geometry-and-trigonometry

[grave pond]

Fine, keep it secret.

[mystic vector]

You said sin(x) cannot, then what is this

[grave pond]

That's not in the form $a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$.

[somber coyote]

Troposphere

[grave pond]

It's an infinite series, not a polynomial of finite length.

[mystic vector]

Then what is the prove of this

[grave pond]

In any case, this is definitely not "geometry and trigonometry". It should go in #calculus or perhaps even #real-complex-analysis.

[mystic vector]

ok

[coarse saddle]

how can we prove BC>BD?

I feel like we have to use the triangle inequality somehow

[neat moth]

You need more information

[dark sparrow]

yeah this is NEI as stated

it is possible to make a triangle like this in which BC < BD

[coarse saddle]

oh true

how about BC<AB and we want to prove BD<AB

[neat moth]

if <BDC is greater than <BCD, then BC > BD

I would have to think about how to prove that though

Wait that’s a super common rule of triangles I think

[coarse saddle]

this?

[neat moth]

I think

[grave pond]

First find a relation between, say, b and d.

Yes.

Now now, they went to such an effort saying "explain how you know it" such as not to scare you with the word "prove".

Yeah.

But how did you get the idea, then?

Supplementary angles will give you d+e=180°.

Then you just need an argument that b=d.

[mint bridge]

good night... thanks to @dark sparrow for the support with the material/exercises as well as the resolution of them, it is appreciated that you take the time to help... you are one of the reasons why I am going to pass my exam extraordinary.

[runic beacon]

Which one you have trouble with??

[spare palm]

Oh sorry I turned off discord

But if at all possible then just one

It's the 3rd

[hollow plume]

The hypotenuse is equally divided by 2, so is the horizontal leg

That should help

[hollow plume]

It’s from the tick marks

[wicked otter]

I currently have 8 days left with this trig course, who can help me with it?

(It’s for credit recovery)

[safe mauve]

Is it trig exact values? Finding angles using trig or using trigonometric equations and their inverses to solve integrals?

[upper karma]

Can someone explain to me how a tangent line multiplied by a normal line would give you - 1?

I know it is because Mnormal = -1/Mtan

But i can't really visualise it

its because the angle between thangent and normal is 90 so by this m1m2=-1

ig

tantheta=(m1-m2)/(1+m1m2)

[upper karma]

Ohh interesting

Didn't think of it thid way

[lime yew]

Does anyone have any good resources to get ahead of high school geometry?

[lone river]

sin(some number), will this return the y-coordinate of the triangle at the given angle?

[frozen lotus]

Can anyone give me a clue on how to solve this pls?

[mild briar]

Well, is there any formula you'd think of using on the right side?

Something which would split αt and φ ?

[frozen lotus]

cos(A ± B) = cos A cos B ∓ sin A sin B ???

this one?

[mild briar]

yeah

that's a good start

If you carry out this computation, you'll have cos(φ) and sin(φ) appear

[frozen lotus]

yeah

[mild briar]

A way to simplify these would be to find some formula f such that cos(x)=f(tan(x))

Then cos(φ) = f(-3/4) which is easily computed

[frozen lotus]

oh ok

thanks

i ll try to solve it

[mild briar]

I'll let you mess around with this in mind

[frozen lotus]

yeah tank u

thanks for your help i think i can solve it by myself now

[mild briar]

This is a common trick and helps to simplify expressions like f(g(x)) where f is a trig function and g an inverse trig function

It usually works because there are all sorts of relations between trig functions

[valid harbor]

with $y = 2\sqrt{3x+1}+4$ the expression of x would be the value of f(x)

[somber coyote]

Pealover

[sage ermine]

So what would the answer be? What is the inverse of the equation because I am beyond confused right now.

[valid harbor]

as I said

the expression of x

in term of y

an then you will switch x and y each other

*and

[sage ermine]

So

y = 4 - 2√3x+1?

[valid harbor]

that's definitely false

i think you didn't get what i meant @sage ermine

[valid harbor]

according to this equality

you give the expression of x

x would be replaced by y and vice versa

[sage ermine]

Ok i've tried again an I'm very close to the actual answer but it's not quite right so there's something I'm doing wrong.

1. x-4 = 2 √ (3y+1)
2. (x-4)/2 = (2 √ (3y+1))/2
3. x/2 - 2 = √ (3y+1)
4. (x/2 - 2)^2 = (3y+1)^2
5. x^2/4 +4 = 3y+1
6. (x^2/4 +4)/3 = (3y+1)/3
7. x^2/12 + 4/3 = y + 1
8. x^2/12 + 4/3 -1 = y + 1 - 1
9. y = x^2/12 + 4/3 - 1

What am I doing wrong here then. Because even after all of this, when I graph my answer, it is not the inverse of the original equation.

[lime yew]

You still make the same mistake on step 5

(a+b)^2 ≠ a^2 + b^2

$(a - b)^2 = (a)^2 - 2(a)(b) + (b)^2$

[somber coyote]

TheBlocker66

[sage ermine]

Ok so instead of (x/2 - 2)^2 = x^2/4 +4
(x/2 - 2)^2 = (x/2)^2 - 2(x/2)(-2) + (-2)^2

Like that?

[lime yew]

Almost!

Since the identity I showed is (a - b)^2, your b term would be positive instead of negative

If you used $(a + b)^2 = (a)^2 + 2(a)(b) + (b)^2$

[somber coyote]

TheBlocker66

[lime yew]

Your b term you would use to expand would be negative

[upper karma]

You play egg inc?

You watch BCS?

Omg

This is the perfect server

[sage ermine]

Do you too?

[upper karma]

Yes

[sage ermine]

Caught up?

[upper karma]

Yes

So soon

Finale

😭

[sage ermine]

I'm terrified for kim

[upper karma]

Yeah i got all kinds of theories

[sage ermine]

Then there's the whole thing of Lalo surviving or not

[upper karma]

My trig teacher last year was a bb and bcs fan and we always talked about it before class

True

[sage ermine]

I'm yet to meet a fellow fan in real life

[upper karma]

I guess i never thought about it that way
I always assumed he would die but its possible maybe he just disappeared

But for so long is unlikely

[sage ermine]

Ok I think this is right now?

