geometry-and-trigonometry

[simple turtle]

Ah I get the picture here

[dusk stag]

this is right

X=70
because X+110= 180
then X=70

[obsidian harness]

How

[dusk stag]

any squre's oposite angle's addition = 180

oh w8

i did wrong

[upper karma]

Of course not.

[golden fiber]

Could you expalin it more to me???

I mean this task

[upper karma]

Which one?

[upper karma]

Where are you having problems? In proving that it's a parallelogram?

[golden fiber]

Yes

[upper karma]

How did you go about doing it?

[elder halo]

is there a program where I can draw euclidean geometry

[simple turtle]

GeoGebra

[golden fiber]

I dont know how to prove it

[upper karma]

Hmm, wait don't disappear again.

Join O_1O_2 and X to O_2. Do you see congruent triangles?

@golden fiber Bruder!

[golden fiber]

I see

But whats next

[golden fiber]

@upper karma

[main fiber]

I’m lost on finding our points to plot I can find period and quarter period just fine

[gray pivot]

can you use calculator

[main fiber]

No

Not allowed

My teacher also teaches ap stats and ap calc so he extra hard in normal trig

I know first I need to isolate theta so I divide by 1/3

And new problem is now y=4tan(theta+15pi/6)+2

[gray pivot]

would it be possible to break up tan into simpler functions

[main fiber]

Which can be y=4tan(theta+5pi/3)+2

[main fiber]

I don’t know

[gray pivot]

you sure thats correct?

[main fiber]

Yes I know it is correct

[gray pivot]

so 6/3 is 3?

[main fiber]

2 sorry

[gray pivot]

so its not correct

[main fiber]

Yes I wrote 2 on my paper tho

But that is barley the start of the problem

Quester period for tangent is pi/b

Which is pi/1/3

[gray pivot]

good luck idk how to help you from here

dont think my teacher has covered this

or if he will this year

[main fiber]

Dang

My class isn’t even honors or nothing

Let me start the problem and show where I get stuck

I will make more neat

[gray pivot]

well im in ap precalc and doing quite well but we havent covered any trig yet

so i only know the basic stuff learned in geometry

[main fiber]

I’m usually really good at math but I’m not sure what to do now

I’m stuck here I don’t know how to find points and graph the function

[gray pivot]

maybe try just plotting the points where y does not equal undefined

you can approximate pi for the y value

as 3.14

its just visual

shouldnt really matter

nvm all the y values are integers

so i dont really see the problem

your x axis is alreadyin terms of pi

[main fiber]

I’m supposed to use quarter period and phase shift

To find points

And use amplitude

For y axis

Ahh I’m cooked

[ivory hill]

Looking for a synthetic proof of this

[golden fiber]

@upper karma and whats next?

[broken hare]

Guys i have a doubt in geometry question

Can anyone explain me how to solve this

Please if u tell solution ping me otherwise i wont get notified

👽

[prime schooner]

HUH

https://tenor.com/view/calculation-math-hangover-allen-zach-galifianakis-gif-6219070

[broken hare]

U want figure

.?

@prime schooner

.

?

[elder halo]

I can't figure out on how to find BQX

[upper karma]

good enough?

[obsidian harness]

angles in the same segment, angle BXQ = 24

[upper karma]

tried my best handwriting lol

[elder halo]

Oh alright, thanks ❤️

[upper karma]

no probs

[golden fiber]

Hello guys could sb explain and help me with this?

[crude compass]

cuz they intersect

intersecting lines are congruent

[golden fiber]

Which lines?

[crude compass]

the one that passes through circle 1 is congruent to line xy, and line xy is also congruent to the line that passes through circle 2

so ax = by

by definition that intersecting lines are congruent

[golden fiber]

Could youwrite it down in more detail for me?

[golden fiber]

@crude compass

[outer wadi]

what does a more steep rectangular hyperbola mean

need it for chem rn we dint study it in maths yet

[main fiber]

I went into office hours either my teacher and finally got it

He explain so bad

[gray pivot]

rip

[main fiber]

It’s ok now though

[subtle onyx]

how do i solve this

pls i have quiz tomorrow and i wasnt here

[obsidian harness]

The correct similarity equation

Is 1/(2x +3 + 1) = x/(x + 2)

Draw out big triangle ADE

And label the sides

[subtle onyx]

how do you label the sides

[obsidian harness]

What's the length of AD?

[subtle onyx]

2x+3+1

[obsidian harness]

Exactly

That's why

Same for finding AE

[subtle onyx]

x+2

would that mean that ac=x

and ab=1

[obsidian harness]

Yes

[subtle onyx]

ty

[obsidian harness]

Np!

[ivory hill]

Anything at all?

[golden fiber]

@crude compass

[hard grotto]

Can someone solve those please?

[upper karma]

For q3 isnt the answer A? -1/2 ?

The question bank's answer key says its B??

question bank is wrong

[golden fiber]

@crude compass

[somber schooner]

how can I simplify this?

[ivory hill]

Isn’t it sin(2θ)

Cuz sin(x+1/2pi) is cos(x)

[faint pasture]

its just 0

its just cos(A-B) expanded

apply that

and you get cos(pi/2)

[ivory hill]

Oh yeah right

[somber schooner]

this is what my tutor did

but I dont understand how

[ivory hill]

Formula for cos(A-B)

They just subbed in the formula

[somber schooner]

oh right

ic

ty guys

[ivory hill]

Cos(A)cos(B)+sin(A)sin(B)=cos(A-B)

[somber schooner]

🙌🙌🙏☝🏿

[fervent timber]

Can anyone help me solve this please

[upper karma]

@fervent timber any tries?

