geometry-and-trigonometry

[upper karma]

how do i do it?

[dark sparrow]

first off: ^

use this symbol for exponents

x^2 - 2xy + y^2 + 2x - 6y + 3 = 0

[upper karma]

yeah, that looks better

[dark sparrow]

anyway, consider making a change of variables

[upper karma]

what do you mean

[dark sparrow]

or rotate your coordinate system

[upper karma]

i have learnt none of these

[dark sparrow]

idk how you're expected to do it then

[upper karma]

well i gotta do it

are any of these complex things?

[dark sparrow]

are there any worked examples you have access to of problems like this

bc if you can't do changes of variables then i don't think this is possible at all

[upper karma]

how hard is changing variables?

if i knew, how many lines would it take to do that?

[dark sparrow]

well it amounts to a bunch of algebra

[upper karma]

ah i see

[dark sparrow]

idk how many lines

but if i was held at gunpoint for a line count, i'd say no more than 20 for all

[upper karma]

i have a similar problem

but the focus is given

focus point

i cannot even do that one

another with the directrix

another with the equation but only asks for thr latus rectum

but this one, if i knew this one, I could've done all of them

[inland kiln]

Can someone walk me thru this

what 1 angle measure is

[vale nimbus]

half of 94° i assume

[inland kiln]

yep

ty

how about this

is it 70

because of similar thingy theorum

[vale nimbus]

they face the same arc so yeah should be 70

[inland kiln]

tysm

[vale nimbus]

[inland kiln]

Please help wtf is this 😢

[inland kiln]

tysm

wait what

[vale nimbus]

thats the standard form of the equation for a circle

[inland kiln]

Ohhhh

[upper karma]

Anyone here with knowledge in algebraic topology?

[inland kiln]

Look at curry man

@vale nimbus how do i convert it to standard tho

[vale nimbus]

uh havent done this in like a year

gimme sec

is that for a test btw?

[inland kiln]

no why?

[vale nimbus]

ah dw then

[hallow rose]

@inland kiln do you know completing the square for the quadratics?

[inland kiln]

No never heard of the term

[tame field]

someone help me with this?

[hallow rose]

@tame field this channel is occupied.

[tame field]

ok

[hallow rose]

It’s basically the idea of grouping the “x/y” terms together to form a nice square identity

@inland kiln https://youtu.be/WrH1Z-SuhhQ

[inland kiln]

so i group them in parathesis

paranthesis

[hallow rose]

See this^^

Yeah you group them, exactly.

[vale nimbus]

found the solution but idk how i did it lmao so ima just let the other dude explain

[inland kiln]

I THINK I GOT IT

(x+3)^2+(y-4)^2=49

[vale nimbus]

thats also what i have yeah

[inland kiln]

LETS GO

tysm guys

[vale nimbus]

good shit bro

[jade sand]

I’m doing hw rn. Idk how to start the problem at all he never really taught us. Can someone help me please)

[silent plank]

look up tangent-secant theorem

[jade sand]

Thank u!

[twin monolith]

Can anyone help me find the perimeter of this please?

[late current]

The whole green shape?

[twin monolith]

Yeah

[late current]

I'm sorry in advance if this doesn't help; but you could try to use the perimeter formula 4a. Multiply 44 by 4. 4 sides.

Or are you meant to use trigonometry ?

[silent plank]

what are you told about the shape?

is it a parallelogram?

[late current]

Looks like a rhombus to me

[silent plank]

not really

shouldn't make assumptions like that

[late current]

True

[silent plank]

anyway, assuming its a parallelogram (since that's reasonable)
use trig to find the length of the slant edge

[late current]

Think its cosine

[silent plank]

try not to spread misinformation

[late current]

Sorry I'm a bit rusty

[acoustic jungle]

you can but then you would need to use 60 instead of 30

[rose sphinx]

Well why to assume its a parallelogram ?

@silent plank

if its a parallelogram ur answer will be 164 @Nikolas#3331

[coral zealot]

can somoene help me reald quizk

please

can someone help me find the surface area

ok

[silver radish]

How do you do this

<@&286206848099549185>

[raw tiger]

@silver radish do you know the formula for the volume of a cylinder?

[silver radish]

@raw tiger yeah

V=pie r2 h

[iron rain]

Find the hight

[acoustic jungle]

@silver radish this might help https://www.google.com/search?q=volume+of+slanted+cylinder&source=lmns&bih=977&biw=1920&rlz=1CARWXF_enCA890&hl=en&ved=2ahUKEwiop_a539noAhXD_KwKHaO8DWgQ_AUoAHoECAEQAA

ok I understand it now

volume = base * height

all you have to do is use tan and find the height

you already know the base

[silver radish]

Would it be 106.8

@acoustic jungle

[raw tiger]

Try making a triangle where the hypotenuse is AB. You can use a trig function to find the missing side. After finding that side imagine pushing the cylinder back upright. Then just plug in the values you have

@silver radish

[acoustic jungle]

perhaps

[unreal nest]

stuck on these

[raw tiger]

What is your question, trying to find h?

