geometry-and-trigonometry

[solid needle]

cis(a+b) = (ax + iay) * (bx + iby)
cos(a+b) + i sin(a+b) = (ax + iay) * (bx + iby)

now expand the right side

and then the only way two complex numbers are equal to each other is if the real parts are equal and the imaginary parts are equal

so set the real parts equal to each other (this gives you the identity for cos(a+b))

and the set the imaginary parts equal to each other (this gives you the identity for sin(a+b))

lemme know if you have any trouble with that

[upper karma]

a(x + iy) * b(x + iy)
(ab)(x² + 2ixy - y²)??

[earnest yarrow]

In this question, the last part " Prove that M moves along a circle"
Does this mean you have to prove M is concylic?

[torn ridge]

guys when you are working on egmo/any geometry book do you take time to rewrite the theorem on paper or you just use paper to do the problems?

[worn scroll]

pretty sure solvable even without knowing any lengths

first step easy find yellow angle

then this angle in terms of alpha

this other angle

and this last angle

[solid needle]

ax was a single variable

you cant split it into a * x

ax is a single name

try again

[upper karma]

Oh mbad

axbx + iaxby + iaybx - ayby

[solid needle]

yes, now do this

[solid needle]

set real parts equal and imaginary parts equal

[proven lantern]

Pspsspps

Anyone here?

[solid needle]

hi

[proven lantern]

Hi, can u help me understand a trig question?

[solid needle]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[proven lantern]

Ok but like u could just be a student wanting help

Anyway, it'd be nice if u can

Should i post here or in #help

Nvm

[proven lantern]

I'm in #help-15 🙏🏼

[upper karma]

Im not sure

axbx - ayby + i(axby + aybx)]

[solid needle]

dont forget the other side of the equation

cis (a+b) = cis a * cis b

cos (a+b) + i sin(a+b) = axbx - ayby + i(axby + aybx)]

[sturdy gyro]

ty ill look into it

[versed pagoda]

okay so I know there has to be a formula that lets you calculate the lengths of the sides of a trapezoid if you know the 4 angles and the length of the median

i mean if you know the angles you know the shape

and if you know the shape just knowing the median or any other length will give you exact dimensions

so what is the formula?

[dense carbon]

By median what do you mean

[dense carbon]

I mean yeah just split the whole thing into a square and triangles

Ok

Assume a trapezoid is two triangles and a square

Line AB is the shorter length

Line CD is the longer length

And they are joined by lines AC and BD

AB is parallel to CD

And each of the points have an angle

If we know all of the angles

Then

Hold on I’m thinking

Ok assume we know the length of AB

As x

You can do sin c to find the height

But

That’s assuming the diagonal of this triangle is 1

Uhhh

I’m thinking

Oh

I figured it out

Split the square into triangles

Four triangles

Not 2

the base of each triangle is x/2

Actually two triangle can work

No wait it can’t

Maybe

I’ll go with four

Four triangles

Line drawn from A to a point on line CD that’s directly under point B

And the same with point A

I’m going to draw this and find a result because this isn’t working

Ok so it’s not possible to use the length of AB to predict the rest of the trapezoid

Since it is just the distance between the triangles

And has no effect on the dimensions of the triangles

[versed pagoda]

sorry i realized you need more than just the median

but what about having the legth of the median

and also the length of the line connecting the mid points of the two parallel sides?

along with the 4 angles

[dense carbon]

I did some stuff

Also you only need angle C and D plus either line AB or line CD

This is the formula that probably works

But that’s assuming you know AC

I’m going to work on if you know BD

Ok I searched up what a median is in a trapezoid

There’s another 3 formulas for me to make now

Oh wait if you do know the median

Nvm taht doesn’t work

The median is just simpler to use

But you still need one more length

[dense carbon]

Yeah that works

[upper karma]

Oh, isnt isin(a+b) = i(axby + aybx)?
Then the other one should be cos(a + b) = axbx - ayby

Then this

Just substitute

Right?

[solid needle]

yep!

sorry this has been so fragmented

hopefully you're still following

now that you know the derivations of the cos and sin addition formulas, make sure you review them

do this derivation front to back a couple times to make sure you really understand the idea

and then you'll never have to memorize these ever again

[upper karma]

Thanks

[versed pagoda]

@dense carbon sorry I meant knowing the median
and the other bimedian or basically the line that connects the midpoints of the parallel side (the median connects the non-parallel sides)
and of course the 4 angles

[frosty kraken]

Then it'll be in terms of alpha not a constant

[dense carbon]

Welp I made a formula using only the median

The bi median is basically useless for me

[versed pagoda]

but it requires knowing one of the sides along with the median right?

if you just know the median and the other bimedian is what i am curious about

[dense carbon]

Ohhh

That’s what you meant

There probably is a way but I’m too lazy rn

[upper karma]

@solid needle I know you already told me a simpler way to prove what I was trying to prove

But, I rechecked my previous lomg message and redid it again

cos(a-b):

cos(a-b) = e^[i(a-b)] + e^[-i(a-b)]/2

[e^(ia-ib) + e^(ib-ia)]/2

cos(a)cos(b):

[e^(ia) + e^(-ia)][e^(ib) + e^(-ib)]/4
[e^(ia + ib) + e^(ia - ib) + e^(-ia + ib) + e^(-ia - ib)]/4
{e^[i(a+b)] + e^[i(a-b)] + e^[-i(a - b)] + e^[-i(a+b)]}/4

e^(ix) = cos(x) + isin(x)
e^(ix) - cos(x) = isin(x)
e^(ix) - {[e^(ix) + e^(-ix)]/2} = isinx
[2e^(ix) - e^(ix) - e^(-ix)]/2 = isinx
e^(ix) - e^(-ix)/2 = isinx
-i[e^(ix) - e^(-ix)]/2 = sinx

sin(a)sin(b):

{-i[e^(ia) - e^(-ia)]/2}{-i[e^(ib) - e^(-ib)]/2}
-[e^(ia) - e^(-ia)][e^(ib) - e^(-ib)]/4
[e^(-ia)-e^(ia)][e^(ib) - e^(-ib)]/4
[e^(-ia+ib) - e^(-ia-ib) - e^(ia+ib) + e^(ia-ib)]/4
{e^[-i(a-b)] - e^[-i(a+b) - e^[i(a+b)] + e^[i(a-b)]}/4

cos(a)cos(b) + sin(a)sin(b):

