geometry-and-trigonometry

[wintry tundra]

theres kind of a theme going on in ur assignment if u havent noticed

[jaunty wren]

Yeah ik

This

So it looks like the non shaded are fourths of circles

[wintry tundra]

yep

and there are 4 fourths of circles

and what is 4 4ths

[jaunty wren]

Og

Uhh

1

I was thinking of halves

[wintry tundra]

yeah

so the radius is half of the square side

so just do 36pi

[jaunty wren]

So I was actually able to do it

Thank you so much @wintry tundra I really appreciate your help

[wintry tundra]

welcome

later in math you will begin realizing that critical thinking is needed for complicated problems

especially for geometry

but also in algebra

[stuck barn]

hi, i have this problem i can’t think of a solution to. being only able to cut, rotate and translate (no scaling), you can easily arrange two equal sized squares into a bigger sized square (length and width are bigger by root 2) like shown in the diagram.

is there a way to do something this simply but with two equally sized rectangles, where the final rectangle has the same proportions?

i can’t think of any ways

[supple onyx]

You can make a rhombus but not a rectangle

[stuck barn]

i should have made it more clear that you can cut as many times as you want, and that the proportion of the rectangle was 16:9

but i found this solution

it works but if anyone has a more elegant solution please lmk

i believe with the current solution, if the rectangle proportions was any different you’d just have more of the big yellow squares

[silent plank]

reduce it to atoms and reform a rectangle

[stuck barn]

honestly the first method i tried had no rotation and quickly realised with no rotation it’s a never ending amount of rectangles since yk, the root of 2 is irrational

[supple echo]

is there an easy way to solve this one?

[upper karma]

why so many geometry students dis year

most of them are 9th graders who skipped a grade in math

but I am in 8th and I was 2 grades ahead since the start

[pure cape]

because why not

[humble pulsar]

dont remember asking

[upper karma]

anyone know how to verify this/show that this is true

What have you tried so far & where are you stuck

I think I'm stuck on just not knowing which identity to use or what relationship i can find to verify this

i guess i dont know my starting point

i think beyond that ill be able to figure it out

oh wait i might have something

let me try

[wintry tundra]

@stuck barn u will never guess what 16:9 approaches

[green osprey]

damn dont need to call me out like that

[upper karma]

bruh

made me think she dmed me back

it's been 20 HOURS BRO

and she's ONLINE

her status is CHANGING

@upper karma first step is put a few numbers in to see if there's a really obvious counter-example

I AM GETTING GHOSTED

second step would be to find a list of commonly used identities and just have a look through to see if any look useful

[deft bear]

stfu

[outer latch]

YOU GUYS COULD BE MY SAVIORS

I know nothing about trig help a brotha out

[silent plank]

lol due Nov13?

[outer latch]

BRO

i know its bad

I procastinated and its the end of the semester i have no clue what to do with this

[silent plank]

watch some vids on it

we're not here to teach you trig from scratch

[outer latch]

pain. okay okay im gonna go do that

[outer latch]

I LEARNED IT

[outer latch]

I got a 76 on it. Could be better but i know trigonometry now

[stuck barn]

i don’t get it, explain

[upper karma]

Let AD be the bisector BAC of the triangle ABC, D ∈ (BC), (AD) = (DC) and AB + BD = AC. Calculate the measures of the angles of the triangle ABC

with trigonometry pls

[upper karma]

Looks tough

:))

[upper karma]

maybe cosine rule on 3 triangles ?

[pure cape]

hmm im think of using cosine rule but only on 2 triangles

we can notate all angles present in the triangles in terms of alpha, or any preferable variable

i think what we can do is to maybe set up a ratio to get rid of all the sides in the cosine rules

[upper karma]

thanks

[upper karma]

hi, exist a site about geometry for olympiad?

theory and solved problems

AOPS

thanks!

[main lintel]

There is a discord server Mathematic Olympiads

[wintry tundra]

hmm

i wanna do the math olympiads but idk if id be good enough

i probably shouldnt doubt myself, after all i bet it just takes studying and memorization

[plain wyvern]

studying and memorization? well, if that was the truth, everyone could ace olympiads like school exams

[west basin]

listen it takes a lot more than just that

you have to be EXTREMELY good at problem solving and have good intuition

theres literal child prodigies that would shit on all of us and they dont even get all the questions right sometimes

[upper karma]

i think geometry is the most hard part for olympiad

[wintry tundra]

Yeah

I was gonna guess its those 2 things and problem solving

Idk

I'm not amazing at math comps even tho people say I'm good at math

[polar summit]

It' algebra+combinatorics imo

[olive cove]

@wintry tundra the only thing you can do to prepare for the Olympiads is answering questions of previous years. But if you like math just participate. The worst thing that can happen to you is that you become better at solving problems. I participated in my country not expecting much and became 60th.

[upper karma]

:)))

[upper karma]

can some one help me

@upper karma csn you help me

no

i came here for help

can some one else help me

wait 5 more minutes and then you can ping @ helpers if nobody else has helped you by then

ok

wait i mean 10, not 5

lol

[silent plank]

ask for help 1 question at a time,

instead of expecting someone to explain 8 simultaneously

[upper karma]

ok can you help ?me with this

[silent plank]

lets start with Q1

[upper karma]

ok

[silent plank]

what seems to be the issue you're having with it?

[upper karma]

i don't understqand it

i have to put the eqaution for it and then the ansew

can you just go to the voice chat

[silent plank]

no mic, don't want to use my phone

[upper karma]

i am in the mathamatics vc

ok

[silent plank]

like the best i can do is listen

and respond in text

[upper karma]

ok thats fine with me

ok join

[silent plank]

moving to #mathematics-voice repost and image there

[upper karma]

can i get some help on my trig homework

i understand it up until i have to use sohcahtoa and set up the equation

[dusky surge]

Hello! @upper karma

Would you post your question here, so that we can have a look at?

