geometry-and-trigonometry

[dire moth]

Oh

Did you figure out my problem?

[forest drift]

I got the answer

[dire moth]

I need help

Oh

Thank you

So much

[forest drift]

I'll send

[dire moth]

Do u have the work too

[forest drift]

after some time

[dire moth]

So I can see how u did it

[forest drift]

yeah

[dire moth]

Ok thank you

[dire moth]

How long, sorry if I’m being rude

[forest drift]

10 minutes

[dire moth]

Ok

Thx again

[hoary mango]

did i use the law correctly

[forest drift]

theta = 45 degrees, w = 19 root 2, h = 9.5 root 2

x = 9.5
area = 3/4 x 361 = 270.75 m2

[hoary mango]

non-right triangles arent bad, it's just way too complicated

[forest drift]

its not complicated

just its weird

[dire moth]

Is it possible if you can do it on a screen or draw it?

Would be easier to understand

If it’s no trouble

[forest drift]

see

it is a square

[dire moth]

Ye I can kinda see it

[forest drift]

so just solve it

diagonals bisect angles so theta is 45 degrees

[dire moth]

Oh

[somber coyote]

officalzenv2

[hoary mango]

let me just remember this for later

[dire moth]

Can you type it one more time

If possible

[forest drift]

what

the answer?

[dire moth]

Yes

Like the work

So I can understand well

In case a question like this pops up in an exam

[dire moth]

If possible

[forest drift]

i'll send it

[dire moth]

Ok

[sturdy forge]

it's a square

[dire moth]

Can you send a picture of it if possible

[sturdy forge]

use the properties of a square and you'll get it easily

[dire moth]

Ye so 19 is the x as a diameter right since it’s a square?

So I go 19/2

Which is 9.5

Which is X

[sturdy forge]

so x = 1/2 x 19, theta = 45 degrees, w = 19 root 2(diagonal of a square) and h = 1/2 x diagonal

[sturdy forge]

yes yes

[forest drift]

thats the same thing i did

[sturdy forge]

bruh

i taught you how to do @forest drift

[forest drift]

partly yeah

I had an idea

[dire moth]

You actually explained it well

Thx

[forest drift]

he's my close friend

[sturdy forge]

no problem

[dire moth]

Wait Is that everything I need to do?

[forest drift]

ye

[sturdy forge]

yeah

[dire moth]

Ok

[dire moth]

Wait so W is the radius?

Right

Wait and to find theta

[sturdy forge]

you mean diagonal?

[dire moth]

Is that theta?

[dire moth]

Cause I thought I do coz-^1 = (9.5/19)

Do I do that to get theta

[forest drift]

you have to know the property of parallelogram which says that diagonals bisect angles of parallelogram. The square is a modified parallelogram so yeah 2 times theta=90 and thus theta is 45 degrees

[sturdy forge]

exactly

if you want to find theta using trig, then wait, i'll send solution

[dire moth]

Oh

Ok

[forest drift]

with trig also you'll get 45 degrees vsg

[sturdy forge]

arccos(19/w) = theta

[dire moth]

Wait so since it’s a square I just divide the 90 degree angle by 2

[sturdy forge]

ik that

[forest drift]

yeah yeah

[dire moth]

Which is 45?

[forest drift]

exaclty

[sturdy forge]

yes yes

[dire moth]

Ohhhh

That’s so much easier

[sturdy forge]

yeah

[dire moth]

Ok

Thx so much

[sturdy forge]

np np

[forest drift]

any other doubts?

[dire moth]

No i think I’m good for now

Thanks so much

[dire moth]

Can I add you in case I need more help some time

[sturdy forge]

sure

[dire moth]

Thx

It said I can’t add you

[forest drift]

add me also

[forest drift]

hehh?

https://tenor.com/view/kuch-to-gadbad-hai-daya-acp-pradyuman-shivaji-satam-cid-gif-21295070

[sturdy forge]

wait

i've sent

[dire moth]

Ok

Thank you

Both of you

[hoary mango]

non-right triangles are so shit

why couldn't trig just stick to advanced functions and right triangles

im tired of this

im skipping to calculus

[forest drift]

If you need help in calculus please let me know

[hoary mango]

its a swallowing choice but i've already dug up enough fucking dirt to know the basics of calculus

but trigonometry is just shitty

its like mining obsidian with a tree branch

[forest drift]

nice - so u know differentiation , integration and limits?

[forest drift]

umm

ok

[hoary mango]

yeah i could probably do some riemann sums in my head

[forest drift]

that is a good level

[hoary mango]

im used to limits

[forest drift]

btw which grade/year are you in?

[hoary mango]

but im kinda only on the basics of differentiation

rather not say

[forest drift]

ok

thats fine

I respect your privac

y

[deep trench]

any software for create maths notes or maths books?

[dim hatch]

latex @deep trench

[grizzled kindle]

@fiery solstice any three points on a sphere has a unique circle passing through them

Find its centre and then find it's normal line

Do it for any three faces of tetrahedron and then solve the system of equation

Coz it's normal lines will intersect

[median fern]

Hi

So

Can you tell me how do I find

BF

When AD // BC and AD =16cm and BE = 10 cm

EF perpendicular BC

[grizzled kindle]

BE = CE ?

If yes then BF = CF

BC = AD = 2BF = 10

So... BF = 5 cm

[median fern]

But BC = 16

[grizzled kindle]

Huh?

[median fern]

Ye

[grizzled kindle]

It's a parallelogram

[median fern]

Hm

[grizzled kindle]

Opposite sides are equal is in its definition

[median fern]

Ye

AB = BC AND AE = BF BUT

AB = BC is true but AD ain't 2BF

Besides AB = BC = 2BF = 16 cm not 10 cm

[grizzled kindle]

Perpendicular in isosceles triangle are perpendicular bisector

[grizzled kindle]

AB = BC ??

