geometry-and-trigonometry

[subtle palm]

needs clarification

i mean it could be the wrong triangle sets

i think it was referring to pairs PXY'-PX'Y and PY'X-PYX'

that's speculation though

[exotic linden]

help please (ping when you answer)

[silk patio]

Triangles have an order

[pure bronze]

You need to understand it yourself, we’re only here to give you a fraction of the solution.

And by the way as a step, sketch it and show your progress

[exotic linden]

im not sure if the double marks are right

im just not really sure how to prove how the internal angle bisector of BAC is perpendicular to AD

[pure bronze]

BE = AB

DE = AC

[wise pawn]

the way I'd do it is represent the points in the complex plane with $z(\theta) = ae^{i \theta}+b e^{i \theta}$ but I think then solving for the real numbers a,b would again boil down to linear algebra, but I think you really only need 2 points not counting the origin

[somber coyote]

Merosity

[wise pawn]

I guess that's pretty much exactly what you didn't want

well I said 2 points not counting the origin

still 3 points

oh typo, missed a negative sign $z(\theta) = ae^{i \theta}+b e^{-i \theta}$

[somber coyote]

Merosity

[wise pawn]

with 4 points maybe we can do something like draw a quadrilateral and some diagonals or something lol

that's the same as the exponential thing I wrote

if you expand the e^{ix}=cosx+isinx parts and recombine

a+b is the radius of the green circle, a-b is the radius of the blue circle

[silk patio]

You can use equations of lines to find it

Wlog suppose you’re at the origin

Given the 4 points ABCD, consider the equation given by (AB)(CD) and the equation given by (BC)(DA)

The notation means find the lines and then multiply their equations

Then you know the ellipse is a linear combination of those lines

But you know it’s the linear combination with no linear terms as it’s centred at the origin

[vernal fog]

does anyone here have a geometry final exam study guide? my finals are tmr and my teachers forgot to add one

[sly iron]

Can someone tutor me geometry?

[keen wind]

If I'm given the formula, "r = 2" and I wanted to find all possible values of x when y is 0 or 1, how would I do that... you know since it's a circle?

[verbal beacon]

the equation of the circle centered at origin is given by x^2+y^2=r^2

use this to find 'x' since you already know 'y' and 'r'

[keen wind]

Alright, thank you!

[blissful bay]

Hi is anyone here good with chapter 3 (parallel and perpendicular lines) for geometry bc I have an upcoming test n I’m struggling

[surreal bolt]

What do you want to know,@blissful bay

[blissful bay]

Like I have trouble finding angles and differentiating between interior, consecutive, and all those other angles

Like angles that are on the same side of the transversal

[surreal bolt]

Sigh one sec

Okay do you have a picture handy?

[blissful bay]

Like there’s a bunch of examples that I just can’t seem to understand

Yeah ik that but ty

[surreal bolt]

What do you not understand about them?

[blissful bay]

like I can’t differentiate between them

[surreal bolt]

So tell me about this picture.

Do you know what interior means?

It means the two angles are BOTH between the parallel lines. (It is like the parallel lines are protecting them)

A pair of exterior means neither angle is inside the parallel lines.

c, d, e, and f are all interior.

a, b, g, h are all exterior.

Alternate signifies two angles are on opposite sides of the transversal.

a, c, e, g are all on the same side.

b, d, f, and h are all on the same side as well.

Um, if you can’t read and understand the exam questions I dunno how you are gonna get an A.

That’s great for you but I am guessing noob is wanting to do well 😛

Take a calculus exam without knowing ‘ means derivative …

Fine without knowing what limit or lim means.

Nope. It is a definition silly.

Lol there are various definitions of a limit. 😛

Lim x —> infinity

[blissful bay]

The transversal is highlighted.
-a is congruent to d since they t vertical angles
-c is congruent to b because they are vertical angles

[blissful bay]

E is congruent to h because they r vertical

• g and h are consecutive angles because they add up 180

applying it to actual problems

,rotate

[blissful bay]

Like this one when I look @ it my mind goes blank

[plush cliff]

pls help me out in this -_-

[rugged lark]

IGCSE huh?

[plush cliff]

ya

[rugged lark]

w = 26 because of the 6th teorem

x = 128 because of the 5th teorem

[plush cliff]

thts why i dont like math

[rugged lark]

y = 52, because x = 128 and TPO = 90, then according to the 3rd teorem TQO=90.Then 360-128-90-90 (because it is a quadrilateral) you get y=52

[rugged lark]

But here you go

[plush cliff]

thanks bro :)

[rugged lark]

x = 128, w = 26, y =52

[rugged lark]

ok

[wintry tundra]

Or just be the ultimate mathematician and just hate geometry

[plush cliff]

screw tht gonna just drop math :D

but for now i gotta suffer

this channel shall be helpful for the long term

[full briar]

help

how would you guys solve this? :0

also the answer needs to be exact

like ik how one half, rad three over two, and rad two over two works

but wtf is this

[devout bolt]

note that sec(sin^-1(x)) = 1/cos(sin^-1(x)) = 1/sqrt(1-sin^2(sin^-1(x)) = 1/sqrt(1-x^2)

so there I've used sin^2+cos^2=1, but you could also derive it from using triangles

and that holds for any -1<=x<=1

[full briar]

ohhhhh

okok thank you very much!

[devout bolt]

np!

[wintry tundra]

Imagine dropping something

[storm portal]

Can't help but notice that if f(x) = arcsin(x), then d/dx f(x) = sec(f(x))

How?

[devout bolt]

Nice spot! So arcsinx is the unique solution to the ODE dy/dx=sec(y), y(0)=0

I cannot think of a nice deep explanation why that’s the case though

I mean you could also think of it like f(x)=arcsinx => sin(f(x))=x => f’(x) cos(f(x))=1

Which is kinda cute

[exotic linden]

<@&286206848099549185>

[plush cliff]

I would definitely be happier that way

[calm orbit]

any tips on how to find the solution set for x?

