geometry-and-trigonometry

[dense ibex]

struggling so hard

[lime crown]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, and no one person can be prioritized over other people, so please patiently wait. Anyone who chooses to help you is a volunteer who is doing so out of their own kindness.

[dense ibex]

oh

[hard garnet]

anyone ever seen this "Find all the solutions to the equation a^2 +(b^2)+c^2+(d^2) = (e^2), in integers a, b, c, d, and e."

[lament tree]

was cavilieri's principle rigorously proven before calculus?

[lament tundra]

[idle thicket]

I forget how to do this stuff

[tepid oyster]

I cannot read anything can you get better quality

[timber cargo]

Get a phone that isn't made in China

[smoky jetty]

yk iphones are mostly made in china

[timber cargo]

They clearly shouldn't be

[smoky jetty]

eh, apple's business strategy

[strong vault]

Hey could anyone help me

[dark sparrow]

[strong vault]

@sullen nimbus I’m doing geometry and having to find the value of angles without a protractor

[sullen nimbus]

Wrong Ann

;-;

But I mean I can help you

[queen knoll]

hi does anyone want to help

i can't find x after 11/12x=11/x

😭

[smoky jetty]

u can think of point M as an enclosed point of rays GE and OE. So if you draw ray EM, u can try and see their relationship regarding angles

[limpid oracle]

i dont get it

i know that multiplying a complex number by exp(I * theta) rotates its theta but it still doesnt make sense

[lime summit]

it makes sense, but in complex plane

use the fact that e^(i * theta) = cos(theta) + i sin(theta)

it this case it is 1/2 + i sqrt(3)/2

[limpid oracle]

It seems that the book did make a typo where z3 * z1 was supposed to be written as z3 - z1 and likewise for z2 * z3

[pliant current]

I'm trying to learn complex numbers what books recommend?

[merry parrot]

#book-recommendations

[upper karma]

why does geometry suck

worst math class i ever took

why tf i gotta prove a triangle is a triangle when its got 3 sides

anyways i need help

i got a fat test on monday on transformations and i still dont know the pattern for rotations

[pearl moon]

what year are you in

[upper karma]

like grade?

[pearl moon]

yeah

[upper karma]

10th

[foggy parcel]

That's why geometry is unique bud

[upper karma]

geometry is pointless

[lyric sonnet]

not true

[upper karma]

if ur getting a job that needs geometry then sure

otherwise it rlly is pointless

[autumn thistle]

the triangle has points.

bah dum tss

anyway

my problem

is this

individual question:

2. Theorem Reflexive Property of Congruence for Angles:
   Given: ∠A
   Prove: ∠A ∼= ∠A

im like

oh ok

well the measures are equal so there

NO

gotta do it like this

1: Theorem All right angles are congruent:
Given: ∠A is right, ∠B is right
Prove: ∠A ∼= ∠B
Proof:
m∠A = 90º (definition of right)
m∠B = 90º(definition of right)
m ∠A = m∠B (substitution property of equality)
m ∠A ≡ m ∠B (definition of congruent)

so

yep

yeah

[weak oar]

If i say some shape is a rectangle given some statement. Then say that because this shape is a rectangle by definition, another statement is true. Are any of these converse statements?

[gentle haven]

In what sense

[pseudo nebula]

if the "another statement" can be used to say the shape is a rectangle then probably

[weak oar]

Ok

So would I say converse of a definition a rectangle

?

[pseudo nebula]

I have no clue what the hell you're trying to do

[weak oar]

mb. I was just trying to use a few given statements in a proof, to prove that a shape was a rectangle. Then using the fact that the shape was a rectangle, I was going to prove something else. It's fine though. I got it. Thanks for answering.

[pseudo nebula]

Honestly not sure the wording

I always just wrote rectangle implies ...

but ig that doesn't make sense

definition of rectangle implies?

[opal girder]

hey can anyone help

i asked in a competiiin math amd channel 49 so please help

[grave pond]

The central angle theorem gives you angle ECD, and angle CED is right by Thales's theorem

From there it seems to be just a matter of chasing angles in the various triangles.

[opal girder]

could you show me work please because idk what thales theorem is is

[grave pond]

Please don't crosspost in mutiple channels, though.

[opal girder]

ok i didnt know

sorry

[grave pond]

https://en.wikipedia.org/wiki/Thales's_theorem FWIW.

[opal girder]

oh thank youi get it now

[lapis vault]

is trig after algebra 2 or after geo

[slow spruce]

algebra 2

But you'll probably learn Pythagoras' Theorem and SOHCAHTOA in Geo

[lapis vault]

k

[fickle rose]

can anyone help in my channel?

[snow crystal]

guys what are the properties of a triangle overscribed on a circle?

if I do something like this, in what ratio are the equal sides divided by the tangent points?

[grave pond]

If you also draw lines from the center to the vertices of the triangle, you get some right triangles that you can do trig on. (Note that those new lines bisect the angles of the triangle).

[snow crystal]

hmm, I'm not even sure if I'm asking for the right thing

basically I'm trying to find the equation of the circle inscribed in this triangle

I know that the triangle is isosceles and I also know the coordinates of its vertices

but idk what to do next

[grave pond]

So what you want is to find the incenter of the triangle. You do that by intersecting two of its angle bisectors.

Alternatively, as a shortcut for this particular situation, you know the incircle will be tangent to the short side at its midpoint -- that is where the bisector you already have meets it.

So the center of the incircle must be at the point on y=2-x/2 whose distance from (2,1) is the same as its distance from the line y=3.

You can write both of these distances as a linear function of, say, the x-coordinate and set them equal to each other.

[snow crystal]

But how am I supposed to use this information

[snow crystal]

ok so now I know the radius which is basically the distance from (2,1) to the incenter of the triangle

it's equal to $\frac{5-\sqrt{5}}{2}$

[somber coyote]

IamMax420

[snow crystal]

but I still don't know how to find the coordinates of the incenter

[grave pond]

How did you find that? Not saying it's wrong, I haven't done the calculations myself, but I don't see how you got it.

If you have the right inradius, then you know the incenter is on both of the lines y=2-x/2 and y=3-r.

[snow crystal]

I calculated the area of the triangle and it's perimeter. We can evaluate the area of a triangle overscribed on a circle with the formula A=(half of the perimeter)(radius of the inscribed circle)

that's how I got it

[grave pond]

Huh, I didn't know that formula, but it looks legit now that you state it.

[grave pond]

The numerical value of your radius looks like it's in the right ballpark for the diagram, too. And with a slope of -1/2 involved, having a sqrt(5) somewhere is par for the course.

