geometry-and-trigonometry

[sonic sphinx]

OK so I'm pretty sure AF, AE and EC are the same length

[gray osprey]

AE can’t equal EC

AE and Af are the radius and extend past AC

The midpoint of the whole circle is A

[sonic sphinx]

AC is shorter than AE and AF m8

[sonic sphinx]

The hypotenuse is alwyas the longest side dude

[gray osprey]

Yeh but AC is shorter than the radius of the circle?

[sonic sphinx]

oh sorry i read it wrong

ur right

whoops

[gray osprey]

All good bro

[sonic sphinx]

AF and the A to the circle are the same length

[gray osprey]

Correct

[sonic sphinx]

There is somehting you can do with that to find BD I forgot what it was tho

[gray osprey]

Pythag?

Ratio of x and y are inequivalant and don’t relate though

[sonic sphinx]

That might be part of it, consider ED into the equation

[gray osprey]

ED is 1/5 greater length than CE

[sonic sphinx]

hang on give me a sec keep trying to solve it

[gray osprey]

All good take ur time

[plucky flicker]

Is answer approx 90 or 91?

I might be wrong

[gray osprey]

How’d you get there though?

Idk if that’s right?

[plucky flicker]

Net is slow image is uploading

Might be wrong I have to check again

[sonic sphinx]

AF:FC = 10:7, 10x^2+7x^2=17x^2

[plucky flicker]

Is it wrong?

[gray osprey]

Idk if it’s that complicated lm in Year 9

I think it’s a bit simpler than that though

[sonic sphinx]

DE:EC = 2:3, 2y^2+3y^2=5y^2

@gray osprey do you know trig

DE:EC = 2:3, 2y^2+3y^2=5y^2
AF:FC = 10:7, 10x^2+7x^2=17x^2

use that

[gray osprey]

No lm learning it next term

[sonic sphinx]

Ok cause you dont need it for this problem

[plucky flicker]

@gray osprey some of the stuff I wrote are unnecessary

[gray osprey]

Yeh you was using shit I’ve never seen

[plucky flicker]

@sonic sphinx am I wrong or right?

[sonic sphinx]

Im not sure im still doing this m8

[sonic sphinx]

Solve for x and y using this info DE:EC = 2:3, 2y^2+3y^2=5y^2
AF:FC = 10:7, 10x^2+7x^2=17x^2

[gray osprey]

xy=480

Diagonal is root x squared +y squared right?

[plucky flicker]

@sonic sphinx ping me after you have solved

Can you?

[sonic sphinx]

sure

P=2(AB+BC)

[gray osprey]

2(Ab+BD)?

Bc is greater than Bd since it’s pythag

[sonic sphinx]

No

Ummm

hold up

[sonic sphinx]

NO NO NO

P=2(AB+BC)

Cause AB+BD wouldnt be the perimter of the rectangle m8

[gray osprey]

Oh shit my bad l wrote my diagram with C and D in the wrong spot

[sonic sphinx]

lol

[gray osprey]

I can’t find the side length of AB or BC

[sonic sphinx]

AB*BC=480

Right

[gray osprey]

Yes

[sonic sphinx]

ok

Use the ratios to solve for AB and BC...

You know when multiplied they equal 480 @gray osprey

[gray osprey]

Yes but l don’t know how to use the ratios to solve a side length

DE is 2:5 rationed to DC length and AF is equal length to AE 10:7 ratio there for AE to AC

We are just finding the perimeter so we just need one full length?

[sonic sphinx]

one more minute

[gray osprey]

All good bro

I got my answer idk how though lol

[sonic sphinx]

what is it

[gray osprey]

92?

Is that what u got?

[sonic sphinx]

Thats wrong

[gray osprey]

How?

[sonic sphinx]

did you get an irrational number for AE and AF?

I mean AB BC

[gray osprey]

Yes

[sonic sphinx]

Actually I think thats right lol

Whoops

ok yeah thats it

[gray osprey]

Yeh thanks for your help bro appreciate ur time

[sonic sphinx]

@plucky flicker 92

[pearl osprey]

guys, when adding vector graphically, is there only one answer? If it's so, how to know where to start and where to finish? Btw, on the algebraically way, actually, it is.

[obsidian harness]

yes, the answer comes out to the same no matter the order you add them

[obsidian harness]

guys, interesting thing I thought of today
SAS, SSS are cosine rule congruent
ASA, AAS are sine rule congruent

[dry shuttle]

Can I ask about grade 10 (or 9) geometry here?

[exotic yarrow]

Yeah

[dry shuttle]

If HP/HQ = (CP/CQ)^2, will CH be the symmedian of ∆CPQ? (I'm at the beginning of grade 10)

I read somewhere that it required EP/EQ = (CP/CQ)^2 so that CH or CE could be the symmedian of that triangle

[quartz minnow]

Where do we actually use sin cos and tan

Like if there's a right angled triangle how can we find the measure of perpendicular, hypotaneus, base by trigonometry

I mean the side length

How to find the length of any side by trigonometry

[obsidian harness]

you need the sine rule and cosine rule

for triangles in general, that may not be right-angled

[quartz minnow]

All right

[obsidian harness]

https://www.youtube.com/watch?v=PUB0TaZ7bhA
this video is for right angled trig

or just look in your textbook

or Khan Academy

[quartz minnow]

I know about soh cah toa

But how do you find the actualy value of a side ? By using the value of that degree of sin?

