#geometry-and-trigonometry
1 messages ยท Page 163 of 1
woah where did we get the 25
typo
Aquiver when you're finished with the homework I can show you why sohcahtoa is a thing
is that the < measure?
yeah
why 180/pi?
like
cuz 180 is half 360 and pi is whats used to find circles or whatever
we havent done circles
only triangles
triangles are boring :(
yesyesyes
i see
fun
So there's 2pi radians in a circle and 360 degrees in a circle
we want a way to convert from radians to degrees
We just multiply the radians by 360/2pi
which is 180/pi
holy shit
I just realized why its called arctan
so interesting
Math is awesome
but so weird
its the length of the arc for the tangent
https://www.maa.org/external_archive/devlin/LockhartsLament.pdf if you ever get the time, this book changed my life
gilderoy lockheart?
sin = soh?
@upper karma everything makes sense
SohCahToa : Sin (opposite/hypotenuse), Cosine (adjascent/hypotenuse), Tangent (opposite/adjascent)
dude even i could see that lol
biology makes things so much easier
meiosis one meiosis two
literally what theyre called
=tex \sin(x)=\frac{\text{Opposite}}{\text{Hypotenuse}}
$$\sin(\theta)=\frac{\text{Opp}}{\text{Hyp}}$$
^
@crude kraken u beat me to it
why a zero
thats theta
in sin(_)
Theta
ฮธ
It's a symbol generally used to represent angles
just convention
but I mean, you saw me use sin(x), it doesn't really matter
hell, I'd be fine with $$\sin(\int)$$
AAAAAAAAAAAAAA
TOO MANY SYMBOPLSSS FOAJSDKL;F
blasphemy
@restive void draw a circle with a radius of one and plot the center point
=tex \sin\left(\int \sin(x)dx\right)
:<
smile
so im assuming in cos the adjacent angle cant be the hypot
adj is never hyp
what is the function
cos
Of
of what angle
Mk
Cos is cah, so for x this is just 13/26 or .5
for y this is 13sqrt(3)/26
or sqrt(3)/2
cos(x)=13/26=1/2
mhm
@restive void Geometry Dash? Lmao
sir
Funny seeing you here
explain who you are
Your know Cemplix?
:0
Wow really
You are popular in osu taiko community?
Hahaha
Are you in the discord server? For taiko community
vladyushko?
I haven't talked to them, i just know he's top canadian, not sure but top 3 def :p
Oh he is really good
Watching his cam videos are crazy
i guess people may recognize me if we played multi together a lot
i remember being in a game that was super out of my league haha
like 10 ppl, all others from japan
and all in the top 100 ๐
spooky but cool
fail every map
Hahaha
sightreading 8* ๐
Yeah crazy stuff like I remember in a lobby with top players and for example they played uhhh
Gypsy tronic DT and it was omg
As in musical instrument @waxen gorge ?
More specifically, violin
For solo ensemble?
Ah
What grade you doing?
Yep
Well I finished la follia/rigaudon and stuff and I wanted to do praeleddium but I'm doing Mozart concerto no. 3
.-.
So
La follia is grade 6
so idk
Is it?
It's Suzuki book 6
Oh nice nice
I've heard of Mozarts concerto
I like suck at sight reading
I'll soon be playing grade 6s but not yet
I had an allstate audition
Apparently grade 5-6 is a big jump
and I did the excerpts perfect
And then flunked the sight reading
So I got alternate :(
Middle school allstate
You're in middle school?
Ya 8th grade
Wtf and you are playing grade 6 already
...
My senior year I'll do a grade 6 but thats the highest Ill go
So India? Lol
Ah
The grades for music is graded 1-7 or 8
Like I barely go outside xD
When doing for auditions or solo ensemble
So bleh is already up there as an 8th grader
Wait omg
One more grade
Till Vitale chaccone
:o
I can't wait
Ur probably mixing it up with bach's
O
It sounds so gud
.-.
Flight of the Bumblebee is overrated
Not that hard
Idk what its like for strings
But for Trombone its not hard to double tongue and move the slide haha
O just realized I practically skipped grade 4
Yep
But with wind instruments don't u need circular breathing
Ahhh I am right below
Fotb?
