geometry-and-trigonometry

[kindred ocean]

once you figure that out, you should be able to draw it

[grave pond]

It might be a trick question. 🙃 When nothing else is specified in ATC communications, a turn means a "standard rate turn", in which it takes 30 seconds to fly through a quarter-circle turn. A DC-9 needs to be going at least 100 knots to stay aloft, which means the radius of a standard rate turn is at least half a (nautical) mile. This is significant compared to the explicit figures of 3/4 and 1 miles in the problem, so assuming that the plane immediately starts flying in heading 138° is going to give wrong results.

Don't mind me, I'm just criticizing the problem author for imagining impossible situations.

Okay. Take a piece of graph paper and declare one of the grid points to be the airport.

[blissful sonnet]

wasnt you though, give the ai credit

[grave pond]

And what I was complaining about is that a flying aircraft cannot just turn on a dime like that diagram pretends it does. In reality it would look more like this. But you don't have enough information to take that into account, since you don't even know how fast the plane is going.

[blissful sonnet]

yeah thats kind of funny

do they even plane?

[grave pond]

(That mock-up looks vaguely like the turn radius doesn't influence the final bearing, but I think that's just an effect of my imperfect image sketching.)

[blissful sonnet]

for sure

[blissful sonnet]

we pointed at what you need to do my friend, read up on nautical bearing, make a drawing, do trig
https://en.wikipedia.org/wiki/Bearing_(navigation)

[true void]

Is there a thread here that teaches geom and trigo from noob to pro

[nimble cypress]

idk

maybe

[nimble cypress]

https://tenor.com/view/pinguim-madagascar-se-escondendo-hiding-gif-27503349

[worthy eagle]

do you know what those terms (complementary, supplementary..) mean?

and do you still need help with that?

[vague dawn]

uhh mb I already got it figured out

[worthy eagle]

okay

[south barn]

Someone here has try a beautiful joruney through meth Olympiad geometry?

[cosmic sluice]

Where do I go to find AoPS books?
Specifically AoPS: Intor to Geometry

and lets assume im "on a budget"

[faint pier]

dm

[tall pasture]

@rotund drum i got 100% on my geometry test

[rotund drum]

dang congrats!

[tall pasture]

Thx

[granite phoenix]

https://tenor.com/view/lordzebax-gif-18360353

[rain scaffold]

Problem statement: 2 perpendicular lines (L1 and L2) are drawn and point O is marked at the intersection of these 2 lines. 14 points are then marked on both of the two lines (excluding O). Any two consecutive points are separated by 1cm. For both lines, these 14 points are marked on the same side of O. These points are numbered 1,2,3,...,14 on both lines where 1 is assigned to the point closest to O on either line and so on. Now point 1 of L1 is joined to point 14 of L2; point 2 of L1 is joined to point 13 of L2 and so on. A curve is obtained, what is the equation of this curve? Take suitable axis to solve this problem.

We can take L1 and L2 as X and Y axis (WLOG). This will setup the coordinate system. Sketching out the curve, it looks beautiful. Feels as if it is a rectangular hyperbola. However, I'm not so sure about that.

[naive bay]

hey, im tryna learn trig and i know the basics, can someone guide me on where to go after this?

[rain scaffold]

This is what the curve looks like.

[rain scaffold]

What have you learnt in the basics?

[naive bay]

the functions and simple questions like-

find the angle

[rain scaffold]

So, like basic cos, sin, tan definitions?

You might give a go at trig identities.

[worthy eagle]

you should also learn the trig values of some important angles like 0,30,45,60,90 degrees after this

[rain scaffold]

The basic ones like cos^2(x) + sin^2(x)=1

[rain scaffold]

You can just use a calculator for that purpose.

[worthy eagle]

oh

[rain scaffold]

And derive this, for instance this trigonometric identity can be derived through Phythagoras theorem.

[naive bay]

yes

i learnt that identity

cant it also be proved algebraically

[naive bay]

where i live calculators arent allowed in exams and stuff so i think we're expected to memorise it

should i learn this-

[rain scaffold]

Oh well, memorize them in that case.

[rain scaffold]

Yes, that is basically counts as a algebraic proof.

[rain scaffold]

If you remember sin and cos for a certain angle, you can compute tan by sin/cos

So, you might as well avoid learning the tan ratios. (In case you're not comfortable with memorizing)

[worthy eagle]

,tex .rocket trig

[worthy eagle]

I will recommend learning sin then reverse the order and you get cos then divide sin and cos values to get tan. Learning sin is must, learning tan after sin is very useful as derivating takes time, learning cos is very optional, you can find cos of any term in mind by thinking of opposite sin value in no time
A trick for sin values, they follow a pattern: √0/2, √1/2, √2/2, √3/2, √4/2

[grave pond]

Huh. I vividly remember drawing such diagrams back in grade school, before we had the knowledge to do anything with them other than "look, that's a pretty curve". :-)
I don't think it's a rectangular hyperbola, actually -- it reaches the axes at (0,15) and (15,0), and if you continue the pattern with negative coordinates for one of the points, the curve continues inwards in the first quadrant. My immediate guess would be that it's a parabola with its axis pointing 45° upwards.

[faint pasture]

i dont see any way you could do it without calculus actually

[rain scaffold]

Can we rigorously prove that its a parabola?

[rain scaffold]

How will we do that?

[faint pasture]

gimme a minute

like uh we can write the general equation of the tangent as x/t + y/(15-t) = 1

so we get slope of the curve in terms of a random variable t

[rain scaffold]

Yes

[faint pasture]

lemme see if and how we can eliminate it

[grave pond]

So far it's just a gut feeling.
An approach could be to assume it's a parabola and figure out a parametric equation for it, something like t -> (3.75-t, 3.75+t) + at²(1,1), with the constant a chosen such that the curve goes through (15,0) and (0,15) at t = ±7.5.
Then (using calculus) compute the tangent to the parabola at an arbitrary t, find its two axis intercepts and if we're lucky they might turn out to sum to 15.

[rain scaffold]

Can we use (15,0) as 14 points are marked in the original problem?

[grave pond]

Or perhaps start by choosing a different scale such that the curve touches the axes at (4,0) and (0,4) instead -- then the symmetry center at the midpoint between (2,0) and (0,2) will have nice coordinates at (1,1).

[faint pasture]

i mean uh there is a way? solve this for t in terms of x and y and put that in dy/dx = -(15-t)/t i dont think the eqn would end up pretty though

[grave pond]

I'm imagining continuing the already given pattern with lines between (0,0) and (15,0, and between (0,15) and (0,0).

[rain scaffold]

Alright.

[grave pond]

In general we have lines between (s,0) and (0,15-s) for s = 1,2,3,...,14 but the pattern really ought to continue if we pick arbitrary s, including 0 and 15.

And I'm hoping those lines will all turn out to be tangents to the parabola I'm betting on.

[faint pasture]

i dunno doesnt look parabola-ey to me

,w graph y = x^2/10

[somber coyote]

Results provided by WolframAlpha

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Requested by enderdoesntmind

[rain scaffold]

I think it should be a parabola

[faint pasture]

well only way to know for sure is solve the diff eqn

[rain scaffold]

Here's what I did:
Equation of tangents can be written as : x/t +y/(15-t)=1. Now let us assume some point (h,k) on the curve so h/t + k/(15-t)=1
Simplifying gives a quadratic in t and its roots should be equal! Because, for (h,k), there should be only one tangent?

