator actually does when you put them into it. e.g if you put cos(30) in it gives 0.15.... but what is the calculator actually doing to get from 30 to 0.15...

[storm portal]

I think the calculator uses what is called a Taylor series or a Taylor polynomial.

A Taylor series/polynomial is essentially an infinitely long polynomial that represents sin(x) or cos(x) in a polynomial form

Addition, subtraction, multiplication, and division are much easier for a computer to handle, so I think that is what calculators use to computer things like cos(30) or cos(30˚)

[small trail]

How would you calculate the sides of an rectangle if you know the following:

1. The area of the rectangle.
2. The aspect ratio (e.g. 16:9)

[onyx cloud]

what does the aspect ratio tell you?

[silk patio]

Socrates?

[small trail]

Already solved my problem

[fervent valley]

What does the working of c mean?

[olive cove]

there are 12 months in a year

The compounding interest is 3%

So x¹² = 1.03

And then you solve that equation

And then substract 1 of course for the interest percentage

[upper karma]

,rccw

[upper karma]

what you wrote is good so far

do you have issues solving the equation?

okay, do you have any ideas on what our next step can be? @upper karma

note that we have on the left side a x+2x and a 34+120 which can be simplified.

34+120*, but yes.

which gives us 154.

and now anything else that you think we can simplify?

do you not know what x+2x is equal to?

great, yes

so we have $3x+154=154$.

[somber coyote]

Al𝟛dium

[upper karma]

do you have any ideas at all on what step to do now?

by what precisely?

okay we can do it, but you can also substract 154 from both sides

[upper karma]

x is just 0 right?

if we divide by 3, we get $\frac{3x+154}{\color{green}{3}}=\frac{154}{\color{green}{3}}$

[somber coyote]

Al𝟛dium

[upper karma]

yes, if you notice we can substract 154 like the following: $$3x+154{\color{green}{-154}}=154{\color{green}{-154}}$$

[somber coyote]

Al𝟛dium

[upper karma]

simplified further, 3x=0

why not

@upper karma please stop spoiling stuff, i know your intentions are good but i want them to solve it

oh

sorry

i didnt realise sorry

x can perfectly be 0, yes

[upper karma]

but most importantly, do you understand what i said here?

what about it you didn't understand?

did you learn to solve equations with the vague "move to the other side with the opposite operation"?

okay then what is causing your confusion?

[upper karma]

oo are those the commands of the bot

$/frac{2/3}

$\frac{2/3}

o>O

[fervent valley]

How do u derive this?

[pure cape]

the non-existent part a, your question is a very interesting one

anyhow, whatever part a asks, you are supposed to be using one of the cross section of the pyramid that includes h as a side to find h

[fervent valley]

Oh ok

[feral musk]

can someone solve this please?, i don't really get it

[silent plank]

what specifically don't you get?

do you know your trig ratios?

[ancient tusk]

idk how to approach this sum can anyone help me

[onyx cloud]

have you heard of the double angle formulas

[ancient tusk]

no

[onyx cloud]

its a good place to start

[dark sparrow]

@ancient tusk sin(2x) = 2sin(x)cos(x)

does this sound familiar to you?

[feral musk]

which is the correct answer

[ancient tusk]

yes

[light loom]

The answer is A since arcsin of 2/3 is angle a which simplifies the expression to tan(a) which you get option A

[feral musk]

thank you so muuuuuch, i hadn't understood arcsin of 2/3 before

[upper karma]

can someone help me prove excenter property

[silk patio]

Which one

[quartz goblet]

Central line of triangle is SD/2 - diameter. so circle have a radius so the square of circle is (SD/4)^2*pi and you need just the half of (SD/4)^2pi

[umbral snow]

SAD

[upper karma]

wait wut

sorry im lost lol

oh wait

nvm

wait wut

no

wait wuts

[umbral fossil]

By diameter of 170feet is it talking about the maximum height?