{e^[i(a+b)] + e^[i(a-b)] + e^[-i(a - b)] + e^[-i(a+b)] + e^[-i(a-b)] - e^[-i(a+b)] - e^[i(a+b)] + e^[i(a-b)]}/4

{e^[i(a-b)] + e^[-i(a - b)] + e^[-i(a-b)] + e^[i(a-b)]}/4

{2e^[i(a-b)] + 2e^[-i(a-b)]}/4
{e^[i(a-b)] + e^[-i(a-b)]}/2

{e^[i(a-b)] + e^[-i(a-b)]}/2 = cos(a-b)```

I was just getting confused with a lot of signs

Is this right

[solid needle]

that is a lot to check

unfortunately i dont have time to check

it is far easier to just substitute b= -b

cos(a+b) = cos a cos b - sin a sin b

we know that cos x = cos -x and -sin x = sin -x

so

cos(a-b) = cos a cos -b - sin a sin -b

= cos a cos b + sin a sin b

done

[solid needle]

but if you really really really want to check, i recommend shoving each line into wolfram alpha

every line is a true identity, so any line you paste into wolfram alpha should output "True" if its correct

[upper karma]

Hmm alright, when I get home, Ill do

I did the same with sin(a+b) = sin(a)cos(b) + cos(a)sin(b)

sin(a) = [-i(e^(ia) - e^(-ia)]/2
sin(b) = [-i(e^(ib) - e^(-ib)]/2
cos(a) = [e^(ia) + e^(-ia)]/2
cos(b) = [e^(ib) + e^(-ib)]/2

sin(a)cos(b):
[-i[e^(ia) - e^(-ia)][e^(ib) + e^(-ib)]/4
[-ie^(ia) + ie^(-ia)][e^(ib) + e^(-ib)]/4
{-ie^[i(a+b)] - ie^[i(a-b)] + ie^[-i(a-b)] + ie[-i(a+b)]}/4

cos(a)sin(b):
Eh, just copy but flip the variables

{-ie^[i(a+b)] - ie^[-i(a-b)] + ie^[i(a-b)] + ie[-i(a+b)]}/4

Add:

{-ie^[i(a+b)] - ie^[i(a-b)] + ie^[-i(a-b)] + ie[-i(a+b)]}/4 + {-ie^[i(a+b)] - ie^[-i(a-b)] + ie^[i(a-b)] + ie[-i(a+b)]}/4

{-2ie^[i(a+b)] + 2ie^[-i(a+b)]}/4
-i{-ie^[i(a+b)] - e^[-i(a+b)]}/2

-i{e^[i(a+b)] - e^[-i(a+b)]}/2 = sin(a+b)```

[lavish lotus]

im still in yr 10 and learnign like basic trig functions like tan theta = sin theta over cos theta is ther any other stuff i should be aware of or??

[worn scroll]

i'd say understand what sin cos and tan represent in the unit circle

and some identities

would be nice to already know some values too for some angles

[lavish lotus]

thanks!!!

[misty topaz]

5sin2x - 11cosx = 11sinx -7
Gentlemen, can you tell me how such problems are solved?

[proven lantern]

What's the question

You just look out for identities you can match it to, to get your desired answer

[misty topaz]

I want to know what x is equal to

[broken harness]

guys when using the cosinesentence and the degree is over 90 degrees do i need to do something to it or am i remembering wrong? can i just put -2abCos(120*) for example

[worn scroll]

no need to do anything to it

you will end up adding stuff because Cosine between 90 and 180 is negative and negative*negative in fact which makes sense because the angle is so large

[broken harness]

thanks @worn scroll

[bleak blade]

https://cdn.discordapp.com/attachments/1026131411813531708/1300687888219901992/image0.gif

[summer ferry]

From here i guess you’ll be able to solve further and find x

You’ll have to find the solutions from the two sinx+cosx equations separately**

[misty topaz]

thanks

[worn scroll]

wym

oh

well if i just know the lengths i'd probably find one angle

otherwise if i know coordinates you can either use vectors or matrices

sad

[sage radish]

bro doesnt know

[somber coyote]

Vanellope von Schmugz

Vanellope von Schmugz

[worn scroll]

hold on gotta order some food

alr

i cant express in terms of just 1 variable right

do i need to use the cosine thing

[worn scroll]

$\cos{(180-c)}\cos{c} - \sin{(180-c)}\sin{c}$

bro is tweaking

[somber coyote]

Ryse

[sage radish]

where is a and b 😭

[worn scroll]

i replaced them with 180 - c is that wrong

[sage radish]

oh no clue idk if its right

[worn scroll]

cos c-180 is just -cosc

sin c - 180 is just uhh yeah - sin

$-cos^2 c - sin^2 c$ erm what the sigma

[somber coyote]

Ryse

[worn scroll]

oh i just want to know sin

but in terms of a and b so nvm

[worn scroll]

just wanna use cosine rule

$cos{c} = \frac{a^2 + b^2 - c^2}{2ab}$

[somber coyote]

Ryse

[sage radish]

i did get a little confused there lol

[somber coyote]

Vanellope von Schmugz

Vanellope von Schmugz

[upper karma]

trigonometry makes me wanna kms

[zealous wasp]

Someone help me in private for a simple trig problem please

Ok

Ambiguous simple trigonometry porblem the teachers says I'm wrong how to counter argue

The problem is the following
The angle of elevation when 50 meters of rope have been released is 37 degrees, which is the height at which the kite is located.

I asked ai and got the addressed pictures as answer

Teacher says 50 meters should ve adjacent side

[somber coyote]

Vanellope von Schmugz

[zealous wasp]

Real is what is

Should take both as what is

It's exactly what I did but the teacher said that I was wrong

Then how do I convince a teacher

Because she says the problem is wrong

My answer

It's all good I did the calculations with very accurate results

[slender rain]

Why did they assume that the triangle with side of r, r, and 50√2 is a right triangle?

[trail tendon]

is there more context...?

just from the context here it doesn't seem like they would be able to assume that

[light fable]

is it not just given info for the problem?

[slender rain]

I need to find the area of the shape. And the book said that to find the area above the rectangle I need to (¼ × area of circle with radius of r minus the right triangle with side of r and r)

I don't understand why the isoceles triangle in that problem is considered a right triangle

[trail tendon]

it just gave a picture and told you how to find the area and thats it?

[light fable]

They told you it's a right triangle

[slender rain]

Yeah sadly

No context provided

[trail tendon]

so like there was no initial problem

[slender rain]

It's told only in the answer sheet

[shadow mist]

what is rhis?

