geometry-and-trigonometry

[quaint tide]

nice

wait how old r y'all that's the main q

cuz

grades are diff in some countries

I'm 15

[smoky jetty]

same-

[gilded cove]

Well I'm 16

[quaint tide]

cool cool

yeah no I was feeling hella optimistic when I started this

but these equations hella complicated

bruh this is gonna be a biquadratic equation wth

[zealous meadow]

same

[quaint tide]

niceeeeee

[smoky jetty]

people at #competition-math might solve it in <10mins lmao

[quaint tide]

Damn fr?

I ain't gon touch that place then 💀

I wouldn't say I'm excellent at maths I'm just interested

[zealous meadow]

imma

post it there

too

💀

[quaint tide]

Aight lessgo

[smoky jetty]

alr then

[zealous meadow]

watch them

speed run

in 2 mins

bro

they solved this q

[quaint tide]

wut

WHO

WHAT

[zealous meadow]

competiotion meth

💀

wht

evn

[quaint tide]

OH YEAH JUST SAW THAT

[zealous meadow]

is tht

wqefq

ewfwq

im ded

[quaint tide]

I first saw the last question there

and was like eh easy stuff

then I decided to scroll up a lil

💀

[zealous meadow]

big mistake

[smoky jetty]

wait till u join the math olympiad server linked there

[zealous meadow]

💀

[quaint tide]

fr

[quaint tide]

hell naw

[smoky jetty]

dont feel demotivated tho

[quaint tide]

my mental math sucks man

[zealous meadow]

i legit
evn mailed this
question
to some math ytber

💀

[quaint tide]

YOOOOOOO

[smoky jetty]

they really just do such maths in a deeper level

[quaint tide]

Which one

[quaint tide]

I c

[zealous meadow]

numberphile
n mind ur decisions
💀

[quaint tide]

Sheeeeeesh

W channels

[zealous meadow]

fax

[quaint tide]

ig mind ur decisions will prolly reply faster

seems like a video worthy question if it's even solvable

[zealous meadow]

i hav an eng exam tom
and here i am doin
a hard ass
geometry q
👍

[quaint tide]

LMFAOOOOOOO

I almost failed my social studies exam

cuz there were some pages in between

that didn't have lines

[zealous meadow]

i hate sst
so much
hist cant get into my head

[quaint tide]

and I kinda discovered a new way

to do this trigo question

So I started doing tht instead

[zealous meadow]

💀

[quaint tide]

OMG SAME

so much memorizing

I got 0.5/20 in history

but I managed to pass thanks to the other 3

[zealous meadow]

bro
💀

[quaint tide]

like at one side I have 27.5/80 in sst and one point I have 75/80 in maths

[zealous meadow]

fr
i have 45/80 in sst
and 80/80 in meth n sci

tht too 45 cuz

eco carried

hist i got 5 or smthin

[quaint tide]

brooo fr

eco helped me a lit as well

[quaint tide]

I messed up science again man

just can't get myself to memorize bio

I can't memorize anything

on the brighter side if we just compared physics I got the second highest in my class

one mark difference between first and me

[zealous meadow]

ye
bio
pain
esp some of the long complex diagrams

[quaint tide]

cuz physics actually requires a lot more thinking and that's what I actually enjoy doing

not memorizing diagrams n shit

[zealous meadow]

||hardcore math ppls de wonderin y we talkin abt bio n sst here since 10 mins||

[zealous meadow]

fr
muggin up not my thing

[quaint tide]

LMFAO

should've switched to #discussion

lets just go there

[zealous meadow]

too late
but sure

[crimson pawn]

is this problem possible???

i’ve been trying to figure it out for a bit. i don’t believe it’s possible

too many unknowns

[quaint tide]

I'll try it out

[crimson pawn]

appreciate it

i tried to make some kind of systems of equations

did not work

[exotic yarrow]

$\overrightarrow{OP}=(3,-1) \implies \overrightarrow{PQ}=(1,3)$

[somber coyote]

Civil Service Pigeon

[heavy stream]

op^2 + pq^2 = oq^2

use distance formula

u shud get value of t

[quaint tide]

oq isn't given tho?

[foggy parcel]

No need of that bth

O is (0,0)

[quaint tide]

huh

[foggy parcel]

use distance formula

[quaint tide]

OH

aight I'll try again

[foggy parcel]

make an equation containing the variable 't' using this

[crimson pawn]

sorry for the bluriness

[quaint tide]

idk what I did but I got 2 as well so

i'mma see mine after eating

[crimson pawn]

i just figured out how to it💀💀

[smoky jetty]

wow thats more complex than i'd imagine

basically, since RQPO is a rectangle, then OP is perpendicular to QP.
so, (m_OP)(m_QP) = -1; where m = slope
from origin (0,0,) we can find the slope of line OP, giving us -1/3. Substituting to the equation earlier, we get:
(-1/3)(m_QP) = -1
(-m_QP)/3 = -1
(3)(-m_QP)/3 = -1(3)
-m_QP = -3
m_QP = 3. So, from P (3,-1), we can use the slope of QP to find the coordinates of Q. Q (x_P +1, y_P+3) = Q(3+1, -1+ 3) = Q(4,2) Hence, t = 2

[smoky jetty]

is the p/q thing also part of the semicircles problem?

[tight ivy]

yeah

[smoky jetty]

what do they stand for?

[velvet pumice]

how can i study trigonometry properly?
been trynna understand trigonometry for a decent amount of time, i think im pretty decent with the simplification but when it comes to questions with unit circle i literally cannot move my pancel, any tips or tricks?

[upper karma]

Mark the centre of the largest semi-circle (midpoint of the radius 2). Recall that in two tangent circles, the line joining the centres and the point of tangency are collinear. Hence, join the centre of the third circle with the other three centres. Then you get a triangle with side lengths: 3, 2+r, 1+r. Then join the centre of the complete circle with the centre of the largest semi-circle. Since these two circles touch internally, the distance between their centres is just the difference of their radius. Then you can apply stewarts theorem. This yields the quadratic equation $$ 6 + 3(3-r)^2 = (2+r)^2*(2) + (1+r)^2(1) $$, this giving $$ r = \frac{6}{7}$$

[somber coyote]

Barycentre

[languid cloak]

this is the millionth time I'm asking this, but is there a standard name for the property of the height, angular bisector and median of an isosceles coincides / being the same?

[shut quartz]

nice

[spark urchin]

Im just beginning geometry. For x I got 8 square root of 5. And somehow for y I got 96 square root of 5. Please help

[foggy parcel]

!show

[lime crown]

Show your work, and if possible, explain where you are stuck.

[foggy parcel]

For x, u r right
But for 'y', u have to show that KNL is simillar to KLM

[red sorrel]

M

[full pivot]

can I get help please

[hoary panther]

now try to find the side lengths now that u have θ

Let $\theta_1$ be one angle in a right triangle and $\theta_2$ be another angle in a right triangle. $\theta_1=90-\theta_2$ or $\theta_2=90-\theta_1$

[somber coyote]

saddayyy_

[hoary panther]

This is because the sum of interior angles will add up to $180^{\circ}$ in a triangle and with a right triangle, you already know that one angle is 90 so $\theta_{a}=180-90-\theta_{b}=90-\theta_{b}$

[somber coyote]

saddayyy_

[silent plank]

why bother explicitly getting the third angle when you could use the given 67° angle directly

[silk frigate]

Heya, does anyone have an Texas calculator

Can someone do cos(30)*6

[languid cloak]

is there a proper name for the property of isosceles triangles having height, angular bisector and median to the base coinciding / being the same?

[heavy stream]

do you have the answer for this ?