[median fern]

Ye

Not

Not

Sry

AD

It's AD

Here's the full question

[grizzled kindle]

Oh sorry here is messed up I meant 16 not 10

[median fern]

So who is 2BF = 16

Anyway

[grizzled kindle]

BC = 2 BF coz isosceles triangle property

Then that would be equal to AD coz parallelogram

Then solve it

[median fern]

Does EF divides the BC into 2 halfs

[grizzled kindle]

Yes

[spark moon]

Anyone understand this

[dusk charm]

opposite sides are equal in a parralelogram, so equate both the sides and solve for t

[spark moon]

ok guys im taking a quiz right now

and its very hard for me

and i really didnt understand the class

and the teacher kicked me from multiple classes for not understanding the material

[dusk charm]

ayo, we need to teach u parralelograms?

do uk wht supplementary and complementary angles r?

[spark moon]

supple

is like 180

complementary

are like 90

[dusk charm]

yes now tell the complementary of 30

[spark moon]

uh????

whaat

[dusk charm]

wht is the complementary angle of 30 degrees?

[spark moon]

uhhhhh

[iron spoke]

if complementary means x + y = 90º

if x is 30

what is y?

[dusk charm]

the complementary of an angle is 90 degrees - the angle
while supplementary is 180 degrees - the angle

so angle + complementary = 90 degrees

[spark moon]

60

OOOOOOOOOOOOOO

i get it nowww

ooooooo

[dusk charm]

and supplementary of 30?

[spark moon]

180-90?

[harsh heron]

Why is 6 afraid of 7?
-0.89594417

[fast ledge]

Angles: SBC = 30, BCA = 50

what is the angle of ABC

[grizzled kindle]

Coz 7 ate 9

[deep trench]

challengue: resolve next in 3 minutes

[dreamy ridge]

So if x is a positive integer, x(x-1)(x-2)!/ -x! = 120(x-2)!

The (x-2)! cancels

So x(x-1)/x! =120

Then write the x! as the numerator times (x-2)!, things will cancel

And then you can go from here I’m p sure

[buoyant oyster]

Is the denominator (-x)! (Which are singularities at negative integers) or something else

[dreamy ridge]

I think it’s -(x!) as the lack of brackets implies that it isn’t the factorial of negative x but rather the negative of the factorial of x

Same sort of logic as -x^2, not (-x)^2

[buoyant oyster]

so it’s -1 = 120(x-2)!

The gamma function over the positive reals is always nonzero

[grizzled kindle]

Because it used factorial rather than gamma notation ig x is a integer so... I'd say no solution

[ivory wyvern]

yo can someone help

[buoyant oyster]

hmm

WAIT

make a triangle from the center of the circle to the secant line

[ivory wyvern]

alr

[buoyant oyster]

the angle of the angle at the center of the circle is 360-267

Wait

yeah

I’m gonna guess here

the triangle is an isosceles triangle

[ivory wyvern]

ok

[buoyant oyster]

Thus the other angles are equal, and equal to the angles next to them

So decide the remainder of the triangle’s total angles (180°) by two

(90-(360-267))/2

[ivory wyvern]

alr alr thx

[median fern]

So hi

Can anyone help me

How do I do it

Ping me

[royal citrus]

traingle BEC is half of abcd (triangle in same base as parrallelogram)
and trangle bec = efc (divided by midpoint)

[hoary mango]

where should i insert the multiplication part?

use law of sines

nah law of sines is overly confusing

can i get some help with it

alright that wasn't really contextual

can i get some help with inserting sides with the law of sines with 2A1S and 2S1A?

[hoary mango]

plug in law of sines?

bro thats a shit ass tip

i am AWARE you need law of sines

i asked for help with PLUGGING the sides in

[dusk charm]

sorry

i didnt read the question properly

[hoary mango]

its okay though

[dusk charm]

1 sec

[dusk charm]

first u have to find B
so since c/sin C = b/sin B
10/sin 59 = 14/sin B
so sin b = (sin 59/10)*14

and B = arcsin(sin B)

i should learn latex

After solving B u should be able to find A similarly

[hoary mango]

oh nice

[dusk charm]

and then i am sure can find the other length

[hoary mango]

im already used to law of sines

next is the cosines

[dusk charm]

law of cosines is annoying

[hoary mango]

i mean

the pythagorean theorem equation part is nice with a calculator

but do teachers even give you calculators when doing these shit?

[dusk charm]

most probably

how the r u supposed to find sin(59)

[hoary mango]

yeah

i'd understand why calculators are used in trigonometry

because no mfs gonna wanna solve the sine of 40 degrees

[dusk charm]

u should be fine

[hoary mango]

since it has a whole damn decimal section, i'd say you'd be dead af without a calculator

[dusk charm]

these dont come for me lol

[iron spoke]

expand 19 terms of the sine taylor series in the exam :^)

[modest birch]

Is the answer 52?

[dark sparrow]

no

[modest birch]

What is it

[silent plank]

show what you did that led to 52

[dark sparrow]

we don't give out answers here

how did you get 52?

[upper karma]

Where does your 5 come from ?

[modest birch]

which 5

[silent plank]

at the very bottom

[upper karma]

You put sqrt (57^2 + 5^2) but I don’t understand where you even get the 5 from sorry

[modest birch]

it was meant to be 52 squared

[upper karma]

But that makes even less sense

[silent plank]

wheres 52 coming from

[upper karma]

It means you made a hypothesis that it was 52 and tried with that 💀

[silent plank]

your attempt at applying the sine law doesn't make much sense either

[modest birch]

i split it into 45 on both triangles

is that even correct

[upper karma]

Nooooo

Ok that’s the problem

[modest birch]

yea xd

[upper karma]

So what you want to do for looking for that angle is starting by getting angle C thanks to either arcos either arcsin either arctan

With the sides of your big triangle

how do i answer for part B

[grave pond]

I think doing trigonometry here is a detour. With a height in a right triangle, the small triangles are similar to the large one, and since we have all the sides of the large one now, no trig is needed.

