geometry-and-trigonometry

[real sentinel]

Simplify it with 1st bracket

And u will get there

[upper karma]

a + b = 10
ab = -11d
c+d = 10a
cd = -11b

Integrate (a+b+c+d) dx when a,b,c,d ≠ 0 and are not equal also.

[No ai Allowed, just try]

[storm ore]

Thanks guys

[whole flame]

So now

Is it smth

1210x+c??

[neat pivot]

anyone?

[final ferry]

X=40 y= 180 - 65 -40

[whole flame]

X is the corresponding angle of 40 so x=40°... y+65°=180° so y=115....

The angle other than x y in the triangle formed be t

Then t and z are vertically opposite angles

So... x+y+t= x+y+z=180°

Z=25°

@neat pivot

Is this the answer?? Bcz i consider 65° wrt the line drawn from transversal

[visual flume]

yooo i did the EXACT same thing lmao

tysm

except the wording

i kinda coord bashed rly hard

ahhh

[neat pivot]

i also thought x was 40 but i kept telling me it was 65 so i gave in tbh

[whole flame]

Nahh ur right... It's property... Corresponding angles

[whole flame]

Is that u in pfp??

[neat pivot]

yea

[whole flame]

U look pretty

[summer cradle]

np

[simple vigil]

Which class question is this
Probably 6-7th

[marble pond]

Frfr

[slate cave]

sqrt

[worldly zephyr]

How to solve 2nd one?

[summer cradle]

use Cosine Rule to find c

the apply Sine Rule to find the others

[worldly zephyr]

Thx

[summer cradle]

np

[earnest cedar]

I have a few conceptual doubts about conic sections:

1. When we talk about conic sections — parabolas, ellipses, and hyperbolas — are they always formed from the same cone (just at different angles of intersection), or can they also come from different cones altogether? If two such curves come from the same cone, is there something common or shared between them that tells us they originate from that same cone? Or in coordinate geometry, does it not really matter whether they’re from the same or different cones?
2. In problems like “find the locus of the midpoint of a chord of the parabola y^2 = 4x that passes through the focus” or “find the locus of the point of intersection of normals to the parabola,” we often get another parabola as the final locus. In such cases, is this new parabola geometrically related to the same cone as the original one, or is it just a mathematical curve in the coordinate plane with no direct connection to the original cone?

[faint pasture]

does someone have a formal proof that 3 planes cant enclose a volume, like i can imagine it but it feels wrong that they cant so a proof would help

[icy merlin]

I think you could look at Euler’s polyhedron formula

if the planes do enclose a volume, then that volume will be a polyhedron

[faint pasture]

yea so that gives edges = vertices + 1 all in all

max number of edges is nc2

nc2 > n+ 1 this has solutions

[icy merlin]

or

faces + vertices = edges + 2

That’s the polyhedral formula

[faint pasture]

yea i skipped steps, do faces = 3

[icy merlin]

But yeah nc2 is number of edges

[faint pasture]

max number but hmm same thing,

[icy merlin]

Sure yeah

[faint pasture]

hmmm so that doesnt prove our initial hypothesis wrong

[solar dew]

How can I find the cost of carpeting needed based on the room shown?

[icy merlin]

Well we know there’s up to 3 edges

A face has to be enclosed by at least 3 edges, so there’s up to 1 face

then we have <= 2 vertices

but then there’s only 1 possible edge, which means no faces at all

[faint pasture]

makes sense, thank you, took me back to the imagining but oh welll its geometry what did i expect

[fallen monolith]

how does this happen I am cooked

[leaden prawn]

Do you know soh cah toa

It doesn’t matter

So you have the angle right

[indigo nexus]

use tan

[leaden prawn]

And the adjecent

Ok nvm

[indigo nexus]

u will do tan (37)= x/8

[leaden prawn]

So you have the angle and you have the adjecent and you’re trying to solve for the opposite dos that make sense

[pearl sable]

do you know how to label each side of the triangle

[indigo nexus]

do you know sin cos and tan ?

[leaden prawn]

Yes

[pearl sable]

nah

it's not

[leaden prawn]

Yeah it’s not

Other way around

adjacent means next to

Yes

Cuz 5 is opposite of the angle and x is next to it

[pearl sable]

yes, why? the adjacent is the same side as the angle you are either looking for or already have

[leaden prawn]

So like you just do tan of the angle is equal to opposite over adjacent

[indigo nexus]

k you have to learn how to use this and maybe if you write a test you have to study p.t. to

k send the pic again

i can help

[leaden prawn]

Bro remember soh cah toa

[indigo nexus]

k here is betterr to use tan

tan is tha oppostite side / adjustment side

[pearl sable]

show your work

[indigo nexus]

$$\tan(39) = \frac{5}{x}$$

[somber coyote]

ponkur

[pearl sable]

we'll see where you went wrong

[indigo nexus]

dont type everything

untill 4.048 the most

do you know the pythagorium theorim ?

not suure

will you use a calc on the test ?

[pearl sable]

show how you got to here

[indigo nexus]

k then untill the hundredth dont write everything and dont change

[pearl sable]

what did you type bro

ok

that is not correct

[indigo nexus]

$$x = \frac{5}{\tan(39^\circ)} \approx \frac{5}{0.8098} \approx 6.17$$

[somber coyote]

ponkur

[pearl sable]

yes

your method of isolating x was incorrect

[indigo nexus]

try it

tell us your anwser and we will help

greek srry it was accident

idk try it we wil help you

[pearl sable]

so how did you get to that answer

[indigo nexus]

ye tell us how you got there in a paper on smthing

you cant just givbe an anwser lke that

mm

wdym

k you do

$$
\sin(70^\circ) = \frac{12}{x}
\quad \Rightarrow \quad
x = \frac{12}{\sin(70^\circ)}
\approx \frac{12}{0.9397}
\approx 12.77
$$

[somber coyote]

ponkur

[indigo nexus]

sure

[pearl sable]

and you know why you were wrong?

it was the same error as last time

[indigo nexus]

do you know anything about square rooots ?

try it tell us and we will help you

[pearl sable]

please show your working out so we can clearly see where you went wrong if you do

and point it out to you

[indigo nexus]

write it in paper or latex send pic

can you give us a definition of what din cos and tan is ?

wich one is it ?