[unreal nest]

yes, that is height. supposed to be an A=B*H problem

(Area = Base x Height)

[raw tiger]

Are we assuming they are parallelograms?

[pastel zenith]

Someone help plz

[unreal nest]

@raw tiger yes

[versed river]

@pastel zenith if <a and <b have equal measures, and <c has a measure 8 less than each of those, than could you set up an equation with that information, given that the interior angles of a triangle always add to 180?

kayden only 2 of the angles are equal

in daemons picture

[unreal nest]

yeah I misread that

[versed river]

np

@pastel zenith you could start by letting angles <a and <b equal y, and then work from there

[pastel zenith]

So c is 8 less than both combined?

[versed river]

nope

c is 8 less than just 1 of them

[compact spire]

What formula would I use to solve this?

[pastel zenith]

Ok that is what is was Herron wrong i was doin it where c was less than both

[raw tiger]

@unreal nest The first thing I would do is solve for the missing side of the triangle with the dotted line. The you know the length of all the sides of the triangle so you can find the angles of that triangles. From there you can start filling in the sizes of the angles in the parallelogram. Hope this gives you a good start

[unreal nest]

that still doesnt help me find h, @raw tiger

I can find Area, not h

[devout harbor]

@compact spire

https://api-project-1022638073839.appspot.com/questions/how-do-i-find-the-cube-root-of-a-complex-number#612158

[compact spire]

@devout harbor thank you

Im going to read that now

[devout harbor]

np

[raw tiger]

@unreal nest if you know the Area you just solve solve for the height, A=bh so h=A/b

[unreal nest]

@acoustic jungle square root to find h?

[acoustic jungle]

Sorry let me just verify that this is true

or false

ok it is false

that is not correct

so I don't know if it's supposed to be this complicated

but I have this

[acoustic jungle]

,rotate

[digital gulch]

Nice music

[acoustic jungle]

I'm 98% sure this is correct. but there might be an easier way.

[digital gulch]

Hm

I’m stuck on this

[acoustic jungle]

@unreal nest

[unreal nest]

yes @acoustic jungle

[acoustic jungle]

It's right?

is it the easiest way?

Doesn't seem like so

[unreal nest]

I’m not sure what you got

[acoustic jungle]

h=sqrt 51.84

[unreal nest]

7.2

?

[acoustic jungle]

ok?

[digital gulch]

This?

[acoustic jungle]

Yes.

[unreal nest]

yes

[digital gulch]

You can use similar triangles

[acoustic jungle]

I am so stupid

[visual mist]

can co interior angles prove lines are parallel?

[acoustic jungle]

yes

You don't need to separate that

god gamer

[digital gulch]

You can use Pythagorean theorem to fine AE

Then say that the smaller triangle is similar to the larger triangle

[unreal nest]

I’m finding line h

[acoustic jungle]

7.2

You don't need pythagorean

[unreal nest]

I’ll let you know if it’s 7.2

[acoustic jungle]

yep

I also did it right my way

the complicated one

but it's easier to do it similar triangles

[digital gulch]

Then find the length of A to B and subtract that by 6.5 and then use Pythagorean theorem again to find h

[acoustic jungle]

$6/h = 6.5/7.8$

[somber coyote]

Fishraider:

[acoustic jungle]

God gamer AEP is already similar to that triangle you don't need to separate it

[digital gulch]

What point is P?

[acoustic jungle]

top right

[digital gulch]

Yeah you could also do that I guess

[acoustic jungle]

You haven't labled it

yes.

[digital gulch]

But they both work

[unreal nest]

ok I have this other question kind of similar

same thing, find the height (h)

[acoustic jungle]

Godgamer, yes, that works but finding BE takes effort.

You could use area to find BE, but it's much simplier knowing that those are already similar

Similar triangles again.

@unreal nest you would use similar triangles

[digital gulch]

Is it given that they are parallel?

[acoustic jungle]

Yes.

He said that some time before

[digital gulch]

Oh ok

So yeah use similar triangles

[acoustic jungle]

@digital gulch what unit are you on?

I am not sure how to solve your question

[digital gulch]

We’re doing 3D stuff

[acoustic jungle]

what grade is it

[digital gulch]

9th

[acoustic jungle]

yeah Canadian education is stupid

[digital gulch]

Eh

I’m doing honors stuff

[acoustic jungle]

Honors?

We only have pre ap what is honors

[digital gulch]

Idek

But this problem is so hard

[upper karma]

same thing basically

[digital gulch]

The formula for volume of a pyramid is v= one third times area of base times area of height

[upper karma]

what question is hard?

[digital gulch]

The volume of the original pyramid is 1000/3

This one @upper karma

[unreal nest]

@acoustic jungle I have no idea what similar triangles means

[digital gulch]

Volume*

[acoustic jungle]

ok you should search that up

[upper karma]

well what is half the volume?