[dark sparrow]

no, the max height is not 170ft

the bottom of the wheel isn't on ground level

[umbral fossil]

@dark sparrow Could u help me

Ik since one trip is three minutes

that should be the frequency

or one full cycle

[dark sparrow]

i mean okay sure but we weren't talking about the period yet

are you able to identify the heights of the peaks and troughs of your sinusoid?

[umbral fossil]

Sorry the what?

[dark sparrow]

the min and max values

[umbral fossil]

Well since it's 5 feet above the ground

and diameter of 170 feet

Max value 175 min value 5?

[dark sparrow]

correct

[umbral fossil]

Amplitude would be 85 then

[dark sparrow]

yes

[umbral fossil]

vertical displacement 90

Now for the period

would I display this is 2pi/b?

so 2pi/3?

[dark sparrow]

you're confusing the period and the coefficient on x

[umbral fossil]

So it takes 3 minutes for one cycle

[dark sparrow]

if you're measuring time in minutes then yes the period will be 3 and you'll have cos( (2π/3)x ) pre-vertical transformations

yes the period is 3 minutes

[umbral fossil]

Right

[dark sparrow]

you need to decide if you're measuring time in minutes or seconds here

[umbral fossil]

Which would be easier

[dark sparrow]

unless you're allergic to the number sixty, they're both equally easy

[umbral fossil]

We could just do it in minutes

Would there even be a phase shift?

[dark sparrow]

i mean, sure, if they really want there to be a phase shift, you could give them a phase shift

since our function 'starts' at a trough rather than a peak as would an un-shifted cosine function

[umbral fossil]

It's asking for one but I don't see a possible phase shift

Is this correct so far?

[dark sparrow]

yes this is correct (so long as you put a y= in front of it) but it may not 100% fit the form

they might want you to write $y = 85\cos\paren{\frac{2\pi}{3} x + \pi} + 90$

[somber coyote]

Ann

[dark sparrow]

to be in full compliance with their asking for a phase shift

and nothing more

[umbral fossil]

the phase shift is pi?

[dark sparrow]

yes

[umbral fossil]

how come

[dark sparrow]

i know this makes it sound like -85 cos(2π/3 x) + 90 is illegal or something but it isn't. this is just an unfortunate side effect of strict grading criteria and bureaucrat teachers

well $\cos(t + \pi) = -\cos(t)$ doesn't it?

[somber coyote]

Ann

[umbral fossil]

I'm pretty sure my teacher accepts it either way

[deep zodiac]

Hi, I’m having some trouble finding an exact value for cos15 by using the diagram of this equilateral triangle inscribed in a square. Any help would be greatly appreciated

[silent plank]

@deep zodiacyou messed up your length for AK

[deep zodiac]

In what way?

Yep I see it now

Thank you

Still doesn’t give me the value for cos15 though

[silent plank]

show your updated work

[silent plank]

expand and simplifiy

[silent plank]

yes

[deep zodiac]

And then factor

[silent plank]

don't really need to

[deep zodiac]

Thank you very much

[silent plank]

its fine as it is

[deep zodiac]

Alright

[hasty mesa]

How do I find the centre(point equidistant from every vertex) of a N sided convex polygon?

[dark sparrow]

it's not guaranteed to exist, is it?

@hasty mesa

like, the way you're defining it, there's no telling whether your polygon even has a point equidistant from every vertex

unless N=3

[hasty mesa]

Shit that's why I didn't get anything when I searched for it

[onyx cloud]

is this possible for any N-sided polygon in N-1 dimensions?

[hasty mesa]

Okay, I need a point as demonstrated in pic such that if we extend lines from that point through every vertex and join the ends we get a up-scaled version of polygon. Will centroid do the job? (sorry for bad diagram)

[onyx cloud]

so you want the side ratios to remain equal and the angles to stay the same too?

[hasty mesa]

Yes, the perpendicular distance from every side should also be the same

[onyx cloud]

what do you mean?

