geometry-and-trigonometry

[dark sparrow]

what is this for

[strong chasm]

I can work with geometry

Are you talking about finding the endpoints?

[daring zealot]

Sorry for the confusion, I think I've figured it out.

[strong chasm]

Well I'm curious... What were you looking for?

[daring zealot]

Originally I was trying to project that yellow curve around the circle given the angle and radius.

Basically trying to identify the 3 borders of that yellow slice

So I could then fill the inside yellow

Instead what I'm doing is calculating the two lines on the sides and filling between them so long as the point is closer (or equal to) to the center of the circle than the radius

[daring zealot]

Okay, I have one more question. How can I calculate the purple point's (x, y) values given the length of the yellow line is 50, the slope of the black line is 1, the angle between the two lines is 25 degrees?

Also the yellow and black line intersect at the origin

Using this to rotate the point:
x′=xcosθ−ysinθ
y′=ycosθ+xsinθ

Would θ have to be -25?

[umbral snow]

@daring zealot
Yes that works

[daring zealot]

Thanks for checking, one more question. How can you calculate a point's coordinates given only its slope and distance from the origin?

Let's say... length is 50 and slope is 2/3

[umbral snow]

A line with slope m is:
y = mx

Length away from the origin is given by
d = √[x² + y²]
d = √[x² + m²x²]
d = x√[1 + m²]
x = d/√[1 + m²]
@daring zealot

So in your case, x = 41.6

y = (2/3)(41.6)
y = 27.7

[daring zealot]

Great! Thank you !

[storm rover]

how do I price that two lines are parallel

[storm rover]

Prove*

[sand marsh]

this book any good on planar geometry? http://djm.cc/library/Plane_Geometry_Wentworth_Smith_edited.pdf

[trail minnow]

@storm rover if two lines have equal slopes then they're parallel

are you doing analytic geometry proofs?

[sand marsh]

any good books for planar geometry?

[stoic arrow]

Can someone identify this shape for me?

And how to find the volume

[sand marsh]

^ think you have to put that in questions sections. this discord's rules pretty strict lol

im not sure though im new here

[trail minnow]

@stoic arrow @sand marsh we have 10 separate questions channels under the category "math help: open"

you have to find one that no one is using and post your questions there

if no one responds after 15 minutes you can ping helpers

[storm rover]

@trail minnow oh no

I just was curious

[trail minnow]

analytic geometry is pretty fun

but anyway yah prove that they have equal slopes

two lines are perpendicular if they have slopes which are each others negative reciprocals

and they bisect each other if they share a midpoint

[sand marsh]

@trail minnow thanks. im just trying to suck all the info I can from this discord. it's a means to an end. :p

[trail minnow]

[spring wave]

hello

there is this confusing problem

"The tangent of the parabola y = ax^2 at x=p intersects the x axis at ________"

[half gull]

whats the equation of the tangent?

[spring wave]

it actually looks easier now that I translated it into English

@half gull that's all that's provided in the problem

[half gull]

you have a formula for that haven't you?

[lavish phoenix]

So how would I go about finding x here? I know the area is 168cm^2

[half gull]

@lavish phoenix what's the arena in terms of x?

[trail minnow]

dfd;flg

lol this isn't where you put questions either haha

[lavish phoenix]

blz maan I rlly need to know

[trail minnow]

lol its ok but in the future

[lavish phoenix]

Thank you sexi boi

[trail minnow]

put them in one of the questions chat

anyway you can split the shape up into two smaller rectangles

one with sides x and 4

and another with sides x+6+2 and 6

you can rewrite this so that 4x + (x+8)6 = 168

10x + 48 = 168

then solve for x

[lavish phoenix]

Thank you hegel you are the fat sexy right now. big thank

[trail minnow]

[spring wave]

@half gull i am sorry but I don't get it

I don't have any formulas

[half gull]

does y=f'(p)(x-p)+f(p) ring any bell?

[spring wave]

nope

first time seeing it most likely.

@half gull but wait a second

first of all

according to my textbook, the general form of a parabola's equation is this:

(y-k)^2 = 4p(x-h)

(or the other one for that matter)

I have seen "p" being labeled "a" in other sources

so I don't know what "a" stands for in this problem

[half gull]

a doesnt means anything

[spring wave]

ah good

sorry for interrupting you then

[half gull]

its just the letter one actually uses in the formula

[spring wave]

so

y=f'(p)(x-p)+f(p)

[half gull]

i wrote it with p here

[upper karma]

hi

[half gull]

but most of the time the formula is written with a instead of p

[spring wave]

so you don't mean the coefficient of 4 in the general equation?

[half gull]

so this is the equation of the tangent of f at a

i never saw this general equation before

[spring wave]

for a parabola ?

really ?

[half gull]

for me the general equation of a parabola is y=ax^2+bx+c

its maybe an equivalent definition

[spring wave]

then they are equivalent

I am pretty sure mine is correct as well (it's the same in my textbook and in another algebra book as well)

[half gull]

yes they surely are

but I never saw it before

[spring wave]

oh sorry

I should have mentioned both

the one i mentioned is not a function

there is this one too:

(x-h)^2 = 4p(y-k)

[half gull]

so what are x, h, k ,p and y?