[snow crystal]

Ok I figured it out, tysm

though I didn't use the information you told me

I basically substracted (radius) from 3 to find out on what height is the incenter situated

and then set it up to -x/2 + 2

[grave pond]

Which is what I said here, right?

[snow crystal]

oh right lol

[opaque rock]

I want to clear my doubt

In q3 Because if pq ll ab then by bpt aq/qc=bp/pc .....(i)
And rp ll ac then again by bpt br/ra =bp/pc ....(ii) then by i and ii aq/qc=br/ra and by converse bpt bc ll rq but the diagram doesent agree

[upper karma]

it depends if you are talking about coordinate geometry or linear algebra

[snow crystal]

geometry ofc

but it's all solved now

[bright isle]

man that's easy

in india a 10 grade student can do it

[merry parrot]

so?

[bright isle]

nothing.

[bold whale]

what channel

should that be in

is it in the right one

[hasty stratus]

Can someone help me with Geometry (Segments and Congruence)

[late kraken]

Maybe

Idi

[hasty stratus]

It's my first time learning Geometry and I don't understand anything.

[late kraken]

Idk

Geometry sucks ass

[hasty stratus]

I know

[late kraken]

Ok lets see if i know what im doing

[hasty stratus]

It pisses me off and I am on my 3rd lesson starting Tuesday. I have a test tmrw on Segments and Congruence.

[late kraken]

Oof

So whats the specific problem

If there is one

[hasty stratus]

Problem: On RT, RS=2x-8, ST=3x-10, and RT = 17. Write an equation in terms of x. Solve the equation. Then find RS and ST.

[late kraken]

Ah

So what do you know about this problem

[hasty stratus]

I am so confused lol

[late kraken]

Itll be helpful to write what you know on the diagram

So write above RS, 2x-8 etc

[late kraken]

Dont forget the 17 below (or above either way, i prefer below so it doesnt get crowded)

[hasty stratus]

Where did you get 17 from lol

[late kraken]

RT=17 u said

[hasty stratus]

Oh yeah

Ok I placed it below RS

[late kraken]

Good

Now

We know S is the middle point correct?

[hasty stratus]

Yes

Can we go in a call?

[late kraken]

Uhh sure

[hasty stratus]

Ok

[late kraken]

Idk where the call is tho

[hasty stratus]

Oh DM’s maybe

[late kraken]

Sure

[warm goblet]

calculate the gradient by y2-y1/x2-x1 then u see the y intercept is -6 bc the x axis is set at 0 so then just substitute the gradient and x axis into the equation

[upper karma]

For finding a midpoint between two undisclosed variables on a number line, can I use (b-a)/2 or still (b+a)/2

[little owl]

(-1-1)/2 =/= 0

b=-1, a=1

[upper karma]

Ohhh ty

[toxic pier]

i need to copy the exact intersecting line segments

[little owl]

for the third time?

draw line, make a mark
copy one segment 1 arc
copy one endpoint of other segment 2 arcs
copy other endpoint of other segment 2 arcs

[onyx relic]

step 1: find slope of the line (y2-y1)/(x2-x1)
replace in
(y-y1)=m(x-x1)

where m is slope

[crisp pawn]

quick question cos^2(2x) = 1/2(1 - cos(4x)) half angle formulas?

[azure anchor]

first expan right side

[crisp pawn]

cos^2(x) = 1/2(1-cos(2x) cause half angle formula

was wondering what happens if it is cos^2(2x)

[jolly slate]

Making equations for desmos is very difficult isn't it? I can't even figure out how to make a stretched semicircle

[coarse fossil]

yall how do i find weighted averages

[naive jackal]

can anyone help me with these questions?

[upper oracle]

I think the for the first one, x would be 93°

For the first blank haven’t gotten to that yet lol

2nd question is ez lol I’m literally doing hw on that exact thing rn

1. Corresponding
2. Vertical
3. Alt interior
4. Alt exterior
5. Supplementary

[jaunty spruce]

can someone help me with this cos graph function? i don't understand how is the amplitude -6

[mystic charm]

are you sure the amplitude is -6? an amplitude can not be a negative number

[wind tree]

hey I have a question. is cos^2(2x) equal to (cos^2 (x) - sin^2(x)) ^2?

[mystic charm]

yes

[visual totem]

Can someone solve this

[upper karma]

Have u heard of de moivres throrem

[naive jackal]

thank you so much

[olive condor]

ive just started geometry as someone in 8th grade, and i need tips on how to memorize what sin, cosin, tan is easiler

tysm, have a great day

ill remember that

[maiden brook]

aops!

[unkempt siren]

Let O be an octagonal prism defined such that each of its faces contains regions both transparent and opaque. For each adjacent pair of octagonal faces:

(i) An arc originates at the bottom vertex of one face, and concludes at the top vertex of the diagonally opposing corner on the adjacent face.

(ii) Symmetrically, from the second face of the pair, an arc commences at the top vertex and terminates at the bottom vertex of the diagonally opposing corner on the first face.

This configuration produces a series of parabolae, each spanning two contiguous faces. The region situated above each such parabola is opaque, while the region beneath is transparent.

Let the octagonal boundary of the prism's height be denoted by P, with its variable radius as r.

Problem 1: Deduce a function, f(P), that determines the set of points on P from which an observer, looking perpendicular to P, perceives the opaque regions of the prism exclusively through the transparent sections within a 120-degree field-of-view.

Problem 2: Investigate the behavior of f(P)with respect to variations in the radius r of the octagonal boundary P. How does f(P) evolve in response to changes in r?

[unkempt siren]

Made this puzzle up myself it’s not homework but I can’t do it and spent too long to leave without knowing the answer

[halcyon glen]

Could someone please explain?

[trail mirage]

Pls Someone help me I’m lost and stuck

Ik it’s probably ex and I’m over thinking it

[smoky jetty]

u have any idea regarsing law of sine and cosine?

[upper karma]

SohCahToa

yes there isn't a different slope

just a different translation

[trail mirage]

I do I solved it dw

[crystal marsh]

Can someone help me with these 3 questions im stuck on?

[hasty ravine]

Did you try swapping the the x & y values of each point?