[hasty mesa]

Do you see x? that's the distance between every side in the initial polygon and the up-scaled polygon

[onyx cloud]

that won't always be the case

i don't think there's a point where that will remain constant across sides

let me rephrase that

i don't think there's a point where that distance will remain constant across all sides for every convex polygon

as a quick example, take a very stretched rectangle

nevermind I confused myself

let me give it some more thought

[hasty mesa]

I think it'll work

[onyx cloud]

hm i don't think it would

the distances to the sides are much larger than to the top and bottom

[hasty mesa]

And all the sides moved parallel to their respective perpendicular vector away from the centroid??

[onyx cloud]

yes, while maintaining the side ratios

obviously this image isn't perfectly centered or perfectly scaled

but it should give you an idea of scenarios where this isn't possible

[hasty mesa]

Let's not worry about the maintaining the side ratios

[hasty mesa]

This is important

Every side moves parallel to their respective perpendicular vector away from the centroid. And I just need the area of the final polygon

[dark sparrow]

so you want to push each side 'out' by x units?

[hasty mesa]

Yes

[dark sparrow]

i'm not sure the resulting polygon would be similar to what you started with

it may even have less sides if some of the original's sides are too short

[hasty mesa]

I forgot all the co-ordinate geometry that I studied in high school

[hasty mesa]

wot?

[onyx cloud]

that's why i wanted to maintain the side ratios lol

[dark sparrow]

maybe im wrong about that

[onyx cloud]

no you're correct ann

unless we're strictly talking about convex polygons

[hasty mesa]

The polygon is convex

[onyx cloud]

do you have a more contextualized example of the question>?

it could give more variables to work with and a more concrete question

[hasty mesa]

Like this?

[onyx cloud]

i mean, what are you applying this question to? where is it coming from?

is it a random convex polygon?

[hasty mesa]

yes

Wait can I dm you? (if you don't mind)

[onyx cloud]

i'd rather keep it in here. does the question come from a private project or something?

[hasty mesa]

Yes something like that

[onyx cloud]

then that's alright

i just want some more context

[deft veldt]

What's the best book/resource for learning euclidean geometry from the ground up?

[upper karma]

Euclid's elements themselves

I guess :D

[stray void]

-W
Wonder Book Of Geometry might be also helpful...

[deft veldt]

Thanks looks good. Will check it out

[brazen pendant]

any chance this can get pinned?

[graceful spear]

can you also pin how they got derived , so im not clueless

[shadow shadow]

why does this need to be pinned?

[graceful spear]

they have this pinned

i mean i dont see those on the list unless im blind

[shadow shadow]

it can pretty easiely be derived

and also

i see a problem

xD

for the periodicity one it should be

$for all n \mathbb{Z}^{+}$

[somber coyote]

Elonmosqito96

[shadow shadow]

eww thats gross

[upper karma]

No, it works just fine for Z

[shadow shadow]

no?

oh

im an idiot

xD

i read Z though R

xD

also no idea why i thought Z+

[onyx cloud]

$\forall$

[somber coyote]

maximo

[onyx cloud]

$\forall n\in\mathbb{Z}^+$

[somber coyote]

maximo

[shadow shadow]

tbh $\forall$ kinda just flexing math notation... and most people in geo/trig wont understand

[somber coyote]

Elonmosqito96

[onyx cloud]

you're right

i thought for a sec this was latex-help

[shadow shadow]

xD

[pastel moon]

lol

[brazen pendant]

its derived from sinx^2= (1-cos2x)/2

i think thats the right formula

[upper karma]

Can someone help with this one?

which theorem is to be applied

[velvet arrow]

hmm did you figure that one out yet?

[upper karma]

no i dont even know if there is theorem

for this

[velvet arrow]

ptolemys theorem?

I think it might work

I will also try to solve it with you

[upper karma]

let me read about this theorem, then i'll be back

[velvet arrow]

I think it can work but, seems to be very complicated if we wish to find the angle

It is better for lengths

It may be possible to try some type of law of cosines

[upper karma]

dude,

i dont think its that complicated,

but i just want to know where to start

[velvet arrow]

Well when i think about it

if we shift P around, that has no effect on angle AOB

Maybe it cannot be determined?

