geometry-and-trigonometry

[small raptor]

When you square, you should get a positive number. It is never negative

[weary drift]

(-2-0)^2 is not -4

[signal locust]

positive 4?

[small raptor]

Correct.

[signal locust]

oh and then that would be positive 144?

[small raptor]

Indeed.

I mean, you still get an ugly number under the radical, but it'll be a real number

[signal locust]

and then i need to convert that to a decimal so i can round it? if thats the right number

[small raptor]

That is the exact answer, yeah. You can approximate by putting it into the calculator, and rounding.

[signal locust]

im not really good with decimals is this rounded ok? to the hundreths place

[small raptor]

Depends on the direction. If it says hundredth, then it'll be 12.17

[weary drift]

,w sqrt((-2-0)^2+(-3-9)^2)

[signal locust]

ooh i thought so

[somber coyote]

Display results online and refine query

Requested by RokettoJanpu#6002

[signal locust]

so my final answer then is 12.17?

[weary drift]

if it requires rounding to 2 decimal places then yes

[signal locust]

i finally got it thanks so much :p

[small raptor]

Just keep practicing to really engrain it. Before you know it, you'll be solving problems about boats in a river going east, while the river is flowing north.

[weary drift]

or solving a wave equation in 2d space

[signal locust]

math has never been my forte lol this is only the third lesson and i am already struggling

[small raptor]

The trick of math is practice.

You now became familiar with the distance formula. Keep putting it to practice.

[weary drift]

practice, plus come on this discord more often if you need further help

[signal locust]

i do have one more question im a little confused on, its like this but with a graph

[small raptor]

Doesn't hurt to graph it too. But that may be hard to think about without knowing vectors. 🤔

[small raptor]

Same concept. Just apply the distance formula.

[signal locust]

would i just take the x,y of the 2 points then use the formula?

[small raptor]

Correct.

[signal locust]

alright awesome thank you!

[small raptor]

If you want a way to visualize what you're doing @signal locust

[weary drift]

are my eyes bleeding a bit?

[small raptor]

Yes.

[signal locust]

thanks, i'm slowly understanding lol

[signal locust]

ive seriously been stuck on this question all day, its the only one i need left to get right on this last assignment and ive tried everything, isnt the formula x+4=1.4x? but i cant seem to get it right

[weary drift]

you set up the equation right, you just gotta know how to "isolate" x

[signal locust]

ive been trying to solve it over and over and i cant get it

idk how im supposed to get x alone in that equation

[weary drift]

you can subtract a certain value from both sides of the equation to get ride of x from one side

[signal locust]

the way i was doing it was subtracting 4 and bringing it to the other side

[weary drift]

ok you can subtract 4... that gets you

$x = 1.4x - 4$

[somber coyote]

RokettoJanpu:

[weary drift]

you still have x's on both sides... what can you do to make x show up only on one side?

[signal locust]

subtract the x on the right?

[weary drift]

remember, what you do to one side, you also have to do to the other side

[signal locust]

oh i mean left

so then it would equal to 0

[weary drift]

subtract x from both sides?

what equation does that get you?

[steady umbra]

X = 10

[signal locust]

idk how you got 10

[signal locust]

ohh i think i tried doing it that way and i didnt get it right

yeah i got that answer before and it wasnt right i think? i can try reentering it

[signal locust]

oh yeah 10 was the right answer

idk how i got it wrong then bc i got 10 before lol

thanks for the help! im finally done lol

[steady umbra]

Glad to help @signal locust

[signal locust]

(:

[dark sparrow]

what ways do you know of, in general, to prove two lines are parallel?

is there any info you could gather here that would let you employ any of them?

[sly marlin]

So what does that tell you?

[dark sparrow]

i don't know

can you

if yes, then do it
if not, then what's stopping you?

ok

what's making you doubt

[sly marlin]

what does SAS similarity say?

Yeah, it is.

So what does applying it tell you?

[dark sparrow]

i was under the impression that he was hesitant as to whether he could apply it at all

[sly marlin]

@upper karma what happens when you apply it?

Which triangles specifically?

Then you are done?

[coarse bronze]

how would i go about memorizing some of the more complex identities

mainly these https://gyazo.com/f9bdb07ea03d95e3c1156bf65555ab81

[flint osprey]

by understanding why they're true

[coarse bronze]

besides school itself where would i go to learn why

[teal sonnet]

need help solving

[weary drift]

have you tried getting cot(x) by itself? (and maybe if you go one step further you can turn cot(x) into another trig function that's easier for you to manually take the inverse of)

[teal sonnet]

ok so I would makie cot^2(x)=1/3?

[weary drift]

that's a step in the right direction

[teal sonnet]

and then make it sec^2(x) -1 = 1/3

[weary drift]

ok, sure

[teal sonnet]

what would I do with the ^2?

turn it into a quadratic and plug it into the quad formula or no

[weary drift]

$sec^{2}(x) = (sec(x))^2$

[somber coyote]

RokettoJanpu:

[weary drift]

do you know how to undo the 2 in the exponent?

[teal sonnet]

take the inverse or is that not allowed>

?

[weary drift]

inverse of what?

[teal sonnet]

of $sec^{2}(x)

[weary drift]

you have this

$sec^2(x) - 1 = \frac{1}{3}$

[somber coyote]

RokettoJanpu:

[weary drift]

get sec^2(x) by itself first

[teal sonnet]

ok umm so sec^2(x)=4/3 ?

