geometry-and-trigonometry

[dense escarp]

Yeah maths is surprisingly very fun

I started 4 months ago because I could hardly divide and I wanted to know what people were talking about with all the X and Y and Z and symbols

[twin depot]

Idk why but trig has been fun for me even when I hated geo out of all classes

[dense escarp]

I’ve gotten into high school geometry so far

[twin depot]

ahhh nice

Geometry is kinda a bummer

[dense escarp]

It’s application to the real world for me personally

You can use it in aviation to find distances and slants

[twin depot]

Yeah it’s interesting

[dense escarp]

Yep been struggling with it

[civic rock]

what did you say? 😡

[twin depot]

It was not bad but fucking quadrilateral messed me up😭

[civic rock]

what's wrong with quadrilaterals

[twin depot]

You had to remember stuff like how it’s a rhombus or if it’s a parallelogram🫩

[civic rock]

oh I thought you're talking about random quads lol

or cyclic quads

[twin depot]

No their fine

[dense escarp]

Cyclic quads?

[civic rock]

quadrilaterals that are inscribed inside a circle

[twin depot]

A quadrilateral inside of a circle iirc

[dense escarp]

Ah

[civic rock]

mmmhm

I remember one time getting scolded because I didnt remember the property of intersecting circles

not the regular perpendicular bisector bullshit but it was a special theorem that was key to solving a difficult problem

[dense escarp]

What makes math interesting for you guys?

[twin depot]

I basically find it like a sandbox. It’s interesting seeing how things work like I would spend an hour on desmo and calculator seeing how theorems work

[dense escarp]

Me too,alongside how well it can describe things in the world,and it’s applicability

It never feels like a waste of time to do mathematics

It’s also like a puzzle and it feels prestigious to do higher level stuff knowing your hard work learning the building blocks got you up to that point

[dire jackal]

the puzzle solving part of it. it feel great to apply basic knowledge and concepts and mix them together to do difficult questions

[dense escarp]

Yes,and finally nailing some problem or concept you’ve been wrestling with feels amazing,like it’s trying to call you dumb and you just proved it wrong by sheer willpower and mathematics

[clear sleet]

thanks!

[agile glen]

CAN ONE GIVE ME TIPS TO TAKE MY TRIGONOMETRY AND CONIC SECTION FROM BASIC TO ADVANCE

[pale sentinel]

DRAW BIGGER CONES

[In seriousness though that's far too broad a question]

Assuming you're a high-school student (by this I mean not a university student), you can probably look up textbooks

[dire jackal]

yo so you prepping for jee right

id assume you know hindi?

theres no other way other than watching lectures and doing problems

there are a lot on youtube

see which teacher you like the best and watch the lectures

and try to find questions

[agile glen]

THANKS BRO

[agile glen]

OK BRO THANKS

[pale sentinel]

[Also - is your CAPS lock broken or something?]

[upper karma]

I THINK SO

[civic rock]

his shift key is locked

[pale sentinel]

His key lock is shifted

[civic rock]

[visual flume]

Lock's shift is keyed

anyways help for 1000000 robux

\

ok i got it nvm

wait does this mean i get 10000000 robux

[prime gyro]

What am i forgetting about this?

[civic rock]

for angle b you can use sum of angles in a triangle

[prime gyro]

buut how do i figure that out when all i have is the 90

[civic rock]

no look at the big triangle

[prime gyro]

OH CRAP

I SEE IT

thank you so much

[quiet herald]

6 x 1.41

[quiet herald]

6 x 1.41

[quiet herald]

I don't know if it's correct but I think it's likely

[civic rock]

your solution is def wrong

[quiet herald]

I tried it xd}

xdxd

[quiet herald]

what is the answer

[summer cradle]

4sqrt(6) ?

[west narwhal]

Nice problem 🙂

[alpine cargo]

its a basic pythagorean. c^2=a^2+b^2

[dark sparrow]

!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.

[dark sparrow]

for future reference...

[dark sparrow]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[alpine cargo]

idk why but ts pmo now.

[deft sail]

Guys I just thought of something

Why have so many product to sum identities when you can use a few

Example cos(9x)sin(8x)

We can change cosine into sine

sin(π/2 -9x)sin(8x)

Or sin(3x)-sin(x)

We can make it positive

sin(3x)+sin(-x)

I feel like the other identities are very useless(kind of)

[upper echo]

Taking your argument to the extreme, I could say that all trigonometric identities are useless, as the definitions of sine and cosine alone are enough 🙂 But we don't do that, because it's much faster to use some suitable identity rather than deduce it every time it is needed. So, it is just faster and more convenient to remember as many identities as you can.

[deft sail]

Yeah I guess you’re right

[dark sparrow]

it's not about anything conceptually new but about practical convenience

[karmic rover]

Was doing a physics question and it says "inclined plane with inclination 1 in l." What does this mean? Does this refer to sin(x) or tan(x)?

[grave pond]

Usually it would mean the slope is 1/l.

[upper echo]

well, something went wrong 🙂 bc the chord AB is definitely supposed to be shorter than the diameter DC which is 12.

[upper echo]

This is exactly the case when a coordinate solution is fairly short. ||If the larger circle is given by x^2+y^2=6² and r=AB/2, then the x-coordinates of the intersection points of the smaller circle with the x-axis satisfy the equation x²-x+(6²-2r²)=0 (by Pythagoras) and we know that x1=-3 and x2=4, so r^2=(6²+3*4)/2=24 and AB=2sqrt(24)=4sqrt(6).||

[visual flume]

lmao why so many replies

but anyways i also got 4√6

thanks guys

[brittle sluice]

In convex pentagon ABCDE, AB = 1, BC = 2, CD = 3, DE = 4 and
m B mC mD = 135 . If x = AE, compute the ordered pairs of integers (p, q) such
that x^2= p+qroot2
can u help me solve it please !
btw i am new here 😊

[upper echo]

||Draw line BF parallel to CD. The rest is easy by Pythagoras + law of cosines ||

[civic rock]

||Construct a rectangle like the following. The rest is Pythagoras||

[brittle sluice]

@civic rock i didn't quite get it dude

[torpid current]

just a curiosity, in trig if we get confused from a question, could we make the trig functions a variable such as x or y?

nvm i dont think its a really good idea if we do that all the time

[gentle canyon]

guys yk when they say in a question that circles are touching does that mean they are tangent or can they intersect

[knotty quiver]

my first thought would be tangent

[grave pond]

Generally tangent.

[gentle canyon]

o ty

just making sure

coz there was this question i did where there were 2 circles with the same centre but different radii and they touched another circle

[grave pond]

Sounds like the latter circles lies between the two concentric ones.

[rancid dragon]

I need some easy to follow trigonometry books, that makes me understand concepts, and has cool and challenging problems, and also covers the high school trigonometry syllabus

[hearty wagon]

hi

[brittle sluice]

hello

[obsidian harness]

yeah, 'touched' is the common English translation of 'tangent'

cause tangent is a Latin word

in Latin tangent literally means touching
||(it's the accusative of (linea) tangēns for all you linguistic nerds)||

[gentle canyon]

the question was smc 2020 i think

lemme check

yeah question 18 of SMC 2020

[visual flume]

!status

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[visual flume]

i js wanna verify, is 55º correct?

