geometry-and-trigonometry

[dull ice]

There is no triangle

Look at it this way

88° is proportional to 44 length

What angle would be proportional to 62

62 is greater than 44

But 44 gives nearly 90° angle

So 62 would give greater than 90

Much greater

And 88 plus whatever that would be, would be over 180

So no triangles exist

[oblique bough]

o_O

so i have 3 problems

[dull ice]

Ok

[oblique bough]

and they all say sinC is greater than 1

so that means there is no triangle

[dull ice]

All of them are no triangle

Yes

[oblique bough]

o_O

[dull ice]

In any problem, if you get sinC= >1

No solution

Or sin any angle

But double check your math whenever you get that

[oblique bough]

im pretty sure i used law of sine correctly in that photo right?

[dull ice]

Yes you did

a/SinA = b/SinB = c/SinC

Or

SinA/a = SinB/b = SinC/c

[oblique bough]

alright

thanks so much lmao i was stuck for a while

[dull ice]

Lol

Np

[granite atlas]

Didnt know where else to ask this extremely basic question.

Q) Does a rectangle HAVE to be 1x2 in dimensions? So next would be 2x4... 4x8 etc

Or does a 1x10 still count a rectangle

[dark sparrow]

no, a rectangle does not have to be 1 by 2

it just has to have right angles, as its name suggests

so yes, it can be 1 by 10

[granite atlas]

Cheers

Google wouldnt answer my question haha

[frozen kraken]

If I have a helix wrapped around a cylinder, why does the helix angle change if I reduce the circumference of the cylinder but maintain the length along the x axis?

[faint solstice]

any germans here? got a task i need to solve

[mint token]

here is my geometry question: when labelling a quadrilateral ABCD, do you label it clockwise or counterclockwise if it's not specified

[astral hornet]

doesnt matter as long as the letters are in consecutive order

[inner lantern]

Heres how we do it

Normal people: rectangle ABCD
Genius: rectangle alpha beta gamma zeta
GOD: rectangle A_1 A_2 A_3 A_4

[dark sparrow]

alpha beta gamma delta, you mean.

learn your greek alphabet smh 😤

[inner lantern]

Delta is for triangle my man

Or lowercase delta, thats fine

😉

[dark sparrow]

the next letter down is epsilon

[inner lantern]

Thats how to call !!!

Is also hard to write, rite?

[dark sparrow]

?

[inner lantern]

no, i will stick with zeta !

whats ur favourite greek character

[dark sparrow]

xi tbh

[inner lantern]

Psi for me

I dont even know how to pronounce that

Ps-ai?

[spark stag]

sie

technically og greek pronounced it psee

but who cares

[dark sparrow]

they still pronounce it psee

[inner lantern]

Ill just pronounce it "pssst"

[snow socket]

https://gyazo.com/7bc56e266694c1a6341e0473e2535357

any help ?

[twin prawn]

What happens when x = 0?

[snow socket]

not so sure

[dark sparrow]

when x = 0, what is y?

based on the graph

[snow socket]

wouldnt it also be 0

[twin prawn]

Look at the graph. Does the function value ever become 0?

[snow socket]

no

[twin prawn]

Exactly

So what's the value at x = 0?

[snow socket]

2

then

[twin prawn]

Correct

And what happens if you plug in x = 0 in your formula?

[snow socket]

2sin

right

[twin prawn]

Not quite

sin(0) = 0, so your formula would yield 1, and not 2, like we just concluded

It doesn't have to do with what number you multiply sin with (also called amplitude), in fact, that's the part you had correct

[snow socket]

im confused

: P

its a problem thats not gonna be on my test but she put it in my homework

[upper karma]

https://gyazo.com/19b59a506a2c895127c11b56bdaf6d58

I need help asap

is this correct

anyone??

[dark sparrow]

we'd rather you'd have shown your work instead of just hitting us with an answer

also, i'm really suspecting this is from an ongoing test

[halcyon coyote]

looks like one of those online tests tbh

[upper karma]

V = r² π h + 1/2 · 4/3 r³ π

[halcyon coyote]

also why do you need help asap?

[upper karma]

because Im tryna get this hw done

lmfao

i got work today

im falling behind bc my teachers don't care about me

[halcyon coyote]

i mean

the formula is correct

[dark sparrow]

,calc (22/7) * 8^2 * 172 + (2/3) * (22/7) * 8^3

[somber coyote]

Result:

35669.333333333

[halcyon coyote]

22/7

[upper karma]

fudge

[dark sparrow]

he was told to use 22/7 for pi

[upper karma]

^

[halcyon coyote]

oh

[dark sparrow]

i mean you dun goofed somewhere in the arithmetic ¯_(ツ)_/¯

your formula is ok

[upper karma]

https://gyazo.com/ba5a36bac7508a1ae0a6a2efc0ed3a70

how do i solve this

i forget the formula

:/

[dark sparrow]

...volume of cone minus volume of gum ball?

[upper karma]

C

0.07

[dark sparrow]

that's way too small

[pure warren]

I suppose this counts as geometry, this popped up and I cannot for the love of me figure out how to solve it

I know the formula but I just don't know how to apply it to this problem

[teal axle]

I’m getting 110/9, idk if it’s right

[mint token]

I got 8, from similar triangles ADC and ABE

[narrow jewel]

Hello

Can someone tell me how you'ed go about truning a basic Right angle equations awnser into one of the awnsers it gives

it's 45/36

but my only awnsers are : 4/3, 5/3, 5/4, 3/5

[dark sparrow]

how did you get 45/36?

that is definitely not the cosine of anything, since it's greater than 1

but in any case you should reduce the fraction

also, answer*

[narrow jewel]

I Was doin fine untill the awnsers changed, Cos is top over bottom

but it appers they want me to do something more

[dark sparrow]

"top over bottom"?