[weary drift]

ok

now try taking the square root of both sides

to cancel the 2 in the exponent

[quick anvil]

@coarse bronze I have found Coursera courses to be an approachable system to use. Some of the courses are truly excellent for explaining the why.

[coarse bronze]

ok thank you

[quick anvil]

Also, whenever possible find a relevant 3blue1brown YouTube video.

[teal sonnet]

ok so i got sec(x) = sqrt(4/3)

[quick anvil]

They show animations while also explaining the "why" of things.

[weary drift]

@quick anvil @coarse bronze not here

we're doing hw

[teal sonnet]

and then i took the inverse of
1/ cos(sqrt(4/3))

[weary drift]

@teal sonnet what does thr right side simplify to? the sqrt(4/3)

[coarse bronze]

i asked before he posted it so

[teal sonnet]

4/3^(1/2) or no?

i'm a bit confused

i'm lost on the sec(x) = sqrt(4/3)

[weary drift]

you got the fraction 4/3

you're taking the square root of the fraction

take the square root of both the top and bottom

[teal sonnet]

oh i see

so 2/sqrt(2)

sorry

i mean 2/sqrt(3_

)

and then now I can get that value in radians

for sec

im getting a domain error

did i make a mistake?

ok nvm i think i got it. I just took the reciprocal of 2/sqrt(3) which is sqrt(3)/2 and then I took the cos inverse of it. which got me to 30 degrees

[weary drift]

do you know what 1/sec is?

[teal sonnet]

it should be cos?

and the answer in radians should be pi/6

is that right?

[weary drift]

ya

$sec(x) = \frac{2}{\sqrt{3}}$

$sec(x) = \frac{1}{cos(x)}$

[somber coyote]

RokettoJanpu:

RokettoJanpu:

[teal sonnet]

ok thank you. can you keep going to see if my work was correct

because i think its asking for more rads than just pi/6

[weary drift]

yes

between 0 and 2pi

what other angles if you plug into cosine will get you sqrt(3)/2?

[teal sonnet]

umm does this involve using reference angles?

I'm very confused on this part

i tried 5pi/6 and that is neg

sqrt(3)/2

[weary drift]

@teal sonnet ya you can think with ref angles if that helps

You just need to find other angles x where cos(x) = sqrt(3)/2

[teal sonnet]

11pi/6?

are there any more?

that is all i got

11pi/6 and pi/6 within the range [0, 2pi]

ahh

i forgot about all students take calc

ok since it is cos pos

it should either be quadrant one and four

[weary drift]

@teal sonnet ya

Pi/6 a͙n͙d͙ 11pi/6 is good

[teal sonnet]

ok thank you so much

[upper karma]

How would you do any of them?

[fringe dirge]

By finding out what corresponding angles are and what consecutive interior angles are and etc

[devout shell]

Starting with definitions is what you do. From there the question guides you as to what to look for.

[upper karma]

well I know how to do them like this:

[devout shell]

Doesn’t need to be parallel lines, though parallel lines accentuate the point they want to make.

Again, read the given definition, and decide for yourself what angle(s) meet the condition(s) of the definition.

[upper karma]

ok

[dark sparrow]

no

@steady umbra don't just give out answers

[steady umbra]

O sry

[silent plank]

what's your definition of the orthocentre?

what was your concern about the 1st image?

(the representation in the 2nd is wrong btw)
what is an altitude?
in 1) which lines are the altitudes?

[quiet mason]

not all perpendicular lines to a side are known as altitudes of the triangle

the ones originating from the vertices are

[silent plank]

pink line doesn't involve a vertex

extend the left leg downwards and draw a perp line from the bottom right corner for the 3rd altitude

do you still have an issue understanding the orthocentre in 1)?

[dark sparrow]

extend the green and yellow lines

both downwards

[silent plank]

^

[dark sparrow]

wdym

https://en.wikipedia.org/wiki/Altitude_(triangle)#Properties

be grateful i didn't hit you with a lmgtfy link as i otherwise might have

don't get what

what do you not get

idk maybe a problem can somehow involve two altitudes intersecting ¯_(ツ)_/¯

you seem to want some sort of instant gratification in the form of a problem which admits an exceptionally elegant solution that makes use of the orthocenter

i don't have any of that

in right triangles, it coincides with the right-angle vertex

i mean yeah it kinda is counterintuitive that the orthocenter can end up outside its triangle

but then

so can the circumcenter

which in fact also happens iff the triangle is acute obtuse (derp)

[silent plank]

*orthocentre is inside in acute
outside in obtuse

[dark sparrow]

er

yeah sorry that was a mixup

@upper karma you're confusing the circumCENTER and the circumCIRCLE

the circumcircle is what you're describing

the circumcenter is its center

uh

what's the question

uh

...

i don't feel like i can answer that in any constructive way at all sorry

[silent plank]

they'll have a common side?

[dark sparrow]

almost nothing, save for angles ADE and FDB being equal

ED and DB happen to lie on the same straight line but they're not known to be equal

...

[silent plank]

ABH is not a right triangle so it won't be H.
which lines are its altitudes?

yes (but not the ortho of ABH)

BH isn't perpendicular to AH and hence neither AH or BH are altitudes

looks a bit better,
but look at the diagram on the left.
you already have the altitudes of ABH there

[dark sparrow]

look at them

what line goes through A and is perp to BH

[silent plank]

note that your drawing on the right somewhat resembles the one you already have

[dark sparrow]

AC maybe?
so is it AC or is it something else

what's making you doubt that AC is perp to BH

@upper karma

ngl it makes me wonder how you didn't see that

that symbol

[silent plank]

that is also on the diagram
when the altitudes are extended, where do they meet?

the altitudes that meet at H are from triangleABC

which are your altitudes for ABH

do they or don't they?