[upper karma]

Then once you’ve found angle C you want to add 90 to it because you obviously have angle D a right angle and substrat the sum to 180

That gets you angle E let’s call it

The subtract angle E to 90 because you have a right angle

I’m not saying that helps your problem but that’s answering to your second question lol

[modest birch]

Whats arcos, arcsin, arctan ?

[upper karma]

can someone tell me if this is right

[upper karma]

I think it is needed it’s just about the “aggrandizement” in Thales theorem (that’s what we call it in France I ain’t gob lie idk what it is in the US)

[silent plank]

don't worry about the trig. consider similar triangles

[upper karma]

The inverse of the sine tan and cos you normally have a tab on your calc that says that

[upper karma]

Yes that’s what you call it I think

[upper karma]

I know RQD is 46

but what's the other one

RQD is 36

[gritty hemlock]

same logic as in how you got RQD

alternate segment theorem

BPR = ( RQP ) = 180 - ( CQP ) - ( RQD ) = 180 - ( PRQ ) - ( RQD ) = 180 - ( APQ ) - ( RQD ) :o

[upper karma]

And this one

It's a simultaneous right

X+Y = 15

what's the other equation

[silent plank]

X+Y = 15
where's that coming from

[upper karma]

So Pythagoras

AC = 15

[silent plank]

X and Y are points (of tangency) and don't represent any lengths or whatever

[upper karma]

And then if you draw the radius lines you get something like this

Both the tangents should be the same right?

[gritty hemlock]

Theorem: Suppose that two tangents are drawn to a circle S from an exterior point P. Let the points of contact be A and B, then PA = PB.

Also, use pythagoras for AC = √(12² + 9^2)

[gritty hemlock]

Idk what you're trying to find coz you just gave the figure, but I think might help!?

[upper karma]

anyone plz tell how the formula given in the solution is derived?

[dark sparrow]

weird how this highly nontrivial formula is presented as just something to invoke without justification

cause i personally do not see immediately how to derive it

all that's coming to mind rn is that maybe drawing the common tangents of the big circle and each of the three small circles might help?

perhaps something could be said about the areas of AER, BGJ and CLP

[upper karma]

ya i tried

[dark sparrow]

you tried this?

[upper karma]

i am unable to do anything with this actually

[dark sparrow]

...

[upper karma]

i was trying it by similarity

[dark sparrow]

ok so i said this a million times already but please do not reply-ping me so often

it's very annoying especially after seeing it for the 20th time

[upper karma]

ok fine

actually i dropped perpendiculars from the centers of the incircles to the tangents

[dark sparrow]

so you dropped perpendiculars from the centers of the three green circles to T, U and V?

(referring to my picture)

[upper karma]

no actually H I M N D S points

[dark sparrow]

ah

[upper karma]

what is ur approach

[dark sparrow]

let s = area(ABC) (lowercase s so as not to confuse with the point i named S)
then area(AER) = s * rA^2/r^2, area(BGJ) = s * rB^2/r^2, area(CLP) = s * rC^2/r^2

i'm not sure if this leads anywhere, but this is what i have in mind right now.

there should be a nice way to derive the area of that hexagon in the middle

[upper karma]

how

[dark sparrow]

similarity

[gritty hemlock]

Uhh

[dark sparrow]

ah. wait.

[gritty hemlock]

Do you know half angle formulas?

[dark sparrow]

AER isnt similar to ABC is it

[gritty hemlock]

the sin(A/2) and the likes in terms of s, (s-a), (s-b), (s-c) :o

[dark sparrow]

@gritty hemlock does this have to do with the problem at hand

..

what

[gritty hemlock]

XD

[dark sparrow]

does this have to do with the problem at hand y/n

[upper karma]

nope

[dark sparrow]

first time seeing that

also thank you for not answering my question directly

[gritty hemlock]

Hmm oh y!

[dark sparrow]

im not convinced the formula you shared (and now deleted) is true

[gritty hemlock]

Not sure tho 👀 coz maybe O,O_A isn't really colinear with A

wait lol it's trivial so yeah the formula must be true

Hmm

[dark sparrow]

what formula

[gritty hemlock]

the sin(A/2) one... just connect the two centers of the big circle and any short circle given, drop perpendiculars from it to any one corresponding side and use the fact that inradius bisects the angle

[upper karma]

yes that formula is true

i have the proof

[gritty hemlock]

wasn't sure but this is trivially true as A has tangents to both C_a and C, so OA and O_AA are both bisectors

[dark sparrow]

@gritty hemlock do you mean dropping perpendiculars $O_AD$ and $OF$ in this picture?

[somber coyote]

Ann

[dark sparrow]

if so, i fail to see how this proves sin(A/2) = (r - r_A)/(r + r_A) nor how we could even use that to solve the original problem

[gritty hemlock]

O_A S and OQ, joining OO_A and dropping a perpendicular from O_A to OQ

[upper karma]

this is the proof

[gritty hemlock]

A + B + C = pi ?

:o

[gritty hemlock]

Lol that also works but ... okay

[dark sparrow]

...

[upper karma]

yep

[gritty hemlock]

Ann

[upper karma]

hey but we were trying to prove this formula

[dark sparrow]

ok yes i guess i can see ansh's formula

but not how to apply it to chgupta's original problem

[upper karma]

actually i know the solution with this formula

[dark sparrow]

what did i say about the reply pings

[upper karma]

ok sorry

actually i want to know another solution with the formula given in the image i shared at first

[dark sparrow]

isnt that what we've been trying to derive all this time

[upper karma]

i don't think so

[dark sparrow]

then what the fuck have we been trying to do

[upper karma]

actually we are doing that ques with another formula

[gritty hemlock]

Are we =_= ? I just introduced a formula in an attempt to approach ur formula...?

