geometry-and-trigonometry

[dawn adder]

hello i need help asap

w trigo

[clever viper]

can someone help me with this

[silent plank]

what have you tried

[clever viper]

nvm im good

i figured it out

[gusty wind]

help with these 2

[silent plank]

what have you tried

[gusty wind]

i didnt try anything

bc i dont know

[silent plank]

are there any words in the problem you don't understand?
have you made any attempts to draw a diagram

[gusty wind]

no and no

[silent plank]

so what exactly don't you know

[gusty wind]

nvm i got it

[hoary sapphire]

im new to geometry so

i was wondering how you would solve this

This is my attempt

[dark sparrow]

they are alternate interior angles

so they are equal, not supplementary

[hoary sapphire]

How would you find x if it was written as equal to each other?

[dark sparrow]

well you have the equation 5x - 18 = 3x + 42

you would solve it just as any other linear equation in one variable

[hoary sapphire]

This is my second attempt but I understand that it isn’t correct, I’m not sure what my mistake was.

[mellow vector]

You need to add 18, not subtract

[dark sparrow]

that and also subtracting 18 from 5x-18 would not have given you -5x anyway

and also the last step is just nonsense - why divide one side by 8 and the other by 13?

[hoary sapphire]

Why do we need to add 18? I know that subtracting will lead me to the wrong path but I dont understand the concept

[mellow vector]

You add 18 so it will cancel out with the -18 on the left side

You want to isolate all of the variables on one side, and the constants on the other side

So you move the -18 to the right, by adding 18 and cancelling it out.

After that you should subtract 3x to cancel it out with the 3x on the right side

And you will be left with only x's on the left, and only constants on the right

[hollow plume]

You might get the answer if you subtracted, however it is best to get rid of the negatives

ASAP

So adding will get rid of them

Will clear lots of confusion

I would say add 18 instead of subtract 42

Both are correct ways, however adding 18 gets rid of negatives

[hoary sapphire]

hmm i see

thank you i will attempt it again

[wispy pecan]

hi

could someone help me with angle congruence

[hoary sapphire]

practice makes perfect

[spring scarab]

did I do something wrong... :с

[gleaming portal]

<@&286206848099549185>

[sharp hare]

Please note that this question doesn't fall under either geometry or trigonometry

It should be asked in #proofs-and-logic

[inland stone]

also, to ask for help use the help channels and don't ping helpers before waiting 15m after you posted your question

[gleaming portal]

it's a conditional statement

in geometry

[gleaming portal]

ok

[sharp hare]

I still feel like this falls under "logic"

[gleaming portal]

conditional statement is under geometry

but wtv

[finite idol]

☠️☠️

[hollow plume]

You learn quickly don’t you

[finite idol]

😅

I only missed one problem in that whole class

Out of like 500

[hollow plume]

Take AP calc lmao

[finite idol]

I will as a junior

[hollow plume]

O snap

You’ve done this 1 year b4 I did

[finite idol]

☠️

I’m taking Linear Algebra and Multivariable Calculus Senior year

[hollow plume]

Here’s the thing

[finite idol]

Maybe Discrete on top of that idk

[hollow plume]

My school only offers calc for discrete math

[finite idol]

I’ll run out of math classes at my high school

[hollow plume]

You should’ve distributed them over the course of 4 years

[finite idol]

Ik

Cuz after BC my school offers 1 term of multi variable and 1 term of linear algebra

[hollow plume]

How lucky you are

[finite idol]

Indeed

[hollow plume]

Both aren’t offered here

Might need to start petition

[finite idol]

Lol

[undone coral]

<@&286206848099549185>

[clever viper]

is the last question just asking for area of rectangle minus area of circular ride

[stray cliff]

Hi , I know math

[upper karma]

can somebody help me with b

[dark sparrow]

@upper karma do you still need help with this

[upper karma]

yes

is this right?

[dark sparrow]

multiple things wrong with that

1. specifying a height for the prism is inappropriate - what if the relation between volumes were different if the height wasn't 50?
2. even if you are going to specify a height, you have to include the units - otherwise, it's not clear if you mean 50 cm or 50 in or what
3. the values you gave for either prism's volume either don't have the units at all (1500) or have the wrong units (21,600 cm).
4. and even the numbers themselves are wrong: the volumes of the prisms so constructed would not be 1,500 and 21,600 cm^3, but 30,000 cm^3 and 43,200 cm^3.

the volume of a prism is the product of its base area and its height, so if the heights are the same, then the ratio of volumes equals the ratio of the base areas

[grand horizon]

Pls help

@dark sparrow

@dark sparrow

[dark sparrow]

who are you

[weary drift]

@grand horizon pls dont ping specific users for math help

[sharp hare]

Have you got a table with frequent tangent values?

[undone coral]

noo

[upper karma]

can somebody check these for me

[misty needle]

Check the volume of pyramid in 1st question again

Also the cube

[sharp hare]

In that case draw a right triangle with base 2 and height 1

Then you can see what the angles should be

And you can calculate the sine and cosine by the lengths

[shy mural]

i have a question thats not particular a problem itself
but in triangle abc, if I is the center of the incircle of the triangle (the intersection of 3 angle bisectors), and A is arbitrary, but B,C, and alpha isnt (alpha is angle BAC), then is I arbitrary?

[upper karma]

i think A will move on the circumcircle of ABC with BC as a chord

then only alpha would be const

BC as a fixed chord

[shy mural]

I is the center of the incircle btw

[upper karma]

so I will be arbitrary

[shy mural]

do u want to see the original problem

[upper karma]

lets see

[shy mural]

Acute triangle ABC has B,C fixed, A is arbitrary such that $\angle A = \alpha = const.$ Let I be the center of the incircle of ABC. E,F are the altitudes from I to AC,AB, respectively. M is the midpoint of BC. Prove that the distance from M to EF is constant.

[somber coyote]

blueberry faygo

[shy mural]

i constructed some additional points

let H be the altitude from M to EF, which means MH is the distance from M to EF

[upper karma]

I should be arbitrary

[shy mural]

Let Z and Y respectively be the altitudes from B and C to EF

when i construct the figure this way, MH = (BZ+CY)/2

because MH is the midline of the trapezoid BZYC

also i figured that triangles AFI and BZF are similar as well as AEI and CYE

so $BZ = BF \cdot \frac{AE}{AI} = BF \cdot \cos \frac{\alpha}{2}$

but the thing is if I is arbitrary then shouldnt BF be arbitrary?

or do we just want to prove that BZ+CY is arbitrary instead of specific BZ and CY

[somber coyote]

blueberry faygo

[shy mural]

@upper karma should BF+CY be arbitrary?

i mean

BF+CE

[upper karma]

hu idk

[shy mural]

is it arbitrary

if I is the center of the incircle ABC and IF and IE are altitudes dropped from I to AB and AC

[upper karma]

yeah maybe

i dk the proof just intuition

[shy mural]

oh

it definitely should be fixed

or else the problem is wrong

@upper karma

heres the figure

[upper karma]

ohh

[shy mural]

how to rotate

[upper karma]

i ma doing something else now

[shy mural]

oh

ok

[rapid flare]

Can you solve this question:

∆ABC is right angled at B.
A semicircle is drawn with BC as diameter and touching AB. If the radius of the semicircle is 10.

