geometry-and-trigonometry

[obsidian thicket]

I no understand

[upper karma]

<@&268886789983436800> ^^ asking for help on a quiz

[obsidian thicket]

Idk if Asia is your name but it's cute

[upper karma]

awww and yes it's my preferred name

[obsidian thicket]

Nice

Damn a lot of helpers on but I got no help

[upper karma]

well it is because people here don't help on quizzes/tests

[humble pulsar]

^

[spiral hawk]

Well, Now that your test is probably over, the length of YZ is 12 cm, as angles X and Z are the same 77 decrees

Then, u would take 77+77=154, then take 180-154 = 26 degrees for angle Y

[oak stag]

^

All triangles equal 180 degrees. So Y is 26

Ughh I did this last year and my brain is stuck

I’ll look in a textbook later, but currently it’s 6am😂

[inland condor]

could anyone help with this

[grizzled lantern]

If they are congruent dont their angles equal? Set the equations equal to each other and solve for x @inland condor

[inland condor]

ty was able to get it.

[obsidian thicket]

Is there anybody I can dm to help me?

[main meadow]

Is this correct?

[upper karma]

is that b x pi or is it something else

[main meadow]

10

10pi

[upper karma]

aight

its correct

if Im right

[main meadow]

Uay

I got it correc5

@upper karma

74 bby

[grizzled lantern]

@obsidian thicket angles for rt and st are equal so set the equations equal

[remote mortar]

i*

[magic wasp]

yeah sure

with what??

[remote mortar]

one sec

[magic wasp]

alright

[remote mortar]

for this do u do y-(-9) = -1/3(x-(-1))?

and then for the perpendicular do u just replace -1/3 with 3

[magic wasp]

oh

lol gimme a sec

[remote mortar]

okay

[magic wasp]

what grade u in?

cuz I don't think I know this

[grizzled lantern]

Parallel means the slopes are the same, perpendicular means the slopes are flipped

[magic wasp]

yes

[remote mortar]

oh

[magic wasp]

lol

[remote mortar]

yeah but like

[grizzled lantern]

For instance, a perpendicular slope of -1/3 means -3

[remote mortar]

how do u find the equation for the line

[magic wasp]

oop

don't ask me ask that guy lol

[remote mortar]

i thought for -1/3 it was 3

[magic wasp]

idk

noo

[remote mortar]

oh

[magic wasp]

its -3

[remote mortar]

lol

wht

[grizzled lantern]

Yeah its positive 3 my apologies

[remote mortar]

oh

wait so do u know how to find the line

[grizzled lantern]

https://youtu.be/Ua_6WPdq1M8

https://youtu.be/13VQU4RAyAs

[remote mortar]

thank u!

[grizzled lantern]

Yep, youtube is clutch

[obsidian thicket]

Ichiro what do you mean

[grizzled lantern]

@obsidian thicket wassup?

[obsidian thicket]

@grizzled lantern here

[grizzled lantern]

Oh you have two sides which are equal which means two angles are equal

[obsidian thicket]

But how do I do the problem

[grizzled lantern]

@obsidian thicket do this, 9x+2=13x-8

you solve and get x=5, plug that bag into both equations and they should both equal 47 degrees,

so you have 2 angles that equal 47 degrees... 47*2 = 94

plug 5 into your 17x+1 and get 86

86+47+47= 180

yayyy

[obsidian thicket]

Damm you mad smart thank you

[grizzled lantern]

Not smart, just in high school

[obsidian thicket]

Damn I'm in highschool as well but I struggle with gwo

The only reason I struggle is bc I'm not doing it in person

[obtuse hornet]

Anyone able to guide me through how to do these proofs

[grizzled lantern]

@obsidian thicket I feel you dude, im a senior this year and doing ap bc calculus which is two ap classes in one

@obtuse hornet I think for 10) <A=<B=<D=<E

because their sides are congruent

For 11) you have alternate interior and alternate exterior angles

[obtuse hornet]

Ahh i shouldve been able to do 10 lol

Where does alternate exterior angles come in?

[grizzled lantern]

for problem 11

[obtuse hornet]

Is that a pair of parallel lines cut by a transversal tho

Cause theres no exterior angles, no?

[grizzled lantern]

yeah might just be interior angles then

[obtuse hornet]

Ah ok

What about 12?

[grizzled lantern]

my brain died on 12

doing calculus atm

[magic wasp]

hey guys

:D

[obtuse hornet]

Anyone able to guide me thru how to do the proof on number 12 up there ^

[silent plank]

where are you stuck?

[indigo chasm]

?

@silent plank

[obtuse hornet]

@silent plank Just dont know how to prove it lol

[tulip grail]

The equation for line a is given by y=-3/4x-2 support line a is parallel to line B and line T is perpendicular to line A. Point (0,5) lies on both line B and T

Part A Write an equation for line B.

Part B Write an equation for line T.

Im a little lost

[silent plank]

@obtuse hornet do you know what's needed for congruency proofs? AAS, SAS, RHS?

[obtuse hornet]

Sorta?

@silent plank

[obtuse hornet]

anyone ;-;

Is it ASA? <P and <T are congruent, <R is congruent and sides Q and S are congruent?

[proud arch]

For anyone that is available I have a question my little sister gave me for geometry and i cant find the right angle to it...sorry for the pun,
anyway, they are given a right triangle with two missing angles and are given the three edges, the question requires you to solve the missing angles without using sin, cos, or tan, is this possible?

[silent plank]

depends

[proud arch]

what will it depend on, i have the edges as a = 12, b=9 and c(hyp)=15

[upper merlin]

uh

what

idk how you would do that

without trig

[silent plank]

depends if you have a special triangle where you technically don't need to involve trig

[proud arch]

Thats what im thinking..