[dark sparrow]

???

[upper karma]

i know

i tried it

[gritty hemlock]

sry that'd still be brute forcing but hey, at least you have one way of proving it!

[somber coyote]

Ansh

Ansh

[upper karma]

ok

[upper karma]

and sorry again

Hey,
I have two Cartesian equations of plans : plan P and P’
I have to verify that they are parallel
I know that for two plans to be parallel they have to have a normal vector in common. The question is how do I find it

[next remnant]

Does anyone know the formula

[grave pond]

If your equations are in the common form ax+by+cz=d, then (a,b,c) is a normal vector. You then just need to check whether the two normal vectors are parallel.
(by the way, it is "plane", not "plan").

[grave pond]

Pythagoras. When (x,y) is on the unit circle, then x²+y²=1.

[next remnant]

ohhh ok

thanks my teacher made it seem so compucated

and went way to fast

[upper karma]

Nice thank you so much I don’t study in English so when it comes to mathematical terms I know nothing so thank you for teaching me that today 👍

[upper merlin]

https://www.youtube.com/watch?v=VociZ23pA0w

some smart person in the comments who finished the solution

I later heard that this was a very well known question in comp math tho

tbh deriving this in exam is just not even close to being possible (at least for me)

[gritty hemlock]

thx lol

ohhhhh yeah right smh, the 1-sin / 1+ sin = (1-sin / cos)² smh

n yeah, we had the same in our Solution of Triangles module in hs

[upper karma]

thanks

[gritty hemlock]

@upper karma Yeah I'm from India

[upper karma]

me too

[upper karma]

r u giving ioqm this year

[gritty hemlock]

can't :o I'm already in Uni 2nd yr smh

passively sat for RMO and INMO but that's the farthest I could go

[upper karma]

ok👍

[upper karma]

actually i have one more question

[gritty hemlock]

Hmm?

[upper karma]

do u know how to find the sum of the distances of the fermat point from the three vertices

[gritty hemlock]

Nope... I vaguely remember using something like that during my +2 coordinate geometry assignments and tests but I always figured a workaround so never learned what the correct method actually was 🤦‍♂️

[upper karma]

ok no prob

[gritty hemlock]

https://math.stackexchange.com/questions/1695823/fermat-torricelli-minimum-distance

[upper karma]

thanks

[upper karma]

but how can we prove this formula R= √Ra.Rb + √Rb.Rc + √Rc.Ra ?

[gritty hemlock]

@upper karma

[upper karma]

ok

[eternal locust]

Can someone help me with this exercise? I feel like I'm missing a step

[mystic pond]

This might be a really dumb question but is we have sin(xc) can we pull out the c to get csin(x) ?

[gleaming nova]

no

[mystic pond]

ty

[upper merlin]

tf are you guys awake at 2 am?

[humble pulsar]

Timezones exist.

[gleaming nova]

no they don’t

[upper merlin]

I also live in India 🙃

[humble pulsar]

That has nothing to do with timezones not existing

[upper karma]

😅

[little terrace]

can someone help with this pls🙏

[gritty hemlock]

,texsp $\sin 2x\cdot\frac{\sqrt{3}}{2} - \cos 2x\cdot \frac{1}{2}= \frac{1}{2}$

[somber coyote]

Ansh

[balmy vale]

I need help with this

[balmy vale]

<@&286206848099549185>

NEVERMIND

[slate night]

i don’t know how to find tangent for this

this is what i got

[grave pond]

Divide by tan(75°) on both sides.

[sturdy spindle]

do you guys know this

[humble pulsar]

yes

[warm eagle]

Did you Type the dollar signs too or wat

Can someone help

[gleaming nova]

pythagorean theorem

[warm eagle]

I forgot this

This is all old homework and my marking period is done by this Friday

I’m so fucked cause I forgot this shot

And I need help

,

[warm eagle]

,

[stark briar]

do you know the pythagorean theorem

[warm eagle]

No i forgot

[gleaming nova]

when in doubt look it up

[warm eagle]

Ok

[stark briar]

you'll need to modify it a bit

[warm eagle]

Wym

I am slow dumb that down

[stark briar]

nvm

just look up the theorem

[warm eagle]

Ok

Nothin

I’m failing

[grave pond]

Are you saying you can't find what the Pythagorean theorem says? Textbooks, notes, google, everything is failing you?

[warm eagle]

Yes

But is this not it

[stark briar]

a^2 + b^2 = c^2

[warm eagle]

I know

But like

[stark briar]

a and b are lengths of legs

[warm eagle]

You cannot multiple

[stark briar]

c = length of hypotenuse

[warm eagle]

17

And only 8

So 64 - 17

Is 47

[stark briar]

then what?

wait

[warm eagle]

I multiply it

[stark briar]

why 64-17?

[warm eagle]

Bc I multiple 8 x8

[stark briar]

but why minus

[warm eagle]

And I can’t multiply 17 cause on the

[warm eagle]

Are you not supposed to minus it

[stark briar]

no

[warm eagle]

After I multiply

HUH

[stark briar]

a^2 + b^2 = c^2

[warm eagle]

So what’s the answer I’m confused

[steady sinew]

wtf are you talking about

[warm eagle]

Cauee i

DAWG

I am lsot

Bro

[steady sinew]

so am i

[warm eagle]

17 is on the left side

[steady sinew]

where did the 8 come from

[warm eagle]

So it stays the same does it not

[warm eagle]

The 8 is on top

[grave pond]

Slow down, breathe.

[warm eagle]

Okok

[steady sinew]

oh

[warm eagle]

I just type fast

[steady sinew]

its a diff question

[warm eagle]

Wait hol up

[stark briar]

the lengths of the legs are 8 and sqrt 17.

