geometry-and-trigonometry

[upper karma]

Since he broke the rules?

[tawdry pivot]

how is Barry breaking the rules? unless there exist take home math exams for some reason

[upper karma]

No asking for help during exams

[spark stag]

It literally says "final exam"

On the bottom left

Probably smuggled in a phone or smthn

[tawdry pivot]

ah, so they might've taken the pic in the bathroom?

or that

[upper karma]

Anyways

Can someone help with 822?

I tried this but didnt go well

I have one small mistake in my attempt i think

come on

the exam is over

bruh

[tawdry pivot]

let $\alpha=\beta=\gamma=90^\circ?$

[somber coyote]

EpicGuy4227:

[upper karma]

I just took the paper wih me

Next time say that the exam is over

ok

So we don't assume that you broke the rules

You caused confusion here

how on earth do you think I can get the paper without taking the exam lol

[tawdry pivot]

@upper karma

[upper karma]

?

[tawdry pivot]

did you read what I said?

[upper karma]

Oh yes

But

I dont know what let means

[tawdry pivot]

for a contradiction

let [whatever i said]

[upper karma]

Ok but that didnt solve anything

I need to prove equality

Between the two sides

[tawdry pivot]

and I have shown that the equality doesn't hold

[upper karma]

Please explain how

[vagrant elk]

Well, it doesn't hold since i have a sneaking suspicion there's a requirement that alpha, beta, gamma be angles in a triangle

[tawdry pivot]

oh

XD

[vagrant elk]

y e a h

smfh

[upper karma]

Still dont understand

[tawdry pivot]

are you tying to prove that the equality holds for all alpha, beta and gamma?

[upper karma]

Alfa beta gamma are in a triangle

Their sum is 180

And i still dont get why doesnt the equality hold up

And how did you come to that conclusion

Just by looking at my equation

[tawdry pivot]

because I thought we were proving for all alpha, beta, gamma, because you didn't say they were in a triangle

[upper karma]

Oh

Sry in hs assignments, we all assume alfa beta gamma are in a triangle

Since all our assignments are based on angles in a triangle

Im sending it again so you dont have to scroll

When I try to do this with my calculator with
alpha = 8 (not degrees)
Beta = 12
Gamma = 14
Left side is -0.53 and the right one is 0.55

So it's a mistake with the assignment

And you can't prove equality

[tawdry pivot]

8+12+14=34? don't you need the angles to add to 180 degrees?

[vagrant elk]

^

[upper karma]

O

It works now

Anyways

How do I solve this

@tawdry pivot

[tawdry pivot]

managed to get it. I set $\gamma=180-\alpha-\beta$ and the equation becomes $\sin\alpha+\sin\beta-\sin(\alpha+\beta)=4\sin\frac\alpha2\sin\frac\beta2\sin(\frac\alpha2+\frac\beta2)$

[somber coyote]

EpicGuy4227:

[tawdry pivot]

then I used double angle formula for sin and $\cos\theta=1-2\sin^2\frac\theta2$ on the LHS, and angle addition formula for sin on the RHS

@upper karma

[somber coyote]

EpicGuy4227:

[peak mesa]

How to get the opposite side by adjacent and angle?

[dark sparrow]

tan(angle) = opposite/adjacent

[peak mesa]

I want to use in progamming but i want to put the size of the opposite side by ajdacent and angle

Math.tan(angle) * img.getHeight();

[dark sparrow]

well you can multiply both sides by adjacent

[peak mesa]

i'm doing this but the values aren't correct

[dark sparrow]

to get opposite = adjacent * tan(angle)

can you show a picture of what you're doing?

[peak mesa]

img.getHeight() = ajdcant

its a random picture xD

[tawdry pivot]

are you using the right units for angle

[peak mesa]

its dirt

its for a game..

[dark sparrow]

ok but can you show the angle you're using

[peak mesa]

right now i'm using 45

[dark sparrow]

to make sure that what you're saying is the adjacent side actually is the adjacent

i don't care for its value

i want to see where it's located

[peak mesa]

right now is not a variable

[dark sparrow]

no, i don't care for the value of the angle

i want to see, geometrically, where it's supposed to be

on the image

[tawdry pivot]

Math.tan(x) gives the tan(x) where x is in radians

[dark sparrow]

also that might be another issue yes

well

try having your program print Math.tan(45) to the screen

[peak mesa]

i dont understand what u asking for

ok

[dark sparrow]

ok let's deal with the radian vs degree issue first

[dark sparrow]

ok

so that means your tan function accepts input in radians

as one would expect, anyway

because if it were degrees, you'd get 1.0

[peak mesa]

ok

[dark sparrow]

so you should replace that 45 with π/4

[peak mesa]

i have Math.toRadians

[dark sparrow]

or that, sure

[peak mesa]

These values ​​are always appearing

0.9 and 255.9

[dark sparrow]

ok the first question is why is it getting printed so many times

[peak mesa]

But the problem is:

size of image on left

its a loop

[dark sparrow]

and second... you're multiplying it by a length in pixels so the result is also going to be in pixels

right

[peak mesa]

yes

[dark sparrow]

you may as well round to the nearest integer while you're at it

[peak mesa]

..

[dark sparrow]

however i'm kinda shooting in the dark here

[peak mesa]

45 degrees should give a lower value

[dark sparrow]

since i don't yet have a picture in mind of what you're trying to do

like

do you have a sketch or something

because that'd probably make stuff clearer

as of now it's kinda hit or miss

[peak mesa]

I'm trying to make a test, conver a flat image to a 3d format

But I am trying from an idea that came to me

[dark sparrow]

wdym by "convert a flat image to a 3d format"

can you like... at least hand-draw what you want the result to be

[peak mesa]

ok wait

The idea may seem a bit sneaky but it's the following, I have an image, that image after going through a function will come out 3d-looking from the angle I insert. What I want to do is take the amount of pixels from the opposite side that I have to remove in the image to be able to give a 3d look

i dont have a example.. i still haven't finished

like that

[mighty narwhal]

kinda shooting in the dark

So you want to rotate a 2d image?