[silent plank]

which means your orthocentre would be:

[midnight hearth]

Hi guys

I'm not very good at maths but i'm trying to code something in C#

And it involves trigonometry i think

Basicly, i have two vectors

Let's say VectorA = (1,0) and VectorB = (0,1)

What i want is the vector between these two vectors such as it would have the same lenght as the previous ones

I know it would be (0.707..., 0.707...)

But now if i have two vectors wich are VectorA(1,0) VectorB(0,3)

How can i find that?

[formal bolt]

It must have the same lenght of what vector?

A or B?

[midnight hearth]

Basicly

This is what i'm looking for

[formal bolt]

Oh on the ellipse

We can parametrize it

Let's say you have V_A=(a,0) and V_B=(0,b)

Then the ellipse is given by
(a*cos(t), b*sin(t))

So we need to solve for t now such that we get the point on the diagonal

b/a = b/a tg(t) -> tg(t)=1

You're point is
(a sqrt(2)/2, b sqrt(2)/2)

[midnight hearth]

Thank-you 😄 I might take a moment to understand that :p

[formal bolt]

I took a long path probably lol

Basically the ellipse is a stretched circumference

[midnight hearth]

Oh no, it's just that i really really suck. I'm an infographics and had really low level of maths at school, and now i'm getting into game developement and i'm a long way from catching up 😄

[formal bolt]

So it's as if all the plane was transformed by
x->ax and y->ay
That's why your point is
(sqrt(2),sqrt(2)) -> (asqrt(2),bsqrt(2))

[midnight hearth]

Mhh 😄

I'm long, sorry 😄

[midnight hearth]

I don't rly understand what is t

[formal bolt]

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

@midnight hearth move the slider of T

To visualize it

[midnight hearth]

@formal bolt Ow Thank-you

That helps a lot 😄

[formal bolt]

you're welcome, if you need some other graph just let me know!

[midnight hearth]

One thing i might ask is... Where to start to learn all this?

Considering my math level is... quite far

[formal bolt]

You could probably start from trigonometry

[coarse bronze]

would anyone mind explaining to me how these particular identities are true https://gyazo.com/567f6be9d34a3d1f1f685fbc11cc1b05

i am struggling to memorize the identities but maybe itd be easier if i understood how they worked

[coarse bronze]

<@&286206848099549185>

[trail minnow]

there are a lot of identities there

which ones do you want to go through?

[coarse bronze]

idk but since theyre related my guess is that some of them can be derived from others

[trail minnow]

do you know the proof that sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

[coarse bronze]

ive seen that identity yes

[somber coyote]

hegel:

hegel:

[coarse bronze]

o

[trail minnow]

ye

[coarse bronze]

yes

[somber coyote]

hegel:

[trail minnow]

applying our formula

[somber coyote]

hegel:

[trail minnow]

does this make sense?

[coarse bronze]

i dont see how it simplifies to that

oh nvm

[trail minnow]

ha

*ah

[coarse bronze]

yes it makes sense

[somber coyote]

hegel:

hegel:

[trail minnow]

do you understand this formula?

[coarse bronze]

i dont understand it but i memorized it since its one of the simpler ones

[trail minnow]

uh

okay the proof is a bit annoying so I guess that's okay for now

so we can basically just do the same thing

[somber coyote]

hegel:

[coarse bronze]

o ok

that makes sense

[trail minnow]

and we can simplify this further

[somber coyote]

hegel:

hegel:

[trail minnow]

oh fucc

that's supposed to be cos^2(theta) - sin^2(theta)

wait nvm im literally stupid

i forgot what I actually wrote like 12 seconds ago

[coarse bronze]

hmmmm ok

[trail minnow]

ye

if you want you can change that to 1 - 2cos^2 (theta) instead

[coarse bronze]

i dont see why all of those identities are necessary tbh

[trail minnow]

because often that's a nicer representation of cos(2x) than the cos^2 - sin^2

like its just useful to know

[coarse bronze]

what about 2cos^2 - 1

[trail minnow]

why do we have it?

[coarse bronze]

no how to get there

[trail minnow]

oh

$\sin^2(\theta) + \cos^2(\theta) = 1$ right

[somber coyote]

hegel:

[coarse bronze]

yes

[trail minnow]

so $\sin^2 (\theta) = 1 - \cos^2 (\theta)$

[somber coyote]

hegel:

[coarse bronze]

yes

[fallow edge]

just complete the substitution with cosine instead of sine

[somber coyote]

hegel:

[trail minnow]

that's a pretty general rule

in most cases when someone goes from an equation involving sin to an equation involving cos (or vice versa)

its Pythagorean identity

[coarse bronze]

hmm that one sort of confuses me but oh well

[trail minnow]

again its just rewriting sin^2 in terms of cos^2

and subbing cos^2 in

[coarse bronze]

oh i think i get it now

ok now for the half angle identities

[trail minnow]

ok back

[coarse bronze]

wb

[trail minnow]

uh for those

just rewrite theta as 2 * theta/2

and apply double angle identities

i think

[coarse bronze]

oh ok

[trail minnow]

ye

[somber coyote]

hegel:

[trail minnow]

uh oops

lol

[stuck torrent]

Imagine x=sin(x

[trail minnow]

stop

no

anyway

this can be rewritten as $2\sin^2 \left(\frac{\theta}{2} \right) = 1 - \cos(\theta)$

then you just sqrt both sides

wait oops im stupid

[somber coyote]

hegel:

[trail minnow]

yeah sorry about that

i forgot the 2 : p

[coarse bronze]

u mean divide both sides by 2

[stuck torrent]

Wait, what is cos(θ)? I’ve only learned cos(θ ?