[digital gulch]

Half the volume is 1000/6

[acoustic jungle]

or 500/3

[upper karma]

right

so consider the top half

[digital gulch]

Yeah

[upper karma]

that also would be a pyramid

[digital gulch]

The top half is also a pyramid

Yeah

[upper karma]

which sides that are proportional to the larger pyramid

[digital gulch]

Yeah

The base of the top half times the height of the top half is 500

The area of the base*

[upper karma]

rihgt

so now we need to find a relation between the base and the height

[digital gulch]

Yes

[upper karma]

it seems like the base length is equal to the height

[digital gulch]

Yeah

[acoustic jungle]

WOAH

[upper karma]

so the base area would be height squared

[digital gulch]

So base length cubes is equal to 500?

Cubed*

[upper karma]

yes

or height cubed

thats what we're trying to find

[digital gulch]

So the new height would be cubed root of 500

So to find the difference

It is

10- cubed root of 500

Ok

[acoustic jungle]

Very nice

Very nice.

Thank you sanath1237

[digital gulch]

Yeah

Thank you

[coral zealot]

yall green how come when i asked for help no one helped me but when someone else ask they get instant assistance

[devout harbor]

ask again

[digital gulch]

I’ll help @coral zealot

[sly marlin]

could be the helpers didn't see you at the right time

[acoustic jungle]

I am pretty sure someone asked you if you know the formula for area and surface area of a cone

that is what you will need

Here is his question https://discordapp.com/channels/268882317391429632/326138757474680852/697519445232451604

[digital gulch]

Surface area of a cone

[sly marlin]

@coral zealot imagine slicing the cone up by cutting a line from the apex to the base, what would you end up with?

intuitively it's a flat shape, but what sort of flat shape?

you can try drawing lines from the apex to the base as well as circles at different heights to figure out

[bitter moon]

how does one go about solving $\sin(x)-.15x=0$?

[somber coyote]

Jeffrey:

[crisp phoenix]

guys someone pls help me w this,

how do i even find it?

[bitter moon]

are those $x-5$ and $x+5$?

[somber coyote]

Jeffrey:

[bitter moon]

@crisp phoenix

[golden panther]

Guys can rectangular form look like: 3-2i or does it always have to be positive sign

[bitter moon]

not exactly certain, but i'd imagine so because i remember writing that and it's the same as 3+(-2i)

[umbral snow]

Think of it as 3 + (-2)i

[bitter moon]

also @crisp phoenix https://www.youtube.com/watch?v=qhtOBRIEPXc

that should answer your question

[golden panther]

Thx u

[bitter moon]

wow a section i can actually answer questions in XD

[dusky surge]

@bitter moon Jeffrey is the best!

[bitter moon]

uwu

ty

but not rlly, just took trig before is all

lol

[upper karma]

uwu

[dusky surge]

uwu

[tawdry pivot]

uwu

[visual mist]

Is angle y and x the same?

[silent plank]

what do you think?

[visual mist]

Idk

[silent plank]

do you know your circle geo theorems?
could one of them be applied here.

[visual mist]

theorem 3?

[dark sparrow]

nobody knows your theorem numbering, sushi492

[visual mist]

circle theorem 3

[high zephyr]

what's the name?

[dark sparrow]

nobody fucking knows what you mean by theorem 3

nobody knows which theorem is numbered 3 in your particular course

[dusky surge]

Or the whole thing, like you can take a photo or something @visual mist

[visual mist]

ok

[dusky surge]

So it's a question about tangents and circles

[visual mist]

yes

[dusky surge]

Which question do you not understand?

[visual mist]

f

[visual mist]

？

[silent plank]

how far did you get?

consider the size of the major/reflex angle/arc

[visual mist]

I don’t really know where to start

[upper karma]

guys is there a relationship between sphere's volume and a cone's surface area

[bleak jewel]

yo

[upper karma]

@visual mist try to divide into triangles

[visual mist]

wdym

[upper karma]

In the area of the angle given

Like

Where they give you the angle

Try to form triangles with that angle

Do you follow me?

[visual mist]

not really

[upper karma]

Hmm

With the angle given, try to form a triangle, which has at least one of the 3 angles, the one you are given, 117°

[visual mist]

oh wait

Like this???

[upper karma]

Yes

Now where do you go

To get angles, almost always search for triangles

[visual mist]

okay

but where

[upper karma]

Well look at the final triangle

What type of triangle is it?

[visual mist]

which one

[upper karma]

The final one

[visual mist]

the one the two tangent makes?

[upper karma]

The big one with alpha

[visual mist]

oh uh isosceles?

[upper karma]

Yes

So both sides must be equal

[visual mist]

yea

then?

[upper karma]

Umm

I was thinking to make to equal triangles again on the sides

But i think that goes nowhere

Wait

[visual mist]

ok

[acoustic jungle]

You know those angles are 90 degrees

and you use the inscribed angle theorem to find out that is 234 degrees, so the center point is 360-117*2 degrees

and the center point + 90 + 90 + theta = 360

@visual mist

also if you don't understand give me a diagram with points.

