geometry-and-trigonometry

[verbal tapir]

the first expression is not an equation

[pearl bolt]

Then what is COT^2 + Tan^2

[obsidian harness]

it's not an equation cause it's just an expression

[obsidian harness]

where's the right hand side of this?

[pearl bolt]

RHS is 0

[obsidian harness]

knew it

[obsidian harness]

set cot A = u, so you have a quadratic

$\sqrt{3} u^2 - 4u + \sqrt{3} = 0$

[somber coyote]

south

[obsidian harness]

then $\tan^2 A = \frac{1}{\cot^2 A}$

[somber coyote]

south

[obsidian harness]

multiply both sides by cot^2 A to check, if you are skeptical

[pearl bolt]

Thank you very very much mann

[obsidian harness]

no worries

[twin spindle]

How would you show your work for this?

Cause this is all I'm getting

[obsidian harness]

this?

cos(alpha + beta) = cos(alpha) cos(beta) - sin(alpha) sin(beta)

oh wait sin(beta/2) you mean

[twin spindle]

Yeah srry

Shoulda circled it mb

[obsidian harness]

yeah I'd write $2 \sin(\beta/2) \cos(\beta/2) = \sin \beta = +\sqrt{1 - (\sqrt{5}/5)^2} = \sqrt{\frac{4}{5}}$

so if $\sin(\beta/2) = x$, square both sides to get $x^2 (1 - x^2) = \frac{4}{5} \cdot \frac{1}{2^2} = \frac{1}{5}$

two of the roots are negative so we exclude them, cause beta is in the 2nd quadrant and so sin is positive

[somber coyote]

south

[obsidian harness]

jesus christ that is one nasty trig problem

[twin spindle]

Tbh idk if im doing it right

Cause this is what I'm doing 💀💀

Oh shoot I forgot to write the square for sin(sqrt(30)/5)

[obsidian harness]

$\sqrt{\frac{1}{2} + \frac{\sqrt{5}}{10}}$ is correct

I have zero idea how you got there

[somber coyote]

south

[twin spindle]

What 😭😭

[obsidian harness]

and you simplified it wrong

this is correct

[twin spindle]

Ohhhhh

Wait im stupid

It's because I wrote cos(-sqrt(5)/5)

And that confused me since I thought I couldn't do anything

But it littrially showed the cos in the question

😭😭😭

[obsidian harness]

do you know why you choose the positive sign, so not $\sqrt{\frac{1}{2} - \frac{\sqrt{5}}{10}}$?

[somber coyote]

south

[twin spindle]

Yeah because -(-)

Right?

[obsidian harness]

it's cause we also got the solutions to cos beta = sqrt(5)/5 when we squared both sides

[twin spindle]

Since in the formula has a - and your cos is also -, they cancel eachother out making it a positive

[obsidian harness]

wait how did you get there

no but your quadratic equation you got has two values for x^2 = ... right?

[obsidian harness]

this is just wrong

you can't square the inside of sin

you can't get from sin(sqrt(30)/10) to sin(30/100)

[twin spindle]

Tbh idk how you got to the quadratic equation 😭😭

[obsidian harness]

oh you're allowed to use the half-angle formula?

[obsidian harness]

after what I wrote just sub u = x^2

[twin spindle]

Yeah

That's what we're supposed to use

[obsidian harness]

oh that's nice

I didn't know you were taught that

[twin spindle]

😭

[twin spindle]

Yeah

[twin spindle]

Wait, so this is right, right?

[obsidian harness]

right then I get what you mean now

[twin spindle]

😭

I was so confused abt the x^2

💀

cause I didn't notice any variable

[obsidian harness]

then what you did originally is correct

except it doesn't simplify to what you circled

[obsidian harness]

simplifies to this

[twin spindle]

Yeah thats kind of the problem I was having

I didn't know how to simplify it

[twin spindle]

And I still dont know how to simplify it to look like this 😭

I js knew how to set it up

[obsidian harness]

you know that $\frac{\frac{a}{b}}{c} = \frac{a}{bc}$ right

[somber coyote]

south

[twin spindle]

Yeah

[obsidian harness]

cool

then you should be able to see it

you should go back a step to the $\frac{1 + \frac{\sqrt5}{5}}{2}$

[somber coyote]

south

[obsidian harness]

then it's just $\frac{1}{2} + \frac{\sqrt{5}}{5 \cdot 2}$

[somber coyote]

south

[twin spindle]

ohhh

So we jus multiply by 1/2 first

I kinda did that last

Or I multiplied 5 by the 1, to get 5/5 + sqrt(5)/5 and then multiplied 1/2

Nvm I got it now

I jus did sum stupid

Thank you 🙏

Wait a minute

Since the whole thing is being square rooted, wouldn't there be a way to simplify it a little bit more?

Like this?

[obsidian harness]

unfortunately no but you're onto something

[twin spindle]

Sry I didn't mean simplify any roots or anything, but rationalizing it

[obsidian harness]

if you have $\sqrt{2 + 3 \sqrt{5}}$, you can test if $(a + b \sqrt{5})^2 = 2 + 3 \sqrt{5}$

[somber coyote]

south

[obsidian harness]

or if $a^2 + (b \sqrt{5})^2 = 2, 2ab \sqrt{5} =3 \sqrt{5}$

[somber coyote]

south

[twin spindle]

Ohhhh I see

So you would jus square it

[twin spindle]

Then it would become this

Would you have to square "sin(sqrt(30)/5)"? So it would be sin^2(sqrt(30)/5)

Right?

[obsidian harness]

no

why would you square it

if you have $\sqrt{(a + b \sqrt{5})^2}$ that's just $a + b \sqrt{5}$, so we've denested the square root

[somber coyote]

south

[twin spindle]

Causeee, you have to so it on both sides right?