[light fable]

I assume the little symbol thing in that angle is indicating it's a right triangle

[trail tendon]

oh, so there is just a picture in the book and then an answer sheet that says that^?

did it even number this as a problem?

[shadow mist]

maybe the angle between r and r is 90

[trail tendon]

i ain never seen that symbolize right triangle but 🫠

[slender rain]

Its only that...

[shadow mist]

dá de forçar a aparecer um triangulo reto ai não?

translate from portuguese

[shadow mist]

what language is that?

[slender rain]

Yeah that what I confused about

Why did they suddenly assume that the isoceles is a right triangle

[shadow mist]

well, it was the only way i think

[trail tendon]

i would say they can't, unless they give information in allat text but i can't read it so

[shadow mist]

fr

is this question solvavel?

[slender rain]

The text only said the function of the window

[shadow mist]

whats the anwser

[trail tendon]

thats a lot of text for just saying "the function of a window"

[trail tendon]

the actual answer, or the answer they gave?

[shadow mist]

ohh

te actual

[trail tendon]

the actual answer is that the problem is not solveable since you don't know the curvature of that arc 👍

unless its drawn to scale in which case i guess you could do it analytically but :l

measure it LOL

[shadow mist]

talvez no enunciado esteja falando sobre

translate from portuguese

oh

[shadow mist]

what language is that?

oh never mind

indonisian

a few months ago

i had an exam

and on the exam had a question like

how much is $\sqrt3+\sqrt2$

[somber coyote]

Argand

[shadow mist]

and the options was

[trail tendon]

yes

[shadow mist]

a)2 b)sqrt5 c)3 d)1,73

anyone of this is right

[trail tendon]

did it say to estimate?

[shadow mist]

but the "anwser" they give was b

[trail tendon]

or ask which one is closest?

[light fable]

me either but then, what is that thing doing there? And there's no way to solve the problem without knowing that angle, so

[trail tendon]

💀

[shadow mist]

Like

what the hell

[slender rain]

I feel scared because this question is in the previous year entrance examination in my country

[shadow mist]

well

indonisia lore:

well

we can respond this

[slender rain]

Why did they make such a unsolveable wuestion in the entrance exam 💀

[trail tendon]

if that exact question with no extra context was given that means it is almost certainly correct which would imply that the symbol in the corner does actually mean its a right triangle, like manifold was sayin

[shadow mist]

usando uma resposta que varia de acordo com a altura do "semi-circulo" da parte de cima da janela

[trail tendon]

i'm just sayin like it would be inaccurate for me to call it "not solvable" if its drawn to scale but like you would have to measure it 😂

[shadow mist]

porque o valor varia de acordo com a altura acima do lado da janela lá

[trail tendon]

in school? yea they wouldnt make you do that
in practice? sure :D

😂

well no not a ruler

you'd measure it with some other instruments i'm sure but 💀

[slender rain]

assume the top side is 50√2 then use the non-intersecting segment formula (2/7.r²)

then the total area = L of the rectangle + 1/2 of the segment area
L total = 50√2.60 + 1/2.2/7.(50√2)²
L = 300√2 + 1/7.5000
L = 1138.5 cm²

How about this one? Is this correct?

[weak agate]

I thought the corners would divide to 45 degrees

[weak agate]

So the diagonals don't always divide both sides equally?

It's just that in no. 16

[maiden brook]

just curious why u r pinging? does the classic area argument still work? I think it should

…

A = ab sin C = bc sin A = ac sin B

I was talking about the proof of the trig addition. I think that’s a bit more difficult to show but not so hard

?

[upper karma]

hello people

How are you all?

What are you all discussing

[maiden brook]

I see

I mean in some sense u need to be able to consider the sin of an obtuse angle. otherwise it’s impossible

but yeah

look at the image I sent before

I think sin and cosine are basically useless in triangles (change my mind?). So I don’t see why u would be so fixated on that

what’s good about it in right triangles

or from the right triangle definition

exactly

better when the triangles are not corrupted by trig

yes

nah but actually like trig doesn’t rlly help u solve like nearly any geo problem that u can’t solve with other tactics

wdym?

[light fable]

there are many situations that require trig. Like knowing one side length and one angle of a right triangle (other than the right angle) and needing to know the other sides. You know, the most basic application of trig

[burnt otter]

60 30 triangles are also essential for trig

[obsidian harness]

<@&268886789983436800>

[sudden harbor]

That looks so beautiful yet so brain wrecking at the same time

[graceful talon]

i thought moderators were called on this

[faint pasture]

"How dare they draw this profanity and throw it in a public channel"

[graceful talon]

the Chinese characters add to its complexity

and the last thing that shouldn't be forgotten: the low quality of the image

it shows the immeasurable number of times it has been screenshot to be displayed and appreciated

[worn scroll]

i mean yea 1/2 ab sin in between

and i get the angle in between using cosine rule

maybe can link to sine law

[earnest yarrow]

Why sin theta=P/H

[tight vector]

I can literally derive it [only if that's a legit question]

[graceful talon]

why not we look up its definition

,w define sine

[somber coyote]

Results provided by WolframAlpha

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

Requested by lkjsfkjsiu

[graceful talon]

so you're basically asking "why is it that for a given angle in a right triangle, the ratio of the length of the side opposite of it to the length of the hypotenuse equals the ratio of the length of the side opposite of it to the length of the hypotenuse?"

maybe your real question is its reason for existence

[graceful talon]

can you do it here?

[tight vector]

holdup

ima do it on a paper and send a pic here

sounds good?

[graceful talon]

ok. thanks

yes

afk in the meantime

[fading dove]

This looks interesting

What exactly do we have to do here tho ?

[quasi sun]

what if the oppiste is missing and adjacent is missing so i am forced to use sin

[maiden brook]

wdym? give me an example?
And don’t give me an example where u have to use a calculator to approximate the sides

[maiden brook]

I mean that’s just approximations of sides so not rlly useful

I think it’s just more useful in something like physics

[meager glacier]

When plotting a function for sine or cosine i have a question about the “phase shift” =C/B

is the phase shift always shifting on the x axis?

[fading dove]

Oh ok

[maiden brook]

I mean if it’s a completely random triangle it isn’t so useful u won’t rlly get an exact answer anyways without trig ratios. if it’s a particular angle that’s nice there are things u can do

[maiden brook]

but u can find those with “pure geometry”

Nothing that was mentioned shows how trig is useful.
Random example but if I said like Γ(a) is the real root of x^3 + x^2 + a and I claimed it was very useful cause it could help me find the real roots of x^3 + x^2 + a, I don’t think anyone would start adapting my notation. In like Euclidean geometry I don’t think it’s so helpful
I think it basically never simplifies/helps solve a problem u can’t do otherwise

[light fable]

they’re useful for real world type applications when you actually want to know how long some sides are or what some angle is, to an approximation.

