bc "share a height" sounds weird af to me
#geometry-and-trigonometry
dark sparrow
somber coyoteBOT
euler/fermat:
upper karma
@pastel swift i meant the midpoint of HG :/
pastel swift
silent plank
what have you tried?
silent plank
do you know how to calculate the area of the curved surface?
upper karma
No
I am guessing it’s Pythagoras
Theorem that I have to use
But I am probably wrong
silent plank
no, pythag isn't needed here.
you should probably review some formulas and come back if you still have issues
dark sparrow
it's pi, not pie.
upper karma
Don’t care I was right
stuck torrent
eager yew
Anyone know the steps to getting from part 1 to 2. I been at this for a while now and I got no idea
terse sage
np
somber coyoteBOT
euler/fermat:
dark sparrow
no
as you can plainly see they don't even come close to C or D respectively
so how can they be "from" C and D
yes that's more like it
slate bay
How would you explain the effects of changing the coefficients of a linear equation in two variables, that is ax+by+c=0.
It's pretty evident that if b=-1, then tweaking a is changing the steepness of the line, but changing them simultaneously does what?
I'd like to describe this without needing to refer to vectors, dot products, and orthogonality.
fringe dirge
I mean it obviously depends how you change them
slate bay
I mean, yeah, but like, it also depends on what the other coefficients are.
Like, if b=-1, I could just say if you increase a, then you get a steeper line, and if you decrease it, you get a flatter line. With a>0, ofc.
Eh, nvm. I'm not even sure what I want myself.
fringe dirge
The slope of the line is -b/a
dark sparrow
that's the problem, innit
fringe dirge
That's really all that needs to be said
slate bay
Hm, that's a start.
eager yew
If I had 9/2Cos^2A and I wanted it to write it in the form 9(x + xtan^2A) Do I use x as 2. Or rewrite as 4.5/Cos^2A then 4.5(1 + tan^2A)
sly marlin
9/(2cos^2(A))?
eager yew
Yes
sly marlin
$\frac{4.5}{\cos^2A}$ then $4.5(1 + \tan^2A)$ should be what you are looking for
eager yew
Am I not able to leave the 2 on bottom line and multiply by 9
somber coyoteBOT
Element118:
sly marlin
What do you mean?
sly marlin
yeah, $\frac{9}{2\cos^2A}$ is fine
somber coyoteBOT
Element118:
eager yew
How could I make that into tan^2(A) ?
sly marlin
You just need to keep your 9/2 to get $\frac{9(1+\tan^2A)}{2}$
somber coyoteBOT
Element118:
copper girder
need assistance with this please
fringe dirge
what have you tried
copper girder
not much im having troubl
spark stag
note that that right triangle has a 45 degree angle
do you know what that means for this question?
copper girder
sin45=1/√2 ?
spark stag
i was looking for something a bit more simple
the right triangle is isosceles; ie, its two "legs" are the same length
copper girder
the other angle in the triangle is 45?
spark stag
yep, that too
copper girder
ohh i see
somber coyoteBOT
Namington:
spark stag
do you see where that comes from?
copper girder
yup i do
dark sparrow
$\frac{x}{x+100} = \frac{1}{\sqrt{3}}$
somber coyoteBOT
Ann:
copper girder
hmm so rationalise, then cross multiply? after that im not sure
dark sparrow
what do you get once you do that?
i'm pretty sure you are overthinking this already
copper girder
3x=√3+100√3
dark sparrow
are you sure you didn't miss anything there
copper girder
yes, correct me if wrong please
dark sparrow
$3x = \sqrt{3}x + 100\sqrt{3}$
somber coyoteBOT
Ann:
copper girder
3x-√3x=100√3
x(3-√3)=100√3 ?
i ended up getting 50(√3+1), no solution sheet to this but it looks right. thanks all.
quiet mason
nicee
dark sparrow
not everything has to be
i mean is there really that big a search space when it comes to trial & error
dark sparrow
........why did you delete your problem again?
quiet mason
lmfao
dark sparrow
i had to disappear and now i don't remember what the problem was!
quiet mason
i thought what was he even talking abt
dark sparrow
sometimes you may need to do some trial & error
quiet mason
euler / fermat youre sys
dark sparrow
nothing wrong with it per se
dark sparrow
as time goes on you'll get a feel for what works when
and that'll let you cut down on the trial and error
but don't try to pursue that directly
you'll just end up wasting more time
eager pendant
wait can you give a diagram to illustrate what you're talking about
idk what you mean by angles inside of each other
silent plank
think about what you need to get the answer
then try and find that info
dark sparrow
just post a diagram
quiet mason
no seriously we wont dox or hack
dark sparrow
AND DON'T DELETE IT MINUTES LATER FOR CHRIST'S SAKE
quiet mason
just post
eager pendant
ok what's the actual problem
(this is why deleting questions is bad)
and re: "the problem really doesn't matter" - context helps because it means people can gauge what you know, etc. and then appropriately help you. it's the entire reason why math.se wants questions to have details of what you've tried, etc. (and while that level probably isn't needed here, i think you can see why not including it makes it harder to help)
silent plank
questions with different given info will require different approaches
eager pendant
@night karma are you sure the blue angle is always the same?
