The following problem was presumably solved by me and @summer mason using just; rules of similarity in shapes and Pythagoras Theory. However, we're not highly confident with our answer. We want someone to make an attempt on solving this question, as a different opinion or approach from someone more familiar to this type of questions will be quite helpful.
#geometry-and-trigonometry
upper karma
fair wigeon
A different approach to this problem would be a bunch of Pythagorean theorem
But I think that your approach of similarity is most efficient
But I'll definitely solve this using pythag
final gorge
Say a eq of line is ax+by+c=0 so can we say slope= -a/b? ..Edit- I tried googling but can't find any source to verify.
covert idol
yes
final gorge
okay thanks.
covert idol
Not really
stark bluff
Yeah does it have any particular uses
inland steeple
devide their dot product by the product of their lengths, this wil give you the cosine of the angle between them
errant ether
just set y to 0 and take their dot product
errant ether
the dot product of two vectors is the product of their magnitudes and the cosine of the angle between them
normalizing them makes that dot product of the normalized vectors equal to just the cosine
take the inverse cosine to find the angle
why are you doing A - B?
inland steeple
if you want to ignore their height just drop the y
errant ether
anyways you have to either change the vectors, projecting them onto the xz plane by setting y to 0 or write a custom dot product that ignores the y component of the vectors
yes, they're equivalent
inland steeple
if it's set to 0 yes
inland steeple
btw A - B isn't the vector you drew there
errant ether
ok, then before normalizing the vectors, set their y components to 0
doesn't matter, normalizing it will always make its length 1
dropping y makes its length less than 1
so normalizing after that makes it 1 again
np
limpid path
hi
gritty flare
Helu
limpid path
helo
gritty flare
Uhm
Similarity
12x=15
Where x= similarity constant and a scalar
That x scales 12 to 15
And it also scales the are
So you know:
12x=15
400x=Y
You need to find Y
K go ahead and solve it now
limpid path
OK
gritty flare
Tell me the answer :D
limpid path
3?
gritty flare
3 times 12 = 15?
charred spearBOT
limpid path
oh
limpid path
ops
gritty flare
limpid path
emm
charred spearBOT
limpid path
you made me even more comfused
gritty flare
12x=15
Divide both sides by 12
12x/12=x
15/12=5/4
So x=5/4
Wat so hard about simple algebra
"Isolate x for equation 12x=15"
buoyant oasis
Where do you find that emoji
gritty flare
limpid path
on emoji bar
buoyant oasis
*forget what I said y'all are learning k"
limpid path
lol
gritty flare
So you understand?
limpid path
yep
limpid path
a question
gritty flare
Lol no
covert idol
gritty flare
Ok so
Figure A's height was scaled 5/4 times to achieve figure B's height
Thus meaning
Figure A's area scaled 5/4 times will give us figure B's area
So
gritty flare
=tex \text{Area}{A}\times x = \text{Area}{B}; \text{ where } x=\frac54
charred spearBOT
gritty flare
Lets substitute in the values
charred spearBOT
charred spearBOT
250
gritty flare
250cm^2
limpid path
ohh
upper karma
law of cosines
nvm
i forgot what it was called
1/2 ab sin c is the right thing
use 1/2absinc
how did you get to this from 12x=15
limpid path
ohh
upper karma
$$ A = \sqrt{s(s - a)(s - b)(s - c)} $$
charred spearBOT
upper karma
a,b,c are the side lengths of triangle and s is tje semiperimeter
he doesnt know all three sides
He can from law of cosines
Once you have the 3 sides apply the formula above
huh well you can do that
but 1/2absinc is more simple lol
upper karma
no thx
upper karma
youre using radians
change settings to degree
nvm
im wrong
where are you getting the cos from?
if you're solving the area problem then you're using cosine instead of sine
inland steeple
Oh right it should be sin
upper karma
Law of cosines
inland steeple
so sin(105°)
inland steeple
I switched up my x and y
upper karma
oh lol
umbral snow
@limpid path
Construct that line. That is the height of the triangle, and since it forms a right triangle, you can find it with right triangle trig. Then A = 1/2 bh
upper karma
both ways work
limpid path
10x13 sin(105)
upper karma
not 1/2absinc?
limpid path
Mm yeah forgot about that
upper karma
lol
lavish haven
What you finding
limpid path
10x13÷2 sin (105)
lavish haven
Don’t you just use 1/2absin(c)
upper karma
yes
umbral snow
I guess you kinda do, lel
lavish haven
Oh are you good then
lavish haven
clap
limpid path
upper karma
unfortunately reasoning doesnt work too well
we have a 10 and 13 length triangle
i forgot a lot from precalc
so heres this
gleaming citrus
Hey guys what kind of stuff should I know/study before going into polygons and polytopes
upper karma
thats a really broad of a question
do you mean regular polytopes in n dimensional space?