[spring wave]

@half gull holy shit I think I solved the problem

x = p/2

and I knew that was the answer but I actually did it myself!

without your help it would have been impossible * 100 tho

@half gull (h, k) is the center of the parabola

p is this number you add to either "h" or "k" depending on the type of the parabola to find the focus

(x, y) is a point on the parabola

[half gull]

yep the answer is correct

x=p/2 is right

[spring wave]

I am God!

lol

@half gull thank you so much man

You have no idea how much I appreciate it

[half gull]

np

[spring wave]

this seemed so impossible

[half gull]

xD

i know how it feels like

i remember trying to solve a problem for 12 days

and finally succeeding

[spring wave]

did you do it all by yourself or did you have help ?

[half gull]

it depends on what you call an help

it was most like a tiny clue

[spring wave]

what was the problem

?

[half gull]

let f be a continuous function such as:
$$\forall x>0, \lim_{n \to +\infty} f(nx)=0$$. Show that $$\lim_{x \to +\infty} f(x)=0$$

[somber coyote]

Key 🍆:

[spring wave]

hmmmmmm.

[half gull]

you won't prove it in 5 min x)

it uses non trivial theorems about R

[spring wave]

@half gull i couldn't do it in 12 days even.

[delicate lotus]

prove that f(nx)=f(n)×f(x) and there u go

[floral dirge]

I dont know how to do any of this

Can someone please help

[trail minnow]

@floral dirge i can help but we should move to #help-2

why do they never read the directions and FAQ sigh

[devout shell]

why do students never read their textbooks, smh

like everyone one that asks me for help says they don't ever use the book

[trail minnow]

no one reads the textbook in hs

like

literally no one

[devout shell]

if only they did...then they wouldn't be asking the very, very simple questions lol

[upper karma]

hi can anyone help me with geometry

[devout shell]

you need to ask a question before anyone can help lol

[upper karma]

I just got the answer, thanks

[devout shell]

ok then lol

[upper karma]

hey, I'm new here, could somebody help me with some geometry work later?

[devout shell]

why not now though? lol

[upper karma]

have somebody else helping me rn

but they'll be logging off later

I'm pretty okay in algebra but garbage in geometry

[devout shell]

what is the question about then?

I still might be on later

[dark sparrow]

post the question when you're ready to receive help with it

[cloud merlin]

is anyone awake?

can someone help me out with quadratic equation

[devout shell]

just post it

[devout shell]

3-6?

or 3x-6?

[dark sparrow]

this belongs in #prealg-and-algebra

and you should check whether you copied the question correctly before posting it there

[cloud merlin]

nvm its suppost to be xto the power of 3 plus 3x - 6 +0

[trail minnow]

uhh

AcE??

[turbid pulsar]

oh mb didnt know i had to post in question channel

[upper karma]

@half gull but how is it non trivial

[gritty siren]

why would you think it's trivial?

[upper karma]

lol i was applying an incorrect argument

assumed f was a function of n lol

[gritty siren]

[upper karma]

imagine if you do x = 1 in lim n->infty f(nx)

but i see why you need baire's theorem now

[half gull]

yes

[upper karma]

bare

[upper karma]

If you guys aren’t allowed to help with hw that’s fine but I’m just rlly stuck

[half saffron]

So create an equation for the volume of the pool

[vague field]

hello

i have a triangle with short sides 3 and 4

for the long side i get 7 because a + b = c

but the answer is 5

???

[dark sparrow]

because a + b = c
🤔

[vague field]

ya

pythagorem theorem

[hard gale]

tfw a²+b²=c² <=> a+b=c

[dire rampart]

how bored are u lmao

[vague field]

what does that mean

so how do i get 5

@dark sparrow

[dark sparrow]

$a^2 + b^2 = c^2$

[somber coyote]

Ann:

[dark sparrow]

this is NOT the same as a + b = c

[vague field]

ohhh

i thought u could square root both sides to simplify

[dark sparrow]

$\sqrt{a^2 + b^2} \neq \sqrt{a^2} + \sqrt{b^2}$

[vague field]

so 3^2 + 4^2 = c^2

[somber coyote]

Ann:

[vague field]

so c is 5

thanks bro

[sand marsh]

any pre-requisites to learning geometry and trig? geometry seems pretty basic and only needing arithmetic. Im not sure about trig.

[dire rampart]

just make sure ur algebra is strong I guess

[sand marsh]

thanks. seems like algebra is the link to a lot of different mathematics branches from what im gathering

[spark stag]

notions of quantity and size are very useful in most things

numbers and relations are the language of quantity and size

elementary algebra is the art of manipulating those relations

so its natural that it comes up everywhere

[sand marsh]

^ well said

[bronze elbow]

shouldn't relations be closer to 'operations'?

[vast jackal]

can someone explain to me what im doing wrng

[bronze elbow]

this needs some rotation magic

[vast jackal]

.rotate

[vast jackal]

oo thank :D

[bronze elbow]

are A1 and A2 areas?

[vast jackal]

yea

the two triangles

[bronze elbow]

what is it you're actually looking for?

[vast jackal]

i dont really know the words in english but i was trying to find the central point of a trapezoid by using this:

[bronze elbow]

centroid?