[narrow jewel]

simplify i guess

my teacher didn't explane this to me

[dark sparrow]

sin = opposite/hypotenuse
cos = adjacent/hypotenuse
tan = opposite/adjacent

commonly known with the mnemonic "SOH CAH TOA"

[narrow jewel]

i know im trying to remamber it

top over bottom is the newb way

[dark sparrow]

"top over bottom" says exactly nothing

cos = adjacent/hypotenuse.

what is the length of the side adjacent to angle X?

[narrow jewel]

36

i mean

hyp

hypotenuse

[dark sparrow]

no, the side with length 36 is not adjacent to X.

the side with length 36 is opposite X.

[narrow jewel]

adjacent whoops

next too

[dark sparrow]

read what i'm asking you.

generally, read what people say. it's a useful skill.

[narrow jewel]

X is next 27 witch is adjacent / hypotenuse

correct?

Witch are 27 & 45

[dark sparrow]

which $\neq$ witch

[somber coyote]

Ann:

[dark sparrow]

yes, 27 is the adjacent side and 45 is the hypotenuse.

[narrow jewel]

yes and angle X is 45

[dark sparrow]

no

you don't care what angle X is. all you're finding is its cosine

[narrow jewel]

Ahh my bad sorry man, ive been doin this for too long

The cosin is wasn't 45/36?

well if X was both adjacent & hypotenuse

then i guess they are the cosin

[dark sparrow]

no

X is an angle. it is not a side.

[narrow jewel]

is this right?

[dark sparrow]

adjacent = XY = 27
hypotenuse = XZ = 45
cos(X) = adjacent/hypotenuse = 27/45

yes that's what the mnemonic stands for.

[narrow jewel]

Okie im seeing it, but 27/45 isn't an option

[dark sparrow]

are you not able to reduce the fraction

[narrow jewel]

Well i've never done that before, do you mean simplify it by dividing by 5?

Doing that I got 5/9

[dark sparrow]

what do you mean you've never done that before. how are you doing trigonometry having never worked with fractions?

also, 27 is not divisible by 5.

divide by 9, not by 5.

[narrow jewel]

I know it was exactly 5.4

Can you go though the process and how me how you'd do it?

It might ring ma bell

Ooof

I probably know what it is but forgot it along the way up here

but hey it's clear what i gotta do now, i have a lead

So thanks for your time

Unless i reduce the fraction and there are more steps afterwords, i yi yi

[dark sparrow]

27/45 = 3/5.

[upper karma]

Can anyone help me on anyone of these?

<@&286206848099549185>

[dark sparrow]

,rotate 90

[hallow smelt]

can anyone explain how this factors? its a trig problem https://imgur.com/nuZh3UA

[spark stag]

do you mean how to get from the first line to the second?

$\frac{h-1000\cdot\tan{32}}{\tan{32}} = \frac{h}{\tan{32}} - \frac{1000 \cdot \tan{32}}{\tan{32}}$

[somber coyote]

Namington:

[spark stag]

do you see what we can do from there?

[hallow smelt]

ya thanks

[hallow smelt]

can anyone explain this https://imgur.com/nVQ9hwD

wouldn't dividing everything by h/100 leave the left side equal to 1

[cold plaza]

h divide by h/1000 is same thing as h multiplied by 1000/h, which gives 1000 @hallow smelt

[hallow smelt]

@cold plaza I dont understand how we go from line 1 to line 2

line 1 being 1000ft/H = 1/tan32 - 1/tan35

[cold plaza]

take the reciprocal of both sides

u get h/1000 = 1/(1/tan32 - 1/tan35), then multiply by 1000

[hallow smelt]

why are they doing that though

[cold plaza]

to solve for h

[hallow smelt]

H/1000 = tan32 - tan35 right?

[cold plaza]

no u cant flip it like that

it would be that if the RHS was 1/(tan32 - tan35)

but its 1/tan32 - 1/tan35, which is completely different

(try turning it into a single fraction to see for urself)

[hallow smelt]

@cold plaza sorry im just not understanding this LOL

are you able to write it out maybe and post a picture?

if you have the time I'm not understanding

[cold plaza]

imagine the first line was h/1000 = 1/(tan32-tan35). after flipping, what do u get?

,, \frac{1000}{h} = \frac{1}{tan(32) - tan(35)}

[somber coyote]

layovah:

[cold plaza]

if that were the first line, then after taking reciprocal of both sides, u'd end up with (h/1000) = tan32 - tan35 (what ur saying)

but thats not the first line. in this case we have smthn different

[hallow smelt]

wouldnt taking the reciprocal of 1/tan32 and 1/tan35 wind up the same?

[cold plaza]

,, \frac{1}{tan(32)} - \frac{1}{tan(35)}

[somber coyote]

layovah:

[cold plaza]

is not the same as

[dark sparrow]

\tan

[cold plaza]

,, \frac{1}{tan(32)- tan(35)}

[somber coyote]

layovah:

[cold plaza]

gget it?

basically, (1/x) - (1/y) is not the same as 1/(x-y)

[hallow smelt]

what is the recip suppose to be then?

