geometry-and-trigonometry

[upper karma]

btw where is D?

hmm ill get my pen one sec

[graceful trail]

My progress so far

D is missing. Probably a mistake or something

[eager pendant]

these type of problems are well-known

especially this particular case, which was known as Langley’s Adventitious Angles

@graceful trail

[graceful trail]

My teacher gave it to us

He told us to solve it

[eager pendant]

ok?

[graceful trail]

I was wondering how to solve it

And I am not sure what the answer is.

[eager pendant]

google what i told you above

if you want to find a solution online

[graceful trail]

Alright

[tawdry pivot]

I saw a method which relied upon BCF being iscoceles, I imagine this problem would be even harder if it wasn't.

[upper karma]

Finally solved it

jesus

I forgot what pairs of angles are smh

@graceful trail got it?

[graceful trail]

Lol nope

I am having a hard time solving the other angles

[upper karma]

Okay my solution was simple as hell

No need really

You just need to know about pairs of angle rule

[graceful trail]

Wdym?

[tawdry pivot]

The solutions I found online were hard 😦

[upper karma]

corresponding angles

Mine is very simple

[graceful trail]

Ah, I know corresponding angles

But there are no parallel lines..

How did you do it?

[upper karma]

there are parallel, it's just hard to see

That's how i solved it

[graceful trail]

Oo

[upper karma]

BC and FE are two parallell lines

[graceful trail]

I didn’t notice that..

[upper karma]

BE is the transversal

yeah

[graceful trail]

Is FC a transversal too?

[upper karma]

yes, but you only need BE

Then it's easy to solve

[graceful trail]

Are you saying x is 60?

[tawdry pivot]

BC is not parallel to FE

[graceful trail]

That’s what I’m thinking

[upper karma]

???

[graceful trail]

Oh wait

[tawdry pivot]

btw the answer is 30

[upper karma]

It is unless this guy's drawing is deadly

[graceful trail]

You got the angel near F wrong. The one that says 60

[tawdry pivot]

hodgy, ABC is an iscoceles triangle, and you have lines rising from B and C with different altitudes, there is no way BC || FE

[upper karma]

60?

oh i thought it was a right triangle

[graceful trail]

lol

[tawdry pivot]

cifer, your teacher gave you hard problem for your level

[graceful trail]

I need to figure out y

He said it’s not necessarily have to be solved because it’s hard

[upper karma]

how did you get A = 20?

[graceful trail]

The whole triangle

[upper karma]

smart!

[tawdry pivot]

ABC

[graceful trail]

But the 20 seem useless

[tawdry pivot]

it is

[graceful trail]

lol

[upper karma]

just draw as many angles as possible and pray to god xD

or learn trigonometry

[graceful trail]

I think for this type of problem, you need a protractor

[tawdry pivot]

This problem seemed too hard, requiring some tricks, so I just looked at the solution straight away. The general case would be even harder (BCF not iscoceles).

[upper karma]

?? how

then it would have parallels

[graceful trail]

No, you need protractor to draw some lines

To actually solve it

[tawdry pivot]

No, the method I followed (Wikipedia) required an additional 2 lines drawn

[upper karma]

dm me answer epic

yeah

i was thinking that

[tawdry pivot]

I already stated x=30. You can look on wikipedia for a derivation.

[graceful trail]

so, y is 80

[tawdry pivot]

Yes, it would seem that it is.

[graceful trail]

I am wondering how to solve y without drawing extra lines

[tawdry pivot]

You would have to find a different method to the standard, doing it without extra lines may help solve the general case.

Well, the adventitious quadrangles problem seems to be research level rip

[graceful trail]

I found a solution where you have to guess

[upper karma]

guessing is not valid

[graceful trail]

It is tideous, but it will give you a correct one

[upper karma]

doesn't matter how tedious it is if you guess lol

[graceful trail]

Lol

[tawdry pivot]

I don't see how your picture explains any guessing

[graceful trail]

Actually nvm, there could be multiple answer to that guessing

I actually give up lol

It’s 11 PM

[tawdry pivot]

I gave up as soon as I found all the angles in terms of x.

[upper karma]

same

this isn't worth it

[graceful trail]

Yep

[upper karma]

Langlands program

I will study geometry for my PhD then solve this problem

Lmao

@upper karma but why

why would anyone spend their life doing such stuff

haha

[tawdry pivot]

from https://www.mathpages.com/home/kmath734/kmath734.htm

$f$ is the solution to the equation $\sin(e+f)=2\cos\left(f+\frac{a-e}2\right)\sin\left(\frac{a+e}2\right)$

so there's the solution for the general case xd

[upper karma]

yeah but that's cheating using trigonometry

we want a general strictly geometric/algebraic method

@upper karma you should ask every researcher in math :D

[somber coyote]

EpicGuy4227:

[tawdry pivot]

it doesn't introduce any extra lines 😃

[sharp cape]

hey can i get some help

i cant visualize this can someone explain or draw it out

https://gyazo.com/1171e5311d156e8011e647adbf3b7cc4

[upper karma]

i should've used more colors

sry

[tawdry pivot]

my eyes

[sharp cape]

alright ty

[upper karma]

yeah

https://www.mathsteacher.com.au/year7/ch08_angles/07_bear/bearing.htm

[sharp cape]

wait and how do i find one of the interior angles to solve it

[upper karma]

u got urself a triangle

[sharp cape]

wait im stupid i got it

nvm

[upper karma]

yeah

u just apply cosine theorem

cuz u have 2 sides and an angle between them

[fiery badge]

that diagram is atrocious lol

[upper karma]

it is

it just works xD

[glad ocean]

does it?!