[ancient crescent]

Give me like 3 minutes and I gotcha

So @visual flume your trying to find Angle RPK correct?

So you have 2 known angles, which are KMR which is 134° and the other is QMK which is 20°

[visual flume]

wait what

so what i did is draw the center O

and kom is 134º

and qom is 20º

wait

i js got 56º

i never saw the mr=mk thing lmao

well based on iscoceles kmr rpk would be 56

but if i use inscribed angles

its 55

[grave pond]

Due to isosceles, the arc RQM is 134°, so we have angle MKR = angle MRK = 67°­, right?

Bu angle sums, then angle RMK = 46°, and angle RPK is 20°/2 more than that.

What was the argument that gave you 55°?

[visual flume]

i used inscribed angles to get 55

but i also got 56 from iscoceles

wait have i seen u before

anyways

if you have circle center O, ROM is 2 times RKM

and draw ROM and its 20+y

its kinda confusing

and then ROK would be 206-y

so RKO would be 103-y/2

and if u add RKO to OKM u get 126-y/2

and u get the equation 252-y = y+20

y= 116

but its prob wrong

[grave pond]

Well, your y should be 114 because the chords MR and MK are equal, so they span equal arcs.

[ancient crescent]

Aren't you just trying to find the angle RPK?

[grave pond]

Currently I'm trying to understand how No Stop got 55° for angle RPK.

[ancient crescent]

Me too 💀

[ancient crescent]

May I dm you?

[grave pond]

Can't you say what you have to say here?

[grave pond]

Wait, why would angle RKO be half of angle ROK? It is RMK that is half of ROK.

[ancient crescent]

It was about the actual answer. But sure I can just say the equation.

[grave pond]

I agree with 56° as the actual answer.

ROK is 92° and RKO is 44°, close enough to half of 92° that the mistake might only make a small difference at the end.

[zealous trout]

Hey can somone help me with mid point theorem

[visual flume]

me too 😭

[visual flume]

wait

hold on

i think im tweaking

[visual flume]

oh sorry i meant (180-260-y)/2

OH

wait thats where i got it wrong

i forgot to subtract from 180

dang i lowk sold that

yeah so basicallly i forgot to subtrac t180 and got 55

its 56 either way then

thanks for helping

[visual flume]

bro

darkest diagram ever

anyways if u look closeley u can see some lines and stuff

so in this diagram ce = eb, cd = ad

de = 1/2 ab

thus de || ab

[zealous trout]

Ye i appreciate the diagram but i understand the basics but got a super complex midpoint theorem question i opened ticket in help so just waiting there ig i could upload it here if you want

[visual flume]

sure

ill try but fail

[zealous trout]

a) Calculate length of CE

[zealous trout]

hahaha

[visual flume]

60?

cuz

if u take

ehm

[zealous trout]

ye

[visual flume]

triangle bfd

u see that it has a midpoint thing

[zealous trout]

omd

[visual flume]

so bf is 15*2 = 30

[zealous trout]

my eyes have been opend

[visual flume]

yessss

wait dy play doom

[zealous trout]

Tysm

[zealous trout]

nah

[visual flume]

oh

[zealous trout]

currently 1am and got school tmr

[visual flume]

cuz thats one of the speech lines in it lmao

[visual flume]

rip

[zealous trout]

lmao

[zealous trout]

yeppp

[civic rock]

<@&268886789983436800>

[jagged geyser]

So for Self Study, Should I use one of the AOPS intro to Geometry books to learn the geometry curriculum but be well prepped for competition math?

[twin depot]

I would recommend to use the state standard curriculum textbooks first.

[upper karma]

BF=2GE,CE=2BF
=>BF=2(15)=30
=> CE=2(30)=60

[upper karma]

I found the answer, it is so fun

[upper karma]

i mean, it took me more than an hour

[maiden brook]

I think that book is very good, yes

[jagged geyser]

So like glencoe?

[jagged geyser]

Thank you for your response actually TBH It was kinda hard to determine which geometry book to get or find a pdf of and read as I wanted it to be rigourous yet still follow a curriculum so thank you so much.

[twin depot]

If it follows your state curriculum then yeah. The only reason why I’m recommending it to follow your state curriculum as you’ll be knowing the necessity of what someone your school have already learned.

Haven’t seen AOPS book yet might look at it later

[jagged geyser]

So if you want the information I can give it but essentially the AOPS textbooks are like harder versions of your typical geometry textbook and introduces more harder problems seen in competitions in math such as AMC for example. These do cover a geometry curriculum though unless you look at their TOC the website doesn't really explain the TOC that well of them.

[twin depot]

Ahh then yeah if you like those advance stuff then go for it. Since it’s going in depth then yeah it will likely cover everything you usually need

[meager flax]

Here's a hard math question

[grave pond]

Very not to scale.

But simple enough if you start by drawing DEF and then extend EF to AC approporiately and note down the various angles that are forced by the given information.

[iron orchid]

hey guys im new to trignometry and i have a question

i was calculating sin of A and uhh sin is perpendicular/hyp but i need to do sin of A so how do i know whats the perpendicular relative to angle A?

what am saying is that perpendicular for the sin of a will be the side adjacent to a or the side opposite to a

[knotty quiver]

what is A? angle?

[iron orchid]

yeah

[knotty quiver]

is A and a the same?

[iron orchid]

angle A

[cosmic rover]

bru im retareted how tf u do ts

[iron orchid]

yeah sorry😭

[knotty quiver]

the only formula for sin i know is opposite side / hypotenuse

[cosmic rover]

CDE and ADE are equal

[iron orchid]

ohhh tysm

[knotty quiver]

idt angles can be perpendicular. only lines do

when lines are perpendicular, they form a 90° angle

[knotty quiver]

,, x+y=360^\circ-2(\bullet)-2(\measuredangle) \
z = 180^\circ - (\bullet) - (\measuredangle)

[somber coyote]

glass

[knotty quiver]

,align x+y &= 360^\circ-2(\bullet)-2(\measuredangle) \
&= 2(180^\circ-(\bullet)-(\measuredangle)) \
z &= 180^\circ-(\bullet)-(\measuredangle) \
2z &= x+y

[somber coyote]

glass

[knotty quiver]

quadrilaterals always have a total internal angle of 360° and for triangles 180°

[upper karma]

Let <BAE=a and <CDE=d
=> a+d+z=180 => 2a+2d+2z=360
=> 2a+2d+x+y=360 => 2a+2d+x+y=2a+2d+z=>
2z=x+y

[potent furnace]

guys check out my geometry-based mario game i made in desmos

https://www.desmos.com/calculator/c4fa6c8d1a

[dire jackal]

respectfully i hate you

but good bait

[graceful crag]

Was genuinely excited 😭

[mystic umbra]

never trust links 💔

[cosmic rover]

oh thank you

[cosmic rover]

ty

[edgy osprey]

Nah man, i suck at geometry

Am i weird that i can do trigs but can't do this kind of geometry😭

[knotty quiver]

what geometry

it's not weird I'm kinda like that

[edgy osprey]

Like this kind of geometry

[knotty quiver]

maybe trig is actually is easier than geometry bc trig is limited to triangles and circles

[edgy osprey]

Yes

like geometries in like andy's video (andy math)

[fresh birch]

Well that is applied trigonometry within geometry

[knotty quiver]

it looks easy

[edgy osprey]

I just need to learn angle theorems to improve my geometry i guess

[edgy osprey]

For me it's hard💀

[fresh birch]

It's about learning a little bit of everything. I recommend that you don't be mechanical and that whenever you see a theorem, axiom, etc. try to demonstrate all those properties. That way, you will hardly forget those theorems and you will progressively improve in geometry.

