geometry-and-trigonometry

[rocky plinth]

HELP MEEEE

[sturdy gull]

Hi

[obsidian harness]

find z first (consider triangle PSB)

then the alternate segment theorem allows you to find x directly

then you should know what to do to find y

[blissful viper]

How?

[obsidian harness]

get a list of circle theorems online and look through it

[blissful viper]

nvm I got it, thx

[sturdy gull]

(a)
x = 35°, y = 70°, z = 55°.

(b)
In triangle OQB, OQ ⊥ AB (radius to tangent).
∠PQB = 55°, so ∠QOB = 70° (central angle).
∠PRB equals ∠PQB (alternate segment theorem), so lines RB and PQ are parallel by corresponding angles.

(ii)
[No inequality given, unable to solve.]

[blissful viper]

you meant OB ⊥ AB right?

[obsidian harness]

please don't give full solutions

unless the person has shown their working

also the alternate segment theorem does not say that PRB is equal to PQB

[rocky plinth]

ohhhhhhh i didnt consider that i got it G thanks

[obsidian harness]

no worries!

[rocky plinth]

yesssir thanks homie ily

[viscid gorge]

thanks!

[upper karma]

hi

[pastel sedge]

PBS and sbr are triangles with one side substended by the diameter you can use this to solve the rest of the trinagle

[pastel sedge]

I don't think so

[obsidian harness]

yeah their reasoning for (b) is completely wrong

every line has an error

[rocky plinth]

are the answers for the values of x y and z wrong too?

because i got the same answers

[obsidian harness]

should be x = 35, y = 70, z = 35

[rocky plinth]

yeah i got x as 35 with the alt segment theorem
how would we prove it?

[obsidian harness]

the diagram already has a proof of the alternate segment theorem

given that PSB is a right angle, you can deduce that angle SPB = 90 - PBS = 90 - (90 - z) = z

but then by angles in the same segment, angle SPB = angle SRB = z

[rocky plinth]

ahh

[young canopy]

a

[pastel sedge]

Yeah same

[jade sierra]

anyone know?

i honestly been on this for like 10 mins by now

[wanton yacht]

What's the question/

[grave pond]

Just one pair of congruent triangles will do that.

[jade sierra]

which is?

[deft bluff]

Well, what do you think? We want to specifically prove that BE = CD, right?
What triangles contain those sides?

[jade sierra]

dbf cef

[grave pond]

Neither of those contains the entire BE.

(Or, for that matter, CD).

[jade sierra]

?

[grave pond]

Do you disagree?

[jade sierra]

i dont disagree but i dont see how that would help me to an answer

[grave pond]

I mean, there is only one visible triangle in the diagram that has BE as one of each sides.

[deft bluff]

If two pairs of triangles are congruent, that means each part of the triangles are congruent!

The point is, if we want to show that BE = CD, we should find two triangles that contain those two sides. That way, if we can prove the two triangles are congruent, it follows that these two sides are also congruent.

[jade sierra]

ye i think it was like cpctc or sum like that

i think the last part im suppose to be looking for is a side

but im not too sure which one

or the reason for it

[jade sierra]

since already got 2 angles

[grave pond]

Did you see this?

[deft bluff]

Ignore the question for a moment, we just want to know what are the two triangles.

If you can figure out what the triangles are, the reasoning/statements should follow.

[grave pond]

Neither of those two triangles has BE as a side.

[jade sierra]

oh

i just see it now

abe

aeb

[deft bluff]

Good! Well, AEB contains BE, but now we want another triangle that contains CD!

Any ideas?

[jade sierra]

acd

[deft bluff]

Good! Now if you want to be consistent, you should write it as ADC :)
Because AEB goes from A to E to B and ADC goes from A to D to C.

Now, are the triangles AEB and ADC congruent?

[jade sierra]

looks congruent

[deft bluff]

Sure but I don't think your teacher accepts the reasoning "it looks congruent" 😭

Is there a specific postulate used for proving congruency?

[jade sierra]

you mean like the sss sas all that?

[grave pond]

Yes.

[jade sierra]

fianly 😭

i got the rest

thanks

i originally thought i would have to prove dbf and ecf are congruent first then say cpctc

[deft bluff]

Yeaaaaa the problem is that these two triangles don't contain BE or CD, as Troposphere mentioned awhile ago.

Usually the first step when doing a proof question is, you need to sort of look for ways that directly relate to the thing you are finding.

That's why I said you need to look for triangles that contain BE.

And in general, if you have a question asking you to prove something = something, a good start would be to find a triangle or shape that actually contains that thing.

[hot apex]

👀 Hello, Im weak in geometry, i want to start from scratch but i don't know from where to start, can anybody help me to create a road map? my final Exam will be in Feb

[finite drum]

could any one explain to me the geometric mean theorem and how to solve problems with it 🥹 i think i know how to find y but i'm not sure about x

[vapid gust]

I’d use Pythagorean theorem idk

If you know y u can find x

[finite drum]

yeah i did but i have a quiz tmr and the teacher wants us to solve it using the geometric mean theoreom 💔

[vapid gust]

oof

lemme try to solve it

well i mean

geometric mean theorem is when you find the geometric mean of 5 and 12

so u solved that

I don’t think you can do anything else

[finite drum]

oh

that's it

😭 ???

[vapid gust]

ig

wait

yea

just do Pythagorean theorem

u used geometric mean theorem already

5^2 + sqrt of 60^2 =x^2 so x^2= 25 + 60

X = square root of 85

That’s it

[finite drum]

OHH okay then i see

THANK U

[vapid gust]

yw lol

[exotic fog]

how do i find the circumcenter of a triangle on a graph? i know it has to do with midpoints and perpendicular bisectors of each side intersecting in the middle to find the center, but each time i try to do it without a graph i seem to get it wrong.

[whole flame]

Find equation of any two perpendicular bisectors

Using slope intercept form

N solve 2 equations to get circumcentre

[exotic fog]

thanks

[obsidian harness]

there's also another way where the equation of a circle is $x^2 + y^2 + Dx + Ey + F = 0$

[somber coyote]

south

[obsidian harness]

for each of your three (x, y) coordinates, sub them into this equation

you will get 3 simultaneous equations involving D, E, F

then by completing the square, you will find that the centre is (-D/2, -E/2)

[whole flame]

Yepp considering the circumcentre of circumcircle

[whole flame]

👍

[slim plinth]

Also it will be $(x-h)^2+(y-k)^2=r^2$

[somber coyote]

Jake

[elder ice]

Hey guys from where to get good level questions for geometry class 9

Except rd

[limber atlas]

#1436381906965889035

[viscid gorge]

any ideas ?