[dark sparrow]

this feels like there might not be enough info to solve for x

[keen geyser]

I agree, without information on the pairs of equal length sides i don't think you can get a unique set of solutions

[storm portal]

Sounds good - makes sense to me 🙂

Appreciate that

[mellow canyon]

Can anyone help me with this pls!! With proper explanation

[wise pawn]

you can plug in $a \sec \theta$ directly to the other formula where $(a \sec \theta)^2$ appears to get $(1-b \tan \theta)^2 = 5+b^2 \tan^2 \theta$

[somber coyote]

Merosity

[wise pawn]

then keep playing with it and hope you can cancel things out

[wintry tundra]

How come

[warm storm]

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[upper karma]

can someone confirm i did this correctly

[mellow canyon]

I did it till here and almost got the answer
As I said before The answer is a^2b^2+4a^2=ab^2

But it got 9b^2 instead of ab^2 on the LHS
Can you please tell me where I’m going wrong … it would mean a lot

[tepid crow]

You made a mistake when subtracting the two equations here

You should get 6 - 2b tan theta = 0

Wait...

Nvm

[mellow canyon]

R you sureee I don’t think it works like that

[tepid crow]

Yeah you're right

Are you sure that the answer is ab^2 ?

The RHS I mean

[mellow canyon]

Tats what the answer key said and I’m confused
I think wht I’ve done is correct ryt??

And theta is also eliminated!!

[tepid crow]

Yeah pretty sure it's wrong

[mellow canyon]

Ok then !!!
Thanks anyways

[acoustic jungle]

the answer key has a 9. It's not an a, it's a 9.

it's just poorly written.

also I'm not sure about the specifics, but that relation satisfies a^2 sec^2 = (1 - btan)^2 && a^2sec^2 = 5 + b^2tan^2. It does not necessarily satisfy asec = 1 - btan && second equation. Could anyone confirm this?

for example, if I had x = y and you squared it, you will get the relation |x| = |y|, which is wrong because that would mean x = -y is a solution.

[tardy lion]

find the number of solutions of cos(cos(x))=x/10
Is there a simple way to see this without plotting arccos(x/10)?

[edgy juniper]

this topic is graphs but didnt know where to put it, so i thought here, anyway just a simple question: how come:
2x^2 (where x is -3) = 18 and (when x is 3) is also 18

i thought 2x-3^2 = -18?

(and they same goes for -2 and -1)

[tardy lion]

wdym? When x=-3, 2x^2=18

[edgy juniper]

doesnt 2 x -3^2 = -18?

[tardy lion]

no 2×(-3)^2=2×(-3)×(-3)=18

[edgy juniper]

oh so i put whatever x is in brackets

[tardy lion]

Yes

[edgy juniper]

thanks

[mellow canyon]

Oh ok thanks for confirming!

[winged valve]

i dont know if this is where you put this but I need help on this

[acoustic jungle]

what can you say about UY and SY.

or XS and XT.

[fading zinc]

hm

[rich wolf]

tangents are perpendicular to the radius

take that as you will

[low estuary]

3-ABCD square
DEC right triangle
it asks the value of x

6-ABC is a triangle
it asks the value of x

5-ABC is a triangle
it asks the value of x

i need help 😬

[upper karma]

@low estuary are you still looking for help?

[low estuary]

yess

[upper karma]

okay, so let's look at our square at the top, are you able to maybe draw 2 new right triangles (by tracing new lines) considering we know that DE is 8 and CE is 2?

give it some minutes of attempts, these 2 triangles will help us solve for x

[low estuary]

i tried to do the same thing too many times

but i am not good at geometry, i am trying to be, so i couldnt do it

[upper karma]

it's okay.

so let's look at our point E, it's going to important, we can trace a horizontal line that meets E, so that it creates 2 new triangles

can you try drawing this triangles in your paper? so that i can see what you are doing

[low estuary]

of course

[upper karma]

great but extend the horizontal onto the other side, so that it touches the right side

[low estuary]

okayy

done

[upper karma]

now you have those 2 upper triangles, CEZ (Z is just the intersection between that horizontal that we drew and the right side of the square) and DEX (X is just the intersection between that horizontal that we drew and the left side of the square)

can you see those 2 triangles?

[low estuary]

yess

[upper karma]

okay great, now if we label the side DX to be z, can you guess what CZ will be as well?

[low estuary]

yes of course

again it will be z

[upper karma]

great. now for the whole side of ZX, the whole horizontal, if we know that ZX is 2sqrt(17) as you well figured out, and if we call maybe EX to be just y, can you guess what EZ will be?

[low estuary]

2sqrt(17)-y

[upper karma]

great job.

so now we have 2 right triangles, with 2 variables, z and y, can you try solving for them by creating a system of 2 eqns?

[low estuary]

i already tried this way too

i just need someone solve this on paper and examine it

[upper karma]

yeah no, i'd rather not to, seeking a solution won't help you as much as you yourself solving it with your intuition as we are doing right now.

so unless you want to wait if someone else "gives away the full solution", we can continue the problem

[low estuary]

ok

i did the things you said

but i am not sure if i could create the right eqns

[upper karma]

okay the equations were made correct, the issue is that you did $(a-b)²=a²-b²$ which isn't correct.

[somber coyote]

Al𝟛dium

[upper karma]

remember that $(a-b)²=a²{\color{green}{-2ab}}+b²$

[somber coyote]

Al𝟛dium

[upper karma]

so instead, you should have $\ \begin{cases}\overbrace{({\color{green}{68}}-{\color{blue}{y}})²}^{({\color{green}{a}}-{\color{blue}{b}})²={\color{green}{a}}²-2{\color{green}{a}}{\color{blue}{b}}+{\color{blue}{b}}²}+z²=2²\ y²+z²=8\end{cases}$

[somber coyote]

Al𝟛dium

[upper karma]

now try redoing this system of eqns to solve for z and y.

[low estuary]

ohhh

i just did all of them again

and i found out that x equals 10!!!

thank you so much

can you also take a look at other questions?

i'd be pleased if you do

[surreal bolt]

Sorry may I comment on problem 3?

You can drop the perpendicular to DC from E.

Call the length there h. Area = 0.5h(2rt(17)) = 0.5(2)(8). So we can find h.