What is sin theta now

[obsidian harness]

that's impossible

didn't you learn about congruence criteria?

you need one side and two angles

or you need all three sides without any angles

or you can have two sides and the angle in between, that also works

[quartz minnow]

Sas sss

[obsidian harness]

yes exactly

[quartz minnow]

Aas

[obsidian harness]

that's what I'm saying

[obsidian harness]

so yeah when we say theta, we don't call the right angle theta

cause we know the right angle

what we don't know is one of the other 2 angles

[quartz minnow]

Oh okay

1 (ii) sin C

@obsidian harness

Oh I solved it

Can we find AC by trigonometry

[obsidian harness]

yes it's literally just Pythagoras

[quartz minnow]

I know but what will be the trigonometric approach for it

[obsidian harness]

why the hell are you using trigonometry to find AC

AC is 25 by Pythagoras

then SOHCAHTOA
sin C = opp/hyp = 24/25

[quartz minnow]

I got it

To find the actual value you need to put those degrees there

I mean without knowing any value you could use trig

[grave pond]

If you insist on calling it trigonometry, then say the law of cosines applied to angle B.

[obsidian hornet]

No, in this case we have less information and cant proof that sin(PHE)/sin(QHE)=1

[obsidian hornet]

Will denote the distance between some point P and some line XY as d(P;XY ).

Let the point S lie on the BC side of the triangle ABC with sides AB = c,BC = a,CA= b.

(1)AS-simedian
(2)d(S;AB)/d(S:AC)=c/b
(3)S(ABS)/S(ACS)=c^2/b^2
(4)d(B;AS)/d(C;AS)=c^2/b^2
(5)BS/CS=c^2/b^2
The equivalence (1), (2), (3), (4) is fulfilled if the point S lies inside the angle BAC, and not only on the side BC.
So, in your question u asked aboout fifth condition and it has no equivalence inside the angle, only on [BC] for the above reason.

[dry shuttle]

Thank you for the explanation

[obsidian hornet]

Glad to help

[bleak blade]

doot sent me here

[stray oxide]

for the polyominoes whats a good way to represent all of them in writing without having to memorize random names like R T etc. etc.

[lucid panther]

in my opinion it's easier to memorize and learn lol

[warm shuttle]

can't you just draw them in text?

[upper karma]

Ah, NCERT...

The first 2 exercises of that chapter are extremely easy. The third one is pretty hard.

[obsidian harness]

not really

there's a way with SOHCAHTOA instead of using the formulas sin^2 + cos^2 = 1 and 1 + tan^2 = sec^2

[upper karma]

Yes, 8.3 is comparatively very hard to the rest of the chapter.

[obsidian harness]

oh wait not the questions ah I see

what's in 8.3 btw?

[upper karma]

Uh, trig identities... It's pretty confusing for a beginner but like everything else, it gets better with practice. 😉

[quartz minnow]

@upper karma @obsidian harness You guys are from India?

[upper karma]

I am.

[quartz minnow]

What r u preparing for

[upper karma]

I'm in the same grade as you, so for now, I'll say board exams.

[quartz minnow]

I'm in 9th

[upper karma]

You're already learning Trig in 9th? Nice.

[quartz minnow]

Ioqm

[quartz minnow]

Will u give ioqm?

[upper karma]

Nope, but I have some mates that are going to give that.

[quartz minnow]

Nice

[obsidian harness]

nahi

[quartz minnow]

Nice

[quartz minnow]

Which grade

[granite fiber]

does anyone know how to rotate a cone about an axis on desmos? Im trying to make it so that a cone follows a point on a circle

slightly urgent

[obsidian harness]

Uni

[upper karma]

hmm

[upper karma]

that goes hard man

[upper karma]

polymonieos?

[warm shuttle]

https://en.wikipedia.org/wiki/Polyomino

[upper karma]

$c^2 = a^2 + b^2$

[somber coyote]

Madhur

[upper karma]

@ashen canopy

[upper karma]

C - Hypotenuse
A & B - The other two sides

[obsidian harness]

what's the question?

[obsidian harness]

it looks like number 5 is an arithmetic sequence

so (n + 2) - 3(n + 1) = 3(n + 1) - 3n

wait do they mean 2(n + 1) instead of 3(n + 1) lmao

number 7 is geometric with common ratio 2

$\frac{1}{2} 2^n, 2^n, 2 \cdot 2^n \cdots$

[somber coyote]

south

[prime stratus]

whattt i dont get itt

[obsidian harness]

you need to review your laws of indices

https://www.youtube.com/watch?v=Zt2fdy3zrZU

[prime stratus]

how do you get their common difference though?

[obsidian harness]

7 doesn't have a common difference cause it's a geometric sequence

it has a common ratio

$\frac{u_4}{u_3} = \frac{u_3}{u_2} = \frac{u_2}{u_1}$ and so on, if $u_n$ is the $n$th term

[somber coyote]

south

[obsidian harness]

the ratio between consecutive terms is constant

[prime stratus]

OH I SEEEE

i love math,,, but math is so complicated

[obsidian harness]

yeah ah so yeah definitely practice more on geometric sequences

there are harder questions than this so I wouldn't start off with anything that requires a lot of simultaneous equations for now

once you feel more confident on questions at a similar level to the #7 you just showed me, then you can move on to harder ones

[prime stratus]

we didn't start geometric lesson yettt😢😢

only arithmetic

[warm shuttle]

take log and it turns into an arithmetic sequence

[upper karma]

I like triangles

[north kindle]

please god can someone help me understand nets i cant visualize them at all

[ocean lava]

thanks

[sterile beacon]

how to prove ABC = ABD

aA = aB
AC = AD
aC = aD

ig

[snow karma]

is there like any info

u can give out

cause

[sterile beacon]

a1 = a2
a3 = a4

[snow karma]

thats all?

