geometry-and-trigonometry

[trail tendon]

doesn't look solveable with elementary functions

no those are not the same for all theta

[faint pasture]

Definitely not with elementary functions yea but wolframs answer makes me doubt if it is even solvable with non elementary ones

[trail tendon]

yea it was probably done analytically

[faint pasture]

Ig youre given the first result and you have to show the second one

[faint pasture]

Mmmm

[trail tendon]

yea but i don't think one follows from the other...

[faint pasture]

,w sinx + cosx = sqrt(2) sinx

[somber coyote]

Results provided by WolframAlpha

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

Requested by enderdoesntmind

[faint pasture]

,w sec(3pi/8)

,w 2 sqrt(2) sin(3pi/8)

Looks like it does

[somber coyote]

Results provided by WolframAlpha

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

Requested by enderdoesntmind

Results provided by WolframAlpha

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

Requested by enderdoesntmind

[faint pasture]

It is true

Idk how youd arrive at that

But its true

[earnest musk]

Bro is this the answer to my question?

Never seen anything like this lol

[trail tendon]

huh?

[faint pasture]

💀 i just subbed the solution to the given equation in what we have to prove

[faint pasture]

Unless wolframs tripping hard then yea it does follow from that one way or the other

[trail tendon]

cos(theta)+sin(theta) = sqrt(2) sin(theta)
cos(theta) = (sqrt(2)-1) sin(theta)

sec(theta) = (sqrt(2) + 1) sin(theta)
sec(theta) = 2sqrt(2) sin(theta)

(sqrt(2)+1) sin(theta) = 2sqrt(2) sin(theta)
sqrt(2) + 1 = 2sqrt(2)
sqrt(2) = 1

um

ignore me

LOL

[faint pasture]

Sec theta = (sqrt(2) +1) / sin theta

[trail tendon]

yea

[faint pasture]

Mmm

[trail tendon]

but then

csc^2(theta) = 2sqrt(2)/( sqrt(2)+1) ?

sin^2(theta) i mean

bro im trippy rn LOL

[faint pasture]

Mahn what are we even doing just weirstrass the first eqn thats given to us

And then get tan(theta/2)

And use that to prove the second eqn

Idk the elementary method

But this works

[faint pasture]

Yea

Actually no

That give sin theta>1

[faint pasture]

You were right when you have csc theta xD

[trail tendon]

i was?

[faint pasture]

Yea

😭 were you

Idk

[trail tendon]

no 😭

[faint pasture]

Listen just apply the weirstrass and its easy

[earnest musk]

Ended up in it the same place

Just took less time then last time

[faint pasture]

Google sin(x) half angle identity in terms of tan(x/2)

And same for cos(x)

Then solve

[trail tendon]

you just did the same thing...

[earnest musk]

I know I just followed your thing and ended up in the same place lol

[earnest musk]

I don't think half angles are allowed in my level i don't even know what they are sadly

[faint pasture]

What all is allowed?

[trail tendon]

i'm not getting anywhere

[earnest musk]

I think there might have been a mistake no way they gon give us this hard stuff I'm in secondary school

[faint pasture]

From the given relation you cant get to a eqn in terms of sin and sec

Obv

Like jjst by algebra i meant

[trail tendon]

boutta throw hands

[earnest musk]

lol I though this was easy at first bout to start crying now

[faint pasture]

I mean its not tough yourr tools are restrictted that is the main issue

[earnest musk]

Is it even possible with the restricted tools I have?

[faint pasture]

Id be suprised if it even was

[dire mirage]

May I ask for assistance on this particular problem?

[earnest musk]

I think Using 3 perpendicular mirrors a light ray with an incident angle of 55° reflects back in the opposite direction, keeping the final incident and reflected angles at 55°.

[faint pasture]

Just draw the diagrans and angle chasing

[faint pasture]

The que is ambiguous tho? Does it want us to give it the total number of mirrors (including the original one) or just how many extras we used

[earnest musk]

I did it guys (I think) can you see if it's correct and I know I have horrendous handwriting

@faint pasture

@trail tendon

[faint pasture]

Cool

It is all correct from what i can see

[earnest musk]

Aight thanks for the help

[obsidian rock]

Why is it sin A?

(unit circle)

[obsidian harness]

that's impossible if you don't know anything about the radius

use the circle theorem that the angle at the centre is 2A

then split the isosceles triangle into two congruent right triangles

you get $2R \sin A$

[somber coyote]

south, just south

[obsidian harness]

https://www.youtube.com/watch?v=wGDUe-psTaM

[obsidian rock]

Wouldn't this mean that, since r in unit circle is 1, then a = 2sinA?

[obsidian harness]

yeah

[upper echo]

yep, it is simple just write the left side as sin(pi/2-t)+sin(t) and use sum of sines.

[dreamy vessel]

Suppose a cube has side lengths of a. What is the area of the region of points that are closer to the center than any one of the vertices?

This problem got a crazy solution and it was in an anime

[upper karma]

assination classroom ?😏

lmao i got the spelling wrong

assassination

[upper echo]

Its formed by inetersections of mid-planes between the center and a vertices. So we get 8 faces. I guess its an octahedron with side 3a/sqrt(8).

[dreamy vessel]

Not quite

[upper echo]

why not?

[dreamy vessel]

The mid points of those lines are good but it's not an octahedron and there's a much more elegant way

[upper echo]

If you take the whole solid whose parts are outside of the cube it should be octahedron

[upper echo]

for some reason you have taken only those points which are inside the cube

so, yes, if you cut out only the part inside the cube you will get the solid on you figure

[dreamy vessel]

Oh man

That's my fault it was supposed to be strictly inside the cube

Whoops

[upper echo]

never mind. the problem and its solution is clear to me.

[dreamy vessel]

Like making more points around the cube?

[upper echo]

I mean, an equation of the midplane is easy: you take a vector from center to vertex: say (1,1,1) and you immediately get an equation (x-1/2)+(y-1/2)+(z-1/2)=0 And the same with two more non-parallel planes. So, we solve a system of 3 linear equations and find coordinates of just one vertice in that octahedron. The rest is easy because of symmetry.

And for cube centered at O with vertices like (1,1,1), (1,-1,1) we get vertices of the octahedron like (3/2, 0, 0). Just exactly as seen in the figure. So the side of the triangle is 3a/sqrt(8).