[humble umbra]

||the angle at the center is twice the angle at the circumference or whatever it's called||

[upper karma]

yeah this is the theorem,

[humble umbra]

should be a direct application of that

[upper karma]

yeah

[upper karma]

Archimedes - measurement of a circle - Prop. 3

The translation by Thomas Heath in english is easy to find.

[velvet arrow]

135?

[upper karma]

yes

[humble umbra]

looks like it

[velvet arrow]

This dude Archimedes had a lot of time on his hands

[upper karma]

I believe you might also find an answer in Euclides' elements

[velvet arrow]

Idk what else I would have done in 240BC

[upper karma]

There must be a section which lists how to construct different shapes.

with compass and straightedge

Look for how to construct polygons of the same angle and sides

in this case the angle belongs to an octagon

Again, Mr. Heath seems to have covered it for us

yeah thx

i will look it up

https://en.wikipedia.org/wiki/Euclid's_Elements#Contents
book 3.
Book 4 even deals with circumscribing regular polygons (inside a circle) specifically.

Another topic. I have the sines of 30, 45 and 60 memorized. What is a good proof to help me memorize other common values like sine of 20, 40, etc..?

[upper karma]

ok i will check it thx man

[upper karma]

this is out of my league , sry man,

there is no pattern in sines or other trig functions, they're just values , i just use a calculator

you can use pythagorean theorem to work out 45º for example.

sine(22.5) seems to be 1/sqrt(5)

calculator disagrees 😦

[upper karma]

sin^-1(n1/n2 sin(th1)) = th2

does that make sense?

[fierce tendon]

no

[upper karma]

I applied the inverse sine,

Maybe it can be simplified to something like n1/n2 th1= th2

[slender wasp]

sin theta2 = n1/n2 sin theta1

[upper blaze]

If I have 2 fixed points and 1 arbitrary point in plane. Can I somehow look at formula for area of triangle that has no absolute values or roots?

[dark sparrow]

i doubt it

you're going to at least need the distance between the two fixed points, and that is impossible to calculate without access to square roots

@upper blaze

also i'm assuming you have those points given in terms of their coordinates, is that correct?

[upper blaze]

yes

Actualy I am doing something for my thesis and roots must disapear

so I used formula with absolute value but even that isnt that pleasent to have but probably better then roots

[dark sparrow]

why must the roots disappear?

do you have any more context for this?

wait hold on i just remembered something

if you really are given the coordinates

ah yes

the shoelace formula will work

https://www.wikiwand.com/en/Shoelace_formula

three points in your case

my apologies for misleading you earlier

[upper blaze]

Oh I have to check this 😄 Well I am doing my master thesis and its from algebaric geometry and I deal with polinomials so if I have root of x... you know 😄

Oh this might solve my problems hehe thanks a lot I never heard of this though

Oh this still has absolute value but that I can manage and this actualy may solve things in generalization hmm

[short iris]

how would i find an area of a circle thats circumference is 1/2 pi squared?

[pure cape]

Find the radius, then find the area

[short iris]

how do i find the radius

like

wait nvm

[smoky flower]

Just that 10th question

[vestal lily]

@short iris radius = C/2pi
you have C = 1/2 pi squares
Therefore you got the radius now
Now you could find the area easily

[vestal lily]

@smoky flower

[smoky flower]

Ohhhh understood thx

[short iris]

thasnks

[brazen pendant]

does anyone have any tricks to remembering the product to sum and sum to product formulas?

[dark sparrow]

rederive them from angle sum/difference formulas

[brazen pendant]

how?

