geometry-and-trigonometry

[fast geode]

Love these problems

Easy

[obsidian harness]

ezpz

[fast geode]

mhm

[weary shoal]

Anybody know the name if the geometry bc i got a test on monday and i do NOT understand this

[jade sierra]

what do i do 😭

i only get to the part where you add u and v but not find the rest

[faint hollow]

so there are two formulas you can use

1. $|u+v| = \sqrt{u^2+v^2+2uvcos(\theta)}$

[somber coyote]

Black_Gold

[faint hollow]

where $\theta$ is the angle between two vectors

[somber coyote]

Black_Gold

[jade sierra]

I’m sorry but I came from alg 1 last year 😭

[jade sierra]

Ik I probably understand it but I feel like I don’t

[trail tendon]

on the right side, every time he says u and v he means the magnitude

[jade sierra]

How did I go from special right triangle a day ago to this

This feels like a whole new class 😭

[trail tendon]

lol

wait lemme fix

[faint hollow]

now to find the angle use --> $tan(\alpha) = \frac{usin(\theta)}{v+ucos\theta}$

where $alpha$ is angle between v and u+v

[somber coyote]

Black_Gold

[faint hollow]

these formulas are derived from that right triangle only

[jade sierra]

what does the weird 0 mean

[somber coyote]

Black_Gold

[faint hollow]

theta, it represents angle, its a greek letter

[jade sierra]

Well as it happens I have 4 angles and I don’t know which to pick

[faint hollow]

do you want the derivation?

[jade sierra]

I just started trig idk what’s derivation

[faint hollow]

do you want to know where these formulas came from??

[jade sierra]

Sure

[faint hollow]

the second one

sorry for the poor camera quality

[toxic linden]

can someone pls explain how the formula of the cone works?

[hoary totem]

vastly oversimplified:

imagine a right pyramid
turns out 3 of these can come together to form a rectangular prism

so the volume of a right pyramid is 1/3 of a rectangular prism

turns out this is true for any base shape though, not just squares (pyramids; rectangular prisms)

hence the volume of a cone is 1/3 of the volume of a cylinder

[toxic linden]

aight thank you

[broken nova]

can anyone send class 11th cbse trignometic funtions questions

[minor pawn]

theta? what is the correct name T^T

[bright pecan]

why does this not work?

thank u

we r year 9s

[minor pawn]

ty

[minor pawn]

im year 9 but we didnt take that yet

[bright pecan]

basically u need to do some intergrals stuff

beyong my capabilities

[minor pawn]

alright.

[faint hollow]

If you are taking the side of triangle 'r'(same as cone's radius) then at that point it's no longer a triangle, it's side on circumference would be arc

[jolly shore]

This may help you guys

[plucky cargo]

moew meow help with this one one

[obsidian harness]

try squaring both equalities and then adding them together

[plucky cargo]

thxx

[faint hollow]

Dayum, I just solved this as a subpart of physics proof i was doing

[wintry yacht]

Q3

[obsidian harness]

by angle chasing, triangle GEA is similar to BFC

so BF/BC = GE/GA

or that BF/1 = 1/GA, as GE = EA (isosceles) = 1

[plucky cargo]

hahaa

[wintry yacht]

Yup

[obsidian harness]

now since GA = x, that implies BF = 1/x

[wintry yacht]

Ohh kaaay

Thankyou

[obsidian harness]

npnp

[dire mirage]

Help pls

If its an isosceles then I can split it up into 2 right triangles right?

And since its congruent on all sides that must mean the angles are the same right?

But somehow my answers are wrong

[silent plank]

can you show your work/calculations

[dire mirage]

Phone is dead so I cant at the moment

[viral mauve]

we integrate

[bright pecan]

i meant taking the area of the triangle and multiplying it by the circumference, my idea was wouldnt the area of triangle just rotate a whole circle around, thus getting us the volume? the cone is straight and not tilted

[faint hollow]

Yeah I get that

Wait let me think

[vagrant snow]

Notice the equilateral triangles

And each angle of an equilateral triangle has the same angle which is?

[worthy tendon]

You are right in your thought process, but if you observe carefully, you can notice that only the tip of the triangle travels through the disstance 2 pi r, while the side opposite to that vertex does not get to move at all. So it is unfair to multiply the parts of the triangle with the distance they dont cover

so you have to kind of cut the triangle into pieces that travel the same distance and multiply the area of that piece with the distance it travels... until you are done with multiplying all the pieces

[jade gyro]

the middle angle is 60 degrees. since AB is straight line, the angle to the right is 120 degrees.

since the triangle with 120 degrees is isosceles, x = 30

[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.

[jade gyro]

ooh i thought we were supposed to give out solution

[undone quest]

Ar(ABC)

[faint pasture]

Uhhhh one way is to apply the cosine rule in ADB and CDB and solve for CD and BD

Now you have three sides

And you can just use the herons formula

Tho i think there should be a better way

[polar cypress]

Can anyone help 😭

I’m dumb

Kay

Second….

ANSWERR PLS

😔

I will after I sleep oh yeah!

Get out

🧌

….

Is it dead

Weird cat…

Mine sleeps like a normal cat

Lalalalalalalala

I choose no thanks

Sigh fine

Buh bye..

[spiral hare]

[undone quest]

@faint pasture I actually dont know law of cosine

Though that way is quicker

I figured it out

[faint pasture]

Uh i dont really think you can do it with just geometry as it involves angles and stuff

[undone quest]

We can assume h

Height

[faint pasture]

Ye

[undone quest]

In base of triangle

And then pythagorus

Eqn1 and eqn2 subtracting

But thank u

[faint pasture]

That sounds kinda like the derivation of the cosine rule

I mean not exactly

[undone quest]

Yup tup

Premath uploads these type of questions

But his question solving is complex

[toxic linden]

Why is the volume of a cone half of the volume of a sphere. We are in 9th grade

[faint pasture]

it is not

[toxic linden]

no like

of the same radius

[faint pasture]

1/3pi r^2 h is not the half of 4/3 pi r^3

[toxic linden]

what does this mean then:

[toxic linden]

thats what i though

[faint pasture]

they let the height = 2r

for almost no reason

and yea

its just a coincidence

if you want a proper proof of the volumes formulas thats gonna be #calculus

[cunning lion]

https://www.mathsisfun.com/geometry/cone-sphere-cylinder.html

[faint pasture]

its still just a coincidence though, right?