[obsidian harness]

like what are you talking about now

[twin spindle]

Or does that "what you do on one side you do to the other" jus bs

[obsidian harness]

that's correct

that's the first rule of algebra

[obsidian harness]

I don't understand where you pulled sin(sqrt(30)/5) from

[twin spindle]

That's jus what I got for sin(B)

But what I wrote from that image is wrong

[obsidian harness]

oh

[ripe oak]

what is n plus 1k plus g+gf+edf+++gtf+fg+ghfg+hfg+hgf+h+fghfz+fh+y+rtgse+td+bin

[bright kite]

hi

[fringe steppe]

Find the least positive integer n such that there are at least 1000 unordered pairs of diagonals in a regular polygon
with n vertices that intersect at a right angle in the interior of the polygon.

[exotic yarrow]

https://math.stackexchange.com/questions/4762892/number-of-perpendicular-diagonals-in-any-regular-polygon#:~:text=In a regular 2n,(n−1)22

[unique nacelle]

im not your friend

[hazy yew]

We can be

[unique nacelle]

ok

are you doing geometry this year?

[unique nacelle]

hi

[hazy yew]

Hey

[unique nacelle]

wat math are you

[hazy yew]

Yeah, i will do in my college entrance exam

[unique nacelle]

ok

so friend?

[hazy yew]

Ok

Yeah

[unique nacelle]

I sent

[hazy yew]

See ya then imma sleep

[unique nacelle]

are you india

[hazy yew]

No

[unique nacelle]

oh

[hazy yew]

Turkiye

[unique nacelle]

ok

I am the united geometry

[hazy yew]

Aight

[unique nacelle]

so sleep

accept friend request

[hazy yew]

That's lovely

[unique nacelle]

ok

[trail tendon]

the entirety of geometry united and became oofdude

[unique nacelle]

goodbye

[hazy yew]

I already did

[hallow ridge]

how would i find the intersection when i’m give the p(a), p(b), and the union

[onyx anvil]

HELLO

sorry caps

[unique nacelle]

I had U and upside down U on my test. completly blanked out but passed.

[hallow ridge]

😭😭

but my teacher is lowkey bad

[unique nacelle]

lol

[signal field]

When I know the angle and opposite of a right triangle I use sin function but if I know the angle and hypotenuse I would use tangent?

[north kindle]

for what

[glad plume]

any tips (and/or practice materials) to master trig, my brain always forgets it

[glad plume]

It depends what you want. Tangent doesn't go with hypotenuse.
When you have hypotenuse and angle:
If you need adjacent then cosine
If you need opposite then sin

[maiden brook]

whats angle DEC

[maiden brook]

also the setup doesnt seem to be accurate...

[oblique widget]

The angle e

The one in the middle

[maiden brook]

what about it?

there’s something off with the setup though, tan(65) != 3

[pseudo nebula]

,w tan(65 degrees)

[somber coyote]

Results provided by WolframAlpha

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

Requested by someone1010

[pseudo nebula]

yea this problem is bad

[oblique widget]

So how do I do it

[broken garnet]

my hands hurts

but I did all of these out of pure boredom

ignore the scratched out part I got bored I wrote random stuff down

[broken garnet]

wait how do you factor using a TI84 calculator

[north kindle]

you dont? it doesnt have a CAS

[fringe steppe]

thank u

[upper karma]

help, i dont get this topic at all

its symmetry btw

i get that symmetry is dividing a shape into 2 equal parts and all

and rotational symmetry is when u rotate a object the amount of times it looks the same

[obsidian harness]

oh bruh you're here as well

btw this question has been answered in another server

but you are welcome to try!

[upper karma]

bro i am desperate

be fr

[broken garnet]

I figured it out dw

[unique nacelle]

GUALA

[upper karma]

Pls help

[grim scaffold]

Guys, I CANT do trig, hold up

[grim scaffold]

Regarding the second image and first problem, the two angles displayed are equal because of where they are placed. I think the word for the position of the two angles is concurrent but anyways, you would just set them equal and solve from there.

So it would be:

10x - 17 = 8x + 1

And you’d solve for X

:Break:
Anyways, I am going to be VERY active here cause I don’t know anything abt trigonometry

[oblique widget]

How do I explain that is a rhombus?

[unique nacelle]

IT SAYS ITS A RHOMBUS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[oblique widget]

ignore that part

[unique nacelle]

DIAGONALS BISECT BRO

SO 16!

[oblique widget]

I know

[unique nacelle]

EASY

ok

[oblique widget]

🗿🗿🗿

I’m saying like

WHAG DOESNIT MEAN

[unique nacelle]

what doesnt it mean?

[oblique widget]

DIAGONALS BISECG

[unique nacelle]

DIATGONAL BISECt

MEANS

IT SPLITS IN 2

BRUUUUUUUUUUH

[oblique widget]

WHAT OMG

HOW DORS IT PROVE

ITS A RHOMBUS

[unique nacelle]

WHAT?

YOU

ARE NOT

DUMB

?

PROVE

FIRSt

PROVE

[verbal tapir]

,

[unique nacelle]

IT IS PARALELLOGRAM

RIGHT?

THEN YOU CAN FIND OUT IF DIAGS ARE PERPENDICULAR!!!!

OR IF DIAGONALS BISECT OPPOSITE ANGLES!

OR THAT ALL 4 SDIES ARE SAME

[verbal tapir]

ABCD is a rhombus because it is a truth that is given by the question

[unique nacelle]

4 SIDES SAME IS RHOMUS BROOO

[verbal tapir]

if the question says its a rhombus, it's a rhombus

[verbal tapir]

you are required to use the conditions given in order to solve it

[unique nacelle]

he said ignore that part

[verbal tapir]

oh

ok then there is no way to prove it

[unique nacelle]

he wants to know properties of

[verbal tapir]

because there is no other given condition

[unique nacelle]

rhombus

IO think

[unique nacelle]

l

[oblique widget]

Oh yeah the properties

[unique nacelle]

IM NOT YOUR FRIEND

d

wd

awd

a

ADW

AW

DAWD

[verbal tapir]

the properties of a rhombus are as follows:
all properties of a parallelogram
all sides have the same length
the diagonals perpendicularly bisect each other

[oblique widget]

So is this like

10

[unique nacelle]

get a macbook

[oblique widget]

Uhm it’s school Chromebook’s sir

[north kindle]

the diagonals of a rhombus arent necessarily congruent

[unique nacelle]

no one said that

[verbal tapir]

side AB is 26 and BE is 10

[oblique widget]

Yes

[unique nacelle]

veo who said they are congruent?