You can't practice circular breathing for more than about 5 minutes or you will pass out
Flight of the bumble
Bee
O
Wait bleh
Are you going to the Troy honour band?
In Florida we have an anual thing where people are nominated to go to Troy and do band stuff for like 3 days
Annual*
It doesn't apply?
We have an honor orchestra...
I would assume they have an orchestra category
Oh ok
It's like grade 2-3
Last yr I got 2nd violin 1st chair (5th best) but prob gonna do worse this yr
Cuz allstate is right before honor orch and I spent a bunch of time on that
But honor orch was delayed so audition is in two more weeks
wha
You play osu as well?
yep
Lol
not that good
Taiko?
Oh standard?
yessir
Lol recognizing everyone
Ahh haha
i am so popular :0
xD
fuck
well
shameless advertising
:)
XD
should I sub.. hm
Btw what ever happened to renevant?
i started hating gd
Yes he posts quality content
i quit for 5 weeks about
Oh ok
Ooh u play csgo~
Did you see that Rafis is new #1?
What rank r u
o I used to be gn3 then quit and stupid mm system put me in s2
ya i saw!
I mostly play community servers tho
Do you still talk with gd people?
weren't u going to bed @waxen gorge
Lol
Oh wow
I barely practiced my Mozart and my rhythm sucked so my violin teacher was like: no..no..no
gd ~ geometry dash?
yessir
Then he played the accompaniement
O k
Amplitude is the absolute value of a term
So the peak is y=4
Oops forgot the vertical shift
+3
To find the period its 2pi/b so its simply just 2pi as the period
Doesn't look to have a c value so no horizontal shift
So the peak is at y=7 trough is at y=-1
Period is 2pi
And to find the yintercept set x=0 -4cos(0)+3
Giving you (0,-1)
Got that
brb, breakfast
๐
just gonna give things names for ease of reference now
AC = x cot(23.5ยฐ), BC = x cot(46ยฐ)
x cot(23.5ยฐ) - x cot(46ยฐ) = 100
=tex x = \frac{100}{\cot(23.5^\circ) - \cot(46^\circ)}
@keen aspen
sure ๐
yeah this looks correct
Sweet imma remember to do yours though much more simple
#11
I have the answer but idk which angle
I thought it was the top angle but apparently its bottom and its confusing me
Unless the triangle is orientated differently
the verb is orient ๐
A Russian teaching an American English lel
186.60254038
0.53012074
...wait that doesn't make sense hang on
31.8mph for reference
== 186.6025/4
46.650625
this is the train's speed in ft/s
oh yeah
46.6506 ft/s * 1/5280 mi/ft * 3600 s/h
== 46.6506 * 1/5280 * 3600
31.80722727
there
yep thanks! ๐
o/
BE splits <ABC in half
AE splits <BAD in half
I need to proove that <BEA = 90 degree
How to?
is ABCD a parallelogram
Yes.
If you mean
AB || CD and BC || AD
@dark sparrow @unborn bane
Sorry I haven't mentioned that
if you let angle ABC and angle BAD to be x and y
then x and y are supplementary
=tex x+y=180
=tex (1/2)x + (1/2)y = 90
1/2 of x and 1/2 of y are simply angles EBA and EAB
That's what I did ayy but why = to 90 tho
what?
oh I know why
=tex <EBA + <EAB = 90
I'm pretty sure that ain't enough :/
and you know that the sum of the interior angles of a triangle is 180
\angle
oh ok
=tex \angle EBA + \angle EAB = 90^\circ
so I need to say that The sum of all the angles in a triangle is 180 degree?
yes ofc
How to parallel helps me?
Because the sum of two angles between parallel are equal to 180?
uh
I didn't parse it correctly
In parallelogram the sum of 2 angles next to each other are = to 180
Yea, that's what I meant
Thanks.
@unborn bane But I don't know if the third triangle angle must be 90 degrees
I'm pretty sure I'm not allowed to use the thing to prove (That <AEB = 90) to know things.
Anyone else? ๐
can you repost your question?
Yes sure.
ABCD a parallelogram
BE splits <ABC in half
AE splits <BAD in half
I need to proove that <BEA = 90 degree
How to?