@silent birch

[upper karma]

did you mean cos 30 or cos 30 degrees

[silk frigate]

Yes

30 degrees

[upper karma]

Recall cos(30 degrees) = cos(pi/6) = sqrt(3)/2

I suggest you read up https://en.wikipedia.org/wiki/Radian

[silent birch]

no

[languid cloak]

yeah just read the proof of how 30 degrees right triangles have hypotenuse = 2 opposites (slice equilateral triangle in half)

[ember lily]

is it better to answer 138 degrees or 222 degrees for this question: "what is the standard position, 0 < θ < 360, if 42 degrees is REFERENCE angle."

would 138 be a better answer or 222 im so confused both have same reference angle

[silent plank]

is that the full question/exact wording?

[remote herald]

anyone please help me with that

[timber cargo]

What you can do is take the arc cos and arc sin to find the angles of the enscribed triangle and find the missing side length of that enscribed triangle. Then when you get your angles you can apply even more trig to find the adjacent sides to get the length of AD

[upper oracle]

Did this for a project, all I needed to do was the streets and place buildings on angle pairs and color it, but for some reason I decided to go overboard and make a whole island lol

[smoky jetty]

I found this exchange online https://math.stackexchange.com/questions/3575601/if-the-median-and-bisector-of-one-of-its-sides-of-a-triangle-coincide-then-the So I think if the triangle's median and bisector coincide, then it's an isosceles. Could the proper name be "isosceles triangle's property"?

[languid cloak]

interesting

[languid cloak]

I think the question lacks conditions, the shape isn't fixed. You can rotate the inner triangle and get a different square, or since the included angle by 12cm side and 13cm side isn't given, you can easily get different squares from that.

e.g. I can literally just stretch the two sides flat and make it almost a degenerated triangle (a straight line) and make a square much larger

[languid cloak]

could be good source for my fantasy world building, yoinked (jk)

[halcyon yew]

wot

[languid cloak]

||people make fantasy maps for fun||

[halcyon yew]

o

im in the middle of my last question or two for geo hw rn

its 2:51 and i got math class tmr lmao

[languid cloak]

? what

it's weekend

[halcyon yew]

2:51 am

[halcyon yew]

extra outside of school math classes

[languid cloak]

holy hell

not that

was pain

[halcyon yew]

I got that 2 times a week

2 hrs each class

and I also gotta go to english for a hr 2 times a week

I dont even have english or math first sem

[upper karma]

Hello @everyone

I am new here

[honest loom]

Yo am I correct on this question

[remote herald]

bro i dont get your thing and i havent studied trignometry

[heavy stream]

uh i got 64.87 degrees idk if i did something wrong

I just used the fact that triangle HAF is isosceles

[woven shoal]

It is 64.87

I think you just use the sine law

[honest loom]

Bro i forgot about the sine law

[woven shoal]

For a triangle

$\frac{\sin a}{A}=\frac{\sin b}B=\frac{\sin c}C$

[woven shoal]

Where side A is opposite to angle a, and so on

[honest loom]

As i remember, there needs to be 1 angle which is a known value for the sine rule to work

Whereas there are no known angle values in the question i had

[woven shoal]

The fact that the triangle is isosceles and you know all sides

I think that's enough

Also how did you get that arccos expression

[honest loom]

Whats arccos

[woven shoal]

cos^-1

[honest loom]

Oh thats from the cosine rule

[woven shoal]

How is your other side 6

?

@honest loom

What triangle did you apply the cosine rule on?

Ping me when you're here

[remote herald]

ig triangle FBA

[woven shoal]

WHY

You're finding angle x, no?

[remote herald]

oh sorry i dint see that

[remote herald]

anyone help

[smoky jetty]

how'd u get 5?

and is this angle given as = 90deg or u assummed it?

[royal bear]

looks like a G

no im wrong sorry

[remote herald]

given

[remote herald]

its a G

[royal bear]

why is analytic geometry so easy

[sonic willow]

Bc you just apply formulas lol

[tawdry gate]

if I know:
its a triangle
b=2
c=2
B=75.52248781407
C=75.52248781407
what is the formula for a? (small letters are sides and big letters are angles)

nevermind I got it

[tawdry gate]

is law of sine this unaccurate or did i mess upp something?:
for a I got 0.691302064481663
I did sin(B)/b-sin(A)

[glad hornet]

I'm in college and doing gcse maths in one of the lessons, we were doing geometry and one of the questions was like this

I didn't really understand the theory

I hadn't done stuff like this in years and I may have already forgot some of yesterday

I got some of the stuff like uuuh this would be 5 x 2 = 10 and you divide that by 2 and you get 5

This should be an obvious one but what is the term to refer to what you're calculating here?

Something like area right?

[smoky jetty]

maybe put labels for the vertices to get those side values clear, and also properly distinguish what that "?" is for

[smoky jetty]

calculating its area yeah

[glad hornet]

what's the vertices again?

[hazy star]

area units

[smoky jetty]

vertex (single), vertices (plural form). Each vertex of a polygon e.g triangle refers to its corner or point of intersection of the lines or the polygons' edge. Thus, enclosing the figure

[glad hornet]

so is vertex a side?

[hazy star]

it's a vertex, lol

[smoky jetty]

nope, its corner basically

[hazy star]

look it up

[glad hornet]

so it's a corner

[smoky jetty]

yeah

[glad hornet]

so on that rectangle there is 4 of em

[smoky jetty]

u can label each of those to make the measurements more distinguishable

yep

[glad hornet]

the question was the line with the ?

Idk how to judge that one

The shapes weren't specific, just generic and given numbers

So we weren't using equipment to find out

[smoky jetty]

sorry but I didnt get what u asked

[glad hornet]

we have to guess that number

[smoky jetty]

which?

[glad hornet]

the line in the triangle

[smoky jetty]

can u share it again, but with labeled vertices? So we can distinguish what u're tryna find

[glad hornet]

like this?

[smoky jetty]

like put A,B,C,D etc to name those corners

[glad hornet]

?

[smoky jetty]

you're missing 1 vertex in there

[glad hornet]

huh?

but there's 3 corners

[smoky jetty]

try to observe the figure

hint: there are in total three triangles, but this line divides the biggest into two smaller ones.

[glad hornet]

Three triangles?

[smoky jetty]

again, observe the figure and try to see it yourself

[pliant nest]

It is not a triangle, it is a pyramid (3 dimensional), try to look into it deeper because you are viewing it as a polygon and not a polyhedron

[agile spear]

is it a isosceles triangle?

[glad hornet]

is the line meant to indicate that?

[smoky jetty]

couldnt be too sure, he hasn't stated what description the triangle has

[glad hornet]

I just drew that on paint, it was what the question looked like

the numbers aren't exact either

[smoky jetty]

honestly tho, u should try looking up geometry lectures first

coz u seem to have forgotten the basics in geo

[glad hornet]

got any recommended sources?

Yeah I have

[smoky jetty]

khan academy is a good way to go

[glad hornet]

I don't really know what else I could say, there were triangles with marked corners and numbers on the sides

[pliant nest]

Honestly, try redrawing it lol

First draw the triangular base

[glad hornet]

sorry... but which is the base?

the top?

[pliant nest]

No

The base is the base, it is what it stands on

The rest are parallel faces

[glad hornet]

this is sounding like a 3d image

it was drawn as a 2d one

[pliant nest]

It is a three dimensional figure on a two dimensional plane

The black is the base, the red are the parallel faces.
It becomes much easier to identify them if you draw them properly

Although my drawing is horrendous too

But as you can see, there are 3 parallel faces and 1 base

4 vertexes in total

Vertices*

[smoky jetty]

i dont think these directions would help him understand the basics of geo well enough

especially when "parallel faces" could have more formal terms

[pliant nest]

Of course not, but I think if he realizes what he needs to look for then he can engage with future material better and know which direction to go in

Best would be to do Khan Academy or formal resources

[smoky jetty]

yeah, but not when he doesnt even know what terms/concepts we're plugging into him

[agile spear]

@remote herald The answer is : (AD=12.6cm). Hints: find congruency in triangle GAE and triangle GDF and also use trigonometric ratios and Pythagoras theorem. I 🙂 encountered with geometry after 6 yrs, still fun.