[dreamy vessel]

https://www.youtube.com/watch?v=64ZjFXrFBqE&t=161s&ab_channel=Phanimations Skip to 1:15 for the problem and from there watch and you'll see two solutions. One of them is basically bash, and the other one is the one

That's not very nice

[vital rain]

Yes

[dreamy vessel]

YOu can't whip this devious stuff out

[vital rain]

My bad let me put that back in

Can you do me a small favour tho

[dreamy vessel]

What is it

[vital rain]

Could you solve my homework

It has a weird name

[dreamy vessel]

Send a picture and then I will decide

[vital rain]

Birch and Swinnerton-Dyer Conjecture

Is that sweedish

[dreamy vessel]

oh hell no

[vital rain]

Pretty please

[dreamy vessel]

I'm a freshman rn

[vital rain]

Good enough

[trail tendon]

assination
lets assinate some peeps

[dreamy vessel]

Wait how did bro know

[trail tendon]

that aint your homework

[trail tendon]

maybe...he watched it :O

[upper karma]

Maybe I tried to solve it

[trail tendon]

no, could never be

[vital rain]

I solved it

[dreamy vessel]

what you get

[vital rain]

The intregal stuff basically

It connects

Source : don'ttrustmebro.com

[cloud sonnet]

Sometimes I’m glad I’m in foundation maths and still in secondary school cause what is this 🤣

[upper karma]

Lol

[cloud sonnet]

It’s hardly even maths it’s just pure letters 🤣

I know they have values but still

[vital rain]

Yh I just made that up lol

[cloud sonnet]

Oh aha

So annoying that we don’t even learn what sin cos and tan is

[vital rain]

You do

[cloud sonnet]

We was told that we don’t need to learn it for our exams even though the last question was what is sin 30

Only higher tier does that

[vital rain]

Is it non calc?

There is an easy way to do it tho

You just need to remember sin that's it

[dreamy vessel]

1/sqrt2 is bad

[vital rain]

Hush

Root2/2 same thing

[dreamy vessel]

that's better

[vital rain]

I picked the 1st one I saw

[dreamy vessel]

Who ever made that

[vital rain]

So many are like that

[cloud sonnet]

Yeah

[dreamy vessel]

petualant

[vital rain]

So for sin you just remember the degrees at the top

The put 0 1 2 3 4

Then root them

And divide by 2

[slim vigil]

Just use a calc (slang for calculator)

[vital rain]

Cos is the opposite of sin

And tan is sin/cos

[vital rain]

There are calc and non calc test

[cloud sonnet]

Yeah my maths calc is tmr

[slim vigil]

Gl

[vital rain]

Uh

[cloud sonnet]

We have 1 non calc and 2 calc

[vital rain]

Oh you do gcse

[cloud sonnet]

Yeah

[vital rain]

I did that a year ago it was pretty calm

[cloud sonnet]

Ahh yeah I’m assuming u did higher then aha

[vital rain]

I kinda flopped didn't get the grades I wanted

[vital rain]

Yh

[cloud sonnet]

What did u get

[vital rain]

8

[cloud sonnet]

That’s really good

[vital rain]

Eh

Wanted a 9

[cloud sonnet]

Only 1 grade off

[vital rain]

Why are you still up isn't it like 00:14

[cloud sonnet]

Good point

Idk tbh I’m just watching maths revision videos

Mostly on simultaneous equations

[vital rain]

The thing is if you revise just before you sleep you won't remember much

[vital rain]

3 or 2 unknowns

[cloud sonnet]

Yeah that’s very true

[cloud sonnet]

I get how to work out x and y it’s the part when you make them the same that gets me

[vital rain]

Better thing to do is sleep early and wake up at like 2 or 3 and revise

[cloud sonnet]

Oh damn

[vital rain]

I got like 3 or 4 hours average sleep during my gcse lol

Had to revise last minute

[cloud sonnet]

that’s pretty bad

[vital rain]

Start you revision Like a month or two before the test and you are set

[cloud sonnet]

Yeah I did kinda do that but I’ve been revising mostly stuff I already know like Pythagoras and angles and stuff

Stupid move

[vital rain]

Yh

[cloud sonnet]

So what do u do at college

[vital rain]

Is college a levels?

[cloud sonnet]

Yeah

[vital rain]

Ah

[cloud sonnet]

Well

[vital rain]

I do maths furthermaths Chem and physics

[cloud sonnet]

Yeah loads of different routes to take at college but A levels are the main ones

[cloud sonnet]

Wow

[vital rain]

5 hours sleep lol

[cloud sonnet]

That is pretty bad

[vital rain]

Mostly cause I leave my homework until the last day

[cloud sonnet]

I always do that

[vital rain]

I wake up at 5 normally arrive to school at like 7.30

If I have homework mabye I wake up at 3-4

[cloud sonnet]

Bro just do it through the day 😭

[vital rain]

Yh I do some at school

[cloud sonnet]

Fair enough

[vital rain]

Our school has like study peroids instead of free peroids

So you can like do homework/revise

[cloud sonnet]

My school does stuff like that

Period 7s

Absolute pain

[vital rain]

Bro what

[cloud sonnet]

It sounds stupid

[vital rain]

When is that 530

[cloud sonnet]

Basically it’s another lesson and you revise for exams and stuff

[cloud sonnet]

My school days are very different from yours I’m guessing 🤣

My school starts at 8:20

We typically leave school at 2:45

[vital rain]

Oh we leave at 3

[cloud sonnet]

And then for year 11s like myself if we have period 7s they are between half an hour to 50 minutes depending on the teacher

[vital rain]

Mist I had was peroid 6 during gcse

Most*

People who didn't meet their target had to go for a 7th one

[cloud sonnet]

yeah that’s basically period 7 at my school

[vital rain]

Peroid 6 was an hour

[cloud sonnet]

My period 6 used to be an hour

[vital rain]

Isn't a peroid an hour

Oh ok

[cloud sonnet]

Well at my school it’s 50 minutes

And period 5 for some reason is 1 hour

Don’t ask why even I don’t know 🤣

[vital rain]

We also had Saturday classes

[cloud sonnet]

Bro did u go my school 🤣

[vital rain]

Can't lie they did not help

I just went for the sake of it

[vital rain]

Prob not

[cloud sonnet]

Fair enough I might go this Saturday

[cloud sonnet]

Yh ik cause our schools are very similar

In this context

[vital rain]

Oh

[cloud sonnet]

Ngl foundation maths has ruined me in what I can do

I wanna do engineering

I can do it but I won’t be able to a uni

[vital rain]

My friend moved up from foundation like a few months before the test

[cloud sonnet]

That’s what I’m trying to aim for

[vital rain]

Only some unis look at gcse

[cloud sonnet]

I’m getting a maths tutor and everything lol

[vital rain]

Mostly a levels and stuff

[cloud sonnet]

oh

Wait what

Seriously

[vital rain]

Yh...