@dark sparrow
Yes!
Yes I did that one
=tex \angle BAE = \frac{1}{2} \angle BAD; \quad \angle ABE = \frac{1}{2} \angle ABC
is that clear?
Yes
=tex \frac{1}{2}(\angle BAD + \angle ABC) = 90^\circ
is that clear?
=tex \angle BAD + \angle ABC = 180^\circ
I'm not allowed to guess that <AEB is 90 because the actual question
we haven't invoked AEB yet
anywhere
=tex \angle BAD + \angle ABC = 180^\circ \ \frac{1}{2}(\angle BAD + \angle ABC) = \frac{1}{2} \cdot 180^\circ
@unborn bane Said that we are comparing it to 90 because the third angle
ok, it's clear now @dark sparrow
How do we invoke AEB now?
Yea. ofc
=tex \frac{1}{2}(\angle BAD + \angle ABC) = 90^\circ \ \angle BAE + \angle ABE = 90^\circ
is that clear now
Yes
=tex \angle BAE + \angle ABE + \angle AEB = 180^\circ
@dark sparrow what did you use to make this figure? https://cdn.discordapp.com/attachments/326138757474680852/405745656011948033/unknown.png
Yes, but we don't know it's 90
we do actually
since the other two angles add up to 90
and all three angles must add up to 180
Wrong mention :/
I didn't ask that but uh okay
lol I forgot we have shape tools in paint
oh, those are just straight lines
@dark sparrow In the trapeze the diagonals split the angles in half?
in a trapezoid? no
ok
ABCD is a parallelogram
AE = ED (I know it doesn't look like that but...)
<BED & <BEA = 90 degree
I need to prove that BD = CD
I know the image is so fucked up but please try to imagine it's perfect.
I want to prove that <DBC = <BCD
To get BD = CD
But how can I do that?
Anyone ?
E?
This one @final prairie
No. they need to be congruent to say that AB = BD
๐
Useful tip: if you draw a line as you keep pressing shift in MSPaint, it'll automatically adjust it as horizontal, vertical or 45ยฐ line.
No problem uwu
so its tan D
well
wait
yes
opposite adjacent right
i lost notes
:(
so 80/18
i think
Sorry to interrupt, but is (x-h)^2 + (y-k)^2=r^2 the equation for a circle?
unsure
@upper karma yes
Thank you!!
to find x, i do tan(32) = x/12?
Yes
Yep
so tan(32)*12 = x?
Yep
is the answer 7.5?
==tand(32)
0.62486935
yes
ty
this is what really confused me
you get tan(40) = 13/x
and all i know is you cant have x in the denominator
like he was saying switch the thing multiplied by tan (40) and x?
well you can during the working
you just need to get x out of the denominator
yep
you can do that
so im so lost on what to do :(((
tan(40) = 13/x
that's correct
now you can do the swap thing
x = 13/tan(40)
it's really just a "shortcut"
OH
you basically multiply through by x, then divide through by tan(40)
i move the tan with it
im so dumb
lol
is the answer 15.49
or did i do it wrong
looks right to me
im not the best at math lol
ill be here a lot
afte all this chat saved my first semester grade lol
isnt it base/height times two
yeah
you don't have enough information to use pythagoras
think about how you solved for x before
well if i solve for x
so i use tan(38) = 11/x?
yep
so now i have x = 13/tan(38)
yeah
now
it depends on how accurate you want to be
you could approximate here
or use the exact value
into the next step
(base*height)/2
aaaaaaaaaaaaaaaa
i forgot
wait so x = 16.639?
just making sure i did that right
yep
base * height is same as height * base
area = 91.5145?
yep
yep
yes
looks right
side+side+side?
yes
yeah
lol
free answer
probably best to do it by hand so its been awhile since i used pythag but its 6^2 + b^2 = 11.65^2?
check it first
could be a first class troll
you can
soh cah toa
if that was written in
tan(31)=6/x
so i should assume thats correct
find x
oh wait
lets see
nvm
wait thats not right?