[deft walrus]

If anybody can help me on my trig study guide much would be appreciated

I don't want the answers but to understand it so I can pass my test tmr

I'm dealing with graphs rn (my weakness)

[stoic dragon]

Can someone explain radian in circles for me?

[fervent timber]

Can anyone help me with this please

[upper karma]

,rccw

[upper karma]

Try it now

[tall coral]

I don't think this is right..
Since theta must be bigger than 45 degrees, that would imply that the shadow would be smaller than the height, but that's not what I'm getting...

[viral lagoon]

ngl, screw math 🤣 Business >

[tall coral]

This is not helpful but sure

[viral lagoon]

yeah sorry thats too complicated for me lol

[acoustic field]

are m<3 and the angle adjacent to it supplementary

[upper karma]

Yes

I don't think adjacent is the right term for those angles

[acoustic field]

im only in middle school, still sped

[upper karma]

oh

:P

I'm not sure

[upper karma]

i’m not sure how you got 17.2

try it on a calculator or somewhere

[main fiber]

Hello I am also doing this now

But I am doing it with tangent and co tangent

Find period and quarter period

Quarter period +c= starting point then just keep adding quarter for your next point

[crude compass]

its an arc

that has a length of r

[upper karma]

sin, cos, tan, cot, sec, cosec in triangles is just a top of the iceberg, agree?

[obsidian harness]

agreed

just knowing what they are doesn't tell you anything about how you can use that knowledge

[upper karma]

I just learned sine rule and found out that sin, cos, tan and cot are not only in right triangles

[obsidian harness]

indeed

time to look into sine and cosine rules

[abstract plaza]

Can someone help me in this?

[obsidian harness]

what even is that

[crude compass]

is that an ellipse

or a football

[abstract plaza]

Ellipse

(1,9)

[crude compass]

whut r u even lookin for

[abstract plaza]

Equation of the ellipse

[crude compass]

u need value of a and b

or at the very least coordinate of a vertex and a covertex

[obsidian harness]

or the two foci

and a point on the ellipse

[abstract plaza]

I have been given just the coordinates (1,9) and the white line is the center of the graph

[crude compass]

your professor is wrong

[abstract plaza]

Why is that sir?

[crude compass]

u cant find equation of an ellipse without values a and b sir

so we can only work with

[abstract plaza]

I see

This is also what he left us with

[crude compass]

(x-1)²/a² + (y-9)²/b² = 1

[abstract plaza]

He wanted us to identify the a and b in this drawing

[dense smelt]

Can I find Y using similar triangles like this

[tall coral]

Not sure what happened either tbh, I think I was tired and put the wrong trig function in.

[hard coral]

Idk if this is useful to anyone, but I found two new identities

The first one looks like tangent but with scale and translation transformations

but I haven't been able to figure out how to scale it properly

This is all I know it has the form of

idk what k,m, and h are so

[fervent geyser]

Can someone help me with a proof

[sleek ledge]

oh no, this is gibberish to me, that's concerning because I'm going to uni for engineering next year

[hard coral]

It equates to the second image

[elder halo]

How do I find x and y

[upper echo]

draw diagonal from 95 to y and use the fact that the quadrilateral is cyclic.

[upper karma]

wow

[upper karma]

We use tg instead of tan, ctg instead of cot in Russia

[lyric halo]

https://www.youtube.com/watch?v=piJkuavhV50

The Twirling Tiles problem is so trippy to visualize

Check it out

Very, very trippy

[cold fulcrum]

what do you call this shape?

[obsidian harness]

right-angled trapezoid

or right-angled trapezium in British English

[obsidian harness]

no wonder

[upper karma]

Congratulations on being the first person to post this on this server

I was beaten to the punch

[late stone]

yo so i have a stupid question, in questions for coordinate geometry (slope) whats the exact difference between x/a + y/b =1 and y = mx + b /y= m (x-a)
Like when do you use the first one and the second one. I dont understand the difference between slope intercept form and jus intercept form of an equation. because im getting confused when to use the first one and when to use the second one

[lyric halo]

You use what is convenient

[lyric halo]

[stray junco]

So in later years do you actually have to use the distance formula or can i continue using the pythagorus? Asking for a friend

[obsidian harness]

the distance formula and Pythagoras are one and the same

[stray junco]

But do they ask to use the formula is my question

[obsidian harness]

well it depends on what the question is asking you

[late stone]

oh so you can use either of them ?

[eager ether]

what is the formula for a spiral on a parabaloid in terms of radius and height, if its helical pitch is equal to R/H.

[bleak roost]

hello! new here, i've just entered college and my first math test is upcoming

but regarding math, i'm absolutely freaking cooked

i always have this issue where i don't know how to start the work, and then the rest is easy

and for starters the material that will be included on the exam is of these themas/subjects

Elements Of trigonometry
Additional formulaes
Complex numbers- algebric trait of complex numbers
trigonometric trait of complex numbers
Matrices - actions with matrices
and then determinants which i'm going to learn this week

if anyone is very good at these and wants to help a re#ar@d out, please reply to my msg here or even dm me 😵‍💫 🙏

[trail tendon]

well yk if you have a question on problem you're working on you could just ask lol

also its probably better in #precalculus

[bleak roost]

rn the issue is more like, how do i create the logic to start the question itself, idk what it is that i'm lacking, just my brain goes poof blank

when faced with it

[trail tendon]

XD

[bleak roost]

for example there was this work i tried with chatgpt earlier

sin square x +1 =2 cos x

so i did just the first two lines right

1-cos square x +1=2 cos x

then substract 2 on both sides

-cos square x =2 cos x - 2

and that's when i started needing help from chatgpt cuz besides making left side positive

what the heck else can we get out of it

[trail tendon]

thats just a quadratic lol

its a bit hidden

[bleak roost]

.. n then we put all in one side and did quadratic

[bleak roost]

SEE

[trail tendon]

yea

[bleak roost]

how you knew that it would be quadratic

how do i make myself create the logic to know it owuld become that

[trail tendon]

you have cos squared term, a cosine to the first power term, and a cosine to the 0th power term

[trail tendon]

honestly now that you know that CAN happen you're more likely to spot it in the future

yk?