[dark sparrow]

here's an example

$\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y) \ \cos(x-y) = \cos(x)\cos(y) + \sin(x)\sin(y)$

[somber coyote]

Ann

[dark sparrow]

adding these two gives $\cos(x+y) + \cos(x-y) = 2\cos(x)\cos(y)$

[somber coyote]

Ann

[brazen pendant]

oh cause the sins cancel

thx

how about the other way though

@dark sparrow

from product to sum

[dark sparrow]

this is product to sum

you meant sum to product

i... guess those are just product to sum but backwards

[brazen pendant]

oh yeah i meant that

[dark sparrow]

like if you take u = x+y and v = x-y then you have x = (u+v)/2 and y = (u-v)/2 and you can put them into the above and get the corresponding s2p identity

[brazen pendant]

what does s2p stand for?

oh wait its sum to product

@dark sparrow i was going through the formulas and im stuck on sin(x+y)+cos(x-y)

its not symplifing

[dark sparrow]

nyeh? how did you get that

[brazen pendant]

so sin(x+y) = sinx cosy + cosx siny

and cos(x-y) = cosx cosy + sinx siny

[dark sparrow]

yeah but then why would you multiply these

what formula are you trying to derive here

[brazen pendant]

i meant sin (x+y)***+***cos(x-y)

sry type

*typo

[dark sparrow]

ah

sin(x+y) + cos(x-y)?

[brazen pendant]

yes

[dark sparrow]

why that would just be (sin(x) + cos(x))(sin(y) + cos(y)), no?

[brazen pendant]

wait how did you do that

[dark sparrow]

sin(x)sin(y) + sin(x)cos(y) + cos(x)sin(y) + cos(x)cos(y)

[brazen pendant]

oh then you factored

i see

ok imma keep working through the rest of the variations

well with a bit of algebraic manipulation i got em all!

🥳

[upper karma]

Can anybody help me here. Obviously to calculate the area of a triangle we do base * height over 2. Somehow he calculated the height as the magnitude of B * sine theta and I'm a bit unsure how that equals the height of the triangle? If anything, I would have done (A/2)^2 + C^2 = B^2 and took the square root to find it.

[silent plank]

consider treating the length of the base as |A→|

the altitude of the triangle is represented by the dashed line

which given |B→| and theta, can be determined using basic trig

[upper karma]

Ah, okay I figured it out. Sine theta = opp/magnitude of B. He's multiplying it by B to get just the opposite.

Doh! Thanks a lot.

simple stuff lol

Mhm, here's another one actually, how is this true exactly?

[storm portal]

cos(π/2 - θ) = sin(θ)
You can use the cosine subtraction formula

[dark sparrow]

@upper karma make a right triangle in which one of the acute angles is θ, then the other will be θ'

cos(θ') and sin(θ) are the exact same ratio of sides

[upper karma]

Ah okay, I get that now. That's interesting.

Thanks a lot.

[brazen pendant]

Express this product as a sum

i got sin(5x)+sin(-x)

am i right?

[dark sparrow]

think you might be missing a 1/2 factor...

[brazen pendant]

hmm

well i got sin(2x+3x) + sin (2x-3x)

[dark sparrow]

sin(u)cos(v) = 1/2 (sin(u+v) + sin(u-v))

[brazen pendant]

ooo

what i did is times 2

thx

express this product as a sum

1/2[sin(5x)-sin(-3x)]

is this right @dark sparrow ?

[dark sparrow]

yes but it can be simplified a little

-sin(-3x) = sin(3x) after all

[brazen pendant]

oh yeah

[brazen pendant]

not like that i swear

[scenic oxide]

I was thinking that since each angle in the regular 10-gon case is 144, you can play a game of give and take with the angle degrees. You can make one angle acute, say 89, by giving another angle (144-89=55), but that would make it obtuse so you have to give it to two angles (27.5 each). You end up only being able to do this 4 times.