[upper merlin]

its not like a handy 30 60 90 or 45 45 90 triangle

im also pretty sure the angles are irrational

[proud arch]

nope with trig, the angles dont come out to those special

[upper merlin]

idk im guessing they are

[proud arch]

lol i mean i used trig and it didnt come out to beginner friendly numbers

[upper merlin]

using arcsin you get the ugly number of 0.204832765pi for one of the angles

yeah

if you cant use sin cos or tan use sec csc cot 😳

[proud arch]

well...i mean none of the trig functions

I think my sister just missed a lecture and maybe they went over the functions without her realizing but also there could be some geometry i just dont know..

[obtuse hornet]

Are rhombus diagonals always congruent

[heavy haven]

is number one bak?

and is number two b?

[summer spire]

there's more to an intersection of planes than just one point

[heavy haven]

how do i do this?

[untold flame]

uhhhh

I don’t really know but I would recommend using IXL because they have similar questions like that

[upper karma]

@heavy haven

it's easy

do you understand these hints?

[silk patio]

It’s easier to do a parallel through V

[upper karma]

🤔

that could work, yes

[silk patio]

Or a perpendicular to the left of U and W and use angles in a pentagon

[upper karma]

||180 - 138 = 42||

||angle UVW (marked in blue) is a vertical angle to the sum of the angles on the right||

(the red line is parallel to the line segments at the top and bottom, btw)

[silk patio]

I didn’t mean like that

I meant using corresponding and alternate angles

The blue angle is split directly

[next jackal]

does $\sin^2(-x)= -\sin^2(x)$?

[somber coyote]

lazypawtato:

[dark sparrow]

no

[earnest echo]

Think about it\\

$\sin²(-x)=(\sin(-x))²$

[somber coyote]

The Godfather:

[next jackal]

okay it just means sin^2 (x) then

[upper karma]

yes

[next jackal]

Is this correct?

[upper karma]

yes

convert cos^2(x) in terms of sin(x) to get a cleaner form

[next jackal]

alright

$3\sin(x)-4\sin^3(x)$?

[dark sparrow]

that 2 is sus

[somber coyote]

lazypawtato:

[next jackal]

now?

[dark sparrow]

yes

[next jackal]

Okay

sin(5x) seems too big

[livid moss]

Too big for what?

[next jackal]

to derive, for now. Feels like busy work

[upper karma]

tbh sin(2x) is already too much work for me

[next jackal]

okay... I understood how to derive them though, may I skip it?

the sin(5x) part?

[upper karma]

okay... I understood how to derive them though, may I skip it?
we are not your teachers

if you understand the mechanism that's good enough

[next jackal]

Okay thanks

[lyric light]

i need help

[dark sparrow]

what is this

[lyric light]

geometry

[dark sparrow]

no, what is this file

[lyric light]

its the problem

ik it seems sus

[dark sparrow]

i've never seen a .HEIC file before

[supple onyx]

HEIC files are image files

[lyric light]

does anybody know how to do it

[supple onyx]

though ive never seen anyone use them

can you put up the screenshot again i didnt have time to look

[lyric light]

ok

[dark sparrow]

what program even uses that file extension

[silent plank]

angle sum on a line

is what

[supple onyx]

I dont know but apparently it's supposed to be more efficient at compression than jpeg

[lyric light]

its because

i airdroppes it from my phone

[supple onyx]

i mean the screenshot you sent earlier

dont wanna go around opening it in windows, much easier to do on mac i believe

[dark sparrow]

i don't trust this image format, can i have a .jpg or a .png please

[supple onyx]

what are you meant to find?

[lyric light]

x

[supple onyx]

and what are the labels pointing to?

and what are the horizontal lines for?

[lyric light]

lol you think i know

[supple onyx]

id expect so since you probably drew that

do you have the original question

[lyric light]

ye

[silent plank]

apply angle sum on a line

[lyric light]

so do i make =180

[silent plank]

make what =180

[lyric light]

x+77+x+177=18-

180*

[silent plank]

yes

[lyric light]

thanks

[supple onyx]

oh the labels are for angles right

[silent plank]

do i dare apply malicious compliance here

since there's no ° those values are clearly in radians

[lyric light]

i got -37

is that right

wait so

how do i know whem to make it equal 100

[silent plank]

100?

[lyric light]

180

[silent plank]

properties of a straight line

[upper onyx]

can somebody help me with 26 27 and 28?

"ĐS" are the answers btw, i want to know how to solve the equations

[supple onyx]

so first for the sin part. $\sin^2(\frac{x}{2} - \frac{\pi}{4}) = \sin^2(\frac{1}{2} \cdot (x-\frac{\pi}{2})) = \frac{1-\cos(x-\frac{\pi}{2})}{2} = \frac{1-\sin(x)}{2}$

[somber coyote]

Little Narwhal:

[supple onyx]

$\cos^2(\frac{x}{2}) = \frac{1+\cos(x)}{2}$

[somber coyote]

Little Narwhal:

[supple onyx]

so we have $\frac{\sin^2(x) - \sin^3(x)-\cos^2(x)-\cos^3(x)}{2\cos^2(x)} =0$

[somber coyote]

Little Narwhal:

[supple onyx]

$\cos^3(x) + \sin^3(x) + \cos^2(x) - sin^2(x) = 0$ iff $\cos^2(x)\cdot (\cos(x) + 1) + \sin^2(x) \cdot (\cos(x) + 1) - 2\sin^2(x) =0$ iff $\cos(x) + 1 - 2(1-\cos^2(x)) = 0$ iff $2\cos^2(x) + \cos(x) -1 =0$

[somber coyote]

Little Narwhal:

[supple onyx]

$2\cos^2(x) + 2\cos(x) - \cos(x) - 1=0$ iff $2\cos(x) \cdot (\cos(x) + 1) -1 \cdot (\cos(x) + 1) =0$ iff $(2\cos(x) - 1)(\cos(x) + 1) = 0$ iff $\cos(x) = \frac{1}{2}$ or $\cos(x) = -1$

and so i must have made a mistake somewhere since that gets you the wrong answer damn it

[upper onyx]

[somber coyote]

Little Narwhal:

[supple onyx]

it's probably just a minor mistake somewhere that you should be able to fix though

i have to go

hopefully you can find the mistake

[upper onyx]

oki

[main totem]

Guys quick question. How can i say that if a point in a rectangle divides the side into a 1:3 ratio then on the other side it will also be 4 units

[dark sparrow]

depends on what you mean by unit

and what do you mean by "how can i say"

didn't you just say it?