[upper karma]

i gotchu

sorry for the late response i tried doing some mysel

*myself

[dusky surge]

Okay, so... Questions?

[upper karma]

@upper karma :
q1, AB=DC.
q2, <T=<I.
q3, <A+<D=180.
q4, KC=CR.
q5, <E=90.
q6, AC=BD.

@dusky surge here yu go

[dusky surge]

Nice!

These are fun

[upper karma]

fun you say

[dusky surge]

😂

[upper karma]

q7, m(<AEB)=90.
q8, LG=GH

[dusky surge]

What's your approach then?

[upper karma]

i would label it first

using sohcahtoa

hyp, opp, adj

[dusky surge]

That's a good move

[upper karma]

all that good stuff

[dusky surge]

So, for question 7, it's given oa right?

[upper karma]

whats an oa?

[dusky surge]

Opp adj

[upper karma]

opposite adj

oh ok

yeah it looks like it

so i would use the tan solution

[dusky surge]

Yep

[upper karma]

this is the part where i mess up

i can put it down fine, opp over adj

but the solving is where i mess up

[dusky surge]

I see

Write the original equation out first

[upper karma]

okay

[dusky surge]

$\tan(75^o)=\frac{17}{x}$

[somber coyote]

Biscuit

[upper karma]

oh wow i didnt know i could do that

[dusky surge]

Is that what you got?

[upper karma]

yeah

[dusky surge]

Good!

[upper karma]

so after this would i cross multiply

@upper karma try with all trigonometric functions at q9 and q11

use tan?

[dusky surge]

Cross multiply is a good move

[upper karma]

so tan75/1 x 17/x

[dusky surge]

I think niuton means you can use all trig func for q9 and q11 for exercise

[upper karma]

oh okay

[dusky surge]

Don't use x as multiplication sign

[upper karma]

my fault

[dusky surge]

tan75/1 = 17/x is that what you mean?

[upper karma]

yeah

thats what i have down

[dusky surge]

Then how do you do cross multiply

Let's see if you did right

[upper karma]

so i would multiply tan75 by x

and 17 by 1

[dusky surge]

That's good!

[upper karma]

17/tan75 is what i got

[dusky surge]

Yea, isolate the x is the next step and you got x=17/tan75 which is correct

[upper karma]

i have to put these in decimal form too

[dusky surge]

Seems that if you don't skip any step, you'll be fine

Oh, that's great!

[upper karma]

i think my biggest mess up is i have trouble finding the equation

like finding tan75=17/x

i have trouble with that part

[dusky surge]

sohcahtoa is what you need to remember then :)

I know you're good at labelling those O A and H

[upper karma]

yeah that parts easy

[dusky surge]

After labelling, you just grab the corresponding one of sohcahtoa

Usually, I will remember sine as opposite

And remember that hypotenuse one is the longest, and the special one, so it's always the denominator for sine and cosine

And well, cosine is like 'co'sine, is just modification of sine

And finally, I will treat tangent as sine/cosine

That's how I remember sohcahtoa

[upper karma]

thats really helpful

i appreciate that

i got 4.5 as my decimal

does that look right

[dusky surge]

,calc 17/tan(75 deg)

[somber coyote]

Result:

4.5551362713291

[upper karma]

ayy

[dusky surge]

If you round it off, won't it be 4.6?

[upper karma]

oh youre right

good catch

[dusky surge]

Careful 😁

[upper karma]

Much appreciated my friend thank you

[dusky surge]

Wanna continue?

[upper karma]

sure

[dusky surge]

So, this time the x is the angle

What's given? O A H ?

[upper karma]

the o and a

[dusky surge]

Then we use?

[upper karma]

tan

[dusky surge]

Good

Equation time

[upper karma]

just for easier access

[dusky surge]

Nice!

[upper karma]

12/9 right

[dusky surge]

Nope

It's toa

Which one is the opposite?

[upper karma]

12 no?

[dusky surge]

No

9 is the opposite

For me, it's easier to recognize which one is adjacent

Because adjacent means the neighbour, which is like the side which sticks with angle

[upper karma]

so the closest one to the angle is the adjacent

[dusky surge]

And I know that there are 2 sides sticks with the angle

But hypotenuse is special

So it's easy to spot out

[dusky surge]

Yep

[upper karma]

i seeee

so 9/12

[dusky surge]

So for q9, my first spot out is adjacent

Yea

tan(x°)=9/12

[upper karma]

then i cross multiply

[dusky surge]

Not this time

Because x is already nearly isolated.

[upper karma]

so when x is the angle i dont have to cross multiply

[dusky surge]

Yep

So what should we do now?

[upper karma]

we just solve then right

[dusky surge]

Yea

Need a calculator for the tan-1

$\tan^{-1}$

[somber coyote]

Biscuit

[dusky surge]

You know how to do it right?

[upper karma]

yeah i got it from here

[dusky surge]

Let's check the answer

[upper karma]

alright

36.86

[dusky surge]

,w arctan(9/12) in degrees

I got 36.87

[upper karma]

i gotta round it anyways so

[dusky surge]

Ok

[upper karma]

so 36.9

[dusky surge]

Good

Q11

My first move would be label the adjacent

[upper karma]

27?

[dusky surge]

Yep

[upper karma]

so h and a

[dusky surge]

Yea

[upper karma]

cah riht

[dusky surge]

Yea

[upper karma]

so cos

cos=27/30

[dusky surge]

Nice

cos(x°)=27/30

And now the work of calculator

[upper karma]

perfect

25.84

[dusky surge]

,w arccos(27/30) in degrees

[somber coyote]

Results provided by WolframAlpha

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

Requested by Biscuit#5208

[upper karma]

yessirrr

[dusky surge]

Good!