[verbal tapir]

because AE and BE are perpendicular

you can use the pythagorean theorem

[unique nacelle]

JAMES

[verbal tapir]

to get AE

[unique nacelle]

STOP

[oblique widget]

OHHHHHHHHHH

I forgot

[unique nacelle]

fuc

I jsut died to guardian

[north kindle]

gulp did

[verbal tapir]

specifically here

[unique nacelle]

I hate ocean monument

[oblique widget]

So 90

[unique nacelle]

NO SHITAKE

OBVIOUSLY

[oblique widget]

Be nice

How would I know

It doesn’t say

[unique nacelle]

ok

it does say

[oblique widget]

Where

[unique nacelle]

it says its a rhombus

DIAG

[oblique widget]

Ojhhhh

Smart

[unique nacelle]

ARE PERPENDICULAR

[oblique widget]

Guy

[unique nacelle]

guy what

[oblique widget]

Does perpendicular form a right angle

[unique nacelle]

are you fuc kidd me

bUTT

IT DOESNt

IT DOES

IT DOEs

IT DOEs

DO YOU KNOW WHAT PERPENDICULAr

PERPENDICUKAR

PENPENCDILAPENCULAR

[north kindle]

Wtf is wrong with you

[north kindle]

Is it your first time seeing a rainbow?

[unique nacelle]

I don't think its june anymore

[north kindle]

Im not sure what this has to do with geometry anyway

[unique nacelle]

ok

perpendicular means right

[north kindle]

And yes that’s what perpendicular means

[unique nacelle]

yes

[oblique widget]

Ok uhm calm down next problem

Whag would be the way to solve this

[north kindle]

Find some congruent and isocoles triangles

[unique nacelle]

no

use properties

[north kindle]

Just rederive them to use

[unique nacelle]

NOOO

STOP

NO

DIAG OF RHOMBUS BISECT OPP ANGLES

SO IT IS HALF OF THE OTHER ONE

SIMPLE

2(2y+10)=6y

!

[north kindle]

<@&268886789983436800> can you get this guy to calm the hell down

[unique nacelle]

im am autism

sorry

im right tho right?

[oblique widget]

Yeah but I’m slow

I don’t get it

[unique nacelle]

ok

I explain

the diag.

bisect opp angles

oh shopot

I got it wrong

wait

so

bisect angles right?

RIGHT?

so

the angle will equal

180-something

1/2 of that

can we bring to help channel?

[neon prairie]

It might be better to move to a help channel, but this channel is also okay for help.

[oblique widget]

What….

[unique nacelle]

im mad at the

people

[neon prairie]

Being generally hostile, flooding the chat or spamming caps lock doesn't help

[unique nacelle]

ok

I was just joking

but not joke

because geometry

Geometry

3

honor

[oblique widget]

Ok so how do you do the math

[unique nacelle]

ok

cna you repaste teh picture bro?

Im not scrolling that far

[unique nacelle]

ok

look

[oblique widget]

Oh shoot

Wrong one

[unique nacelle]

BRBUUUUUUUUUUUUUUUUH

[unique nacelle]

OOK

I HAVE ALGEBRA TO DO

LETS MAKE IT QUIZK

[oblique widget]

Ok

[unique nacelle]

BCD

IS

180-ADC

AND 1/2BCD is ACB

ACB = 2y+10

AND

CDB IS 1/2(ADC)

bruh

its easier to think on paper

CDB IS HALF

OF THE BIGGER ANGLE

kaer!x can you hlep?

[hallow blaze]

Could someone please help me out? I got x = 29 degrees and y = 119 degrees but i'm not sure if i did anything wrong :(

[unique nacelle]

oh hell nah

ok

I dont kow

glup are you still there?

[oblique widget]

Yes

So triangle BCD

right

[north kindle]

Seems right to me

[unique nacelle]

aradia pelase help gulp

[oblique widget]

Sorry for the hazard 😞😞😞

[hallow blaze]

let me see hang on

[unique nacelle]

help gulppoo

[oblique widget]

I don’t understand the concept

Blame my teacher

[unique nacelle]

yueaj

did u go over htis in class

[north kindle]

16y + 20 = 180

[oblique widget]

Nvm

[unique nacelle]

aRADIA

HELP

[oblique widget]

I got it

[unique nacelle]

GULPYOUROWNPOOOOOOOOOOOOOOOOP

[oblique widget]

I just did 2y+10+6y=90

And I got 10

[unique nacelle]

no

taht is not not correct

[north kindle]

is it not

[oblique widget]

Yes it is

[north kindle]

its what i got as well

[unique nacelle]

bro

[unique nacelle]

I said it was not not correct smart

[oblique widget]

I
Photo math it its correct

[unique nacelle]

I love photo math

ARIA MATHJ

[north kindle]

im pretty confident its correct

[unique nacelle]

are you not smart?

[unique nacelle]

not not = yes

[oblique widget]

Ok two negatives makes positive

[unique nacelle]

aradia Im surprize you dont knows that

[oblique widget]

Is this like

Undefined

Or what

[unique nacelle]

NIOO

[neon prairie]

Being condescending about dumb things like this falls into the territory of being hostile to other users imo.

[unique nacelle]

what?

[unique nacelle]

bro

BRO

ALL SIDES OF A RHOMBUS

ARE EQUAL?

[oblique widget]

Oh oops

[neon prairie]

I.e. the thing I told you to stop doing just a bit ago

[oblique widget]

So I just set them equal to each other

[unique nacelle]

I didnt use caps locks I used shift

[unique nacelle]

correct! 🤟

[neon prairie]

I genuinely do not care what key you used.

[unique nacelle]

Got it

[unique nacelle]

What

[neon prairie]

?