[upper karma]

[gloomy slate]

Hi, I had this problem on my test review and Im not even sure if i wrote it down correctly because I couldnt find how to solve it anywhere, if anyone could help I'd appreciate it

[dire rampart]

that bracket is just 1

and you can write 2sincos as sin(2theta)

and then solve

@gloomy slate

[steady sleet]

That's not even necessary

[gloomy slate]

You're saying the part after the 1 isnt part of the problem?

[steady sleet]

Leaving it factored here is fine actually

It is part of the problem :l

What does the left side equal?

[gloomy slate]

I am not sure

Im supposed to prove this problem but Im not really sure how

[steady sleet]

What does $\sin^2(\theta)+\cos^2(\theta)$ equal?

[somber coyote]

Simple_Art:

[gloomy slate]

1

[steady sleet]

Okay

So what does the left side equal?

[gloomy slate]

1

[steady sleet]

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

[somber coyote]

Simple_Art:

[steady sleet]

Can you solve this then?

[gloomy slate]

Im not sure what Im supposed to solve, I wasnt there during this lesson :/

[steady sleet]

???

[gloomy slate]

I dont even know man sorry

[steady sleet]

There's only one variable to solve for wtf

[gloomy slate]

The problem was to prove that thing that I drew, is that the same meaning as solve? lmao

If so then oops

[steady sleet]

what

The left and right sides are definitely not equal for all theta

Take theta = 45° for example

You get 1 on the left side

And...

== 1+2 * sind(45) * cosd(45)

[charred spear]

2

[steady sleet]

you can't prove something that's false

[gloomy slate]

So I mustve wrote the problem incorrectly then

I think the answer for it was 1 + 2sin(theta)cos(theta)

[dawn stone]

you know, your problem was probably prove [sin(theta)+cos(theta)]^2 = 1 + 2sin(theta)cos(theta)

[sick veldt]

https://cdn.discordapp.com/attachments/377497721583697930/534433681553620992/Screen_Shot_2019-01-14_at_1.01.56_PM.png

[supple grove]

Would anybody like to help?

[earnest plover]

so im working on geometry right now and some guy is representing length in a fraction

and it doesn't really make sense

there is a side called x

and the fraction is x/2 but I don't know what the fraction represents or how he got the 2

does it represent length or something?

[thorn talon]

what's the full context?

but yes, usually it would still represent a length

half of side x

[thorn talon]

which two?

well

well they both have a right angle right?

and the vertical side is equal right?

goes up and down i guess

now

what about the bottom side

both of the triangles share it right?

?

ah yeah

i'm dumb

do the angle instead

you see the isosceles triangle formed?

you know the thing where

equal angles opposite equal sides

?

so then you have two angles and a side

yep

exactly

AAS

i honestly consider it the same thing

but ehh

if i had to

hmm

well

figure out the two angles

angle sum the third angle

the one at the top left and top right

ASA

but that's memes

[upper karma]

is there a formula or method to find the perimeter of any polygon (simple and non simple) given vertices?

[astral hornet]

if you can calculate perimeter and apothem using coords of vertices, you can use (ap)/2

[vernal tusk]

@cobalt sun mention me when you solve it? curioud

curious*

[umbral snow]

The angle is better stated as 18.33 because nobody uses minutes

[clear haven]

no such thing as a noobish question here unless u're asking us if the natural numbres are finite

also yes make sure u convert dms to degrees or radians if u want to use stuff like calculators

[umbral snow]

You have a right triangle where the opposite is 70m. You want the adjacent

[clear haven]

so which of the trig operation thingies do u want to use

[upper karma]

helppp

how do u solve

3tan^2 x - 1 = 0

[woven wigeon]

add 1 to both sides

then divide by 3 on both sides

then take the sq root of both sides

answer should be -rad3/3 and rad3/3

@upper karma im guessing this is correct but im still rarted

lo late reply f

[upper karma]

@woven wigeon thank you so much and early happy birthday!!

[glad ocean]

@upper karma see 18°20? It is part of a right or 90 degree angle

so in the bottom small triangle the angle is 71°40" or 71.6666°

That is adjacent to a side, which is 70

[upper karma]

@glad ocean seems like you pinged a wrong person

[glad ocean]

so tangent of x =

Oh heck srry

tan (71.667 deg) = x/70

x = 70 tan(71.6667)

x = approximately 211.25

[woven wigeon]

@upper karma thanks, rly appreciated

[cobalt sun]

@vernal tusk sure I'll let you know when I solve it

[sick veldt]

I have the math all on paper done

but I need confirmation

I started off with 2x + 6/x + 6 = 10/8

and cross multiplied

did basic algebra and got x = 2

<@&286206848099549185>

[steady sleet]

#❓how-to-get-help, rule #4: If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once.
Please read the rules: #❓how-to-get-help!

[sick veldt]

sorry

[unreal flame]

is that a mistake?

cause when i check

(cos(x))'

im getting sinx

and not -sinx

cause im doing

cos(x)' = sin(pi/2-x)' = cos(pi/2-x) = 0cosx + 1sinx = sinx

[hard gale]

ach the chain rule

what's the derivative of pi/2 - x with respect to x ?

[gritty siren]

also, bad notation

[hard gale]

(ofc)

[unreal flame]

i dont understand

I made cosx'

by all the rules of derivative

i made everything exactlly how it should be

but i got that cos'x = sinx

and not -sinx

cos(x)' = sin(pi/2-x)' = cos(pi/2-x) = 0cosx + 1sinx = sinx

[hard gale]

"what's the derivative of x -> pi/2 - x with respect to x ?" there's a reason i'm asking you that q

[unreal flame]

but i dont understand the question sir

my english is not very well

it should be -1 right?