[warm eagle]

What does that mean

I don’t multiple 8

So I just subtract 8 -17?

[grave pond]

If you just keep posting in panic and never even wait to let anyone write anything to help you, then what did you expect?

[warm eagle]

True

[grave pond]

Please give people time to answer instead of just flailing with random guesses.

[warm eagle]

Idk I like to get it done bc I am very stressed in all my classes

And I have midterms

And all trust

Tamhat

[warm eagle]

Yes yes you are right that is my fault

[grave pond]

It's alright if you don't want to listen to us, but in that case why are you posting here?

[warm eagle]

No like I am

Confused

But I like to kinda learn fast

My teachers teach us the hard way

And I know some easy ways

They complicate it more

I hate it

But yea my school really does suck ass I’m not gonna lie

[grave pond]

You won't learn anything by being here unless you sometimes stop typing and wait for an answer. And then react to that answer instead of continuing your own train of thought.

[warm eagle]

Okok

I just overthink a lot

And slot of possibilities come in my head

[grave pond]

You're still doing it!

[warm eagle]

Okok so

MY FSUKT

Okok

Restart

What do I do

[grave pond]

You got pointed to the equation a^2 + b^2 = c^2.
Write down what you know using the letters in the equation.

[warm eagle]

Wait

Sorry but

17 is a and 8 is b? And I’m looking for c square correct?

[grave pond]

b is 8, yes. But a is not 17, it is the square root of 17.

[warm eagle]

OH

OHHHHHHH

[grave pond]

You're eventually looking for c.

[warm eagle]

Okok

[grave pond]

So the next step is to insert your known values of a and b into the equation.

[warm eagle]

8x8?

Or no

[grave pond]

Just take "a^2 + b^2 = c^2" and insert a=sqrt(17) and b=8.

[warm eagle]

So this is right correct

[grave pond]

No.

[warm eagle]

Dam

[grave pond]

You've made several mistakes already by getting ahead of yourself and skipping steps and getting things wrong because you lose track of the big picture. So I'm really going to insist that you take it one step at a time and for now do nothing except insert a=sqrt(17) and b=8 into the equation a^2 + b^2 = c^2.

[warm eagle]

4

[grave pond]

What.

[warm eagle]

Sorry for asking bht what do I do with the 8

[warm eagle]

My fault

[grave pond]

Once more: The ONLY thing I'm asking for at the monent is to insert the values sqrt(17) and 8 into the appropriate places in the equation a^2 + b^2 = c^2. Do not attempt to solve it yet. Do not attempt to skip all the way to a final answer. Just insert the values, and show what the equation looks like with the known values inserted.

[warm eagle]

Okok

17

And basically

[grave pond]

Are you even listening to what I'm saying?

[warm eagle]

Wait

Hol up

I am doing it

I am just typing some there here to remember

[grave pond]

It might be better to use a separate sheet of paper for your personal scratch work ...

[warm eagle]

okok

I

Is

9?

[grave pond]

Sigh.

How many times do I have to say it.

I am not askign for a number here.

[warm eagle]

o

[grave pond]

You're still attempting to jump all the way to an answer.

[warm eagle]

WAIR so I wasn’t right

I did it

I’ll tell you

So bad

Basically

17 stays the same right

[grave pond]

I am asking this and ONLY THIS: What does the equation a^2 + b^2 = c^2 look like when you have inserted the value sqrt(17) for a and the value 8 for b?

[warm eagle]

YE

8*8=

I’m starting to remember how to do it

[grave pond]

Do not do ANYTHING with the equation except inserting the values for now.

[warm eagle]

..

So basically

I found out how to do it

Like

I know how I was just having problems with the 17

Cause that was just hurting my brain

[grave pond]

Your problem is that you're not listening to ANYTHING I'm saying!

[warm eagle]

yes yes but I found out a way with what you said before

[grave pond]

So again, why are you even posting here if you're just going to ignore everything I or anyone write to you?

[warm eagle]

I went on a separate piece of paper

No no

I went what you said

I wrote it down

Wrote the values

And boom

I got my answer

[grave pond]

(Hmm, it actually turns out the final answer is in fact 9).

[warm eagle]

Yes

I told you🤪

[grave pond]

But that was not what I was asking.

[gleaming nova]

can’t blame troposphere though

[warm eagle]

True but I wrote what u said in a paper

And I did it from my self from there

[warm eagle]

Yea that’s my fault tbh

I’m just a very weird learner

But I will save ur messages so I can rem

Remember for other stuff later on

I’m getting them all right @grave pond :D

[barren musk]

What are proof Theroms

[sleek isle]

How does a 5 point and 3 lines look like?

[gleaming nova]

your question makes no sense

[dusk charm]

wha?

[zinc mural]

Would someone help in applications of logarithm derivatives?

[silent plank]

not geo/trig
also post a clear question in an appropriate channel if you want help
#❓how-to-get-help

[visual basalt]

?

[gleaming nova]

apply the formula for interior degrees in an n sided polygon

[visual basalt]

Idk how to do that ...

[dark sparrow]

do you know what formula quantum is talking about?

[hoary karma]

help ples

[dark sparrow]

@hoary karma are you asked to write the proof in two-col format

[hoary karma]

yes

i also got this one in 2 col format

[hollow oxide]

what am i doing wrong? 😦

[dark sparrow]

it's just y = (7/3)x

[hollow oxide]

oh, thanks

[hoary karma]

can i get help with these

2 column format

[knotty delta]

solution

[hoary karma]

thanks

i appreciate it

can i get help with the other one also?

[knotty delta]

solution

[hoary karma]

bro you are the best i couldnt thank you enough

[knotty delta]

np!

[iron walrus]

Can anyone tell me how I'm supposed to find the errors in the triangles?