[pearl sable]

yeah you're good, but can you send us a photo of you doing the question lol

but the notes are fine

[indigo nexus]

k do you want me to give you a problem ? solve it

and tell us the anwser

let me find one

[pearl sable]

send us a photo of your work

[indigo nexus]

if ΕΓ=5CM and ΒΖ=6cm find the area of the 2 tgrianglesw and the square

solvbe the one i gave to you

if you solve it you are good

and i need ecxplanation to everything if you can do this you are fine

ik its basic but its good for a start

yes thts an easy example

[pearl sable]

i mean you're js skipping a few steps so

[indigo nexus]

if you started yesterday that means yk everything abut finding areas and stuf

wait listen to me

you learned yester ay rite ?

right *

do you know anything about square rooots pythagorium and areas ?

if you dont know these you cant do trigonometry

the problem i gave to you has everyithng

[pearl sable]

yeah it's wrong also

[indigo nexus]

yes

[pearl sable]

let me break down what you should do into steps

when you see a trigonometry question:

• label the sides of the triangles
• note the information you were given, and what you need to find
• based on the information given, you will select one of the formulae: sin, tan or cos
• you will sub in the values or lackthereof into the formula and solve for the unknown

let me do an example from the most recent question

[leaden prawn]

tan

[pearl sable]

kinda ran out of space but here

[indigo nexus]

@true chasm

[pearl sable]

flipped my bad

[indigo nexus]

if youre teacher say to you no calc and only gave you these

can you solve ?

and all the angles are 30 45 and 60

[pearl sable]

bro he doesn't need to know allat 😭

not now

[indigo nexus]

still is trigonometry thats the basics

like if thats the test that easy

[pearl sable]

he js needs to understand the fundamentals w these questions

and how'd you get that

do you see the photo i sent

can you show your work like that

okay

[indigo nexus]

@pearl sable what grade are you ?

[pearl sable]

no idea

but i'm not that far in maths

[indigo nexus]

what are you doing now in maths

[pearl sable]

linear acceleration

[indigo nexus]

i can see nice

am doing geometry trigonometry algebra

not that good

[pearl sable]

convenient for nate

yes that's good

because now we can see exactly where you made a mistake

this line

is where the mistake arises

you see the fraction 10/x

you want to take
sin37 = 10/x
xsin37 = 10
x = 10/sin37

so basically what you did in the question

but you did it the wrong way around

it's always the denominator of the fraction that you do this with

any issues?

yes

yep

oh yes, your formula was wrong

see how you wrote

yeah, the algebra was correct

but the formula itself you miswrote

i never checked that that's mb

it's sinθ = opp/hyp

you wrote sinθ = hyp/opp

you see the issue?

yeah

accidently

let me show you something about fractions to help clarify

want me to do the question, but this time we don't mistype the wrong formula

yes

here you go

okay 👍

well here x represents the opposite

so we just subbed in x for the opposite

for sinθ = opp/hyp

sin37 = x/10
because
θ = 37
opp = x
hyp = 10

yes

nope

wrong

show me your work

ik what you did wrong

it's the fraction still

yep

there are two ways, only one is correct

yes

do this:
x/6 = 9

find x

you understand what i mean when i type x/6 right?

1/2 = one half

okay

i want you to give me a value for x

yes

now i want you to do this

correct, so x = ?

yes

how did you do it

9 x 6?

and do you know why that works?

it's the fundamental concept of this that you need to know for these trigonometry questions

so you're saying that if you have the x on top, multiplying the bottom number by the other side of the equation gives you a value for x?

okay great, can you hold off on the trigonometry question for a second and do this question

you're good you'll get the hang of it really quick if you know this

a = c ?
but a/b = c

nope

think back to what you did with my previous question

x/6 = 9

x = 9 x 6 right?

so now we have
a/b = c

you could interpret it that way

a doesn't equal x, but a is in the same position as x.

yes

what's the opposite of division?

this might help you visualise it

1/2 = 0.5 yes?

1 = 2/2

if you take the bottom number, 2 in this case, and multiply by 2, you get the top number on its own

so 2(1/2) = 1

2(0.5) = 1

1 = 1

or like you did with x/6

6(x/6) = 9(6)

x = 9 x 6

so it's 1 divided by two
if you want to get one, you multiply by two

1/2, i want 1
1/2 x 2

x/6, i want x
x/6 x 6

what could be one?

when i ask for the value for a, i am not looking for a number

i want you to write a = ?

it's not a number

x divided by 6, multiplied by 6 will give you x

yeah

my next question will tie directly into trigonometry i just need you to understand how to find a

if a divided by b equals c

a/b = c

then a = ?

if x divided by 6 = 9

x/6 = 9

then x = 9 x 6

x = 54

1/2 = 0.5

1 = 2 x 0.5

1 = 1

a = ?

it functions the exact same way

there doesn't need to be a number

i'll tell you

yes

correct answer

[pearl sable]

yes

that's for finding the angle, if you have two of the sides

can you do this

x = 5? no

oh excuse me x does equal 5, i wasn't expecting you to put it into the calculator

but can you write x = 10sin30

yeah

yeah so you understand

how that is the exact same thing as

yes, but i'm not too concerned about the answer

i want you to understand the theory

to then do a/b = c

a = ?

oh yeah

let me ask you this

if x/6 = 9
can i say that x = 9?

yes but with the new thing

[pearl sable]

^-1

nah you can't, we've already figured out that x = 54

if you want to know why

if we let x = 9

then it becomes
9/6 = 9
1.5 = 9

which does not make sense

click 2nd function

i'll show you how to do it

yep

it is

send photo of your calculator

try pressing shift cos

see if it shows cos^-1

if it doesn't try alpha cos

the answer is 53, 9.999 is wrong

you know how to do the question?