[cold plaza]

it would be

,, \frac{1}{\frac{1}{tan(32)}- \frac{1}{tan(35)}}

[somber coyote]

layovah:

[cold plaza]

which is what they're writing in the second line

[hallow smelt]

okay

at least i see what they did now

thanks

[cold plaza]

np

[hallow smelt]

so you cant just cross multiply on line 1

tan32 * 1000 and tan 35*1000

to be clear

[cold plaza]

yes

(u cant do that)

[hallow smelt]

LOL

remembers how I failed to understand the recip

clarifies

thanks alot though

[cold plaza]

np

[upper karma]

just curious

if i have a cylinder but the height goes above the normal top (like the circle on top) and instead goes all the way up to a sort of 'arch', is there anything different i'd need to do there?

this is for finding SA and volume

[spark stag]

i have trouble visualizing what you're saying

do you mean its bent like an arch?

do you mean it caps off with some shape other than a circle (eg a dome)?

[dark sparrow]

@upper karma got a picture?

[upper karma]

Unfortunately not. But let me try and draw one

kinda badly drawn but

this http://prntscr.com/n4hppu

the arch extends upwards

it just cuts off

@dark sparrow

[dark sparrow]

...what's the shape of the arch?

is it a spherical dome or what

[upper karma]

Yeah

[dark sparrow]

...well then there you have it don't you

you've got a cylinder with a hemisphere stuck on top

[upper karma]

Oh now I feel stupid

Lmao, thanks

[hallow smelt]

why cant i say y = cos45*600

then sin 45*y = H

[keen aspen]

what

cos(45) would be (600-d)/y

[dire rampart]

600 is the base of the whole thing

that triangle isnt right angeled

[hallow smelt]

so you can only SocCahToa right angles?

[keen aspen]

yes

[hallow smelt]

oh dam did not tell us that

[keen aspen]

since its not right angle you have to use law of cosines or sines

with the information given, you can find y and x with the law of sines

[hallow smelt]

does the big triangle have to equal 180?

to get the angle for the 600

[keen aspen]

I dont know what you mean by that

[hallow smelt]

I mean the Y-X-600M triangle

the third unknown angle is it just 180-45-75

[keen aspen]

yep

now apply law of sines

[hallow smelt]

ehhh

got the right answer =]

H = 473

[keen aspen]

epic

[hallow smelt]

can i assume that on the other side of the dotted red line at the bottom

is another 90 degree angle

[keen aspen]

other side?

[hallow smelt]

the H dotted red line

you see how one side is 90 degrees

can i assume the other is also 90

[keen aspen]

tyes

[hallow smelt]

thanks

I should try and find some time to retake geometry

do you see a simple way of solving this?

I already solved it correctly, but i did SohCahToa

and it was along process..

[frozen kraken]

180-35 is 145
So first angle is 3

Then with 1000 feet you can find the side of that triangle

If 1000 feet equals 3 degrees difference, then d is 11666.6667

Well I guess that’s of angle of h/d is 90

[vernal tusk]

i don't have a question, but i did something neat that one of you might find cool

had a geometry quiz where it asked to find the equations of all the tangent lines of a circle

like, every y =

which is impossible, because there's infinite

and the teacher said nvm only do 4

but i thought it meant a general rule, so i did one before erasinf

wasn't actually right, but i fixed and completed it this morning

[fleet wolf]

you can find the slope of the tangent line at every point on the circle

[vernal tusk]

y = ((h-x1)/sqrt(r^2-(x1-h)^2))(x-x1)+k+sqrt(r^2-(x1-h)^2), then you just negate both sqrts to get the negative

yeah, that's what i did

[fleet wolf]

then use where it meets the circle to find y intercept

[vernal tusk]

exactly

i'm not asking a question, again

i solved it

it works

just thought it was interesting

[fleet wolf]

ok

[vernal tusk]

used slope and point slope form

[supple abyss]

it's good that you're having fun

[vernal tusk]

get to drag it around and it's fun

yeee

[dire rampart]

no fun in maths

[fleet wolf]

shh before i wack u with baguette

[frigid wasp]

i havent learned about hyperbolic functions but it tanh(x) =

[hallow smelt]

When I plot the function 400sincos that shit looks crazy as fuck and not like this picture

why does theirs not look insane

[spark stag]

@hallow smelt window settings?

also make sure you're in radians

if you're graphing in radians, on the interval [0,pi/2] with range [0,300] it should look the same

as such

[supple abyss]

this might just be my eyes being fucked up

but their pictures does't look symmetrical to me

[spark stag]

ah yeah, good point

fsr it also goes above y=200

i think it was hand-drawn

still looks the right general shape though

[hallow smelt]

Oh shit how did you decide to use range 0 to 300?

I wasnt using desmos was using my ti-84

[spark stag]

trial and error works, or you can reason

at pi/4, both sin and cos have the same value: 1/sqrt(2)

multiply those together and you get 1/2

[hallow smelt]

is there a way to get the ti-84 to give me pie/4 as a value instead of decimals?

[spark stag]

so you need to have, at least 1/2*400

which is 200

so i overshot that a bit to 300

again, early estimates arent always perfect; use trial and error

no, the 84 won't give exact values when you trace a graph, im afraid

but you can find the y coordinate (200) and reverse-engineer it

200 = 400sin(theta)cos(theta) =>
1/2 = sin(theta)cos(theta)

which you might notice works if theta = pi/4

although whenever you get a weird decimal in a trig problem, its probably worth a quick check to see if it's pi/2, pi/4, 1/sqrt(2), etc

[hallow smelt]

ya that makes sense

the whole point of this is so I can get the dimensions of a cross section

is everything i colored a cross section?