[unique nacelle]

you mean button? what is a key

[oblique widget]

So would I just 42/2

[unique nacelle]

obviously

its a square

[oblique widget]

Just making sure

[unique nacelle]

HEY

GUlp

I will

RECOOMEND

you something alr?

its a list of properties

[oblique widget]

A better teacher

Yeah

I agree

[neon prairie]

Are you trying to argue about the difference between a button and a key with me?

[oblique widget]

Bro calm down

Everybody

[unique nacelle]

TAKE THIS

[unique nacelle]

I jsut don't know what a key is

[oblique widget]

You’re so helpful

[unique nacelle]

👍

[unique nacelle]

I was in your shoes so I know how it feels

[unique nacelle]

I passed my acceleration exam

[neon prairie]

Okay, now you know I suppose.

[unique nacelle]

what is caps locks?

[neon prairie]

Okay, last chance for you to stop interacting in bad faith here.

[oblique widget]

Calm down @unique nacelle

Let’s just do math

Peacefully

Can you help me or no

@unique nacelle

[unique nacelle]

wait

did I already said ?

yes

its square

divide by 2

[oblique widget]

Diffferent problem

[unique nacelle]

nevner posted bro

[unique nacelle]

bro

can you tell?

rectangle bro

[oblique widget]

How

I wasn’t here for the lessons

Don’t blame. Me

[verbal tapir]

oofdude i recommend not being so aggressive to other people

[verbal tapir]

step 1 is to graph the points

if you graph them, you can see the slope of the line from point A to point B is 1

the slope of the line from point B to point C is -1

the slope of the line from point C to point D is 1

and the slope of the point from point D to point A is -1

now we know that sides AB and CD are parallel

and sides BC to DA are parallel

this guarantees a parallelogram

but then, we can tell that sides AB and BC are perpendicular (if the product of the slopes of two lines is -1, the two lines are perpendicular)

which makes it a rectangle

and finally, we can see every line is the same length,

which confirms it as a square

[oblique widget]

I photo marh it

[verbal tapir]

also squares are rhombuses

so

[verbal tapir]

?

[mighty jay]

what is this???? how to do

im clueless

fr

[verbal tapir]

i don't think this is geometry but ok

|x-a-c| is either x-a-c or -(x-a-c)=-x+a+c

depending on the value

thus,
x-a-c<b (i)
and
-x+a+c<b

multiply both sides of the bottom inequality by -1

x-a-c>-b

which can be rewritten as

-b<x-a-c (ii)

therefore
-b<x-a-c<b from (i) and (ii)

[mighty jay]

idk im learning this with "parabolas" and other stuff that i dont know so i guessed

[mighty jay]

ok thanks!

[opaque heath]

For #27 am I solving for 180?

[faint hollow]

what??

[vagrant snow]

If you mean
Equating -14x-4x to 180 then yep

[opaque heath]

Just what I was typing

Thank you

[eternal yoke]

u just set the sum to be equal to 0

[formal geyser]

Hello guys, can anyone explain to me how to solve both these equations, but can we begin with the x) one first?

These are the hardest equations on the website i am currently scrolling and it seems like photomath overcomplicated the solution, so maybe there are more straightforward ones

[obsidian harness]

wouldn't it be nice if you could change the $1 - \sin 2x$ to $1 - \cos 2x$?

[somber coyote]

south

[obsidian harness]

we can actually do this by $1 - \sin(2(\pi/4 - u)) = 1 - \sin(\pi/2 - 2u) = 1 - \cos 2u$

[somber coyote]

south

[obsidian harness]

so the right hand side becomes $\cos(\pi/4 - u) - \sin(\pi/4 - u)$

$= \cos(\pi/4) \cos u + \sin(\pi/4) \sin u - \sin(\pi/4) \cos u + \cos(\pi/4) \sin u$

$= \sqrt{2} \sin u$

[somber coyote]

south

[obsidian harness]

so we just have $2 \sin^2 u = \sqrt{2} \sin u$ or that $\sin u = 0, \frac{1}{\sqrt2}$

and don't forget to change $u$ back to $x$

[somber coyote]

south

[grave mulch]

I want to know what is the slope if the equation is y=18 and x=0

[faint pasture]

0 nd not defined

[plain ravine]

https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_24

really fun geometry problem

tell me if you get it

[faint talon]

hello

[obsidian harness]

in the actual competition I would just make AB equal to CD so we have 3.5^2

[worthy tendon]

I was curious https://artofproblemsolving.com/wiki/index.php/2024_AMC_10_Problems/Problem_21

[nova kraken]

What textbook are you using?

[faint pasture]

💀 i was taught this as like a formula lmfao-

Like length of common tangrnt of two circles touching each other = 2sqrt(r1 r2)

[faint pasture]

How did you do that btw

[worthy tendon]

I didnt

I was curious to see the latest version and the brainrot was already there

[lost widget]

wait wtf lol

[faint pasture]

Lmfaoooo

[grave turret]

can anyone help me with Trignometry i am in grade 10th CBSE ik the basics but idk how to solve the identities

[faint pasture]

We'd need a question

[formal geyser]

Hi, so how do i solve it?

[formal geyser]

Sorry man, you lost me at the first step

I dont understand your solution

[formal geyser]

I would appreciate it

[obsidian harness]

cos(pi/2 - x) = sin(x), sin(pi/2 - x) = cos(x), those don't ring a bell?

then you need to revise your trig identities

I already deliberately avoided writing cos x - sin x in the form R cos(x + alpha)

[faint pasture]

Sin2x = 2 sinx cosx

And 1 = sin^2(x) + cos^2(x)

So the lhs = sin²(x) + cos²(x) - 2 sinxcosx

Which is just

[obsidian harness]

oh that's clever

[faint pasture]

(Sinx - cosx)²

Ig you can do it from here

[opaque heath]

Trigonometry Lial 11th edition

[obsidian harness]

I mean you still have to solve cos x - sin x = -1

cos x - sin x = 0 is the easy part

so you need the R cos(x + alpha) form as I said

[formal geyser]

And then it will be variable substitution right? (sin x + cosx) = t

Thanks

t^2 - t = 0

[faint pasture]

Are there any solutions not of the form 2npi+ pi/2-

[faint pasture]

Mm

[obsidian harness]

yes

[faint pasture]

Oh

2npi + pi

No other right💀

[obsidian harness]

,w 1 - sin(2x) = cos(x) - sin(x)

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by sooouth.