[hard gale]

ye

[unreal flame]

pi/2 gone and -x becoming -1

there is no step that including (pi/2-x )'

its sin(pi/2-x)'

[hard gale]

cos(x)' = sin(pi/2-x)' = cos(pi/2-x) this step clearly does

[unreal flame]

i dont understand

that is the step i made

I said that

that what i did

here you are not doing derivative of pi/2 - x

"derivative of pi/2 - x" is not included

cos(x)' = sin(pi/2-x)' is just an identity

and then you do derivative on sin(pi/2-x) and not on pi/2-x

I dont understand how your step is included in the proccess

like, its not

Onion_ArtToday at 10:35 PM
cos(x)' = sin(pi/2-x)' = cos(pi/2-x) this step clearly does

this step clearly doesnt.
the first is the identity of cos(x) = sin(90-x) and the second is derivative on sin(pi/2-x)' that is giving cos(pi/2-x) and NOT -1 .

@hard gale

im sorry but i think you are wrong dude

@gritty siren hi man, you said he is right but he is wrong if you want i can explain you why you both were wrong

[gritty siren]

wut I only pointed out the bad notations

[unreal flame]

where are the bad notations here?

[gritty siren]

one shouldn't write (cos x)'

[unreal flame]

why

?

[gritty siren]

you put the ' on a function, and cos x is no function

cos is a function, cos x is a number

[unreal flame]

o_O

sir it can be cos'(x) and also cos(x')

these are different things

[gritty siren]

no, nooo

[unreal flame]

😐

cos(x') = cos(1)

[gritty siren]

x' isn't something to be written

[unreal flame]

😐

[gritty siren]

x isn't a function

[unreal flame]

dude how old are you?

[gritty siren]

why does it matter?

[unreal flame]

cause maybe you have not learned that yet

[gritty siren]

x isn't a function

$x\longmapsto x$ is a function

[somber coyote]

Tuong:

[unreal flame]

like you say that cos'(2x+1) and (cos(2x+1)') are same?

[gritty siren]

$\text{id}_\bbR$ is a function

[somber coyote]

Tuong:

[gritty siren]

they are not the same, and the 2nd one doesn't make any sense

or at least, it doesn't have the meaning you want it to have

[unreal flame]

i meant to write cos'(2x+1)

[gritty siren]

this is correct

[unreal flame]

you dont know that
sin'(2x+1) is cos(2x+1)
and
sin((2x+1)') is sin(2)
?

these are different things

[upper karma]

How do I simplify a ratio ._.

[gritty siren]

sin'(aaaa) = cos(aaaa) is alright
but sin((2x+1)') doesn't make any sense

[upper karma]

How do I simplify a ratio?

[gritty siren]

maybe try one of the questions channels, Roide

[unreal flame]

but I didnt had it in the question

[upper karma]

I was hoping you could give me the answer since you're typing all these chinese equations

[unreal flame]

i just asked what is cos'(x)

sinx or - sinx

[upper karma]

You guys arguing over maths

IQ: Over 9,000

One of you tell me how to simplify ratios

I need help with my work

[unreal flame]

actually im pretty stupid

i dont know even what is ratio

radious?

[upper karma]

Nice sarcasm

I'll send a screenshot of the question

[unreal flame]

no really

my english is bad

[upper karma]

It a simple question but my dumbass doesn't know how to simplify ratios

I don't evne have a formula to begin with

https://gyazo.com/886fce61c60d24c1f4801fb8b2da6ace

[astral hornet]

To get simplest ratio, divide everything by lowest number

[elder oak]

can anyone explain to me why geometric series are named as they are

like I have a few ideas of like nested infinite squares inscribed inside of each other but idk

I’m talking bout like 1.2.4.8.16.32

[vital frost]

help

[astral hornet]

if angles 3 and 4 sum to 90, then its linear pair, composed of angles 5 and 6, sum to 90 too

use angle 5 to find angle 6

since angles 2 and 6 are formed by the same lines and share vertices, they're congruent

use angle 2 to find angle 1

[upper karma]

Is trigonometry possible to master by self teaching

[desert ruin]

good evening could yall help with dis one

[winged matrix]

got one question

if the adjacent is x and the hypotnuese is 7 would it still be adj/hypo to use tan?

[desert ruin]

could someone help me

[desert ruin]

could i have some help with this to

[random sable]

Have you tried using congruent triangles?

[glad ocean]

45°

[heady juniper]

@desert ruin

opp angles are congt in paralleogram

at least thats what im assuming it is

wait

thats not it

wait

its just corresponding angles in congruent traingles

thats even easier

angle sum of bottom triangle for 3rd angle

then just say it equals X

[hexed island]

ya i did the problem aswell, it equals 35

since all angles of a triangle equal 180, just say that 16+129+x=180

x being the undefined angle in triangle BAD

and if you say that the x you are supposed to find is equal to the angle we just solved since both of them are alternate interior angles, then x = 35

[elfin moon]

Can I use the determinant to determine which side of the triangle a given point is in 3D?

[elfin moon]

I guess another way to see this is to use the triangle to get the halfplane that splits the space into two parts. I then want to check which space it is in (possibly on the plane too)

Oh, just dot product, perhaps

[upper karma]

When I type in 5 x tan(25), exactly like shown in the problem, I get a different answer

From the shown answer, on the computer

[cobalt sun]

@vernal tusk I solved that 😈 question

[upper karma]

Nevermind, my calculator was set to radiant and not degrees

[sick veldt]

are my box answers correct?

I'm not sure

oops 18 supposed to be 8

Is this one also correct?