[hallow smelt]

can you cut ABE in half to make a 90 degree angle and cut 2m to 1?

[flint compass]

I was watching a video titled: Sine & cosine identities: periodicity on Khan Academy. In the video, he draws a right angle triangle in a unit circle and then makes another one by adding pi/2 radians to the angle theta. He then says how Cos theta = Sin(theta+pi/2) . I don't understand how the purple line is equal to Sin(theta+pi/2)

[upper karma]

Given area of rhombus and the angles how do u find the side lengths?

[supple abyss]

@flint compass

also this

[flint compass]

@supple abyss I can't read the first drawing and I don't understand what the second drawing proves.

[supple abyss]

how about now

[flint compass]

How’s it equal to sin of theta plus Pi/2 tho? I still don’t get why it works out that way.

[supple abyss]

congruent triangles

[flint compass]

Right but they’re both theta not theta + Pi/2. How does that relate?

[supple abyss]

the blue angles is pi/2 -theta because it's on a straight line with pi/2+theta

making the blue angle pi-(pi/2+theta)=pi/2-theta

the yellow lines are both of length 1 as they are radii of the circle

and the red angles are both pi/2 as they are right angles

this is enough for the triangles to be congruent by Angle Angle Side

[flint compass]

What do you mean by “the blue angles is Pi/2 - theta because it’s in a straight line with Pi/2 + theta?

[supple abyss]

in this picture for example

ignoring the degree sign

a=pi-b

due to this

[flint compass]

Yeah

So is the x axis the line?

[supple abyss]

yeah

[flint compass]

Ok, so then how does that make the line sin( theta + Pi/2)?

Wouldn’t it be sin(Pi/2-theta)?

Oh, I get it

[supple abyss]

well it's actually both

[hallow smelt]

How did they get 138.2 degrees as a second answer???

[vagrant elk]

well, it kinda uses how sin x is periodic

If you were to plug that number into your calculator, you'd see it ends up right about 2/3. Just looking at a graph for a sine wave, you see it's symmetrical around the theta = 90* line (if you consider just from theta = 0 to 180) and it uses that symmetry to go backwards across in a sense

[hallow smelt]

so how do i easily find the 138.2 answer?

[vagrant elk]

you use sin ^-1 as provided by your calculator

and then 180-that because it's symmetrical you just do that to "work backwards"

[hallow smelt]

why does the following question only have 1 answer?

[vagrant elk]

Cosine doesn't have the same symmetry

since it starts at one and goes to -1 in that same interval

[hallow smelt]

so then sine and csc are the only ones who are going to have two values 0-180

or 0 to pie

[vagrant elk]

Cos definitely doesn't have the same niceness

[hallow smelt]

is there a reason they chose to use COT instead of Tan ?

[vagrant elk]

not as far as I can see at first glance other than to introduce them all

[hallow smelt]

sometimes they do random stuff and i have no idea why

let me see if i can find an example from earlier today that confused the shit out of me

[dark sparrow]

mb because they consider cot^-1(s/50) to be less intimidating than tan^-1(50/s)

[vagrant elk]

i don't get it

[hallow smelt]

how was i suppose to know to add K there

[supple abyss]

ye

[dark sparrow]

proportionality constant

[supple abyss]

[hallow smelt]

I have never used that explicitly in my life

that i know of

[supple abyss]

although the k makes the 8000 redundant

[hallow smelt]

so K and 8000 are the same thing?

[supple abyss]

no

an arbitrary constant multiplied by 8000 is still an arbitrary constant

[hallow smelt]

is that something most teachers would want us to signify is there?

the K

[supple abyss]

if you don't put it there it's just wrong

nothing about your teacher

[hallow smelt]

oh

ya so then it does matter

this is what im taking away from this whenever I see the keyword proportional there is a constant there

[supple abyss]

ye

[hallow smelt]

I just went through thermodynamics in chem a shit ton of proptionalities

I dont think we signified there being constants in them

like Delta E = Delta T but no constant there

[supple abyss]

$\Delta E∝\Delta T$ or $\Delta E = k\Delta T, k \in \mathbb{R}$

[somber coyote]

CaptainLightning:

[hallow smelt]

LOL

what is the constant then

do you just throw planks in there

lmao

[supple abyss]

specific heat capacity and mass

product anyway

[hallow smelt]

the mcdelta t thing?

[supple abyss]

ye

[hallow smelt]

i hope this is not fake news

or im gonna sound dumb as shit on monday

-_-

hahaa

thank you though good to know

[hallow smelt]

My Answer for A ) is ASA, SSA and SAS, My Answer for B) is SSS

is that right?

[dark sparrow]

is it possible to have two triangles with the same side lengths but different angles?

[upper karma]

I need halp

Help*

You'd know the answer if you knew what a rigid transformation was. Google it.

im slow with alot

Alot isn't a word.

[hallow smelt]

21 savage said it was

[hallow smelt]

on the last line why is cos^2 negative??

[dire rampart]

sin^2-1=-(1-sin^2)=-(cos^2)

[hallow smelt]

oh dam I knew it was going to be something obvious like that fuck

1-sin^2 = cos^2 not sin^2 - 1....

thanks

[raw prism]

<@&286206848099549185> could I get some help with question 9

[dark sparrow]

what's that pencil-drawn squiggle on top meant to be

[dark sparrow]

well what have you tried so far for that one?