[marsh glade]

is there any way to take the inversecosine of 6/10 without a calculator?

start by identifying the adj, hyp and opp i think

[pearl sable]

i wouldn't know

[marsh glade]

np

[pearl sable]

do the exact same thing as you did for every other question

and show me how far you get to

it?

no

i don't want an answer

i want a line from your work

yes but you realise if you get the answer wrong and you don't show any working out then it's the exact same as just plucking a number from your head in the eyes of the examiner

you need to show your though process behind the answer

go line by line

how you would do this, let the angle = x

send the lines

who said anything about a calculator

i don't want you to get an answer

i want you to show me what you do

[marsh glade]

im sorry to butt in, but do you think this would be ok? Im learning the samething coincidentally, and was wondering whether this would be correct? the image has my work

[pearl sable]

i didn't get 39.79

[marsh glade]

i rounded a lot, do you see any errors in my work?

[pearl sable]

your methodology is overly complicated, but it should work let me do it myself your way and see where you went wrong

[marsh glade]

thank you!

i rounded it a lot in random places, 😓 i didnt see that

[pearl sable]

yeah it was purely your rounding

it works otherwise

although inputting what you did would get 39.72, not 39.79

[marsh glade]

maybe my error in rounding or writing

[pearl sable]

probably

[marsh glade]

where did you end up rounding?

at the very end?

[pearl sable]

look bro you gotta go line by line

[pearl sable]

i just wrote hypotenuse = 25.612, instead of 26

[marsh glade]

ohh

i see

[pearl sable]

so it was more precise

label the sides of the triangle

figure out which formula you can use with the sides you do have values for

then i want you to write out the equation

example: cosx = 20/25.612

nah

bro stop trying to skip to the answer you need to know how this shit works 😭

[marsh glade]

maybe grab a paper and pencil?

once you get one question down, the rest will be easier

[pearl sable]

exactly

i want you to know how it works so then you can do all of them

show me what you've done

i don't want a random number i want to see your work

[marsh glade]

we believe innyou nate!!!

[pearl sable]

let me know

honestly you're moving too fast, you shouldn't be finding the angle on your first day of trigonometry

but it is what it is and you're here now so

[marsh glade]

i would cry 🤑🤑💔💔💔

[pearl sable]

show me plssss 😭

idk why you're so aversed to showing me

[marsh glade]

https://tenor.com/view/madden-giferator-madden-giferator-madden-15-homework-gif-9437354540607036120

[pearl sable]

i don't want you to give me an answer i want you to show me what you've done

[marsh glade]

YAY

[pearl sable]

you know the theorem of pythagoras ?

js stick to what you've been doing this entire time

there's no need

[marsh glade]

oh no

well

at least yoy showed work… 😥 its progress

[pearl sable]

broo pls just treat this question the exact same way you've treated every other one

no.

you won't get an answer

i just want you to show me your work

idgaf

? no

[marsh glade]

show MOREEE work

[pearl sable]

bro 😭

well duh

yep

[marsh glade]

what did you type in the claculator

you can probably express that as an equation (?)

The work you shwoed shows how you got the hypotenuse

In order to see how you got to the answer

[pearl sable]

bro he has no idea what the theorem of pythagoras is let's stay away from that it's just piling more and more onto him 💔

[marsh glade]

the steps in between is what we need(?)

[marsh glade]

💔

[pearl sable]

@true chasm do the question like this

but this time you have two sides and let the angle = x

and i want you to get to a line akin to this

[marsh glade]

||Now you have hypotenuse 👅👅👅based on this image you know which trig function the angle x is, cos. Cos is the adjecent/hypotenuse and you have those values, now you can make a equation akin the the equation above! In the upper example, the angle is known, (37) and we are finding sin. Instead, we put the angle as x and we are finding the cosine ||

[marsh glade]

explanation(?) not very good

[pearl sable]

well yeah

it is tan

what's your point

ok bro let's restart

i don't get the confusion

you sent this image right?

these are the formulae if looking for an angle

you literally js sub in the values and you're done

yes we gave up on that you spent like 20 minutes and are still yet to show me any SHRED of rough work

[marsh glade]

I see how i overcomplicated it 🙏🏽🙏🏽now

[pearl sable]

broo idgaf you're going to get it wrong i want you to just show me what you did

it's okay ik you're js gonna tell me caculator

so we move on

let's just sub in values for a formula

• label the sides of the triangle
• find the right formula
• sub in the numbers
• input into a calculator

ok great

[marsh glade]

After this homework please come back and do pythagoras 😍😍

[marsh glade]

pythagoras my goat 🤤🤤🤤

[pearl sable]

see this is what i wanted

this is the incorrect line

so basically

you js split up the tanx into x and tan-1

move the tan-1 onto the other side of the equation

i'll call it tan-1, but js keep it in mind it's ^-1

and then

you put tan^-1(20/16) into your calculator

and you have the angle

done ✅

yes 51.34 😄

exactly bruh i told you ts like 10 months ago

it's right tho so idk what to tell you lol

[marsh glade]

whats the ans help

It doesnt tell u the correct answer?

thats wonderful

idk is it correct?

[pearl sable]

lemme do it rq

it's 67.38

show me your work

it's 12/5

[marsh glade]

so close

[acoustic walrus]

Can anyone help with my geometry

[pearl sable]

what you need help w

[acoustic walrus]

This

On question 4 and 5

[pearl sable]

ok

yes

[pearl sable]

i can't really see the sstreet lines

[marsh glade]

can we have the points a and b?

[pearl sable]

yhou are correct

🎉

[acoustic walrus]

What is -800 100 and point B is 100 -400

I also think the answer is 450. I just don’t know how it is for a question five

[pearl sable]

i can't see line l

[acoustic walrus]

It’s the white line

[pearl sable]

it can't be

oh it's just any line

np

your midpoint ain't working for what i've plotted

green dot is the midpoint

if A = (-800,100) and B = (100,-400)

[marsh glade]

the distance of these points is this i think (?)