[smoky jetty]

and our own interpretation of the terms might confuse him in other stuff

[pliant nest]

Yeah, he definitely needs to go back to Area and improve those foundations

[glad hornet]

I don't really understand all the lingo

I forgot pretty much all of this stuff I learned in hs

[pliant nest]

Why did I say parallel faces

I meant to say lateral lol

I just realized

[agile spear]

yep

[smoky jetty]

hence why he should go to khan first, lol

[pliant nest]

Yeah, but I'll try giving him a rundown because I know when I first learned it I could not even understand how to tell pyramids from prisms apart

[glad hornet]

https://tenor.com/view/milk-gif-15983354

[pliant nest]

So, the Base is the base (shocking). Think of a house. It has a floor (the base) and a ceiling [also a base, they are parallel (parallel means is that they are opposing one another) and congruent (which means equal)]. The lateral faces would be the four walls of the house, assuming the house is cube shaped.

Another example: In a square pyramid, the square would be the base and the lateral faces would be the triangles. There would be a total of four lateral faces (in this case, in pyramids, triangles) because a square (the base) has four sides. You identify pyramids by the base, by the way.

I think these two terms are pretty self explanatory though, some figures have two parallel and congruent (equal) bases such as prisms (like the house example above).
Some do not, like pyramids, they only have one base. As used in example above.

A way to identify whether a solid (a three dimensional figure) is a pyramid or a prism is pretty easy.
If the lateral faces are parallelograms (a type of quadrilateral, which means a four-sided 2D figure), it is a prism. If the lateral faces are triangles (three sided 2D figures, not 4), then it is a pyramid.

Holy shit explaining this is so hard

I can try help answering some of your questions if you have any btw. Just ping me, but just like the others above definitely check out Khan Academy, or The Organic Chemistry Tutor (a YouTuber who is awesome).

[agile spear]

still you made it like a prof.

[pliant nest]

Thanks 🥺

[glad hornet]

I'm sorry if this is a silly question but is there like a test paper where it explains the theory of the following questions?

You know how like a test would in school?

[pliant nest]

I always employ the Feynman Technique while studying, even in math. So I try to simplify it as much as possible to be as digestible as possible

[pliant nest]

And remember dude, mathematics builds upon itself. If you don't know how to add or subtract (the theory), multiplication and division is very hard. So since you struggle with 3D figures, go back to 2D such as identifying them, areas etc.

[glad hornet]

I don't think there was any implication of 3D shapes

[pliant nest]

The problem you had involved a three dimensional figure

[glad hornet]

It was a triangle with a line in the middle, all of the prior questions were squares/triangles with numbers on the sides until that point and you calculate the area

[pliant nest]

Yeah, it was asking about the Base

Two Dimensional figures do not have bases, only 3D Figures do

Not sure what exactly it was asking about, but it was definitely the base

[pliant nest]

Many books, Khan Academy and YouTube videos. I cannot really recommend a text-based one as I used a book which I assume you do not have access to. I would definitely check out Khan Academy and The Orgo Chem. Tutor as I said

[glad hornet]

It was like, still drawn as 2D like the others

It just had the line

[pliant nest]

Can you resend it

[pliant nest]

Oh, yeah. Definitely 3D. You just cannot draw a line connecting the third vertex because it overlaps with it

I am not sure whether it is asking about like the...area of the base or the length of that side though. But it is definitely three dimensional

[glad hornet]

The rest of the questions were about area so this has to be area too for sure

[pliant nest]

Then it is asking about the area of the triangular base

But the shape itself is three dimensional

[agile spear]

add more details

[pliant nest]

^ like I do not know what the problem is. Are you asking about how to calculate the area of a triangle or?

If it is to calculate the area of that triangular base, you just do 1/2bh

[glad hornet]

Yeah it's the area

[pliant nest]

Okay, so you don't know how to calculate Triangle Area, that's the problem right?

[glad hornet]

The others went like this
50 x 25 halved makes the area

I did the other triangle areas fine it's just that ONE question had a line in the middle with a ?

So I figured I gotta get the value of that line

[pliant nest]

Not going to lie dude, you cannot get either the line or the area of that problem because you do not have any values

Do you know anything about that pyramid? You need some actual numerical values to figure out the rest

[glad hornet]

You mean like the numbers in the question I'm talking about? I was throwing around generic ones (because I couldn't remember the exact numbers)

If I had to think real hard...

[pliant nest]

Woah, my internet is fucking up what the hell

[agile spear]

@pliant nestwhere are you from?

[pliant nest]

Bro, why does the figure change every time? It is still a 3D figure, but now it seems like you are supposed to find C. For which you could use the Pythagorean Theorem (a^2 + b^2 = c^2) as opposed to finding the area of the base or something

[pliant nest]

I live in Europe

[glad hornet]

is the figure the number? last time i just threw 50 50 together just cuz, i think the exact numbers were 50 and 45

i dont have the exact triangle from the document so i draw what it looked like in paint

[pliant nest]

Nvm

You cannot use the Pythagorean Theorem on that triangle, that is not a right triangle

But now it seems like a different problem altogether nonetheless

[glad hornet]

the shape of the triangle doesn't really matter that much (as far as I know), the document just used generic shapes

[pliant nest]

You need some form of values. Length of sides, angles etc. to actually do anything with the image

So knowing that 1 side is 50 units, the other is 45 units

Does not really help because now it does not seem like you are supposed to figure out the area of the figure (to me at least). But the line that runs through the middle is there to represent that it is not a triangle (2D figure), but a pyramid (3D figure)

[glad hornet]

Maybe...

Apologies for all the trouble mate I know this must be frustrating

[pliant nest]

Idk how to help you dude. The formula to figure out the Area of a Triangle is 1/2(B)(h) or 1/2Bh

[glad hornet]

Khan Academy would probably help with this right?

[pliant nest]

Do what you can with that knowledge

[pliant nest]

Definitely

[glad hornet]

The line might actually not mean anything at all, but I saw it and I thought "wait... how do you calculate the line?"

[pliant nest]

Who knows, good luck though

[glad hornet]

most questions with triangles looked like this

no line in the middle

it's just one had that line in the middle that confused the shit out of me

[pliant nest]

Well, dude, if all the questions were about area then it was definitely to solve the area using the known values

[burnt arch]

okay im kinda struggling here, when im converting radians to degrees and vice versa
and i have a radian greater than like 2pi or degrees over 360 the formulas just kinda fall apart for me
like im trying to convert 14pi/3 to degrees and i get 840 degrees
but the correct answer is 120 degrees, and im clueless on how to get from 840 degrees (my answer) to 120 degrees (the correct one)

so i just simply subtract 360 from my answer until i get an angle under 360?

because that gives me the answer but i feel like im doing it in a wrong way

[obsidian mesa]

well consider what 14pi/3 radians really means, it's like doing 2 full turns and a bit

that "bit" just happens to be 120

for most intents and purposes you tend to want to find that angle, as telling us that the angle is 2 full turns and 120 is redundant when we can just say its a turn of 120

lmk if that doesn't really make sense

[burnt arch]

well i know what that means in the bigger picture

but when actually calculating that angle

you get 840, but 840 isnt really saying much

is it as simple as just subtracting 360 from 840 until it gets to 120?

[obsidian mesa]

basically yeah

[burnt arch]

oh, okay then lol thank you

[obsidian mesa]

all good!

i get what you mean though it doesn't seem particularily mathsy to just subtract until it works

[burnt arch]

omg yeah that always makes me overthink things

when you have a complicated thing you expect a complicated way of solving it

so when it's simple it just feels wrong at first

[obsidian mesa]

true true

[rain leaf]

Does anyone know why or how there are so many different forms of the angle bisector theorem?