[upper karma]

@acoustic jungle what

Could you like develop it

[acoustic jungle]

I don't have any points

[upper karma]

Oh wait

[acoustic jungle]

The center angle is 360-117*2

and the center angle + theta + 90 + 90 =360 because quadrilateral

[upper karma]

Lmk if you need more

@acoustic jungle

||Sorry for my ass circle||

[visual mist]

wouldnt the centre be 117*2

[silent plank]

are you referring to the major/reflex or minor angle?

[upper karma]

Isn't phi equal to 31'5? Bc 180=117+2*phi

[acoustic jungle]

what's with all these triangles

Major arc AB is 117*2

ADB = 360 - 117*2

angle DAX= angle DBX=90

[upper karma]

Major arc AB is 117'2 is bc of the theorem?

[silent plank]

• for multiplication (not ')

[acoustic jungle]

because of inscribed angle theorem

[upper karma]

@silent plank is my thing correct

[silent plank]

your images?

[upper karma]

No, this

Isn't phi equal to 31'5? Bc 180=117+2*phi
@upper karma

According to my last drawing

[silent plank]

no

[upper karma]

Why

[silent plank]

you made some bad assumptions

[upper karma]

Umm

Which one?

[silent plank]

triangle ABC bein isosceles

there isn't enough info to determine if it is

[upper karma]

You mean there isnt enough info to see that phi is twice?

[silent plank]

wdym by phi is twice?

[upper karma]

Like 2phi

Is it a wrong assumption to make both sides of 117 angle, to be the same angle?

The phi angles are the side ones

@silent plank

[silent plank]

whats your justification that the red and green lines are equal?

[gray marten]

oo look an icecream cone!

[upper karma]

Lol

@silent plank that they are isosceles

Right?

[silent plank]

how do you "know" though

[upper karma]

Bc both have 31'5°

Or was that wrong

[silent plank]

why are you using ' for a decimal point

how do you "know" that its 31.5?

[upper karma]

@silent plank its used in my country

Look

Like this???
@visual mist

Oh

https://cdn.discordapp.com/attachments/326138757474680852/697769274437271653/image0.jpg

Is this wrong

[silent plank]

yes its wrong

[upper karma]

Oh

Lol

My bad

So there is not enough info to determine that they are isosceles

@silent plank

[silent plank]

correct. and that has 0 effect on how the problem is solved

[upper karma]

Lol

So can be only solved using the theorem

Lol i didnt know of that theorem

@silent plank

So can be only solved using the theorem

?

[silent plank]

no, there are other theorem(s) that can be applied

[upper karma]

Oky

Ty

[silent plank]

such as angle in alternate segment, angle sum on a line, with a triangle that IS actually isosceles

[dark sparrow]

how did you arrive at 90°

[void sigil]

It's about the arcs

Not the lines

try to draw it out first

[tiny halo]

how do find the are of the triangle

im confused

[silent plank]

the green line would have a length of 28m

and there are various methods you could use

splitting it into 2 smaller triangles, w/ pythag to determine the altitiude
trig
heron's

are some that could be applied here

[tiny halo]

which one would be easier

[silent plank]

depends which of those you know and are most comfortable with using

[rose sphinx]

Hey

I need a detailed explanation of this

DM me if required

[onyx basin]

first off those pings dont work

[acoustic jungle]

imaging pinging the whole server

[onyx basin]

secondly, do you know what:

• a circumcircle means
• an internal bisector means

[rose sphinx]

Yeah I know

[onyx basin]

so given triangle ABC

and you have the circumcircle Ω for triangle ABC

the internal bisectors of angles A, B, C intersect Ω at A1, B1 C1, respectively

you following along so far?

[rose sphinx]

Yeah

[onyx basin]

so a1, b1, c1 make a triangle

the internal bisectors of angles A1, B1, C1 intersect Ω at A2, B2, C2, respectively

then that forms another triangle

[rose sphinx]

K

[onyx basin]

and we know that the smallest angle of triangle ABC is 40deg

[rose sphinx]

Yeah

[onyx basin]

so we need to find the magnitude of the smallest angle of triangle A2B2C2

do you know what magnitude is

[rose sphinx]

Well come to the point pls

[onyx basin]

well i just laid out the steps to find the answer

[rose sphinx]

I hv come through these but couldn't really think any further

[onyx basin]

did you draw it out

[rose sphinx]

Well I drew something

🙄

Ar u gone ?

[onyx basin]

busy

can i see what you drew?

[upper karma]

If y=4, then s=(1;0), k=0, and a=0°
If x=-8, then s=(0;-1), k=-1, and a=-45°
Are my assumptions right?

[rose sphinx]

I assumed that the lines don't coincide

[upper karma]

No, two different graphs

Book says that in the case of x=-8 there is no k and a=90°, but i think it's a printing error. Can you confirm?

[grave fulcrum]

Is there a formula or property I can use to solve this?

[silent plank]

yes. properties of parallelograms

[grave fulcrum]

Sorry my b

ig a better question would be
what's the value of AE and which property am I missing?