Find the minimum area of ∆ABC

[hollow plume]

First draw the figure

[rapid flare]

Why? Isnt the question clear

[hollow plume]

It’s better to draw it out

That way you clearly know what you’re finding

[rapid flare]

Ok i will provide my high quality image

@misty needle

[misty needle]

What is the question?

Wait what?

You said BC as diameter

[rapid flare]

Ok leave it. Its bit too easy for jee aspirants like you

Come back to #calculus

[rapid flare]

Correction.
AB is diameter.
And it touches AC

[vital maple]

if s is the semiperimeter, a, b, c are side lengths then BF=s-b, CE=s-c, so BF+CE=a, which is constant

[shy mural]

how'd you get BF=s-b and CE=s-c?

is it some formula related to the inscribed circle

wait

BF=s-b = a+b+c/2 - 2b/2 = a+c-b/2

where is this from

[vital maple]

it's a well known formula

[shy mural]

may i have the name

[vital maple]

there is no name

[shy mural]

how do you derive it

[vital maple]

you can derive it by noticing AF=AE, BF=BD, CD=CF where D is the foot from I to BC

then writing systems of equations

[shy mural]

what D

oh

thanks

let me try deriving

wait you dont even need the semi perimeter

its just BF+CE=BD+CD=BC

thankyou

[rare kelp]

Bruh I just understood exactly how tangents work

if you multiply f(x)=x by the tangent of a constant in a graph, like desmos, the line will smoothly turn while changing the constant

[rare kelp]

Example: https://www.desmos.com/calculator/m7xhiqd27n?lang=fr

[chilly whale]

Hello, in a problem I am supposed to simplify $\sin x -\cos x$ into $\sqrt{2}\cos \left(x-\frac{3 \pi}{4} \right ) $, where as a hint it was said it might be helpful to use the addition formula later on.
\

How would one go about this? I tried googling about the derivation for this identity but with no avail. Thanks.

[royal raptor]

does anyone know descriptive geometry?

or is it too advanced for you

[somber coyote]

Aslan

[worthy sapphire]

@chilly whale cos(x-3pi/4) = cos(3pi/4)cos(x) + sin(3pi/4)sin(x)

[chilly whale]

I see, how would one come up with this?

Take note that the answer I provided was not listed in the problem, but only for me to clearly show what the answer was supposed to be.

Is it just intuition ? Can't think of a systematic way of seeing that just from sin x - cos x alone, which was the only thing available in the problem

[glacial urchin]

Can someone please help me with this?

[verbal gulch]

Can u use any theorem?

[verbal gulch]

@glacial urchin learn about Euclid's Theorem 1 & 2

[worthy sapphire]

you can work backwards from sqrt(2) cos(x - 3pi/4)

you want to split it into sin(x) and cos(x) somehow

and with the hint of 'addition formula' this was the only logical option

oh wait

i didn't read the note

hm

@chilly whale what was the question then?

i guess it is to simplify to the form a*cos(x - b) or smth

in that case, you want to use cos(a+b) = cos(a)cos(b) - sin(a)sin(b), and you can note that this looks incredibly close already like sin(x)-cos(x), just that you have these 'annoying' cos(b) and sin(b)

you want to make them equal to each other so you can factorize

[chilly whale]

The original question was to solve the equation sin(3x) - cos(3x) = 1, and the same hint was specified but for sine.
So it was a slight modification on my part because I only wanted to know how to derive the identity. Because the cheat sheet simply glossed over that part but took use of it.

But thank you, i think I'm more comfortable now if i ever encounter a similar problem again

[worthy sapphire]

ah ic

[upper karma]

would it be a AA

[hollow plume]

AA

[worthy sapphire]

what is your question?

[upper karma]

i solved it

[worthy sapphire]

ah nice :D

[upper karma]

but thx for trying to help

[worthy sapphire]

this is the angle bisector theorem if you're wondering

there's also a nice other proof, do you wanna see it?

[opal hare]

Its not E. And I'm betting its not D, and possibly a, anyone got ideas

[abstract mesa]

I got C(20 degrees

[opal hare]

Do u have paper to show me

Or not really

[abstract mesa]

Yes

[opal hare]

The answer is C, but how did you get it

[opal hare]

angles of triangle are 70 80 80..?

what?

how? angles add to 180

@abstract mesa

[abstract mesa]

No

20 80 80

It says in the problem isosceles and the ABC is 20

So (180-20)/2= angle A= Angle C

[opal hare]

oh lol

couldnt read writing

[opal hare]

Why -160?

u add the angle?

[abstract mesa]

Because we're solvint for angles b

Yes

A + B +20=180

But because it's isosceles by def sides AB and BC Respectively are equal and the angles that corresponds to them from across are also equal

So angle A and angle B are equal

So in reality you get A+A+20=180

And solve for A which also gives me out B

You

[opal hare]

i see

sso therefore 160=2a

[abstract mesa]

Yes

[opal hare]

isnt the point between CA D

Not E

[opal hare]

You write E

[abstract mesa]

It doesn't matter whether it's changed or it's symmetric in that sense

[opal hare]

True

But better notation still :/

[abstract mesa]

But no I did it right

[opal hare]

U could've tried lol

[abstract mesa]

It says point E is on side AB

And D at the base or AC

I did it right

[opal hare]

🗿

y u keep deleting ur own messages lol

[abstract mesa]

Lol

[opal hare]

ah im done with math, im literally going to use very few math modules

[abstract mesa]

Why did you send me a request you perve

[opal hare]

rt

https://tenor.com/view/penguin-pirate-edit-walk-silly-gif-5875569

why not

[spiral bough]

Not sure if it’s the same in English, but in my language it’s “nedian”

[round cedar]

Can someone please check if my answer is correct

Question

Answer

[round cedar]

I did all of this but forgot to out the value of y can't be more angry at myself

What other variable is their it doesn't tell you what z plus x is

[round cedar]

Yeah I was thinking about it do you think they are gonna cut marks for it?

[upper karma]

does anyone know how to solve this ;-;

[upper karma]

can somebody check this for me

[hoary sapphire]

Seems correct to me

[eager berry]

is there such a thing as arccot? desmos calculator isn't acknowledging it

[dark sparrow]

the graphing calculator acknowledges it just fine.

[ocean pond]

can some1 help pls :D

[worthy sapphire]

hint: let the line through Q perpendicular to LI intersect LI in X. then prove triangles QXL and IML are similar :D

[upper karma]

thank you!

[worthy sapphire]

https://cdn.discordapp.com/emojis/589062916372824085.webp?size=48&quality=lossless

did you get it?

[upper karma]

i understood! though I haven't solved it yet right now

[worthy sapphire]

alright

let me know if you need another hint :p

[upper karma]

use this bro

[polar plaza]

help pls T-T

idk anything abt trigonometry

[rapid flare]

Then learn it

[hollow plume]

4)sin is opposite side over hypotenuse

5. calculator question

6)sin(36)=x/8

7)sin(46)=8/x

For 6 and 7 solve for x

The trigonometric symbols will be done on calculator

[upper karma]

can somebody check this for me

[hollow plume]

Looks good

[upper karma]

thanks

[upper karma]

can somebody check this

[upper karma]

can somebody check this for m

[onyx onyx]

this question makes no sense

[upper karma]

Hi! Does anyone know how to solve this?