[obsidian harness]

the two quadrilaterals are similar to each other (both have two right angles; corresponding angles)

oh in fact the largest quadrilateral is also similar to the other two (it's not, I got the wrong orientation)

[viscid gorge]

how to do it with triangles ?

we shouldn't use similar quadrilaterals

[queen venture]

yeah what are similar quad

[obsidian harness]

it's the same as any similar shape

the criterion we are using is that the 4 angles are the same in both quadrilaterals

[obsidian harness]

why not?

[viscid gorge]

cause we only got taught similar triangles

[obsidian harness]

is this a textbook problem then?

or did you randomly find it on the internet

[viscid gorge]

internet

[obsidian harness]

the internet doesn't limit itself to the school curriculum

in fact that's the whole point of competition maths and so on

[viscid gorge]

i mean my friend sent me it and said it's a good one

so i assumed it's based on knowledge we know

[drifting thorn]

Probably it just isnt posible to get the value of x without more information

[queen venture]

oh his friend trolled him

[obsidian harness]

one sec I'll draw it out

[obsidian harness]

this gives the answer away but

reflect the bigger quadrilateral from left-facing to right-facing

and then I just rotated the smaller quadrilateral counter-clockwise

[queen venture]

https://tenor.com/view/сигма-gif-3883804279153061520

[eager brook]

@everyone recommend me a book that has a butt ton of problems for me to solve and practice for trigonometry

[grave pond]

The horizontal line from the midpoint of the right side divides the slanted 10+4 line in half. So the green triangle is a 3:4:5 triangle.

[plucky nova]

how to find area of right angled triangle where one of sides is 31

[grave pond]

Knowing one side length is not enough to find the area of the entire triangle.

[hearty kite]

I came until here and I am stuck here :p

turns out I was going the wrong way

(I think)

[plucky nova]

ik but teacher told me that its posible

[plucky nova]

this side is not hypothenes

[grave pond]

Either you have additional information that you have not told us here, or the teacher is wrong.

[grave pond]

So the side length of the square is 14·(4/5), and we get
x = 14·(4/5) - 4·(4/5) - 10·(3/5) = 2

[plucky nova]

the best part that he told me that i dont even need to use trigonometry

[grave pond]

There must be information you have not told us about.

[plucky nova]

no

thats all information

[grave pond]

A right triangle with sides 31, 2, sqrt(965) has area 31.
A right triangle with sides 31, 4, sqrt(977) has area 62.
These triangles both match the single bit of information you have; so it is not possible to find the area knowing just the length of one of the sides.

[hearty kite]

If you know the angle, you can find it

31 = Hypotenuse * sinx
Adjacent = Hypotenuse * cosx
(31 * Adjacent)/2 = Area

If you don't know anything other than "One side is 31" you can't find it

[plucky nova]

one angle is 90

thhats it

[grave pond]

!xy

[lime crown]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[hearty kite]

So this iis the triangle we know?

[plucky nova]

yes

[hearty kite]

You can't find it

you need other information

(i.e.) is this triangle inside of a rectangle
are there any other triangles we know

[grave pond]

You can get it to have any area you want by setting the other leg to be (desired area)·2/31.

[hollow sand]

hey everyone, what is the best way to understand geometry? I'm trying to learn by textbook's

[hearty kite]

2 or 3 Dimentional understanding of objects/shapes

espacially triangles

[hollow sand]

I was think about triangles, has one post-it front of my desk saying "every kind of triangles"

thank you!!!!

[hearty kite]

you're welcome

[plush sorrel]

how do i know which one is a and b

[hearty kite]

you don't need to

[hearty kite]

c^2 = 9+100-3*10*2*cos34

cosine theorem

[plush sorrel]

ohh ok thanks

[hearty kite]

you're welcome

[plush sorrel]

im at 40.622 square root = c square root

i got 6.4 rounded and its wrong

[grave pond]

It would need to be at least 7 in order to satisfy the triangle equality.

Check your arithmetic that gave 40.622 again.

[hearty kite]

c^2 = 109-60*cos34

cos34 is approximately 0.8

(8/10)*60 = 48
109-48 = 61
61 is approximately c^2
c is approximately 8 (c < 8)

[slim plinth]

3-4-5 triangle

[verbal matrix]

Does anybody have any good websites/sources i can use to study for my trig final? Cumulative. Im already doing everything my prof has up, but I want more.

[hearty kite]

angles for a 3-4-5 right triangle is (in degrees) 37-53-90

[grave pond]

Approximately.

[slim plinth]

Khan

[slim plinth]

Really

[hearty kite]

Approxmately, yes

[slim plinth]

C ≈7.69

[rare agate]

does anyone know any good resources that are free for trigonometry and in depth for self learning?

[whole flame]

Which level?

[slender dragon]

A level

(if ykyk)

[real sentinel]

Books , you can google for one mate!

ask doubts here !

[slender dragon]

bro I was just trying to make a joke

i'm not even doing A level

[slender dragon]

joke for this

[real sentinel]

Not a joke channel brota ... Use #chill

[slender dragon]

bro, is it that deep...

[whole flame]

Use other channel

[next mantle]

@spring lion You do 🤨

[spring lion]

I do?

Nope it doesn't appear

[next mantle]

Bro you are literally in the channel

[spring lion]

No like

On my own

[pale sentinel]

it's also a shitty joke

[spring lion]

Okay I got access to pre uni

[ebon ice]

anybody know this?

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[ebon ice]

1

[real sentinel]

hmhm

[earnest wraith]

if u try drawing it it helps

[molten lichen]

Σ

[dim grail]

Did I draw it correctly?

[slim plinth]

Use V=Bh

[dim grail]

or are the edge lengths the edge of the diagonals?

[grave pond]

The slant edges are 12 m too (otherwise there's too little information to find the requested angle).

[dim grail]

I see

Thank you for the info!

Imma update my MS paint drawing

[grave pond]

(So it's really half of a regular octahedron, which if you squint the right way gives you the angle it asks for immediately without any calculation. But you're probably supposed to calculate).

[dim grail]

Yeah it does look like it

Isn't an octahedron extended though?

[grave pond]

You get an octahedron by taking two of these pyramids and gluing them together along the square sides.

[dim grail]

yeah

[grave pond]

The problem didn't ask for the volume, though.

[slim plinth]

What do they want us to do

[grave pond]

"Find the angle between a slant edge and a base diagonal".

[dim grail]

They want us to find the angle the diagonals that converge make

[slim plinth]

Is there a formula

You use slanted height formula

[grave pond]

We should probably see how @ebon ice himself reacts to the drawing before going too far ahead.

[slim plinth]

Ok

[dim grail]

I guess we can look at it from top down actually

it will be the midpoint of the square

[slim plinth]

Use $l=sqrt(h^2+(b/2)^$

[grave pond]

Not quite.

[somber coyote]

Jake
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dim grail]

Is it the exterior angle? or the alternate interior near the converge?