[crystal marsh]

i already figured it out lol

[hexed raven]

Cant use trigo ratio if not right angled triangle

[elder quartz]

uhhhhhhhhhhh

nah

i dont think so'

they are curved

honestly idk

if 2 curved lines like that are parallel

[little geode]

could you guys help me

[elder quartz]

#❓how-to-get-help

[visual totem]

yeah but how do i use that here

[upper karma]

Express the left hand side

In terms of sin n theta

Where n is an integer

[visual totem]

oh right

thanks bro

[frail frost]

really failing to grasp angles. i passed alg1 with ease but i’m struggling with geometry

mainly question 7

[delicate loom]

Read Euclid's "Elements"

Pretty good basis, should still be required in public schooling

[cursive agate]

I don't think reading a 404 page book for one problem is really assisting him lol

[upper karma]

🙄 didnt notice

[cursive agate]

Geometry is tricky! I sent you some dms that (hopefully) help guide you a bit :D

[delicate loom]

For all future problems lol

[cursive agate]

Fair lmao

[unkempt zodiac]

Can somebody privately dm me and help me? I am currently a sophomore and we are doing two column proofs and I don’t rlly understand.

[unkempt zodiac]

Anybody?

[fickle rose]

on one side you write down a step towards proving what the problem tells you to do, and on the other side you write down the theorem that allows you to do that step

thats the simplest way i can explain it

[fickle rose]

think about the 2 smaller angles inside angle lpr. you already know the value of one of them so use that to find the measure of angle lpe

we know that angle tpa is vertical angles with angle lpe, which means that they are the same measure. remember that apr is a linear pair with epr, and we know the value of angle epr, and angle tpa. use this to find tpr

i explained tpa above.

angle tpe is vertical angles with apl. you have angle tpe by doing simple addition.

[unkempt zodiac]

Nah Ik that but the main problem is utilizing the evidence and using the proper evidence

[fickle rose]

do u get a theorem sheet

[unkempt zodiac]

Yea but we don’t get to write all of them

[fickle rose]

what do u mean

[fickle rose]

give me an example problem and i can walk u step by step. lets move this to one of the help channels

[unkempt zodiac]

Let’s say we learn ten theorems we can only have eight in the theorem sheet

[fickle rose]

oh really

[unkempt zodiac]

Ok can u open one

[fickle rose]

alright join

well talk in it

[unkempt zodiac]

K

[halcyon glen]

Could someone please explain?

[merry chasm]

I got a decimal is that my fault??

[zealous vessel]

does this mean the angle at A is 3x+5 and angle at B is 2x and sides AB and BC are equal?

[merry chasm]

Yes???

Uhm well so I put

3x + 5 + 3x + 5 + 2x = 180

[zealous vessel]

,calc 170/8

[somber coyote]

Result:

21.25

[zealous vessel]

was that your answer?

[merry chasm]

Yeah

But I don’t think I am supposed to have decimals

[zealous vessel]

was there some constraint saying you can't have decimals?

[merry chasm]

No but

We also haven’t had decimals

[zealous vessel]

I didn't get what you mean

Well from what I can see, 21.25 is the answer, it's mathematical so you can prove it : )

[merry chasm]

Okay!

Well I was just unsure because we haven’t had any problems which had have decimals

[flint marlin]

Help?

[frozen ocean]

,rccw

[frozen ocean]

do you know what is the area of trapezoid?

[flint marlin]

(Ab+dc)/2 × h

[frozen ocean]

[A=\frac{a+b}{2}h]

[somber coyote]

Akira 🍉

[maiden brook]

coordinate geometry looks easy for this

[teal talon]

2(4)^2 = AB^2

32 = AB^2

[lament tree]

I think it isn't cus our geometry is based on hilbert, not euclid
and the lack of algebra gives a really big headache

reading old math texts is really confusing cus of the lack of recent notation

[royal bear]

oh god...

geometry, my enemy

[violet kiln]

Same

[carmine ore]

when solving for theta,
1 = tan theta, doesn't this have multiple solutions for theta?

err

actually isn't is just root 2 / 2 / root 2 / 2 = 1?

so tan is 1 when cos and sine are root 2 / 2

[lyric sonnet]

yes

[silent plank]

that's only one case, and even then there are infinitely many values of theta where sin and cos theta are sqrt(2)/2

[zealous vessel]

So does cot, their period is just pi instead of 2pi like for other functions.

Wait even though sin and cos have period of 2pi they also have two solutions for any given value right?

[young kestrel]

Yeah dawg i need help bruh

[shut barn]

hello

[frozen ocean]

#❓how-to-get-help

[vagrant tendon]

what is the answer to this

pls help

[vagrant tendon]

help me please

what is answa

[frozen ocean]

we don't give out answers for you since this is against the rules

but we are here to help you

what's the question asking?

[vagrant tendon]

Idk they werent specific

All i csn rlly say

Somehow

It adds to 180

But x aint 60

Im working it rn so ill ask if i need helo but thx

[frozen ocean]

isn't the question asking to find the angle x?

try adding 180 and subtract the other angles on the angles given

you should be getting your x

[icy cobalt]

Concave Quadrilaterals add up to 360 degrees on the inside "(n - 2) * 180, where n =4 (the number of sides)"

[smoky jetty]

u can extend the lines and do angle chasing

[zealous vessel]

$$sec(x) = \frac{tan(x)}{sin(x)}$$ right?
But $$tan(0) = 0$$ and $$sin(0) = 0$$ and $$\frac{0}{0} = undefined$$
and yet $$sec(0) = 1$$
How?

[somber coyote]

Navee

[storm escarp]

secant is defined as 1/cosx

it is the same as tanx/sinx everywhere except for where sinx is 0

[zealous vessel]

Why is that, like algebraically that seems contradictory

[storm escarp]

I don't have a good answer for why secant is 1/cosx instead of tanx/sinx. You are right that it seems wrong because secant is supposed to be the length of the hypotenuse of a triangle with 1 as its base and tanx as its height, so it follows that sinx=(tanx/secx) or secx=(tanx/sinx)

[zealous vessel]

I think I should post it as a question in the forum but I don't see any chance of attention being drawn to that 😐

[timber cargo]

Well technically that would be tangent(gerine)

[ember urchin]

Three people are pulling on a stuck object at the same time. Once the object gets unstuck, draw the vector path that the object will follow. Below is the diagram of the forces the people are placing on the object

[frozen ocean]

!memes

[lime crown]

memes go in #chill

[frozen ocean]

do you still need help with this

[smoky jetty]

if in intercept form (i.e (x/a)+(y/b)=1 where a= x intercept; b=y intercept) y/b is negative, does it automatically means y intercept is negative? tho its slope intercept form says it is negative

[frozen ocean]

depends on equation if it's positive or negative

but yea since you're saying it's negative then negative

[smoky jetty]

alright thanks!