[grave pond]

Would it sound less daunting if instead of "identify the error" it said "for each of the diagram, explain why it is impossible"?

There's not any "the" error to identify in the sense that you can know which measurement is wrong; all you can show is that they can't all be right.

[near sand]

What does it mean when two angles are a linear pair?

Say I got angle QMN and angle QMP. And QMN = 40

Oh nvm

They're supplementary, so they add up to 180

I just solved my own question

[fiery jolt]

wait i thought only triangles can be congruent

[grave pond]

Any kinds of geometric figure can be congruent.

[near sand]

Can someone help with #8?

So far, all I know is angle QMP is 120 and QMN is 40

[fiery jolt]

oh

ok

@near sand still here?

[near sand]

Yes

[fiery jolt]

now firstly u need to know what is bisect

do you?

@near sand

[near sand]

Yeah. If we have an angle ABC, and a ray BD. Then the measure of the angle ABD = measure of angle DBC

[fiery jolt]

yea exactly

[upper karma]

Can you apply standard form for the expression 3c+2a+s?

[fiery jolt]

uou know that angles QMN is 40 therefore QMX is 20

because it bisects there

QMX+XMN=40
QMX=XMN
therefore QMX=XMN=20

[near sand]

Ohh

And angle QMY would be 60 since QMP is 120

[fiery jolt]

same as QMP, PMY and QMY are the same so they are all 60 degrees

yea

exactly

[near sand]

Ah, thank you

I guess I over thought it

[fiery jolt]

yea np have a nice day

[near sand]

You too

[tight wagon]

I need help with this

[gleaming nova]

distance formula

[gritty hemlock]

How the heck are lines 2 and 3 parallel? two lines need to have the same slope for them to be parallel

[upper karma]

i didnt mean to click that

im very confused

[gritty hemlock]

Just check the slopes of each line equation, equal slope mean parallel lines, product of slope = -1 implies perpendicular lines

[wheat onyx]

Im asking for help on how to answer this, thanks in advance

<@&286206848099549185>

[slate wren]

use the similiarity of triangles

[wheat onyx]

Okay so...
9/3=4+x/4
x=8

But
9/3=5/y
y=1.67 which is different from the correct answer

Are there any mistakes made in my solution?

[upper merlin]

BDC, CEA and AFB are equilateral. We have to prove AD, BE and CF are concurrent

preferably using ceva's and menelaus'

[willow vessel]

Hello everyone

Anyone who can help me to test tge independence of each axioms of three line geometry?

[hoary karma]

again, my dumbass cannot figure out these problems, you are supposed to find x

[storm heath]

Why and how is this so?

[humble pulsar]

Euler's Identity

[somber coyote]

Pebble

[grave nova]

How do I solve this?

[grizzled kindle]

$sin( arccos(4/5) - arcsin(4/5)) \newline = Sin( arcsin(3/5) - arcsin(4/5)) \newline = sin( arcsin( 3^2/5^2 - 4^2/5^2) \newline = - 7/25$

[somber coyote]

∆ddY

[grizzled kindle]

@grave nova

[grave nova]

Woah, where did you get arccos from?

Wait was it the thing with sin(x+y)=sinxcosy+cosxsiny?

[grave nova]

The question was to compute the ordered pair (A, B)

I can do the I write left side but idk how to solve for both variables

[dark sparrow]

@grave nova still here?

[grave nova]

No I’m supposed to be sleeping sorry

[grizzled kindle]

arccos(x) + arcsin(x) = π/2

[grizzled kindle]

Yes

[frigid zodiac]

i dont understand this

like

in the tan 20 = x/5

he multiplied both sides by 5

ok i understand that

but for the cos 20 = 5/y he multiplied both sides by Y

can someone help?

[dark sparrow]

...isn't that the exact same principle tho

[storm heath]

Oh.

[storm heath]

INTELLIGENTTTTTTT.

[grave pond]

Yes? Did you want to say something about that exercise?

[silk jacinth]

if I have a triangle with side lengths of 105, 220, 320 and I want to solve angle X that's opposite to 105 through law of cosines, shouldn't the equation be:

105^2 = 320^2 + 220^2 – 2 * 320 * 220 * cos X ?

and if that's true, should X be ~6.92° ?

[dark sparrow]

,calc (320^2 + 220^2 - 105^2)/(2 * 320 * 220)

[somber coyote]

Result:

0.99272017045455

[dark sparrow]

,calc acos(0.99272017045455) * 180/pi

[somber coyote]

Result:

6.9177052509777

[dark sparrow]

seems to check out @silk jacinth

[silk jacinth]

yeah but is the original form of the equation correct?

[dark sparrow]

yes

[silk jacinth]

okay, so by that logic, shouldn't the area of the triangle be 0.5 * 220 * 320 * sin 6.92° ?

which would be around 4200 ?

[dark sparrow]

,calc 0.5 * 220 * 320 * sin(6.92 * pi/180)

[somber coyote]

Result:

4241.0145956378

[dark sparrow]

4241 but close enough i guess

[silk jacinth]

yeah the computation checks out but is the formula for the area correctly determined?

[dark sparrow]

yes

though i'm a little bit weirded out at your wording

[silk jacinth]

sry I'm tired and english isn't my first language

what I mean is

have I properly written the equation of law of cosines and the formula for triangle's area prior to solving them if I want to determine the area of a triangle with sides of 105, 320 and 220

[dark sparrow]

yes your formulas are correct

[silk jacinth]

then goddammit, the textbook I'm supposed to be using seems to be just full of errors

this is like the 3rd time I've seen an error

the "correct answer" according to the textbook should be ~8800

I'll need to run these assignments by other teachers at my school and let the writers of the textbook know that they should really be more careful bc I've just wasted so much of my time just checking my (apparently correct) answers with the answers insisted by the textbook

[trail musk]

Someone

Can someone help

[raven hearth]

hey can someone help me with my geometry

[weary drift]

[upper karma]

can somone help me with problems 2-4 ? i am stuck and i cannot solve them i need to fix them by tomorow for credit

[upper karma]

hey guys doyou mind helping me with this

[dark sparrow]

@upper karma do you still need help with this?