[grave pond]

We want the angle between the two green lines.

(The thin green line I've added is the "base diagonal").

But perhaps that was what you meant too?

[dim grail]

yeah

thats what I meant

btw nice drawing!

this is now a 2d problem btw

[grave pond]

Yeah, those two lines lie in a plane.

[grave pond]

It's even easier if you don't split it into two triangles by the altitude. :-)

[dim grail]

I think we can just use Pythag

and then just solve for all lengths

since sqrt(12 squared + 12 squared) = missing length that cuts through our square pyramid

and then just use sin/cos quotients to get the angles

which should lead us to 30/60 or 45/45 but I gotta go eat lunch at this point

[grave pond]

The shortcut is to notice that the red and blue triangles here are congruent by SSS.

[tight girder]

hello

I am in need of assistance

could someone be of help?

[dim grail]

I can try 😄

I can try*

[tight girder]

sure uh ill post it

i had to use google translate, the angle is v (not y). I need help with b)

The answer is -a/b, but why?

[stoic hearth]

cause the angle after 270degrees drops to the quadrant iv wich x in positive and y negative so tan becomes negative

[tight girder]

As I understand tan is y/x so normally it would be b/a

But why is it a/b? I also realize that the y is negative

But the flip I Dont follow

[stoic hearth]

i think it has to do for the (a,b) look at the exercise at the circle i think you flip this

so tan(270+u)=-a/b

[tight girder]

I think I get it now

[stoic hearth]

am not sure about the flip though

oh maybe because its getting under it

[tight girder]

The a becomes the y value for the "flipped" version

if that makes sense

[stoic hearth]

does that work for any value of a,b ?

[tight girder]

Im not sure

But i think im good now

Thanks

[stoic hearth]

np

[slim plinth]

What is A=1/2 aP

As a polygon

[wanton yacht]

What?

[dim grail]

triangle?

[slim plinth]

It’s a polygon

It’s a a something height x perimeter

[wanton yacht]

A = 1/2 aP?

I don't know what that means, lol

[dim grail]

height * perimeter

is wire frame cube

essentially

imma draw it in ms paint rn

[slim plinth]

The one that spells like Apocalypse

Same stem apo but I forgot the word

[dim grail]

this is height times perimeter essentially but the inside is empty

so a cube but not filled in my material

[slim plinth]

Ok

Ths

This

[trail tendon]

<@&268886789983436800>

[dim grail]

oh then just mult it by the number of lines of the polygon 4 * (19.6 in * 15 in)

[slim plinth]

And then I saw that person using Pythagorean theorem and sin,cos, tan

[rare agate]

just resources for trig, that cover precalc as well, I am self learning so there isn't a level I can say.

[plain bison]

how to solve

[dull blade]

Ooh that one's fun!!! I'm gonna ask a bunch of leading questions to let you figure it out yourself.

[plain bison]

no gng js give me the answer

[dull blade]

So first, what are you looking for? What are some things that would prove it's a rhombus?

[plain bison]

bruh

[dull blade]

Nah I'm not giving you the answer you have to figure it out yourself. I'm just giving you guiding hints

Step by step

✨

Fine. To prove it's a rhombus the sides all need to be equal length. Idk if you've learned any other properties yet

Draw the diagram out and mark the congruent angles

What triangles are congruent? What angles are congruent?

What triangles are right triangles, etc.

[plain bison]

ok

that works

[dull blade]

So you see two congruent right triangles, correct? And you know that the overall shape is a parallelogram

Prove the right triangles are congruent

Do that by finding congruent angles and side lengths.

Then you know that TY is congruent to TW

And since opposite sides of parallelograms are congruent, TW is congruent to YX and YT is congruent to XW

So all 4 sides are congruent and it's a parallelogram

Do you need any help proving the two right triangles congruent?

You use ASA

With the base of the right triangles, the right angle, and the shared angle

[dull blade]

So do you know how to solve the question now?

[frosty crypt]

Holaaa

Alguien que hable español?

Hellooo, something speak spanish? I need help!

[young crow]

hi I have a probably easy problem that I can't be bothered to solve myself

what's the circumradius of an n-simplex with unit inradius

[trail stump]

1/2 apothem times perimeter = Area

[polar crow]

yesterday i taught myself trig with triangles, i am not in a trig class (i'm in advanced algebra)
should i move to unit circles next? after that i don't know what

[polar crow]

sorry if that was a dumb question but i'm genuinely curious because this is my first time i'm learning a math topic for fun

[silent plank]

yes, you should learn unit circle

[modern dune]

help

eb is throwing me off

[whole flame]

Ok so the triangle is an isosceles triangle so use it's 2 angles congruent properly

[modern dune]

oh right

[whole flame]

Also use vertically opposite angles to find BCE

[modern dune]

thats just 44 right

[whole flame]

And find other angles of triangle by sum of angles= 180

[whole flame]

Yepp

[modern dune]

ohh

[whole flame]

And sum of ABE and EBC angles is 180 as they r complementary angles

[modern dune]

well how do i know that BCE goes onto the other side since EB is like at a weird angle

[whole flame]

I can share some documents related to it

[whole flame]

????

Whatt

[polar crow]

thank you

[modern dune]

broo idk how the 44 goes onto the other side

is it because it's a isoceles triangle?

[whole flame]

No

[whole flame]

This is a property

[modern dune]

no yea i know that but what am i supposed to do with the vertical angles

if it goes onto that side then how does it help me find the answer

[whole flame]

Consider CEB and EBC both x as they r equal

N find them

[modern dune]

is this because theyre isoceles

[whole flame]

Use sum of angles of triangle is 180

[modern dune]

68

[whole flame]

Yess the triangle is isosceles as 2 sides r equal

[modern dune]

oh so then

[whole flame]

N if 2 angles r equal then sides r also equal

It's the other way around the property

[modern dune]

oh ok

thank you

[whole flame]

Yepp

[whole flame]

👍

[modern dune]

i just have a lil question what do the letters mean

[polar crow]

side length

[modern dune]

like they're equal to each other?

[polar crow]

yeah, t=t, y=y
it's just telling you that the sides of the rectangle are equal

[modern dune]

ok thanks

[polar crow]

np 👍

[whole flame]

I believe these r not ur hw questions

What's ur age??

[modern dune]

i'm 15 dawg

[whole flame]

Okk

Any other doubts then do show ur work

Check the rules

[slim plinth]

I saw some the question is word problem

That they are a rectangle

But their length and width are the same

[fallow talon]

Hyperbolic geometry attack in Fist Of The North Star 2:

[dark sparrow]

where is the hyperbolic

[fallow talon]

Look at the pink net that the evil guy shoots from his hand.