[silent plank]

look up the properties.

there's a concise list

note that the values in the diagram aren't entirely accurate. (a few errors from rounding)

[grave fulcrum]

oh alright

7?

Because it bisects

[silent plank]

yes

[grave fulcrum]

Thanks mate!

[inner cave]

hello is this the right adress to ask for help with accelerated movement of an object?

[cold tangle]

When using this formula, if I need to find the SinA for ex. after I solve for SinA I need to apply Sin^-1?

What I got so far is SinA = Sin25 * 23 / 12

[onyx cloud]

@inner cave try one of the #❓how-to-get-help channels

the SinA will be whatever is on the right of the equation

the angle A will be what you get after applying the inverse sin function

[cold tangle]

Oh alright

Thx for clarifying

[grave fulcrum]

Last question

[foggy iron]

how i do this

[junior thunder]

Please

[acoustic jungle]

@junior thunder tan

[junior thunder]

What degree should the calculator be @acoustic jungle

[acoustic jungle]

you need to find the height of the first triangle on top, then find the height of the bottom one and add the mup

@foggy iron just type 8788 * pi/3 in your calculator and round it

[junior thunder]

Should it be in DEG?

[acoustic jungle]

@grave fulcrum I'm not sure the properties for trapzoid, but if you got exercise 24 by doing (57+76)/2, you can do 25 by (10+AB)/2=7

it should be in degree mode yes

[junior thunder]

Thank you

[upper karma]

Hello guys sorry to bother you

Does someone have the time to take a look at this

Is that in a different language?

@upper karma

Yes

The first one (18) u need to find how much is sin4α+cos4αcotg2α, IF tg2α=4

The second (19) and the third (20) u need only to simplify

[acoustic jungle]

whoever wrote this is so bad

sin sin (x)

sin^2 (x) ??

cos cos (x)

wth

@upper karma are you supposed to just write is simply like sin (4x)+cos(4x)/4 or do you have to expand it out

[upper karma]

I guess sin sin = sin ^ 2

[grave fulcrum]

Thanks man!

[lyric summit]

i need math problems that involve pythagorean thm 😔

[onyx basin]

https://www.google.com/search?q=pythagorean+theorem+practice+problems

[weary drift]

missed opportunity for lmgtfy link

[onyx basin]

true 😔

[lyric summit]

what

[onyx basin]

lmgtfy stands for "let me google that for you"

[lyric summit]

oh

im sorry 😔

i was lookng for lharder problems that in some way incorporate the pythagorean thm but i cant find any

[acoustic jungle]

have you tried googling "hard Pythagoras questions"

[onyx basin]

here i'll give you a hard pythagoras question

in what city was he born and in what year

[versed river]

probably the 60's

[onyx basin]

nah today's actually his tenth birthday

[versed river]

oh wow

kid genius

[onyx basin]

yeah ikr

[summer spire]

@lyric summit do you have a textbook to do problems from?

[upper karma]

hey can anyone help me understand something?

?

its abt trig

[lyric summit]

i do have a textbook

but those are too simple for the assignment

[summer spire]

@upper karma can't help you with a question if you never ask it

[upper karma]

Ok

so my question was

For example

Lets say

you have to find tan(135) degrees

w/out a calc

you would draw it on a coordinate plane

and then use the reference angle (45 degrees)

to make a 45-45-90

and then use the unit circle to map out the lengths for that triangle

and then find the tangent

but its weird to me that you use the reference angle

to make those ratios

how does 45 degrees correlate to 135

besides being supplementary

[lyric summit]

unit circle

i am having trouble making up problems similar to this one

[versed river]

@lyric summit

[lyric summit]

oop sorry

and thank youu :)

[twin shard]

anyone know how to do this

[high zephyr]

do u know how to find the radius based on that eqn?

[round anvil]

Anybody got good recommendations for geometry books?

There's no geometry section in #books-old

[tame ore]

sucks dick

[versed river]

?

[high zephyr]

...

[lyric summit]

i need more math problems

[versed river]

can you give an example of the type of problem?

[humble pebble]

can someone help me

[high zephyr]

which one?

[humble pebble]

how do i do 6, wouldn't it be diameter if it goes to the middle?

[versed river]

it doesnt mean the chord passes through the centre of the circle, its asking about the chord's distance from the centre

[humble pebble]

oh im dumb lmao thanks

[versed river]

dw i thought the same thing when i saw the question

[humble pebble]

w8 but how do i solve it doe

cause i have only one smaller side

which is the 8 bisected from 16

[high zephyr]

@versed river im not that sure but isnt it 4 and 6cm?

cuz perpendicular distance

[versed river]

yeah

i would think it cant be greater or equal to the radius

[high zephyr]

yup

there u go brisk

[versed river]

because like if it were it wouldnt be in the circle

and the 8cm one is a tangent not a chord, i think. not too familiar with circle geometry

[high zephyr]

8cm is tangent yes

[upper karma]

is there any way beside the one where i will have to solve for line AB and then solve for B from the two line equations to solve for the coords of B?