[toxic linden]

yeah we did the integrals for proofs of the formulas

[faint pasture]

yeps (which high school is teaching integrals to 9th graders damn)

yall are lucky, i was introduced to calc for the first time in 10th grade and had to mostly self teach back then too

[toxic linden]

wait can you help me in #precalculus

[faint pasture]

ask away i can try at least

[faint hollow]

I was introduced in 12th, but i read pretty much on my own from 10th aswell

[faint pasture]

mhm, like in here it is properly taught from 11th grade cuz of physics so yea

[faint hollow]

yeah but for physics, our physics teacher gave us same basic knowledge, so that we can do the integrals in physics in 11th

[faint pasture]

💀 they were like you gonna have to learn this properly in 12th anyways, so we might as well go a little deeper

[cunning lion]

depends on what you mean by "coincidence". archimedes used this relationship between the volumes of spheres, cones, and cylinders to discover the formula for the volume of the sphere

relevant video: https://www.youtube.com/watch?v=5q_sfXY-va8

[marsh elk]

Useless

[queen coral]

Do you Have the Resolution of the 1 problem?

[covert mango]

i dont have

@queen coral

[queen coral]

Hm, ok

I liked this question

[faint hollow]

can you translate and send again

[pearl compass]

how do you solve this?

[faint hollow]

make a straight line equation with the help of given info

like length of first and last rod, spacing b/w holes

also it would help to take the bottom of first rod as origin

[pearl compass]

Alright, I will try that now. Thanks.

[formal geyser]

Hello

How do i solve this problem

[vernal pilot]

This is really interesting. I would start by finding the side lengths of the triangle.

[formal geyser]

Of what triangle

[formal geyser]

I have found the rectangular

Both triangles have the area of 1 cm^2

[formal geyser]

These are the sides, since the side of the square is 2 cm

[vernal pilot]

When it comes to geometry puzzles it helps to annotate as much as possible and remember you can draw lines as you want.

[formal geyser]

I have just solved it

4 cm^2

These are all the sides i have found

Then i calculated the area of upper small triangle (1/5 cm^2), added it to the 3 cm^2 of trapezoid (given in the instructions), calculated the area of the big square and substracted the are of trapezoid + triangle from it

The sides of the small triangle i have calculated using the similarity of triangles

The answer is right, but correct me if i overcomplicated the solution or did unnecessary steps

[flint marlin]

Can you please translate?

[old haven]

Sorry to bother you, I wanted to ask. I have the intention of joining the Mathematics Olympiad but I'm confused about where to practice. Do you have any suggestions for websites/YouTube channels for practicing math questions, guys?

[undone quest]

Grade ?

[mild smelt]

Website of the Olympiad? Problems from previous years

[cunning lion]

#competition-math may be more suited (see also the discord server in that channel's description)

[gentle wren]

yo how to do vector shit

im so confuse

like how to find the ratio of a point on a line doing vectors

[round pulsar]

Can someone give me a thorough explanation on how to do this please

[cunning lion]

use the pythagorean theorem

[maiden brook]

In triangle ABC, segment BD is an angle bisector of the triangle. suppose point E lies on AB. why does the reflection of E over BD have to lie on line BC?

[formal geyser]

hi

can anyone explain why if we have two triangle and all their corresponding angles are the same, then the triangles are similar and the ratio of their sides is the same? Is it because when the corresponding angles are the same, then the "shape" of triangles is the same while the size may differ

[cunning lion]

resizing a triangle doesn't affect the angles, so any two similar triangles can be thought of as the same triangle just moved around and enlarged/shrunk

and the size change affects all of the sides equally, so their ratios are preserved

[formal geyser]

okay, got it. By the way, the angles ratio doesn't equal the sides ratio?

@cunning lion

[cunning lion]

we have a theorem relating ratios of angles and sides which is the law of sines

[formal geyser]

i know it, just wanted to clarify whether it works with just angles

[cunning lion]

it does not

[flint marlin]

It doesn't?

Could you verify your question?

[faint pasture]

Actually it does

[faint pasture]

I got kind of a coordinate geometry based proof if you want it

[flint marlin]

Ok I didn't get the question then

[faint pasture]

Or nvm it isnt really coordinate

[flint marlin]

Could you explain the question?

[faint pasture]

Ah i need a diagram to explain the solution

[faint pasture]

Oh thats just like if i have a triangle ABC

Then i draw the angle bisector through A

Then the mirror image of B about the angle bisector

Will lie on AC

They just kinda changed it up a bit

Instead of B they took another point on the side

The solution doesnt change tho

[faint pasture]

Oh i get what you mean

They wanted to say BD where they said AB im kinda sure

[flint marlin]

Oh yeah

Got it

[faint pasture]

Reflection of a point about a line passing through it doesnt make sense and they defined BD

So ig yea they meant BD

[flint marlin]

Did you prove it?

Ok I got the proof nvm

[faint pasture]

I know the proof but i need a diagram to explain it properly and i dont have anything to draw it on

But basically let the side AB and angle ABC be fixed, ie all we can do to the triangle is adjust AC

So if we choose AC such that ABC is isoceless the mirror image of B will lie at C

However you can extend the side AC much as you want the mirror image wont change up

So we again fix up AC to be the original length

And the mirror image still lies on there

[flint marlin]

Hmm

Basically we can always find a point let say E such that AE=AB

On AC

[faint pasture]

Yea in short that

[flint marlin]

Yeah

Are you into olympiad geometry?

[faint pasture]

Oh hell naw

[flint marlin]

Oh..

[faint pasture]

I just had to do this proof for coordinate geometry

[flint marlin]

Oh

[faint pasture]

Which is in my syllabus

So yea

[flint marlin]

Mmmm

If you get any good questions , ping me please I would like to try

Because im preparing for the 2nd stage for representing my country in IMO , I just cleared the first stage

[mild smelt]

ABC isosceles triangle with angle A = 80°, base BC.

Inside of the triangle is point M such that CBM = 30°, BCM = 10°.

Find angle AMC

😊😊

[formal geyser]

Hi, how do i solve this one

I have already calculated the area of the left right triangle

But what about something that is similar to a circle's sector, but is not

I mean, the sector where "Blue Area" is written

I know it is not a part of a small circle, so i dont know how to calculate its area

[mild smelt]

Look at this triangle

[flint marlin]

Thanks

[formal geyser]

Okay, so it has 4 cm side and arctan(3) + 60° angle

[mild smelt]

Uhmm one of its angles is 30° actualy

[formal geyser]

How did you figure it out?

[mild smelt]

So the red triangles angle is 30.. so now it's easy to find it's area...