[peak mesa]

i want to convert a 2d image to 3d image

[mighty narwhal]

You want to view a 2d image from an angle

[peak mesa]

yes

[mighty narwhal]

Take a piece of paper

[peak mesa]

the original image

[mighty narwhal]

Take a piece of paper
If you look at it from and angle it is the same as rotating it

[peak mesa]

But where can this help me?

[mighty narwhal]

Well

Rotate the paper
You are seeing less of it

Delete some of the lines when rotating
The amount you need to delete increases by the angle of rotation

[peak mesa]

yes..

but what i want is get the opposite site by adjacent and angle

but this dont work

[mighty narwhal]

oppoite side by adjacent ang angle?

[peak mesa]

and*

[mighty narwhal]

and*

tan angle=opposite/adjacent
opposite=tan angle * adjacent

[peak mesa]

https://cdn.discordapp.com/attachments/326138757474680852/590461365181874186/unknown.png

dont work..

img.getheight = adjacent

Math.tan(Math.toRadians) is tan angle..

[mighty narwhal]

What is the expected output?

[peak mesa]

look the values

[mighty narwhal]

Also

[peak mesa]

The value on the right is lower than the value on the left

[mighty narwhal]

the two legs or a 45deg right triangle are equal

they should be the same

[peak mesa]

left value = size of image

right value = output

[mighty narwhal]

tan(45) is 1

So essentially you would get the same

[peak mesa]

It may be another value that does not fit the same

[mighty narwhal]

by size you mean width or height?

[peak mesa]

yes

As you can see the orange side is bigger than the green side

and that's what I want, understand?

[mighty narwhal]

Okay let me think

Something really isnt adding up
The output should be less, as you said

Why is is greater

what are the outputs when a=20deg?

[mighty narwhal]

is this the output?

????

[peak mesa]

yes

but if i put in radians

[mighty narwhal]

Show me the code

Something is really not adding up

[peak mesa]

but the problem is:

40 degrees output:

[mighty narwhal]

Erm this makes sense

[mighty narwhal]

This also makes sense

[peak mesa]

goes up too fast

[mighty narwhal]

?

[peak mesa]

at 45 degrees should give about half of 256

[mighty narwhal]

No

tan 45 is 1

If you have a right triangle with angle 45

it is the half of a square

the two legs are equal

[peak mesa]

yes

but

but look at the difference

ignore de size of adjacent xD

opposite face is less by what? by angle..

[mighty narwhal]

the one on the right doesnt look like a triangle with angle 20

[peak mesa]

its from paint '-'

Try to understand my point of view xD

[mighty narwhal]

The 20deg output is the half of the 40deg output

Almost the half

It makes sense
I understand that it looks weird

But it works out?

[peak mesa]

nope

[mighty narwhal]

Scale the two triangles so that the bottom leg is the same size

[mighty narwhal]

Erm yes

[peak mesa]

look

[mighty narwhal]

Yes?

[peak mesa]

look at size

[mighty narwhal]

NowBro

[peak mesa]

size of opposite side

[mighty narwhal]

the upper one should be 40 deg

not 30!

[peak mesa]

i will work with this angle

not the 70º

[mighty narwhal]

The upper triangle should have angle B = 40

you are making it 30

we calculated it with 40

:C

[peak mesa]

same thing

look at angle of B

40º

but

look at the opposite face

its bigger than the other

[mighty narwhal]

ofc its bigger

It should be bigger, by the math

your code says its bigger

I dont understand the problem

[peak mesa]

Yes and I want to get this side from angle B

[mighty narwhal]

,calc 93/215

[somber coyote]

Result:

0.43255813953488

[peak mesa]

oh wait

[mighty narwhal]

Look at the two images
The side is a little shorten than the half

IT WORK OUT

Your code works

:C

[peak mesa]

problably i will use the hyp side

-.-

omg..

[mighty narwhal]

Oh deer .-.

[peak mesa]

subtracting hyp side with the adjacent side

-.-

[mighty narwhal]

hyp=cos angle * adjacent

[peak mesa]

Thank you anyway, I had not remembered this problem

[mighty narwhal]

Your welcome

[peak mesa]

I'll try later

[mighty narwhal]

And good luck with your code!

[peak mesa]

Thks

[peak mesa]

@mighty narwhal https://gyazo.com/8fbbcf402a3e4e818fe094f97f7bd596

xD

[mighty narwhal]

ERrrrmm

[peak mesa]

:v

[mighty narwhal]

Well umm

The top triangle is correct

if the top is the fix point and we rotate it towards us

[peak mesa]

My idea was to make an image later rotatable

360º

[mighty narwhal]

Well

Its half-way done in one direction

then you can just do another iteration in another direction

you get the point

[drowsy estuary]

Hello there, is someone can help me with perspective depth ?