[bleak roost]

0.0

i feel like i do, and i feel like even though i'll encounter the same problem i'll be like

this can be solved quadratic... but then after it what happens

cuz even on this one

[trail tendon]

well then you solve for the possible x values

you solve for cos(x) = blank and then you can find the x values

[bleak roost]

we had cos x1 =-1+ square root /3 and cos x2= -1 -square root 3

so does the work end there or do i gotta do more

[trail tendon]

well it depends on the question

did they ask "solve for all possible x values" or...

[bleak roost]

holdon let me check

Practice Problem for You:
Let’s give you a quick practice equation to work on. Try solving this one, and use the steps we talked about:

sin⁡ square 𝑥+1=2cos𝑥

chat gpt gave me like this

so we went through the steps

[trail tendon]

ah

well then it probably wants you to solve for x

[bleak roost]

and then solve for x it seems

damn you know everything

lmao don't blame me for being surprised you can guess my math questions

i'm being honest i'm in college but my math development was stunted because of bad teachers and not being to catch the elementary logic of stuff in it

and now i'm grinding my brains out trying to understand stuff out to make sense to me

[trail tendon]

yea thats fair 💀

[bleak roost]

deada$$--my high school teacher was so bad

[trail tendon]

😭

[bleak roost]

in 11th grade we like, learned trigonometry... she just wrote the formulas on the blackboard and said who wants to do any exercises

[trail tendon]

💀

[bleak roost]

and my class had math talents so yea... they would do and teacher wouldn't explain jacksh

and then 12th grade she got cancer

and we didn't have a teacher all throughout it

[trail tendon]

br thats crazy

[bleak roost]

very lovely story to show how i am a math illiterate

for college level stuff

now

[trail tendon]

im ngl tho now is your time to grind fr 😂

[bleak roost]

EXCEPT IDEK WHERE TO STARTTT

😵‍💫

[trail tendon]

with da homework? :>

[bleak roost]

i joined this server now to maybe try and somehow broaden my knowledge

[trail tendon]

thats what i did lol

[bleak roost]

ahh yeas, homework which i don't even know how to start them

[trail tendon]

whats the first problem :>

[bleak roost]

🥰

we got like these pdfs

holdon

for example this one

we have like the solutions

but i got no idea why or where a thing popped up from

[trail tendon]

don't look at the solutions

for a sec

[trail tendon]

do you have any idea what to do as a first step?

[bleak roost]

i just snipped it, should i send it or na? didn't look at them

[bleak roost]

literally no.

[trail tendon]

you dont need to send the solution yet no

[bleak roost]

okay

let me think

to like prove both sides are equal

it should be that kind of work

[trail tendon]

what about
x/4 = 5/(3x)
would u know how to solve for x

[bleak roost]

... idk how to start it. this either.

even tho that's probably grade school level

holy damn i'm getting more and more disappointed with myself by the second

math is very humbling

[trail tendon]

naw bro don't be 😂

[bleak roost]

first 1-cosa would equal=sin a?

at least i learned that earler

oh nah i mixed it up

[trail tendon]

when you have fractions on either side of an equal sign, you could try to multiply both sides by the denominators just to get rid of the denominators

[bleak roost]

it has to be sin squared a

[bleak roost]

what would be the right value

[bleak roost]

let's try the work here

[trail tendon]

ok

[bleak roost]

should i go with /* cos a on both sides?

[trail tendon]

so lets say you don't want 1-cos(a) in the denominator anymore. what could you multiply by both sides to get rid of that denominator?

[bleak roost]

it would have to be 1-cos(a) itself right?

[trail tendon]

right

what do you get when you multiply both sides by 1-cos(a)?

[outer wadi]

Just cross multiply

[bleak roost]

so left side would become

sin(a)

[trail tendon]

mhm

[bleak roost]

and right side would become 1+cos(a)/sin(a) * 1-cos(a)

or is that entirely wrong

[trail tendon]

yeah no its correct

do you know how to simplify (1+cos(a))(1-cos(a))?

[bleak roost]

1+cos(a)/sin(a) * 1-cos(a) simplify this value right here?

[trail tendon]

specifically the numerator but yeah

[bleak roost]

mm i'm trying to think if there are any identities for that but i can't recall any

[trail tendon]

not even identity

just like distribute

(a+b)(a-b)

[bleak roost]

i'm trying to think but... yea yk what i'm going to say

so it would have to be

a* a + a*(-b) and such?

[trail tendon]

sort of

[buoyant sedge]

wait if you just multiply (1-cosa)

[bleak roost]

first with first, first with second, then second with first and second with second

[buoyant sedge]

if that's what you want

[trail tendon]

so if you have
(1+cos(a)) * (1-cos(a))
this means
1 * (1-cos(a)) + cos(a) * (1-cos(a))
right?

hold up

fixd

[bleak roost]

wait wait, so now we multiply the value

let me send a pic of the work rn

just so we're clear on stuff.. sry for taking so much time

[buoyant sedge]

also you can do sum to product formula if you want to make this more complex

[bleak roost]

so like so far the work is looking like this

[trail tendon]

oh, you should multiply (1-cos(x)) in the numerator rather than the denominator

[trail tendon]

like this

[bleak roost]

hell i'm tryna make it easy for a 1st grade kid to understand

[bleak roost]

oh? so if a value is multiplied like that it would be added on the numerator?