In the show it is called Anryu Tenha. Anryu Tenha is a technique of the North Star Lapis Lazuli martial art that distorts the physical space around the opponent, causing the opponent to become disoriented.

[cunning lion]

it seems to be non-eucludean but idk about hyperbolic specifically

[winged mountain]

I'm about to do that lesson soon too!

We have the same textbook.

[hasty crescent]

Sorry wrong send

[upper karma]

Could somebody help me with some trig study I’m doing for the act?

[jovial tendon]

what should i study for my final for geometry

[grave pond]

Old exam sets are very often the best advice.

[slim plinth]

Yes

[obsidian harness]

hopefully when you were practicing through each topic, you only did all the odd-numbered questions or something

to refresh your memory, practice the questions you didn't do while you were learning that topic

start from the oldest topic and work your way forwards

once you have a decent grasp of all of the content, then you can go into practicing previous years' papers

otherwise those exam questions will seem like "class: 2 + 2 = 4; exam: if Johnny has an apple, what is the mass of the Sun?"

[grave pond]

About 10^29 times the age of the shepherd, of course.

[vagrant shard]

hello how does putting a one over a trigomentric sign like cos switch the adjacent and hypotenuse

[obsidian harness]

cos = a/h

so $\frac{1}{\frac{a}{h}} = \frac{1}{\frac{a}{h}} \cdot \frac{h}{h}$

[somber coyote]

south

[obsidian harness]

pls show your work

[queen venture]

my work? all i was able to find was the locus of the centre of circle C3 by using conditons of orthogonality

[obsidian harness]

right, what did you get for that?

[queen venture]

its shortcut is to find the radical axis of both cirlcles , so js subtract both the circles' equation

[obsidian harness]

no like do you have your work on paper

so that I can see the equations

[queen venture]

man its so negligible

[obsidian harness]

send it anyways

[queen venture]

js why

[obsidian harness]

so that it's easier to check

otherwise you're not going to get much help

[queen venture]

hmm i cant show my work on paper as i dont have my phone with me rn

is there some software to show calculations and stuff?

[obsidian harness]

https://www.desmos.com/calculator/irsxk8baxl

okay hopefully you got something like this

[silent plank]

multiple options,
latex if you know it
graph/type stuff up in desmos or geogebra
or draw in paint

[obsidian harness]

using $2g_1 g_2 + 2f_1 f_2 = c_1 + c_2$

[somber coyote]

south

[silent plank]

or a combination of the above

[queen venture]

it was an exam question so i couldnt graph it, i was trying to find as many relations as possible. I did make a rough sketch but i got stuck when i tried to use chord of contact with given midpoint(assuming h,k for the locus) formula, T = S1 as i didnt have the centre of circle C3

[obsidian harness]

https://www.vedantu.com/question-answer/locus-of-the-midpoint-of-the-chord-of-contact-class-12-maths-cbse-5f04962498c19458ec0ea637

[queen venture]

ty

[wary elm]

What is the minimum value of sum of intercepts made by a variable tangent of ellipse with semi major axis a and semi minor axis b on the coordinate axes

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[obsidian harness]

!help

[vague pivot]

​No Category:
  help Shows this message

Type !help command for more info on a command.
You can also type !help category for more info on a category.

[lime crown]

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

[obsidian harness]

you should show your work if you want help

[wary elm]

Oh

[obsidian harness]

let's say the tangent passes through (p, q)

where (p, q) is on the ellipse

immediately there's an equation for this tangent (think back to the tangent equation for a circle)

actually there's likely no closed form for this

https://math.stackexchange.com/questions/3167873/minimum-value-of-length-of-tangent-of-the-ellipse-fracx2a2-fracy2b
it's probably intended to be this

[elder ice]

Hey guys any suggestions on what questions to do for geometry which are good level for class 9th

[summer oracle]

can someone help me with this

[flat fox]

Just do
1/2(x+15)=3x solve for x

How well we know that 3x is half of x+15 based on the tick marks

[summer oracle]

wait im kinda confused on that

doesn’t that mean it’s congruent

[grave pond]

No, they're just similar (by a version of the SAS theorem for similarity).

[grave pond]

Since angle BAC = angle DAE and AB:AD = 1:2 = AC:AE, triangles ABC and ADE are similar.
So in particular BC:DE = 1:2 too.

[queen cliff]

could someone help me with trig identities

I hate it!!

Well I love it but it’s a little confusing

[lime crown]

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

[queen cliff]

How do you solve those type of equations tho?

[upper karma]

any1 recommend some books for inmo im beginner in geo

[queen venture]

do you have sl loney trigonometry?(its a book)

[obsidian harness]

it's easier if you send a question from your book

[wanton sinew]

the value of x is 3

well its pretty easy to solve these type of questions

[plucky sail]

basic proportionality, as the 3x side divides the traingle proportional we know its parallel to the third side therefore 2.3x = x+15, x=3, similarity not required

[silent plank]

that proportionality comes from similarity

[slim plinth]

What’s this formula

[bright grotto]

does anyone have any tips and tricks to be able to quickly prove trig identities

[plucky sail]

u dont need toprove it while answering if u just write the theorem, thts wht he asked

[pallid flower]

@low ocean sorry bit late but i can't get how will this prove bisection? What do we use?

[pallid flower]

oh wait, triangle BAD ~ BEC, if so then BAD = BEC. Hencee AD parallel to EC

[low ocean]

yup, also BAD=BEC

[pallid flower]

Then ACE = DAC

DAC = BAD

there we go

[low ocean]

yeah

[pallid flower]

thank you a bunch once again, i prolly wouldnt figure this all out

[low ocean]

np

[main sphinx]

screw radians

radians were made by the devil

[slim plinth]

Hello

[sharp wigeon]

excuse me how do I know whether to put the 2 in front of k"pi"

whether I do 2kPI or just kPI

and whether it has a second solution?

or just one

[silent plank]

depends on what trig function you have

[sharp wigeon]

yeah thats kind of the question

how to I know which to use it on

or a rundown of there variables would help like why or why not put the 2k instead of just k

[proven parcel]

is this a good place to ask a question about area moment of inertia?

[wanton yacht]

Area moment of inertia?

That's more physics, lol

Wait, haven't I talked to you before?

Yeah, weren't you the guy asking about integrals?

[slim plinth]

How about conservation of momentum

[wanton yacht]

What about it

po = p, Lo = L

[slim plinth]

I learned that btw

[wanton yacht]

Nice

[slim plinth]

That states that momentum cannot be created or destroyed

That was part of my teachers notes

But what’s the original

[wanton yacht]

The original?

I mean, the statement is just that if there are no external forces on a system, then both linear and angular momentum is conserved.

[slim plinth]

Yea

So then the statement for energy is just that if there are no external forces on a system, then both linear and angular energy is conserved.