[sterile beacon]

But I think this is the solution

[snow karma]

u js answered ur question

a3 = a4

[sterile beacon]

you just compare from the postulate what's equal than something and you replace it by the stuff given directly from the exercise

[sterile beacon]

Nope

that's the info from the exercise

[snow karma]

oh wait

do u mean

OH

THE TRIANGLES

ah ok

well

[sterile beacon]

do you play chess?

[snow karma]

yes

[sterile beacon]

rating?

[snow karma]

like

1000 smt

[sterile beacon]

my rating is 1200 but my puzzles rating is 2200 😂

[snow karma]

i havent played

[sterile beacon]

let's play one day

[snow karma]

chess drives me

crazi

and drunk

[sterile beacon]

same

but it's good

[snow karma]

anywho, to answer ur question

a1 = a2

a3 = a4

and they both share same length

AB

[sterile beacon]

yeah

[snow karma]

so theyre congruent by ASA

[sterile beacon]

and same angles

[sterile beacon]

ig lol

yes

[snow karma]

i miss this type of geometry man

now i barely see triangle congruence or similarity

😔

[sterile beacon]

It's not that important either, the important thing is to know when it's equal and when it's not

I don't intend to become a mathematician, I just care about understanding the proof, having visualization, and understanding the logic

[sterile beacon]

for engineers! 😂

[sterile beacon]

I do want to know some calculus 1 2 3 and to solve hard logical questions

[snow karma]

calculus

is

a type of hell

i cant even describe it

😭

[upper karma]

it never ends.

[stuck sleet]

Do all of the trigonometric identities (e.g. sum and difference identities) also work on oblique triangles?

[warm shuttle]

sum and difference do

but there exists an identity that doesn't work

(is it even an identity then?)

[snow karma]

its more of a 'guess the right way to do this' in calculus and me no like 😔

and there are ten million ways to do it

[sinful flame]

Hi would like to ask if somobody could help me with this geometry tas k

Construct a rectangular trapezoid ABCD given:

α = 90°; a = 6 cm; b = 4 cm; c = 4 cm.

what would be the
construction procedure

I did something by myself but i still stuck on how do i find point C and D

i had idea with placing parrael but idk

thank you for anyone that has tip to help me out 😄

[snow karma]

whats a, b and c?

[sinful flame]

those are sides of the trapezoid

[snow karma]

yeah but

they correspond to which lengthes

like

[sinful flame]

so basically |AB| = 6cm

Points ABCD
sides abcd

[snow karma]

so AB = 6

[sinful flame]

ys

[snow karma]

BC = 4

[sinful flame]

mhm

[snow karma]

DC = 4

[sinful flame]

yeah

[snow karma]

and angle alpha is angle DAC?

[sinful flame]

no angle alpha is DAB

[snow karma]

oh

ooh yeah mb

[snow karma]

ok thats way better

so u can actually find side AD

using pythagorean theorem

[sinful flame]

well the think is i need to do construction procedure

so basically step by step draw it

example

1. |AB| ; |AB| = 6 cm

i sent similar thing earlier but it was rectangle i think

or idk

[snow karma]

so like u cant do any calculations?

[sinful flame]

im not sure but my goal is just draw it

and write step by step contruction

and if its the only answer

yk what ima draw what i mean

and send it

wait

when i check it again

all i have to do i just find point C and then make parrel to connect it with perpendicular prime from A

since this is rectangular trapezoid

[snow karma]

or just simply calculate AD 😦

[sinful flame]

lmk i said something that doesnt make sense cuz geometry aint my sh

[sinful flame]

okey sure but the thing is i NEED to draw i

it

[snow karma]

calculate it then draw it

[sinful flame]

so u say pythagorian right

[snow karma]

yes

[sinful flame]

wait how tho

since its A2=B2 + C2 right

but in this case i dont know D

[snow karma]

consider a point H on AB such that CH is perpendicular to AB which makes CHAD a rectangle

then u have right triangle CHB

do pythagorean on it

[sinful flame]

https://tenor.com/view/say-that-again-say-that-say-that-agian-marvel-fantastic-gif-12441585058103544841

[sinful flame]

alright ima do it hol on

[sinful flame]

wait

you mean point A ? or side a

[snow karma]

a point H

on AB

such that CH and AB are perpendicular in H

will look smt like this

[sinful flame]

Oh okey now i get it

[snow karma]

now

notice how CHAD is a rectangle?

[sinful flame]

wait so H is point on AB right

[snow karma]

right

we added that

[sinful flame]

OH

YEAH I SEE IT NOW

[snow karma]

alr

tell me whats the mength of AD when u get it

length*

[sinful flame]

ight

so this is the part where i calculate it right

[snow karma]

mhm

this whole thing is just to find the length AD

like that we have all the lengthes

[sinful flame]

okey nevermind i think i might be aucustic right now and can form single way how to put this

if CHAD is rectangle

[snow karma]

mye

ye*

[sinful flame]

do we know lenght if AH

of

[snow karma]

yeah

[sinful flame]

whats lenght of AH

[snow karma]

in a recatngle

[sinful flame]

yeah

[snow karma]

opposites lengthes are equal

so AH = CD

=

4

[sinful flame]

oohh okey cool

so we know basically either A or B

[snow karma]

still stuck?