[knotty quiver]

for that particular problem i used facts like supplementary angles and complementary angles and a triangle always have a total internal angle of 180° and special triangles like 30°60°90° triangle

a flat line has an angle of 180°. if a line divides it, the resulting two angles will have a sum of 180° (supplementary angles)

[edgy osprey]

Yeah, i just need to do a lot of practices, but currently i am just learning calc 2, is it wrong that i learn calculus without even mastering geometry (and i left behind linear algebra [like matrices, Eigenvalue])

[knotty quiver]

idt u have to really master geometry

u might feel the need to go back to some concepts ig

[edgy osprey]

Yeah, you are right

[fresh birch]

Geometry doesn't have much to do with calculus, but if you're learning Calculus 2, you should master functions, limits, series, and more advanced trigonometry.

[knotty quiver]

maybe geometry like conic sections

[fresh birch]

clear geometry of space

[knotty quiver]

but mastery isn't mandatory

[fresh birch]

the calculation is applicable within that

[edgy osprey]

I haven't master series yet, currently i just want to integrate everything😂, like i still learning to derive the integral of arc hiperbolic functions

[knotty quiver]

i haven't mastered series either

i just know taylor/maclaurin series and geometric series

but I don't fully grasp the concept of convergence/divergence

[fresh birch]

It is not that difficult either, summative, productive, convergence, divergence, etc.

[knotty quiver]

or at least I don't, calculationwise idk

[edgy osprey]

I don't even know what taylor series and mclaren series are

[knotty quiver]

basically any function can be written in taylor series form

it's fun to do

[edgy osprey]

Ahh, so you can convert them back and forth

[knotty quiver]

I've forgotten how to write it lowkey

,w taylor series

[somber coyote]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by usrcard

[edgy osprey]

I learned a bit about the convergence and divergence tests from the OG chemistry tutor guy

[knotty quiver]

it shows how any function can be written as a series of polynomials

,,\sum_{k=0}^\infty\frac{(x-a)^kf^{(k)}(a)}{k!} ?

[somber coyote]

glass

[knotty quiver]

maclaurin series is taylor series where a=0 (radius?)

[knotty quiver]

that's neat

[edgy osprey]

Yeah i think, i'll learn those two series rn, thanks bro!

and the chemistry tutor guy already has videos about it👍

[knotty quiver]

idk if i wrote it right. lemme check by trying jt on sine or exp. i kinda remember their series expansions

-# it's correct

[knotty quiver]

i think it's fun if you try to derive it yourself

i forgot the concept behind it tho. like u start with linear approximation oh yeah

basically taylor series is a series of approximations of the function

starting with linear, then quadratic, then etc. to infinity. The more terms, the better the approximation is

#calculus

[maiden brook]

that’s not true

[knotty quiver]

ah yeah, probably non differentiable ones

since the series involves derivatives

[potent furnace]

e

[meager flax]

You're correct

I have more and harder ones

[meager flax]

U should've said lim x---->♾️ (1+1/x)^x

[upper karma]

where the hell is geometry problem

[meager flax]

Bro it's in persian I need time to translate it

Do u want matrix or just shapes problems?

[fresh birch]

send them so people can try them

[meager flax]

Hm sure

Everyone

I spent a lot of time translating this question to English so u better solve the shit out of this question

Btw 18 is a wrong option it was supposed to be 10 gpts mistake

[obsidian harness]

triangles ABF and BEC are similar so $\frac{BF}{CE} = \frac{AB}{BE} \implies \frac{x}{\frac{2}{3} s} = \frac{s}{2x}$

[somber coyote]

south

[obsidian harness]

hang on, but isn't the question at fault cause it implies AB = AE?

[meager flax]

Noo it doesn't say AB=AE

Lemme give you a hint

You must draw a segment from a point to another

[obsidian harness]

oh that's not a right angle, my bad

[fresh birch]

Sometimes the figure looks like something else, don't be fooled by the graph.

[fresh birch]

xd

[upper karma]

an inscribe right triangle has area π, prove that it is 3-4-5 triangle

[fresh birch]

ehm a triangle could not have area Pi

[upper karma]

i mean a circle inside triangle has area of π

[fresh birch]

ohhh

[upper karma]

So... do you have an idea?

???

DAMN, i am stuck, should i just answers it is not enough information?

[fresh birch]

Poncelet's theorem

[fresh birch]

Ok, when you tabulate a value of a and find the value of b, the relationship a+b = 5 will always be maintained; so with that you would demonstrate the triangle. What happens is that a and b can have infinite values, but their sum will always remain constant at 5.

Another way to demonstrate it would be to state the statement the other way around, that is, if we have a right triangle of 5-4-3, prove that the area of the inscribed circle is 1

That way it would be easier

[maiden brook]

or non-analytic functions

[knotty quiver]

what is analytic function

[maiden brook]

I think it’s basically just a function where it’s Taylor series converges everywhere to the function

[knotty quiver]

i see

alright that's enough #calculus or #real-complex-analysis right there

in this world of #geometry-and-trigonometry

[maiden brook]

But the point is that not every smooth (infinitely differentiable) function necessarily converges to its Taylor series. usually they don’t bring that up

[knotty quiver]

yea i could see that

like 1/x?

or

maybe u could pick another radius of convergence (idk if that's the term)

[maiden brook]

No

[real scroll]

hello, im just starting to learn geometry, what are some tools that i can use to learn it, other than khan academy, which is what im using now.

[spiral lodge]

Geogebra and/or Desmos

[quaint chasm]

this is probably very well known, but a cool property i found of 'skewed' (i.e effectively rotated) grids is that they have large gaps orthogonally and diagonally between any two points on the grid, despite there being points close by

[meager flax]

Ok guys I'll send the answer now

There are 2 super ways to solve this 1 long way whoch came to mind of people and 1 is a cooler and faster way

Omg

I can't take screenshot of the question cause the site policy

This is gonna take long

[fresh birch]

Lol

[meager flax]

Yo guys I just realized tex it is a girl cause it's pronounce is she/her

[analog maple]

Hi, when a formula says: a sen α. It means a * sen α; or it says the sen α of a? Ty

[upper karma]

quadri oabc is a kite because angle o=b=90, and C+A=180
So you can tell the length of CO=3,OA=4, and OA is radius, and the formula of the quart circ is πr²/4

[visual flume]

wait for that to be a kite then oca and abc would be the same

which idk how to prove

[upper karma]

You don't need to prove by length.