[trail minnow]

you divide by 2 and get $\sin^2 \left(\frac{\theta}{2} \right) = \frac{1 - \cos(\theta)}{2}$

[somber coyote]

hegel:

[trail minnow]

this makes sense so far right?

[coarse bronze]

i think but i never knew that u could divide the inside of the () by 2 to get 2sin^2

[trail minnow]

you can't

i messed up writing original formula

actually i'll rewrite

$\cos(2 \theta) = 1 - 2\sin^2 (\theta)$

[somber coyote]

hegel:

[trail minnow]

right

[coarse bronze]

o

[trail minnow]

so we just do a nice trick and do $\theta = \frac{\theta _1}{2}$

[somber coyote]

hegel:

[trail minnow]

and get $\cos(2 \cdot \frac{\theta_1}{2}) = 1 - 2\sin^2(\frac{\theta_1}{2})$

[somber coyote]

hegel:

[coarse bronze]

ok that makes sense

[trail minnow]

so we can simplify this to $\cos(\theta_1) = 1 - 2\sin^2 \left(\frac{\theta_1}{2}\right)$

[somber coyote]

hegel:

[trail minnow]

and from here we simply rearrange our equation

$2\sin^2 \left(\frac{\theta_1}{2}\right) = 1 - \cos(\theta_1)$

[somber coyote]

hegel:

[trail minnow]

so here we just divide by 2

$\sin^2 \left(\frac{\theta_1}{2}\right) = \frac{1 - \cos(\theta_1)}{2}$

[somber coyote]

hegel:

[trail minnow]

this makes sense, right?

[coarse bronze]

yes

[trail minnow]

yep

[coarse bronze]

i just need to remember all the double angle ones to derive these

[trail minnow]

yes

[coarse bronze]

but then i remember the double angle ones by deriving from other ones

[trail minnow]

really you can just remember sin(a + b) = sin(a)cos(b) + sin(b)cos(a)

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

for the half angle formula for cos you do the same thing

[coarse bronze]

ok ty for helping me

[mighty gull]

cos(a+b)= sin(pi/2 - (a+b) )= sin((pi/2-a)-b), so no need to remember both tbh

[trail minnow]

but this time write $\cos(2 \cdot \frac{\theta}{2}) = 1 - 2\cos^2 \left(\frac{\theta}{2} \right)$

[somber coyote]

hegel:

[restive wren]

What next?

[somber coyote]

187:

[restive wren]

So the Answer is +-omega and +-omega^2

Oh ok

Square root of unity....

[upper karma]

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[frozen bluff]

which is which?

[limber whale]

I know this is an easy question

But how do I solve

What is the side of a square whose diagonal is a?

Also: Find the area of an equalateial triangle if the side is a

[silent plank]

what have you tried?

[frozen bluff]

if anyone can do this

i will love u forever ;-;

[limber whale]

@silent plank for the first, I have no clue

for the second, I tried splitting the base in half and finding the area

[silent plank]

they're both doable using pythagoras
they can also be done using trig but i don't think you've learned that yet

[limber whale]

I have not done trigonometry yetg

[silent plank]

just pythagoras is enough

what is angle AOC in terms of AOB and BOC? @frozen bluff

[frozen bluff]

uh

i gueesed on test

ill just retake

;-;

[limber whale]

@silent plank I'm still having trouble

[silent plank]

let the side length of the square be 's' and the diagonal be 'a'
what's your equation when applying pythagoras?

[limber whale]

,tex $b^2 = (1/2a)^2 - a^2

[somber coyote]

dscrdjm:
Compile Error! Click the reaction for details. (You may edit your message)

[limber whale]

@silent plank

[upper karma]

We have 7x+30 + 9x + 42 = 104

[silent plank]

are you using b as the side or the diagonal

[limber whale]

Oh sorry that is the one for question two

for question one I get

,tex sqrt(a^2-a^2)

[somber coyote]

dscrdjm:
Compile Error! Click the reaction for details. (You may edit your message)

[silent plank]

type it out normally

[limber whale]

sqrt(a^2-a^2)

[silent plank]

howd you get that? that's 0

[limber whale]

I know

[silent plank]

whats the formula for pythagoras?

[limber whale]

a^2 + b^2 = c^2

[silent plank]

lets use d for the diagonal and s for side lengths
so we don't get mixed up

substitute those into the formula

[frozen bluff]

can anyone solve that?

[limber whale]

s^2 = d^2 - a^2

[silent plank]

what are the properties of a square?

[frozen bluff]

idk tbh

im a sophmore 2nd week of school idk this ;-;

All four sides are congruent. Opposite sides are parallel. The diagonals bisect each other at right angles.

[silent plank]

you are able to substitute both edges for s

[frozen bluff]

there u mean that?

[silent plank]

that was aimed at dsc

[frozen bluff]

by s?

oh

[limber whale]

Oh okay

s^2 = d^2 - s^2

[silent plank]

you weren't supposed to move an s^2 to the other side

what is that shape squidward?

[frozen bluff]

is the asnwer 87?

a Square?

