geometry-and-trigonometry

[naive scarab]

yeah then this is not a good question

[upper karma]

i think our class got cheated out of 2 points

[naive scarab]

but I can tell you how to answer it if you have the axioms and definitions

[upper karma]

dang

[naive scarab]

if you want, you can take that to your teacher

[upper karma]

Yeah

[naive scarab]

“you didn’t define this thing, so it’s not really fair”

i’m a math phd student too if that gives you more credit lol

[upper karma]

first i wanna see if anybody got it right

[naive scarab]

ok so to be fair, I don’t think euclid defined “same side” either, though he uses it a lot

(this is still bad)

here are all the original axioms and definitions if you want to check it out: https://mathcs.clarku.edu/~djoyce/elements/bookI/bookI.html

[upper karma]

k

[naive scarab]

to give you a bit of the historical context, as far as we know elements by euclid is the first mathematical text that had rigorous proofs in it

like there is older stuff that had true mathematical formulas and stuff, but there was never a development from axioms or proofs

the greeks started coming up with this stuff around 800 BC if i remember right, studying classical geometry in particular

they didn’t have cartesian coordinates or equations like we do today

everything was described geometrically

so for example, the pythagorean theorem wasn’t “a^2 + b^2 = c^2”

[upper karma]

it's funny seeing the wording of these axioms and stuff compared to the wording of the stuff i do in class it's the same thing but worded very differently

[naive scarab]

it was “if a right triangle has legs a, b, and hypotenuse c, the squares on the lines a and b sum to the square on line c”

yeah it is worded really differently these days

[upper karma]

mhm

[naive scarab]

a lot of the modern axioms weren’t developed until the 20th century

so back to euclid: it’s thought that elements was basically a compendium of known geometric proofs that the greeks figured out

like euclid probably compiled all the results and put together the presentation

[upper karma]

it went from "a line is breadthless length" to "a line is infinite"

[naive scarab]

yeah 🙂

although some of that is a quirk of the translation from ancient greek

[upper karma]

yeah

[naive scarab]

i think the translation most people use was written by this specific guy in the 20th century

in english i mean

[upper karma]

Yeah

[naive scarab]

the last fun historical note: for basically all of history, if you were to become an educated person, one of the first things you’d learn is geometry out of euclid

cause that taught you how to think, make specific and rigorous arguments

it’s kind of incredible that this one book has been used for 2500 years

[upper karma]

Yes

Also

[naive scarab]

oh so if you want to see what the modern axioms would look like for planar geometry, they’re basically this: https://en.m.wikipedia.org/wiki/Hilbert's_axioms

[upper karma]

K

[naive scarab]

in the 20th century mathematics became a lot more formal and rigorous, things started getting built on the foundation of set theory

it went from 5 axioms in euclid to 20 here

plus the axioms of set theory lol

for you, i’d focus on doing things with the 5 original axioms

they’re not perfect but you can do a lot with them, and you’ll learn a lot

[upper karma]

I really only do stuff rn with the things I do in class

[naive scarab]

yeah that’s totally fine

[upper karma]

we have a different postulate called the parallel postulate as well

[naive scarab]

ok cool, what does that say

[upper karma]

but it's totally different

Lemme pull it up

[naive scarab]

it’s probably equivalent in some way, we’ll see

[upper karma]

I doubt it

[naive scarab]

well it is called the parallel postulate....

there are lots of ways to write it

some that don’t seem equivalent on the surface but actually are

[upper karma]

through a point not on a line, one and only one line can be drawn parallel to the given line

definitely not the same

[naive scarab]

i think it actually is the same

[upper karma]

???

how

that's all it says

[naive scarab]

yeah i know lol

so we can prove with that axiom that if two lines are parallel, then the interior angles on either side sum to 180

[upper karma]

I guess

[naive scarab]

so if we draw a line with different interior angles, we know those lines aren’t parallel, they intersect somewhere

[upper karma]

yeah...

But

[naive scarab]

you probably need the axiom “lines intersect at 1 point at most”

ok so on one side the angles are less than 180

and on the other side they’re more than 180

the intersection can’t be on the more than 180 side

[upper karma]

k

[naive scarab]

because you’d be able to draw a triangle with more than 180 degrees

so therefore, they must intersect on the less than 180 side

also, if you have the original form of the parallel postulate axiom, you can prove there is only one parallel line

[upper karma]

yes

[naive scarab]

so they’re equivalent axioms

so here’s maybe how i’d do this problem as rigorously as I can:

step 1: prove from the axiom you’re given the original version of the parallel postulate

meaning the lines will intersect on the side where the interior angles are less than 180

step 2: use that to show p and q are not in the same side as I outlined above

[upper karma]

yeah

[naive scarab]

imma be real though, this question is way too subtle for people trying to learn this for the first time

[upper karma]

the only thing he told us was that it took him 10 steps but that's with the things I've learned so far

[naive scarab]

especially if you don’t have the original form of the parallel postulate

i bet you $5 i can poke a hole in his 10 step argument lol

[upper karma]

he won't show anybody

[naive scarab]

but i would need a list of his axioms and definitions

[upper karma]

he did it two column

[naive scarab]

maybe you can like, draw a parallel line in the particular diagram you have and work with that, might make it easier in this specific case

i’m gonna play around with this and see what i can come up with

[swift copper]

Hey all! New member here! I'm helping my partner complete a Grade 11 Math course for his admission to college. We're both "well past our prime" lol - him being 30 and I'm 34 -- so it's been a while since either of us have had to look at angles, triangles, SIN COS TAN, etc.

I'm helping him with his homework and studying, and am currently stuck on a problem regarding calculating the length of a staircase (I know how tall it is and what it's slanted height is).

Before I dive into the question -- is this the right room to post my question in? (Not looking for hard answers, but hoping to show what I've done and see if I'm on the right track lol) THANKS EVERYONE!

[naive scarab]

yeah this room is fine, you can also use one of the question rooms further down

[swift copper]

Fantastic! Thank you! I'll ask in here right now, but make note of the questions rooms below for the future. THANK YOU!

[upper karma]

K

[naive scarab]

it’s good to go to those when this channel is actively being used. if nobody answers for 15 minutes, you can ping the helpers

#❓how-to-get-help has more info

[swift copper]

Ah yes! Thank you! I was reading through that and noticed the 15 minute bit 🙂

Actually, I'll throw the quick question here + the solution we worked out

a = c x sin(a)
a = 325 cm x sin(22)
a = 325 cm x 0.3746
a = 121.745
a = 122 cm

Therefore, the lean-to roof’s height is 122 cm.~~~

[upper karma]

aight

why'd you cross it out

[swift copper]

a = c x sin(a)
a = 325 cm x sin(22)
a = 325 cm x 0.3746
a = 121.745
a = 122 cm

Therefore, the lean-to roof’s height is 122 cm.```

Haha it's been a while since I've been on discord too -- wanted quotes not strikeouts lol

[naive scarab]

yeah that looks right

i’d recommend using * instead of x to not get confused with variables

[upper karma]

yeah

[naive scarab]

or you can use $\LaTeX$

come on bot

you can do it

forgot the weird capitalization lol

[swift copper]

Hahah, thank you!
Going on 18 years since I last took a high school math class (Ontario) - so that's a pretty neat ego boost that we did it right lol

hahahah thank you!