[carmine abyss]

hii can someone please help me!!! ive been stuck on this for a while

[sweet minnow]

quick question. Is this trig identity valid? If so, how or why is the cos negative?

[upper karma]

Ig, it is like:
$sin (2t - π/2) = -sin (π/2 - 2t) = -cos (2t)$

[somber coyote]

Subrat Panda

[upper karma]

Pls check this, and if anything is wrong in it, do tell

[sweet minnow]

yeah that makes sense, thanks!

[worthy sapphire]

uhm... there is a theorem that will let you solve this in one step, i'll just give that theorem i guess

[hollow plume]

It says use special right triangles and not trigonometric functions

Though you may have gotten the right answer, you have to do it they way the question asked

[hollow plume]

That is correct

[floral moon]

NH is a chord and HR is a tangent, there is a theorem that says "the measure of the angle formed by a chord and a tangent is equal to half the measure of the intercepted arc". Since HN is 114, <NHR will be 57 degrees because 114/2 is 57.

[night cape]

are there any good resources for learning trig?

[hollow plume]

SOH CAH TOA

There’s the universal acronym

[hearty pivot]

does anybody have resources for learning proofs?

[night cape]

wait thats it O.o

[hollow plume]

Yes

Universal acronym

[night cape]

Ohh

But I mean like

a complete learning of trig

[hollow plume]

There’s khan academy

[brave rock]

bruh tf is this

[hollow plume]

So the hypotenuse is 6

[brave rock]

sorry, i forgot to delete because i solved that one

[hollow plume]

You already did 👍🏼

Good job

[steep flax]

Hi is anyone here

[timid fulcrum]

ye

[steep flax]

I have a treat for any anime lovers in here

[timid fulcrum]

ok.

[steep flax]

Do you like anime

[timid fulcrum]

sure, but i don't watch very much

[steep flax]

It’s sending

Why did it send like that

It’s a funny anime triangle video

[timid fulcrum]

maybe review the bisecting diagonals theorem and other theorems, because i proved it as: PX = SX because TX = AX, thus AT must bisect PS. PAST is a parallelogram because of the bisecting diagonals theorem.

You could also first prove that TP also = AS since TX = AX and they're inside a quadrilateral.

after that state that ST = and || AP because there's a pair of opposite equal and parallel sides. And then prove that it's a parallelogram because of the opposite pairs theorem

[timid fulcrum]

the question is basically saying: out of these 5 rectangles, create one that is 15 by 11. You can only place them next to each other but use take multiple. I think it's best to first eliminate some choices. And i feel like the best way to do this is copy the image into something like mspaint and copy and rotate the pieces around to see if the sides add up correctly.

[timid fulcrum]

main thing here is to just focus on arc HN since it's only needed for solving. If you draw lines JN and HJ, you make an isosceles triangle since JN = HJ because they're the radius of circle J. Use what you know about angles in isosceles triangles to find the two smaller ones. Since line LR is a tangent, < JHR has to be 90 because HJ perpendicularly bisects LR. then just find the complementary angle of < NHJ

[carmine abyss]

thank you so much this really helped!!😇

[upper karma]

can somebdoy help me with this

[dark sparrow]

(question is being addressed in #help-9 )

[silent plank]

are you omitting any information?

[upper karma]

Well

BEC is similar to DAE

And BAE is similar to DEC

[carmine abyss]

im not sure what my next step would be for this prooff

[timid fulcrum]

you may want to review the opposite sides theorem because since KL is congruent to LM and JKLM is a parallelogram, then all sides of it must be the same length (equal sides being the main thing for rhombi)

[foggy osprey]

hi can somebody help

[hollow plume]

The center is the numbers in the parenthesis times -1

Radius is determined by the square root

(x-h)^2 + (y-k)^2 =r^2
Where the center is (h,k) and the radius is r

[foggy osprey]

ok thanks

[upper karma]

hii, does anybody know how to make the stem-leaf plot for this?

[opaque ibex]

i have a question

is the circumference of the inner tangent circle equal to the tangent triangle?

[dark sparrow]

what's a "tangent triangle"?

do you have a diagram?

[opaque ibex]

hold on

a + b + c =? 2rpi

[dark sparrow]

no, not even close

the perimeter of a triangle is in fact always longer than the circumference of its inscribed circle

[opaque ibex]

is this true: ((a+b+c).r)/2 = A(ABC)

[dark sparrow]

if A() means area, then yes

[opaque ibex]

and how can i proof

it

prove it

i dont want the proof just what can i try

okay thank you

[spiral bough]

If you name the incenter I, consider the areas of triangles ABI, BCI and CIA

[bitter lodge]

Is it proven that it is not possible to construct Pi as a line segment? (using only a compass and straightedge ofc, and no approximations)

[dark sparrow]

yes but the proof goes way way way beyond the extent of highschool geometry

[bitter lodge]

Ok thanks, do you know what it's called so I can look it up.

[dark sparrow]

https://en.wikipedia.org/wiki/Lindemann–Weierstrass_theorem#Proof

this is the more general theorem

[latent leaf]

would this be a enlargement? just double checking

[soft sail]

yes.

[hollow plume]

The blue figure is a dialation of the black figure

Therefore the blue is the image

You are therefore incorrect

[silent plank]

<@&268886789983436800>

[hollow plume]

Ping doesn’t work lmao

[worthy igloo]

I wonder why the bot didn't delete it

[worthy igloo]

I hope you realize these are bots

[hollow plume]

👍🏼

[zealous hollow]

Hi guys! Hope everyone's doing well. If anyone is able to give me guidance or some help on this question. It'd be much appreciated <3

[worthy sapphire]

label everything like this

[smoky jetty]

thats a lot of composite figures (for my level atleast)👀

[hearty pivot]

could someone help me with this?

nevermind, i got it lol

[zealous hollow]

Thank u bro! I understand now💯 💯

[last oak]

Yo

Can someone help me figure smth out

How did they go from i to 2 and sqrt 44 to sqrt 11i

How did they simplify it

[zealous hollow]

Do u know what i is equal to?

They just simplified the expression that's all

44 is the same as 4*11

[zealous hollow]

Not me bro

Alg haha

[hollow plume]

If there’s a negative in the root, there will be an i

$\sqrt{44}=\sqrt{4}*\sqrt{11}$

[somber coyote]

% Openglobe %

[hollow plume]

Square root of 4 is 2

[hollow plume]

I can’t believe it

I can’t believe I replied to the wrong person

[zealous hollow]

XD happens mate

[inland vale]

do yk how

would i

what do i do to find

tangent

[clever viper]

just use a calculator

[hollow plume]

Sin/cos =tan

In this case degree number stays the same

[upper karma]

i nee help

[inland vale]

how come

nvm

figured it out

[valid musk]

It is tan 20, I’m not too sure how to explain it

[hollow raft]

sin = y
cos = x

tan = y/x

In unit circle terms

That’s a way to explain it?