[faint pasture]

,w graph y=x^2/10 and (y-sqrt(2))^2-x^2 = 1

[somber coyote]

Results provided by WolframAlpha

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

Requested by enderdoesntmind

[rain scaffold]

Does this work?

[faint pasture]

could be either tbh

[grave pond]

Consider this, fitted by eye

[rain scaffold]

,w graph (y-x-15)^2=60x

Does this work

[faint pasture]

damn looks identical so i guess thats correct

[rain scaffold]

This is what I got

[faint pasture]

not a parabola that i can say lol

but should be correct

[grave pond]

Hmm, looks wrong to me that it doesn't seem to be symmetric in x and y.

[faint pasture]

your method sounded about right

[rain scaffold]

Yeah

What curve is that?

[faint pasture]

bro but look at his method i dont see a flaw

[faint pasture]

lemme simplify the eqn and check it

its an ellipse

[faint pasture]

click the more thingy

[rain scaffold]

h^2-ab<0 is the condition right?

[somber coyote]

Results provided by WolframAlpha

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

Requested by leplusgrand_leader

[faint pasture]

yeps

[rain scaffold]

Parabola-..

[rain scaffold]

Well?

[faint pasture]

🤡 why the fuck is h^2-ab<0 in that case

[rain scaffold]

@grave pond Your guess was correct

[rain scaffold]

Calculation error?

[faint pasture]

real

,w simplify (y-x-15)^2 = 60 x

[somber coyote]

Results provided by WolframAlpha

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

Requested by enderdoesntmind

[grave pond]

Oh, and with a slider in Desmos, the tangent seems to follow beautifully. Not a proof, but encouraging.

[faint pasture]

huh? 1^2 - (-15)(-15)<0 is true bruh

[grave pond]

(Pity one cannot share Desmos constructions without being logged in anymore).

[rain scaffold]

a,b are coefficients of x^2 and y^2?

so h^-ab=0

[faint pasture]

there was nothing here tropo dont ban me

[rain scaffold]

Why lmao

[faint pasture]

😭 why did i go for g and f

i need a recap of coordinate geometry fr

[rain scaffold]

rip..

[rain scaffold]

Such a shame.

[grave pond]

Actually, plotting my parabola and your (y-x-15)²=60x shows they seem to overlap precisely, even though your expression doesn't look symmetric around x=y at first.

Rewriting to y=x+15-sqrt(60x) should at least make it simple to get an equation for the tangent :-)

[earnest harbor]

how a white boy get goated wit the sauce

[dark sparrow]

go to #chill

[sly urchin]

I'm taking IGO tmr, any advice?

My expectations aren't too high, if i get 2 questions right, that'd be a great success for me

[tropic blaze]

@sly urchin hey bro wanna play chess

[sly urchin]

Sure, gotta say i'm quote rusty, haven't been training in months

Let's take it to the DM

[tropic blaze]

@sly urchin what's your rating

[rain scaffold]

Is it necessary for a parabola to be symmetric about y=x? Anyways, I think we got the correct expression.

[obsidian harness]

there's nothing that states that a parabola needs to be symmetric around any particular line

of course all parabolas have exactly one line of symmetry

[obsidian harness]

from the diagram, you can see that at least all of the intersection points are symmetric across y = x

[rain scaffold]

Yes, so does that make it necessary for the curve (which turns out to be a parabola) to be symmetric about y=x?

[obsidian harness]

I have no idea

you can reverse-justify that because there is a unique conic that passes through any 5 (non-collinear) points

like, that specific parabola satisfies the envelope condition, so it's the only one that works

[rain scaffold]

What's an 'Envelope' condition? I haven't heard about it.

[obsidian harness]

a curve formed like this is called an 'envelope'

so, the condition for the envelope to occur

[grave pond]

A random parabola doesn't need to be symmetric about that line in particular -- but since your original construction is symmetric about it, the curve it converges to should also be.

[west depot]

how would i solve this

[dark sparrow]

let x be the number of 7cm pieces and y be the number of 11cm pieces

then 7x+11y = 200

and x and y are both natural numbers, so you need to solve this diophantine equation

[grave pond]

You can somewhat wing it: You want to use as many 11cm pieces as you can, so first try 18 pieces, leaving 2 cm, which cannot be filled by 7cm pieces.

Then try 17, 16, 15, 14, .. pieces of 11cm each, until what is left over can be covered exactly by 7cm pieces.

[faint pasture]

What special characters does a parabola have to posses to be called an envelope

[grave pond]

The leftovers are 2, 13, 24, 35, 46, 57, 68, ..., and we look for the first of those numbers that is a multiple of 7.

[grave pond]

https://en.wikipedia.org/wiki/Envelope_(mathematics) is a better explanation than I'd be able to type up.

[winged oyster]

11x+7y=200

i mean youd have to just guess with numbers

[faint pasture]

Damn it could be extended to other curves as well.. that's kind of scary lol

[winged oyster]

but it would be 20

15 11 cm and 5 7 cm

[grave pond]

There are more systematic procedures based on the extended euclidean algorithm and Bezout's identity, but they are not worth the trouble when the numbers are as small as 11 and 7.

[granite peak]

how do i know which values are the A,B and C

When using cosine and sin rules

[sturdy oriole]

did anyone take the igo already and wants to discuss

[dapper blade]

Does anyone have a solid book/pdf for mensuration of volumes which V=1/3bh

[slate goblet]

Hey I got a question can someone help me out pls?

[knotty quiver]

!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/

[spiral lodge]

Wdym? Are you referring to the angles or to the vertices? Cause the vertices are already written there

[obsidian harness]

angle A is always opposite to side a
angle B is always opposite to side b
and angle C is always opposite to side c

so it's always (opposite side to the angle you want)^2 = (adjacent side)^2 + (adjacent side)^2 - 2(adjacent side)(adjacent side) cos(angle you want)

[upbeat crystal]

https://tenor.com/view/burdgis-landsat-remote-sensing-gis-geemap-gif-22770097

[carmine jetty]

whatever's across from that little bendy straw looking thing (called angle) is gonna be your opposite, and whatever's next to it is your adjacent. although this is a little weird since it's not a right triangle

I've never had to use the law of cosines formula

but its this apparently? $cos A = (b² + c² - a²) / (2bc)$

[somber coyote]

drew

[carmine jetty]

never knew TeXit makes ^2 into arrowed down s'es

thats a weird little nook and cranny detail

apparently a is your opposite (the one that goes straight across from the angle), and b and c are your adjacent angles for the equation

this would give you 45/60

which is the same thing as asking 4.5/6

which is 0.75???

i mean 3/6 is 0.5 because half of 6 is 3.. wait

3.5 would be some irrationally long number

4 would be as well

ahh im not gonna sit here and act like ik what im talking about

dippin peace

wait is it the same way that quarters of time work?

like 15, 30, 45, 60?

so technically 45/60 is a set of 4th's

i guess if you look at it like that then obviously the 3rd/4th would be .75

but who tf automatically thinks of quarters of time when they see 45 and 60

[cunning lion]

use ^2 instead

,, \cos A = \frac{b^2 + c^2 - a^2}{2bc}

[somber coyote]

κλαουντ ☁ (cloud)

[kindred elk]

hey guys

[winged oyster]

hi

[fading mist]

Anyone have an efficient way to do this

[summer cradle]

Let length = x and width = y
x² + y² = 8² = 64
(x-2)² + (y-2)² = (4sqrt(2))² = 32
=> x² - 4x + 4 + y² - 4y + 4 = 32
=> (x² + y²) - 4(x + y) = 24
=> 64 - 4(x + y) = 24
=> 4(x + y) = 40
=> x + y = 10
(x + y)² = 10² = 100
(x² + y²) + 2xy = 100
64 + 2xy = 100
2xy = 36
Area = xy = 18 (E)

[wicked garnet]

insane problem with not much data

[naive bay]

guys im learing trig for no reason, my brother just taught me a trick forvalues of sin/cos/tan of 0,30,45,60,90-

like choose one finger and take the no. of fingers above/below that

[rain scaffold]

-1=<sinx=<1. That means the range of sinx is [-1,1]. Let t be some real number in [-1,1]. Can we always find a 'x' such that sinx=t?