[pearl sable]

why

what's so different

nothing

why

❔

[marsh glade]

Try flipping the triangle

or turning it

[acoustic walrus]

Yea i don’t think this is right but thanks

[marsh glade]

same triangle btw

look familiar…

whatever you used for this

Hmm

[pearl sable]

i got my answers, they line up with the coordinates you gave for A and B

Q4
m = (-5)/9
|AB| = 943.4m (rounded)
Q5
perpendicular of -5/9 is 9/5, m = 9/5
5y = 9x + 2400

[acoustic walrus]

So it is 450?

Buy I don’t know how to shoe the work

That’s what my teacher wants the most

[pearl sable]

wym by 450

i can show you the work

i can run you through it

[marsh glade]

wrong img

[acoustic walrus]

No but we did it with no graph

[acoustic walrus]

As the y intercept

But I put 750

Making it wrong

[pearl sable]

480

lemme do it rq

wait nah send me the question again i'm lost

@true chasm

[acoustic walrus]

Wait, so can you work out the problems like, what formulas did you use?

[pearl sable]

step 1: calculate slop of AB
A = (-800, 100)
x1 y1
B = (100, -400)
x2 y2

m = (y2-y1)/(x2-x1)
m = (-400 - 100)/(100 + 800)
m = -500/900
m = -5/9
step 2: find the distance between a and b
|AB| = sqrt (x2-x1)^2 + (y2-y1)^2
|AB| = sqrt (800)^2 + (-500)^2
|AB| = 943.4

step 3:
find the perpendicular slope (inverse fraction), 9/5
test to prove that they are perpendicular
-5/9 x 9/5 = -1, correct they are perpendicular

step 4:
find the midpoint of AB
((x1 + x2)/2 , (y1+y2)/2)
( -350, -150)

step 5:
find the equation of the line
(-350, -150) m = 9/5
x1 y1

y - y1 = 9/5(x - x1)
y -(-150) = 9/5(x - (350))
y + 150 = 9/5(x + 350)
5y + 750 = 9(x + 350)
5y + 750 = 9x + 3150
5y = 9x +2400
( y-intercept = 2400/5 = 480 )

Answers:
Q4
m = (-5)/9
|AB| = 943.4m (rounded)
Q5
perpendicular of -5/9 is 9/5, m = 9/5
5y = 9x + 2400

what's the question gang

so label the sides

adj and hyp

cosA = adj/hyp

cos x = 10/14

x = cos^-1(10/14)

it's simple

[acoustic walrus]

Wher did u get 9/5 the slope of AB is -1/2

[pearl sable]

slope of AB is -5/9

i showed you my working

come on you know how to do it

yes

fire

can i go to sleep 🥺

[whole flame]

Cos^-1 5/7

[thorn pier]

Does anyone know how to solve this? Prove the ABC triangle (lhs=rhs)

[hollow lily]

,rccw

[hollow lily]

Ehh

Sine rule, Cosine rule, Area formula A=abc/4R and heron fomulas

Or just use A+B+C=180 and do a bunch of algebra

[thorn pier]

i used a+b+c=180 on lhs but im not sure on what to do on rhs.

Maybe move the 2 to the other side to then divide by -2, but after that its gonna be -cosAcosBcosC, not sure what to do after that

[hollow lily]

I don't like algebra

Doing all of that seems boring, I think you'll find a way it's just hard to see

,w expand (a+b+c)(a+b-c)(a+c-b)(c+b-a)(a^2+b^2+c^2)/4

[somber coyote]

Results provided by WolframAlpha

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

Requested by fionna_fx

[hollow lily]

Eh that doesn't seem easier

[exotic yarrow]

You can Google this (https://math.stackexchange.com/questions/4011258/proving-sin2a-sin2b-sin2c-2-cos-a-cos-b-cos-c-2?noredirect=1). If you don’t want to look at the full thing, the tldr is ||double angle|| then ||sum to product||

[upper karma]

ig I have this question in my tb

start taking it from rhs

and prove it

gradually you will come till LHS and you find that LHS=rhs

and taking from RHS is easiest all the time

[thorn pier]

Tried from rhs but its not possible

This is from lhs but i got stuck

Ignore the A+B+C

[upper karma]

what well

[mystic shoal]

Huh

[thorn pier]

thats what im saying

[mystic shoal]

Upper one or lower?

[thorn pier]

wdym?

[mystic shoal]

Prove is it?

[thorn pier]

yeah prove

[mystic shoal]

The top one is the question?

Gimme 5 min

Hm just change sin^2A to cos2A

Use the formula

Cos2A = 1 - 2sin^2A

If you still can’t do it i would solve it

[thorn pier]

solve it if you dont mind, my trig midterm is tmrw 🥲

[thorn pier]

If anyone else solves it id be grateful 🥹

[hollow lily]

Solution is right here

[mystic shoal]

Sry bro i had a lecture

Do you still need it?

@thorn pier

[thorn pier]

All good tyt

if you’re fine w it then yeah

[mystic shoal]

Alr gimme some min

[mystic shoal]

@thorn pier

Let me know if there is any step you don’t understand

[thorn pier]

i love u bro 🥹 tysm

[mystic shoal]

In which standard are you studing rn?

[thorn pier]

Wdym?

[thorn pier]

Also how does cos(A+B) become cos(C)?

[mystic shoal]

-cosc

[whole flame]

A+B+C= π so A+B=π-C .... Cos (A+B)= cos (π-C) ...... We know cos(π-O) is -cosO

O is theta

[thorn pier]

Ahh ok

[mystic shoal]

Class

[thorn pier]

12th

[mystic shoal]

Oh

Nice

[thorn pier]

imo its not that hard but the teacher gives us ridiculous questions

[obsidian hornet]

There is a geometry problem that, in my opinion, pretty hard to solve by synthetic methods, this solution is the only thing i came up with help, so mb someone can solve it in a shorter and simpler way possible ?

[hollow lily]

!nopdf

[lime crown]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[hollow lily]

Sorry pal, can you provide a pic

[obsidian hornet]

yes, no problem

[hollow lily]

What the hell am I even looking at

Where you get this problem from

[obsidian hornet]

Someone posted this problem on this forum, without a source, I tried to solve it, but nothing worked, later I returned to it several times over the year and tried to solve it, but it didn’t work, then I turned to the forum and one person gave me these lemmas, I proved them, but I couldn’t complete the solution, then he helped me and in the end I got this, so far this is the most difficult and longest problem for me, I think it’s from some Chinese Olympiad, judging by the resulting solution, so I believe that a more natural solution can be found.