You're getter better and better

[upper karma]

very much appreciated thank you

would you mind checking this answer for me

@dusky surge

[dusky surge]

Yea?

[upper karma]

pretty sure i did something wrong i got 0.034

just dont look right

[dusky surge]

Question please

[dusky surge]

Okay, so we have A and H

[upper karma]

using cah

[dusky surge]

So, we have cos(32°)=x/14

[upper karma]

yup thats the equation i put down

[dusky surge]

Did you accidentally input cos^-1 instead of cos in your calc?

[upper karma]

i had cos

[dusky surge]

Okay, so we have x=14cos(32°)

[upper karma]

wait so when i cross multiply how should i write my answer

like what would do put down first

[dusky surge]

Is it like this?

[upper karma]

i wrote the opposite down

[dusky surge]

Ah, usually we write the number before the cosine

[upper karma]

ohhhh

[dusky surge]

Because if we didn't have the brackets, the calculator might got it wrong

And think that the 14 is with the degrees

[upper karma]

yeah i got a whole different answer

11.87

[dusky surge]

,calc 14*cos(32 deg)

[somber coyote]

Result:

11.87267334619

[dusky surge]

That's better :)

[upper karma]

alright

still got some more to do

thank you sir

[dusky surge]

Good luck!

[upper karma]

ill need it

when i have the x isolated already which function should i sue

use*

the regular cos/tan etc or the one with the -1 at the end

[dusky surge]

If x is in degrees and it's inside the function sin, cos or tan, use the corresponding ^-1 (inverse function) to get the x out of the function

[upper karma]

so if i have

tan(x)=54/22

i would use -1 right

[dusky surge]

$x=\tan^{-1}(\frac{54}{22})$

[somber coyote]

Biscuit

[dusky surge]

Yea

[upper karma]

alrighttt

thank you sir

sorry for the bombardment of questions

[dusky surge]

Remember to use brackets :)

No problem! Learning and asking are always related

[upper karma]

i got 67.83

[dusky surge]

,w arctan(54/22) in degrees

[somber coyote]

Results provided by WolframAlpha

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

Requested by Biscuit#5208

[upper karma]

alright good

$x=\tan^{-1}(\frac{43}{31})$

[somber coyote]

fuego

[upper karma]

,w arctan(43/31) in degrees

[somber coyote]

Results provided by WolframAlpha

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

Requested by fuego#2724

[upper karma]

,w arccos(11/35) in degrees

[somber coyote]

Results provided by WolframAlpha

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

Requested by fuego#2724

[dusky surge]

You can type in #bots channel to check your answers

[upper karma]

oh my fault

[strange gulch]

can someone help with my geometry hw? <@&286206848099549185>

[humble pulsar]

@strange gulch just post your question, dont ask to ask. and you ping helper after 15 mins of not getting help

[strange gulch]

ok

[humble pulsar]

so if triangles are congruent, what does that mean about their sides?

[strange gulch]

their sides are congruent

[humble pulsar]

which means what?

[strange gulch]

that SR is congruent to KJ

[humble pulsar]

yeah, I was looking for side lengths are equal

[strange gulch]

its just that when i try to solve i cant get a whole number

[humble pulsar]

You got to 14w+4v-6=13w+6v-33 and w+v+23=16w+2v-37?

[strange gulch]

yes

[humble pulsar]

w-2v=-27
15w+v=60

You got this @strange gulch ?

[strange gulch]

i got it

v=15 and w=3

[humble pulsar]

yes

[ivory fractal]

help guys im dumb and dont know this

please help asap

[humble pulsar]

@ivory fractal why asap?

[strange gulch]

x=15

[ivory fractal]

family things

okay but what would i have to do to get 15

like 4x - 10 + 2x + 20 = ?

[strange gulch]

no

[ivory fractal]

how would i get x?

[strange gulch]

2x+20=4x-10

[humble pulsar]

What does it mean to bisect an angle?

[strange gulch]

then solve for x

[humble pulsar]

@strange gulch try not to just give answers

[ivory fractal]

midpoint,

[strange gulch]

ok, sorry

[ivory fractal]

cut in half

[humble pulsar]

Bisect means split in half

[ivory fractal]

yes

[humble pulsar]

so the halves of the bisected angle are equal

[ivory fractal]

jesus thanks you guys you saved my ass

[strange gulch]

np

[royal citrus]

Hey how do u do this

when i use a^3 - b^3 this is what I get,

[wintry tundra]

Idk how to solve cuz I'm not a NERD haha gotem 😎

[dense vine]

@royal citrushave you learned about complex numbers?

[royal citrus]

i have not

[dense vine]

oh ;-;

[royal citrus]

because of nerds u are using discord, so be greateful

[plain wyvern]

you don't need complex number to solve that

using a^3 - b^3 is the right way but the expression you ended up with here #geometry-and-trigonometry message

is dead end

[wintry tundra]

@royal citrus a complex number is a number that has $sqrt{-1}$

[somber coyote]

hiidostuffmc

[wintry tundra]

Welp

That's not how u do it i guess

But it has the square root of negative 1

[royal citrus]

well i got it

thx for your help though

[wintry tundra]

Eh just thought I'd inform u on the magic of math that doesn't exist but does

[upper karma]

• (Assume straight angles) Solve for x and y -
  can someone help me?

[west basin]

just make two systems and solve them

[silent plank]

apply properties of straight lines and ^

[upper karma]

im dumb please help

[plain wyvern]

@upper karma that equal sign with tilde above itbetween triangles is the sign of congruence, I think that is enough for you to understand what is going on here

[neon hamlet]

Or nvm

Do you know how you could do this?

How would you solve for x in this case

[plain wyvern]

the measure of angles of triangle have been given to you

what is sum of the angles of triangle?