1. x-4 = 2 √ (3y+1)
2. (x-4)/2 = (2 √ (3y+1))/2
3. x/2 - 2 = √ (3y+1)
4. (x/2 - 2)^2 = (3y+1)^2
5. (x/2^2) -2(x/2)(-2) + (2)^2 = 3y+1
6. ((x/2^2) -2(x/2)(2) + (2)^2)/3 = (3y+1)/3
7. x^2/4 - 2x/3 + 4/3 = y + 1
8. x^2/4 - 2x/3 + 4/3 - 1 = y
9. y = x^2/4 - 2x/3 + 4/3 - 1

[lime yew]

There is still one more mistake you made. In step 7, you didn't divide every term by 3

[sage ermine]

Oh yeah the first one. So it'd be y = x^2/12

Ok so

I've got it

but

How come when they're graphed they aren't inverse of each other

[lime yew]

It's because a quadratic function maps multiple values to one, so if you took the inverse of that, you would get a function that maps one value to many values, but is that a function?

[sage ermine]

This was the original question so it is a function

It's for a test that I'm currently writing

[grave pond]

You need to determine the appropriate domain of f too.

[lime yew]

Right

[grave pond]

(Which is of course the same as the range of f^{-1}).

[valid harbor]

yeah that is what I was about to say

[sage ermine]

When I add the ^-1, it no longer appears on the graph

[lime yew]

I am pretty sure desmos can not calculate inverse functions

[lime yew]

There's one more thing too. When you divided (3y+1) by 3 on step 6, you need to divide both 3y and 1 by 3

[sage ermine]

so that makes it -1/3 correct?

[lime yew]

Yes

[smoky raptor]

Can somebody pls help me get the formula for these tops pls

[hollow plume]

What shape do you need for surface area

[smoky raptor]

um idk but i think this

[hollow plume]

I thought you meant surface area for solids

Not plane shapes

[smoky raptor]

oh sorry about that but stuff like this

[hollow plume]

Prisms can have rectangles and not squares

[smoky raptor]

yea it mainly triangles squares and hexagons but the main Focus is the shape itself and how many triangles are in it but I don't understand like the formula and how to like solve with it and also solve for surface area

[hollow plume]

Please elaborate

[smoky raptor]

i need to know the formula to solve for surface area

understand what i just said?

[sage ermine]

Can someone help me with this? I'm completely lost I've never seen the transformations isolated before.

[wicked otter]

Who can vc and help

[supple kiln]

It is function stated by intervals

From what i can see assume part modular,quadratic and linear

[wicked otter]

@brittle onyx

[twin oar]

I made this visualisation of circular functions on the unit circle for my students
https://www.desmos.com/calculator/qt4nghrl3n

[warm iris]

trig problems are discussed here?

[upper karma]

yes

[bleak blade]

No

Equidistant from the sides

Circumcenter is equidistant to the vertices

[fallen sleet]

how to solve LOGARITHMS L3.ii

[dire blaze]

guys

how do i do trigonometry

i dont get how this dude got 35 to 0.57

[dark sparrow]

$\log_a(b) = \frac{1}{\log_b(a)}$

[somber coyote]

Ann

[fallen sleet]

$\log_a(b) = \frac{\log_b{b}}{\log_b(a)} = \frac{1}{\log_b(a)}$

[somber coyote]

alshfik

[civic yew]

will someone help me with my math? i will pay

[bleak blade]

ok

[surreal trout]

can somwonw help me with this

anyone?

[supple copper]

https://docs.google.com/forms/d/e/1FAIpQLSfixcUBqLqqSQl3gHl6mV_5BOQD88dNN8Dt8H3OBslj3K3BpQ/viewform?usp=sf_link

[crystal quail]

@surreal trout If you have two lines (FB and EC) that are intersected by a 3rd line (FD), they are parallel if they have congruent corresponding angles. Basically angles on the same side. Draw this picture with just FB, EC, and FD (you can extend them beyond where they are here if that helps). If we get 2 angles on the same side to be congruent, these lines will be parallel. Angles EFB and DEC measure corresponding angles. So if they were congruent then these lines would be parallel

[surreal trout]

ok tysm]

im starting to get it

[vestal smelt]

Can we find a & b with just this picture?

[dark sparrow]

no

[fallen sleet]

you can't give an exact value for a or b (not enough info) but:
a=110-b
b=110-a

both equations derived from a+b+70=180

[oak ferry]

can anyone help me with this question

A riding lawn mower has wheels that are 15 inches in diameter, which are turning at 2.5 revolutions per minute. How fast is the lawn mower traveling in miles per hour?

[dark sparrow]

where exactly are you stuck?

@oak ferry

[wraith pumice]

Well wouldn't both be equal to 55?

[silent plank]

no

there's nothing to indicate that the angles are congruent

[fallen sleet]

indeed, the angles could be, for example, b=60 and a=50, or b=109.99 and a=0.01, plenty more

[dark sparrow]

@upper karma this sounds more like #prealg-and-algebra than geometry

[upper karma]

oh sorry

ill post it there

[dark sparrow]

but also you are apparently supposed to drag-and-drop the reasons into these blanks and you havent shown us what the things are

[upper karma]

can anybody check if this is right or not

[dark sparrow]

definitely not

your triangle should not change shape after this transformation

[silent plank]

looks like you messed up the transformation of the point (-2,-2)

[dark sparrow]

not just that, surely?

oh

i see what you're getting at

[upper karma]

did i get it right?

[dark sparrow]

much better

[upper karma]

thank you

[surreal trout]

hwy can anyone help me on this?

[bleak blade]

are you old enough for discord

[surreal trout]

yes

nvm its 126

im so dum

wait nvm

actully it is

[bleak blade]

the problem is elementary

[hoary folio]

If I’m in geometry rn, how long do you think it will take for me to speedrun all the way to calculus?

