geometry-and-trigonometry

[earnest yarrow]

I know for orthocentre but what does baricenter mean??

[obsidian harness]

barycenter is the center of mass

so just the centroid for a triangle

[earnest yarrow]

Oh

For orthocentre you can use radical axis and for centroid use cevas theorem

[bold pond]

Does anyone have tips for geometry in currently failing the class because the tests are waited and idk how most of this stuff works

[vale oasis]

im sorry but ,i have question too
How does the people call the Similar Triangles but not accroding rule( like if there is Δ　A B C traingle on koridnate A(-4,0) B(-1,2) C(0,0) and the similar traingle is A1B1C1 A1(8,0)B1(2,2)C1(0,0) )

[hybrid belfry]

Didn't get an answer on the
math-discussion channel so i'll try my luck here then.

(Please feel free to prove it in ANY way you like, i don't care if it's not being taught in highschool)

Can we somehow RIGOROUSLY prove that rectangles exist?

To be more specific,
let's say i draw 2 parallel straight lines and then i draw another 2 parallel straight lines that are perpendicular to the other 2.

How would we RIGOROUSLY prove that all opposite edges of the shape are equal? 🤔

I couldn't prove triangle congruence theorems before proving that rectangles exist... 🤔...
(surface area is measured using square unit, i used surface area to prove the theorems so that requires me to prove that rectangles/squares exist first 🤔)
Can you? 😃

I guess we gotta pick a very good definition for parallel straight lines in order to prove this 🤔...

[hybrid belfry]

What do you mean?
The lines are paralleled already,
that's in the assumption.

I'm trying to prove that the opposite edges are equal

I'd like to see your
definition of 2 parallel straight lines
first, we need that, i think

You said "draw 2 parallels" without saying what the definition of 2 parallels is

That works for me,
anything else?
I mean is there any other definition of 2 parallel straight lines that might work in this case?

No no, i meant
is there any other definition of
2 parallel straight lines that we may use in this case instead of the
"there's also the definition of parallels as lines of fixed distance"?

Just wanna know the options 😄

Ok,
so now we gotta prove that it's possible to
draw 2 parallel straight lines according to the definition
" there's also the definition of parallels as lines of fixed distance"

[hybrid belfry]

How can we RIGOROUSLY prove that it's possible to draw such lines

I'm not sure if that's an axiom

Is this another way of defining parallel straight lines?

"parallels as lines of fixed distance".
This is a DEFINITION.

[hybrid belfry]

Such thing exists or not is not in the definition 🤔

It's like defining a term that might not exist.
Now we gotta prove that it exists.

Well then there's nothing to prove here if we accept the assumption that such thing does exist an axiom.

Which is not what i wanna ask.

I'm not convinced that there's no other way but accepting that it's an axiom 🤔

If i did then i would have shared it already instead of asking 😂

What if we pick another definition of parallel straight lines?
Can we work with that?

@strange narwhal

[tardy ravine]

all i know is parallel lines aint gonna intersect 😭

[hybrid belfry]

If i choose this as the definition, can you prove it now?

Please use the Direct Message,
i'm going to sleep now, it's pretty late here 😄
@strange narwhal

[compact trellis]

Please, someone tell me what C is

i did this but idk if its correct

no answer key eiter

[solid needle]

youre cutting it in a way that doesn't make it very easy

imagine the cross section that contains angle BHE

in other words, draw the ||rectangle BCHE||

[compact trellis]

ohh

thanks

i think i can take on from here

nvm i use pytahoges

@solid needle Check my answer please, i got 68

[solid needle]

,w arcsin(5/sqrt(29))

[somber coyote]

Results provided by WolframAlpha

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

Requested by cosmovibe

[solid needle]

3 sigfigs

[compact trellis]

wowwww...???

[solid needle]

you only gave 2

[compact trellis]

oh cmon

68.2

i don;t understand what this thing spewed outtt 😭

[solid needle]

dont come on me, im just following the instructions on your assignment 😂

[compact trellis]

oh is this the answer

[solid needle]

you can just ignore anything you dont need

just find the thing you do need

[compact trellis]

wow it is

thank you

im in grade 10

lol

[solid needle]

yeah learn to use wolframalpha as a tool

very useful

dont overrely on it obviously since it is a calculator

but do know that its there to help you check your work or provide insight when you get stuck or if you want to play around

[compact trellis]

I’ve used it before, and it’s probably one of the most powerful calculators

Dude, are you familiar with

Calculators, like physical

the 991cw in specific

[solid needle]

unfortunately not, sorry

[compact trellis]

Alright, that’s fine, thanks for the help I didn’t think of taking a cross section like that

Oof,’only thing left to study is constructions 😢

[tepid violet]

ADB is a right angle triangle, AB is 3, AD is 2 and BD is x (say), use pythagores, x^2 = 9 + 4

x = rt 13

[tepid violet]

for part 2, you can say tan theta = ad/db => tan theta = 2/3 and theta = tan inverse (2/3)

you can convert it into sin using identites

[teal island]

has anyone seen functions like this before? I found these two functions while solving motion equations

these appear to be trig functions that "phase" into each other very nicely and are solutions to the time it takes to reach a maximum velocity given an initial velocity and acceleration vector

[maiden brook]

do u have the equation for them?

[teal island]

they are solutions to a quadratic equation in terms of theta

[fading dove]

can you send the original question ?

this graph looks interesting

[fading dove]

the red curve is a sin curve where for sin(ax), ax= -ve value

trying to find the amplitude and translation

and for sin(ax) its the same which is blue

[maiden brook]

wdym?

what would a trigonometric function mean without trigonometric functions

[maiden brook]

(i think u can see how u finish from here)

[maiden brook]

the area of a triangle in three ways

[bold pond]

For the second problem how do I solve this?