night karma
I tried to make is so it is, by adding the 180 degrees and that extra extention with 90 deg.
eager pendant
i don't think that ensures that it's constant
sly marlin
The extra extension of 90 degrees doesn't really do anything
night karma
Mhm.. you can extent it, right?
sly marlin
It could be 91 degrees and it wouldn't make a difference
The vertical line could be further to the right and the angle would get smaller
night karma
If the 90 deg. was further away to the right and the point with the missing corner was higher that'd be possible too?
sly marlin
No, the angle would get smaller.
night karma
That's what I mean haha
sly marlin
It's underdetermined.
night karma
Aight, I'll try to improve it again. Thanks! 😉
upper karma
For trig is it n/x or x/n
zenith ember
Context?
sly marlin
Might be sin rule, but I have no idea.
dark sparrow
@upper karma you're going to have to provide context
what is n? what is x? what is their ratio supposed to represent?
left folio
How would I find the base of a triangle inscribed in a circle where each vertex touched the perimetre of the circle?
Well not base, the value of each side
eager pendant
well it depends @left folio because every triangle has a circumcircle
left folio
Oh yes sure, the triangle would need to have all 3 sides equal in length
night karma
Would you say this triangle problem is consequent?
gritty flare
it seems doable
night karma
That'd be amazing if it is!
Yet, I'm having a hard time creating the shape in Geogebra; to see if it's possible, without calculating any of the missing angles.
night karma
What would that mean for the problem? 🤔
gritty flare
idk just asking
okay anyways write out as much angles as possible
the missing angles
night karma
Well, I made the problem myself, but I'd like to know if (this time) the problem is actually consequent. Already doubting it since I am unable to draw it myself if I were not to know all angles
Didn't experience any flaws while solving and no corners are e.g. 60.893 deg. So the problem has been made exact, I hope
rapid depot
@night karma are you asking if its doable?
gritty flare
it did look doable at first
but im having troubles now
it may actuslly be rly ez and im just dumb
night karma
Yeah, asking if the question is consequent and that there's ONLY ONE solution.
night karma
Yep! It's meant to be. Nice find. Hope that's consistent too 😬
gritty flare
the drawing is not
but u can just put to scale i think
drawn to scale if im not wrong
night karma
How'd I do such a thing?
gritty flare
no
night karma
Oh, I did the same
No, it's meant to be 50. Sorry for any spoilers, but I need to check
rapid depot
how did u learn to do math.....
O.o
that looks way more complicated then it needed to be
night karma
Yeah, haha. I'd like to believe that
gritty flare
uhm lemme try something here
lets name ? - x
then in quadrilateral one angle is 180-x
night karma
Look at this cut-out piece (from a previous try), the ? can be ANYTHING (well.. in between certain values of course). That's basically what I want to avoid with the problem I sent a moment ago
But.. you think the one I sent (not the one right above this message lol) is consequent? 😅
night karma
That'd be sick!
gritty flare
lemme actually get to it and see if it is
night karma
Yeah, it's inspired by this page: https://eylemmath.weebly.com/
You may or may not have seen it before.
night karma
You don't?
gritty flare
uh cant find any and its like 2am
night karma
You're not to blame for needing pen/paper by the way. It's too complicated to even try in your head
gritty flare
i labeled 20 degree angle as x
and followed thete
idk if its rly doable but it seems so
mighty gull
uh let me try. is this problem independent of the one youposted above?
night karma
This is the problem. Tell me if it's consequent!
mighty gull
oh i see. let me try solving this one then
night karma
👍
sly marlin
I feel that maybe we are missing something here.
night karma
As information to solve the problem
As information for the problem to be consequent
Other subject..
?
sly marlin
Let me construct it properly
night karma
All right!
sly marlin
Yes, we are missing something in the question
Is the big triangle equilateral?
If yes: Add in that condition properly and you are fine.
If no: There's still a way to change that angle we want to find.
night karma
Hmm.. following my own calculations, the biggest triangle indeed turns out to be equilateral. If I were not to add that to the 'question', how'd someone be able to misuse this?
night karma
What exactly would be unknown yet?
sly marlin
Proof: Fix the position of the 100 degree, 60 degree angle. Then the top vertex is fixed. For any 70 degree angle you can draw, you can always move the vertex up and down that side. That gives the intersection point in the triangle and the length of the line segment. With a lower intersection point and longer line segment, the triangle would not be equilateral.
night karma
And the equal lines (marked with 1 'marking') would still be equally as long?
upper karma
n is the number that you know; x is the number you are trying to find
dire fog
I need help with this question
spark stag
do you know what the internal angles of a triangle add up to?
steady forum
silent plank
what did you get when you evaluated it?
