geometry-and-trigonometry

[quaint raven]

Da hell

[dense wren]

https://tenor.com/view/rocket-science-complicated-difficult-challenging-math-gif-23793443

[floral shale]

It is geography idk what you mean

Can't you see Germany on there?

Set any side equal to any other side (that has a different variable)

You'll have a system of equations

But it doesn't look possible lol

I mean there is a solution but it wouldn't make a square lol

[dense wren]

https://tenor.com/view/ok-its-gonna-be-gif-23529916

[floral shale]

Atheists be like

[opaque plume]

Add 3 to 3, 84 times

[eager bolt]

👍

[silent plank]

not quite

you're only supposed to do that 83 times

[earnest phoenix]

[upper karma]

Can I have help

[twin plover]

does arithmetic sequences questions asked here?

[surreal meadow]

you mean trig sequences ?

[twin plover]

ye

[surreal meadow]

wait

they are literally just formulaes

kinda difficult to memorise tho

ill still send them tho

@twin plover

here h is the common difference

what grade are you in tho? you prolly might not use this

[twin plover]

8

[teal kayak]

On geogebra how do I measure the distance along a circles surface between two points on that same circle (e.g the path an ant would take, NOT the straight line distance)

Surely it has a built-in way to do this and I won't have to die manually calculating it

[tulip shadow]

i need help

(like almost everyone else here)

i need to "determine 2 coterminal angles, one positive and one negative for each angle. give answers in radius." problem is for -20

what am i supposed to do

[dark sparrow]

does it say "give answers in radius" or "give answers in radians"? @tulip shadow

[tulip shadow]

apparently radius, but im really not sure

perhaps its a typo

[dark sparrow]

sure does sound like one

talk to your teacher and refrain from doing this problem until you receive clarification

[tulip shadow]

ok

[wheat dove]

https://cdn.discordapp.com/attachments/888753723189383234/984716136333598740/unknown.png

[loud shard]

can someone please help me with this question?

[dark sparrow]

@loud shard do you still need help with this?

[final yew]

it is literally tan(8x)=0

[dark sparrow]

not quite

tan(3x)+tan(5x) is not tan(8x)

[supple dune]

yea this is def wrong u can

can't add trig functions like that

wow this is actually really nice

its not tan(8x) but it is sin(8x) with both cos(5x) and cos(3x) defined

so basically u use def of tan

and get

$\frac{\sin(5x)}{\cos(5x)}+\frac{\sin(3x)}{\cos(3x)}=0$

[somber coyote]

JustKeepRunning

[supple dune]

multiply through by $\cos(5x)\cos(3x)$ (you can assume this is nonzero as otherwise the origianl equation would not have been defined, but u have to account for this at the end)

[somber coyote]

JustKeepRunning

[supple dune]

so u get $\sin(5x)\cos(3x)+\cos(5x)\sin(3x)=0$

[somber coyote]

JustKeepRunning

[supple dune]

and then use sin sum identitty to get solutions are just

values of x such that sin(8x) is 0 but neither cos(5x) nor cos(3x) are 0

and from here its easy to just compute the solutions

does that help

@loud shard

[twin plover]

can anyone help with sequences

[dark sparrow]

[final yew]

well , i can def eat up some steps right? tan(8x)=tan(5x)+tan(3x)/(...)=0 as tan(5x)+tan(3x)=0 , so we eventually end up at sin(8x)=0 keeping cos(8x) non zero , right?

[dark sparrow]

well you could do $\tan(5x) + \tan(3x) = \tan(8x) (1 - \tan(5x)\tan(3x))$

[somber coyote]

Ann

[final yew]

yes with $\tan(5x)\tan(3x)$ not equal to $1$

[somber coyote]

Eren Yeager

[supple dune]

wut do u mean by "eat up some steps"?

if u mean skip some of the steps i did yea sure thats fine

i was just going really in depth cuz like i was explaining it

[simple rain]

pls need quick help

[modern olive]

do you know which coordinate represents sine?

x or y?

[simple rain]

it was D

i made a triangle and it worked xd

[modern olive]

nice, it's good to know where the triangle shows up in the unit circle

[onyx hollow]

question is kinda wrong

sin(200) is a number on the y line

not a coordinate

[dark sparrow]

the coordinates of a point on the unit circle are (cos(theta), sin(theta))

this is in no way "wrong"

[onyx hollow]

exactly

[dark sparrow]

there is nothing wrong with saying that one of the coordinates of one of these points equals sin(theta)

[onyx hollow]

(x,y) cant be equal to a number

[dark sparrow]

one of the coordinates

[onyx hollow]

oh

[dark sparrow]

look at the answer options... they say "x coordinate of A", "y coordinate of B" etc

[onyx hollow]

i didn't read the answers 💀

[onyx hollow]

i was reading "coordinate" as a whole point

thanks for correcting me @dark sparrow

[night inlet]

Im a little stuck on the following problem:

Given an irregular triangle ABC and lengths A (the hypotnuse) and B, find C.

Ignore that sticker lol

Im just not sure how to move forward without knowing any angles and not sure how to derive the angle. (It seems like the shole problem amounts to deriving the angle between A and B)

[dark sparrow]

can you show the problem exactly as it was stated?

[nocturne remnant]

Not enough info unless it is specified to be a right triangle

[silent plank]

the seeming disregard for upper/lower case is questionable

you mentioned that the triangle was irregular. was that just suppose to mean that its scalene?
also you used the term hypotenuse. are you explicitly told that you have a right triangle (or are you improperly using that to refer to the longest side)

[night inlet]

I was improperly using the term hypotnuse to refer to the longest side. And yes by irregular I meant scalene.

(Geometry was a while ago, thanks for the terminology check)

[night inlet]

I now see that it is possible to draw more than one triangle with 2 fixed sides, even with the knowledge that one of them is the longest.

Thank you!

[night inlet]

So I was exploring this problem as a lemma to another problem, which is stated as follows:

The three sides of a triangle are of lengths l, m, and n, respectively. The numbers l, m, and n are positive integers l <= m <= n. Find the number of different triangles of the described kind for a given n [Take n = 1, 2, 3, 4, 5, ... .] Find a general law governing the dependence of the number of triangles on n.

[dark sparrow]

right, so you want to find the number of triples (l,m,n) such that l+m>=n and l<=m<=n

[night inlet]

I see. It seems to me that l + m must be greater than n (if they're equal then all points would be on the same line, no?).

[mighty hare]

Hello

ah nothing I am going to try it myself

[lofty tangle]

I don’t have a problem to ask help for but I’ve lost all my papers on radians and I need to re go over it before finals tomorrow, could someone give a quick summary of like circumscribed angles and inscribed angles

[supple dune]

just remember that 180 degrees = pi radians

inscribed angles is half the arc it inscribes (the angle formed at the center)

and that holds for edge caes as well (including tangents)

[lofty tangle]

Got it

Ty

[supple dune]

for circumscribed angles

ima assuming u mean the angle formed by two tangents

u don't really neeed anything fany

*fancy

just remember its 180 - central angle it circumscribes

but u don't really need to remember that u can probably just see that using like commn sense

[dense wren]

khan academy? xd

[simple rain]

ya

[tepid junco]

Can someone help me

With a question

Nvm I figured it out

[mighty hare]

Hello I need help

A bit ugly

AC=CB=CD=DE

BCE is 30degrees

Find BCD

[mighty hare]

No?

Since BC is not equal to BE

[jaunty girder]

oh my bad

sorry

[mighty hare]

so anyone knows how to solve it

[wanton prawn]

e=160 and f=80 and i need to find how long the side of the square is. inside the Rhombus

I can find a with pythagros

pythagroas

a= 40(5)^0.5

[strange hatch]

Are the "square's" end points on the middle of the sides of the rhombus?