[faint pasture]

I just meant the cosx - sinx =-1 thingy but sure

[obsidian harness]

(sin x - cos x)^2 = (cos x - sin x)^2 = (cos x - sin x)

ah so cos x - sin x = 1 not -1

[celest radish]

Yeah, I got the same values for x). You can get it down to (cosx-sinx)(cosx-sinx-1)=0, so then either cosx=sinx, which is x=π/4 + πn, n is an integer (since sin x and cos x are both positive in quadrants 1 or 3), or cosx-sinx=1, which holds on the unit circle for x=2πn, n is an integer, or x=3π/2 + 2πn, n is an integer.

The other one should be x=πn, n is an integer or x=3π/4+πn, n is an integer, which you would be able to extract once getting to (sinx+cosx)^2(cosx-sinx)-(sinx+cosx)=0, then (sinx+cosx)((sinx+cosx)(cosx-sinx)-1)=0, where you have a difference of squares that ends up simplifying to the double angle identity cos2x.

I was really confused at the start though because I thought it was asking to prove, and it clearly does not work for the same values of x when comparing the LHS and the RHS

(This is a foreshadow to all of you planning on taking post secondary math: You will look at problems for sometimes hours and realize you missed the smallest of details that end up changing everything. Trust me, it happens more than you think)

[wise zodiac]

Hello, how to prove b)?

[naive cove]

maybe we should say that the sum of angles in ABCD equals 360 degrees, and as we have two pairs of equal angles, it will be 2*(A+B)=360; A+B=180, but they are one-sided angles at AD, BC and secant AB, so AD||BC. the same way A+D=180 and AB | | CD as well. so we have two pairs of parallel straight lines and ABCD is a parallelogram

[unique nacelle]

stop bro lol

[naive cove]

but im not sure that we can use the sum of angles

[celest radish]

The main thing to keep in mind with b) is that if you extend all of those lines, what you really have are sets of parallel lines on transversals. Once you make that realization, you can then use the property of co-interior angles to show that the diagonals are equal (you would need a couple steps, but it’s similar to what geshka was trying to mention)

You could also use the property of alternate interior angles and supplementary angles to help with the claim (again, with support of the transversal lines)

[unique nacelle]

are you not smart lol: JK

bro

rhombus diag. are perpen

rectangle doens't always have perpen diags bruv.

[verbal tapir]

this can be done with alternate angles

if you're asking about how to prove that b) is correct, when the fact that the shape is a parallelogram is given

if you're asking about how to prove the quadrilateral is a parallelogram, when b) is given

oops

fixed image

if B and D are set as θ, A and C must add up to 360 deg - 2θ
which means A and C are both 180 deg - θ

if we extend any side (for example, AD), we can see that due to the rules of corresponding angles, the other two sides that meet (AB and DC) must be parallel to each other

[kind quail]

Hello I was wondering if someone would help me. I am not sure what to do with X.

I know how to find r but how do I find x?

[trail tendon]

find r, and see if you can solve for x

[trail tendon]

you forgot to square the entire (x-0.9)^2 and (x+0.9)^2

also you shouldn't have canceled out the xs anyway, they were both positive

[kind quail]

ohh i see! I understand

[wise zodiac]

but u can't use that?

you should be using <A = <C and <B = <D

we don't know they are parallel right off the bat

[wise zodiac]

oh yep nvm thanks i get it now

[celest radish]

You’re told it’s a parallelogram from the beginning. Anything the question directly tells you is something you can use. If it just said “quadrilateral”, it would be a different story

[oblique widget]

yo someone check if this right or not

[maiden brook]

looks good to me

[unique nacelle]

wrong the isosceles trapizoid at the bottom is inaccurate

[oblique widget]

the base angles are congurent?

[balmy bolt]

i wish the geometry im doing was actually like applicable in life

physics is interesting ig

[trail tendon]

its probably the basis for something that you'll learn later that you can use irl

[balmy bolt]

yeah

its not really applicable to real life
i guess its just preparing you for a field of study in the future

that can be

[trail tendon]

what is it?

[unique nacelle]

no the drawing is worng the legs arent congruence

[oblique widget]

I didn’t even draw anything…

[wraith kiln]

Help!!!

[wraith kiln]

<@&286206848099549185>

[upper karma]

uh

[maiden brook]

who is this guy..

[bright kite]

Whoever is giving ioqm

good luck

[fervent lava]

can anyone help me?

the question is"if sec theta +tan theta=p,obtain values of sec theta,tan theta,sin theta in terms of p

<@&286206848099549185>

[kind brook]

use chat gpt

[fervent lava]

preparing for mid term exams

[marble isle]

Not enough information to solve it in my opinion. Do you have a picture to go with the problem statement? I can provide this diagram for you to steal for a cheat sheet regarding the laws of sine and cosine…

[kind cairn]

Does anybody have any good references about perpendicular conics passing through cyclic quadrilaterals? That is conics whose asymptotes intersect at a right angle? Is this conic unique? Does it always exist? Any help would be great. (By conic I am referring to a hyperbola)

[nocturne remnant]

i think the word youre looking for is "rectangular hyperbola"

in general, a conic passing through A,B,C is a rectangular hyperbola if and only if it also passes through the orthocenter of triangle ABC

[kind cairn]

Oh so a rectengular hyperbola through ABCD would pass throguh the orthocenters of ABC, BCD, ACD and BDC?

Thats what Im trying to prove so

[nocturne remnant]

yeah

https://artofproblemsolving.com/community/c2927h1273728_rectangular_circumhyperbolas

you may find this interesting

[kind cairn]

Oh great thank you.