[somber coyote]

doubledual:

[swift copper]

Super appreciated! I'll be hanging around here for the foreseeable future lol

[naive scarab]

if you go to the #bot channel, you can play around with how to do that

put math stuff in between dollar signs and the bot will do things for you

you can use more advanced math symbols that way too

$\int_a^b f’(x),dx = f(b) - f(a)$

[somber coyote]

doubledual:

[naive scarab]

$\sum_{n=0}^\infty r^n = \frac{1}{1-r}$

if you hang around you’ll pick up the syntax fairly quick, and it’s not to hard to google, or just ask how to do something

[somber coyote]

doubledual:

[naive scarab]

@upper karma i think i may know what your teacher had in mind

or at least a fairly straightforward proof

[upper karma]

yeah?

[naive scarab]

it’s a proof by contradiction

if P and Q are on the same side, either P is between B and Q, or Q is between B and P

this is by the definition of “same side”

consider triangle APB

[upper karma]

K

[naive scarab]

we know angle APB is 90,

so in particular angle ABP is < 90

[upper karma]

k

[naive scarab]

same reasoning tells you QBC < 90

[upper karma]

yeah?

[naive scarab]

but since P and Q are on the same side, <ABP + <QBC = <ABC

which is a straight line, so <ABC = 180

contradiction

[upper karma]

so like what

An indirect proof?

[naive scarab]

yeah it’s a proof strategy called proof by contradiction

you want to prove a proposition P

assume not P

derive something absurd

therefore P has to be true

here we got that <ABC < 180, but also <ABC = 180

we got that by assuming points P and Q were on the same side

so they can’t possibly be on the same side

that make sense to you?

[upper karma]

yes

[naive scarab]

you could definitely make this into a two column proof if you had to

[upper karma]

we learned that by the name indirect proof

[naive scarab]

ah ok

most people say proof by contradiction

so if you see people say that, you know what it means

[upper karma]

blame my teacher

[naive scarab]

yeah lol

for the record, the parallel postulate is involved in this proof, subtly

the “triangles add up to 180 degrees” thing requires that axiom

there are non-euclidean geometries where that isn’t true

for example, geometry on the surface of a sphere

triangle ABP here has interior angles adding up to 270 degrees here

in this geometry parallel postulate isn’t true, technically no two lines can be parallel

if you define “line” to mean “a great circle on the sphere”

a great circle is a circle that evenly splits the sphere in two, like the equator in earth

also lines intersect twice

also the notion of “side of a line” is not well defined

so this is like a really weird geometry

[upper karma]

So

[naive scarab]

way advanced stuff but fun to think about

(the reason the lines should be great circles is because that is the fastest way to travel between two points)

[upper karma]

Wait I'm writing it down as a two-column

[naive scarab]

ok sure

also angles on a triangle adding to 180 degrees is apparently logically equivalent to the parallel postulate, according to wikipedia

[upper karma]

Hmm

[naive scarab]

the “at most one parallel line” axiom is also equivalent, and it’s usually referred to as playfair’s axiom

[upper karma]

So how do I prove that angle ABP is less than 90 degrees

Wait nvm

[naive scarab]

just to prove i’m not crazy for honing in on that axiom lol

[naive scarab]

i wrote out a 2 column proof and got it in 11 steps, you can probably condense it a little and get 10

so i bet this is exactly what he did

[upper karma]

@naive scarab

[naive scarab]

hi

[upper karma]

what is the thing called that a line creates a straight angle

or how we got that <ABC = 180

[naive scarab]

uh i think that’s an axiom?

[upper karma]

idk

what is it

[naive scarab]

eh maybe not, but you should be able to reference something, i’m sure you use that a straight line is 180 all the time

off the top of my head i don’t know what the name for it is

[upper karma]

Me neither

I'm looking at my notes

[naive scarab]

it’s proposition 13 in the original euclid

book 1

[upper karma]

k

[naive scarab]

https://mathcs.clarku.edu/~djoyce/java/elements/bookI/propI13.html

for the two column proof I might just write “angles on the same side of a straight line sum to 180” as the reason

it’s a proposition that you’ve probably already done, or been implicitly using a lot

if you want to be fancy you can reference euclid book 1, prop 13 lol

[upper karma]

Hmmmnnn

f

[naive scarab]

btw every time you use the pythagorean theorem, you should be citing euclid book 1, proposition 47

really i think putting “angles on the same side of a straight line sum to 180” is perfectly fine

i wouldn’t stress about it

like it’s how you prove opposite angles are congruent for example

and i’m sure you have that

[upper karma]

I dont need to

we can do proofs somewhat informally

idk what I was trying to say

[spark totem]

what exactly is hypotenuse leg?

[upper karma]

k

take a right triangle

so like this

                              /|

hypotenuse / | leg
/__ |
leg

[spark totem]

right

[upper karma]

the other side is the hypotenuse

[spark totem]

correct

so what exactly is the theorem

[unkempt nest]

if a hypotenuse and a leg of a right triangle are congruent to the hypotenuse and the leg of another right triangle then the two triangles are congruent

[spark totem]

ah

i see

thanks fellas

[unkempt nest]

nw

can someone help me with this question

[spark totem]

proofs are horrible

[unkempt nest]

fr I hate it

[upper karma]

in this quadratic formula 4x^2-5x=6 can i move the 6 over and make the equation = 0?

[naive scarab]

@upper karma absolutely. whenever you have an equation, if you do the same thing to both sides the equality is still true. so you can subtract 6 from both sides of the equation

also 2 column proofs aren’t real proofs lol

they’re what happens when attempting to begin real math hits the american education system

[unkempt nest]

lol

[brisk holly]

what would cos theta be?

im thinking its r/r but then arccos(1) is 0 so that doesnt make sense

[slender nacelle]

It will depend on the length AB

There's no right angle triangle here

[brisk holly]

wdym

[slender nacelle]

I mean, this triangle is an isosceles triangle. So there are not base or hypotenuse defined

[brisk holly]

so how do you get the trig functions for an isosceles triangle

[slender nacelle]

There are different ways. What have you been taught?

[brisk holly]

idk

can you just show me one way

[slender nacelle]

Yes, but the angle will depend on the length AB

Unless there's anything given more information

[brisk holly]

show me for an isoceles triangle where you have all 3 lengths

[slender nacelle]

There's something called cosine rule

$\cos C = \frac{a²+b²-c²}{2ab}$

[somber coyote]

oscillatingEquilibrium:

[brisk holly]

how do you know which are a b c

[slender nacelle]

a is the side opposite to angle A, b is the side opposite to angle B, c is the side opposite to angle C

For this question,

$\cos\theta = \frac{r^2+r^2-AB^2}{2r^2}$

[somber coyote]

oscillatingEquilibrium:

oscillatingEquilibrium:

[brisk holly]

thanks

[slender nacelle]

You're welcome

[brisk holly]

this is the full problem

im still kinda lost

[slender nacelle]

Oh you don't need to find theta here

[brisk holly]

yeah i wasnt

i was trying to just get sin

thought maybe that would lead me somewhere

[slender nacelle]

Do you know formula for area of triangle?

