geometry-and-trigonometry

[upper karma]

was I wrong, @silent plank ?

[silent plank]

extremely

[upper karma]

where was I wrong?

[silent plank]

pretty much everywhere

[upper karma]

I see where I was wrong 🙂
you can't just "add 1.0 to the unit circle", that doesn't give you a full rotation, pi is a full rotation (3.1415...)

[silent plank]

(the last few steps are mathematically valid, but has nothing to do with the original question)

[dark sparrow]

you have a bad case of everything-is-linear disease

[acoustic jungle]

you need to apply the angle addition formula if you want to do that.

that is just over complicating the problem but at least it'll be right.

also you can't add pi to that thing, what?

you have arcsin of something that is outside the range

[latent iron]

I don't get this

@acoustic jungle ^

[upper karma]

better now?

[silent plank]

no

you have a bad case of everything-is-linear disease

[latent iron]

O

Im

[silent plank]

are you implying that:
0 = sin(180°) = sin(90° + 90°) = sin(90°) + sin(90°) = 2
therefore 0 = 2?

[acoustic jungle]

honestly I can't believe you got the right answer by doing all the wrong steps.

[silent plank]

one of the solutions

[latent iron]

reposting

[silent plank]

the 2 errors cancelled each other out

[latent iron]

oh

Is anyone able to help

[acoustic jungle]

ryan is yxb = yxa = axb

I do not understand what those hexagon angles are

[serene rose]

probably just right angles

[acoustic jungle]

bruh..

[latent iron]

THEYRE RIGHT angles

[acoustic jungle]

ok my bad.

[novel flax]

it's 3d

look at it that way

[serene rose]

exactly

[latent iron]

i dontget it omg

[serene rose]

its not a triangle its a 3d figure xy is 1 dimension, xa and xb are the others

[silent plank]

consider what information you can get from angleYAX = 45°

(and that angleAXY is a right angle)

[upper karma]

well, that means that angle YXA is also 45 degrees
(correction: angle AYX)

which means that 2 x 45 degree angles + 90 deg angle = 180

it's an isosceles triangle

it means XA = XY

but how does that relate to 165?

is AB = YA = BY?

[acoustic jungle]

The way I would solve it is representing XB in terms of XY using tan 35,

[silent plank]

its not mentioned in the info
do you have any justification for that?

[acoustic jungle]

it's probably not the best way.

but that's how I would solve it.

[crisp bramble]

The width of a regtancular painting is eight inches more than the length. A frame that is4 inches wide goes around the paiting. Write an expression for the combined area of the painting and frame. Anyone know who to solve this. Thanks

[silent plank]

i used ||congruency to determine angleABX and/or YB and then used sin.||

[upper karma]

congruency with what?

[serene rose]

is abx 35 degrees too?

[silent plank]

the vagueness is deliberate to make you think

left something out of that spoiler

[acoustic jungle]

That is a better way than mine.

[silent plank]

would've needed more complicated stuff if that angle wasn't 45 degrees

[serene rose]

so is the answer 94.64 degrees?

[acoustic jungle]

ah yes the length would be measured in degrees.

[serene rose]

ah yes im smort

meters*

[upper karma]

4 inch frame means that it increases both the width and the height by 4 inches

[acoustic jungle]

4 inch frmae?

shouldn't it be +8.

You are increasing 4 on each side.

[upper karma]

Guys I’m really confused how do I do this

@acoustic jungle yeah, that makes sense

[acoustic jungle]

perpendicular means the 8 is bisected into 2 equal parts

[upper karma]

@crisp bramble add +8 to both width and height instead of 4

[acoustic jungle]

and the radius in uniform around the circle.

[upper karma]

Pythagoras's triple

3, 4, 5

because any chord on a circle is perpendicular to the radius

and that splits it in two

this is also a radius:

so if you find that (using Pythagora's theorem), you're set

ah I got the answer it's 5 thanks for the asnwer

[torn linden]

How would you find YZ?

[upper karma]

, rotate

[upper karma]

WZ = 7
XZ = 12
is that right?

do you know the formula for the area of a triangle?

[torn linden]

Yes

Also that is right

[torn linden]

Ohhh

Thank you

Lol

Well actually

I know that you use that angle

[bitter jetty]

An angle with its vertex at the center of a circle intercepts an 80° arc of that circle.

What is the measure of the angle?

its 80 right

or am i trippin

[upper karma]

Umm

I dont think so, at a quick glance

[torn linden]

So once you found xw you would use tan theta? @upper karma

[bitter jetty]

how come

[upper karma]

Wait

[bitter jetty]

it wud be

right

[upper karma]

I may be tripping too, wait a sec

@torn linden sohcahtoa

SOH - CAH - TOA

you need CAH TOA, yes

for angle theta

Yeah you are right @bitter jetty

My bad

[torn linden]

O

[bitter jetty]

how do i do this again

[upper karma]

Do you know the equation of a circle

If not, google it and come back

[bitter jetty]

yes

[upper karma]

Oh ok

[bitter jetty]

x^2 +y^2 = r^2

right

[upper karma]

Ew nope

Look it up

[bitter jetty]

o u mean (x-h)^2+ (y-k)^2 = r^2

[upper karma]

I have never seen h and k, but yeah

Its the same ig

[night summit]

h represents the x or axis of symmetry

k represents a vertical shift

https://gyazo.com/bca26ea79d43caa7e3f51f227b046f02

i need help finding out what it wants from me

[upper karma]

@torn linden tangent, sorry 🙂 you were right

to find the angle

[night summit]

what angle?

[upper karma]

In the first case, theta

For example in the third, A

[silent plank]

@bitter jetty completing the square

[night summit]

@upper karma so for example, in #2 i have to solve for theta?

[silent plank]

its not asking you to solve for theta

you need to show LHS = RHS

[upper karma]

Its what Dspider said

I was out of context

[night summit]

wtf is lhs=rhs

[silent plank]

use algebra to show that the
left side is equal to the right side

[night summit]

oooh

wait are we doing the same thing?

i use the triangle to solve?

https://gyazo.com/bca26ea79d43caa7e3f51f227b046f02

this?