Are the triangles themselves (formed by the bisector) similar?

and also does anyone know where BD = ab/(b+c) come from?

$BD = \frac{ab}{b+c}$

\begin{align*}
\frac{BD}{DC} = \frac{c}{b}
\frac{AB}{BD} = \frac{AC}{CD}
BD = \frac{ab}{b+c}
\end{align*}

\begin{align*}
$$\frac{BD}{DC} = \frac{c}{b}$$
$$\frac{AB}{BD} = \frac{AC}{CD}$$
$$BD = \frac{ab}{b+c}$$
\end{align*}

\begin{align*}
\frac{BD}{DC} = \frac{c}{b}\
\frac{AB}{BD} = \frac{AC}{CD}\
BD = \frac{ab}{b+c}\
\end{align*}

[somber coyote]

donut123

[burnt arch]

Just made this to help me remember the unit circle (excuse my handwriting)

[grave pond]

An interesting mnemonic pattern I've seen used to remember the coordinates of the 0°, 30°, 45°, 60°, 90° points is:
(sqrt(4)/2, sqrt(0)/2) .. (sqrt(3)/2, sqrt(1)/2) .. (sqrt(2)/2, sqrt(2)/2) ... (sqrt(1)/2, sqrt(3)/2) ... (sqrt(0)/2, sqrt(4)/2)
but the precise angles that go with each of those would still need to be memorized separately.

[glad cloak]

guys i made parabola go spinny

[remote herald]

Thanks a lot for your help

[remote herald]

but i havent studied trignomerty

[smoky jetty]

idk why but I simply just do some calculations to fill up the unit circle, lol. Like i never memorized the values at all

like from the special right triangls in the 1st quadrant, i can just find the corresponding angles in the other 3 quadrants and get the coordinates

[woven shoal]

I had something for you

Let me host it up wait

[woven shoal]

Alr ping me when you're here

[grave pond]

I hope that's what everyone does. But at least the magical angle 30° for the one of the special triangles that's not diagonal is arguably something of a memory item too. Once you remember that the 30°-60°-90­° triangle is the one whose ratios can be expressed with square roots, you can easily calculate what those square-root expressions are.
I suppose the physical work of drawing up the entire circle neatly with pen and paper can possibly be valuable for recalling the relations later on. The actual paper can be thrown out as soon as you can get it over your heart, but having done it helps remembering what it was you did.

[high glen]

I don't understand why i had to put my calculator in radians for this question.

[grave pond]

Because that's the usual convention for angular velocities.

[high glen]

how is this angular velocity

id appreciate any help on this

[grave pond]

The comment under the formula helpfully describes the cosines as $\cos(\omega t)$, and $\omega$ is the usual symbol for angular velocity. So I'd conclude that the $25\times 10^{-3}$ factor is a time in seconds (they speak about "25 ms" later), which would make the $200$ an angular velocity of 200 radians per second.

[somber coyote]

Troposphere

[high glen]

so is if it was in degress would it have the degree sign and when not it is radians

[grave pond]

It more like it's conventional not to use degrees for angular velocity at all. If you want to describe the speed of rotation in something where whole revolutions are nice numbers, you'd be working with frequencies (f) instead of angular velocities (omega).

[high glen]

ok, I'm digesting what you said and how to apply it what I am learning. Thank you for your help!

[grave pond]

It's not unlikely that somewhere above what you showed, there's something that makes it more obvious that the number 200 is given in radians per second rather than degrees per second.
However also perhaps not, since this seems to be electronics. Generally numbers you find by calculating with capacitances/inductances will naturally come out in radians rather than degrees.

[high glen]

This is the original question

Unfortunately I’m weak it my understanding of some math concepts. I’m just beginning circuits 2. So I don’t understand when it’s in radians or why

[grave pond]

Okay, then your only hope seems to be to know the convention.
In general, any number that you multiply by an amount of time to get something you then stick into a trig function is an angular velocity, and the convention is that such numbers are measured in radians per unit time.

[agile spear]

No issue if you haven't studied trig. ratios, just use AAA congruent property and pythagorean theorem. tada

[fallen vector]

Can you use Cosine to find the area of a triangle when two sides and an angle are given or do you just have to use sine rule everytime?

[grave pond]

Depends on whether your two sides and an angle are SSA or SAS.

[fallen vector]

its SAS

[heavy stream]

yea i mean u have to use cos rule here to find cb and then u can apply heron's formula

right?

or if u have calculator then you won't even have to use the rules

because you could just draw a line perpendicular to ab joined to c

then just using trig ratios u can add the areas of the 2 smaller triangles u have

[fast cedar]

Someone knows about vectors?

[heavy stream]

no ;-;

lmfao bruh

[burnt arch]

Sorry I was sleeping lol

[smoky jetty]

I see, on the basic trigo identities tho, i just glimpsed a bit of it and realized that many of them were pretty much algebraic manipulation. Soon, I might see if memorizing it would be better or just some manipulation stuff would do

[drowsy gust]

I’m horrible at geometry help

[steady mason]

the volume of a cylinder is unaffected by the lid right?

[steady mason]

(3x + 48) + (x + 12) = 180

solve for x after that

since q and s are parralel these 2 angles are the same

[steady mason]

actuallu an easier way to visualize it is

This might make it more intuitive

[remote herald]

thanks alot

[smoky jetty]

correction: AAA is not a congruence property. 2 triangles with the same angles "can" have different yet proportional corresponding side lengths

[woven shoal]

Aight when you're here ping me

[smoky jetty]

e.g

[woven shoal]

Wut you doing

[burnt arch]

kinda here (I need to get ready for school(

[smoky jetty]

huh?

[woven shoal]

Just ping me when you're here. Not kinda here

[burnt arch]

Uhh alr well I'll ping you in like 9 hours lol I got school

[woven shoal]

9 hours?

You're in a very bad timing conflict with me

[burnt arch]

Yeah school and possibly homework

[mental pelican]

Can someone please check what I have done wrong:
Extend DE and AB to meet at a common point Y. Now triangle ABC ~ triangle YBE
This gives BY/BE = 5/4 . Also triangle DBE ~ triangle BYE so BY/BE = DB/DE which leads to DE = 52/5 . So the ratio is 52/65

[vast saffron]

Can someone explain this, here the author is deriving 'law of cosines' but from what I see, shouldn't it be a•cos(theta) + b squared? Why is there a minus inbetween?

[slate pewter]

okay i am not that good at trig but i didnt understand how does the line that goes through D , that is parallel to AC meets cb

something i am missing ?

[mental pelican]

I didn't say it meets cb , I said extend AB

Oh you meant in the question

[mental pelican]

Why wouldn't it?

Erase the line be and de

And extend CB

[slate pewter]

oh i didnt know you can just extend it :D

[mental pelican]

<@&286206848099549185>

[burnt arch]

Lol I'm in American central standard time zone

[drowsy gust]

Thx!

[smoky glacier]

Is this PDF drunk? It says WX is a line, then goes on to say that 3 points make a plane as long as they aren't on the same line, so WXB is a plane?
Or am I misunderstanding something

Or am I to read that as the line doesn't exist when they are defining plane and to just consider the points WXB

[charred cosmos]

Could someone please solve this problem?

Given a pyramid A-BCD, AB=CD=AC=BD=a. AD=BC=b, if A, B, C, D are all located on the surface of the same sphere with radius of 2, determine the range of a+b.