Base of h to D is Pythagorean using h and 8. Call it r.
Then x is a Pythagorean with r and 2rt(17) - h.

[gusty bay]

Just wondering how others would address this question, I can't seem to find the "Correct" answer though I have about 5 different ones haha.

Thanks.

[livid moss]

Show your attempt

I would draw vectors, break them into components, add up the components then find the magnitude and angle

[mild briar]

complex numbers

[gusty bay]

There is a whole lot of crazy on this page. Be warned.

A had a whole nother page of crazy but I scribbled htem all out so that when i solve it, I don't think that those were the possible answers haha.

[livid moss]

Yeah, I have no idea what's happening

You don't seen to have used vector components

[gusty bay]

I actually haven't learned about vectors yet haha.

[livid moss]

Oh

[gusty bay]

That's something we get to this semester haha.

But in you saying that means i'm going to go look up vectors and figure out how to use them for this question. Which is helpful in and of itself.

[livid moss]

Can you draw one single picture? Start with a starting point, then draw an arrow to indicate moving 20 paces south, then where that arrow ends, draw an arrow to indicate the next movement, etc.

Until you are done.

Then draw an arrow from the starting point to the tip of the final arrow.

The length of that arrow and its direction is what you are after.

[gusty bay]

Yeah, that's the conclusion I ended up coming to, but using my limited knowledge of trigonometry, sin, cos, tan, their inverses and trigonometric value. I wasn't able to conclude how I would solve for x or the degree as the result isn't a triangle, unless manipulated in some way.

Which I'm assuming means I can use vectors to solve it? Thanks again @livid moss

[livid moss]

Yeah

If you don't formally know vectors, you would still basically do the same thing just except without calling it vectors

Here is a picture. I took each arrow and split it into a triangle with a vertical and a horizontal side

Then the vertical side of your final arrow, is just adding up or subtracting the lengths of the vertical sides of your other triangles.

So you go down 20 paces, then with the next arrow you go down ? paces, and then you go back up ?? paces

[gusty bay]

oh beautiful, i see it now.

[livid moss]

Then you can do the same thing for the horizontal sides

I didn't split the first arrow into a triangle since it's just straight down so its effect is obvious

[gusty bay]

Thank you so much hahaha. I was so close to it earlier, I had the bottom right, right angle and instead of 2 of the left top portion, I had 1 that spanned the whole left side. But that doesn't work because the angles were different.

[livid moss]

Glad I could help. Ask if you get stuck using this method.

[gusty bay]

Will do, thanks again! I was super nervous about asking the question as I have stewed on it for a very long time hahaha. Built it up, ya know?

[livid moss]

Aww, don't be nervous to ask for help. But it's good to try first as much as you can. Sometimes you just can't do it unless you learn the method though.

[gusty bay]

Thanks again @livid moss I managed to get there in the end, it looks like something out of a scribble book but its correct haha.

[upper karma]

@low estuary okay, are you still looking for help?

[low estuary]

yes, for the 6th. the other one did i handle

[upper karma]

Ok nerds I need help

How do I determine or how can I cut circle so that contact point always stays in center line?

I mean that it stays in center line when you rotate whole thing

[meager holly]

hello, is anyone able to help me with this? thank you.

[meager holly]

<@&286206848099549185>

[mental plaza]

wich one of those or in general

[meager holly]

in general

[mental plaza]

So you probably learned congruence properties or whatever you call them like side angle side and stuff like that

[meager holly]

yes

[mental plaza]

and you get every time a different set of given things

and you have to see if you can apply those

or if there is not enough information

do you understand or do I have to show the first one as an example?

@meager holly

[meager holly]

ohh okay, can you give an example please?

[mental plaza]

sure so I'll show you how 4 goes and you try the rest

[meager holly]

okay thank you

[somber coyote]

VincentBH

[meager holly]

yes

[somber coyote]

VincentBH

VincentBH

[meager holly]

yes

[somber coyote]

VincentBH

[meager holly]

ohh okay

[mental plaza]

and you keep doing this

[meager holly]

okay but im confused on the part of the theorem and postulate part

[mental plaza]

wait where are you confused

[meager holly]

i understand the AAS part but im confused on what postulate and theorem to use

it says i have to write a theorem or postulate

[mental plaza]

are you confused by what the difference is between theorem and postulate?

[meager holly]

no, im confused on what theorem or postulate i am supposed to use like vertical angle theorem or like segment addition postulate

[mental plaza]

1 sec

[meager holly]

okay

[mental plaza]

basically you have to write any theorem or postulate you use

you only used AAS here, so that's what you gotta write down

[meager holly]

oh okay

[mental plaza]

if you use smth else than you gotta say what you use

[meager holly]

ohh okay

thank you

[mental plaza]

np gl with the others

[meager holly]

okay thank you

[upper karma]

how am i going to explain these two parts are congruent

theres no point there

ik that but idk how im going to show ABC and ABD are isosceles

theres no name for the point in the middle

[jaunty scaffold]

Assume a name for the point yourself

[upper karma]

that isnt allowed

[jaunty scaffold]

Why not?

[upper karma]

so far in the entire book every proof was done without assuming a points name on ur own

and they never taught that

[jaunty scaffold]

$c = \frac{2}{\frac{1}{a} + \frac{1}{b}}$
$\frac{c}{a} + \frac{c}{b}$ = 2$
$\frac{c}{a} - 1 = 1 - \frac{c}{b}$
$\frac{c-a}{a} = \frac{b-c}{b}$

Y it no work

[upper karma]

they never told me it was allowed so im going to assume its not allowed

if i do assume its allowed

thats a gamble

[jaunty scaffold]

Imma just get a pen and paper and send. 😂 Well basically you can manipulate (c-a)/a = (b-c)/b to c = 2ab/a+b

Yup

Afaik, it is just another way of defining Harmonic Mean of 2 quantities

Oh. I see. You're confused over how they achieve that expression and not about why that's HM.

They used the property of similar triangles that the ratio of corresponding sides of similar triangles is proportional in this diagram

Yes I am writing it out, one moment please.