You just need to proof by angles

and that is enough

[visual flume]

i meant angles oca and abc

[upper karma]

You mean, oca and acb?

You don't need to

[knotty quiver]

-# fun fact: there used to be none

[visual flume]

bro

wait

i think

ocba cant be a kite

[upper karma]

Go back learn about the property of a kite

[visual flume]

bro

[upper karma]

Go!!!

[visual flume]

it cant be a kite

look

[upper karma]

IT IS!!

[visual flume]

if it is a kite

oa is 4

[upper karma]

EXACTLY

[visual flume]

then ob would also be 4 because it isa also the radius of the circle

so oba is equilateral

which means ANGLE boa is 60º

[upper karma]

YES

[upper karma]

EXACTLY!!! WHY???

BECAUSE IT IS A KITE

[visual flume]

and due to the fact that coa is 90º, cob would be 30º, and ob would be perpendicular to ca, meaning oca would be 60º

so now, we have a 30 - 60 - 90 tirangle, and skince "ocba is a kite", oc would be 3

meaning the intersection point between ob and ca , may we call it d, cd would be 3/2, resulting in od being 3√3/2, and ob SHOULD be 4 if it is a kite, but we find that 2od = ob => 2(3√3/2) ≠ 4

so it is not a kite

…

[upper karma]

why would cd be 3/2

Why??

[fresh birch]

[upper karma]

Just where you got cd=3/2

there must be a reason

[visual flume]

no way bro

how old ru

search up 30-60-90 rule

[upper karma]

i got the wrong question to say you are wrong, how it is perpendicular?

[civic rock]

one of the ways I see is to use Ptolemy's theorem

[civic rock]

4CO + 3R = 5R
-> R = 2CO
-> R^2 + R^2/4 = 5^2
-> R = 2√5

[upper karma]

how can it be.... Okay, i give up, i used the wrong method.

[visual flume]

what even is ptolemey's theorem 😭

imam search

[civic rock]

if you don't want to use Ptolemy's, you can extend BC and OA so that they intersect at D

And D will lie on the big circle by Thale's theorem

[visual flume]

wait what this is what google showed

[civic rock]

yep

and if you look at OCBA, you see the opposite angles add up to 180°

so it is a cyclic quad

[upper karma]

Hhmm... my memory messed up avain

[civic rock]

so Ptolemy is applicable here

[junior tiger]

pretty good theorem btw

[visual flume]

ohhhh

[civic rock]

I believe it's the fastest way to solve this problem if you don't want to deal with a gazillion similar triangles and ratios

[visual flume]

ohhhhhh

i see

tysm!

i was trying to do it with inscribed angles all the time lol

[junior tiger]

also if it is a problem related to sum of opposing angles in a quadri then pitots theorem is the solution

[upper karma]

I see.

[civic rock]

anyways

wait I found an easy way without Ptolemy

CA = CE = 5

then EO * EA = EC * EB
or 2R^2 = 5(5+3)
-> R = 2√5

[upper karma]

the chord theorem?

[civic rock]

sorta

you can see ∆EOC and ∆EBA being similar

if we don't want to bring any cyclic quad shenanigans here

[upper karma]

Yep, i see that.

[hollow finch]

chat im going into geometry next year is it easier than algebra 🙏

[round beacon]

Eh. For me about the same

but for others idk

[hollow finch]

alright

[upper karma]

This is easier.

[shut pebble]

"Use your straightedge and compass to construct a triangle, square, pentagon, hexagon, and heptagon. How many steps did each take?" i'm stuck

(regular)

[civic rock]

Impossible for heptagon

do you have an idea for a triangle?

[shut pebble]

yes i did the first 4

i was stuck on heptagon

thanks

[maiden brook]

btw, if u r curious which polygons are constructabld

First a Fermat prime is a prime of the form (2^2^m) + 1, where m is a nonnegative integer

Eg m = 2 gives 17 so this is a Fermat prime

[shut pebble]

i know

[maiden brook]

ah

cool

[fresh birch]

XD

[analog maple]

Hi, is this exercise correct? I drew the triangle with the data, so idk if it's correct, the idea is to calculate c and Â. I think C is correct but idk Â. Ty!

In fact the angle is wrong, because I'm calculating the difference between CA and H, but idk how we can calculate te angle

[knotty quiver]

it's hard to differentiate your H and A handwriting

also why not just use c for hypotenuse since it's the variable for it instead of H

[analog maple]

It's better like that?

[west narwhal]

Yes, I think so.

[signal karma]

a^2=b^2+c^2-2bccosa

[analog maple]

¿250^2 = 308^2 + 396^2 -2(308*396) (cos a)? How does this gives the angle of A? It's going to give a huge number

[weary haven]

Is there a special term for an angle θ that are defined such that it is defined for all real numbers θ

[grave pond]

What?

[weary haven]

I mean

in simpler terms is there a term for an angle that allows for measures greater than 360° or less than 0° (for all real numbers essentially)

unlike in elementary mathematics where angles are just defined to be not less than 0° and not greater than 360°

[obsidian harness]

it's the reverse: an angle between 0° and 360° is called a principal angle

when we say angle, we mean that the angle can be any real number

[weary haven]

oh

[grave pond]

That depends a bit on context: in elementary geometry it will usually be implicit that the angles that are being asked about are to be beteeen 0 and 360⁰ -- or even between 0 and 180⁰, ignoring direction.

[obsidian harness]

yeah we don't really need a specific name

it's pretty clear when a trig context will differ from a Euclidean geometry or an angle between two planes context

[grave pond]

It's good to be aware of these implicit differences in convention, despite the lack of generally understood terse words for stating them.

[warped sail]

I am gonna loose it

[grave pond]

Hint: Lots of ||central angle theorem||. All five points ||lie on the circumcircle of the square||.

[warped sail]

Yea i am just plain stupid, so can you explain it to me ?

[grave pond]

Do you know the central angle theorem?

[warped sail]

yea but i cant see any of it here

[grave pond]

Suppose ABCD is already drawn, then the central angle theorem tells you where to find all the possible locations for a point E where the directions to C and D differ by 45°.

Namely, since E is to the left of CD, it must be on a circle that passes through both of C and D, and whose center sees C and D separated by 90°. And that circle can only be the circle that passes through all of A, B, C, and D.

[warped sail]

is that what you mean ?

[grave pond]

Yes.

[warped sail]

oh. yea i get it now

[grave pond]

It doesn't look very circular, but then again the square doesn't look very square to begin with ...

[warped sail]

I drew that myself so its not perfect

Thanks for your help btw

[grave pond]

It's not any worse than the original diagram, is what I said. :-)

And now you can find angle DEA, and then you're almost done.

[warped sail]

the rest is easy, I just didnt see that inscribed angle

[obsidian harness]

yes, it's because angle DEC = angle DBC = 45 degrees, so the converse of the angles in the same segment theorem implies that D, E, B, C are concyclic

and you can similarly observe angle DCB + angle DAB = 180

oh yeah and there's also a unique circle through 4 points if they are cyclic, so the two circles are the same

[warped sail]

no need for all that, since I got angle DEA, and DAE is 90+25 I can minus all that from 180 and get angle ADE

[obsidian harness]

I know just re-explaining why exactly you can draw a circle through all these points

[warped sail]

oh, ok thanks !