Need to complete algebra 2 and pre calculus to do that

[storm portal]

@hoary folio Depends on how many hours you spend per day

[silent plank]

depends how well you understand stuff

there's like a 10hr course on YouTube

[sudden tundra]

If they add up to 185 instead of 180, can we still use the AAS theorem and say they are congruent?

[silent plank]

no

if the triangles can't even exist it doesn't make much sense to talk about congruency/similarity

[sudden tundra]

Ok, I see

[grave pond]

Are you sure the black numbers are angles? They have no degree signs.

[crystal quail]

Anyone know a decent proof of the existence of triangle similarity? So far I've gotten two triangles that have 3 pairs of congruent angles, I just have to show that these triangles have proportional sides. Couple explanations I've seen online just hand-wave it and don't really give anything that rigorous. Everything else I find is similarity tests. I want to prove that similarity exists. That two triangles with congruent angles have proportional sides

[grave pond]

http://aleph0.clarku.edu/~djoyce/elements/bookVI/propVI4.html ?

[silent plank]

Are you sure the black numbers are angles? They have no degree signs.
didn't catch that

@sudden tundra the 65 in black wasn't an angle but the length of a side so you shouldn't be adding that to the 85° and 35°

[grave pond]

On the other hand, if they're side lengths, then it's pretty unclear which of the sides is 85 long.

[silent plank]

yeh, it's poorly labelled and looks very deliberate

[sudden tundra]

Hmm, it's asking me to "Determine if their is enough information to determine if the triangles are similar by using the Angle Angle Similarity Theorem. Then write a Similarity statement."

[silent plank]

since it isn't explicitly shown whether the required angles are congruent, you should start by determining the third angle in each triangle

[sudden tundra]

Alright

[bleak badger]

Hey, I'm having a hard time understanding sinusoidal phase shift equations, does anyone know a helpful video that explains it?

[storm portal]

I'd guess that you can probably find one on YouTube from The Organic Chemistry Tutor

[bleak badger]

Thanks

[crystal quail]

hey sorry for the late response. I was thinking about this problem and just passed out 😅 . Thanks for the link though, the site looks very useful. I am curious as to this proof though. 4 lines from the bottom in the proof, Euclid states that, since FACD is a parallelogram, these 2 proportions (BA : AF = BC : CE) must hold. I'm finding that a little hard to justify. It doesn't quite click with me.

Oh I didn't realize he included Lemma's in the side of the proof. I found the one referring to that. I'll check that out first

[crystal quail]

Alright I'm stumped again. I have this proposition in Euclid's Elements that I'm trying to wrap my head around. Am I misunderstanding something? Triangles ABC and DBC are not equal in any way I can gather. In GeoGebra I formed two triangles on the same base, and on the same parallels, and none of the angles, sides, or even areas were congruent. What does he mean by saying they equal each other? Am I misunderstanding?

Link to it is here
http://aleph0.clarku.edu/~djoyce/elements/bookI/propI37.html

[dark sparrow]

they are not equal indeed

they are equal only in area

[crystal quail]

ah fuck. when i did the areas on my diagram I looked at the wrong triangles. Thank you 😦

[upper karma]

can soembody check if this si right

points are r=2,0.333 s=2,-4.667 T=-3,-4.667

[upper karma]

can somebody check if this right'

[crystal quail]

looks good to me

maybe step 4 might be looking for something else though? Maybe something that says that bisecting means angle ADC is chopped into equal halves? I'm not good with writing proofs so don't take my word for it. But that'd be the only thing I can think of if this isn't correct as is

[upper karma]

thanks!!!

[cold meadow]

I think it's C.

The original statement:
A quadrilateral can only be inscribed in a circle if it is a square
From which, follows:
Non-square quadrilaterals that can be inscribed in a circle do not exist

Statement from C:
All quadrilaterals can be inscribed in a circle
From which, follows:
All non-square quadrilaterals can be inscribed in a circle

It is also given that non-square quadrilaterals exists, therefore non-square quadrilaterals that can be inscribed in a circle exist
This is a contradiction, proving the original statement false.

[jolly grotto]

Thanks a lot, i will think about it.

[upper karma]

i've been trying to get the line of best fit, but it always gets me undefined

can anyone help

[upper karma]

oh this deosnt work on a set of colinear points, if only they wrote it down

wasted 2 hours in this

[surreal trout]

is this correct guys?

[sharp delta]

yeah it is

[hollow plume]

FH and DH are perpendicular to each other

That should give you a hint

[upper karma]

how can i calculate trig functions that are 90 degrees+ ?

[raven verge]

@upper karma you just keep going around the circle, basically

[upper karma]

instruction unclear

[raven verge]

so theres the pythagorean identity (sin(x))^2) +(cos(x))^2)=1 which is like pythag on a normal triangle

so its sort of like sine is the height and cos is the width of the triangle, if the hypotenuse(going around the circle) is set to 1

[upper karma]

cos = x while sin = y

[raven verge]

if you make the hypotenuse point around to 180 degrees, then you find the length of the width, (though not in absolute value)

yes

[upper karma]

how can i prove that cos = x?

[lime yew]

On a unit circle, the hypotenuse is the radius of the circle, which is one

[upper karma]

yes

oh so hypotenuse is given

i seee

x = cos data because

adj/hyp = adj

ncie

[lime yew]

Right

And you can see why $\sin^2(x) + \cos^2(x) = 1$

[somber coyote]

TheBlocker66

[upper karma]

👍 thank you so much guys

[lime yew]

You're using Khan Academy right?

[upper karma]

yeah i have some spare time during the summer so i decided to learn something new

[lime yew]

You should probably watch their video on the unit circle, considering how it is important in trig

[upper karma]

i did like 3 times 😭

it shown an example of a something that is less than 90 degrees

i got confused when it showed me something greater than 90

[lime yew]

Hmm

[upper karma]

i dont know how you can make that blue line when its like yk greater than 89

[lime yew]

Have you tried watching other videos?

[upper karma]

yes and it showed me "all students take calculas"

can somebody help me with this

the center of the dilation is 2, -2

[raven verge]

the blue line just goes to the top left quadrant, then around

[lime yew]

Does this help?