[cloud sonnet]

Stupid question I know aha but I thought it cause u need good GCSEs to get into certain A levels

[vital rain]

They like gaslight you too much at gcse

[vital rain]

Yh

Depends on the school

[cloud sonnet]

Fair enough

[vital rain]

If you get 6+ you are set

7+ grades even better

I would show you my gcse certificate if I had them rn I just got them earlier today

[cloud sonnet]

Yeah with foundation I can’t cause max u can get is a 5 and the grade boundaries for edexcel are ridiculously high

[cloud sonnet]

Yh

[vital rain]

Yh

[vital rain]

It's in my files I'm too tired to get them

[cloud sonnet]

Fair enough

[vital rain]

I can tell you off the top of my head what I got if u like

[cloud sonnet]

Can do

[vital rain]

My grades aren't like crazy

I got 6 for physics, bio, French, business, both English

I got 7 in chem

I got 8 in maths, furthermaths , re and geography

That's it I think

[cloud sonnet]

Why do u do so many GCSEs

[vital rain]

I did triple science

So I got 3 science grades instead of 2

[cloud sonnet]

Ahh

[vital rain]

And furthermaths was optional

Ngl it was way easier than the actual maths higher last year

[cloud sonnet]

how 🤣

[vital rain]

The test I mean

[cloud sonnet]

Ah

[vital rain]

It easy too

You should look at a past paper they are doable after you learn a few things

[cloud sonnet]

For someone who got a grade 8 it probably was easy

[cloud sonnet]

Yeah for these exams there mocks and old exam papers and I’ve found them on maths genie and Ive predicted which paper it’s gonna be

[vital rain]

Eh I only revised for it the day before the test

[vital rain]

But do you learn how to do them

[cloud sonnet]

Yeah what I normally do is I watch a video of a guy explaining it and if I don’t get a topic I try and rewatch if not I got on for example maths genie or dr frost and find the type of questions I don’t get from the exam and attempt them

I did that today and I did meh

[vital rain]

Mabye lives streams may help

[cloud sonnet]

Yeah I’ve been watching some on TikTok

It’s crazy I’ve been revising purely maths for the mocks aha

[vital rain]

Ah ok

[vital rain]

Might not be a good idea

[cloud sonnet]

Yeah it isn’t

Atleast these are only practice exams

[vital rain]

What subjects do u d

[vital rain]

No mocks are just as important

[cloud sonnet]

I do history Re stats and engineering

[cloud sonnet]

Ik they send these to colleges

[vital rain]

English?

[cloud sonnet]

Yeah I do the usual 3

Maths English and science

Science it’s combined

[vital rain]

Oh I also somehow got destination for that gcse speaking thing

I just waffled

I thought I was getting a merit

[cloud sonnet]

Ahah

ima go sleep now I’m shattered

Night mate

[vital rain]

K good night

[narrow crescent]

the question:
Write the corresponding triangle congruence theorem (SSS, SAS, ASA, AAS) and complete the congruence statement

i wrote " triangle STA is congruent to triangle RTA becaus eof the AAS postulate"

she then marked my paper with

Angle STA, and Angle RAT "Do you have 2 sides"

basically what it looked like

[exotic yarrow]

You have:

• ST congruent to RA
• Angle STA congruent to angle TAR
• Side TA is congruent to itself

Since you have a two pairs of congruent sides and the pair of angles included b/w said sides, it should be SAS.

[narrow crescent]

thank you 🙏

wait so

would i do

triangle(STA) ≅ triangle(RTA) because SAS

or angle(STA) ≅ angle(RAT)

[exotic yarrow]

the order of the letters should match up w/ the congruent parts

[narrow crescent]

because sas

[exotic yarrow]

ex. the angles at T and A are congruent, so those should match up in your names

$\triangle STA \cong \triangle RAT$ is one way to write it

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

there's 5 other ways though so

[narrow crescent]

ight

thanks

[devout whale]

By reading some messages I come to conclusion, that I'm so fucked up in school

What de hell I was thinking to go into high school

[trail tendon]

isn't it kind of mandatory...

[rich knot]

don't think this needs a whole forum post, just going to ask; Trigonometry, unit circle, yeah? am I right in thinking every point on the circle's x is the cos to that angle/y is the sin to that angle?

[obsidian harness]

yes, this is the unit circle definition

unit circle, as in the hypotenuse has to be 1 for this to work

otherwise it will be $(x, y) = (r \cos \theta, r \sin \theta)$

[rich knot]

awesome, that's all I needed

[somber coyote]

south, just south

[obsidian harness]

np!

[hidden willow]

hii ive got a question

[obsidian harness]

ah so let s = side length of the red square
since the side length of the squares that make up the cross = 1

[hidden willow]

ye but i dont know where to go from there

[obsidian harness]

then by Pythagoras we observe $s = \sqrt{3^2 + 1^2} = \sqrt{10}$ hmmm

[somber coyote]

southlander!

[obsidian harness]

ah so that must mean that we have a pair of similar right triangles

[hidden willow]

hmmm

where do we go from there

?

[obsidian harness]

these 2

so comparing side lengths we have:

$1, 3, \sqrt{10}$

$?, 1, ?$

[somber coyote]

southlander!

[obsidian harness]

it follows that the short side = 1/3

and the hypotenuse of the small triangle = sqrt(10)/3

this is enough information to find the perimeter of the blue pieces

also note that 1 - short side = 1 - 1/3 = 2/3 appears multiple times

[hidden willow]

So we got that this length is 1/3

?

is this it

and then

1/3×12 + 4/3

[obsidian harness]

no, this length is 1

the squares have side length 1

[hidden willow]

oh yeah wait nvm

[obsidian harness]

also I see 4 * sqrt(10)/3 in there

since it's the perimeter not the area I think you can't join up the three pieces btw

that works for the area ofc

[hidden willow]

i think i get it

ill have a look again tysmmm

[obsidian harness]

np!!