[brisk holly]

1/2bh?

[slender nacelle]

When you know two sides and angle made by them

[brisk holly]

no

[slender nacelle]

Area = 1/2 bc sinA = 1/2 ab sin C = 1/2 ac sin B

It's same as 1/2 bh, just a more general way

[brisk holly]

can you format that

[slender nacelle]

$A= 1/2 bc \sin A = 1/2 ab \sin C = 1/2 ac \sin B$

[somber coyote]

oscillatingEquilibrium:

[brisk holly]

so AB would be c ?

[slender nacelle]

Yes

[brisk holly]

so area = (r AB sinA)/2

[slender nacelle]

No

[brisk holly]

what then

[slender nacelle]

I mean, yes

But you have to write it in form of theta

[brisk holly]

area = (r^2 sin(t)/2

is that correct

[slender nacelle]

Yes

That is area of triangle

You need to find the shaded area

[brisk holly]

so I need the total area

[slender nacelle]

Yes

[brisk holly]

of the triangle plus that curve

how do I get that

[slender nacelle]

It's a pizza slice

ie. Some fraction of the full circle

How is area of this pizza slice is related to total area of circle?

[oak knoll]

so uh

you guys done with your question?

don't wanna be rude and just bump in while you guys are still working on a problem

[slender nacelle]

You can ask on a math help channel

[oak knoll]

oh ok

I'm new so I don't really know much

alright thanks

[slender nacelle]

Np

[oak knoll]

@slender nacelle

Do we just go to a random channel that isn't occupied and just post your question there?

or something else?

[slender nacelle]

Like question alpha, beta,..kappa

Any channel that is free

[oak knoll]

ok

thanks

[slender nacelle]

Np

[molten bronze]

I need help anyone free

#geometry-and-trigonometry

[hushed spear]

any help is appreciated

@silent plank can you please help when you can

[brisk holly]

@slender nacelle So I solved it but I just get down to theta = pi/3

that doesnt prove anything

[slender nacelle]

You can't get theta

[brisk holly]

yeah i think i made 1 grave mistake

[molten bronze]

Can anyone help me

With geometry

[brisk holly]

whats the question

[slender nacelle]

You have to figure out the area of cicular segment.

[molten bronze]

Did I do this right ?

It looks wierd so I wanna make sure :?

[slender nacelle]

Yes it looks correct

@molten bronze

[molten bronze]

Kk

Wait I have one more now this ones harder

[brisk holly]

i solved it, lets go

[slender nacelle]

Great

I am not exactly sure about this notation

Does that rotate 90 degree counterclockwise about origin?

And reflect about y axis

[molten bronze]

Yes

[slender nacelle]

Okay so first you have to reflect it wrt y axis

[molten bronze]

Yea and then rotate

[slender nacelle]

What will be new coordinate after reflection?

[molten bronze]

The x negates ;?

[slender nacelle]

Yes

Your answer looks correct

[molten bronze]

Thanks 🙂

[slender nacelle]

You're welcome

[molten bronze]

I have a question about the format of this question

[slender nacelle]

Sure

[molten bronze]

Just with a lot of words :?

[slender nacelle]

Yes

[molten bronze]

I have another question about formatting

[slender nacelle]

Ok

Yes

Option 3 is correct

[molten bronze]

Okay these word prom lens are confusing for me thanks so much 🙂

[slender nacelle]

You're welcome!

[molten bronze]

Oh wait one last question

Doesn’t this question have 2 answers ?

[slender nacelle]

Which?

[molten bronze]

Wouldn’t rotation also work in this context or am I stupid

[slender nacelle]

Rotation changes the orientation

Like a square would look like a diamond

[molten bronze]

For a polygon to?

Ohh I see

If it’s not 72 degrees then it would be different

[slender nacelle]

72 degrees?

[molten bronze]

It would be rotated on top of itself

[slender nacelle]

For a pentagon?

[molten bronze]

360 divided by 5 =72 so if I rotated 72 degrees orientation is same

Yee

Oh that’s polygon

W

[slender nacelle]

Yes

[molten bronze]

Ahh thanks anyway haha

[slender nacelle]

You're welcome!

[molten bronze]

I got those right :?

[slender nacelle]

Yup

[molten bronze]

Yay

[slender nacelle]

Haha

[still venture]

Hey I need some help

[lyric stream]

does anyone know this?

[vague crypt]

Move each coordinate of each vertex of the triangle down to points, and reflect that over $x = 1$

[somber coyote]

LifeSource:

[tropic bobcat]

Helppp

Help

Anyone online?

[eager kraken]

how tf does this round up to 99d 39m

[west basin]

where else would it round to

[eager kraken]

99 38

[upper karma]

@west basin it should round down to 99 38

[west basin]

sorry i only do ceiling

no floor

[upper karma]

@tropic bobcat dont ask to ask, just ask

[steep temple]

i want to find a linear equation such that it passes through the top right "corner" of the left curve and through the "middle" of the two curves on the right, something like below:

$$ f(x) = \frac{x-2}{x^2-2} $$

[somber coyote]

dlp:

[steep temple]

I'm aware not every equation of the form $$ f(x) = \frac{x-m}{x^2-n} $$ produces a graph like this but preferably I'll work on that later

[somber coyote]

dlp:

[slender shoal]

Hey y'all I'm new to the server and need some help with an algebra 2 assignment, I was given a 9th degree polynomial and used a graphing calculator to find three zeros, I then divided these out to get a 6th degree polynomial. I was given 2 imaginary zeroes which equates to 4 zeros and asked to find the last 2 zeros. I turned my 4 zeros into two trinomials, and attempted to use long division to remove each trinomial from the 6th degree polynomial like I was taught. However I was unable to obtain a rational number. I checked with a calculator and it could not find a solution. Anybody know what I can do/what I might be doing wrong?

[silent plank]

can you show your work

you last 2 zeros would be complex

[slender shoal]

Yeah sure, sorry my work is so sloppy it's late and I got stressed out

Of course it's sideways

In the top left are my 4 given zeros that I simplified and combined

At the bottom was my first attempt to divide which failed and so I attempted to use the other trinomial

In the center is just some incomplete idea I had

[silent plank]

so that 6th degree is what you had after dividing your 9th degree?

[slender shoal]

Yes

I synthetically divided out all real zeros

[silent plank]

,w expand (x+1-2i)((x+1+2i)

[somber coyote]

Results provided by WolframAlpha

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

Requested by ramonov#0921

[silent plank]

-2i*2i is 4

not -4

[slender shoal]

Isn't i times i 1?

Oh

Well shit

That sounds exactly like the kind of thing I'd do

[dark sparrow]

no, i * i is not 1.