[maiden brook]

edited

so then how do u do it?

[restive heron]

Got a study question. I'm taking this trig course that doesn't have any lectures. It's basically read the chapter, then do homework problems. Then there's a huge playlist of trig videos. I don't know if this is normal for online school.

Anyways. Would it be more efficient to work on the problems as I'm reading? How do other folks study?

[mild smelt]

Everyone studies differently

But my personal approach is... go straight to the problems. and when you encounter some term your don't understand, find it in the lectures 🤣

[restive heron]

Lol. Thank you! I think I'm going to take this approach to the next section.

[hot latch]

can someone help

[cold crane]

does anyone know how to explain geometry proofs to me😔

[hot latch]

show me notes and i probably can

[jolly hinge]

help

😭😭

ive been on ts for 2 days😞

[bitter granite]

if you are allowed a calculator you could try an exponential regression

doubt it would work with only 2 points

yea idk then

[hot latch]

chat gpt it

[trail tendon]

Do you know the exponential model general function?

[jolly hinge]

mb gng i turned ts in idgaf anymore😭😭😭

[trail tendon]

☠️

[jolly hinge]

yeah like i got notes from class but like i would keep applying it and it would say its wrong😭

[trail tendon]

br

[hot latch]

SO like please

im struggling

[trail tendon]

is this the "figure" they gave you?

[hot latch]

yes but i redrew it cuz she put the wrong degrees in hers

[trail tendon]

how do you know she put the degrees wrong?

[hot latch]

she told us and i didnt wanna have to explain it so i just redid it

she put 60 instead of 45

[trail tendon]

oh

[hot latch]

do yk how to do it

[trail tendon]

i mean you know how a 45 45 90 triangle has side lengths 1:1:sqrt(2) where the hypotenuse is the sqrt(2)?

[hot latch]

yes

[trail tendon]

can you see that your diagram is a 45-45-90 triangle?

i mean the triangle created between the person on the boat and the tower

[hot latch]

oh so would i be like tan45=5280+x/1454

[trail tendon]

because the angle of elevation is 45 degrees, and the tower goes straight up

[trail tendon]

yeah lol

that works

[hot latch]

okay so im just slow

let me work it out

[trail tendon]

nah that was fast 😂

[hot latch]

i just didnt even consider that until you reminded me 45-45-90s exist

[trail tendon]

😂

[hot latch]

hold on theres another i cant figure out

if u dont mind

[trail tendon]

what is it?

[hot latch]

one sec

wait it doesnt work

tan45= 1 right

[trail tendon]

yeah

[hot latch]

so id be getting a negative if solving for 4

x*

[trail tendon]

uhhh

wait wth

[hot latch]

she somehow got this

[trail tendon]

she did that with 60 degrees tho?

[trail tendon]

thats also negative

[hot latch]

which if i used the 60 instead of 45 id have 1454 sqrt3

[trail tendon]

i guess the boat is on shore LOL

🤣

[hot latch]

imma DIE

hold on

[trail tendon]

that is actually very weird

[trail tendon]

is this from a textbook or did your teacher make this problem?

[hot latch]

she made it

[trail tendon]

💀

[hot latch]

but shes like hella smart so im lost

[trail tendon]

i think the answer is fine to be negative

logically it doesnt make sense but im not sure if she thought that through

like her answer is negative as well

[hot latch]

if she wouldve said 30 degrees it would make sense

[trail tendon]

i guess the boat is just on the shore 😂

[hot latch]

here lemme show u what she drew

[trail tendon]

ok

[trail tendon]

XD

[hot latch]

im soooooooooooooooo lost lol

like that makes sense right??

where does she get the denominator 3

[trail tendon]

yeah, that makes sense, except in the sense that the distance is still negative

[trail tendon]

you mean the square root 3?

[hot latch]

her answer is 1454sqrt3/3 - 5280

so like

[trail tendon]

oh

[hot latch]

i have that but just no demon

denominator**

[trail tendon]

she used 30 degree elevation that time ☠️

🤣

tan(30 deg) = sqrt(3)/3

tan(60 deg) = sqrt(3)

tan(45 deg) = 1

💀

[hot latch]

right thats exactly what im thinking

bro whatever hopefully i get partial credit for setting it up right

[trail tendon]

i bet you'd get full credit if the problem is screwed up 💀

if you answer the question as is

[hot latch]

i mean if theres one similar on the test

[trail tendon]

correctly

[hot latch]

here this was the other one

thats the answer but coukd u explain why i use tan -1 cuz she never even showed us that

[trail tendon]

ok so tan(x) = 9.3/7.4, that makes sense right?

[hot latch]

YES

[trail tendon]

thats the inverse tangent function

[hot latch]

sry caps

[trail tendon]

do you know inverse functions?

or like what they are?

[hot latch]

i mean isnt it cotangent

[trail tendon]

no, thats the reciprocal

[hot latch]

oh

[trail tendon]

of tangent

[hot latch]

i thought tan-1 was cot

[trail tendon]

nah they're different

tan^(-1) is arctangent

[hot latch]

okay so how do i know to use that?

[trail tendon]

or inverse tangent

basically, if you apply an inverse function to a function of x, you get x back

so arctan(tan(x)) = x

so, if tan(x) = 9.3/7.4, you can take the arctan of both sides
arctan(tan(x)) = arctan(9.3/7.4)
x = arctan(9.3/7.4)

does that make sense or 💀

[trail tendon]

like tan(45 degrees) = 1, so arctan(1) = 45 degrees

because arctan(tan(45 degrees)) = arctan(1)
45 degrees = arctan(1)

[hot latch]

LMFAO wtf is an arctan is that just the inverse thing

bro 💀

im cooked

[trail tendon]

arctan is the inverse of tan

arctan(x) = tan^(-1)(x)

so if y = tan(x), x = arctan(y)

(theres domain restrictions but that doesnt matter for triangles here)

[trail tendon]

it just does the opposite of what tangent does

[hot latch]

okay so how do i know like when to use that? like when i am trying to find the angle of depression?

[trail tendon]

yeah, when you know tan(x) = something and you need to find x

you want to know the angle

then you can use arctangent

x = arctan(something)

and then you can probably just plug into a calculator lol

[hot latch]

omg ur lying so tanX is tan^-1??

times whatever

so like tanX=(7.4/9.3) so i can do tan-1(7.4/9.3)

[trail tendon]

no what?