If they are that's not a square.

[boreal bloom]

could anyone help me with this?

[strange hatch]

Well, I'm assuming that the arrows are indicating that the sides are parallel therefore ABCD is a trapezoid, and since KF is also parallel and spliting the sides in half it must be the midline (I hope that's how you call it). So, KF = (AB+CD)/2, and from there, by pluging in the values, you get x=3 I believe.

[supple dune]

are u sure this problem is well defined

i feel like theres not enough conditions to fix everything

[boreal bloom]

oh i see thank you!!

[strange hatch]

No problem 🙂

[humble stump]

what do i do next with 4x+18=(8x+12x)2

do i combine like terms

[strange hatch]

yeah

You should get x=2

No, its x=3, sorry

[wanton prawn]

they are not

[strange hatch]

still working on the problem?

[wanton prawn]

yeah kinda. The problem is I am not allowed to use "cathetus theorem" not sure if its the right english word. With them its a simple problem. but without its kinda challenging. Cause I dont know how the corners of the square sperate the sides of the rhombus.

[strange hatch]

What is the "cathetus theorem" exactly, I'm not familiar (I may know it I just don't speak english as a mother tounge).

[real wyvern]

In a right triangle a leg is a proportional mean between the hypotenuse and its projection on it.

a is the hypotenuse
b and c are the legs
m is the projection of leg b on the hypotenuse
n is the projection of leg c on the hypotenuse```

[strange hatch]

Thanks!

[real wyvern]

yep

[rocky wigeon]

A geometry problem
"Let ABC be an acute triangle
BE & CF be the height of the triangle, they cut each other at H
M is the mid-point of BC
On the line EF, construct a point X such that the angle XMH= HAM
Proof that AH split MX in half. In other words, call the intersection AH at MX is I, prove that MI=IX"
I put it in #help-0 but it didn't work out due to time out

[steady spire]

I tried to explain Geometric Algebra in this Twitter thread: https://twitter.com/nileshtrivedi/status/1536599538991087618
It's well within the reach for any high-schooler. I will be very disappointed if physics textbooks don't switch to this in next 10 years.

Nilesh (@nileshtrivedi)

@ravihanda The idea is actually really simple and well within the grasp of a high-schooler.

We created complex numbers a+bi, where i^2 = -1. This gives us a world where multiplication = rotation.

Why stop here? Let's also create j where j^2= +1 and k where k^2 = 0.

Twitter · Jun 14, 2022, 6:40 AM

[wise pawn]

Prepare to be disappointed. Not just because Hestenes has been beating that horse for years, but my understanding is historically after Hamilton discovered quaternions they were used for a while but just too cumbersome in practice, so Gibbs came along and started popularizing things like the cross product and dot product to not deal with quaternions anymore.

[steady spire]

True. But as a working programmer, there is a lot of practical utility in GA for me: https://twitter.com/nileshtrivedi/status/1536611520561414144

Nilesh (@nileshtrivedi)

I try to explain Geometric Algebra in this thread. If you loved high-school physics, you gotta out GA.

I once wrote a 2D physics simulation engine. Took me a month and ~5K lines of code. With this algebra in place, it takes less than 60 lines of code. And works in 2D,3D,4D etc.

Twitter · Jun 14, 2022, 7:28 AM

[rocky wigeon]

any ideas?

[untold willow]

are any helpers avaliable for vc for geometry

[trim breach]

You are given sides that are congruent. You know the end reasoning can either be SAS or SSS. I would look at what you know about intersecting lines.

[eager bolt]

,rotate

[elder geyser]

jn = mn ( given )
ln = kn ( given )
=> jn + ln = mn + kn
=> jl = km
angle nkl = angle nlk
kl = kl ( common side )
triangle jlk = triangle mkl ( sas congruency )
=> jk = ml ( cpct )
=> triangle jnk = triangle mnl ( sss congruency )

[cinder scaffold]

can someone help me on how i find out x ? (the two lines are parallel btw)

[dark sparrow]

use similarity of triangles

500/(500+10) = x/(x+13) or whatever other similarity statement you find most intuitive

[cinder scaffold]

ahh okay thank you

[sinful quartz]

can anyone help me with this 😭😭

<@&286206848099549185>

[supple dune]

its the first one

thats the only biconditional statement there

biconditional means it has to contain a phrase of the form "if and only if" or smth analogous to it

[sullen sequoia]

ive tried isolating var dont think im right tho

[nocturne remnant]

can you show us what you have tried

[shrewd heron]

I’m ashamed to say 8th grade math is making me struggle

[supple dune]

XY+ZY=XZ

so (2+3x)+(3+6x)=5+9x=23

which means that x=2

and so ZY = 3 + 6 * 2 = 15

[light rune]

Use SSS

[clever maple]

can anyone help me with a vectors question involving lines and planes

[light rune]

Yes

[signal herald]

May i get some help with a problem i have

[real wyvern]

X=(81/45) x40

[pearl ravine]

https://cdn.discordapp.com/attachments/625408817412440109/986748918043922472/IMG_1819.jpg

Can someone help me out here? I forgot I had to do this last thing for geometry and I’ve been procrastinating it but now I forgot how to do it ;-;

[sullen sequoia]

thanks

[misty wing]

somebody

anybody pls help :')

zoomed in

[glossy stream]

hello i have a question please why is i=i'. Thank you

[nocturne remnant]

There are 4 angles in the figure, i, r, r’ and i’
What relationships do you know between these angles?

[glossy stream]

I know that r=r' and that inside the triangle II"I', the angles I I' are respectively i-r and i'-r'

[nocturne remnant]

Is there anything you know between just i and r

[glossy stream]

only i and r ?

[nocturne remnant]

Yeah

Is this optics?

[glossy stream]

yes 🥲

[nocturne remnant]

Yea so snell’s law or sth

[glossy stream]

i is inside of air so n=1 so sin(i)=n2sin(r)

ahhhhhhhhhhhhh

[nocturne remnant]

Yeah then you can eventually get i=i’

[glossy stream]

ah ok now i understand it 😆

thank you very much

https://tenor.com/view/the-office-bow-michael-scott-steve-carell-office-gif-12985913

i was starring 1 hour at this exercise

[violet badge]

Does this look right? Doesn't seem like much for 8 marks from b) onwards, so I'm not sure if I've gone wrong. Don't have a markscheme either

Thanks ^^

[stark snow]

sorry whats x1 pls

can't see it

is it 10 or 16

[violet badge]

x1 is 10, x2 is -6

[stark snow]

tks

[violet badge]

sorry i have terrible handwriting

[stark snow]

ur not the only one

[silent plank]

there are some notation issues

you should have () around your 2,12 for you midpoint

[violet badge]

oh yeah, coordinates

[silent plank]

and near the bottom you shouldn't be using x for multiplication

[violet badge]

() again instead?

[silent plank]

multiplication dot would be ideal

using () to imply multiplication also works

[violet badge]

fair enough

is the answer alright though? i feel like i mix up the equations alot

[silent plank]

end result looks ok

[violet badge]

awesome, thanks

[robust crane]

if you have 2 polygon shapes - is there a way to know the maximum number of intersecting points the edges between the two shapes would have?

this is what i mean by intersecting

ignoring parallel overlapping edges which would have infinite number of points

[north heart]

depends on a few of factors

i very very very very much doubt there's a general answer

well

there probably is

but it would just be very tedious

[wanton merlin]

how do i do this?