[blissful rover]

no

i havent in fact

[young helm]

can someone help me, i just started geometry unit and like im confused on the SSS ASA stuff i dont understand them 😭😣

and im confused because sometimes u need to visually write the angle itself and that makes no sense because clearly other sides are also congruent, why do we NOT need to also write them down??

[solid needle]

can you see an angle bisector in this diagram?

[solid needle]

are you around? if you are, i can walk you through this

[neat wigeon]

opposite angles in isosceles trapezoids r supplementary
so 4x + 36 = 80
x = 11
perimeter is 76

[obsidian harness]

It's by symmetry; sum of angles in a quadrilateral is 360

By symmetry, angles F = G and E = H

F + E = G + H

So each side is half of 360 = 180

[vernal pilot]

@trail tendon

[wise zodiac]

that makes no sense

[obsidian harness]

indeed

have you seen this proof?

WLOG let side c be the base and you can do that
so it also holds pairwise, e.g sin B / b = sin C / c if a is the base, and hence the law of sines follows

[proper root]

silly little question but are the cartesian spatial axes not considered "concurrent"?

[obsidian harness]

depends on your exact definition of "concurrent"

two non-parallel lines always intersect at a single point

which isn't useful

so many people say that concurrent is when 3 (or more) lines all intersect at the same point

[dark sparrow]

what's your definition of "concurrent"?

[obsidian harness]

also like are we talking in 2D?

[proper root]

that's what i was unsure of T . T

in some places they use "plane" in the definition, in others they don't

[obsidian harness]

what's the exact definition you got?

[proper root]

that 3 or more lines in a plane are considered concurrent if they intersect at the same point

[obsidian harness]

ah, that makes more sense

[proper root]

though in some youtube videos there was no mention of "plane"

[obsidian harness]

yeah they're considering 3D geometry then

like that's again another subjectivity in what the word even means

[proper root]

so in 2D geometry the definition is exact (since all lines would be in the same plane by default), but in 3D it depends?

[obsidian harness]

yep

[proper root]

why have it be that way though

[obsidian harness]

personally speaking I would say that 2 lines in 2D absolutely do concur, if they aren't parallel or the same line (they need to meet at a single point)

[obsidian harness]

people are fallible

they don't often clarify what they mean

also more symbols doesn't mean more clarification a lot of the time

words just work better

[proper root]

wait coinciding lines aren't considered concurrent?

[proper root]

i guess so

[obsidian harness]

yeah everything seems to say that they must meet at a single point

[proper root]

i'll try to figure out with questions from the exam i'm preparing for ig then

the definition that fits my use-case

thanks for answering!

[obsidian harness]

np!

[sturdy surge]

It made me so FUCKING happy when I saw it come out the same as when I had solved it manually. I wanted to cry joy but I'm in public rn lol. I'm just amazed at how far tech has gone

Made a program for the calculator

I just enter whatever sides or angles are avaible

And boom

I get all I need

9th grade math was boring so I had to teach myself this stuff lol

[solid needle]

congrats, very cool

[sturdy surge]

Thank you. I could probably program some way to go to the moon lol

There's 230-ish lines of code available lol

[dark sparrow]

HP15C HOLY FUCK

IS THAT AN ACTUAL HP 15C

i used to have one of those...

[sturdy surge]

11c. I got a 15c ordered

77 dollars for a 15c is epic

They're great calculators

Simple to learn yet does complex stuff

[young helm]

hey

[solid needle]

out for dinner atm

[young helm]

alr

[solid needle]

will be home in about an hour

[young helm]

ok ill come back

[solid needle]

here now

[young helm]

OOPS

[young helm]

u there? 😅

[solid needle]

yeah

so you said you wanted help with triangle congruence theorems?

do you have anything specific you want to go over?

[sturdy surge]

231 lines of program is such a bottleneck. Instead of buying a new calculator, I'll do it the German engineering way and buy 10 hp11cs

I'm gonna be honest you can probably calculate a trajectory to the moon with 10 hp11cs

That gives 2000 some lines of program

[nocturne garnet]

Can somebody help me realise where we got this from😭 the only numbers that are as such is that CN:NS = 1:3

[sturdy surge]

For number 6 would you use pythagorean or sum however it's spelt to find the shared side and then figure it out from there?

[upper echo]

Because of the picture to the right. JN=CN, so MJ=CF, so DC=2CF, so DF/CF=3 and that equals KF/XF.

[haughty basin]

Can someone hop on a call, I really need help. I got a test tomorrow and don't know wtf to do.

[upper karma]

what's the test over?

[upper karma]

rahhh

[haughty basin]

My homework sheet says unit 6. On here is logarithms, graphing functions and interest rates, both quarterly and annually. If you could help that would be amazing.

[dawn hearth]

Who can give me a hard math prob

[haughty basin]

Riemann Hypothesis

Idek what that is but if you can solve it, apperently you get a $1mil reward

[neat wigeon]

find x and y
x + y = 10
xy = 50

[crude quiver]

guys

solve this

its easy

[haughty basin]

Can someone tutor me bro I'm so slow

[crude quiver]

me

but

who wants to solve my question

[obsidian harness]

stop spamming

[crude quiver]

how

[obsidian harness]

also it's already solved

[crude quiver]

im not

it isnt

find the final answer

it isnt yet

[haughty basin]

That's cool, but I'm still tryna figure this interest rate bs out for my test tmr, I need heeeelp

[crude quiver]

this is the final answer

[obsidian harness]

it's on the calculator screen

yeah so 0.35 is one of the answers, approximately

[crude quiver]

but how about number 2

[obsidian harness]

then you just need to calculate $\frac{-6 - \sqrt{(-6)^2 + 4(1)(2)}}{2}$

,calc (-6 - sqrt((-6)^2 - 4 * 1 * 2))/2

[somber coyote]

Result:

-5.6457513110646

[haughty basin]

can somebody be my tutor for a min please?