[trail tendon]

yes

x = tan^(-1) (7.4/9.3)

and then you plug that into a calculator because no one in their sane mind would manually estimate that

[hot latch]

ohhhhhh i see omg okay i get it

[trail tendon]

this is a joke btw

[hot latch]

also i figured out the other one

no u right tho

those ppl scare me

[trail tendon]

oh cool

[hot latch]

i did adj/opp instead of opp/adj lol

whoops

[trail tendon]

oh LOL

and its still negative 😭

[hot latch]

dumb mistakes will be the death of me

[trail tendon]

nah i fell for it too

😭

just blame the brain not functioning correctly at that moment 100%

[hot latch]

ive been mixing up cos and tan for like the past 2 days and its actually driving me insane

true not my fault

[trail tendon]

get sleep, drink water, practice 😔

try to use brain more ✅

[trail tendon]

this is actually top tier tip for not making mistake

use brain more

[hot latch]

the practice makes me wanna die cuz i panic so much abt this stupid class

[trail tendon]

😂

[trail tendon]

funny i feel like practicing would make you panic less...

[hot latch]

idk i like over practice and then blank out when its infront of me

[trail tendon]

yea honestly thats relatable XD

[hot latch]

whatever hopefully that doesnt happen tmrw cuz it caused me to fail my first one

[trail tendon]

just think through everything if you can

[hot latch]

ill try my best thank u for the help ill probably be back

i say give it 5 min

[faint hollow]

@trail tendon , Bro how did you get the "very active tag"?like i live in India so i can't help when its night here and thats when most of the convo happen because of people living in USA, also where are you from

[trail tendon]

i dont know how i got it, it just appeared one day probably cuz i talk a lot 💀
im from USA yea

[oblique widget]

@trail tendon

i dont really understand the pattern

the triangle similarities

[trail tendon]

well, you can see that MNL is 90 degrees, and NPL is 90 degrees, right?

[oblique widget]

i do see that yes

[trail tendon]

and if NPL is 90 degrees, NPM is 90 degrees as well...

right?

[oblique widget]

i mean like the order

of the letters

[trail tendon]

ohh

go from left to right

[oblique widget]

ok so if i go left to right

pln

[trail tendon]

so MNL means if you start at M, then go to N, then go to L

[oblique widget]

is similar to

mpn

[soft vector]

Does anyone have any trigonometry tips and tricks to help with memorizing values on the unit circle

[oblique widget]

right bro i need that

oh i know how to get to MNL just NPL and MPN

[trail tendon]

no

[oblique widget]

becausae if i find NPL and MPN i could find MNL

[soft vector]

I have some holes in my understanding I feel like im so close 😭

[oblique widget]

i dont understand lowkey dont see the vision

[trail tendon]

if you have triangles LPN and NPM, the corresponding sides have to match up. so LN is the hypotenuse, just as NM is the hypotenuse. LP is the long side, just as NP is the long side. PN is the short side (of triangle LPN), just as PM is the short side (of triangle NPM)

[trail tendon]

wait i read what you said wrong

[oblique widget]

ok that i see the vision

LN is hupotenuse and MN is hypotenuyse

[trail tendon]

yeah

[oblique widget]

yeah i see the right angle for both triangles

that lowkey make sense

[trail tendon]

mhm

[oblique widget]

so if i start at the right angle

for example i said

triangle pln

it would be

npm?

[trail tendon]

PLN goes along the long side and then it goes along the hypotenuse
NPM goes along the long side and then it goes along the short side.

so those wouldn't correspond, right?

but if you went like PNM

then you go along the long side and then along the hypotenuse

so that would correspond to PLN

[oblique widget]

UHH

[trail tendon]

XD

[oblique widget]

sorry would it be like

NMP

[trail tendon]

NMP goes along the hypotenuse and then the short side, right?

(i'm leaving out the last but technically then it goes along the long side because its a triangle)

[oblique widget]

it does the hypotenuse

NM is the hypotenuse

HOLD ON

TRIANGLE PLN

WOUD BE SIMILAR TO

NPM

DID I GET I

[trail tendon]

PLN would be congruent to PNM

they both go through the long side, then the hypotenuse, then the short side

[oblique widget]

NO 3HQW6

[trail tendon]

huh?

💀

[oblique widget]

let ,e function rq

i kinda see it

but

let me show you an example

[trail tendon]

mhm

[oblique widget]

so

if it was

PLN

wouldnt the second triangle start with N

[trail tendon]

where?

[oblique widget]

im talking about the example i sent earlier

because based on this example i js send triangle KPM is similar to MPL

and it starts with M

[trail tendon]

PLN -> goes from P to L, then L to N

[oblique widget]

since its the last letter of the first triangle

[trail tendon]

then N to P but just considering the first two constructions is enough to identify it

[oblique widget]

lets say i do triangle NLP

dont my similarity triangle has to start with P

it would be triangle PNM would it

[trail tendon]

then you go from N to L, and then L to P

[oblique widget]

hypotenuse to the right angle

[trail tendon]

so NLP would be going along the hypotenuse, and then going along the long side

[trail tendon]

first NL goes along the hypotenuse

[trail tendon]

first PN goes along the long side

so those wouldn't be the same

however, if you did like MNP

it goes from M to N , and then N to P
first it goes through the hypotenuse MN, then it goes through the long side NP

[oblique widget]

this is confusing

[trail tendon]

similar to how NLP goes through the hypotenuse NL, and then through the long side LP

[trail tendon]

😔

i dont know the greatest technical terminology for it lol

im saying "goes through" i don't really know how ur supposed to say it

but like thats the order

[oblique widget]

so does NPL similiar to MNP

[trail tendon]

if you walked from N to L to P, you went along NLP
first you went along NL (the hypotenuse route) , then you went along LP (the long side route)

[trail tendon]

no, but NLP is similar to MNP

[oblique widget]

oh shoot yeah

but looking at triangle LPN the long side for it is LN

[trail tendon]

but its not similar to LMN XD

[oblique widget]

because angle LPN is a right angle

[trail tendon]

the long side is LN yeah...

sorry uh

the hypotenuse is LN

the "long side" i was using was the longest side that wasnt the hypotenuse lol

[oblique widget]

oh

[trail tendon]

oh right its called a leg i think

long leg 💀

[oblique widget]

i thought hypotenuse are the longest side

[trail tendon]

they are

my bad

[oblique widget]

😂

that was a little confusing

[trail tendon]

i was just calling it long side because it was longer than the short side and the hypotenuse was diffferent

[trail tendon]

my bad i shoulda clarified 😭

[oblique widget]

so lets say PNL

it would be PMN?