[upper karma]

are those reference angles?

[lime yew]

Yes

[raven verge]

there's also this one: https://www.mathsisfun.com/algebra/trig-interactive-unit-circle.html

[upper karma]

OMG

my brain

OHHHHHHHHHHHHHHH

SO YOU DRAW THE LINES

ON THE NEXT

OK

the points next to it are the ones that i used

idk if teere right

[raven verge]

if the center of dilation is at S, shouldnt the corner of the bigger triangle also be at S? I'm not that familiar with how they set those kinds of problems up though

[upper karma]

thank you this helped tremendously

this is the original question

[upper karma]

should the lines be 5 units long or 5.5

[raven verge]

what would you do if the scale factor were 6/3, which is to say 2

[lime yew]

The length of the new base is 5 units

[upper karma]

yeah i just realized this as well

desmos weird

[lime yew]

I am pretty sure they should be 5 units though

[upper karma]

yes its 5

this is not right?

[lime yew]

It is

When you dilate a shape around a center, you scale the distance from the center, not the locations of the vertices

[upper karma]

the problem said the center of the dilation is at (2, -2)

so im pretty sure you don't change the location for point "S"

[lime yew]

Yeah

T is 3 to left from S, so T' should 5 to left, or at (-3, -2)

[sharp wagon]

Is anyone on

I need help on my homework

I need answers

and I have a time limit

[topaz vector]

Can someone help me with this?

My eyes genuinely hurt looking at it

[dark sparrow]

i suggest giving some names to points

this looks like a 3D diagram btw

[topaz vector]

i have like

weird eyes

and it's making me strain them

geometry <

trigonometry<

[topaz vector]

i don't rly know. I just need to find x but i'm not really sure how to since hte lesson didnt explain it well

[dark sparrow]

...again, i suggest that you give names to the 4 relevant points in this diagram

[topaz vector]

x is the length

[dark sparrow]

reproduce it on paper

and give names to points like A, B, C etc

so that we can talk about things more clearly

[topaz vector]

it doesn't tell me... >.>

[dark sparrow]

you have the freedom to choose your own names

it doesn't matter which ones you go with

[topaz vector]

Oh okay

[dark sparrow]

use whatever four letters you like most

[topaz vector]

Like that?

orr

o u meant the points

LMFAO

[dark sparrow]

i meant points and i said points

[dark sparrow]

ok

consider finding angle DAB and from it finding DA

then AC by using the right triangle DAC

[topaz vector]

how do I do that...

oh wait

so this is easy just

180 - 51 - 65 yes

64

[dark sparrow]

yes

[topaz vector]

DA is a 90 degree angle yes?

wait what is DA referring to

[dark sparrow]

DA is a line segment

[topaz vector]

63 ig??

[dark sparrow]

where did you get that from?

[dark sparrow]

that does not sound right.

...

what

[topaz vector]

i thought that it's a 90 degree angle?

[dark sparrow]

i asked you to find DA. the length of DA.

[topaz vector]

how do I do that

I actually have no idea what i'm doing ;/

[dark sparrow]

like, sure, angle DCA is 63°, but that's barely relevant to us

[dark sparrow]

law of sines to the rescue

[topaz vector]

...what is the rescue

[dark sparrow]

"___ to the rescue" is a fixed expression lol

[topaz vector]

i feel really stupid right now.

[dark sparrow]

apply the law of sines to triangle ABD.

||DA/sin(DBA) = DB/sin(DAB)||

[topaz vector]

so

225 x sin65/sin64

226.9

x/226.9

[dark sparrow]

do not use the letter x for multiplication

[topaz vector]

x=115.6

asterisk makes it bolded

sometimes

so I avoid it on discord

[dark sparrow]

\*

[topaz vector]

*

How u do that

[dark sparrow]

also if you put spaces around asterisks it never gets read as formatting

[dark sparrow]

monospace stuff is `backticks`

[topaz vector]

\*

115.6 seems to be our answer

I don't like using diagrams

[jolly grotto]

hello all,

lets consider two spheres in the space

respectively S1 radius = 6 and S2 radius = 3 and center distances is 3.

Then, Are they intersect in a plane or have a common point?

If they are circles exactly would be intersect in a plane, but what about spheres?

[pliant roost]

two distinct spheres can intersect in a circle, point, or not intersect at all. in this case the intersection must be a point, because the center distance + S2 radius = S1 radius

if you have two distinct circles, they can intersect in two, one or zero points, so I don't understand what do you mean by saying "circles exactly would be intersect in a plane"

————————————————————
found a problem online, kinda stuck on it

Opposite sides of an inscribed quadrilateral ABCD intersect in points E and F. Prove that the A-Humpty point of the triangle AEF lies on the circumscribed circle of ABCD.

I've approached it in two ways: trying to prove, that if P is the Humpty point of the triangle AEF, then it must lie on the circle, or trying to prove that intersection of a median an the circumscribed circle is the Humpty point of the triangle AEF. the second way seems more promising, but I am not sure

[jolly grotto]

oh sorry my bad, i thought r1+d<r2

[pliant roost]

also you need to review cases when angle BAD is obtuse and acute, and I have stopped at trying to prove the obtuse case first

[pliant roost]

no problem

[jolly grotto]

ok in this case one common point, not intersect in a plane

[pliant roost]

https://www.geogebra.org/calculator/m2musaak here is the geogebra of the task, I've used to make a normal drawing of what's going on

[pliant roost]

yes

[jolly grotto]

Thank you very much

i appreciated

[pliant roost]

please try it, I am really stuck on it. for the second way I tried different ways like using Ceva, seeing that a rotational homothety must appear (I proved one angle for the triangle similarity, but couldn't prove the other one) and stuff like that. I still didn't understand how to use that we have a median, so it is possible to focus on that

here's the most minimalistic drawing without any more drawn structures

[pliant roost]

for the first way I tried to prove some angles, but got stuck and thought that this was not a good approach. I am not very sure it isn't, so you can try doing it first too

also, I don't want a full solution, but some good ways to go and things to prove first at least, would be cool if you can find multiple ways to solve this task

[upper karma]

can somebody help me with this

[hollow plume]

Act each step out, then you can put them in order.