[slender shoal]

Yeah like I said thats the kind of mistake that I'd make

Thanks for the help @silent plank

[cinder nacelle]

For RHL can I prove triangles congruent if I have a right angle and the hypotenuse congruent

[silent plank]

you need a leg/side too

apply angle addition posulate

[cinder nacelle]

What’s that

[silent plank]

that's for the question above yours

[cinder nacelle]

Oh

[silent plank]

also that question seems to be missing some information

[gray whale]

Could someone be so kind to explain this question?

Ive been stuck for a little while. If you can @ me that would be wonderful

[late monolith]

YO guys

I have to turn this in by :45 and im stumped

i have 13 mins LOL pls help me im begging

if anyone can help me fill this out quickly it is greatly appreciated

[urban knoll]

I’ve been sitting here going insane over triangles🤦🏼‍♂️ do the boxes look good

[dusky surge]

The boxes look nice, they are green and shady, with numbers
Yea, look good to me

[urban knoll]

The little triangle in the top box was the only thing wrong

I’m so mad

It’s only hw so it’s not a big deal

[dusky surge]

Ohhhhhhhh you mean I have to check the yellow boxes too?

Awwww, my bad

Yea, the middle one of the acute triangles box is actually a obtuse triangle.

[dense crypt]

Does anybody know how to do this? I would like an explanation and an answer bc my teacher doesnt explain

[earnest echo]

Those marked angles are corresponding Angles

[gray whale]

https://cdn.discordapp.com/attachments/326138757474680852/779366372940841010/unknown.png

If someone could explain i would absolutely give you a ❤️

been struggling with it for a while please @gray whale me

[hollow raven]

You can use the identity sin(x+y) = sin(x)*cos(y) + cos(x)*sin(y).

oh hmm

you can isolate one of the trig ratios for each equation and just sub those resultant values in for the identity

[dark sparrow]

you can square both sides of the first eq to get 9sin^2(x) + 16cos^2(y) + 24sin(x)cos(y) = 25

and likewise with the other

and add them

@stark nebula @hollow raven

[hollow raven]

I see, thank you

[neon hamlet]

Anyone know how you can solve this?

Just 4 and 5

[earnest echo]

If you have 1 already then you can easily find 3

And 4

[neon hamlet]

How would you get 4?

[earnest echo]

Angle on a line

[neon hamlet]

3 in this case is: 46

OH

Right they all add up to 180

[hollow raven]

think of supplementary angles

[neon hamlet]

Right?

[hollow raven]

yeah

[neon hamlet]

Triangles I know add up to 180 but how would you also account for the x and y?

[upper karma]

okay

our goal is to make a system of 2 eqn with 2 variables

can you think of a first eqn?

[neon hamlet]

Right.

[upper karma]

considering what you just said

[neon hamlet]

y+ 25 + 30 + 65= 180?

[upper karma]

almost

[neon hamlet]

Oh the x

[upper karma]

i know you are looking at the big triangle right?

yes

[neon hamlet]

Right so x + y = 60\

Then 4x + y = 180

[upper karma]

yes

wait

,calc 180-(25+30+65)

[somber coyote]

Result:

60

[upper karma]

okay yes

$\begin{cases} x+y=60 \ 4x+y=180\end{cases}$

[somber coyote]

Al𝟛dium:

[neon hamlet]

Right so - the first equ

[upper karma]

this is the system we want to solve

[neon hamlet]

then add

yeah

[upper karma]

yes

[neon hamlet]

I got it thanks!

[upper karma]

yw!

[neon hamlet]

x was 40 annnd

y is 20

[upper karma]

hold up

[neon hamlet]

?

[upper karma]

yes, that's correct

i was checking

[neon hamlet]

ah okay

Alr thanks once again!

[chrome wagon]

could someone help me with this problem i’m stuck

[limpid gust]

can somebody help me with what theorems i need to prove this statement?

[pallid cloud]

since l parallel to m, so angle 1 equal to angle 3, alternate exterior, or think of it as corresponding then vertically opposite
since angle 1 equal angle 2, and we proved that angle 1 equal angle 3, so angle two equal angle 3, so equal corresponding angles, so a parallel to b

@limpid gust

[limpid gust]

thankyou

[pallid cloud]

not necessarily

you should here add the condition that a parallel to b

here he has to prove that

so indirectly he has to prove what you said

since in your case he has to prove the supplements of the angles 1 and 2 to be equal

which is equivalent to proving that angles 1 and 2 are equal

[cedar oriole]

for Q8 i keep getting an incorrect answer

what i did is:

i have a+pb = (3,2)
so i calculated (3,2) - a = (1,2)
p(3,1) = (1,2)
and I cant solve it

[humble pulsar]

a+pb = k[3,2]
[2,0] + p[3,1] = k[3,2]
[2+3p,p]=[3k,2k]

[cedar oriole]

@humble pulsar oh yeah ofc, thanks

[humble pulsar]

👍

[sick ember]

can any one help me though this work sheet i am new to all of this stuff

[hollow raven]

Idk if its just me but it looks like a quiz

[sick ember]

nah its review

i read the rules

[lyric stream]

does anyone know this?

[vague crypt]

I literally helped you with this

Cmon

Move the vertices down two units and reflect the new triangle about $x = 1$

[somber coyote]

LifeSource:

[mossy valve]

k

ive done number one but for number two i dont remember how to find the adjacent and opposite legs

how would oyu find the adjacent and opposite legs

i know hypoteuse but i dont know which one is which

[lyric stream]

@vague crypt thanks, i'm in a few different math servers so I didn't remember which one i asked it in...

[vague crypt]

Its fine!

[sick ember]

well hello there @lyric stream

[fickle sleet]

@mossy valve Do you understand the definition of an adjacent and opposite angle?

[mossy valve]

no not really

[fickle sleet]

It's fine if you can't exactly explain it, do you have a general understanding? Anything?

[mossy valve]

uhhhh

theyre both next to the 90 degree

just idk what sides theyre on

[fickle sleet]

mmhm

Ok cool

So you know the hypotenuse is always the longest side of a right triangle, yes?

It's a pretty absolute rule. Even without corresponding angles, you would know what the side is immediately just by seeing its length.

Adjacent and opposite sides can't be found using length comparisons, unfortunately.

[mossy valve]

y

e'

[fickle sleet]

In that case, they are found using relative angles. The hypotenuse is always opposite to the 90 degree angle. The adjacent and opposite sides are found relative to another angle given on the triangle apart from the 90 degree one.

The adjacent side is the side that is right next to the angle. The opposite side is the side that is across or furthest away from an angle.

Your question doesn't exactly specify any angles to work with, so it doesn't really matter what is adjacent and what is opposite because there is none to choose from in the first place.

When the second question asks you for acute angles, it is asking you to find the two other angles in the triangle apart from 90 degrees. You do not need to know which side is opposite or adjacent, because you don't know the angles corresponding. You already have the values of each side, 8, 15 and 17.

[lyric stream]

hey lol

[fickle sleet]

Now if you're asking how to solve those two angles, then you would probably need to use the sohcahtoa formula.