[trail tendon]

yup

[oblique widget]

NO

what

i was taught that the middle letter of a triangle is always a right angle

[trail tendon]

if you start at P, go to N, then go to L, you traverse the short leg and then you traverse the hypotenuse

[trail tendon]

if you start at P, go to M, then go to N, you traverse the short leg and then the hypotenuse

[trail tendon]

you should probably add arrows i dont know which one was first 💀

[trail tendon]

no...

if you're considering an angle, <PLM, then you are talking about <L

but theres no guarentee of a right angle and we're talking about triangle orientation and not the angles

[oblique widget]

uh oh

[trail tendon]

what 💀

[oblique widget]

well this is my problem

[trail tendon]

yea

[oblique widget]

why are they flipped

[trail tendon]

wdym?

[oblique widget]

back to this problem

my teacher took the first letter of the first triangle

and the last letter of the second triangle

then filled in the middle letter

i dont se that pattern for the answert i put

[trail tendon]

what does "the first letter of the first triangle" mean

[oblique widget]

you see triangle KPM

the first letter is K

and the last letter of triangle KML is L

and for the middle letter you just fil in whats missing

[trail tendon]

oh i see

[trail tendon]

im pretty sure that only worked here because you happened to pass through your last destination 💀

[oblique widget]

bruht

she literallyt said that

why is she giving me problem that ion know how to do

[trail tendon]

did she say that for in general, or for that problem

[oblique widget]

in general

[trail tendon]

💀

[trail tendon]

did you go through these and you think the answer is D or you're not sure?

[oblique widget]

i have mp odea

no idea

[trail tendon]

ok

for LMN, first you go along LM, and then you go MN, does that make sense or nah?

like you go from L to M to N

if you start at L, then go to M, then go to N
its the same as starting at L, walking the distance of LM, then walking the distance of MN

[oblique widget]

i mean i get what you['re saying

[trail tendon]

so for LMN, first you walk the hypotenuse, then you walk the shortest leg, right?

it makes sense so far yes?

[oblique widget]

consider

ablly

next qauestion

[trail tendon]

did you get the last one?

[oblique widget]

i dont wanna spend too much time on one, i partially understood it but i need to move on

i still got a lot

[trail tendon]

well the answer you selected was correct so 😂

[oblique widget]

well is it what is it

nextone

[trail tendon]

can you create ratios for this one? (based on similar triangles)

[oblique widget]

no i cant

i dont know ho

ww

[trail tendon]

:<

[oblique widget]

don t blame me my teacher dont teadch

im trying my best

[trail tendon]

do you know that the ratio sides of similar triangles are the same?

[trail tendon]

like this?

[oblique widget]

oh yes

i know that'

[trail tendon]

i just drew each triangle i saw within this triangle, and you can tell they are similar because of the angles

[oblique widget]

so how do i find the variables

[lilac nebula]

Can anyone figure out the longest pole that can fit through this corridor without using a calculator

[hoary totem]

r u asking for help or r u just.. asking

[lilac nebula]

Helpp btw

[hoary totem]

do u know how to find a?

[hoary totem]

good

so now look at this triangle

u know the angle and the opposite side

do u know any relation that can give you the length of the hypotenuse y?

[hoary totem]

oh ur wondering how to find the sine value without a calculator

[lilac nebula]

yeaahh

[hoary totem]

generally this is impossible

unless u have a table of sines

which im guessing u dont

so if u dont have a calculator u can leave it as sin(37.5)

[faint hollow]

well sin(37 degree) is approximately equal to 3/5 or 0.6

[faint hollow]

you got the answer right? i just got here

[graceful talon]

,,\sin^{-1}\theta+\cos^{-1}\theta=\frac{\pi}{2}

[somber coyote]

0_א

[lilac nebula]

I'm not sure, and I think the maximum value I got might be wrong

[faint hollow]

can you resend the question

or tag it maybe

[faint hollow]

all the info shown is given right??

[lilac nebula]

yeahh

[faint hollow]

yeah the answer seems to be correct, but i might have a way to prove it

i'll need some time to write it down

[faint hollow]

is this a software you can type in??

[lilac nebula]

I can prove it with a trig identity, but it's so complex

[lilac nebula]

Goodnotess btw

[lilac nebula]

Thanksssss

[faint hollow]

what was your idea regarding the proof??

[lilac nebula]

I just manually proved sin 37.5 using double-angle and half angle trigon identities

[pulsar prawn]

3:4:5 triangle

[faint hollow]

Yeah exactly

[marsh elk]

Express an angle of 1.26c/rad in degrees & minutes.

Before converting the answer to degrees and minutes, is it 72.18 Degrees or 72.19?

[worldly iron]

hello

[marsh elk]

ridiculous

[obsidian harness]

,w 1.26 gradians / radians

[somber coyote]

Results provided by WolframAlpha

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

Requested by sooouth.

[obsidian harness]

do you have more digits of what you got?

if the digit after the 8 is 5 or higher, round up, so 72.19

if it's 4 or below, round down, so 72.18

[marsh elk]

go and check the info first instead of just rounding it off. Are you so high up that you cannot bend your knees...

[obsidian harness]

I don't even know what c is

why the heck are you so rude

[marsh elk]

rad

[obsidian harness]

so 1.26 radians right

[marsh elk]

do not mistake the truth for rudeness

[obsidian harness]

,w 1.26 * 180/pi

[somber coyote]

Results provided by WolframAlpha

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

Requested by sooouth.

[obsidian harness]

well I should ask you why you are rounding early then

it's not equal to 72.18 degrees or 72.19 degrees exactly

[marsh elk]

ok u r dumb and will cease all communication w/ u)

[marsh elk]

just wanted to confirm, tnkz

[dense wren]

"ok u are dumb, thanks"

[narrow olive]

How are they not banned yet

[broken dock]

<@&268886789983436800>

[broken dock]

gotta ping mods sooner

[light dock]

please don't be rude to others.

[marsh elk]

the discussion here tends to prioritize superficial intelligence focusing on surfacelevel interactions rather than engaging with the profound themes and complexities that this channel aims to explore

[copper comet]

homie if you cant understand the surface, good luck with the profound stuff

[tough jolt]

i got sniped

[marsh elk]

maybe I should come back in 5000yrs when intelligence has increased
ciao ciao ^^

[dense wren]

I can see your point, but the emphasis on surface interactions is merely a byproduct of recursive semiotic structures, where profundity dissolves into performative cognitive oscillations, rendering thematic depth an illusory construct within post-structural discourse.

[mighty dawn]

?