Actually i try to find the mathemathical formula to find the distance between (IO, LP, JN, KM) to make my figure look like a cube, not a rectangle

[peak mesa]

sqrt(LxL+PxP)
@drowsy estuary

[drowsy estuary]

ah sorry, i'm not understanding. or maybe i haven't provide enough information. at the begining i only 1 square (IJKL) and the W point. and with amthemathics (or geometry) i must find where to place the distant square (ONMP). So by emample, let's say i have a square (IJKL) of 1 meter of each side & the W point is at 2 Meter height from K and 1 meter on the right of K... at this point i dunno where to place O,N,M,P to get a cube (or maybe i haven't understand your formula =/ )

[foggy nimbus]

hello

can someone here help me with my geometry work

it's graphs and such

would be great if you could

thanks

[drowsy estuary]

ah! i think i found the solution. i just need to one angle by 2. by example, if i divied the angle (ILW) by2 and draw a line, it should go straight to O and my guess is that it respect the Cube proportion ^^;

i mean to be more precise, the line that i draw start from L and cut the IW line. That's giving me O. and then from O i just need to draw straight line to get my ONMP square to form a perfect cube (and not a rectangle)

from here, using Sinus Cosinus tangent should do the job 😃

thanks for your interest, have a nice day everyone

[foggy nimbus]

Can someone help me solve this?

[devout shell]

JKM and m4 form what kind of angle?

[foggy nimbus]

To answer that: Not sure. I'm terrible at Geometry.

[devout shell]

I encourage you to put a piece of paper on the screen and just trace those angles out

you see what you get and then I think you'll see what I mean

[foggy nimbus]

Alright

I have a few other questions, hopefully it isn't considered spam if I keep asking here

[devout shell]

did you get the answer to the previous one?

[foggy nimbus]

I'm still working on that one

[devout shell]

do that one first, then we can move on

[quiet mason]

prove that length of arc of a circle=radius of circle*angle described by arc

[devout shell]

which formula in particular do we need to prove?

S = r * θ formula?

or the proportion one?

[quiet mason]

yea s=r*theta

[astral patio]

Hi guys. Does this look right? <@&286206848099549185>

[minor arch]

https://i.imgur.com/nzAifKJ.png

Can anyone explain to me why for the sin/cos, we're using POQ.

But tan/sec, we're using triangle VOR?

I get the logic behind the calculation. not getting the logic behind changing the triangle.

[fringe dirge]

It's just so one of the sides has length 1

Like try using tan on the POQ triangle, neither of the sides in the ratio will have length 1

[minor arch]

Oh I see.

[pulsar dirge]

What does question 8 mean

[dark sparrow]

you are asked for the lowest solution of 4sin(3θ) + 2 = 0 in the interval [2π, 3π]

[upper karma]

Can anyone help me out w/ this?

The exercise asks for <Y, but how?

oh

wait

🤔

X = 180-123 = 57

Z = 180 - 97 = 83

Y = 40????

😮

@upper karma ;#

yes

@upper karma ok, also help

Doesn't one postulate say that the sum of the 3 exterior angles = 180?

But that clearly isn't true here? It's 123+97?

[vagrant elk]

What's the sum of your exterior angles?

Actually just add them up

[upper karma]

Uh😳

123+97 isn’t 180

@vagrant elk

[dark sparrow]

[upper karma]

😳😳😳

What am I doing wrong

[vagrant elk]

I'm saying

add all three exterior ones together

[upper karma]

360

[vagrant elk]

you'll notice it's clearly not 180

but rather 360

[upper karma]

Yeah 😛

[vagrant elk]

because for your three angles X,Y,Z

you have the exteriors, leaving 3*180 - (X + Y + Z)

which you should know the right side is 180

2*180 = 360

boom done

[upper karma]

Nice.

I have read before it was 360, but I saw a Reddit post saying 180, so I was confusion.

@vagrant elk exterior angles need not be > 90, right?

[vagrant elk]

what

the exterior angle can do whatever it wants

[upper karma]

😮

Word.

Yeah, because if <X = 110, its exterior angle is 80

which is, as a matter of fact

let me check

,ask 80 < 90

[somber coyote]

Display results online and refine query

Requested by rudy#0446

[upper karma]

It happens to be less than 90.

[worldly junco]

sup

[manic osprey]

4sin^(3)xcosx+4cos ^(3)xsin x=0 which is the real answer: πk/2 or πk? My friend showed me how can you get that but I say because the equation is homogenic, its πk

[dark sparrow]

$4 \sin^3(x) \cos(x) + 4 \cos^3(x) \sin(x) = 0$

this?

[somber coyote]

Ann:

[manic osprey]

yes

[dark sparrow]

this can be factored to $4 \sin(x) \cos(x) (\sin^2(x) + \cos^2(x) ) = 0$

[somber coyote]

Ann:

[dark sparrow]

and in turn to $2 \sin(2x) = 0$

[somber coyote]

Ann:

[dark sparrow]

so yeah $2x = k\pi$ so the sol is $x = \frac12 \pi k$

[somber coyote]

Ann:

[twin shard]

im not sure how to do d

i have 4(xy) + x^2

thats for surface area

for volume i have the formula its 16000/x + x^2

im not sure how to get the formula for surface area

[dark sparrow]

what do you mean by "normal area"

[twin shard]

area in general

not surface area

[dark sparrow]

what do you mean by the "area" (as opposed to surface area) of a box, then

[twin shard]

sorry?

oh

so the area of a box is l.w.h

in this case its l^2 . h

[dark sparrow]

that's called volume

not area

[twin shard]

oh yeah

my bad

i meant volume

[dark sparrow]

and no, the volume is not 16000/x + x^2, it's 4000.

you're literally told it's 4000.

[twin shard]

yes

but that is the formula for the volume in terms of x

is it not?

[dark sparrow]

no, it is not

[twin shard]

oh

do you know how to find the formula for the surface area?

because i dont know how to do it than

ahhh

its 2lh + 2lw + lw

there fore it would be 2xh + 3x^2

so sa/h = 2x + 3x^2?

im not sure if thats in terms of x

fuck idk how to do it

[craggy night]

"Assuming alpha = -3pi..."

The answer section says the result is -2, but no matter what I try it's always 1/0

Can somebody please confirm I'm not insane

[jaunty plume]

definitely not -2

[dark sparrow]

the denominator is not 0

[craggy night]

thanks

[jaunty plume]

yes it is?