[upper pebble]

How would you find x in this triangle? I used geogebra and found out it's 10° but I don't know how to prove it. It seems that every angle works when plugging them in.

[limpid gust]

Am I right so far?

@pallid cloud

[pallid cloud]

1=3 alternate exterior

[limpid gust]

oh 1=2 isnt

[pallid cloud]

1=2 given

[limpid gust]

omg lol i forgot

[pallid cloud]

3=2 transitivity property

QED

[limpid gust]

whats qed mean?

[pallid cloud]

Quod Erat Demonstrandum

that which had to be shown

forget it, don't put it

[limpid gust]

yeah im in geometry

[pallid cloud]

just signaling that this is the end of the proof

[limpid gust]

my teacher is weird with the ends

we just kinda leave it

[pallid cloud]

you can use it everywhere when you write a correct proof, geometry or the rest

😛

[limpid gust]

alright thankyou, but may i ask if that reply feature is new, I have never seen that

[pallid cloud]

yeah it is new it seems

[limpid gust]

How are the 2 things provided similar, the angle b is different in both

I can do most of this myself

[pallid cloud]

b is 58

alternate angles

or alternate interior angles

both are correct, but there are alternate interior and alternate exterior

[limpid gust]

oh those are 2 different problems

im stupid

not what i mean lol

[pallid cloud]

x=82+24=106

[limpid gust]

I thought I needed both for b

[pallid cloud]

construct the parallel to (PQ) passing through B

[limpid gust]

so x is 106

and b is 58

thats simple

I dont know how to show work though

[pallid cloud]

you'll have that the angle from above is 82 (corresponding) and the angle from below is 24 (corresponding), so add them

[limpid gust]

yeah so x = 106

[pallid cloud]

ye

[limpid gust]

But i am wondering if i have to show it

[pallid cloud]

1

[pallid cloud]

2

gonna go to sleep now, it's late and I have a class to teach tomorrow in the morning

[limpid gust]

ok have a good night

[pallid cloud]

thanks👍

[sick ember]

i think i can get though this just need a little help

[dusky surge]

Hint, draw a line on z parallel to AB.

[slate blade]

@limpid gust if this helps IDK if you've already solved it

[dark sparrow]

@novel quiver

[novel quiver]

BEC a right triangle on E, such as, CE = 2 and BE = 8
Let O be a point in the segment [BE] such as OE = 2
We give A the symmetric of C from the point O and I the middle of [AE]

Prove that vector BA . vector BC = 28 , use that to deduce the distance of AB

Solution: Drop a perpendicular from A to BE, and denote the intersection of the line with BE by D.
Then triangles AOD and COE are congruent. (AAS via angle ADO = angle CEO = 90 deg, angle AOD = angle COE (vertical angles), and AO = CO (given, since A is symmetric to C about O)
Hence AD = CE (corresponding sides of congruent triangles) = 2, and DO = EO (corresponding sides of congruent triangles) = 2.
Then you have the lengths of BD (being the complement of DE in BE) and AB (via Pythagoras' theorem), hence you have the sine and cosine ratios of angle ABD.
Then you have cos(angle ABC) via compound angle formula (angle ABC = angle ABE + angle EBC = angle ABD + angle EBC).

Be it S = (point M such as MA^2 + MC^2 = 32)
Verify that A ∈ S

I posted the solution to the first question

could maybe help for some context

[dark sparrow]

uhh

what plane

wait, is this a 2D or a 3D diagram

[novel quiver]

2D*

[dark sparrow]

[novel quiver]

sorry

it's 2D

ehh just forget the plane part

i guess

it's just that we have to prove that for every point M, MA^2 + MC^2 = 32 is verified

i guess

[dark sparrow]

MA^2 + MC^2 = 32 does not seem true for all M

[novel quiver]

hm, wait I probably translated the question wrong lol (the original version is in french)

Be it S = (point M such as MA^2 + MC^2 = 32)
Verify that A ∈ S

@dark sparrow

it should make sense now

what do you think?

[dark sparrow]

bah

il faut seulement montrer que AC^2 = 32

c'est tout

[novel quiver]

mdrr t fr

ok att

pk AC^2 ??

comment ta eu le AC^2

[dark sparrow]

M ∈ phi <=> MA^2 + MC^2 = 32

[novel quiver]

phi?

[dark sparrow]

le nom de ton ensemble

la lettre grecque phi

[novel quiver]

ah ouais dacc

[dark sparrow]

c'est à dire que M est dans l'ensemble phi si et seulement si la somme des carrés des distances de M à A et à C est égale à 32

si tu veux vérifier que A ∈ phi

ça devient

AA^2 + AC^2 = 32

où AA note "la distance de A à A"

qui est évidemment 0

[novel quiver]

ahh je vois, donc le M c'est juste pour donner l'example
et si on veut voir si un point verifie MA^2 + MC^2 = 32
on remplace M par ce point

merci <3 @dark sparrow 💪

tu peux m'aider avec cette question stp?

"pour tout point M" - je dois faire quoi exactement? est ce que je dois montrer que 2 MO^2 + 16 = 32 ??

[main lintel]

lul, speak english please

[novel quiver]

ADO et OCE sont identiques (ASA) du coup AO = sqrt8 et OC = sqrt 8
donc AC = 8 (2sqrt8)

AC^2 = 32
8^2 = 32
64 != 32

@dark sparrow pk c'est pas juste?

[dark sparrow]

$2\sqrt{8}$ n'est pas égal à 8 wtf

[somber coyote]

Ann:

[dark sparrow]

t'as vraiment pensé qu'AJOUTER deux copies de qqch c'est pareil que les MULTIPLIER ?

ou que la RACINE CARRÉE est identique à LA MOITIÉ ?

[novel quiver]

mdr g confondu le * 2 et ^2
ok dsl

MA^2 + MC^2 = 2 MO^2 + 16
on remplace M par A

AA^2 + AC^2 = 2 AO^2 + 16

0 + (2sqrt8)^2 = (2sqrt8)^2 + 16
32 = 32 + 16

qu'est ce que j'ai fait de mal? @dark sparrow

[dark sparrow]

[soupir]

[novel quiver]

lol

[upper karma]

can we non-french humans be part of the conversation?

[dark sparrow]

premièrement : $2 MO^2 \neq (2MO)^2$

[somber coyote]

Ann:

[dark sparrow]

mais le plus importamment : ça dit cette fois de le montrer pour TOUT POINT M DU PLAN

ce qui n'était simplement pas le cas pour le problème précédent

tu peux oublier les B, D et E

y a que les points M, O, C et A

AO = OC = sqrt(8)

M se situe peu importe où

[novel quiver]

pourquoi doit-on les "oublier"? pck ils sont sur la meme droite?

[dark sparrow]

non

c'est pour que ton diagramme ne devienne pas un bordel complet dans lequel il est impossible de voir ce qui se passe

[novel quiver]

hm, dacc

[novel quiver]

du coup dans ce cas on fait quoi?