[trail tendon]

yes, when you multiply by a value it gets multiplied into the numerator

if you divide it goes into the denominator

[bleak roost]

oh? OH?

wait what

i felt like i was about to have a groundbreaking enlightening

so if we multiply by both sides there would be the value on the numerator

but if we divide the both sides

[trail tendon]

here ill draw out the thing

[bleak roost]

the divided thing would be added to the denominator?

[trail tendon]

here's the original right?

[bleak roost]

yup

[trail tendon]

we wanna multiply both sides by (1-cos(a))

[bleak roost]

sina/1-cosa=1+cosa/sina

yep

[trail tendon]

and these multiply to the numerator right?

they don't go in the denominator because that would mean we're dividing by them

[bleak roost]

yup, it seems so, i didn't know of this, i was using them without ever knowing the logic behind them

[trail tendon]

😭

[bleak roost]

thank you, so multiplication numerator, dividing below

[trail tendon]

like 4/3 multiplied by 5

you multiply it by the numerator so its 20/3

[bleak roost]

💀 please, my brain feels like a cancer to my own skin at this stuff

[trail tendon]

rather than going in the denominator where it would be 4/15

[bleak roost]

yep, it makes sense to me now

[trail tendon]

alr

so

[bleak roost]

so now 1-cos a went ot the top

so would that make the square binominal or how was it spelled in english

[trail tendon]

yeah, like this right?

[bleak roost]

yup

[trail tendon]

now we wanna distribute (1+cos(a))(1-cos(a))

because it allows us to simplify

u remember how to distribute that?

[bleak roost]

it goes like this right?

[trail tendon]

yeah

so what does (1+cos(a))(1-cos(a)) become?

[bleak roost]

let me write it down real quick

[trail tendon]

ok

[bleak roost]

1-2cosa-cosa

?

wait wait no

why has my brain shut down

[trail tendon]

close

you'd get

1 + cos(a) - cos(a) - cos^2(a) right?

[bleak roost]

wiat wait ama write it again

[bleak roost]

would'nt the first cos be negative?

1 times 1 =1

1 times (-cos a)= -cos a

[trail tendon]

it doesn't matter which one is first lol

[bleak roost]

then +cos a times 1=+cos a

[trail tendon]

you could also write 1 - cos(a) + cos(a) - cos^2(a)

[bleak roost]

then +cos a times -(cos)a

would equal -cos square a

[trail tendon]

oops

the last one is negative yeah LOL

💀

[bleak roost]

would be like this right

[trail tendon]

yeah

and you see how the middle terms cancel out?

-cos(a) + cos(a) = 0

[bleak roost]

yep it becomes

1-cos square a

[trail tendon]

yes

[bleak roost]

which makes... sin square a

[trail tendon]

yes

so then what u have?

[bleak roost]

so now we got

sin a = sin square a / sin a

[trail tendon]

yeah

[bleak roost]

wtf do we do now...

[trail tendon]

can u simplify sin^2(a) / sin(a)

[bleak roost]

how would i be able to do that?

[trail tendon]

sin^2(a) = sin(a) * sin(a) right?

[bleak roost]

oh shi

so

oh god

[trail tendon]

mhm

[bleak roost]

what the hell is hte solution of this work tho

the one they put up on the platform

did they use some other work?

[trail tendon]

they did a very similar thing lol

they just manipulated one side

theres multiple ways of doing it

you could also cross multiply

[bleak roost]

i'm wondering for example here what the hell happened to the sin a on the right side

did it turn into 1+ cos a out of thin air

[trail tendon]

wdym?

here?

[trail tendon]

those aren't the same

they were only using the left side

and they multiplied by 1+cos(a) on the numerator and denominator

like

they were trying to turn the left side into the right side

[bleak roost]

hmm, like they do to rationalize when there's square root on the denominator?

[trail tendon]

yeah, except here they multiply by the conjugate

[bleak roost]

opposite sign of the denominator, multiply both numerator and denominator

god, this shi makes my brain tweak, how would i be supposed to know we gotta multiply it with that instead of trying to prove that they're the same?

thank you, i understand the process behind this work and how we solved it

it is a rather disgustingly easy thing to do

if you know the grasp of the functions ofc

lol

thank you for your time

[dreamy vessel]

The six trigonometric function: son(x), con(x), tall(x), cotton(x), seenot(x), seenot+(x)

[quaint marsh]

can someone

get on a call and help me with my geometry

ive cheated my way through online school

im in 10th grade

[daring bronze]

how to i find the slope dotted lines and the rise over run for it

[errant lance]

The slope is right there???

Or are you specifically talking about in general?

[daring bronze]

in general

[errant lance]

You can kind of count it out

Like look at the dotted line

And look for a point where the line crosses through

Now, look for the next closest point where the line crosses through going up

Then do rise over run, which is 3/1, simplifying to 3

That's for the dotted line only In this sample

[daring bronze]

look

[errant lance]

I think it's y= 1/2x + 3

For the b part

[daring bronze]

it worked how was it 3 wth

[errant lance]

There is a little number right there on the y axis

Or the bold line going up and down

[daring bronze]

ohhh cause its going through the 3 right?

[errant lance]

Yep.

[daring bronze]

ohh okay thank you for your help bro

[errant lance]

You're good bro

I'd probably go to algebra for this though.

[daring bronze]

i got sent here

by a mod

[errant lance]

Hm.

Were you kicked out?