[grave pond]

(I didn't go that way, by the way -- instead I looked for a point O on the perpendicular bisector of CD such that angle COD is 90°).

[honest pumice]

Am i wrong or is this exercise is wrong? where x+2w = 500 feet of fencing and my area is x*w but since x+2w = 500 <-> w = (500-x)/2 then A = x(500-x)/2 and 0 < x <500

the author's answer assumes we also add fencing to the river side, right?

[silent plank]

you are wrong

the question says that

x is the length of the side perpendicular to the river

[honest pumice]

ahhh, my bad

[silent plank]

you have the wrong location of x in your diagram

[honest pumice]

i assumed it was parallel, lol

thank god i'm reviewing my knowdledge of calculus before taking the real analysis course

[honest pumice]

i'll assume i'm doing something wrong again.

my cylinder will be composed of 2 circles and a rectangle, the two circles are 2(2*pi*(r^2)) and the rectangle is 2*pi*(r)*h so the total are of material is 100 = 2(2*pi*(r^2)) + 2*pi*(r)*h, if i solve for h i get h = (50-2*pi*r^2)/pi*r so my volume is V = (2*pi*r^2) (50-2*pi*r^2)/pi*r = 100r - 4*pi*r^3

but the result is half that, what am i doing wrong?

maybe i have to assume it's not a rectangle but a square?

[silent plank]

you double counted the circles

[honest pumice]

don't i need two circles to make a cylinder?

[silent plank]

yes, but you doubled it twice

area of one circle is pi * r^2
for two, it's 2 pi r^2
but you had **2 ** (2 pi r^2)

[honest pumice]

ahh, right, thank you, yeah, it's just that forgot the formulae for the area of a circle

i did something like

and counted two triangles instead of one

the area was 2(pi)(r^2)/2 <-> (pi)r^2 not 2(pi)r^2

thank you

[cedar mist]

Is there anyone available to aid in my understanding of trig func verification?

It is part of an exam, so I undestand if it crosses moral boundaries; however, we are allowed to seek help as long as we give credit.

[short osprey]

write sin2x as 2sinxcosx and then write tan as sin/cos then you get 1/cos^2 = sec^2

[cedar mist]

Man, I just realized that I posted the wrong problem, already got the answer SEC^2(theta)

Thank you so much for helping!

[short osprey]

no worries

[mystic umbra]

8 pts is a bit much for a two to three line solution, but it could be just me

[cedar mist]

I dont understand how you know so much, when you have such little context.

[mystic umbra]

w-what

?

[short lynx]

how do u find the coordinates of point k

[grave pond]

It asks for approximate coordinates, so you're just supposed to eyeball it using the grid lines.

[short lynx]

the x and y int right?

[grave pond]

They shouldn't be integers, if that’s what you're asking.

[short lynx]

ye so how am i suppose to find the decimals

do i js guess?

[grave pond]

I don't immediately recall a more formal English word for it than "eyeball", sorry.

The x-coordinate looks to me like it's about a third of the way from -0.8 to -1.0.

The y-coordinate looks about halfway between 0.4 and 0.6.

[short lynx]

Ok

[cedar mist]

Do you have your unit circle?

Cos = x

Sin = y

You can plug them into your calculator for the approximate values

(-0.866,0.5)

[obsidian harness]

yes

[cedar mist]

Sorry, ill stop now

[obsidian harness]

no need to apologise btw

[cedar mist]

I know, I want to help but don't want to give inaccurate info, so I kept sending before I checked, lmao.

[short lynx]

Oh

[short lynx]

Ahh thx

[grave pond]

But doing that with the calculator was the next part of the exercise, so the initial question that only asks for approximations must be about just reading the coordinates off the grid by eye.

[short lynx]

cos = x, sin=y?

ye u find the second and third part to find the x and y coordinates

without them, i cant answer the first qs

[short lynx]

u cant find the points by eye, would've gotten it wrong

it would be like (-0.8,0.5)

idk why its rounded to 3 decimal places

-0.866

[grave pond]

You're not being asked to find the point to 3 decimal places by eye.
That's the follow-up questions where it says to use a calculator.

[short lynx]

To find the intersection you would make the equations of both graphs equal each other, such that sqrt(1-x^2) = tan(-pi/6)x
Which would just simplify it sqrt(1-x^2) = (-1/sqrt(3))x . Is this right way to find it asw?

[grave pond]

The first question is just to approximate the coordinates, and (-0.8,0.5) would be reasonably good answer to that -- although the x-coordinate is obviously a bit to the left of the -0.8 grid line, so I would say something like -0.9 or -0.85 or -0.8666.

[short lynx]

oh alright

[grave pond]

The exact coordinate is -sqrt(3)/2 = -0.8660254... so neither -0.85 nor -0.87 would be that much off for an approximation by eye.

[short lynx]

ohhh i get it

[grave pond]

The only way the entire exercise makes sense to me is that it wants you to have some experience both with reading approximate coordinates out of a random graph by eyeballing the distance from the gridlines, and with computing them to greater precision with a calculator.

Both these skills are important; they can serve as checks on each other.

[swift ibex]

Can anyone help me understand how to apply trigonometry formulas while solving problems?

Anyone?

[worthy eagle]

@swift ibex u mean these?

[swift ibex]

Hii sorry for replying late

[swift ibex]

Yep

[worthy eagle]

You will be given an angle say 30° and a side's length say 4 cm and will be asked to calculate the length of remaining sides

Now, firstly you will need to look if the given side is opposite to the given angle or adjacent or hypotenuse

[swift ibex]

Okay

Actually I did some of trigonometry but as lack of practice I forgot how to solve

[worthy eagle]

suppose you find the given side is adjacent and you are asked to find the opposite side, just look for the formula that includes adjacent and opposite sides,i.e., tan and use it

you write tan 30° = opposite/4 cm

then either you remember that tan 30° is 1/√3 or we can say √3/3 and substitute for it there and get the answer or you can use calculator to find tan 30° if it's allowed

[cedar mist]

Approximate does not mean to read by "eye"

When you use pi, to get an answer you are only using a very small part of the actual number. I believe that's what it means by approzimate.

[swift ibex]

Okay

[worthy eagle]

you will be given 2 of any three things in formulas mentioned here and you will be asked to calculate the third which i told you how to do

[worthy eagle]

aah yea tropo's right first one is supposed to be done by looking, i think goal is to verify that sin, cos give coordinates of a point on unit circle, and even better exact coordinates which we can't find by just looking

[grave pond]

It's the only possible meaning that makes sense to me in this context where they ask for "approximate" coordinates as a separate question before asking for 3-decimals approximations by calculator.

[swift ibex]

Here is problem I got in my book

A ladder is placed at an angle of 60° with the ground in the window of our house. If the ladder is 2√3 meters long, draw a picture and calculate how high our window is from the ground.