[gleaming nova]

the two triangles i’ve highlighted are similar, with the highlighted angles equal

this will be helpful

[mint bridge]

I know that this exercise is written in Spanish but I would still like to solve it
translated says
5.If the circumference is inscribed in the triangle PBC. Calculate θ

help please

[dark sparrow]

there is no theta in the picture what the fuck do they want you to calculate

i see an alpha and a beta but no theta

also there are no point labels

[mint bridge]

that's exactly what's funny
take beta as if it were theta
It is an exercise that my teacher gave us as a challenge and if there are no labels, it is even more complicated to know what the angle is
I was hoping I could get some help ;-;

[dark sparrow]

well the only correct course of action is to throw this problem back in the teacher's face and ask them to label all the points

[mint bridge]

xd
I will take that into account

[dense bough]

lazy diagram as you just have to assume that the intersection points are where the circle is tangent to the lines

[humble bolt]

maybe if you express each line algebraically, and see where the lines intersect

[supple dune]

ok if u are solving for \beta heres a sol

green stuff is stuff u find red stuff is given

so basically from exterior angle you get that \angle AEC = \alpha + \beta

and then cuz AE and AD are tangent AE = AD and so \angle ADE = \alpha+\beta

and then it follows that \angle EAD=180 - 2\alpha - 2\beta

and then just sum the angles in \triangle AEC

so you get \alpha + \beta + 180 - 2\alpha - 2\beta + 24 = 180

which yields \beta = 24^{\circ}

[rocky wigeon]

@supple dune you seem good at geometry
Maybe you can help me with this problem?

[robust crane]

yeh but i am wondering if there is a way to know the maximum intersections before having to find out manually

[dull adder]

do u guys know what is 2+2

[silent plank]

its the kinda thing that could potentially get you kicked for suspected intentional trolling

[floral dew]

Is it possible to make a sine function that falls between two lines

right now I'm looking at X and 2x+1

So the troughs fall on the line X and the peaks are on 2X+1

[dull adder]

2+2 4 eh

[floral dew]

?

[upper karma]

can i get help in this channel?

[mint bridge]

thanks you are great

[upper blaze]

I love this hahaha

[floral dew]

I've come up with (x/2)*sin(x)+1.5x

x is tangent to the bottom of this while 2x but not 2x + 1 is tangent to the top

[floral dew]

figured it out completely:https://www.desmos.com/calculator/vsgmhuj87u

[cerulean scaffold]

A triangle ABC is given . Let Am and Ap respectively be the foot of the median and the height passing through A. Finally, let B' and C' be the feet of the perpendiculars drawn from By C to the bisector of angle A.

Theorem: Points AM, AP, B' and C' are concyclic.

Theorem: The previous circumference and the other two that are obtained in a similar way, have a radical center that coincides with the incenter of the given triangle ABC. The centers of these three circumferences are on the circle of nine points.

[supple dune]

i've been trying a bunch of different things but nothing seems to be working

i have a feeling u might have to use a projective sol

but my projective is a bit rusty rn

the other thing I was thinking about was using humpty dumpty points

but idt theres enough to actually prove that H acts as one, and in fact, ima pretty sure it isn't even true

do u have any hints or anything like that

[rocky wigeon]

Unfortunately, I also don’t have any idea
I’ll let you know if I have something

[olive stone]

I am see an alpha and beta but I dont see theta.

[desert steppe]

No specific homework help here but I've been having a really hard time working with radians.

I understand everything about what they are and how they work but I just can't work with them, I always convert the radian values into degrees (unless a formula requires radian values).

What has helped make reading and working with radians more intuitive for you guys?

[supple dune]

just keep working with them

u will eventually get used to them

especially in calc because radians are used all over the place

and conversion to degrees make calc really annoying

trust me u will get used to radians in no time just keep using them and they'll become like second nature to u

[desert steppe]

I get you, but I've been doing it for some time and still even if I don't directly convert it to degrees, that's where my mind always tries to do. Never really overcame that, and don't really see how to atm

[supple dune]

ok so for me

one thing that was really helpful

was htinking of the unit circle and the points on it in terms of radians

and i slowly got used to using them in general

[upper karma]

Hey y’all!!!

Any book recommendations for starting of with Geometry?

I wanna be the best at it, I didn’t come across a book that was worth my time. That’s why 😂

[thin holly]

Geometry by Jurgensen Brown King should be nice

as a book

there is a full course from scratch on Khan Academy

[grave nova]

If I have a pole that is rotating from the bottom (like the hand of a clock) and I know it rotates at some theta per second, how can I determine how fast the tip is moving given the length of the pole?

[nova bridge]

The distance moved is the arc length $r\theta$. Assuming length is fixed, then the speed is given by $r\dv{\theta}{t}$, i.e. the angular speed times the length.

[somber coyote]

PhenomPlasma

[real wyvern]

If you were told to prove that the area of rectangle is A x B then how would you do it?

I did by using herons extended formula

[dark sparrow]

youd probably start by asking for the definition of area on which to base your proof

almost all area formulas rely on that of a rectangle, so what you did is very likely a circular proof

[real wyvern]

Well I did lie saying sqrt( (s-a)(s-b)(s-c)(s-d))

Is herons formula

Just extended it to quadrilaterals

[dark sparrow]

did you prove that this extension works?

because i think it doesn't.

[real wyvern]

No but I think it does

[dark sparrow]

the area of a quadrilateral is NOT a function of its side lengths. there exist quadrilaterals with the same side lengths which nonetheless have different area

[real wyvern]

Well didn’t mathologer made video abt it

[dark sparrow]

consider these two shapes

they have all side lengths equal to 1

any formula you make using only side lengths will give the same value for both of these

but the area is different

do you agree or disagree?

[real wyvern]

I agree

Hmm

Well isn’t Brahmagupta’s formula just extended herons?

Which works

[dark sparrow]

In Euclidean geometry, Brahmagupta's formula is used to find the area of any cyclic quadrilateral

it requires an assumption to work

granted, all rectangles are cyclic,

HOWEVER all of this is beside the original point i was making

[real wyvern]

No I see where you’re coming from

[dark sparrow]

ALL area formulas in geometry, with the exception of a select few if any, rely in some way on the fact that the area of a rectnagle is the product of its dimensions.

[real wyvern]

I see

[dark sparrow]

until you have a rigorous definition of area (which you might not as of yet), you may be unable to actually prove in any meaningful way that the area of a rectangle is what it is.

[north heart]

area of triangles?

[dark sparrow]

really now? can you show me a triangle area formula that does not eventually rely on rectangles?

[north heart]

hm

could you not instead argue that every rectangle area formula relies on triangles

[dark sparrow]

well idk about you but from my perspective it's part of the definition of area itself that the area of a unit square is 1

whence one can derive (with some difficulty) that the area of a square with sidelength s is s^2

whence one can derive that the area of a rectangle is width times height

[north heart]

huh, is that really doable?

hm

[dark sparrow]

well it's easier for rational lengths i suppose

[grave pond]

For irrational lengths we'd have to do some very tedious definition-chasing (as in: what is an irrational number anyway?) if we're not satisfied with appealing to intuition and say rational arithmetic extends continuously to irrational and therefore our intuition should be good.

[keen condor]

hello im new here

[upper karma]

can someone explain me why in sinusoidal function we divide 2*pi By the period

[nocturne remnant]

The “default period” of sin(x) is 2pi if you’re wondering what’s the significance of the number 2pi

[upper karma]

Yea I know that

but I am confused as to why we divide it by the period of a changed wave (stretched or shrunked to be more specific)

[grave pond]

You want an increase of the variable of "one period" to become an increase of 2pi in the number you feed into the sine function.

The division lets you compute the scaling factor that achieves that.

[upper karma]

thanks got it

[tiny flicker]

Hello

I'm new can someone teach me some trignometry

[wise pawn]

You might have better luck trying to follow something like khan academy, a book, lectures, etc and trying to follow that. Then when you get stuck on stuff you can post what you've tried here and ask for help on a specific problem.