[lusty maple]

They look right

but not sure if first one is 16/8 = 5/2.5 or is it 16/5 = 8/2.5

former is correct I believe

[thorn lichen]

it's the same thing

$$\frac{16}{8} = \frac{5}{2.5} \Leftrightarrow \frac{16}{8} \cdot \frac{8}{5} = \frac{5}{2.5} \cdot \frac{8}{5} \Leftrightarrow \frac{16}{5} = \frac{8}{2.5}$$

[somber coyote]

Metal Santa:

[lusty maple]

good to know thanks

[thorn lichen]

🍻

[lusty maple]

https://math.stackexchange.com/questions/1925552/solving-for-the-unknown-magnitudes-of-two-vectors-when-given-the-equilibrant-vec how do I go about applying this to n vectors?

There would be multiple solutions and I'm stuck on multiple parts lol. For example, my vectors have a maximum amount (they are the torques of forces from rocket engines acting on a mass, which I am trying to balance) and they are scaled by a factor f (between 0 and 1, this is the throttle for that specific engine). I would need to find the solution where the sum of all factors is greatest. Any idea where to start with this?

[thorn lichen]

not really😂

sorry I don't know much on linear algebra

[lusty maple]

no worries somebody else may be able to point me in the right direction

[keen tangle]

@tawny river you can repost here

[upper karma]

can someone explain taylor series

like i mean i get that their approximates of trig waves

but like how do u find the taylor series for a wave

[umbral snow]

@upper karma
Still looking for it?

[steady sleet]

Requires calculus

[glad ocean]

yeet

[lusty bronze]

A central angle θ in a circle of radius 7 m is subtended by an arc of length 8 m. Find the measure of θ in degrees.

idk how to do this if anyone can help please let me know

[clear haven]

s=rtheta

[upper karma]

@umbral snow yea

[proper citrus]

I need help nerds

Say we have a cone

Base diameter is 3.5m

And the slant height is 3m

The cone is open at the base

The cone was made using a circle with a sector cut out

What is the angle of the major sector?

That's what I tried to find out

The curved surface area is given by πrl

Where l is the slant height

So

πrl = (x/360) * (πr^2)

Sound about right?

Solving this doesn't give the correct answer tho

Pls fix

I get 204 degrees

But the answer's supposed to be 210°

Uhh

Anyone willing to take a gander or sth?

Sigh

Just ping me

[umbral snow]

@upper karma
Taylor series are an infinite series method to calculate any analytic function, not just trig functions. You calculate them with their definition

[proper citrus]

And here I thought you were typing to help me XD

[upper karma]

uhuh

[sly charm]

i have an exam tomorrow: anyone interesting in helping me review some trig stuff? teacher doesn't let anyone us calculators 🤐

[graceful trail]

What Grade are you? @sly charm

[leaden moon]

What do colinear vectors mean if we are dealing with complex numbers

If I have colinear vectors x and lamda*x

Is lambda now complex or still real?

[upper karma]

Are distances between figure/objects from their centers or their edges

[glad ocean]

Idk but imo it should be centers

edges*

[keen tangle]

the distance between sets is usually the length of the shortest line segment with one endpoint in one set and the other endpoint in the other set

e.g. between two points, point and line, two lines in space

[glad ocean]

yep

think of it this way

A car is in front of another car

To measure the distance you would calculate from the front of car 2 to the back of car 1

obviously it might get a bit more complicated than that

But I believe that's the basis for measuring distance between multiple figures

[mighty hull]

hello

okay so,,, i know its probably not hard for you guys but can you help me with special triangles?

im having a lot of trouble grasping the concept

[devout moon]

hi guys can someone help me with relative rotations with Euler Angles

[jaunty rover]

so like

[vernal tusk]

@cobalt sun how did you solve it?

[zinc grail]

hello

so I get confused by this trig rule

so its cos is = to x-90?

[umbral snow]

sin(90 - x) = cos(x)
If that's what you're trying to say

[glad ocean]

yeah

So arccos(12/13)

[zinc grail]

so what would I do

so it be the same?

so the answer would be A?

[umbral snow]

@zinc grail
That is, cos(x) = 12/13
Adjacent = 12
Hypotenuse = 13

By pythag, opposite = 5
Then, sin(x) = opp/hyp = 5/13

[zinc grail]

oh crap

ok thanks

[glad ocean]

Yeetus

[vernal tusk]

you could even just use sine

by saying that the unlabeled angle is 90-x, so 12 is the bottom side there and 13 is the hypotenuse by soh(cahtoa), then pythag theorem to get left side is 5, then soh angle x to get 5/13

which is literally rhe same thing Kaynex said, just without cosine

lol

[lament bay]

I’m lost

[worthy harbor]

Think about what happens to the angles as BM increases

BM > AM means B is an acute angle

[lament bay]

Idk

<@&286206848099549185>

[worthy harbor]

Think about this, if A + B + C = 180

and you know that B is less than 90 what does that have to mean for A and C?

[lament bay]

A+C>90

[worthy harbor]

so everything hinges on proving that the angle B is less than 90 degrees right?

[lament bay]

uh huh

[worthy harbor]

Now they tell you that the length of BM is greater than the length of AM

[lament bay]

ok

[worthy harbor]

That's all you need to prove your statement, that's why I said think about BM getting realllllllyyy long and AM staying the same

can you imagine the angle B getting closer and closer to 0?

[lament bay]

uh huh but how would I write this in proof format?

I think the question wants proof

[worthy harbor]

Well, I don't know exactly what you're required to do, is it like a euclidean proof?

In your book you should have identities that express these relationships, I've been trying to give you a sense of direction more so than just an answer

[lament bay]

https://www.google.com/search?q=paragraph+proof+example&safe=strict&hl=en&authuser=0&tbm=isch&source=iu&ictx=1&fir=1_nBXitRugOG9M%253A%252C8hfF0VUl2eq6CM%252C_&usg=AI4_-kSyxvmLo7opLppg5_E_iljAUp7vZw&sa=X&sqi=2&pjf=1&ved=2ahUKEwiJm6rktvbfAhVKkSwKHZsFCzgQ9QEwAHoECAAQBA#imgrc=FGl20AwHEw58sM:

Supposed to be somewhat like that

I'm trying to understand it but its hard

[worthy harbor]

Look at the Triangle Inequality theorem

[lament bay]

The sums of two sides are greater than 3rd

im struggling too hard 😦

[worthy harbor]

ok, so if AM > BM what does that mean about the angles opposite those sides?