So yeah y(20)/x(20) would simplify to y/x(20)

[dark sparrow]

tan(x) is literally defined as sin(x)/cos(x), nothing else to it

[glacial urchin]

Bet thanks

[nocturne remnant]

Fix your language dude🗿

Let angle A be the smallest angle of triangle ABC. BC divides the circumcircle of the triangle into two arcs. Let U be a point on the arc BC not containing A. The perpendicular bisectors of AB and AC meet AU at V and W respectively. BV and CW intersect at T. Show that UA = TB + CT.

I assume you meant perpendicular bisectors because I have no idea

[stray wigeon]

#help-3 help pls

[floral moon]

basic trig question: will the initial side always be the positive x axis

[low sky]

what's the initial side?

[stray wigeon]

#help-2 round 2

[low sky]

how would i go about showing that the red line is parallel to AC? Here G is the centroid and D is constructed by drawing a line through the midpoint of the median

[signal peak]

Bruh I just joined this server as a 10th grade student looking for enlightenment but instead I got blinded by this geo prob💀

[low sky]

heres the drawing

[stray wigeon]

** Below is where reking's question comes from**
“Here is my attempt at solving this geometry problem:
(1) Draw the median of side BC and name its point of intersection with DC as K,
(2) We know that K is the centroid of triangle ABC, so 2 * DK = KC,
(3) We also figure out that DK = 2m, KF = m, and FC = 3m,
(4) Connect points K and E, and show that KE is parallel to AC in order to get FE / AF = 1/3.
I am simply stuck on the “showing that KE // AC is true” part.”
(Solution: #help-3 message)

[low sky]

yeah i just took the part we were stuck on

[silver blaze]

Can anyone help me prooving this by formula

[somber coyote]

Pebble

[heavy mortar]

hey guys i have a dumb question that im stuck on

[teal kayak]

Hey, do you guys know of any programs like geogebra I could use that can like, record values as they change and automatically graph them

[heavy mortar]

dw about the french text its useless

[teal kayak]

Or just better programs than geogebra idk it doesn't seem like The best

i think you are supposed to use homework help section

#❓how-to-get-help

[teal kayak]

update: nvm

[spiral bough]

There’s an instrument called locus in geogebra, that can help with your problem I think

[smoky jetty]

desmos?

[teal kayak]

not quite bc it was geometrical values i wanted to graph and i was hoping for something cleaner than just graphing a bunch of points but i figured it out

[teal kayak]

i just made a point in the second graphics view with its x and y values corresponding to the thing i wanted to graph, and then made geogebra trace the point

very clean uwu

[pure void]

how to do this

[silent plank]

angle chasing, angle sums on lines and of triangles

[low sky]

anyone think they can figure this out?

[final yew]

Find the value of $\tan \frac{\pi}{15}+\tan \frac{2\pi}{15}+\cdots+\tan \frac{5\pi}{15}$

[somber coyote]

Eren Yeager

[floral moon]

Rewrite $\sqrt{cos^2\theta + sin^2\theta + cot^2\theta}$ in nonradical form without using absolute values for $\pi < \theta < 2\pi$.

[somber coyote]

Chrovo

[floral moon]

ok so I solved most of it and got up to | csc theta |

but the answer says its -csc theta

oh wait nvm

I ignored the pi < theta < 2pi part

[dense bough]

dude is this a competition problem or something?

it was just asked here: https://math.stackexchange.com/questions/4505640/show-that-frac-sin3-beta-sin-alpha-frac-cos3-beta-cos-alpha/4505684#4505684

[eternal pier]

Hello?...

I just need help for number 1 and 8...

(I just started this Trigo stuff)

[silent plank]

were you able to solve 2 to 7

[eternal pier]

Nope... Not any...

[silent plank]

consider sohcahtoa

[eternal pier]

I also don't know "sohcahtoa"...

[silent plank]

do you know any trig at all

[eternal pier]

Nope... Just started...

[silent plank]

I'd recommend you start with something like a lesson on khan

[eternal pier]

Khan Academy?

[silent plank]

yes

[smoky jetty]

welp, u gotta know the fundamentals first. From triangles

[low sky]

that was me who asked it on stackexchange!😅

[worthy sapphire]

it's certainly not the way you wanted to solve the problem, but vectors work (sadly)

(youll get the midpoint of the median from C is 1/4a + 1/4b + 1/2c, D = 1/3 b + 2/3 c, G = 1/3(a + b + c))

(so vectors GD = 1/3 AC)

i'll look for a synthetic solution

i think it's a very nice result though

[low sky]

someone else helped me out on that one, and figured out a nice way to show it

if you want i can show you

[worthy sapphire]

@low sky ah got it!
important step is to prove that TAB~TDE with factor 3:1 (where E is the intersection of BT and AC, and N is the midpoint of AB, and T the midpoint of CN)
do this by defining E' and D' such that AE':E'C = 1:2 and BD':D'C = 1:2, and then that T', the intersection of AD' and BE', is the midpoint of CN [footnote 1] -- this will imply T = T' and since there is only one possible intersection for AT with BC, we must have D' = D and E' = E analogously
then since NG:GC = 1:2 so NG:GT:TC = 2:1:3, we know TGD~TCA by sas, so AC//GD

footnote 1: see second picture

:p i was typing while you said that

i wanna know tho!

[tame pulsar]

geometry

:(

[worthy sapphire]

please do btw

[low sky]

#help-19 message

[worthy sapphire]

that one's nice -- should've thought of that because of all the ratios (also kinda the reason why i did harmonic mean because of intersecting at the median)
i don't do menelaos often enough anyway--

[silver blaze]

Nah that wouldn't work i found easier way

[low sky]

in this proof, can you explain how you get that T' is the midpoint of CN from footnote 1? all i can gather from this is that we now know the length of the line PQ... PQ is not visible in the first image right?

[upper karma]

can somebody pleas ehelp me with this

[worthy sapphire]

if you let P,Q to be on AC,BC respectively such that PQ//AB and P,T',Q collinear, then |PQ| = 2/(1/|AB| + 1/|DE|) = 2/(1/|AB| + 1/(1/3 |AB|)) = 2|AB|/(1 + 3) = 1/2 |AB|
by PQ//AB we know that a homothety from C with scale factor 2 sends CPQ to CAB, so also T' (the midpoint of PQ) to N (the midpoint of AB), so T' is the midpoint of CN

i did leave that as a big gap to be checked, so i'm glad you asked me about that

oh, and read T as T' lol (edited)

if you don't like random people just citing homothety, you can also note CPT~CAN by PT//PQ//AB//AN so CT/CN = PT/AN = (2PT)/(2AN) = PQ/AB = 1/2

this is also the reason why T lies on EM, where E and M are the midpoints of CA and CB respectively, and then you can also apply ceva on triangle MCA and point T

but ceva is basically menelaos

[low sky]

okay i get that part then! now in the original proof you then show that AE:EC = BD:DC = 1:2, but why does it follow that TAB~TDE with factor 3:1? are you using SAS here?

[worthy sapphire]

yeah, CDE~CBA by sas implies DE//AB implies TAB~TDE by aa

and AB/DE = CA/CE = 3

this is very similar to showing that the centroid lies on ratios 1:2 on a median

actually just the same

[floral moon]

is csc the same as sin^-1

I meant csc

[somber coyote]

potato

[slender wasp]

If you need more detail I can explain

[floral moon]

yea I know that but can't sin^-1 be rewritten as 1/sin

[slender wasp]

(note really we have to restrict sin to an interval)

[floral moon]

and if not why

[somber coyote]

potato

potato

[slender wasp]

As a similar example, we can take the function f(x) = x^2 defined on positive numbers. Then the function 1/f would send x to 1/x^2, but the function f^-1 sends x to its square root.