[terse scaffold]

pandaThink

[eager hatch]

I have no clue where to start

[exotic fjord]

uhm

[earnest echo]

Can you use vectors

Then it's just one line

[eager hatch]

I have no clue what that is so probably not

[earnest echo]

Fuck vectors

[eager hatch]

Alright

[earnest echo]

I believe you have already determined that it's an isoceles triangle

Right?

[eager hatch]

Ye just now

Cuz the base angles

Oh wait yeah I’m stupid

Thx for the help

[upper karma]

If you can call and help me with geometry dm me

[upper karma]

😳

[kindred rapids]

Hi can someone help me..?

This is solving trigonometry equations

[upper karma]

sure

post

[kindred rapids]

I've got these two...but neither me nor my friends know how to do the questions...

Okay

Thank you

[upper karma]

first is a quadratic in the variable cos(theta)

[kindred rapids]

Yes but it couldn't factories...

Factorize*

It's a quadratic yes...but then what do I do..with the quadratic?

I pulled out the common factor of that and I got (cos Ø) (3cosØ + 4)

But then when I try to find theta with the alpha thing, I'm getting error in the calculatator

So I'm not sure what to do...

And the second question...I got a quadratic but it cannot factorize...

[dark sparrow]

3 cos(theta) + 4 = 0 has no solution

also can you show your quadratic for c?

[kindred rapids]

sin^2 Ø + sinØ - 3 = 0

I replaced cos^2 Ø with 1 - sin^2Ø

Then worked it out and ended up with that quadratic ☝️

[dark sparrow]

please don't use Ø for theta

[kindred rapids]

Sorry

[dark sparrow]

anyway ok

so substituting x = sin(theta) we have x^2 + x - 3 = 0

no we don't

we have sin^2(θ) + 2sin(θ) - 3 = 0.

[kindred rapids]

I dont see how we have 2sin(theta)

[dark sparrow]

2cos^2(θ) = 2(1-sin^2(θ))

wait shit fuck

put the two in the wrong place

it'll be on the squared sine not the plain sine

[kindred rapids]

but we will get 2 - sin^2(theta)

[dark sparrow]

no

2(1-a) is 2 - 2a not 2-a

[upper karma]

Hi I need help

I need he,o with geometry

Help

just post

[cinder nacelle]

The sum of angle and 4 times it’s complement is 20 degrees greater than the supplement of the angle ?

Please help

[glacial dawn]

Solve

[cinder nacelle]

X+4(90-x)=20+180-x
x+360-4x=200-x
-3x=-160-x
-2x=160
X=80?

[glacial dawn]

Lemme see

My equation was bad sorry

[cinder nacelle]

No problem

[glacial dawn]

I think what u did is right

80 + 4(10) = 20 + 100

120 = 120

[cinder nacelle]

80+10(4) =120
180-80+20=120

Yeah thanks

[glacial dawn]

Np

[random flume]

@cinder nacelle

[cinder nacelle]

Normal font

[pearl lava]

Is number 1 and 3 no

[supple onyx]

Consider the first triangle in 1. Then make the angle between the two sides with dashes in them bigger. Youll get a triangle with two sides the same but obviously not congruent triangles

As for 3, are the angles indicated equal?

[zenith dew]

I dunno if this goes here but any help?

[limpid gust]

can someone help me with this proof

[fallow tusk]

hey guys
so when doing cast diagram
the formula for positive value is thetha, 180- thetha, 180+theta , 360 - thetha
what about negative value
what is the formula for cast diagram
i know that it is a bit different

[kindred rapids]

@dark sparrow Hello...you were helping me, please can you continue to help me?

I got through....yes it will be 2 sin^2(theta)

Then we will put those two brackets = 0 and then find for theta in both

But when I put into the calculator...I'm getting error for one of the equation =0

What do I do..?

[upper karma]

What is the problem?

[kindred rapids]

Go further up....in the chat

We talked a while ago

The question and everything is there...

[upper karma]

So uh

$2\cos²(\theta)+1-\sin(\theta)=0$?

And you tried to substitute (1-sin²(theta)) by cos²(theta)

Wait for the bot

[kindred rapids]

The bot..?

[upper karma]

Texit is a bot that renders code into mathematical language

[kindred rapids]

Oh okay

[upper karma]

Dang

It's taking it's time

Anyways

[kindred rapids]

Also I didn't substitute....i factorised

It was a quadratic

[upper karma]

Yeah but you should've substituted before hand right?

[kindred rapids]

Actually I did

Sorry

I didnt read that well

[upper karma]

And after that you get a quadratic which you can factor

[kindred rapids]

Yes

[upper karma]

Can i see what your quadratic looked like before failing to factor?

[kindred rapids]

No I factorised it fine...Ann corrected me and factorised it properly this time...

I*

(2sin(theta) + 3 = 0 ) (sin(theta)) - 1 = 0 )

Those are my two brackets when I factorised...

Quadratic = 2sin^2(theta) + sin(theta) - 3 = 0

[upper karma]

Yeah so

[kindred rapids]

But now...I have to find the values for theta in both brackets
First bracket = Second bracket =
sin(theta) = (-3/2) and sin(theta) = 1

Oh

So what's my next step?

[upper karma]

That's why i ask for how your quadratic looked like before factoring

[kindred rapids]

Okay

[upper karma]

Yeah okay it was a misunderstanding

So everything is fine BUT you have to notice that the domain of arcsin is (-1,1)

So sin(theta)=-3/2 won't give real solutions to our problem

[kindred rapids]

....so what do I do with that sin (theta) = -3/2..?

[upper karma]

That equation does not give us any real solution

So we are only interested in solving sin(theta)=1

Which should be very remarkable

As it is known which angle gives 1

Do you recognise it?