[somber coyote]

Result:

15.549703546891

[obsidian harness]

ah so it might be easier to think about it if you join the 3 blue pieces together, which has a perimeter of 3 * 4 = 12

then you have to add on the 4 edges which each have length sqrt(10)/3

,calc 12 + sqrt(10)/3 * 4

[somber coyote]

Result:

16.216370213558

[obsidian harness]

@hidden willow

[chrome token]

does anyone know a synthetic proof of existence of isogonal conjugates? (so no trig ceva stuff)

[compact current]

want to torture yourself ?? Have some fun dealing with some trigonometry problems
|| translate the question from french to english ||

[vital rain]

Crazy ritangle

[obsidian harness]

the question says it's made from squares actually

nice problem

[vital rain]

Na it's a maths challenge

I see the question

[trail tendon]

why does that look like the python symbol...

💀

[vital rain]

Huh

It's because there is a plus with blue and yellow

[vital rain]

Uh

Ping the other guy I arl solved it

Thx anyways

[neon ledge]

ok sry ill delete my message

[cobalt pecan]

guys rate these questions

[obsidian harness]

nice if you know your stuff

$\sec^2 \theta - \tan^2 \theta = 1 \implies \sec \theta + \tan \theta = 1/x$

[somber coyote]

southlander!

[obsidian harness]

should be clear what to do from here

[cobalt pecan]

This is wht i did

[obsidian harness]

yeah I think the other question with tan and sin is harder

since it gives you a choice I'd just do the easier question, haha

[cobalt pecan]

yea fr

This is wht I did for or

Question

which grade question u guess it would be?

[obsidian harness]

I'd say 11th

[cobalt pecan]

its 10

cbse

[obsidian harness]

ah nice

[cobalt pecan]

you are which level?

[obsidian harness]

uni

[cobalt pecan]

oh damn

which one

[obsidian harness]

oh I can't say but I study in Australia

[cobalt pecan]

oh srry

nice

which course u took?

im planning to take engineering

[obsidian harness]

I switched from maths to sociology

[cobalt pecan]

but that itself is a vast field

[cobalt pecan]

I see

so u studied in australian curriculum?

[obsidian harness]

yeah

[cobalt pecan]

nice

CBSE is rough

purely theoretical till 10th

like actually

i can't do a hydrogen pop test

[obsidian harness]

oh lmao sciences

[cobalt pecan]

or study refractio thru glass prism

[cobalt pecan]

oh for us stream is only at 11th

[obsidian harness]

ah right

I heard they removed the periodic table from 10th.......

[cobalt pecan]

so we have to study science math and sst all the way till 10th

[cobalt pecan]

yea they did

periodic classification of elements

[obsidian harness]

yeah RIP

that's embarassing

[cobalt pecan]

in a way its good

but also bad

[obsidian harness]

cause now they'll shove all the info into 11th instead

[cobalt pecan]

yeaa

[obsidian harness]

so I think making 9th and 10th a bit more rigorous is good

if it makes 11th and 12th more manageable

[cobalt pecan]

actually it is rigourous enough

[obsidian harness]

true

[cobalt pecan]

we have board in like 2 months 😭

*boards

bcs this i didnt open discord for a year

like discord is literally dying

[tawdry arrow]

I had a trig test yesterday… I studied heavily the week prior but when I got to the big questions I blanked on part of the formula

Tan ( x) side/side

Forgot what x was and where to find it

Teach said there would be a retake but I dunno how the scoring will work. Hopefully I can raise my grade

[pale orbit]

they should introduce analysis in 9th, the lack of rigor in calc is quite concerning

[maiden notch]

when you realize math gets easier and easier

[trail tendon]

where you get this idea from

[limpid tapir]

....💀 harder if you have a weak foundation, but in advanced topics its tricky

Like quantum mathematics as an example

[trail tendon]

thats physics

[limpid tapir]

You study it in mathematics to use it in physics its not physics

Quantum mechanics is physics

[trail tendon]

oh i thought you said mechanics LOL

[obsidian frost]

Prove That A F > D B
Btw F is the one in the red circle

[merry knot]

Split AF into FS and AS

[keen lake]

are there anything else?

[maiden notch]

when you draw that line it gets proven, since in a right triangle the longest line is always hypotenuse

[faint pasture]

is 3pi/2 considered a solution for tan(x) + sec(x) = 2cos(x) or no

like desmos says yes, but undefined =/= 0 though

[upper echo]

nope, tan(3pi/2) and sec(3pi/2) are undefined.

[faint pasture]

but like if i take the limit it does indeed say its equal

[upper echo]

yes the limits are equal. But you cannot plug in x=3pi/2 directly into this equation. It has no sense.

[faint pasture]

mmm

thanks

[rich scroll]

Find the apothem and the area of the given regular hexagon with radius of 3 cm. Draw the figure using the exact measurement of the given radius.

How do i find it with only the radius given? What i know ks the radius+ midpoint of a side= the apothem. But i only have i given PLS HELPPPP

[faint pasture]

Radius + midpoint of a side?

Length + point

Uh

What

[undone quest]

Yo can anyone tell me how to do approach in complex problems like of advanced level

[frosty rose]

can anyone explain this formula? what is m and where does this formula come from

[undone quest]

@frosty rose It is like point slope form as far i have read

[frosty rose]

oh

its for finding 2 tangents outside a circle connected by 1 point

[undone quest]

Brung that x thing to y and u are left with m

[frosty rose]

what is m?

[undone quest]

M is slope

Of a linr

[frosty rose]

ohh okay

thanks brother 🫶

[undone quest]

These are various ways in which u can write an eqn for a line

If you are new to cood

[trail tendon]

"complex problems like of advanced level"

[hybrid cosmos]

Love it that y=x intersects the unit circle at irrational points, but when u shift it one unit to the left – it intersects only rationals

(yes, I was lazy to do it by hand lol)

[obtuse tartan]

hi guys

i just started trig

can anyone help me?

[pulsar vale]

someone help

pls

[vital rain]

There is also a quick way to find the equation from a tangent to a circle if you have the equation of the circle and the coordinates of intersection:
the lets say the equation of thr cirle is x²+y²=64
Let's say the point intersected is(a,b)
x²+y²=r² is the same as xx+yy=r²
You sub in the x and y values so the equation of the tangent would be ax+by=64

This also works if the equation is something like this (x-1)²+(y+2)²=16 and let's say the point of intersection is (c,d)
It would be (x-1)(x-1)+(y+2)(y+2)=16
you then plug in the x and y coords and you get (c-1)(x-1)+(d+2)(y+2)=16 this would be easier to expand with actual values, try it see if it works out

[pulsar copper]

can somebody explain how this is correct

[soft linden]

what all should i study for 10th grade geometry

[frosty rose]

i will try it in class today and see how it checks out, thanks!