[neon hamlet]

180

aaa

so they add add up to it

[plain wyvern]

yes

[neon hamlet]

Alright thank you!

[neon hamlet]

@plain wyvern
Any idea how you would solve for this?

I know CP is 2/3 of the total line

[upper karma]

you actually dont trig

you can easily get the coordinates of P || (coordinates are means of the coordinates of the 3 vertices) ||, then find CP using distance formula

[wintry tundra]

@upper karma u do what u always do for 2 variable equations and isolate a variable and substitute to find the other

And apply the straight line properties like @silent plank said

[echo veldt]

Need help with the next step

Don’t know how find R or a from here

But if I find 1 finding the other is easy

[upper karma]

@echo veldt don't multipost.

[echo veldt]

Apologies

[pure cape]

You can divide everything by 2

And then apply angle sum to get sin(some angle +/- x)

[cursive hazel]

does anyone know this

[upper karma]

What do you think?

What are your best thoughts on this so far

[cursive hazel]

a and p are parallel

[upper karma]

That's correct.

[cursive hazel]

thanks

[arctic cobalt]

How would I solve this

It’s asking if the three sides make a right triangle

[upper karma]

Ok what have you tried so far

[arctic cobalt]

Pythagorean Theron inverse trig

[upper karma]

inverse trig
?

[humble pulsar]

which of those sides has to be the hypotenuse?

[arctic cobalt]

Trig but when your trying to figure out an angle instead of a side

The largest is the hypotenuse right

[upper karma]

Yeah but why would you need trig here

[arctic cobalt]

Idk just trying it to see if it worked

[humble pulsar]

so which measurement is the hyp and which are legs

[arctic cobalt]

26 is the hypotenuse and 25 and 10 are the legs?

[humble pulsar]

yes

is pythagorean theorem satisfied with those?

[upper karma]

sigh

[arctic cobalt]

No 25 squared plus 10 squared is bigger than 26 squared

Is that it is it really that simple

[humble pulsar]

if pythagorean doesnt work, it's not right

[obtuse hornet]

Could anyone help me w 5

[short ravine]

Can you use one of the theorems you know to create an equation for the angle in question

[upper karma]

Are we supposed to assume ABC is a right triangle?

[short ravine]

I believe we can solve it without making that assumption

[upper karma]

I need to find the perimeter of the kite and the value (s) of x. Can anyone help me?

[arctic cobalt]

Do you remember the rules of a kite

[upper karma]

I know.

[obtuse hornet]

Umm tbh i just dont rly know how to solve this type of problem

I was hoping someone would be able to walk me thru it

[upper karma]

What do you know about the angles in a triangle

[obtuse hornet]

Huh

[upper karma]

How am i supposed to interpret that

Do you know anything about the angles on a triangle?

[obtuse hornet]

Well what kinda things

[upper karma]

Do you know what does the angles of a triangle add up to?

[obtuse hornet]

180

[upper karma]

Are you able to make an eqn considering that

To find the measurement of <ACB

[obtuse hornet]

uhh

idk if can lol

[upper karma]

Give it a try.

You won't know if you don't try

[obtuse hornet]

i cant rly figure it out besides knowing B is 21 but idk how to figure out A

[upper karma]

And please don't be lazy and don't tell me "i've tried but idk how to" having not actually tried at all

[upper karma]

What A?

A isn't an angle

[obtuse hornet]

CAB

[short ravine]

@upper karma could you shift to a questions channel

[obtuse hornet]

Like iv tried but i still cant figure out how you find the measure

does the measure of arc AB or ADB have anything to do w it

[upper karma]

Yes

Okay let's move slowly

What did you got for the eqn

And don't worry right now of having 2 unknowns on it

[obtuse hornet]

Tbh i still dont know that cause i dont really have a good understanding on the arcs stuff yet

[upper karma]

The eqn given by using that the sum of angles on the triangle ACB is 180°

[upper karma]

It's ok as long as you are cooperative with me

[obtuse hornet]

Alright

[upper karma]

Are you able to make an eqn using this

Don't worry about having 2 unknowns right now

[obtuse hornet]

I am not :I

[arctic cobalt]

Quick question 4square root3 Times Square root of 6 would it be 4square root 18?

[obtuse hornet]

Unless its like ABC +CAB +BCA = 180 or something

[upper karma]

Ok what's one angle on the triangle ACB

Yes

[short ravine]

@arctic cobalt you could simplify that tho

[upper karma]

(answering to ShongyPonga)

[obtuse hornet]

Oh ok

[upper karma]

So okay

[obtuse hornet]

I'm just not really sure how to find all those measures

[upper karma]

We know ABC, don't we?

We are given m<ABC

Wait not CAB

[obtuse hornet]

Wait what

ABC right

[upper karma]

Yeah

[obtuse hornet]

yeah yeah

[upper karma]

So far we have the equation $m\angle{ACB}+m\angle{CAB}+21=180$

[somber coyote]

Al𝟛dium

[upper karma]

m<ACB is what we want

[obtuse hornet]

Yep

[upper karma]

Ok now for the m<CAB

Consider this circle theorem

Look at the first and last one for our case

You may want to save this somewhere.

[obtuse hornet]

hm

So how do i apply these

[upper karma]

Look at the arc ADB

And at the angle CAB

Hold up a sec

[next jackal]

What's m in mADB?

[obtuse hornet]

Sooo tbh im still not exactly sure how to apply the theorems with that

[upper karma]

Ok what we want to use here is that we know the measurement of the arc ADB, and the measurement of that arc is equivalent to the angle corresponding to that arc located in the center of the circle, and that angle in the center is twice the angle CAB. In conclusion, The measurement of the arc ADB is twice the measurement of the angle CAB due to the theorems i posted above

[upper karma]

measurement

[next jackal]

Okayy

[obtuse hornet]

Ohhhh so CAB would be 110 and ACB would be 49?