[thorn rivet]

yoh bro why this math so difficult

[weary citrus]

Yes, this is what stated in Intermediate Value Theorem

[weary citrus]

You're not gonna believe how simple the problem is

FBD,ABC and EDC are congruent triangles

[wicked garnet]

may I ask how and what can I use it to prove the parallelogram?

[weary citrus]

FB=DE=FA and FD=AE=AC if you could prove they're congruent

that makes FDEA parallelogram

[wicked garnet]

but how can I prove it congruent though

[weary citrus]

I could send the whole sol here

If you want

[wicked garnet]

yeah tysm

[weary citrus]

Wait

[wicked garnet]

can you explain more on how to prove bca and dce is congruent?

[sinful violet]

first 2 lines should be angle DBA

[wicked garnet]

yeah indeed

[sinful violet]

not BDA

[weary citrus]

Oh yeah sorry

my bad

[weary citrus]

It's the same process really

[wicked garnet]

ohhh

i think I see it

thx

[rain scaffold]

I'll look that up.

[oblique sinew]

Is this rectangle even possible? The two diagonals should be equal but here it looks like one diagonal is 8cm and the other diagonal is (4√2+√2+√2)cm or 6√2 cm. And 8 is not equal to 6√2. Am I missing something?

[sinful violet]

becuase the 4rt2 line isnt aligned with the actual diagonal

and adding the rt2+rt2 wont get you a straight line

[thorn rivet]

guy can i ask some thing how to get good score on math exam

[grave pond]

Start caring about the course early enough that you know the material once the exam happens.

[novel ingot]

sm1 teach me the basics

[silent plank]

of?

[novel ingot]

coordinate geometry

[silent plank]

for the basics, you could try something like khan academy

[novel ingot]

good idea

[wet pecan]

Are you new started geometry?

[weary citrus]

No, I've done geo for a while now

[wet pecan]

Oh nice

Im bad geo

I new started

[weary citrus]

What kind of geo are you studying

[strange horizon]

One message removed from a suspended account.

[grave pond]

Hmm, if you assign names to the angles marked with one, two, and three ticks, then together with alpha, beta, and C you have six variables.
Angle sums in triangles ABC, AEC, AEF and DEF gives you four equations in the six variables, and if you eliminate the three first variables, you'd have a single equation left after all the algebraic dust settles, which will hopefully rearrange to the one you're supposed to prove.

[strange horizon]

One message removed from a suspended account.

[grave pond]

Which result do you actually get, rather than something with 900°?

I get the desired conclusion by adding up ||the angle sums of 8(DEF)+3(ABC)-4(AEF)-2(AEC)||.

[strange horizon]

One message removed from a suspended account.

[strange horizon]

One message removed from a suspended account.

[grave pond]

8+3-4-2 = 5

[strange horizon]

One message removed from a suspended account.

[strange horizon]

One message removed from a suspended account.

[honest shore]

could i use geometry or trig to determine the ideal racing line

if so, is there like any other concepts that id have to use?

[storm sparrow]

Guys, I know it is not geometry by itself, but I'm studying applied geometric algebra, does anyone wanna go with my now?

[storm sparrow]

Me*

[obsidian harness]

you should ask in #advanced-algebra

go to #get-advanced-access to get access

or if you're not talking about exterior algebras such as Clifford algebra, please clarify what exactly you mean by 'geometric algebra'

[rigid torrent]

Hello, everyone!
What are some book recommendations for AMC10? I aim to make USAJMO this year but have my 8th grade (next year), my 9th grade, and my 10th grade years.
I use all books on the AOPS Book Store, handouts, AMC books, formula books, and other resources found on https://mathematics.gg/books.

[storm sparrow]

I mean this yeah

Okok

[quasi hound]

wht grade r u in rn

[west narwhal]

I think 7th probably..

[west narwhal]

Well I think you can do a lot of practice tests and mocks for the AMC 10, aside from the books. For books, you can do Mastering AMC 10/12 by Sohil Rathi and other contest preparation books. There's also Alcumus and MathDash for practice.

[quasi hound]

7th? so hes 12

[obsidian harness]

he can be 13

[quasi hound]

yea

[night cape]

Hi im doing this quiz and from what i understand U and N are not on the same line, so how is it they are collinear?

[tall pasture]

My geometry got pissed af today from a student he did 8+6=2

[summer cradle]

a set of points are collinear if you can draw a single line which passes through all those points

[night cape]

ooooooooooh so its a question of IF you can draw not its its already drawn through

that makes sense thank you 🙏

[summer cradle]

you're welcome!

[tall pasture]

Guys what are adjacent angles?

[native pier]

angles that are, quite simply, adjacent

[smoky stream]

Chat dose anyone know how to answer this

[rigid torrent]

7th

[rigid torrent]

We have $\frac{pi\cdot d}{2}=28$. Then $\pi\cdot d = 56$ and $d=\frac{56}{\pi}$. $d$ is the side length of the square part of the figure, and there are 3 sides of the square part which are also part of the perimeter. Therefore, there is $3d = 3\cdot\frac{56}{\pi}=\frac{168}{\pi}$. Adding the $28$, the total perimeter is $\frac{168}{\pi}+28$, which is $81.476...$ and rounds up to $\boxed{81.5}$

[somber coyote]

C++
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

C++

[smoky stream]

Yo do this

[terse terrace]

That's easy

[rigid torrent]

We have $\frac{pi\cdot d}{2}=26$. Then, $\pi\cdot d = 52$ and $d=\frac{52}{\pi}$. $d$ is the side length of the square part of the figure, and there are 3 sides of the square part which are also part of the perimeter. Therefore, there is $3d = 3\cdot\frac{52}{\pi}=\frac{156}{\pi}$. Adding the $26$, the total perimeter is $\frac{156}{\pi}+26$, which is $75.656...$ and rounds up to $\boxed{75.7}$.

[somber coyote]

C++

[rigid torrent]

Real.

[smoky stream]

I already did it

[maiden brook]

!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.

[rigid torrent]

Okay sir.

[opal dawn]

Add 10 and 13 alternate exterior angles?

Also I’m does 12 and 14 add up to 180

[trim abyss]

Bruhhh dude just left from the help chat

Is the Y axis right?