Angle A is larger than Angle '1' in the diagram

I mean BM > AM

AM = MC

so BM > MC too

So also Angle C is larger than Angle '2' in the diagram

[lament bay]

uh

[worthy harbor]

That's almost every statement you need to construct the proof

the final thing is to show that B is = to Angle '1" + Angle '2', then just show that A + C > Angle '1' + Angle '2'

[lament bay]

uh huh

[worthy harbor]

A > Angle 1, C > Angle 2, therefore A + C > Angle 1 + Angle 2

Angle 1 + Angle 2 IS B

[lament bay]

Its hard to put it all together cause I dont get this stuff

i hate geometry

[woven wigeon]

same

[lament bay]

I can’t even do a problem lol

[woven wigeon]

it's alright, eventually you'll get it

i had trouble with geometry too

still confusing lol

[lament bay]

@worthy harbor gave me hecka tips but I still don’t get it

[woven wigeon]

dont rely on one person to help, they have lives too ya koe lol. there's plenty of helpers

not me ofc coughs

[lament bay]

What do I do? Tag em again? Ok

[woven wigeon]

be patient

[worthy harbor]

lol, I'm just playing chess poorly

[woven wigeon]

f

[worthy harbor]

The shortest side is always opposite the smallest interior angle

https://www.mathopenref.com/trianglesideangle.html

[lament bay]

Uh huh

[worthy harbor]

Once you understand that fully, you'll be able to prove it

[lament bay]

Ok

I understand that one but idk how to write it :/

[woven wigeon]

what does ur current lesson cover? proofs?

cuz im bad at it

[lament bay]

Nah

Just doing some early practice

trying to do some early practice LOL

[woven wigeon]

f

welp gl

[lament bay]

i dont know anything man lol

[woven wigeon]

I feel u

cuz rn I’m studying for a test tomorrow, and I don’t get anything lol

[lament bay]

lol

[woven wigeon]

just need help with problems 25-27. I’m having trouble setting it up

and there's a typo on 27

actually i got 27 down lol

[lament bay]

im dumb

[woven wigeon]

ur not the only one

who's dumb here

i need help with 25 & 26

plws

<@&286206848099549185> f

i had to xd

welp going to fail either way

[worthy harbor]

#25 you use SSA

26 as well

like this https://www.triangle-calculator.com/?what=ssa&a=20&b=30&b1=70&submit=Solve

the bottom shows all of the formulas for how it's worked out, use it to help you learn, not to do it for you 😛

[woven wigeon]

thanks, but a friend helped me already xd

used law of sines to solve it eee

[supple haven]

yo

"match the trigonometric expression with one of the following

and they gave me

sec(x)cos(x)

what would that go with?

<@&286206848099549185> can anyone help me out?

[plucky marlin]

sec(x) = 1/cos(x)

[supple haven]

its sec(x) times cos(x) tho so what about in that situation?

[plucky marlin]

replace sec(x) with 1/cos(x) and see where that takes you

[supple haven]

1?

[plucky marlin]

yeah

[supple haven]

thanks man you're a god

[plucky marlin]

[supple haven]

sin^2 + cos^2 = 1

but what about the inverse with sec and csc and all that

[shadow anvil]

urm wot

[supple haven]

uh like

[shadow anvil]

You mean like sec^2(x)=tan^2+1?

[supple haven]

I think

sorry i have a hard time explaining stuff

0 iq on my part

[plucky marlin]

u divide both sides by cos^2 (x)

[supple haven]

thanks man

[plucky marlin]

np

[supple haven]

tan^2(x) + 1 = sec^2(x)

so I checked online and saw that

but like

if you divide sin^2 + cos^2 = 1 by cos^2

wouldnt you just get tan^2 = sec^2

im a lil lost on that

now to wait 10 minutes so I can @ the helpers

[narrow sleet]

U need to divide all three terms by cos^2

[supple haven]

ahhhhhh I just realized what I did tysm

I forgot that when something cancels it leaves 1 behind as a remnant

I thought it vanished

thanks dude

[narrow sleet]

Haha 👍

[austere glen]

From a square piece of paper Martha cut out some same circles. Sum of their circuits is bigger than the circuit of the square. How many circles did Martha at least cut? (I got the answer is 4 but i want to check)

[cobalt sun]

@vernal tusk

[marble perch]

can someone help me

these are based upon the properties of a square - if you need them I can send them

(parallelogram properties)

Forgot something was cut out

Given: Square MNPQ has a perimeter of 140

[astral hornet]

hint: the diagonals form 45-45-90 triangles

[glad ocean]

yeah

Cause the diagonals bisect the 90° angles

So it's 4x - 15 = 45

so x = 4, just solve it from there

[vernal tusk]

why does it even have the perimeter?

also how can one assume those 2 angles are congr. in the first step @cobalt sun? that's interesting af also, ty

[narrow sleet]

So u can find MQ, that why they gives u the perimeter

[lapis geyser]

whats the method for solving this called, i want to do practice questions?

[narrow sleet]

Because this is a square, so PQ = MQ and this shows that triangle PQM is an isoceles triangle where <QMP = <QPM. Since this is a square, < PQM = 90°. So I think u can think why <MPQ = 45°

@vernal tusk

[umbral snow]

@lapis geyser
Check for similar triangles
(x + 4)/6 = x/3

[lapis geyser]

@umbral snow thanks a lot!