It's completely analogous to that, though the notation is a bit confusing potentially, because of course with numbers we expect a^-1 = 1/a

[somber coyote]

potato

[slender wasp]

Or as another example, sin^-1(1/2) is just 30 degrees, whilst 1/sin(1/2) is some weird decimal

[floral moon]

oh so sin^-1 is kind of an exception to this rule

[slender wasp]

I'd actually view it as being the other way round aha - sin/cos etc are the exceptions :)

Typically (in my experience at least) if f is a function then f^n is what happens when you apply f n times

So f^-1 is when you apply it '-1' times i.e. inverse

[slender wasp]

It's just that with sin, cos etc typically we mean this

It's just a weird thing I guess here

Hope that clears it up?

I remember asking my teacher the exact same thing a few years ago

[floral moon]

oh

yeah I that makes sense

If I'm understanding correctly

Since $f^-1(x)$ implies inverse function, in an ordinary function, you would have to do stuff to find its inverse.(swap the x and y, and solve for y). On the other hand, $f^n(x)$ is different, it is $(f(x))^n$.

[somber coyote]

Chrovo

[slender wasp]

That's only strictly true for sine and cos, but correct for them

[somber coyote]

potato

[slender wasp]

(ignore the either)

[floral moon]

wait so you can't do tan^2 theta

[slender wasp]

I'd say it depends on context but wouldn't worry about it too much

Oh sorry well yes tan, csc, etc

Any standard trig thing.

[floral moon]

wait so which trig functions were you referring to? vers theta, covers theta, etc..?

[grave pond]

If the reader recognizes it as a trigonometric function in the first place, they will probably use the general convention that positive exponents refer to exponentiating the function value rather than iterating the function.

Always remember that this is about notational conventions rather than deductive facts that follow crisp rules.

[slender wasp]

Can't say I've ever seen anyone use vers for non meme purposes aha

[floral moon]

Oh alright

[upper karma]

can somebody pls quickly tell me what yd^3 means

i need to turn this in in 2 hours

[grave pond]

Probably the volume of a cube with a side length ot 1 "yard", which is an archaic unit of length still used in the USA.

I think there are 3 "ft" to one "yd" and therefore 27 ft³ in 1 yd³.

[upper karma]

ooooh ok thanks

i think i get it

[wanton merlin]

what is 8-10?

can you use the properties of rotations to show what the measures of the length of diagonals are equal?

[worthy sapphire]

does everything make sense now? sorry for skipping over some similarity stuff--

[worthy sapphire]

the harmonic mean argument was not even needed btw

you can work with ratios on the median CN and show that T'=T from that

so it'll become completely elementary

[weak smelt]

8. you can see that the shape you're given is nothing, but 4 triangles and a rectangle, now do something with triangles to show that the given shape is indeed a rectangle

9. notice what triangles the bisecting lines form within the main triangle, find a way to apply pythagoras once or twice and deduce the lengths from that

10. for this one you just need to draw these points and then remind yourself what a square is

[upper karma]

inverse sin and sin to the power -1 are different

i remember when i was dumb and thought inverse sin was csc

lol

[weary drift]

@upper karma cut it out

[upper karma]

Ok mate. Chill. It was just a joke.

😉

[dapper sable]

[red bluff]

part i) is straight forwards

hw would i do part 2 and 3?

[worthy sapphire]

for part 2, visualize cos(k θ) on the unit circle and use the fact that the centroid of an inscribed regular 7-gon is (0,0)

and for part 3, look at what you've done at part 2

[spare stump]

hey this is a pretty basic concept but im confused, can anyone tell me the difference between aa similarity and aa corollary of triangles, like i just dont get when to mention which one of these theorems

[grave pond]

I don't think "AA corollary" is a standard name. Can you quote (or post screenshots of) the exact statements of the two theorems you're talking about?

[naive rivet]

For Part 3: You can have a completely algebraic solution if you know this fact:
for a complex number z
$$z^7-1=(z-1)\prod_{k=1}^{3}{\left[z^2-2z\cos{\frac{2k\pi}{7}}+1\right]}$$
dividing by z-1 both sides
$$\sum_{k=0}^{6}{z^k}=\prod_{k=1}^{3}{\left[z^2-2z\cos{\frac{2k\pi}{7}}+1\right]}$$
dividing by $~z^3~$both sides we have :
$$1+\sum_{k=1}^{3}\left[z^k+\frac{1}{z^k}\right]=\prod_{k=1}^{3}{\left[z+\frac{1}{z}-2\cos{\frac{2k\pi}{7}}\right]}$$
Now you could substitute $~z+\frac{1}{z} =t~$.
You'll get RHS as (t-a)(t-b)(t-c) = some polynomial now just use the fact that a,b,c are the roots of the polynomial in the LHS, and use vietas directly.

[somber coyote]

mkay

[worthy sapphire]

[worthy sapphire]

hm, how to see the factorization in the first line algebraically (geometrically with vieta it's alright)

cool sol though!

[naive rivet]

z^7-1= (z-1)* product.. that line you get from geo yeah

[worthy sapphire]

right

sort of a translation :p

[turbid berry]

Hey idiot engineer here.
I look for midpoints on lines using x = ym + b (Most lines are close to vertical) on a screen, and I find the midpoint of these lines using

m = (x1 - x0) / (y1 - y0)
b = x0 - y0*m 
line_midpoint_on_screen = (SCREEN_HEIGHT*0.5 * m + b, SCREEN_HEIGHT*0.5)

But if the offset goes beyond the screen width or below 0, I get a problem finding this midpoint. My idea so far is to just isolate y and find the midpoint by

y = x/m - b 
y = SCREEN_WIDTH*0.5/m - b

But is there perhaps a more clever way of doing it? A bit annoying to have special cases everytime b is outside screen space or m goes to infinity

[nocturne remnant]

Isn’t the midpoint just

https://tutors.com/math-tutors/geometry-help/midpoint-formula

Wait no, I read your question again

[upper karma]

How can I prove that |sin x| <= |x| (the absolute value of sin x is less than or equal to the absolute value of x)?

[spiral bough]

If $|x| \geq \frac{\pi}{2}$, then $$|x| \geq \frac{\pi}{2} > 1 \geq |\sin{x}|.$$

[somber coyote]

DVD_Koce_DVD

[spiral bough]

Therefore you only need to look at the case $|x| < \frac{\pi}{2}$

[somber coyote]

DVD_Koce_DVD

[spiral bough]

But then if you look at the unit circle and build both angles x and -x, they cut off an arc AB, which length is bigger than the length of the chord AB

Or in other words $2|\sin{x}| < \frac{2x}{2\pi} \cdot 2\pi$

[somber coyote]

DVD_Koce_DVD

[upper karma]

Thanks. That was really useful.

[upper karma]

yeah

Xm [ (X1 + X2)/2 ] , Ym [ (Y1 + Y2)/ ]

[amber condor]

How do you do that

[nocturne remnant]

Hint: First prove that BE and CF are angle bisectors of D1E1F1 and reduce the problem

[rapid elm]

Can someone help me to find the area of the rectangle ABCD

[worthy sapphire]

think back on the definition of a circle and draw in another radius

[rapid elm]

How i'm supposed to find the value of r?