well thats a subset of topology
runic rock
https://i.imgur.com/0uOb4vK.png
how would this be solved? I know that you start from the left hand side.
umbral snow
16/13 = x/8
By similar triangles
frosty flame
how do u know they are simliar?
umbral snow
They are similar if they have the same angles
upper karma
AAA
valid pike
I thought it had something to do with the Pythagorean theorem and I ended up getting x=9.08
Am I right
What should I do
drowsy spoke
The smaller and larger triangles are congruent
gleaming citrus
@upper karma yea regular polytopes in Rn
deep sedge
1/2
wild hamlet
whats the half angle formula?
wild hamlet
=tex sin(\frac{\theta}{2}) = \sqrt{\frac{1+cos(\theta)}{2}}
charred spearBOT
deep sedge
so do i put 1/2 where cos is
wild hamlet
no
deep sedge
oof
wild hamlet
cos(150) is not sin(150)
cos(x) = sqrt(1-sin(x)^2)
knowing that, sub sin(x) for 1/2
=tex sin(\frac{\theta}{2}) = \sqrt{\frac{1+\sqrt{1-sin(x)^{2}}}{2}}
charred spearBOT
wild hamlet
you there?
deep sedge
yesss
wild hamlet
what do you think we do now?
deep sedge
now do i put 1/2 where sin is
wild hamlet
yep
clear haven
~~derive 1/2 ab sintheta from cross product ~~
silent sparrow
helllo guys
and girls
does anybody know how to graph sin cos tan cot csc sec ?
if so message me
spring apex
@silent sparrow
how can you not know how to graph sin and cos but are needing to graph cot, csc and sec?
🤔
you generally learn how to graph sin and cos atleast a couple years before those other ones
so im not sure where you're coming from with this
silent sparrow
im in trg
we went over sin and cos
but i forget fast lol
trig
this is in college
spring apex
what kind of fucked up education system do you have lol im sorry
upper karma
Can someone help me with radian angles?
I'm struggling with an exercise about Chasles relation
covert idol
upper karma
Ok lol
It's a french one though
I don't have all the vocab for it rn
I'll need to look it up
median holly
What makes cosine and secant even functions?
dire rampart
a function is even if $$f(-x)= f(x)$$
charred spearBOT
dire rampart
ie: its is symetrical about the y axis
winter ingot
Does anyone know how to solve this
wild hamlet
yes
area = 144 = area of triangle
the legnths of the trapezoid are 12,6, and sqrt(153)
winter ingot
Could you explain how you got that please
So since BCDE=144, doesn’t that mean that ACD would be 288
wild hamlet
i'll use heron's formula
s = (a+b+c)/2
a,b,c are the triangles sides
A = sqrt(s(s-a)(s-b)(s-c))
a=b,
c=12
A = 144
144 = sqrt((2a+12)(a+12)^3)
solve for a
a=b due to it being an isosceles triangle
it doesnt have to be, but itll make things simpler
144^2 = 2(a+6)(a+12)^3
grand yacht
cold plaza
if u know how y=sinx looks, u can vertically stretch it up and down by a factor of 3
silent sparrow
just use the base function
cold plaza
huh
cold plaza
yep
drowsy spoke
it'll be kinda confusing in the beginning, as you learn more you'll see that it's pretty easy, usually there's hardly any effort involved - we usually just stretch, squeeze, or flip the original graph along the xy axes to get a graph that satisfies our equation
covert idol
Yeah its pretty straight forward once you know what putting numbers where does
And they're all pretty intuitive
flint depot
surreal bolt
hey
flint depot
hi
surreal bolt
so 1) distance formula with Pythagorean theorem
2) perpendicular lines have negative inverse slope
3) circumscribed circles of right triangles have a center on the hypotenuse.