If you’re talking about more pure geometry kind of problems like proving theorems, then you may be right that they don’t come up very much.

[maiden brook]

yeah I mean in pure geometry they aren’t that helpful, especially compared to like similar triangles, Heron’s formula, etc etc

[warm robin]

Goodmorning

[compact ocean]

is this supposed to be easy 😭

[maiden brook]

there’s a theorem that gets u the answer immediately. I’m assuming u don’t know it
construct segments so u can make use of the angles given

[compact ocean]

i asked ai and it said 70+40=100/2=55

is that correct?

[maiden brook]

yes but the answer isn’t important

[maiden brook]

^

[compact ocean]

is m>1 55

thats what its asking

[compact ocean]

no i dont know it

[maiden brook]

wdym? yes I said u don’t know the theorem but find the angle nevertheless given what I said. it’s a good thing to be able to do yourself

if you ever forget this theorem, especially on a test, u can quickly rederive + it helps u understand it better

[chrome token]

do you know the circle angle thing? that central angle/2 thing?

[obsidian harness]

Construct a line connecting the top two endpoints of the lines with the circle

By the angle in the centre principle, the base angles are 40/2 and 70/2

Then instead of doing 180 - (180 - x) you just have x, where x = 20 + 35 = 55

Well I mean you posted a lot about the theorem already

Once they understand it they can now look at my message

[upper karma]

@strange narwhal

Guys
Why is arcsin (2x/1+x²) for x>1 equal to π - (2 arctan x)

[obsidian harness]

wait

https://math24.net/weierstrass-substitution.html

ah yeah it's this

[upper karma]

Uhhhh

[sudden harbor]

It's a property ain't it?

2arctan x = arc sin (2x/1+x^2)

Idt there's a pi there

Besides when u put a negative value in arc of sin and tan it'll just put the negative outside
But for arc cos (-x)= pi-arc cos(x)

As cos is an even function but sin and tan are odd so they aren't affected by the negative value

[obsidian harness]

there is for x > 1

[obsidian harness]

okay, $\sin u = \frac{2 \tan(u/2)}{1 + \tan^2 (u/2)}$

now sub in $u = 2 \arctan x$ and you get $\sin(2 \arctan x) = \frac{2x}{1 + x^2}$, because $\tan(\arctan x) = x$

[somber coyote]

south

[obsidian harness]

now, you have to be careful with arcsin(sin u), because that is definitely not u

same for arccos(cos x) and arctan(tan x)

(in fact, the period of arcsin(sin x) and arccos(cos x) is 2pi, and for arctan(tan x) it's pi; why?)

the range of u will be form 2 arctan 1 to 2 arctan infinity
pi/2 to pi

for these values of u, you have arcsin(sin u) = pi - u

because pi - u will be in [-pi/2, pi/2], where [-pi/2, pi/2] is the domain of arcsin

that completes the expanation, given what we defined u to be

[median plaza]

about to take my test, someone bless me please

red is wrong green is right

how does this make sense

[obsidian harness]

ah, the triangle with the 30 and 60 is a right-angled triangle

you know this because the angles on a straight line add up to 180 degrees

so 90 + x = 180, x = 90

[median plaza]

so equalateral triangles are 60 60 60 and right triangfes are 30 60 90?

[obsidian harness]

yes, well the right triangle made in a equilateral triangle is 30-60-90

okay so you've divided the hexagon into 6 equal parts

so angles around a point = 360 deg

each central angle must be 360/6 = 60

then you break one of the 6 triangles into 2 equal halves

so 60 breaks into 30 and 30

[obsidian harness]

the blue triangle is equilateral so the base angles are 60

but you could also do 180 - 90 - 30 = 60 too

[obsidian harness]

there's also a 45-45-90 right triangle which is very common

and a lot of other right triangles with random angles

[median plaza]

so an isoleces

[obsidian harness]

mhm!

an isosceles right triangle

[median plaza]

im so cooked

[obsidian harness]

it's actually just area = 1/2 * n * s * a

n is the number of sides

[median plaza]

so

1/2 x apothem x permiter

[obsidian harness]

yep

number of sides * the side length of each triangle

that gives the perimeter

in fact it's just 1/2 bh = 1/2 * s * a

and then you multiply this by n, cause you have n triangles in total

[median plaza]

oh yeah this is easy

[obsidian harness]

5 triangles here

all the same shape and size (congruent)

[median plaza]

oh yeah

[obsidian harness]

yeah the apothem is a fancy Greek name for the height

the side length is the base

that's how

[round leaf]

Guys why is the bgtan (-V3) = -π/3 and not the supplement π-π/3

I know its about the range of the bgtan but like is 2π/3 inside the interval -π/2,π/2?

[somber coyote]

Vanellope von Schmugz

[round leaf]

Yes

Which means 2pi/3 should be the answer

Right

Bec the range of bgtan is ]-π/2,π/2[

[noble cedar]

HEY GUYS QUICK QUESTIONS DO I ALSO CALCULATE PI OR LEAVE IT THE PI LIKE ITS A UNIT

[round leaf]

2pi/3 is not in there

So 2pi/3 should be the answer

[silent plank]

it's not in there so it isn't the answer

[round leaf]

But its an open interval

Meaning everything but the things inside the interval

[silent plank]

no

that's not what open interval means

[round leaf]

What abt [-1/2,1/2] meaning everything between those two numbers and those two numbers right

?!

[silent plank]

yes

[round leaf]

Thus ]-1/2,1/2[ means everything but these numbers

[silent plank]

open means those edge values are excluded

no

it'd mean values between -1/2 and 1/2
but not including -1/2 and 1/2

[round leaf]

No?

[maiden brook]

no

[round leaf]

Hold on

The bgtan isnt even defined outside that interval

That’s impossible

Thats why there’s two asymptotes

Okay mb

!!!

Okay thanks!!!!

This was kinda embarrassing

Im supposed to know this

[maiden brook]

you would point out that this is relying on trig ratios for obtuse angles

[lone pike]

If abe and cde are congruent, then ac bisects bd and it's a parallelogram. Why don't we know that ad is parallel bc?