[wanton yacht]

Well, yea

I mean

Yeah

Check this otu https://www.youtube.com/watch?v=vWVZ6APXM4w

[slim plinth]

Conservation of energy

[wanton yacht]

Yeah? So what's the answer?

Same height, higher, or lower?

[slim plinth]

Higher

[proven parcel]

Yeah I just haven't found any physics/engineering discords that are really active like this one, and no I've only ever asked one question in this discord before and it was in #multivariable-calculus about maple and partial derivatives

[wanton yacht]

Same height

[slim plinth]

Ok

[wanton yacht]

Oh, maybe I'm thinking of someone else then.

Well, I would probably ask about area moment of inertia in either #multivariable-calculus or #calculus

Definitely not here

[proven parcel]

okay thanks

[past path]

What should I study to understand this better? The book isn't really helping me much. 😔

Like what specific lesson is this

[obsidian harness]

are you having trouble with the algebraic equations you get?

[past path]

Not really

I like algebra

[obsidian harness]

or do you literally not know that $m(CD)$ is angle $COD$

[past path]

It's geometry I'm not that good with

[somber coyote]

south

[past path]

I know that too

[obsidian harness]

questions A, B, C only use "sum of angles on a straight line is 180"

[past path]

Okay thank you

[obsidian harness]

no worries!

[vestal comet]

angle segment rules and circle theorums

[topaz hill]

hi

[upper karma]

.

[obsidian hornet]

Books:

1. Lemmas in Olympiad Geometry (Sam Korsky...);
2. Euclidean Geometry in Mathematical Olympiads (Evan Chen);
3. Geometry revisited (Coxeter...).
   Sources:
4. AOPS;
5. Sharygin geometry olympiad (geometry.ru);
6. IGO (Iranian Geometry Olympiad);
7. GoGeometry.

[upper karma]

ty

[toxic anvil]

Bruh, I need help on this

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[toxic anvil]

1

[exotic yarrow]

By definition, what is perimeter? Use that to set up an equation (note they seem to give you lots of information about triangle DPH).

[toxic anvil]

Ok

[wraith meadow]

Can anyone help me solve for angle A? I am lost.... Both sides are symetrical. Don't know where to start.

[grave pond]

You can find two sides of this right triangle and use trigonometry.

[wraith meadow]

I'll give it a shot

Bro.... that made every so easy.... That was the only triangle I didn't try or even see. THANK YOU! I was going crazy

[summer plaza]

solve this for me twin

i was using ai ngl and i dont think ts right

gotta be a different way to type ts

[midnight sable]

💀 u=16

[summer plaza]

do i js put 3 for the v 2nd part?

[midnight sable]

yea

[summer plaza]

ty man 😭

[tropic fox]

Dude no?

Shouldn't u be 16/3*√3

Or 16/✓3

[midnight sable]

no

[summer plaza]

this mf gon know i used ai 😭

[feral lichen]

Wym it is 16

[tropic fox]

How

[feral lichen]

It's 8/cos(60)

[midnight sable]

16/8 = sin90/sin30

[tropic fox]

Shouldnt be be u sin60 and u cos 30

[midnight sable]

or this

[tropic fox]

My bad

[summer plaza]

yall recommend any way to get better at geometry

?

[tropic fox]

V is 8√3 right?

[summer plaza]

this was right

[summer plaza]

and yes

[feral lichen]

Practice

[feral lichen]

So you can see patterns between questions easier

[tropic fox]

Yes before i messed up the angle i wrote 60 as 30 and 30 as 60

Ai must have done the same mistake

[summer plaza]

why are yall not teachers 😭

[feral lichen]

No where near qualified

[summer plaza]

yeah sure......

[tropic fox]

Real

[summer plaza]

in what way

smarts wise or like , age n shit

[feral lichen]

Prolly both, depends on what level but teachers gotta have some sort of degree in maths before teaching it inni

Like one of my maths teacher got a masters and the other got a PhD in physics

[summer plaza]

oml bro get out of here

[harsh hamlet]

All you need is a degree that is somewhat mathematical and then get the teacher's stuff done

My former maths teacher has a degree in engineering, and she teaches up to higher maths. So you don't really need a PhD or Masters

Although you didn't say one was needed

[elfin trail]

Isnt this just tan(60) = 8/v ?

I mean

use calculactor?

[feral lichen]

As long as they can actually teach then it's ight

[feral lichen]

😂

[ebon ice]

So I did 12^2 + 12^2=288 and then square rooted is 16.9 cm

now im stuck so help

[wary summit]

Is chat gpt trippin?

[ebon ice]

yea sometimes it makes stuff up

[wary summit]

Its the pro one too

Second time it messed up and I caught on today

[tender arch]

bro forgor how to math

[nova spruce]

he's been through much, cut him slack

[orchid rampart]

Physics pov

[slow pelican]

I suck at geometry as the fornules dont make any sense

[jade sierra]

is this possible to prove

i dotn think so

i think its AAS or ASA but im not sure where the side is suppose to be

might be this

not sure

[grave pond]

It's not even true -- for example if MATH is a square, MT is sqrt(2) times the length of HT.

Wait a minute, the diagram doesn't show MT and HT intersecting at B like the text claims; they intersect at T!

Perhaps there's a typo in the statement to be proved too.

Probably the goal should have been |MH| = |TA|.

[cold spear]

Huh

How do they intersect at T

[grave pond]

T is the endpoint of each of them.

[cold spear]

Oh nvm

Im dumb

[cold spear]

Yeah

[cold spear]

But they do intersect at B

[grave pond]

"MT bisects HA" so |HB|=|BA|.

[cold spear]

If it gets touched it is a intersect

AH NO

NVM

I WAS MISSUNDERSTANDING

[grave pond]

B does not lie on HT.

[cold spear]

With A

[cold spear]

IK

MB

I tought it were MT and HA

But nvm

[grave pond]

That must be what they meant, but it's not what they wrote.

[cold spear]

So it aint true

Its a tricky question

[upper karma]

Is it asking to you prove MB + BT ≅ HT?

[jade sierra]

mt congruent to ht

[upper karma]

Yep but MT is MB + BT

[upper karma]

I'm with you

[upper karma]

Every time I try to create a form to solve my brains melt. Have you figured it out?
It's really seems to be wrong 😕

[jade sierra]

I think it is AAS and then just cpctc

You have vertical angle and alternate interior angles for sure

And since mt bisects ha, hb and ba is congruent sides

[jaunty jasper]

Could I get help with this? I genuinely cannot figure out where to go with this diagram

[exotic yarrow]

recall that kites have perpendicular diagonals

this should clue you into the ||pythagorean theorem||

[jaunty jasper]

oh my gosh i just realized this whole time ive been looking at it and havent seen the 10 😭

[exotic yarrow]

rip

[wicked garnet]

if cosx+cos²x+cos³x=1, what is sin⁶x-4sin⁴x+8sin²x?