[gentle haven]

Is #3 asking for the midpoint of the first two things as if they were in a number line

Like the average

And the 2nd part of it makes no sense to me

[pearl elk]

did i do it right or am i missing something

sorry for bad hadnwriting

[maiden blade]

I’ve never been so confused in my life

[strong gorge]

sin(q)/cos(q)

[weak oar]

You know that the whole angle of should be 180 right. So you are given that there is a 90 and 46 degree angle. Together they make a 136 degree angle. Meaning you have q-3=180-136. This simplifies to q-3=44. Add 3 you get q=47.

[zealous vessel]

Is the fact that all trig functions can be represented as slopes of lines a trivial fact?

[mortal sundial]

i an not sure how to answer these questions would love it somebody could help me answer them

[lime crown]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[zealous vessel]

!help

[lime crown]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[fickle rose]

46 + 90 + (q-3) = 180

[mortal sundial]

!help

[lime crown]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[zealous vessel]

Nah bro read the channel and ask for help according to that.

[mortal sundial]

ok cool cheers mate

[upper karma]

hi, does someone know in which yt video I can learn this?

or the name of the topic

and also things like:
sen80º = cos10º
I get why, but I want to watch a video about it and know the name of the topic to save the notes in a page with that name

[zealous vessel]

what a coincidence, I just watched this video : )
https://www.youtube.com/watch?v=f2pLhomzFXc&ab_channel=EddieWoo ( forgot to paste link out of excitement : )

• minor correction, it's not sen but sin. As in sine
• what you're looking at for sin and cos is them complimenting each other

[upper karma]

oh that's great, thanks (:

and in spanish is seno, or sen
yepp

[zealous vessel]

ah ok, I didn't account for the language part : |, anyway happy to help

[vague vortex]

so close to crying can someone help

[granite hollow]

m<D = 108

[foggy parcel]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[fickle rose]

it’s a parallelogram

use parallel lines to figure it out, or use any of the properties of a parallel organ (opposite angles are congruent and consecutive interior angles are supplementary are some that should work)

[frail frost]

was not in school when we learned this and i have a quiz today on it

please break it down and tell me if my answer to 20 was right

if 20 is right i understand it and only need to know how to do 21

[silent plank]

q20 as written doesn't make much sense

for q21, you can apply stuff like pythagorean theorem and/or distance formula
(effectively the same thing)

[smoky jetty]

how did u get 135 already in your equation?

idk if these are shortcuts but they're unclear

[fickle rose]

ur wrong on q20

135 is the measure of one interior angle

ur adding the measures of 2 interior angles which should equal 270 not 135

[smoky jetty]

let him cook lmao

[fickle rose]

?

[fickle rose]

how to get help in this server. been asking for help requests periodically for the past 4 hours for the same math question and not one response, meanwhile the person who comes in right after me gets a response instantly from someone else

is there some sort of way where u can get help when no one has responded for several hours

[smoky jetty]

the #help channel?

u could ping @ helpers there once, after 15 mins of no response afaik

[regal scarab]

We have started with Unit circles, but I am extremely confused. I simply cannot understand what the numbers mean and the textbook did not provide any background information, simply threw me into the deep end. Does anyone have any good sources that explain (preferably visually, fine if not) from the ground up, the knowledge i need to start learning about Unit Circles? I do not know how to research it and the videos/articles I have found go off of the same assumption that I simply know some information before-hand

[zealous vessel]

Try 3b1b trig lecture, Eddie woo's vids (on YT) they show it visually.
But unit circle is just a circle with radius 1, and it's diameter 2π. Also try to get what is radians, that'll give more insights.

[regal scarab]

Many thanks! i will check them out

[deep ruin]

An equilateral triangle ABC with side 3 is inside in a circle point D lies on the arc AB doesnt have the too find BD and CD.

[frail frost]

i missed school the day my teacher went over it so i was looking on youtube for a breakdown and they said 180-360/n lmao

[frail frost]

are vertical angles sometimes never or always supplementary

[smoky jetty]

afaik vertical angles are formed by the intersection of 2 distinct lines, so they are never supplementary?

[frail ridge]

I have this open box

How do I solve for the middle square in the cross figure?

I can find the length of most edges of the cross figure, except for the furthest edges

[smoky jetty]

i kinda not get your diagram, but maybe try pythagorean

[inner vapor]

someone help me with this geometry question

[smoky jetty]

which site do u make those symbols stuff on an image?

[smoky jetty]

hey guys so I was learning how area of triangle on a coordinate plane is derived, and I was able to understand the processes and concepts involved. upon simplifying the terms (encircled green), I got the ff. terms on the 2nd image. However, idk how i could simplify it further to get the final terms (encircled red) below the 1st image. I tried observing it and got a sensible idea, but idk the formal name of the process(i kinda suck in algebra oof)

so uhh how do I simplify the 2nd image further by following a formal process (that idk rn) to get the encircled (red) temrs below?

I found another quicker method to derive the area of triangle, called determinant and thought it was quite similar to shoelace method, but I wish to know this one first, lol

[slow spruce]

What axiom or theorem do I need to prove if the vertex is on the edge of the circle, the angle is half the arc?

[smoky jetty]

tl;dr how can I simplify further the 2nd image to get the bracketed terms in encircled (red) RHS expression (1st image)? Dont mind the (1/2)

[smoky jetty]

nvm it was a simple factorization that i didnt know, lmao

[round remnant]

Demonstration of the half-angle formula in trigonometry

[upper karma]

Are there any of you that have learned Trigonometry by using predominantly the degree measurement ? Anyway, that's what I've seen in a High School near, and thought it quite bizarre.

[surreal escarp]

Does my math make sense

[foggy parcel]

U r right

@quartz horizon

Opps

@surreal escarp

[fickle rose]

good

[junior verge]

hello

i had a question on a quiz yesterday that told me to give a counter example to this conjecture

a^2 < b^2

a<b

so i said

3^2 < (-5)^2

3 > -5

did i get it wrong or right

[silent plank]

you'd probably need an additional line and some words to go with what you're writing for full marks

3^2 < (-5)^2
according to the conjecture,
3 < -5
which is false since 3 > -5

(also not geo/trig)

[junior verge]

thanks for telling me

i didnt put words cuz i thought counter examples needed tp be very specific and limited to numbers

im stressing about that question ngl

[jagged venture]

Need some help with this one guys

The equilateral triangles ADF and BCE are constructed outside the square ABCD. If DE intersected by CF = G show that triangle ABG is equilateral

Pretty stuck

[gray girder]

because the height of an equilateral triangle within a square where the triangle and square share a side is always going to be 7/8ths of the side length of the square?

idk if that really shows it though

cuz you don’t actually know any values

[modern thicket]

I think its a mistake in this line, because if we dont know a is more or less than 0, we should write sqrt. a^2 = module of a

[modern thicket]

module of -5 is more than 3

[junior verge]

wdym

im kinda stupid can you explain it simpler lol

[modern thicket]

before i write why its a mistake in question, because sqrt. -5^2 != -5, right?