[tidal crest]

@dark sparrow hello 👋

Can you help me with a problem

[dark sparrow]

who are you and why are you pinging me personally

[tidal crest]

Mb

[dark sparrow]

instead of posting your problem and waiting for someone to come along to help

which may or may not be me

[gleaming nova]

[rugged cargo]

It's a square with side 10 . There are 2 semi circle in figure and a quadrant find area of shaded part

[upper merlin]

I feel like this is definitely featured in mind your decisions

[dusk charm]

lmao

[civic shoal]

how would I determine the value(s) of theta to the nearest degree 0˚ ≤ θ ≤ 360˚ for sin theta = 0.4526 and cot theta = 0.5814 and sec theta = 1.4526?

[trail musk]

3C. Point D is on side BC of equilateral ▲ABC. From point D, perpendicular line segments with lengths 4 and 8 inches are drawn meeting sides AB and AC at points R and T. Find the number of inches in the height of ▲ABC. [Hint: Draw a line segment from A to D.]

[trail musk]

Can someone do without trig

[candid gyro]

hii

I'd appreciate if anyone of you can help me on a Rotation project

If you have a pdf lesson or some resources that can help please share it with me.

[upper merlin]

without trig we can follow their hint and use area of ADC+ADB=ABC

[distant inlet]

Let H, a hyperbola be the graph of $x=\sqrt{y^2+1}$ and let P be a point lying on H. Prove that for any point P and some line l that passes through P, the probability that line l intersects H exactly twice is 1/2.
Proposed by Yuya Shan

[somber coyote]

AeroBennu

[dark sparrow]

question is not well-posed until a probability distribution is specified for the set of all lines passing through P

[silent plank]

have you tried anything at all?

[mild birch]

exactly one possible line l would be a tangent. since there are infinite lines possible, it makes the probability of this happening 0. and the normal would usually intersect twice too except for exactly one point in H, the turning point and this has a probability of 0 too because there are infinite points. as you’d know, $\sqrt{y^2+1} > y$ so rotate the tangent upto 90 degree one way and you’ve got two intersections, rotating it the other way, you’ll only cross at that point because x of H will always be greater than x of l

[somber coyote]

Rylo

[mild birch]

idk how to construct a formal proof for this tho

i could prolly give it a try

maybe try construct the equation of a line l

[upper karma]

Hello, I need some help. I'm studying for my math exam and there's one task that's bothering me. I have marked three procedures and they all gave the same and correct answer. The first one made my teacher but she haven't explained me why she did it like that and I don't understand it. Third one I did few days ago but now I don't understand it and second one I did now and it makes sense to me. I'm wondering which way is the best to solve other tasks like that cause my teacher told me my way can result wrong answer at some other tasks.

[silent plank]

option 2 is the simplest

the issue with option 3 is that x is being used to represent different things

[upper karma]

ooh okay thank you

[silent plank]

i.e. x is somehow 6.4 but the real x you want is 6.4-4

to set that ratio and equation up properly you can start with
$$\frac{x+4}{4} = \frac{4}{2.5}$$

[somber coyote]

ℝamonov

[upper karma]

alright so thats the first option

[silent plank]

its the reciprocals of the first option, but yeh pretty much the same

[upper karma]

so you said the second option is the simplest, will it always result the correct answer in other taks?

[silent plank]

considering that the 4 in 4/2.5 was a result of simplification of 1.5 + 2.5
$$\frac{x+4}{4} = \frac{2.5 + 1.5}{2.5}$$
$$1 + \frac x4 = 1 + \frac{1.5}{2.5}$$
$$\frac x4 = \frac{1.5}{2.5}$$

[somber coyote]

ℝamonov

[silent plank]

is a way of justifying why that simplified ratio works

[upper karma]

okay, thank you for your help

[harsh heron]

animation I made in Manim

[upper karma]

whats the length of the sides of a

right angle triagnle with theta of 60 degrees?

[queen drum]

30 60 90 triangle

[tulip shadow]

i need help, part a is simple, part b i have no idea how to get it to prove its supplementary

[stiff robin]

Since AED = ACB, AB/BC = AD/DE is what you would use.

\U+1F601

:U+1F601

[tulip shadow]

omg thks so much

[valid chasm]

Guys can anyone tell me where people discuss about 8th maths

[serene harness]

can someone help me?

[deft veldt]

Need a geometry textbook with loads of questions that start off relatively simple and get progressively harder

[forest zodiac]

dumb question but

https://cdn.discordapp.com/attachments/937423899346481155/937684409874792508/ces.png

how do you do this on calculator

[upbeat charm]

According to what you've been taught I presume... angle PQT should be 90° right??

[runic kraken]

anyone can answer?

[thick jay]

It's an inscribed angle, https://www.khanacademy.org/math/geometry/hs-geo-circles/hs-geo-inscribed-angles/v/inscribed-and-central-angles

[thick jay]

A lot of people can. How far do you get when you try, what's the stumbling block?

[runic kraken]

i dont even know what to start with lmao

i did all other questions they were ez, but this is the only complicated one

@thick jay

[thick jay]

So, from the diagram, what is the angle of CBA?

[runic kraken]

thats what i cant find

do I have to add 66 and 26

[thick jay]

I have brain fog and the word I'm looking for is escaping me, but on line NS, all angles that connect to a point on that line at point B, what do they have to add up to?

[runic kraken]

bro idk

just answer it

[thick jay]

Like, angle CBN, CBA, ABS, what do they have to add up to

[runic kraken]

ooh

180

so 180-(66+26)?