[fervent stump]

I would try finding sin^2 x first. Think about how you can do this from the given equation

[wicked garnet]

nvm I solved it

😅

[cold spear]

did you resolve it?

[digital stone]

:0

geometry dash

[cold spear]

.-.

[digital stone]

i havent learn what is on the image thing yet

WHY IS THEREE APPROXIMATE SYMBOL BUT ONLY ONE LINE

gone D:

geometry crash

[cold spear]

it means its the same as the top one

what grade are you in btw?

[digital stone]

im on grad 7

[cold spear]

you like math?

[digital stone]

YA

[cold spear]

alr

[digital stone]

i did able to solve a 9th grade math

[cold spear]

you want any help on somth?

[digital stone]

no

[cold spear]

alr

[digital stone]

but i like math and geometry dash

:)))

gone again

[slim plinth]

Will u use sin^2(x)+cos^2(x)=1

[civic cobalt]

Can anyone help me

[solid sorrel]

With ?

[civic cobalt]

Trigonometry ig

Or geometry

Idk the diff

[solid sorrel]

Uhh ok

So trigonometry involves right angled triangles and canculation of various things based on that

Geometry is just , all the shapes in general n calculations related to that

[civic cobalt]

Theres a exercise i have to understand for exam tmrw

Its cooking me

Can u help me its not a very big one

[solid sorrel]

I can but i am only in high school , so i gotta know the level of difficulty

[civic cobalt]

Im in highschool too

Last garde

[solid sorrel]

So u gotta tell me /show me the q first

[civic cobalt]

Can u go dms

[solid sorrel]

Aight!

[plucky kelp]

INSIDE/ON a unit circle:
Sin is the absolute value of the vertical line the goes from a point on the x axis to a point on the circle's perimeter directly above.
Cos is sin but flipped 90°
Tan is sin/cos
In radians, pi is half a circle. 1/n of the circle's perimeter is 2pi/n in radians.
That's all i know about trig from 3b1b currently. Can anyone correct me?

[jaunty jasper]

yes and I got 6

[worthy eagle]

sin is the value of vertical line that goes either above or below. It can be positive when above and negative when below.

It's better to understand cos as the value of horizontal line that goes from the vertical line's base to the origin. Cos is sin but flipped is also correct.

1/n of a "unit" circle's perimeter is 2pi/n in radians. Unit circle means the circle whose radius is 1. If radius isn't 1, then 1/n of circle's perimeter is 2pi/n × radius in radians.

[queen shadow]

they are like very different but id say that geometry helps you understand trigonometry well

[dim grail]

good job!

That looks tough

[dapper blade]

How does the figure for this one look like?

The slant height is the radius right?

[worthy eagle]

yes

[spring lion]

Can someone help me with rotating shapes?

I'm doing high school geometry on Khan academy right now

And I have no idea what to do

[hollow lily]

!da2a

[lime crown]

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

[spring lion]

Oh

Okay my bad

Let me ask

Basically questions like this. It could be 180°, 270° etc etc.

I did watch the explanation video but I'm still confused

[obsidian harness]

is this 90 degrees clockwise or anticlockwise?

[spring lion]

Counterclockwise

Right?

[obsidian harness]

I don't know actually

I assume they mean counterclockwise but this is bad wording

[spring lion]

Well

When it's +

It's counterclockwise

[obsidian harness]

yes that's the convention

anyways, here's the idea

[spring lion]

Hmmm

[obsidian harness]

you swap the x and y coordinates

so (x, y) = (-2, 1) would become (1, -2)....
except you have to check the quadrant, so that's why it's (-1, -2)

[spring lion]

OH WAIT

Does it apply to

180-270 too?

[obsidian harness]

not 180

but 90 and 270

[spring lion]

Oh

[obsidian harness]

180 is the easiest one: (x, y) goes to (-x, -y)

[spring lion]

Okay

What about -180?

[obsidian harness]

180 is the same as -180

[spring lion]

Huh

Oh

YEAHHH

I understand

Okay I'll try to do the question now.

[obsidian harness]

no worries! have a try

[spring lion]

90 degrees

So

(x, y)->(y, x) but it's first quadrant so

X is positive y is negative?

Is that it?

🧐

(-y, x)?

I'll try!

Damn I failed but that's ok I'll try again.

[obsidian harness]

(x, y) to (-y, x) yes

or it can also be (-x, y)

[spring lion]

Ohhhh okay I realized why I got it wrong

In the last point I made a small mistake

Ok ok

I got it right!!

😭 THANKSSSSS

[obsidian harness]

no worries!

<@&268886789983436800>

[spring lion]

I did a 4/4 god bless you bro 😭

[digital stone]

:0

[crude rain]

Geometry dash???

[fallen burrow]

Hi, I'm currently in upper high school and we're doing algebraic trigonometry.

I understand most stuff, except demonstrations or comprovations.

They give me an expression like this and I need to simplify it and express it in similar ways until I find that both sides are equal (1=1)

[warm shuttle]

I also wonder that ^ I have no clue how to do these systematically

I suppose besides just applying complex exponential definition, and hoping algebra takes care of the rest

[tight grove]

if you're just trying to prove the left = the right try using the cosine sum and difference identities to expand the left hand side and then recognize the difference of squares

i can write it out in a sec

[warm shuttle]

Like sure, you can show how to solve a particular example

But that doesn't give me a method

[unique lynx]

so first of you want to take it apart so:
(cos(a)+cos(b))(cos(a)-cos(b))=cos²(b)-sin²(a)
that would be in other words:
(cos(a)cos(b)-sin(a)sin(b))=cos²(b)-sin²(a)
and that would again be:
((1-sin²(a))(1-sin²(b))-sin²(a)sin²(b))
and that is
1-sin²(b)-sin²(a)+sin²(a)sin²(b)-sin²(a)sin²(b)
and that is:
1-sin²(b)-sin²(a)
and that is
cos²(b)-sin²(a)

cos²(b)-sin²(a)
-# I hate my life

-# if there is an issue my explanation is that I am in 9th grade

[tight grove]

most of these in my experience are using other identities or occassionally its easier to use the unit circle or a triangle to prove them

[warm shuttle]

I think trig might just be the first non-trivial piece of mathematics taught in schools? Like that's the image I'm getting at least

[tight grove]

its definitely the first one that most people get stuck on

[warm shuttle]

maybe square roots. But somehow schools bypass those being an absolute pain in the quack

[tight grove]

i never really had to prove identities aside from a math competition when i was a senior in high school but trig as a whole is pretty easy to get stuck on. combining all the algebra in with the geometry makes it easy to trip up on

[grave pond]

It would feel more direct to me to start by applying the product identity to the product of cosines on the left.