[upper karma]

Yes.

[upper karma]

Save this, i believe it'll be useful

[next jackal]

Weren't we supposed to add a point at the centre of circle?

I don't remember learning geometry

[upper karma]

I mean you can, but i simply just verbally stated it instead of taking the time to draw it, and do a couple more steps

Assuming that's what you are saying

[next jackal]

yeahh

[mossy valve]

how would you dilate a prism without any values?

example:a rectangular prism with a volume of 2 dilated by 3

[slate blade]

The volume increases by the cube of its scale factor

@mossy valve

[mossy valve]

i thought every side length or whatever would increae

not the total volume

[slate blade]

Huh? Why

[mossy valve]

because thats what ive been told

and im confused on how i would dilate something without side lengths

i cube it and it goes to some massive number

i dont think thats right

[green osprey]

is this in use right now?

[mossy valve]

is what in use

i still dont know what he meant

[silk crown]

why is $(\vec{b} - \vec{a})\cdot (\vec{b}\times\vec{a}) = 0$?

[somber coyote]

What does "friend" mean?

[weary drift]

@silk crown by definition axb is orthogonal to a & b, ie its dot product with a & b are 0. this fact is applicable after distributing the sum over the dot product

[analog cloud]

https://catgirls.wtf/19/brl6224x5e.png

[pure cape]

wrong channel

and easiest way is to just plug in the numbers

harder way is to solve for m

and then compare it to all 4 answers

[analog cloud]

its trig?

[pure cape]

doesnt seem like trig to me

well anyway, like i said, hard way is to solve for m, which is not that hard with a calculator

and thats also the quickest way

easier way is to plug in the numbers to check

[analog cloud]

oh okay

https://nigga.eu/19/bdkjbuilgj.png

[pure cape]

still isnt trig

...

[polar summit]

log is not defined for negative values over real numbers(don't know about complex numbers)

so your answer is x=1

@analog cloud

[analog cloud]

its algebra 2, which is trig over here

[pure cape]

there is a reason its called algebra and not trig

[upper karma]

is this a test?

No

I know the rules

Just say the answer and explain iirc

we don't give away answers here either.

anyways

have you tried anything

or what have you tried at least to try to get it

[shadow wadi]

Hi guys how to calculate this without a calculator

[upper karma]

Yup

[shadow wadi]

The result is 0.5

But I cant get it through transformations

[upper karma]

can you show it

@shadow wadi please move to a different channel, this one is occupied.

[shadow wadi]

Bro which one

[upper karma]

there are a plenty of #question channels unoccupied, choose one.

kappa for example.

[odd blade]

Find the length of a djogonal of a square where the side of the square is root 6

May I get some help with that ?

[silent plank]

have you drawn a diagram?

[silent plank]

label your diagram

[odd blade]

I just add root 6?

[silent plank]

no,
the diagonal splits the square into two right triangles

[odd blade]

I get that how do I label it

[silent plank]

indicate which sides have what length

[odd blade]

Isn’t the square all equal ?

[silent plank]

the diagonal will be the hypotenuse of your triangles

[odd blade]

Do I delete one side

[silent plank]

wdym by delete?

[odd blade]

And it makes a right triangle

Then I put the root 6 to the side

[silent plank]

well you only need to focus on one of the triangles if that's what you're wondering

[odd blade]

Yup so is that correct so far ?

[broken cargo]

hello

does anybody know how to do translations

[silent plank]

mark all your sides

[broken cargo]

in geometry

[odd blade]

What do you mean by mark ?

[silent plank]

write. sqrt(6) on all sides that have length sqrt(6)

[silent plank]

and you can determine the length of the diagonal using something like pythagoras or trig or special triangles

[odd blade]

So it’s a 45 45 triangle now basically ?

[silent plank]

yes

only just realised you had sqrt(6)*sqrt(2) = sqrt(12) written

was that your own work?

[odd blade]

I searched up how to find the length of. Digabal it said multiply by root 2

So I did that got root 12

But i need simply it

And it is equal to 2root 3?

[silent plank]

yes

[odd blade]

Does it simplify more ?

[silent plank]

no

3 doesn't have any non-1 square factors

[odd blade]

I see thanks ramannov

I am dead stuck now

How do I get half of 4root 15

Divide by 2

So it’s 2root and the 15 won’t divide I am stuck

The question is find the length of the altitude of an equalteral triangle where side of the length is 4 root 15

@silent plank

[upper karma]

HOW

[wintry tundra]

@odd blade half of 4 root 15 is just 2 root 15

[odd blade]

Oh I see and how can I multiply 2 root 15 by root 3

[wintry tundra]

So $4 \sqrt{15} = 2 \sqrt{15}$

[somber coyote]

hiidostuffmc

[wintry tundra]

4 root 15 divided by 2 i mean

Sry forgot to write that

[odd blade]

It’s okay

But how do we multiply 2 root 15 by root 3?

[upper karma]

\frac{}{}

[wintry tundra]

Thx

$\frac{4\sqrt{15}}{2}=2\sqrt{15}$

[somber coyote]

hiidostuffmc

[wintry tundra]

U can multiply roots togethsr

[odd blade]

So it be 2root 90?

[wintry tundra]

$\sqrt{15} * \sqrt{3} = \sqrt{45}$

[somber coyote]

hiidostuffmc

[odd blade]

I see thanks I tusgjt it was a 30

What about the outside numbers

Do they just get added

[wintry tundra]

Yeah

Roots are like parentheses

But they actually do something besides group numbers

[odd blade]

So by long leg be

6root 45?