hypotenuse can't be adjacent
you can find the other adjacent and then use pythogoras
yes
yes
so then pythag
though, if you round every step, it can get quite far from the actual answer
6^2
yes that's what I said
oh
misunderstood you
sorry
hey how the hell do you write a decimal root as a simplified root
so root 134.01
how the hell do you write that as a simplified radical
calculator not allowed?
my teacher likes simplified radicals
wait
im so dumb
i dont need to do that
i just need the perimiter lol
im so silly sometimes
wait no i still have to right
cuz c^2 is still there
== root 134.01
Failed to parse equation: Invalid syntax at position 1
root 134.01
^
bruh
== (134.01)^(1/2)
11.57626883
=pup square root of 134.01
Query made by @upper karma
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=square+root+of+134.01
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/
ill just do that lol
lol
my teacher didnt show me how to do any of this in class..
not even gonna bother with the next two lol
yes y is 60
yea
i dont even see where to start
tan(42) = z/60?
yes
i dont have a reference angle
or you can do
so do i multiply by 60
seems good
so now i can use 48 as a reference angle
yes
unless my calc fucked up
or i messed up doing tan(42)
yeah probably that
it did
hey
Wait im gonna go up and see the full problem
Is there an image you can post?
@upper karma is that what you got
its like
way up
im kinda not good at math lol
@restive void yes
be optimistic
indeed
Tan 45 = 1 yes
it just is
imagine you have a square, with side lengths one
for my brain
and then cut through the diagonal
yessir
that's like one way to visualise it i guess
are you sure math is the cause
to me
uh
probably more me not understanding
but its probably math
lol
well that doesnt make sense
anyway
yeah ill be back here maybe later
idk
do you need to get variables out of denominators for sine
what do you mean?
so i have sin(48) = 11.1/a
look
you have
sin(x)=8โ2/c
solve for c using pythogoras
then solve for sin(x)
i figured it out, for some reason my mind switched to having to use sin, cos, or tan with a number
happens
does it actually require you to solve for the angle or nah?
true
How?
what've you tried so far and where did you get stuck?
well i first descided to calculate the area of the sector
then minusing from the area of the triangle
so area of sector is 1/2*r^2 * theta
and area of triangle is 1/2*absin theta
i still end up getting it wrong tho
okay, can you show your work?
...
okay first off, that doesn't sound right at all
second, did you really use a calculator to compute sin(3ฯ/4)?
ye
are you sure it's not in degree mode
Can I get some help?
I need to find the values of x and y so that the quadrilateral is a parallelogram. I found x is 27 but I'm having trouble with the y. The equation is (1/3)y=y-60. What do I do first?
you need to group up the y
because you want it to be alone
so
(1/3)y - y = -60
(-2/3)y = -60
y = 90
Okay thanks ๐
How do I find the center of.. a 3d circle
(English is not my first language)
And the radius
a 3d circle is called a sphere
you want to "complete the squares"
here's the idea. if you have an equation like
x^2 + bx = c
you can express the left hand side as a square
like so
x^2 + bx + (b^2)/4 = c + (b^2)/4
(x+b/2)^2 = c + (b^2)/4
this will help you, in your problem, to put the sphere in its standard form
as (x-x0)^2 + (y-y0)^2 + (z-z0)^2 = r^2
okay, thanks! ๐
ok someone pls help i missed 10 days worth of school and i have no idea where to start
well, start with a
draw the graph (with the x, y coordinates) and locate the points
then draw the circles
okay
try to draw it to scale
wym draw the circles? like how do i know how big to make them or how many?
my teachers making me use this app, but i dont even know if im using it right bc i missed her class
help me understand translation and enlargment... plz!
multiply the coordinates of the figure by -1.5 and then translate 3 units right 6 units up
Assistance porfavor
so im starting from the bottom and working my way up
but these triangles are very confusing
Except I'm not at home :/
:(
if at some point in time u make a calculation error gg
And I'm.not gonna do cos(28ยฐ)
you have a fun calc smile
u also have to consider tons of significant figs
or if u can store the ans that would be useful
which is why im here because im not fucking good at this lo
Just don't do the sig figs
And do everything in a row
like keep going on ur calculator
from the bottom triangle, u can try finding the hyp
@runic jackal i got a negative value using cos