[vast jackal]

and k realised that even though A1 +A2+A3 should be equal to A it wasnt

yea i think so

the result i got from this process was different from the actual formula that gives the centroid

[bronze elbow]

never actually had to look for that, lemmy look it up :D

in the mean time a thought:

[vast jackal]

and i realised it probably had to do with the As being wrong

[bronze elbow]

a system of equations for finding b,b1 and h, based on the areas

[agile bridge]

hello

can someone help me with a problem

[vast jackal]

its not about finding those so much as the numbers not matching up when they should

[bronze elbow]

@agile bridge you need to be a bit more specific xD

[agile bridge]

ABCDA'B'C'D' is a cuboid . The angle between planes (A'BD) and (C'BD) = 90 degrees. The angle between planes (AB'C) and (D'B'C) = 60 degrees. Find the angle between planes (BC'D) and (A'C'D).

it's a big one

[vast jackal]

can anyone help me?

[bronze elbow]

@vast jackal for starters this is what google threw at me

[vast jackal]

yea but that should match up with the result i found also

why doesnt A1+A2 +A3 =A

[bronze elbow]

how did you decide its not =?

@agile bridge please hold on, can't mutitask

[vast jackal]

because when i equated them it came up with 2b1=b

[agile bridge]

ok man

[vast jackal]

when it shouldnt

sorry aurel

[agile bridge]

no no it's fine. keep going 😄

[vast jackal]

its a 3

A3=h(b-b1)

basically my question is what am i doing wrong since they should be equal

[bronze elbow]

wait, what's wrong with 2b1 =b?

that just suggests b1=b/2

it wouldn't be a trapezoid if they were equal

[vast jackal]

in my original problem b1 js something else

i dont want them to be equal i was expecting them to 0 out

0=0 type situation

[bronze elbow]

ok, just gonna solve it myself, sec xD

[vast jackal]

lel thats probably best

[bronze elbow]

A3=h*b1
for starters

[vast jackal]

lol omg

[bronze elbow]

you can have the rest :D

[vast jackal]

tfw retard

thanks

NEXT!

@agile bridge

[bronze elbow]

ABCDA'B'C'D' is a cuboid . The angle between planes (A'BD) and (C'BD) = 90 degrees. The angle between planes (AB'C) and (D'B'C) = 60 degrees. Find the angle between planes (BC'D) and (A'C'D)

[agile bridge]

yes

i can;t even visualise this

with draw in my face

[bronze elbow]

its just a cube

[agile bridge]

cuboid

[bronze elbow]

that has non 90 angles

this needs some drawing, brb

[agile bridge]

what do u mean by "cube"

[bronze elbow]

a 3d square

(that's not quite square)

[agile bridge]

a ok

😃

[bronze elbow]

I didn't find a good drawing tool..

[agile bridge]

paper :D? old style

[bronze elbow]

ignoring the part I haven't put in the walls..

(nor are the angles correct..

[agile bridge]

uhh..

what next 😄

[bronze elbow]

find the parallels and transfer angles in 3d..

[agile bridge]

you are right

i think this is a cube

what tool is that?

i need some aditional drawings 😄

[bronze elbow]

yup better tool..

its a cuboid, as mentioned

just a bit stretched cube

parallel walls, parallel bottom and top, rest is variable

[agile bridge]

can u tell me the name of tool u draw?

@bronze elbow

[bronze elbow]

closed it already, try desmos, bit better

still looking for even better one xD

prolly easier to just draw by hand

[agile bridge]

ty

i think i basically solved it

and that's a cube

100%

[bronze elbow]

@agile bridge https://www.math10.com/en/geometry/geogebra/geogebra.html decent drawing tool

(phew took a while to find it xD)

[agile bridge]

oh thank you so much

[bronze elbow]

a bit fiddly, but at least it has most the bells and whistles

[agile bridge]

i found the angle

ABCDA'B'C'D' it's a cube and because of symetry the angle between (AB'C) and (D'B'C) = 60 degrees

[bronze elbow]

well it was bound to be either 30, 45 or 60 :D

[agile bridge]

yeah...

it's hard to visualise this problem

[bronze elbow]

that's 3D for ya :D

it would've probably been easier to come up with a matrix to calculate every angle in the cuboid instead xD

[agile bridge]

yes probably

I considered BC to be 1 unit, AB = a and AA' = c

then if : P = pr_BDA => m((A'BD),(ABC)) = m(A'PA)

if P' = pr_BDC => m((C'BD),(ABC) = m(C'P'C)

and we know that triangle A'PA is congruent with triangle C'P'C

=> m(A'PA) = 1/2 (180 - 90) = 45 = >

[bronze elbow]

tbh I have trouble thinking about 3d geometry in letters..

[agile bridge]

$\c = AA' = AP = AB * AD / BD = \frac{a}{sqrt(a^2+1)}$

[bronze elbow]

so I rather just look at pretty pictures and mark angles on them xD

[somber coyote]

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

[agile bridge]

i thought that it must be a cube

😄

[bronze elbow]

its a cube with sides a b and c as opossed to normal cubes in which you only have a's :D

[agile bridge]

😄

[dark sparrow]

not a cube then

[bronze elbow]

almost a cube?