[sinful flame]

yeah

[snow karma]

so

can u find length HB?

[sinful flame]

well my thing is

oh

well yeah thats

2cm right

[snow karma]

there we gooo

now isnt triangle CHB a right triangle

[sinful flame]

yeah it is

[snow karma]

u have the length CB = 4

and HB = 2

how to find the third length?

[sinful flame]

pythagor

OHH

damn im slow asl 😭

[snow karma]

its oki

js tell me the length

to confirm

[sinful flame]

and the final lenght basically CH = AD

[snow karma]

YEP

WE GOT AD

[sinful flame]

4.5

cm

?!???

[snow karma]

what

no

[sinful flame]

oh mb wrong way

[snow karma]

so?

[sinful flame]

im so fucking lost even tho im soo close same time

this sh simple af but still

[snow karma]

write me the pythagorean formula for triangle CHB

ill help

[sinful flame]

CH 2 = HB 2 + BC 2 ?

[snow karma]

pls write using CH CB and HB

[snow karma]

me too autism to understand this 😔

[sinful flame]

like this ?

[snow karma]

ok yeah u did it wrong

the length that is isolated by its own is always the hypotenuse

[sinful flame]

right

[snow karma]

u isolated CH

meaning

ur saying CH is the hypotenuse

which is wrong

[sinful flame]

but its BC

thats the hypotenuse

[snow karma]

exactly

[sinful flame]

wait

[snow karma]

so u wrote the formula wrong

[sinful flame]

wait wait

okey i need to switch the CH

wait no whats weird

or it isnt

[snow karma]

huh

no

thats what ur supposed to do

CH is in the wrong spot of the formula

[sinful flame]

so i switch CH with CB ?

[snow karma]

exactly

always the hypotenuse is left alone on one side of the formula

[sinful flame]

so its gonna be 4 (2) = CH (2) + 2 (2)

[snow karma]

perfecto

solve for CH

and bam u find AD

[sinful flame]

i got 3,5 again tho

wait no

i said 4.5 before

so its 3.5 cm ?

[snow karma]

no

how did u get that

[sinful flame]

i might be mentaly challanged

im send

I used X for CH

[snow karma]

alright so

theres only one solution correct

which x2

cause in geometry

lengthes arent negative

[sinful flame]

right

so it should be x2

to make sense

[snow karma]

yes

so AD is 2sqrt3

do u know how to construct a length that has a sqrt3 in it?

[sinful flame]

so the alternavice form doesnt count ?

[snow karma]

the what

[sinful flame]

the one under it

[snow karma]

well its an approximation

if ur teacher allows u to approximate roots to build lengthes

go for it

but

[sinful flame]

hear me out the thing is

[snow karma]

hm?

[sinful flame]

i have to find point C wich i can get with circle k which has radius = BC

ima draw it for you

[snow karma]

alr

[sinful flame]

this is what i can do with raw data

or just lenghts

but u see i need to find point C

so i can find point D

[snow karma]

wait

which center did u take

to build that arc

[sinful flame]

B

[snow karma]

with radiu 4?

[sinful flame]

since B has lenght of 4 cm

[snow karma]

radius*

[sinful flame]

yeah i need make point at one place

[snow karma]

alr but u need to find the correct angle

[sinful flame]

if I had an AC angle prick

and knew lenght of BC

actually yeah i would need angle

[snow karma]

exactly

[sinful flame]

okey well if u look at photo how would u draw it

[snow karma]

all what i js said

is how ill draw it

[sinful flame]

so i have to find angle for DAB ?

wait nah

DAB is 90 deg

since its rectangular trapezoid

[snow karma]

u need mABC

[sinful flame]

but that would leave me with

[snow karma]

U CAN DEFINETLY FIND IT

with trig

but

[sinful flame]

Right-angled trapezoid
Two opposite sides are parallel, two divergent. The sum of the interior angles is 360˚. It has two interior right angles. Interior angles at the bases are not congruent.

[snow karma]

ye

[sinful flame]

okey well

if ABC is 90 that means CXA is 90 too

[snow karma]

ABC isnt 90

[sinful flame]

my bad i meant DAB

XAB*

[snow karma]

mhm

[sinful flame]

okey well that leaves me with another 180 deg right

[snow karma]

yes

[sinful flame]

dont tell me i can use pythagor to find ABC

actually tell me 😭

[snow karma]

pythagorean isnt used for angles

[sinful flame]

what used for angles then

[snow karma]

most likely

in american schools

trigonometry

but

in order to do trigonometry

u gotta do the whole shit i did

[sinful flame]

WAIT

I GET WHAT U MEAN

u mean

TAN
COS
TAN

this type of sh

oh my i god then thats my biggest flaw rn

[snow karma]

yea but

[sinful flame]

cuz i never understood this

[snow karma]

welp

not like ur gonna use it rn

u have to draw the whole thing i drew

and find the angles

also the angle is gonna a bit

weird

[sinful flame]

is it gonna be like 45,23 deg type of sh ?