[crude quiver]

me artin

[haughty basin]

I need iiiiiit'

[obsidian harness]

we don't tutor here but you can go to #❓how-to-get-help

[crude quiver]

what you learning

[haughty basin]

trig

unit 6

[somber coyote]

south

[crude quiver]

trigomontry ?

[haughty basin]

yea

[crude quiver]

oh thats kinda easy

[haughty basin]

yea

[crude quiver]

i learned it in sumer

summer

[haughty basin]

I learned it never cuz Im a bum and watch ig reels in class

I added you

if you can vc it would help tremendously

[low lark]

guys how do i become better in geometry

im very bad at it

[silent plank]

practice

[tardy ravine]

@silent plank im not sure if this person is intentionally trolling or not because this person asked this yesterday, and around 5 other people spent about on hour with this guy #help-43 message from this point on, going through it step by step and saying either that the answer is wrong or some other nonsense to people helping him even when it wasnt 😭

[dark sparrow]

practice, practice, and practice.

but one big thing i see people not do, and which gets in the way a lot, is not drawing diagrams.

whenever you see a geometry problem, it is never a bad idea to make a diagram for it

[flat orchid]

hey

cos a / 1+sin a + sin a/ cos = ??

[crude quiver]

?

[flat orchid]

wait i will send u

[green jetty]

now try to understand where exactly there is a /
maybe this way?

[obsidian harness]

okay what have you tried?

common denominator?

[crude quiver]

,rotate

[crude quiver]

,rotate

[crude quiver]

,rotate

[crude quiver]

,rotate

[obsidian harness]

you can press the emojis next time

[flat orchid]

Ok

[crude quiver]

oh sorry

[flat orchid]

Can anyone solve for me 🙏🏻

[obsidian harness]

you'll learn more if you try a bit yourself

[obsidian harness]

.

[flat orchid]

Np

[dark sparrow]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[flat orchid]

Ok

I tried to solve but couldn't get it

[obsidian harness]

yeah so what have you tried so far

[flat orchid]

So I came here

[dark sparrow]

show us one of your failed attempts

[obsidian harness]

I think their image is loading Ann

[obsidian harness]

yeah 1 + sin A is definitely not -cos^2 A

[dark sparrow]

did you try to replace 1+sin(A) with -cos(A)?

or that...

mega illegal

[flat orchid]

Ohh don't worry I got the answer

[obsidian harness]

ok...........

just saying, common denominator and the numerator and denominator will cancel partially

[green jetty]

Well, if you first bring everything to a common denominator, and then use the fact that the sum of the squares of the sine and cosine of one angle = 1, then the problem is literally solved in 2 steps

[dark sparrow]

denominator is written incorrectly

should be (1+sin(A))**×**cos(A) not +

[crude quiver]

Hello ?

[obsidian harness]

just ask

[crude quiver]

?

[neat wigeon]

he's saying that if u have a question just ask, don't say "hey guys can i have some help"

[vague badger]

i need helo in a math

help*

[dark sparrow]

you should just post your question upfront

[vague badger]

post where?

[dark sparrow]

here. or open a help channel instead if you want.

[vague badger]

this

[dark sparrow]

ok

any progress so far?

[vague badger]

i did but i dont think i did anything useful

i need someone to solve it

[dark sparrow]

well show it anyway maybe there's something good in there

[dark sparrow]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[vague badger]

i erased it..

[dark sparrow]

blech.

ok

let's start with some thinking then

first, do you know what a kite is?

[vague badger]

yep

ik

[dark sparrow]

ok, and do you know how to find the area of a kite?

[vague badger]

yeahh

[dark sparrow]

can you say how?

[vague badger]

product of diagonal divided by two

[dark sparrow]

ok good

right, so the diagonals of our kite are AB and XY. we now have two goals: find each one of these.

do you agree?

[vague badger]

yeahh

[dark sparrow]

ok right

now, to find the length of AB, the easiest way is to find the coordinates of A and B themselves.

do you see how to do that?

[vague badger]

we solve equation of the line and equation of circle simultaneously?

[dark sparrow]

yes, exactly.

[vague badger]

ohh

waitt

le me tryy

[dark sparrow]

ok. do that, then send your work here & ping me.

[prisma gyro]

??

[dark sparrow]

?? indeed

how on earth are we supposed to read the coordinates of any points from these shoddy ass diagrams? 💀

[tawny pier]

What the shit

How tf you get the coordinates

[prisma gyro]

LMAO

idk

[tawny pier]

What the shit are the numbers even representing

[prisma gyro]

wtf

[neat wigeon]

63. 0 slope
64. -n/2m slope
65. (2d-b)/(c-2a) slope

[dark sparrow]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[neat wigeon]

ah ok

[acoustic terrace]

ik thats x is 18 but how

[solid needle]

!status

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[acoustic terrace]

1

[solid needle]

is there any other information from the diagram you can see that helps you?

[acoustic terrace]

no thats all that is given

[solid needle]

ok hint: what do you know about OA, OB, OD?