[kind cairn]

Do you know if this is a known fact?

Like that ABCD and the orthocenters pass through one conic?

[nocturne remnant]

it is somewhat known(?)

not too widely known

fyi the center of that hyperbola is called the poncelet point of ABCD

[kind cairn]

Yeah, I know, Im just interested in it cause Im writing a paper about a generalization of the Euler Line for cyclic n-gons, and came across this fact while trying to generalize the center of the nine-point circle, playing around in geogebra. Wondering how easy it it to prove. Do you know maybe any papers talking about it?

I mean the hyperbola

[nocturne remnant]

hmm

that sounds very interesting

https://www.jstor.org/stable/3614382

i looked on the internet and found this

altho i cannot access it :p

[obsidian harness]

$\sec^2 \theta = 1 + \tan^2 \theta = p^2 - 2p \tan \theta + \tan^2 \theta$

so $p^2 - 2p \tan \theta - 1 = 0 \implies \tan \theta = \frac{p^2 - 1}{2p}$

the rest is straightforward if you know your identities

[pulsar prawn]

“If you know”

[somber coyote]

south

[fervent lava]

guys can u help me in example 36

please

[pulsar prawn]

The solution is below

You want another way than this?

@fervent lava

[grave sigil]

seems like they do

[pulsar prawn]

Probably the only way

It is identities

[faint hollow]

Is this class 10 NCERT??

[fervent lava]

yas

[faint hollow]

I tell you I never did ncert examples, it's good to know people are doing it now

[kind quail]

Hello I was wondering if I could get some help understanding my Trigonometry. I asked about this question the other day and I thought I understood but then I realized I did not.

[frigid fog]

!nogpt

[lime crown]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[frigid fog]

It wants you to use the Pythagorean theorem, and algebraically manipulate it in order to find formulas for a^2 and b^2 or in your case the short leg and long leg

atleast im fairly sure thats what its asking, I'm about to cook dinner so i unfortunately cant help with the specifics rn

havent seen a problem like that before XD

[kind quail]

Yes my teacher never taught us in class anything like this. It's not even in the text book so I was confused. I've just been guessing while trying to use the Pythagorean theorem

[sleek sonnet]

hi

someone

can help me with this question?

in a rhomboid ABCD, AC=30 and BM=9 where M is the midpoint of CD, calculate the maximum value of AD

[rough coral]

have you tried taking the square of both terms?

[pseudo nebula]

hi im someone

what's a rhomboid?

[sleek sonnet]

like this

[sleek sonnet]

hi

[pseudo nebula]

isn't that just a parallelogram ? lol

[sleek sonnet]

in my country

we call like this

[sleek sonnet]

is similar

you help me?

[sleek sonnet]

xd

[pseudo nebula]

been too long since my geometry days, im sorry, my brain doesn't wanna work

[sleek sonnet]

😭

[rough coral]

this is most likely your illustration

[sleek sonnet]

now?

yessssss

but i don´t know i have to do later

[pulsar prawn]

!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.

[pulsar prawn]

!nogpt

[lime crown]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[kind quail]

i keep confusing how to do that

[hoary prism]

(a+b)^2
= (a+b)(a+b)
= a^2 + ab + ba + b^2
= a^2 + 2ab + b^2

its just distributive property

some may call it foiling

[hoary prism]

so due to the pythagorean theorem the setup here is

(x-1.3)^2 + (x+1.3)^2 = r^2

[warm shuttle]

also commutative ab + ba -> 2ab

[upper karma]

sirf value hi to put krni hai

[fervent lava]

@upper karma calculation is lenghthy

[faint pasture]

Well the question is lengthy nothing you can do bout that

Shit probably cancels out

[wise zodiac]

Hi, I proved this but just a little query. Which definition of isosceles are they using?

1. Exactly two sides of a triangle are equal

2. At least two sides of a triangle are equal?

I feel like depending on the definition the answer can change?

If they're using definition 1) then we can't tell? Maybe AB = AC = BC and it's equilateral...

[warm shuttle]

1 would be weird

[wise zodiac]

It is still a definition though

so for the A <- B direction, we can say that by the angle bisector theorem CB/BM = CA/AM and since AM = BM we have CA = CB

but can't CA = CB also perhaps equal AB?

[warm shuttle]

Of course it can

and you are right

What i'm saying is that if they used 1 definition, then the answer would be unsatisfying

[wise zodiac]

Yeah but like it's still a valid definition, shouldn't they specify what definition of isoceles they're using?

[warm shuttle]

So from the view point of making the exercise definition 2 is more likely

If you are so unsure just add a

note: assuming regular triangle is also considered isosceles```

[wise zodiac]

Lol i'm self studying

[north kindle]

is this not just a diagonal of the rectangle?

[warm shuttle]

no

I saw this exact problem in one of help channels as well. Maybe with different numbers

[north kindle]

can you help me

i cant figure it out

im trying to find x here but i cant do it

[obsidian harness]

the perpendicular bisector of the fold must pass through B and D

cause it's a reflection, so the reflection direction is perpendicular to the fold line

[warm shuttle]

what he said

then by similar triangles 8:12 = b:8
b = 64/12 = 32/6 = 16/3
x = sqrt(b² + 8²) = sqrt(256/9 + 64/1) = sqrt(256 + 576)/3 = sqrt(832)/3 = (8/3)sqrt(13) i think?

if no mistake

[obsidian harness]

wow

[wise zodiac]

How would one start this out?

[fervent lava]

i have a doubt ,x sin3theta+ y cos3theta= sin theta cos theta

x sin theta= y cos theta

to prove-x2+y2=1

[faint hollow]

Can you send photo of the question

It's hard to understand if it is power or coefficient

[obsidian harness]

if the trapezoid is isosceles, then the diagonals are congruent by SAS:

both triangles share the blue angle, the base, and the same diagonal sides

the other way is harder, starting out from if the diagonals are the same length

[wise zodiac]

Yep i solved this exactly like you did it

I’m talking about the “only if”

#help-42

[fervent lava]

can i get another help?