[silent plank]

how are you using triangles?

[night summit]

wdym

[silent plank]

misread the question

[night summit]

i just want to know

[silent plank]

you apply appropriate trig identites

[night summit]

what the question is asking

so how do I solve it

[silent plank]

i'll guess i use the first question as an example

[bitter jetty]

how do i do this one?

[silent plank]

fk typo

[night summit]

lmfao

[upper karma]

@bitter jetty are you here

[somber coyote]

ramonov:

[upper karma]

Lmao this is way too funny

@torn linden

Why is there 3 questions being answered here at the same time?

[torn linden]

Thanks @upper karma

[night summit]

@silent plank so what do I have to reach at the end? what is the answer that I have to box?

[upper karma]

@bitter jetty hey?!

[bitter jetty]

@upper karma

ye

[upper karma]

Oh

Lets move to another free channel

[bitter jetty]

dm?

[silent plank]

the answer would be the entire thing that i typed up showing that the left side and right side have the same value

the question isn't asking for any numerical values of any sort

[night summit]

well dam i dont know how to prove it lol

i havent done proofs in such a long time

and I havent really been following that well since online learning started

do you think you can show me how to prove #2?

https://gyazo.com/bca26ea79d43caa7e3f51f227b046f02

[silent plank]

now that you have an idea of what you're supposed to actually do,
give it a go first

start from which ever side seems more complicated

and apply trig identities and/or definitions

it may help to convert to sin and cos first

[night summit]

i dont know how to put it in a triangle

im terrible at trig

[silent plank]

you don't really need triangles

[night summit]

i dont even know where to start

[silent plank]

i'm not going not repeat myself
read my recent posts

[night summit]

i read it, i just dont understand it

i dont know what you mean by start from which ever side

[silent plank]

start from either the left side and do manipulations to show that its equal to the right OR
start from either the right side and do manipulations to show that its equal to the left

[night summit]

ooooh

the right side seemsharder

[silent plank]

for q2, i would recommend starting from the left side

combining terms is generally easier than splitting them

[night summit]

i just dont know trig identities well enough to make them equal

i bet its straightforward and easy but I just dont know the relationships between them

[silent plank]

do you know the definitions of tan, cot, sec, cosec?

[night summit]

only tan sin and cos

oh wait

no i do

they are the inverse of them

so yes

i do

[silent plank]

and the pythagorean trig theorems?

[night summit]

a^2+b^2=c^2?

[silent plank]

the ones involving trig

functions

[night summit]

oh no

i dont

[silent plank]

sin^2(x) + cos^2(x) = 1
1 + cot^2(x) = cosec^2(x)
tan^2(x) + 1 = sec^2(x)

the difference of 2 squares is also quite common in these types of questions

and that's pretty much all you need to solve 95% of these

[upper karma]

tan (x) = sin (x)/cos (x) too right?

[night summit]

i think so

[upper karma]

In case he forgot that one

[silent plank]

i'll give a general outline for Q2.
start from LHS
express it in terms of sin and cos
combine fractions (common denominator)
apply an appropriate trig identity
apply definition of the reciprocal functions

[upper karma]

tan (x) = sin (x)/cos (x) too right?
@silent plank this one too right?

Or am i tripping

[silent plank]

i mentioned that already and it was confirmed that it was known

[night summit]

wait but there is no sin and cos in q2

oooh

tan is sin/cos

and cot is cos/sin

?

do i have to plug those in instead of tan and cot?

nvm i got it

[dark sparrow]

you never """have to""" do anything

[night summit]

ty for the info

[gritty sail]

<@&286206848099549185>

[upper karma]

yo i tried doing this problem, i think my answer is right but im not sure

i got 20.8

52/130 = 0.4

[gritty sail]

uh

[upper karma]

0.4*52 = 20.8

[gritty sail]

it is 39

[upper karma]

why is it 39?

[gritty sail]

for ur question

(130-52)/2

formula

u can find the formula online

[upper karma]

what is it called

[gritty sail]

can someone answer my q

[twilit zenith]

JDP, the outside angle theorem for a circle states that if you have an angle outside the circle that creates two arcs inside the circle, then its value is equal to half the difference of the angular size of the arcs.

For example, in the third drawing (the right one), if the angular size of the arcs is 100° and 60°, then the angle is equal to:
(100° - 60°) / 2 = 20°.

Jbao, I'm not too sure about it, but it seems to me that you may be able to use m, n in order to express sin (a) through sin (b), cos (b), and cos (a) with the tan equation, but I'm not too sure, sorry.

[fervent thistle]

can someone point me in the right direction on how to do this "wxyz is a kite. Angle WXY has a measure of 133 degrees and angle ZWX has a measure of 60 degrees. Find the measure of angle zyw

ZYW*''

[gritty sail]

search

properties of kite

[twilit zenith]

This drawing may help you visualize it.
(The drawing is not to scale.)

[fervent thistle]

thanks

[fervent thistle]

what does "classify triangle CAC' according to its side length" mean?

[twilit zenith]

Probably tell whether CAC' is an equilateral triangle, an isosceles triangle, or a scalene triangle.

[fervent thistle]

how wuold you solve "Triangle DEF is formed by connecting the midpoints of the sides of triangle ABC. the length of the sides of DEF are shown. What is the length of AB?"

[acoustic jungle]

similar triangles.

notice how the small triangle formed by the vertex of the large triangle and the two midpoints is similar to the large triangle.

try using SAS porportion.

[fervent thistle]

They didn't provide an image so it is confusing

[acoustic jungle]

then draw it.

I had to draw it.

[fervent thistle]

ill try

how do i find the length when the numbers are not shown it says length of the sides of def are shown but they aren't

oh nmv

[acoustic jungle]

nemindver.

[acoustic jungle]

the circle has radius of 2.

you would have to split it into a cylinder and a cone.

and use those formulas.

[silver radish]

Oof

Anybody know how to do this problem

[acoustic jungle]

I just told you how to do it.

[silver radish]

Oh

Yeah I dont understand

Use formula for circle and cylinder?