[exotic yarrow]

$l$ and $m$ aren't given to be parallel, so you can't necessarily say this.

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

get your protractor and measure those two angles

you'll see what I mean

[exotic yarrow]

10 and 15 are. Not 10 and 13 though.

[opal dawn]

What is 10 and 13 then

[exotic yarrow]

none of these

[opal dawn]

HELP ASAP

Which of these are both interior and exterior

I’m so confiseeddd

[exotic yarrow]

wdym

an angle pair can't be both

[opal dawn]

Wdym

Which angles are interior exterior etc

[echo schooner]

Yo can u help me

[opal dawn]

I don’t understand it

What makes an angle interior or exterior

Aren’t they all interior?

[echo schooner]

Alternate exterrior angles are angles outside the 2 lines interrior is inside

[opal dawn]

Which two lines

[opal dawn]

Help

[opal dawn]

?

Are 9 and 10 consecutive exteriors?

[toxic shard]

Like the 2 lines that are crossed by a transversal if the two angles are somewhere between the 2 lines its interior and if its outside its exterior you cant have one angle on two sides of the same line

[fossil tree]

help im crashing out

[outer raven]

the triangle with angles 5 and 6 is an isosceles traingle

so 5 is 68 degrees and 6 and the other angle r the same

[fossil tree]

i got it thx

[winged oyster]

i think 6 is the mistke

mistake

not sure abt what it is

[obsidian harness]

6 is the base angle, and 5 is the apex angle

Only the two base angles are equal

[vague dawn]

https://discord.com/channels/268882317391429632/1430047214436094063 can anyone help?

[obsidian harness]

(solved)

[worldly zephyr]

Someone pls help with this problem

"Plot the following points with polar coordinates (r, 0):
a) (2, pi/4)
b) (3, pi/6)
c) (1, - pi/4)
d) (2, - 5 pi/6)"

[spiral lodge]

Are you sure it's (r, 0)?

!original

[lime crown]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[worldly zephyr]

It's (r, theta) but Idk how to write

Screenshot doesn't work on my tab coz the volume buttons are destroyed

[worthy eagle]

You know what's formula for x,y coordinates

in terms of r, theta

[worldly zephyr]

Ik how to find (r, theta)

But not the opposite

Finding (x, y) from (r, theta)

[worthy eagle]

(r, theta) means a line of length 'r' units pointing at theta degrees from positive x-axis with one end at origin

it represents the point at which the other end of the line points

[worldly zephyr]

So r is on the unit circle and theta is the origin?

[worthy eagle]

no not on the unit circle

on unit circle means r=1.

It means on a circle having radius r units

[worldly zephyr]

Then r is the hypotenuse

[worthy eagle]

yes

[worldly zephyr]

?

And what is theta?

[worthy eagle]

now you know r is the hypotenuse in the circle of radius r units and theta is the angle that the hypotenuse makes with the positive x axis

[worldly zephyr]

Doubt here

[worthy eagle]

yes ask

[worldly zephyr]

Why is theta a coordinate?

Isn't it an angle?

[worthy eagle]

theta isn't a coordinate
hypotenuse is a 'line' that has one end at origin and the angle between this line and the horizontal is theta

so like i can send a diagram

[worldly zephyr]

So r is radius of circle and theta determines where on the circle its end point is

Right?

[worthy eagle]

yes

this point which this r touches the circle is the point that (r, theta) refers to

[worldly zephyr]

Ohhh

[worthy eagle]

you have to find coordinates of this point

now, if you draw a perpendicular line from this point on the x axis, you get a nice right angle triangle where the perpendicular line represents the y coordinate of the point and base represents the x coordinates of the point

[worldly zephyr]

Oh OK

So then how do we find (x, y)?

[worthy eagle]

think how can you find the x, y from the triangle, use trigonometry

[worldly zephyr]

Ik the angle and the hypotenuse so...

[worthy eagle]

try anything sin, cos, tan and see what works

[worldly zephyr]

Tan won't work

No hypotenuse

Ohhhhhhh!

Thanks!

[shut thorn]

(cosθ , sinθ)?

[worthy eagle]

so can you write formula for x,y in terms of r, theta just so I know you got this?

@worldly zephyr

[worldly zephyr]

Um

[worthy eagle]

no

[worthy eagle]

no

near

end it man, i will just send the formula in few min

[worldly zephyr]

sin theta = opp/hyp

So if theta = pi/4 then sin theta = 1/sqrt 2

So if hyp = 1

[worthy eagle]

you know hypotenuse=r

[shut thorn]

Wait. Isn't cosθ the value of x?

[worldly zephyr]

So then r sin theta = opp

[shut thorn]

Is it not a unit circle?

[worldly zephyr]

Not always ig

[worthy eagle]

in this case rcostheta

[whole flame]

Nahhh for unit circle r=1 unit

[worldly zephyr]

But rsintheta also works no?

[worthy eagle]

yup
x=r costheta
y=r sintheta

[shut thorn]

Oh

r cosθ

And
r sinθ

[shut thorn]

Yes I thought it was a unit circle.

[whole flame]

Okayy

[worthy eagle]

no x≠r sin theta as then it would mean sin theta = x/r = base/hypotenuse which is wrong

[worldly zephyr]

For y

[shut thorn]

Yeah it does because when r = 1 then it's just sinθ

[worthy eagle]

so you get it right? so you can solve those questions

[worldly zephyr]

Thank you so much

[whole flame]

X= costheta y = sintheta for unit

[shut thorn]

Yep.

[worldly zephyr]

@worthy eagle sry if I was a little thick skulled

[worldly zephyr]

👆

[worthy eagle]

np bye

[shut thorn]

Yep this question.

Wait lemme get it brb.

if x = cbrt(28) and y = cbrt(27) then find the value of (x + y) - (1) / (x² + xy + y²)

How do I do this.

I think it would be useful to say that it's about cubes? Like x³ + y³ = (x + y)(x² + y² - xy)

Or the identity of difference of cubes? x³ - y³ = (x - y)(x² + y² +xy)

[whole flame]

Ok so

[whole flame]

Use this

[shut thorn]

?

[whole flame]

But what u can do here is

[shut thorn]

?

[whole flame]

Let me complete😅

[shut thorn]

Kk.

[whole flame]

-1/x^2+xy+y^2

Multiply and divide this part by x-y

So in denominator u will get x^3-y^3

[shut thorn]

Ye because difference of cube

[whole flame]

N the value of it is 28-27

[whole flame]

Yepp

So the remaining question is

[shut thorn]

So it becomes x-y/28-27

[whole flame]

28-27

1

In denominator

[shut thorn]

So it becomes x+y - (x - y)

Which is 2y.

[whole flame]

So u only have to solve x+y-(x-y)

[shut thorn]

Ohhh

[whole flame]

Yepp

[shut thorn]

Thanks.

[whole flame]

👍

[humble knoll]

Hey! Guys can you help me in Trigonometry?

[honest dock]

you can send your question right away no need to ask

[dark sparrow]

!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/

[humble knoll]

Actually! I wanna know how someone think while solving a question. How does it comes in mind just after seeing a question? I am just curious.