[silent plank]

nope, 87 deg angle. clearly not a square

[frozen bluff]

RECTANGLE

IDK ;-;

[silent plank]

same applies ^

[frozen bluff]

oh

a rhombus

[silent plank]

yeh

[frozen bluff]

LMFAO

fr

damn

so whats the asnwer?

[upper cosmos]

isnt it just 87

[silent plank]

yeh. but it seems that was guessed without proper justification

[upper cosmos]

opposite sides are congruent so opposite angles have to be as well

[limber whale]

Wouldn't the other angles be 93

also @silent plank I thought you saidthat both edges are s

[frozen bluff]

ill just put 87

;-;

[silent plank]

yes.
so you would have
s^2 + s^2 = d^2

[limber whale]

Okay

[silent plank]

simplify and write s in terms of d

[frozen bluff]

IT IS 87 I GOT IT CORRECT u know me

[limber whale]

wdym s in terms of d

Like

sqrt(s^2+s^2)

[silent plank]

that would be the other way around,
d in terms of s

(your s^2 terms can be added together)

[limber whale]

sqrt(1/2(2s))?

[upper cosmos]

what r u trying to solve

[silent plank]

side length of square in terms of diagonal

[limber whale]

@silent plank

[silent plank]

from
s^2 + s^2 = d^2
2*s^2 = d^2
how would you continue to isolate 's'

[limber whale]

S = sqrt((d^2)/2))

also is there a way to do square roots with TeXiT

[silent plank]

\sqrt{}

can that be simplified?

[limber whale]

,tex \sqrt{(d^2)/2)}

[somber coyote]

dscrdjm:
Compile Error! Click the reaction for details. (You may edit your message)

[limber whale]

Um

yes

[silent plank]

use dollar signs on the edges instead of .tex

[limber whale]

$\sqrt{(d*2)/2)}$

[somber coyote]

dscrdjm:

[limber whale]

@silent plank is the simplification d over sqrt(2)

[silent plank]

it was still correct before but now its wrong

[limber whale]

Is there a way to do fractions? and is there a menu for all this?

[silent plank]

just type normally and learn tex later

[limber whale]

kk

Isn't It d over sqrt(2)

[upper cosmos]

yes

[silent plank]

do you know how to rationalise that?
people don't like radicals in the denominator

[upper cosmos]

S=(D*sqrt2)/2

$S=\frac{D\sqrt2}{2}$

[somber coyote]

MineClipper:

[limber whale]

Is that it simplified?

$S = \sqrt{\frac{d^2}{2}}$

[somber coyote]

dscrdjm:

[limber whale]

@upper cosmos @silent plank

wait

[upper cosmos]

no

[limber whale]

I remember how to rationalize now

[upper cosmos]

the d wouldnt be squared

actually thats technically right

just u need to rationalize it still

[limber whale]

$S = \frac{{2}d}{2}$

[somber coyote]

dscrdjm:

[limber whale]

@upper cosmos

[upper cosmos]

no

$S=\frac{D\sqrt2}{2}$

[somber coyote]

MineClipper:

[limber whale]

okay wait

[upper cosmos]

u want me to work it with u?

[limber whale]

so you multiply it by two

[upper cosmos]

no

[limber whale]

to make the denom ration

[upper cosmos]

well kinda

yeah

[limber whale]

Yeah

So then d^2 * 2

2d^2

[upper cosmos]

let me do my work and take a pic

[limber whale]

Oh okay

I see it now

So it's like:

$S = \sqrt{\frac{d^2}{2}}$

[somber coyote]

dscrdjm:

[limber whale]

then:

$S = \sqrt{\frac{d^2}}{2}$

[upper cosmos]

take a look at that

[limber whale]

Yeah

[somber coyote]

MineClipper:

MineClipper:

[limber whale]

okay so then to turn that to a

$a=\frac{a\sqrt2}{2}$

[somber coyote]

dscrdjm:

[silent plank]

s=

[upper cosmos]

$S=\frac{D\sqrt2}{2}$

[somber coyote]

MineClipper:

[silent plank]

or x if you prefer

[upper cosmos]

thats our equation

we had this before

[limber whale]

It says the answer is $1/2\sqrt{2a}$

[somber coyote]

dscrdjm:

[limber whale]

In the book

@silent plank @upper cosmos

[upper cosmos]

s^2 + s^2 = d^2
2*s^2 = d^2

this is the orginal?

[limber whale]

The original is:

What is the side of a square whose diagonal is a?

@upper cosmos

[upper cosmos]

so its $A^2 + B^2 = C^2 right?$

[limber whale]

Y?

*?

[somber coyote]

MineClipper:

[upper cosmos]

its a square right?

[limber whale]

yes

[upper cosmos]

and we r tryign to find the diagnol

[limber whale]

No the side

[silent plank]

question says diagonal is a

[limber whale]

fromthe diagnoal

[upper cosmos]

ooh crap my b

$D^2 = S^2 + S^2$

[somber coyote]

MineClipper:

[upper cosmos]

and diagnol is a?

[limber whale]

Yes

[silent plank]

can you show a pic of the answer?
not sure if you texed it properly

[silent plank]

oh

yeh thats the same as what we had

[upper cosmos]

yeah

[limber whale]

It is?

[somber coyote]

MineClipper:

[silent plank]

sqrt doesn't extend to the a
so its just a * sqrt(2)/2

[limber whale]

Oh yeah because youcan move it in cuz the commutative property

Oh wait, so @silent plank is the answer in the book wrong?