[upper karma]

what edits should i make or am i fine

.

bruh

ok

im doing the indirect proof righ

right

they want me to do it like this

ok

would this work then

this is a direct proof btw

oh ok

i think i have to put that in

bc i cant think of any other reason

[jaunty scaffold]

Sorry to keep you waiting, multiple ratios here are tripping me too. 🤦

hope this helps @upper karma

Most likely there's something easier which I am unable to see right now

[opaque crater]

Can anyone help pl

pls

<@&286206848099549185>

[sly marlin]

this isn't geometry btw, it's closer to #probability-statistics

though it's pretty basic (compared to other topics in probability)

yeah so this is just applying the definition of expected value @opaque crater

[fiery solstice]

how do I write the equation of the line BC to include the y-coordinate of the point C? I got the slope

[trim breach]

You mean like point-slope form?

[fiery solstice]

nvm I had it all wrong

[upper karma]

is this correct

[wise pawn]

yup

[upper karma]

right?

[upper karma]

yes

[upper karma]

explain this somebody?

[untold cosmos]

some one plz help on this

[slender oyster]

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

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

For 27 Just Apply Pythagoras Theorem

sin R = TS/RS

Find TS using Pythagoras Theorem

or u can find cos R using b&c and using sinx = sqrt(1- cos^2 x) convert it to sin

[storm heath]

Can I get proofs that SAS, SSS, ASA and AAS are congruent triangles.
And SSA is not.

KhanAcademy proofs is not very good for that.

[slender wasp]

(ASS is not sufficient I think for congruence I believe anyway?)

[storm heath]

Sorry, it was a typing error.

I meant SSS.

[upper karma]

Arc Length = 2 x pi x r (theta/360)

Theta is the angle Subtended by Arc of Arc Length

[supple talon]

theta= l/r

[blissful bay]

Does anyone know like how to start this off? I don’t want the answer but I’m just confused as to how I would solve it

[silent plank]

37. outside angles theorem (for circle)
38. properties of tangents + angle sums

[blissful bay]

Thank you!!

[silent plank]

inside angles theorem (circle)

[unique badge]

@fringe basin ask here

[upper karma]

how do you do number 6

[storm heath]

How do I rigidly transform the 1st figure into the 2nd one?

Or the 2nd one into the 1st one?

[storm heath]

Ehhh anyone?

[crimson sluice]

do you mean change the positions of vertices so they coincide?

if yes, then the first triangle should be positioned as def, the points d, e and f being positioned anticlockwise

[slender wasp]

What have you tried?

[upper karma]

Wdym?

[slender wasp]

Have you attempted the problem / what ideas do you have

Feels less helpful just giving the answer

[dark sparrow]

we don't give out answers here anyway

@upper karma are you here to learn or did you just want someone to do this for you?

[upper karma]

I have the concept of doing and plus I already did that answer so no

And I can't find a pencil

[dark sparrow]

...

[slender wasp]

yeah thought so, sweet

[mystic coral]

how do i find the length of a non right angle triangle

@dark sparrow

[smoky owl]

Hello

[mystic coral]

hi

[smoky owl]

can i get some help in this topic?

im doing preparations for my next year courses

[mystic coral]

yes

[smoky owl]

and i dont really got anyone to teach me

[mystic coral]

so what degrees do u have if u have a flat line

[smoky owl]

180

[mystic coral]

so see that 105 if u take that flat line how to u find the angle conected to it

[smoky owl]

uhm

[mystic coral]

so u have a flat line

an ammount of it is 105 degres

[smoky owl]

d to c?

[mystic coral]

what is the missing angle

[smoky owl]

that flat line

[mystic coral]

ye

[smoky owl]

okay

[mystic coral]

il give a hint substraction

is needed

[smoky owl]

dfc

missing

oh i think i get it

d to c is 180

f d c is 39

and f c d is 39

so dfc is missing?

[mystic coral]

yes

are u trying to find out every angle or what?

[smoky owl]

trying to find x

the blue angle

is what im trying to find

[mystic coral]

ye

so if u do 180-105 angle D I J is 75

[smoky owl]

ohhhhhhhh

so the mid angle is what you subtract from the straaight angle to separate the triangle

[mystic coral]

since is it a flat line

and u know 1 angle and u know that a flat line is 180

u substract that angle u know from 180

[smoky owl]

oh ok

so than i find ijd

than its the same as x

39+75 = 114 - 180 = x

[mystic coral]

why 114-180?

[smoky owl]

because the 2 angles that we found out than we minus to find the j angle

[mystic coral]

isnt it 180 -114 because 180 is the sum of the angles of a triangle

[smoky owl]

oh yes

i forgot

66

x = 66

right?

isnce j and x are the same

since*

[mystic coral]

yep

[smoky owl]

tysm

i get it now

B)

[mystic coral]

ok

[smoky owl]

the answer for this one would be 80 right?

because if 14 outer = 50 than the other side 14 is 50 as well, and the total angle is 180 so 180-100 = 80

@mystic coral

[mystic coral]

yep

[smoky owl]

ok ty

[upper karma]

plan for proof?

[smoky owl]

yo momma

[thorn rover]

so far, the I found the angles 47 and 90. Also, for b, the value I believe is 50. How would I determine the river width? Would I need to determine each side's length first?

i found c = 46.63 and a = 68.37

[thorn rover]

anyone there? i could really use your help. this is the only problem I have left.

[halcyon mango]

is the river's width the actual question? is that what it's meant to be?

or the distance from A to B

[thorn rover]

the width of the river

[halcyon mango]

the first sentence makes me think otherwise

[thorn rover]

I already solved for each distance between each point: C - A = 50, C - B = 68.37, and A - B = 46.63

[halcyon mango]

how would the river's width affect any of the numbers given or calculated?

i mean sure if it exceeded them

but within the margins

[thorn rover]

I asssume it wouldn't, but I just want to know how wide the river is, I thought I would need to find the angles and sides first though.

[halcyon mango]

it's not like you're given any info about the river

you'd need to know at least one more side of the river

[thorn rover]

you mean like in between A - B?

[halcyon mango]

well the either the width itself, or the line segment of the hypotenuse that crosses the river

there's other grammar mistakes in the problem which is why i think it was miswritten

[thorn rover]

is there anything I could at least to get the one side of the river?