[dark sparrow]

y a que les points M, O, C et A
AO = OC = sqrt(8)
M se situe peu importe où

dessine le nouveau diagramme

[novel quiver]

ouais mais

ok

[dark sparrow]

ça peut rendre les choses plus claires

[novel quiver]

tu veux que jenleve les points B, D et E? comme ça?

[dark sparrow]

non mdr

[novel quiver]

mdr

[dark sparrow]

c'est tjrs bcp trop compliqué

[novel quiver]

ok donc tu veux que jenleve quelque droites qui sont lies aux points dont on en a besoin?

[dark sparrow]

voilà

[novel quiver]

a we donc on a carrement changer le tout

ok

AO = OC = sqrt(8)

MA^2 + MC^2 = 2 MO^2 + 16

pour prouver que tout point va valider MA^2 + MC^2 = 2 MO^2 + 16

on doit faire quoi?

[dark sparrow]

je suis un peu plus occupée maintenant, donc désolée, je peux pas donner une solution détaillée

[novel quiver]

ok np

[dark sparrow]

tu peux probablement utiliser le théoreme de pythagore ou qqchcomme ça

[novel quiver]

nn mais att

MA^2 + MC^2 est l'équation d'un cercle nn?

lol

quelque chose au carre + qqch au carre

ca y ressemble

[humble pulsar]

baguette

[novel quiver]

@dark sparrow

[dark sparrow]

MA^2 + MC^2 est l'équation d'un cercle nn?
non

la semblance n'a aucune signifiance

[novel quiver]

donc comment on peut savoir si c lequation dun cercle ou pas?

pck si ca se trouve c leq dun cercle

nn?

[dark sparrow]

non...

[novel quiver]

ok 🤔

et pourquoi tu en es si sur?

ca maiderait la prochaine fois

pck jai cru que la ressemblance etait assez pour dire que c leq dun cercle

lol

[dark sparrow]

sûre*

en tout cas ... les x et y dans l'équation d'un cercle en coordonnées cartesiennes

ce sont des coordonnées

MA et MC ce sont des distances

elles sont pas liées à un système d'axes

[novel quiver]

donc si on pose O(0,0) et A(-sqrt8,0) avec C(sqrt8,0)
on pourra dire que MA^2 + MC^2 = 32 est leq dun cercle?

juste pour savoir

[dark sparrow]

bruh

c'est qu'une coïncidence que ça se simplifie en une équation d'un cercle ...

[novel quiver]

bon

[main lintel]

guys, this is an english speaking server, you can as well use private messaging to speak in french, or otherwise you are uselessly filling up the channel.

[pure cape]

Well ann is helping someone, and maybe its easier to use french since its easier for them

I wouldnt call that filling up the channel

[dark sparrow]

yeah there was actually math help happening here even if it's in baguette

[upper karma]

🥖

[silent plank]

baguette: a gem, especially a diamond, cut in a long rectangular shape

[upper karma]

baguette: long, thin loaf of French bread that is commonly made from basic lean dough

[novel quiver]

so if we wanted to convert this MA^2 + MC^2 = 32 into the formula of a circle, we would switch the M with numbers (that would represent the coords of one of the points of the circle) @dark sparrow

[dark sparrow]

bruh

why are you so obsessed with circles

[novel quiver]

i am trying to find a way to distinguish the equations that look like the equation of a circle and the ones that are the equation of a circle

well, my exam's on that 😂

ok i got it actually, ty

well, I still didn't get the solution of the question though lol

[main lintel]

Maybe if you spoke in English someone else might help to clarify 😛

[novel quiver]

Prove that for every point M of the plane, MA^2 + MC^2 = 2 MO^2 + 16

we give MA^2 + MC^2 = 32

@main lintel

[main lintel]

What this q signals. That OE and EC is equal?

So this second formula is also given?

Then the first statement is not for every point of the plane.

But for every point M that satisfies the equation MA^2 + MC^2 = 32 ?

You should at least copy the question as it is stated.

[novel quiver]

BEC a right triangle on E, such as, CE = 2 and BE = 8
Let O be a point in the segment [BE] such as OE = 2
We give A the symmetric of C from the point O and I the middle of [AE]

Prove that vector BA . vector BC = 28 , use that to deduce the distance of AB

Solution: Drop a perpendicular from A to BE, and denote the intersection of the line with BE by D.
Then triangles AOD and COE are congruent. (AAS via angle ADO = angle CEO = 90 deg, angle AOD = angle COE (vertical angles), and AO = CO (given, since A is symmetric to C about O)
Hence AD = CE (corresponding sides of congruent triangles) = 2, and DO = EO (corresponding sides of congruent triangles) = 2.
Then you have the lengths of BD (being the complement of DE in BE) and AB (via Pythagoras' theorem), hence you have the sine and cosine ratios of angle ABD.
Then you have cos(angle ABC) via compound angle formula (angle ABC = angle ABE + angle EBC = angle ABD + angle EBC).

Be it phi = (any point M such as MA^2 + MC^2 = 32)
Verify that A ∈ phi

We replace every M by A, which valids the equation.

Prove that for every point M of the plane, MA^2 + MC^2 = 2 MO^2 + 16

I'm stuck on the last question @main lintel

I have sent the whole problem - every question and its answer

[main lintel]

Oh, the green q is a 2?

[novel quiver]

yes, everything written in green is just a 2

there is no q

@main lintel

[main lintel]

You are missing I on the picture

[novel quiver]

there is no I?

[main lintel]

Why is it given in the task, seems irrelevant

and I the middle of [AE]

[novel quiver]

oh, yeah it's missing
but we didn't use it at all

so it's irrelevant, i guess

[main lintel]

This . is scalar multiplication vector BA . vector BC = 28 ?

Dot product?

[novel quiver]

it's a scalar multiplication

yes

yeah

👍

[main lintel]

Prove that for every point M of the plane, MA^2 + MC^2 = 2 MO^2 + 16

[novel quiver]

what's the solution?

[main lintel]

I don't know, have to think about it.

[still venture]

Hey anyone know how to do proofs

[main lintel]

#proofs-and-logic @still venture Just ask the question there

@novel quiver My first idea would be to imagine point M, and to draw its projection on the extended [AC] line.

But this is without checking any of the previous questions, which might give you some hints.

You know the [AO]=[CO]=sqrt(8)

[novel quiver]

doesn't MA^2 + MC^2 = 32 kinda look like the equation of a circle?

[main lintel]

Perhaps visually, but also looks more like the pythagorean theorem.

[novel quiver]

how can we know if it's the equation of a circle (or not)?

[main lintel]

Or the equation of the elypse 😄

@novel quiver You have to carefully examine what each component of the equation represents

What would you take as circle center, what as radius and what as points on the circle

I mean it is an ok way of solution to consider, just wouldn't be my first attempt.

[novel quiver]

aigh, but like - is it okay to actually draw point M?

isn't point M supposed to be "everywhere"

if you see what I mean

[main lintel]

Yes, you draw it, but you keep in your mind that it is not fixed, but can be anywhere.

[novel quiver]

alright, and now that we have it - what do we do?

[main lintel]

You need to be make sure that every conclusion you make further can be generalized on the case that M can be anywhere.