[kindred rapids]

90°

[upper karma]

Yeah π/2 in radians

[kindred rapids]

Yes

[upper karma]

Is it asking for all the solutions or just on the interval between 0 and 2π

[kindred rapids]

...my teacher styled the question herself.....we are instructed to just get the values of theta

We just have to get the answers in degrees....not radians yet

[upper karma]

Can you copy the exact statement of the problem without skipping any words

Okay yeah so theta=90°, it does not matter

[kindred rapids]

Okay ...so the first question I sent...could that be worked out..?

[upper karma]

Wdym?

Which first q

3cos²(theta)+4cos(theta)=0?

[kindred rapids]

Yes

It was part b ...I worked it out but it just like this question we just did...

Only one set of brackets worked out and the other didnt...which means, just as you said that it didnt give a real solution...

[upper karma]

The only way i can see if you are correct or not, is if you show me your workout

Or your final answer

[kindred rapids]

Okay

I'm sorry but it's on scrap paper so...it might be a bit messy

Sorry those scratches at the underneath is theta

This is the way my teacher taught me btw

I'm sure I went right...doing this....everyone else got this.
And I guess it's similar to the previous one....where one doesnt give a real solution...

[upper karma]

Yeah arccos(-4/3) isn't defined over the reals

So yeah that's correct

(assuming you want the solutions under such intervals)

[kindred rapids]

Alright thank you...so much for your help...

[upper karma]

Yw!

[kindred rapids]

I really am grateful...

:)

[cinder nacelle]

The bisectors of adjacent angles are opposite rays

Always true
Sometimes true
Never true ???

[wide venture]

what is next step

[cinder nacelle]

The bisectors of adjacent angles are opposite rays

Always true
Sometimes true
Never true ???

[polar locust]

hey guys

can you help me with this question?

Explain how the graph of y = sinx, over [0 º, 360 º) shows that every sine ratio in the interval has two angles that share the same ratio.

[upper merlin]

draw horizontal line through graph, it always intersects graph twice

[hidden crescent]

Yes

[upper merlin]

this means that for the constant y represented by the horizontal line, there is always 2 x values that satisfy y=sinx

[hidden crescent]

this is why there is ambiguous cases for triangle with SSA

[upper merlin]

^

unless its right

[hidden crescent]

yee

[upper merlin]

wait i just realized 1 or -1 dont have 2

they only have pi/2 and 3pi/2

[tired knoll]

can somone help me with this problem

The ray of light, emitting from point M, reflects the line AB at point C∈ segment AB, and then goes through point N so that m∠ACM=m∠BCN. Prove that the bisector of the ∠MCN is perpendicular to line AB.

<@&286206848099549185>

[obsidian thicket]

Somebody help me

[tired knoll]

um I can try if you would like

@obsidian thicket so what are you confused about

oh wait btw we dont help people with quizs cus thats cheatin

[obsidian thicket]

Its not a quiz

That's what my teacher puts everytime but we get unlimited chances etc I can show you

[tired knoll]

ah no Ill just trust you

so what are you confused about @obsidian thicket

well I have to go to, if your still confused I recommend drawing it out (and dont forget you can use algebra when trying to solve a geo problem)

[obsidian thicket]

I just dont know how to do it

It's ok bye for now

[tired knoll]

have you drawn it out?

I need to go soon but im still here for the next min or two

[obsidian thicket]

Idk how to draw it out I do school online and its muy difícil para mi

Very hard for me to do

[tired knoll]

well it says triangle DEF so you should draw a triangle and lable its vierticies D E and F

if a problem has two letters together it means a line segment for instence segment/side DE

[fresh pike]

i dont quite understand how to figure out which optiont he information is trying to prove

[livid moss]

8 and 4 are alternate angles between AD and BC with AC as a transversal

So if they are equal, then the lines are parallel

[fresh pike]

so which option would i choose to show that?

[silent plank]

its not really about showing
its applying the theorem(s) concerning alternate angles on parallel lines

[muted spoke]

find the signs of a,b,c like - or + and -b/2a ,,, -D/4a are coordinates of the topmost point

anyone

[silent plank]

consider sum and product of roots and concavity

[muted spoke]

i didnt understand

[pale dew]

how can i do this without CosA - CosB = 2cos(A+B)/2* cos(A-B)/2

[next jackal]

@pale dew I might be wrong here but can you see 13π/24 as 12π/24 +π/24?

[limpid gust]

"write the reason for each conclusion", does that mean the property of equality?

[humble pulsar]

@limpid gust what do you mean?

[limpid gust]

@humble pulsar

is this what I am supposed to do

[humble pulsar]

Yes

[limpid gust]

alright is the last one the subtsitution property?

i dont understand that one

[humble pulsar]

Maybe? I'd personally say it's the same as whatever you put for 10

[limpid gust]

yeah i was thinking that to

[humble pulsar]

wait no scratch that

complementary doesnt mean equal

[dark sparrow]

#13 is not even true

your answers to 11 and 12 are wrong

[humble pulsar]

oh shit yeah

[dark sparrow]

12 should be multiplication property of equality or something similar, 11 i'm not 100% sure on the exact wording but i can tell you with certainty that transitive property ain't it

[limpid gust]

how is 11 not transitive

[dark sparrow]

transitivity has precisely nothing to do with addition

it simply is not what the statement is talking about

[limpid gust]

Oh 12 is multiplication

[dark sparrow]

maybe additive property of lengths?? i don't know what name your teacher expects you to use.

[limpid gust]

probably addition property

[humble pulsar]

probably, we dont know what your teacher uses terminology wise

[limpid gust]

Then what about 13

[humble pulsar]

13 isnt true so it's likely a typo

angle 2 = angle 4

not angle 1 = angle 4

[limpid gust]

is this the same thing?, are the conclusions properties again or something else

like would 15 be midpoint formula

[humble pulsar]

no clue cause you can make multiple conclusions

[next jackal]

how did we get those highlighted terms?