[upper karma]

Q. 6 plzz anyone?

[storm heath]

Can someone explain this to me please?

[storm heath]

Nice drawing!

Ohhh so that is the criterion.

Thanks!

[soft laurel]

hey does anyone know how to solve this ? if so could u show how u solved it so i could understand ?

[trim breach]

Do you know what a full rotation is in radians?

[fervent valley]

How does it turn into 2x - 4?

[silent plank]

wdym turn into?

both sides of your initial equation were differentiated to get that result

[fervent valley]

oh no like how x^2 - 4x + 5 change into dy/dx = 2x - 4

I forgt

Just that part

[silent plank]

both sides of your initial equation were differentiated to get that result

(using something like power rule)

[fervent valley]

wait wdym

could u walk me through an example?

[silent plank]

look up differentiation (using power rule)

[fervent valley]

This rule?

[silent plank]

yes

[blissful bay]

<@&286206848099549185>

[upper karma]

what theorem/property is this

[upper karma]

well you can start by noting that the diameter of both spheres = 18cm

since if you put a line through the middle of the rectangle and cut it in half, it'll be the diameter of the circle

[rare sierra]

I am confident that this is the answer-

34.83 sq cm
Somebody correct me if I am wrong

[blissful bay]

Oh so it basically splits it in half so the radius is 4.5

[blissful bay]

Thank you!

[rare sierra]

Yea what he said

[upper karma]

I think the rest is fairly simple?

[blissful bay]

Yupp thank you! I think u are correct, I got 34.77cm^2

[upper karma]

perfect

[blissful bay]

Yeah I just had to do it on paper

[rare sierra]

Yep, I only went to 3.14 so you are right

[blissful bay]

thank u guys so much!

[rare sierra]

Np

[upper karma]

my pleasure

[upper karma]

is there a name for it? I just know they are parallel cause my teacher said so 😎

[rare sierra]

I think there is a theorem

I am going through his steps to see what he proved so far

[blissful bay]

I think that it’s either angle side angle or just transitive property

cuz the given has AE is congruent to AD

And then the statements are that the angles are similar. Which is why they are parallel

[rare sierra]

All A D C angles are congruent is basically what he said so far

with some additional angles

And since we can picture

well

Assuming that they are parallel

We can usesegment DC as a transversal I suppose

[blissful bay]

Well didnt they prove that both triangles are congruent

[rare sierra]

and we know that D and C are congruent

[blissful bay]

Cuz it’s an isosceles triangle

So if they proved that they are the same due to sas you can say that they r parallel through substitution/transitive property

[blissful bay]

^

[rare sierra]

You may be right

[rare sierra]

I am gonna check something though

[blissful bay]

aight bc I’m not 100% sure

[blissful bay]

For this what formula would I have to use?

[rare sierra]

Hm

Lets see

We know every segment length

My guess would be to make a lot of triangles

I am not that far into geometry tbh

[blissful bay]

no worries!

[rare sierra]

5x5x8.4

[blissful bay]

I appreciate your help

[rare sierra]

I think I may have got it

sec

so we got 2 triangles that are congruent there

each are about 23

hmmm

and then thats 4 triangles

so

92 for all triangles

then 100

and another 100

then a 30

85

Alright my best guess is

about 407

I honestly just messed around with sides and found the areas of them

I dont know the formula if there is one

but the again I am not far into geometry so dont take my answer as yours

very well may be wrong

[blissful bay]

Yeah same I’m mixing them up. But I think that what u have is close

[rare sierra]

Yea I know its probably close to the answer if not the answer, but I moved some things around that I don't know if I was allowed to

[blissful bay]

Bc it’s a worksheet with answers attached n I just cant seem to get the answer

[rare sierra]

Do you know the answer?

[blissful bay]

I’m not sure but thank you

Yeah

[rare sierra]

What is it

[blissful bay]

One sec

104.76mm^3

[rare sierra]

104

huh

well I got the answer 104

[blissful bay]

Right that’s what I’m thinking

[rare sierra]

for one part of it

that was

lemme check

nvm it was 102

well I can look at a formula

ill brb

[upper karma]

add up the area of both trapezoids and the 4 rectangles

[rare sierra]

Well thats what I tried

So where did I go wrong

[blissful bay]

Thank u

[blissful bay]

Not sure but I’ll try again

[upper karma]

can a rhombus be a rectangle

[upper karma]

Depends on your definition of rhombus

But id say yes

[azure matrix]

Yes, I'm pretty sure it can.

[upper karma]

ok

did i get it right

[blissful bay]

For 14 which formula am I supposed to use bc I just don’t know what or how to start

[halcyon mango]

try to see if you can decompose that shape into simpler shapes for which you know surface area formulas

[upper karma]

D

[true gate]

what is the volume

pls help

[olive cove]

Google formula for volume of a sphere

[fickle mantle]

if you morph a triangle ionto a circle, at what point does it go from 180 degrees to 360... and why does it seem like 360 degrees if i imagine walking on a triangle path and using a compass?>

[wise pawn]

sounds like you're conflating interior and exterior angles

[fickle mantle]

if i walk on the inside opf a circle its 360 right?

but your right, i would turn 180-60 degrees

for qual triangle

equal

[wise pawn]

if you're morphing a triangle into a circle, then I'm imagining you're blowing it up like a balloon or something

the angles are going to get larger as you do that

but those angles are not going to turn into the 360 degrees of rotation in a circle though, the 360 degrees is measured from standing at the center and rotating around

not the same as an angle measured from a vertex of a triangle

a circle doesn't even have any vertices to measure this way

[fickle mantle]

im thinking of walking on a line

[wise pawn]

as you walk around a path and measure your change in angle, you'll get 360 for both yeah

[fickle mantle]

thank you

[wise pawn]

yeah you're welcome

[brazen topaz]

Hii! I need help..

In the attached figure, CH is the height of the equilateral triangle ABC, BP ​​is to PC as 1 is to 2 and Q is the intersection of the lines AP and CH. The value of area of ​​triangle AHQ / area of ​​triangle ABC is

[woeful finch]

can anyone help??