[dark sparrow]

cuboid

or box

[bronze elbow]

I like box

lets paint it blue and hop through time and space

[agile bridge]

enough math. i think i'll move to physics. bye @bronze elbow

[bronze elbow]

time and space fits right into that :D

[agile bridge]

😄

[bronze elbow]

gonna check your answer (slowly)

or not.. gtg

[slate pewter]

hello my friends i did find the answer but i want to be sure is it right answer

in this circle

the point from the center to B is 18 cm

and the center to A is 24 cm

how much the area in the black spaces

what i did is a little bit complex and long but i will say what i did do

from center to B is 18

18 squared is 324

324 * 3 = 972

972 is the area of the smaller circle

from center to A is 24

24 squared is 576

576 * 3 = 1728

1728 - 972 = 756

so the second circle without the red circle is 756 area

756 / 8 = 94.5

each one is 94.5 area

so 94.5 * 4 = 378

378 is the answer

(3 = Pi

in the quistion they did say pi is not 3.14

thats is see ya

[celest moon]

can someone help simplify this?

it wasnt shown what the end result should be

aka not solving for an identity

[vagrant elk]

Try multiplying by 1-cos(x) on both the top and the bottom (good ol conjugate trick)

[celest moon]

i tired

but it just ends up messier

[vagrant elk]

well, doing that gives you (sin+tan)(1-cos)/(1-cos^2)

so you can then use that to simplify the bottom

$\frac{(\sin +\tan)(1-\cos)}{1-\cos^2} = \frac{(\sin + \tan)(1-\cos)}{\sin^2}$

[somber coyote]

Darkrifts:

[celest moon]

yeah

[vagrant elk]

which, while that looks a bit nastier, you can separate the sin and tan because of distributivity of multiplication

[celest moon]

to sin+sinsec

[hard gale]

das pretty convoluted tho

[vagrant elk]

$\frac{\sin*(1-\cos)}{\sin^2}=(1-\cos)(\csc)$

[somber coyote]

Darkrifts:

[vagrant elk]

it's maybe not the best way to go about doing this, but it certainly is a way

then you work with that tan

h m m

[hard gale]

oh man i should go to sleep

[celest moon]

i mean

the farthest i got was

[vagrant elk]

So you end up with $(1-\cos)(\csc) +\frac{\tan(1-\cos))}{\sin^2}$

[somber coyote]

Darkrifts:

[vagrant elk]

which I'm fairly sure can be simplified further

just a hunch

[hard gale]

wait actually my thing wasn't bullshit

[vagrant elk]

it probably wasn't

[celest moon]

changing tan to sin/cos?

[hard gale]

yeah

[vagrant elk]

That would certainly help

Since you have (1-cos)

$\tan(1-\cos) = \frac{\sin}{\cos}-\sin$

[somber coyote]

Darkrifts:

[vagrant elk]

if i'm not retarded and multiplied wrong, that is

[hard gale]

yas

[vagrant elk]

Then you can use your sin^2 on the bottom

and yeet those away

[hard gale]

well should be 1+cos but we

[vagrant elk]

why 1+

[hard gale]

cuz that's what's given

[vagrant elk]

I multiplied by 1-cos to get rid of 1+cos on the bottom

conjugate boy

[supple abyss]

oshit

[hard gale]

$$\frac{\sin(x)+\tan(x)}{1+\cos(x)} = \frac{\sin(x)+\frac{\sin(x)}{\cos(x)}}{1+\cos(x)}$$ $$\hspace{7em} = \frac{\frac{\sin(x)+\sin(x)\cos(x)}{\cos(x)}}{1+\cos(x)} = \frac{\sin(x)(1+\cos(x))}{\cos(x)(1+\cos(x))}$$

[somber coyote]

emeric75:

[hard gale]

boom

[vagrant elk]

yeah, but that's ugly as fuck

so therefore b a d

[hard gale]

you ugly af

[vagrant elk]

shit u rigte

[celest moon]

could you cancel the two 1+cosx?

[vagrant elk]

ye

[celest moon]

damn alright

thanks for the help

this problem was bugging me

[trail minnow]

i have the IQ of a nugget i don't remember the centroid theorems for triangles

isn't it like

the segments from the midpoint of every side of a triangle to the centroid are congruent

or smth

[devout shell]

they are proportional to each other I think

let me find the theorem

[trail minnow]

i already solved it lol

[winged saddle]

nut

[trail minnow]

$SR^2 = ST * SF$

[somber coyote]

hegel:

[trail minnow]

iirc

you're setting the radius as x

right

so you would get

$15^2 = 9 * (9+r)$

[somber coyote]

hegel:

[trail minnow]

so r should equal 16

with Pythagoreans you said

$16^2 + 15^2 = (9+16)^2$

[somber coyote]

hegel:

[trail minnow]

wat

that works too

is it possible that whoever wrote this like

fucked up and chose nonsense values

but why the fuck this discrepancy is occurring?

$x^2 + 15^2 = (9+x)^2$

[somber coyote]

hegel:

[trail minnow]

$x^2 + 225 = x^2 + 18x + 81$

[somber coyote]

hegel:

[trail minnow]

$ 18x + 81 = 225$

[somber coyote]

hegel:

[trail minnow]

$18x = 144$

[somber coyote]

hegel:

[trail minnow]

yep you solved this properly

i'm honestly unsure

it's possible they just chose ratios that don't make any fucking sense?

this is definitely very confusing

yeah it's fucking weird

[somber coyote]

Syn:

[bronze elbow]

@upper karma two solids can have the same area and volume, yet be different, as long as they're not of the same form

on the other hand https://www.quora.com/Can-different-shapes-have-the-same-volume-and-surface-area

[acoustic vector]

$9 \cdot 4$

[somber coyote]

iiTrap_Housell:

[acoustic vector]

$\dfrac{9}{4}$

[somber coyote]

iiTrap_Housell:

[acoustic vector]

$\text{test}$

[somber coyote]

iiTrap_Housell:

[acoustic vector]

$\sigma{1}^{4}$

[somber coyote]

iiTrap_Housell:

[acoustic vector]

$\sum{1}^{4}$

[somber coyote]

iiTrap_Housell:

[acoustic vector]

$\sum{1}{4}$

[somber coyote]

iiTrap_Housell:

[hard gale]

$\sum_{1}^{4}$

[somber coyote]

emeric75:

[acoustic vector]

$\infty$

[somber coyote]

iiTrap_Housell:

[hard gale]

there's a #bots chan btw

[devout shell]

can you show me what the polygon looks like?