[snow karma]

idk

didnt do it

[sinful flame]

okey i guess trapezoids are my biggest opps

[snow karma]

also u probably cant

u dont have

restrictions for ur angle

cos(ABC) = 1/2

yeah nah

[sinful flame]

So this cannot ve drawn?

Cuz that can also happen

[snow karma]

it can definetly be drawn

i literally told u what to do lmao

[sinful flame]

Shitt

Im thankfull u doin allat

[snow karma]

its alr

[sinful flame]

So wait u telling me

If I find out angle beta

I might be saved

[snow karma]

thats

useless

and takes time

[sinful flame]

Ima just skip if I see this kind of question atp 😭😭

[snow karma]

and requires u to draw the right angle

[snow karma]

i have an idea

continuing on what u drew$

drew*

send the photo again

[snow karma]

alr

now on that huge ass straight line that u drew

with 90 degrees

[sinful flame]

yeah right

[snow karma]

draw a perpendicular line with it

until u find one that touches the circle u drew

in a length

4

cm

[sinful flame]

well

i already drew it

perpendicular line

it has x lenght

or not x

[snow karma]

bruh

[sinful flame]

hold on

yeah i let u cook since u know better than me

[snow karma]

uh

so

i have another idea

hopefully ur able to do this

first of all make that circle intercept with AB

in a point

we will call it H i guess

[sinful flame]

so i make circle with radius of 3 right

since AB is 6

[snow karma]

what

no

continue on the drawingg

witht he circle

of radiu s

4

[sinful flame]

wait second my compasses tweakin out

[ocean venture]

Hi .
Let me know if anyone wants to have 1:1 sessions for maths/ competitive maths/ programming

[snow karma]

ok

[sinful flame]

wait

i can see it

i think

so the circle i made nost intercepts with k+

k1

and that gives me point C ?

[snow karma]

bruh

im sad now

thats not what i said to doooooooooo

dioùBIOEZJFNIJEZµFNEZµF

😔

[sinful flame]

i have failed u master

[snow karma]

this is what i wanted u to do

continue drawing that one arc

til it intercepts

with AB

we will call it H

then draw D on that huge ass line 90 degrees such that HD is 4

then draw the radius that is parallel with HD in that circle

cause that radius makes the point C

bam

ABCD

trapezoid

[sinful flame]

omfg

yeah i wont draw allat straight away but now i read

it makes sense

[snow karma]

yeah

cause

[sinful flame]

omfg

[snow karma]

rhombus

yk

makes rhombus

[sinful flame]

ight man thanks

i will save this

[snow karma]

oki doki

[sinful flame]

cuz in like 2 weeks i will be doing tests cuz i failed math this yeart

(but u already noticed it )

[snow karma]

damn

[sinful flame]

in my schoo if u fail from certain thing

u do test at end of summer break

and if u get anything better than F

that will be on your report card updated

so lets say i had F

and if i get C from this test

i will have C on final report card

[snow karma]

ur overcooked

[sinful flame]

im dust.

[unique nacelle]

hey

I also takign the test

but I didnt fail

its geometry honors acceleration exam

our school only allow algebra 1 in 8th grade

so yeah

[upper karma]

Lol

[undone quest]

I can help with construction

[unique nacelle]

help

area

how?

is there like a formula?

@everyone

[odd zodiac]

I have a question.
Isn't it the case that when you want to prove a statement about triangles that involves the sine function, it's sufficient to consider acute angles because of the identity sin(180°-x) = sin(x)?

[north kindle]

im having a lot of trouble visualizing what shape this is

i got the actual problem but im just curious about what this shape is

[grave pond]

It's called a "cuboctahedron".

[north kindle]

very interesting shape

[distant frigate]

Can some one help me with conic sections

[unique nacelle]

circle elipse parabola and hyborfefkowekfobla

circle is horizontal cut

elipse is angled

and parabola is like vertival or something idkl?

[plain horizon]

am i legit, the only person who's tried to cut a square with angles and stuff?? 😭

[lucid panther]

nah i'm with u lmao

[plain horizon]

like, how do i find the angles to cut it up eqaully

[lucid panther]

i use khan academy

[plain horizon]

i see
i know how to find angles and what not
just i don't know how to set up all the like, brute forcing and what not

[acoustic snow]

is richard ruscyzk's introduction to geometry a good book for an accerelated geometry class

like will they be up to a high difficulty

[pallid reef]

Hi, any recommendations to understand solid mensuration???

[warm shuttle]

any line through the center will cut it into 2 equal parts. Any line not through the center will not cut it into 2 equal parts.

[plain horizon]

😱 NO WAY

[warm shuttle]

If you want to cut off 1/3 of the area, then for every point on the perimeter there is a line (i think there is exactly 2) that cuts off 1/3. And every line that cuts off 1/3 will pass through some point on the perimeter. You get partition into 3 equal parts by cutting 1/3 2 times in such a way that cuts don't intersect in the square

4 parts is complicated tho

[plain horizon]

whar???

i got it divided into thirds, idk how to find the angles

[warm shuttle]

where do cuts intersect?

[plain horizon]

center

[warm shuttle]

oh so you don't cut it with straight lines

but with rays instead

then there are still infinitely many ways to do this

does one of the cuts go through the corner or through the center midpoint of the side?