upper karma
im pretty confused i tried a lot of different stuff
and tried just dumping it in a calculator
dark sparrow
$\frac{2\pi}{24} \cdot (-10) = -\frac{5\pi}{6}$
somber coyoteBOT
Ann:
dark sparrow
$\sin\left( -\frac{5\pi}{6} \right) = -\frac12$
upper karma
😂
somber coyoteBOT
Ann:
upper karma
hang on i need to give it another shot cus i just realised i messed up the arithmetic
in the problem it was originally (t-10) for the phase shift and its basically trying to get you to find the Y value when t = 0
i did mess up my arithmetic but still confused
been on it for a few hours but think im missing some understanding somewhere
upper karma
how is sin of -5pi/6 = -1/2? why do i get something different if i just dump than in a calculator
silent plank
what is the relative angle of -5pi/6
and what quadrant is it in?
is your calculator set to radians
night karma
That'd make sense
upper karma
well that explains that 👍
im just learning thats a thing and now everything makes way more sense
😂
autumn kestrel
why are these two both true?
mighty gull
thats just wrong then
oh they said for some value of x
so they arent saying its true for all x, but that there exists some value of x for which it is true
autumn kestrel
yeah
mighty gull
so try solving all 3 to see if theres some value of x for which it is true
autumn kestrel
it would be 0/360 degrees right?
mighty gull
for which one
autumn kestrel
both 2 and 3
mighty gull
yeah should be
so there is an x for both II and III, and obviously none for I so thats your answer
autumn kestrel
so is it just as simple as plugging in? or is there something conceptual to it
mighty gull
i mean if you stretched it i am sure we could find something conceptual, but it seems like they just want you to solve the eqns using sin^2+cos^2 = 1
small raptor
Third one, I think you can just divide out the 2, and you'll get cosx^2 + sinx^2 = 1 which is the Pythagorean identity.
autumn kestrel
what about dividing the second one then
mighty gull
theres no 2 on the sin^2 neveza
small raptor
Yeah, was thinking that would violate the whole (a+b) relationship after I said it.
mighty gull
np
dark sparrow
why would you need to remember any at all
what side lengths
sounds like another futile pursuit
you'll waste more time memorizing triples than you'll gain being able to "instantly recognize" a right triangle by its sides if they do happen to form one, and you'll eventually be able to do the latter anyway with enough exposure and practice
i mean if you wanna waste however many hours of your time memorizing and memorizing and memorizing then go ahead really
eager pendant
wait why do you want to remember
dark sparrow
the rule that lets you generate pythagorean triples given two arbitrary integers?
yes, i'm familiar with that
eager pendant
if it's for recognising them in tests, idk, but if you want to generate lots, use the parameterisation yeah
try and derive it yourself!
it helps you remember, (im sure you could find a proof online though)
dark sparrow
are you asking why $(u^2 - v^2)^2 + (2uv)^2 = (u^2 + v^2)^2$?
somber coyoteBOT
Ann:
dark sparrow
you are perfectly able to check this yourself
it's nothing more than a little bit of algebra
eager pendant
i think what ann said above is pretty clear lol
dark sparrow
sounds like another futile pursuit
you'll waste more time memorizing triples than you'll gain being able to "instantly recognize" a right triangle by its sides if they do happen to form one, and you'll eventually be able to do the latter anyway with enough exposure and practice
i mean if you wanna waste however many hours of your time memorizing and memorizing and memorizing then go ahead really
what places?
are you saying that because you saw them in a lot of practice problems?
the only reason for that is that it's easy to use them to create geometry problems with "nice" numbers, which may even be necessary in a setting where the only communication channel between the student and the teacher/grader is through an answer sheet
or a textbox that only accepts integer or at best terminating decimal or fractional input
dark sparrow
huh
dark sparrow
can you show some exercises from that chapter
eager pendant
(actually i lied, this one is marginally easier if you know some pythagorean triples. the main idea is influenced by the parameterisation - which is infinitely more worth learning than arbitrary triples )
the point of this exercise isn't to memorise the triples with 97 on them and go down a list
it's to use the parameterisation to find all the triples, which isn't bad since 97's prime
dark sparrow
i mean
honestly
the idea of memorizing triples THAT FAR just sounds like idiocy to me
eager pendant
btw what book is this
by aops i thought you meant the community, not the actual company who writes the books
which is why i was taken aback at first
tawdry pivot
proving that a,b,c must be in the form of euclid's formula is a little more subtle than proving euclid's formula is a pythagorean triple
eager pendant
well if your impression of the chapter on pythagoras is to memorise triples, i feel like you've misread or missed something
(also, what's true of most aops books is that you're not bound to a particular reading order. if you feel like this thing isn't useful, just move on to a different chapter or exercise, etc.)