[junior verge]

no i put paranthesis

then

(-5)^2=25

[modern thicket]

ok, u right, but because its incorrect line in question

[junior verge]

incorrect line?

im so confused sorry

[modern thicket]

sqrt from negative number in square is equal to module of this number, may be its my mistake, but i know that sqrt can be only positive

[junior verge]

yeah

the sqrt i put was equal to a positive number

which was 25

[modern thicket]

but why than you put a negative 3< -5?

[junior verge]

no i said 3>-5

look

[junior verge]

@modern thicket

[modern thicket]

bro, can i describe it by photo in your dm?

[junior verge]

yeah

[upper karma]

can someone help me w this

[smoky jetty]

Ever heard of the cosine function? Cos x = (adjacent side/hypotenuse)

[upper karma]

how do i find the hypotenuse side

[dire snow]

In the trapezoid ABCD, from the midpoint L of the diagonal AC draw the parallel to the bases, which meets the diagonal BD in P and AD and CB in E and F respectively.

Prove that LP (AB-CD).

I don't know how can I find it lol

[smoky jetty]

u can plug in the values in that equation and do some manipulation

[round remnant]

How much geometry would you learn from reading Euclid's elements?

[grave pond]

It would get you all the way up to the state of the art as of about 300 BC.

[upper karma]

like?

[smoky jetty]

you're given angle 71, so as a reference to its cosine function we get the ratio = 6.3cm/CB
cos 71deg = 6.3/CB(isolate CB)
CB cos 71deg = 6.3
CB (cos71deg/cos71deg) = 6.3/cos 71 deg
CB = ?
from here u can put it in a calculator to find the approximate value of CB

u can also refer to its secant function

[upper karma]

question

Consider the rhombus $ABCD$ with centre $O$, with $\angle DAB < 60°$ and the equilateral triangle $ABE$ so that the points $E$ and $D$ are on either side of the line $AB$. The center circle $E$ and radius $AB$ intersect the line $AC$ at the points $F$ and $A$, and the line $BD$ at the points $G$ and $B$. Prove that the centres of the circles circumscribing triangles $EBA$, $FBD$, and $GCA$ form an equilateral triangle.

idea

We see that $\angle BGA*2=\angle BEA=60$, which means that $\angle BGA=30$

Because $ABCD$ is a rhombus, the diagonals are perpendicular $=> \angle GOA=90$

Using the 30-angle theorem we can simply show that $OA*2=GA=CA$

Because $OG$ is the perpendicular but also the median of the triangle $GCA =>CG=GA $

From the congruences, we can show that $CGA$ is an equilateral triangle.

WE can show that $B$ isn't the center of the circumscribing circle, because $OA^2=OB*OG$, which makes $BG=3k$ and $OK=k$

We also know that triangle $DFB$ is isosceles.

ALSO $BFAG$ is an writable quadrilateral

I don't know what to do forward. Hope one of you can help me! Thank you!

[somber coyote]

Ionela

[upper karma]

Shouldnt LP=(AB-CD)/2???

[dire snow]

Yeah it's the same

[upper karma]

This us like a formula. You don't really have to demonstrate it

[upper karma]

I know how to demonstrate it. Do you wanna know how?

[dire snow]

Ook

[upper karma]

You just have to use the theorem of the middle line

EL*2=DC

[dire snow]

Name of the th?

[upper karma]

PF*2=DC

[upper karma]

Middle line theorem

Do you understand?

[upper karma]

Help

[dire snow]

Hmmm

Wait

[frail frost]

how do i get better at geometry when we dont have a textbook and my teacher hates teaching?

[polar jewel]

hey

[brittle fiber]

can anyone help me in this

[gentle haven]

Try online courses

[orchid forum]

why doesnt geogebra give me an answer

the syntax looks correct

[slate pewter]

$1+\dfrac{2}{3}\simeq 1.667$ test

[somber coyote]

Yui Hirasawa

[slate pewter]

oh nice it worked

SOOOOOOOOOOO

i got sick and for a week and missed school when i came back they were doing this kind of dark magic
i am in a turkish school and tbh the turkish lessons arent helping so i was thinking if yall can just tell me where i can learn this or , what is the name of this subjects the only thing i know is that its related to trig

$\dfrac{\sin ^{2}\left( x\right) }{1-\cos \left( x\right) }\Rightarrow \dfrac{1-\cos ^{2}\left( x\right) }{1-\cos \left( x\right) }\Rightarrow \dfrac{\left( 1-\cos \left( x\right) \right) \cdot \left( 1+\cos \left( x\right) \right) }{1-\cos \left( x\right) }\Rightarrow 1+\cos \left( x\right)$

[somber coyote]

Yui Hirasawa

[slate pewter]

or something like this

$\dfrac{\dfrac{1}{\sin \left( x\right) }+\dfrac{\cos \left( x\right) }{1}}{\dfrac{1}{\cos \left( x\right) }+\dfrac{sin\left( x\right) }{1}}\Rightarrow \dfrac{\dfrac{1+\cos \left( x\right) \cdot \sin \left( x\right) }{\sin \left( x\right) }}{\dfrac{1+\sin \left( x\right) \cdot \cos \left( x\right) }{\cos \left( x\right) }}\Rightarrow \dfrac{\cos \left( x\right) }{\sin \left( x\right) }\Rightarrow \cot \left( x\right)$

[somber coyote]

Yui Hirasawa

[smoky jetty]

there are many resources online, and perhaps more enganging lectures available for u to enjoy learning geometry

u can try khan academy in particular, but i suggest referring to an online textbook as well for supplementary definitions/concepts

[dire snow]

Hmm I don't understand why el*2=cd but I understood the theorem

[orchid forum]

why does this happen

[errant beacon]

1 + cot^2 if sin=1/3 is 13/9 right?

that's what i got

[surreal escarp]

https://cdn.discordapp.com/attachments/882996198397599744/1158212463892893726/Screen_Shot_2023-10-01_at_3.22.01_PM.png?ex=651b6ce9&is=651a1b69&hm=e6ba18c2096d33c6941aeec8ebd82aef7b5267796fdb7946fd409f66312cf88b&

Would the answer be x = 5?

[upper karma]

yes

[upper karma]

9

[upper karma]

what is the answer

[hollow osprey]

how would you solve for z?