[upper karma]

can somebody cehck if this is what its telling me to do

[valid harbor]

trollish troll

no me

who do you think i'm talking about

[slim grove]

How do I find the radius of a circle

[knotty plover]

what information do you have about the circle

Diameter = 2r
Circumference = 2pi*r

if youre given one of those you can find "r"

[bleak birch]

how do we get RHS for cos(arcsin(u/2))?

oh nvm

[wanton merlin]

identify all the numbered angles that are congruent to the given angle, justify your answer

[median shale]

/..

[brazen stratus]

use vertical op property and supplementary ez

[wanton merlin]

is this right for question 1.

[hollow plume]

You forgot to add something:
Angle CAB is congruent to Angle DAB
And Angle ABC is congruent to Angle ABD

[pulsar garden]

Hi to everyone, basically im trying to learn geometry from myself and i wanted to ask you: is there only a way to demonstrate the problems?
Basically i just have to demonstrate that the angle aOc is congruent to the angle bOd.
i already know that i could just demonstrate that with the addition and the difference of the angles, but i would like to know if even this answer is acceptable

where i wrote "oppure"(basically at the bottom of the paper) there is another way to prove

[wet mortar]

can someone help me with trigonometry

[limpid flint]

Yes

[lament kite]

Anyone know how I might go about finding a hyperbola with just the foci?

complete nvm i am actually blind

asymptote was right there on the page 😅

[zealous rapids]

Hi, can I solve this equation for θ, θ - sin(θ) = 2*π/n?

n is constant here

[upper karma]

can somebody help me with this

[rocky cave]

https://media.discordapp.net/attachments/993867856624230504/997233545091219456/unknown.png

can someone help me with this

[near crescent]

C=(a, a)
The slope of Segment AC, when simplified, is equal to 1
The product of the slopes is equal to -1

[hollow plume]

I know the answer, it’s just I don’t know how to help him get to the answer

[near crescent]

lol

same

[upper karma]

whats b

is it a?

a,0

[near crescent]

to help you, try factoring the expression into 2 more smaller expressions and then try, I guess, using the factors as the length and height of the traingle multiplied by 2

or smth: im not very good at explaining whatsoever

[near crescent]

you filled out b?

it is (a,o)

becuase all sides of a sqare are equal by the definition of a square

[rocky cave]

https://media.discordapp.net/attachments/997225358162149478/997249778356334602/unknown.png

this was the answer ;-;

[near crescent]

yeah...................

did u want the answer?

or help

........

sry-

[rocky cave]

help

i just wanna understand

but its so hard

[rocky cave]

i didnt try the multiply by 2 part

lemme see if that works

[upper karma]

can somebody chekc this

[hollow plume]

You’re correct

[exotic linden]

hii I need help with the diagram

[upper karma]

can somebody check this for m

[rocky cave]

someone please

im so desperate

[rapid flare]

27

[rocky cave]

my lord and savior

how did you get 27?

[rapid flare]

b=10
So 7b²-31b-20=370
lb=370
So l=37
37-10=27

[rocky cave]

thank you so much

[rocky cave]

where does the 370 come from?

[rapid flare]

subsitute b=10

[rocky cave]

so do 7(10)^2-31(10)-20?

[rapid flare]

Yes

[rocky cave]

i got 1010

[rapid flare]

700-310-20

[rocky cave]

aaaaah

yep, got 370

imma try it for another question

thank you so much

actual life saver

[rocky cave]

one last question

where did you get 37 from?

370 divided by 10?

im guessing

[rapid flare]

Ofc

[pure granite]

any good book for geometry?

[peak fossil]

For Euclidean geometry, imo Khan Academy is the best

[grave nova]

I haven’t a clue on how to solve this. I have a vague idea of what a locus is but idk how do this

[grave nova]

My guess is that it’s a circle defined by the distances from A and B to P…?

Ok I figured it out

It’s a circle

Not too hard actually

I was overthinking it

[idle chasm]

Can someone help me?

Please

<@&286206848099549185>

[hollow raft]

locus meaning plural locations?

find the place(s) where p is located?

Wouldn’t you just plug in each x

[idle chasm]

?

[hollow raft]

2p^2 + 0p^2 = 20

so 2p^2 = 20

p^2 = 10

p = sqrt of 10??

[hollow raft]

oh wait you aren’t the other dude

[idle chasm]

No

[hollow raft]

Arcs are the same as their inscribed angles

If I’m not mistaken?

I’d use that here to start

[idle chasm]

U done yet?

Wumpus

@hollow raft

[hollow raft]

Nope

I’m providing help not doing it for you

[idle chasm]

Where

I am new so I don’t know where to ask

[hollow raft]

Do you know what an arc is?

[idle chasm]

I just need help setting the problem up

[idle chasm]

Yes

[hollow raft]

Do you know what an inscribed angle is?

[idle chasm]

No not really

[hollow raft]

Do you know what equal means?

[idle chasm]

Yes

[hollow raft]

Inscribed angle means angle inside the circle

[idle chasm]

Ok

[hollow raft]

So the inscribed angle here would be 2x + 17

[idle chasm]

Yes ok

[hollow raft]

using the three definitions you know what I said earlier

They are equal if im not mistaken

[idle chasm]

?uh

[potent garden]

i can help with apex learning geometry if someone needs just dm me and i got you

[stark sentinel]

who cares about trig bro just remember that sin(x)≈x 👍🏻

and of course pi≈e≈3

[near crescent]

....bruh

[near crescent]

You have to first know that the measure of the arc is 2 times the measure of the inscribed angle

Try to solve from there

[wet mortar]

pls help

[silent plank]

is there more information

[wet mortar]

nope

only this

i think the question is invalid

[vital maple]

turn the tan into sin and cosine

[silent plank]

then there would be two possible values

[vital maple]

yes

i guess find both

[wet mortar]

tan=sin/cos???