I did use this
Perpendicular= Ab , land=Bc, hypotenuse=Ac
Sin60°=Perpendicular/hypotenuse
√3/2 = Ab/2√3
Abx2 = 2x3
Ab = 2x3/2
Ab = 3cm

@worthy eagle

[cedar mist]

I understand why you would see it that way.

I really think the problem is poor due to the it causing such confusion.

[worthy eagle]

oh u did some work let me check

[worthy eagle]

yes good

it's correct

[swift ibex]

I'm glad I could solve one

I can solve this but some struggling with harder ones

[worthy eagle]

I don't understand why did you use Ab, Bc, Ac for perpendicular, base and hypotenuse?

is there any reason or you just picked it randomly

I mean I would normally use variables like A, B, C here or just P, B, H or something like that

[grave pond]

I think he meant AB, BC, AC -- he has named the corners of the right triangle A, B, and C.

[worthy eagle]

oh

[grave pond]

(Where B is the right angle, and C is where we measure the varying angle).

[worthy eagle]

ok

[grave pond]

This goes to show that the "draw a picture" is an important part of the exercise, and you need to show your drawing together with the rest of the work in order for readers to make sense of it.

(It doesn't quite sound right that the window is only 3 cm off the ground, however).

[worthy eagle]

oh units

[worthy eagle]

@swift ibex read this and remember units, ladder's length was in meters not cm

sry for ping(tropo), mb

[swift ibex]

My teacher actually wrote them to make me understand better so I kept using them (sorry for bad handwriting)

[swift ibex]

Mb

[swift ibex]

Yep

Can you help me with this ones

If xsin45° cos45° tan 60° = tan²45° - cos 60°, find the value of x
If x sin 60° cos² 30° = tan²45° sec60°/cosec60°, find the value of x.

@worthy eagle

[worthy eagle]

do you remember the values of sin45°, tan60°...?

@swift ibex

[swift ibex]

Yep

[worthy eagle]

then just write their values there and then solve for x

it will become a linear equation of form
ax+b=0

[swift ibex]

Okay I'll try solving it

[hollow pawn]

who can tutor me basics of trigo

Dm

[grand condor]

can anyone take a look of this problem, i already solved it but i'm still looking for better and simpler solution.

[worthy eagle]

can you show how you solved it

[grand condor]

ok hold on

honestly, it's not that easy

[grand condor]

see

[worthy eagle]

this was one more way to do it but both ways look equally time taking

[grand condor]

what about problem2

the first one is actually easier to handle

[pale linden]

if we let 5 points make a pentagon then does there always exist a triangle with corners on these 5 points, such that the triangle at least has half the pentagon area?

[visual flume]

help pls

im cooked

i got angle bad is 45 😭

tryna use angle chasing to solve for x (which is dp probably)

[civic rock]

notice anything about ∆ABD and ∆APC?

[visual flume]

uhhhhhhhhhh
UHHHHHH
UHMMMMMMMMMMMMMM
ERMMMMMMMMMMMMMMMMMMMMMMM
theyre similar?

idk

im tryna use cyclic quadrilaterals to make abpc

[civic rock]

why are they similar then?

[visual flume]

thtas how i got bad is 45 😭

[visual flume]

well all i know is angle bad = angle pac

dang it my pens broken

[civic rock]

what about angles ABD and APC?

[visual flume]

hold on imma try to fix my pen 😭

[pale linden]

yo we left angles in 2024

[visual flume]

ok finally fixe dmy pen

[pale linden]

answer my question instead gng 🥺

[civic rock]

dell

[pale linden]

it’s not what im initially trying to solve but I think that should be easier?

[civic rock]

no, take the regular pentagon for instance

[pale linden]

well I have this question where you place 5 points and the 10 triangles that u can make all have area > 2

and u gotta prove at least one has an area of 3

[civic rock]

?

!xy

[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.

[pale linden]

if u place them in a pentagonal pattern u know the pentagon will have an area 9>=A>6 cuz it can be made from 3 triangles and u don’t assume a triangle has an area that’s greater than 3 initially

if its minimal then a triangle must be at least half

if the thing holds right

so if this doesn’t hold then the problem doesn’t hold?

@civic rock

[visual flume]

hell no.

every time i understand it now, i have to go somewhere

like i JUST finished getting EVERY single angle and i had to go cut my hair

and i js came and got 3/5

is it ✅ 😀 or ❌ 😡 👎

[fresh birch]

The way you did the 2nd statement seems the fastest to me, using the law of sines and expressing the angles in terms of C, you could also derive the function to find its minimum

[short lynx]

is cos always x and sin always y?

[short lynx]

would u find qs like this on the SAT

[fresh birch]

sorry?

I didn't understand what you mean

[short lynx]

like would you find a similar question like that on the SAT math section?

oh wait nvm

its alr

[fresh birch]

:V

[civic rock]

just use ∆ABD ~ ∆APC

[visual flume]

yipeee

[grand condor]

i see, thx

[fresh birch]

Thx master

[cedar mist]

Mathster*

[fresh birch]

jajajaja

[tender leaf]

Hey everyone, im gonna be taking geometry this year and im pretty nervous, i did well with Algebra1 and i take algebra2 after geometry, anyways i havent really done anything geometry related sense like 8th grade and im gonna be in 10th and id like to start the year off without feeling like im going into geometry blind😅 if anyone is willing to dm me and give me some pointers or a rundown of geometry and some advice id greatly appreciate it 🙂

[upper karma]

My advice is... 🤔🤔🤔

I don't know.

What??

[civic rock]

think of illuminati every time a triangle appears

[upper karma]

Well, let me give you formula of to learn smoothly:
Success=RCD/S

R is for Reading C is for Confident and D is Don't give a damn about fail.

And S is S**ess

So... the more you don't give a damn about being fail, the more you will be succeed at Something.

But still you need to read and confident (trust yourself)

Good luck buddy 🤞

[grand condor]

can anyone help me on the problem,pls, finding the maximum of y

[upper echo]

Rewrite it as this. ||Then use sqrt(13)<4.||

[obsidian harness]

here is another general method

https://en.wikipedia.org/wiki/Tangent_half-angle_substitution

but yes the way above is particularly neat

[upper karma]

derivetive?

[steep granite]

can someone explain to me the sine sum and sine difference formula

i dont understand where cosine comes from

[steep granite]

why does cosine appear

[worthy eagle]

Do you want derivation?

[steep granite]

im

in only alg 2

I never learned calc

[worthy eagle]

I can send you a derivation by diagram video, you will understand how cos came, wait a minute.

[steep granite]

so u can only understand where cosine comes from if u know derivatives?

[worthy eagle]

https://youtu.be/_RteYGcHmM0?si=2-7R8N-GpdlnDzH_

[worthy eagle]

Derivation, not derivative

[steep granite]

theres a difference?

what is derivation?

oh

is derivation

still in calc

?

[worthy eagle]

"Derivation" of a formula refers to the process of deducing or obtaining a formula from a set of established principles, laws, or simpler equations using logic and mathematical steps.

I hope you understand now, watch the Vedio and I think it will clear your doubt

[steep granite]

wait I have one question

is the two definitions of deriviation

something u learn in calculas

or is the

second meaning

to the thing

the proof thing

something u learn

earlier

[worthy eagle]

The process of deducing something from already known facts is called derivation. This meaning is not at all related to calculas. So don't confuse yourself.