[brave crane]

You can prove that <5 + <6 = <1. The proof goes as follows:
We know that <1 + <4 = 180, so <1 = 180 - <4.
But <4 = 180 - (<5 + <6) since <4, <5, and <6 are the angles of the triangle WUV.
Now, replace <4 = 180 - (<5 + <6) in <1 = 180 - <4; this will give:
<1 = 180 - (180-(<5+<6)) = 180 - 180 + <5 + <6
Thus. <1 = <5 + <6.
Now... since <5 + <6 = <1, then both <5 and <6 must be less that <1.
Think about it in that way: 4 + 5 = 9, both 4 and 5 should be less than 9. In general, if x + y = z where x, y, and z are positive, then x and y are less than z.

[upper karma]

How do we prove the statement "If two lines are perpendicular then the product of their gradients is -1" ? i forgor

[obsidian harness]

Dot product?

[upper karma]

but we dont know the magnitudes?

oh

[obsidian harness]

WLOG you have (a, b) and (kb, -ka)

[upper karma]

WLOG?

[obsidian harness]

Without loss of generality

[upper karma]

how is it (kb,-ka)?

dot product between what btw ?

south?

[obsidian harness]

Those two

Basically you want any two vectors whose dot product is 0

[upper karma]

😅 im still lost

[compact trellis]

how can i shorten these statements and reasons?

[faint pasture]

Wasnt there a proof without this

Yea

Like we know angle between two line is

arctan((m1-m2)/(1+m1m2))

And if this is pi/2

Then 1+ m1m2 = 0

And m1m2 =-1

@upper karma

[upper karma]

brb

[merry knot]

you can also do this with just pythagorean theorem

$|BC| = \sqrt{m_2^2+1},;; |AB|=\sqrt{m_1^2+1}\implies (m_1-m_2)^2=m_2^2+1+m_1^2+1$

[somber coyote]

daniel

[merry knot]

so rearranging gives the desired result

[upper karma]

back

[upper karma]

ah yes

thanks!

[upper karma]

how?

[merry knot]

look at what i posted

[upper karma]

m1 and m2 are derivatives right?

how can they be lengths of sides?

[merry knot]

they're the slopes of the lines

[faint pasture]

Well technically they are also the derivatives

[merry knot]

because BD is 1, AD must be m_1 in order for the slope to be m_1, same for m_2

[faint pasture]

😭 the tilted axis made it look weird

[upper karma]

which ones are the tangents again?

[cyan hamlet]

Can someone solve this for me? It’s beginner physics PLEASE

a cannon ball is launched off of a 28 m cliff. it lands 22 m away and it takes 5.5 seconds. what is impact velocity with angle

[merry knot]

there aren't any tangents?

AB is perpendicular to BC

[faint pasture]

The horizontal velocity is 4 cuz 22/5.5

For vertical just use the second equation of motion

To get the initial vertical velocity

And then the first equation

To get the final velocity

[upper karma]

OHHHH

lol

[cyan hamlet]

I did all that and I think I got 32 m/s [7 degrees from the vertical]

Idk if it’s right

[upper karma]

the tilted axis did confuse me

[upper karma]

i get it now thanks

[faint pasture]

I so dont wanna do the calculations so you are on your own

[random otter]

can anyone help with me the value of k

,rccw

[tawdry arrow]

i could not do proofing

that stuff is wack

[serene pond]

Guys I'm stuck on something, can any line. intersect on any point on a line and it could be it's bisector?

technically if a line extends both sides infinitetly, then any point on the line could be a point of bisection right?

or is there some weird caveat?

[trail tendon]

I would think yes

i dont think theres any caveat

but idk

[serene pond]

i'm considering the length of expansion of the line given a point in time as the line extends at a constant rate

so one side of tge bisector could be like 3m and the other is 4m

but thats just a given point in time

as time resumes, the lengths will change

but im not smart so idk

[trail tendon]

like you have a given parameterized line and a given point, and you want a function that tells you the distance from the given point to any point on the line?

or idk what ur tryna do

[serene pond]

well say the line extends infitely

but before it does

the given line on the sheet of paper

which demonstrated finite in length

has line that perpendicularly intersects it

but to the observer it doesn't bisect the finite line in half

so in the beginning

the line starts with unequal length sides

you know what nevermind

I'm thankful I got that out

cuz now I look poopy

doesn't matter if both directions have infinite length

any point on a line will be a point of bisection

[rustic plover]

As I was saying what math is happening here 10th grade or 11th grade

[serene pond]

highschool geometry

so prob 9th grade

[rustic plover]

and I assume the people doing this is middle schoolers

[rustic plover]

oh same thing

[serene pond]

well its still pre-uni

[rustic plover]

Well I did it last year too

[serene pond]

did what

9th grade geometry?

[rustic plover]

Pre calc and geometry

[serene pond]

oh

[rain summit]

Can anyone help me with this im lowkey lost

[ornate path]

Your answer is the 1st part

Use pythagoras theorem

Since it's a square, DBC becomes a right triangle upon adding a diagonal

and since it's a square, the sides are equal

if the sides are x, then the diagonal, by pythagoras theorem will be:

x2 + x2 = DB2

2x2 = DB2

DB = x multiplied by square of 2

root of 2

which is in the 1st part

[dire mirage]

Pls help I've been solving for the past hour and I'm genuinely lost

[ornate path]

what do you have to find?

[dire mirage]

The area of the rectangle minus the area of the 3 circles

[amber cobalt]

The rectangle side is 4

Is it a square?

Ah yes it's a square

[dire mirage]

Nope its a rectangle

If it were a square I would've solved it hours ago

[flint ingot]

4(sqrt3 + 2) is the area of the rectangle then subtract area of circle which is just 3pi

[gilded forge]

are there any formulas for calculating the area of a regular polygon with side length and number of sides given

cus i can't find any online

[flint ingot]

Probably not

[gilded forge]

:(

i'll try to find one with my friends then

[flint ingot]

If there was then they would output something if you put in 1 as the amount of sides

Which doesn’t make sense

Idk

[gilded forge]

oh we found one

the output is undefined with 1 side and 0 for 2 sides

[random otter]

can someone tell me the addtion part ans

[upper echo]

2nd one is 4

[random otter]

k

[limber cliff]

can someone do the trig part

[obsidian harness]

start by writing the LHS of 4i in terms of sin and cos only

then use the Pythagorean identity

[steel hinge]

It is a rectangle. You can tell because the top of the bottom circle goes slightly above the bottoms of the top circles. As for finding the missing height, here's what to do. If you look carefully, you can see that if you draw lines from the center of each circle to the contact points of the other circles, you make an equilateral triangle. You can then cut the triangle in half for a right triangle to be formed, with a hypotenuse of length 2, and a short leg of length 1. You then do pythagorean theorem, to find the long legs length is sqrt(3). You also have to account for the upper radius of the top circles and the lower radius of the bottom circle. This ends up being 1+(sqrt(3))+1. This equals 2+sqrt(3), this being the height.
For the original question, you first multiply the height and width, being 4(2+sqrt(3))=8 + 4(sqrt(3))
You then subtract the area of the circles, adding up to 3pi:
8 + 4(sqrt(3)) - 3pi
That is your answer.