[narrow sleet]

Guys, any hint how to start the proof for this?

[narrow sleet]

<@&286206848099549185>

[twin prawn]

Hint: what kind of triangle is ABC?

[narrow sleet]

Isoceles triangle so angle ABC and angle ACB are the same

Then?

Oh I get it....

Thx @twin prawn

[twin prawn]

🍮

[red sierra]

How to prove sin2A = 2sinAcosA

[junior lake]

sin 2a = sin a * cos a + sin a * cos a (because of the double angle identity) = 2 * sin a * cos a

maybe?

if the task is to prove that the double angle identity is true i have no idea though

[light finch]

It's to prove double angle identity.

[somber coyote]

Penguinlay:

[wild hamlet]

You can do it geometrically.

[vernal tusk]

@narrow sleet did you figure out that angle inequality problem? i like almost have it but not quite

[ocean nebula]

Do you guys know how to find all the heights of a triangle if you have all the side lengths and all the angels

[narrow sleet]

@vernal tusk Same, I found out OC > OB. Then probably use the SSS inequality theorem to solve it?

[limpid basin]

Find the area using herons, then use A = bh/2 to find each height

[buoyant oasis]

Cone I :
r = 18, h = 21
Cone 2 :
r = ? , h = 28
If both cones are similar, find r of cone 2!

r1h1 = r2h2
18.21 = r2 28
24 cm = r2

Is this correct?

[fleet wolf]

it should be r1/r2=h1/h2

not r1h1=r2h2

@buoyant oasis

[buoyant oasis]

e.e oops ok thanks!

[buoyant oasis]

@fleet wolf but the result is correct tho?

[fleet wolf]

oh nvm

i tought you took r2 as 18.21

no 24 is correct youre right

[buoyant oasis]

ok thanks!

Cone 1 :
r = 9
h = 15
Cone 2 :

r = 20

if both cones are similar determine the height

I used the r1/r2=h1/h2 eq. and got 33.3 as an answer.

The number is a bit ugly so I just wnated to make sure i'm in the right path before I move on.

Is 33.33 correct for the height of cone 2?

[fleet wolf]

yup

[buoyant oasis]

Thanks!

[vernal tusk]

@narrow sleet...i feel like they should be congruent by SAS and CPCTC

[narrow sleet]

I think the converse of hinge theorem should work

[upper karma]

can someone reteach me trig

I have a pretty basic understanding but now I'm in high level math classes I realized it's pretty dang important

[dire rampart]

lots of good resources online

khan academey for example

[rapid hatch]

How do i show that a triangle ABC with A (-2,5) , B(0,-1) and C (-3,-2) is rectangular?

i missed one week school idk anything

@twin prawn can you help me out fam

oh fam this ez pez mb, ill come back later with otehr questions tho

[astral hornet]

rectangular?!?!? triangle?!?!?!

[fading bough]

I'm completely lost.. Idk what to do

[astral hornet]

using vertex and roots can get equation

then height when tracking starts can get coords

roots can find distance

[austere jasper]

hi guy can someone help me with this please? if cosx = -sinx --- 3cosx=?

[upper karma]

Never seen - - - before

[vernal tusk]

@astral hornet i think he meant right by rectangular

[astral hornet]

Oh

@rapid hatch then just find perpendicular slopes

[primal tide]

I'm not sure whether this goes in #geometry-and-manifolds or this one

but how do I determine the shape of the following relation:

[somber coyote]

MrSpaceCow:

[primal tide]

<@&286206848099549185>

[toxic pumice]

hello, could anyone help me with this problem

y=2x+5, x is restricted to values greater than or equal to 0 and less than or equal to 100, what is the y range

[wide echo]

Try finding the maximum and minimum values of y=2x+5 with those restrictions on x

$0\le x \le 100$

[somber coyote]

stephen:

[toxic pumice]

y range would be greater than or equal to 5 and less or equal to 205

[wide echo]

Yep

$5\le y \le 205$ in other words

[somber coyote]

stephen:

[toxic pumice]

thank you

[wide echo]

[mellow scroll]

Directed angles are quite counter-intuitive but I think I'm getting the hang of it. Proof for existence of Miquel point btw.

[half yarrow]

How would I prove and solve for X?

[plucky marlin]

well ACB = BCD

ABD = 90 = CDE

and if all of the angles of a triangle add up to 180

so the other angle must be same

thus they are similar

in similar triangles ratio of the sides opposite to the same angle are same

so $\frac{5}{9} = \frac{x}{7}$

[somber coyote]

soap:

[half yarrow]

Oh thank you, how would I solve for X?

[dire rampart]

isolate x

what can you do to both sides

so one side has just x

[half yarrow]

Do I just find the difference between 9 and .5 and apply it to 7 and X?

[dire rampart]

multiply both sides by 7

what do u get

[half yarrow]

3.5 cm and 7x and 63

@dire rampart

[dire rampart]

ya bur right side is 7x/7

which is just x

[half yarrow]

That brings me back to X, doesn’t solve it though?

7x/7 = x

[dire rampart]

you get x= left side

so you have ur x

[half yarrow]

Wait I’m a little confused, X= 7 even though there different sizes?

[hybrid wolf]

Would this be correct?

[elder oak]

[hybrid wolf]

Adjacent line is 210
Opposite is 44
210^2 + 44^2 = c^2
c = 214.5
opp/hypo = .205
arcsin (.205) = 11.8
round to 12 degrees ```

^ that's what I did for my answer

ok I suck at league, but I can find more fun in it with math

[elder oak]

🤔

[glad ocean]

oml

I guess it's right but WHY

[hybrid wolf]

Math > League

I play it with my friends and suck at it, but maybe some math can help B)

[neon stag]

Would I have to solve two equations to find two x's and then add them to find the real x?