[maiden brook]

No it doesn’t

taking the derivative is called differentiation not derivation

[worthy eagle]

oh man, when he got confused with this. My brain also confused 'derivative' with 'derivation'

[worthy eagle]

when he used 'derivatives' big shame

[dark sparrow]

derivation is a set of steps to arrive at some identity, formula or other result

[grave pond]

Though, just to keep everyone on their toes, a "derivation" can also mean a linear operator satisfying f(ab) = f(a)b+af(b).

[mossy mirage]

yess sure

[knotty quiver]

terms confusement

[fresh birch]

[warped pendant]

how can i do this problem?

[upper echo]

A triangle's area is maximal when its height is maximal, i.e. it is a right triangle and its side is a diameter..

[warped pendant]

okayy thanksss

[zealous trout]

Hey could someone help me with this. Ik im probably smooth braining it it's been a long day but I have to prove that ABCD is a parallelogram

[worthy eagle]

||you start by proving AQB congruent to CSD then prove PBC congruent to RDA||

[cinder ledge]

||Try proving that PBC is congruent to RDA, then use CPCTC.||

What how to mark spoiler

[worthy eagle]

use || on both ends of your text

[cinder ledge]

Oh thanks

Also is the CPCTC part correct?

I am a bit rusty

[worthy eagle]

yes

[cinder ledge]

Ok thanks!

[orchid tendon]

anyone here read or is reading the book "Basic Mathematics" by Serge Lang

[zealous trout]

Thanks so much

[cerulean hearth]

Hello, how are you? . Where on the server do I ask about books on Euclidean geometry? I'm looking for it for the first time.

[left mauve]

#book-recommendations

Welcome to mathcord Stiven

[lofty spear]

@cinder ledge welcome to the mathcord c:

[cinder ledge]

Ty

[night cape]

hii im a tad confused

if the circle is used for angles, why is the triangle important and how is it related to this idea

[knotty quiver]

idk but i know u can form triangles in the unit circle

[noble garden]

yo guys help i cant beat this level on geometry dash

[mystic umbra]

[little tendon]

Hey guys, can someone help me with this question, greatly appreciated

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[pearl hinge]

Can u draw the figure

[keen shuttle]

Hey guys these two problems of my trigo homework (10th grade) are really challenging and absolutely beats me, any kind of help will be highly appreciated

[void drum]

Hey

[keen shuttle]

Hello

[void drum]

n or o?

[keen shuttle]

Both actually

[void drum]

ok. we can use some rules with the sin: sin B - cosB = sen A while sin B + cosB = cosA

[void drum]

This is a proportion

So its just multiplying

[keen shuttle]

But sinA/cosA is jus TanA

[void drum]

Yes, but that doesn't mean that sinB - cosB = sinA (i got confused)

[keen shuttle]

What have I done?!

[keen shuttle]

Still couldn't solve the second one :/

[spark yacht]

Give me a sec

[haughty root]

I thought tanα=tan(β-π/4), then cosα=±cos(β-π/4).

while sinα+cosβ=√2cos(β-π/4).

oh no

[haughty root]

sinβ+cosβ=√2cos(β-π/4).

[summer cradle]

k sin(a) / k cos(a) = (sin(B) - cos(B)) / (sin(B) + cos(B))
sin(B) - cos(B) = k sin(a) ...(²)
sin²(B) + cos²(B) - 2 sin(B) cos(B) = k²sin²(a)
1 - 2 sin(B) cos(B) = k²sin²(a) -> (1)
sin(B) + cos(B) = k cos(a) ...(²)
sin²(B) + cos²(B) + 2 sin(B) cos(B) = k²cos²(a)
1 + 2 sin(B) cos(B) = k²cos²(a) -> (2)
(1) + (2)
2 = k²sin²(a) + k²cos²(a)
2 = k²(sin²(a) + cos²(a))
2 = k²
k = ±sqrt(2)
sin(B) + cos(B) = ±sqrt(2) cos(a) (Proved)

[keen shuttle]

Thanks y'all

[little tendon]

idk how the polygon even is whether its a regular polygon or some other type of polygon

[night cape]

ohhhhhhh

[night cape]

oh so they are related

[knotty quiver]

ya like what frances said

[river sand]

guys i have two exams for both geometry semesters tomorrow whar do i do 😭

[knotty quiver]

learn from your mistakes

[simple vigil]

how to do trigonometry easily i m in class 10

help me

i have problem with algebraic identity in trigo

[silent plank]

show the specific problem

many of those trig identity proofs involve the use of
conjugates and pythagorean trig identities

[obsidian harness]

!show

[lime crown]

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

[simple vigil]

i want to learn how to solve a question in trigo

like how to make a path

to convert ratios

[silent plank]

too vague, the topic is too broad
show specific questions

[spiral lodge]

What would that mean? You should realize by yourself that your questions are extremely vague...

[simple vigil]

overall i m stuck in remembering ratio

i know sin and cos

[obsidian harness]

which ratios do you need help remembering?

[simple vigil]

cosec cot and sec

there name are very confusing

[obsidian harness]

ah okay, well for cosec and sec

they start with C and S respectively

swap them around, so S and C

so then cosec = 1/Sin
sec = 1/Cos

cot = 1/tan cause both cot and tan have the letter T

hope these mnemonics help!

[simple vigil]

but there value like if somebody ask randomly

tell TAN 30

[obsidian harness]

well from the definition

you know that must equal sin(30)/cos(30)

[simple vigil]

then i have to think for a second

sin 30 = 1/2

[obsidian harness]

the more you practice non-calculator the easier it will become

just trust

[simple vigil]

thanks bro

[obsidian harness]

yeah all of the tangent values (and thus cotangent) don't have any 2s

for tan it's 0, 1/sqrt(3), 1, sqrt(3), infinity

so tan is increasing

[simple vigil]

i know this

[obsidian harness]

(cotangent is decreasing, the reverse order, but don't worry about that for now)

[simple vigil]

sin increase

cos decresde

[obsidian harness]

yep for first quadrant ofc

[simple vigil]

but i have a doubt in identity sinA=sin(180-A)

[obsidian harness]

oh, have you learned the unit circle yet?