[hollow lily]

ehhhh, I feel like I'm gonna waste serveral hours of my life solving it

Sorry

I can't help u

[obsidian hornet]

It's okay, I would never have solved it myself because I'm not a good geometer and I don't study modern methods of solving problems, so let someone who wants to figure it out try to solve it.

[zenith stirrup]

Guys who created trigo

[obsidian harness]

https://en.wikipedia.org/wiki/History_of_trigonometry

welcome to the history of maths

the answer is usually some culture / civilisation for something as old as trigo

[whole flame]

It's a Greek word ig

[zenith stirrup]

I think aaryabhatta gave the foundation

[simple vigil]

Aryabhatta gave SIN also

I think

[obsidian harness]

the law of cosines is in Euclid's Elements

but you're right, Aryabhata and other Indian mathematicians made significant progress in trigonometry

[simple vigil]

Wait let me share my book page

Idea of sin was of Aryabhatta

[obsidian harness]

correct

but I mean if you can trust Wikipedia,

The first trigonometric table was apparently compiled by Hipparchus of Nicaea (180 – 125 BC), who is now consequently known as "the father of trigonometry."[17] Hipparchus was the first to tabulate the corresponding values of arc and chord for a series of angles.

[simple vigil]

It's debate on who started
Because it developed over time

[obsidian harness]

that's my whole point

[whole flame]

In Greek tri means 3 gono means sides and metry is to measure or work on

I guess the word comes from greek

[obsidian harness]

it's a bit silly cause Greek influenced European languages

so there are plenty of other mathematical developments that were not known to the West for a long time

[whole flame]

Yep

[spice pasture]

hello

what is formula for slops of a line

[summer cradle]

m = (y2 - y1)/(x2 - x1)

where (x1, y1) and (x2, y2) are two points on the line

[dark sparrow]

angles actually not sides

[elfin cosmos]

I'm new to trigonometry and can anyone help me learn it from the beginning

🙏🏻

[hollow lily]

The Organic Chemistry Tutor

[elfin cosmos]

I was talking abt maths u sent science

[hollow lily]

He cover everything

Math Physic, Chem, Bio

[upper karma]

he covers everything

hes mah pookie

[elfin cosmos]

Can u tell name i don't click to random links

[hollow lily]

ehh

https://www.youtube.com/watch?v=g8VCHoSk5_o&list=PL0o_zxa4K1BVCB8iCVCGOES9pEF6byTMT

[elfin cosmos]

Thanks

[whole flame]

Difference of y upon difference of x

[whole flame]

Okkkk... But Indian scientist Aryabhatta is also considered to be a Pioneer of trigonometry

[dark sparrow]

...and?

that contradicts me how?

[whole flame]

Nahh

It was the argument going above

I just stated the meaning of trigonometry word

Where it came from

But i wasn't sure

I mentioned that

[knotty quiver]

I've never asked myself why's it actually JG

the Jorganic Ghemistry tutor?

[marsh glade]

Its his name

julio gonzalez

whats why

[dark lily]

Do you guys have any lectures/books etc on stuff for conic geometry

Especially regarding the properties of parabolas, hyperbolas and ellipses

Also question
How do you find the length of a common chord between two circles without going through the whole find eqn of chord by radical axes, find where it intersects the circles and then find its length

[real sentinel]

U can simplify find PDFs of notes on Google

Best is youtube

[dark lily]

Ig I'll just have to go through questions on my own

[real sentinel]

Yea

start from basics

cuz properties are vast

u may feel low solving heavy questiona

properties are too much to mug up

[dark lily]

I'll have to speedrun that shit 🤧

[coral salmon]

İm kind of bad at trigonometry any ideas on how i should go about learning it ?

[cloud laurel]

Learn circles and triangles and how they relate

Learn identities and try to prove them

[coral salmon]

i know triangles

well then ill try to do circles then

thx

[cloud laurel]

np!

[visual flume]

aaaaahhhhhhhhhh

[exotic yarrow]

In the future, please show what you’ve done so far when asking for help - it gives us more context of where to help you (and saves our time from explaining things unnecessarily). \ \

Let the line $x=k$ intersect $BC$ at $E$ and $AB$ at $F$. Then, $\triangle BEF$ is a right triangle with an area that is half of $\triangle ABC$. What I’d do is start by finding the coordinates of $F$ (in terms of $k$).

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

Further hint: ||A = bh/2 on both BEF and ABC||

[visual flume]

wait in ur solution are there similar triangles included

cuz i was doing that before i asked and i got a number greater than zero

[summer cradle]

BD = 3 - (-5) = 8
Area(ABC) = ½×8×2 = 8
Area(ECB) = 8/2 = 4
½×(k+5)×h = 4
h(k+5) = 8
h = 8/(k+5)
ECB ~ ADB
h/2 = (k+5)/8
(8/(k+5))/2 = (k+5)/8
4/(k+5) = (k+5)/8
(k+5)² = 32
k²+10k+25 = 32
k²+10k-7 = 0
k = (-10±sqrt(128))/2
k = (-10±8sqrt(2))/2
k = -5±4sqrt(2)
k < 0 (since to the left of the origin)
thus k = -5-4sqrt(2)

[fresh birch]

is it your solution?

[fresh birch]

Is this the problem you need to solve?

[whole flame]

X=k drop a perpendicular to the base and consider distance k from point b or drop another perpendicular from A to BC and consider a trapezium

Apply the given coordinate to find k

[real sentinel]

Find area of triangle

Find equation of line AB

Put x equals k

Find y coordinate in terms of k

Now find the area of the new triangle and equate it to half of the area of the triangle ABC

[shell otter]

To find k, note that the total area of triangle ABC is ½(base)(height)=½(5+3)(2)=8, and the line x=k cuts the base from x=-5 to x=3, creating two smaller triangles that share the same height-to-base ratio since the top vertex is at (3,2). The area from x=-5 to x=k is proportional to (k+5)² because area scales with the square of the horizontal distance when the vertex is not centered. Setting half the total area equal gives (k+5)²/(3+5)²=½ → (k+5)²/64=½ → (k+5)²=32 → k+5=√32 → k=-5+4√2 ≈ 0.66.