[ionic bluff]

what 2d shape will the crosssection of the solid look like through the plane of the diameter?

[true knoll]

1:6 according to me, tho i simply didn't apply formula but the fact that nature loves symmetry, so i divided the whole triangle into 6 smaller triangles of equal area, now we know that area of an equilateral triangle is root3/4 a^2, where a is the side of the triangle and now as i said that i divided triangle ABC into 6 equal parts then the area of triangle AHQ world be 1/6th of the triangle ABC

i mean i just applied logic here

🥲

[true knoll]

umm use the formula for frustum?!

[rich wolf]

,w 20 + 2sqrt(32^2-30^2)

[somber coyote]

Results provided by WolframAlpha

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

Requested by AMD#4434

[rich wolf]

,calc sqrt(32^2-30^2)

[somber coyote]

Result:

11.13552872566

[obsidian spindle]

how does one find the centre of a hole?

[dark sparrow]

context?

[obsidian spindle]

I work with low cost machine vision systems. manufacturers want to detect drilled holes in plane wings

the problem is how do you approximate the centre of a hole to a high precision.

[dark sparrow]

is the hole known to be circular?

[obsidian spindle]

so far the computer can detect the x,y coordinates that build up a contour

[dark sparrow]

hm.

are those like, pixels?

the ones highlighted in green on the right

[obsidian spindle]

thats the problem, the hole is never circular in practice, it is an ellipse mostly and always has outliers

yes

[dark sparrow]

why not take the arithmetic mean of the green points then

[obsidian spindle]

this gives a centre that is prone to outliers

[dark sparrow]

can you show an example with a lot of outliers

[obsidian spindle]

ya

this hole is at an angle, and the lighting is not optimal which confuses the camera

[dark sparrow]

hmm

these outliers seem clustered

maybe try to separate them off somehow?

[obsidian spindle]

im think this, consider the contour of the hole,

if we draw a line between every point

then draw a perpendicular line towards the middle,

[dark sparrow]

sounds kinda overcomplicated

if you can draw a line separating the big contour from the two small ones, you can just ignore points belonging to the two small ones

[obsidian spindle]

no no, these are not part of the contour of the hole

this is considered a different contour

in practice, I discard these, the focus is on the ellipse/circle in the middle

[dark sparrow]

so why not take just the hole contour

if you discard them already

taking the mean should work fine

[obsidian spindle]

I'll try this way thanks

[brazen topaz]

Thankss!! I did the same

[true knoll]

ah no all cool XD

[modern heart]

Heyyy

What would you call an area enclosed by 2 concentric arcs?

Found it, it's called an Annulus Sector
if you guys have a better name for it, lemme know

[untold cosmos]

plz

help

im in need

[stoic chasm]

Alright

So for question 1)

@untold cosmos I'd like to know that you're still here btw

[untold cosmos]

ye

[stoic chasm]

Before I start the explanation

Ok.

[untold cosmos]

im here

[stoic chasm]

There are two ways you can do this.

you can use the Pythagorean theorem

or the distance formula.

Which one would you like me to use?

I think that the distance formula would be better.

[untold cosmos]

ok

[stoic chasm]

Since that's more applicable..

alright!

Would you be fine if i screenshare?

You don't ahve to talk.

[untold cosmos]

ok

[stoic chasm]

You can just join teh voicechat.

👍

ummmm

bruh we have like 2 vc

join this server

https://discord.gg/homework

real quick

they have a lot of voice channels

@untold cosmos

[untold cosmos]

ok

ill join

[stoic chasm]

UGH IM TERRIBLY SORRY!

MY DISCORD CRASHED

Join again @untold cosmos

(the vc)

Hello?

@untold cosmos

Are you there?

[untold cosmos]

yes

i typed their

[stoic chasm]

ok join teh vc again

im really sorry about that

[untold cosmos]

its ok

[humble pulsar]

have you drawn it?

[stark apex]

Its A?

[trim breach]

No.

[stark apex]

how is it not?

[upper karma]

anyone know how to do this

[quick cedar]

I can help!

first find the co-ordinates of the 3 initial points

then reflect them in the given line by swapping their x and y values

and then dilate them as given be leaving A' the same and finding the midpoint of P' and A' for P'' and the midpoint of L' and A' for L''

do you get why those are the steps?

[upper karma]

its abit blurry so idk what the coordinates are

for L i got (2,4) for A I got (-7,-1) and P i got (-4,7)

idk what to do after

tho

@quick cedar

[quick cedar]

i think P is (-3, 7)

so then what you do is to flip the points in the line x=y, we swap around the x and y values of every point

so L' (4,2) A' (-1, -7) and P' (7, -3)

[upper karma]

ooo why do we switch the coordinates for geo

[quick cedar]

well if you think about flipping a point in the line x=y

youre drawing a line from that point to the line and then extending it by that same length

the two points need to be symmetric in the line

and as such (a, b) flipped in x=y is (b, a)

[upper karma]

ooo

yeah that makes sense

i forgot how to do dilate by a scale factor so now coordinate geo is biting me in the ass

[quick cedar]

oh thats all good

well if you imagine something dilating by a factor of it

2

its getting twice as big, however point A' doesnt move

[upper karma]

ah so the prime shape will still be within the original?

[quick cedar]

in some way yes. We are basically taking the triangle and making the sides twice as long

keeping the angles the same, and the position of A' the same

so we know that the midpoints of the sides of the new triangle will be the old triangle

[upper karma]

ah so does only L' and P' change

[quick cedar]

yes!

[upper karma]

huh

[quick cedar]

ok so if you think about it, P'' is twice as long as P'

but since the line hasnt moved, only gotten longer that means that if you find the midpoint of P'' and A'' it must be the old point of P, P'

[upper karma]

yesh

ohhhh

so u just ahlf the points of P

*coordinates

[quick cedar]

no you need to make it longer

and also it needs to be relative to A'

[upper karma]

ooo

[quick cedar]

so this is a bit confusing but

do you know the midpont formula?