[upper karma]

Z is 112

180-x bcoz it's a cyclic quadrilateral

t = 44 btw

kite property or sm

[smoky jetty]

u can refer to central and inscribed angles theorem, or property of cyclic quadrilaterals

or possibly other ways

[night mist]

how does one calculate sin(2x) = 0.2197?

this is part of a physics problem but i cant for the life of me remember how to solve a simple thing like this

i suspect i can just do:

sin2θ = 0.2197
arcsin(0.2197) = 2θ
2arcsin(0.2197) = θ```

[sinful forge]

Can somebody lend a hand with this question

[dark sparrow]

last step is wrong

[raw tusk]

Hey everyone, I want to know why in an equation sinx-sin3x=0 I can't just do:
Sinx=sin3x
X=3x

I know this is wrong, but why exactly?

[night mist]

oops, should have divided 🙏

[toxic edge]

Is x 50m or 121m? Or none?

[somber coyote]

Personal LaTeX configuration.

​ ​ ​ ​ ​ ​ autotex: Disabled (may be overriden by guild settings)
keepsourcefor: Don't delete source (may be overriden by the guild)
​ ​ ​ ​ ​ ​ ​ colour: Using the grey colourscheme
​ ​ ​ alwaysmath: Disabled
​ ​ ​ alwayswide: Disabled
​ ​ ​ ​ namestyle: DISPLAYNAME
autotex_level: WEAK

To see more detailed information use ,texconfig <option>.
To set an option use ,texconfig <option> <value>.

Preamble

No personal or guild preamble, using the global default preamble.
Use ,preamble to view or modify your preamble!

Oct 3, 2023, 11:11 AM

You have switched to the darkgrey colourscheme.

[cedar flower]

Sorry, my mistake, I thought I was writing in my private chat with TeXit

[smoky jetty]

i forgot how to solve this one easily, but I did some angle chasing and extended the segments' length and got an answer quite far but <50m

[smoky jetty]

||~33.32m||

[toxic edge]

Regarding my problem?

[smoky jetty]

yeah

[toxic edge]

Can you show the solution? This was a test task

And we (the class) are quite divided between 50 ish and 121

[smoky jetty]

i assumed that this side is equal to 15m + x, otherwise my solution is nothing but fallacy

[smoky jetty]

wdym?

[toxic edge]

Yes you are right on that

[smoky jetty]

yall got diff answers?

[toxic edge]

Yeah we quite divided

[smoky jetty]

yall haven't gotten the aswer key yet?

[toxic edge]

My priv teacher says its 53 ish

[smoky jetty]

oh

lemme grab my notes real quick

im certain theres a much easier way to find that

realative to what i did which was just overkill i'd say lmao

[toxic edge]

Oh wait i think im dumb

[smoky jetty]

there prob is, but i couldnt find a similar problem lmao

[toxic edge]

Just get the hypothenuse of small triangle using theorem of sine and then just sin (58deg)

sin 58deg = x/hypothenuse

[smoky jetty]

whats your hypotenuse's value?

[toxic edge]

Im on train rn

But lemme figure, 1 sec

[smoky jetty]

alr alr

[opal ermine]

im supposed to find the value of NP, does my math check out? just started learning abt this so not sure and its messy

[toxic edge]

Around 62.6

[opal ermine]

that’s what NP equals?

[smoky jetty]

mind to share?

[opal ermine]

could u explain to me how u got it

oh mb, wrong person

[smoky jetty]

oh wait, on which problem? lol

[toxic edge]

The hypothenuse

And then my x came out to be around 74

I am beyond confused

[royal bear]

where do we get the reasons for Euclidean geometry

[toxic edge]

Sorry for bad writing, gggyeah, not easy to write on a lap while in train

[royal bear]

example In ∆ABC III ∆ADE
 =  (given) <- this

[smoky jetty]

it's ok, wait a few mins as I try to make mine understandable, lmao

[smoky jetty]

whoops I had a small but impactful arithmetic error earlier lol

but heres my improved/more accurate solution

and yeah the answer is ~53@toxic edge

[toxic edge]

Ah rip

Dont got the perfect score :(

[smoky jetty]

i hope u dont get overwhelmed by the amount of equations there, I just elaborated them to show how I approached the problem. Should be easy if u know properties of quadrilaterals

[smoky jetty]

is it the only problem u didnt get right?

[toxic edge]

Yeah most likely

[smoky jetty]

ohh dang

does your teacher have a simpler approach to it?

[toxic edge]

No idea, will find out on thursday

[smoky jetty]

i see

what grade u on?

[toxic edge]

11th

School edu consists of 12 grades

And then uni obv

[smoky jetty]

oh, im in 10th

[toxic edge]

Yeah, education systems differ a little bit

There should be a united one

[smoky jetty]

lol

[toxic edge]

Still hoping for a 9/10 tho

Or at least 8

A mini test task had the same points as this one

[smoky jetty]

what topic was it in?

[toxic edge]

Trig

Quite basic stuff

I was prepping like 2 hours previous day

And i overprepped

[smoky jetty]

u should practice problems with increasing difficulty

[toxic edge]

I did that

Here we have a website with like tasks

An actual cheat code

And i went thru ALL the tasks

Got 89%

100 ish tasks

[smoky jetty]

wow, may I see the problems?

[toxic edge]

They differ in difficulty

Some take 10 seconds

Others like 3-7 minutes

https://www.uzdevumi.lv/p/matematika-pec-skola2030-paraugprogrammas/matematika-i note: you will need a translator. The tasks are at VII

[smoky jetty]

i see, yeah seems like introductory lessons to trigo

[toxic edge]

Yeah this is Maths I

Non intro would be probably Maths II

[smoky jetty]

I wish u good luck!

[toxic edge]

thanks!

[upper karma]

i would like a beginner textbook on geometry, with pdf

[upper karma]

geometry like 3D or 2D?

[upper karma]

2d

[upper karma]

try cengage

or

there is an author named rd sharma

his book for maths 2d is good too

[upper karma]

they have a textbook?

yeah

textbook with theory

solved examples

and practice problem

alright

thanks

u can get the pdfs online easily

for free

so yeah

[stable jay]

Are there any convenient ways to find the radius of a circle given 3 points?