[tiny flicker]

Oh ok, thanks

Is there a learning section here though

Do you have any book recommendations for torsion homology?

[gritty kestrel]

torsion homology
I think Khan's academy videos on torsion homology is pretty good. So is 3b1b's video.

[real wyvern]

Hey Ann this would work though for area of quadrilateral always?

[dark sparrow]

maybe it would, i don't know

unlike what you tried to appeal to earlier this one involves (some of) the angles

[amber locust]

I'm trying to prove a similar proposition to Euclid I.1 ("On a given finite straight line (call it AB) to construct an equilateral triangle") but for isosceles triangles. What I'm thinking is (1) describe a circle with center A and radius AB; (2) select any point C from AB that not A or B; (3) describe a circle with center B and radius BC (here there'll be a point of intersection D of the two circles); (4) produce two straight lines AD and BD.

Most of the steps are similar to Euclid's construction of an equilateral triangle and I'm comfortable proving them, but I'm wondering if I'm justified in (2): there isn't any definition or postulate literally saying "for any finite straight line, there's a point between its extremities," but maybe it's a direct implication of a definition or postulate that I do have? If so, I'd think it'd be either post. 2 ("To produce a finite straight line continuously in a straight line") or def. 4 ("A straight line is a line which lies evenly with the points on itself"). Am I in the right direction?

Also, I know Proclus devised a construction for this that doesn't require selecting a point in AB, but it seems a bit over-complicated (maybe I'm wrong and it's actually the simplest solution):

[trim needle]

How do you make a sine wave more extreme? Similar to x^2 -> x^4?

[grave pond]

for any finite straight line, there's a point between its extremities
It's not a postulate, but proposition I.10 will produce such a point.

[amber locust]

Hm, that just shows that there's a point that divides some straight line AB into two equal straight lines, right? But it doesn't follow that there's some point that is in AB but still doesn't divide it equally, correct?

Also, I think I found a possible error (?) in my method: it fixes the height of the triangle, so it won't work to construct all possible isosceles triangles for that given straight line. Perhaps I'm missing something, maybe you can confirm. (Also, thanks for the answer.)

[grave pond]

Well if AB has a midpoint, then in particular that shows that at least one point between A and B exists.

If you want the legs to have a particular known length, you can use postulate 2 (and the implicit assumption that you can extend lines arbitrarily far) to extend AB to be long enough to use I.3 to set of C at the desired distance.

[amber locust]

So one point is enough?

[grave pond]

That was what you seemed to think needed an argument -- once you know for sure there is a point to pick, it's definitely acceptable to say "pick a point", no matter if the point actually picked is the particular one you're sure exists.

[amber locust]

Oh, ok. I wasn't sure if once you know for sure there is a point to pick, it's definitely acceptable to say "pick a point" was justified, but since it doesn't matter if it's not the one I know exists, then alright. Just to see if I understood this properly then: we know, from I.10, that a straight line can be bisected, producing two equal straight lines, which again can be bisected and produce two equal straight lines and so on. Then this proves that there is a point in the original line that does not bisect it; therefore, I'm allowed to say "pick any point of the original line."

[willow anvil]

Let ABCD be a convex quadrilateral such that AB = CD. If the points E, F, G and H are the middle of the lines AD, BC, AC, BD, respectively, show vectors EF and GH as linear combinations of vectors AB and DC, and then prove that the vector EF is normal on the vector GH.

[grave pond]

ABCD doesn't need to be convex.

If you translate C and D by some vector v while leaving A and C in place, then E, F, G, H will all be translated by ½v. In particular they keep their relative positions.

In particular, we can translate CD such that the midpoint of CD coincides with the midpoint of AB.

Then ABCD is a rectangle, and EF, GH connect midpoints of opposite sides, so they are necessarily normal to each other.

There's a degenerate case where AB and CD are parallel. In that case either E=F or G=H, so depending on whether we consider the zero vector to be normal to everything or nothing, it's either a trivial case or an exception to the claim.

[narrow stump]

are the points on the unit circle they make you memorize

ie $(0,1), (\frac{\sqrt3}2, \frac12), (\frac{\sqrt2}2, \frac{\sqrt2}2)$ etc

[somber coyote]

coseen™

[narrow stump]

are they the only points $(x, y)$ at $a\pi$ degrees around the unit circle

[somber coyote]

coseen™

[narrow stump]

such that $x, y,$ and $a$ are all algebraic

[somber coyote]

coseen™

[dense wren]

I didn't understand your question so imma be quick, these points are points for pi/3, pi/2,pi/4,pi/6, but you can actually memorize/prove points which are like 3pi/12, 1pi/12, 5pi/12 and so using trig identities. But these 3 you've showen are the main.

• there are points in different quadrants

[upper karma]

Hey y’all? Hope you are doing well?

If a straight line through C(-√8, √8), making an angle of 135° with the x-axis, cuts the circle x = 5 cosθ , y = 5 sinθ , in points A and B, find the length of the segment AB.

I am having trouble solving this, and understanding this as well? It’ll be a great help if you could explain this problem in steps.

[narrow stump]

yes but are those 16 the only points on the unit circle with both algebraic coordinates and an algebraic angle

(algebraic angle as in $a\pi$ where $a$ is algebraic)

[somber coyote]

coseen™

[grave pond]

All rational multiples of pi have algebraic sines and cosines.

[narrow stump]

wait rly

so like $sin(\sqrt\pi)$ is algebraic?

[somber coyote]

coseen™

[narrow stump]

o shi

alr

[grave pond]

sqrt(pi) is not a rational multiple of pi.

[narrow stump]

oh sry misread as algebraic

but $sin(\frac\pi8)$ is algebraic?

[somber coyote]

coseen™

[grave pond]

Yeah, since cos(pi/8)+i·sin(pi/8) is a root of z^8-1.

(In this case, you can use half-angle formulas to get an expression in radicals, given the known values for pi/4).

[narrow stump]

ohhhhhh that makes a lotta sense ty

[grave pond]

On the other hand, the Gelfond-Schneider theorem implies that an irrational-algebraic multiple of pi cannot have algebraic sine and cosine.

[amber locust]

Question about Euclid I.3: is it really a necessary proposition? Is it not just an instance of I.2? That is, given two finite straight lines AB (the greater) and C (the lesser) -- as stated in I.3 -- wouldn't I.2 be enough to cut AB at some point D such that AD = C?

[grave pond]

The proof is indeed by appealing to I.2, so you're free to think it is a fairly obvious corollary. But Euclid had no special word for "corollary", they're all just propositions.

[amber locust]

I see, that makes sense. Thank you.

[clever fable]

Hi, can someone explain the solutions for this question pls, it doesn't make sense:

how do they get $\lvert a \rvert^{2}$ = $\lvert b \rvert^{2}$ and $\lvert c \rvert^{2}$ = $\lvert a \rvert^{2}$

[somber coyote]

LordEmrys

[upper karma]

by pythag0rus theorem

[clever fable]

pythagoras of which triangle?

[clever fable]

is it pythagoras of triangle ABO and ACO?

[upper karma]

yesss

[clever fable]

kk

[tranquil grove]

hey can someone help me with the volume of a coin

i dont know how to work out the area of dodacagon and get the apothem

and then get the volume out of that

i know its pretty basic but my assignment is due in 8 and a half hours and i havent slept yet

[dark sparrow]

what's the coin

[tranquil grove]

50 cent coin

its an australian coin

[upper karma]

think of the coin as a cylinder

[tranquil grove]

ok

then what

[upper karma]

just tried, it wont work

[tranquil grove]

ok

i need to know how to work out the area first

then i think i remeber the volume

[upper karma]

Any more information about the coin?

apothem is rcospi/n

r=circumradius

[tranquil grove]

no like easy words

not smart yet

i know radius and apothem though

[upper karma]

then use the radius and apothem to find the area of a single triangle

since it's a dodacegon it must have 12 apothems

[upper karma]

yep

[tranquil grove]

arent the raius and apothem the same

[upper karma]

nah

[tranquil grove]

o

so which is which

radius is the one from the center going to the outer edge?