[split raven]

How can I simplify sin(270 + A)?

[dire rampart]

sin(270+A)=sin(180+A+90)=-sin(A+90)=-cos(A)

[honest bay]

$\frac{cos(2a)+cos(2b)}{sin(a)+cos(b)} = 2cos(b) - 2sin(a)$

[somber coyote]

mohad12211:

[honest bay]

how do i prove this?

$\frac{2(sin^2(a) + cos^2(b))}{sin(a)+cos(b)}$

[somber coyote]

mohad12211:

[honest bay]

I tried this but i am stuck now

[dire rampart]

what

how did u get that numerator @honest bay

[honest bay]

$cos^2(a)-sin^2(a)+cos^2(b)-sin^2(b)$

[somber coyote]

mohad12211:

[honest bay]

then

[dire rampart]

ok then?

[honest bay]

$1-sin^2(a)-sin^2(a)+cos^2(b)-1+cos^2(b)$

[dire rampart]

[somber coyote]

mohad12211:

[dire rampart]

I'm not even sure the original thing is true

[honest bay]

well

is this the question

[dire rampart]

,w is ((cos(2x)+cos(2y))/(sinx+cosx) equal to 2cosy-2sinx

[honest bay]

lol

[halcyon coyote]

[dire rampart]

gimme sec lemme try the question

[honest bay]

$-2sin^2(a)+2cos^2(b)$

[somber coyote]

mohad12211:

[honest bay]

$\frac{-2(cos^2(b)-sin^2(a))}{sin(a)+cos(b)}$

[somber coyote]

mohad12211:

[honest bay]

should be like this

[halcyon coyote]

tho now you can use

(a²-b²) = (a+b)(a-b)

[honest bay]

Ohhhh

[dire rampart]

that gives the minus of what u want tho

[honest bay]

well it is solved now

[dire rampart]

no what u have is -2(cosb-sina)

[honest bay]

yeah

[dire rampart]

u want 2cosb-2sina

theres a minus that shouldnt be there

[honest bay]

hmmm

yup

[dire rampart]

oh u factored wrong

[honest bay]

yeaaah

i see

[dire rampart]

rip

[honest bay]

lol

thanks

[dark sparrow]

$\cos$, not $cos$

[somber coyote]

Ann:

[dark sparrow]

[dire rampart]

[golden arrow]

Hi I'm trying to solve for alpha in the equation I have just sent, I feel like I've come up with a solution but the values arent checking out, can anyone help?

[carmine pasture]

000hubibibhiuhiuh

[upper karma]

I need help

7-9

@lusty kernel

[knotty falcon]

I may be misreading it but 7 is just simply divided by t

And the other two are just rearranging the equations

C= 5/9(F-32) becomes $9C/5 = F-32$

[somber coyote]

Jimbo:

[knotty falcon]

And then you add the 32 to both sides and then you can write the equation as

oof, sorry I'm new to this. xD

[somber coyote]

Jimbo:

[raw wren]

Anyone here good at verifying identities? I’m desperate

[umbral snow]

Nope. Not a single person here

We just don't know high school math sorry

[raw wren]

I’m in college :(

[umbral snow]

@raw wren
Or college!
Feel free to post it I was kidding. In the future just ask, don't expect people to prompt you.

[raw wren]

Oh I’m stupid lol thank you

Okay I’m stuck on the first question. I know the answer but I’m confused on the steps. It’s verifying identities

[supple abyss]

use $\sin^2\theta + \cos^2\theta \equiv 1$

[somber coyote]

CaptainLightning:

[raw wren]

When do I use that formula? I tried working on the right side first

[supple abyss]

use it on the left

[raw wren]

Okay thank you

[supple abyss]

np

[spark stag]

Why do you use \equiv

[eager mesa]

identity

[spark stag]

[slate pewter]

hello friends

i didnt cant get that

[eager mesa]

what you solving

[slate pewter]

you can know the both in the tringle

but i need to know (D) (F) and (E) Angels

[eager mesa]

cant be determined i think

angle FDE

can vary depending on lengths FE

even when angle DFE is 126

think about it if point E is close to F, the angle FDE will be very small

E goes to very far, angle FDE will be bigger

same goes for angle FED i thnk

[slate pewter]

i think its like this

becuse FDE

is cute it by the half

so its need to be the other angle is will be equal to it

so i think its like this

[solemn harbor]

Geometry is fascinating, I mean, at least the very basic Euclidean plane geometry course that I took. But surprisingly simple things are tedious to prove.

Not as fascinating as abstract algebra though.

[lofty scaffold]

I need help at #help-2 thank you.

[upper karma]

Stop breaking rules

Geometry is fun

Without geometry we'd likely have no calculus and then we'd be boned

Don't remember what an orthocenter is tho

[spark stag]

Hey I need help in #changelog

Thanks

[upper karma]

https://i.imgur.com/8Y2kDyO.png

https://i.imgur.com/D4xL5jI.png

So, this might be a dumb question, but up to this point, I have been factoring to find the correct phase shift, but when I factor this equation I get 1/2(4x-pi) not pi/4, am I factoring incorrectly?

[spark stag]

why would you factor out 1/2?

you want the coefficient on the x to be 1

pi is a constant, it can be whatever

[upper karma]

I see, thank you.

[spark stag]

yeah, so you factor out 2

not 1/2

[swift thistle]

whats the logic of ?