[upper karma]

yep ig

maybe if u show me figure

i can help

or else general method is if u have like three points

make sure one lies on the centre

and other 2 lie on the circumference

then u use distance formula

and find out the lengths

then u can see if ur getting the radius as same

[upper karma]

i made this math thing that uses cos and sin after learning about them

it uses scratch

[upper karma]

I am trying to solve a problem, this... A is a variable point belonging to P, where P is defined by (x-1)^2 = 8(y+1), with abscissa λ, λ belonging to real numbers. Line S is parallel to the x-axis through point A, and line T is defined as x = λ-1. The intersection between T and S is M. Find the locus (geometric locus) of M." how I should start it, I am two hours on in it and I don't understand it

[lavish dew]

Planter Box Project

You are building a rectangular planter box. The designer gave you the following instructions:
a) The top of the box should have sides parallel to 2x+3y=6. Since it is a rectangle, you know
that the other sides should be perpendicular to the given line
b) Two of the vertices are given as C (7,0) and E (3,7)
c) The rectangular box has a height = 3 feet.
d) Each side is made of concrete with thickness of 2 inches
e) All length measurements are in feet
f) The volume of a rectangle =lengthwidththickness

1. Find the equation of the lines that represent the sides of the top portion of the planter box
2. Find the missing vertices
3. Draw a sketch, by hand, of the top rectangle on a coordinate grid and another sketch of the box
4. Find the perimeter of the top rectangle. Round to the nearest hundredth.
5. Find the volume of the concrete in each side of the rectangle
6. If concrete costs $5.25 per 6 cubic feet, how much does the box cost?
7. It was mentioned in the instruction that the diagonal of the top rectangle makes an angle of
   2x+3 with the shorter side. Write an equation to represent the measure of the angle formed
   by the diagonal with the longer side.
8. If the angle formed by the longer side and the diagonal is 52 degrees, find x
9. Use Google Slides to represent your work and post it on your e-portfolio.

someone please help this is due tomorrow as an exam grade

[frail wind]

hey, my geo teacher doesnt rlly teach and he doesnt give hw, although i find it to be frustrating i helps a lot, any websites to use for preactice? or mabey a book with problems to do?

[lavish dew]

youtube

[frail wind]

true

[unkempt finch]

Can someone lend me a hand with this question

[shut quartz]

which one

[unkempt finch]

Both

[shut quartz]

ok um i can barely read it that and the fact that i need glasses

[solar heron]

A ship sees a whale at 10 degrees. After 10 minutes the ship sees the same whale at 50 degrees. The height of the boat is 100m. Find the mph of the whale.

[woven shoal]

Is the ship at rest?

[solar heron]

Yes

im stuck on this probkem for 2 hours

pleae help

[woven shoal]

Well, draw a diagram, first

[solar heron]

i tried

here ill show a pi

[solar heron]

@woven shoal what u think

[woven shoal]

...

The 10 comes after.

The 50 degree angle comes before.

[solar heron]

?

why would that be

[woven shoal]

You see, logically speaking...

That 10 degree angle you made.

Is bigger than the 50 degree angle.

[solar heron]

it was not meant to be drawn to scale

[woven shoal]

Well still.

[solar heron]

the ship sees the 10 degree angle before the 50 degree angle

[woven shoal]

If you see something at a 10 degree angle, and then at a 50 degree angle

Did it come closer

[solar heron]

no it went furtthe

thats what im asying

cause the whale went further from the ship

[woven shoal]

It didn't go further.

[solar heron]

yes the whale went away from the ship

[woven shoal]

Well then your angles are drawn wrong

[solar heron]

and we r supposed to find the mph of the whale

[woven shoal]

Its not written so

[solar heron]

so how would the diagram be

could u draw it then

[woven shoal]

The whale is coming towards the ship from what I understand

[solar heron]

no

i dont think ur understanding the question

the whale is going away from the ship

regardless the question is not regarding that

it says

after 10 minutes

it sees a 50 degree angle

therefore we can conclude that the whale has moved away from the ship

right... @woven shoal

that 10 degrees is not to the whale

thats the degree to the base of the ship

we arent trying to find that

[woven shoal]

Degree to the base of ship?

The observer would sit on top of the ship

So he would look down

[solar heron]

yeah look to do the whale not to the base of the ship

[woven shoal]

Well wait

One min

[solar heron]

hold on

[woven shoal]

mb

Now I don't know if it's 10 degrees from the horizontal or the vertical

[solar heron]

hortizontal

because these are equal

[woven shoal]

Alr so still.

Where would the 50* be

[solar heron]

ill show u

[woven shoal]

Nearer to the ship, or farther.

[solar heron]

farther

because the angle has increased

ok how about this

like ill give u a real life example

if you and ur friend were standing on ground level

and you guys are looking straight at eachother

[woven shoal]

Bruh.

[solar heron]

and then ur friend climbs up a hill

and you look at him

he is farther from you now

does the degree from you looking at him increase or decrease

[woven shoal]

Seriously.

[woven shoal]

Look here.

You're measuring the angles from the horizontal

The ten degree angle is smaller

So the line at a 10 degree angle is gonna be going further ahead

[solar heron]

which side represents the ship?

[woven shoal]

Left.

[solar heron]

let me edit ur image

it dosent matter which angle is bigger @woven shoal

first you see the 10 degree angle

and then you see the 50 degree angle after 10 minutes

[woven shoal]

It does matter for your understanding

But yeah okay

[solar heron]

yeah

so continue

how do i find mph

[woven shoal]

Now use some trigonometry

To find the bases

Of both right angled triangles.

[solar heron]

i already did

ill just give u the angles

[woven shoal]

The difference will give you the distance travelled.

[solar heron]

i mean

why is the difference

shoudnt it be the sum

[woven shoal]

Why the sum?

[solar heron]

because the total distance travlled was the 10 degree and the 50 degree

[woven shoal]

It was FROM the 10 degree TO the 50 degree (according to you)

[woven shoal]

This diagram is gonna give me a headahce.

Its so wrong

[solar heron]

ok then u do it

[woven shoal]

Whats that 576 128

[solar heron]

the side length

[woven shoal]

128 is what

[solar heron]

the decimal

567.128

[woven shoal]

,w 100 tan(40)

[solar heron]

pi?

whya re u using pi

[somber coyote]

Results provided by WolframAlpha

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

Requested by muhammadhussaini

[woven shoal]

...

Now

What do you think.

[solar heron]

i got them right

no?

[woven shoal]

WHy did you put it as 839

No.

83.9

[solar heron]

i put

83.9

🤦‍♂️

[woven shoal]

So tell me now

Oh boy

How do I make you understand

Yk what

Lets make a whole diagram

As per your way.

[solar heron]

ok...

[woven shoal]

I even lost my mouse now

[solar heron]

..

[solar heron]

perfect

[woven shoal]

Lol.