[wintry tundra]

So $2\sqrt{15} * 2\sqrt{3} = 2\sqrt{45}$

[somber coyote]

hiidostuffmc

[wintry tundra]

Hmm

Wait i might have done that top one wrong

Lemme think

Yeah no disregard the one where it has coefficients of 2

[odd blade]

So we got half of 4 root 15 it be 2 root 15 then we times 2root 15 by 3 we get

[wintry tundra]

The long leg is 2 sqrt 45

Which simplifies to 6 root 5

[odd blade]

2root 45 correct?

[wintry tundra]

So put 6 root 5

[odd blade]

So

[wintry tundra]

Are u just trying to find the long leg?

Yep

Thats right

[odd blade]

Thanks so much !

[wintry tundra]

Welcome

Radicals can be a bit tough to simplify, but remember to always think if radicals have square factors

And then keep the postulates and theorems in mind so u can apply them

[odd blade]

Ye it seems easy I think I just get confused when I see radicle sign

[wintry tundra]

Mhm

Its different, it has all the properties of parentheses tho

So thats something to also remember

But it has the property of finding the root tho so

[odd blade]

Thank you

Sorry guys one more thing I need help dividing 4 root 6 by root 2

I got 4 root 3

Is that correct?

Nvm I got it

[upper karma]

How can I show it’s not a test

It’s legit a homework

I wasn't asking for that

Oh

I was asking what had you tried so far

[unique flower]

i cant help you, as you are doing the test rn

[languid wharf]

good luck

[earnest echo]

sin(30°)=sin(3×10°)

Now we know identity of sin(3A)

It still requires some heavy duty computation

(you'd still need the calculator to solve the cubic)

[west basin]

just use maclaurin series expansion

and guess

[plain wyvern]

better write those other language sentences and words in english, so that someone can help you easily @lilac pine

[lilac pine]

everything needed is shown in the figure

there isn't anything written in text that isn't shown in the figure

it is asking for tan a

i did found out what the answer is, but i didn't actually calculate it

[upper karma]

@shadow wadi don't multipost...

[wintry tundra]

Isn't sin 60 just pi/3

So like just go from there

Sin 10 is pi/18 probably

[upper karma]

I don't know what logic to make it there but you can't just do sin(60°)=π/3 then sin(10°)=(π/3)/6 or whatever you did

It actually is approximately π/18 because it's a relatively small angle

[wintry tundra]

Hm

I mean since apparently sin 90 is pi/2

Sin 60 is pi/3

And sin 30 is pi/6

So it'd be weird if the pattern didn't follow

[humble pulsar]

That's called a coincidence

[wintry tundra]

Worst coincidence ever

[west basin]

yes remember that coincidence is not the same as causation

[humble pulsar]

Correlation is not causation*

[wintry tundra]

Still thats just a sad coincidence

[humble pulsar]

not really, but ok

[wintry tundra]

I mean it's confusing for people like me who only knew the pattern

[humble pulsar]

also sin30 != pi/6, sin30 = 1/2

[wintry tundra]

Oh

[west basin]

30 degrees = pi/6

its just degrees and radians

[slim saddle]

the answer is 4

i am not sure how i got it tho

please @ me if you can explain

[livid moss]

@slim saddle can you see the length of their diameters is the width of the rectangle?

So their radii is half the width

[slim saddle]

yea

the radii is 3.5

[livid moss]

Yes

[slim saddle]

but they overlap

[livid moss]

The centre of each circle is 3.5 away from one of the sides

Because it's a radius from the centre to the side

[slim saddle]

ohhh

i get

it

11-7=4?

[livid moss]

Yes

[slim saddle]

oh ok thanks!

[livid moss]

Np

[thorn finch]

Any1 know why i dont get the same Segment Area

[plain wyvern]

@thorn finch (just the area of sector - area of the triangle) is the area of segment. How do we know why you don't get it? Instead you must ask what you can't understand in it

[molten cloak]

Can someone explain that step plz

[silent plank]

seems there's a typo

they wrote an extra 2

[molten cloak]

ohh thanks , let me see if i can get the answer without that 2

[burnt halo]

can someone help me prove the equation of a circle

or like explain it

seems like i need to do it in functon of it radius

[heady summit]

It's an expansion

a^3+b^3= (a+b)(a^2-ab+b^2)

[heady summit]

See these

And then there is an expansion a^3-b^3 =(a-b)(a^2+ab+b^2)

[steep temple]

Geometrically, what does $$ F_2\sin\theta_3 $$ symbolise?

[somber coyote]

dlp

[steep temple]

It just forms some new triangle with some new sides right

[upper karma]

it seems like so from a quick look

[humble tundra]

One possibility is this:
Call the three vertices of the triangle A, B and C (anticlockwise from top).
Draw a line through C parallel to AB (call that line CX).

Then

F2.sin(theta3) = (Perpendicular distance from point A to line CX)

[fast tide]

I have a book of trig identities. I was thinking if I gave each one a person's first name that might help to remember them all. or name them after foods maybe that way when I eat I can think of the trig identity again. what do think? I thought about numbering them but that might not be the best idea. I have heard some people have a method where they visualize putting numbers on shelves in the home.

[upper karma]

nicest way to remember formulas / identities is knowing how to derive / prove them

[fast tide]

do you think the derivations are all about geometry? or can they be algebraic?

[upper karma]

mostly algebraic

but also geometrically

[fast tide]

I would think that it is all geometry

I prefer algebra

but then that brings me back to the identities again

[upper karma]

combine

[fast tide]

yeah so there must be some primitives (no offense herambe) that I can use to make the others

[upper karma]

yeah sure

like sin^2 + cos^2, sin(a+b) and cos(a+b)

these are the most imporant

[fast tide]

ok

[plain wyvern]

just practice lots of questions on them, they will stay with you for life

[upper karma]

whole life? lol

[cunning pagoda]

Watching a video and I'm a bit confused. Shouldn't the bottom edges have the same length? I know the left one is normalized, but if v and n are the same value for both triangles, why is one (v dot n) while the other is (v dot n)*n?