tawdry pivot
idk if it's necessary to actively memorise triples, to do well in competitions
dark sparrow
it is most definitely not
but some people think MEMORIZE MEMORIZE MEMORIZE CRAM CRAM CRAM is the solution to everything ¯_(ツ)_/¯
tawdry pivot
ye I don't recall seeing anything much other than 3 4 5 and 5 12 13 in problem solving
cinder portal
foggy lance
I’m not sure exactly where to put this as it’s a mixture of trig and vector algebra, but I’ll put it here
dark sparrow
holy FUCK what a mess of technicolor on a transparent background
foggy lance
This was a guide I created to describe the relationships between the î and j vectors of a magnitude
Yeah, I tried color coding it in medibang. Discord doesn’t do well translating it, probably need to go back and redo it over white background
spark stag
i feel like I'm in Vegas at midnight
trail minnow
you are
foggy lance
^^
foggy lance
Only spawns of Satan use light mode
And while I’m diabolical, I’m not yet at that level
devout shell
This is what it looks like
foggy lance
That hurts
simple rune
@autumn kestrel you can subtract the equation sin²(x)+cos²(x)=1 from both II. and III., so you get:
II. sin²(x) = 0
III. cos²(x) = 1
and if you solve these, you get the same result: x = π/2 + πk
dark sparrow
the derivation i've seen involves complex numbers
are you ok with that
why what
because it's sometimes useful to consider the plane as a complex plane
and this happens to be one of those times
all i'm asking is whether you're ok with me talking about complex numbers; if you aren't, i'll need to somehow find a way around mentioning them explicitly
you don't need complex analysis
you just need a basic understanding of complex-number arithmetic
ok good that's all you need really
so then
consider the complex plane
and specifically, consider the "lattice points" on the plane - i.e. those numbers whose real and imaginary parts are both integers
now, as you (hopefully) know, the absolute value of a complex number is precisely its distance from the origin (a.k.a. 0), and can be computed like so: if $z = x + yi$, with $x$ and $y$ real, then $|z| = \sqrt{x^2 + y^2}$
somber coyoteBOT
Ann:
dark sparrow
now the thing is
each lattice-point number whose absolute value is an integer corresponds to a pythagorean triple
specifically, if we know that $z$ is a lattice-point number and $|z|$ is an integer, then the corresponding pythagorean triple is $(\Re(z), \Im(z), |z|)$
does that make sense?
somber coyoteBOT
Ann:
dark sparrow
real part and imaginary part respectively
Re(x+yi) = x, Im(x+yi) = y
i didn't get to the 'meat' of my derivation but ok w/e
devout shell
Show your counterexample then
Because I was thinking the only possibility was the one given in your provided answer
dark sparrow
you are indeed not
devout shell
Not pairs of points. It is asking as individual points, where can you place them so that they are the same distance away from both lines
dark sparrow
you want the set of all points P such that the distance from P to one line is equal to the distance from P to the other line
no
terse sage
@dark sparrow Could you tell me about the derivation of pythagorean triplet formula, or link to somewhere about it. I know about complex numbers
Thank you
dark sparrow
tl;dr Pythagorean triples correspond to complex numbers whose real part, imaginary part and magnitude are all integers
all such numbers can be obtained by taking complex numbers with integer real & imaginary parts and squaring them
quiet mason
X and Y
you can prove it yourself
yes
if its the perpendicular bisector
no
AAS
wont work then
the 90 degrees wont be eq
but thats only one point
any point on the perpendicular bisector
is equidistant
from X and G
Y*
ofc there are other cases but they dont satisfy this
look if youre asked write down two points from which all points on the perpendicular bisector are equidistant
you definitely cannot write only one point
i.e. the one with the midpts of the line segments on either side
but X and Y have all points equidistant
yes
dark sparrow
...
EULER WHAT THE FUCK.
@upper karma what did i say about deleting your messages in bulk like this??
flint osprey
@terse sage there is a 3Blue1Brown video about exactly that that you could look up, it's really nice
trail minnow
lmao lionel talking to himself 
olive solar
mighty gull
well how does it feel now euler
devout shell
no one is judging
mighty gull
looks like you were talking to yourself which is wierd isnt it
dark sparrow
@upper karma yknow there's this thing called dialogue
formal bolt
Is it some sort of ecological internet behaviour?
upper karma
I NEED HELP
silent plank
what have you tried?
upper karma
i don’t know which step to start with @silent plank
i’ve been trying to figure it for hours now
silent plank
what would be required to find the coordinates of a point on a circle?
upper karma
having an angle and put it in a cos and sin in ur calculator
silent plank
good.
in the question it tells you that the 5 arcs are the same length. what info can you get from this?
upper karma
they r all the ‘same’ as A than
zenith ember
How many degrees in a circle?
silent plank
A is a point. please clarify
zenith ember
So if you divide it into 5 equal arcs?
zenith ember
How big are the arcs?
upper karma
72 degrees each
zenith ember
So you can get angles for all the points.
upper karma
i’m very confused
i learnt about the unit circle today and we got homework for it including that exercise but i have no idea on how to apply it
silent plank
angle between OA-OB is 72.