Okay

[solar heron]

so what does BDC stand for

[woven shoal]

Now tell me

Wait wiat

Dont rush

Tell me

[solar heron]

sry

[woven shoal]

What did you find BD as

[solar heron]

567.128 ft

[woven shoal]

Good

BC

?

[solar heron]

83.9

[woven shoal]

Okay

Now for God's sake tell me, if BC is bigger how's it only 83

and BD which is supposed to be a part of BC is 567

[solar heron]

um

thats a good question

shouldnt 83 be just DC

not Bc

[woven shoal]

Well you found BC using a right angled triangle

Where angle C was 50 deg

Doesn't it tell you that you flipped the angles?

[solar heron]

but how would that make sense

[woven shoal]

Or you're still very stubborn

[solar heron]

??

[woven shoal]

I mean just think

[solar heron]

what does stubbornous have to do witht his

[woven shoal]

Wait a min I got something for you

[solar heron]

clearly ur older than me so u understand better

im still learning

[woven shoal]

Wait a minute.

I gotta use Desmos now

You have the undergraduate role.

I'm still preuni

[solar heron]

@woven shoal im in 10th grade

g

[woven shoal]

It took me 10 mins but I hope it helps you

See as that angle gets bigger the base of the triangle gets smaller and smaller

@solar heron

[solar heron]

ok how about this

we can do two different triangles

it dosent need to be on the same triangle

clearly its a problem

[woven shoal]

I mean you need to think

Is a 10 degree angle bigger

Or a 50 degree angle

[solar heron]

i get that now

@woven shoal ok so continue

how do we find the distance

and why do we subtract

[woven shoal]

Okay now tell me

We have to find the distance.

Distance means: Final position minus initial position

[solar heron]

yeah and isnt that just the value of the side lenght of 10

so 83

[solar heron]

how is the inital the end of the whale

bro yeah im done ur confusing me

[woven shoal]

...

[solar heron]

DonutDude | ZT5 Player profile

[proper quarry]

i just joined and im very dumb but hi

[zealous meadow]

do yall know how to solv this

[smoky jetty]

u can find the value of c using the 2nd triangle, and then the value of d using the 4th triangle. Substitute them for sides c & d of the 3rd triangle and try heron's formula?

[heavy stream]

u could use heron's formula but it would be very time taking

[smoky jetty]

same thinking, lol

[quaint tide]

how would you find C???

[smoky jetty]

oh right, lemme think

[quaint tide]

maybe it can be solved with a system of equations

something makes me feel like a = c

[zealous meadow]

how will i find c
💀

and herons
will giv
5 variables
and 4
quartic equations
tht i aint evn gunna try to slove

[quaint tide]

oh

well shi

[zealous meadow]

prolly some elegant way to slove
tht i dont kno

[smoky jetty]

time to be creative, nifty

[quaint tide]

hm

print out the question

and

cut out the triagnles

then put each equal side

on itself

then create a quadrilateral

idk

[smoky jetty]

ikr, transforming the first triangle to make a parallelogram w/ the 2nd triangle seems sensible.. i dont got proofs tho

[zealous meadow]

if

[quaint tide]

[zealous meadow]

a=c

area of

triangle 1

shud be = to

[quaint tide]

oh yeah

[zealous meadow]

area of triangle 2

by sss

[quaint tide]

ye

[zealous meadow]

fr
me rn

[quaint tide]

h m m m

[smoky jetty]

however from the reflection itself, b's heightto the base of 1st triangle is lower than the 2nd's

so a is not equal to c

[quaint tide]

ye

ok we can find the height of each triangle

hmmm

[zealous meadow]

ofc
the height has to be greater
so tht the area is greater
since the base is alr same
💀

[zealous meadow]

o

ye

after tht

[quaint tide]

💀

[zealous meadow]

um

[smoky jetty]

try making a composite shape

out of those shapes, and see what u get

[quaint tide]

lets atleast just note down the heights man

gotta do something

wth those are some pretty goofy heights

[zealous meadow]

triangle 1 : 4/3
2 : 5/3
3 : x/3
4: 13/3

[quaint tide]

Hmmmmmmm

@zealous meadow @smoky jetty technically we can join these two triangles by the 90° angles to make a new isosceles triangle

Nvm

I'm wrong srory

[zealous meadow]

im

tryna

think out

of the box

so hard rn

help

[quaint tide]

💀

I'm trying

dude where did you find this question this is hella interesting

[smoky jetty]

is the 2nd triangle a right triangl?

[quaint tide]

Unless there isn't sufficient information then i'mma be hella mad

[quaint tide]

not rlly specified so idts

[smoky jetty]

a lil' more interest and i might dive into it with my notebook, lmao

[quaint tide]

if it was the question would be a 100 times easier

[quaint tide]

lessgo

[zealous meadow]

my friend sent it

i searched it on google

and only 1 site

has the answer

but i need to pay

[quaint tide]

how do you search geometry questions on google-

[smoky jetty]

yall doing competition math?

[zealous meadow]

image search

💀

[quaint tide]

OH

SHI

[zealous meadow]

no rn
he sent it for fun

[quaint tide]

THAT WOULD'VE BEEN SO HELPFUL

[smoky jetty]

don't, we brainstorm here

[quaint tide]

In the past obv

[quaint tide]

aight

brain has stopped working after seeing this question

[zealous meadow]

fax

[quaint tide]

I'll just try the herons formula man

[zealous meadow]

best of luck

[quaint tide]

maybe it works

[quaint tide]

thx man

gonna be tough

[smoky jetty]

lmao

we missed the 4th triangle

[quaint tide]

wdym

[zealous meadow]

how

so

:sk

[quaint tide]

[smoky jetty]

nvm

[quaint tide]

LMAO

Aight I'll go onto herons

[smoky jetty]

eyes fumbled

[quaint tide]

I c

saw another 90°?

[smoky jetty]

do the rigor! lol

[quaint tide]

what's rgor

rigor

[zealous meadow]

aint no way u dont know wht rigor means

[quaint tide]

nope....

pls explain

Guys the herons formula is √[s(s-a)(s-b)(s-c)] where s is half of perimeter right?

[zealous meadow]

yes

[quaint tide]

been a while since I used that

aight

[zealous meadow]

wat grade r yall even in, tryna help me
idek
💀

[smoky jetty]

we need to have the value of all sides of a triangle

[quaint tide]

10th

[quaint tide]

we can just make 4 linear equations

[zealous meadow]

trust me

it aint

linear

its quartic

[quaint tide]

idk if that stuff is even possible

[zealous meadow]

i tried

[quaint tide]

OH SHIT

IT IS

[zealous meadow]

o same

[smoky jetty]

same