[vital maple]

yes, then rearrange that

[wet mortar]

ooo

[silent plank]

consider where tan is positive
and do the two cases separately

[wet mortar]

aight

[silent plank]

draw your reference triangle and apply the appropriate signs for sin(x) and cos(x) for each case

[wet mortar]

mm ok

[royal citrus]

What would be the angle alternator to theta

[royal citrus]

got it

[near crescent]

Gj!!!

[winter scaffold]

I'm stumped on why the bottom step is shorter than the rest of the steps in the following picture.. Could someone explain the reason?

I'm building stairs, in this case this is called a Stair Stringer.. and I'm wondering why this Stair Calculator suggests I use a smaller "Rise" calculation of 6.25" for my first step riser..

I would prefer all the step risers to be the same going up to prevent trip hazards obviously. The rest of the upper step risers are all 7.75" ..

My total rise is 69.5" and total run is 84.5625"
Any help explaining the reason would be appreciated.

(Hopefully this math channel is the correct one)

[languid ore]

I can't help but notice that 6 and 1/4 inch is 7 and 3/4 inch minus the height of the "treads"

Is it possible that it is in fact 7.75 inches high when you factor in the tread?

[winter scaffold]

First step as shown to my view says that the first step in whole including the riser and tread, it would be a total of 7.75 yes.. but the rest are already 7.75 riser alone.. plus the tread it'll be 9.25"

[languid ore]

The diagram seems to indicate that on at least one step that is the height of the step plus the tread

[nocturne remnant]

The steps appear to be all of the same size if you include the treads

[nocturne remnant]

For the rest of the steps if you put in treads that will raise the height of both the lower and upper part of your step so the height you step actually doesn’t change

If that made any sense

[languid ore]

And if you multiply 7.75 times 9 you get your total rise, approximately. 69.75, as opposed to 69.5

I notice that the decimal column gives more precise measurements

Which may or not be practical to cut, idk

Haven't done woodwork of that sort since I was a teenager

[winter scaffold]

Erm no, but the first step is still 1.5" shorter.

[winter scaffold]

I did not notice that, thanks!

[nocturne remnant]

Yeah my point is that the rest of the steps are 7.75 regardless of the treads

But the case* is different for the first step only since you’re not also putting treads on the floor

[winter scaffold]

Well that's true, so the question is.. why would the first Step Riser have to be shorter by exactly 1.5" (Size of a tread). It would simply make that first step shorter, making the next step a trip hazard thinking it could be the same height that you'd have to raise your foot in when in fact its not.. Get it?

I'm confused by that. lol.. i dont understand the reasoning why this calculator did it like that.

[nocturne remnant]

The calculations are correct

I’ll try to draw a diagram or sth

[languid ore]

it's pretty handy that they have a nice canned up tool for this though

[winter scaffold]

OHHH!!!!!! I just understood it!! My god.. it took me hours to try to figure it out. @nocturne remnant , @languid ore . You both made me think more than i have before and finally get it.

Let me explain and see if it makes sense.

The bottom riser only needs to be 6.25 because the tread will make it 7.75.. so the next riser comes and its cut a 7.75 height.. but since the tread is taking up 1.5" space out of the 7.75".. then we are left with a 6.25" riser to work with.. the then the TOP tread on that second riser raises it again to 7.75"!!!

I was forgetting to add that top riser and subtract from the rest.. that's all..

[languid ore]

Yeah it took me a minute or two to visualize that as well

I spoke up before I really saw that

[winter scaffold]

I didn't want to cut an expensive 2x12 first because it didnt make sense to me.. but now it does. It makes sense

[winter scaffold]

I figured it out, but I'd still be cool to see your diagram if you ended up doing it

[languid ore]

Is this staircase going to descend onto dirt? (or grass, etc)

[winter scaffold]

No it will be sitting on a concrete step that is 8" high. So it's 1/4 higher than these steps

The stairs will never be in standing ground/water

[languid ore]

Ah. I was just fretting in my head over any error involved. I.e. how much wiggle room you have if the stairs descend too far once you cut them

[winter scaffold]

I see i see, well after looking at this more and finally understanding the reason why the first riser is smaller it makes sense now. I think I will just double/triple check my Total Rise and Total Run to make sure I'm good to go before i begin to cut

[languid ore]

But I suppose it's not that big of a deal. It could be worked around

[winter scaffold]

Should be fine as long as my run/rise is accurate. The concrete step is already in place

So I just have to be on the money with those numbers

[languid ore]

Yeah

[nocturne remnant]

I think you’re doing great without the diagram i was going to draw

Nice

[upper karma]

you guys mind if i ask really simple questions

[nocturne remnant]

Go ahead

[upper karma]

😔

sure

[winter scaffold]

Well thanks again guys, off to bed. @nocturne remnant @languid ore

[languid ore]

☮️

[upper karma]

im 8th grader starting 9th grade math so

xD

[languid ore]

Is that in Vietenamese?

[upper karma]

yes

how to calculate AC

😩

[languid ore]

Been a while, but I think they want to use the fact triangle ACH is similar to triangle ABH

And by similar I mean they have the same internal angles

Which means that their sides are proportional to one another

[upper karma]

oh

[languid ore]

Is that 19.9 (19,9) on one side?

It's a little hard for me to tell for sure from the photo

[upper karma]

14,4

[languid ore]

ah

[upper karma]

10,8

18

so i guess AC^2 = BC × CH

but i don't have BC yet

[nocturne remnant]

You can use another pair of ratios
Consider AC/AB=HA/HB

[languid ore]

^ that

It may be helpful to redraw the triangles ACH and ABH on their own, taking care to match up identical angles within them

You know at least that they're both right triangles. That said, you don't need the Pythagorean theorem to solve this.

And you don't have to solve for the other two angles either, just identify which ones match up

[nocturne remnant]

Interesting
I used triangles ABH and ABC

but both comparisons arrived at the same result

Very cool

[languid ore]

Yeah i didn't mean to give the impression that there was only one approach, if I did

[violet raft]

I’m going into Geometry next school year, anything I should do to prepare?