[dusky surge]

I guess it's from the "intuition" of v²-u² =(v+u)(v-u) =2xy

[next jackal]

ohh

thank you

[dusky surge]

😁

[next jackal]

wait that didn't clear my doubt

[limpid gust]

[dusky surge]

Like $$v^2-u^2 = 2xy$$
$$(v+u)(v-u) = \sqrt2 \sqrt2 xy$$
$$\frac{1}{\sqrt2}(u+v) \cdot \frac{1}{\sqrt2}(-u+v) = x \cdot y$$

[somber coyote]

Biscuit:

[next jackal]

Oh okay got it

thanks a lot

wait how did the book factor between x and y?

@junior light

[junior light]

Which portion is not clear?

[next jackal]

wait how did the book factor between x and y?
this one

can you once check the screenshot?

[junior light]

I'm looking at it but seem to be missing the context.

[next jackal]

okay, do you have the copy?

[junior light]

Anyway, I think you should check the angles and draw a picture.

I think I have, let me see.

Oof I have so many legitimate books it'll take time browsing through them.

Nvm I'll get it.

[next jackal]

Oh okay

[junior light]

Alright, so it's the minus sign in front of u that is bugging you?

Basically, the second part(the one you highlighted) is about proving that the curve v^2-u^2=2, when rotated by -pi/4 counterclockwise, i.e., pi/4 clockwise gives you xy=1.

[next jackal]

not just that, how did x and y get factored like whatever it is in the screenshot?

[junior light]

Mmmm check the rotation matrix?

[next jackal]

okayy

[junior light]

I don't exactly understand why Lang is trying to prove the converse, the first part was sufficient to justify the curve P' is obtained by rotating the curve P by pi/4.

[next jackal]

Me neither 😕

[dark sparrow]

what are we proving exactly

[junior light]

Let me snip out the theorem.

[dark sparrow]

okay

do we know that the image of (x,y) under G is ((x-y)/sqrt(2), (x+y)/sqrt(2))?

[junior light]

Yes.

[next jackal]

yes

[dark sparrow]

okay,
do we also know that the image of (u,v) under G^-1 is ((u+v)/sqrt(2), (u-v)/sqrt(2))

[next jackal]

wait I got it

just had to rotate it back again

I'll do it

Thanks

[silk patio]

What’s the topic exactly?

[next jackal]

rotation of hyperbolas

[pseudo quiver]

o

[junior light]

just had to rotate it back again
@next jackalYes, that was the idea, but I'm still not convinced why it is needed.

[silk patio]

Here’s a cool thing, the discriminant is invariant under rotation

[next jackal]

Yes, that was the idea, but I'm still not convinced why it is needed.
me neither

[silk patio]

You rotate xy and get x^2-y^2, they both have discriminant -1

In fact any special linear transformation will preserve the discriminant

[dark sparrow]

you are proving a two way set inclusion

G(H) = H'

you need to prove G(H) subset H' and then H' subset G(H)

the latter can be more easily proved as G^-1(H') subset H

[junior light]

Owww, set inclusion makes sense.

I was just perceiving it as rotation of a curve and couldn't see why the curve should be anything else.

[silk patio]

You can apply G like 3 times instead of doing the inverse

[dark sparrow]

7

[silk patio]

Rotational symmetry

[dark sparrow]

yeah but in general G^-1 is G^7 not G^3 lol

[silk patio]

Yeah but as a set it’s embedded when you apply it 3 times

[junior light]

How is it 7 in general?

[silk patio]

G^8 = a full turn

So it does nothing

G^4 is a half turn which also leaves the set invariant

[junior light]

Only if G is rotation by pi/4, right?

[silk patio]

Yeah

[junior light]

Okay

[grizzled lantern]

E is a 90 degree angle

[arctic cobalt]

Can someone help me with this the topic is rhombuses and finding the angles and sides. This worksheet asked me to find the measure of angle BCD and the measure of DE

[grizzled lantern]

180-(53+90)=????

thats your third angle

[arctic cobalt]

Why is it 180 for Angle BCD I thought the angle was equal to 90

[grizzled lantern]

Ok so think of that rhombus as 4 triangles

look at triangle AEB

AEB is the same as the rest of the other 3 triangles that make up the rhombus

so AEB's angles equal all the others

[arctic cobalt]

So angle BED is 37 90-37 is equal to BCD right

[grizzled lantern]

BED is a line

I think your angle is supposed to be in the middle

lemme send you a photo

[arctic cobalt]

Sorry Angle BEA is congruent to BEC

[grizzled lantern]

does that help?

[arctic cobalt]

I think I get solving angles but how do I find side DE

[grizzled lantern]

pythag

CE is equal to 9 and your hypotenuse is 10

9^2+b^2=10^2

[arctic cobalt]

How did you get 9

[grizzled lantern]

whats half of 18?

so look, AC=18 EC is half of AC

[arctic cobalt]

Oh ok that makes a lot of sense thank you

[grizzled lantern]

yep

[odd blade]

I need some help with a word problom

Michele phone bill for 30 days was 350 dollers how much should her bill be for 16 days

I did 350/30 x/16

Is it right ?

[grizzled lantern]

350/30= #'s

16*#'s gives u the answer

[orchid hearth]

Hi! I would really appreciate some help this 'prove that' question involving the addition formula

[odd blade]

So not set up like that ?

Then I just gotta multiply divide and round

[grizzled lantern]

@odd blade probably more like 350/30 =16X

[odd blade]

And how whould I go to solve it ?

Because when I add the fraction I can cross multiply then divide then round

[grizzled lantern]

you could find what 350/30 is and then divide by 16 if your teacher lets oyu

you

do you need it in fraction form?

[odd blade]

Nope just need to round it

[grizzled lantern]

ok

[odd blade]

I just have trouble setting up word problems the solving is easy

[grizzled lantern]

so 350/30= 11.67

[odd blade]

So her phone bill for the 16th day will be 11.67 dollers ?