[upper karma]

the radius is the distance from the centre of the dodacegon to the point which intersects a circle, in simple words it's the distance from the centre to the edge

[tranquil grove]

ok

then whats the apothem

[upper karma]

the line from the centre that intersects the side of the dodacegon( or any polygon)

it creates a 90 degree angle

[tranquil grove]

uhhhhh

like this?

[upper karma]

let me see if I can find an image

[tranquil grove]

ooohhhh

so its the side x the apothem?

or am i stupid

then x 12 aswell

[upper karma]

yes

[tranquil grove]

OKIE

thankyou

and what would the formula be?

[upper karma]

for what?

[tranquil grove]

finding out the area for a polygon

[upper karma]

since u know the radius and the apothem, use that to work out the area of one triangle

a dodacegon has 12 sides right?

[tranquil grove]

ye

[upper karma]

so it must have 24 right triangles

[tranquil grove]

no

12 triangles i think

\

[upper karma]

two triangles for each side

[tranquil grove]

no only one side

but yes

but no

idk how to explain it

[upper karma]

I am talking about a right angled triangle

[tranquil grove]

right angle triangle.....

like from the apothem and radius?

then yeah

[upper karma]

yea

[tranquil grove]

so its (apothem) a X s (side) X 12

cuz teacher said something about 12 but not 24 i think

[upper karma]

strange

[tranquil grove]

what?

i found an image

that

i remember that

[dark sparrow]

do you have a physical 0.50 AUD coin on hand that you're expected to measure?

[tranquil grove]

ye

[upper karma]

yea if u know the formula for the area of an isosceles triangle

[dark sparrow]

right

[upper karma]

use that

[dark sparrow]

do you also have a ruler?

[tranquil grove]

ok

ye

[dark sparrow]

right

in that case, you should measure two lengths on the coin

[tranquil grove]

ok

which lenghts

[dark sparrow]

the edge to edge distance, which is 2a in the picture you just sent

and the length of one side

[tranquil grove]

ok

the side is

a cm

[dark sparrow]

both lengths would probably make most sense to be given in millimeters

[tranquil grove]

so 1cm

ok

[dark sparrow]

...are you sure it's 1cm tho

[tranquil grove]

so 10mm?

[dark sparrow]

that feels a little too big

but what do i know, i've never seen any australian coinage

what's the edge to edge distance?

[tranquil grove]

well the measurements dont have to be correct he said you just have to show working out

[dark sparrow]

it's better if they are correct

[tranquil grove]

but i already wrote a bunch of stuff down ok fine

[dark sparrow]

hm, 10mm might be okay actually

[tranquil grove]

Yay

[dark sparrow]

what'd you get for the thickness?

[tranquil grove]

like the apothem

[dark sparrow]

er

sorry

had things mixed up

the edge to edge distance

which is double the apothem

[upper karma]

dont u think u can just google this stuff, I mean u would be more accurate

[tranquil grove]

cant

he say no

[dark sparrow]

i think it's better if OP just measures it themself

[tranquil grove]

hey i know op means me but is there a reason why people say that?

[upper karma]

Oh ok

[dark sparrow]

it stands for "original poster"

[tranquil grove]

ok

oooooooooooooooooooohhhhh

[dark sparrow]

i personally say it out of habit

[tranquil grove]

yeah

[dark sparrow]

so yeah did you measure the red distance and if so what did you get

[tranquil grove]

doing it now

2.9cm

so

29mm>

i think

[dark sparrow]

29mm alright

looks close enough

[tranquil grove]

ok

[dark sparrow]

ok so you wanted the volume right?

[tranquil grove]

ye

[dark sparrow]

that means we need one more measurement

the thickness

[tranquil grove]

so measure hieght

[dark sparrow]

now for this

[tranquil grove]

THICCC

ok

[dark sparrow]

do you have multiple coins on hand?

as in multiple 50c coins

[tranquil grove]

YES

[dark sparrow]

if so there's a way to make your measurement of the thickness a little more precise

[tranquil grove]

WHY THE TEACHER SAY THE SAME THING
t

WTF

[dark sparrow]

oh, did your teacher also mention stacking several 50c coins?

stack them on top of each other and measure the thickness of the whole stack, then divide by how many coins were in it.

[tranquil grove]

he said bring 5 50 cent coins then figure out what they are there for

thanks

[dark sparrow]

alright, so you brought five of them?

i mean like, the more the better

[tranquil grove]

ye

[dark sparrow]

up to a point

[tranquil grove]

hold up let me turn on my light

[dark sparrow]

ok

measure the stack and report the measurement & the number of coins to me

[tranquil grove]

ok

14mm

5 coins

[dark sparrow]

great

that means each coin is 2.8mm thick

[tranquil grove]

so do i now like divide by 5?

[dark sparrow]

yes

[tranquil grove]

wait

[dark sparrow]

as i did for you just now

[tranquil grove]

oh

ok

[dark sparrow]

alright so we have our measurements

edge to edge distance = 29mm
side length = 10mm
thickness = 2.8mm

those are the figures you got, right?

[tranquil grove]

yea

wait

no

[dark sparrow]

oh?

[tranquil grove]

thickness is 2.8mm

[dark sparrow]

ah

mb

alright

[tranquil grove]

wait no thats good cuz your like testing me

makes me think im smarr

[dark sparrow]

it wasn't intentional on my part but good on you for catching the error.

[tranquil grove]

ok

[dark sparrow]

ok so we're picturing the coin as basically a dodecagonal prism. which means that to find its volume we find the area of its base (i.e. the face of the coin) and then multiply it by the thickness

so to find the area of the face, look at this diagram

the area of this orange triangle is $\frac{1}{2}as$, and the coin is comprised of 12 such triangles (because it is a regular 12-gon)

[somber coyote]

Ann

[dark sparrow]

does that make sense?

[tranquil grove]

uh what

hold on

imma process

takes a while

[dark sparrow]

half base times height, if that helps

[tranquil grove]

so like its a wierd triangle

but its still a triagnle

[dark sparrow]

it's not a weird triangle

it is a triangle just like any other

[tranquil grove]

oh

[dark sparrow]

maybe it's positioned in somewhat of an unorthodox manner

[tranquil grove]

yeah that word

[dark sparrow]

but we still have the right data to apply the half-base-times-height formula to it

[tranquil grove]

what does formula to it mean?

[dark sparrow]

...?

by "it" i mean "the triangle"

[tranquil grove]

your last three words

oh

[dark sparrow]

we apply the formula to the triangle

to get its area

[tranquil grove]

smacks face

[dark sparrow]

and then to get the area of the whole coin we take twelve times that

[tranquil grove]

wait

might be big brain move her

is it

base X height X 12

[dark sparrow]

don't use X for multiplication

also no not quite

[tranquil grove]

oh

[dark sparrow]

the area of a triangle is half its base times its height.

[tranquil grove]

so

[dark sparrow]

so the area of the whole coin will be 6as

a is our height and s is our base

[tranquil grove]

ok

what do i use for multiplication?

if not X

[dark sparrow]

with spaces around it if you dont want it eaten by discord

[tranquil grove]

that work?

[dark sparrow]

yes

[tranquil grove]

ok

so its

a * s / 2 x 12

because its a triangle

thats why half it

[dark sparrow]

it's (a * s)/2 * 12 yes

which is the same as 6as

as i wrote

[tranquil grove]

so im right?

or am i stupid again

[dark sparrow]

you are right

[tranquil grove]

wait

[dark sparrow]

you're just being a little inelegant about it

[tranquil grove]

for the stupif or smart

[dark sparrow]

.