[next gazelle]

Anybody knows the proof to the fact that
If there's a circle C1 then when we drawn another circle C2 through 3 points:
A random point P outside C1,
Two points of contact of tangent from P onto the circle C1
then C2 also passed through centre of C1?
(I know Thales theorem but that doesn't prove this cause I don't know if there's only one such cases possible or multiple)

[steady mason]

np

[burnt arch]

@woven shoal alright im here

[lime crown]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[lapis moon]

\colorlet{a}{blue!40}
\colorlet{b}{red!40}
\begin{tikzpicture}[scale=3,c/.style={a},d/.style={b},column 1/.style={anchor=east},column 2/.style={anchor=west},row 4/.style={font=\Large},row 8/.style={font=\Large}]
\pgfmathsetmacro\aA{20}
\pgfmathsetmacro\aB{60}
\coordinatelabel=left:$O$at(0,0);
\coordinatelabel={[c]\aA:${C =(\cos A,\sin A)}$}at(\aA:1);
\coordinatelabel={[d]-\aB:${D =(\cos(-B),\sin(-B))}$}at(-\aB:1);
\coordinatelabel=right:$E$at(1,0);
\draw(O)circle(1);
\foreach\V\colo in{A/c,B/d,E/{}}{\draw->,\colo--(\V);}
\draw->,carc(0:\aA:0.3);
\draw->,darc(0:{-\aB}:0.25);
\node[c]at(\aA/2:.4){$A$};
\node[d]at(-\aB/2:.4){$-B$};
\node[sloped,above,c]at($(O)!.5!(A)$){$\vec{a}$};
\node[sloped,below,d]at($(O)!.5!(B)$){$\vec{b}$};
\matrix[matrix of math nodes,anchor=north]at(current bounding box.south){
{\color{a}\vec{a}}\times{\color{b}\vec{b}}&={\color{a}|\vec{a}|},{\color{b}|\vec{b}|}\sin((-{\color{a}A})+(-{\color{b}B}))\
({\color{a}\cos A}\hat{\bf i}+{\color{a}\sin A}\hat{\bf j})\times
({\color{b}\cos(-B)}\hat{\bf i}+{\color{b}\sin(-B)}\hat{\bf j})&=
({\color{a}1})({\color{b}1})(-\sin({\color{a}A}+{\color{b}B}))\hat{\bf k}\
\sin({\color{a}A}+{\color{b}B})&=-({\color{a}\cos A},{\color{b}\sin(-B)}-{\color{a}\sin A},{\color{b}\cos(-B)})\
\sin({\color{a}A}+{\color{b}B})&={\color{a}\cos A},{\color{b}\sin B}+{\color{a}\sin A},{\color{b}\cos B}\
{\color{a}\vec{a}}\cdot{\color{b}\vec{b}}&={\color{a}|\vec{a}|},{\color{b}|\vec{b}|}\cos((-{\color{a}A})+(-{\color{b}B}))\
({\color{a}\cos A}\hat{\bf i}+{\color{a}\sin A}\hat{\bf j})\cdot
({\color{b}\cos(-B)}\hat{\bf i}+{\color{b}\sin(-B)}\hat{\bf j})&=
({\color{a}1})({\color{b}1})(\cos({\color{a}A}+{\color{b}B}))\
\cos({\color{a}A}+{\color{b}B})&=({\color{a}\cos A},{\color{b}\cos(-B)}-{\color{a}\sin A},{\color{b}\sin(-B)})\
\cos({\color{a}A}+{\color{b}B})&={\color{a}\cos A},{\color{b}\cos B}-{\color{a}\sin A},{\color{b}\sin B}\
};
\end{tikzpicture}

[somber coyote]

vin100

[silk dock]

hi

does anyone know how to solve this one??

<@&286206848099549185> ?

I can't think of a single start

[burnt arch]

!help

[lime crown]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[silk dock]

oh

[silk dock]

what does this do?

[burnt arch]

read the message 😭

[burnt arch]

look

[silk dock]

oh

my bad

sorry!

[burnt arch]

youre good 🙏

[silk dock]

can you solve it though?

help 48 btw

[frozen ocean]

#❓how-to-get-help

[woven shoal]

Are you right now

[burnt arch]

yeah

oop

[woven shoal]

https://muhammadhussaini.github.io/pages/desmos/trigwheel.html

Here

[burnt arch]

woah

[woven shoal]

I saw that circle kinda thing of yours so

[burnt arch]

thats nifty as fuck

imma bookmark that

[woven shoal]

Lol

[burnt arch]

yeah no thats super useful

[woven shoal]

I mean there should be others available online

[burnt arch]

wowie

[warped stone]

Does anyone know what is Pythagoras related to real-life application

[dark sparrow]

did you maybe mean "pythagoras"? as in the pythagorean theorem?

@warped stone

[warped stone]

Yep

In my lang is known as theorem pythagoras

Sorry for confussion about "tha"

[dark sparrow]

"phythagoras" and you are absolutely sure it starts with a ph and not just p?

anyway like

i guess the most "real world" that you're gonna get is calculating distances

in various circumstances

like... idk if you were designing a sloped roof on a house and you know how tall it is and how far it needs to extend horizontally

you are gonna need the theorem to calculate how much roofing you need

[warped stone]

Anyway in graphic designer

Or geometry related

All 2D-based object triangles then combine two 90-degree triangles and here the box
...

So in this topic, I will guess the basics of all shapes is triangle

I want to be an architect btw

[bleak trout]

i gotta do something

[opaque sluice]

Indeed. I discovered it in my programming language theories

A square is a combination of 2 triangles, a cube 2*6 = 12

A circle, many triangles

And even though, triangle is made of a single triangle (wow)

[proper eagle]

I have been working in this problem for some days and still don't have any solution.

It ask for the triangle with the biggest area that you can find with the vertices on the circumferences.

I had some ideias, like use the cosine law with the three central angles, what's a way to found the sides in function of the radius.

[grave pond]

I don't have a full solution, but you'll want to have OA perpendicular to BC, OB perpendicular to AC, OC perpendicular to AB.

(If they're not, moving just one of the points so its radius becomes perpendicular to the opposite side will make the triangle larger).

[grave pond]

Hmm, and we can then derive using the law of sines that the cosines of the three central angles stand in the same ratios as the three radii.

[high glen]

how do you plot by hand w(t)=.3125(1-cos(160t)/2) I have T=12.5 ms and the wave peaks at 312.5mW.

[woven shoal]

I mean that's a part of a larger line so

Locate the larger line it's part of

[flat oar]

Hey can anyone help me with this, its angle relationships

[silk basalt]

SR?

[woven shoal]

Yeah

[silk basalt]

thanks 8 more questions

I wanna say RQN

[woven shoal]

You're doing a test?

[silk basalt]

but I’m not sure

It’s not a test it’s a practice test for this week

It’s worth 25 points

[woven shoal]

It's worth points in your Grade????

[silk basalt]

Or homework I got told to work on it had home in my free time

yea

[silk basalt]

is this one correct? @woven shoal

[fickle rose]

no

point n ain’t on the plane

[upper karma]

How

[grave pond]

Angle sum in a triangle.

[upper karma]

The angle is outside the triangle

not inside

[grave pond]

Compute the inside angles first.

[upper karma]

Then what

[grave pond]

Once you know the inside angle next to the x° you want to find, it is easy to get the last bit of the way.

Does "vertical angles" match something you've learned?

[twin lotus]

its like linear pair postulate

[slender kelp]

I think

[grave pond]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[grave pond]

And it's not the right answer anyway.