[obsidian harness]

let the height of both triangles be h

then you can set up simultaneous equations for both the big and the small triangles, in terms of both x and h

using tangent

[mighty dawn]

how do i set up the simultaneous equations with tan?

[obsidian harness]

well tan(angle) = opp/adj right

so can you set up an equation like this for the triangle with 43 deg?

[mighty dawn]

tan(43) = 2?

[flint marlin]

SOHCAHTOA

Dont forget this

[obsidian harness]

no

what's the side opposite the 43 degree angle?

[mighty dawn]

h

[obsidian harness]

yep!

okay what's the side adjacent to 43 degrees?

so the adjacent side is not the opposite side, and it's also not the hypotenuse

[mighty dawn]

2

[obsidian harness]

cool

putting it all together

tan(43) = h/2

[mighty dawn]

for the other triangle would it be tan(31) = h/x

[obsidian harness]

nearly......

draw the big triangle out

the adjacent side is not x

[mighty dawn]

is it not? it can't be O, would it be H?

[obsidian harness]

no, the adjacent side is x + something

[mighty dawn]

x + 2?

[obsidian harness]

yep

[mighty dawn]

so tan(31) = h/x+2

[obsidian harness]

yep!

now from tan(43) = h/2

you get 2 tan(43) = h

can you do something similar to tan(31) = h/(x + 2)? so that you get rid of the fraction?

[mighty dawn]

2x tan(31) = h ?

[obsidian harness]

no

you should get (x + 2) tan(31) = h

[mighty dawn]

oh ok

[obsidian harness]

and then you can combine 2 tan(43) = h and (x + 2) tan(31) = h !

||2 tan(43) = (x + 2) tan(31)
2 tan(43) = x tan(31) + 2 tan(31)
2 tan(43) - 2 tan(31) = x tan(31)
(2 tan(43) - 2 tan(31))/(tan(31)) = x||

[jade sierra]

quick

need help

i got 239.1 cuz the teacher taught us to subtract the quadrant degree with the one you get from doing tan^-1 (y/x)

[cunning lion]

if the x component is negative then you should add 180° to the angle

[jade sierra]

oh

💀

i thought you always subtracted

god this delta math took me a hour 30 to do prob

and i got a test on this in 2 days

[gilded bronze]

Finally finished this code, yall might find this very useful. Or not lmfao

https://bit.ly/HolyTrinityTriangleTool

[flint marlin]

Cool

[high solar]

can anyone help me?

i sent question in help

[flint marlin]

Send here

[flint marlin]

.

[zealous pike]

It is an exercise that got in my mind. Given the right triangle ABC, two triangles can be formed by drawing a line that bisects the right angle and intersects the point N, a line perpendicular to AB that intersects point N and a line perpendicular to AC that intersects point N. Show that the figure formed by the points AMNO is either a square or a rectangle.

Given that AMN is a triangle and that line AN bisects the angle BAC, the angles of the triangle AMN are 90, 45 and 45, because 180 = 90 + 45 + x.

That being the case, side AM is equal to side MN.

Angle NAO is equal to the angle MAN, AN is equal to AN, angle ANO is equal to angle ANM, therefore the triangle AMN is congruent to the trianglr ANO by ASA, hence, figure AMNO is a square

Is this right?

[winter dagger]

Hii I have a question abt trig:

if sin^-1 x = arcsin x rather than csc x, is sin^-2 = arcsin^2 or csc^2 x?

I know that trig function to the -1 makes it an inverse, but I was just wondering abt all other negative numbers

[obsidian harness]

$\sin^{-2} (x)$ is bad notation and should be avoided

[somber coyote]

south

[obsidian harness]

but it would be csc^2 x

[unkempt gulch]

I need help finding h.

[obsidian harness]

let the distance of B from the base of the height be x

so you have h/x = tan 60

can you set up a similar equation for the 45 degree angle?

(in fact the 45 makes x really easy to figure out, since you have a right triangle haha)

[kind rapids]

$\sin^{-1}(x)$ is for inverse function only. You can write $(\sin(x))^{-2}$ to mean $\csc^2(x)$ or write $(\sin^{-1}(x))^2$ to mean $\arcsin^2(x)$

[somber coyote]

҉C ҉l ҉ø ҉s ҉e ҉r

[unkempt gulch]

I already did that but I don’t know what you mean by setting up a similar equation for the other angle

[obsidian harness]

h/(x + 10) = tan(45)

[unkempt gulch]

Ok

Thanks

[obsidian harness]

np

[unkempt gulch]

I got h = √3 * (x + 10) but I forgot to mention that the result needs to be a rounded off number in meters. How do I get that from this result?

[obsidian harness]

tan(45) = 1

[obsidian harness]

that's weird cause h = x + 10 so you went wrong somewhere

so (x + 10)/x = sqrt(3)
x + 10 = sqrt(3) x
x - sqrt(3) x = -10
x = -10/(1 - sqrt(3))

and h = -10/(1 - sqrt(3)) + 10 so round that to the nearest m

[unkempt gulch]

You’re right. I messed up the value of tan(45)

[winter dagger]

Okay, thank you! So should I only use the notation sin^n (x) when n>0 and use (sin x)^n instead when n<0?

[obsidian harness]

yep

I mean $\sin^1 (x)$ looks weird

[somber coyote]

south

[obsidian harness]

but yeah apart from that, yes

[old elbow]

Does anyone know how to solve this problem?

[formal geyser]

The first one:
Minute hand rotates with 6° per minute (360°/60 minutes), hour hand rotates with 360/ degrees per second (24 hour is 1440 minutes). So we need to determine the time when two hands meet each other, so the angle between 12:00 and the minute hand equals the angle between 12:00 and the hour hand. X is this time.
x * 6 = 150 + x*360/1440. The answer is x equals 26 minutes and around 5 seconds

05:26:05

[hard smelt]

I was following along until it says divide the interval between the two vertical asymptotes. Where on earth are they getting those numbers from?

[cunning lion]

the length of the interval is (pi/2 - 0) = pi/2

each subinterval should have 1/4 of that length

therefore each subinterval should have length (1/4)*(pi/2)

[hard smelt]

I'm not sure I understand can you show me how you would setup an equation for a couple of them and mark where you're getting the numbers from?

[cunning lion]

if you want to divide something into 4 equal parts, mathematically that corresponds to dividing by 4

[hard smelt]

I understand now and I feel a bit stupid but why on earth would they draw the number line so inaccurately?