[cedar stratus]

id start with trig identities

and see that 1 + tan^2 = sec^2

[humble knoll]

Then should I take lcm? or convert that 1/sec^2 to cos^2

[cedar stratus]

is it asking to evaluate (sin^2 (1) + 1)/(1 + tan^2 (1))?

the formatting looks weird

[humble knoll]

Yeah 🧐

[cedar stratus]

oh

then just put it in a calculator

[umbral bane]

What is maths even about

https://tenor.com/view/jew-jews-jewish-jump-happy-gif-1531468640197245258

[weary citrus]

Oh wao what math course is this?

Or program

[umbral bane]

HSC extension 2 Mathematics

[umbral bane]

No bro 💀💀💀💀💀💀

[tall spruce]

so how can u understand me 😂

[whole flame]

What is to be found

X=?

[summer cradle]

what are you asked to find?

if x then there's not enough info cuz you need the height of the cross section from the base

[umbral bane]

It’s a silly question

[umbral bane]

Cos I searched it up

we have functioning internet here

[pearl hazel]

does anyone have a good strategy on memorizing the unit circle

[grizzled dust]

what?

[shut thorn]

just memorise the values of cos and sin 30, 45, 60, 0 and 90

[grizzled dust]

or do a logic asosiation with pithagoras, to understand why it happens

[shut thorn]

and then

[grizzled dust]

??

[shut thorn]

in I quadrant. sin and cos arent negative.
in II quadrent. cos is negative. sin isnt
in III quadrant. cos and sin are negative
in IV quadrant. sin is negative. cos isnt

[shut thorn]

yeah that also works but he did want help im memorising their values

which can be confusing considering cos60 and sin30 are equal

and that you could accidentally interchange those values

[grizzled dust]

no bro, like, you just relate it in the diferent dot, and, doing it, you memorize it cause i want (if you understand, you memorize)

[shut thorn]

i saw that 🤨

[grizzled dust]

what? i do not do anything 🤨

[shut thorn]

🤨

[grizzled dust]

who is hungry in bread thinks

🤨

[whole flame]

What's the problem

[grizzled dust]

XD, you don't catch it

[whole flame]

U need help then ask otherwise u can chat in #chill

[grizzled dust]

what? my english isn't pretty well

i was helping someone with trigonometry

[whole flame]

Alr

[tall spruce]

ooooo

[thorny nova]

Can somebody help me with dis sum?

Find the minimum value of 2^sin²x +2^cos²x

[shut thorn]

Is it a multiplication sign or an addition sign?

[thorny nova]

To the power

[shut thorn]

No like it is 2^(sin²x) + 2^(cos²x) or 2^(sin²x) × 2^(cos²x)

?

[thorny nova]

No it's plus

[shut thorn]

Oh

[thorny nova]

It's an addition

[summer cradle]

can you use derivative?

[shut thorn]

I think it uses the Pythagoras identity?

[thorny nova]

Nah

[shut thorn]

sin²x + cos²x = 1

[thorny nova]

Nope

[shut thorn]

Oh it autocorrected 😭

[thorny nova]

Have u heard of arithmetic mean

[shut thorn]

Pythagorean*

[thorny nova]

Or geometric mean?

[shut thorn]

I mean no but I can try.

To atleast simplify.

[thorny nova]

Wait

Any problem if I dm u?

[shut thorn]

Dms are open.

[north kite]

whats trigonometry

[whole flame]

Tri gono metry

[simple vigil]

Measurement of triangle

[upper karma]

Three gooner meter?

[wintry ibex]

bruh

[wintry ibex]

triangle 📐

[spice hazel]

this guy can't be serious right 🙏

[winged oyster]

brainrot 😂

oh my god

[shut thorn]

ratios of the sides of a right angled triangle

[whole flame]

Ik

[shut thorn]

oh wait

[whole flame]

That guy was just fooling around

[shut thorn]

wrong ping 😭

[whole flame]

Yeahh

Gooner

[kind dawn]

Hello

No 3 goon meters 😞

Nnn is soon

What is bread?

[kind dawn]

This has 1 bread

[shut thorn]

now it has none

[kind dawn]

😭who stole my bread

[real sentinel]

Dude wrong channel

[kind dawn]

Oh

Which one then

[shut thorn]

not me definently

[whole flame]

#chill

[latent finch]

can soimeone solve this this might be in my test tmrw

solve for x

im going to sleep

im cooked

[obsidian harness]

you have a right triangle with hypotenuse = 15 cm and base = 12 cm

yeah if you're sleepy it's not a good idea to solve this

wake up and try again

actually, x could be anything I didn't realise it's a square, so here you go

[upper karma]

isn't the height of the rectange = 9? Because you can draw a diagonal in the middle of the rectangle and the length of that diagonal is equivalent to the hypotenuse of the triangle on the top. Therefore, you can form pythagoras theorem with 12^2 + h = 15^2 where h is the height of the rectangle.

[obsidian harness]

how do you know that the length of that diagonal is equivalent to the 15 cm?

[upper karma]

I mean if you turned the top triangle into a rectangle with the height of h as well, wouldn't the diagonals be congruent? This is at least how I think of it...

[obsidian harness]

that's only true if the height of the triangle is equal to the height of the rectangle

[shut thorn]

If I can add on.
If either the base or the perpendicular are shown to be equal.
Then x = 15cm through converse of midpoint theorem.

Question,
Is it supposed to be a square?

[obsidian harness]

no, not a square

[shut thorn]

Kk.
I'll try my best. Brb.

Can't do it.

For me there's not enough information.

Sorry.

[obsidian harness]

exactly, I concluded the same

draw the small right triangle and the rectangle

let the rectangle have any height

then you can always draw the diagram, so x can be anything

[kindred ocean]

If it's not a square, then I think its impossible

However, there are ticks on all 4 sides of the rectangles, so doesn't that mean all 4 sides are equal?

[shut thorn]

how in the hell do i do this

One thing i didnt add is that DO is the altitude

[simple vigil]

Wait I can try

[simple vigil]

Which class question is this ?

[shut thorn]

ninth

[novel stone]

What is the question?

[shut thorn]

ABCD is a rhombus. DO is an altitude from point D which bisects AB

find all angles of the rhombus.

this is the original question. I made that one myself

[simple vigil]

Do you have answer key ?

[shut thorn]

nope

[civic rock]

well inspect ∆AOD and see what's special about it

[shut thorn]

AD = 2OA?

And the fact that its right angled

[civic rock]

yea

actually another way to look at it is that DO is AB's perpendicular bisector so DA = DB

[winged vessel]

hello can someone help me :(?

[shut thorn]

just send the question.

[winged vessel]

In the Oxyz space, given a cylinder with axis Ox, radius equal to 2. Given point A(0; 4; 4), call M a point on plane (Oxy) so that AM touches the cylinder and creates plane (Oxy) at an angle of 15 degrees. Determine the abscissa xM (xM>0) of point M?

( I use google translate so it may have some mistakes)

[shut thorn]

got it thanks

only that part.

that part was the only thing i needed, Thanks/.

[obsidian harness]

wait, I didn't spot that

the markings were too faint

[shut thorn]

is it an isoceles triangle too?

tbh it shouldnt be one

why am i even asking

[upper karma]

ofc I am not how could a three gooner meter major be serious

[young ermine]

@muted osprey

[muted osprey]

Hii

[upper karma]

any1 suggests good book s for beginning geo for maths oly

like evan chan doesnt have good theory

[spice hazel]

[upper karma]

Could anyone explain to me how the sides of a triangle are connected to the angles? Such as how is sin(30) = opposite/hypotenuse, how is the acute angle connected to the ratio of the opposite to the hypotenuse?