[trail tendon]

💀

[daring bronze]

no

idk why he sent me here

😭

[trail tendon]

cuz its a geometry problem-

[errant lance]

It ain't that funny

[trail tendon]

it uses algebra

but is geometry

😔

[errant lance]

Somewhat, but usually they bring you into geometry at the end of Algebra 1

So there's a chance.

[daring bronze]

pls guys

i dont get this shit

[errant lance]

Alright, first, the dotted lines are all parallel, so you can choose any one you want, and the reflection is just the bolded line.

The one going between the two figures

Pick any point by which one of the dotted lines crosses through

Then, look for the CLOSEST point next to it still crossing through that line

This one is a little bit trickier, so it will be a negative slope.

Do you get this so far?

[daring bronze]

no..

[errant lance]

Oof

[daring bronze]

bro im cooked

can u show me on stream?

or ill stream bro pls

[errant lance]

I can't stream, as it is late and I don't really want to wake anyone up

I'll try to make a drawing to show you

[daring bronze]

okay bet

[upper karma]

bruh

[errant lance]

This is for reflection only

It's a little confusing

[errant lance]

You can choose any point you want where the reflection line is crossing through

Like (0,2) for example

Then, look for the CLOSEST point with the same requirements (line crosses through the point)

Next, look at the formula I drew

On the top part of the fraction, you want to put the distance between the height of those two points I mentioned

On the bottom is the distance right or left depending on where the line is tilting to.

The distance up and down of the points is 3, while the distance left or right is 1, as you are literally going 3 points up from the point you chose originally, and 1 point right after that

And that's basically it.

Do you get it a little better now?

[daring bronze]

yeah

i get it now

is it the same for the dotted line?

[errant lance]

Kind of, but it is slightly different

If you go down in the rise over run formula, then it is negative

Say you went down 3, then the top part of the fraction would be -3

Or if you went LEFT 1 point on a line, then it would be -1

This one would go down 1, making the top part of the fraction -1

Then, it would go right 3, making it -1/3

That would be the slope

As if it goes right three, then it is positive, but if it goes left, then it is negative

And that's it for the dotted line.

You got it?

[daring bronze]

yup yup

thank you so much bro

i really appreciate it

[errant lance]

You're welcome, but I have to get off, as my battery is low and I may not have enough to stay on much longer.

[daring bronze]

have a good night bro, and thank you again.

[acoustic field]

Can i send a problem here? Or do i send it in a help channel

?

[trail tendon]

sure u can send a problem here

[acoustic field]

Thats a quick diagram i made

[weak plaza]

how would you find sin 58 without a calculator?

[vernal pilot]

you dont

[trail tendon]

sin(58) is sin(58) ✅

[weak plaza]

my teacher told me to do this without a calculator

wtf 😭

[vernal pilot]

then either you have done something wrong or your teacher is evil.

[weak plaza]

leaning towards evil

imma just use a calculator

[flint marlin]

@sooouth.

[unreal storm]

Hello guys can anyone explain i have done the solution for this question right or wrong:
1 -since OQ and OR are 2 units each QR will be 4 units and since it is an equilateral triangle QP=PR=4units
2- I proved the triangles congruent by SSS axiom and proved the angles QOP=ROP by CPCT
3- I used pythagorus theorem in triangle QOP and got the lenght of OP as 2root3 units then i got the co-ordinates of point P as 2root3,0 answer is right but i did the method correct right because everywhere they mentioned the angle QOP and ROP as 90 without proving since we its not mention OP is perpendicular so thats why i found 90 by congruency

[stray junco]

do the midpoint stuff have ties with the pythagorus like distance

[unreal storm]

that one is right?

which i did

[ember ore]

use pythagoras

[stray junco]

I learned the formula do never mind that

[unreal storm]

i did bro.

[ember ore]

and?

[unreal storm]

uh and hwat

what*

[ember ore]

oq and or are 2 units

[unreal storm]

bro read the whole thing i mentioned the steps i did.

[ember ore]

so take op as x

[ember ore]

hold on

[ember ore]

yeah it is correct

[unreal storm]

ok thanks

can we do like also this way that since op line is drawn from vertex A of equilateral triangle it is perpendicular and bestics qr

[ember ore]

bisects qr but yeah we can

[unreal storm]

oh yea mb

[ember ore]

also the x and y axes are perpendicular to each other by default

so

[unreal storm]

oh

[ember ore]

there's ntg to prove abt right angles

but yeah

what you did was correct

[unreal storm]

ohhhhh okaayy

[ember ore]

which grade r u in btw

[unreal storm]

9

[ember ore]

class 9?

ah nice

[unreal storm]

yea

how did u know;-;

oh ya i mentioned

[ember ore]

lets just say i also did it when i was in grade 9

[unreal storm]

oh u indian? just asking

[ember ore]

huh i didnt see it anywhere

[ember ore]

yeah

[unreal storm]

which board

am icse

[ember ore]

cbse

[ember ore]

noice

[unreal storm]

dam

btw do u think cbse people should hate icse people?

coz i recieve hate that icse papers are easy .

[ember ore]

no why that doesnt make sense

[ember ore]

thats just shi-

icse board is harder that cbse especially science

[unreal storm]

fr

it is coz syllabus is more

[ember ore]

yeah the amount of topics u poor kids have to learn is insane 😭

[unreal storm]

cbse paper is so less and some chapters are taught 1 grade after than that of icse so i think thats wrong as for competetive exams its bad if u gotta learn so less

[unreal storm]

fr

brb

[ember ore]

ok i see why icse kids get hate coz they brag abt how tough their subject is

and although cbse topics are limited, the depth of concept in every topic is just 💀

try solving cbse grade 10 maths it will drive you nuts

[unreal storm]

they dont brag that much;-

[ember ore]

u just did lol

[unreal storm]

am in 9 cant do 10th.