I think 2 is the easiest in this case
flint depot
ok
surreal bolt
so what is the slope of LM?
surreal bolt
kk. What slope should MN have?
flint depot
4/3
surreal bolt
and what is the expression for MN's slope using p and the other three numbers?
flint depot
idk that
surreal bolt
NM. What two points are on it?
flint depot
oh
surreal bolt
😃
surreal bolt
it's not "again" 😃
surreal bolt
you find two representations of the sane thing 😃
and set them equal 😃
same*
okay sorry it was technically "again"
flint depot
is p=8
surreal bolt
yes
flint depot
help
chrome fiber
there's one obvious transformation which you should be able to catch right away
what's that?
candid harbor
?? i think its the one with the line by it
gritty flare
Each letter corresponds to each other
Like
△△
M P
N Q
O R
M congruent to P
N to Q
O to R
MN to PQ
NO to QR
MO to PR
So a) can not be proven
candid harbor
ok i got the answer right...
vivid ridge
Hm, the way I'm thinking about this might be way too complicated, so try doing it step by step: what are the other angles of the roof?
I was thinking getting the area using absinθ then comparing that with 1/2 b*h
but I realised it's way simpler if you just make a right angled triangle with the middle of the roof
you can get those from the sine rule, sinA/a = sinB/b = sinC/c
where A, B, C are the angles of the triangle
we do
it's 30 and 30
yeah that's what I assumed
vivid ridge
you don't need it, you can use tan(30)=h/6
oh wait
right, that's the next question?
in that case use cos(30)=c/6
jaunty horizon
!15m
thin hound
?
jaunty horizon
tf y no work
vivid ridge
Sorry, what do you need help with?
thin hound
2 and three
vivid ridge
... Could you be more specific? What part do you not understand, what did you try?
fluid hawk
@copper valve guess what my prof (the one who told me he would help me do independent study on calculus on manifolds) said
he says "CoM is just rigorous revision of your multivar class, we'll skim it through and you'll have to read it yourself. The main focus on your study will be A comprehensive intro to diff geometry book"
lol
median holly
Why is the root mean square in the circumference of an ellipse?
fluid hawk
@copper valve y u ignore me
thin hound
😂
fluid hawk
i dont understand
copper valve
@fluid hawk maybe Lee's smooth manifolds would suit better?
glacial cedar
sick veldt
I have my answers filled in
check if they are right please
I have to drag my answers into a box
<@&286206848099549185>
vivid ridge
Looks okay I think
royal dew
@sick veldt why is point A equidistant to the sides PQR?
sick veldt
because both sides are equal
royal dew
why?
royal dew
so, is the length from point A and point R the same?
vivid ridge
No I'm pretty sure it means that the distance from the sides of PQR (i.e. QP and QR) are the same, i.e. XA and YA are the same distance
royal dew
I apologize, I meant to ask him if the distance between the points A and R are the same as A to Q
vivid ridge
Right, I see.
vivid ridge
I suppose, yeah
sick veldt
So could it be to place the point of the compass on point c and draw and arc?
vivid ridge
There is no point C
vivid ridge
Yep, which is the 3rd one on your list
sick veldt
this is what it should look like now right?
or wait
should the 4th one replace the 3rd?
Since you have to do D as well?
vivid ridge
no, from your first image it's the 3rd one that goes first
which is "Place the compass on the point B and ..."
vivid ridge
... wait, you still don't have that first question correct
sick veldt
oh
vivid ridge
vivid ridge
looks about right
vivid ridge
no
Well, let's think about it - out of all the steps in there what doesn't rely on any previous steps?
I can see at least 2 of them needs to be done first before anything
sick veldt
number 3 doesn't
vivid ridge
No, number 3 needs an arc
vivid ridge
Yeah, that looks about right
vivid ridge
what do you mean?
sick veldt
This was my original answer
these are the remaining answers
some have a triangle on the left
nvm
vivid ridge
I don't know what SAS and AAS Congruence Theorem means in this case but I'm pretty sure this is okay
vivid ridge
I'm gonna take a break, good luck with your assignment
sick veldt
Okay
cold plaza
are the branches of hyperbolas parabolas?
wild hamlet
if im thinking what you're thinking, thne no
x^2 - y^2 = 1 is the most basic one
y = sqrt(xx-1)
=pup draw xx-yy=1
cold plaza
if they were, then parabolas would have oblique asymptotes
but parabola don't (correct me if im wrong)
so they arent parabolas
rustic cairn
How do I solve 10 & 11
astral hornet
complementary are 2 angles summing to 90
supplementary are 2 angles summing to 180
kindred bronze
Can anyone explain this to me? I have to solve for x
@rustic cairn Find out what other angle measure do you need to get 90°/180°
wild hamlet
do you know what sin(30) is?
more importantly, do you know what the ratio of sin is?
kindred bronze
Not really. I'm in geometry. But I'm trying to do this for extra credit; which is going to be counted as an exam grade
I would be able to do this if I was given an actual value rather than an expression
meager acorn
i have a question
complimentary angle is 90 degree ; pi/2
supplimentary angle is 180 degree ; pi
?
meager acorn
thanks
so let say you have
pi/12 and find comp angle it would be
pi/2 - pi/12 ?
supp angle it would be
pi - pi/12?
comp = 5/12 pi
supp = 11/12 pi?