Holy sh1t i js spent 10 minutes on that

🤣

[obsidian hornet]

Yes

[fresh birch]

Really? And what problem do you need help with?

[obsidian hornet]

Looking for different solutions

More natural solutions

[fresh birch]

Mnh...

[fresh birch]

and the resolution of this problem?

[obsidian hornet]

I mean, I'm looking for a more optimal and suitable solution for this, maybe using some kind of plane transformation or projective approach. I don't have a lot of knowledge in this area, but I could understand the proposed solution, even if it's a simple coordinate count.

[fresh birch]

lemma 1: Let E,F be the other intouch points, let D' be the reflection of D over EF then A,D',Q colinear

Lemma 2: Let M be the midpoint of minor arc BC, let I' be the reflection of I over BC, then A'I'M colinear

Now let G be the midpoint of DD', foot from A to EF. Project D,D',G,inf onto (ABC) thru A
U get (AD int (ABC),Q,AG int (ABC),M)=-1
Project this through D onto (ABC)
Let N be midpoint of arc BAC , then easy angle chase/well known that ND and AG meet on (ABC)

Now project this bundle thru D onto (ABC).
U get (A,AQ int (ABC)=X, N,sharky)=-1
but now if we project (I,I',D,inf) onto (ABC) from M we get (A,A',sharky,N)=-1
Thus from (1) and (2) we have AQ int(ABC)=A'||

[obsidian hornet]

Thank you, that’s what I meant, i will analyse and ask you if smth is missing for me

[fresh birch]

okay first of all...

Are you sure these are your solutions?

[obsidian hornet]

Yes, this guy helped me to solve this and provide the picture, i wrote about it above

[fresh birch]

Ehh well, Master Kerem Karam is a friend of my own teacher

The solutions there are from Master Kerem Karaman.

[obsidian hornet]

Ok, that sounds interesting, he’s really good and experienced in geometry

Without him I would never solve this problem

[fresh birch]

Okay, but I don't see the part where you sent any of your own resolutions.

only the solutions of kerem karam

[nimble lodge]

Do you guys have book recommendations for Algebra and Trigonometry ?

[obsidian hornet]

I wrote my resolutions in post with problem, but none of them are actually were applied in the solution, In fact, the solution is completely his,
But I didn't expect anyone to be particularly interested, so I wrote it for simplicity's sake

[fresh birch]

Okay, but you just told me that the solutions are yours...

[fresh birch]

.

[obsidian hornet]

Yes and I'm not making excuses. I fully acknowledge the authorship of this solution with respect for the person, and I haven't shared it anywhere other than this channel. I decided to write it this way for simplicity, so I don't have to explain the entire situation again.

[fresh birch]

So you're telling me that you decided to say that the solutions were yours (and you didn't attach any of your own solutions) just to avoid explaining who the real author was....?

I want you to understand that this is not the first time that solutions have been "stolen" from Master Kerem Karam.

[obsidian hornet]

Actually, I didn't originally claim his solution, but I said that he helped me solve the problem completely, and I wouldn't have solved it without him. However, when you asked me a second time, I thought that since the person hadn't seen the original post, I would simply tell them that it was my solution without implying any copyright verification.

[fresh birch]

Well, that's a bit malicious, to be honest... I guess if I ask Master Kerem Karam if that's true, there shouldn't be any problems...

[obsidian hornet]

You could, i can share our chat in reddit if u need

[fresh birch]

And I reiterate, this isn't the first time that Master Keren Karam's solutions have been intentionally stolen, which is why I'm taking protective measures...

[fresh birch]

Okay, fine, I won't go into that further, just always be careful to give the respective credits to the authors of those problems.

🫡

[obsidian hornet]

Ok, still thank you for your solution

[fresh birch]

No problem

🫡

[obsidian hornet]

By the way, do you know where i can get better in geometry, you said that Master Kerem Karam was your teacher’s friend and actually you know geometry on a high level, so, I am really interested where to study like this or at least some sources if you don’t mind to answer

[fresh birch]

My teacher and Master Kerem Karam have many years of experience in that type of geometry, so it is with patience my friend

[obsidian hornet]

Coz I don’t really have teacher or tutor to experience geometry

[fresh birch]

I don't consider myself to be at such a high level in geometry either, that solution is partially mine, since a friend and I did it a while ago

[fresh birch]

I didn't have one at the time either, I met my teacher by chance and he agreed to teach me.

@obsidian hornet A good way to start is by studying the Clark Kimberlin Triangle Encyclopedia.

👍

[obsidian hornet]

Ok, thank you, it would be nice if i’ll meet someone like your teacher one day

[fresh birch]

And above all, practice. The lemmas used are little known. With practice, you will encounter situations like this and from there the lemmas arise. You can learn through practice.

[obsidian hornet]

Practice makes perfect

[fresh birch]

🫡

[obsidian hornet]

Additionally, Have you tried using the method of moving points to solve problems?

[fresh birch]

The translation method is useful for certain cases, although I prefer to solve using strokes and lemmas.

Also if you learn about special triangles and special complementary triangles, they can be a good tool.

You can also find some geometric olympics tricks, especially from Russian, Turkish or Chinese publishers.

[obsidian hornet]

Yes, thats true, but to make sure that you understood me in the right way i meant such like this methods: let we have ABC triangle, I - incenter, there is a well-known fact for point F on the picture is that FG = FH + FJ, but its a trashy thing to solve it using bisector properties and etc. Instead lets set up a linear function f = ax+by+c and using lemma thats if f(a) = f(b) = const => f(c) = const for any c on AB line, omitting details, thats enough to check this for two any special cases and we can choose D and E points instaed F and for them this is obvious so the problem is done, is this called translation methods?