[upper karma]

Yes

[silver radish]

oof

[upper karma]

GIVENS:

AC is a diameter.

BZ is tangent to circle O

Measure of arc BC = 90 degrees

Measure of arc CD = 30 degrees

Measure of arc DE = 20 degrees

On a separate sheet of paper, set up an answer column as follows:

1. m<1

2. m<2

up to

10. m<10

I am completely lost on this entire problem.

[signal harness]

Honestly, I have no idea how to solve these problems

[acoustic jungle]

inscribed angle theorem.

[acoustic jungle]

is this a test.

[signal harness]

yes

i've no idea how to do them

[acoustic jungle]

Ok just leave them blank then.

[signal harness]

i'm not looking for you guys to give me answers

just how to get them

and i'd rather not leave them blank since it depends on my grade

[acoustic jungle]

BONUS

[upper karma]

@silver radish you need to find the volume of the cylinder, and volume for the cone
the cylinder has a radius of 2 units and a height of 2 units
while the cone also has a radius of 2 units but a height of 3 units
then you add the volume of the cylinder with the volume of the cone

[real lily]

Is this remotely close? Its for my online school quiz

[wheat lichen]

they dont help w quizzes here

how do i find the tan/cot/cos/sec with just sin and cos?

[real lily]

Oh really? didnt realize

I just dont understand this question at all really lol, got everything else down

[acoustic jungle]

You wrote the answer.

[signal harness]

you mean TOA?

[wheat lichen]

oh crap im dumb nvm lmao

[signal harness]

for the second one?

[pallid cloud]

I have two lines $(d)$ and $(d')$ such that the angle between them satisfies $((d);(d')) \neq \frac{\pi}{4} \mod{\pi} \ C$ is a point oustide $(d)$ and oustide $(d') \$ I am asked to construct a triangle $ABC$ so that $A \in (d)$ , $B \in (d')$ , $CA= \frac{CB}{2}$ and $( \overrightarrow{CA} , \overrightarrow{CB} )= \frac{\pi}{4}$ , how should I do it?

[somber coyote]

Patrick Salhany:

[real lily]

wait so what i have is right or wrong?

Is this correct?

more specifically cos S and cos R

[pallid cloud]

The sines are correct

The cosines aren't

[upper karma]

another quiz? 🤐

[real lily]

nope, new assignment for today

[rapid temple]

SOH CAH TOA

[pallid cloud]

<@&286206848099549185> my problem is above, so please whenever someone wants to answer me, feel free to ping me

[real lily]

Well yes, except for cos S when i do cos(16/26) it gives me 0.81655147937, which is none of the selectable options (in fractions)

[rapid temple]

why would you do cos(16/26)?

[real lily]

sorry if this is super easy to see for some reason

[upper karma]

@real lily you need to reduce 16/26 to their lowest form

[real lily]

oh god

[rapid temple]

do you know what SOH CAH TOA means?

[real lily]

ok that makes sense

yeah, i was just doing it horribly wrong apparently

[upper karma]

16/26 is the same thing as 8/13

it's just divided by two

but the result is the same

[real lily]

yeah i get that now lol, dont know what i was doing before

[rapid temple]

what do you have for cos S?

[real lily]

should be this?

[rapid temple]

looks good

[real lily]

Alright, thank you guys

[upper karma]

@upper karma open youtube, search "circle theorems"

right off the bat, 1 and 2 are 90 degrees

so that's angle #3

and you know all angles from a triangle need to add up to 180 degrees

so that means angle #4 is... ?

@upper karma aaaaand you're offline

nice

[upper karma]

Angle 4 would be 65 degrees then?

[acoustic jungle]

Yes.

[upper karma]

ah okay

thanks guys !

geometry has been a pain since learning through voice calls dont really teach me anything, but the yt vid i watched helped a lot so ty @upper karma

sure thing

[topaz panther]

i need help

im super bad in math

[upper karma]

@topaz panther post your question

[topaz panther]

Jan has a rainwater collection catchment system. The main storage cistern is 250 meters from his garden. The pipe
used to transport the water has a slope of 0.05 degrees. How high off the ground is the output value on his cistern?

[upper karma]

well, when you hear slope, that means "rise over run"

the rise (or vertical) is your "x" that you wanna find

slope is best defined with tan

since that's also "rise over run"

tan(0.05) = x / 250

[topaz panther]

do i use pythagoras theorem?

[upper karma]

SOH - CAH- TOA

250 is the adjacent

x is the opposite

so it's TOA

that means tangent (like I said before)

tan(angle) = opposite / adjacent

tan(0.05) = x / 250

x / 250 = tan(0.05) (same as above, just switched places)

x = tan(0.05) * 250
x = 8.726 * 250

do i use pythagoras theorem?
@topaz panther no, you dont have 2 side lengths, listen to what DSpider is telling you

,w x=8.726*250

that can't be right...

,w x=8.726*250

So it is this one

[somber coyote]

Display results online and refine query

Requested by Al3dium#4084

[upper karma]

it can't be that high up...

Ikr is weird right?

@upper karma its impossible if its a 0.05 degree wtf

"0.0500417..." is in radians

the calculator needs to be in degrees

Weird problem

[acoustic jungle]

e-4

bro.

[upper karma]

@upper karma now i put in the calculator and it says 8.726*10^-4

[topaz panther]

im lost already

i gotta do 25 pages of it

[upper karma]

Bro you missed the exponent

@upper karma

what does that mean? 😳

It isnt the same 8.726 than 0.0008726

@topaz panther why are you lost

He only used a simple trig definition

[topaz panther]

not that simple tbh

[upper karma]

Do you know trig definitions?

so the "e-4" at the end means you add 4 zeros in front?

8.727 bcomes 0.0008727?

The e-4 means 8.726 multiplied by 10 and THAT 10 has an exponent of -4

$8.726e-4=8.726*10^-⁴$

Wtf

[somber coyote]

Al3dium:
Compile Error! Click the reaction for details. (You may edit your message)

[upper karma]

Now lol

[topaz panther]

huuuuh??

whaaaat

can someone do it step by step so i can understand it lol

[upper karma]

Dont overthink

But

Do you know trig definitions?
@upper karma .