[slender kelp]

Mb

Oh I get it

[dry bough]

I know my hand writing sucks but am I right on this problem
tan(x-pi/4)+2

https://cdn.discordapp.com/attachments/476221292341886979/1161743259417182350/20231011_151208.jpg?ex=653968b9&is=6526f3b9&hm=673c76f8c4d4eeb09f7e2d5c4c6c0520fb6f123115d627cdd6b45abd7dfe791f&

[jagged venture]

Also posted this in a help channel

In the right-angled triangle ABC with angle ABC = 90, let's denote by D the foot of the height from A and by M the middle of the leg AB. The bisector of angle ABC intersects the bisector of angle DAC at point E. Show that:

a. Triangle AME is isosceles
b. ME parallel to BC

[verbal sluice]

Can someone help me on trig real quick

[hoary panther]

I can try if u give an example on what u are confused on or a question from hw

[burnt arch]

@stark vapor

like i'll draw one of these

and from it i can just start pulling identities out

i know how to create the identities from this circle at least lol

[stark vapor]

what identities, give me an example

[burnt arch]

sin x = sin (pi-x)
-sin x = sin -x
cos x + pi = -cos x
cos x = -cos (pi-x)
tan x+pi = tan x
tan (pi-x) = -tan x

and so on

[stark vapor]

ah

personally i just remember these:
sin(x + pi/2) = cos(x)
cos(x + pi/2) = -sin(x)
sin(-x) = -sin(x)
cos(-x) = cos(x)

and then tan = sin/cos

[burnt arch]

well imma take your word for it lol and just remember what i need

also i have sin^2+cos^2=1 already memorized along with the cofunction identities and yeah

it's just the symmetry identities or whatever they're called are a little tricky

[burnt arch]

but these are the most useful out of all of them i assume

[stark vapor]

yeah maybe also memorize cos(x + pi) = -cos x and sin(x + pi) = -sin x

[burnt arch]

oh yeah ofc

[stark vapor]

technically you can rederive it from sin(x + pi/2) and cos(x + pi/2) but it’s not hard to remember adding pi

then one has cos(pi - x) = -cos(-x) = -cos x

since cos(-x) = cos x

[burnt arch]

yeah

cuz the x coord will still be the same

it's just sin that becomes negative

[stark vapor]

saying cos(2pi - x) = cos x is like, a geometric way to state it, too, but i prefer just saying cos(-x) = x

since cos(t + 2pi) = cos t for all t

[burnt arch]

yeah

cuz 2pi just returns you to the original spot so nothing really changed

being able to imagine the unit circle makes comprehending and memorizing all of this so much easier

thats why im always randomly constructing unit circles and writing literally all the information i see from that circle

to test myself to see how much i've memorized

and each time i do it i can write a little bit more on my paper so yeah thats fun

[jovial bough]

help

sos

[neat python]

Point O is the center of the circle, point A is inside the circle, and point B is the outside point. OA=6.7 cm; OB=7.5 cm. Find the length of the radius of the circle if it is known to be expressed as an integer in centimeters

<@&286206848099549185>

[gusty kernel]

Ang, QPR Is half QSR because inscribe angle is half of central angle standing om same arc

So

As QSR = 88° SO, QPR = 44°

[cosmic cedar]

bit confused with this one. i tried using the area=1/2absin(c) but got mixed up. help

[upper karma]

Hi!

Use this Area=1/2 x r^2 (2πθ/360∘−sinθ)
11m^2 = 1/2 x r^2 (2π(75)/360∘−sinθ)

This is the area of the minor segment

Once you have the radius, you might want to use the formula for the area of the triangle as 1/2 × r^2 × sinθ

[cosmic cedar]

not sure ive learnt this sorry 🥲

[upper karma]

What formula do you use?

[cosmic cedar]

wait wait actually i mightve learnt it

hold on im gonna try the question again

[upper karma]

1/2absin(c) is the Sine Area Rule

The area of the triangle is actually taken from the Sine area rule

[cosmic cedar]

nevermind i still havent got it. how should i go about solving for the radius

[cosmic cedar]

mhmm i tried that but i mightve done smth wrong because it didnt match the answer in the answers sheet

ill try again

[upper karma]

give me the answer in the answer sheet

[cosmic cedar]

radius 8.008m, triangle 31.0m^2

[upper karma]

Ok I'll check if it works

[cosmic cedar]

it probably does, thanks

[upper karma]

Area=1/2 x r^2 (2πθ/360∘−sinθ)
11m^2 = 1/2 x r^2 (2π(75)/360∘−sin75)
11m^2 = 1/2 x r^2 (1.309 - 0.966)
11m^2 = 1/2 x r^2 (0.343)
11m^2 = 0.343r^2/2
11(2/0.343) = r^2
√64.14 = r^2
r = 8.008m

got it

[cosmic cedar]

ahh i see now

[cosmic cedar]

i was trying to use this rule to find the radius haha

[upper karma]

use the formula for area of the triangle that I gave you. Might work.

[upper karma]

there is no radius on your sine area rule

[cosmic cedar]

yup i kinda figured that out. i forgot the other rule u wrote existed

only started learning this today so im still trying to get my head around it

[cosmic cedar]

alrighty

i got it

30,97, rounded to 31.0

thanks a bunch :) couldnt have done this without your help

[wicked narwhal]

Good evening, sorry to bother you. What is the meaning of the double arrow? Does it mean the QS vector?

[hoary panther]

the line going through the points "Q" and "S"

[wicked narwhal]

Thank you I did not know this notation, is it the equivalent of the following image?

[hoary panther]

it is not. usually, that notation is used for vectors and rays depending on the context

[upper karma]

Ray

[wicked narwhal]

Oh thank you for the explanations ^^

[hoary panther]

if it is AB with the arrow to the right, usually it is talking about a ray

[dry bough]

https://media.discordapp.net/attachments/476221292341886979/1162117144985751642/20231012_155814.jpg?ex=653ac4ee&is=65284fee&hm=f4b6afc10051a293549099e545c52f882be92da3a455edf3988a340ce2c88abd&=&width=507&height=676

costheta = -8/17 where pi/2 lessthan = to theta < pi
find tan2theta
I got 240/161
does that look correct?

[upper karma]

can barely see the writing

[hoary panther]

I can see it

[upper karma]

i can but it ain't clear

[hoary panther]

ahh

[sudden schooner]

Hello is it correct to say that the curvature of a curve represents the difficulty of approximating this curve by a polynomial application?

[dreamy hornet]

guys

does anyone knows a good geometry book??

[lofty terrace]

for learning geometry or for practicing it?

or both?

[hoary panther]

most books have exercise questions

[narrow marsh]

https://cdn.discordapp.com/attachments/801167302913032213/1120545784853700618/image.png?ex=65373095&is=6524bb95&hm=dcfc33cf88fdc0a1ce97999b93c9a9864dfa087f9876c5254797926705b7329b&
i forgot how to do this, can someone explain?

[wicked narwhal]

do you know similar triangles ?

[lofty terrace]

Yeah but there are books with only practice questions

And then there are textbooks

With a mix of learning material pairs with examples and except uses

[narrow marsh]

yeah is the equation like 11/4x = x+12

[wicked narwhal]

yeah can you solve it ? (x is to the numerator in 11/4x)

[narrow marsh]

how would i solve it that way?

i can solve it 4/11x = x/x+12

[somber coyote]

Primordial

[somber coyote]

Primordial

[remote herald]

which country's question is this

i anit even getting a clue

[nocturne remnant]

Hint: ||add two more triangles to make a square||

[spring plume]

for the derivation of eqn of hyperbola why is b^2 = c^2-a^2? for ellipses theres a proper geometrical way to determine b^2 = a^2-c^2 but i cannot find why this equation holds for hyperbolas

i saw a video and it said that b^2 = c^2-a^2 is just an arbituary constant thats meant to represent c^2-a^2 after deriving the formula to x^2/a^2 - y^2/c^2-a^2 = 1

??

[hearty crest]

Hi

[lethal moss]

Welcome 😀

[hearty crest]

I'm taking trig rn, started graphing and... proofs

[lyric sonnet]

do you have more recaptcha puzzles

[lethal moss]

Where did you find them though? It was a long time ago

[lethal moss]

Did you really dig through the entire server?