[solid needle]

is there like a *yet implied somewhere here due to context?

because you totally can show they are parallel, but it requires more steps

using CPCTC, you can get enough info to show that triangles ADE and CBE are congruent

and then you can get that angles ADE and CBE are also congruent, which proves AD || BC

[lone pike]

Uhh, the original question was something like, which to triangles, when congruent show that abcd is a parallelogram

[glad apex]

Need help on this one <@&286206848099549185>

[solid needle]

ABD and CDB

[opaque wadi]

chat

how do yall draw shapes

globular and cone

[upper karma]

for the second one
arctanx + arctan1/x
arctanx = y
for y >=0
x = tany
arctanx + arctan1/x
y + arctan(-tan(y - pi/2)) = pi/2

for y < 0
y + arctan(-tan(y + pi/2)) = -pi/2

is this looking right

[solid needle]

yep!

here are two more ways of understanding this

method 1:
arctan a + arctan b = arctan( (a+b)/(1-ab) )

this is easy to derive from the tan sum formula

the denom 1-ab is zero, and the only way you can take the tan of an angle to get no run is if the angle is +/- pi/2

method 2:
think of the tan function as a function that converts an angle direction into a slope direction, so the arctan function converts slope to angle

imagine a line with slope x and a line with slope 1/x

if you can visualize this, you should also be able to visualize why they are complementary (in terms of their ref angles)

[upper karma]

[west knoll]

Conatuct a Rhombus PQRS such that PQ = 60 mm and diagonals PR=9 mm.Measure and write Down the Size of Qp̂S.

Construct*

<@&286206848099549185>

@lime crown

[tender jetty]

did bro just ping the fucking bot role

[upper karma]

Yes

[west knoll]

what does that mean

bot role ?

ping ?

[upper karma]

Lmao must be new to discord

Aren't you

[west knoll]

yes i am

[upper karma]

Anyways let's help you out

What grade are you in

[west knoll]

8

[tender jetty]

help me out here i forgot the 15 minutes command can you do that rq

[upper karma]

Can I send pics in here

Of the answer

[west knoll]

yeahhh !

[upper karma]

@west knoll what do u want

The area of the triangle

Or the perimeter

The question is not well given

It's pretty easy

Ill let u in on a hint

[west knoll]

Conatuct a Rhombus PQRS such that PQ = 60 mm and diagonals PR=9 mm.Measure and write Down the Size of Qp̂S.
Construct*

it hust it

its just it

[upper karma]

Ok

Then

Both the diagonals are equal

Should be enough for u to solve the question?

Or do u need more

The diagonals are always equal whether in a parallelogram or in a rhombus

Sorry for the bad handwriting man

Should be simple enough

@west knoll is your problem resolved?

[upper karma]

I pinged you because of your about me

[dusky pelican]

what are identities?

i don’t understand them

and i’m in last year of high school

[solid needle]

identities are simply statements that are always true

for example:
x+1-1 = x

that one is kinda silly, but examples like

x^2 - y^2 = (x+y)(x-y)

would be more useful

but nonetheless they are all considered identities

identity in math refers to "itself". saying something is equal to itself is obviously true, but not so obvious when they look very different, like in the example i just gave

so we value those identities to be more useful because it allows us to substitute those expressions freely

does that help?

[dusky pelican]

yes thanks you!

[tardy ravine]

not sure whats going on here

it says its a regular pentagon

so 360/5 = 72 degrees internal angle

and A is at 5,2 so it has a lenght of sqrt(29)

[tardy ravine]

how to draw these polar curves without suffering

[rain sparrow]

Sqrt21

Sqrt(5^2-2^2)
Sqrt(25-4)
Sqrt(21)

[tardy ravine]

wait hold on

[rain sparrow]

Wish I could help but I can't even remember 😭

[tardy ravine]

how?

if im finding OA

[rain sparrow]

Oh wait

Oops yeah 😭

[tardy ravine]

isnt it sqrt(5^2 + 2^2) = sqrt 25+4 = sqrt29

[rain sparrow]

You're right ignore everything I said

I was tripping 😔

[tardy ravine]

that book giving me diff answers for that question so im skipping it 😭

[tardy ravine]

its alr ill eventually find a way 😭

[rain sparrow]

Wdym different answers

[rain sparrow]

Do you mean finding the points?

[tardy ravine]

i mean actually sketching it by hand

[rain sparrow]

Oh 😭

[tardy ravine]

so when i use cosine rule to get OB using 72 degrees i get like 6.3 something

[rain sparrow]

Plot the points and draw it I don't think it matters if it looks rough

[tardy ravine]

book is saying its 8.3

[rain sparrow]

Let me try it

[tardy ravine]

alr

[rain sparrow]

Is OB the angle??

[tardy ravine]

OB is the length from O to B

[rain sparrow]

Where did you get 72° from

[tardy ravine]

regular pentagon

[rain sparrow]

Just dividing by five?

[tardy ravine]

so internal angle is 360/5 = 72

yea

[rain sparrow]

Oh I've got it ok

Let me draw it

[rain sparrow]

Should be 540° that's the problem

[tardy ravine]

ahhh

tysm that makes alot more sense S:ob

[rain sparrow]

Yeah no problem

Lmk if you get the answer

[tardy ravine]

LETS GOOOOOOOOOOOOOOOOOOOO

.racw

,racw

[tardy ravine]

,rccw

[tardy ravine]

first ever actual polar curve thats not a circle done by hand lets goooo

now i gotta do this 10 more times 😭

[rain sparrow]

Good job 🙏

[rain sparrow]

Practice makes perfect

[tardy ravine]

the thing is

i hate drawing as well 😭

i was so bad at it they barred me from taking it as a subject 💀

(art)

[rain sparrow]

Wow 😭😭😭

Usually what I would do for the ones on the right is just do it in one go

Loop loop loop loop

[pine thunder]

can every polygon with number points of form 2^2^n +1 be drawn with a straight edge or does it hold for the primes only

[brave dirge]

Does anyone have the PDF file to the "Geometry and it's Applications" textbook? Any edition is fine, lol

[ionic hawk]

If someone wants to try

$\forall x \in [-1;1] \medskip$
$\sqrt{1-x}=2x^2-1-2x\sqrt{1-x^2}$ find every solution

[somber coyote]

nayuzz

[upper echo]

you can put x=cos(t) and then it is factorized into simple factors.

so, there are 3 real roots: -√3/2 and ±√(10+ 2√5)/4

[oblique dock]

hello

[rain sparrow]

@verbal quail did you ask the teacher

[rain sparrow]

Hello

[tulip birch]

i need help finding x on this question

i thought about "cos(30)15" but that doesnt work

[dark sparrow]

write out the trig ratio in full, and do it carefully!

don't try to immediately guess at the answer.