you told us you would be able to find the coordinate with this info
silent plank
that will first get you point B
upper karma
OH SH*T
THANK U
@zenith ember i calculated all of the points and A too to check and when i calculated (1,0) came out as the coordinates, do i have to multiple all the other points with 2 now because in the figure point A is (2,0)
or am i using bad logic..
zenith ember
That's correct
upper karma
thank u
silent plank
note that it is symmetrical and that E has the same 'x' but negative 'y' value of B.
same with C and D.
upper karma
i paid attention to that
flint pelican
upper karma
I need help with 48,50,52. The graph is for a different problem
weary drift
,rotate -90
somber coyoteBOT
weary drift
@upper karma Not sure this is how it's "meant" to be done, but for 48 I would calculate AB, the vector from A to B, and since we know B is halfway between A and C, we just add AB to B's position to get C
upper karma
Yeah that’s not what she wants us to do. It’s like she didn’t even explain it
weary drift
How about we call the first number in each coordinate x, and the second coordinate y
Calculate the difference in x distance between A and B. We know that B is halfway from A to C, so what we do is add our x difference to B's x coordinate to get C's x coordinate
And we'll do the same thing with A and B's y coordinates
wide onyx
doing physics but stumbled on an equation I couldn’t solve
180 = 250tanx + 49/cosx^2
thanks!
dark sparrow
i hope that by cosx^2 you do not mean cos(x^2)
if that is the case, then you can write $\frac{1}{\cos^2(x)} = \tan^2(x) + 1$ and get a quadratic in $\tan(x)$.
somber coyoteBOT
Ann:
somber coyoteBOT
crypthes:
dark sparrow
$180 = 250 \tan(x) - 49 (\tan^2(x) + 1)$
somber coyoteBOT
Ann:
wide onyx
wait I’m confused on how 1\cos^2x became tan^2x + 1
oh wait pythagorean identity thing
thanks!
dark sparrow
1**/**cos^2(x).
ionic delta
weary drift
@ionic delta do you know how to identify relative extrema?
btw this question feels like it goes in #precalculus ... 
weary drift
if a point has a lower y value than all of its neighboring points, then it is a relative minimum
if a point has a higher y value than all of its neighboring points, then it is a relative maximum
spark stag
"let that triangle be not a right triangle, therefore the pythagorean theorem will not work"
You would have to prove this statement
but if you do so (probably defining the "Hypotenuse" equivalent as the "longest side"), then sure
not necessarily; consider the theorem "polynomials of degree <5 always have solutions in the radicals"
this doesnt mean, however, that its converse is true
consider, say, x^5 - 1 = 0 having solution 1
(and others)
you can't just say
"by saying if it isn't x, then y theorem would not work"
you'd have to prove that y theorem never works
unless it meets that specific criteria
if you do that, then sure
the proof is valid
in the case of pythagorean theorem
you can quickly sketch a derivation of the cosine law
and its easy to tell using that (and the fact that internal angles of a triangle add up to 180) that pythagorus only ever works for right triangles
in other words, you have to prove that it's an $\iff$ relationship
somber coyoteBOT
Namington:
spark stag
if we have $A \iff B$ and we show $\neg B$, that implies $\neg A$
somber coyoteBOT
Namington:
spark stag
and same with $\neg A \implies \neg B$ in that case
somber coyoteBOT
Namington:
spark stag
i mean, I guess you could just say
"we know pythagorean theorem EXCLUSIVELY holds for right triangles, so..."
basically we know $$right \implies pythagorus$$ my above statement asserts $$pythagorus \implies right$$ so if we have $\neg right$, that implies $\neg pythagorus$
somber coyoteBOT
Namington:
spark stag
technically, ONLY the "pythagorus implies right" part is necessary
in this case
since $A \implies B$ means $\neg B \implies \neg A$
somber coyoteBOT
Namington:
spark stag
so the statement $pythagorus \implies right$ can be rewritten as $$\neg right \implies \neg pythagorus$$
somber coyoteBOT
Namington:
spark stag
so if you show that pythagorus actually does hold
theres your contradiction
but we can be more direct about it
because pythagorus implies right
you have to make that statement either way; your proposed proof goes down the path of proof by contradiction, but that isnt actually necessary here
and in many ways, it's "cleaner" to just quickly show "if pythagorus holds, then a triangle is right"
but your method should work too as long as each step is decently justified
it's just inserting an argument by contradiction when one isnt really necessary
which is ¯_(ツ)_/¯
dark sparrow
@upper karma don't you even DARE delete your part of this dialogue several hours after this convo is over.
worthy root
Lol
fringe dirge
Law of sines
dark sparrow
or law of cosines
tawdry pivot
law of cosines seems a bit more straightforward
dark sparrow
arctan(3/4), arctan(4/3) and π/2
upper karma
you shouldn't. arctan is just the inverse of tan
angle being arctan(3/4) means that tangent is 3/4
dark sparrow
@upper karma why would you
the fact that your first response to most everything is "should i memorize this" is a bit disturbing
small raptor
Maybe he should pursue Biology. They do a lot of memorization
formal bolt
Following corollary: π is approximately 3.1416
small raptor
It's okay. I enjoyed my time in bio I
small raptor
I never take into consideration of the teacher. I enjoyed just learning it on my own. My teacher was decent though. She seemed to have a proper understanding, and even experience in research. Even my lab teacher did the teaching as part time and worked in a lab for her actual job.