[ember ore]

the weird thing is

[unreal storm]

i am not bragging... like i did in response when a cbse kid said to me

[ember ore]

cbse grade 9 maths is harder than grade 10

idk why

but yeah

[unreal storm]

bc the papers are easy chapters are less so i replied in that perspective.

[unreal storm]

10th is easier than 9th

coz 9th is base for 10th

thats what i have heard people say./

[ember ore]

so you should be able to solve the papers then

[unreal storm]

idk.

[ember ore]

ok lol

[unreal storm]

ur thinking worng am not tryna brag or hate xd

leave this topic

[ember ore]

yk what forget it lol

[unreal storm]

💀

yea

welp ima go do construction of quadrilaterals 💀

[ember ore]

have fun

[upper karma]

what on earth

[unreal storm]

no

[unreal storm]

no fun.

[ember swan]

It’s ok I’m doing sphere geo 🌐

Luckily google search engine is goated by already including a sphere surface and volume calculator in the top result when searching for“sphere” in the address bar

But seriously relearning polygons but on different radius spheres is really easy.

Best visual example is how like a circle can e split into fourths a sphere can be split into eighths

[lyric marlin]

can anyone help me with this

[obsidian harness]

this should help

[obsidian harness]

for question B, 1 is PS for example

it's just asking you this

[lyric marlin]

it includes tangents it's frustrating me

[obsidian harness]

angle between tangent and radius is always 90 degrees

also you need this

[lyric marlin]

that one is kind of easy but finding the measurements kills me

[lyric marlin]

I don't even know how to start Letter C

[obsidian harness]

which question in C?

[lyric marlin]

1

I'm crying😭

[obsidian harness]

well the angles on a straight line always add up to 180 deg

their angle notation is also used for the measure of an arc of a circle

but for question 1 they just mean the angle on the straight line

[lyric marlin]

can you try do the B 2 cuz our teacher teaches easy example and gives hard example

[obsidian harness]

right, if C is the centre of the circle, that means angle PCM = 157

so you have two right angles

and sum of angles in a quadrilateral is 360 deg

[limber gust]

where'd u get c from tho

[obsidian harness]

that's what the arc measuring 157 degrees means

[limber gust]

here?

[lyric marlin]

I wouldn't bother doing that problem😭

[limber gust]

well if your looking for C. 1, u can see its an inscribed angle

since the angle of the intercepted arc is always half

150x2 is 300

since now u got the measure of that arc, all u need to do now is to to find x. u just have to subtract the measure of the circumference (360) to find arc BA

360-300=60 (measure of arc BA)

[obsidian harness]

oh wait 150 is the measure of the arc

wow

that's terrible placing

[limber gust]

then u can now get the measure of the angle (x) which is 30

[limber gust]

REAL

[lyric marlin]

omggg

I didn't know what theorem i'm suppose to use

[jade sierra]

secant , cosecant , cotangent is screwign with me 😭

[edgy wraith]

can someone help me with this

i dont really understand the question

[blazing rampart]

i think its 4th option

[obsidian harness]

what's the surface area of a hemisphere?

[edgy wraith]

my teach said that i gotta use the formula of sphere to get the hemisphere but i dont even know what a hemisphere is

[obsidian harness]

remember to add the base

[obsidian harness]

it's half a sphere

and the base

[edgy wraith]

yes and its volume

[edgy wraith]

ohh alr thanks

[edgy wraith]

thanks

after calculating it the answer was indeed 4th

[weak plaza]

can someone explain how they got B

[faint pasture]

By sine rule

a/sinA =b /sinB

We know a,b,A

So we can find B

(Although i personally prefer to use cosine rule in that type of questions instead of sine rule)

[weak plaza]

so like
7/70=5/sinB?

[faint pasture]

7/sin(70)

Yea

[weak plaza]

im sorry im still a bit lost
so I do 7/sin(70) which is 7.449
then what do I do with that ?
like my mind cant proccess the correct way to go about it

[faint pasture]

7/sin70 = 5/sinB as you wrote earlier

,w (5h)^2-(4h-24)^2-300^2=0

Tf

[weak plaza]

[somber coyote]

Results provided by WolframAlpha

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

Requested by enderdoesntmind

[faint pasture]

Nvm that was unrelated to your question

[faint pasture]

This is what you gotta know though

[weak plaza]

holy shit I got it, I just thought it was more complicated than that

overthinking it as usual

sry about that, thanks

[static imp]

guys I need help

pretty please?

[buoyant sedge]

can you identify the coordinates of point A, B and C?

[lyric marlin]

just did 2 pages 😭

can someone help me do just the G. ik it's easy but I get frustrated

[buoyant sedge]

draw radius

from the center of the circle to the points of tangencies

then draw line from the center to point U

and then you should be able to figure it out

[static imp]

I already got it thanks for your help though

[buoyant sedge]

always good to figure it out by yourself🤩

[static imp]

which one would it be?

[trail tendon]

do u know what image means?

[static imp]

I know its not A

I think its B but I'm not sure

[trail tendon]

why?

[static imp]

I got the x and y coordinates mixed up

its def B

yep it was b

[wind bear]

help - Given a function f(x) = 2 sin (3 x) cos (3 x).
Understanding the expression for the given function gives f (x) =
The derivative of the given function f' (x) =
cos
(x).

[buoyant sedge]

In general, sin(2a)=2sin(a)cos(a)

So you can notice that

2sin(3x)cos(3x)=? Then take derivative

[austere mortar]

is this alright? I am to find the sum of the first 100 solutions to sin 2x = cos x

also in the end instead of "3/pi" it shouldve been "3pi"

sorry for that little mistake

[hearty walrus]

If i’m doing geometric column proofs, does the order of the steps matter?