[muted bloom]

Shoelace mydude

[devout shell]

you can form right triangles using the sides of the polygon as hypotenuses

then find the area of that rectangle and subtract off the area of the four right triangles you made

yes

you see how you would do that with triangles?

I'll have to show you then, let me make drawing in geogebra

this is a pretty cool way to go about finding the area lol

better scale

[upper karma]

in the equation sin(x/2) + cos(x) = 0, why aren't x= π/4 and x= 7π/4 valid solutions? it seems like it works if you factor it out.

[devout shell]

factor?

[upper karma]

Here, I'll send a picture of my work

[spark stag]

erm

[upper karma]

I guess the proposed solutions would be pi/3 and 5pi/3

[spark stag]

$\cos{\frac{\pi}{4}} \neq \frac{1}{2}$

[somber coyote]

Namington:

[spark stag]

same with 7pi/4

[upper karma]

see earlier message I guess the proposed solutions would be pi/3 and 5pi/3

but the answer key in the book says that the only answer is pi

and graphing the same equation only gets roots of pi

[spark stag]

hold on, lemme look it over

ah

you squared both sides

which introduced extraneous roots

theres nothing wrong with that but, whenever you square both sides

you have to check all your roots at the end

by "plugging them back in" to the original equation

[upper karma]

oh, forgot about that

[spark stag]

because squaring sometimes introduces additional "solutions"

since (-2)^2 = 2^2, for example

checking the equation with pi/3 or 5pi/3 gets you answers that certainly arent 0

so those roots are extraneous

[upper karma]

yeah, that makes sense now, thanks

[stoic nova]

hi can any help me with my geometry homework please

[trail minnow]

[stoic nova]

can i dm you

[trail minnow]

better to post it here lol

other ppl more competent than me can intervene

[stoic nova]

its multiple questions tho

is that fine

[trail minnow]

yes

post 1 at a time tho

[stoic nova]

oops

[trail minnow]

oh if it's like that then it's fine lol

what are you solving for?

in 11?

[stoic nova]

1 sec

area

[trail minnow]

ah

BC = 10 and DE = 6

[stoic nova]

yessir

[trail minnow]

well you know the formula for the area of a paralleogram right?

wait nvm it's a rhombus

much easier then

$\frac{d_1d_2}{2} = \text{ the area of a rhombus}$

[somber coyote]

hegel:

[trail minnow]

correct

uhhh you there lol?

[stoic nova]

yeah sorry i was trying to do it my own first

[trail minnow]

ah okay

still need the help?

[stoic nova]

i think i just need help 13 and 14

[trail minnow]

okay

13 is actually very easy

you have a square with side length 4, correct?

[stoic nova]

yeah

[trail minnow]

so the area of the entire square is 16

from there

each quartercircle has an area equal to

$\frac{\pi r^2}{4}$

hm

one moment lol

[somber coyote]

hegel:

[trail minnow]

ere we go

so we know that the radius is 2 for each right?

[stoic nova]

yeah

[trail minnow]

$2^2 = 4$

[somber coyote]

hegel:

[trail minnow]

so you can simplify this equation to

just

$\pi$

[somber coyote]

hegel:

[trail minnow]

and that's the area of one of the small circles

you have 2 of them, so multiple by 2 to get 2pi

[stoic nova]

ahhh

[trail minnow]

and then your final area is the area of the entire square minus the unshaded sections i.e the area of the 2 circles

yes

not a difficult problem haha

[stoic nova]

do you think you can check some problems ive done

[trail minnow]

sure

do you want help with 14 though?

[stoic nova]

my teacher doesnt post key 😦

[trail minnow]

same here lol

[stoic nova]

i think i got that

[trail minnow]

mmh

if you know the formulas it really isn't all that complicated

[stoic nova]

oops

[trail minnow]

okay

6 is correct for sure

hm

i don't think 4 is correct though

do you know the formula for the side lengths of an isosceles right triangle?

[stoic nova]

no

[trail minnow]

if you have a right triangle with a leg of length x

then the hypotenuse is equal to $x\sqrt{2}$

[somber coyote]

hegel:

[trail minnow]

in this case we know that $x\sqrt{2} = 10$

[somber coyote]

hegel:

[trail minnow]

so you can solve for the height of your right triangle by dividing 10 by root 2

so $x = 5\sqrt{2}$

[somber coyote]

hegel:

[stoic nova]

oh ok i get it

[trail minnow]

yes

24 * 5 root 2 lol, not hard math

[stoic nova]

im really bad at geometry

😦

[trail minnow]

haha dw about it

you just need more practice

and potentially a better teacher

[stoic nova]

how about 1 and 5

[trail minnow]

if you could repost 1 that would be helpful i can't really see the whole problem

[stoic nova]