[safe maple]

Can someone verify if this is right

@neon stag That’s how you get the answer

[elder oak]

,rotate -90

[elder oak]

nani!

,rotate 90

[safe maple]

,rotate -90

[hybrid steppe]

How do I solve this? I don’t understand the -1 thing for any of them

[astral hornet]

^-1 means inverse function

inverse of tan is arctan

it'll return the angle that has that ratio

[frozen saffron]

can anyone help me?

here's the problem

A new lookout tower is to be located equidistant from three ranger stations. If the stations are at positions A(0,-5), B(8,3), and C(6,5), find the coordinates of the point where the new tower should be built.

can anyone help me with this one?

[astral hornet]

Circumcenter

[frozen saffron]

i think so

might be centroid

i'm a bit confuse

[astral hornet]

Circumcenter is center of circle that contains the verticies

[frozen saffron]

so can you help me with this question then?

[astral hornet]

Do you know how to find circumcenter?

[frozen saffron]

find the perpendicular bisector

AC AND AB

[astral hornet]

Yea, get the equations of those and solve in a system for coord of intersection

[frozen saffron]

do i still need to find the midpoints?

[astral hornet]

You need midpt

[frozen saffron]

2 of them right

?

[astral hornet]

That should be enough

[frozen saffron]

i have a question again

how would you know if the word problem is asking for a centroid or a circumcentre?

[astral hornet]

Question is asking for equidistant. Circumcenter is center of circle that contains the verticies, meaning the verticies are a radius distance from circumcenter

[frozen saffron]

so basically if the question have "equidistant" word on it or "vertices" automatically it's a circumcentre question?

[astral hornet]

Equidistant to verticies yes

[frozen saffron]

ok thank you for your assistance

[sick veldt]

hey can someone confirm 4 answers for me?

[sick veldt]

<@&286206848099549185>

hello...?

I really wanna submit this quiz in soon

[keen tangle]

The first one's wrong

[sick veldt]

oh

I did 168/95.2 = x/34

[keen tangle]

The bisector theorem (or whatever it's called) says that KY/YV=168/95.32

Your equation calculates KY

[sick veldt]

so what should i do with the x and 34

should it have been x/34 - x?

[keen tangle]

I mean, that would give you the answer but it's pointless calculating

Your method gave that KY=60

And x is just VY+YK

[sick veldt]

yeah idk

Ik that when you add VY and YK you get x

but that doesn't help

and ik my method gives me 60

but what should I do

[keen tangle]

Your method gives YK=60 and we know VY=34

So x is just their sum

[noble marsh]

hello I need help

what does it mean when there are lines on the triangle?

im not sure how to solve this

[keen aspen]

That means the side lengths are the same

If on both triangles they have the same number of nodges on a side then the side is congruent

[noble marsh]

so what if one side has 2 lines and the other has 1 line

what does that mean?

[keen aspen]

That means they arent congruent

Rather the two sides with the 2 lines are congruent and the two sides the with 1 line are congruent

[noble marsh]

so in the picture above, how would I determine if its SSS,ASA,or SAS?

[keen aspen]

If you are given three sides being congruent then SSS, if two angles and a side in the middle then ASA and two sides and an angle in the middle its SAS

[noble marsh]

im not sure what my homework means

in the picture above

[ocean nebula]

Do you guys know how to draw in the 3 heights of this triangles. I already have the lengths

[keen aspen]

Wym?

[astral hornet]

perp to a side through opp angle

[noble marsh]

I got it wrong

can someone correct me

[marsh spruce]

Hey guys I just wanted to share with you guys what I've just found and it's very useful to memorize the unit circle
https://m.youtube.com/watch?v=LE6dmczMc68

[upper karma]

#dontmemorise

[dry glen]

yes

memorizing is good

is da key for learn

lmao

[vernal tusk]

@noble marsh 1 is SAS, 4 is impossible (AAS)

[molten sail]

"To find x, make y = 0"

Doing this and not getting the right answer.

It only works on a few question

(Linear systems)

On another question I had to make y = -2

How do I know what to make y?

[keen aspen]

Can I have an example

[lament bay]

What theorems would I use?

[glad ocean]

Substitution

Substitution

Subtraction

[vernal tusk]

first is partition, not substitution

also feel like there should be a reflexive b4 subtraction but hey

btw, not to guilt or anything, just some advice: study your theorems, postulates, and corollaries

cause those are pretty much the very basics of geometry right there

although it does say work with a partner so

[sand wharf]

you cant do trig if you eat all the pi

[glad ocean]

sin(90°)

That doesn't require pi 😉

[hazy granite]

nooo

not the pi

[rustic cairn]

How do i do this?

[upper karma]

is this a parallelogram @rustic cairn ?

[upper karma]

NQP = 29deg?

did you put in the numbers? or are they part of the question?

QLM=100deg?

LQM = 180 - 100 -29 = 51

LN=14

QP =8.2

[upper karma]

what is the line we reflect inverse functions over called?

I know its y=x but does it have a specific name?

I feel like I want to say axis of symmetry but that doesnt feel correct.

[upper karma]

<@&286206848099549185>

Also, what is the "power" of a root called?

like in the root x^1/7 what is the 7 called?

Is it called a "root index"?

[astral hornet]

Yea

[upper karma]

Hell yeah

what about the inverse functions?

[astral hornet]

Wdym

[upper karma]

what is the line we reflect inverse functions over called?
I know its y=x but does it have a specific name?
I feel like I want to say axis of symmetry but that doesnt feel correct.

[astral hornet]

Y=x

[upper karma]

is there a name for this line

I undrestand its just a line with a slope of 1

but it seems like one of those things that would have its own name

[upper karma]

What are some 'useful' transformations of the plane which preserve the incenter of every triangle?