[topaz panther]

wait hol up, if its a right angled triangle it means all the degrees inside of it add up to 180 degrees, so 90 plus 0.05 is 90.05. 180-90.05=89.95

so we know all the angles \

or idk

[upper karma]

Idk what you did in

180-90.05=89.95

Oh wait mb

yeah, then you can use something else other than tan

Yeah thats correct

[topaz panther]

but how do i find slope

i need to find hypotenus and opposite

[upper karma]

Can you answer to my 3rd attemp of asking you if you know the trig definitions?

[topaz panther]

no

[upper karma]

Look up for trig definitions on google

And come back.

We'll explain what you dont understand

[topaz panther]

oh you mean the COS TOA ...

[upper karma]

Yeah

[silent plank]

(the hypotenuse isn't required)

[topaz panther]

yea ik that

[upper karma]

Oh then

If you know that, honestly i have no idea of what you dont understand by

tan(0.05) = x / 250

Yeah i saw slope and i confused my variables lol

[topaz panther]

i just dont know what to do with those numbers. like ik i gotta divide the x by 250 and put the 0.05 in my calculator but thats it

[upper karma]

@silent plank can you take over me? I gtg sleep

[silent plank]

first, solve for x like you would for any other question

[topaz panther]

idk how

[silent plank]

can you solve:
x/2 = 2?

[topaz panther]

lol no

[silent plank]

fml

[topaz panther]

yea fml

[silent plank]

this is very basic algebra

do the same thing to both sides to work towards isolating x,
in these cases, it only requires 1 step.

[lament warren]

He doesn't need you to be condescending to him, Ramonov.

[silent plank]

I'm not trying to be condescending.

[topaz panther]

ight whatever peace out

[upper karma]

x = tan(0.05) * 250
x = 0.0008.727 * 250
x = 0.218175

...right?

[silent plank]

no

[upper karma]

0.05 degrees is a very tiny incline

[surreal snow]

May someone explain how does one know what “opposite” and “adjacent” is? In theory any side could be opposite or adjacent of any angle and it gets me really confused..

(right triangles)

[upper karma]

no, only one side can be opposite of the angle

[surreal snow]

which angle are they usually referring to

[upper karma]

and the adjacent is the side that's not the opposite and that's not the hypotenuse either

[silent plank]

1. degrees isn't mentioned in the angle (otherwise it is implied its in radians, especially for low numbers)
2. multiple decimal points in that number
3. you multiplied the rounded number instead of the unrounded value causing your final approximation to be inaccurate
4. use of equal sign instead of approximation and/or stating that its an approximation

[surreal snow]

Ah, so when someone does sin(x), it means adjacent of that angle x?

[silent plank]

no

[surreal snow]

huh

[upper karma]

@surreal snow there's a mnemonic called "SOH CAH TOA"

[surreal snow]

Yeah I heard of that

I know what sin and cos do and all

I just have trouble understanding which side they are referring to

[upper karma]

https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles/intro-to-the-trig-ratios/a/opposite-adjacent-hypotenuse?modal=1

[surreal snow]

I’ve actually watched that video before

for our class

[upper karma]

and did you complete the practice questions?

click the above link

[surreal snow]

Yeah I have

We did the whole lesson for it

[surreal snow]

Ah so the adjacent opposite thing is relative to the angle

[silent plank]

yes

[surreal snow]

alright thanks

[upper karma]

how about now?

[silent plank]

missing degrees symbols and you're off by a factor of 10

[upper karma]

😐

it's 3:25 am

going to sleep now

[silent plank]

(and your x look like *)

[upper karma]

can't think straight

yes, it's a shorthand for multiplication

you've never seen this symbol?

[upper karma]

that's an "x" my dude...

[silent plank]

$x *$

[somber coyote]

ramonov:

[silent plank]

looks a lot more like an asterisk which is my concern

x x

[upper karma]

https://i.pinimg.com/originals/5f/13/ab/5f13ab0b3cdb4a17c91162a164227b25.jpg

[high zephyr]

would look like x if u exclude the dash in the middle

[acoustic jungle]

what is that alphabet.

that r looks like an n.

[lament warren]

What is that, Portuguese?

[tight pebble]

Can I have some help with my maths homework plsor is that not cool?

This is it btw

The last one is a real pain for me idk why

<@&286206848099549185>

[umbral snow]

Check #❓how-to-get-help for the rules on the helper ping

[tight pebble]

ok thx

got ya

[rich wolf]

you must wait exactly 15 min

[umbral snow]

If v is perpendicular to both x and y, then anything perpendicular to v is of the form ax + by for some a and b

Or, that (1,1,p-1) would need to lie in (-1,1,1) and (1,p,p²)'s plane

[tight pebble]

So what i've done is I've crossed the two vectors given to get an expression for the perpendicular vector

[modest spear]

Need some help

I'm kinda stupid when it comes to math.

Especially trig...

[twilit zenith]

Try to look at the triangle the following way. Does it help?
And yes, C is right.

Nah, it's okay. 😁

Sorry, xi. 😢

[modest spear]

Yes that helps

SOH CAH TOA right?

[upper karma]

Yes

[modest spear]

Ah okay

So for the enxt question, I would just do the same thing?

So it'd be

answer B?

[upper karma]

@modest spear nah I don’t think so

Oh wait

Ya it’s D

Sec = Hypotenuse/Adjacent

I don’t think you knew that

But if you get c by itself

It would be D

@modest spear

@quiet mason yes what

Nvm

[modest spear]

Ah okay thank you brother

[rich wolf]

@upper karma bro what is that username

[upper karma]

Idk

It’s cool tho

Very long

This name always get me in trouble tho, they usually tell me to change the name/nickname or just kick me out

[rich wolf]

Gee wonder why

[twilit zenith]

AMD, could you please pin this? I think it could be useful.

[upper karma]

That’s pretty cool

[burnt shale]

Question.

Graph

Answer.

How the heck do you tell the difference between the two, they look so similar!