A regular polygon has 16 sides find measure of one exterior angle

[trail minnow]

ah

i see

i think you confused the formula for the interior angle of a regular polygon with the exterior angle of a regular polygon

they're pretty similar by exterior is a little bit simpler

$\text{the interior formula is } = \frac{n-2(180)}{n}$

[somber coyote]

hegel:

[trail minnow]

while the exterior is

[stoic nova]

damnit

[trail minnow]

$\frac{n-2(180)}{360}$

[somber coyote]

hegel:

[stoic nova]

yeah i did mess it up

[trail minnow]

it's no big deal

if you can see where you went wrong you can correct yourself next time

[stoic nova]

is it formula 360/n

[trail minnow]

ah wait yea

i messed up my formula too >.>

too much cosine work in #prealg-and-algebra

sigh

[stoic nova]

so it would just be 360/16

[trail minnow]

yes

[stoic nova]

22.5 ?

[trail minnow]

yeah

[stoic nova]

how about 5

[trail minnow]

for 5 you were basically right, i think you just accidentally wrote down 24 instead of 8 because you added the perimeter of the triangle rather than the base

[stoic nova]

isnt it supposed to be perimeter tho

[trail minnow]

no, area of triangle is $\frac{bh}{2}$, remember?

[somber coyote]

hegel:

[stoic nova]

where did i get 1/2Pa

[trail minnow]

that's the area of a regular polygon probably

but the height isn't the apothem of the triangle

the apothem is the perpendicular segment running from the center of a polygon to the midpoint of one of its sides

ah nevermind @stoic nova !

i see

your answer was correct, your diagram just confused me

or

it may or may not be i have to check

you elongated the apothem to the point that it is too long

you drew the height, but you were only given the apothem

hmm

you could do

$\tan{30} = \frac{x}{4\sqrt{3}}$

[somber coyote]

hegel:

[trail minnow]

from there simplify

so x = 4

yes you were correct, but your diagram is wrong >.>

[stoic nova]

dont worry about that lol

[trail minnow]

also the method you used to solve was incorrect

hmmm

[stoic nova]

oh

well

😦

[trail minnow]

hmm

wait no

this is actually a rather odd problem...

[stoic nova]

is my number 2 correct

[trail minnow]

you didn't fully distribute the 180 i believe

it should be $156n = 180n-360$

[somber coyote]

hegel:

[trail minnow]

also i'm pretty sure your teacher messed up creating #5

its certainly doable but i don't think you know law of sines

$\frac{\sin(30)}{A} = \frac{\sin(120)}{8}$

[somber coyote]

hegel:

[trail minnow]

$8\sin(30)=A\sin(120)$

[somber coyote]

hegel:

[stoic nova]

this is for 5?

[trail minnow]

yes

$4=A\sin(120)$

[somber coyote]

hegel:

[trail minnow]

whoever made this problem is dum

why r they trying to give hs freshman/sophomores something like this

this is just annoying

[stoic nova]

is number 2 15 sides?

[trail minnow]

ye

ok so you can confirm it is an equilateral triangle

therefore

[stoic nova]

i still have one more page if you dont mind lol

[trail minnow]

lol

one moment please

ok so you were right

basically

[stoic nova]

no im just saying after this problem i have more

[trail minnow]

yeah

well

[stoic nova]

no disrespect lol

[trail minnow]

depends on how much it is

its 12:00 rn in my timezone and i have ap world lmfao

[stoic nova]

its 6 questions 😮

[trail minnow]

hmmm

i might have to pass on that one

i really do have to finish this hw lol

[stoic nova]

thats fine

you have helped me alot

[trail minnow]

[stoic nova]

can someone assist me with 6 geometry questions please 😃

[upper karma]

i will try

[upper karma]

Pogi I'll try.

[rustic cairn]

How do i solve number 8

[spark stag]

Use the principle of similarity

The larger triangle is just the smaller one scaled up

So we have to find out by what factor it's scaled up

To do that, we can take 1 side of the larger triangle and divide it by the same side of the smaller one

So, 50/25 or 80/40

In either case, we get 2

So, to scale up 32, we multiply it by 2

[rustic cairn]

So the answer is c.64

[spark stag]

Yes

[rustic cairn]

Thank u

[stoic nova]

I just need help on 18 and 19

[paper mauve]

@stoic nova first draw it out

[muted bloom]

For 18: Find the area of the circle. Subtract the area of the circle from the area of the square to find the shaded area

The divide shaded area by entire area

19: draw it out before asking

[storm rover]

If I have sunlight going into a mirror how would I calculate what angle the light would go into a box or what angle should the mirror be at compared to sun

[umbral snow]

@storm rover
I imagine you have a source, like a window or something

Draw a vector from the bottom of the window to the bottom of the mirror. Reflect the angle based on how the mirror is positioned. That's the lower ray of light off the mirror. Repeat for the upper

[stoic nova]

My bad

[storm rover]

I have a few questions btw

[visual fulcrum]

Anyone willing to help me?

Anyone

[storm rover]

I will ig

[spark stag]

area of whole circle - area of inner circle

[storm rover]

^

[visual fulcrum]

Thanks, I just need help with like 17 questions

I need the area of outer circle

The shaded one

[storm rover]

_thats how you get it

[visual fulcrum]

C = d(3.14)
C = (2r)(3.14)
R = (C/3.14)/2
D = (C/3.14)

Heres some stuff to find the circumference

But I need the area

And I dont know how to find area

[storm rover]

It’s pi not 3.14

[visual fulcrum]

You multiply by pi, I use 3.14 as a reference

[storm rover]

Always use pi unless it says to approximate

[visual fulcrum]

But this is my problem

And around 17 more

After around 7 of these I can do some probability questions

Anyone able to help?