[humble pulsar]

So n itself is a vector with magnitude 1. $v \cdot n$ is a scalar, so $(v \cdot n)n$ is a vector with magnitude $v \cdot n$

[somber coyote]

moshill1

[humble pulsar]

Both "horizontal" sides are length v . n, one just has the unit vector n drawn and one has the scaled up vector

[fast tide]

I started working on this

\def\arraystretch{4}
\begin{tabular}{|c|c|c|}
\hline
\raisebox{0.3 em}{\Huge $\sin{\theta}=\frac{1}{\csc{\theta}}$ }&
\raisebox{0.3 em}{\Huge $\cos{\theta}=\frac{1}{\sec{\theta}}$ }&
\raisebox{0.3 em}{\Huge $\tan{\theta}=\frac{1}{\cot{\theta}}$ }\
\hline
\raisebox{0.3 em}{\Huge $\sin{\theta}=\cos{\theta}\cdot\tan{\theta}$ }&
\raisebox{0.3 em}{\Huge $\cos{\theta}=\frac{\sin{\theta}}{\tan{\theta}}$ }&
\raisebox{0.3 em}{\Huge $\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}$ }\
\hline
\raisebox{0.3 em}{\Huge $\csc{\theta}=\frac{1}{\sin{\theta}}$ }&
\raisebox{0.3 em}{\Huge $\sec{\theta}=\frac{1}{\cos{\theta}}$ }&
\raisebox{0.3 em}{\Huge $\cot{\theta}=\frac{1}{\tan{\theta}}$ }\
\hline
\raisebox{0.3 em}{\Huge $\csc{\theta}=\frac{1}{\cos{\theta}\cdot\tan{\theta}}$ }&
\raisebox{0.3 em}{\Huge $\sec{\theta}=\frac{\tan{\theta}}{\sin{\theta}}$ }&
\raisebox{0.3 em}{\Huge $\cot{\theta}=\frac{\cos{\theta}}{\sin{\theta}}$ }\
\hline
\end{tabular}

[somber coyote]

galois

[fast tide]

is that correct?

[limber sinew]

looks right

quite a lot to remember though, you don’t need to keep that much in your head

[weary drift]

see pins for trig identities

[fast tide]

I think the symmetries make it easy. that is why I had to retype everything

without the symmetry it looks incorrect to me

[plain wyvern]

the table can be made smaller, too much redundancy

[misty girder]

what is 2+2?

[burnt halo]

anyone online?

[burnt halo]

ok so i need to set the limits of the integrals of this oval

it has 3 equations wich are the 2 small circles and the big circunference

my integral limits for the big circunference AKA the curve on top and the bottom of the oval are from 0 to point D

and i dont know how to set the other limits

<@&286206848099549185>

[fervent obsidian]

So what exactly do you want to calculate?

is it the area of the ellipse?

@burnt halo

[burnt halo]

nah its an oval

[fervent obsidian]

Well? What is your goal? You want to calculate the area?

Also if thats supposed to be an oval then its too symmetrical.

[burnt halo]

thats the way yo do it

i swear

its supposed to be like that

but anyways i think i got it

thx tho

[long light]

how can i find the union of those two equations:
sin(x) = 0
cos(x) = 0
?

[patent plume]

Uh

They don't exist?

at least, not in the real plane

[long light]

i will explain my self:
sin(x) = 0 => x = pi * k (k -> Z)
cos(x) = 0 => x = pi/2 + pi*k (k -> Z)
how can i find the union of those two

[silent plank]

well the solution set for the the first one are integer multiples of pi, (or even multiples of pi/2)
the set for the second one are the odd multiples of pi/2
so their union would be the integer multiples of pi/2

[long light]

so it could be (k*pi)/2

thank you

[patent plume]

Oh that's what you mean

[slate blade]

How does one find the magnitude of AP?

I get the direction part

it's just $$AP = k(\frac{b}{|b|} + \frac{c}{|c|}) $$

[somber coyote]

Doodaide

[slate blade]

but what is the general form of the "k"

[slate blade]

<@&286206848099549185>

[polar summit]

What is the resultant vector of b and c? How do you represent it?

[slate blade]

you mean like b-c?

bruh u still there?

[polar summit]

parallelogram law of vector addition

I am not sure if it's applicable but that should give you some insight.

[slate blade]

...ok I'll think on it

[green osprey]

quick question, how important are unit vectors

[polar summit]

very important

[humble pulsar]

If you mean i j k, then very very important

[green osprey]

oh

how do i do this

[humble pulsar]

but if you mean like the normal unit vector in cross product,

[green osprey]

im so confusedd

[humble pulsar]

what's the magnitude of a unit vector?

[green osprey]

im learning this for machine learning, but im thinking about dropping it

the magnitude is the length right

[polar summit]

EXAM PROBLEM

[green osprey]

its kahn academy

[polar summit]

Umm then okay

[humble pulsar]

man was prepared

[livid moss]

Way to dox all the weird servers you are in

[green osprey]

lol

[polar summit]

$|v| = \sqrt{(-7)^2+6^2}$

[somber coyote]

usernamephobic

[green osprey]

isnt that the magnitude

with the pythogream therom

[humble pulsar]

yes..

[green osprey]

what even is a unit vector

isnt it one unit

[humble pulsar]

a vector of magnitude 1

[green osprey]

im so confused then whats the answer

[humble pulsar]

Basically you scale a vector up / do via scalar multiplication to get a vector with mag 1

$1 = |kv| \implies k = \frac{1}{|v|}$

[green osprey]

ohhhhhh

[somber coyote]

moshill1

[green osprey]

so a?

[humble pulsar]

yes

[livid moss]

You really need to drop the defeatist attitude though. This is going to be the smallest bump you'll encounter if you're learning ML. You can't want to drop it everytime you struggle.