Funnily enough, the one with the doctrate doing all the research was the lousy teacher. 😛
worthy root
coarse bronze
would anyone teach me how to evaluate sin, cos, tan, csc, sec, and cot without a calculator
i know the ratio for each of them already
spark stag
what do you mean "evaluate" @coarse bronze?
you cant easily calculate them for all angles, but there are certain "special" angles that students are generally expected to memorize at some point
those being 0, 30, 45, 60, 90, and the angles elsewhere on the unit circle that correspond to those
(I'm writing them in degrees, but you should recognize them in radians too)
"memorization" of the values at those angles can be greatly aided by just knowing the construction of two "special" triangles, though
to the point where you dont really need to "memorize" specific values, just be able to draw a triangle
spark stag
ah, ok
coarse bronze
thanks for tryna help
spark stag
all of those are equivalent to finding the length of the entire line
except one
which one is not the entire length?
@wide ferry
i'm not sure what it means by "both" answers, admittedly
somber coyoteBOT
Lucid:
spark stag
ah, sure
yeah, in that case, not sure
ah maybe it means
find the answer to both parts
the multiple choice, and the fill-in-the-blanks
and for the fill-in-the-blanks, it just wants you to calculate the lengths
using the diagram
the "different" one would be the length of BC - AC, and the "same" would be the length of AB
wide ferry
ok thanks
spark stag
grats
glad ocean
@spark stag the different is 4, the same is 10
somber coyoteBOT
mesmore:
bold folio
queen python
worthy root
copper girder
i need help with this
dark sparrow
what have you tried and where are you stuck?
copper girder
so far ive found that the unknown angles in the top two triangles are 50 and 60
i dont know how to find the angles on the ground though
dark sparrow
well
okay, so for ease of reference
let's call the point with the three right angles O and the point on the top C
so what you're saying is you don't know how to find angles OAB and OBA
is that correct
copper girder
yep
dark sparrow
well
if you can't find them
give one of them a name
like θ
and try to express some things in terms of θ
copper girder
∠OAB = θ
OB = sinθ x 400
my guess is you would need to find either OA or OB and then apply trig ratios to find the height
dark sparrow
OB = 400 sin(θ), OA = 400 cos(θ)
OC = OB tan(30°) = OA tan(40°)
400 tan(30°) sin(θ) = 400 tan(40°) cos(θ)
@copper girder does this make sense to you?
copper girder
i dont understand:
400 tan(30°) sin(θ) = 400 tan(40°) cos(θ)
dark sparrow
OC = OB tan(30°) = OA tan(40°)
do you understand this
copper girder
oohh i understand now
copper girder
yes, thanks for the help. appreciate it ❤
dark sparrow
nope
devout shell
going to pull up some circle theorems
mild cargo
Notice that AF and AE are tangent to the circle and then use the known angle + the inscribed angle theorem
dark sparrow
@upper karma want a hint?
there's a circle here
what happens if you consider its center?
in general, when you have a circle, its center is almost guaranteed to be involved somehow
so mark off the center on the diagram
give it a name
maybe connect it to some other points
see what you can get from that!
what are those
are those your diagrams?
okay so you got angle EOF = 108° it seems
...wait, you said you were not supposed to use the inscribed angle theorem or sth?
(O is what i'm calling the center btw just for refs)
okay alright
the center of the only circle in the picture
okay so
another hint i suppose
extend the radius DO to meet the circle again at another point
call that point K
(so that DK is a diameter)
consider triangles DOE and DOF
and also angles FOK and EOK
opal sonnet
zenith ember
You have to ask a question before we can help.
mental wharf
they still have yet to actually ask a question
umbral snow
We still haven't gotten that question
tawdry pivot
RemindMe! 1 day
weary drift
And we never will get the question... he left the server
mental wharf
whats wrong with these people
formal bolt
Maybe he just turned caps on by mistake and didn't know how to go back
flint osprey
plot twist
he thought this was #methematics
torpid harness
Hello! I am a 11th grader with a learning a disability (Comprehension ) , I've been having a hard time getting time to get work done. I am lost and I get stuck and i'm getting overloaded with work, I really want to catch up with my class and not fall behind. I hope I can get some help. My teachers don't seem to have time for me. ):>
upper karma
teachers are mean 😦
dark sparrow
csc(θ) is not 1.7 and cot(θ) is not 1.33; the other four values happen to be correct but only because they're terminating decimals
fervent oracle
yup
dark sparrow
5/4 is not 1.666666667 either
if you meant 5/3, still not the case
5/3 != 166666666667/100000000000
4/3 != 1333/1000
no
weary drift
either keep it in fraction form or write it as a repeating decimal
mental wharf
just write is as a fraction
dark sparrow
let me say this clearly: 4/3 HAS NO REPRESENTATION AS A TERMINATING DECIMAL
weary drift
$\frac{4}{3} = 1.\overline{3}$
somber coyoteBOT
RokettoJanpu:
dark sparrow
why do you hate fractions so much, @upper karma?