Well fuck, then I fail the school year...

[winged saddle]

hi

ok i can help

like he said

so

how about you start by finding both of them

and then you can substract

so i will start by finding the area of the big circle

so the formula for area of a circle is pi times radius times radius

or pi r^2

so

radius is 11

so pi times 11 times 11

which is 121 pi

and then

the area of the small circle is

pi times radius times radius

which is pi timmes 3 times 3

which is 9 pi

so now

121 pi - 9 pi is 112 pi

sooo

about 351.68

@visual fulcrum

oh wait

351.848

round to the neareest hundreth

i would say 351. 85

[visual fulcrum]

@winged saddle nevermind but thanks tho xD
I figured out I was already finished with my school year and I was doing the wrong school year math

[winged saddle]

BRUH

[covert orchid]

how could i calculate the volume of this shape ?

[devout shell]

Composite shape of triangular prism and rectangular prism

To help you get started, the dimensions of your rectangular prism would be:
40x20x4

It’s up to you to find the dimensions of the triangular one

[trail minnow]

[upper karma]

[maiden stag]

Can any1 help me with my geo final

[devout shell]

we can't do a test for you lol

[maiden stag]

My teacher said use what ever

It's also a take home exam

Created with You Doodle

What's the volume and area of the 3

[devout shell]

there are three different solids lol, how could we know which one you are referring to?

[maiden stag]

Of all 3

[devout shell]

since this is open resource test, you can just look up the formulas lol

[dark sparrow]

^

[maiden stag]

If u double the dims of a rectangle wouldn't it double the perimeter and the area?

[dark sparrow]

perimeter yes, area no

area would be quadrupled, not doubled

[maiden stag]

This is not from a book or a paper my geo teacher made it in a period

[supple gyro]

i can help youw ith the trig ratios

but

cant find x

[twin ore]

just wanna make sure

sinx = 3 / 5, where is an obtuse angle

what's the value of tanx?

i got the answer, jsut wanna make sure it got it right

[supple gyro]

3/4

[twin ore]

no negative?

[supple gyro]

depends

on what quadrant

1st or second quadarnt

must be one of those two

1st is positive

2nd is negative

[twin ore]

coz that's the only question. x is an obtuse angle such that sinx=3/5, find the value of tangent

[supple gyro]

oh

[twin ore]

im thinking tanx should be in the second quadrant

[supple gyro]

its negative

obtuse

yeah

[dark sparrow]

x is in the second quadrant yes

[twin ore]

thanks

[upper karma]

hi there i have a tube with two curves and i would need the total lenght of it , i have no idea how to calculate it with the angle of the curves to get the complete lenght but i have a sktech of it i tried to solve it but my solution if 588,6 mm is wrong (thats what my teacher told me )

[mortal socket]

is sin y axis and cos x axis? on the legendary unit circle??

[upper moat]

sin = opp / hyp = opp / 1 = opp => the y value
cos = adj / hyp = adj / 1 = adj => the x value

[mortal socket]

ok

thanks

@upper moat what is hyp and adj tho

[upper moat]

opposite, adjacent, hypotenuse

Since it's the unit circle, radius = hypotenuse = 1

[mortal socket]

yah

[dusk brook]

Half of 2 is 1 which is the length of the overlapped region square, 1^2 is 1

[maiden pawn]

Ahh... My first Proof by Contradiction.

FeelsGoodMan

(Only exception is 90 degrees, to which construction of a right angle, an equilateral triangle & a line of symmetry for the triangle suffices)

[upper karma]

Hi, I was wondering for a hyperbola in a conic section, does the law: eccentricity=(point to focus)/(point to directrix) of a focus on one end apply also to the graph on the other end?

what i mean is if the below picture is true

ignore the AP=ePQ written above, and also the line that P lies on is the diretrix

[upper karma]

<@&286206848099549185> ?

[upper karma]

@upper karma I didn't learn this in geometry...

Try putting this in precal

hm, its not calculus though so idk

but i solved it so its all good

Aight.

I'm sorry I'm wasn't much help.

[twin ore]

I think im doing something work. I kept getting far out andwers for theta

*wrong

it got 0.5 r^2 costheta sintheta for its area

*i

then I equate that to 3/5 pi r^2

am i doing something work?

*wrong

oh gosh too much typo

[upper moat]

$0.5r^2\sin{\theta}\cos{\theta}$ is 3/5 of the sector, not of the entire circle

[somber coyote]

Hello World:

[twin ore]

oh. so can I just equate that to (3/5)(0.5)r^2 theta <--- area of the sector using radian

[upper moat]

Yep

So you get $0.5r^2\sin{\theta}\cos{\theta} = \frac{3}{5} \cdot \frac{\theta}{2\pi}\pi r^2$

Whoops

[twin ore]

thats area of triagnle

[somber coyote]

Hello World:

[twin ore]

will 0.5 theta r^2 works the same?