[sick vine]

,rotate

[sick vine]

what part do you need help in?

[molten gulch]

have someone here the list of all points that can be represented by the notation X_i (for example : point X_25 (center of the homotethy ......))?

[signal field]

C is considered a side on a right triangle for law of cosines right?

Y would be my third angle

[cunning lion]

the law of cosines can be applied to any triangle

[signal field]

Yea I meant is Y an angle but C is a side

I might have to show the problem

Nvm I got it

[signal field]

Anyone have tips on when to use law of sine vs cosine

Like when to differentiate when I need to use cosine vs sine

[warm shuttle]

You are asking this like they are similar

Oh i guess they are related actually nvm

But just try both and see if one of them is applicable

I personally always think about it like you could change the measures of stuff. And then it's like, oh but if these 3 values are fixed, then this other value also has to be fixed. And then i figure out what the formula for it is.

[quartz minnow]

Q1- (i) (we have to find angle AOB and OAB)
Q3 (i)(ii)(iii)(iv)

[faint hollow]

what is P in angle APB??

btw if i consider the angle shown as 50, then angle AOB = 130, and angle OAB = 25

3. i) x = 22.5

3. ii) x = 50

3. iii) x = 60

[faint hollow]

,rotate

[faint hollow]

,rotate

[faint hollow]

,rotate

[faint hollow]

3. iv) x = 230

[faint hollow]

btw class 9 RD SHARMA?

[stuck sleet]

In the trigonometry book I was reading it is stated on law of cosines that:

Angles A and B may be found by the law of sines, the smaller of these angles should be found first, for if the larger is obtuse some "CONFUSION" may arise.

what is that confusion, and how can I prevent this confusion, also is there an ambiguous case for the law of cosines?

[vagrant snow]

There is no ambiguous case for law of cosines

[verbal tapir]

that confusion can arise because of how the value of sine is

for angles between 0 and 90 degrees,

and for angles between 90 and 180 degrees

[upper karma]

why are so many people cooked?

[quartz minnow]

How?

[quartz minnow]

How ?

[quartz minnow]

Yep

I want to know how, answers are written in the book

[nova kraken]

What textbook do you use?

Does anyone have recommendations regarding books on Geometry? Introductory or intermediate level.

[fervent lava]

its rd sharma

can i get some help

[wicked mist]

can sm1 solve this please?

[faint hollow]

I'll be able to send in half an hour

Its easy

First try reading the properties thoroughly

[faint hollow]

Divide both sides by cos^2 theta

[fervent lava]

okie

then?

[faint hollow]

Write sec^2 x as 1+ tan^2 x

[fervent lava]

done.

[faint hollow]

Then let tan theta = x, which will give you a quadratic in x

Solve for its roots

The values of x obtained will also be values of tan theta

Tell if any problems arise

[fervent lava]

what if we dont let tan theta x?

and just write
2tan^2theta-3tan theta+1=0

doneee

[faint hollow]

I was gonna say it, but more often people have a little trouble dealing with quadratics this way(without letting x)

[faint hollow]

just square and add the 1st and 2nd equations

then find mn

by writing tan^2(x) = sin^2(x)/cos^2(x)

[lament trench]

sanskriti mam are u indian @fervent lava i think so by your name

[faint hollow]

names can be misleading, argue that she's Indian on the basis of the NCERT and RD questions, its much more objective

PS : its a joke

[chrome depot]

is x^2=y^2 considered a hyperbola?
something involving this appeared on a test earlier today and idk if i got it right

[verbal tapir]

it is called the degenerate hyperbola

a special case of the hyperbola

[chrome depot]

alright, thanks :)

[trail tendon]

have you ever thought maybe the hyperbola doesn't LIKE being called a degenerate? 😤

[kind rapids]

Yeah, who came up with the name degenerate for conic sections 😔

[fervent lava]

correct

10th grader

[fervent lava]

[wise zodiac]

if 0 < a < b < c then prove that 1/2 ac > area of triangles with side (a,b, c)

Does this work:

for a< h = b < c -> 1/2 ac > 1/2 ab and it works for this case too 
i'm guessing these are the only two cases
well a = h case is trivial```

Btw i'm looking to do this geometrically not with trigo, thanks

[signal field]

How could I solve an unknown triangle with trig if I only have 2 angles and 1 side given

Idk how I could apply law of cosine for instance

[trail tendon]

well if you have 2 angles of a triangle, you have 3 angles of the triangle lol

you can apply law of cosines if you have only two sides and one angle, but you can use law of sines if you have two angles and one side (or also two sides and one angle)

[rough crater]

*solved

[devout pecan]

i just noticed there doesnt seem to be a simple way to write the infinite solutions of sin(x) = 1/x

i challenge anyone here to find a way to calc them

[hoary prism]

Yeah I don’t think there’s a closed form solution

You’d wanna use newtons method or something

[lost wyvern]

How

[obsidian harness]

What info is there in the previous q?

[worthy tendon]

Just go on finding the locations of every point. G -> E -> F -> I -> A -> B
The problem is just too much work

[lost wyvern]

None

[lost wyvern]

But how so you find their locations

[worthy tendon]

G from being the midpoint of two known points and so on

each of the points has special properties, which you can use

[obsidian harness]

yep midpoint formula, finding the perpendicular then intersection for point E is another procedure

then FI has the same slope as AB

so then you just need to equate the line AB (slope AB, passes through A) with the perpendicular at point C

[wise zodiac]

does A_1 > A_2 suggest that s_1 > s_2

where A_i is the area of a triangle and s_i is the semiperimeter of the respective triangle

I used this fact in one of my proofs and now it feels iffy lol

[dense smelt]

I’m not understanding the symmetry thing

So does that mean I can just look at something declare it’s symmetrical and call them equal?

South you get this?