0+2*0.5-1

[dark sparrow]

wait crap

😂

i thought sin(-3pi) = 1 😂

brain is not running at full capacity today

[jaunty plume]

[mighty narwhal]

You have a regular hexagon with each sides colored to be a different colors
I rotate your hexagone (always by 60deg) to the left twice, then I flip it along the vertical axid
I then rotate it again to the right 5 times, I flip it again, then I rotate it to the right 1 more time

Your task is to figure out how you can get to the current state of the hexagon from the start with the least amount of transformatinos

transformations*

[subtle gate]

wait, what you do to the hexagon is given, or what exactly do you mean by transfromations?

[dark sparrow]

wdym by rotating "left" and "right"

cw and ccw respectively?

or the other way around

[fair wigeon]

my iq is hexagone

[mighty narwhal]

left is ccw
right is cw

Transformation is either a rotation or a flip along the vertical axis

[subtle gate]

wait

[narrow sleet]

Reflection?

[subtle gate]

so are we working with your transformations, or do we need to choose them ourselves?

[mighty narwhal]

with my transformations

rotating cw and ccw
And mirroring against the vertical axis

[subtle gate]

each side has a diffrent color?

[mighty narwhal]

So like
Simplify the following steps:
rotate 2 times ccw, mirror, rotate 5 cw, mirror, rotate 1 cw

Each side (edge) has a unique color

So every state of the hexagon is unique

[narrow sleet]

Is there a front and a back face?

[mighty narwhal]

No, you just mirror it

Rotate it cw or ccw by 60deg
Mirror on the vertival axis

[narrow sleet]

Ok

[subtle gate]

||you just leave the hexagon if i didn't miscalculate if you want to replicate the effect||

[mighty narwhal]

|| what u mean?||

||wait||

||you rotate it two times ccw||

[subtle gate]

||The transformations you described leads the hexagon to it's starting point, so if we want to save on transformations, you just do nothing (at least from what i can get, i am gonna put this into geogebra for a sec)||

[mighty narwhal]

||Noooo... you rotate two times ccw||

[subtle gate]

|| start - +2ccw -> 2ccw - mirror -> 2cw - +5cw -> 7cw = 1 cw (since 6 rotations are 360°) - mirror -> 1ccw - +1 cw -> 0 what is wrong with this?||

[mighty narwhal]

2ccw mirror 5cw mirror 1 ccw

if you mirror, rotate, mirror again
The mirroring will be gone but the rotations are in the other direction

So it becomes 2 ccw 5ccw 1ccw

=2ccw 6ccw
6ccw becomes e (identity)
You are left with 2ccw

[subtle gate]

you typed the last thing wrong from where i worked with

if i just change my last step to the one you said, it actually is ||2cww||

||start - +2ccw -> 2ccw - mirror -> 2cw - +5cw -> 7cw = 1 cw (since 6 rotations are 360°) - mirror -> 1ccw - +1 ccw -> 2ccw||

[mighty narwhal]

Why is my step wrong tho

If you mirror, rotate 5cw, mirror again

you are rotating 5ccw

2ccw 5ccw 1ccw

2ccw 6ccw
2ccw

[subtle gate]

i am agreeing with your outcome tho?

[mighty narwhal]

Yay

I did make rules for how you can manipulate them

with the help of abstract algebra

[subtle gate]

i mean you could do it like that

but that is just complicated

[mighty narwhal]

the last two rules is about how you can switch the order of mirroring and rotating,
So you can always get them next to each other and they will kill each other

Yea

But like after you understand it you can do it almost instantly

r^-2 f r^5 f r^1

r^-2 r^-5 ff r^-1
r^{-2-5-1}=r^-8=r^-2

[subtle gate]

okay i modeled it in geogebra

[mighty narwhal]

Cool

Does geogebra agree with us

[subtle gate]

yep it all works out

[mighty narwhal]

Yey

I also thought my friend who is like a normal math student in 9th grade
She could understand it can now use it ._.

I am also teaching her calculus lol xD

[buoyant oasis]

whaaaa

i would've loved to have a teacher to teach me calc at 9th

[subtle gate]

i mean you can kinda teach it yourself at that point

[mighty narwhal]

Well I did teach it myself in the begining of 8th grade

I mean the very basics

Then someone recently taught me integration by parts and stuff like this

[violet tartan]

I need some help on this worksheet, my teacher doesnt really teach people and the final on this

[mental wharf]

if you have 2 lines from the center of a circle to the circle, then what is their length

[violet tartan]

They are equal

[subtle gate]

unless you have any side, you cannot determine a...

[mental wharf]

yeah, so the triangle is isosceles

[subtle gate]

you need a radius and an angle and you only have the angle to solve for the arc

[violet tartan]

Alright

[violet tartan]

Sorry for not responding afterwards, I had to go to class

and thanks for the help

[covert orchid]

hey guys, i'm looking for some trig and geometry books (for someone who has already some knowledge in these areas but wants to go a bit deeper) can you guys give me any recommendations?

[fringe dirge]

There are some nice aops books on the topics

They're geared towards more competitive math students, but they have nice information

[covert orchid]

aops?

[fringe dirge]

https://lmgtfy.com/?q=aops

[tawdry pivot]

thought about doing that lol

[mighty onyx]

Topic 4 ques 2

[rustic dagger]

Pretty cool

Forgot the rotate command and the note thingy is blocking some of the question

[dark sparrow]

,rotate -90

[mighty onyx]

Ok let me remove it

[rustic dagger]

,rotate -90

Omg

!rotate 90

,rotate 90

[rustic dagger]

Mabye b?

Idk acually

I guess find the angle of elevation when reaching the pillar and go from there?