[acoustic terrace]

radiuses all equal

[solid needle]

correct

did you realize that already and just didnt think that was relevant or did you just now see that

[acoustic terrace]

ik that but thought it ws irrelevant

[solid needle]

ok pro general tip

in geometry, we can translate statements about one thing, like side lengths, into statements about other things, like angles

for instance, if you know the triangle congruences, like SSS and SAS, you know that for instance knowing all the sides to a triangle gives you all the angles

and its not just side lengths and angle measures, other geometric concepts like area or arc measure all matter too

just keep that in mind, never think any bit of info is useless

what do you know about triangle OAB now

[acoustic terrace]

angle ABO is 2x

and AOB is 180 - 2x

[solid needle]

not quite

why is AOB that

[acoustic terrace]

because the angles in a triangle add up to 180

oh wait 180 - 4x

[solid needle]

yes perfect

so circles give us radii which are super useful for things such as, but not limited to, finding isosceles triangles

having so many radii all be equal in length is very very useful

so maybe we can draw some missing ones?

triangles are easier to analyze than quadrilaterals after all

[acoustic terrace]

AOE?

[solid needle]

yeah

[acoustic terrace]

would that split the 7x into 3.5x

[solid needle]

good question

if youre not sure, you better verify

geometry is full of "it looks like x" but isnt

see if you can prove it

otherwise if you know what the inscribed angle theorem is, that would be much easier, you wouldnt even need to draw the radii

i gotta dip, if you still need help hopefully someone else can take it from here

good luck

[acoustic terrace]

thank you

[obsidian harness]

don't think about splitting actually

just use the angle in the centre theorem to look at the angles around point O

hence you must have (360 - 4x) + (360 - 14x) + (360 - 12x) = 360

actually, drawing OC and then using the isosceles triangles thing also does work

so if you call OCB = c, then OCD = 6x - c

so you get BOC = 180 - 2c and COD = 180 - 2(6x - c) = 180 - 12x + 2c
and so BOC + COD = 360 - 12x indeed

that's how the circle theorem I showed you can be proved

[potent kelp]

how would you solve this like idc about the answer

[obsidian harness]

that means AE must be the diameter

[potent kelp]

yeah

[potent kelp]

as its perpendicular

[obsidian harness]

yeah so then angle ACE = 90 and hence BAC = 70

[potent kelp]

the 20 degree is useful only for adding up and finding i assume?

[obsidian harness]

now then KAC is an isosceles triangle so you can use that to find angle CKB

[potent kelp]

its called a tangent if i remember correctly?

[obsidian harness]

yeah like it turns out that that one angle determines everything else, given the setup

[potent kelp]

[obsidian harness]

no, a tangent intersects the circle only once

[potent kelp]

[obsidian harness]

tangents and circles come up a lot but not for this question

[potent kelp]

if sides same angles same right

[obsidian harness]

which congruence theorem are you using?

[potent kelp]

actually

its a random question a friend was solving that i screenshotted as it seemed intresting

[obsidian harness]

yeah I know, and the key step is this

[potent kelp]

been a while since i last touched geo so its badge

[potent kelp]

so is that statement correct

[obsidian harness]

it's too vague

[potent kelp]

[obsidian harness]

you need to revise SAS, SSS, ASA, AAS congruence

[potent kelp]

here we go again

nice

[obsidian harness]

look up the ambiguous case for example

(given sides 10, 6, and a 30 deg angle not in between the sides)

[potent kelp]

sorry but

this bounced off

[obsidian harness]

like you weren't very specific so I had to keep asking you to clarify

I think you're actually asking about two congruent triangles

corresponding parts of congruent triangles are equal

so then you can just compare the same sides in both triangles, and they'd be equal

same with the angles

[potent kelp]

my bad

[mellow merlin]

a

[mellow merlin]

When youre trying to solve for x in a sinusodial equation for example
6sin(8x) + 4 = 5

which equals
(sin^(-1)(1/6) + 2πn)/8 = x

why does x have multiple solutions?

the context is this video hold on

https://www.youtube.com/watch?time_continue=381&v=0_4uTKBI99o

its like rlly weird to me cuz why did he only tried finding the intervals from [-π/2, 0]?

i think i get why x have multiple solutions but i still dont get the interval

x has multiple solutions since something like cos(1) and cos(361) are the same

[steady wyvern]

Hi guys

[steady wyvern]

Why sin(90-x)=cosx and cos(90-x)=sinx

[steady wyvern]

?

[worn scroll]

think about the unit circle

sin goes 0 1 0 -1

cos goes 1 0 -1 0

by making x negative sin does loop other way around

0 -1 0 1

add 90 i start from 1

1 0 -1 0

that's cos

[steady wyvern]

Than you

[knotty socket]

I don't know if I'm asking in the right channel but does anyone know what the name of this is called, I have come across this motion pattern while programming tween related stuff (accidentally) many times before, and it's really beautiful

[digital moss]

please help me solve

[cunning lion]

<@&268886789983436800> crypto scam/ad

[neon prairie]

Poor texit

[mellow merlin]

why tf is it -180

[trail tendon]

are you familiar with sin(x) = sin(180-x)?

[mellow merlin]

yes

but why -180 😭

[trail tendon]

would you agree that sin(x) = sin(x - 360)?

you just go once around the circle and end up in the same spot

[mellow merlin]

mmm

yes

[trail tendon]

so then
sin(x) = sin(180-x)
could be
sin(x - 360) = sin(180 - x - 360) = sin(-180 - x)
right?