[naive tapir]

represent each segment as a vector, then:

• take the angle between each vector and ensure its <180deg
• take the cross/dot product to find the orientation of the vector

[lyric marlin]

how do i start this one

[frosty rose]

when do i know if a vector is a parabola?

[obsidian harness]

wdym?

[wild veldt]

Whaaaa

Wdym 😭

[frosty rose]

uh

one sec

ill send smth

[wild veldt]

The vector can’t be a parabola, a relation between vectors can be

[frosty rose]

oh fr ?

[obsidian harness]

yeah a vector is a straight line segment

[frosty rose]

hold on

[wild veldt]

Like how (a)+k(b) where a and b are both vectors is a plane

[wild veldt]

Could’ve used texit but I’m sigma 😼

Ight imma dip I got chem to do

[frosty rose]

trajectory vector r(t) = {x = cos a +3}
{y = sin a}

this is a system idk how to type it tho

[obsidian harness]

okay

then that means $(x-3)^2+y^2=1$

[somber coyote]

south's secret twin brother

[frosty rose]

yes

[obsidian harness]

you're looking for a parametric equation

and you wanted to convert that into Cartesian form

[frosty rose]

ya

[frosty rose]

that isnt cartesian right?

[obsidian harness]

it is

it's a Cartesian relation not a function

[violet bronze]

a and b are generic variables

[frosty rose]

oh

[violet bronze]

How do I show the ratio AB:AC

I don't know the values of a, b

AB = B-A =b
AC = C-A = 4a-2b

[frosty rose]

so in here the trajectory is a circle-ish?

[obsidian harness]

if you have A = (1, 2), B = (-1, 3) and C = (5, 0)

[violet bronze]

Nope

[obsidian harness]

can you do the problem

[violet bronze]

a and b dont denote i,j

I emailed my lecturer about that even

[obsidian harness]

well but can you do the problem now

[violet bronze]

No

You can't assume a=i, b=j

[obsidian harness]

cause it's very similar

[violet bronze]

He explicitly stated that

[violet bronze]

.

If a=i and b=j then its easy but its not meant to be

[obsidian harness]

a line through B and C is $5a + (-6a + 3b)t$ where $t \in \mathbb R$

[somber coyote]

south's secret twin brother

[obsidian harness]

sub in t = 1 to see how

[violet bronze]

yh I did that

[median harbor]

anyone here knows grade 10 math

[violet bronze]

Wait I'm being dumb lol

You can assume a=i and b=j

Give them both the same length and the ratio stays the same lmao

My bad

[violet bronze]

Yh what's up

[obsidian harness]

yeah then you just need $5a + (-6a+3b)t = a + 2b$

if it exists there will be a real value of $t$

[somber coyote]

south's secret twin brother

[obsidian harness]

no worries

[median harbor]

alright sick, wait

[obsidian harness]

but this is the proper logic with a, b

in case you wanted to know

[violet bronze]

If what exists? I showed colinearity already with the proper logic

A is at t=⅔

t in a vector line equation can be used for ratios?

actually ig yeah lol

[obsidian harness]

cause we found CA:CB = 2/3 or AC : BC = 2/3

[violet bronze]

B is at t=0, A is at t=⅔

[obsidian harness]

this is defined so that when t = 0 we get C, when t = 2/3 we get A and when t = 1 we get B

[violet bronze]

Oh i used B

Not C

[obsidian harness]

yeah so it asks for AB:AC

[violet bronze]

For t=0

But its the same

[obsidian harness]

but yes t = 2/3 here

[violet bronze]

I used C too mb

But yea

[obsidian harness]

ahhhhh so yeah AB : AC = 1/3 : 2/3

[violet bronze]

Wait

I understand that im being really really incredibly stupid rn

but how does that give AB:AC = 1/3 : 2/3 lol

Δt from A to B is 1/3 lol

I realise now

I'm dumb

dw

[obsidian harness]

the length of a line segment is proportional to the change in t

[violet bronze]

yh I'm just being dumb dw

I'm usually not this shit at maths

But I come here for help only when I'm being stupid

sooo

I appear as such lol

[obsidian harness]

it's okay

like you just need time to understand

[violet bronze]

Thing is I've known vector stuff for years and years

Like plane geometry and etc

I just never considered λ (I use λ not t)

In terms of line segments

Cuz I only ever had to think about them in context of like distance between a line and a plane or intersections of lines and planes

[median harbor]

can you teach me how to get the answer to this?

[rare furnace]

hi chat can someone explain similiarity and congruence or whtever this shit is i havnt been paying attentuon in class

[median harbor]

so basically

[median harbor]

in similarity, same shape, different size

in congruence, theyre exactly the same,

to get the " ° " of an triangle with an extension, add a and b together.

a tringle's angles must always be equal to 180 if you add all of it up,

a quadrilaterals angles must be equal to 360.

[median harbor]

i cant really read the other letters in the 2nd image, but if you need to know how to solve, i can also help

[elder halo]

How do I solve this?

[native viper]

can someone help me🥹🥹

[austere mortar]

if Im understanding these funny arrows on the lines correctly try using this as a hint

[austere mortar]

try using this theorem, itll help tremendously

@native viper especially since we dont know if |BC| = |AC| and thus y doesnt necessarily have to be equal to 43°

why is this wrong? Im supposed to find all of the values of m for which this function has any solutions in the real numbers

[elder halo]

is y here 90 degrees

doesnt look like it

[austere mortar]

nope, however

[elder halo]

How do I find it?