[astral hornet]

@upper karma you just call it by its equation

[upper karma]

any ideas <@&286206848099549185>?

[worldly glen]

If Sin^4(x) + Cos^4(x) = 1/3 what is sin(2x)?

[wild hamlet]

uh cos^2(x) = 1-sin^2(x)

cos^4(x) = (1-sin^2(x))^2

that would seem to be the easiect way to do things

or to at least get sinx

[worldly glen]

fixed* this is what I've tried:

[dense garnet]

Guys
Stupid question
If I want to multiply a line(?)s length on a sphere with X, what should I multiply the R with?

[loud zodiac]

What do you mean?

[dense garnet]

Like you have a circle

A unit circle

And you want to make the "quarter" of it X cm

By multlpying the R with x

[loud zodiac]

So you're trying to change the size of the circle by multiplying the radius with some value x?

Sorry, not quite following

[dense garnet]

Yup that is it

What I'm initially trying to do is to spreading points evenly on s sphere with golden spiral method

But I want to decide the distance between the points

[loud zodiac]

I'm getting a call. I'll try to come back and help in a bit if no one helps you out before I get back

[dense garnet]

Alrighty

[dense garnet]

Ok so another question

nvm

[loud zodiac]

Did you figure it it, @dense garnet

[dense garnet]

Nope

Been 6 hours since I started to solve the initial problem

Literally nothing so far

[loud zodiac]

😭

Sounds like you're looking for arc length?

Also, are you talking about a sphere or circle?

[dense garnet]

A sphere

Yeah I am wondering if the every arc grows with the same rate when the radius changed

And if yes, what is that rate

Anyone? lol

I've read about 20 different papers on this

NOTHING

[loud zodiac]

Sorry, not sure how that would work in 3d

[cobalt sun]

@vernal tusk idk how...

I just know that it's solved

Oh, they are corresponding angles between two parallel lines

[sick veldt]

I think its HL

cause if you split the triangles

you'd get an HL no?

hello?

anyone?

:/

[supple bramble]

Just check if your answer is correct and if it isn't try another one

[sick veldt]

I can't check it

[supple bramble]

Then you're probably supposed to do it on your own without help

[sick veldt]

its a practice lol

the real quiz

is after this

<@&286206848099549185> Been 15 minutes

can you confirm 3 of my answers?

I used the pythagorean theroem

so

sigh

or tan

[magic chasm]

Can someone help me with the last question on ex 04

I know I'm supposed to use the sin 2x, I did x= arctan 1/3, but then what?

[drowsy spoke]

Translate

[zenith ember]

Verifier = Verify. Montrer = Show. Deduire enfin = Finally, deduce

[vernal tusk]

@sick veldt 40 degrees

what looks like the hypotenuse (cant be a right triangle tho) is 15 as given

tree is 18

tan(x) = 15/18

(x = the abgle)

inverse tan it

atan(tan(x))=atan(15/18) (atan is just tan^-1())

x = atan(15/18)

x = 39.80557

x = 40

...wait, no

sohcahtoa

opposite over adjacent

so atan(18/15) = 50.1944

x = 50

dorry

sorry

[verbal jolt]

Any trig tutors out here? I don't necessarily need help right now. But I am looking for someone to get help from throughout my term. ($$$)

[umbral snow]

Feel free to ask any questions on the server, whenever you may want

[upper karma]

@sick veldt the first problem cannot be HL because the hypotenuse (AB and AD) aren't equal in this case, which is a requirement for HL. So your answer of SAS is the correct answer

[weak adder]

i've been stuck on this problem for a while now, i have 2 quboid shapes defined by 8 points each, i only know that they overlap somehow, and i need to find the optimal transformation vector to seperate them

[tawdry pivot]

,rotate -90

[dim flare]

ok so i've got this problem
triangle solving
i'm given 12.9/sin30=b/sinB=25.8/sinC
i got C=90 degrees, is that correct?

[vital ermine]

I'm stuck on part D. I have the equation y = -(3/b)x + (12/b) but not too sure how to solve for b given a different gradient.

[dire rampart]

do you know how the gradients of 2 perpindicular lines are related?

[vital ermine]

ah shoot I meant part C, and yea if the product of two gradients = -1 they are perpendicular

[real sparrow]

@vital ermine so basically you have the slope from that line and we know that the product of the gradients of two perpendicular lines is -1 , you'll get the slope then and use the points to find the intercept.

[dire rampart]

@vital ermine the gradient is -3/b which you are told equals 1

so -3/b=1

solve for b

[vital ermine]

-3 oof i'm dumb

I didn't understand the question I think haha. thank you!

[dire rampart]

no worries

[blazing kelp]

could anyone help me

Prove that: 2/cos2 - 1/sin2+2/sin4 = cot4 -sin4

[keen aspen]

cos^2(x)?

Or cos(2x)

[blazing kelp]

cos to the power 2

like cos square

[keen aspen]

Ok

[blazing kelp]

I am sorry sir

one change

edited

please check now

[keen aspen]

$ \frac{2}{\cos^2(x)}-\frac{1}{\sin^2(x)}+\frac{2}{\sin^4(x)}=\cot^4(x)-\sin^4(x) $

[somber coyote]

PJS:

[keen aspen]

This?

[blazing kelp]

yes

[keen aspen]

Okay

so try getting a common denominator with the sin^4(x) and sin^2(x)

[blazing kelp]

sir I got 2sin4 + 2 cos2 - 1/sin2

I don ot now how to proceed form there

[keen aspen]

Actually move everything with sin to one side

[blazing kelp]

but we cant use rhs ?

[keen aspen]

We cant?

[blazing kelp]

we have to prove it

[keen aspen]

Ugh okay