[mighty onyx]

Ok

Ans is d

[hasty mesa]

@mighty onyx you must be from a familiar place

[mighty onyx]

?

[hasty mesa]

those are prev. year jee papers aren't they??

[mighty onyx]

Yes

Have u given jee

[hasty mesa]

are you preparing for Jee??

[mighty onyx]

Yes I am

Come In dm

[hasty mesa]

lol I used to ask jee questions here once upon a time now you continue

[novel pecan]

I'm trying to find the angle between the base edge and one of the other two in a tetrahedron. I keep getting pi/3 no matter what values I use for the edges. Does someone know why this is wrong?

[spark stag]

so we know the entire angle JWL = 87 degrees

so, when we add the two smaller angles together

we get 87 degrees

in other words, (2x+24) + (3x-12) = 87

can you solve for x?

[hallow token]

https://tenor.com/view/head-scratcher-confused-perplexed-huh-clash-with-cam-gif-10198268

[frail rock]

maybe start off by adding like terms?

btw very nice gif. quality is great

[hallow token]

Got it its 36

[olive sinew]

I know it’s wrong

Can anyone explain to me how can you break the interval

[fringe dirge]

could you split the interval in half?

[dark sparrow]

can you split the interval [0, 12] into 6 equal parts?

[olive sinew]

2

I guess

?

[spark stag]

what would those intervals be?

[olive sinew]

2,2,2,2,2,2

[]

[dark sparrow]

what would their endpoints be

surely they're not all 2?

draw a number line on which you split [0, 12] into 6 parts

[olive sinew]

I did the number line and everything but nothing kicks in yet

Can you show me an example please

[fringe dirge]

take a picture of your number line

[olive sinew]

-.-

[fringe dirge]

You're not labeling the numbers correctly at all

How does the interval go from 2 to 2?

The very last number on the right side should be a 12 since you're ending at 12

First of all, Ann's not a guy

[upper karma]

Uh, oops

How do we know this rhombus forms a 45 degree angle?

From DC to AC

[upper karma]

Nvm— apparently, the question was worded in a way that tou can arbitrarily set the degree;c

[tawdry pivot]

we can readily see that you can just cut off overlapping ham and place them over uncovered bread. if we were trying to get the least number of cuts then it'd be interesting

[barren ibex]

do it with the least amount

and how would you cut the sausage

[left rose]

Does anyone know some resources that could lead me to figuring out the points on this irregular shape? The instructions say to use a ruler and protractor.

https://i.imgur.com/fLQMYGi.png

[fringe dirge]

Just measure the x coordinate and the y coordinate of each point separately

@left rose

[left rose]

Not sure how to. Here is an example from the class.

https://i.imgur.com/fl54j3Z.png

@fringe dirge

Not sure if there is some correlation. Or just measure in cm or something. So lost. haha

[fringe dirge]

it doesn't really matter what you measure with

[left rose]

Ah, okay. It just flipped my brain upside down with the numbers being so huge.

[left rose]

@fringe dirge just to make sure. I can just do cm to measure?

[lime flame]

how are these two triangles congruent?

[cinder portal]

Two things

KE is equal to EI

and angle EKA is equal to angle EIT

knowing that KE and EI are the same, that means the ARCS KE and EI are the same

by the inscribed angle theorem i think? angle A and angle T are congruent them

since angle EKA and angle EIT are the same, that means the chords are also the same AK and IT by some theorem i forgot

and then u have some triangle theorem

and shows they are congruent

ASS? AAS? i forgot o god

[dark sparrow]

@upper karma that isn't even a rhombus

[upper karma]

Oh my bad Im dumb

Its a trapezoid @dark sparrow

Im srry😟

[elfin scarab]

Hello,
here's a really easy highschool-level geometry-related problem for you... So, my textbook has this question:

The centre of a circle drawn around a triangle lies on one of the sides of the triangle. Prove that the triangle is right-angled.

Now, after thinking about this for some while and using the assistance of some geometry programs, I've come to the conclusion that the triangle does not need to be right-angled, therefore making the whole question false.
Have I reached the correct conclusion? Is there a way to prove this?
My refutation is as follows:

The circle drawn around a triangle is defined by the fact that the distance to all the corners of the triangle is the same from the centre of the circle.  (The radius) Therefore we can just construct a right-angled triangle, draw a circle around it and then just change the position of the point at the right angle so that it keeps the same distance to the centre, still keeping the hypotenuse and the circle in place but disrupting the right angle. The centre of the circle still lies on one of the sides of the triangle. (The hypotenuse)

[elfin scarab]

Ah yes, the diameter 😃

Do you have a continuation? @glad quarry

[glad quarry]

Yeah, I thought you understood..

Waiy

[elfin scarab]

But yeah, isn't the radius more fitting there?

[glad quarry]

The diameter always subtend 90 at the circumference at the circle

It's a very basic property of circles

[elfin scarab]

Okay, and?

[elfin scarab]

Okay, but doesn't my refutation still hold?

[glad quarry]

I don't think so because when you change the position of the point on the circle it's angle still remains 90 we have done a general proof the point 'A' could be any point on the circle

[elfin scarab]

Aaaaah wait

Well okay I didn't really get that explanation but now it seems logical 😄

[elfin scarab]

Yeah, I got it now. Thanks 😃

[glad quarry]

You're welcome.

[elfin scarab]

It's interesting though how the constant need to translate blurs the actual meaning of the text so well...

[nocturne tapir]

Circular theorems lmao

I'm getting flashbacks of highschool

oh my god

The alternate segment theorem

[frail rock]

Is it proper to say that congruent shapes are kinda like perfect replications in a way?

[lusty quest]

how would i find the radian of this? it doesnt even give me degrees to work with

[devout shell]

is there like a radius or something given?