[plain horizon]

midpoint, cause it looks better on a graph

[warm shuttle]

so you have a right trapezium where one of the bases is 1/2

[plain horizon]

basically yeah

[warm shuttle]

and the height is 1/2, and the other base is x such that the area A = height x (base_1 + base_2)/2 = 1/3
(1/2) (1/2 + x)/2 = 1/3
1/2 + x = 4/3
x = 4/3 - 1/2 = 8/6 - 3/6 = 5/6

then you can drop the height from the center to the other base and get a right triangle

with sides 2/6 and 1/2

so you can get the angle there to be toa tan(angle) = (2/6)/(1/2) = 2/3

by calculator ≈ 33.69

so the whole angle is about 123.69

same for the symmetrical angle

and the third angle is 360 - 2(123.69) = 112.62

[plain horizon]

dude, wtf? am i tweaking??

i got something else 😭

[warm shuttle]

what did you get?

[plain horizon]

nvm, i'm tweaking

i did arctan(1/3)

i got it lol

i got the same thing

[warm shuttle]

now i have a question tho

what are all the ways to cut a square into equal parts with lines

if you have 1 line it's easy

if you have 2 lines there is one way for each point outside the square

fix your lines to pass through that point and sweep them inwards

by intermediate value theorem there will be a point at which they cut exactly 1/3

so what about 3 lines

[plain horizon]

gang, i can't visualize, wtf? 😭

pls give diagrams 🙏

[viral mauve]

that’s so real 😭😭

[formal geyser]

There are two equal squares

How do i find this red angle?

[upper karma]

depends

[formal geyser]

Quadratic? You mean for quadratic equations?

[upper karma]

tokyo is trolling you

extra information is required to solve this problem

[formal geyser]

I changed the description, two squares are congruent

[upper karma]

depends how you cut it

[upper karma]

extra information is required

[formal geyser]

That's it, i dont have more info

[upper karma]

what is the question source

wait

i did not see a key point

ok it’s solveable

i did not realise the red line passed through both corners

drop a perpendicular from the highest point to the bottom

you need a triangle with hypotenuse rt(2) whose side lengths differ by one

[formal geyser]

Highest point is the one containing red angle?

[upper karma]

yes

and extend the bottom blue line

you get an isosceles right triangle

[formal geyser]

I thought about it before but i still dont understand how it helps

[upper karma]

call the length from the shared corner to the base of this perpendicular x

you get that x, x+1, and sqrt(2) forms a right triangle

(assume blue and purple are unit squares)

[formal geyser]

If it is isoceles triangle why side lengths are different then?

[upper karma]

the isosceles triangle is the one from the bottom left corner, base of perpendicular, tallest point

this is a different right triangle from the shared corner, base of perpendicular, tallest point

[formal geyser]

You mean these triangles?

[upper karma]

not quite

orange triangle is yes

draw the other triangle from the corner which is blue and purple

[formal geyser]

Ehat is shared corner

[upper karma]

the one where the squares meet

[upper karma]

yes

call the length of the horizontal black line x

[formal geyser]

Okay

Then isoceles triangles side lengths are x+1

[upper karma]

the lengths of the orange triangle are x+1 and x+1

so the horizontal black line is x and the vertical orange line is x+1

[formal geyser]

And what is next

[upper karma]

the diagonal black line is rt(2)

it is the diagonal of the purple square

it is also the hypotenuse of a right triangle with side lengths x and x+1

[formal geyser]

We need to find x, right?

[upper karma]

it would help solve the problem

[formal geyser]

X = (-1+sqrt(3))/2

[upper karma]

sounds about right

let the smallest angle in the black triangle be t

[upper karma]

sin(t) = x/rt(2)

which is (rt(6)-rt(2))/4

incidentally equal to sin(15)

so t=15

the red angle is 75

[formal geyser]

Okay, but how did you understand that the red angle is 75

[upper karma]

the angle between the red and black diagonal lines is 45-15=30

the angle between black vertical and purple right is 45

adding gives 75

[formal geyser]

I don get this part

[upper karma]

the black diagonal line is the diagonal of the purple square

so it makes an angle of 45 with the square

[formal geyser]

Oh yes, thanks

I am quite bad at geometry

[formal geyser]

But you explained it very well

[upper karma]

thanks

i enjoy geometry

[formal geyser]

@upper karma how do you study geometry

Do you use websites, books?

[upper karma]

idk

[upper karma]

aops introduction to geometry

[formal geyser]

Okay, thanks

[upper karma]

add 180

square roots of complex numbers are always diametrically opposite each other

[sinful flame]

Hello, I have a question about the rhombus, if the diagonals are different for example : |AC| 8cm and |BD| 6cm does it mean its not rhombus anymore? Because all sides of a rhombus should be the same length

[kind rapids]

But its diagonals can be different

[sinful flame]

i might find it

Area of a Rhombus Formula = 1/2 × d1 × d2 where, d1 and d2 are the diagonals,

[sinful flame]

.

wich gives me side or area is 24 cm

what make sense right

or is it too big

wait no i might said somethin

[kind rapids]

24 cm^2 to be precise

[sinful flame]

so wait

how do i find out side

if this is area

then

[kind rapids]

You need to use properties of rhombus

[sinful flame]

maybe this ?L

if im right i should get that a = 5cm

and perimeter is 20

[sinful flame]

wdy think

[kind rapids]

Yeah

They use formula: $a = \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2}$

[somber coyote]

Closer

[kind rapids]

Use this formula and you will get a = 5

[sinful flame]

oh thanks

well yeah my goal was to calc it and then draw it

[twin spindle]

How do you get 256.7175 degrees? Ik that by adding 180 from -26.5651 and dividing by 2 gets you 76.7175 but im not sure hou you're supposed to get the other angle?