[austere mortar]

use this as a hint

and that gamma + 22° = 102°

[elder halo]

I see what I did wrong

[median harbor]

what equations should i use for the first and 2nd one

i need to find the whole radius of the circle

• and i need to find the possible measures of angle P and R in the second one

[austere mortar]

@median harbor this is surely for the first

[formal geyser]

Can we say that the big red triangle and the ACB triangle are similar because the base of big red triangle and AB line are parallel and therefore i can perform further calculations?

I just cant prove that the base of big red triangle is parallel to AB line

[upper echo]

The first figure implies that the base of the red triangle is sloped 45 degrees. So you can take it parallel to AB. Though strictly speaking the initial figure doesn't have all needed data for the problem. because it might be something like this. And that changes the area.

[covert moss]

15π/4 how can I know what quadrant this angle is without evaluating it. One method is subtracting revolutions but I am just wondering if theres any other way

[trail tendon]

idk if you count this as 'another way' but go around the circle 1 and 7/8 times LOL

[upper karma]

not enough information

[trail tendon]

agreed

[upper karma]

you could bring it to the nearest multiple of 2π (16π/4 = 4π) i suppose

that’s largely the same thing though

[faint pasture]

What if you assume its isoceless

Looks like enough info in that case

Like i can definitely do this with coordinate bashing

Without it uh

[finite glen]

how do you solve this, the asnwer is 2/5 btw

[upper karma]

yes, if you assume it’s isosceles

[upper karma]

find the other three triangles as a fraction of PQRS, then subtract them off from the whole rectangle to get ABP

[finite glen]

I found the other 2 triangles, but how do you find BAR?

[upper karma]

you’ll need AR/QR (and therefore QA/QR) as well as BR/SR (so find BS/SR)

both should be findable with the information you’ve been given

[finite glen]

2/3 and 1/2?

[upper karma]

second one is correct but i don’t think the first one is

[finite glen]

wait 3/2?

[upper karma]

what ratio is this?

[finite glen]

oh wait sorry wrong ratio

3/5?

[upper karma]

that looks good

now find the area of BAR as a ratio of the area of PQRS

[finite glen]

oh okay thx i got it

[stray junco]

how do you divide line segments again

[proper otter]

!xy

[lime crown]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[stray junco]

Point
[A] is at
[{(1, 5)}] and point
[C] is at
[{(-5,-7)}].
Find the coordinates of point
[B] on

[\overline{AC}] such that
[AC] is
[6] times as long as
[AB].
[B=\large(]
[,]
[\large)]

[somber coyote]

Baramsniper 산토끼토끼야

[little pawn]

Help pls

❤️

[upper echo]

Denote AZ=x. Then XM=2x, then write Pythagorean th. for MAW:
WA²=MA²+MW²:
8²+x²=((8-2x)²+(4-x)²)+(4x²+4²)

[little pawn]

is it 15?

[upper echo]

It's up to you to decide 🙂

[little pawn]

pls i jjust want a confirmation

i htink i decided 15

[upper echo]

Rely on your own reasoning. I don't have the answer

[little pawn]

🙁 ok

[jolly zinc]

js found out that cos^-1(-1) in radian mode is pi

thought it was cool

[frosty rose]

whats the difference between the equations uw + vy + w = 0 and y = ax+b of a straight line?

[obsidian harness]

(u/v)x + y + (w/v) = 0 can be rearranged into y = -(u/v)x - (w/v)

just a different way of representing the line

[frosty rose]

ohh

what do u, v, and w represent?

[obsidian harness]

it's hard to say

[frosty rose]

oh

[obsidian harness]

w controls the vertical translation

u, v control the slope

the issue is that you have 3 variables controlling two unique pieces of information

[frosty rose]

its for more precision?

[obsidian harness]

a line can be uniquely defined by a slope and a point on the line, or two points on a line

[obsidian harness]

no

my point is that one of the variables has to be redundant

[frosty rose]

oh

[frosty rose]

ah

yea i kinda get it

thanks brother

[obsidian harness]

no worries!

[soft ermine]

if you have any problems, plz dm me anytime.

[upper echo]

The difference is that y=ax+b misses vertical lines, while the first equation can give vertical lines if v=0. And also there many equations of the first type giving the same line. Unlike y=ax+b which is unique for any given non vertical line.

[obtuse stag]

PLEASE someone im rly confused bcz hypotense + perpendicular be the total hieght... I HAVE MATHS exam tmr please

[obsidian harness]

(this q has been answered already)

[tardy loom]

Anyone know where I can find a visual, geometric proof/explanation of the Dot Product equation?

$\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|cos\theta$

[somber coyote]

Plentimon

[tardy loom]

Draw and label a diagram, then use the pythagorean theorem. You've got this!

[tardy loom]

Think I found what I was looking for: https://www.youtube.com/watch?v=Sl4lr8tATXE

[gray pivot]

Confusing wording but you just use 30-60-90 triangle. You can assume that the tree is straight.

[sand zinc]

Say you have a triangle with angles ABC, and sides abc. You have the angles C and B, side length b, and want to find side length c.
Set h = sin(C)*b
sin(B) = h/c
sin(B) * c = h
c = h/sin(B)

I figured this out before learning what the law of sines is

Maph

[faint pasture]

Isnt that the definition

[trail tendon]

intuitionally yea is just a definition
but there is geometric interpretation like between the different ways of computing it lol

thats all the video was talkin about :l

oh i see yea u meant specificially u.v = |u||v|cos(theta)

[faint pasture]

Yee

[main fiber]

how to solve y=4tan(theta/3+pi/2)+2 no calculator

[trail tendon]

what are you solving for exactly