[rich wolf]

180*8=1440

90 * 10 = 900

A convex polygon may not have any angles greater than or equal to 180

1440 - 900 = 540/0 = ∞ (not possible)
1440 - 810 = 630/1 = 630
1440 - 720 = 720/2 = 360
1440 - 630 = 810/3 = 270
1440 - 540 = 900/4 = 225
1440 - 450 = 990/5 = 198
1440 - 360 = 1080/6 = 180
1440 - 270 = 1170/7 = 167 + 1/7

So at most 3 acute angles. In the process above I subtracted the maximum sum of the non-obtuse angles from the total interior angle sum and then divided that by the number of remaining angles, repeating the process by removing one non-obtuse angle at a time until the average measure of the obtuse angles was less than 180.

(1440 - 90n)/(10-n) < 180

In fact ANY convex ngon may only have 3 acute angles

Let me actually make this a bit more rigorous. The interior angle and exterior angle at any vertex of a convex n-gon form a linear pair that sums to 180 degrees. This means that 180 - interior angle = exterior angle at each vertex. So the sum of the exterior angles must equal 180n - 180(n-2) = 180(n-(n-2)) = 180(n-n+2) = 180(2) = 360.

Let us represent the average measure of the acute angles in the polygon by 90-ε, for 0<ε<90. Then each corresponding exterior angle to the acute angles is 180 - (90 - ε) = 90 + ε. For some number k of acute angles, k(90+ε) must be less than 360. For k = 4 we have the sum of the exterior angles as 360 + 4ε which is not possible as the exterior angle sum must be exactly equal to 360. For k = 3 we have the sum of the exterior angles as 270 + 3ε, which depending on the exact measure of the acute angles allows for the other non-acute angles of the polygon to be less than 180, avoiding the degenerate case. Therefore the maximum number of acute angles in a convex n-gon is 3.

[rich wolf]

You can extend this to show that in a convex polygon with three acute angles the average of the acute angles will be greater than or equal to 60, with equality occurring only in the case of a triangle.

[scenic oxide]

OK that makes sense, thank you

[idle bobcat]

can someone tell me why this is wrong?

or i guess "partially" correct

[umbral snow]

Consider instead x² - x = 0

There's two solutions to it, x = 0, and x = 1

If you divide both sides by x, then you lose x = 0 as a solution

The better method is to factor:
x(x - 1) = 0

Same with yours:
cos(x)(cos(x) - 1) = 0

That is, the incorrect solution above is missing π, (3/2)π

@idle bobcat

[idle bobcat]

oh wow

that makes a lot more sense

i need to make sure i dont make this mistake on a test cause i feel like i would make this mistake 9 times out of 10 lol

[umbral snow]

Basically, never solve by dividing. Polynomial problems are solved with factoring always @idle bobcat

[idle bobcat]

gotcha thanks

[ionic bluff]

yes

unless you have something that can only be solved with division

take for instance

sinx = cosx

there is no way to solve this without saying that tanx = 1

[idle bobcat]

makes sense

also is it just me or is the answer key wrong here

shouldnt it be neither

-(-x)^3 should be -x

[ionic bluff]

-x^3 = (-x)^3

-(xxx) = (-x -x -x)

lol

[slender wasp]

tbf there is a way to solve sin x = cos x more nicely in that way lol just say that if cos x = 0 then |sin x| = 1, so if x is a solution then cos x is non-zero

[ionic bluff]

you have to divide and take the arctangent

[slender wasp]

yh

[idle bobcat]

for the domain how come the constraint it greater than or equal to

why isnt it just greater than

[slender wasp]

It's because sqrt(y) is defined for all y >=0

[idle bobcat]

why is it not defined for y=0?

[ionic bluff]

it can be 0

just not LESS than 0

[idle bobcat]

oh sorry my brain was broken earlier lol disregard

[peak mesa]

hey guys, what is the name of this geometric shape?