[obsidian harness]

,w (pi - arctan(2/4)) * 180/pi

[somber coyote]

Results provided by WolframAlpha

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

Requested by sooouth.

[obsidian harness]

so 153.43 is the same angle as 153.43 + 360 = 513.43

,calc 513.43/2

you shouldn't get 256.7175

,calc 153.43/2

[somber coyote]

Result:

76.715

[twin spindle]

So the hw assingment is wrong?

[obsidian harness]

wait sorry lol

[twin spindle]

Oh wrd

All good

[somber coyote]

Result:

256.715

[obsidian harness]

yes that's how you get it

so you can always add 360 to the angle then divide by 2

that's the same as dividing by 2 first and adding 180 deg

in fact, if you keep on adding 360 and then dividing by n
that's the same as dividing by n and adding 360/n each time

[twin spindle]

Wait so basically after you do the arctan of some number related to this question, you can add 360 and divide by 2, or divide by and then adding 180?

[twin spindle]

How did you get 256.715? because 153.43/2 is 76.7175

[obsidian harness]

yes

[obsidian harness]

I rounded a bit

you can just do 76.7175 + 180

I was just making sure you got the correct angle

[twin spindle]

Oh wrrddd

I see I see, thank you!

[obsidian harness]

np

[upper karma]

https://cdn.discordapp.com/attachments/238956921581862913/1273356990046932992/image.png?ex=66be519c&is=66bd001c&hm=c68a66fc7cdb7369603e149f55a6bb6fe4dfd0bb0a1f9faf91aef14f00f9bfce&

{k(n^2 + 1}, k(n^2 - 1), 2kn} is still a pythag triple

then what? map 1 to n and m to n? is that even legal? Lol

[maiden brook]

$(2mn)^2 + (m^2 - n^2)^2 = (m^2 + n^2)^2$

[somber coyote]

hockeydude85

[bleak blade]

Yeah

And then there’s a scale factor

Of k

[upper karma]

yes clearly it is a pythag triple

i was asking how you go from {(n^2 - 1), (n^2 + 1), 2n} to that

can you map 1 to n and m to n? and then scale factor it by k?

[maiden brook]

imo that’s a very weird way of going about it I don’t think so? why not just say it just satisfies the Pythagorean theorem and that’s good for u

[odd thistle]

Anyone know an easy way to remember Sum to product

And product to sum trig formulas

[faint pasture]

for the sin ones, if you are addding the sines, the arguement in the sine got a + and the cosine -, ie the sign between the sines is same as the sign in the arguement of the sin

for subtraction

the same

for cosines i dont really have a tip

[half snow]

Smn pleaee help

@here

[violet cairn]

is it a regular tetrahedron?
if not then i think there's not enough information, as you could extend the truncated part to arbirarily big one

[half snow]

Regular one

[violet cairn]

nvm, even if it's regular you can apply the same reasoning

as long as we have no information on the truncated piece

[half snow]

So cannot be determined?

[violet cairn]

i'd say a and c cannot

b is doable

(calculate area of a equilateral triangle then (area * height)/3)

[half snow]

Aight yea

I got b

[north kindle]

this problem confuses me, what does it mean by "1/4 inch thick"? is the diagram to scale?

[obsidian harness]

the length of the red line on the left and the height of the red line on the right
are both 1/4 inch

also the diagram isn't to scale

if you think about it, it's just similarity

6/4 = 3/2 so (3/2)/12 = 1/8 of the volume is reduced on the left block

so 7/8 is remaining

[north kindle]

yeah thats what it seems like but im trying to show it for the right one

[obsidian harness]

ah consider similar triangles
the large triangle is 10 by 6
the smaller triangle is 10/4 by 6/4

and you want the ratio fo the areas

[north kindle]

yeah

i got that 15/16 is remaining on the right wedge though

which im not sure is correct

actually it might be

it makes sense to me that the left one has less remaining area

[ornate pasture]

54

[obsidian harness]

yeah that's what I get too

bruh I'm wondering how are you doing this level of problem while not understanding MATHCOUNTS

I guess you're following the book but

[north kindle]

yeah the book is a good teacher

that one wasnt too bad

[obsidian harness]

okay

[maiden brook]

^^

[outer wadi]

how's the slope of this graph -sqrt3 shouldn't it be -1/sqrt3

[warm shuttle]

the angle in picture is not 60°

if it was accurate, then it would be much clearer

[outer wadi]

isnt slope just y2-y1/x2-x1

[warm shuttle]

$\frac{y_2 - y_1}{x_2 - x_1}$

[somber coyote]

CyclicTree

[warm shuttle]

yes

100√3 is wrong if you are wondering

or at least contradict the 60° angle

[outer wadi]

oh my bad got it

[warm shuttle]

Notice that 60° > 45°

[outer wadi]

thanks @warm shuttle

[warm shuttle]

angle in the picture is less than 45

[bleak blade]

Ok this excersise is actually super easy tho

Because the parallelograms

[warm shuttle]

you can nuke this problem a bit

Problem is stated only using affine geometry, so it's enough to consider the case of an equilateral triangle with side length 1.