@steady hull
meager acorn
so let say in test or any quiz let say they give you pi/3
pi/2 - pi/3 for comp angle
pi - pi/3 for supp angle
steady hull
ok
meager acorn
to find angle x 180/pi ?
radian x pi/180?
bitter viper
Hi, is there a chance someone could double check my answer? I'm doing online homework and only am able to submit once
i got answer -tan(x)
slender gyro
@bitter viper that's almost right, but i think you forgot a factor of 2
The answer Being -2tanx
bitter viper
thank you!
cloud cave
i just realized something
why is tan 0 zero when sin 0/cos 0 = 0/0 (which is undefined)
flint depot
dire rampart
@cloud cave sin0/cos0 is 0/1 not 0/0
dire rampart
lol dw bout it
steep nexus
can someone explain when to use sin, cos and tan in a right angled triangle ?
cloud cave
my brain stopped working after preparing for my calculus midterm tmr lol
dire rampart
wdym use?@steep nexus
steep nexus
um like
dire rampart
as in, calculaationg sides and angles?
dire rampart
steep nexus
SOH CAH TOA
steep nexus
i do know the mnemonic
but it gets confusing when to use sin cos tan whether to find a side or an angle
like if i have to find the adjacent side, that is;
CB
what do i do ?
considering AC is xcm and BA is y cm
dire rampart
so you want the adjacent side?
steep nexus
hm
dire rampart
thats just pythagoras
steep nexus
ya but
our questions strictly mention: no use of pythagoras
we use a^2 + b^2 = c^2 in mathematics B
dire rampart
are you given an angle?
dire rampart
im not sure what your exact question is
trig can be used given you know 2 sides or angles
steep nexus
pythagoras is used to find only the hypotenuse right ?
dire rampart
no
it can be used to find any side
provided you know the other 2 sides
ok lemme help
if BH splits it into 90 degrees
then AH=HC
steep nexus
ight i'll look into it
dire rampart
you can work out x using a mix of trig and pythagoras
trim breach
You can find angle ABH and use any trigonometric function to find all values of triangle AHB. That should make the problem easier.
Because then you know AH and can then find BH.
trim breach
Yeah, my bad.
But I meant triangle AHB.
Because nothing indicates line BH bisects line AC as far as I know.
fluid hawk
@copper valve can u tell me why that suits better
upper karma
When a person asks 'why' 
upper karma
I have No idea what you guys are discussing though 
fluid hawk
differential geometry book
upper karma
👀
copper valve
what's up
oh Lee's smooth manifolds Vs calculus on manifolds
CoM is super non handholdy I guess
like it goes through a decent amount of stuff, but also skips out on a little bit of stuff
and either includes it as exercises or is just omitted
it's a really good book tho
take a look at both of them
at some point I was considering of using both books for my reading group, but around that time ppl were losing interest anyway :p
bbl shower
fluid hawk
@copper valve i thought u were doing lee vs diff geo vol 1
oh wow u have reading group? o.O
upper karma
What happened to my comment: Lee > CoM?
copper valve
@fluid hawk had
basically my calc 2 class was a trainwreck
2 semester course
during the last month or so of classes I made a small group to go thru first 2 chapters of CoM to study kinda
and we succeeded !! woohoo
My mark went from 50% to 68% from the final
copper valve
nah it was passing
upper karma
fluid hawk
@copper valve lol and in my college 90 is B+
another proof that it's shitty
it's funny that you read CoM for your Calc II class....
uh why tho
copper valve
it was recommended to me
and I already had the book
the first 3 chapters is calculus not on manifolds
so it was still very applicable @fluid hawk
fluid hawk
yea but its hard
like u dont need compactness for calc II
maybe ur uni too advanced
idk
anyway, yea my prof wants me to do spivak vol I 
I'll prolly fail, but im doing it anyway
@copper valve
left folio
Uhm i seem to have another problem i don't understand too wel unfortunately
" Use an angle sum identity to compute cos(π/12)"
@flint depot Have you tried sketching it?