[dire rampart]

24=4 times 6

[swift thistle]

yeah but why 2 sqrt 6?

what is 2 doing there

[dire rampart]

$\sqrt{24}=\sqrt{4 \cdot 6}=\sqrt{4} \cdot \sqrt{6}=2\sqrt{6}$

[somber coyote]

lemon catto:

[swift thistle]

oh...

thanks a lot @dire rampart

[upper karma]

https://gyazo.com/1ab762da04309a00fb2c8eb423a17d55

Is my explanation correct?

Please @ me 😄

[limpid basin]

You need to explain it in terms of math

[dense tundra]

does anyone know how to do this?

i solved for y and got y + 4 = t then i just used x and substituted y + 4 for t in x

[upper karma]

looks right

unless they want you to solve for y= something

and expanding it would help

[dense tundra]

yea i got it, i just had to expand it and set it equal to x = (answer)

[heady juniper]

Hi all

i am trying to study for my maths exams

and i am in need of circular geometry questions

if anyone has some good ones

please message me

[flint compass]

Is there a formula to calculate cosine?

[gritty siren]

adjacent / hypotenuse?

[flint compass]

Isn’t that only for a right angle triangle?

When I do cosine of angle 45 how does my calculator know the ratio/correct answer?

[gritty siren]

ah, outside the context of elementary geometry, there is more stuff

there are algorithms to approximate the values of cosine and other stuff

[flint compass]

Ok.

[gritty siren]

t!wiki cordic

[loud cedar]

📖 | ** https://en.wikipedia.org/wiki/CORDIC **

[gritty siren]

found it

so, this set of algorithms is probably the one that most calculators use

[flint compass]

Ok, I’m also confused on why cos(x) has less oscillations then cos(4x)

[dark sparrow]

cos(4x) oscillates four times as fast

[steady sleet]

lul

Taylor tho

[celest wave]

how do i figure out c when i know a and x?

[dark sparrow]

you know a and x but nothing else?

[celest wave]

yes

[dark sparrow]

write out the pythagorean theorem for every right triangle in this problem

see what you can get

[celest wave]

i cant really do much with it as i only know these two

[dark sparrow]

no i mean you've got all those other variables to play with

in the picture

do it in terms of those

[celest wave]

i dont understand what im supposed to do with the unknown variables

[vague pagoda]

trig

[dark sparrow]

express them in terms of known variables, and potentially each other

this can be done trig-free

$\begin{cases} a^2 + b^2 = c^2 \ x^2 + h^2 = b^2 \ h^2 + y^2 = a^2 \ x + y = c \end{cases}$

[somber coyote]

Ann:

[dark sparrow]

@celest wave does this make sense to you

[celest wave]

i understand these

but do you want me to make up values for the unknown variables?

[dark sparrow]

no

i do not

immediately you can see that you can take the first and second equations and combine them into one: $a^2 + x^2 + h^2 = c^2$

[somber coyote]

Ann:

[dark sparrow]

and since you don't care about b you might as well just forget x^2 + h^2 = b^2

$\begin{cases} a^2 + x^2 + h^2 = c^2 \ h^2 + y^2 = a^2 \ x + y = c \end{cases}$

[somber coyote]

Ann:

[celest wave]

i cant really use any of these yet as i dont know h or y

[dark sparrow]

yeah, but again, you can eliminate another variable you don't care about

in this case, h

h^2 = a^2 - y^2

$\begin{cases} 2a^2 + x^2 - y^2 = c^2 \ x + y = c \end{cases}$

[somber coyote]

Ann:

[celest wave]

i still dont know what i am supposed to do with this as i dont know y

[dark sparrow]

you don't care about y, so eliminate it

y = c - x

[celest wave]

i mean i do care about y as y is needed to figure out c

[dark sparrow]

you don't care about y in that you don't need to find the value of y

and you can eliminate it

$2a^2 + x^2 - (c-x)^2 = c^2$

[somber coyote]

Ann:

[dark sparrow]

et voilà, an equation in a, x and c only

[celest wave]

lets say a = 9 and x = 3

then would something like 14.316 sound reasonable

[dark sparrow]

over 10000 might be just a little too much

[celest wave]

those are decimals

[dark sparrow]

but why do you need to replace a and x with specific values and give me approximate solutions when it's possible to solve for c directly?

[celest wave]

those values are the ones i had

and i dont know what you mean by solving c directly

[dark sparrow]

$2a^2 + x^2 - c^2 + 2cx - x^2 = c^2$

[somber coyote]

Ann:

[dark sparrow]

$c^2 - xc - a^2 = 0$

[somber coyote]

Ann:

[dark sparrow]

$c = \frac{x + \sqrt{x^2 + 4a^2}}{2}$

[somber coyote]

Ann:

[dark sparrow]

like that

[celest wave]

that results in 19.562

[dark sparrow]

,calc (3 + sqrt(3^2 + 4*9^2))/2

[somber coyote]

Result:

10.624143795447

[merry rapids]

Is anyone able to help me with this question please? We are meant to find the surface area however I can't seem to do it. The answer is 156cm however I somehow get 149cm doing: (7x3x2)+(4x3x2)+(7x4)+(1/2x7x5x2)+(1/2x4x5x2)

[dark sparrow]

ok first off don't use x for multiplication

second, 5 m is not the height of any of those triangles

[merry rapids]

Is 5m the slant height?

[dark sparrow]

no

[merry rapids]

The perpendicular height of the pyramid?

[dark sparrow]

yes

[merry rapids]

I want slant height right? So would I do Pythagoras (5^2+3.5^2...square root ans = 6.1 ) to get the slant height?