[novel quiver]

Prove that for every point M of the plane, MA^2 + MC^2 = 2 MO^2 + 16

[main lintel]

Now you compute MA MO and MC which are parts of your equation.

[novel quiver]

N is the middle of AO right

[main lintel]

MA^2 = MN^2 + AN^2 MC^2=MN^2 + NC^2 MO^2 = MN^2 + NO^2

[main lintel]

No

M can be anywhere, so N can by anywhere on the extended AC line

Now, not saying that this will give the solution, but this would be my mental process, that is my first attempt

Now, I put this into the final equation, and see which parts are missing

MN^2 + AN^2 + MN^2 + NC^2 = 2*(MN^2 + NO^2) + 16

AN^2 + NC^2 = 2*NO^2 + 16

Hmmm, does this give me something 😐

[novel quiver]

well, how did you go from MN^2 + AN^2 + MN^2 + NC^2 = 2(MN^2 + NO^2) + 16*

to AN^2 + NC^2 = 2*NO^2 + 16

[main lintel]

We cut the MN^2 from both sides.

[main lintel]

We have at least reduced our equation to the segments on the same line [AC]

Now we have further as we calculated at the beggining that [AO]=[OC]=sqrt(8)

[novel quiver]

I thought we'd get 2 MN^2 + AN^2 + NC^2 = 2*(MN^2 + NO^2) + 16

[main lintel]

And now cut 2MN^2 from both sides

[novel quiver]

how so? do we square root both sides?

[main lintel]

No, multiply right side with this 2 in front of the parentheses.

You will have 2MN^2 on both sides, so they can be cut away.

[novel quiver]

2 MN^2 + AN^2 + NC^2 = 2*(MN^2 + NO^2) + 16

2 MN^2 + AN^2 + NC^2 = 2 MN^2 + 2 NO^2 + 16
= AN^2 + NC^2 = 2 NO^2 + 16

ok gotcha

[novel quiver]

can we use that in the new equation we got?

[main lintel]

Now you have these segments on the same line in the equation.

[main lintel]

Yes, you need to think how to utilize this simple conclusion on these segments on the same line that we got.

Think if you can progress further.

[novel quiver]

which segments are you referring to?

[main lintel]

To the segments from the new formula that we calculated.

[novel quiver]

AN^2 + NC^2 = 2 NO^2 + 16

oh right, I see what I mean

all of the segments in the equation are on the same line

alright

[main lintel]

I said before, not sure that we have the solution.
But it seems we have made progress, because we reduced our very general task of M in entire space, to these segments on one line.

And moreover we have this sqrt(8) distances on the same line.

Seems like we could merge these two facts to finalize the solution.

[novel quiver]

this is probably the answer AN^2 + NC^2 = 2 NO^2 + 16

meaning that for every point M we draw, its orthogonal projection will always result in all of the segments being on the same line

@main lintel

[main lintel]

No.

This is not the answer.

@novel quiver

[novel quiver]

did you find the answer?

[main lintel]

You need to prove that the let side is equal to the right side.

You need to understand how the poofs work in mathematics.

You have been given the task with some "definitions" and the statement to prove.

[novel quiver]

so we have to substitute for the values of the segments and validate the statement, right

[main lintel]

We have went with a way of proving something by going from the reverse. This is common procedure in mathematics.
We have assumed that the final equation holds, and we substituted stuff in that equation to simplify it. We have to simplify it enough so that we get for sure that it holds.

We don't know that it holds, because it is not obvious.

We simplified it by having M (any point in space) replaced by the simpler O (any point on the AC line). But it is still not obvious that it holds.

We need to simplify it further, with the known "definitions" (given in the task), to be certain that it holds.
One of such given "definitions" is that AO=OC and from OE=EC=2 we know that AO=OC=sqrt(8)

[novel quiver]

AN^2 + NC^2 = 2 NO^2 + 16
AN^2 + NC^2 = AC^2 + 16

[main lintel]

Yes, that is an example of the "simplification", but seems like it is wrong one.

[novel quiver]

we can probably say that AN^2 + NC^2 is the same as AC^2 ?

[main lintel]

No, this is not correct.

[novel quiver]

why is that?

[main lintel]

You need to be 100% certain in the simplifications you make, this is rule number 1 in mathematics.

If you are not certain in some simplification you cannot make it.

[novel quiver]

isn't vector AN + vector NC = vector AC

[main lintel]

Yes

[novel quiver]

if we square both sides

[main lintel]

But you cannot square each term.

[novel quiver]

vector AN + vector NC = vector AC

oh

[main lintel]

You can square both sides, but not each terms by itself.

[novel quiver]

yeah and if we square both sides vector AN + vector NC = vector AC

it becomes (vector AN + vector NC)^2 = (vector AC)^2

[main lintel]

You are missing a lot of practice to be honest. You shouldnt make such "simple" mistakes.

You need to practice a lot.

[novel quiver]

what are these simple mistakes I made?
My thinking was just:

AN^2 + NC^2 = AC^2 + 16

and if we square root both sides we get:

AN + NC = AC + sqrt16

[main lintel]

You understand now the simplifications, and you are trying things in the correct direction, but you need a lot of practice.

[novel quiver]

is this correct?

AN^2 + NC^2 = AC^2 + 16

and if we square root both sides we get:

AN + NC = AC + sqrt16

[main lintel]

You can square root both sides, but again, not each term by itself

This is a simple mistake you shouldnt make.

Indicates that you need a lot of practice.

[novel quiver]

oh, so the answer would be

sqrt(AN + NC) = sqrt(AC + 16)

[main lintel]

This is one way to proceed, but you need to progress further to see if this is the correct simplification

Or perhaps it is a complication, not a simplification.

[novel quiver]

yeah, likely 😕

[main lintel]

When I did the M point projection on the line, I'm not sure that was the good simplification. I'm still not sure because we dont have the solution.
But seems like a move in the correct direction because we ended up on the segments of one line, instead of the whole space

[novel quiver]

should probably go back to this step AN^2 + NC^2 = AC^2 + 16

[main lintel]

This is not the step we got.

AN^2 + NC^2 = 2 NO^2 + 16

This is what we got.

And we also got before that AO=OC=sqrt(8)

[novel quiver]

Yeah but I simplified it into AN^2 + NC^2 = AC^2 + 16 , which iirc was correct?

[main lintel]

I dont remember this step

Here you say that 2*NO^2 = AC^2 right?

[novel quiver]

Yeah

[main lintel]

You used this fact to make this simplification.

[novel quiver]

Yes

I did

[main lintel]

Why you think this is the true fact

As I said, what is the rule number 1. Never make a simplification you are not sure it holds.

[novel quiver]

hm, yeah :/

Well, we can probably isolate the 16 in ** AN^2 + NC^2 = 2 NO^2 + 16**

and then substitute for every value of the segment, and see if it is equal to 16?

[main lintel]

Not exactly sure what you mean. But let me give you a small hint.

We have this equation, which is good. But what is the other thing that we have?