[grizzled lantern]

hold up

[odd blade]

When I did my method I got 186.666

Since it’s money I need round that but my rounding is terrible

[grizzled lantern]

im so sorry

your method works

yeah so keep your method and cross multiply

you should get 30x=5600

[odd blade]

Yup thanks

Now I am completely stuck on this one

[upper karma]

https://prnt.sc/vesdse

[chilly moss]

Confused

[next jackal]

Are you allowed to use calculator for this?

Do you know the definition of sine?

[urban knoll]

Anyone active?

[chilly moss]

I do, and I think I got it

5740 feet

[urban knoll]

Can someone help me w these?

[chilly moss]

90-64=?

[urban knoll]

Thanks

Ur smart

[chilly moss]

Np haha

[urban knoll]

Got that in like 2 seconds

[upper karma]

Test?

[chilly moss]

Nope homework

[urban knoll]

Nah just homework

[chilly moss]

Same

[upper karma]

Micheal try to look up definition of sine and cosine

[urban knoll]

I have 14 more questions and I don’t know how to do them

[chilly moss]

I have one more question

How do you find h?

[upper karma]

AD = AB + BD

[chilly moss]

Ah

[upper karma]

You can express BD in terms of h

Also AD

[urban knoll]

How would I find x would I just do some sort of 2 step equation?

[grizzled lantern]

112=4x

[hollow raven]

opposite angles are equivalent

[urban knoll]

112 isn’t a choice tho

[upper karma]

?

[hollow raven]

You would need to isolate for x

[urban knoll]

So I would need to do a 2 step equation?

[hollow raven]

If that's what you need to do then yes

[grizzled lantern]

lol

[upper karma]

dude just 4x=112 and x=112/4

its not magic

[limpid gust]

Can someone help me what I have circled is what I have done so far, its a 2 column proof

[upper karma]

How would I find x would I just do some sort of 2 step equation?
@urban knoll idk if im right but just do 112/4

since opposite angles are congruent or equal

[limpid gust]

if AB is between AC and AT to form CAT, will CAB be congruent to BAT. explain

[upper karma]

could you show a picture if any

[limpid gust]

Thats the problem

in words

theres nothing else

@upper karma

[upper karma]

oh

then im guessing u got to make the triangle urself

im just confused on whats T

[limpid gust]

I dont think you get this

[upper karma]

ye i dont

[inland canyon]

bruh

can someone help me rq

ill type it out

Billy is building a frame for his fathers garden in the shape of an isosceles trapezoid. The lengths of the parallel sides are 10m and 14m. The preimeter is 40m what are the lengths of the nonparallel sides?

[silent hollow]

am i?

[upper karma]

lol

[hollow raven]

@inland canyon the nonparallel sides are equal because it is an isosceles trapezoid
P = a+b+c+d, P = 40 m
a = 10 m, b = 14 m, c = d
40 = 10+14+c+d
Substitute side D as side C
40 = 10+14+c+c
40 = 24 + 2c
Just solve for c

[rotund current]

I'm given this, does anyone know how to solve this?

[lime hawk]

im so lost

[young knot]

I have an exam on Tuesday and we have to understand this but I have 0 clue. I tried it once and my diagram was wrong on so many levels and I can’t work out how to draw this or do the questions.

[silent plank]

what was your attempt at a diagram?

@young knot

[young knot]

I only managed the girls part but got it wrong one sec I’ll screenshot

This is all I had but it’s badly wrong

That’s only for the info given on the girls

I think I put the 25 degrees in the wrong spot

[silent plank]

consider starting with something like this

involve the bearings later

[young knot]

I’m just confused at the right angles because I would’ve thought those spots would be 40 and 130 degrees but I’ll try it thanks

[silent plank]

one sec i'll give something more detailed.

your diagram should somewhat what's happening in 3D

B,O,G are on the ground,
T is the top of the tower and extends upwards.

[young knot]

That makes more sense now, thanks a lot

The questions should be easy enough, thanks for your time and help

[mystic pebble]

I think this is the correct channel to ask

[upper karma]

@mystic pebble what have you tried

[mystic pebble]

@upper karma This is my working so far

Let me put it into words so you don't have to look sideways

[upper karma]

,rccw

[mystic pebble]

wow

[upper karma]

Yeah no i have no clue on what you did

That is ilegible

[mystic pebble]

I tried to multiply both sides by sin^2(theta)

So I got = 2

then I tried using sin^2(theta) = 1-cos^2(theta) on the numerators on the left

So I ended up with (1-cos^2(θ))/(1+cos(θ)) + (1-cos^2(θ))/(1-cos(θ)) = 2

Which I think is true?

But I didn't know where to go from there or if I was even going down the right path

cc@upper karma

[upper karma]

Alright

When doing proofs as these, we don't want to modify both sides, but instead only work with one side

But the idea is definitely that

Try to only apply common denominator on the LHS

i.e express the LHS as one fraction

[mystic pebble]

Okay, so it would be come:

(1-cos(θ))/(1+cos(θ))(1-cos(θ)) + (1+cos(θ))/(1-cos(θ))(1+cos(θ))

Which can be turned into

(2)/((1+cos(θ))(1-cos(θ))

And from there

[upper karma]

$\frac{1-\cos(\theta)}{(1+\cos(\theta))(1-\cos(\theta))}+\frac{1+\cos(\theta)}{(1+\cos(\theta))(1-\cos(\theta))}$

[mystic pebble]

Why is it 1-cos(theta) as the numerator for both of them?

[somber coyote]

Al𝟛dium:

[upper karma]

Typo

[mystic pebble]

Oh

[upper karma]

But yeah

[mystic pebble]

(1+cos(θ))(1-cos(θ)) becomes sin^2(θ)?

Thank you very much 🙂

[upper karma]

Yes, good job

Yw

[mystic pebble]

So can you always do these identify proofs by changing only one side?