[silent plank]

its the same

[upper cosmos]

no

its the sme thing

i have a multiplied by the sqrt of 2

[silent plank]

they just wrote it in a different order/format

[upper cosmos]

the book as the sqrt of 2 multiplied by a

[limber whale]

ohhh

okay I see it now

Thanks 🙂

[upper cosmos]

👍

[limber whale]

Theres another problem similar to this one

That I also need help with

[upper cosmos]

what is it?

[limber whale]

Whats the area of an equalitateral triangle if a side equals a

[silent plank]

formula for area of a triangle?

[limber whale]

a = 1/2hb

[upper cosmos]

no

[silent plank]

use parentheses

big A for area

assuming you mean (1/2) base * height

[upper cosmos]

$A=\frac{\sqrt{3}}{4}a^2$

[somber coyote]

MineClipper:

[upper cosmos]

little a is side length

[silent plank]

we're going through the process to reach that

(w/o trig)

[upper cosmos]

oh my b

[limber whale]

Yeah I mean (1/2b * h

[silent plank]

parentheses should be around the (1/2)

[somber coyote]

dscrdjm:

[silent plank]

or even use \frac12 if you're using tex

[upper cosmos]

$A = \frac{1}{2}hb$

[somber coyote]

MineClipper:

[silent plank]

you know the base which is a
how would you find the height?

[limber whale]

Pythagoras

Theorem

[silent plank]

what lengths are you using?

[limber whale]

A

The length is a

@silent plank

[silent plank]

and the other length of your right triangle?

[upper cosmos]

its not a right triangle

its an equilateral

u have to use trig

[silent plank]

ur drawing one to use pythag

(trig is just 1/2 a^2 sin(60)

[limber whale]

other length is 1/2a

[silent plank]

pls parentheses or just write a/2 for that

[limber whale]

a/2

okay so then you get

$b = \sqrt{a^2 - \frac{{a}{2^2}}^2 $

[silent plank]

{}{} for frac

you mean h for height

[stuck torrent]

close the }

You don't need another {} around the frac

[somber coyote]

dscrdjm:
Compile Error! Click the reaction for details. (You may edit your message)

[stuck torrent]

$h = \sqrt{a^2 - \left(\frac{a}{2}\right)^2} $

Is this what you want?

[silent plank]

its (a/2)^2

and h =

[somber coyote]

Whoever:

$h = \sqrt{a^2 - \left(\frac{a}{2}\right)^2} $

[stuck torrent]

This?

[limber whale]

yeah

How did you do that?

@silent plank

It's that right?

[stuck torrent]

...

I wrote it

thx

[limber whale]

Yeah

Thank you

SO then you multiply it by the base and you get @silent plank

$A = {a}\sqrt{a^2 - \left(\frac{a}{2}\right)^2} $

[somber coyote]

dscrdjm:

[silent plank]

and then simplify the root

[limber whale]

so it's

$A = {a}*{a} - \frac{a}{2}$

[somber coyote]

dscrdjm:

[limber whale]

@silent plank

right?

[silent plank]

nope

subtract the terms inside the root first

and i gtg someone else take over

[upper cosmos]

u lost me when explaining all this

so i cant help at all

@limber whale did u end up getting it

[limber whale]

no

wait yeah

[cinder portal]

could just find the area of the whole thing

then subtract small traingles

basically shoelace

[coarse bronze]

if it makes it easier for you there should be 4 right triangles and 1 rectangle that you subtract from the entire grid

[cinder portal]

don't assume anything, only use what you know

if i was u, i'd subtract 3 triangles and like 1 trapezoid

or 4 triangles + 1 rectangle works

[upper karma]

damn i dont understand how to do this

[tawdry pivot]

do you see that QR=RS?

[upper karma]

combine like terms?

sorry im not v good at math

ohhhh okay yea i remember doing this

okay thank you 😃

This should probably be in algebra but it on my geometry paper so why not put it here...

if EB = 4y - 12 and ED = y + 17, find the value of Y.

oh thats easy

im doing that now

wait no i thought of the other one

thats kinda like what i needed help with

yea I completely forgot how to do it

like I did it in school the other day no problem

but like

im just blanking rn for some reason

yea

ok

ok thanks

[radiant nimbus]

Ok if I have a unit vector AB = i+7j-2k and B = (8,3,6) how can I find A?

[weary drift]

unit vector AB = i+7j-2k
Doesn’t seem to be a unit vector, its magnitude isn’t 1. So instead the vector we have is just the vector from A to B.

If you also know point B and if you remember how to find the vector from one point to another, how can you find A?

[radiant nimbus]

hmm

@weary drift to a vector AB from A -> B I would subtract the A values from the B values. So do I just Add AB and B to get A?

like i Know that if A (x1,y1,z1) and B (x2,y2,z2) I would do (x2-x1,y2-y1,z2-z1)

[weary drift]

Ya that’s how you’d find vector AB... but your conclusion as to how to find A isn’t quite right

[radiant nimbus]

could you tell me the rule/process to find A or point me in the right direction? I don't want an answer, just wanna know how to do it.

if it's not B + AB... then maybe it's AB - B? I dunno.

[weary drift]

So as you said already, you’d find the difference of A and B’s components to find vector AB (let’s call the vector U), so...

B - A = U

Rearrange to isolate A:
B = U + A
A = B - U

[radiant nimbus]

hmm that's the associative property right?

[weary drift]

Didn’t apply associative property anywhere

[radiant nimbus]

jk jk <.<;

ok I got it wrong in this case is i 0 or 1?

like if it's i+2j+3k is it <1,2,3>?