[fresh birch]

I know what you're talking about, but as I said, I'm more of a geometrician. For example, what you mentioned isn't difficult to solve geometrically... it's just a generalization of Viviani's theorem; it can be demonstrated with areas...

[obsidian hornet]

okay, i understand

[fresh birch]

And those translation methods aren't always going to work for higher-level problems; that would be pushing it too far...

As I told you, if you want to learn about it, in Clark Kimberly's encyclopedia they are detailed, including the translation methods.

[obsidian hornet]

Sure, thanks

[quaint chasm]

has anyone tried representing the net of a cube this way before? it's pretty simple but i don't think i've seen this done

4-0-1-5-4
| | | | |
6-2-3-7-6

i was trying to write out a representation and realized you can make it into a tiling or a sort of loop
when tiled as a grid of squares, each vertex has one duplicate connection since otherwise you'd need to make connections bend or cross

i also then tried doing it differently and got the graph in image 4 which is functionally the same

[grave pond]

That seems to be a cube without top and bottom sides.

[quaint chasm]

effectively yeah
but those faces are implied since it's cyclical

no clue if this is useful at all anyway but i figured it would be best to show it anyway
im sure theres some graph theory to this but i haven't really studied graph theory much so idk

[grave pond]

It's not what I would understand a "net" to be -- e.g. Wikipedia's definition specifies "an arrangement of non-overlapping edge-joined polygons in the plane that can be folded (along edges) to become the faces of the polyhedron", so if some of the faces are missing, we don't have a net.

[quaint chasm]

for the tiling yeah

since you can't have 'curvature' or whatever it'd be called if it's a plane

[hollow lily]

<@&268886789983436800>

[real sentinel]

Dude wrong channel

[quaint chasm]

?? how
its geometry

[open vapor]

its meant for simple geometry, not geometry on steroids

like sine law and stuff

[grim valley]

Coordinate lines making my head hurt🥀

[true chasm]

can anyone help with this?

[dapper grove]

hi

[true chasm]

hi

[true chasm]

can u help?

[dapper grove]

ok

[true chasm]

ty

soooo idk how to work it out

@dapper grove hello?

[dapper grove]

your answer is 0.525, but is 001 if its approximated

[true chasm]

ok thx

how u got that?

[last flume]

hi

[true chasm]

hi

u good at trigonometry?

im having trouble with this

@last flume?

[last flume]

what have you tried?

[true chasm]

https://tenor.com/view/monkey-thinking-monkey-thinking-think-money-gif-9734025314392564509

i tried using my caculator

[last flume]

That's not what I meant

You are given one side and 2 angles and you are required to get another side. How do you do this?

[true chasm]

uh

i get the answer to b

then i get c using both sides

[whole flame]

Calculator allowed??

[whole flame]

Just do sin 55/8=sin 43/C

Just remember that the side opposite to angle A is a

N B is b

C is c

So i used sinA/a = sinC/c

Pretty easy

[quasi garnet]

can somoene help me with this

[runic bay]

So AC T BE means that they intersect at a 90 degree angle, same with AC T CG

So using that info then CG is parralel with BE

And if BE and CG are parralel then AF crosses through both at the same angle

[real sentinel]

Just plug in the data in the sine rule formula, you will get it

Exclude the B part cuz we don't have data for B

[real sentinel]

Easy,

BE and CD are parallel

FA is transverse

So by property of corresponding angles angles BEA and CDA are equal

And angle CDA And FDG are equal by vertically opposite angles

Use transitivity, hence proved

[whole flame]

Prove the triangles ADE & ACD similar.... Angle ACB equals ADC by similarly.... ADC and FDG are equal by vertically opposite angles... That means FDG equals BEA

[real sentinel]

Incomplete question

[knotty quiver]

show also Fig. 15.4

i forgot the standard equation for ellipse

,,\frac{(x-h)^2}a+\frac{(y-k)^2}b=1

i forgot the squares

[somber coyote]

glass

[grave pond]

Perhaps fig 15.4 is the classical proof that a cut-off cone is an ellipse.

[somber coyote]

Fionna The Unemployed

[hollow lily]

you may ask if I'm okay, I'm not please send help

i think the sols of mine have some flaws, ehh whatever

ahh I did the mapping wrong , ehh I'm too lazy to keep latexing

nvm I think the mapping is right

[sterile heron]

can someone help meh

[hollow lily]

yeh, but for these problem why not open a help channel #❓how-to-get-help

helpers will explain things more detail here

[whole flame]

Take lhs

Or rhs

Convert tan into sin /cos

Cross multiply

But I think SMTH is crazy abt this question.... It's equality doesn't exist

[hollow lily]

divide both numer and denom in lhs by cos(x) we have the rhs=-lhs that's truthly weird I agree

[summer cradle]

it's not true

[real sentinel]

Something's wrong abt the question

[cunning lion]

<@&268886789983436800>

[true chasm]

can someone help me?

[knotty thunder]

using the law of sines:

sinx/8 = sin43/7 (notice the angle is always opposite the side)

sinx = (8sin43)/7

x =arcsin(8sin43/7)

[true chasm]

so....

38.20?

@knotty thunder

[knotty thunder]

Yep I think so

[true chasm]

can someone give me the answer to this pls?/

[visual flume]

sin(x)/32 = sin(84)/34

sin(x) = 32*sin(84)/34

so ~0.99

around 0.99

wait oops mb i forgot to arcsin

[true chasm]

lol

almost put it in

another me?

[visual flume]

natebate lol

[runic bay]

so im just in grade 9 but i thought that Sine, Cosine and Tangent were only for right triangles

[visual flume]

nope

[runic bay]

huh

[visual flume]

theyre for all triangles

[runic bay]

ok 👍

[visual flume]

however

you can like

sort of

find the ratios and stuff with right triangles

[runic bay]

ok

[visual flume]

they start off trig with right triangles because it clearly represnts

uhhhh

the stuff

also nate its around 58.82

[true chasm]

ty

[runic bay]

so they still kind of work with the other ones but it's easier and nicer to see with right triangles

[visual flume]

yeah

[true chasm]

it said it was wrong

[visual flume]

wth

wait lemme check

so you have arcsin(34*sin(84)/32)

uh

oh shoot

flip the 32 and 34

[runic bay]

arcsin is sin^-1 right

[visual flume]

yup

[true chasm]

https://tenor.com/view/benjammins-sad-chair-sad-chair-meme-sad-in-chair-depressed-in-chair-gif-9576715358533884604

[visual flume]

i typed it wrong

but my work was right

aughhh what lemme lock in for ts

[runic bay]

I saw some of the math work for the grd 11s on the whiteboards in my math class and saw some stuff that basically said a fraction to the -1st is just switching the numerator and denominator. Is this correct?