[twilit zenith]

Interesting question. You'd have to look at their domains.
Notice that the function tan is defined whenever input ≠ 0.5π + πk (the zeros of cosine), which in your case would be:
(0.5x) ≠ 0.5π + πk x ≠ πk + 2πk,
which would mean the vertical asymptotes differ by 2π each, which isn't the case.

The function cot however, is defined whenever input ≠ πk (the zeros of sine), which is the same as where the function shown is undefined.

[upper karma]

so i saw a graph like this the other day on the internet

and had something to do with trig functions

anyone know what it is?

[dark sparrow]

this looks like the graph of the fractional part function

[eager pendant]

a sawtooth wave?

[dark sparrow]

if you wanna talk waves it's a sawtooth

[upper karma]

i am just curious which function it represents

in equation form

because we were given an electrical wave such as this

but i had no idea it was a trig function grapjh

had i known, i couldve answered the question

can anyone provide me the general equation of graphs such as this?

and is it actually a trig function?

[dark sparrow]

it's not a trig function per se

but i guess you could write down its fourier series

so long as you give me some actual scale on those axes

[upper karma]

ah i see

well no, i dont think we are expected to divulge into Fourier transformation

i was wondering if it is a function which is summation or multiplication or whatever...maybe a series made out of trig functions

which WONT need fourier transformation

thanks

[gray marten]

looks like floor

y = x - floor(x) or y = x mod 1

[somber coyote]

Lang:

[gray marten]

$$y = \frac{\pi-2\arctan\left(\cot\left(x\pi\right)\right)}{2\pi}$$

[somber coyote]

Lang:

[gray marten]

all of these will give you that sawtooth wave

sawtooth waves can be for electrical, musical, etc a lot of stuff.

[fleet wolf]

well its no even a function per se since it maps one x onto mulitple y values

[dark sparrow]

wut

arctan is not multivalued, rps

@fleet wolf

[fleet wolf]

the sawtooth image

when it goes down there is a line implying that the function has a value everywhere on that vertical line

[dark sparrow]

maybe that's just a graphing artifact

[olive scarab]

Give elipse (E): $ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 $ and the foci F(c,0) $\in $ straight line d. d intersects (E) at M,N.
Prove that $\frac{1}{FM} +\frac{1}{FN} $ is constant

any idea to do this :(

[somber coyote]

.,.:

[upper karma]

what is the length of red line

[pallid cloud]

@olive scarab are you allowed to use polar coordinates?

If yes, then I have a proof, I can help you

Ping me when you answer

[upper karma]

@upper karma wdym by slope of red line is 8%

[versed river]

when slope is a percent it refers to rise over run, essentially

like for every 8 metres vertically, theres 100 metres horizontally

[upper karma]

yes

[versed river]

so try 8%*x=200 to get the bottom side

[upper karma]

Ik what a slope is, just never heard of it by a percentage

Ok

m2 xd

so I wanted to ask ^^

and then pytahgoras

I can try it

[versed river]

ja

[olive scarab]

@pallid cloud yes, but I rarely use it though, can you so me the way ?

[upper karma]

so the length is 2508?

[versed river]

approximately, yes

[upper karma]

can I say thanks or something to give you a +1 or something like that for your help @versed river and @upper karma

[versed river]

its all good mate. Have a good one

[upper karma]

ty guys for your help ^^

:)

[pallid cloud]

@olive scarab give me some time to write it, cause it might take some time
When I post it I will ping you

[olive scarab]

thank you!

[pallid cloud]

.,. The proof took me five pages I'll post it now

[rich wolf]

,rotate

[pallid cloud]

@olive scarab

[rich wolf]

What do you need help with patrick

[pallid cloud]

No lol this is a problem I solved for @olive scarab

Thanks

[rich wolf]

Mind posting the original problem? Cant really read your handwriting

[pallid cloud]

Posted by @.,.
Suppose we have an ellipse $(E)$ of equation $\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1. \ $
A line $(d)$ passing through the focus $F(c;0)$ intersects $(E)$ at $M$ and $N. \ $
Show that $\frac{1}{FM}+ \frac{1}{FN}$ is a constant.

[somber coyote]

Patrick Salhany:

[pallid cloud]

Couldn't figure it out in cartesian coordinates
I used polar

Btw @olive scarab I used a>b
The case a<b is obtained by analogy

[olive scarab]

oh yes, I used a>b also

[rich wolf]

Same

[pallid cloud]

I used polar
There must be a way using cartesians
But I can't figure it out rn

[rich wolf]

Doesnt really matter in the context of this proof anyway which one you choose

[olive scarab]

I'll read your solution rn, thank you very much

[pallid cloud]

No problem
If anything unclear ping me

Nice problem in fact

[rich wolf]

Very fun

[pallid cloud]

This shows that the harmonic mean of FM and FN is the parameter of the conic
Here ellipse

Parameter: k=b^2/a

[olive scarab]

oh, I understood it very ckearly

thanksssss

[pallid cloud]

👍 👍 👍

[upper karma]

@pallid cloud that's some good dedication

[pallid cloud]

🙂

[novel flax]

i found CD and the length of the tangent already, but i don't know how to find BC

[novel flax]

<@&286206848099549185>

[pallid cloud]

@novel flax

Have you used similar triangles to find your results?

[novel flax]

i let the intersection of AD and BC be E, and so ABE is similar to CDE but i don't know where to go from there

[acoustic jungle]

Have you tried drawing the diagram?

could you post that here.

[novel flax]

how do i do that

[acoustic jungle]

could you post the diagram.

[novel flax]

ok

[acoustic jungle]

Nevermind I will draw it.

[humble quail]

ADX is similar to CBX so it's not hard to find everything in these two triangles

(As soon as we found CD ofc)

[acoustic jungle]

oh yeah you are right.

that's also how the secant proof was done.

[humble quail]

Yep

Tangent is found even more easier

[pallid cloud]

You don't need that E intersection

[novel flax]

so 9/CB = 10/7

[acoustic jungle]

,rotate

[humble quail]

Lmao everybody draw it

[pallid cloud]

You don't need E ...