That separates the a's and b's immediately.

[warm shuttle]

I think trig is just more algebra? But it's dealing with special functions, so it gets tricky

[unique lynx]

I will try to do that the next time

[fallen burrow]

exactly

that's where the problem is

I learn about specific examples but nothing that puts it together

[warm shuttle]

I finished school ~6 years ago, but I still have no clue

[grave pond]

There's not supposed to be a "method" as such other than trying things to see if they work (and having enough experience to be able to tell quickly what is likely to work for an expression that looks like such-and-such).

[fallen burrow]

and I end up resolving for a specific set of values rather than equate two expressions

[fallen burrow]

LOL

[warm shuttle]

Like really? I know there are some situations in mathematics that really do require unique insight, but trig of all things?

And it's not like you are given a week on a problem in a test

[grave pond]

Unfortunately school tends to spend so much effort on teaching methods that it's easy for students to leave with the impression that mathematics is about knowing all the cookbook methods and identifying which one to use for which problem.

[wanton yacht]

this is the problem with math classes

[unique lynx]

-# I have no experience in that section (I just did the expression with sin cos tan the first time but I know the base of trigonometry and the technical aspect wich is why I could even do that)
-# still took me forever

[fallen burrow]

the problem is what he says: they don't give you a week to slowly think about it, they give you an hour and with pressure

[wanton yacht]

It's not about problem solving skills at all

I mean, maybe some. Good ones will be

[fallen burrow]

so often the solution ends up being teaching cookbooks

But anyway, I made this document kind of explaining it to myself (it's automatically translated to English), but I think there are more identities that I must know?
If you have time, would you please look it up?

[grave pond]

This one doesn't take a week to try a few different things. An hour should be more than sufficient unless you squander all of it by panicking about not having a cookbook method to follow.

[warm shuttle]

Like four function algebra over rationals is a cookbook pretty much. There are certainly some questions you can ask, which don't have cookbook answers, but those are also quite recognizable.

[fallen burrow]

oh the translation fucked up the document

Is this all?

[warm shuttle]

In my experience there are about way too many identities to remember

[grave pond]

Yeah, it's useful to have a table of them to consult.

[warm shuttle]

Something like 200

[grave pond]

(And I'll stick my neck out and say that school systems that expect students to remember all the standard identities are Doing It Wrong).

[warm shuttle]

Obviously infinitely many, but like you get what I mean

[fallen burrow]

well, not all the 200 but knowing some 10 or 20 helps

specially the 5-8 most important ones

if it's teached well, of course

[warm shuttle]

The stuff I know is like cos^2 + sin^2 = 1, e^(ix) = cos(x) + i sin(x), tan = sin/cos

I knew more in school tho

[grave pond]

Yeah, agreed. Anything beyond that it seems to be a better use of one's time to look up as needed rather than deliberately memorize.

(If looking it up frequently means you end up remembering some of them, then fine, that's a bonus).

[warm shuttle]

Still, I am really surpriside there isn't an algorithm for trig identities

I mean there are clear algorithms for integration and combinatorial identities

It bothers me greatly

I have a feeling it exists, and I just don't know it.

[grave pond]

Algorithms for integration? Huh?

[fallen burrow]

this is something I experienced while learning programming as a hobby! You end up memorizing things, when you continuously look them up and use them. Not as a deliberate exercise!!

[warm shuttle]

Well, there are commonly taught "methods", which are practical by hand. But beyond that there is Risch's algorithm

Obviously methods of integration aren't algorithms, but they are at least somewhat systematic

[grave pond]

Yes, Risch's algorithm exists but runs to hundreds of pages.

[warm shuttle]

But it exists

That's... kind of a big deal

[grave pond]

If that's the limit for what you'll accept, than undoubtedly one could concoct something for bringing trigonometric expressions to a standard form that works within such generous bounds. But for the kind of identities that show up in "prove this" exercises, it's unlikely to be an improvement over trying things by hand -- especially since the point of those exercises is not to find the answer but to train willingness to try different things and see what works.

[fallen burrow]

Dalton is in the chat? Oh wow that's such a pleasure to meet the founder of the modern Atom model...

[tight grove]

even if there was a "method/algorithm" id have to imagine due to the natural of transcendental functions that it would be so complicated it would be far easier to just recognize the patterns from the handful of identities you know off the top of your head or from ur reference to a formula sheet

[warm shuttle]

Yes, but it really does bother me that there apparently isn't, when there are methods for integration and combinatorial identities

[warm shuttle]

it is to find the answer tho

It even says that in the problem statement

(usually)

[grave pond]

That's what the problem statement says, but it is not the purpose of assigning the exercise.

[tribal rose]

I'm confused by this. -0.23 can't be an answer because it's referring to time, so which part of the equation do I look at?

[grave pond]

"After she starts measuring" must mean that we're looking for the smallest positive t that gives D(t)=50.
So -0.23 is not a correct answer even though D(-0.23) is indeed 50.

[tribal rose]

Hmmm...

Let me think...

Okay, so sin(pi) = 0, so maybe t = 0.77?

Someone correct me on that if I'm wrong.

Nevermind, that's right. Thank you so much, @grave pond !

[fading plover]

Bro I thought this would help with my geometry bit I swear I'm seeing matrix codes rn

[vital knoll]

Can no 11? Idk 11 means

[fading plover]

B

I think its 7

[vital knoll]

How u know?

[fading plover]

I don't I just think it's 7 cause like 8 is too high and d/e are the same

[vital knoll]

Oh sorry d 0.5

[digital stone]

GEOMEYRY DASHH

[whole flame]

B)7. Just a hinge based on observations

[tired bramble]

guys is there any VC channels for geometry?

[cunning lion]

there are no public VC channels in this server.

[pearl osprey]

ppl can you explain me why the area of this triangle is always the same while the little square is less or equal than the big one? without limits pls i'm asked for euclidean geometry

[grave pond]

O lies on a line through B that is parallel to AC, so taking AC as the base always gives the triangle the same height.

[vital knoll]

High of lamp its 1.5m (6 floor from lamp-4.5 shadow)
So, lamp <1m (asume) and 0.5m huonge. I don't know how next step

[dark sparrow]

<@&268886789983436800>

[civic trout]

hello, i have an ICGSE exam in 2 days and im kinda cooked. Some topics are easy but others are a bit harder for me. Dose anyone know if i could get some help

[silent plank]

just ask #❓how-to-get-help

[tribal rose]

Does this look correct? I'm usually worse at graphing cos(x) than sin(x).

[exotic yarrow]

,w 3cos(2x+6pi)+4, x=pi/2

[somber coyote]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[tribal rose]

Oh... so it has to equal 1 in the green area.

I tried moving the maximum point, and it seems to equal 1 at pi/2.

Wait... it doesn't...