[steel hinge]

.

(I forgot to reply to the message)

If you're more of a visual learner, here's pictures:

[steel hinge]

While I was able to help with this, I am currently in algebra 1, so don't expect me to help here often.

[dire mirage]

U r a blessing from the lord above

[steel hinge]

You're welcome.

[steel hinge]

The first and last answers are correct. I will give an explanation shortly

[steel hinge]

Proof for the first answer

Another easier way to prove this, is by just multiplying the sides by sqrt(2) to see if that equals the length of the diagonal

[steel hinge]

Proof for the last answer

[obsidian harness]

you have good maths skills for only knowing up to algebra 1

keep it up bud

keep the effort and the energy up!

[steel hinge]

I am in algebra 1

I know a bit of trigonometry and geometry

[obsidian harness]

yeah like some people are a right mess, I'm guessing many others in your class

[obsidian harness]

damn

[steel hinge]

Like how cos is adjacent over hypotenuse

Sin is opposite over hypotenuse

[obsidian harness]

ah yeah you do cover that in alg 1 though

[steel hinge]

Ye

I'm also pretty good with i

And have a full on desmos account

[chrome token]

can someone tell a nice book for synthetic geometry? (or will this go in book reccs)

[smoky apex]

can anyone teach me this

[faint pasture]

Looks like inconsistent data or not enough info

[smoky apex]

oh

thats all my teacher gave

[faint pasture]

Yea id say inconsistent

Sum of all angles in a triangle = 180

150+120 + angle 2 = 180

Angle 2 = -90 degrees

Which doesnt make sense

[smoky apex]

oh damn

okay ill try and ask for more data

[dark sparrow]

What’s the formula for area of an irregular hexagon?

[merry knot]

Extremely ugly

It's probably easiest to just use the shoelace formula

[lyric notch]

Chat i fucking hate trigonometrics

I need help with this ( em means in)

[merry knot]

maybe the 150 and 120 are outside

i.e. the two red angles aren't the ones that are 150 and 120

[faint pasture]

Then it doesnt satisfy the isocelless triangle condituon

Actually wait

[merry knot]

one is one side and the other is on the other

[faint pasture]

Yea it does

My bad

Still tho

Ambiguous

If you gonna mark the angle and put a value nezt to it

Thats not the marked angle

Yea thats ambiguous

[merry knot]

yes

[faint vine]

Find the polar coardinate of (3,3)

[merry knot]

one of the many problems with geometry problems that don't define everything in words

[faint pasture]

I dont mind just figures but at least do it properly

[blazing raft]

@oblique pagoda

[oblique pagoda]

Hello
So uhhhh
My exercise is to derive the trigonometric theorems with the Euler relation
Everything on the top is…bullshit, don‘t bother with that because I think it was a mistake
I very feel like the result is just in front of my nose, but at the same time, I‘m not moving at all and wasting time
So it would be nice if you helped me out

[blazing raft]

Whoa that does look bullshit

@manic tusk help my gang member is struggling

[oblique pagoda]

I‘m not your GIRLFRIEND BROOO

[blazing raft]

LOL

[manic tusk]

Wtf

[merry knot]

You should use that $e^{ix}\cdot e^{iy}=e^{i(x+y)}$

[somber coyote]

daniel

[oblique pagoda]

That is what I did

[merry knot]

Sorry I didn't look at your work

[trail tendon]

e^iy = cos(y) + i sin(y) tho

i think is missing the i

[merry knot]

Yes there is a missing i

If you add the missing i and recalculate both sides, then equate real and imaginary parts, you get the result

[oblique pagoda]

Oh

I wish I had eyes

Thanks a lot! That must be revolutionary for my answer

[trail tendon]

I think you really wish you had... i's

[oblique pagoda]

BWHAHAAAHAHAHHAA

[blazing raft]

XDDDDDDDDDDDRIZZ

[light dock]

happens to everyone

[oblique pagoda]

You‘d be surprised how unconcentrated I can be
But yeah, you‘re right, thanks for the reminder

[dusty junco]

I can help u. Is it urgent? because I have to review trigonometry

[manic tusk]

,rotate

[somber coyote]

Couldn't find an attached image in the last 10 messages.

Nov 26, 2024, 11:09 PM

[manic tusk]

,rotate

[manic tusk]

@lyric notch whatve you tried

[junior wigeon]

how is k expressed in the format of x/k + d when the k in the function is a fraction

[manic tusk]

you divide by the fraction, so multiply by reciprocal
if k = 1/2 for ex, you have x/(1/2) = 2x

[junior wigeon]

so I have the function 7sin(1/5x) + 2, that would mean it would be 5x + 0?

[manic tusk]

x goes to 5x yeah

so you can do mapping rule

[junior wigeon]

okay thanks for the help

is there a way to solve this algebraically or should I graph it?

[manic tusk]

for a), the easiest thing you can do is find points a, and b where the function f is equal, so then the avg rate of change [f(b)-f(a)]/[b-a] = 0. like, you could find a point on the midline, and then add the period to it, and the function would be the same, so the avg rate of change is 0
then for b) you can go from the max to the min (or from the max to the middle), and for c) you can do the opposite

if thats confusing, this would be the visual of what im saying for each subquestion

[vapid gust]

hi

[ripe widget]

Bro this si brutally upsetting

[blazing raft]

Hi

[weak plaza]

can someone explain the steps for me?

[trail tendon]

do you know the dot product?

specifically, the two ways to compute it?

[weak plaza]

vw/|v||w|

problem is I cant figure out how to get |v| or |w|

it doesnt do a good job explaining

[neat yarrow]

pythagorean theorem. if you draw the vector out on a coordinate plane you'll see what i mean

[weak plaza]

yep that did it

[weak plaza]

can someone help me with this one too

im supposed to convert it to 11cos270 and 11sin270 right

[near topaz]

I can help you

[weak plaza]

i figured it out I was applying tan accidentally

[drifting zealot]

Does anyone have any tips for Proofs?