[acoustic jungle]

oh that's what the tangent means, I thought it was trying to find the length of x from the circle.

well I drew it anyway.

[pallid cloud]

The 2 secants determine 2 similar triangles

And the 1 secant and the tangent too

[acoustic jungle]

How would you find the distance from x to the circle?

[pallid cloud]

You don't need the distance from X to the circlw

[humble quail]

secant theorem too

[pallid cloud]

You want tge length of the tangent

[acoustic jungle]

no I was asking a different question.

[pallid cloud]

Ah ok

[novel flax]

oh

you can draw a right triangle and then do hypotenuse - radius

because the hypotenuse is from center of circle to X, the other vertex is the point of tangency

@acoustic jungle

if you wanted to find the distance from x to circle

[pallid cloud]

Yes and use Pythagoras to find the hypotenuse

[humble quail]

what is distance from x to circle? You mean to the center?

Cuz if not there are infinity of distances from x to the circle(if it's not tangent)

[pallid cloud]

He meant the smallest length XM where M describes the circle I guess

[acoustic jungle]

right.

hmm.

I'm not sure how to find the radius of the large circle. @novel flax

[humble quail]

The shortest distance from x to the circle will lie on XO, where O is the center

It's quite easy to prove anyway

[acoustic jungle]

I meant the edge of circle O.

on the circumference.

oh I got it.

the radius of large.

I just had to solve with two equations

n is the radius of the large circrle, k is the radius of small circle on X, and t is the tangent, which is known.

(2n+k)k=t^2 and t^2=(n+k)^2-n^2

wait don't tell me these are the same thing.

whoops.

[humble quail]

What are you trying to find...

[acoustic jungle]

I'm dumb.

the radius of the large circle.

[humble quail]

What u call large circle?

I lost it

[acoustic jungle]

the bigger circle.

I make another cricle on X to find the radius of X to the circumference to the large circle.

I'm not sure how I would find the radius of circle X.

orange said find the hypotenous - radius of large.

but I don't know how to find the radius of large.

[novel flax]

oh

what

[acoustic jungle]

Could you elaborate on how you could do it?

[humble quail]

Something like this?

[novel flax]

to find radius of large circle?

or distance from X to large circle

[acoustic jungle]

yep @humble quail

[humble quail]

If we draw another circle with center in X it will tangent big circle

[acoustic jungle]

well I need the first to find the second.

[novel flax]

there's a formula

[humble quail]

And the shortest path from x to big circle will be XM

Hmm

[novel flax]

circumradius = abc/(4A) where A is the area of the triangle

and a b c are the side lengths of the triangle

[humble quail]

Ye it's easy to prove with sine theorem and knowing that A=absinγ/2 for ex

But will it help

[novel flax]

alternatively, the diameter is a/sinA

[acoustic jungle]

What triangle are you referring to?

triangle ADC?

[novel flax]

ABC

[humble quail]

Well I probably figured out how to calculate XM but for this I'll have to use cosine theorem 2 times and then use secant secant theorem...

Painful calculations

[acoustic jungle]

delta phi, how would you solve it with oranges way?

I have no idea what he is saying

[pallid cloud]

https://www.google.com/url?sa=t&source=web&rct=j&url=http://2000clicks.com/MathHelp/GeometryTriangleCyclicQuadrilateralCircumscribedCircle.aspx&ved=2ahUKEwjV8avUkITpAhUQ1xoKHcfcCd8QFjACegQIDhAI&usg=AOvVaw23CQi89K--CSq7ciIy_A1U

That's what I thought about

Héron's theorem for quadrilaterals, cyclic

Aka Barmaghupta's theorem

[acoustic jungle]

Thank you.

[pallid cloud]

And other relations obviously

[humble quail]

Well it's just too painful to calculate this

I'll try again

[acoustic jungle]

that link was very helpful, but I think I'm gonna stick to the cosine law.

what value did you get for XM, delta phi?

[humble quail]

Nothing. I think I can end my calculations, but this way I'll need calculator and I don't rly like when geometry becomes algebra :<

Wanna find beautiful solution if there's one

[acoustic jungle]

Man, I used cosine law of ADX to find angle DAX then I did 1/2 * ab sinC to find area of ADB, then I applied the formula thing provided by orange to find the radius, then I used pythagorean theorem to find XM.

[humble quail]

But to use that formula you have to know BD, so you must've also find it

[acoustic jungle]

wait I had to use cosine law twice.

[humble quail]

Or not

[acoustic jungle]

yeah I use it to find angle DAX then i apply it again to find BD

this is such a brute force way.

it sucks.

[humble quail]

Yea

Don't like it in geometry

[acoustic jungle]

this is what small brainers do.

[humble quail]

It's like to shot birds with a huuuge gun xD

[upper karma]

Its what it saves you valuable time from an exam

[novel flax]

oh

[acoustic jungle]

orange do you have a better way.

[bleak flint]

Hey, I was wondering if someone would be able to go over some practice questions with me? I'm trying to get back into school and turns out trig is hard when you haven't done it in 8 years 😛

[upper karma]

Me

[novel flax]

@bleak flint sure

[bleak flint]

i'm clueless once a graph comes up on screen

[upper karma]

Same I’m also clueless

Goodbye

[novel flax]

use best fit of sinusoidal equation

[bleak flint]

googling what that is now

[umbral snow]

Which question are you looking at?

And can we get a zoom in on that one lol

[bleak flint]

I'm really confused by all 3, I read through the manual and it didn't make sense so i started watching youtube videos on it and now i'm even more confused.

but the first question is asking me where the rate of change zero is

[novel flax]

oh ok

[bleak flint]

i wanna say the first questions answer is A

[novel flax]

rate of change means slope

[bleak flint]

C**

[novel flax]

yes first one is C

[bleak flint]

for b i'm inclined to say D since the slope changes

[novel flax]

yes

[bleak flint]

but it could also be B and D since they both change

[novel flax]

for B the slope stays constant so it's not changing

[bleak flint]

okay

so then for C) it should be A and B

[twilit zenith]

Orange is right. Straight lines have a constant slope, and thus the rate of change is constant. (but not necessarily zero)
It's a bit hard to see it from this image, so I'll give you that.