[high zephyr]

you can use a formula to find the perpendicular distance from a point and a line

[upper karma]

ah man

that makes it even complex

i was looking for an one liner of some sort...

thanks, though

[high zephyr]

when u mean solve AB u mean finding the length right?

alrighty

[upper karma]

nope

the coords of B

the ax+by+c line is given

and the coords of A is given

i need to find the coords of B when AB is perpendicular on the line

[high zephyr]

ah ok

[upper karma]

i know i can do so by finding the AB line equations

ans then solving for x, y using both line equations

but i was looking for a more compact method

[hollow tiger]

hey um, I kinda need some help with this equation, I can't really figure out a way to solve it, other than just expressing everything through x and then solving for x, which is already looking super difficult

$a_{k} = -sin(x)

a_{k+1} = 4sin(x)ctg(2x)

a_{k+2} = cos(x)

a_{7} = \frac{1}{5}$

[somber coyote]

Behemoth:

hey um, I kinda need some help with this equation, I can't really figure out a way to solve it, other than just expressing everything through x and then solving for x, which is already looking super difficult

$a_{k} = -sin(x)

a_{k+1} = 4sin(x)ctg(2x)

a_{k+2} = cos(x)

a_{7} = \frac{1}{5}$
```Compile error! Output:

! Missing $ inserted.
<inserted text>
$
l.57

I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

[dark sparrow]

this isn't an equation

[hollow tiger]

equation system** yeah my bad

[dark sparrow]

this looks like an incomplete problem statement

do you have a screenshot or something

[hollow tiger]

yeah, but it's in Russian, I'll translate it real quick...

[dark sparrow]

i speak russian

[hollow tiger]

oh

ok

[dark sparrow]

а, так это арифметическая прогрессия

[hollow tiger]

да, я поэтому и записал формулы, думал с помощью них как-то надо)

[viscid ginkgo]

why is the period of -3sin-3x pi/3 instead of 2pi/-3?

[dark sparrow]

@viscid ginkgo this channel is occupied, please move!

из того, что ты записал ранее, не было понятно, что речь идет именно об арифметической прогресии, если что

а так, наверное надо не морочить себе голову и просто записать уравнение, гласящее, что разности соседних сленов совпадают

то есть $\cos(x) - 4\sin(x)\cot(2x) = 4\sin(x)\cot(2x) + \sin(x)$

[somber coyote]

Ann:

[dark sparrow]

решить его все равно придется, т.к. спрашивается значение x

а когда уже его решишь, найти k не так уж и сложно

[hollow tiger]

ага, понял
Спасибо 🙂

[crystal iris]

somebody please help

i have an assignment due in 20 minutes and i don't understand remaining problems

[acoustic jungle]

👍

[dark sparrow]

and you decided not to even post it, wasting 10 min of your own time?

[acoustic jungle]

use secant theorem for the first one

[silent plank]

and intersecting chord theorem for the second

[silent plank]

you do have access to a list of circle theorems right?

[crystal iris]

no

i'm running out of time

how do you do the intersecting chord theorem

[silent plank]

i'm more concerned about why you're doing these questions without the required knowledge

[crystal iris]

i've missed lesson time

a lot of it

i know

[silent plank]

do a quick search on circle geo theorems

and then match which theorem is applicable

is probable more efficient than me explaining from scractch here

[gray marten]

Hi, I literally know only basic trig

I'll see if I can do it

Hmm

[acoustic jungle]

no you need to know circle stuff

[gray marten]

So xD and BA are two secants that intersect at point C

[acoustic jungle]

CD is not a secant

it's tangent

[crystal iris]

right

[gray marten]

Rip so it's not secant theorem?

[silent plank]

||BA isn't a secant either||

[gray marten]

Ooh wait.

https://en.wikipedia.org/wiki/Tangent-secant_theorem
https://en.wikipedia.org/wiki/Intersecting_secants_theorem

[silent plank]

use < >

[gray marten]

These articles seem useful.

[silent plank]

instead of deleting the http

[gray marten]

Oh okay

Huh cool didnt know that.

OOoh wait

Maybe

Btw I'm trying that last image.

CB * CA = CD * CD

[acoustic jungle]

yes

[gray marten]

Meaning: 9 * 12 = x^2

Damn thats easy

lol

Cool theorem tho.

I guess its same idea for the 1st image too.

[upper karma]

is geometry fun

Depends

No

On solver and question setter 😂

[upper karma]

@silent plank why you not answering my dm's

[silent plank]

because i prefer not to answer in dms,
and you weren't entirely specific with your request

[upper karma]

Okay

I repeat my question

Could you give me a clear example where C is "shift up" (as you said) and SC and ST aren't equal?

@silent plank

(Sorry for the last tag)

[silent plank]

that is the example

[upper karma]

Like an image

Bc i dont see it clear

[silent plank]

also did you mean CT and CS

[upper karma]

Yeah

[silent plank]

just move the point C a bit higher on the circumference

like maybe extremely close to point T

[silent plank]

and its extremely obvious that CT and CS aren't equal

[upper karma]

Oh

I thought you meant shift up as moving it closer to P