[weary drift]

Ya

[radiant nimbus]

hmmm how did I get it wrong then

[weary drift]

Show us your work and we’ll see

[radiant nimbus]

Nvm I made a typo 😦 it worked.. A = (7,-4,8)

thanks for the help

[silent plank]

start by constructing the centre O
and lines OE and OF
what are the properties of the incircle?

the perimeter isn't needed to find the angle here

what are special about the points D,E,F?

[dark sparrow]

i remember giving you some hints on this matter...

[silent plank]

wasn't this already explained a few days ago

[dark sparrow]

to elaborate: what little you do know about FOK and EOK will be sufficient to find EDF

triangles DOE and DOF both come from a class of triangles with a special name

and keep in mind that D, E and F are all points on a circle with center O

i'm running a bit dry here

i feel like i've done almost everything short of just spoiling the solution

which involved carefully dancing around the invocation of a circle theorem you're "not supposed to" use...

and that's what we are doing

so what do you know about isosceles triangles

I mean obviously the base angles are equal
ok great

give me some pairs of equal angles in your diagram then

no

look

triangle DOE is isosceles

with DO=OE

this means that angle ___ = angle ___

fill in the blanks

...no?

cmon

you're overthinking again

how did C even get involved

I???

there isn't even a point called I on the diagram??????

...oh. you're calling the center of the circle I instead of O.

...

be consistent, please

and call it O, like you did in your first response

anyway yes

angle ODE = angle OED

mark those off on your diagram

now

triangle DOF is isosceles with DO=OF

this means that angle ___ = angle ___

ok

great

now

consider angles ODE, OED and EOK

what can you say about these

no

not necessarily

as in it might happen but it's such a niche special case that it's not worth considering at all

ok

good

now

consider angles FOK, FDO and OFD

what can you say about those

okay good

so now

given that angle ODE = angle OED, and EOK = ODE + OED

can you express ODE in terms of EOK

then what is ODE in terms of EOK

ok great

now

analogously to what you just did

express ODF in terms of FOK

[buoyant idol]

fok is my bios teachers name

[dark sparrow]

ok @buoyant idol thank you for injecting this completely irrelevant detail into this conversation

[buoyant idol]

lmao

[dark sparrow]

ok great

now

what is ODF + ODE

no

ODF + ODE is not FOK

geometrically, what angle is literally the sum of those two angles

no, that's not what i'm asking you

look at those two angles, ODF and ODE

when you put them together

what angle do you get

don't overthink it

exactly!

and now you can express it in terms of FOK and EOK.

and, surprise, you know their sum!

is WHAT not 54 degrees

[silent plank]

you can try to prove that yourself

[dark sparrow]

is this unrelated to what we were doing

[dark sparrow]

ok fine

fine

fine

fine

fine

fine

if you didn't want to go along with my solution then why didn't you say that right away

are you asking me why two tangents to the same circle originating from the same point are the same length

bc you are most certainly capable of proving that yourself

i don't know what the fuck you're gonna use

and i'm too busy trying to not lose my mind to care

[fleet wolf]

yes you can use congruent triangles

[dark sparrow]

YOU'RE FUCKING WELCOME FOR NOTHING.

i did exactly nothing worth thanking me for

[vale bobcat]

hoho, I learned how to convert a decimal into a fraction, kono Dio da!

[balmy pelican]

lmao

[stiff bolt]

guys when you want the exact value of cos, and you have cos(3pi/4 + 4pi/3) then you add first and then find the value right?

You cant take cos 3pi/4 and cos 4pi/3 and then add the values

that's cheating correct?

[dark sparrow]

it's not "cheating", it just doesn't work. $\cos(x+y) \neq \cos(x) + \cos(y)$ in general.

[somber coyote]

Ann:

[stiff bolt]

@vague light

what now

Thank you Ann

[vague light]

Okay

No

I used the double angle identity

[stiff bolt]

No you didn't cause I told you to use it dingus, and it still doesn't work

[vague light]

@dark sparrow

[dark sparrow]

,w cos(4π/3 + 3π/4)

[somber coyote]

Display results online and refine query

Requested by Ann#0413

[dark sparrow]

checks out

[vague light]

Thank you

[stiff bolt]

No I dont wanna hear it, it's a singular case where it works

In general compound angle 60-45 should be used

[vague light]

Dude

Instead of using the sum

I just use the two angles in the sum

Cause it’s easier

Than 25pi/12

[dark sparrow]

this is not geometry

[upper karma]

What would be then?

It is in my geometry class

[weary drift]

looks like #proofs-and-logic ?

[upper karma]

Ah ok

Thanks

[finite frost]

Wouldn’t this been just Pythagorean theorem backwards? Its been too long sense I’ve done this mess

[weary drift]

still use pythagorean theorem, yes

[finite frost]

It’s a backwards version though, yes?

[weary drift]

in a sense, yes

[finite frost]

Sense it’s provided with c^2

[weary drift]

where c is the length of the hypotensure, ya, i guess you're "working backward" to find the length of one of the legs

[finite frost]

So like b^2= c^2-a^2

What I’ve been messing with just feels weird bc it comes out with radicals

[weary drift]

if c = hypotensure, a = length of BC, b = AC, then ya you got it set up right

,w sqrt(40^2-12^2)

[somber coyote]

Display results online and refine query

Requested by RokettoJanpu#6002

[weary drift]

you got it

[finite frost]

Guess I’ll just round off with a whole and run with it

Thanks bud ^~^

[weary drift]

np man

[frozen bluff]

pls help

[formal bolt]

Have you tried something?