[dark sparrow]

the pyramid doesn't have a square as its base, and so does not have a single "slant height"

you need to find the height of each one of its triangular faces

[merry rapids]

Sorry my bad, but wouldn't I still do Pythagoras in order to achieve that?

[dark sparrow]

i mean yes you'll end up using the pythagorean theorem

[merry rapids]

Thanks

[vital frost]

🙏

Need help

[upper karma]

@vital frost

[vital frost]

yo

[upper karma]

In [3] solve for the other angle of that triangle

First do that

[vital frost]

could u please do it

im completely lost

[upper karma]

You first need to find what t is

[upper karma]

Hmm

@vital frost I think I found a way

[vital frost]

ok

@upper karma what is it

[upper karma]

@vital frost angle b is a right angle because it intercepts a semicircle

So t=8

Then plug 8 into (4t+1)

[vital frost]

how do u know t =8

[upper karma]

Because angle b is a right angle

So (7t+1)+(4t+1)+90=180

Then you do the algebra and get t=8

@vital frost

So then you plug 8 into (4t+1) to get 33°

And since the central angle is twice the inscribed angle, the central angle is 66°

[vital frost]

so 7t+1 would be 57

[upper karma]

Yes but for the measure of AB you only need 4t+1

Which is 33°

The angle measure of the arc equals the central angle, which is twice the inscribed angle @vital frost

So the angle measure is 66°

[vital frost]

is that what the question is asking me>

[upper karma]

I think so

Wait

[vital frost]

Solve for the arc length AB is segment AC is a diameter of the circle measuring 8 yrds

[upper karma]

We found the angle measure of the arc so now we can calculate arc length

So the radius is 4 yards

[vital frost]

mhm

[upper karma]

Use the arc length formula for 66° and 4 yards

@vital frost got it?

[vital frost]

uh no

[upper karma]

So you have the measure of arc AB, right?

[vital frost]

yes

66

[upper karma]

And you have the radius

Take 66/360

Multiply it by 2×pi×4

[vital frost]

to get this?

[upper karma]

Yes

Now round to whatever it tells you to round to

[vital frost]

doesnt say

[upper karma]

I would round to the hundredth

Now for [2]

@vital frost

[vital frost]

yo

[upper karma]

Now to solve [2]

[vital frost]

ok

[upper karma]

Notice how LMOP is a quadrilateral?

[vital frost]

yes

its a kite

[upper karma]

And quads have a total measure of 360°?

[vital frost]

yes

[upper karma]

So rays LP and LM are tangent to the circle and segments MO and OP,respectively, are perpendicular to these rays

If an two lines are perpendicular, the angle measures are 90°

So 360-165-90-90 will give you angle L

L is 15°

[vital frost]

ok

is that all u ahve to do ?

[upper karma]

Yes

[vital frost]

how about the circumscribed angle method

[upper karma]

Okay

So the measure of a circumscribed angle is supplementary to its arc

So the arc MP=165°

180-165=15

[vital frost]

alright

on to [1]

[upper karma]

I helped with what I could

My geo skills are getting rusty

[vital frost]

lol

it's alright

[upper karma]

Fixing them isn't my top priority right now

My top priority is calc and linalg atm

[gloomy python]

I need help finding x and y

I know what y is

I put y=45 is that right

[astral hornet]

as per markings all sides are congruent?

if y is the measure of that angle then yes its 45

and 45-45-90 ratio to find one leg

and since all sides are congruent you now have the length of all sides

[upper karma]

Yeah u use 45-45-90

[upper karma]

Hello

I'm new.

So, I am currently trying to learn Geometry

Please tell me topics that I might see in Geometry class, 😄

t!yt What is geometry

[loud cedar]

📽 | ** https://www.youtube.com/watch?v=9dNIrj7w-LE **

[upper karma]

Thank you!

😄

[upper karma]

So

who can solve this:

Hint:Below 100

Hint: 180-(38+71)=x

Hint:180 - 109=x

[dark sparrow]

this is triangle

[upper karma]

yea

[dark sparrow]

what do you know about the angles in a triangle

[upper karma]

Hint: 71=X

[dark sparrow]

why are you posting these hints

what do you want

[upper karma]

I'm bored

[dark sparrow]

go shitpost elsewhere

[upper karma]

it is not shitposting Btw

[eager pendant]

which geometry are we talking about

[upper karma]

basic

Solve this

[supple abyss]

I didn't know the answer was considered a hint

[upper karma]

Lul

I just got bored

[dark sparrow]

i don't think i'd be lying if i said 95% of the active population of the server would consider this too trivial to be of any entertainment value whatsoever

[upper karma]

Capitalize your "I" "smart person" and also add a punctuation mark.

[supple abyss]

[eager pendant]

just 95?

[dark sparrow]

look who's talking, mr/mrs/mx doesn't-use-commas

[mild cargo]

this is math server not english grammar server lol

[upper karma]

*This is a math server not a english grammar server, lol.

[dark sparrow]

*This is a math server, not an English grammar server, lol.

[upper karma]

C RAP

[dark sparrow]

i can outgrammar you just as easily

[upper karma]

ok I'm a 10th grader

JESUS

[dark sparrow]

if you wanted to flex you should've prepared to get flexed at

[gritty siren]

is math an actual word ?

[upper karma]

yes

it is a shortened word

for mathematics

but yes'

it is a real word

[eager pendant]

maths

[upper karma]

@dark sparrow *If you wanted to flex, you should've prepared to get flexed at.