Okay i get that now lol

[gray marten]

Oh lol its the ice cream problem again

Is O, in the middle of the circle?

[upper karma]

But in the problem he showed it was really clear that TC and TS were equal lol

[silent plank]

which was an issue

because that wasn't necessarily true

[upper karma]

Oh

[gray marten]

If O is the middle of the circle CT or CS.

[upper karma]

Yeah

[gray marten]

This doesnt' mean TC and TS are equal. though.

[upper karma]

No

[gray marten]

There may be cases where TC and TS are equal

[upper karma]

If C

[gray marten]

But if it is CS, CT and TS must all be equal.

Meaning it will be equilateral triangle in circle.

[upper karma]

But HOW can we exactly know that the problem wants us to go through that assumption (that CS, ST and CT are equal?

Is there any key in the question that may give it to us?

[gray marten]

From the image, I would assume TS and TC are not equal

But I haven't seen the question so idk.

[upper karma]

I would

[silent plank]

TS was a typo

[upper karma]

Well yeah

[silent plank]

probably

[upper karma]

But

But HOW can we exactly know that the problem wants us to go through that assumption (that CS, ST and CT are equal?
@upper karma

[gray marten]

But yeah CT and CS are most definately equal.

Given O is the middle of the circle.

[upper karma]

?

Oh

[silent plank]

if the given diagram doesn't explicitly give certain properties, try not to assume anything

[upper karma]

So if it says that O is the middle of the circle

CT and CS are equal?

[onyx basin]

CT = CS when O is center of circle and OP is angle bisector

[silent plank]

IF it lies on PO and PT and PS are tangents

[gray marten]

If O is the center CT and CS are guaranteed to be equal.

From the image it looks like CT and CS are equal.

[onyx basin]

if O is the center and that's all we're given, we can't assume ct and cs are equal

because we dont know that OP is an angle bisector

it looks like one, but we can't assume that it is

[upper karma]

Yeah

So thanks

[silent plank]

well it is actually reasonable to assume that PT and PS are tangent

[onyx basin]

ah wait

i just noticed the PT and PS tangent lines

[upper karma]

I have definitely got to see a bunch of geometry theorems

[onyx basin]

then again, it's not given that they're tangent

[silent plank]

and from theorems etc, PO would actually be an angle bisector

however, lets make a small adjustment to that image by removing that line PO

like i said, reasonable assumption based on the context

[onyx basin]

yeah

if this question is asking to prove that ct=cs, it's kinda poorly formatted

[silent plank]

now, if line PO was removed, you have no helpful information about where C actually is (other than its on the circle)

[gray marten]

Then TO and ST are probably the same.

[silent plank]

probably?

[gray marten]

I would assume it is given O is in the center

[silent plank]

assuming its a circle and O its its center

dammit

[gray marten]

Yeah, discords being weird for me rn

[onyx basin]

discord has been shitting itself every week

[tropic stirrup]

OS = OT, PS = PT are all you can gather

[gray marten]

Yeah

[tropic stirrup]

usually you arent to assume things that have its own designated symbol

like the right angle, same length, parallel

tangent line is not one of them so you can at least comfortably assume that

[graceful rivet]

i need math homework help

trigonometry is my weakness

[onyx basin]

just ask

[lyric summit]

is the answer to this problem 3+root57 over 2

[versed river]

,w solve -2x^2+6x+24=0

[somber coyote]

Display results online and refine query

Requested by Sneaky#0928

[lyric summit]

thank u

[silent plank]

parentheses

when transcribing in plain text

[lyric summit]

ok

[silent plank]

and sqrt(57)

to make it extra clear what's actually being sqr rooted

[lyric summit]

ok owo

[upper karma]

UwU

[upper karma]

The angle θ=π6 is in standard position and has a terminal side that coincides with the graph of a proportional relationship represented by y=kx.

so angle theta is 30 degrees

Do I make a 30,60,90 triangle

and see what points that lines up on?

or

do cos30, sin30 = x,y

becuase my math teacher said (cos theta, sin theta) = (x,y)

if its on the unit circle then yes

x^2+y^2=1

It doesnt specify if its on a unit circle or not

holdup (costheta, sin theta) = (x,y) only works on the unit circle?

yah

at the point (0,0) if u connect the orgin to any point on teh circucfrence (x,y)

that angle formed

sin of that angle is the y coordinate

and cos is the x coordinate

[abstract scarab]

Yaaa

[upper karma]

holdup

so lets say i have circle O where O is the origin (0,0) and circle O has radius 1

x^2+y^2=1

and the point (1,0) is point B on the circle

mhm

lets say we have a random point on the circumfrence A

angle AOB=30

ya

sin 30 is the y coordinate of of A

and cos 30 is the x coordnate of A

The angle θ=π6 is in standard position and has a terminal side that coincides with the graph of a proportional relationship represented by y=kx.