[novel flax]

correct

[bleak flint]

i'll try and zoom in more for the next one if you guys have time to help a brotha out

[twilit zenith]

Oh, we can see it, it's just that it takes a second until you actually notice it.

[bleak flint]

I'm assuming for D) it's asking me which slope has the greatest incline

which i think would be B

[novel flax]

yes

[bleak flint]

the rate of change is the greatest in B I guess would be the way to word it correctly

[twilit zenith]

Indeed. 😁
In B, the function "increases the fastest," and that's why the rate of change is the greatest.

[bleak flint]

I know the answer to B) is that it's not linear since the rate of change isn't constant

the general trends on the other hand have me scratching my head

[umbral snow]

It's approximately constant

Close enough that I might just say it is

[twilit zenith]

Generally speaking, in statistics, if the data looks "close enough" to being linear, then we use linear estimates for it, even though the relationship isn't truly linear.

[bleak flint]

okay

[twilit zenith]

I can definitely agree with you that it isn't a perfectly linear graph, though.

[umbral snow]

But you're right, maybe they want you to say it's slowing down.

[twilit zenith]

Fish, your bread reactions are the tastiest.

[bleak flint]

I'm just confused as to how I would word that

I'm thinking something like:

[acoustic jungle]

Log x

[bleak flint]

The relationship does not appear to be linear due to the fact that the data shows the rate of change varies

english the language i learnt math in so this is super tough

I mean it probably wouldn't be easier in french but i like to give myself an excuse

[twilit zenith]

I think you're correct.

However, I think that what they're asking in "does the relationship appear to be linear?" is whether, if you discount the fact that there could have been mistakes here and there, that the overall rate of change is linear?

[bleak flint]

okay so I think i was wrong in thinking it wasn't linear since overall the rate of change is the same

at least in a broad sense

So i'm gonna go with "The relationship apears to be linear since the majority of the discuss throws have a posittive rate of change

[twilit zenith]

Sorry for responding late.
I guess I'd say that the "general behavior" is "close" to being linear.
I know it's not really an exact explanation, but I don't think they expect you to give them advanced definitions like R².
(R² is a statistical measure that says how accurately an approximation is to the actual data. For example, if you consider the line approximation in the graph I sent above, then the R² measure would measure how "closely" the line matches the data.)

[bleak flint]

no worries i'm just happy I'm getting some help homie

I'm completely clueless on the last 2 questions though

[twilit zenith]

Seems to me that question C wants you to explain how come the Olympic distance results increased over the years [not really a Math question, I suppose], and in D, you're supposed to use the trend line (the linear approximation of the graph) to guess what the result at 2020 will be.

[full ruin]

i already missed it but can someone explain what the answer is?

i got .4 rounded to the nearest tenth

[acoustic jungle]

isn't it given in the question.

[full ruin]

wdym

i’m pretty sure that’s algebra not geometry 🙄🖐

[bleak flint]

i'm pretty sure i'm in the wrong channel

[upper karma]

@full ruin it is given in the chart, as Fishraider said.

Although this isnt geometry neither trigonometry

[full ruin]

oh well i’m in geometry so this is where i went

and i got it wrong. saying it’s in the chart isn’t gonna help me

@upper karma

[upper karma]

This isnt still probability channel

#probability-statistics @full ruin

[surreal roost]

I guessed 57.9, but thats not right

[upper karma]

Can someone help me find x

And explain

[twilit zenith]

Distance, this drawing (not of a 7-sided polygon) may give you a bit of an intuition as to how to go around calculating the area.

Also, if I may ask for clarification, Calcfer: Is the point inside the circle the center of it, and also - what's written on the exterior angle?

[upper karma]

Exterior angle is 40 degrees

The point is the center of the circle

@twilit zenith

[twilit zenith]

If I may give a hint: Consider the following construction:
EDIT: Apologies. No need for the DE line. Could be solved another way.

Do you want the full answer?

In case you do: [BTW, I call O the center. Sorry I forgot that in the drawing.]
Hint 1: ||Consider the triangle ABC. In the said triangle, the angles sum up to 180°, causing ABC + BCA = 140°.||

Hint 2: ||BO and CO are angle bisectors of DBC and BCE, respectively.||

Answer: ||Knowing the two above, we get that DBC + BCE = 220°, and so, due to the property of angle bisection, we have OBC + BCO = 110°, and considering the triangle BCO, we have the sum of the angles being 180°, causing COB = 70°.||

[upper karma]

Ok

Thx

Wait

The angle is 180?

For x?

[twilit zenith]

My mistake, I'm sorry. I meant to write 70°. 🤦‍♂️

[sacred garnet]

Yo i got some simple questions i need the answers too

[gray marten]

Is it just me or is the majority of questions in this channel "inscribed circle/triangle thingo"

[sacred garnet]

I got some circle stuff

Sorry about the quality

Im trash at math

[gray marten]

So whats the area of the circle first

Do you remember the formula?

[sacred garnet]

Is it 452.39?

[gray marten]

No I don't think so.

Do you remember the formula?

[sacred garnet]

No lol

[acoustic jungle]

You used the diameter for the radius as a mistake.

[sacred garnet]

Damn this quarantine really makes math hard

Ohhh

[gray marten]

So its πr²

r means radius, which is half of the diameter

[sacred garnet]

Ye

[gray marten]

so r = 6

So π6²

And so you now know the full area of the circle

[sacred garnet]

113.1

[gray marten]

Yep

Then there are 6 pieces

So you just divide it by 6

[sacred garnet]

18.85

[gray marten]

Ye

[sacred garnet]

Then just 2x

[gray marten]

Ye

[sacred garnet]

Aight thanks

[sacred garnet]

Uh i need help on 2 more questions

I always make the most simple mistakes ;-;

[silent plank]

what mistakes are you making?
how would you approach this?

[white dew]

https://media.discordapp.net/attachments/701211089488183376/703780326207389816/image0.jpg?width=702&height=936

am I doing this right? it feels wrong. I solved a problem like this yesterday, but it was Sin not Cos

I was told by someone that I should not use decimals for trig, so do I just stick with square root 3?