I proved the one after this but by re-arranging both sides into another identity

So maybe this is the wrong way of doing it? I'll send a picture

@upper karma

[upper karma]

I'd recommend only to one side

I don't know if your teacher enforces you to do so

[mystic pebble]

Would you be able to quickly go through how you would prove 1/sin(x) - sin(x) = cos(x)/tan(x)

[upper karma]

Yours isn't wrong but just not recommended

[mystic pebble]

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

I managed to change the right side to this

[upper karma]

First choose one side

Yes the RHS looks better to choose

[mystic pebble]

1/sin(x) - sin(x) = (1 - sin^2(x))/sin(x)

[upper karma]

Yes

[mystic pebble]

dsdasd

Okay yeah I got it

Then I just simplify the fraction on the right

[upper karma]

Yes

[mystic pebble]

You just being here makes me solve it 😂

Ty man, you have been a big help

[somber coyote]

Al𝟛dium:

[upper karma]

Yw!

[tired knoll]

hey can someone help me with this problem

The ray of light, emitting from point M, reflects the line AB
at point C∈ AB, and then goes through point N so that m∠ACM=m∠BCN. Prove that the bisector of the ∠MCN is perpendicular to line AB.

[upper karma]

not drawn to scale

We are given that m∠ACM = m∠BCN and that there is a bisector of ∠MCN.

Since the bisector bisects ∠MCN, ∠MCD ≅ ∠NCD.

Notice that ∠ACD and ∠BCD are a linear pair.

m∠ACD + m∠BCD = 180°

m∠BCN + m∠NCD + m∠BCN + m∠NCD = 180°

m∠BCN + m∠NCD = 90°

I hope I answers your question correctly.

[haughty karma]

Does anyone know how to do this I'm stuck

[upper karma]

hint: AXY and ABC are similar

[haughty karma]

10/2?

XY = 5cm @upper karma

[upper karma]

yes

[haughty karma]

Alright thank you

[hot rain]

@upper karma too much proof. could have been shorter

[upper karma]

You're right, thanks for telling me.

[haughty karma]

Help

Please

[olive cove]

The area of a triangle is base multiplied by its hight divided by 2

So if the base is twice as small the area should also be twice as small.

[haughty karma]

Ok thank you @olive cove

[upper karma]

Im confused when it comes to proving congruency in triangles. What order would you look at the letters, for example if there was HJK or something

which one would i count first

the angle or the side?

Depends on the location of the third side or angle.

which order would i look at them in?

We are trying to prove the triangles are congruent, right?

yeah

Look at BF. What can you say about the line?

both triangles share line BF

Correct, meaning we should have two congruent sides and one congruent angle.

so whenever you see a line like that there's always gonna 2 congruent sides and 1 congruent angle?

oh wait no

Since the two triangles are sharing the same side, that shared side is congruent to itself.

but then how would i order then

i cant do ASS

SSA?

I think so.

Is SSA a valid option?

wait no

so that leaves sas

im confused

Sorry, I didn't type fast enough.

SSA is not a valid condition for them to be congruent.

yeah so i thought that SAS was the only option

but i think thats where the order comes in

and im just confused on how to order them

Well, I like to look at it in counter-clockwise or clockwise, depending on the triangle.

Let's look at the bottom triangle going clockwise starting with an unmarked side.

Does it help to identify if it is SAS or SSA?

tilting my head clockwise it makes it look like a regular triangle but i still cant figure it out

Oh, sorry, Ima clarify.

Just look at the unmarked triangle side BC.

https://media.discordapp.net/attachments/326138757474680852/774740847521693716/unknown.png

Only take note of congruent parts.

Going clockwise, we see BF, FC, then angle C.

Thus, SSA.

ohh

i can see that

but then why did the question say the answer was "These triangles cannot be proven congruent."

Lemme list of the conditions for two triangles to be congruent.

SSS
SAS
ASA
AAS
HL

It specifically has to satisfy one of these five.

oh ok

However, we got SSA, which isn't in the list.

We can't make anymore conclusions about the other sides and angles being congruent, meaning SSA is all we have.

Therefore, the triangles are not congruent.

I hope this helped you.

yeah it did a lot tysm

[night basin]

Find a point that is equidistant from the following:

( 3, 12 ), ( 5, -2 ), and y = −4

i dont have graph paper so this is diffucult for me to visualize

[upper karma]

my mans on that ixl

IXL pog

[unkempt wasp]

so I have a function of two parallel lines. How would I get two separate functions for each line?

can't really find any information on this on the internet

is this even geometry?

[upper karma]

Looks like a conic @unkempt wasp

[unkempt wasp]

it's 2 parallel lines

4x^{2}+4xy+y^{2}-12x-6y+5=0

[upper karma]

Yep, a degenerate conic

[unkempt wasp]

ohhh ok

I know what to look for now. thanks

[upper karma]

No prob

[weary barn]

where can i find an exhaustive list of trig identities
for example : cotx - tanx = 2 cot(2x)
another example : product of sin((pi k)/n) from k=1 to k= (n-1) is n/(2^(n-1))
I mean the rare identities you don't see usually

pls ping

[dark sparrow]

@weary barn if you're looking for an exhaustive list you're out of luck

[weary barn]

oof

they turn up in my test

and I am supposed to somehow know them beforehand

like 5-6 new things every test including 1-2 new rare indetities

then during the test discussion, the prof says u are supposed to know them before giving the test

[summer spire]

what textbook are you using?

[dark sparrow]

and I am supposed to somehow know them beforehand
??

why

[weary barn]

what textbook are you using?
@summer spire none

Nothing except class notes

and a lot of practice problems (from weekly tests)

btw I am in senior year of my HS (I learnt trig last year tho, only tests this year)

and the tests are very tough
shall I give some examples of problems so someone can suggest me how to prepare?

[silk patio]

Show me problems where you need rare identities

[plain wyvern]

@weary barn Ah! I would like to see, how tough are these problems really!

[sinful jay]

does anyone know how to match graphs and equations for sec csc cot and tan? iv been trying to look up things to understand but just dont get it

[humble pulsar]

Probably asymptotes @sinful jay

[stark marsh]

and the tests are very tough
I smell JEE

[upper karma]

Nothing unusual

Product of sines is not something you memorize

Not everything can be taught or compiled in a book

Also "nothing except class notes" is risky unless you are in Sankalp or similar batches in other institutes

[weary barn]

I smell JEE
@stark marsh yes

Also "nothing except class notes" is risky unless you are in Sankalp or similar batches in other institutes
@upper karma I am in top batch in ALLEN

[upper karma]

Good

You shouldn't be asking that, then

[stark marsh]

lol

[weary barn]

lol ik

[dark bison]

how do you get pi/6 from arcsin(0.5)?