[upper moat]

Yeah it's the same thing once you get rid of the pis

You're left after simplification with sintcost = 3/5 t

[twin ore]

cos theta sin theta = theta * 3/5

[upper moat]

Yea

[twin ore]

i end up having sin theta = 6/5 theta

*sin 2 theta

[upper moat]

Yep

[twin ore]

then just find the intersection

i dont know how to do it manually

i got approx. 0.83

[upper moat]

Hrm I'm not too sure on how to solve sint = 6t/5

[twin ore]

sin2t = 6t/5

[upper moat]

yep my bad

[twin ore]

i just relied on my calculator

[upper moat]

Other than the obvious 0 rad, but that's not valid

[twin ore]

is there even a way to do it ?

[upper moat]

Probably is

[twin ore]

Thanks for helping @upper moat

[upper moat]

I haven't done anything yet 😅

[twin ore]

lols. its nice to have someone to at least tell me if my line of thoughts are corrects

sometimes my mind is just messed up

lols

[hard gale]

tbh i think you can't solve it analytically

so yeah just calc ig

[upper moat]

Welp

[twin ore]

btw, how I used the Textit Bot

[upper moat]

How to use it?

[twin ore]

yes.

oh gosh too much typo

[upper moat]

Go to #bots and type ,help

Join the support server 😄

[supple goblet]

Hello

I have a short geometry question, can anjine assist

I need to find the alpha

[upper karma]

the alpha male?

The alpha angle at B?

isn't that just 60degrees?

honestly I can't see much in that image (might be cus i have a low resolution)

[supple goblet]

It may be also due to my bad hand writing. The left corner B is with the angle alpha

Blue markings are the original question, other is my addition for the solving

D is 70 degrees

And as dc is same

AD DC

[devout shell]

@violet summit what circle THMs can you remember?

[cyan violet]

Would you like help on this.

It doesn't seem that arduous to be honest.

Yes, there are theorems that can be utilized to derive an answer.

[devout shell]

to be honest, it is not lol, yet I'm not going to pass judgement on someone who may be struggling, there just isn't a question, but a picture with some problems on it

need the person to respond lol

[cyan violet]

Yes, simply wait on them.

[upper karma]

these problems look like those 3rd grade chinese homework

sry my brain cant function this far

geometry is a no to me

[devout shell]

I liked it when I took it in highschool

[upper karma]

we never did geometry in highschool

explains why i forgot everything about it

[devout shell]

some people come asking for help on fill in the blank proofs like for real lol? They should make you write two column proofs from scratch

[violet summit]

nvm mb dont know why i posted this i dont need the help no more

i pisted the wrong picture anyway

[cyan violet]

Meh, read up and work though chapter 12 of introduction to geometry AoPS and those theorems will be explained and you'll be constrained to prove them.

[violet summit]

posted*

[devout shell]

was about to say, they always come back and say "oh I got it" lol

every time lol

[upper karma]

xD

[devout shell]

It was even funnier this one time

I might have screenshot

I don't

I can probably search for it though

[trail minnow]

@devout shell re proofs they do make u write proofs from scratch

im sure it is on the regents

[devout shell]

it's not that bad honestly

[trail minnow]

and we have one every test

oh no

it's always pretty easy

short answers are easier to fuck up on cause teachers can make u go through many more steps

w/ proofs if there is more than like 20 steps then u won't have time to make a table

:o did u always have honorable btw?

[devout shell]

what if they make you do Reimann Sums from scratch

[trail minnow]

i wish tbh

[devout shell]

Just got the role lol

[trail minnow]

yah

but then

[devout shell]

because of Reimann

[trail minnow]

who lost it

[vagrant elk]

h m m

[trail minnow]

[last pollen]

in an equilateral triangle, for any point inside the triangle, the sum of the heights relative to each side is a constant

apparently, for a regular Tetragon the same holds

does it hold?

[fossil lotus]

What does “the heights” mean?

[last pollen]

pick a point inside the equilateral triangle, the height relative to one side is the distance to the closest point on that side

[fossil lotus]

Ah

[last pollen]

for the tetragon it would be the distance to the closest point in the plane defined by each face

[fossil lotus]

By “tetragon” you mean a quadrilateral?

[last pollen]

nopee

sorry

a pyramid

[fossil lotus]

Or a tetrahedron?

[last pollen]

yeah

a tetrahedron

a regular tetrahedron

[fossil lotus]

-gon = 2D, -hedron = 3D

[last pollen]

yeah smh

this is what I mean if it wasn't clear

a+b+c is equal to the height of the triangle, no matter where you place the center point

if the triangle is equilateral

[last pollen]

ok, Googled a bit. this I called viviani's theorem and holds for simplexes

[dark sparrow]

wow oof

[low herald]

help

[hallow smelt]

Final study guide, not sure how to do this question

[dire rampart]

can u find the sin of tan^-1(9/40)?

[hallow smelt]

9/41 right?

[storm rover]

Can you describe cosine in the terms of sine?

[cyan violet]

@storm rover Are you talking to me?

[storm rover]

No

[cyan violet]

Okay, I was confused.

[quartz edge]

@hallow smelt Use the double angle identity

looks like you found arctan of 9/40 already

[idle bloom]

@storm rover

[storm rover]

Thanks

[idle bloom]

since cosine and sine are just horizontal shifts of each other that's how you get these formulas

[storm rover]

Okay cool

@cyan violet if you got this from AOPS it should have a solution for it

And there is a video about it

https://mindyourdecisions.com/blog/2017/07/30/tricky-interview-question-the-ratio-of-areas-of-squares-sunday-puzzle/

[cyan violet]

It's an alcumus problem.