[green osprey]

alright but like this khan academy course is long asl prolly going to take me a few weeks to get through this alone lol.

[green osprey]

Quick question, what are unit vectors even used for lmao

[polar summit]

google

[humble pulsar]

$a \times b = \left(|a||b| \sin{\theta}\right)n$

where a and b are vectors and n is the normal unit vector

[somber coyote]

moshill1

[weary drift]

there’s no THE unit normal, you must specify the right one satisfies a right hand rule with a & b

[idle bloom]

They're used for alot of things

So like one example would be redefining a shape in terms of a new coordinate space

So maybe instead of the normal xy plane

You want to work in terms of a rotated xy-plane

You can define your new coordinate axis with vectors

And the math ends up alot nicer when you use unit vectors

[green osprey]

oh okay thanks

[fast tide]

is there a method to pull the coefficient out of sin(ax) before factoring it out? like if a is in R?

like how the double angle pulls out the 2 to the outside. but is there a more general solution where the coefficient is any number in R?

[proud mesa]

Meeeh there are not very useful methods to do so

I guess your best option for the general case is using the fact that $\left( \cos (x) + i\sin(x)\right)^{\alpha} = \cos(\alpha x) + i\sin(\alpha x), $\forall \alpha, x \in \mathbb{R}$.

[fast tide]

$2\sin\left(ax\right)\cdot\sin\left(\frac{\pi}{2}+ax\right)$

[somber coyote]

galois

[fast tide]

this works for *2 or /2

for all x without

ignore a

[somber coyote]

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

[proud mesa]

I guess this is cleaner now

Well uhhh

In the general case I guess this is really your best option

[fast tide]

I like what you did there

[proud mesa]

If alpha is natural, then there are formulas for sin(nx) and cos(nx) that you can prove by induction.

But I don't find them useful

[fast tide]

that is interesting

[proud mesa]

Like this one

I've seen a closed formula for sin(nx) that doesn't relly on previous values of n

But like

These formulas are just a summation mess

[fast tide]

sen()?

sin()?

[proud mesa]

Idk why would you find this kind of thing useful

[proud mesa]

Yup

We usually call sin as sen here.

Because it's pronounced "seno" in portuguese.

But anyway

[fast tide]

https://mathhelpforum.com/threads/what-does-sen-mean.185979/

seno en espanol

oh ok

[proud mesa]

https://brilliant.org/wiki/expansions-of-certain-trigonometric-functions/

Brilliant has got other identities

But as I said

These closed formulas for sin(nx), cos(nx) and etc... Are just a mess

And not even useful

These first two identities are simpler to memorize and get useful information about.

[fast tide]

only for sums

[proud mesa]

I mean, the identity for sin(nx) is so complicated it's barely useful.

[fast tide]

it makes sense actually that n is in N

if sin(ax) with a in R then you can't remove it

because you would need some other function to make it periodic

e^(itheta)

that requires i

[proud mesa]

$\sum_{k = 1}^{n} \sin(kx) = \frac{\sin\left(\frac{n+1}{2}x\right)\cdot \sin\left(\frac{nx}{2}\right)}{\sin\left(\frac{x}{2}\right)}$ is just much cleaner and actually applicable.

The first identity I mentioned may be even more useful in most cases.

[fast tide]

what is the interval for this sum?

0 to n?

[proud mesa]

Yeah you are right

Sorry

[somber coyote]

MisterSystem

[proud mesa]

Now it's right

[fast tide]

ty

I saved it

this seems useful in fourier analysis

[proud mesa]

Try proving it with induction

[fast tide]

is it easy or hard?

I know about induction proofs

[proud mesa]

It's not that straightforward I guess, you gotta do some computations, but it's not hard.

Just remember your usual trig identities or just change everything to exponential form.

[fast tide]

I plotted that sum. it is periodic but the function looks strange for large values of n

sine wave for n=1

the equation simplifies to sinx

when n = 1

ok it makes sense now

if you swap it to cos that should look a lot different

this sum starts out with y=x=0

3 different frequencies all phase aligned to p0 at t0

I see the function is continuous for natural numbers only

[upper karma]

Help Plz

[sly marlin]

@upper karma what have you tried?

[upper karma]

Keep in mind that the area of an equilateral triangle is $\frac{s^2\sqrt{3}}{4}$

[somber coyote]

Shen

[upper karma]

Where $s$ is the side length.

[somber coyote]

Shen

[sly marlin]

@upper karma ? Side of what rectangle?

[upper karma]

Triangle i mean

[dull pendant]

hi everyone, it's my first time posting here
i cant seem to understand how to prove 2a + 2b + 1 = 2c at all

i attempted:
c^2 = a^2 + b^2
2c^2 = 2a^2 + 2b^2
but then i got stuck after doing this:
sqrt(2a^2 + 2b^2) = 2c

[silent plank]

use information from both triangles

manipulating

c^2 = a^2 + b^2
by itself won't get you what they want

[sly marlin]

you need to use information from both triangles

[dull pendant]

alright so:
(a+1)^2 + (b+1)^2 = (c+1)^2
a^2 + 2a + 1 + b^2 + 2b + 1 = c^2 + 2c + 1

im guessing i should make 2c the subject here

[pure cape]

you're almost there

[sly marlin]

remember to use both equations

[dull pendant]

ugh yeah

[pure cape]

use pythagoras again

[sly marlin]

hence ||triangle inequality||

Hence 2=3

[dull pendant]

actually, i think i should be making c^2 the subject of that

[silent plank]

not much point

[dull pendant]

oof

[silent plank]

you're making more work for yourself

recall what you got from your first triangle

$\red{a^2} + 2a + 1 + \red{b^2} + 2b + 1 = \red{c^2} + 2c + 1$