[tranquil grove]

YES

LES GO

to i need too make it more elegant

ok

[dark sparrow]

i uh

...

i mean, you do you but thats not what i meant

[tranquil grove]

oh

[dark sparrow]

tbh like, we have everything ready to plug in the numbers

a = 14.5mm, s = 10mm, h = 2.8mm

area = 6as = 6 * 14.5 * 10 = 870mm^2
volume = area * thickness = 870 * 2.8 = 2436mm^3

[tranquil grove]

ok

so

that is the answer?

[dark sparrow]

yes

[tranquil grove]

LESS GOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOl

thank you so much

like you probably saved me 20% of my grade because f that

i wouldnt have failed but

it would have been like a c

now its b+ or better

(in australia)

let me write this down

do i have to pay you for this?

[dark sparrow]

you don't

i can't take payment anyway for reasons id rather not get into

[tranquil grove]

ok

thank you so much though

[sinful quartz]

can someone help pls

[upper karma]

By the Corresponding Angles Postulate m1 + m2 + m3 = m6.
Plugging in m1, m3, and m6 gives m2 + 95 = 135 or m2 = 40.
By the Vertical Angles Theorem m2 + m3 = mQ.
Plugging in m2 and m3 gives mQ = 85.
Note that m1 + m2 + m3 + m4 = 180.
Plugging in m1, m2, and m3 gives m4 = 45.
The vertical angle opposite of 4 (we shall call it Z) is also 45.
By the Corresponding Angles Postulate mZ + mQ = mU.
Plugging in mZ and mQ gives mU = 130.

We conclude mQ = 85 and mU = 130.

[mint bridge]

Does anyone have problems with any of the following topics:
logarithms
exponential equations
trigonometric equations
Trigonometric identities
Law of Sines and Cosines Problems
Please, if you have them, I would greatly appreciate it. I have an exam as an insufficiency... it has to be passed.

[dark sparrow]

you want to be given a set of problems on any of these topics to practice?

[mint bridge]

yes please

[dark sparrow]

how many problems would you like?

[mint bridge]

enough to be prepared
10 per theme obviously in their respective difficulty

[dark sparrow]

10 per topic, got it

i'll get back to you once i'm done composing them

[mint bridge]

Thank you

[dark sparrow]

@mint bridge here's what i've composed so far. going to make something for trig equations and laws of sines/cosines soon.

[mint bridge]

I really appreciate it thank you

[atomic tulip]

Could you ping me, when you do that? Would appreciate that.

[dark sparrow]

sure

[atomic tulip]

Thank you.

[atomic tulip]

Let $\alpha$ and $\beta$ be nonzero real numbers such that $2(\cos\beta - \cos\alpha) + \cos\alpha\cos\beta = 1$. Then which of the following is/are true ?\
A. $\tan{\left(\frac{\alpha}{2}\right)} - \sqrt{3}\tan{\left(\frac{\beta}{2}\right)} = 0$\
B. $\sqrt{3}\tan{\left(\frac{\alpha}{2}\right)} - \tan{\left(\frac{\beta}{2}\right)} = 0$\
C. $\tan{\left(\frac{\alpha}{2}\right)} + \sqrt{3}\tan{\left(\frac{\beta}{2}\right)} = 0$\
D. $\sqrt{3}\tan{\left(\frac{\alpha}{2}\right)} + \tan{\left(\frac{\beta}{2}\right)} = 0$

[somber coyote]

What the hell am I doing here?

[atomic tulip]

I DONT have an issue while solving this.

I solved this and got A and C.

But

If I put alpha or beta (either) = pi.

Let's say alpha

I get cos(beta) = -1

Which has solutions.

While none of the corresponding options will work in that case.

So yeah was wondering how that works.

[dark sparrow]

@mint bridge @atomic tulip

the last one is more conceptual in nature partly bc i believe the knowledge of these concepts to be somewhat more important than the number crunching normally associated with these problems

[mint bridge]

ok I understand anyway thank you very much for your support and dedication

[dire pollen]

Please someone give me 1 hour crash course in math vectors

i am preparing for exams and i studied the topic 5 years ago

if someone could just sit and give me a refresher on all its properties, it would be great

also matrices

[dense wren]

1 hour isn't a lot... I can recommend you Khan Academy -> precalculus -> vectors and matrices

[normal fjord]

what is the formula for the surface area of a cube that was split into two

[inner hemlock]

It should be 8L^2

[dense wren]

What do you mean by "split into two"?

[normal fjord]

split in two halves

[dense wren]

8(a^2) ig. It's like a regular cube but where you "cut" it, there are 2 extra areas. Both are a^2. Regular cube is 6(a^2) and if you add 2(a^2 ) to it you get 8(a^2)

[pliant roost]

$\text{Given }\triangle ABC\text{, bisector }AD\text{ and orthocenter }H\text{:}\
R=\sqrt{AB \cdot AC}\
X=\text{S}_{AD}(\text{I}_A^R(H))\
\text{What is (are the properties of) }X?$

[somber coyote]

DV Game

[pliant roost]

I have only found that X always stays out of ABC's inner part

I also have made GeoGebra graph, if it could help: https://www.geogebra.org/calculator/fua8fa3m

If you wonder, the problem comes from an article, and it was left to solve to the reader. Article is about humpty-dumpty points, and that you can perform the transformation used in the problem to get some nice things to work around

It lists the consequences of the transformation to some related objects (P is denoted as the A-Humpty point of the triangle ABC):

1. Line BC becomes the circumscribed circle of ABC.
2. Circle, circumscribed around triangle APB becomes tangent at point C to the circumscribed circle of ABC.
3. Circle, circumscribed around triangle APС becomes tangent at point B to the circumscribed circle of ABC.
4. Point P becomes the intersection point of tangent to the circumscribed circle of ABC to points B and C.
5. A-Apollonius circle becomes the perpendicular bisector of BC.
   Then it says that you can prove some facts using listed ones, and after determining where does the orthocenter after this transformation, you can prove the rest (as we can probably guess, the orthocenter-related ones)

So, what I want is the facts you need to "determine, where orthocenter goes". I am not sure what does this mean, since it doesn't go to any special point or something like that

Moved this to #help-16 for now

[amber locust]

I'm trying to prove Euclid I.6 directly (in contrast to Euclid's proof by contradiction) and without appealing to posterior propositions. I think I found a way, but I'd like to have a second opinion. Euclid I.6 reads: "If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another." My proof is:

(i) Let ABC be a triangle such that ∠ABC = ∠ACB. [Given]
(ii) Describe a circle A1 with center A and radius AB. [Post. 3]
(iii) Describe a circle A2 with center A and radius AC. [Post. 3]
(iv) It will be seen that the points of A1 coincide with the points of A2; so A1 = A2. [C.N. 4]
(v) Since A1 = A2 [iv], the radius of one must be equal to the radius of the other; so AB = AC.

I see two potential problems with this: in (iv) I'm relying on C.N. 4, which I don't really like, but since it's there I might as well use it; in (v) I'm saying that if two circles are equal, then their radius must be equal. Although this seems rather obvious, there's nothing in the Elements defining equality of circles.

Any thoughts? Is this acceptable (in the context of the Elements, at least)? Is there a more rigorous direct proof of this proposition?

[grave pond]

Your (iv) seems to be assuming your conclusion.

How do you know the points will coincide?

[amber locust]

Yeah, that's exactly what I don't like about C.N. 4. It seems the only way to use it is visually (and so it's not that rigorous). But without it I'm not sure how I can prove the proposition directly.