[tulip birch]

i mean thats what i would usually do when a question like this pops up

[dark sparrow]

well, do it now

what is the equation that you get?

[tulip birch]

hold on

well ofc you have to think about soh cah toa

and seeing where everything is placed i think its cosine

[dark sparrow]

well indeed it is

now write out the actual ratio

there's a detail that i want to see you get right, or correct you if you don't

[tulip birch]

thats adjacent over hypotenuse right?

[dark sparrow]

yes, but again

[tulip birch]

yes yes i know

[dark sparrow]

do it

just do it

[dense carbon]

15 is cos(30) times x

Now find the implications of that to reverse engineer the value of X

It’s because x is the hypotenuse and 15 is adjacent to 30

Cos = adjacent/ hypotenuse

Or which way was it I forgot

[worn scroll]

$\cos{\theta} = \frac{A}{H}$

[somber coyote]

Ryse

[worn scroll]

$\sin{\theta} = \frac{O}{H}$

[somber coyote]

Ryse

[worn scroll]

$\tan{\theta} = \frac{O}{A}$

[somber coyote]

Ryse

[worn scroll]

you have an angle and the side adjacent to the angle

and you want to find the hypothenuse

so we have theta, A and we need to find H

[dense carbon]

$ x= \frac{15}{\cos(30)}\$

Gosh darn it

[worn scroll]

which one of these have an angle, A, and H?

[rain sparrow]

Did you get it?

[cold sun]

how do i find the angle thats on the red

[pale dagger]

180-106

[cold sun]

how did you get this tough

why 180-106

[pale dagger]

a plate is 180°

yk

[cold sun]

a triangle is 180 degrees ik that

[elder flicker]

Omg that is just a straight line

I was trying to find sin and cos —

[pale dagger]

i bet all my money he's 12

[cold sun]

who me?

[pale dagger]

yh

[cold sun]

im just bad at math

[pale dagger]

fr ?

how old r u ?

[cold sun]

17

[elder flicker]

Geometry was at 14 for me

[pale dagger]

dang

[cold sun]

math diffed lol

[pale dagger]

you're american right ?

[cold sun]

nope

[pale dagger]

which country ?

[elder flicker]

Math can be hard

[cold sun]

Aussie

[elder flicker]

Esp when u don’t have good teachers

[cold sun]

my maths knowledge is really incomplete

[pale dagger]

wtf is your school systèm

[cold sun]

i've down calculus before but i could barely understand some aspects of functions

done*

[pale dagger]

in my it were 11-12

[elder flicker]

American

[cold sun]

i still don't understan the logic behind why its 180-106 though

[elder flicker]

180 degrees is a straight line

[pale dagger]

understandable

[elder flicker]

Technically it’s supposed to be
Alg 1
Geometry
Alg 2
Precalc

[cold sun]

meaning

[pale dagger]

yk what's a 360 ?

[cold sun]

circle

[pale dagger]

did you played call of duty ?

[elder flicker]

The 106 lays on the straight line so the other angle must be supplementary, meaning it adds up to 180

[cold sun]

oh

[elder flicker]

But I took accelerated math in middle school so it was
6th and half of 7th
7th and pre algebra
Algebra 1
All in middle school and then
Geometry
Alg 2
Took precalc over the summer
Ap calc + took the ap precalc exam
And then first semester I took calc 2 and now I’m in calc 3

[pale dagger]

i don't even know what's the garde

tell with the ages

[elder flicker]

6th is 11-12yo
7th 12-13
8th 13-14
Geometry was 14-15
Alg 2 15-16
Precalc at 16
Calc 1 16-17
Calc 2 17
And I’m still 17 taking Calc 3

[pale dagger]

goofy school system

[elder flicker]

I made it worse

[pale dagger]

wdym

[elder flicker]

Precalc over the summer, otherwise I’d be in ap calc rn

[cold sun]

pretty much the same here

[pale dagger]

what tf is precalc

i don't live in the same country sis

[elder flicker]

All the math between algebra 2 and calculus

[pale dagger]

ohhh

i see

we did the same at 15

[pale dagger]

in france we did it all at the same time

[elder flicker]

Ah

[pale dagger]

dang

[elder flicker]

Oui oui baguette

[pale dagger]

i just understand

you live in america ?

[pale dagger]

oui oui baguette très bon

[elder flicker]

Ye

[pale dagger]

dang

what's the hours in your country ?

[elder flicker]

Like rn? 8:11pm

[pale dagger]

who tf say pm

why don't u say 20:11 ?

[elder flicker]

That’s time in the us for ya

[pale dagger]

kinda weird

[elder flicker]

24 hr is military time

[pale dagger]

uh ?

military time ?

[elder flicker]

That’s what we call it

[pale dagger]

ok

fr you're on the other side of the sea

so weird to think that it's 20h in your country

[elder flicker]

Mhm

[pale dagger]

what's your opinion on france ?

[elder flicker]

Je taime

No wait that’s ily

J’aime

[pale dagger]

t'aime*

[pale dagger]

what're u trying to say ?

[elder flicker]

I like

[pale dagger]

j'aime

[elder flicker]

Yeah

[pale dagger]

je t'aime mean i like u 💀🙏🏼

[elder flicker]

I know I said that

[elder flicker]

^

[pale dagger]

do you learn french at school ?

[elder flicker]

Last year, I’ve lost most of it from lack of practice

[pale dagger]

like what did you learned ?

[elder flicker]

Just a bit. French 1, there’s 4 years total, 2 required

[pale dagger]

wdym 2 required

[elder flicker]

Liek to get into a good college

[pale dagger]

so you did the 4 years by choice ?

[elder flicker]

No, I did one year, I was trying to help u understand my level. There’s up to 4 years

[pale dagger]

oh ok

so you begin learning french

[elder flicker]

Yeah but I didn’t get to finish cause of my schedule

[pale dagger]

so which language did you learned at school ?

[elder flicker]

Spanish now

[pale dagger]

how many years ?

[elder flicker]

2 semesters im in my second

[pale dagger]

in america you count in semesters ?

for us that's trimestre yk

[elder flicker]

College does

HS is years

[pale dagger]

HS ?

[elder flicker]

High school

[cunning lion]

it depends on the university. most american universities use a semester system but some use a trimester system (although it's usually called a "quarter system" because of the existence of a summer session)

[elder flicker]

Mhm

[pale dagger]

k ty

so they count summer either ?