[hollow plume]

Circles, Proofs, and Basic Triginometry

[violet raft]

Trig is Senior year, unless you Ace geometry

[hollow plume]

I’ve been taught trig in geometry

Well basic trig

Only sin cos tan, and laws of sine and cosine

[violet raft]

gotcha

[hollow plume]

The bulk of trigonometry is in precalculus

[violet raft]

Okay thanks

[upper karma]

whats the difference between a similar triangle amd a Side-Side-side similarity theorem

[hollow plume]

Side side side is when all sides are proportional

Similar triangles is when all sides are proportional and the angles are congruent

[karmic cliff]

side side side implies similarity (and congruence)

[nocturne remnant]

Two triangles being similar refer to the fact that they have the same “shape”.
Meanwhile, the Side-Side-Side similarity theorem states that two triangles are similar if all 3 pairs of their corresponding sides are in equal proportion.

[teal yoke]

Alright can u guys help me on 3 and 4?

I’m kind of confused on those 2

Can you guys also provide a step by step explanation on what to do on those types of problems?

Thank you guys

[upper karma]

3 is 10 i think

can somebody tell me what ~ means i forgor💀

[teal yoke]

That symbol means congruent

So triangle bac is congruent to triangle dec

[silent plank]

similar not congruent

[teal yoke]

My bad

[sharp dawn]

Help

[serene rock]

Anyone can?

[upper karma]

wowthis looks cool

[serene rock]

Yeah ...

[dark sparrow]

whoever wrote this problem confused the words "intersect" and "inscribe"

[spiral bough]

I guess the answer they're looking for is ||4+5=9||, but there is a small problem with that

[serene rock]

What is the problem ?

[spiral bough]

no problem, my mistake

[serene rock]

please tell me details about your answer...

your answer is correct...but i cant understand clearly...

[spiral bough]

Ok, I'll dm you

[upper karma]

a+b should ne 26

[spiral bough]

No, why?

[upper karma]

can u send me the soln

[spiral bough]

Sure, here or dm?

[upper karma]

dm

[lime python]

Hey my friend has been struggling with trig and she has finals coming up. I am going to invite her here.

[undone fossil]

coming up? r u in the us

[lime python]

Yes in California to be exact!

Summer classes though

[undone fossil]

ohh

i was going to say, school ended like 6 weeks ago

[lime python]

School never ends Lolol

[undone fossil]

lol

imma have to start hs in a month 😭

[lime python]

Good luck!! The work is easier than college!

[undone fossil]

I'd hope so

and thanks!

[dawn dirge]

hard problem,can someone solve?

[vital maple]

coordinate bash

[dawn dirge]

CAN YOU?

[vital maple]

[teal kayak]

does geogebra 3d have any way of making a plane tangent to a point on a sphere

[spiral bough]

https://youtu.be/ty_ENfQ8YeI

[gloomy sapphire]

Can someone help? 😁

[hollow plume]

You need the “given” reasons first

[short patio]

Can someone help>

[upper karma]

Altitude

[opaque ibex]

i have a question

A(1,1) and B(4,1) are two points on the coordinate system, and point P is a variable that contains certain distance value from A and B. So, |PA| / |PB| = 1/2 is knowsn and the question is line equation of the possible values ​​of the point P

what can i do

is problem clear

[opaque ibex]

what i'm actually asking is how to find a set of points placed a certain ratio away from two points

[smoky jetty]

I think you're supposed to find the hypotenuse using Pythagorean theorem, wherein c^2= a^2+b^2 for no.3 problem

and for no.4, use the pythagorean theorem as well

[opaque ibex]

frankly i don't know if this set of points forms a circle, a line segment, or an ellipse

[dawn dirge]

can someone solve?

[dawn dirge]

😄

okay check dms

[dark sparrow]

why not send it here

[indigo hinge]

Can someone help me solve this impossible problem?

[dark sparrow]

did someone tell you outright that this is impossible?

it doesn't look that way at all from my perspective

perhaps you could say what exactly about this problem is making you consider it impossible

[indigo hinge]

nAH

[dark sparrow]

right

[indigo hinge]

just sarcasm but i just dont know what to do with only one side

[dark sparrow]

well you're given one pair of sides

[indigo hinge]

yeah

[dark sparrow]

and that's enough to find the scale factor from the smaller shape to the bigger shape

[indigo hinge]

mhm

[dark sparrow]

the perimeter will scale by that same factor

while the area will scale by its square

[indigo hinge]

ok

[indigo hinge]

its the area part im stuck on

fsr idk

[dark sparrow]

have you found the scale factor that i mentioned

[indigo hinge]

ye

[dark sparrow]

and what did you get

[indigo hinge]

2:3

[dark sparrow]

right

[indigo hinge]

OHHHH

[dark sparrow]

is that so? do you have a counterexample?

[indigo hinge]

treue

[dark sparrow]

so what

[indigo hinge]

TRUEW

[dark sparrow]

if you're so insistent, you can split it into two triangles

but area scales with the square of the scaling factor no matter what 2-dimensional shape youre dealing with

it even goes for surface area of 3-dimensional objects, so

i take it you do not have a counterexample to corroborate your "not necessarily".

[indigo hinge]

true

https://tenor.com/view/shaking-dog-white-dog-scared-meme-gif-23585947

[dark sparrow]

??

koce, are you sure we're on the same page here

do you have work to show that confirms your point

show

the scale factor from the big to the small is 3/2
so the ratio of areas is 9/4
which means the small trapezoid has 4/9 the area of the big

that is what you seem to be arguing against

where are you getting it from that the area of the small trapezoid is 5?

you haven't shown any work so far, you know.

you haven't shown any work so far, you know.

you haven't shown any work so far, you know.

[indigo hinge]

AINT NO WAY

[dark sparrow]

i hope it is readable.

i'm starting to suspect that you might be full of shit, but who knows.

[indigo hinge]

omfg

[dark sparrow]

and what have we here?

what are S_1 and S? @spiral bough do you have a diagram in which the areas you denote as S_1, S_2 and S are marked?