[scarlet glade]

x²+y² = r²

Area = xr + r²

y² + (x+r)² = 6²

x²+y²+r²+2xr = 36

2r²+2xr = 36

Area = 18

Hope that helps

[rapid elm]

Ok thanks you

[rapid elm]

I don't understand this

[scarlet glade]

6 squared is the hypotenuse

Pythagorean theorem

One side is x+r, the other side is y

So we get (x+r)² + y² = 6² = 36

[rapid elm]

OH ok i see

[scarlet glade]

I'm supposed to solve for the angle x, but I'm stuck. Any tips?

[worthy sapphire]

let CAB = alpha and ABC = beta, then find as many same angles as you can! :p

[vital maple]

or notice that ABQP lies on a circle with center M the use inscribed angle theorem

[rancid wave]

idk if im over thinking this

but do yall know how i could solve this?

[plush robin]

do i need to know algebra before learning geometry

[covert pine]

No, but if you know algebra, solving geometry would be more easier

[rare kelp]

Brackysowchrom

Whacky stochrome

[valid harbor]

why do you say that M is the center of that circle

actually nvm lol

[floral shale]

Extend this line (shown in red) then use right triangle altitude theorem

[amber quartz]

why doesn't bezout's thereom apply for circles?

[final yew]

isn't bezout a number theoretic stuff?

[amber quartz]

According to the theorem, they can have 2 more points of intersection as well. But in reality, they do not have more than 2 intersection points. why?

[final yew]

i am really sorry to bother but afaik , bezout theorem says that there exist solutions to diophantine eqn ax+by=c iff gcd(a,b) divides c iirc

i am unable to relate it to circles

[slender wasp]

different theorem aha

One is about intersection points of curves

[nocturne remnant]

The NT one is called Bezout’s Lemma or sth

[nocturne remnant]

Found this on Wikipedia

Not that I understand it but seems like it would be of use

[upper karma]

musk? is that u?

[dusk light]

I am super confused on this problem. Can someone possibly help me? (:

<@&286206848099549185>

[floral shale]

You can use special triangles

A regular polygon of $n$ sides has $n$ number of congruent isosceles triangles whose vertex angle is $\frac{360}{n}$ degrees

[somber coyote]

Umbraleviathan

[floral shale]

@sweet orchid#1179

Bruh

Did this person leave the server

[velvet jolt]

I don't understand

Idk wht to do

Teacher didn't rlly help us with it

[austere hearth]

can someone help me find x in fonction of a and b

[plush robin]

thanks im in rational expression want to learn next geometry

[worthy sapphire]

calculate the area of triangle ABC two different ways, and use the similarity AFB~ABC

[fleet cosmos]

I couldn't simply it further

[upper karma]

maybe mutltiply divide by sin(r+2-r)

[sterile sail]

How to find the Surface of this triangle

[slender wasp]

Uhh

I can't tell if this is just a joke, this could be any triangle unless the task is to measure it with a ruler or smth lol

[hollow plume]

There are no labels…

[clever fable]

can someone help pls, I get that the direction changes i.e. it can't be a, but idk how to find which of the other options it will be?

[floral shale]

Measure side lengths with a ruler, use Heron's Formula

[analog breach]

So the main questions that I need help with are questions 2 and 4

[grand olive]

Do graphs of equations fall into geometry?

[upper karma]

yeah

[opaque ferry]

2 is straightforward

my problem is one

[onyx cloud]

have you heard of the law of sines?

@opaque ferry

also, how did you manage to solve 2 without getting an answer for 1

[silent plank]

Q2 can be solved from applying cos law directly. finding other angles is unnecessary despite what the hint says

[onyx cloud]

yeah but im still curious on how they thought of the cos law but not the sin law

[silent plank]

perhaps they did but had trouble using inverses

[runic beacon]

not the right channel for this question, but do you still need help with it?

[last fable]

Yes

[runic beacon]

Okay so what exactly do you have trouble with? And what does the question state?

[floral shale]

Should be sine law

[flint rose]

I guess u can use defactorization formula

2SinASinB = cos(A - B) - cos(A + B)

[fleet cosmos]

well this did the job

[flint rose]

I see great

Hmm was this is a jee level question?

I've seen few with Inv trigo in it

[fleet cosmos]

yup jee adv

[flint rose]

Ic

[jade merlin]

hi

i put 12 * tan (55) into my calculator and I got -542.1970549

what did I do wrong and how did they get 17.1

oh bruh i got it

[floral shale]

Imagine being in radian mode

[quick crater]

Dead

[upper karma]

ig the the formula is derived from like if we keep the largest possible square pyramid in a cube of same side lenght as of the square and then we ahave to find the volume of it

ofc this would require some calculus

right

[stiff bloom]

Is it true that the altitude of an equilateral triangle with an inscribed triangle of a certain radius is just 3r?

*inscribed circle

Not triangle

[upper karma]

yeah its right

[static glacier]

But it seems to easy and I'm worried

[upper karma]

what the hell becoz it is easy no need to worry

just ask

the arc angle

ig

and ef too

which angles u want to find

ig they have calculate angle 140 for the major arc DF

yeah

for the minor arc

it is 160

for major it is 140

yess

but idk about how the hell they got 140

[zenith raptor]

can someone give me the answer

[upper karma]

ye one sec

[zenith raptor]

I heard i could use desmos but I have absolutely no idea how to use it

[upper karma]

my math teacher also told me that when doing calculus

[zenith raptor]

lmao I'll probably watch something

[zenith raptor]

uh

[upper karma]

y=0.4x

[floral shale]

Wel no

It's not that

[floral shale]

Are you able to click on the points and see their coordinates?

[upper karma]

Does anyone know how to solve this?

[fierce shadow]

could u solve it?

@violet nest

@obtuse field

[vital maple]

use complex numbers

[fierce shadow]

it's for me?

[vital maple]

yes

[fierce shadow]

i should't

without it

it must be -0.5, but I cannot solove this

[fierce shadow]

@agile roost

[fierce shadow]

how do I find out which quarter the point is in?

tan(28pi/3)

[empty star]

oh

well

[upper karma]

I have been given a project to make a model on trigonometry, So I wanted a new innovative idea which is not generally available on search engine, like something completely creative.

So can anybody help me out

[lucid sandal]

What kind of model

[upper karma]

math

[opal sedge]

why does sin(x) have an inverse function when it doesn't pass the horizontal line test

[upper karma]

Means?

[lucid sandal]

... a project to make a model on trigonometry

What kind of model

[gusty umbra]

Answer is D

AB/DE = AC/CE = 6/7 and <c is congruent to <c

Two pairs of corresponding sides that form the same ratio and one pair of congruent angles can be found in the two triangles ABC and EDC, therefore satisfying the SAS similar postulate

[upper karma]

But i said they arent similar since their ratios arent the same

[silent plank]

it doesn't have a true inverse function. the domain can be restricted so that it does.