Yeah, I may need to take this to one of the help channels.

Thanks for pointing that out, though!

[dusky granite]

I seek to sharpen my trigonometry skills but lack any practice problems.Any know of a way to generate practice problems?

[spice nebula]

https://mathematicalolympiads.wordpress.com/wp-content/uploads/2012/08/103-trigonometry-problems-titu-andreescu-zuming-feng.pdf

103 trigonometry problems

[dusky granite]

thanks

[spice nebula]

i've never personally used this book but from what i know they are pretty hard, but solving them will definitely increase your understanding

[vapid portal]

titu andrescu it might be cooked 💀

[elfin saddle]

oh i remember being stuck in the sinusoidal graph part for a month

[queen venture]

at (x=0) the value of function would be 3sin 2pi/3

you could also find the value of x for which the sin function becomes zero and shift the graph to that x

horizontally

thats x = -2

so shift the whole graph to the left by 2 units

also to account for the -3, shift the whole down by 3 units

[versed dragon]

Is there an ez way to memorise the cosines and sines plz i need this🥀💔😭

[dark sparrow]

well there is the "trigonometric hand" thing

this what you're looking for? @versed dragon

[versed dragon]

Ye ty

[simple vigil]

Do people need this 😭
Just remember sin values and all done

[dark sparrow]

i mean different ppl remember things in different ways

having it laid out visually helps some

[simple vigil]

Yes it works for some people

[queen shadow]

Tan is infinity or undefined

And can someone tell the difference in the two

???

[warm shuttle]

it's sin(x)/cos(x), so when cos(x) is 0 you are dividing by 0

So it's certainly not a real number, but if you imagine you could "wrap" the graph around, and then it's actually connected

(where it would be projective infinity)

[queen shadow]

Actually some of the textbooks use infinity while some use not defined so which is correct and can both be same ??

[warm shuttle]

I think textbooks saying infinity are more thinking about it in terms of limits

[silent plank]

depends on the exact wording/context

[silent plank]

that's missing a few words/numbers

[queen shadow]

Means??

[silent plank]

Tan is infinity or undefined
makes no sense

[queen shadow]

Tan 90

My bad 😭

[silent plank]

ok, then that's undefined

[queen shadow]

Is it different (infinity and undefined)

[silent plank]

yes

[drifting citrus]

Does anyone have recommendations on resources for trig? I'll be doing calc 1 soon and haven't been in school for years. Doing college algebra next semester so I would like to freshen up on trig during the break!

[nimble cypress]

what the sigma

🗿

[fossil glacier]

What if i spread my fingers too much

then it wouldnt be 90 degrees

jk

very smart

[sly urchin]

Given triangle XYZ in the first quarter of the plane, G is the centroid, X_1, Y_1, Z_1 are the midpoints of YZ,ZX,XY.
Prove 2(OX_1+OY_1+OZ_1)-(OX+OY+OZ)≥3(√5-1)/√2 . OG

Equality occurs when XYZ is an isosceles right triangle

[whole flame]

It's actually undefined

As it's sin90/cos90= 1/0 which is a weird mathematical operation so its not defined like no actual value

But as rhl and LHL or asymptotes tend to - and + infinity

It's actually associated with infinity

[upper echo]

I think that's not correct. Let X=(1;0), Y=(0;1), Z=(2;0). Then LHS<RHS.

[sly urchin]

Well it's my geometric interpretation of an inequality, maybe I did that incorrectly

#help-6 message

Oh I see what i did wrong

X(a,b), Y(b,c), Z(c,a)

[vale iron]

help me with euler lol

like

[elfin iris]

how do i get better at solving trigonometry identities questions, it takes me forever to make LHS = RHS, and that also by trying 50 different things

[wanton yacht]

Kind of just practice

Just try to get everything in terms of one kind of trig function. So if you have a bunch of sines, cosines, or tangents, try to rewrite all in terms of say sine, or maybe all in terms of cosine, depending on which is easier

Do you have any examples you'd want to go over?

[elfin iris]

i will try this, thank you

[elfin iris]

i will let you know if i get stuck on any problem

[silent plank]

most of them involve
conjugates / pythagrean identity
or compound angle

start with the side that looks uglier/more complex

[dense current]

anyone can help me with pre calc 12 trignometry?

how do i remeber all those special angles in trig

[dark sparrow]

there's this thing that someone else asked for just yday

that gets you the specials in quadrant 1

@dense current dunno if you saw this but theres a mnemonic for you if you want

[warm shuttle]

One thing that I noticed is that special angles are exactly angles that are of the form rational number times half-pi, where the ratinal number has denominator at most 3

I think that denominator determines the algebraic nature of the cosine and sine values

Which is bizzare

[willow glacier]

Can someone double check my work?

[dense current]

lol what does those 0123 mean?

[dark sparrow]

didn't i write it in the image

[dense current]

whats sqrt?

@dark sparrow

[dense current]

euler's number?

[dark sparrow]

square root

[dense current]

oh i get it now, wait thats actually genius

did you invent that?

[dark sparrow]

no

i don't remember where i first saw it though

[dense current]

i have a question tho, for 45 degrees, and if you do root 2 over 2, but isnt it 1 over root 2?

[dark sparrow]

same shit

$\frac{\sqrt{2}}{2}$ and $\frac{1}{\sqrt{2}}$ are equal

[somber coyote]

Ann

[dense current]

bruh rly

[dark sparrow]

yes

[dense current]

damn i am stupid

mbmb

[dark sparrow]

$\frac{\sqrt{a}}{a}$ and $\frac{1}{\sqrt{a}}$ are equal

[somber coyote]

Ann

[dense current]

ic

ty

[spring lion]

I'm struggling with these kinds of questions

Every time I look at an explanation it solves it in a way I don't comprehend

Figured I'd ask it here!!

[grave pond]

There are probably several ways through a problem like that, and it's a matter of trying different things until you find something that gets you to an equation you know you can solve.

A base fact that is surely going to be relevant is to know that when you divide a right triangle in two that wall, all the three triangles you end up having (ABC, DAC, DBA) end up being similar. So I'd look for a way to exploit that similarlity.

After several minutes of trying out combinations, I landed at the proportionality 8:x = (12.2+x):8 (by similarity of ABC with DAC) which looks like it would produce at worst a quadratic equation when I multiply across ... and I know I can solve quadratics.

[spring lion]

Okay

[shut sand]

yeah because 8 is the hypo of the small one and x+12.2 is the hypo of the big one

[spring lion]

Wait let me see

[shut sand]

Similarly, x is the smallest side on the smaller one and 8 is the smallest side on the big triangle

[spring lion]

So we take both of their ratios?

Let me see...

8/12.2+x?

8/12.2+x=x/8

[grave pond]

No, 8/(12.2+x)=x/8