[frozen bluff]

bruh idk how to do it

its khan academy

[formal bolt]

I see, there is a formula for the area of a circle

[frozen bluff]

i have 5 questions

mhm

[formal bolt]

Try to search it on google

[queen river]

Hey. I am new to this server and my math needs to improve and I was wondering if anyone would help me.

[river gull]

What grade level

[queen river]

And my class has geometry and I need help with it.

10th

Geometry

[river gull]

@frozen bluff it’s 4 x Pi

[queen river]

What?

Will you help me?

[river gull]

was answering someone else question

maybe, I need to see the problem

[queen river]

Oh ok

You're gonna be mad

[river gull]

okayy

[queen river]

I start school next Tuesday and I start Geometry math 2nd trimester 2nd period

Can we still be friends

[fleet wolf]

triangle EHG has area $\frac{1}{2}\cdot EG\cdot JG$

[somber coyote]

rockpaperscissors:

[fleet wolf]

now what is the area of reactangle EGJI?

yeah

which is twice area of triangle EHG

[dark sparrow]

omg

not this again

@upper karma why did you delete your problem again

[eager pendant]

lmao

@upper karma why

I wanted to read the problem

[small raptor]

The person does that all the time. I stopped caring what problem they have at this point.

[worthy root]

[steep reef]

∫dxdθ

[worthy root]

,w ∫dxdθ

[somber coyote]

Display results online and refine query

Requested by Krishna(stop listening rap)#4971

[jagged lantern]

Hey, i have a question

What's the most absurd SOHCAHTOA/Trig fact you have ever heard

[weary drift]

some people use tg for tangent and ctg for cotangent

[umbral snow]

Some people use sen for sin

[dark sparrow]

sen for sin is because "sin" means "without" in Spanish and Portuguese iirc

[steep reef]

Thas not geo

[dark sparrow]

trigonometry fits in this channel, @steep reef.

[steep reef]

But I speak Spanish

$/4/3/-/44

[jagged lantern]

@umbral snow @dark sparrow I really like that fact lol

[nocturne musk]

hello

I need help, if possible

I just need a.

[quiet mason]

if some line is divided into two at the midpoint or any point on the line
the length of the line=sum of the lengths of the points into which it is divided

[wide ferry]

I just need the y

[olive solar]

don't just give the answer please, pride

[upper karma]

Ok. Sorry bout that

[olive solar]

@wide ferry you would just reverse the midpoint formula for y

it'd be (y+7)/2 = 5

[wide ferry]

ok thanks

and can you help me with one more

[olive solar]

do you know what the distance formula is

[wide ferry]

yea

oh

i did that

and i got the square root of 5.5

[olive solar]

uh

[upper karma]

What scope are yuo asking this in terms of?

sqrt (x2-x1)^2+(y2-y1)^2?

@wide ferry

[wide ferry]

yes

nvm

i got it

[tender igloo]

Is it solvable

[upper karma]

,rotate -90

[upper karma]

A triangle can not be evaluated for only 1 lenght and 1 angle

no such theorem exists

to my knowledge

[tender igloo]

Ok

[tender igloo]

It is fine

,rotate -90

[tawdry pivot]

,rotate

[upper cosmos]

what r u solving for?

[upper karma]

help pls

[worthy root]

Which problem @upper karma

[buoyant idol]

whats sec and csc?

[devout shell]

Secant and cosecant

Secant is 1/cos(x) and cosecant is 1/sin(x)

[buoyant idol]

@devout shell is it possible if u list out all these strange things

if u can, no need if ur busy...:/

[devout shell]

There are six trigonometric functions

The three you are most familiar with are some, cosine, and tangent

[buoyant idol]

ye

[devout shell]

Those are written as sin(x), cos(x), and tan(x) respectively

In terms of the sides of the triangle:

sin(x)=opp/hyp
cos(x)=adj/hyp
tan(x)=opp/adj=sin(x)/cos(x)

[buoyant idol]

oh wow tan(x) is sin/cos didnt know lmao

[devout shell]

The remaining three trig functions:

Cosecant, csc(x)
Secant, sec(x)
Cotangent, cot(x)

Are the reciprocals of sine, cosine, and tangent respectively

[buoyant idol]

wait whats cot?

[devout shell]

Now we know that:

csc(x)=1/sin(x)
sec(x)=1/cos(x)
cot(x)=1/tan(x)=cos(x)/sin(x)

[buoyant idol]

ah okok thx

[devout shell]

You should prove that tan(x)=sine/cosine

[buoyant idol]

yeye the hyp cancels each other out

[devout shell]

Very good

[buoyant idol]

lmao

can u do like inverse cotangent or smth?

[devout shell]

There is an inverse for cotangent, though I forget what the interval for the inverse function is

[buoyant idol]

ah okok

thx tho

[devout shell]

Given some value of cotangent, provided it is in the interval, you can use the inverse function to recover the angle measure

[buoyant idol]

wait can u get cotangent in terms of hyp,opp,adj?

like cot = hyp/opp or smth (its probs not correct but im asking if u can do this lmao)

[devout shell]

Take the reciprocal of tangent

That will be cotangent

[spark stag]

well, $\tan \theta = \frac{\sin \theta}{\cos \theta}$, and $\cot \theta = \frac{1}{\tan \theta}$

[somber coyote]

Namington:

[buoyant idol]

ah okok