[summer cradle]

the largest angle is opposite to the longest side

same goes for the smallest

this is because if you change the size of an angle in a triangle, the opposite side also changes accordingly

for the Trigonometry part I have actually made a video about it

https://youtube.com/shorts/jhffI7LU4UU?si=enq05521ZSV8D1Ds

[obsidian harness]

and just to clarify, this is true for all triangles, not only right-angled ones

for a non right-angled triangle there can be a longest side

but there is no 'hypotenuse'

[obsidian harness]

your best bet to properly understand this is to draw a 30-60-90 triangle yourself (you can construct an equilateral triangle with a pencil, a ruler, and compass, then split it into 2 equal parts)

same for a 45-45-90 triangle

now for each angle, 30, 60, 45, measure the opposite side to that angle and the hypotenuse

then you can plot a graph on paper, with 30, 45, 60 on the x-axis, and the ratio on the y-axis

oh yeah and the graph also goes through (0, 0)

[summer cradle]

I never said it only applies to right triangles

[obsidian harness]

I know

just to clarify, cause you changed the context away from right triangles + they're definitely very new at this

[summer cradle]

oh

[reef wasp]

can someone help me: A parallelogram is drawn in a triangle with sides of 3 cm and 5 cm, and a diagonal of 6 cm. Find the sides of the triangle, if it is known that the diagonals of the parallelogram are respectively parallel to the sides of the triangle, and its shorter side lies at the base of the triangle.

[civic rock]

,rotate

[reef wasp]

damn

[civic rock]

one thing I see is that ∆BAC is just ∆ODK but scaled up by 3

you can use Apollonius's theorem to find the length of OK

[reef wasp]

alright i will just clear my mind for a sec

thanks

[latent finch]

welp im cooked

[marble forum]

does anyone knows how to bring the complex number at hold?...I can't really solve the questions

[cunning lion]

do you have a specific example of a question where you are stuck? that is kind of vague

[pearl hazel]

LOL thank you, 2 days later i managed to memorize it

[wise shoal]

hello, can someone teach me mass point geometry? i really don’t understand

[cloud laurel]

Can postulates be proven?

In geometry?

[cunning lion]

no

there are always some set of statements you must take as true in order to prove further statements

so you can prove a postulate true only if you start from some different set of postulates

[stark pulsar]

Guys what was/is your guyese geometry honors grades

[whole flame]

Use #chill to ask

[whole flame]

Yepp by indirect method

[dark sparrow]

no, a postulate's entire PURPOSE is to be a STARTING POINT for any subsequent proofs.

what you may be able to prove is that some postulate holds in a model of geometry that you're constructing

[dark sparrow]

??

[summer cradle]

to find OK without Apollonius Theorem you can make use of the Cosine Rule

Let <FKD = x, then <EDK = 180-x

Triangle FKD:
5² + 3² - 2×5×3×cos(x) = 6²
34 - 30cos(x) = 36
-30cos(x) = 2
cos(x) = -1/15

Triangle EDK:
5² + 3² - 2×5×3×cos(180-x) = EK²
34 - 30×-cos(x) = EK²
34 + 30cos(x) = EK²
EK² = 34 + 30×-1/15 = 32
EK = sqrt(32) = 4sqrt(2)
OK = EK/2 = 2sqrt(2)

Triangle ABC is similar to triangle ODK with ratio 3:1

So AC = 3×3 = 9, AB = 3×3 = 9, and BC = 2sqrt(2)×3 = 6sqrt(2)

[tough heart]

Can smb help me with this pleassseee

[plucky sierra]

ok

ill try

[summer cradle]

Hint: Divide into smaller shapes then use Trigonometry to find the lengths of the unknowns

[plucky sierra]

yes

listen to him

[tough heart]

Okay I’ll give that a try

[plucky sierra]

make 2 triangles

and also calculate the square's area

[tough heart]

Where 😭

[plucky sierra]

hold on

there

[tough heart]

Ooooh okay

Gotcha

I’m stuck after

I lowk don’t jnkw the next step

Nvm I got it thank you for the help I appreciate it!

[plucky sierra]

wait

nvm

[plucky sierra]

alg

[tough heart]

I might need ur input though just in case if that’s okay

[plucky sierra]

uhh

wait u got all the areas right?

[whole flame]

Ahh

Nothing

I thought he was asking abt theorems

I just read the proof part

Didn't see it's axioms

[plucky sierra]

kk

[shut thorn]

There's actually a proof for this in our book. I can send it here.

Nvm. My network disagrees 💔

[whole flame]

Divide the figure into different shapes

Try to divide into triangles, square and rectangle

Drop perpendiculars to do so

Use basic geometry to deal with triangles

[upper karma]

Could you send it if you can please?

[shut thorn]

Yeah sure. Lemme check if my network agrees.

It was split onto two pages so the photos are kind of janky.

[upper karma]

Thanks so much dude!

[night cape]

hi, why does the unit in "unit circle" imply that the radius will be one?

come to the of it even in unit vectors its implied to be one too

[shut thorn]

The word unit itself means one.

[night cape]

OHHHH

[stark spear]

1 is used as unit probably because its the multiplicative identity element and just makes things easy

[night cape]

That is good to know

thanks guys

[shut thorn]

No problem. Just tell me if you don't understand something.
The way the book writers wrote it is kinda confusing sometimes.

[thorny nova]

Why coordinates are cos and sine of theta and respective values

[real sentinel]

What's the doubt

[thorny nova]

No it's just demonstrative

No doubt

[obsidian harness]

that's the unit circle definition of sin and cos

[thorny nova]

I just thought of posting it

[obsidian harness]

it's a definition

[thorny nova]

Ya

[obsidian harness]

cos(theta) = opp / hyp

if you now let hyp = 1, then the opposite side must be cos(theta)

[wintry ibex]

rs aggarwal

[shut thorn]

yep i have it

having doubt in these two.

and in this question. Its the answer none of them because a square is also a rhombus and a rectangle is also a ||Gm?

[obsidian harness]

that's correct

[obsidian harness]

so they are saying that DB = BC for example

but what do you know about sides DC and BC?

[obsidian harness]

and given that all four angles sum to 360 degrees, you should be able to replace the variables using the equalities, then rearrange for angle DGH

[shut thorn]

ohhh so its an equilateral triangle. Which means C is 60. and A is 60. And similarly ADB is also equilateral. So that means that the angles are 60,120,120,60

[obsidian harness]

yep!

[shut thorn]

got 'd' for the second one

[obsidian harness]

yeah, d is correct

nice work!

oh lmao there's an obvious typo in the question statement

which of the following is true about DGH, it should be

[shut thorn]

yeah

thats like the only reason i got confused 😭

now this question.
First of all, no figure. So i made this figure

you have to show that angle EOF is 90

so do you show angle DEA and CFB 45?
Or do something else entirely

[worthy tendon]

Those two angles would be 45 only if the rhombus is a square, which is a very specific and special case. So that won't work for your proof.
You can also have a situation when those angles are 30 and 60 deg, and so on.
So you shouldn't try to impose additional conditions on the problem

[shut thorn]

oh wait its a rhombus

lemme see

wait

what if i

construct the diagonals?