You just have to know the coordinates of all points though
limpid basin
Find the coordinates of their intersection with that line, then use Shoelace
left folio
@limpid basin Can you help me please
limpid basin
Use cos(x-y) = cos x cos y + sin x sin y
left folio
Well i'm not quite sure what that means since i'm new to this
If cos = π/12, then how do i find the y of cos?
umbral snow
So we're not finding cos = π/12, whatever that means. We're finding cos(π/12)
And "the y of cos"? Huh?
umbral snow
cos is a function, it takes in a value (in this case π/12) and outputs a value (which we want to know)
It is an identity that
cos(x - y) = cos(x)cos(y) - sin(x)sin(y)
So you can use special angles you already should know to find the value of cos(π/12)
Problem is, do you know the special angles?
left folio
Wow i don't get why my book doesn't explain these things and ecpects me to know them
I don't even know what a special angle is tbh
So what i got from this is that these are all functions which i can solve no problem, but how is there no value in the dominum?
I mean how can i find anything like that
umbral snow
left folio
I'm sorry if i am too inexperienced with these instead easy concepts i imagine
steady sleet
🤢
umbral snow
This is the unit circle with all of the special angles on it
Yes, it's gross. But, there's easy ways to cut this down
left folio
Idk how these radiants impact my exercise but cool
umbral snow
I feel your exercise depends on a few things you don't know. You should at least know how to use the special angles
left folio
I don't know what that even means and it's driving me crazy because i don't know what to do and feel stupid
Is there no kinda quick way to calc it?
steady sleet
Do you know of "special right triangles"?
umbral snow
To use that chart, for something like
sin(3π/4)
Go to the 3π/4 angle
Read the y value at that point
Get √2/2
sin(3π/4) = √2 / 2
left folio
I don't know any of that, i'm still far behind probs
umbral snow
Remember sin means y value
cos means x value
steady sleet
Do you know the value of sin or cos at any angles?
left folio
Oh i see, you're saying the point
umbral snow
Yeah! You'd use the x-value instead, and get
cos(3π/4) = -√2 / 2
left folio
Mmmmh i understand
So that's the sum identity rule right?
Adding up x and y of the point, that is
Which is basically adding up two function if i am correct
umbral snow
For an angle θ
sinθ is the y-value of the unit circle at that angle.
cosθ is the x-value of the unit circle at that angle.
This is the DEFINITION of sin and cos. Any time you use them, you're finding values off the unit circle
left folio
This is all like 5 whole steps above what my 3 pages explained lol
Yes i did know this and thank you
Is this any close to correctness? cos(π/12)+sin(π/2-π/12)
umbral snow
We want to use
cos(π/3 - π/4) = cos(π/3)cos(π/4) - sin(π/3)sin(π/4)
cos(π/12)
= cos(π/3)cos(π/4) - sin(π/3)sin(π/4)
These are each special angles, ones you'll be expected to know by heart
left folio
If i don't know these it's because i'm 17 and haven't done them yet but i wanted to do more
But so, haven't we just found the coordinates of the point, and the lenght s of catets?
I might not know what i was going to get as result i guess
@umbral snow Thanks 😃
upper karma
How would I find the zeroes for a function like sin(x) + cos(x)?
Or something like sin(3x) + 2cos(x)
gritty siren
you can often change the thing into 1 cosine
upper karma
I just need some trig identities right?
gritty siren
Yeah
upper karma
cool thanks
a sin x + b cos x = c sin(x + φ)
Seems like I was looking for phasor arithmetic
rain tulip
@upper karma if it's sin(x) + cos(x), use sum to product and find where sin and cos are zero
upper karma
I don't see how I'd use the product-sum identities on that
None of those identities have a straight sinx + cosx that I can convert to anything else
gritty siren
sin x + cos x = sqrt(2)cos(π/4-x) iirc
gritty siren
you can always resort to complex numbers when you're unsure of your trig identities tbh x')
pastel bough
i came up with this sort of conjecture
if you want to know more about it i will consider sending it as a question for challenge
left folio
Do identities in trigonometry have to resort in their sine and cosine form to be verified from a given equation?
steady sleet
wdym @left folio
left folio
I have this "Compute limx_0 (sin(5x))/x" and i don't really know how to compute it, could you help?
steady sleet
=tex 1=\lim_{x\to0}\frac{\sin(x)}x
charred spearBOT
steady sleet
You are familiar with this limit?
left folio
Ok i can understand what that is but no, i'm not familiar with it
I wansn't given an equation, just a computing of limit
steady sleet
You need to know this limit to be able to do limits with trig functions
it's one of those "fundamental limits" that you should've learned first before anything else
steady sleet
You misunderstand my point
in order to solve problems like that
the limit I wrote above must be given
in some form or fashion
unless you're allowed to answer in the form of stuff like "this is the derivative of sin(5x) at x=0"
rather than giving the actual value of the limit
steady sleet
The first 5 questions all require the limit I mentioned
Okay, why don't you tell me what you do know and were taught to do?