[sinful cobalt]

feeling dumb posting this here but im super behind n I gotta lock in but would it make sense to say angle c is reflexive?

[obsidian harness]

no, the reflexive property is that anything is equal to itself

what is true however is that angle ACB = angle DCE

cause vertically opposite angles have the same measure

[weary plume]

The graphs of trigonometric functions always appear to be constituting of consecutive rapid growth and decay. Can I model a given interval using a logarithmic or exponential function? Just out of curiosity.

[obsidian harness]

it is possible given a small enough interval of x

you should probably add together multiple exponential functions and logs though

https://www.desmos.com/calculator/bskysm1m6u

here's a nice example for sin x

[autumn surge]

A clock has 6 hands instead of 3, each moving at a different speed. Here are the speed values for each hand:

1: Moves forward by $\frac{x}{12}$ degrees each minute

2: Moves forward by $x^2$ degrees each minute

3: Moves backward by $x$ degrees each minute

4: Moves forward by $\frac{x}{2}$ degrees each first minute and $2x$ degrees each second minute

5: Moves forward by $x$ degrees each minute

6: Moves backward by $\sqrt{x+y}$ degrees each five minutes

We know that two of these hands are the real minutes and hours hands, but that there is no seconds hand.

$y$ is a prime number that is a possible value for minutes in a clock, e.g.: 59 works, but not 61.

At the start, the clock shows midnight, which is the actual time. After a certain amount of time, 4 hands meet in one one spot, while the other 2 meet in another spot.

Question: What is the actual time?

[somber coyote]

⛊Alpha™

[spark wasp]

Where i can find all of the geometry formulas

[wise pawn]

doesn't exist

[autumn surge]

yeah

[maiden brook]

thankfully

[weary plume]

Here's a better one

Logistic appeared to be the best option

[meager flax]

Wonder how will it be solved

[tame scarab]

Hey help me with this question please

Among all the rectangles with a perimeter of 20cm, determine the one with the maximum area

My result is 32cm

cause 4+4+4+8 =20cm perimeter

4*8 = 32cm²

[bright sedge]

That is not a rectangle though

4x4x4x8 is a quadrilateral but not a rectangle

It would look something like this which makes it a trapezoid

You can let w and l be width and length. The area is wl and 2w + 2l must equal 20 ( the perimeter ) so w + l = 10, so w = 10 - l. If you substitute this into the area, you get l(10 - l) or 10l - l^2

That is a quadratic that represents all l values that work and the y-value is area. If you graph it, it looks like this and you can see area is maximized at the peak

The length there is 5 and so the width is also 5. The maximum area with that perimeter is a 5cm by 5cm square, with an area of 25cm²

[trim ore]

A circle line is drawn around the triangle. One angle of a triangle is 50 degrees. Calculate the narrow angle between the bisectors of the other two angles.

I have been stuck on this problem and I cant figure it out. Could someone please help me with it?

[regal wedge]

draw out a picture to start

[stray cedar]

Someone teach me trigonometry

[regal wedge]

send a screen shot

[regal wedge]

specifics?

[bright sedge]

Do you know what the law of sines is?

What is the problem then?

Yes I am aware what it is. What do you need help with, if you know what it is too? Do you want me to walk you through a question or?

You can solve for AC

Okay so there's your answer

Also it should actually be ~14

24 * sin(22 degrees)/sin(40 degrees)

How did you get 31.6

Did you type it in wrong

[wise pawn]

stop spamming this please

[trim abyss]

hi

[upper karma]

<@&268886789983436800>

[drifting zealot]

Are there any tips for proofs?

HELP ME

[regal wedge]

show smth specific

[drifting zealot]

Are there any general tips or advice for writing proofs, specifically in a form with two columns. Statements and Reasons.

[regal wedge]

angle chasing is fun

congruency is useful

[grand tulip]

in 2 column proofs?

js remeber every little step needs writing so even if basically anybody with any level of math advanced beyond prealgebra can understand it you still have to write it

and dont stress about axioms when substituion can be aplied

[lyric halo]

The circles having radii A and B intersect orthogonally. The length of their common chord is:

@faint pasture

[faint pasture]

What was the condition for orthogonal intersections i forgot

2ff1 + 2gg1 = c+c1?

Actually wait

Im dumb we dont need that

[faint pasture]

C is the length of the common chord

[lyric halo]

niceee

[wise pawn]

,rotate

[lyric halo]

@faint pasture @wise pawn

[faint pasture]

These look oddly add and subtractable cuz of the symmetry but i tried that and it doesnt look like it took me anywhere

So i dont rrally have a clue

Wait a minute

Its just h1+h2= 0

Because of the family of curves thing

Ie all the intersection points of s1 and s2 also satisfy

S1 + Lambda S2 = 0

Lambda is just an arbitary constant

Doing that in here and subbing lambda = 1 gives us a circle as the curve if we consider only the x^2 and y^2 coefficients

And it would be a circle if coefficient of xy =0

I really need to revise circles

@lyric halo

[lyric halo]

i see

[wide lynx]

This identity holds, right?
$\sec(x+\pi\cdot k)=(-1)^k\cdot \sec(x)$

[somber coyote]

horizon2.0

[faint pasture]

Yea

[vestal shard]

Can someone help me to prove the angle ACD is equal to the angle HKB?

[dry junco]

Can anybody help me with this question?

[obsidian hornet]

AK-simedian of angle BKC by defenition, so, BKH=AKC and AKC=DCA couse tangent

[restive mauve]

Is it only me or as a beginner, I find really easy to remember sin, cos, tan?

[obsidian harness]

it's not only you

great, move onto more challenging stuff if you're ready

especially if you've already done a million right triangle find the side / angle problems, and it's making sense

[wise zodiac]

what is the volume of the largest cylinder that can fit inside a closed rectangular box measuring 12 inches by 10 inches by 8 inches?

What’s the justification for why the cylinder can’t be along the space diagonal of the rectangular box? is it just because it intuitively feels like it leaves empty space?

Like relative to keeping it horizontally or vertically

[faint pasture]

Lets see

Mods would like to see this probably

<@&268886789983436800>

[spark stag]

bizarre

[golden fiber]

Hello can sb help me with this task pleas?