[opal sedge]

huh

[silent plank]

i.e. arcsin(sin(x)) = x for -pi/2 <= x <= pi/2

,W graph sin(x) from -pi/2 to pi/2

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ramonov#0921

[upper karma]

https://en.wikipedia.org/wiki/Versine

I found this on the web.

I mean, versine, coversine, haversine are rarely known and rarely used trig functions.

you can do your presentation on this. it will be amazing.

how do you feel about my idea?

[hollow raft]

Could I use this in a lesson on dilation for kids in grade 5?

Billy and his dilating anabolic steroids, although maybe increasing his outside size, don't increase the angles from within. :)

Cause in reality, he's still the same old triangle.

[upper karma]

Anything that is new that I can represent in front of everybody

[upper karma]

Thank you very much this is also very useful

[hollow hazel]

@gusty umbra Not correct. The angle should be between the pair of lines. Here <c is not between AB and AC. So your argument is wrong.

[upper karma]

Can someone help me with trigonometry

If u can even read my handwriting that is

It's due today....

[grave pond]

Is it correctly understood that all you have done so far is copy the problems verbatim to your own notepaper?

[upper karma]

What.

I don't understand what u said

[grave pond]

I'm trying to figure out if there's any work of your own (other than penmanship) in the image you posted.

[upper karma]

My teacher gave it to me. And I don't understand how to work it out, and the teacher won't help.

None of the questions above are from me.

It's from my teacher.

[upper karma]

lol sure

what is that bro lol

...no help?

<@&286206848099549185>

Can anyone help?

[grave pond]

Does your textbook at least contain statements of the law of sines and the law of cosines?

[upper karma]

I'm sure.

It does

Troposphere

Just froegt about it

Nvm

I can barely even understand u

[grave pond]

Your teacher's handwriting is not terribly legible, but in context I suppose the squiggle at the top of (3) is supposed to say "use the law of sines" and (4) says "use the law of cosines".

[upper karma]

I'll just deal with my punishment

[upper karma]

Tyats my handwriting but ok.

[grave pond]

You said your teacher gave it to you?

[upper karma]

She wrote it

On the board

I copied it

Cuz that's what we do.

[upper karma]

Thanks for ur time but...I'll just figure soem5jiyn put

[fierce shadow]

could u solve it?

[upper karma]

-1/2

[grave pond]

Amazing, that is indeed what the calculator says.

[upper karma]

yeah

lol

actually this is from calculator lol

[grave pond]

It doesn't seem at all obvious why it would be exact.

[upper karma]

it looks better if we convert these sines into cos

[sterile sail]

so we are going to make the red one 2:1 to the blue one, is there a quick method to find it?

and the red one has to have the same form as the blue one

[carmine marten]

@grave pond

<@&286206848099549185>

[stark sentinel]

Your first step is to read the rules

!15min

[lime crown]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[carmine marten]

ok

[carmine marten]

<@&286206848099549185>

[upper karma]

Please help ;-;

[grave pond]

Okay, I have it. The bold diagonals that make up this 7-pointed star each have unit length, and the star is oriented such that A points due right. By the central angle theorem, angle EAD is pi/7, so the direction from E to A is pi/14 above the horizontal. Then the direction from E to B is 3pi/14, and the direction from F to B is 5pi/14. This means that the y-coordinate of A is sin(pi/14) above that of E, which is sin(3pi/14) below that of B, which is sin(5pi/14) above that of F.
The sought-for expression is therefore just the difference in y-coordinate between A and F, but by symmetry this is just half the length of CF, which was 1.

[floral moon]

<TSE is 57 degrees, line ED is a tangent, ST is a chord. There is a theorem that says "the measure of the angle formed by a chord and a tangent is equal to half the measure of the intercepted arc". By the theorem, we can write this as 57 = x/2, multiply 2 on both sides, x = 114(A).

[frail karma]

Where do Plane M and line K intersect? I don’t understand

[slim tundra]

E

[upper karma]

Thank u so much!

[floral moon]

np

[floral moon]

The line k cuts through plane M. You need to find the point where the line touches the plane, which would point E

you could think of it like sticking a pencil(line k) through a piece of paper(plane M) and the hole formed would be where it intersects (point E)

[exotic roost]

how do i solve for x

[floral shale]

What do the dots mean

[final yew]

those angles are equal obvio

||use similar triangles DAF~ DBG => 14/(x+5)=x/24|| @exotic roost

[grave pond]

Hmm, they don't look equal in the diagram.

Diagrams are not required to be to scale, of course, but this looks unequal enough to warrant some doubt about the interpretation.

[dark sparrow]

maybe the dots in the angles serve to mark their equality?

[gusty umbra]

For the buidling problem, you are given a rectangle, and on top of that rectangle is an isosceles triangle (if base angles are congruent, 30 degrees = 30 degrees, then the triangle is isosceles). As the base of the isosceles triangle and the width of the rectangle are equal (assumed from the diagram), the base of the isosceles triangle is 30 units. Now dropping an altitude in the isosceles triangle to form two 30-60-90 triangles and bisecting the base of an isosceles triangle (Dropping an altitude in an isosceles triangle bisects the base as the two triangles that are formed are congruent by AAS), it can be derived that the base of both 30-60-90 triangles is 15. Now to find the altitude of the isosceles triangle, the ratio Tan(60)= 15/x (x being the altitude)can be established, which yields 5 root 3 as x. The height of the structure is the height of the rectangle (22) plus the altitude (height) of the triangle (5 root 3), meaning that the answer is 22 + 5 root 3.

[gusty umbra]

For the problem regarding the intersection of perpendicular sectors, the answer would be 14. Using the perpendicular bisector thm which states that the distances from any point on the perpendicular bisector to the line segment the perpendicular line segment lies on is equivalent allows us to distinguish that CD is 14. In this case, DG would be the perpendicular bisector, and the distances would be AD and DC, which are both equal as the thm states. AD = DC or CD would be equal to 14 as well.

[jovial stag]

i’m really confused

[coarse prism]

How is this possible?

Both cos^2(B) and cos^2(A) must be equal to 1

[floral shale]

Difference of squares basically

Its a bit disguised in trigonometry voodoo but its a difference of squares

You see if you let:
$$a = \sin{(A)}\cos{(B)}$$
$$b= \cos{(A)}\sin{(B)}$$

You'll notice that youll get the following phenomenon: $(a+b)(a-b)$. This is what you might recognize as the factored form of the difference between two squares. $(a+b)(a-b) = a^2 - b^2$, so if we just plug in we get your answer

[somber coyote]

Umbraleviathan

[floral shale]

yo @coarse prism you here bud

[coarse prism]

Yh

Everything was answered

I just substituted cos^2(x) with 1-sin^2(x)

[smoky jetty]

Use Pythagorean theorem

[smoky jetty]

try using either law of sine and cosine, or use both.

[upper karma]

..thx ig

[smoky jetty]

i hope it helps..