balmy pelican
lmao
dark sparrow
cot(θ) = 4/3 that's it
just leave it
as
a fraction
again
why would you ever think that fractions weren't an appropriate form for an answer
dark sparrow
consider viewing this sum as (AB+AF) + (AC+AE) + AD
worthy root
3*AD
left folio
I calculated the areas of an equilateral triangle inscribed in a circle and of a square inscribed in the same circle. I found out that the square's area is not twice that of the triangle, can anyone confirm?
upper karma
yeah
steady umbra
@worthy root from unacademy.
buoyant idol
how do u find an angle from lengths?
silent plank
be a little more specific
buoyant idol
wait
is it possible to find an angle from a triangle
that isnt 30,60,90 or 45,45,90?
silent plank
depends how much info you are given
worthy root
@steady umbra nope . I searched for vectors problem pdf found this
buoyant idol
well what kinda info do u normally need?
can u like have 1 length and an arch and find the rest of the lenghts?
subtle gate
can you specifiy
silent plank
arch?
buoyant idol
angle soz
also its not a question (for schl or anything) im j clarifying to see if its possible so sorry, i cant give an example or anything :/
silent plank
that wouldn't be enough
buoyant idol
whats the minimum u would normally need?
silent plank
if you know your triangle identities, sin/cos rules, you would know the requirements
buoyant idol
the ratios?
silent plank
a/sinA = b/sinB etc...
zenith ember
It -is- possible to get an angle from a triangle that is not one of those two special triangles.
buoyant idol
it is?????
zenith ember
Although it is much more unlikely to be easy to calculate.
buoyant idol
eh its not test practice or anything so i guess its fine :/
zenith ember
As it will involve the inverse trig functions.
buoyant idol
sin^-1(x)?
zenith ember
I prefer arcsin, but yes.
buoyant idol
idk the name but at least i got it right lmao
subtle gate
(edit:) some people use sin²(x) and so sin^-1(x) is really confusing if you do it that way
zenith ember
Usually, the meaning is clear, but there is the chance that sin^-1(x) can be confused with 1/sin(x)
So I prefer arcsin.
buoyant idol
im already confused
is arcsin linked with derivatives or no?
cuz the first result came up w derivatives
subtle gate
it's just the inverse of sin
buoyant idol
ah more info to learn
nice
this is all hs level right?
cuz if it is
i have a extremely strong feeling im not gna make it to uni lmao
zenith ember
Trigonometry is typically covered in HS, yes.
But a lot of folks find it difficult.
buoyant idol
FUG
subtle gate
i had problems too, but you just need to practice with it
buoyant idol
eh
i have so many things i want to practice
but trig isnt 1 of them j cuz i like to do challenging stuff
zenith ember
Gotta walk before you can run.
buoyant idol
but i guess i gta do the """""""basics"""""""" (although not very basic) stuff b4
yeye
weary drift
so you got your graph and they want you to find the y value of a certain x value. what x value did they give you?
upper karma
thats the problem
It's unclear
ok fren
what we gotta do here
I fr don't get what it is tryna ask me
rustic temple
basically find the y-value when the x-value is -1
upper karma
so how do I do that
rustic temple
draw the line x=-1 and find where it intersects with the graph
upper karma
so imagine this
btw this has to deal with functions just to let ya know
f(x) = y
f(x) equals some stuff
basically when x = -1
what is y?
f(-1) = y
go to the graph lets take a zoom in 
ok gotcha
so when x=-1
y is -1?
where is the y point in the grap
yea fren 
it's my 5th day of college
bruh moment
i can't beleive it's that ez
EZ PZ
ez cuz u smrtboi
this isn't a function tho right
To be a function, the graph must pass a "vertical line test"
o ik that
ye
so that aint a function
Meaning, for every x value, there can only be 1 y value
in this case when x=2
y equal two number
confirm not function
got it
latent lynx
Using only lines of undetermined length that can be vertical, horizontal, or either of the 45 degree diagonals, is there a short way to produce a 60 degree angle?
I did it in 10 lines, but is there a solution with less work
upper karma
@upper karma can u come in clutch again
Domain: what numbers can x be?
Range: what numbers can y be?
infinite?
o wait
x is infinite righrt
and y is 4
can y be 3?
no
how bout 5?
ye
so yeah x is infinite
buy y is 4, or higher
because it CAN be 4
yeah so (4,infinite)
no
[4,infinity)
hah
bracket matters
oh yeahhh
@upper karma look right?