how bout for that question

specifcally

im guessing we dont use the unit circle

no but it still works for any circle

a. Find the constant of proportionality, k.

 k=

angle θ = 360 degrees

so lets say we have (x-a)^2+(y-b)^2=r^2

awh shit its just 0

yah

pi=2xpi for any radian

what happens if the angle was 30 degrees

how would i approach that problem

well theres 2 ways

if its ust a generallity problem

use WLOG

whats WLOG

without loss of generality

meaning we just say a length or something is a number

so if we assume the origin is (0,0)

actually

here

let (a,b) be the origin of a circle

ok

and r be that radius

ya

so a+rcos30 is the x coordinate

and b+rsin30 is the y coordinate

bc

the origin is at (a,b)

can u voice call

and the rightmost point of the circle is (a+r,b)

no

k

i was just gonna make

a 30,60,90 triangle

hmm do you now the radius

find the point

being

root 3, 1

and the origin of the circle

nope

do u know the coordinate where the 30 degree angel is made

nope

literally

all it says

it only says the radius?

The angle θ=π/6 is in standard position and has a terminal side that coincides with the graph of a proportional relationship represented by y=kx.

a. Find the constant of proportionality, k.

 k=

the angle is actually 30 degrees

my bad

OHH its not 6*pi

ya

k

my copy paste

fucked up

because it cant copy the divide sign

actually do you now tan

tangent

yeah i can do tangent

like sinx/cosx=tanx

well tan is just the slope of the line

no

i didnt know that lmao

so sin30/cos30?

yah

that makes sense

because if this was on a unit circle

that'd find the slope

it could work on any circle

but its more coplicated

thats why we use unit circle bc it has the easiest coordinates to work with

the origin is at (0,0)

and all the intercepts have a 0

and the radius is 1

k but theres a general way to do it though

if u have any circle cenetered at (a,b)

and the radius is r

and theres a point on the circle (c,d)

the angle from the point (a+r,b),(a,b),(c,d) is x

then c=a+rcosx

and d=b+rsinx

ok tyy

[heavy osprey]

I need help with part b of this

In part a I found the big hexagon area and so far in b I found small hexagon area

[upper karma]

do you know about 30 60 90 triangle?

[heavy osprey]

Yeah we learned that but the only value we have is x

Is it still doable with just x

[upper karma]

yah

[heavy osprey]

How

[upper karma]

because if we let the length of the small hexagon be x

then root(3)*x/2 is the apothe

apothem

SCP can u help me with one more question after

k

ty

[heavy osprey]

So what do you mean let it be the length of the hexagon

Do you mean like x is side length

[upper karma]

yah

[heavy osprey]

Okay I’ll try again thanks

[upper karma]

Write an arithmetic sequence that gives the nth positive x-intercept of the graph of f(x)=cos1/8x. Leave your answer in terms of π.

k theres only 2 places where it hits the x intercepts

in one rotation

0 and pi

so its just pi(n-1)

because the period is 2pi

for every 2 rotations it touches the x int right

oh wait

[abstract scarab]

It's 1/8 or 18

[upper karma]

its cos1/8(x)

ya

thought it was just cosx

nonono

[abstract scarab]

Cos x/8 right?

[upper karma]

well in that case its just the same thing but it muliplys by 8 each time

yea cos x/8

0+8pi+16pi...

isnt every nth positive x intercept

just

+8pi

yea

it is

yah thats the domain

so should I say

8pi(n) - 4pi

[abstract scarab]

Correct??

From n=0 to ...

[upper karma]

sequences start at n=1

so its

[abstract scarab]

No sequence starts at n=0

For my answer

[upper karma]

oh

did u make a recursive rule?

nvm u didnt

[abstract scarab]

Used general solution

[upper karma]

k u have the function f(x)=cos(x/8)

using the winding function

the positive x intercept

will always any multiple of 2pi

bc pi and 3pi and 5pi... and so on

have a negative x inetcept

the answer is

4pi(2n-1)

right

assuming n=1

[abstract scarab]

Yaa..

[upper karma]

kk

[heavy osprey]

So back from earlier should my equation look like this

[abstract scarab]

Question??

[upper karma]

bazinga

can i dm u

[heavy osprey]

Wrong one part b of this

[abstract scarab]

@upper karma sure

[upper karma]

no its not 4p(2n-1)

pi*

@heavy osprey do u remember 30-60-90 triangle well

[heavy osprey]

Yes

[upper karma]

since u know the area

let the side length of the smalller hexagon be x

[heavy osprey]

Okay

[upper karma]

and by 30-60-90

the apotom is root(3)*x/2

[heavy osprey]

So I’m pretty sure we have to use the apothem formula where: Area=1/2(Permiter)(Apothem)

[upper karma]

yah

[heavy osprey]

Ooh I see now

So the perimter would be 6x. Then I just solve

[upper karma]

and the perimeter is 6x

yah

[heavy osprey]

Okay thank you

[upper karma]

and then when u get x

times that by root3/2

for the apothem

[abstract scarab]

Cool

[gray marten]

Would it be useful to know all of these:

https://en.wikipedia.org/wiki/Category:Theorems_in_plane_geometry