[indigo hinge]

check

btw the answer i got for the Area was 30

[dark sparrow]

30? sounds more like it should be 20.

[indigo hinge]

y

[dark sparrow]

is that "yes" or "why"

[indigo hinge]

why

[dark sparrow]

the ratio of areas is 4:9 (unless Koce here is about to prove us all wrong with a revolutionary insight which has a chance of being revolutionarily wrong)

okay, hold on

am i to understand the smaller trapezoid is on the left and the larger trapezoid on the right?

ok, let's see...

[spiral bough]

That’s the answer for p

[dark sparrow]

$\frac{45 - \frac{9S}{4}}{S_2} \neq \frac{45 - 9S}{4S_2}$.

[somber coyote]

Ann

[dark sparrow]

you made an algebra mistake.

[spiral bough]

Omfg yes

[dark sparrow]

so in fact this line just is not true

$\frac{180 - 9S}{4S_2} = \frac{9}{4}$ would've been correct here.

[somber coyote]

Ann

[dark sparrow]

and i repeat,

do you have a counterexample?

the last time you claimed to have a counterexample, it was with this problem by MrHippoP. but now, as i hope i've demonstrated clearly, that counterexample turned out to fall apart.

do you have another counterexample?

[dark sparrow]

by the way, if you continue with this, you will find that S + S_2 = 20, exactly as it should be.

well if you can't substantiate your claim with a counterexample then i don't see why any of us should just take your word for it.

[indigo hinge]

true

[spiral bough]

Wait I think I might be wrong

[dark sparrow]

every polygon can be broken up into triangles, and every curved shape approximated by polygons.
and you seem to accept that the square of the ratio does work for triangles, so...

[spiral bough]

Damn

[spiral bough]

Sorry, my dumbass again

[indigo hinge]

can you show why?

[dark sparrow]

the ratio of areas is 4:9

and the big one's area is 45

[indigo hinge]

proofz?

[dark sparrow]

@indigo hinge what do you want proof of

[indigo hinge]

proof to me how this is possible

https://tenor.com/view/gaming-monkey-monkey-gaming-gif-24891666

[dark sparrow]

...congratulations, you picked the least helpful answer possible.

and a gaudy gif to go with it.

[lilac crescent]

S to the I to the P I M P

[dark sparrow]

sipimp?

[storm portal]

simp?

[indigo hinge]

pims

[upper karma]

can somebody help me with this

[plain ingot]

I don’t even understand the question lol

So many ands

[upper karma]

ye

[native moss]

wattahel

[lilac crescent]

puebes

[frozen lotus]

how do you calculate the angle?

you know the coordinates for C, P and L

[dark sparrow]

it's the angle between the vectors CP and PL

[frozen lotus]

but whats the formula?

[dark sparrow]

look up "angle between two vectors formula"

[fast vortex]

How can I solve this with parallelogram law?

Should i add 3&4? To get an angle at 500N

[undone coral]

these ss were taken on my first attempt of the problem btw

[undone coral]

<@&286206848099549185>

[violet pelican]

hii could someone help me on this? its the 3rd one y=a sin x im struggling with

[opaque ibex]

$\cos(x) \leq \frac{\sin(x)}{x} \leq 1$

[somber coyote]

theman

[opaque ibex]

how can i prove this

[spiral bough]

This is only true for values close to $2k\pi$

[somber coyote]

DVD_Koce_DVD

[undone coral]

hi can i pls get some help on this

[spiral bough]

This is a theorem, $GI = \sqrt{FI.IH}$

[somber coyote]

DVD_Koce_DVD

[iron skiff]

1. Find the equation for the perpendicular bisector of a line segment, given that the end points of the line segment are C(4, 3) and D(7, 6).

could someone please help me solve this?

[spiral bough]

1)Find the coordinates of the vector CD, say (u, v)
2)Find the coordinates of the midpoint of CD
3)The equation of the perpendicular bisector is of the form:
ux + vy + c = 0, where you can determine c by knowing where’s the midpoint

[undone coral]

is this right?

[silent plank]

no

[undone coral]

its pi ik now

[undone coral]

pls help

[abstract mesa]

Costheta=1/sqrt65

Plug it into calculation and round up to 4 decimal llaces

Places

[copper nimbus]

Can "sin(4cosx)" be simplified further?

[copper nimbus]

is sin(arccosx) = tanx?

[clever viper]

is that byu

[undone coral]

ya bru

[undone coral]

is this right?

[turbid moss]

no

[versed valve]

Hey guys can you help me with something?

Suppose we have two circles with radius r1 and r2

If we only know that r1-r2=c,cεR can we find the equation for the intersection points relatively to c?

[undone coral]

<@&286206848099549185>

[gleaming portal]

can someone help me with this

[edgy bronze]

if sin35 and H = 12meters then x/H =0.57 how would you get to the 0.57 ? In a written method

[gleaming portal]

<@&286206848099549185>

[versed valve]

<@&286206848099549185>

[hollow plume]

What is the shape for plane R?

For C, a ray has a terminal point and has an arrow

For B, a line has 2 arrows

For D, a point on Plane R is a point inside of plane R

[gleaming portal]

What bout A

[hollow plume]

That is A

[gleaming portal]

no could u just give me the answers

and explain it to me

I want to understand

[hollow plume]

I’m trying to make you understand

I can’t give you the answer

[gleaming portal]

y not

[gleaming portal]

its a quadrilateral (parallelogram)

[hollow plume]

Ok

[gleaming portal]

is that what you say?

@hollow plume

.

it says give another name for plane R

[hollow plume]

actually it’s just a quadrilateral

[gleaming portal]

yea

ty

[hollow plume]

There’s no way you identify parallel lines

[gleaming portal]

yea

there isn't the sign

that shows that

so I guess you are right

and thanks for the help btw

really appreciate it

[hollow plume]

👍🏼

[dark sparrow]

NEI

[undone coral]

huh

[dark sparrow]

not enough info

unless we assume the boxes to actually be cubes, which nobody said had to be the case

[undone coral]

so its none of the above?

[dark sparrow]

would say so, yes.

[undone coral]

ok thanks