[visual flume]

broe what

why is my calculaator tweaking

it said 58.82

but when i used google calculator it says

69.39

[runic bay]

69

[visual flume]

i think i need a new one

[visual flume]

yup

heres a more generalized way to define it

[runic bay]

did you type it out the same?

[visual flume]

yeah

[runic bay]

ok dumb question but i had to ask

[visual flume]

(x/y)^n where n < 0 = y^n/x^n

[visual flume]

no problem

not a dumb question tho

[runic bay]

ok

i think i made sense of that

so if (3/4)^-1 is 4^-1/3^-1

[visual flume]

oh wait im stupid

= y^-n/x^-n

mb

so for example if u had (2/3)^-3

it would be 27/8

[runic bay]

so (3/4)^-1 is 4^1/3^1

[visual flume]

yeah

[runic bay]

so its just the reciprical

[visual flume]

yup

[runic bay]

cool thank you so much

[visual flume]

also can anyone help pls 🥺

im soooooo coooooookedddddddd ahhhhhhhhhhhh

[runic bay]

whaaa

im really confused

[obsidian harness]

complete the square

start from (x^2 - 2x) + (y^2 - 4y) = 0, and then you get the equation of a circle

which translates into finding the rightmost point of the circle

[true chasm]

im back

and i come with problems for you to solve

help

@visual flume

[fresh birch]

XD

[civic rock]

tuah

[dark sparrow]

progress?

actually, do you still need help with this, 2 hours later

[whole flame]

Bro i told u the solution to a similar problem like this yesterday

[whole flame]

..

[real sentinel]

It's an equation of circle

Find its centre, radius

Then just add x coordinates of centre with radii

[real sentinel]

Is calculator allowed

[jade depot]

Yeah u need it for arcsin

[high badge]

im year 1 where can i find trigonometric questions specifically differentiations and inverse questions that i can try to solve

[whole flame]

U want a book?

I can tell u smth of higher level

[high badge]

I'll be glad to take a book aswell

[whole flame]

Black book

I don't know if it's available where u stay

If u want online sources which r paid specifically for math

Download cengage app

Try their questions

Depending on ur level decide

[worldly zephyr]

Someone pls help with 11th and 12th

[whole flame]

Send the complete question

What do we have to find

[worldly zephyr]

We should solve the triangles

Finding all other values

[whole flame]

Show me the main question

[whole flame]

Use sine rule

[split imp]

Using law of sines, c/sinC = b/sinB so you can find the other two angles and find the last one using law of cosines

[whole flame]

Yepp

[whole flame]

@worldly zephyr

It's a similar kind of question

Just use the cosine rule in the last

Cos A= b^2+c^2-a^2/2bc

[true chasm]

njah i goty it and went to sleep

[worldly zephyr]

Thank you 😊

[whole flame]

👍

[grizzled warren]

answer is 1+√5. well treat it like quadratic equations of y and use quadratic formula.

then use the fact that discriminate must be greater than or equal to 0 and then it's over

[jaunty pine]

How do I know that 2 lines are perpendicular?

Is it like a specific shape or angle?

[hollow lily]

Using eye

[jaunty pine]

?

[hollow lily]

It depends on the scenario

In coordinate we have some ways like dot product, product of tangent m1•m2=-1

In Euclidean it is quite a lot

Pythagoreans, 90° angle

Prove that it's square, a rectangle,.....

[jaunty pine]

O

square = 4 sides ez

[grizzled warren]

I mean it's depend on question and situation.

if it's follow the condition of parallel line then line is parallel.

you can also use vectors if you can. I mean dot product ab cos(theta)

[kindred terrace]

bro.. this book is kinda bad tbh

i have solved the questions

many are not correct tbh

[whole flame]

It's the best

[whole flame]

Show the questions

[runic garden]

I am so dumb at geometry and my tutor always rages at me about that...

[tall pasture]

I have struggled with it but @rotund drum helped me with it

[true chasm]

can someone help me out?

[strong quail]

whenever you have 2 sides and know the angle between them, you can use the cosine rule to find the 3rd side

you are given the rule in the question, you just have to plug a,b and the angle C, and square root to find c

[real sentinel]

Dude you have the formula and values so just plug in the data

[true chasm]

pls help 🙁

[tall pasture]

This geometry or trigonometry?

[true chasm]

trig

[tall pasture]

Oh nvm

[true chasm]

im cooked

[real sentinel]

Is calculator allowed?

If yes, dude use the cosine rule given below the question

Where LHS is to be calculated

And cosC is 73°

[true chasm]

yea

im unsure of the way its supposed to be in order

[solar dew]

What can I do to make my scale drawing?

[queen shadow]

Can someone please help with this

[queen shadow]

Well like since you took the scale as 1cm = 3.45 ft , shouldn't the white board be like 6.96 * 1.16 (cm)

This as well

[grave pond]

Hmm, I don't think there's a slick way to do that other than just bashing it -- more algebra than it's geometry, really.
The graph gives you this information:

• f(0) = 0
• f(2) = -3
• the y-coordinate of the vertex is 1.
• b > 0
• a < 0
  The first three gives you three equations in the three unknowns a,b,c that you can solve. (It is simplest to start by eliminating c before you start writing down the other ones).
  You'll end up solving a quadratic, and one of the inequalities will help disambiguate which of its roots you need.

[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

[blissful sonnet]

try drawing i guess

it's angles and lines

you know what the bearing is right?

[fresh birch]

jajajaja

[kindred ocean]

I think the problem becomes much easier once you understand bearings