[earnest echo]

You need to remember that

[dark bison]

cant you get it in calculator?

[earnest echo]

Yes, Calculator can do that for you

[silent plank]

depending on your calculator, it may not output they symbol pi

[arctic cobalt]

The question asks for the length of side RV and SU how would I go about solving this

[earnest echo]

Did you find the length of the diagonal??

[arctic cobalt]

That’s what I don’t know how to figure out

[earnest echo]

Can you see that triangle SUR is right angled?

[arctic cobalt]

Yeah

[earnest echo]

How do you find hypotenuse of a right angled triangle if two perpendicular sides are given?

[arctic cobalt]

I’m pretty sure you use a^2 plus b^2 equals c^2 but I’m not sure how to put the equation together

[silent plank]

plus: +
equals: =

a^2 plus b^2 equals c^2
→a^2 + b^2 = c^2

(where a and b are you legs and c is the hypotenues)

[earnest echo]

First step is to recognise what is your a ,b and c is here

[arctic cobalt]

I’m pretty sure side SU is c because it is the biggest

[silent plank]

if you're familiar with properties of right isos triangles you could potentially take a shortcut

[arctic cobalt]

Wait so if c is SU that means the equation would be 25^2+25^2=su which would give me 1250

[silent plank]

=c**^2** (not c)

[arctic cobalt]

Then what would the equation look like?

[silent plank]

25^2+25^2=(SU)^2

[arctic cobalt]

Oh nvm the variable tripped me up

[orchid perch]

What's the deal with square roots and the plus/minus sign? I never know when teachers want me to give both answers or only the positive answer. I get that 64 = 8^2 and (-8)^2, but i never seem to know when they want the negative answer

[dark sparrow]

-8^2 is -64 not 64

you meant (-8)^2

[orchid perch]

yes

[dark sparrow]

when you write $\sqrt{x}$ it refers to only the positive root

[somber coyote]

Ann:

[orchid perch]

$\sqrt{x}$

[somber coyote]

Yonder:

$\sqrt{x}$

[orchid perch]

babbys first latex

[humble pulsar]

👏

[somber coyote]

Yonder:

thanks for the help, i'm not sure if my question's been fully answered. what would it look like if they were referring to the negative root? $-\sqrt{x}$? or how about both?

[orchid perch]

i've had a pretty good run of just imitating like I know math, but I never seem to know when they want me to provide the positive square root, the negative one or both

it's hard to describe

i'll just get points off and i don't know why

[dark sparrow]

the negative root is $-\sqrt{x}$ yes

[somber coyote]

Ann:

[dark sparrow]

if you want to refer to both at once it's $\pm\sqrt{x}$

[somber coyote]

Ann:

[orchid perch]

ok. thanks for your time @dark sparrow

well wait, is $\sqrt{64}$ = -8?

[dark sparrow]

no

$\sqrt{64} = 8$

[somber coyote]

Ann:

Yonder:

[orchid perch]

well why?

[humble pulsar]

x^2 = 64 -> x = +/- sqrt(64)
but sqrt(64) just equals 8 since that's the convention we use

[orchid perch]

when I see $\sqrt{x}$ should I just assume that they always mean the positive x root?

[somber coyote]

Yonder:

[humble pulsar]

The convention is that square root is positive unless the minus sign is explicitly stated

[orchid perch]

okay ty thanks for your time

[upper karma]

which one is wrong?

[earnest echo]

Check the first one again

[upper karma]

How would Log(A)/Log(B) expanded look like?

I always thought it would be Log(A) - Log(B)

[earnest echo]

log(A/B)=log(A)-log(B)

[upper karma]

oh yeah, right.

[upper karma]

can someone explain how this is wrong

i dont get it

[round isle]

wrong channel

maybe try #proofs-and-logic

[foggy delta]

o

[haughty karma]

Hey guys i need help on these quetions

[round isle]

well, the last question is simple. Two angles on a line must add up to 180

[haughty karma]

but what's the equation

cuz we can't use two variables

[round isle]

ah yes you're right

[haughty karma]

3x + 100 + ? = 180

[round isle]

hmm lemme see

oh right

so, all angles inside the 4-sided shape must add up to 360

get all the four angles in terms of x

[haughty karma]

yes

get all the four angles in terms of x
@round isle

what do you mean by this?

[round isle]

? = 180 - 3x - 100

do the same for the other three

[haughty karma]

oh rigth

[round isle]

you get four angles in terms of x

[haughty karma]

thatnk you

what do you mean by that tho

[round isle]

but I don't understand the remaining questions

[upper karma]

Who understand this like

Explain

[sinful jay]

Does anyone get how this math makes sense?

[upper karma]

I am so confused, would love help

[humble pulsar]

,rotate

[humble pulsar]

@upper karma what have you tried for 23?

[upper karma]

I figured it out thanks though @humble pulsar

[silver spruce]

does anyone know how to solve

[pliant dawn]

What is this triangle looking symbol called and what is it

[round isle]

it is the angle at vertex C with vertex D and E being the neighbors of C

similar for CAB

[upper karma]

bit confused on how to start this

[livid moss]

Well every line has an equation lf the form y = mx + c

Now you are given two pieces of information, a point and that it is parallel to another line

So use that information to find m and c

[upper karma]

thanks

ok so

@livid moss isnt c 5?

I havent learned this formula so im kind of confused

[livid moss]

No

c is 5 for the parallel line

You are working out the equation for a different line

[upper karma]

ohhh

parallel = same slope

so

i got

[silent plank]

that does not have the same slope as y=x+5

nor does it pass through the point (2,3)

[livid moss]

parallel = same slope
@upper karma what happened to this? I thought you were on the right track?