[mellow merlin]

yeah i can imagine it being equal

but how can i know which one to use

my first instict was to do 180+30 since sin = -30*

[trail tendon]

?

i'm not sure what you're asking

[mellow merlin]

when i tried solving this way this wasnt in the solutions

then i skipped and found out they used -180 - x to get a different solution than mine

[trail tendon]

why would you do 180 + 30 since arcsin(-0.5) = -30

you have two solutions for 4x between -180 and 180, and both are below the x-axis (because sin(4x) = -1/2)

[mellow merlin]

it just made sense in my head since wouldnt 180+30 equal to sin -1/2 too if sin -30 = -1/2

[trail tendon]

-30 = -1/2 😭

[mellow merlin]

oh intervals 😭

[trail tendon]

oh you were ignoring the intervals here i see

[silent plank]

supplementary identity is usually presented as
sin(t) = sin(180-t)
180 - (-30) = 180 + 30 = 210
sin of that would be -1/2,
but its outside your interval so not quite what you want
there isn't anything wrong with using the identity that way
but you'd want to then subtract 360

oh, already answered

[mellow merlin]

do all sin equations normally have an interval of 180 and -180 unless stated otherwise

[trail tendon]

no

i mean they have a period of 360 degrees but if its not specified there is usually no domain restriction

[mellow merlin]

ok ty yall 🙏

[somber coyote]

Vanellope von Schmugz

[coarse flint]

Prove LHS plss

And @ me when answering

[dark sparrow]

you pinged everyone in a server of 200k+ members?

you serious?

also we don't do your questions for you

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[coarse flint]

Ok

[hybrid belfry]

JUST TO BE SURE.
This channel is in the Pre-University section.

Does that mean the majority of people here can only answer Pre-University questions? 🤔

[dark sparrow]

can like able, or can like allowed to?

[worn scroll]

what did you try

[visual elk]

HI guys

[flint marlin]

You are literally wolowitz

[visual elk]

hi

[visual elk]

If the bearing of Q from P is 150 degrees, find the bearing of P fom Q

[obsidian harness]

try drawing a diagram

you should get something like this

[dark sparrow]

memes go in #chill

[glacial magnet]

hellou

[silk jacinth]

if I have a family of curves x^2 + y^2 -2ax + 2ay = 8a + 8 (a != 2) with each member being a circle and I've determined a common point (-2,2) by using x=-2 and y=2 to eliminate the parameter a, have I also determined that this is the only common point?

[upper echo]

A short answer is not. In this case it the only point because the centers also lie on a line passing through (-2,2).

[silk jacinth]

why does centers lying on the line passing through (-2,2) narrow it down to one point?

[upper echo]

Because if two different circles have two common points then their centers lie on a perpendicular bisector of the segment between these two points.

but perpendicular bisector does not pass through any of these two common points.

[humble thunder]

can someone help me with a triangle proof problem

please😭

[maiden brook]

send

[lucid talon]

Guys who can do these easy substitutions just for warm up?

sin^2(x) + ? = 1
sin 3x = 3 sin x - ?
tan x = sin x/?

[worn scroll]

why?

[solid needle]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[flint kestrel]

how do you solve these two

[neat wigeon]

equal cords have equal arcs if it's in the same or congruent circle

[last jolt]

why the hell is geometry so overly complicated

[solid needle]

,rccw

[last jolt]

whoops sorry

[solid needle]

this seems like a very straightforward geometry problem to me

what are you struggling with

[wet pond]

Is this not just trig

[last jolt]

the steps

where i even start

i never really got geometry so

idk how it can be straightforward

[last jolt]

im getting in my geometry class so I'm not sure

if this is straightforward i must be a dumbass bc no way

[wet pond]

oh no I wouldn’t expect you to know how to do it with trig

That’s algebra 2 crap

Haven’t taken geometry yet so I couldn’t tell you how to do it the intended way

[solid needle]

have you learned the trig functions yet, like sin and cos

[wet pond]

but this is the triangle right?

[solid needle]

i don't think it is expecting you to do it any other way

[wet pond]

Actually 75 would be in the top right corner but you get the point

[wet pond]

You do any of this yet?

[last jolt]

nope

[wet pond]

Alright then yeah you wouldn’t know how to do it with trig

[last jolt]

well yeah sin cos and tan

but i kinda forgot them

[solid needle]

ok so

that's why

with sin/cos/tan this is really straightforward and simple

do you at least remember the definitions of sin/cos/tan

[last jolt]

tan is opp/adj

sin is i think adj/opp?

[wet pond]

SOHCAHTOA

use that

remember it

[last jolt]

geometry is so hard 😭

i cant memorize all this

[wet pond]

Sin=Opp/Hyp, Cos=Adj/Hyp, Tan=Opp/Adj

[wet pond]

Why memorize

just write notes and study them

[solid needle]

uh

the amount of memorization you have to do here

is not much at all

calm down

[wet pond]

This is an example problem of how you’d solve it

[solid needle]

take a deep breath, relax

so the mnemomic soh-cah-toa makes it really easy to remember the ratios

if you sound out, verbally say "soh cah toa"

do you think you can remember that?

[wet pond]

Nine letters

choshacao :>

[last jolt]

what are these steps ☠️

[wet pond]

Write this down if you can’t remember it immediately

[solid needle]

he's throwing you a bunch of steps at the same time

just focus on one step at a time with me

[wet pond]

It’s algebraic reasoning

You’ll get used to it

[wet pond]

I wasn’t trying to make them understand the example problem, just showing them what it looks like to solve it

🔥

[wet pond]

You could technically use the altitude to determine the value of this but that’s some mid-trig stuff

[last jolt]

so whats the answer

[solid needle]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[solid needle]

we can help you understand the problem and the concepts

we arent going to directly give you the answer

[wet pond]

You don’t learn anything if we tell you the answer directly

[last jolt]

oh yeah thats true

mb

[wet pond]

We can teach you, it’s not really a big deal

Trig is defo one of the more annoying subjects in my opinion

[last jolt]

so if i just have the angles how do i start

[wet pond]

If you’re comfortable with calling in a dm or smth that’d make it easier

No pressure

[last jolt]

not rlly sorry

[wet pond]

Nah it’s totally fine

alright so we should prolly go over just basic trig first

[last jolt]

alright sure

[wet pond]

?