[grave pond]

It goes wrong even before you appeal to the common notion. Whatever it means (I agree that is somewhat murky), it can only be applied once you know that the two things coincide, but you have no argument yet that they do coincide.

[amber locust]

So I'd have to show that the two circles coincide before, correct?

[grave pond]

Yes.

[amber locust]

But to show that I'd have to know the radius of each circle, correct? 😭

And show that they are equal.

Yeah, I don't think I'm on the right path with this proof.

[grave pond]

I don't see an immediate way to complete it, no.

If you could show that the two circles coincide, you wouldn't really need to appeal to the common notion. Because B is on one circle and C is on the other, but they coincide, C is also on the first circle. Therefore the definition of circle (Definition I.15) says that AB=AC.

[amber locust]

Yeah, that makes sense. Well, I'm gonna try out some other methods for a while (probably to no avail). Thanks for the help!

[sturdy hull]

How would one even approach this task?

[silent plank]

you could consider a sub like u = 2^sin(x)

[sturdy hull]

hmm I guess

[fallen pier]

anyone know how to solve c?
Image
i tried to find the slopes, make a line and solve for x2 elimating t but could not figure out how that equates to x2 in the expression for x

[fallen pier]

just derivative and algebra, the 2nd derivative is really nasty :/

[steel ibex]

How do I Start?

[dark sparrow]

you are asked to find the value of x which lies between -90° and +90°, and satisfies sin(x) = -1/2.

perhaps a sketch of the unit circle is in order.

[sturdy hull]

ohhhh smart

[dark sparrow]

did someone ping me?

[ionic cliff]

hello

i nee help on this problem

[strange hamlet]

what is a good site similar to khan academy to explain euclidean geometry but with more in text instead of videos

[wild bluff]

a book

[dense wren]

The first step is to construct circle K with any radius with a center at point E. Then construct circle K2 with the same radius as circle K, with a center at point S. (These 2 circles are already in the picture) Second step is to get the distance between intercepts of circle K and lines EF and ED. Then just construct circle K3 with the radius of the distance you got in the last step and center at the intercept of circle K2 and the non-labeled line. The last step is to construct a line between points S and the intercept of K3 and K2. And you are done :)

[upper karma]

Not really website but sure

[ornate dust]

brilliant

but you have to pay for it tho

that really is the only downside

[upper blaze]

But its worth paying for brilliant :) i paid one anual its good

[strange hamlet]

ok

thanks

[dreamy ridge]

This is TMUA, you’re not expected to differentiate this. Although icl I think that’s far easier than the alternative. I think it’s something to do with the fact that you want the subtraction of 4 lots of 2 to the sin x from 4 ^ sin x to be as little as possible

Sin x goes between -1 and 1

So the subtraction is minimised when the power of 2 is -1

So when sin x=-1

As 4(2^sin x)> 4^sin x for all real x

Ik by the time you do tmua you won’t have done chain rule at a level but tbf chain rule is far easier and methodical to find the maximum point

If you have 2 solutions you can avoid the second derivative by testing the value of the gradient either side

Eg if x=3pi/2 , test x=4 and x= 5

At max point, gradient should go pos to neg ( think the only way from a high point is down lol)

[dreamy ridge]

In reference to this lol^

[sturdy hull]

thanks a lot

[jolly monolith]

someone please help

[wheat fable]

for all parallelogram, two diagonals bisect each other, so lots of information in this question is actually useless lol

@jolly monolith

[fleet fog]

How would I find angle BAC and CAB. I think I can find one of the angles, because The Angles are the Sum of the opposite remote angles, by the Exterior angle Theorem, and because there is a right angle, those 2 angles add up to 90 so subtracting 36 from that amount should get me one of the angles, but I'm not sure which one

[wheat fable]

angle ABC and ADC should be 36 and 38, cuz line above and line BD are parallel

[fleet fog]

Oh Transversals 😦

[wheat fable]

btw angle BAC and CAB are same lol

[fleet fog]

Is Angle B and the Angle from the Triangle and the Transversal Alternate Interior Angles?

[fleet fog]

I don't think thats true because they would have the Same Exterior angle measure

[wheat fable]

sorry i dont get it, do u mean the top line and line BD are not parallel

[fleet fog]

No, I'm just confused lol

I don't understand how to get the measure of angle BAC, ABC, CAD and D

I can find ABC once I find BAC and Same for CAD and D

[wheat fable]

oh, basically u already know the angle ACB and ABC which is 90 and 36, then just use the theorem of sum of interior angle, ACB +CBA+BAC=180

have u learned all of these, cuz tbh knowing proof is necessary

[fleet fog]

Kind of lol

I think Angle A is Supplementary with 38 and 36

So when you divide it and get 53 you can use that to get 37 for Angle B

[wheat fable]

no, AC did not bisect angle BAD

[fleet fog]

Oh ok. But 36, Angle A and 38 all sit on a Straight line

So they are Supplementary right?

[wheat fable]

yeah true if angle A means angle BAD

[fleet fog]

Yes

[wheat fable]

😄

[fleet fog]

I think I overthought it when I saw Transversals lol

I'm horrible with those

Thanks a ton for working me through this!

[wheat fable]

ur welcome😁

[fleet fog]

😄

[fleet fog]

So sorry but one last question If I may.

How do we know AC bisects Angle BAD

[wheat fable]

no it does not

[fleet fog]

:O

[wheat fable]

angle BAC is 54, angle CAD is 52

[fleet fog]

How do you get that

[wheat fable]

angle BAC = 180 - angle ABC - angle BCA

sum of triangle interior angles is 180

[fleet fog]

Ok I understand that part. But from Algebra we need 2 knowns to find another

We only know angle BCA

How did you find both angle values from that

[wheat fable]

transversal bro😅

[fleet fog]

RIP 💀

[wheat fable]

that's ok, just needs some practice lol

[fleet fog]

Yeah

Thanks again for taking your time with me lol

Really appreciate it!

[spiral abyss]

how did cos changed to sine

[dark sparrow]

cos(270°-x) = -sin(x)

[obtuse nova]

can i get some help on this problem

[fleet fog]

180-3a
The equation to get this is: 180= a + 2a + m∠C

That's the best you can get with 2 unknowns

[upper karma]

GUYS

can someone ANYONE

Pls help w a maths question I’ve been struggling w

BDANNANDNWNSNEE

LIKE ANYONE I BEG 😭

[hollow plume]

send

[upper karma]

Question 7 b

And question 12

Yes

[hollow plume]

can you turn that 90 degrees counterclockwise pls

the image

[upper karma]

Okkkk

[hollow plume]

let me be honest with you ive never seen these types of questions

[upper karma]

It’s fine 🥲

[autumn sierra]

yo

can someone please help me with this?

[sinful halo]

I just learned sin cos and tan

[wheat fable]

IK=KL, so angle KIL = angle x

[light holly]

I'm a bit rusty on my math; why are all (three basic) polar equation tests for symmetry inconclusive when they fail? I'm struggling to find counterexamples.

[hollow raft]

again, isn’t this calc?

[gentle plinth]

Can anyone explain how this would be used?

Cause C and F are at two different graidents?

[gentle plinth]

nvm

got it

[cold meadow]

well you're trying to find when this function's derivative is 0
also yeah this is precalc not geometry i guess

[sturdy hull]

Just to be clear here; the notation on the left denotes the angle between the two vector AB and CD, or is it something else?