left folio
If this is readable
left folio
This might do
steady sleet
read the first line of that
left folio
This process is what i should be able to do
steady sleet
or rather, the first 2 lines
left folio
Ok
I think i understood
That part wasn't as clear and it should've been imo, thanks a lot
steady sleet
The entire section before that should've been concerned with the limit I wrote as well
left folio
One would think that but i don't think so
steady sleet
well the top of that page begins with "Finally,"
implying the section before it leads up to that point
steady sleet
yes, those are all parts of the standard approach to proving lim sin(x)/x = 1
left folio
I can see how, but still i don't see it as a nice introductory way of putting it
steady sleet
wdym?
You don't like that they prove the stuff they tell you?
Instead of stating facts without telling you why they're true?
left folio
It's a little too brief for my liking since in school i haven't even done what cos and sin are, but that may make me sound like a dummie so
steady sleet
#❓how-to-get-help rules:
Stick to one channel and don't post the same question in multiple channels. Please don't ask for help in other channels if no one is responding in the one you have posted your question in.
And certainly do not ping mods/admins for help.
austere frigate
what is "below" mean in this graph?
hard gale
Y1(x) <= Y2(x) taht's what it means
hard gale
Reread their whole sentence, they're not saying it's the case for all x
austere frigate
yea, and I don't understand what they mean by "below"
if 0.28 and 0.93 is below, then wouldn't 1.99 be above?
what I am saying is I don't understand what they mean by "above" or "below" .
and looking at the graph doesn't help either
hard gale
What does above and below mean in real life? That's a starting point
hard gale
woeful flame
ABC and DEV are similar triangles(edited)
AA1 , BB1 , CC1 , DD1 , EE1 , FF1 are median lines(edited)
so they are also similar
like:
AA1/DD1 = BB1/EE1 = CC1/FF1
I have to show that AA1/MA1 = DD1/ND1
How can I prove that?
(Please dont send me links for sources because I already leaned that and already have great sources but just this specific question is hard for me and I need some help)
vivid ridge
It's similar triangles, so MA1/ND1 = AA1/DD1 right?
MA1 is a fraction of the line AA1, ND1 is a fraction of the line DD1
but because they're similar triangles that fraction is the same
think about it via ratios - AA1 : DD1 is the same ratio as MA1 : ND1
woeful flame
"It's similar triangles, so MA1/ND1 = AA1/DD1 right?"
why?
thats not abvious, because this is exactly what they asking from me to prove
vivid ridge
It's not exactly what they're asking you to prove
vivid ridge
you have to argue that the line from M to A1, and the line from N to D1, are similar via the property of similar triangles
then, after that you get what I said, MA1/ND1 = AA1/DD1
do some shuffling around to get what you need
woeful flame
wait sir
please
I understand ALL you said
but you said it like its abvious
they want me to prove that
How do you know that M and N
flint depot
help pls
woeful flame
is cutting AA1 and DD1 equally
vivid ridge
... because you claimed it's a similar triangle. It just follow from that, you can even argue from the median line CC1
ALL I've assumed in my proof is that it's a similar triangle
woeful flame
Because they are similar triangles you can understand ONLY THAT AA1/DD1 is like the triangles similarity
becuase median are also have the same propotion
but HOW can you say that AM/MA1 = DN/ND1
by wich criteria you say that
?
vivid ridge
Alright, how about another way of thinking about it: Is the triangle AMC1 and DNF1 similar?
woeful flame
I cant prove that
how can I ?
I need AA \ SSS \ or SAS
to prove that triangles are similar
by wich one you proved that AMC1 and DNF1 are similar?
vivid ridge
It seems like you're only trying to use the fact that AA1/DD1 = ... and not actually use properties of similar triangles
left folio
It's quite easy to prove tbh, if those 3 lines are have the same ration you only need to say one of several ways of proving it
vivid ridge
you won't get anywhere with that
woeful flame
wait, guys,
please just try to tell me how did you prove that AMC1 and DNF1 are similar?
like tell me by wich criteria
wich angles did you find \ wich sides did you find?