[last jolt]

tan is opp/adj sin is hyp/opp cos is adj/opp?

[wet pond]

Read it like ‘ratio, numerator, denominator’

S = sine, C = cosine, T = Tangent

O = Opposite, A = Adjacent, H = Hypotenuse

[last jolt]

alright

[wet pond]

If you can’t memorize it just yet it’ll be fine, I’ll keep writing it down until you can confidently remember the ratios 👍

[last jolt]

oh sin is opp/hyp

[wet pond]

mhm

[last jolt]

cos is adj/hyp

[wet pond]

👍

[last jolt]

oh

ok so how does that apply to the problem

[wet pond]

Alright so

Let me just test your algebra rq

[last jolt]

ok

[wet pond]

Just basic multiplication and division

[last jolt]

15?

for the first one

oh wait no

60

i was thinking about the wrong thing

[wet pond]

Yeah

now the same applies to this

Sin has no definitive value so it can’t be simplified

[last jolt]

60

so whatever sin is is the result of something being divided or

[wet pond]

since sin can’t be simplified

X = 2*sin(30)

sin30 has a definite value

but sine by itself does not

you’d either write X as being equal to 2sin(30) or 1 because 2sin(30) = 1

[last jolt]

how does 2sin(30) = 1

[wet pond]

because sin(30) = 0.5

it’s important to understand that we measure trig ratios in degrees (DEG) and not radians (RAD)

big difference

[last jolt]

oh

[wet pond]

also important

sin(x), cos(x), tan(x) and any other trigonometric ratios are not proportional

sin(15) does not equal 0.25

it’s a little bit off

the reason why it’s not proportional is an entirely different thing that’s difficult to explain

Am I moving too fast?

[last jolt]

no its fine

[wet pond]

Alr

so back to your problem

[wet pond]

This

[last jolt]

yes

[wet pond]

Instead of saying side AD, we’ll call it side X

We want to use the ratio where X is over 85

[last jolt]

so that would be opposite over adj

right

[wet pond]

yes

in the top left corner

We know from the problem that angle ACD, or the top left corner, = 75 degrees

so if we take tan, or opp/adj of 75

tan(75)

[last jolt]

ok

[wet pond]

then we can determine that tan(75) =x/85

because X is opposite and 85 is adjacent

[last jolt]

and you'd do the same thing for both sides?

[wet pond]

well since you already have 85 and tangent gave you the value of X, then you can use Pythagorean theorem to determine the hypotenuse

if you mean the other side of 85 then yes

There’s your X value

So now you just have to find the value of X and add it to 317.2243186

[last jolt]

oh that's how you're meant to type it

oh my god

[last jolt]

so basically the same thing but for the other sidr

[wet pond]

Yeah

Given one side length and and two angles you can find everything in a right triangle

btw

If you have two angles then you know the third

If you have two sides then you know the third for a right triangle

[last jolt]

so 35° tan x/85?

[wet pond]

tan(35) = x/85

so what does x equal?

[wet pond]

that’s in radians :p

use desmos.com/scientific

It’s a good calculator

use DEG

[wet pond]

yep!

:D

I am gonna nitpick a little and say that you should usually put 85 before tan(35) as a general rule of variables

but you did it right

that’s correct

[last jolt]

ok so now i just add both 59.51 and 317.22?

[wet pond]

mhm

[last jolt]

well thats 376.73

so is that the answer?

[wet pond]

That’ll give you the value of side AB

it said to the nearest meter

so I’d assume 377

but yes 376.73 is the correct answer to the decimal point

[last jolt]

that actually wasnt that bad

knowing how to do it

[wet pond]

There’s a lot more to learn

Although I feel like you don’t want to know

[last jolt]

why not

im already here and itll probably help

[wet pond]

I’m not saying I won’t teach you

Just like

You seemed kind of fed up with geometry so I didn’t want to teach you more than you needed to know

didn’t wanna overwhelm you or anything yk

[last jolt]

i get it lol

but after solving this i feel way better

[wet pond]

I should probably explain the trig ratios then

they’re super silly

Sine is actually a very strange function in the sense that it is constant

it’s a perpetual wave, such as cosine is

tangent is scary so I’m not really gonna talk about it

You don’t need to worry about sin/cos/tan graphs for a bit

Sine, Cosine and Tangent aren’t proportional because they’re measured in pi radians

[last jolt]

so they're all like waves

[wet pond]

yes except tangent

we don’t talk about tangent

[left bolt]

can someone help me with a geo problem?

[wet pond]

I might be able to

can I see?

[left bolt]

alr

sure thing

[wet pond]

I’ll teach you law of sines and law of cosines in a sec, just let me do this

[left bolt]

A friend asked me for help, but I lowk can't figure it out

[wet pond]

I see the delta notation

We’re finding the side length of sides AE and EC?

[left bolt]

yup

I understand how to get AB and BC

since the triangles are similar

[wet pond]

so hyp is 20

[left bolt]

yup

[wet pond]

angle

$B_1\left and\left B_2\left\left equal\left 45\left degrees$

[left bolt]

how can u be sure

its just an altitude

the triangles will be similar

[somber coyote]

Stoneheart
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wet pond]

The square at angle B

[left bolt]

yes?

[wet pond]

represents a 90 degree angle

[left bolt]

ik

[last jolt]

alright

[wet pond]

oh I’m dumb

disregard that

[left bolt]

alr lol

[wet pond]

Haven’t done trig in a while I’ve been focused on calculus

lemme get my notes

[left bolt]

real

honestly I don't think it's trig

[wet pond]

we don’t know the angles of A?