[novel quiver]

AO=OC=sqrt(8)

[main lintel]

And what is cool about it is that it is on the same line as our 3 segments.

Which is a pretty good indication that this should help us.

To do further simplification.

[novel quiver]

We can probably replace every N point by O?

AN^2 + NC^2 = 2 NO^2 + 16

AO^2 + OC^2 = 2 OO^2 + 16

[main lintel]

Only if you are 100% sure

(rule no 1 😄 )

[novel quiver]

Yes.

AO=OC=sqrt(8)

AO^2 + OC^2 = 2 OO^2 + 16

After we substitute:

(sqrt8)^2 + (sqrt8)^2 = 0 + 16

which gives us:

8 + 8 = 16

The only thing I'm not sure of is that

whether or not we can replace a point by another

[main lintel]

Yes, this is a good question. Very important!

[novel quiver]

Yeah, are we "allowed" to

Because if we are, then what I just sent checks out

If not, we're still stuck

[main lintel]

Yes. Unfortunatelly I'm not sure we are allowed.

It must be proven that we can, because it is not obvious.

[novel quiver]

Well, they're all on the same line (A, N, O, and C)

so we can probably do it?

[main lintel]

I don't know any rule by which you can do it. At least not immediately.

If you don't know the rule, or it is not obvious you should discard it.

(rule number 1 😐 )

[novel quiver]

for now that's kind of our best bet

but alright

probably better to play it safe

[main lintel]

Yes, rule no 1. The most important of all rules.
If you focus on the image and on what we have:

AN^2 + NC^2 = 2 NO^2 + 16
AO=OC=sqrt(8)

To me it seems that we are pretty close to the solution, and that only a final insight needs to be made.

[novel quiver]

hm..

[main lintel]

So you only have this line AC, and the segments on it. Everything else we can ignore.

It all boils down to this line and the segments on it.

Some of the segments on it we know the distance, like AO, OC, even AC (it is twice, because AO and OC is equal)

Some of the segments we dont know, like AN, NO and NC.

Unfortunatelly, all the once we don't know are in our equation 😄

[novel quiver]

yeah :/

this one question is very difficult

[main lintel]

Can you make any relation between the segments that you are 100% sure about?

Like I gave one relation: AC = 2*sqrt(8)

Because AO=sqrt(8) OC=sqrt(8) and AC is their sum.

And this is 100% sure.

So we can use that in the simplification.

Unfortunately I don't think it will give us progress.

But there should be other relations between the segments, that you might exploit.

[novel quiver]

welp, still no solution I guess..

Did you find it?

[main lintel]

I have an idea @novel quiver But you should try it yourself from this point further

as per my last hint

[novel quiver]

👍

[lilac moth]

Is this channel free?

[round isle]

please post your question

[cosmic pebble]

From

#❓how-to-get-help

Rule 1

The help channels are solely for help with math, so feel free to post your question. Asking whether you can ask a question or if anyone knows about some specific topic is unnecessary, so please try to avoid questions of that nature.

[woeful lagoon]

for part c you should use cosine rule right

because it gives 1 answer instead of 2 w/ sine rule

[plain wyvern]

cosine rule is too cumbersome. Looks like an SAT question. I would first find AB using area of Triangle as area = 1/2 . AB. AC . sin(BAC) then I will find angle B or angle C using sine rule and then again I can find BC using sin rule

[silent plank]

AB from area,
BC from cosine rule
there isn't really an issue with using the sine rule if you know which sides is the longest

but with sufficient info i'd usually go for the cosine rule anyway

[opal slate]

How do i count the area of the figure when R=182mm and r=74mm

[dark sparrow]

you can split it into twelve sectors, six small (radius r, angle 32°) and six big (radius R, angle 28°)

do you know how to calculate the area of a circular sector?

[opal slate]

yea

[dark sparrow]

okay, then does this answer your question?

[opal slate]

(alpha/360)piR^2

and ill take that *6

and same with the other sector *6

[dark sparrow]

yes sure

[unkempt wasp]

how would I get a hyperbola formula if I know two tan line formulas?

[dark sparrow]

(5x-6y-16)(13x-10y-48) = c

c is a constant to be determined

[unkempt wasp]

oh wow thanks. Learnt something new just now, but I also forgot to mention that hyperbolas axes are same as coordinate axes so this doesn't seem to match. I'm really confused about this because I don't even imagine how it would look like

I'm kinda new to this topic

[unkempt wasp]

ok I found the answer through desmos but I'm still not sure how would I calculate this

[unkempt wasp]

Still don't know how. Any hints would be appreciated 😄

[glacial dawn]

What are you trying to figure out

[unkempt wasp]

how would I get a hyperbola formula if I know two tan line formulas and hyperbola axes are same as coordinate axes?

[glacial dawn]

Didn’t u alrdy get the formula?

[unkempt wasp]

I managed to find the end result through desmos but idk how would I do that using formulas

[glacial dawn]

O

I’m not sure tbh. Sorry

[slim warren]

Hi all I've done 3a and 3b, not sure how to do 3c can someone show me what to do?

[plain wyvern]

first find out the general form of the tangent line for the given circle

[pliant dawn]

A circle with a radius of 8 cm is inscribed in an equilateral triangle. Calculate the area between
the circle and the triangle.

[slim warren]

How do u do that?

[plain wyvern]

There is a formula for it but if you don't have then you can find one for a general circle say (x-a) ^2 + (y-b)^2 =c^2 put y = mx+c when you put it up, and when you evaluate it, the discriminant =0 will give you a solution, which you can use to find the equation of tangent line.

[slim warren]

I tried that but i got rly weird (and the wrong) answers

Can u show me how?

[plain wyvern]

@slim warren I just checked my notes there is a very suitable way, the tangent equation will be of the form y-17 = m(x-7) find the distance of this line from the centre of circle, you will get the value of m. From that you can do it easily

[slim warren]

Ahh ok I'll work on it thanks for ur help mate

[plain wyvern]

I remember those days of preparing for JEE for coordinate

[slim warren]

Yh its hard lol

[unkempt wasp]

how would I get the formula of a hyperbola if I know a point and a tan line?

to be more precise the tangent line is 9x+2y−15 = 0 and the point A(squareRoot(6),3)

[soft zodiac]

I’m having trouble trying to set up #6

[exotic talon]

Could you guys help me with dilation

I have this homework could I show you the questions

DONT GIVE ME THE ANSWER

Just tell me how to do it

[blazing coyote]

hello

can i get some help with this

[exotic talon]

@blazing coyote do you know dilation

[blazing coyote]

nope

sry

[lunar sand]

times the scale factor with each coordinate

so for 9, (2x5,2x5)

[exotic talon]

@lunar sand thank you

@lunar sand so for the first one it would be like k’=(1.5,2)

So like 0.5 times 3 and then times 4

[lunar sand]

correct

[exotic talon]

Thank you!

This is starting to get easier

I have a test on this tomorrow

Anyways thanks for looking out!

I have been so stressed on this

[lunar sand]

Np, and good luck

[exotic talon]

Thanks

[main lintel]

@soft zodiac do you understand these golf terms?