[cunning lion]

most people don't take classes during the summer quarter, but it still exists which is why they're called quarters despite most people taking classes for 3 of them

[elder flicker]

I took a summer class

[pale dagger]

you've school during summer ??

like additional school

[elder flicker]

I took precalc over the summer

[pale dagger]

so you sacrifice like 1 weak ?

like you were doing math the whole week ?

[elder flicker]

Two months?

[pale dagger]

you sacrified 2 months ??

[elder flicker]

That’s how long the class was

[pale dagger]

dang

[elder flicker]

I’d do my work in big chunks

[pale dagger]

and what were the hours

when you start and when you finish yk

[trail tendon]

who keeps track of that

💀

[pale dagger]

brain

[trail tendon]

brian

brian keeps track of it

go ask brian

[pale dagger]

k

@rough pier

[trail tendon]

bro 😭

[pale dagger]

what was it ?

[elder flicker]

I was like not in real class so I’d just do it when I felt like it

[pale dagger]

like you goes there when you want ?

[elder flicker]

No it was online

[pale dagger]

why are amercians that weird fr

[trail tendon]

wdym weird

[pale dagger]

idk y'all are just weird

like you take am and pm, online school, weird school system

[trail tendon]

how is that weird

[pale dagger]

kinda weird to say like 10pm

22h is way more easier

[trail tendon]

🤔

[trail tendon]

they're both literally just as easy

💀

[pale dagger]

nah not even

[trail tendon]

its just because you're used to it

i don't see a reason for doing one or the other, as long as you know what day it is and what time of the day

both systems got that so yea

[pale dagger]

nah i use both

[trail tendon]

then how is it weird

💀

[pale dagger]

i use way more the 20h

[trail tendon]

then like i was sayin its because you're more used to it

[pale dagger]

stop yapping pls

i'm rly tired fr

[trail tendon]

[pale dagger]

(3am in my country)

[trail tendon]

3h

[pale dagger]

👍🏼

[verbal quail]

yoyoyoy

[pale dagger]

what's bro saying

[trail tendon]

bro can't read anything longer than seven letters

[pale dagger]

fr

my eyes isn't eyesing

[solid needle]

am/pm is objectively just worse than military time

here's a question to ponder

[pale dagger]

fr

[solid needle]

we "flip" to a new cycle from 12 o'clock to 1 o'clock

because we are no longer adding one, we are "resetting" the hour value

but for some reason

we "flip" from am to pm (or vice-versa) between 11 and 12

why?

if you treated 12 o'clock as 0 o'clock, now the flip is from 11 to 0

and this lines up and makes perfect sense

[pale dagger]

did you know that military times does the same thing ?

[solid needle]

you begin the day at the beginning of the cycle, 0 o'clock

[solid needle]

yeah and this is why military/24h clock is just superior

because it does this

and you also remove the ambiguity of am/pm

because no one ever is going to confuse 3 o'clock with 15 o'clock

but there have been, and there will always be, some people who get confused between 3 am and 3 pm in some contexts

[pale dagger]

@trail tendon he's way smarter that me to explain why military time's better

[solid needle]

nothing the us does makes any sense whatsoever

even our date format is inferior to the EU's

and actually the rest of the world

we use mm/dd/yyyy instead of dd/mm/yyyy

[pale dagger]

fr and why even do you use miles ?

[solid needle]

we organize everything else in order of magnitude and division of units

either from smallest to largest or vice versa

for instance

hh:mm:ss

why would we ever do hh:ss:mm

but that's precisely what you're doing when you do mm/dd/yyyy

and no one ever questions this insanity

[graceful talon]

maybe because mm/dd is seen as a single entity

[pale dagger]

fr that's kinda dmb

[solid needle]

that's the historical reason

but then when you try to normalize date formats

it gets all messed up

you should never allow yyyy/dd/mm nor should you allow mm/dd/yyyy

but im not going to just pick on the US, i can also pick on the entire human civilization too

we should always count starting from 0, and we should use tau instead of pi

those are probably too late to change because it requires the entire planet to cooperate to a new standard, unfortunately

formally, none of these things prevent you from doing proper and correct math in any capacity

but unfortunately humans are biological fleshbags and not perfect computational machines and so design is an important factor

so we don't make mistakes like mislaunching a rocket because we forgot to convert units and killing people

./rant

[trail tendon]

thats cuz they don't know the system
how do i know if 0 o'clock is mid-day or mid-night?
obviously its mid-night but if i didn't know that i would get confused
same thing with am pm. how do i know if 12 am is mid-day or mid-night?
its obviously mid-night but if i didnt know that i would get confused

lol

now yes there are two things to remember instead of one which makes it slightly worse i guess

[solid needle]

we should also never teach the english language then

[trail tendon]

but like

[solid needle]

if they don't know what the word "sub" means

they might get confused when i say that i want to eat a sub

thinking i want to eat a submarine

ignorance is not an excuse

at one point you didn't know what am/pm meant either, but you learned those dumb rules anyways

[trail tendon]

what does this have to do with anything?

[graceful talon]

is am/pm really English tho? isn't it latin?

[trail tendon]

you didn't know what 0 o'clock meant either...

[solid needle]

idk, you're the one arguing that am/pm isn't inferior

you tell me

[trail tendon]

until you learned it

[solid needle]

right, that's MY point

what's yours

[trail tendon]

you're saying you don't know until you learn it and thats true for anything

[trail tendon]

that would argue that it makes no difference which one you use

you have to learn it anyway

[pale dagger]

don't think so i learned latin for 4years and never heard abt that

[solid needle]

GOOD, YOU GET IT

im so happy, we agree

[trail tendon]

so am/pm isn't inferior, they're both like the same

XD

just different systems

[solid needle]

no it is inferior, for the reasons i mentioned

because humans operate under design

not as perfect calculators

i could replace hours 1-12 with the zodiac

that's also the same system right?

hell let's use the chinese zodiac because it's also the same

120 minutes after mouse is tiger

is that just as good of a system?

[trail tendon]

we could have 🤷‍♂️ but we aren't boutta change it now lol

[solid needle]

no the point is

this is clearly a worse system

[trail tendon]

why?

[solid needle]

what is 60 minutes after mouse o'clock

tell me right now

[trail tendon]

unless you make it overly-complicated it doesn't matter much lol

[trail tendon]

what is mouse o'clock?