left folio
Like if a triangle has 180° you should be able to say that if the ratio between all those 3 copies of lines are the same, i think that is enough to state they're the same isn't it
woeful flame
no its not
left folio
Give me a min to think it a little more though
vivid ridge
droreh, we're not just skipping steps, it's what we get from stating that they're similar triangles
woeful flame
They want me to prove that by Angle-Angle \ Side-Angle-Side \ Side-Side-Side
left folio
Well it's not obvious but you have to use only one criteria
woeful flame
thats the only way to prove similarity
vivid ridge
You said that the trigangles are similar, and so are the median lines.
woeful flame
Yes, the median lines are similar too
vivid ridge
So clearly the angle MAC1 and NDF1 are the same.
vivid ridge
droreh please just listen for a second.
woeful flame
its not clearly sir
vivid ridge
IT CLEARLY IS
woeful flame
How?
woeful flame
of?
left folio
If the sum of angles is 180° there is only way that they can both exit with same value in ratio of copies of lines, or i think that's the case at least, they should be the same
woeful flame
@vivid ridge did you find MAC1 and DNF1 are similar by: AA \ SAS \ SSS ....?
wich of them?
vivid ridge
Okay, droreh will you listen to me for a little bit please. Let's take this slowly, no interruptions.
woeful flame
@left folio The only 3 ways they gave me are: AA \ SAS \ SSS
I can't prove by another ways
Do you have any idea?
left folio
Let @vivid ridge explain it, i don't want to interrupt him
woeful flame
to show that MAC1 and DNF1 are similar will be very helpful. but how can I do that?
(By the 3 ways that I got AA \ SSS \ SAS)
"IT SPLITS THE ANGLE IN TWO"
vivid ridge
the angles MAC1 and NDF1 are the same. This is because you said that they're similar triangles so angles CAB and FDE are the same. AND M is on the median of A, so is N on the median of D. And because the median are similar, the angles must be the same. You can apply this to the remaining angles MC1A and AMC1 with NF1D and DNF1
woeful flame
@vivid ridge how can you prove that the middle point is actually split the lines in similar way?
vivid ridge
Hence the triangle AMC1 and DNF1 are similar. In fact, you can apply exactly the same logic to the triangle A1BM and D1NE, which actually gives you the fact that the side MA1 and ND1 are similar.
vivid ridge
droreh, NO INTERRUPTIONS PLEASE.
You're asking way too many intermediate questions
I'm making this complicated because you said that it's not obvious
woeful flame
please wait
you said that
the angles MAC1 and NDF1 are the same. This is because you said that they're similar triangles so angles CAB and FDE are the same.
I cant understand why the reason of
"the angles MAC1 and NDF1 are the same"
is
"This is because you said that they're similar triangles so angles CAB and FDE are the same."
what?
😐
I understand thay the angles MAC1 and NDF1 are same, but from another reason
not the reason you said
the are same because AA1B and DD1E are similar triangles by SSS proof
but I dont see why they are same because that angles CAB and FDE are the same.
left folio
@woeful flame Please do wait
vivid ridge
So we just established that MA1 and ND1 are similar. Which means the fraction MA1/ND1 is the same as the rest of the triangle, namely AA1/DD1. So MA1/ND1 = AA1/DD1. Cross-multiply to get MA1 * DD1 = ND1 * AA1. You divide the whole thing by MA1 * ND1 to get AA1/MA1 = DD1/ND1
woeful flame
thats weird that you allways show that triangles are similar but you never use AA \ SSS \ SAS thats why i dont understand you
vivid ridge
Okay, let's resolve your complaints. You're saying that you don't know why the angles CAB and FDE are the same?
vivid ridge
droreh, you keep saying to wait while interrupting me, I can't take you seriously.
please just wait until I'm clear to talk.
Again, let's resolve your first complaint, one at a time.
woeful flame
Ok but please try to understand my questions
vivid ridge
Angles CAB and FDE are the same.
This follows from the fact that they're similar triangles.
vivid ridge
Okay, next, the medians AA1 and DD1 are similar
you stated that, and again it follows from similar triangles
woeful flame
those are sides, you cant tell about them that they are similar
vivid ridge
I can, I can make a statement about at what angle it cuts CAB or FDE, and its proportion to the overall triangle
vivid ridge
Correct. The lengths AB AC and DE DF are similar in that they have the same proportion as the two triangles.
So are the side BC and FE where the medians are based off of.
vivid ridge
Yes
vivid ridge
Yep. And again we can say the same about BC and EF
woeful flame
True
CB/FE = AC/DF = AB/DE = AA1/DD1 = CC1/FF1 = BB1/EE1
thats becuase of the similarity of ABC and DEF
So I also understand that AA1B and DD1E are similar triangles by SSS
so angle A1AC1 is equal to D1DF1
vivid ridge
We want to establish that the triangles MA1B and ND1E are similar. If they have the same angles, then they're automatically similar.