[gusty umbra]

Here for the 1st problem set up this ratio: Sin80/16 = Sin50/a (Just solve for a and u get c as well as it is an isosceles triangle if base angles are congruent then the triangle is isosceles)

[upper karma]

Thanks

Just didn't answer it

I sent it to my teacher

He was pissed

Who cares

[upper karma]

@upper karma basivally theres smth called the law of sines which lets u find a trianglws sides and angles

so if that kinda problem comes up again or you wanna study for a test

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:trig/x9e81a4f98389efdf:law-of-sines/v/law-of-sines this video could help, if you wanna bookmark it or smth

cuz in trig classes it tends to be used a bit

hope all goes well for u in the class bro

[wooden moon]

yo

gn everyone

i would like to know if in the triangle inequality theorem you can have c=a+x or just c<a+x

[mighty trout]

if you had equality then you'd have a degenerate triangle aka a line

[upper karma]

It didn't but thanks

[pure birch]

#help-19

[orchid pulsar]

<@&286206848099549185>

I understand the fundamentl rules

but these, i just cannot

[gusty umbra]

@orchid pulsar For problem number 18, we must find the m<ACD which is 1/2 of the measure of minor arc ad by inscribed angle thm. Using the inscribed angle thm, we can derive the value of major arc ACD is 244 as the angle the measure of the inscribed angle that intercepts it is 122 (The measure of the intercepted arc is two times the measure of the inscribed angle that intercepts it). We know that the sum of non-overlapping arc measures in a circle is 360, so the equation 360-244 = arc ac can be established. From this equation, arc ac is 116 meaning that the inscribed angle that intercepts that arc ( angle ACD) is 58.

[gusty umbra]

For problem number 19, we know the measure of angle OAR is 90 degrees as the angle subtended by the intersection of a radius and a tangent line (Im making the assumption the line is tangent) is 90 degrees. The measure of angle BAR is 1/2 the measure of arc BA, From the previous problem, we know Angle ACD is 58, so arc AD must be 116, which means that angle AOD is 116. As linear pairs are supplementary, angle AOB is 64 which means arc BA is also 64. As angle BAR is 1/2 the measure of BA, angle BAR is 32. Angle OAB = Angle OAR-Angle BAR = 90 - 32 = 58. So angle OAR is 58.

For problem 20, BAR is 32 as derived in the explanation for problem 19.

[gusty umbra]

The sum of the two sides must be greater than or equal to the third side for a triangle to be a triangle, so both a +x > c and a +x = c would create a triangle

[orchid pulsar]

Thank you @gusty umbra

[somber coyote]

British people

[gusty umbra]

280/30 = x sin(360+120)

= 28/3 = x (Sin360Cos120 + Cos360Sin120)

= 28/3 = x(Sin120)

= 28/3 = x(Sin(60+60))

=28/3 = x( Root 3/2)

= 168root3/27 = x

= 56root3/9 = x

[bright rover]

How do I do this question ?

[dense sierra]

hey hey, can someone explain me why this equalities happen that way by explaining the logic of it?

[candid comet]

Have you heard of the idea of ASTC?

Sine is always positive in A,S (First and Second Quadrants).

Cosine is always positive in A,C (First and Fourth Quadrants).

This may have to do with the explaination.

Also, Sine is an odd function and Cosine is an even function.

So, cos(-x) = cos(x) and then sin(-x) = -sin(x).

[upper karma]

The question asks to find the missing endpoint given that R is the missing endpoint and S is the midpoint is T is another endpoint. For the co-ordinates of R, I am getting (-10, 0) however, the online homework grader is showing that this is wrong. May somebody please be willing to verify? Thanks.

[silent plank]

read the question carefully

specifically about which point is supposed to be the midpoint

[pure birch]

#help-16

[steel trout]

hi im confused on how to find k given these 2 equations can someone pls help

[broken iron]

put s=10 and t= I6 in last equation

since last equation is valid for 16<= t <=30

[gusty umbra]

You can use the interpolation formula, mx2 + nx1/m +n = X and my2 + ny1/m+n = Y for points (5,2) and (12,-5). This will yield answers (9,-2) and (8,-1). Note that you must consider both (5,2) as (x1,y1) and (12,-5) as (x1,y1) as the 3:4 ratio can be split up both ways.

[dense sierra]

Thank you sm

[candid comet]

Has anyone done Euclid's book 7?

[upper karma]

what do mathematicans call the power to ban someone from a server?
ban theta (mispelling of tan theta) LOOOOAOOAOAOOAL

[nocturne remnant]

https://tenor.com/view/damn-you-got-the-whole-squad-laughing-damn-bro-you-got-the-whole-squad-laughing-not-funny-unfunny-gif-21764390

[mossy briar]

[broken nimbus]

4(4 - x) - 2(8 - y) = 8

16 - 4x - 16 - 2y = 8
4x - 2y = 8

what do I do after i've done this

question:
"find the equation of line A under central symmetry in the point (2, 4)"
A = 4x - 2y = 8

[strange robin]

y = 2x-4

(x-2)2=y-4

2x = y

[broken nimbus]

why remove the 8 tho

[strange robin]

i just divided by two

4x = 2y+8

4x-8 = 2y

2x-4 = y

[broken nimbus]

so you just swap them around ?

[strange robin]

um

sure

[broken nimbus]

any idea how to do (v) ?

[strange robin]

ye i dont know what axial symmetry is

lol

y= (2/7)x +2?

idk

[broken nimbus]

did i start the central symmetry part right tho

[strange robin]

wdym

[broken nimbus]

4(4 - x) - 2(8 - y) = 8

is that the right way to calculate iv

[strange robin]

-4x +2y = 8

2x+4 = y

ig not

[strange robin]

-2 * -y

= 2y

[distant scarab]

The two points on the line 2x+3y+4 = 0 which are at a distance 2 away from the line 3x+4y-6=0 are
ans. (64,-44) and (4,-4) obtained thru point-line distance eqn.

When I tried to solve this without using the point-line distance eqn., I used a drawing similar to the one in the pic
Solving for the angle between the two lines using (M2-M1)((1+M2*M1)^-1) returns tan(theta) = 1/18

Assuming that the triangle was a right triangle, tan(theta) = Opposite/Adjacent, and that the opposite side was 2, scaling that would leave me with with a distance d of 36

However when checking against the actual answer (4,-4) with the intersection of the two lines at point (34,-24), I get a distance of 10sqrt13 or approx 36.05551275

Did I just draw the fig wrong or perhaps used the angle between two lines eqn wrong? What wrong assumption did I make? my brain hurts bruh

oh wow nvm guys the right angle was on the opposite line

thx

Anal.Geom. is suffering

[rapid flare]

Guys can anyone tell detailed solution:

How to find the volume of void between 4 unit spheres that touch each other

@supple fern

[supple fern]

I wasn't hired for this, I just redirected you to the more appropriate channel

[mellow copper]

what does this mean? seems like they dont enclose a void.

[mellow copper]

yea so it doesnt seem bounded by the spheres

[sly halo]

Hello Everyone. Can anyone tell me if I'm on the right track with this?

[rapid flare]

It is bounded by spheres

R equal spheres

[mystic grotto]

Why am I not getting the same solution set when substituting 2sinxcosx with sin2x?

[supple fern]

I can't quite read everything, but you can probably prove that both sets are in fact equal

they might not give the same formulas

but the points they generate should be the same

[mystic grotto]

how to prove that both sets are equal?

[supple fern]

can you type out more cleanly what you got for each set?
you usually do that by assuming an element is in one of the sets, and then showing that it must also be in the other and vice versa

so for example if a number has the form x = pi/3 + 2kpi/3 (left set) then it also must be in one of the four other forms from the right set