[lusty quest]

no, this is the whole question

if theres a degree, id figure it out

it looks like a 60 degree

[hard gale]

they just want you to eyeball it ig

yea looks like pi/3 rad or something

ie approx 1

[lusty quest]

its not pi/3

[hard gale]

approximations of approximations wow

[lusty quest]

it said im wrong

[hard gale]

i didn't say it was

they want an integer

[lusty quest]

ohh

so its a 1, because apparently pi == 3

[hard gale]

fundamental theorem of engineering

pi = e = 3

[gritty siren]

sin x = x

= tan x

[hard gale]

g = pi²

hehe

[lusty quest]

i honestly hate pearson

[bright parrot]

is this the discworld definition of pi?

[devout shell]

said use 3 as pi lol

[barren notch]

why is it when finding area of a circle in terms of the diameter you divide by 4?

[mental wharf]

radius squared

d = 2 r

[astral patio]

Anyone able to confirm this please?

[dark sparrow]

is the character i circled meant to be a 7?

also, what interval were you asked to solve over?

@astral patio

[astral patio]

Yes, 7. 0 < pi < 2pi

[dark sparrow]

then why do you have 2pi listed as a solution?

[radiant nacelle]

Should I learn trigonometry through the unit circle method or the right angle method or does it matter

[ebon spear]

Id say unit circle tends to be more useful in my experience but the right angle method is pretty good if you need to remember certain values of sin and cos

[radiant nacelle]

Ah okay, thanks 😃 i remember sohcahtoa from highschool, is that from the right angle approach?

[ebon spear]

Sohcahtoa relates very strongly to that approach yea

The circle is kind of a big extension of the right angle approach

[radiant nacelle]

Ahh okay I see, I’m in a precalc and trig class for the summer and it seems like we’re just going to be doing the right angle approach and was just wondering if I should study up on the unit circle method beforehand

Or afterwards or smth

[ebon spear]

I imagine it'd be taught in class, with a good amount of time spent on it. The right-angle method just quickly gives certain useful points on the unit circle

[radiant nacelle]

Ah gotcha, alright thanks for your help 👌

[ornate lodge]

Is it always true that the longer base of an isosceles trapezoid inscribed in a circle is the circle's diameter?

[dark sparrow]

no

[ornate lodge]

I had a problem where the side edges and the top edge are the same length, and the base edge is double that. Does it always work in that case?

[dark sparrow]

yes because then your trapezoid is forced to be half of a regular hexagon

[pallid narwhal]

could someone give a full walkthrough of this? teacher gave no explanation and I'm fully lost

[spark stag]

we can model the motions of the satellites through sinusoidal functions

can you construct the sine (or cosine) functions for both satellites?

[urban oxide]

halp

14

tf is that

[dark sparrow]

,rotate -90

[urban oxide]

sorry

[dark sparrow]

so what's giving you trouble

[urban oxide]

idk how to start

[night karma]

With?

[urban oxide]

14

i thought of writing tg as sin/cos

[night karma]

Mhm.. probably someone here who can help. I've never seen '1g' or 'tg' in my books before 😂

[urban oxide]

thats lg

log10

and tg is tan

[upper karma]

what's tan 45

or rather i should ask

what's $\log(\tan 45^{\circ})$

[urban oxide]

oh shit

im dumb

[upper karma]

lol

[urban oxide]

thank you

[upper karma]

is that romanian

[urban oxide]

yeah

how did you know?

[upper karma]

i have seen a couple of assignments from romanian people

[urban oxide]

oh

[fervent raft]

@upper karma you can use log rules to expand that

remeber that $\tan\theta = \frac{\sin\theta}{\cos\theta}$

[somber coyote]

agentnola:

[devout shell]

solving right triangles using logs, old technique

[upper karma]

sigh

[devout shell]

no one uses that anymore

[upper karma]

first of all that wasn't my question

[fervent raft]

oh f

sorry

[upper karma]

second of all tan 45 is just 1 so log(tan 45) is 0

[fervent raft]

yes

I dont know the degrees in my head

[upper karma]

[fervent raft]

I was just gonna expand it and sin(45)=cos(45)

so it has to be 0

[mighty narwhal]

,calc arcsin(2)

[somber coyote]

The following error occured while calculating:
Error: (intermediate value)(intermediate value)(intermediate value) is not a function

[mighty narwhal]

[dark sparrow]

,calc asin(2)

[somber coyote]

Result:

1.5707963267949 - 1.3169578969248i

[hard gale]

[mighty narwhal]

,calc asin(2)

[somber coyote]

Result:

1.5707963267949 - 1.3169578969248i

[mighty narwhal]

,calc sin(1.5707963267949 - 1.3169578969248i)

[somber coyote]

Result:

2

[hard gale]

god i didn't see the i at the end

[mighty narwhal]

Person who learned that the output of sine is in [-1;1]:
Impossible

[hard gale]

was like wtf that thing is full rarted

[polar mason]

Can anyone give me a tricky trigonometry question (11th grade)?

[fringe dirge]

Idk what exactly you're looking for but

Show that $\cos(\sin^{-1}(x)) = \sqrt{1 - x^2}$

[somber coyote]

Zopherus:

[devout shell]

[fringe dirge]

@polar mason

[polar mason]

That scared me

But I'll try it

[fringe dirge]

is that the stuff you're looking for

[polar mason]

I don't exactly know

I just wanted a sum

That was tricky

[fringe dirge]

a sum?