[trail tendon]

what part are you talking about, exactly?

since D and E are midpoints of the EQUILATERAL triangle, the line between them creates another similar EQUILATERAL triangle

[dense smelt]

It says DG=EF by symmetry that’s my problem,

[trail tendon]

well

everything on either side of this line is symmetrical

because its a circle, so the circle is symmetrical. and the triangle is equilateral, which is symmetrical in this way

i don't know how to PROVE it cuts line GF in half, im thinking about it, but to me its obvious

intuitionally

and like thats not a good thing i should be able to know why lol

but it does seem like they are symmetrical

so GD and EF would be equal

[dense smelt]

Well I’ve had problems in geometry where I assume to much bc it’s intuitive and then I’m wrong and I was supposed to prove it first n blah blah

But I get what u mean

[trail tendon]

yeah, exactly, thats why im tryna think of a way to prove it 😂

or at least show somehow

oh well we know <BDG = <BEF

[dense smelt]

Yea idk why they just make a blank statement like that I’m ngl it does seem like one of those things u can assume from the pic you showed

[trail tendon]

that doesnt help

[trail tendon]

yea

like we know every object here is symmetrical

idk how to prove the line is symmetrical about the line i draw but

it really makes sense that it is lol

oh oh i know

[dense smelt]

Ima just start declaring things are equal by symmetry now tho, just cuz I said so and bc they look the same

[trail tendon]

the line i drew is a diameter

[dense smelt]

I like this method

[trail tendon]

🤣

[covert mango]

find area

pls

[obsidian harness]

length of AD is equal to length of BC?

this should be the way to go if that's the case

are you sure this is even solvable?

seems like there are infinitely many solutions

[crystal tangle]

Is this channel occupied?

I need some help

[obsidian harness]

you can go ahead and ask

[crystal tangle]

Idk how to do 3b

[obsidian harness]

okay so $\frac{5}{\sqrt3 + \sqrt5} \frac{\sqrt3 - \sqrt5}{\sqrt3 - \sqrt5}$

[somber coyote]

south

[crystal tangle]

Yeah but I need it in this form

[obsidian harness]

do you know what the difference of two squares is?
(a + b)(a - b) = ....

[crystal tangle]

The 3b

[obsidian harness]

yeah

that's how you get there

[crystal tangle]

Idk I'm a new y11

[obsidian harness]

https://www.youtube.com/watch?v=t1Pg0Ar52-k

here are some worked examples

btw this channel, 1st Class Maths, explains really well

[crystal tangle]

Thx

[obsidian harness]

start from around 9:06

or earlier if you don't understand the more simpler examples

[crystal tangle]

That was kinda helpful by the end but idk how to put it in the form 5/2 (-root 3 + root 5)

[obsidian harness]

after you simplify you get $\frac{5 (\sqrt{3} - \sqrt{5})}{-2}$

[somber coyote]

south

[obsidian harness]

so write this as $\frac{5}{2} \cdot -1 \cdot (\sqrt{3} - \sqrt{5})$, so that we can distribute the negative inside the bracket

[somber coyote]

south

[obsidian harness]

doing this gives you the answer

[crystal tangle]

K thx

[obsidian harness]

np!

[restive heron]

My Trig class just opened up yesterday, so I guess i'll be hanging around.

[jade sierra]

same

[tall coral]

Trying to prove Miquel's Theorem. The paper says "Just chase a few angles," but I have no idea on how angles would connect to the point, can I get a hint as to where to start?

[upper karma]

Q.1

Q.2:
Let's say you have been given the angle ABC and the sides AB and AD.
We do not know the lenght of BD so let a line pass through like that.
AD is given so we can draw a circle with center A with radius AD.
The third vertex lies in the circumference of the circle and also on the line BC.
There is only one intersection of theirs that gives the angle ABC the measure that was provided; the other intersection gives a triangle in which the measures are not as such.

So my question is: Isn't ASS theorem non-ambiguous?, i.e., it forms a certain triangle?

Q.3.
Is this proof of mine of the basic proportionality theorem correct considering the fact that I am allowed to use similarity?

EI is parallel to GH.
FJ is the perpendicular to GH from F.
Angle GJF = 90 degrees and therefore the angle EKF = 90 degrees.
Triangle EKF is similar to triangle GJF.
JK:KF = GE:EF (this can be proved using similarity as:
GF/EF = FJ/KF.
(EF+EG)/EF = (FK+KJ)/(KJ)
1 + EG/EF = 1 + FK/KJ
EG/EF = FK/KJ.)
Similarly, IH:IF = JK:KF.
This implies that GE:EF = IH:IF,
which completes our proof.

I have another proof but I am quite certain it's correct.

It utilizes the mid point theorem and it's correct.

Which itself can be gotten by this theorem of basic proportionality.

[lucid panther]

how are u always online

get a life bro

[trail tendon]

im doing homework 😔

[lucid panther]

always lmao

??

[trail tendon]

real 😂

[trail tendon]

this was a genuine question? well no I sleep and go to my classes and occasionally socialize and stuff

but most the time im just at home, chillin, doin homework

[lucid panther]

ok beta whatever you say

[upper karma]

Please answer to my questions.

[trail tendon]

whats ur question?

[upper karma]

above see it

[trail tendon]

oh

[manic swan]

oh same anjali?

[upper karma]

Yes.

[upper karma]

.

[covert mango]

yeah

[marble cliff]

hey i have small question

in regard the following

according to this synthetic proof
I would like to know how OP = BP * CP / AP

this prove that all the alttitude meets at one point

i do not see it

[obsidian harness]

do you get why OP/BP = CP/AP?

[marble cliff]

yes

to there i get it

that all 3 altituds intersect at the same point

[obsidian harness]

then multiply both sides of this by BP

oh hmm

[marble cliff]

sure then you get OP

i get that

[obsidian harness]

it's the reasoning after okay

[marble cliff]

but i do not really see how that will show that all 3 altitude intersect at one point

still

so if you can elaborate

[obsidian harness]

do you have a link to the ProofWiki page?

[marble cliff]

sure

[obsidian harness]

found it

[marble cliff]

https://proofwiki.org/wiki/Altitudes_of_Triangle_Meet_at_Point