[simple vigil]

not learned but know it

[obsidian harness]

okay, so just to recap

[simple vigil]

is this for only sin or all ratio

[obsidian harness]

cos(theta) = x-coordinate
sin(theta) = y-coordinate

[obsidian harness]

sin and cos

but all ratios come from sin and cos anyways

[simple vigil]

but if i put A=180 this is false

bcz cos 180=-1

but in this formula it will be cos 0 = 1

[obsidian harness]

no! that formula does not apply to cos

the correct one for cos is cos(-x) = cos(x)

[simple vigil]

okay

you helped me a bit

thanks

[obsidian harness]

this is the picture proof for sin(180 - x) = sin(x)

[simple vigil]

i still have like 6-7 months before boards

[obsidian harness]

oh it's a bit dark

but yes, same height = same y-coord = same sin value

[obsidian harness]

ahhhh okay

you have plenty of time don't worry

[simple vigil]

but there is like 15 chapter in 10th class

[obsidian harness]

keep making progress on maths, also your other subjects ofc

[simple vigil]

my maths marks are not good or else in othere subject i m in top 3 students

[obsidian harness]

interesting

honestly you don't need to top every subject

[simple vigil]

no but i really like coding AI ML DL which require a lot of maths

[obsidian harness]

I think there's a difference with the maths you need for coding and the maths you're learning in school

not exactly the content, but the way it's taught and how the knowledge gets applied, such that school really kills learning

[simple vigil]

in coding i require like PnC

[obsidian harness]

honestly just try to do well enough to get into those uni programs

[obsidian harness]

exactly and you're never, or rarely, going to put a married couple in some seats right

[simple vigil]

honestly you know IIT

?

[obsidian harness]

but then there might be some coding scenario that is similar

[obsidian harness]

yeah

you aiming for JEE then?

[simple vigil]

not i want to tell i never thought of it

i want to go to

BITS PILANI

[obsidian harness]

oh BITS

@jaunty quail can you help me out with India specific stuff bhai

[simple vigil]

you know there is an institute in india which has only one subject

MATHS

and a student got a package of 60LPA

[obsidian harness]

crazy

[simple vigil]

yeah

its true

its in chennai

[left mauve]

Hello

[left mauve]

Nice

CS?

[simple vigil]

yes CS

and i really like physics too

you know technical guruji?

[left mauve]

A bit

[simple vigil]

he is graduated from this college only

and many other

[left mauve]

Yeah well it's the top private eng college

Getting in is not too easy

What grade are you in

[simple vigil]

you can give BIT SAT

10th

[left mauve]

ik

[left mauve]

Ah, you still have time then

[simple vigil]

what about you

[left mauve]

I'm in uni

[simple vigil]

i think you are in 3rd year

[left mauve]

No, 1st year

[simple vigil]

fresher

[left mauve]

Yes

[simple vigil]

which college

[left mauve]

Why go for BITS then?

[slender flint]

BITS isn't in Chennai 😭

[left mauve]

Not doxxing myself, lol

[simple vigil]

we know

its in GOA and PILANI (rajasthan)

[left mauve]

BITS is always a secondary target

[left mauve]

And hyderabad (and dubai)

[slender flint]

And Dubai

[simple vigil]

why secondary?

[left mauve]

Because it doesn't offer as much ROI as a good NIT or IIT might

[simple vigil]

bye guys let me go i have to do worksheet and work of triangles

[jaunty quail]

[azure rivet]

i knew you were indian

[obsidian harness]

I'm not

[azure rivet]

whats this then

[polar otter]

Hellooo

[dire jackal]

if he was indian why would he ask someone else for help with indian stuff

also having BITS as your goal in 10th grade is stupid

should aim for iits and nits since they are better for job placements which the person cares about lol

[azure rivet]

your point is true tbh

[simple vigil]

the problem is competition

jee given by more than 12 lakhs students

meanwhile bitsat maybe 6-7 lakhs

[left mauve]

Uhh

Bits is a single college

Jee offers admission to multiple colleges

It evens out

[dire jackal]

why are you aiming lower though

if aim for iit, you have a better chance of geting into bits

aim for bits and you would prep less

[pearl hinge]

Yeah but if we drawn PQ and MN is such a way that QM is perpendicular on PQ.i.e. then we can find the R+S easily
I couldnt come up with anything else
Since they didnt specify how the segments would be arranged exactly, we can draw one which can actually give an answer ,i suppose

[worthy eagle]

I came up with something ↓

[worthy eagle]

Begin by drawing a seven sided regular polygon, you only drawing a regular polygon for your reference. Just remember to show PQ parallel to NM in this diagram.

Now, extend MS and QR and let O be the point where they meet.
Extend NM and PQ and draw a perpendicular line from O to the extended NM and PQ.

Now you got 2 right triangles. Mark angle Q = 98° and M = 111° and use 'sum of all angles on a line = 180°' and 'sum of all angles in a triangle is 180°' to find your answer.

you can send a diagram so that I can tell you if you are doing it right

[pearl hinge]

That's a vey cool wayy,thanks!
Also there is already a vertice named O in the polygon,so it should be named as something else

[crimson crescent]

does anyone have any good youtube videos to learn how to do statement reason cus im talking geo next year and I havent done statement reason in a while and I dont know how I should learn it so youtube would be good or any other resources

[solid radish]

Hi

For the vertex form if it’s like x - 3 + 2

Would the vertex be (3, 2)

And if it was the opposite like x - +3 + 2 the vertex would be (-3, 2) right

Sorry I posted it in the wrong channel

[blazing ermine]

Can someone help me..

It's physics..

😓

[dark sparrow]

you should never drop the ()^2

otherwise you'll confuse everybody

[spark yacht]

send ig

!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/

[spark yacht]

cmi?

[simple vigil]

i have problem in such kind of question

especially (viii) and (iv)

[dire jackal]

yo is that 10th ncert by any chance

[dire jackal]

in 8th, you just open the squares and you will see general properties like sin^2x+cos^2x=1 pop up

[worthy eagle]

You can try converting LHS to sin, cos and then multiplying and dividing LHS with the denominator of RHS so the denominator on both sides become same and solve the rest to make the numerators same.

For example in iv,
(1+secA)/secA
= (cosA+1)
= (1+cosA) (1-cosA)/(1-cosA)
= (1-cos²A)/(1-cosA)
= sin²A/(1-cosA)

[simple vigil]

you are using identity (a+b)(a-b) =a^2-b^2

?

[worthy eagle]

yes

[simple vigil]

what are you typing so long

[dire jackal]

discord is bad at showing when someone is actually typing. if i type one letter and erase it it will still show me as typing for some time

[worthy eagle]

but this multiplying and dividing thing doesn't work always, just tried on viii and it's getting very long,

[dire jackal]

viii is pretty simple just expand the square

[worthy eagle]

actually I do solving on mobile by typing, not on paper so it shows me typing for very long

[dire jackal]

ohh ok

[dire jackal]

man these are so nostalgic

things were so much simpler in 10th

[worthy eagle]

yes, converting everything to sin, cos makes it unnecessarily long

[dire jackal]

hmm ok

[simple vigil]

it look simpler now for me as a first time VERY HARD

[dire jackal]

true it wasnt that easy when i first did it

just do some practice and youll start to see certian patterns

[simple vigil]

no like in vii part

i have a solution to divide all by sin A

both numerator and denominator

first we will take common sin from numerator and cos from denominator

then divide

[dire jackal]

instead of dividing take sin common from numerator and cos from denominator and keep it as tanA

[simple vigil]

we can expand this 1-cos^2 theta

should i share what i have done?

[dire jackal]

ye

[simple vigil]

1-2 steps are skipped

[dire jackal]

i mean yeah ur done its right

why are you asking it then

[simple vigil]

no like i found it difficult

i took approx 10 min thinking what to do

[dire jackal]

ohh damn

[simple vigil]

i can't afford this much time on a question of 2-3 marks in boards