[solid swan]

you gotta solve for x

[twilit zenith]

Yes, your general process is correct.
I would actually advise remaining with the √3, as using only a few decimal places instead of √3 could result in a loss of accuracy, which is more apparent when you're supposed to get a specific angle as a result.

For example, let's say that you got the equation sin (x) = 0.5√2. In this case, the answer should be x = 45°, 135° + 2πk, but rounding 0.5√2 to 0.707 could make you lose the exact answer.

[solid swan]

i think 30° and (360°-30°) works

because of how cosine behaves

because of unit circle you know 30° has to be a solution

then you take those 4 angles

α, 180°-α, 180°+α, 360°-α

and check if they work under the CAST thing

[twilit zenith]

Conan is correct about the answer.
If it helps you visualize it, try using this interactive unit circle:
https://www.mathsisfun.com/algebra/trig-interactive-unit-circle.html

[solid swan]

since cosine is only positive in I and IV you know only α and 360°-α work

[white dew]

yea I will need that visualization!
also I apologize if I am not talking as much as I should, its hard for me to sync this kind of stuff in so I am trying to understand what you guys are saying to the best of my capability.

[twilit zenith]

That's completely okay. Please don't feel pressured. I know what it's like learning something for the first time.

[white dew]

Since I keep √3 and not make it decimal
I get left with CosX = √3 over 2

What Conan is saying is different from what I am seeing in my phone.
(I am using Mathway to assist me)

I thinks its primarily because it is using the pi symbol instead of α

[twilit zenith]

I apologize for responding late.
Also, Conan was talking about how you can start with one solution to the equation cos (x) = 0.5√3 and extrapolate the other solutions from it.

He concluded that if α is a solution to cos (x) = 0.5√3, then (360° - α) must be a solution as well, which it is.

And indeed, I suppose your phone found the solutions 30°, 330° in degrees, or π/6, 11π/6 in radians.

[olive scarab]

if you dont mind, I am stuck at this problem :
Give elipse (E): $ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 $, M,N $\in$ (E) that M,O,N is a straight line, foci F(c,0) and P,Q $\in$ (E) that P,F,Q is a straight line and PQ parallel with MN
Prove $MN^{2} = 2a.PQ$

[somber coyote]

.,.:

[olive scarab]

the picture should look like this

[twilit zenith]

I haven't fully tried it myself, but I guess there's the naïve solution of expressing the lines, and thus the intersections, through some parameter like the slope or the co-ordinate and eventually reaching an equality on either side.

[spare heart]

@olive scarab given that both lines are parralel with the x axis, and that f = 0 (meaning lines are on top of each other at origin). Wouldn't that give 44^2 = 44? Thus disproving it?

parralel to the y axis*

64 =/= 16

[olive scarab]

no, if it parallel to y-axis so MN = 8
PQ =6.4 and 2a= longer diameter=10
=>8^2 =6,4.10 = 64 is true

[spare heart]

oh my bad, I thought 2a was a typo/question no.

how is PQ = 6.4? if we make them exactly where M and N lie?

[olive scarab]

the drawing so...I change specific number in for seeing if it was true or not

[solid swan]

ok through using directrix and eccentricity i managed to get the following

[somber coyote]

Conan:

Conan:

[solid swan]

MN = $\frac{2b^2 - 2c^2}{a + c cos\theta}$

[somber coyote]

Conan:

[solid swan]

assuming theta is the angle between positive x axis and the lines

$c = \sqrt{a^2 - b^2}$

[somber coyote]

Conan:

[solid swan]

but the rest doesn't match up

i'm just stuck here

<@&286206848099549185>

[empty matrix]

Can anyone solve this question?

With explanation

[solid swan]

my first thought is

since angle BAC = angle ABC (isosceles triangle)

triangle ABD and triangle CAD would be similar

[upper karma]

Similar triangles

i need help with this too if anyone has time

@upper karma do you expect us to understand that lenguage?

yes

[solid swan]

it's straightforward

[upper karma]

😳

its not that hard i just cant concentrate on any numbers

[solid swan]

where is k though

[upper karma]

k is the similarity thing

[solid swan]

so the ratio of one side to another?

[upper karma]

yes

[solid swan]

assuming that the triangles are similar

[acoustic jungle]

3,25

[upper karma]

hm

1.25/3.25

[solid swan]

oh wait 3.25

sorry

[upper karma]

is ok

so it should be

2.6

[solid swan]

CB would be 15*3.25/1.25

[upper karma]

97.5

[solid swan]

KL would be 130*1.25/3.25

which is also x

[upper karma]

x*2.6=1562,4

how do i use the bot

a ok nbvm

im stupid

[solid swan]

also this hasn't been solved yet: https://discordapp.com/channels/268882317391429632/326138757474680852/703818666613145610

[acoustic jungle]

they lie on the centers of the other

and they are congruent.

You wouldn't need a graph to solve this, you can draw it on paper

[upper karma]

i dont exactly know this subject

but i need to find the outside length of them all

[upper karma]

can someone help me find the equation for that

[hallow rose]

,rcw

[upper karma]

ty

I figured the period is pi/10

For the sake of the problem, lets assume its a sine function

is the vertical displacement at 41?

thats whats bugging me

$8^2+15^2=(8+15-2x)^2

pause what

46 = (23-2x) times 2

46 = 46 - 4x

46 - 46 = -4x

0=-4x

[upper karma]

@upper karma

there

are you familiar with Pythagora's theorem?

Yea

That’s pretty cool

I just didn’t know how to set it up

Because of the circle stuff

all tangents form 90 degree angles with the radius

Ya thanks my dude

Btw I’m not the person who asked the question @upper karma

That was @upper karma

Ya u knew that

Bye

😄

i mean

i am familiar

we havent really studied it

the first one was the only one i really struggled with after a bit of thinking x)

this helped

thank you!!

let me know if you need help determining side "c"

i mean

pythagoras theorem i

s

(I intentionally marked it as "c", btw)

a times a plus b times b