[worthy tendon]

do you have a plan in mind where those diagonals would become useful?

[shut thorn]

showing that AC and ED are equal?

[worthy tendon]

mhm

[shut thorn]

cant think of anything

[worthy tendon]

well, extending your argument from here, you can assume angle DEA = x, and try to prove that the angle CFB = 90-x.

[worthy tendon]

also, for your figure, I suggest making the ticks the same for AB and AD, since ABCD is not only a parallelogram, but more specifically a rhombus

[shut thorn]

yeha

lemme send the new pic wait

oh wait

in this. Since DBCF is a ||Gm, Angle CDB and angle CFB are equal

angle CDB is 180 - (90 + angle PCD)

and DEA = Angle PCD (from ||gm DEAC)

so if i take angle PCD as x. And DOC as y.
90 - x + x + y = 180
-> 90 + y = 180
-> y = 90

[shut thorn]

thanks. I did it

just tell me if theres a problem in my method,

[worthy tendon]

wait a minute

[shut thorn]

because im sure there is one

[worthy tendon]

90 - x + x + y = 180
where did this come from?

like whats 90 - x?

[shut thorn]

thats after i simplify 180 - (90 + x)

it becomes 90 - x

[worthy tendon]

oh, okay, so you are using the result that diagonals of rhombus are perpendicular?

[shut thorn]

yep

[worthy tendon]

well, that works

good job

[shut thorn]

thanks

[worthy tendon]

I was thinking something simpler, like let angle DEA = x, so angle EAD = 2x (since EA = AD makes the triangle isosceles).
That leads to angle CBF = 2x (from parallel lines)
And since triangle CBF is also an isosceles triangle, you can calculate angle CFB = 90 - x, which would lead to triangle EOF be right angled at O

[north kite]

the hell

[honest dock]

[thorny nova]

Guys I can't solve this

Pls help

[exotic yarrow]

In the future, please show what you've done so far when asking for help (it gives us more context to work with and prevents us from wasting our time explaining things unnecessarily). \ \

What I would do here is divide both sides of $a sin^2 \theta+b \cos^2 \theta=c$ by $\cos^2 \theta$ to obtain
$$a \tan^2 \theta+b=m \sec^2 \theta.$$
Consider how you can use this to solve for $\tan \theta$ (which is particularly relevant since we can use that in the third equation $a \tan \theta=b \tan \phi$).

[somber coyote]

Civil Service Pigeon

[remote shale]

Let there be a triangle ABC where cosBAC = 1/7. The tangent of the circumcircle of triangle ABC at point C intersects the tangent of the circumcircle of triangle ABC at point B and point A at point D and F respectively. The line AD intersects the circumcircle of triangle ABC at E. It's known that cosEBC = 1/2. There is a point G that lies on the side that is different to point B facing AC such that AF is parallel to BG and angle AFC = angle FGC. If the value of FG/FA = a\sqrt{b} / c where a and c are coprime natural numbers and b is a natural number that is not divisible by any perfect square other than 1, find the value of a + b + c

can anyone help? im unfamiliar with these types of problems so I havent really made any useful progress on it

[hollow lily]

Do you have a diagram?

[hollow lily]

Btw the problem is kinda heavy you should consider open a help channel

[remote shale]

i tried making it in geogebra however im having trouble with where to place point G as angle afc and angle fgc always looks very different in size

this is what i have

[remote shale]

alright wait im gonna try and open

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

rewriting the question so it reads more easily ^

better diagram: https://www.geogebra.org/calculator/hqpzusnr

In the future, please show what you've done so far (even if you don't think it's relevant, a quick list of your observations can save us a lot of trouble since we don't waste time explaining things unnecessarily).

@remote shale

[exotic yarrow]

see if this helps though

since it looks quite different from your diagram

I gave you the link so you can test if the diagram actually satisfies the diagram if you want to

[exotic yarrow]

oh I should probably give you a hint just in case

prove that ||FG/FA = AB/AC|| and ||AB/AC = BE/CE||

I don't know which of these should go first so

flip a coin ig

[frosty mesa]

Can someone help me verify if this curve drawn in my notebook is a Jordan curve?

And if it is a Jordan curve, can it provide a counter example for Inscribed square problem?

[nimble hearth]

👍

[remote shale]

okay i understand

[exotic yarrow]

[remote shale]

hm yes the point G placement looks VERY different

[fathom kraken]

Is my whole approach to the question incorrect- (i feel its just the way im thinking abt the question)

[elder yacht]

can someone help me with my math homework im confused:

[fathom kraken]

just apply exponential laws

[elder yacht]

ya but i dont know what that is

how do u know

[elder yacht]

oh

wow

thats it?

[fathom kraken]

yup

[elder yacht]

oh nvm thats easy how did u know this stuff bro ur smart

[fathom kraken]

like the first question is a mix of 1 and 5

[fathom kraken]

thxs

[elder yacht]

i see this makes sense

thank you

btw why do these rules work?

[tacit perch]

$a^n = \underbrace{a\times a\times a \times … \times a}_{n \text{ times}}$

[somber coyote]

GigaGoat

[elder yacht]

yes i understand that

[tacit perch]

So $a^n \times a^m = \underbrace{a\times a\times a \times … \times a}{n \text{ times}} \times \underbrace{a\times a\times a \times … \times a}{m \text{ times}}$

[somber coyote]

GigaGoat

[elder yacht]

ohhhh that makdes sense

[tacit perch]

Ikr

Cus it’s like

a multiplied by itself

m+n times

[elder yacht]

r u also in 8th grade learning this stuff??

[tacit perch]

🗣️🔥

[elder yacht]

i am in 8th grade i have 86 rn

im doing pretty good

[tacit perch]

👍

I’m not in 8th grade tho

[elder yacht]

oh ok

[tacit perch]

Anyways

Try to use the above reasoning to figure out why the other rules work

[elder yacht]

ok i will thx

[upper karma]

Oohh, this looks fun

[tacit perch]

Lmk if u get stuck

[kind torrent]

Messi goated

My friend akhilesh is also messi fan and good at math

[whole flame]

Use #chill

[summer cradle]

just curious is this an olympiad problem?

[summer cradle]

which direction is the current flowing?

[whole flame]

The answer is cos^-1 (1/2) ??

[fathom kraken]

answer is 60 degree

[fathom kraken]

uh down? (gravity)

i got my mistake- the two shld be at the slope

[rapid glade]

Hello

Where can I get help for a geometry problem?

[whole flame]

Ohh yeah so cos

Right

[real sentinel]

Shoot

[whole flame]

Actually there is a direct formula for such problems

U can derive it

[rapid glade]

#help-6 its here

[whole flame]

It's cosx = Vr/ V man

Vr is velocity of river

Water

[fathom kraken]

right

oh yea

got it!

thanks

[whole flame]

🙏

[upper karma]

Hey guys, I’m not sure if this applies to geometry and trigonometry but what is a vector in mathematics? I know in physics it is just something that has a size and direction but is that the same in mathematics?

[hollow lily]

kind of

In math vector has direction and magnitude ( length )

So it's kind of the same thing, Finn

[upper karma]

yep it is the same

anyone in 10th rn?