[deft sail]

Do you have to know Half/Double angle formulas

Or do you not have to and do something else

They do feel stupid to me

[cunning lion]

the double angle identities are special case of the angle addition identities

and the half angle identities are a rearrangement of the double angle identity for cosine

[trail tendon]

you don't have to but they're good to know, makes sum stuff easier/accessible later. Depends what u mean by have to.

[limber phoenix]

Can omeoe help me

HOW DO I FIND THE LINE OF REFLECTION

[wheat granite]

I dont know exactly what youre given, but I would guess that you just find the midpoint between two points that are opposite of the line

[limber phoenix]

@wheat granite

Its this

I understand that the line of symmetry has y

But Im not sure if its y=0 or y=-r

[wheat granite]

its y=0

the midpoint between n and -n is 0

[limber phoenix]

Okay thank you so much

[limber phoenix]

Where would y=-r be placed

?

It was y=-r

I got it wrong

THanks anyway

[wheat granite]

oh wait sorry I missed your message and didnt realize you were looking for a 2nd line of symmetry, I thought you mean the x one

[wheat granite]

If I somehow end up with a situation like this in a two-collumn proof, what reason can I give for angles a and 1 being congruent? (the white lines are the same)

[deft sail]

Do you have an example

[cunning lion]

sin(2a) = sin(a+a)

[uncut portal]

Is this where you can get help?

Wait this isn’t a geometry nvm

[old elbow]

Thank you

[runic scarab]

idk if this is the right place to ask about 3d transformations, but heres my question:
assume you have a point P at (1 ,3, 1)
you want to scale it by 2, rotate about the y axis by 45 deg and the x axis by 30 deg, then you to want to transalte it by (2, 0, -1)

[warm shuttle]

that's not a question

[runic scarab]

forget to mention "where will that point be?"

[formal geyser]

By the way, which grade are you in?

[warm shuttle]

You say rotate by 45 degrees, but you don't say in which direction.
(x, y, z) --scale--> (2x, 2y, 2z) --rotate y--> (2x cos(p45°) - 2z sin(p45°), 2y, 2x sin(p45°) + 2z cos(p45°)) := (x', 2y, z') --rotate x--> (x' 2y, z')
(x', 2y cos(q30°) - z' sin(q30°), 2y sin(q30°) + z' cos(q30°)) --translate--> (x' + 2, 2y cos(q30°) - z' sin(q30°), 2y sin(q30°) + z' cos(q30°) - 1)

p and q are +1 or -1 depending on the direction of each rotation

[runic scarab]

oh mb, i apologise 😅 but how have u done this? i was looking at one of the given slides and they have done this

[warm shuttle]

oh with homogeneous coords you just google what the correct matrix for each transformation is and then let computer do its thing

[runic scarab]

whhatt, how? i have a online (open book) test on this soon

[warm shuttle]

which part do you not understand?

you form your vector with your x, y, z and w = 1 (pretty sure, double check with the book)

then you figure out what the matrix for each transformation is

again double check with book or online resources

then you apply them in order

reading from right to left

S - scaling
R_2 - rotate
R_1 - rotate
T - translate

[runic scarab]

um so i understand that the order of caluations go this way but what i dont understand is why the second rotation is calcauted before the first and both the rotaton stuff is inverted

[warm shuttle]

i think that's just a naming thing

[runic scarab]

hm?

[warm shuttle]

R_2 applies before R_1

it seems to say something about inverses

which you left out

but if you wanna do like f(g(x))

and then you want to get back to x you do g^-1(f^-1(f(g(x))))

so that you cancel them out

so if you are undoing transformations then the naming makes sense

otherwise it's weird

[dense smelt]

How is this a parallelogram in the first place?

If I got the props of parallelogram right, then it doesn’t make sense bc equilateral triangle but 8 , 6?

[hard smelt]

how am I supposed to figure this out? I don't have a cot function on my ti84 and even if I did it would give it to me as a decimal

[mild smelt]

Remember the definition of cot

[ivory cloak]

Aider moi s'il vous plaît

[deft sail]

Je vais essayer

[ivory cloak]

Merci infiniment

[deft sail]

Je suis vraiment confus sur cette question ? Je ne l'ai tout simplement pas fait depuis un moment. Je ne sais pas ce que je suis censé faire

[mild smelt]

Reflect one of the points over the line. Shortest path is a straight line.

So angles should be equal

[upper karma]

.

Consider an n-sided polygon.
What is to be found is the number of in it.

Let the first point be A.
We move Anti-clockwise.
From point A, the number of diagonals (on observation) = n-3 (in which we excluded from A to U - the last point and B - the adjacent point).

For this (from point B) there are n-3 diagonals again (without any re-ocurrence of any diagonal).

Things become a little sus for point C.

[deft sail]

I never learned this yet 😭

[upper karma]

This is not a subject to be learnt; it's logic.

[deft sail]

I don’t use logic that much

[upper karma]

No problem.

[deft sail]

Oh it actually makes sense

I never thought of reflecting the point

I thought it was already equal by four parts

[upper karma]

We can see that the diagonal AC has been considered again (it was previously considered for the first case).
So we subtract 1 from (n-3). We get (n-3) - 1 diagonals for this.

Now for point D there are two re-occurrences.

AD and AB.

So we subtract two.

We get (n-3) - 2 diagonals.

[deft sail]

😼👍

[upper karma]

Let the last two points be P_(n-1) and P_(n-2).
We can find the number of diagonals (without coincidence) for S (because for T and U all diagonals will be repeated).
For C which was the P_3 point, there were (n-3) - (3-2) diagonals.
On observing we can generalize this for a give point P_k (considering first point as A).
Number of non-repeated diagonals from P_k in this polygon of n sides = (n-3) - (k-2) = n - 3 - k + 2 = n-k-1
Tada!
S is the P_(n-2) point.
Observing the pattern we get it has (n-3) - (n-2-2) diagonals which equals n-3 - n + 2 + 2 = 1 diagonal.
The number of points (including S and C) between S and C = n-4.

In conclusion the total number of diagonals can be given as follows:
(n-3)(n-2) - (n-4)(n-3)/2 = n(n-3)/2

Here's the breakdown of each part:

For the 'without subtraction' part for the diagonals for a given point we have n-3.
Now it continues till S [which is (n-2)].
So we multiply (n-3) and n-2 to get (n-3)(n-2).