[graceful talon]

i thought am and pm stands for ante meridiem and post meridiem maybe those sound familiar to you

[solid needle]

i just told you, it's ordered by the chinese zodiac

[trail tendon]

i don't know the chinese zodiac cuz thats not a standard way to tell time 💀

not that it couldn't have been, it just isn't

[solid needle]

why can't you just honestly say that assigning arbitrary names to memorize for the numbers 1-12 is dumb

[pale dagger]

yh you're probably right

[trail tendon]

now that we've already done it yes it would be dumb to change it
in the first place though i don't see why it would be dumb :l

actually nvm it would be dumb because how would you express time in between lol

it should be numerical yea

mb

[solid needle]

ok so you see the point of design

and that's my point

you don't even have to change anything because hello the entire world uses the 24h clock pretty much

it already exists as a standard, and it's better

[trail tendon]

i don't think either design is better
but i could agree that if the entire rest of the world uses 24h system then we should too 💀

[paper spear]

i love

[left zealot]

any ideas how to find CAE?

[restive flare]

Is A the center of the circle?

[ivory path]

j

[deft sail]

How do you describe transformations?

Like, saying that one figure slides, flips, reflects to another?

[sudden harbor]

Is BE the diameter of the circle?

[left zealot]

not mentioned in the question

[left zealot]

i dont know actually, question only gives out the diagram and the value of some angles

[sudden harbor]

Wish you'd shared this sooner anyway

So

Using the given info

We know
IBD+DBE+EBJ=180

So we have DBE=58

Now they have given DHC as 68

Which means
Angle DBC will also be 68

Hence DBE+EBC=68
58+EBC=68
hence ebc is 10

And now u have eac

[obsidian harness]

yeah assume A is the centre, that's correct

[sudden harbor]

Yep it has to be the centre

[obsidian harness]

I bet it's the centre so yeah

[sudden harbor]

Cause be is not the diameter fs

[sudden harbor]

Using
2ebc=eac

Hence your ans is 20

[left zealot]

woahh i totally missed that one

ive obtained EBC, but didnt know that u can use the "angle at center is twice the angle at circumference" theorem

[left zealot]

why it has to be centre?

[left zealot]

thanks for this insight btw! appreacite it

[obsidian harness]

super important

[obsidian harness]

it's the question's fault that it didn't say

[narrow sail]

it's degree measure theorem

[obsidian harness]

you can't do the question otherwise

[pale jewel]

can someone help me with chemistry??

[solid needle]

https://en.m.wikipedia.org/wiki/Inscribed_angle

[left zealot]

Ahh I see, thanks

[worn scroll]

intuitively in a quadrilateral the sum of opposite angles is 180

so yeah you can prove that with this the problem is the proof of the other one

[meager glacier]

On question 4 im supposed to use identities to simplfiy but the result i got looks more complex lol does this still count as “simplification” bc i removed the degree of 2?

[silent plank]

no

and they want you to approach this with factorisation

[sudden harbor]

Ywww 😁

[gentle haven]

Are all values of sin or cos of (π/n) expressible by radicals?

Some people are saying pi/11 isnt, but i don't understand why

wolfram gives this for cos and it appparently (according to desmos, chatgpt and copilot) gives a complex number which the real part is cos(pi/11). but when i evaluate it myself i am getting -1?

for sin it gives one which i know gives a complex solution.

[solid needle]

if you mean radicals of only positive real values then no it is not always possible, and the proof is quite advanced and deep if you want to be rigorous, afaik

the reason youre getting weirdness here is due to how complex numbers work

basically, if you take the principal solution of 1^(1/n), you get cos(pi/n) + i sin(pi/n)

so if you do it this way, its almost a tautology

if you know a bit about complex numbers, specifically how rectangular and polar form work, then you can look up either euler's formula or roots of unity

[gentle haven]

ohk

[gentle haven]

where cani find such proof?

[solid needle]

maybe start here: https://mathworld.wolfram.com/TrigonometryAngles.html

but i wouldn't know anything more than that without spending a significant amount of time digging through it

good luck

[patent light]

can someone pls send a parallelogram

[left zealot]

i think it's a similarity problem but dont know where to start

[timber saddle]

Does anyone have the PDF or book about complex number on solving geogemetry problem

[terse sphinx]

hey can someone help me with this problem? i was absent that day

[dark sparrow]

,rccw

[dark sparrow]

https://www.youtube.com/watch?v=g8VCHoSk5_o&list=PL0o_zxa4K1BVCB8iCVCGOES9pEF6byTMT

look through this playlist

[vapid niche]

Given a circle (O) with diameter AD. Choose point B on the semicircle (B is different from A and D). On arc BD, take point C (C is different from B and D). The two chords AC and BD intersect at point E. Draw segment EF perpendicular to AD (F ∈ AD).

a) Prove that quadrilateral ABEF is cyclic.
b) Prove that AE . AC = AF . AD
c) Prove that E is the incenter of triangle BFC.
can someone help me solve c and provide me a detailed and thorough explanation

[tulip kite]

hello can anyone give me the basic lessons of geometry and trigonometry

[solid needle]

we dont do that kind of thing on this server, but there are plenty of free resources online and you are free to ask as many specific questions on this server as you want

[solid needle]

c is annoyingly tricky so far

just checking, you already know how to do a and b? and you know the basic properties of the incenter right?

[upper karma]

okay shortie princess

[tulip kite]

okay, thank you what free online resources can you recommend for geometry and trigonometry?

[silent plank]

Khan academy

[neat wigeon]

khan academy is just the goat

[tulip kite]

okay thank you

[meager glacier]

Oh so 4 would be (cosx)(cosx+sinx)

[meager glacier]

i did that sh way complicated the first time 😭

[left zealot]

anyone can help?

[left zealot]

angle should be equal and length should have the same ratio since they are congruent triangles.
i have no idea what cevins is

my progress so far

[brave dirge]

Heyyy, does anyone have the PDF for this textbook??

[left zealot]

Ohh okay. i dont know the theorem about cevians.

Can you give guide me?

Ohh I see. learn something new today hehe

[left zealot]

Am I on the right track? Still clueless how to find length DX (point to to base BC)

no worries

[patent light]

I have a question

[dark sparrow]

[patent light]

How do you derive the formula for the surface area of a frustum

[dark sparrow]

conical or pyramidal?

well actually no the idea is the same in either case

a frustum is a pyramid with a smaller pyramid chopped off the top

use that to figure out both pyramids' lateral surface areas and subtract them to get the lateral surface area of the frustum

and then add the areas of both bases

[earnest yarrow]

..