[sturdy hull]

this isn't trig or geometry, that's Calc 1

[sturdy hull]

try analysing the function V & finding its maximum on R, then use that last tid bit of info to find the answer

[sinful quartz]

does anyone know the reason for #7

[vast monolith]

Similar triangles

[woeful viper]

hi
our teacher gave us this geometry problem as homework and i wasn't able to solve it after trying for hours
appreciate any help and guidance
also sorry if the graph is a bit off

[neat pasture]

Think x = 3

Reason why I di a^2+b^2=c^2 and on the bottom reduce the 7 to 6 as the 7 is 4 lines between each other as the dotted line is 3 so reduced it to 6 and then plug it fromthere

[north heart]

there

that should be all you need

[silent plank]

Reason why I di a^2+b^2=c^2 and on the bottom reduce the 7 to 6 as the 7 is 4 lines between each other as the dotted line is 3 so reduced it to 6 and then plug it fromthere
diagram isn't necessarily drawn to scale

[neat pasture]

Okay as make sense but still think the a^2+b^2=c^2

It is not easily answerable since I 3squareroot 5

For thata one line but think can be hard since not give another number besides 7 which can only think from the graph is having to reduce it down

lIke 12 takes up the entrire top length and need to split it so then can get 6

[silent plank]

wdym by split it to get 6

[neat pasture]

Because 12 takes up the entire one side of the length and so if want to solve for that one line right next to 3 as need to reduce it or get to. huge of a number for the length then

It got for me 6.7082

[silent plank]

I have no idea what you mean or how you're getting that

[north heart]

^

[silent plank]

and you're overcomplicating the problem

[north heart]

this is basically all you need

and the answer is some simple trig away

[silent plank]

don't really even need trig

[north heart]

makes it easier ¯_(ツ)_/¯

[silent plank]

not really

[north heart]

how so?

[neat pasture]

Sorry

[silent plank]

set up a ratio directly

[north heart]

ohhhh

i forgot about addition lol my b

[silent plank]

what addition

[neat pasture]

orry @ℝamonov about the geometry problem as well as was trying to say with the graph presentation you do not want to say have a length of 15 and with the 12 it would show to be too long

[silent plank]

the only assumption you should be making is that the bottom right angle is a right angle and ignore the grid markings

[neat pasture]

You could also use sah coh toa as a way to solve for the side length as well

As a way to solve for the side length then

[silent plank]

sah coh toa
that is not the correct mnemonic for right triangle trigonometry

and as I mentioned earlier using trig is unnecessary

[neat pasture]

Okaay as do get it

Am trying my best at this as not my day job

Are you trying to solve for x anyways

Yes my assumption is correct as can split to be 6.34 as for the 12

to reduce the length as a way to solve for the other lengths for the triangle

[silent plank]

I still have no idea what you mean

[silent plank]

it's assumed that finding x is the only thing required. finding any other length is unnecessary and isn't required to find x

[neat pasture]

This should be splitted from 12 so then can solve for that one 90 degree any length below it to solve for x

Need two lengths to solve for x

[silent plank]

if you insist...you could use pythag multiple times and algebra bash

if you were to do that you'd want to keep sqrt(40) as sqrt(40) or 2sqrt(10) instead of rounding

but again I must reinforce that is overcomplicating this

why are you directing others here...? @neat pasture !!! I'm NOT the one that needs help

[neat pasture]

Got a person name eric taro who can better help you our

Like said before I cannot help out as better people out on this discord that can help solve the problem

I am also trying to be kind here as well

[silent plank]

I know how to solve it...

the op isn't here

[onyx holly]

Well, if you can't help, then stop trying to drag others to help.

[silent plank]

and from the convo, it seems like "you" might actually be the one that needs help with solving

[safe mauve]

Big triangle, inner triangle, scale factor?

[silent plank]

pretty much

[safe mauve]

Nice 👍 tho @woeful viper (sorry for the ping) did you end up learning how to solve for x?

[floral shale]

Find this side

Then use tangent ratios, then use sine ratios

Tangent ratios to find θ, use θ to find x

Which is a lot easier

[floral shale]

Than whatever this is

[silent plank]

why would you need tan

[floral shale]

That he replied to

3 and the green line

[silent plank]

why would you need to find the green line

[floral shale]

Because you have a right triangle there

[silent plank]

yeh...

and you already know two of its sides...

[floral shale]

And then the big boy triangle shares the same angle as the green-3-7 triangle

[silent plank]

if you wanted to use trig you can consider sine in that small triangle directly

[floral shale]

You could

Yeah

[silent plank]

or you can bypass trig altogether with similar triangles

[floral shale]

But similar triangles lame

[dim plaza]

isn't x an altitude of that triangle?

[somber coyote]

Thunder7

[light holly]

I'm a bit rusty on my math; why are all (three basic) polar equation tests for symmetry inconclusive when they fail? I'm struggling to find counterexamples.

[grave pond]

Can you be more specific about which tests you're talking about?

[light holly]

sure

I am referring to the tests for symmetry about the polar axis, about theta = pi/2, and about the pole

which basically just involve making r and/or theta negative and comparing the original function to the transformed function

[grave pond]

You'll need to give significantly more context for this, I think. What are you testing for symmetry? Which kind of symmetry? How do the tests you're talking about work?

[light holly]

I just mentioned it above; it's basically all I know/remember/could find about it

[grave pond]

I'm sorry I can't be of any help. I don't even understand what you're doing.

[light holly]

this more-or-less covers what I'm referring to: https://math.libretexts.org/Courses/University_of_California_Davis/UCD_Mat_21C%3A_Multivariate_Calculus/10%3A_Parametric_Equations_and_Polar_Coordinates/10.3%3A_Polar_Coordinates

I guess I'm just confused as to what information is missing at this point in order for my question to make sense

(and I know that the link says multivariate calculus, but this was grouped under/a part of trig at my college)

[grave pond]

Your link doesn't seem to make a claim that the tests are inconclusive when they fail.
But I suppose an example could be something like (r=0, theta=7) whose solution is a perfectly symmetric single point, but the equation itself doesn't satisfy the tests for mirror symmetry about either axis.

[light holly]

that's why I'm confused about the inconclusiveness; I couldn't find any material online that states that it is inconclusive, but one of the placement tests at my college required us to answer that the test is inconclusive when failed

I just find it odd how it's not really mentioned anywhere else

I don't think that r=0 fails the test while being symmetrical, but something like r=1 probably would

[grave pond]

In general I think the point must be that the mapping from (r,theta)-space to (x,y)-space is not injective, so a nice and symmetric set in the plane could result from mapping a subset of (r,theta) space that is not itself symmetric.

[light holly]

I wonder if another example would be something that traces at a different rate the same path when r/theta is made negative and/or traces it with a negative radius

though that would likely involve multiple and/or nested trig functions

ah something like r=sin(theta)cos(theta)

it fails all of the tests but is symmetrical about all of them

is there a place I can look for more (preferably rigorous) material about this topic?

[grave pond]

Hmm, it seems to satisfy the theta -> theta+pi test, though.

[light holly]

ah the tests were ever so slightly different for the placement test at my college; they all involved making theta and/or r negative

no +/-pi

[grave pond]

It seems to be tricky to produce a good counterexample that still looks like a single algebraic-ish equation. But a compound criterion such as { (r,theta) | theta = 0 or (|theta|=pi/2 and r>0) } wouldn't pass any of the simple tests, but still produce a symmetric result.

[mystic vector]

I am learning Taylor series, I have some question when I saw the proof
the first step of proof is $Let f(x) = a_0 + a_1 x + a_2 x^2 + … a_n x^n$, but how can we sure f(x) can be written as this form?

I take an example, when $f(x) = sinx$, how can I know it can form as $a_0 + a_1 x + a_2 x^2 + … a_n x^n$ form?

[somber coyote]

HelloWorld

HelloWorld

[grave pond]

Can you show some context for that "first step"? What exactly is being proved? Indeed not all functions can be written as a polynomial, and in particular sin(x) cannot..

[mystic vector]

I just google and don't know whether this is correct, I self study this🥲

[grave pond]

Proof of what, though?

[mystic vector]

it say proof of taylor series