[polar mason]

Nah it'll do

Ty

Mr Zop

[fringe dirge]

Try this one

[polar mason]

Okay

[fringe dirge]

Calculate the sum $\tan^2\left(\frac{\pi}{16}\right) + \tan^2\left(\frac{2\pi}{16}\right) + \tan^2\left(\frac{3\pi}{16}\right) + \tan^2\left(\frac{4\pi}{16}\right) + \tan^2\left(\frac{5\pi}{16}\right) + \tan^2\left(\frac{6\pi}{16}\right) + \tan^2\left(\frac{7\pi}{16}\right) $

[somber coyote]

Zopherus:

[fringe dirge]

This one's a toughie though

[polar mason]

I'll try it

Dw

It's a long one or a tough one?

[fringe dirge]

Just tough, not really long

[gusty crater]

Oww

[polar mason]

Okay

[gusty crater]

Damn these trigs

Hate em

[polar mason]

Ty @fringe dirge

[fringe dirge]

np

[mighty narwhal]

Hmmmm

how do you do the cos(arcsin) one tho

[devout shell]

drawing the triangle is one way

[mighty narwhal]

hmmmmm

OOOOHhhhh

[devout shell]

and I think from that idea you can see what to do

[mighty narwhal]

with hyp 1, side x and the other side sqrt(1-x)

Yea okay I got this

[devout shell]

perfect

[mighty narwhal]

I wont draw or write latex for it

I was just curious

[devout shell]

that's fine lol, you described it perfectly

so clearly you know it

[mighty narwhal]

I thought of substituting on the right side like you would do at integration and simplify and get something?

[devout shell]

oh man lol

[mighty narwhal]

:/ I am dump

[devout shell]

matter of trying out all sorts of problems

[mighty narwhal]

What grade are you in

[devout shell]

going to be junior in college

[mighty narwhal]

doesnt know about american school system grades so pretty much clueless

[devout shell]

third year student at university

[mighty narwhal]

Oh thanks!
Good luck with that

[fringe dirge]

The algebraic way to do it is to use the fact that \cos^2(x) + \sin^2(x) = 1

[mighty narwhal]

How would that help

$\cos(\arcsin x)=\sqrt{1-x^2}$

[somber coyote]

DarK:

[fringe dirge]

Think about \cos^2(\arcsin(x))

[mighty narwhal]

Yea we can square it

$\cos^2(\arcsin x)=1-x^2$, ofc $1\ge x^2$

[somber coyote]

DarK:

[fringe dirge]

yeah then just take square roots

No I mean like

[mighty narwhal]

.-.? I just squared them

[fringe dirge]

Think about \cos^2(\arcsin(x))

[mighty narwhal]

I know what u mean

I have it already

$\sin^2(\arcsin x)=x$

[somber coyote]

DarK:

[fringe dirge]

not quite

remember what sin^2 means

[mighty narwhal]

x^2

[gusty crater]

What's Ptolemy's theorem...can anyone explain?

[mighty narwhal]

I forgot about the stupid ^2, dont harass me Zeph

We get $x=x$, $x\in [0;1]$

[somber coyote]

DarK:

[mighty narwhal]

Which is ofc true for all x between 0 and 1

[upper karma]

sin(pi/3)=sin(8t)

how do i solve for t?

[mighty narwhal]

If you have

Sin(a)=sin(b)

Then a=b+2kpi, where k is an integer

[fringe dirge]

that's just not true

at all

[mighty narwhal]

Why

[fringe dirge]

sin(0) = sin(pi)

sin(pi/4) = sin(3pi/4)

[mighty narwhal]

My book literally says what I wrote

Gimme 5 mins to make a photo

[fringe dirge]

i literally gave you counterexamples to what you said

how can you still think what you said is right

[mighty narwhal]

Please just give me 5 minutes :-:

A=b+2kpi
Or
A=pi-b+2kpi

With the second possibility your counter example is countered

[fringe dirge]

Yeah but you literally didn't say that

[mighty narwhal]

Did u give me time to say that ;-;

But nevermind

You can just take asin() of both sides

And take dont forget that sin(pi-x)=sin(x)

So you have 2*2=4 possible equations
And 2 solitions?

[lusty quest]

if this is an acute angle, then how would you know which angle to put in?

[trail minnow]

you have to remember that $\sin a = \cos b$ if $a$ and $b$ are complementary

[somber coyote]

hegel:

[trail minnow]

so what they are asking is to find some value of a where a and (a + 40) are complementary

so you can say that $a + (a + 40)$ is going to equal 90

[somber coyote]

hegel:

[trail minnow]

and therefore just solve for a

[lusty quest]

oh ok, so a would be 25 then

i see, thanks!

[oak hearth]

https://i.gyazo.com/5c0102f45a69bf7cadc875e984681b37.png

https://i.gyazo.com/7d600fff15b52e3c13b1079cf1df9a51.png

https://i.gyazo.com/ce418ee4c8f854b13412f6e957523eec.png

anyone know answers

[split compass]

How would you find the exact value of csc(8pi/3)?

cosecant of 8pi divided by 3

[dark sparrow]

are you able to find the exact value of sin(8π/3)?

@split compass

[split compass]

yeah i got it now

[dry phoenix]

Hi guys. I need some help with analytic geometry.

I have tangent y=6x-2√37 and a circle equation x^2 + y^2 = 4.

I need to find coordinates of the point where they collide. I tried just implementing tangent equation into circle equation but it gets retarded with numbers. Any other way?

[devout shell]

More like tedious with the numbers, but that’s what I would have done initially. Then the thing is choosing the correct x value.

Unless you get a zero discriminant

Then there’s only one value for x

[surreal bolt]

er are you sure it is a tangent?

The circle is centered at the origin so all points have a tangent with slope = -x/y

y = 6x - 2root(37) ... means the slope is 6 for the tangent

6 = -x/y or y = -x/6 and x^2 + y^2 = 4

[dreamy hollow]

I have 3 points in 2-D (a, b and c), in order, what does this equation do? Does it have a name? Thanks
calculation = a.x*b.y + b.x*c.y + c.x*a.y - a.y*b.x - b.y*c.x - c.y*a.x;

[gusty crater]

What's Ptolemy's theorem?