[left bolt]

No, but we do know that AE is a bisector

as in A1 and A2 are congruent

[wet pond]

So the angles A1 and A2 are equal

[left bolt]

ngl geomentry left my brain, a friend of mine asked me for help on this problem

yes

[wet pond]

Sorry I don’t know notation cus I’m in 8th

Skipped geo and self taught trig

[left bolt]

ur in 8th grade?

[wet pond]

Yh

[left bolt]

dang

im in 9th

[wet pond]

Taught myself trig and differential calculus

Neat

[left bolt]

I took an online geo course so I skipped it in school

now taking ap precalc and ap stat

anyway

[wet pond]

ah

I forgot

Altitude equals geometric mean of the two hyp values

[left bolt]

yeah

[wet pond]

slipped past me

[left bolt]

but, u don't even need that to figure out the altitude

[wet pond]

sqrt 8*12

[left bolt]

just becaus there is an altitude from the right angle to the hyp. it means that the triangles are similar

[wet pond]

Yeah didn’t take geometry

Had no idea

Xd

[left bolt]

from there we can set up proportions to find the side lengths and altitude of the original triangle

I just can't figure out how to find the perimiter of the triangle

AEC

[wet pond]

so

BC = 3.55655882

[left bolt]

yeah

[wet pond]

ABC is a right triangle

[left bolt]

yes

[wet pond]

which means by extension

BEA is a right triangle

[left bolt]

wait I don't think thats right...

[wet pond]

It should be

[left bolt]

hold on

[wet pond]

geometric mean of 8 and 12 is 9.7979

sqrt 9.7979^2 + 8^2 = 3.55655

[left bolt]

but BC isn't the altitude

[wet pond]

yeah ik

I used pythag to find BC

the altitude is part of a right triangle with BC being the hypotenuse

[left bolt]

yeah I'm with u

[wet pond]

so

AB = 15.4919

[left bolt]

yes

√240

[wet pond]

Because pythag with altitude and 12

yes

[left bolt]

ik

i've already solved for those

how do we find AE and EC

I just don't know where to go from there

[wet pond]

ok so

we can find the angle of A

with cosine

12/15.4919

[left bolt]

oh

yeah

waitttt

alr

[wet pond]

either angle of A = 19.61576023

[left bolt]

wait what

I got 22.1905599794

[wet pond]

Eh I messed it up

the angle of A = cos^-1 (12/15.49193338)

[left bolt]

hold on, I think I got it

[wet pond]

Sorry just doing the math to check ur answer

I’ll tell you in a sec

23.575

@left bolt

[wet pond]

I can teach you more now

[last jolt]

oh sorry i didn't get this

are you still there

[wet pond]

Yeah

[last jolt]

ok nice

[wet pond]

Sorry just give me a sec

Sorting some stuff out

[last jolt]

all good

[wet pond]

ok

done with that

next thing you should probably learn is law of Sines

It’s used for non-right triangles

[wet pond]

Sorry I’m too tired to write all that in my notes rn

Use this Google image

make sure you enlarge it

[wintry seal]

Lami's theorem right

[wet pond]

.

law of sines

but yeah

Lami’s theorem

First time I’ve heard somebody call it that

[wintry seal]

Which grade are you in

If you don't mind

[wet pond]

8th

although I taught myself this stuff

[wintry seal]

Great. Limits in 8th

ahead of your peers

[wet pond]

and derivatives

Haven’t gone deep into integration yet

[wintry seal]

Functions are interesting enough to keep your attention

[wet pond]

I like math in general

Integration is just a lil confusing

I’ve learned most of differentiation and limits for the most part

[wintry seal]

Integration has a lot of brainstorming involved some times

[wet pond]

I need to work on continuity and integration

[wintry seal]

They can use whatever principle they want and it becomes quite diffficult to decode that

[wintry seal]

Continuity and differentiability is an important step

[wet pond]

should move to calculus ig

[wintry seal]

Though it can be quite boring at times

and repetitive

[wintry seal]

yeah

[last jolt]

alright

[wet pond]

It’s not very important for you right now but in the case that you go into non-right triangles then you should use law of sines and law of cosines

[wet pond]

oh do you not have calculus channel

[wintry seal]

Precalculus its showingg

That's what you are talking about?

[wet pond]

oh no

There’s a channel for uh

calculus

But it’s in early university section so you don’t have access

you’d have to change roles

[wintry seal]

How to change roles

[wet pond]

Scroll to top and press ‘Channels & Roles’ and pick Undergraduate

augh

no

one lower

Yeah

Undergraduate

[wintry seal]

Wow, a new dimension opened

[glad apex]

Someone help solve me this

[cloud ice]

is anybody taking online geometry on pearson connexus

[broken harness]

?

[broken harness]

i tried swapping 5sinxsinx with 5(1-cosxcosx) but i realise i should divide the whole thing with cosx*cosx ?

but then i lose a solution?

The book did a similiar problem that way, but i dont know how to check if cosx=0 is a valid solution or not at the end

[dark sparrow]

use ^ for exponents

[flint marlin]

Lol

[dark sparrow]

you don't want to have to type out (LONGEXPRESSION)*(LONGEXPRESSION) to talk about (LONGEXPRESSION)^2

anyway can you show what equation you got to just before you divided both sides by cos^2(x) (or planned to)?

[broken harness]

i think i got it but thanks will use ^. I divided by (cosx)^2 @dark sparrow

[dark sparrow]

ok but where did you get to just before dividing by cos^2(x)?