I agree with the answers you've put in there
ty
also since you're using what looks to be my laps plus
@upper karma
make sure you use comma not a period
, .
i've made that mistake too many time
just a warning
I'm using Pearsons MyMathLab
graceful hemlock
Turn the equation to slope intercept form
upper karma
yeah ok
upper karma
but it doesn't have 2 x and y values
graceful hemlock
Mhmm lemme get pen and paper lol brb
upper karma
ok ty
graceful hemlock
Get the opposite reciprocal of the slope. Then, put the slope into point slope form and plug in (-3,-5)
upper karma
can u show me what u did
graceful hemlock
You’ll get something that can be manipulated into slope intercept form
Lol I haven’t done anything yet
paper mauve
Why use reciprocal
upper karma
yeah
graceful hemlock
Point slope is y-y1=m(x+x1)
upper karma
yeah thats the formula I use
paper mauve
So get it into y=mx+b
upper karma
aight
paper mauve
It happens
graceful hemlock
Change the 7/3 to b
upper karma
k
graceful hemlock
So it’s y=-4/3x +b
graceful hemlock
Oh crap oof
paper mauve
Yes it is in point slope form
paper mauve
Now you plug in -3, -5
graceful hemlock
Get your slope and plug in the points to point slope
paper mauve
I trust your arithmetic skills enough to say that this should be right
Is it correct?
upper karma
prob
doesnt tell me till the end
@paper mauve any idea?
ughh
,w graph the point (0,-4)
somber coyoteBOT
Requested by Babowwz#1394
upper karma
A horizontal line goes along the horizon
left to right
ye?
y=mx+b amirite?
yes
oh slope is 0?
so a flat line has no slope
yee fren
y=+b
but what is b
b represents the y intercept
which be -4
ok
,w graph y=-4
somber coyoteBOT
Requested by Babowwz#1394
upper karma
so it's litereally f(x)=-4
So i think we should talk about the Cartesian coordinate system
yes
ok
it's a flat line through point (0,-4)
-1,-4
-500,-4
ok yeah I was never taught that xD
horizontal is left to right
vertical up and down
signal locust
alright so ive just started geometry and i am struggling with the basics already
these types of questions trip me up im not sure how to solve them?
spark stag
Since B is the midpoint of AC
weary drift
so B is the midpoint... what does that tell you about AB and BC?
signal locust
right i understand that but do i need an equation to get x?
spark stag
Yesv
signal locust
like 2x+7+3x-4?
signal locust
ohhh so then it would be 2x+7 = 3x-4?
signal locust
alright thanks lol i get it from there
weary drift
@upper karma did you make sure you're converting correctly between radians and degrees?
the above formula uses 360 degrees
but the answer must be given in radians
$360^\circ = 2 \pi$ radians
somber coyoteBOT
RokettoJanpu:
weary drift
the circumference is not 36
do you know the formula for the circumference of a circle?
no prob
signal locust
i used the x1+x2/2 formula for this and got 8,6 im not sure what to do next? idek if i was supposed to use that formula or if its right
weary drift
why do you believe you have to use that formula?
do you know what that formula gives you?
signal locust
oh damn isnt it the midpoint?
signal locust
how would i find that?
small raptor
I think you use the distance formula, no?
weary drift
$d = \sqrt{\Delta x^2 + \Delta y^2}$
somber coyoteBOT
RokettoJanpu:
weary drift
delta(x) is the difference in x coordinates, delta(y) is the difference in y coordinates
signal locust
so i get the distance of the x and y coordinates and then add that together?
im just confused i dont really see anything in my notes for this
weary drift
you have coordinates N at (4, -3)
and D at (8, 9)
those are (x, y) coordinate pairs
delta(x) is the difference in their x coordinates
delta(y) is the difference in y coordinates
small raptor
Easier to visualize if they were vectors, honestly
weary drift
whatever works for you
signal locust
ive never seen that formula before
somber coyoteBOT
neveza:
weary drift
$d = \sqrt{(x_1 - x_2)^2 + (y_1 -y_2)^2}$
small raptor
yeah, above. I just cannot work the bot
somber coyoteBOT
RokettoJanpu:
weary drift
delta(x) implies you are taking the difference of 2 x coordinates
but if x1 and x2 makes it easier for you to know what's being calculated, that's fine
small raptor
I mostly suggested that format, because most early introductory do not show the delta symbols
So, it may look foreign to the person
signal locust
yeah im not familiar with delta quite yet
small raptor
Delta is just shorthand for difference/change
weary drift
anyway... now that you've gotten to know the distance formula, what have you got as your answer?
signal locust
i tried plugging it in but i dont think im getting it? i got a radical as my answer so either im doing it wrong idk
small raptor
Show us the work by hand?
small raptor
Show us the work regarding your specific problem. 😛
Maybe you put something in wrong, or maybe you got it right, and it's just uncomfortable answer
signal locust
alright this is what i have so far i then just entered into the calculator but im gonna try it again by hand i just dont know if im plugging things in right
this formula is brand new to me idk what im doing
small raptor
And what was your calculation from there?
Is this a different problem? the coords you have listed is different from the one you posted above
weary drift
doesn't look like you plugged in the right values