[quick cedar]

knowing this, we want to find P'' so that MP''A'' is P'

and that L" so that ML"A"

[upper karma]

oh sh

sorry

[upper karma]

yeah

(x1 + x2)/2, (y1+y2)/2

idk how it applies

to da question

[quick cedar]

because imagine a string from point a to b and its 2cm long. If you double its length and put one end on a then the middle of the string will be on point b, because now the string is 4cm long, so 2cm is the middle, and 2cm from a is where b is

so the same works for dilation

[upper karma]

oooo

[upper karma]

cant we just divide each cordinate by 2

or nah

[quick cedar]

no because A is not the origin

[upper karma]

ahhh

so than what do we do

[quick cedar]

so you want to find the coordinates that make the midpoint of P" and A" = P'

and the same goes for L"

its a bit complex because instead of finding the midpoint of two points, you need to find the point which as a midpoint at a point we know already

[upper karma]

wait let me send u the answer key fro the mock exam question im doing

[quick cedar]

ok lets just go from here. Do you get why we have these points here

[upper karma]

ye because erm we had to shift

[quick cedar]

yep so want to find the point L" so that the midpoint of it with (-1, -7) is (4, 2)

try and find it

[upper karma]

yeh tahts what im confused ab it

[upper karma]

so uh 1.5,-2.5?

[quick cedar]

no you worked out the midpoint of A' and L'

[upper karma]

[quick cedar]

we want the point so that M of L" and A" is L'

[upper karma]

im a bit lost

[quick cedar]

ok so we have point (-1, -7) and point (4, 2) and we want to find the point such that the middle of our unknown point and (-1, -7) is (4, 2)

[upper karma]

OHHH

so -1 + x/2 = 4 and 7 + y/2 = 2?

[quick cedar]

yep

[upper karma]

im confused why doe sthat work

and wouldnt that be 9,11

[quick cedar]

yep

[upper karma]

huhhh

[quick cedar]

wait

no

because its -7

sorry not +7

[upper karma]

oo

oops

so -7 + 9/2 = 1 tho

wait no

[quick cedar]

well yeah but 9 is the x value, and -7 is the y value

[upper karma]

-7 + 11/ 2 is 2 yeah

idk why it works tho ahh

[quick cedar]

yep

Ill draw it for u

[upper karma]

alrrr

[quick cedar]

obviously the line is on a plane but in essence this is what we want

we have the points A and L and we want to find the point that has midpoint with A, at L

because we essentually stuck a pin in A, and pulled the line to be twice its lengt

[upper karma]

huh

[quick cedar]

ok so to dilate a line with factor 2 is to make it twice as long

[upper karma]

yesh

[quick cedar]

and the question said we make it twice as long from vertex A meaning that point A does not move

so we double the length, extending the end of the line

notice that if we double the length of the line, the halfway point of the new line is the end of the old line

[upper karma]

ohhh

so from a' we have to make the line twice as long from both vertices?

[upper karma]

btw whats the distinction between '' and '

I kinda understand what u mean now oWO

[quick cedar]

im glad 🙂

[upper karma]

AHH WAIT CAN U EXPLAIN WHY A ISNT CHANGED AGAIN

IM A BIT SLOW WITH GEO

SINCE I JUST STARTED TODAY HJKDEHED

[quick cedar]

its all good!

we are dilating the triangle at vertex A

so we are making the triangle bigger but where? it could be anywhere

so we double the triangle's side legnths, and keep point A where it is thats what it means to dilate from vertex A

[upper karma]

OO THANK U

[grizzled oriole]

hey i need help

imagine you have a triangle ABC and you know all of its corner's coordinates and you also know X

you need to find the coordinates of F and G

given the distance between them. for example 5

any idea?

[idle wadi]

@grizzled oriole X is a vector?

if so then G would be the origin

F would be the unit of that vector * 5

[untold cosmos]

plz

[trim breach]

Use Pythagorean Theorem to solve the triangle, then express sin(θ) as the ratio between the sides.

[grizzled oriole]

X isnt a vector its a point

and XG as a vector isn't perpendicular to BC

[subtle palm]

ok uh um

don't think you can do that

unless ab||fg at the minimum can be done

and/or like before if x vector

[grizzled oriole]

its possible

there is only one solution

[cold totem]

how do you get a^2 + b^2 = c^2 from this?

[pure cape]

you can cancel something out in both sides of the first equation

and b+a=a+b

[upper karma]

You mean how equality arises?

[cold totem]

i got it

i had considered it only in terms of rhs and a^2 + b^2 = c^2 for some reason

disregarding the lhs

thanks

[storm heath]

How is Triangle DAE Isosceles?

[dark sparrow]

ABDE is a square.

[storm heath]

So?

[humble pulsar]

so... AD = AE

cause it's a square

[storm heath]

Yes.

Woah yes!!!!

Thank you!

[quick cedar]

this is not true

[humble pulsar]

DE=AE*

[quick cedar]

yeah sorry

[fallen path]

Can someone help me with these 2 questions

<@&286206848099549185>

[trim breach]

Consider using the Law of Cosines for the top one, and Law of Sines for the bottom.

[thorny terrace]

#❓how-to-get-help

also @untold cosmos , your question isn't really appropriate for this channel. Your question is a probability/stats question and as such, should be in a probability/stats channel.

[sly iron]

can someone tutor me geometry and algebra 2?

[storm heath]

Is not this congruent by SSS?

[dark sparrow]

you have posted a low-res picture, but it looks like GH and IJ are marked parallel

so it is not known whether or not they're congruent

[storm heath]

Ohhh.

So it means parallel.

Thanks Ann!