[thick jay]

yeah

[runic kraken]

thanks

I got 869.1 final answer

its correct right? @thick jay

[thick jay]

I didn't actually do it out

[upper karma]

should I take honors geometry

[upper merlin]

RQT will be half of ROT, then cyclic quad

[wet swan]

https://mathoverflow.net/questions/415077/value-set-of-a-composite-function

I need help determining the value set of a composite function. You can check the link. I appreciate all tips and help

[glass harness]

so im trying to really understand geometry.

I am thinking of just using the elements by euclid to reteach my self.

but i assume a regular text book work be better?

[thick jay]

khan academy?

[upper karma]

can i ask for physics help here?

[abstract pollen]

I'm in calc based physics and I am trying to understand a trig situation.

i have the change in x =sqrt(d^2_1+d^2_1)-d_2

d_2 is the x side d_1 is the y side.

I don't can't visualize the change in x on the triangle

[sharp flame]

is this correct?

[upper karma]

I'm trying to learn calculus, but I don't know trig

What's a good book on trig

[frosty matrix]

sl loney

[manic shell]

just use paul's math notes or something

[deft veldt]

just khan academy

and questions

loads of questions

[upper karma]

Uhhhh

Wait

Uh

I accidentally sent that

Yeah but now I'm not sure

So SL Loney, Paul's math notes, or khan academy

[lone sandal]

Help

[upper merlin]

try to spot some similar triangles

proof of menelaus' theorem

[opaque ibex]

I have a feeling that the sum of the perimeter angles is 180

[upper merlin]

join the centres of the three circles and remember that those lines pass through D,E,F

[opaque ibex]

oh thank you yes

[split drift]

What is the solution?

[tidal hull]

I need help like rn pls

[grave pond]

Fins a pair of known sides in one quadrilateral that have the same ratio as a pair of sides in the other. That will tell you how the sides match up.

[near sand]

Do three noncollinear points always determine a unique taxicab circle?

[tidal hull]

can someone give me the answer to this?

[grave pond]

No -- consider (0,0), (1,1), (-1,1)

[vernal grotto]

i dont understand how exactly you'd find the phase shift of a sine function because aren't they periodic so how would I know which point was actually the original point at x=0 before it got shifted away from it?

[versed mulch]

18

[versed mulch]

add all the values of x and equate it to 180 angles in a straight line make 180

[dark sparrow]

if you're drawing things in paint you don't need to screenshot the entire screen, you know. you can just select the part of the image you want to share and copy-paste it

[vast jungle]

Whats the proof that sum of angle in triangle is 180?

[grizzled kindle]

@vast jungle let ABC be a triangle
Construct a line parallel to BC passing through A , now apply the relation between interior angles of a transversal

[vast jungle]

But isnt this proof cyclic. You are using the fact sum of all angles in triangle is 180 to prove the same? 🙄

[dark sparrow]

no?

@vast jungle they're basically saying this, where angles of the same color are shown to be equal

[vast jungle]

I agree with you but how do we prove those colour are actually the same

[dark sparrow]

derive it from the parallel postulate

the relation between interior angles of a transversal

[vast jungle]

Okay

[upper karma]

hey guys! i need some help in this…

Two parallel lines passing each by one of two opposite vertices of a square divide it into three parts of the same area (two triangles and a parallelogram). If the distance between the lines is d cm, what is the measure square area? When d = 1

[upper merlin]

the triangle's area would be 1/3 of the area of the square and you can use this to find the area of the parallelogram in terms of d

[upper karma]

oh, but we don’t need the area of the parallelogram

we want the square area

[upper merlin]

you basically find x in terms of s and solve area of parallelogram=1/3 area of square

[upper karma]

sorry for the translation of the problem, i just used google translate cause i was so lazy 💀

oh wait

[trail anvil]

what function finds the opposite side length from an angle?

like what function would I use to find AC

[upper merlin]

think about which trig ratio would get you to the answer

you are given the side adjacent and want to find the opposite side

[upper karma]

just memorize SOH CAH TOA

xd

[upper karma]

ok i still can’t arrive there

[upper merlin]

you found x in terms of s?

[upper karma]

like it has to be a enter number and my solution was decimal 🙄

[upper merlin]

send your working

you are not entering the side of the square right?

[upper karma]

no xd

we are like 15 people resolving this

and we can’t do it

[upper merlin]

but do you know that angle is 30 degrees?

[upper karma]

that’s the problem

we r not sure

ok the solution is 13, but how do u arrive there?

[upper merlin]

following this diagram you'd get 1/2 s.x = 1/3 s^2 which gives x=2/3 s

now area of parallelogram would be d.sqrt(s^2+x^2)=1/3 s^2

[upper karma]

ohhh

[upper karma]

ok, after i will check it

cause we had a limit of time to do it and we couldn’t finish this one

[wet lance]

Is there any tricks to doing hard circle questions?

[silent plank]

depends on what you mean by a hard circle problem

[upper karma]

@upper merlin hey! sorry for the LONG respond, but I was able to do it

and thank you

[zinc mural]

No problem

[atomic solar]

I need help😭

It's due in almost two hours and my teacher won't help

It's not even a test

Rip

I give up

[charred nimbus]

what is x

Homie needs help

[fierce lodge]

how do you find Cot 3pi/4

[charred nimbus]

yall are useless

<@&268886789983436800> \

<@&286206848099549185>

[weary drift]

dont ping mods for hw help

[charred nimbus]

answer

question

Goddamn

[weary drift]

ok

[spark stag]

i'm sure that calling people "useless" and demanding an answer will make them eager to help you.

[weary drift]

come back in 24h

[broken dock]

pay for a tutor if you want help

oh they left

[fierce lodge]

<@&286206848099549185>