[thorny nova]

Listen, maths is a tool. Vector is also a tool applied in physics just as u apply calculus in physics as such. Vector is in the Cartesian plane coordinate system

[thorny nova]

U use it as atool in physics

For example

[hollow lily]

Well it's not completely the same thing tho, like how force in physic is fixed in place sometimes

[thorny nova]

In projectile motion

U take u as a vector

And use respective components

Of u sintheta and u costheta

[thorny nova]

U know

Physics is like a sister to maths

Tbh

Force is what

Mass times acceleration

What is acceleration

dv/dt

Or simply, velocity /time

So maths is involved

But u use it as a tool

Like without multiplication or division

None of this would exist

[hollow lily]

math is involved in anything bro

[thorny nova]

Exactly

Thank you

For understanding my point

[shut thorn]

WHY do I have to use the midpoint theorem in questions ABOUT CIRCLES?

[hollow lily]

Math is math

[shut thorn]

Why does this question use the midpoint theorem man 😭

[hollow lily]

Math is math mate, we have anything we make use of it all

Like how coordinate geometry is more algebra than geo

[shut thorn]

💔

[simple vigil]

Because it's property in mid point theorem
That the line which is parallel is half of base

[shut thorn]

Yeah I know.

[simple vigil]

Then what's the problem?

[shut thorn]

Nothing.

But now I am curious.

Os there any other way to do it or no?

Is*

[simple vigil]

What you did ?

Used mid point theorem

[shut thorn]

I did the midpoint theorem one

[simple vigil]

To show AD=DB and CO=OB

[shut thorn]

Is there another way?

[shut thorn]

That's just a thing because OD is perpendicular to AB

[simple vigil]

Yeah perpendicular to chord bisect the chord

And CO = OB both are radius

This is the easiest approach

[shut thorn]

Yeah.

[kind torrent]

Are you from India?

[shut thorn]

?

[frosty mesa]

Is there any constants besides π,√n and golden ratio that can be estimated using geometric construction?

[obsidian harness]

you're looking for https://en.wikipedia.org/wiki/Constructible_number

I notice you used 'estimated'

there are some constructions which require tools beyond a straightedge and compass, for example using origami folding or using a marked ruler (neusis constructions)

these are also exact constructions

and then there are also inexact constructions like this https://www.youtube.com/watch?v=L1EMeqg7TIk

[strange pond]

midpoint theorem isnt entirely necessary here

[frosty mesa]

Like what are the examples of exact constructions?

[obsidian harness]

read this page

[strange pond]

you can use the converse of the corresponding angles property for the first part, and show that the triangles ACB and DOB are similar for the second part

[frosty mesa]

Also what is your point on Rational distance problem for 4 sided polygon?

[obsidian harness]

what do you mean by 'what is my point'?

I never mentioned anything about that

[frosty mesa]

Your point is like your opinion about it

And the topic is about rational distance problem for 4 sided polygons

[obsidian harness]

I don't care

https://math.stackexchange.com/questions/4839673/four-rational-pythagorean-triples-in-a-square

you might be interested in this related problem if you're asking though

[kind torrent]

The chapter’s name is Circles, a chapter in 9th and 10th grade here

And the book appears to be some extra guide book

[obsidian harness]

if you've already proven the midpoint theorem somewhere else, that gives you the result straight away

https://www.youtube.com/watch?v=2TK0DOawX7I

actually this theorem is pretty powerful

I've used it to solve this JEE problem

[shut thorn]

Fr?

I knew there should be another way.

[strange pond]

read this @shut thorn

[shut thorn]

Ohhh.
Thanks.

[obsidian harness]

that's basically how the midpoint theorem is proven

[strange pond]

oh lmao

[obsidian harness]

there's another way where you use the fact that the perpendicular of an isosceles triangle to the base cuts the base at its midpoint (OA = OB so OAB is isosceles)

so then since angle ABC = angle DBO, SAS similarity tells you that AC/DO = AB/DB = 2

[strange pond]

oh yeah perpendicular bisector of chord = radius

[obsidian harness]

mhm

that's just a property of isosceles triangles if you think about it

[obsidian harness]

this one is more elementary

[runic bay]

Guys i just found weird proof about x^-1 = 1/x. I understand it now

note i did not make it on my own i googled it

wait this isnt the right channel my bad sorry

[hollow lily]

maybe in #prealg-and-algebra or #precalculus

[frank pebble]

Do Leibniz's geometrical theorem about the centroid in a triangle, which states:
If M is the centroid in ABC, and P is a random point, then:
PA²+PB²+PC² = MA²+MB²+MC²+3*PM².

And one of the vector equalities, in particular:
If M is the centroid in ABC, and P is a random point, then:
PM = 1/3(PA + PB + PC), where all of those are vectors

have anything in common? (can one maybe be derived from the other, even?). Since they have the same conditions, and in both equalities it can be made so that there is a 3 being multiplied by PM.

[exotic yarrow]

can one maybe be derived from the other, even?
yes ofc

$\norm{p-a}^2+\norm{p-b}^2+\norm{p-c}^2=\norm{m-a}^2+\norm{m-b}^2+\norm{m-c}^2+3\norm{p-m}^2$

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

Standard expansion yields

$$3\norm{p}^2-2p \cdot (a+b+c)+(\norm{a}^2+\norm{b}^2+\norm{c}^2)=3\norm{m}^2-2m \cdot (a+b+c)+(\norm{a}^2+\norm{b}^2+\norm{c}^2)$$

[forest dew]

@old raft @grave pond

[somber coyote]

Civil Service Pigeon

[old raft]

hi

[forest dew]

send question here

[forest dew]

trapop can help

and estelle

[placid star]

hiii

[exotic yarrow]

And hence $(p-m) \cdot (3m-(a+b+c))^2=0$

[somber coyote]

Civil Service Pigeon

[forest dew]

hm if so

[exotic yarrow]

thus m = (a+b+c)/3

[forest dew]

6x3= 18 the white area?

[exotic yarrow]

@frank pebble apologies for the segmented response, I got cut off above

[old raft]

I mean you could go do it

[lime crown]

Please do not ping individual helpers unprompted.

[old raft]

ermmmm

[grave pond]

That is still not how you compute the area of a trapezium.

[forest dew]

I need learn

anyways

[placid star]

you can find both areas and subtract

[old raft]

@forest dew 1/2(sum of parallel sides)(height)

[exotic yarrow]

minor remark, but in the future, please try not to cut ppl off in they're in the middle of typing something (see above)

[forest dew]

alright.

[old raft]

and we're moving to help channel because I'm severely burdened by guilt from interrupting other people working

[placid star]

have you been introduced to areas yet

[forest dew]

valorine i thought this is what we see in math

sorry but this is not what we see in math

i realized now

Ill just watch

[grave pond]

Huh? There's no reason why only one person can be typing at the same time.

[frank pebble]

no worries. So after m = (a+b+c)/3, we know M is a centroid, and from there we derive the vector equality as usual?

[grave pond]

Whoops, I think I misunderstood your point here. Sorry.
Still, having parallel conversations is the kind of things that can happen in the topic channels; that's how they differ from help channels.

[winged oyster]

hi

[dusk cape]

hey!

[stoic sequoia]

Can Somebody help me?

[grave pond]

You'll never find out unless you ask a question.

[dusk cape]

depends... whats the question?