So let's show that they have the same angles
woeful flame
Ok
How?
enough to show 2 pairs of angles
and the triangles will be similar by AA
proof
but how?
I have only 1 pair
AA1B = DD1E
but I need a second pair of equal angles
so I will be able to prove by AA(Angle-Angle)
vivid ridge
Let's consider the median BB1 and EE1. Then, BB1 will cut B at a specific angle, going through the middle of AC. because AC is similar to DF, and E1 is the middle of DF, the side EE1 will also cut the identical angle E exactly the same way. So MBA1 and NED1 are the same.
vivid ridge
Another way to establish this is this: EF/BC = CB1/FE1, so since the lengths EF and FE1 are similar proportions, it must be that the remaining side is also the same.
woeful flame
but this is not obvious for them, they want me to prove that
vivid ridge
Hence the angle MBA1 and NED1 are the same, because we just constructed a median from similar triangles
woeful flame
"because AC is similar to DF, and E1 is the middle of DF, the side EE1 will also cut the identical angle E exactly the same way."
you cannot say that sir.
you have to prove that if you want to say such a thing
its NOT obvious
atleast not for my teacher
this all question is about to prove that
vivid ridge
EF/BC = CB1/FE1. Is that clear?
woeful flame
Yes
vivid ridge
Okay. So we have two sides of the same length proportion, and the angles ACB and DFE are the same
Which means that the triangle B1BC and E1EF are similar
vivid ridge
Hence the angles B1BC and E1EF are the same.
woeful flame
Right
Now you proved
Not I understand
the proof
that A1MB and D1NE are similar
by AA
vivid ridge
Right, you can make the same argument that we've been making
vivid ridge
Not really, I went all this way because you wanted me to define every single thing which could've been shortened by the property of similar triangles
woeful flame
Wait but I still dont get somthing here
ok you show that A1MB and D1NE are similar
but why AA1/MA1 = DD1/ND1
vivid ridge
I really gotta go now, but if you proved that then AM / DN is the same proportion
woeful flame
OK sir thank you very much, but you didnt gave me a correct solution
I hope someone else here will be able to find the solution for my question
Im sorry but you are wrong In2erval
@vivid ridge you cant use transitive in similarity
if triangle X is similar to triangle Y is similar to triangle Z - you CANT say that X/Y = X/Z
you say that if ABC and DEF are similar triangles
and if also A1MB and D1NE are similar triangles
so obvious that A1/MA1 = DD1/ND1 but its not true
because you are using transitive
upper karma
I know how to find the roots for any problem of the form sin(ax) + cos(bx) but where would I begin to find the roots of sin(ax) + cos(bx) + c?
I'm struggling to find an approach
vivid ridge
Can you make it into a quadratic involving sin(x) or cos(x)?
like asin^2(x) + bsin(x) + c = 0?
upper karma
Not for large values of a and b which are the cases I'm interested in
vivid ridge
how large are we talking
upper karma
4 and also problems where a = b are trivial so I'm excluding those
vivid ridge
would something like this work? https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Sine,_cosine,_and_tangent_of_multiple_angles
just a wild guess really
upper karma
Hmm I don't think it would
Maybe I've got something I'll start with sin(3x) + cos(2x) + 1. I know I said a and b > 4 but I cant even do this one yet so here goes
I can rewrite that as
1 + cos^2(x) + 3 cos^2(x) sin(x) - sin^2(x) - sin^3(x)
And then I can use power reduction on most of the terms to get a wall of text I'm not going to write out
vivid ridge
1 - sin^2(x) is cos^2(x) so that's one part you can simplify
upper karma
Yeah true
vivid ridge
maybe take sin(x) out as a factor in some of the remaining terms
upper karma
2cos^2(x) + 3 cos^2(x) sin(x) - sin^3(x)
that seems easier to work with now but ehh
now I'm stuck lol
vivid ridge
you could also try to make it a bit simpler by substituting u = cos(x) and v = sin(x)
make that into a multi-variable polynomial
upper karma
so that would look like
2u^2 + 3uv - v^3
vivid ridge
yep, and... people would use stuff like elliptic curves to study the behavious of that
vivid ridge
oof
upper karma
and some other scary looking roots with all kinds of garbage
and so then I would need to get
2cos^2(x) + 3 cos^2(x) sin(x) - sin^3(x)
to evaluate as 1 , 2 which I should be able to do using inverse sine and cosine
I believe
And then I'd know one of the roots
upper karma
Yeah