I would be very grateful for any tips because Im strugling with this task for very long time

[golden fiber]

Sorry its he right drawing to this task

[upper echo]

This one is still not quite correct. M should be the midpoint of the longer arc BC and it looks like it is AC in your figure.

[upper echo]

If it's still relevant.
First, show that triangles BJN and BAG are similar (J is an intersection of AN and BE). Then EJN is similar to EAG, and then deduce that GAEN is a cyclic quadrilateral. Then similarly for the other side HAFN is cyclic. The rest is easy.

[azure helm]

how can I come up with the identity by myself

sin(2x) = 2cos(x)sin(x)

[cunning lion]

from the angle sum identity

[novel shadow]

Im gonna go insane i believe its 4.5(root thing) 3/9 but idk

[obsidian harness]

$\frac{\sqrt3}{2} = \frac{x}{9}$

[somber coyote]

southlander!

[obsidian harness]

it's just 4.5 sqrt(3)

[novel shadow]

Heck yeah im right then!!
Thx!!!

:DDDD

Found another one im struggling with (sorry)
The adjacent should be x(square root thing) 3
I dont know how to get x in this

The opposite should be just x

[obsidian harness]

$\frac{\sqrt3}{3} = \frac{x}{10}$

oh yes you need to rationalise

[somber coyote]

southlander!

[obsidian harness]

I originally wrote 1/sqrt(3) but that's equal to sqrt(3)/3 by rationalisation

[novel shadow]

Yea

Im so confused on this Q^Q

[obsidian harness]

$\frac{1}{\sqrt3} \frac{\sqrt3}{\sqrt3} = \frac{\sqrt3}{3}$

[somber coyote]

southlander!

[ionic thorn]

How to show Beta = 2 * alpha

so far, I know Beta = gamma + alpha

and Gamma = delta

[merry knot]

are the sun rays parallel to the x-axis?

[ionic thorn]

yes

[merry knot]

then delta = alpha by parallel lines

[ionic thorn]

is that a theorem?

[merry knot]

I guess

Idt it has a name

Corresponding angles are equal

[ionic thorn]

Ohhh I see

you do it from delta opposite and then transition to alpha

thank you!!

[latent crow]

paris pamfilos' lectures on euclidian geometry is ass

[merry knot]

Just read EGMO

[latent crow]

i dont watch elmo

[merry knot]

Your loss

Still, read EGMO

[latent crow]

he fell off after the episodes started repeating the same procedure

what is egmo

[merry knot]

Euclidean geometry in mathematical olympiads

[latent crow]

although my statement was one of general acknowledgement

i will consider your suggestion

[maiden brook]

i wish that book put a bit more effort into explaining things

i think aops does it better but its a lower level

[south elm]

HELP

My teacher sucks at teaching and I’m failing the class with a FINAL of 68

I’ve never failed so bad before

[acoustic field]

Guys help me figure this out: determine cos(2pi/3) using the ratio 30-60-90

[upper karma]

2pi/3 is 30°

[acoustic field]

okay thanks!

[loud ridge]

Do you have ur special angles memorized lol

[acoustic field]

yesnt

[obsidian harness]

assume there exists a triangle XYZ such that XY = 4, XZ = 8, and YZ <= 4

show this breaks the triangle inequality

hence by contradiction, YZ > 4

oh well YZ = 4 is a degenerate triangle, so the statement you have to prove is not technically correct

[upper karma]

I meant 60°

WAIT

NO

What have I done

[thorn cradle]

Does anyone know what this rule is called (if it has a name)

[obsidian harness]

this follows by similar triangles

$\frac{a}{b + c} = \frac{b}{a}$, (short leg) / (hypotenuse) constant

[somber coyote]

south

[obsidian harness]

$\frac{e}{b + c} = \frac{c}{e}$, (long leg) / (hypotenuse) constant

[somber coyote]

south

[obsidian harness]

$\frac{d}{c} = \frac{b}{d}$, (short leg) / (long leg) constant

[somber coyote]

south

[chrome token]

can someone give me a hint?
A circle is outscribed around the triangle ABC. Chords, from the middle of the arc AC to the middles of the arcs AB and BC, intersect sides [AB] and [BC] in the points D and E.
Prove that (DE) is parallel to (AC) and passes through the centre of the inscribed circle.

Ik the angle bisector thing(that it passes through middle of arc)

idk how to approach

[glossy pecan]

@obsidian harnesstysm

[obsidian harness]

no worries!!

[serene bay]

can someone here explain trigonometry real quick?

[obsidian harness]

https://www.youtube.com/watch?v=g8VCHoSk5_o&list=PL0o_zxa4K1BVCB8iCVCGOES9pEF6byTMT

if you have any questions about the videos you can ask here

[acoustic field]

Too late now i failed the test ive worked so hard to prep for (it was mainly on cos and sin, this was in the study guide)

[serene hedge]

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[light dock]

please don't advertise this here.

the server isn't really for advertising your paid tutoring services.

[twilit girder]

hi guys so we did a test recently and I got a different answer then my friend, he got 9 while I got 3.9 for X. I have a feeling I did something wrong can anyone correct me?

[exotic ore]

x= 9

use the side lengths

simply the porpotions

so do 77/21 = 11x +11/30

and then solve for x

77/21 x 30 = 110

110 - 11 = 99

11x = 99

x = 99/11

x= 9

@twilit girder

[twilit girder]

aw man

well thanks anyways

[exotic ore]

np

[twilit girder]

quick question though, how comes its 77/21 and 11x + 11/30 rather then what I wrote?

[exotic ore]

u did 11x + 11 /21

[twilit girder]

yeah

[exotic ore]

i think you kinda got the picture right in your head but you executed it wrong

its because that wasnt the porportional adjacent side

with ur image it kinda looks similar so i can see why u got confused

[twilit girder]

in the question the triangle was the opposite way so I switched it, my friend thinks that could of caused the mistake

[exotic ore]

yh

u confused it around

becuase if they are porportional means it has the same ratio to each other

so u divide it by each other to find x since they are similar

if a and b are proportional a/b is the same value

if u understand

[twilit girder]

yeah I see where its going

thank you 🙏

[exotic ore]

np

[indigo wadi]

Can someone help me with this question? Im not sure if the solution takes the form of a ± or not

sin(arcTan(1/x))

[tiny burrow]

yo what grade are u in

[twilit girder]

10

[tiny burrow]

I am in grsde 9 and I am working on that lol

any tips

It's pretty complicated for me

[twilit girder]

oh trigonometry?

[tiny burrow]

yes