Then just slide triangles UTX and SXR down to the base

you get sum = side length = 1

[north kindle]

youd need to prove UX = BP and RX = CQ first since the book hadnt covered parallelograms by that point

[ripe plaza]

I'm being humbled so bad

I don't get the law of cosines

😭

[bleak blade]

It’s by definition

if it’s a parallelogram

[north kindle]

yeah and the book hasnt taught the definition of parallelograms yet

[opaque acorn]

a² = b² + c² - 2bc * cos(angle)

[bleak blade]

Yeah but

There’s no point in not using it

If you have a tool always use it to your advantage

[north kindle]

its only a hard problem in context because you dont have that tool yet

[upper karma]

Honestly

The hardest part of trig for me is remembering the formulas

Like

Idk they’re so random

[bleak blade]

tbf

if you spend enohgh time with them

you know them by heart

[upper karma]

Ah

I see

[upper karma]

How long did this take you, bubo

[bleak blade]

idk

[north kindle]

just rederive them whenever you need them

[bleak blade]

if you solve enough problems

[upper karma]

Yh practice makes perfect ig

[ripe plaza]

that's it? to find the sides? orr

[vagrant snow]

Find one side given that you know two sides and that Theres a angle opposite the unknown side

[azure helm]

1. Let ( L: X = \lambda(2, -1, -2) + (-6, 1, 1) ) and (\Pi) be the plane that passes through the point ( A = (-6, 1, 1) ) and is perpendicular to the line ( X = \lambda(-1, -2, 1) ). Find two points ( B ) and ( C ) such that the triangle ( ABC ) is right-angled at ( C ), the hypotenuse ( AB ) lies on ( L ) and has a length of 9, and the side ( AC ) lies in the plane (\Pi).

[somber coyote]

938c2cc0dcc05f2b68c4287040cfcf71

[azure helm]

Help

[north falcon]

can someone please explain the math behind all 3 problems and how to apply it to similar questions? I haven't learned this method of math yet but am trying to self study

idk if this is considered geometry srry if its not

[warm shuttle]

For first one, can we assume that the angle is 90° and that the line is tangent?

For second one, can we assume that the right angle is B?

[worthy acorn]

Why do we keep the pi/root 2 when its cos2theta

[warm shuttle]

set $\theta = \frac{\pi}{8}$

[somber coyote]

CyclicTree

[warm shuttle]

and honestly these aren't that easy to answer, so maybe start a help channel

[trail tendon]

because they're using theta = pi/8

so cos(2 (pi/8)) = cos(pi/4) = 1/sqrt(2)

then they should write = 1-2sin^2 (pi/8)

[twin spindle]

Tips on how to verify this?

[worthy acorn]

Ah that makes sense

Thank you

[twin spindle]

I tried multyplying by the conjugates, but then I felt as if that didn't work

[trail tendon]

find a common denominator and then magic happens ._.

[twin spindle]

I also did that, and that got me here 😭😭

[graceful talon]

I got 2/cos\theta

But sec is not sin

[faint pasture]

maybe it is and we just dont know that, yet!

[twin spindle]

I also got 2sec(theta) when I re did it

[graceful talon]

I mean i know trig functions are just different forms of one function

You can choose which should be the basic function

,align \sec\theta&=\frac{1}{\cos\theta}\
&=\frac{1}{\sqrt{1-\sin^2\theta}}\
\sec\theta&\neq\sin\theta

[somber coyote]

רי-ית

[faint pasture]

💀 orr and hear me out just compare ranges

sin's range is -1,1

sec's is everything except that + {-1, 1}

[graceful talon]

Damn

You saying they're equal at ±1?

Wait no

But csc is

[stuck sleet]

What is the geometrical meaning of inverse cosine of an oblique angle that have a negative result?

e.g.

arccos(argument) = theta

where the resulting theta is a negative number

[faint pasture]

nah nah

am just talking about the ranges

[graceful talon]

Ok

[graceful talon]

In a unit circle angles can go negative

When u measure an angle clockwise from the right horizontal line i forgot what's it called

[stuck sleet]

My confusion is when I use law of sines or cosines or getting the angle between straight lines (in analytic geometry)

Is when I get the angle I have the negative result

[graceful talon]

Maybe an error in computation

[graceful talon]

But inverse trig functions have limited ranges

I think arccos range is [0, π]

So there should be no negative

Hmm but it's more about functions, idk

[faint pasture]

for arcsin there can be

[graceful talon]

Yea

[low shuttle]

can someone tell me if my working is correct

[obsidian harness]

bruh you didn't post anything

[upper karma]

what are some physics prerequisites in mathematics?

in trigonometry?

[upper karma]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[dire furnace]

I’m in AP physics one rn, so far it’s mostly simple algebra, but precalc is a prerequisite for it and it’s still early in the year so I assume it’ll get more complicated

[upper karma]

no like just in trigonometry

algebra is ok

precalc has no involvement rn, for me

[upper karma]

except basic t ratios and stuff

[brisk karma]

Looking to get into Geometry, specifically algebraic geometry,just as a fun hobby. If anyone has reccomendations at good introductory books I can find online lmk 🙂

[upper karma]

algebraic geometry?

[brisk karma]

Yup yup

[upper karma]

formulas and shit right?