The subtraction part continues till S (from C).
So we subtract the summation of the natural numbers from k = 1 to k = n-4 (since the number of diagonals to be subtracted for the n-2 point = n-(n-2) - 1 = n + 2 - 1 = 1.
And the difference between the location, aka the number (for example, 1, 2, 3...., etc.) and the second part of the term (n-k) where n-k = the subtraction to be made for the point n-k+2.
_So that thing for S is (n-4).
_So finally the thing that we have to subtract = the summation of natural numbers from 1 to n-4 = (n-4)(n-3)/2.

The final formula for this calculation becomes:
(n-3)(n-2) - (n-4)(n-3)/2
= (n-3)[(n-2) - (n-4)/2]
= (n-3)[{2(n-2) - (n-4)}/2]
= (n-3)(n)/2
Which is the required formula.

So the question I am asking is:
Are all the steps of my monumental proof correct?
I know that all of it is correct but I would like to hear praise from others.

[zealous pike]

Can someone help me please

Well

It is a question that came from my mind

Given a conic section of height h and radios r, what function describes the change of r as h grows or decreases?

[flint marlin]

Damn good

But I would prefer combinatorics here

[mild smelt]

Hmm.. there are n points and each point can be connected to n-3 other points by a diagonal.

n(n-3) will be each diagonal counted twice, so to get the answer just divide by 2

n(n-3)/2

[upper karma]

how

[upper karma]

yes that's better

nC2 - n

[split charm]

Yeah i learned it like this too

[upper karma]

https://www.geogebra.org/geometry/xxhhey7d

in this figure where all the diagonals are joined of a 21-sided regular polygon, how many triangles have been formed in total (every triangle)

inside the polygon, as small as a bacteria and also including triangle formed by the vertices of the polygon

[mild smelt]

Found a pdf about this https://cs.uwaterloo.ca/journals/JIS/sommars/triangle.pdf

[upper karma]

i am just one ahead of them

can anyone calculate this for me

for number of sides = 21

[zealous pike]

i am invisible

[wise zodiac]

does anyone know how to prove the ares of XYW = ZYW USING HERONS

[obsidian harness]

why would you specifically want to do it using Heron's

just note that the perpendicular height of W above XY and YZ is the same

[wise zodiac]

yep i have 3 solutions without heron's

[obsidian harness]

if you insist

[wise zodiac]

but i want to do it with heron's because i'm learning what's going wrong

😭

can u help me

[wise zodiac]

is this where heron's fails or something because not sufficient information?

[obsidian harness]

use the fact that cos(x) = -cos(180 deg - x) and then cosine rule, that might work

[wise zodiac]

i thought of it but like

wouldn't that give negative measure lol

[obsidian harness]

$\frac{XY^2 + YW^2 - XW^2}{2 \cdot XY \cdot YW} = -\frac{YZ^2 + YW^2 - WZ^2}{2 \cdot YZ \cdot YW}$

[somber coyote]

south

[wise zodiac]

nice thing is using that argument you can have sin(x) = sin(180 - x) and yet another proof to show that the area is equal

[wise zodiac]

oh wait

[obsidian harness]

yeah I was thinking along similar lines

[wise zodiac]

i think u got it 😭

[obsidian harness]

basically if we have a way to relate XW and WZ somehow

$XY^2 + YW^2 - XW^2 + YZ^2 + YW^2 - WZ^2 = 0$

$2 XY^2 + 2 YW^2 - (XW^2 + WZ^2) = 0$

[somber coyote]

south

[wise zodiac]

so it's just not sufficient right?

sufficient information*

[obsidian harness]

it's going to be hell algebraically

I think it will work somehow but don't ask me

[wise zodiac]

u mean it's possible?

[obsidian harness]

ye

[wise zodiac]

wow

[obsidian harness]

then I think this form would be the most useful

so if we let a = XY and b = YW
and a' = YZ and b' = YW for the other triangle

we just need to consider a^2 + b^2 - c^2

[obsidian harness]

ah so $XY^2 + YW^2 - XW^2 = WZ^2 - XY^2 - YW^2$

[somber coyote]

south

[obsidian harness]

so $(XY^2 + YW^2 - XW)^2 = (XY^2 + YW^2 - ZW^2)^2 = (YZ^2 + YW^2 - ZW)^2$ hence proven

[somber coyote]

south

[wise zodiac]

thankss

@nocturne remnant (sorry for ping) helped me solve it

😭

[obsidian harness]

nw

[zealous pike]

Is it possible to use the similarity of triangles to find the height of the smaller conic section?

I thought about something like this

The height of the small conic section is 12

[tame oriole]

guys what do i need to know to learn trigonometry

[mild smelt]

Everything that is in your geometry textbook before trigonometry 🤔

[zealous pike]

Isn't just angles and triangles and circles enough

Anyway

[zealous pike]

If I multiply the area of the triangle by the conic base's circumference, what figure will I have

[maiden brook]

geometric trig is 🤮

[kind quail]

Hello I have this in my home work which is just really practice questions. And I can not seem to figure out exactly how to solve this using sine

I've been trying but nothing seems to add up

[acoustic pollen]

i cant try it rn but maybe law of sine and law of cosine

[spark vale]

How can I find the answer of tan(135) ?

let's say I don't know how to find the answer of trigonometry functions when the theta value is over 90 degrees.

[kind quail]

@maiden brook yes I believe it is law of sines

[spark vale]

i know how to solve it.
tan(135) = -tan(45)

but this does not make sense

[acoustic pollen]

why?

[spark vale]

i mean it's 135 degrees

but how can we find the answer with 45 degrees for that?

nonsense to me

[acoustic pollen]

do you know about the unit circle?

[spark vale]

yes

that's what i am using

the radius is 1. that's unit circle

[maiden brook]

why are u pinging me

[acoustic pollen]

because theyre equal

i mean not fully equal since one of them is negative

[spark vale]

thanks for your image

but what's that tangent line

[acoustic pollen]

tan axis

[spark vale]

i am not finding any relationship between Tan func and that line

[acoustic pollen]

ok so in the unit circle, the y axis represents the sine and the x axis the cossine, right?

[spark vale]

yes

because the radius is 1

[acoustic pollen]

all the other trigonometric functions like tan, csc and cot also do have lines that represent them

that one is for tan

[spark vale]

really?

this is the first time i hear

now i think i need to understand why they have those lines.

I will show u what i am thinking of trigonometry funcs

[acoustic pollen]

ok

[acoustic pollen]

yes

but the one i sent you is a way to visualize it

[spark vale]

so u see there's no extra line on the image

only triangle inside the unit circle.

[acoustic pollen]

yeah because usually you don't need it, but that doesn't mean it's not there

wait let me try to find an image of the full unit circle

[spark vale]

thanks