[dark sparrow]

i know how to punctuate Properly™, thank you

this isn't the kind of environment where one would normally put in the effort to do that

[upper karma]

Well, why did you capitalize the "p"and add a TM

[dark sparrow]

for emphasis

in some sense, anyway

[upper karma]

ok I am here for geometry I give up

[dark sparrow]

if you have a geometry problem you're stuck on, post it

[upper karma]

ok

[dark sparrow]

instead of, yknow, being a prescriptivist dick

[upper karma]

ur the dick

[dark sparrow]

[upper karma]

well stop yelling

[dark sparrow]

yelling?

[upper karma]

well shit

JUST GET BACK TO MATH!

[vague pagoda]

@upper karma thats just trig and pythagorean theorem

[dark sparrow]

lmfao what

cotton don't overcomplicate things lmfao

[vague pagoda]

OH LOL

[dark sparrow]

the angles in a triangle add up to 180°

that's it

[vague pagoda]

lmao ik

I thought like

sides

[upper karma]

The missing side it 109

*is

[supple abyss]

U wha

[vague pagoda]

no its not

[upper karma]

oh wait

[vague pagoda]

divide the triangle into 2 right triangles

🤔

71+19 = 90

[upper karma]

its 71

[stuck iris]

what is going on lmao

[vague pagoda]

right triangle

[upper karma]

Its right

[vague pagoda]

and the other right triangle is the same

so 71

[dark sparrow]

cotton

[supple abyss]

You literally assumed it was isosceles while doing that

[vague pagoda]

what

[dark sparrow]

you really are overcomplicating everything

so

so

much

[vague pagoda]

why

[mild cargo]

No @supple abyss

[vague pagoda]

why cant u overcomplicate things

[dark sparrow]

38 + 71 + x = 180; what more does one need?

[mild cargo]

You make a right triangle, and then do 180-90-71 to get the final angle, which just happens to be half of the 38 angle

[vague pagoda]

idk nothing?

[hallow smelt]

why is tan-1(-20) possible?

doesnt that mean the opposite over adj is a negative? how can distances be negative

[plucky marlin]

trigonometric ratios can extend beyond geometric interpretation

it doesnt always have to be "opposite over adjacent"

u can extend the definiton of tan(x) using unit circle

[hallow smelt]

they wanted me to calculator that out and then do 180-that for the angle

isnt 180 - (-w/e tan-1(20) a positive number over 180?

[plucky marlin]

what ?

[hallow smelt]

the value of tan-1(-20) is -1.5

x180/pi = -87 degrees

wouldnt 180 - (-87degrees) be over 180?

thats what im saying

[plucky marlin]

yeah 180 + 87 > 180

[hallow smelt]

but they ignored the fact that its negative

they just did 180 - 87

whyyy

[plucky marlin]

what exactly are you doing ?

context might help

[hallow smelt]

ill ss the solution sec

I get that they even/odd propertied it out right

wouldnt it be 180 - (-tan)

[plucky marlin]

do you know in which quadrants is tan(x) negative ?

[hallow smelt]

2/4

[plucky marlin]

yeah

in this case your theta is between 0 and 180

so only the 2nd quadrant sits in there

[hallow smelt]

so Q 2

[plucky marlin]

so ur theta cant be more than 180

hold up

lemme draw it

[hallow smelt]

ah i see where you're going with this

that was super helpful

[plucky marlin]

https://prnt.sc/n8peql

yeah okay ignore my poor drawing

[hallow smelt]

no man that was really good

wait

but why wasnt it Q4 again?

because starting at 0 we could go negative

and it would be slightly before the -90 degree mark

oh wait

nvm nvm

they restricted it to 0 - 180

[plucky marlin]

yeah

[hallow smelt]

this is not the first problem I have done with this thing where they ask for angles between 0-180

I never was picturing the unit circle

good tool thanks

[plucky marlin]

np

[hallow smelt]

can you show me where I went wrong here

thats better

im suppose to solve for solutions from [0, 2pi)

[young hemlock]

can someone help with the first one

[umbral snow]

The first one is three quarter circles each with radius 7

[little osprey]

Assuming they are quarters

Yeah Baynex got this

[umbral snow]

Implying they are quarters

[little osprey]

lmao

[young hemlock]

lol

[little osprey]

please

ignore me

[young hemlock]

3 quarter circles

got it

[umbral snow]

No. Now you have to have a very large presence here

[young hemlock]

ok

um

[little osprey]

He's joking

[young hemlock]

so is anyone willing to help or

[upper karma]

Well he told you 1

3 is a quarter of 6 and a half each of 3 and 12

And 5 is half of 6

[little osprey]

what?

[young hemlock]

what

[little osprey]

Sorry ignore Kronk rn 😂

So ok

On 1

You have 3 quarters right?

[young hemlock]

yes

[upper karma]

[little osprey]

So to find the arc length what do you need?

[upper karma]

Don't ignore me

I told u answer

[little osprey]

Wait you're confusing me Kronk

[young hemlock]

@little osprey theta?

[little osprey]

Uh yeah, and what is theta?

[young hemlock]

the angle

[little osprey]

What is the angle in between the sides of a quarter

[young hemlock]

90

[little osprey]

Yes

Do you know the formula to find the arc length?

[young hemlock]

um

theta by 360 times 2 pi radius

[little osprey]

yeah, that's for a circle

so in a quarter

[young hemlock]

oh

[little osprey]

you would replace the theta with 90