What?

[fringe dirge]

t!wiki Ptolemy's theorem

https://en.wikipedia.org/wiki/Ptolemy's_theorem

[frail rock]

I got both of these correct. I understand how I got Part A correct but for Part B I just guessed. I'm not sure why Part B is correct though.

[fossil ledge]

So the frame is 2 inches wide

Meaning from the Center it’s 10 inches away

[frail rock]

oh

[fossil ledge]

It’s still a circle, but the diameter increases to 10

[frail rock]

i kinda get it

[fossil ledge]

Diagrams are op op for geometry and trig questions

[frail rock]

sadly they didn't give any

and I couldn't quite picture it in my head

so thanks 😄

[fossil ledge]

Diagrams are half the battle, you only get better by practice, and no worries :)!

[covert orchid]

hey guys, is there any book that you'd recommend to learn trigonometry ?

[olive sinew]

how can i find both measures for a, when i dont even have enough information ? any suggestions ? hints?

[elder aspen]

Well there's 27 of them and they're equally distributed over 360 degrees, no?

[olive sinew]

yes, but i still dont get it, the way i see it is like a whole pizza with 27 seven equal slices with where theta = 2pi or 360dg

should i divide 360 by 27 ?

[covert orchid]

can someone explain me why both circles measure pi/4 for 45 degrees?

[mystic shadow]

the blue dotted line represents 45 degrees from the x or y axis

picture them as extending forever

you can draw a circle around the origin of any size radius but that doesn't change the blue dotted lines

[upper karma]

If u have a triangle like this, is the number on the outside always larger? The 3 in this case?.. sorry for the tard question xd

[tawdry pivot]

,rotate

[tawdry pivot]

3 is the largest out of 1,2,3 no matter the trinalg

(except for degenerate)

[upper karma]

Oh I see, thank you very much for your help!

[tawdry pivot]

wait you could just read the "solve" part

[upper karma]

I Didn't quite understand it, but now I understand it. Thanks!

[tawdry pivot]

np

[covert orchid]

About my photo, but why is it like that? that what i don't understand but myabe it has to do with 's = angle * r' ?

[zenith ember]

It's a -ratio-

The arc length gets bigger as the radius gets bigger, but the ratio stays the same.

[covert orchid]

oh proportionality ?

[zenith ember]

That's what the image says.

[covert orchid]

ooooooooh

okay

❤ thanks

[upper karma]

Just a quick question, why is measure 5 and 3 in part A not less than measure 8? And in part B why is measure 4 greater than measure 2? I though they were corresponding which would mean they’re congruent right?..

[covert orchid]

could someone explain me the t' thing?

that's about coterminal angles

[spark stag]

they're using t' to indicate the reference angle

which is the smallest angle an arc makes with the x axis

do you have a more specific question?

[covert orchid]

yeah, i mean why is that useful

oh i get

it basically helps me to get an angle like the one in Quadrant I even i have one like the one in quadrant IV

right?

[spark stag]

essentially.

the sine of an angle is the same as the sine of its reference angle, except the sign might be different (- instead of +)

same with cosine/tangent.

[covert orchid]

reference angle t' ?

[spark stag]

?

[covert orchid]

i was asking what is the reference angle, is it t or t'?

[spark stag]

your book is using t'.

sometimes its notated as

$t_r$ or $\theta_r$

[somber coyote]

Namington:

[spark stag]

where the subscript r stands for "reference"

but t' is what your book is using

[gentle fog]

Can somebody help me figure this out? I have no idea how to solve this

[spark stag]

so, we know cos(3theta) = -1. what angles make cosine = -1?

[gentle fog]

pi

[spark stag]

right, are there any others?

[gentle fog]

Well the cosine is the X value (and this is based off of the unit circle) so I don't think so

[spark stag]

well, theta has to be in the unit circle

but our angle is given by 3theta

if theta is between 0 and 2pi

what is 3*theta between?

[gentle fog]

0 and 6pi?

[spark stag]

right

so what angles in [0,6pi) make cosine = -1?

[gentle fog]

pi, 3pi, 5pi?

[spark stag]

right

so we know that the stuff in the cosine

3theta

equals pi, 3pi, or 5pi

so set up our equations:

3theta = pi
3theta = 3pi
3theta = 5pi

solve for theta

[gentle fog]

So the answer would be pi/3, pi, and 5pi/3 right?

[spark stag]

yes, that's our solution set.

and you can check these

,calc cos(3*pi/3)

[somber coyote]

Result:

-1

[spark stag]

,calc cos(3*pi)

[somber coyote]

Result:

-1

[spark stag]

,calc cos(3*5pi/3)

[somber coyote]

Result:

-1

[gentle fog]

Oh my god thank you so much. It's the same process for like other trig functions (considering sin and cos, no reciprocal functions)?

[spark stag]

yeah essentially

if you want you can think of it as

substituting x = 3theta

then the interval 0 <= theta < 2pi

x=3theta, so theta = x/3

substitute in

0 <= x/3 < 2pi

and solve for x

0 <= x < 6pi

this is the symbolic interpretation of what we were doing.

its not really necessary to think about it like that, if you can "visualize" it

but it might be helpful

[gentle fog]

Alright, thank you so much lol

[dry phoenix]

Hi

I need number of whole number solutions for this inequation in the interval [0,8]

i dont know how to implement this interval in this problem

When i start doing this inequation, i get that 3pi/4 < x < pi/4

but i dont know what to do after that

[supple abyss]

[dry phoenix]

I do not know how to use anything that could present that better

I m sorry for that.

it should be 135° < x < 45°

[supple abyss]

so you're saying that 135<45