[grizzled galleon]

i have question

why is the matrix for the reflection in the line y=mx

this

[livid moss]

@grizzled galleon because the reflection of (1,0) through the lines is ((1-m^2)/(1+m^2), 2m/(1+m^2) and the reflection of (0,1) is (2m/1+m^2, (m^2-1),1+m^2)

You reflect the basis vectors

So reflect (1,0) and that gives you the first column, and reflect (0,1) and that gives you the second column

[weary hearth]

@grizzled galleon u know about householder matrix?

if $a$ is a vector then reflection matrix along the hyperplane normal to $a$ is given by $$H_a = I -2\frac{aa^t}{a^ta}$

[somber coyote]

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

[weary hearth]

u just put (-m, 1) in place of a

[soft arrow]

Hey
I'm trying to create an object on a cricle with x distance from centre... actually I want to make 8 objects, each 45 deg away from each other, but I'm working on a function to do it... I'm very bad at it... can any1 help ? >
function >

for inx in range(0, 8):
    light = rt.CoronaLight()
    light.width = 2
    light.height = 2
    x = math.cos(inx) * 10
    y = math.sin(inx) * 10
    light.pos = rt.Point3(x, y, 50)

current result :

I also need to rotate them to face centre but thats another problem o.o

uu maybe inx is angle o.o

[upper karma]

Multiply inx by pi/4

[soft arrow]

yeh was just test-printing it 😄

[upper karma]

What library are you using?

[soft arrow]

python, math

aaa it works! 😄

inx = (math.pi/4) * (inx * 45)

[upper karma]

no I meant the rt library

But I’m glad it works

[soft arrow]

oh, rt is 3ds max python stuff

ok, what about next problem, make each object face the centre o.o hmmm

[upper karma]

That depends what way they are facing naturally

If they’re facing up, rotate by pi-inx I think

Or something like that

[soft arrow]

mmm, I think thats easier as I just use auler angles, so each angle is inx *(360/itemCount)

so I think thats ok, but I broke my 1st test now 😄

this >
inx = (math.pi/4) * (inx * 45)

why is math/pi/4?

math.pi/4

[upper karma]

Because a full rotation is 2pi

In radians

[soft arrow]

hmm

[upper karma]

So an eighth is pi/4

[soft arrow]

so math.cos/sin wants radians ?

[upper karma]

Yes

You can transform degrees to radians using math.radians iirc

[soft arrow]

something like this

 count = 8
    degCount = 360/count
...
inx = math.radians(inx*degCount)```

[upper karma]

Yea

Although modifying the iteration variable is generally bad practice

[soft arrow]

count = 8
degCount = 360 / count
for inx in range(0, count):
    light = rt.CoronaLight()
    light.width = 2
    light.height = 2
    i = math.degrees(inx * degCount)
    x = math.cos(i) * 10
    y = math.sin(i) * 10
    light.pos = rt.Point3(x, y, 50)

I broke something 😄

[upper karma]

It would be easier if I could test it

Please send the full code

[soft arrow]

def makeCoronaLight():
    count = 8
    degCount = 360/count
    for inx in range(0,count):
        light = rt.CoronaLight()
        light.width = 2
        light.height = 2
        light.rotation = rt.eulerAngles(90,inx*degCount,0)    
        i = math.degrees(inx*degCount)
        x=math.cos(i)  * 10
        y=math.sin(i) * 10
        light.pos = rt.Point3(x,y,50)

you have max?

or python/pyside2?

[upper karma]

Never heard of it

But ill install real quick

[soft arrow]

nah its big app

hours to install

do you have pyside2?

I can make gui app example easily

[upper karma]

Ok its installed

Send the file

[soft arrow]

ok 2 min

leme mock it up

[upper karma]

I have the libraries I just want to know what the imports look like

[soft arrow]

yeh im making it now

if u dont have max I have to build one using pyside

all good 2 mins

[upper karma]

I have it

[soft arrow]

from PySide2.QtCore import *
from PySide2.QtGui import *
from PySide2.QtWidgets import *
import sys
import math

app = QApplication()
w = QWidget()



count = 8
degCount = 360 / count
for inx in range(0, count):
    light = QLabel(str(inx),w)
    light.setFixedWidth(10)
    light.setFixedHeight(10)
    i = math.degrees(inx * degCount)
    x = math.cos(i) * 100+100
    y = math.sin(i) * 100+100
    light.move(x, y)
w.show()


sys.exit(app.exec_())

just resize the widget

woa hes jumping all over the place o.o

eee

I needed radians! 😄

taadaa 😄

danke!

[upper karma]

yeah you used math.degrees lol

ok great

[soft arrow]

dis amazing o.o

[fervent valley]

hey can anyone explain this

[weary hearth]

do you know wavy curve method of solving this kind of inequalities?

[livid moss]

wavy curve method

It's funny because I think know exactly what it means

[weary hearth]

ya that's a weird name

[placid gate]

help?

[sick sable]

Looks like (m\angle WTL=m\angle QTR), as they're opposite, such that (2\cdot(2x-1)=3x+7\Rightarrow 4x-2=3x+7\Rightarrow x=9). Then, (m\angle PTQ= m\angle LTS=5\cdot9+6=51^{\circ}).

[somber coyote]

SubGui

[vernal quartz]

Does anyone know of a good source of trigonometric proof questions which start at around UK A level level and work their way up to stuff you would see in olympiads?

[earnest echo]

SL Loney perhaps

[vernal quartz]

thanks, this looks good!

[storm heath]

How did he do this?

[dark sparrow]

there is a very thin gap spanning the diagonal in the 5×13 rectangle, and its area is precisely the missing unit

it's a parallelogram whose sides have slopes 5/2 and 8/3, which are deliberately made to look very close together

[valid osprey]

oh fr

[dense marsh]

hey guys ........

m

m quit new up here

[upper karma]

just go ahead and ask ur question

[dense marsh]

been just ooking for some help si i can deal with trigonometry tho ..........

i just

well first when it comes to complex numbers

and write them under the trigonometrical forme .........

and when it comes to integrate well

things like arcsin and arccos

i quit face that a lot so ........

and m 13 btw

m james ..........

[upper karma]

oh darn I don't really know for the complex number one, haven't really delved into it unfortunately

but wdym by integration arcsin and arccos? that should probably go in #calculus if you are talking about purely integrating those two trig functions

[dense marsh]

discord looks quit complicated tbh , it was just the fact that many people use it thta's why i , quit thought i can give it a try and well see how can i get some help

[upper karma]

well there are some nice youtube videos that may be help navigating all of this, so that may be of help

[dense marsh]

how about equations that contain trigonometry functions tho

i really got serious troubles with that ........

like serious ones ........

[dense marsh]

means i can find out how to work it on youtube then

huh