[cloud ice]

How did I get this wrong

Anyone

[dark sparrow]

show your work

[cloud ice]

Sin 52° times 481 equals 0.788 then i multiplied it by 481 and got 379

Am I right

@dark sparrow

[dark sparrow]

so you calculated 481 * sin(52°)?

yeah that's the wrong calculation to be doing lol

this is a side profile of the pyramid

[cloud ice]

It’s 481

[dark sparrow]

typo on my end, my bad.

but my diagram stands.

[cloud ice]

Oh ok

Thank you

[wet pond]

@cloud ice

for the trig problem

You’d want to determine the side adjacent to 52 first

Unless you want to go into reciprocals

[last jolt]

did i do this right

[wet pond]

how do you know the height of the half cylinder?

[last jolt]

well it gives us the height of both

27 feet or 324 inches

volume is L • W • H so you just multiply

[wet pond]

the height of the rectangle + the half cylinder = 2.5

The formula for the volume of a cylinder is $pi\left r^2\left h$

[somber coyote]

Stoneheart
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wet pond]

so

[last jolt]

yeah but its a half cylinder

[wet pond]

$0.5\left pi\left r^2\left h$

[somber coyote]

Stoneheart
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wet pond]

It tells us the height of the entire railing is 2.5

but not the height of each individual part

Unless it says somewhere that the height of both is equal then we don’t know the height of each individual part

[last jolt]

oh

ok i read it again it said the railing is 2.5 inches tall

the length was the 27 ft

whats the equation for rectangular prisms

[wet pond]

Dunno

[last jolt]

alright

@wet pond so i kinda did it right

2025 was a number i needed to get but i fucked the cylinder part up and skipped a step

[wet pond]

Alr

[tribal rose]

Is this normally considered geometry? I thought it was because of ratios in math (so it would relate to similar triangles), but another website says that it's more of an algebra/precalculus concept.

[trail tendon]

it's algebra that is used in geometry so...

[tribal rose]

That makes sense. So it's basically both.

[trail tendon]

yea. i mean without context this looks more like an algebra problem but could be incorporated into a geometry problem

so yea

[dark sparrow]

this is not how the \left command is used.

$V = \frac{1}{2} \pi r^2 h$

[somber coyote]

Ann

[dark sparrow]

is this what you wanted to write?

[frank rivet]

I dont know much, but is latex a software where we can write mathematical equations without having to struggle trying to settle the format in a single line?

[dark sparrow]

LaTeX isn't a software at all. it's a markup language

TeXit, our bot, allows you to render LaTeX code into equations

[frank rivet]

👌 ok

[sonic ferry]

I'm just wondering is this fully simplified?

My own process went to csc^2, but I expected it to finish at 1/sin^2

[dark sparrow]

does it make a difference?

it depends only on whether you consider csc a function deserving of its own name

[sonic ferry]

So it's a personal preference thing?

[dark sparrow]

precisely

[sonic ferry]

There's also this problem I'm wondering about

How do you know what to use to start simplifying this?

Because with my process I just distrubuted sec first which gave me a completely different answer

[dark sparrow]

what answer did it give you

[sonic ferry]

Sec - SinSec
1/Cos - Sin/Cos
(1 - Sin)/Cos

They're squared obv I just didn't bother to type the squares

[dark sparrow]

your thing can be simplified further still

[dark sparrow]

never do this

[sonic ferry]

Oh yeah actually

1 - Sin^2 = Cos^2
Cos^2 / Cos^2 = 1

So they actually do lead to the same thing

[dark sparrow]

yes they do

[fading pewter]

Guys help plz

[obsidian harness]

what have you tried so far?

[eternal shadow]

Can someone give me prooving trigonometry questions for practice for 11th class?

ping me when you send

[obsidian harness]

https://madasmaths.com/archive/maths_booklets/standard_topics/trigonometry/trigonometric_identities_with_solutions.pdf

and https://www.madasmaths.com/archive/maths_booklets/standard_topics/trigonometry/trigonometry_pythagorean_identities.pdf

[eternal shadow]

tysmm

ill try them

[frank rivet]

Whats the apporoach to proving LHS = RHS in trig questions

[trail tendon]

doodle with each side until you get them the same

💀

[steady wyvern]

Hi guys

[steady wyvern]

Does ||A|A|=|A|A||?
In matrix and determinant

[trail tendon]

multiplication between matrices and scalars is commutative yes

[brave dirge]

What do you guys get for this problem? I got about 35 but I think the Quizlet explanation mixed up a term and got something wildly different

[steady wyvern]

Thank you

[sweet spruce]

Anyone could help me with triangles math please??

[dark sparrow]

don't ask to ask

http://dontasktoask.com/

just ask your question(s)

[sweet spruce]

Well I have a test tomorrow about similarity and congruence of triangle where I have to calculate how much a side of a triangle is long based on other sides of a different triangle and I don’t understand well

Also I’m sorry if my English is bad I speak Spanish

Things like this is what I’m given for tomorrow test

[steady wyvern]

Does it true ?
In matrix and determinant

[dark sparrow]

can you write this in a manner that's less purposefully confusing with notation

[wicked bolt]

Say $x \neq y, x,y \in \mathbb{R}^2$. Define the line to be $L = {ta + (1 - t)b : 0 \leq t \leq 1}$. How to show there is $a = (z,w) \in L$, such that $z,w \in \mathbb{Q}$

[somber coyote]

Spice

[sweet spruce]

Any1 know how to get x

[wet pond]

Who pinged me

[wet pond]

whar

I don’t really understand what’s going on here, otherwise I’d help

[obsidian harness]

similar triangles!

well you are missing information, cause the bigger triangle's base is definitely not x

and this is a cursed question cause 36/90 is not 40/80, so it's an impossible setup