[dark sparrow]

looks like it might be an astroid, but it's impossible to tell for sure

it may just be 4 quarter-circle arcs arranged in a pointy shape

[peak mesa]

But I was looking for a specific name, because I'm creating something with geometric shapes and each shape needs to have its own name and I don't know what to name :/

oh lmao

its a astroid yeah

not 100% but yeah

thank you

and this one btw xD??

can i called Star 4?

its because a star 4 normally is something like

[dark sparrow]

idk this one doesn't strike me as one that necessarily has its own name

[peak mesa]

its like ninja star but thats not the correct name i guess?

[dark sparrow]

i think obsessing over names is kinda pointless...

[peak mesa]

i need to have names for all the shapes im using

i can give the name "Ninja Star"

but i think its not the correct one

[dark sparrow]

why do you need names

[peak mesa]

programming - enum

and class names

[dark sparrow]

just give them names you recognize then

[proud merlin]

any software similar to wolfram alpha? WA is really broken , so any similar recommendations

[upper karma]

How is it broken?

I think it generally works pretty well

[proud merlin]

lot of wrong answers , and not good presentation , at least from my experience

like so many errors in integrating functions and solutions of differential equations

[upper karma]

Wolfram alpha is a Computer Algebra System, so you can use that term to do research into alternative ones

Sage math is the most prominent free alternative

But it has a learning curve

[placid gate]

help?

[storm portal]

For this problem, you’ll have to identify what you do/don’t know, and how you can use the given information to find unknown angles. These unknown angles may help you get other unknown angles, or they can help you directly find the angle you’re looking for.

[placid gate]

ik PEF and EPF and FQR and don't know QRT

[storm portal]

How can you use the angles you know to find angles you don’t know?

The new angles you find can subsequently be used to find either one of the following:

1. Other unknown angles
2. The answer: QRT

[placid gate]

?

what other unknown angles are there other than QRT

?

[storm portal]

Well, what angles were you not given?

There is one angle where (if you find it), you can solve the problem

In order to find that angle, you may need other angles

[placid gate]

but what angles do i need?

they gave me all of them

[storm portal]

If they gave you every angle, then you should be able to solve the problem using the following concepts:

Vertical angles are congruent
The sum of the 3 interior angles in a triangle is 180°
A line has an angle of 180°

[placid gate]

r is 180

this is 180

[storm portal]

Correct. Call that angle FRT

You can use the 3 concepts I gave you to find QRT

[placid gate]

but i don't get them

[storm portal]

Vertical angles are essentially angles that are opposite each other

The second one is just that the angles in a triangle sum to 180°

[storm portal]

And I know you know the last one because you made this observation

[placid gate]

we know that is 180

but how do i find the measurment?

[placid gate]

i got it

it was 145

[hoary wedge]

Can someone help? The answer is x= -0.47 but i dont know how

[dark sparrow]

take u := arcsin(x), then your equation becomes u(pi/2 - u) = -1, a quadratic in u

[hoary wedge]

thankyou :)

[silent plank]

<@&268886789983436800>

[spark stag]

sorry, it seems harmless but we dont allow advertising

[versed flame]

what does multiplying the radian with the area of a circle and dividing it by 2pi (360 degrees in radian) do?

i saw it somewhere and im confused

[silent plank]

not sure what you mean exactly

[versed flame]

so like if i have a circle with a radius of 2, does ((pi*2^2)(pi/2))/2pi give me the area of 1/4 of the circle?

[silent plank]

oh 1 sec

[versed flame]

sry idk how to use latex

[silent plank]

$\frac{\theta^c}{2\pi} \cdot \pi r^2 = \frac{\theta^c}{2} \cdot r^2$ \ \
$\frac{\theta \deg}{360°} \cdot \pi r^2$

[somber coyote]

ℝamonov

[silent plank]

this gives the area of a sector with angle theta

[versed flame]

wait gimme one sec to process that

[silent plank]

fraction of a full revolution, multiplied by the area of the full cicle

[versed flame]

wait whats that c?

oh waiut that means radian

right

[silent plank]

c in superscript can be used to denote radians

[versed flame]

ah

wait but what does dividing by 2pi do?