[lyric sonnet]

no

i just remembered

[lethal moss]

oh

surely I'll find something

I think you haven't seen this one yet

[lyric sonnet]

there exists no x:|x| = x ?

[lethal moss]

It was a trick question, try again

[lyric sonnet]

😢

[lethal moss]

hint: ||There are 2 false statements||

[lyric sonnet]

[lethal moss]

Yeah, that was an error on my side, but I turned it into a mechanic now

[lyric sonnet]

i think all are correct

[weary hound]

Statement 2 and 4

Assuming a, b, and c isn't restricted in positive numbers

[lethal moss]

Actually statement 1 too
think about 0,5^2

[weary hound]

ah right

[lyric sonnet]

oh

[weary hound]

How could I not think of this

[wooden wharf]

yesterday my teacher asked me to prove that a triangle is a triangle and i said because it has three sides

is there any other reaoson why tho

[thick fable]

show your diagram.

[thick fable]

hmmmmmm

i think i got it

wait

[hearty crest]

I can't understand, what differences do I have to make when graphing tangent? I can graph sin and cos kind of well

[rain leaf]

Tangent is just fundamentally different

Drawing a bunch of lines as borders starting at pi/2 and repeating in both directions every pi

Then just draw a weird cubic shaped graph between then

[gilded dew]

This sounds silly and I'm probably on the wrong discord server for this, but I'm making a minigame and I want to figure out how to detect if a player is looking at another player based on their yaw & pitch.

For example, I'm staring at this NPC named "Aryn":
https://media.essential.gg/933f89d0-9cf2-49b3-ce6f-adfe3f60ba00

And as you can see Minecraft says my yaw value is 63 & pitch value being 1 (as you can see top left).
Minecraft also has boundaries and limitations on their yaw value when it hits a value of 180, where it becomes -180, same when it crosses 0 (1, to 0, to -1) afterwards. If anyone who plays Minecraft can denote a connection between this reasoning and the real life applications of trigonometry then please help me figure this one out.

tl;dr I know trigonometry is involved so if anyone can tell me how I could use that here that would be awesome thank you

[gilded dew]

help me figure out how to detect if a player is looking at another player using yaw & pitch coordinates

[hot quest]

What coordinate axes does yaw work with? (e.g. looking straight at positive x axis is a yaw of 90)

Either way, for the yaw, imagine drawing a graph of minecraft x and z (east-west and north-south respectively) coordinates. The two players will be two points on those axes, and the yaw will tell you which way one of those players is looking. To check that a player is looking in the direction of another player, you can use an inverse trig function, using the distances between the two players in the x and z axis, to check if the angle between the two players matches the yaw.

For the pitch, imagine a graph of distance between the players-height of players. Using that, you can again use an inverse trig function to see if the angle between the two players matches the pitch.

@gilded dew , does that help you?

[gilded dew]

Sort of, im using ChatGPT to actually help me and he's helped me a lot

Apparently something to do with the dot product formula

Thanks though, I think my work here is done

[frozen ocean]

chatgpt can't do math

[upper karma]

Can someone write all the special triangles?

Like 3-4-5

[nocturne remnant]

if you are looking for all triplets of integers which form the sides of a right triangle, it is infinite and has well-known characterizations: https://en.wikipedia.org/wiki/Pythagorean_triple#:~:text=68%2C 285%2C 293)-,Generating%20a%20triple,-%5Bedit%5D

[upper karma]

it might work for me.

thanks

[coral carbon]

hi

i just joined the server

im in geometry btw

[smoky jetty]

if I dervied a formula from something (e.g property, equations, etc), can it also be a proof to why the formula works?

i.e can it be like proving the formula itself

[gilded dew]

You’d be surprised

[frozen ocean]

I'm not surprised

ik what does it do

[gilded dew]

Nah chatgpt gave me the dot formula and convert pitches and yaw values in Minecraft to their respective unit vectors

[frozen ocean]

gpt 4?

[gilded dew]

It even changed the calculations for arctan based on how Minecraft interprets the yaw range of values

idk but I just used a free version

[frozen ocean]

free version is shit at math

they once gave me chatgpt plus for three months and still found it shit at math

[gilded dew]

Ok

[frozen ocean]

before it goes away they added a image feature

[coral carbon]

gusy

guys

[frozen ocean]

what?

[coral carbon]

i got a queston

guys
can u help me
so when ur trying to prove that 2 triangles are congruent
like triangle ABC and triangle DEC
u use statement and reason format right?
im pretty new to geometry so im kind of confused

[upper karma]

right

[upper karma]

did u get it?

[thick fable]

LMAOOO no sorry

[upper karma]

is the answer 12?

My friend solved it and got the answer as 12.

[hearty crest]

anyone know how to graph trig functions? I'm stuck

[upper karma]

Guys my Yankee ahh teacher asking me what a 1M sided shape is named? I Googled BU nun came up

By shape I mean polygon pentagon yk

[hallow scroll]

almost a circle

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

[timid remnant]

bruh why does this even exist its basically a circle

u cant even see any of the sides lmao

[slow path]

,rotate

[slow path]

I got an answer through brute force but I'm not sure how to do it normally

I got 8cm² which is right but I kinds guessed

Cus what if EB was triple AD or if DF was not half of AD?

Am I meant to guess?

@empty tiger

[celest raft]

Can i give you my answer ?

[slow path]

Sure

[celest raft]

Since ABCD is similar to DAEF

The ratio AB/AD must be the same as DA/DF

[slow path]

Ohhh

I read the question wrong I think

[celest raft]

Same as saying 8/4 =4/DF

[slow path]

I think the values I said were wrong

[celest raft]

The values you said were right

[slow path]

I don't think EB = 8

[celest raft]

But your answer was wrong because you didn't justify

[slow path]

If the longer side to smaller side = 5:2 then I gotta do 4/2.5, no?

To work out DF?

[celest raft]

The ratio is 10:4

[slow path]

Which is 5:2

[celest raft]

Yes

[slow path]

So if the 4cm is the 5 in the 5:2 then surely the 2 is 4/2.5?

[celest raft]

But 10:4 is easier cuz you will get integers

Wait no

My bad

[slow path]

I may have confused you with my original values I wrote on

[celest raft]

U were right

[slow path]

So area of DAEF = 4 × 4/2.5

16/2.5

6.4

Correct?

[celest raft]

That's it

[empty tiger]

Hm

[slow path]

I feel like I'm asking too many questions on this test but the main reason I'm doing it is to see my weak points

[celest raft]

Anyway if you have anything else to ask...

[slow path]

'rotate

[empty tiger]

Love the scale of the rectangles

[slow path]

I'm so lost on this one ngel

[slow path]

,rotate

[slow path]

I don't know where to start, because in my mind I'm blank because I don't know the shape

[celest raft]

Wow i'm trying to figure out what that means cuz im not an english guy and that's blank to me

[slow path]

I can help? What terms don't make sense?

[celest raft]

Surface area

[slow path]

The area of the each sides face added up

So like on a cube you'd do the area of each square and add them up

[slow path]

Google would probably help more than me

[celest raft]

Yeah

I'll try to see

Well

I maybe have a starting point

[celest raft]

The surface can be converted Into a square

[slow path]

Wdym?

[celest raft]

SA of A = SA of B*9/4

But like take thé square root

So sqrt(A) = sqrt(B*9/4)

[slow path]

I'm so confused, I think I'm better off asking my teacher, but I won't be able to sleep without knowing. Can I come back to you in like 20 mins?