$\frac{\pi}{2\pi} \cdot \pi 2^2 = \frac{\1}{\4} \cdot \pi 2^2

it not working

i cri

ok but i kidna figured out

[silent plank]

there are 2pi (radians) in a full revolution

[spare iris]

In a certain right angled triangle one acute angle is double the other . Prove that hypotenuse is double the smallest side.
Can anyone help with this one?

[brave needle]

The conversion factor is rad=deg*π/180 since π=180°

So for i) 10°*π/180°=π/18

[low oak]

<@&286206848099549185>

[nocturne remnant]

There are interesting length properties of equiangular hexagons (hexagons with all angles 120) which may help you.

[nocturne remnant]

Lemma: the side lengths of such hexagons can always be expressed as
a, b, c, a+x, b-x, c+x (in that order)

(Note that x can be negative.)

This should be quite helpful cuz there’s 3 equiangular hexagons lying in your diagram

[upper karma]

Nvm sorry

[nocturne remnant]

By setting up extra variables x,y,z you can find the perimeter of the biggest hexagon

[nocturne remnant]

Proof of lemma: ||Consider the side lengths as vectors. Then opposite sides can cancel each other out. (Such as a and d cancels to d-a which can be negative) Then we have d-a = f-c = b-e, as desired. ||

[lyric kettle]

Can someone explain how to obtain the coordinates of the vertices of this triangle?

[rose karma]

I know it may be easy to some but I don’t even understand what this is asking me and I have been looking everywhere.

[onyx cloud]

an angle ABC means the angle between lines AB and BC

or in other words, the angle created by going from A to B to C

so $\angle AEC$

[somber coyote]

maximo

[onyx cloud]

is

the angle i drew in black (lmk if you can't see it)

$m\angle AEC$ is the measure of that angle (here the angle is measured in degrees)

[somber coyote]

maximo

[oblique tiger]

Can any one help me with that’s

Theses

[onyx cloud]

wdym

@oblique tiger can you be a little more descriptive

[oblique tiger]

In geometry

I fail this and I wanna know what I did wrong

[onyx cloud]

you fail what?

can you show the problem?

[oblique tiger]

Six questions I didn’t get

I would but when I send it

It takes forever

Do you have any other apps

[onyx cloud]

no

[oblique tiger]

Hmmm

Let me try

Again

Hopefully it loads

[onyx cloud]

maybe you could send 1 question at a time

[oblique tiger]

I did

But it’s not loading

[onyx cloud]

could you type out the question?

[oblique tiger]

Yes

It is find the volume

Of

12 is my height

And the

L is 10

W 6

And there is eight too

My only answer choice I got are

288cm^3

360^3

96^3

120^3

Choices -

[onyx cloud]

is there a picture?

do they show you a shape

[oblique tiger]

Yes

[onyx cloud]

what does it show you?

[oblique tiger]

But it won’t load

Is there a way I can show you

[onyx cloud]

you could describe it

[oblique tiger]

There

Are you able to see it

?

[onyx cloud]

yes

what about it is confusing you?

[fervent valley]

I need help boiss

[umbral snow]

Thoughts on the problem? Anything getting in your way?

[fervent valley]

nah

Im just stuck I cant think of a solution

[dark sparrow]

try calculating some other angles

look at triangle ABD for instance

[upper blaze]

Have anyone ever saw any equations that can help determine is poligon positivly or negativly oriented? :/

Like I have triangle with 2 fixed points that I know coordinates A(0,0) and lets say B(1,0) and C is arbitrary in plane so C(u,v), and I wanna determine using equations (not v>0, v<0 couse I know that determins it) but some equation concerning u,v and maybe some other points posibly

Any idea would be wellcome

[wintry harbor]

You can use the determinant of the matrix with columns that are vectors AB and AC

I think that works

[upper blaze]

But determinants still requiers comparing it with 0 no?

I need literaly equation

[wintry harbor]

Yeah

[upper blaze]

like something = something else

[wintry harbor]

I don't follow