[celest raft]

Of course

Hey

[slow path]

hey

[celest raft]

I just found out the SA of a cube of V=8 is the same as the SA of a sphère of V=8

[slow path]

v?

nvm

[celest raft]

Volume

[slow path]

i dont think it is

[celest raft]

I just did it by hand

[slow path]

2x2x6=24 for dube

[celest raft]

But i verified and it is true

I computed thé needed value of r for V=8

[hot quest]

ratio of areas: 4 to 9
ratio of sides: 2 to 3
ratio of volumes: 8 to 27
405/27*8 = 120

[celest raft]

Just destroyed my whole research 'bout that to demonstrate it

[hot quest]

also is that edexcel gcse maths?

[slow path]

2015, yes

[hot quest]

nice, i remember doing gcse a loong time ago. back when maths was way easier

[celest raft]

What is it ?

[slow path]

that is so much simpler than i anticid lmao

[slow path]

exams when youre 16

[celest raft]

Ohh

In France that's different

[hot quest]

tip for edexcel gcse math: practice the vector 5 markers

anyway goodbye

[slow path]

vector 5 markers?

[slow path]

I know this isnt the case but surely this only works for squares? (Someone else can snwer for them)

[celest raft]

Well since all shapes can be converted Into cubes, this do works

For 3d shapes

And since 2d shapes can be turned Into squares it works too

[slow path]

i dont unserstand

[celest raft]

Imagine you have a sphere with a volume of 8 cm^3

This is the same as saying you have a cube with a volume of 8 cm^3

Volumes stay thé same not the solids

That's topology

I can't explain very further

[slow path]

I just fried to prove you wronwrong and instead proved you right, which js good

[celest raft]

That's the same as having a ball with a non-stretchable surface, you Can bend it and change its shape but thé volume and thé surface area will stay thé same

Eventualy you Can bend your ball so that it becomes a cube

Here you are !😄

[celest raft]

Can i add you as a friend so that if we have more things to talk about we Can ?

[slow path]

sure!

[celest raft]

Nice

[empty swift]

Can someone help me with a problem

Im confused how to solve for X

Nevermind I found it out

X=29

[oblique wharf]

Yo i’m kind of stuck on this

I cannot find any way to get past x = 10 / sin(20)

[frozen ocean]

use cos

,tex .sohcahtoa

[somber coyote]

Akira

[copper gorge]

Send help

[stable orbit]

can someone solve this for me ive been waiting for like for over an hour

[gaunt radish]

Add the equations together

[gaunt radish]

with like terms

[mossy estuary]

"Send help" is an interesting way of saying it lol

[vernal vale]

Is this correct??

[fickle rose]

should be because all interior angles add up to 180

u can always check by using an exterior angle as well

[brisk sundial]

bruh IGO 2021 advance

[knotty magnet]

what the hell of a diagram is that

[pliant current]

Idk how to find angle DAB PLS HELP

[old arch]

geometry dash

[frozen ocean]

fr

best game

[heady scaffold]

first of all, you must understand that KM is a line bisector of JL and angle bisector of JKL. therefore, angle JKM and angle LKM are congruent. Therefore, using SAS theorem, we can conclude that both triangles are congruent. Therefore, line JM is equal to line ML, so 3x + 4 = 5x - 16. Using algebra,

3x + 4 = 5x - 16
3x - 5x + 4 = -16
-2x + 4 = -16
-2x = -20
x = 10

Therefore, since JL = JM + ML

3x + 4 + 5x - 16
3x + 5x + 4 - 16
8x - 12
8(10)-12
80-12
68

of course, always recheck my work because i may have made a few mistakes

im doing this all in my head so theres a high chance there might be a few mistakes

[fossil raven]

how are 5 and 8 congruent

[heady scaffold]

,rotate

[heady scaffold]

youre welcome

[fossil raven]

what

[heady scaffold]

no i was talking to the people that are going to try to answer the qudstion

i hate geometry so im not even going to try but i rotated it so its easier to read for those people whoo are going to answer

[fossil raven]

oh alr

[leaden radish]

my brain isnt working what

[timid remnant]

the 3rd one right

bcs if its equilateral its gonna be acute as well bcs 60 degree angles

rights can be scalene if u just increase the lenghts

and iscoseles can be obtuse obviously

[latent pivot]

Da 3rd one coz an equilateral triangle must have three same angles of 60 degrees so it cant be obtuse

[fickle rose]

3

[fickle rose]

is it any special kind of special quadrilateral or not

cuz obviously you can’t say that with 0 information

[teal imp]

need more info, only way I can think of that happening is EBDC is a parallelogram, which basically means that all those triangles are equilateral

Perhaps the question mentions 2ED = AC and the perpendicular line from point B to ED bisects ED?

and ACDE is a trapezium

[honest summit]

has anyone got deductive gemoetry like provin congruency and similarity worksheets and resources i can find

would be rlly helpful

[left stag]

Hello fellow trigonometricians

[thick fable]

what about geometricians

[fickle rose]

ye

[thorn crow]

Good as the old days (From the last 3000 years literally)

[thorn crow]

You can refer "A School Geometry" by Hall and Stevens

For a bit hard, Challenge and Thrill of Pre-College Mathematics

Both the books are available free online

A School geometry is very easy and gives you some good practice while challenge and thrill of pre-college mathematics very good problems on similarity and congruency

[heavy grove]

hello

i need help

@errant viper

[frozen ocean]

poor helper getting pinged for no reason

[timid remnant]

fr

[smoky jetty]

imagine a triangle whose each interior angle is >90 deg when triangles have total of 180 interior angle deg.

[deep pecan]

Could anyone help with this. circle theorems and you need to find x and y, anwsee in cm

[smoky jetty]

i forgot particular theorems but the way I see it is like this

then u can find the hypotenuse i.e the red line, thus x and y

[deep pecan]

Oh i see what you mean

Thanks for the help

[smoky jetty]

np!

btw the red lines are radii so they're of equal lengths

[deep pecan]

Yep thank you

[steel ermine]

Plz

I need help

[snow crystal]

Hi I have a problem, can someone tell me why if I have an equation sinx + cosx = 1 why can't I just square both sides

Bc when I do it I'm getting wrong results and idk why

[tired ruin]

I don't understand why the solution of sin(x) = -(sqrt(3)/2)
is -2pi/3

I made a post about it here: #1164248299197431808

[timid remnant]

a' is the new point and if a is going there then everything else also has to do the same thing, meaning u literally move all the points 5 units to the left

and thats quadrilateral a'b'c'd'

oh wait

u need to rotate it first

90 degree clockwise is gonna be a at (-1, 1), b at (-3, 1), c at (-4, 4), and d at (-2, 4)

abd then u move it all 2 units up and 5 units to the left

[heady scaffold]

yes equilateral triangles always have three 60 degree angles

three

[copper gorge]

Need help

😔

[upper karma]

Help pls 😢

[fickle rose]

do you know the properties of a circumcenter?

if u do, you can very easily solve 3 and 4 with no math at all

not even a. single expression or equation is needed

point P is the incenter for problem 5 and 6

knowing the properties of an incenter will have you easily solve the next 2

[snow crystal]

I know that well

But I'm still getting wrong results after doing it

[foggy parcel]

power of AI

[foggy parcel]

are u supposed to solve for x or what? @snow crystal

[snow crystal]

Yes

But When I square both sides I dont get the correct result

[foggy parcel]

Are u informed of the coorect answers though?

@snow crystal

Anyways

1. sqaure both sides
2. use the identity sin^2 (A) + cos^2 (A) = 1
3. u will get an equation containing sinA × cosA in the left side and 0 in the right
4. if sinA = 0, A =?
5. if cosA = 0, A =?

Hence A =..

[frail temple]

Please anyone help f(x) = 4^x

[thick fable]

what is the full question

[fickle rose]

that’s just an exponential function?

[frail temple]

Yes it is

[fickle rose]

so what do u need help with

[frail temple]

Graph and solutions on how to solve it