geometry-and-trigonometry

[upper karma]

4 squared = 16

[sick hemlock]

now √9 = ?

[upper karma]

25-16=9

[upper karma]

3

[sick hemlock]

do you understand how to apply Pythagoras theorem?

Like, say the sides are a, b and c where c is the hypotenuse

[upper karma]

Yes

[wild rivet]

that's the answer of the sample question

[upper karma]

I thought u meant what is the type

or sth

[wild rivet]

now apply it on the question we had. x^2=?

[sick hemlock]

can you see how you can find the third side in the triangle?

since you already know 2 sides

[upper karma]

yeah what is the x tho

[wild rivet]

we don't know, you have to find out

[sick hemlock]

alright lemme send the figure one sec

[wild rivet]

10^2-8^2=x^2

[upper karma]

solve this?

[wild rivet]

10² - 8² = x²

solve for x

[upper karma]

(10-8)^2 = (10-8)(10-8) = 400-64 = 336

[wild rivet]

no

oh no

[upper karma]

Bro

U confuse me

wdym

[upper karma]

this?

[upper karma]

or this?

[sick hemlock]

Shit forgot to put in G

my ocd is acting up now

argh

[wild rivet]

(10²) -(8²) = x²

now solve for x

[wild rivet]

this is

[upper karma]

i have to solve 2 things in the same time wait

[wild rivet]

the same as this

[wild rivet]

this is the same as

[wild rivet]

this

[sick hemlock]

Find the side length AB@upper karma

You said you knew how to apply Pythagoras theorem

apply it and find AB

[upper karma]

AB = AG - BG = AB = 10 squared (100) - 8 squared (64) = 36

[sick hemlock]

Another thing to keep in mind is that a² - b² and (a-b)² are 2 different things

[wild rivet]

yeah

[sick hemlock]

what you just found was AB²

Not AB

[wild rivet]

exactly

[upper karma]

so 6?

[wild rivet]

YES

sorry, caps was on

[sick hemlock]

Pythagoras Theorem says that AB² + BG² = AG²

[upper karma]

ahh yeah

[wild rivet]

but you can't use squares

you have to use square roots

[sick hemlock]

not getting what you mean

you use square roots to go from AB² to AB

[upper karma]

yeah

[wild rivet]

I mean that taryla is coming to an answer but not square rooting it

[upper karma]

u confused me

[sick hemlock]

I think he misremembered the Pythagoras theorem

[wild rivet]

whom?

[sick hemlock]

which is fair since he was probably taught it 3-4 years back

[sick hemlock]

taryla

[upper karma]

maths is complicating for me

I forget what i was taught

[wild rivet]

umm, you don't want to know about me. Can I talk to you on personal chat? You seem to be nice

[sick hemlock]

sure?

[wild rivet]

ok

[upper karma]

Goodbye guys

[sick hemlock]

you need to practise questions based on what you're taught

[upper karma]

thx for ur help ig

[sick hemlock]

maths in lower grades is basically formulae memorization

[wild rivet]

it is? weird

anyways bye! @upper karma

[upper karma]

Goodbye ❤️

[upper karma]

D

Since it does 8 revolutions, the total distance is 8*22,159km = 177,272km

[frozen ocean]

Nope, I wasn't need help with this because I already know how

It's just an example on how we learned trig in canada

[dark sparrow]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[upper karma]

mb

[violet robin]

Helpppp

[knotty fjord]

help...

[bleak blade]

Should be just 1 cause all the numerators cancel out with denominators

[grave pond]

The first picture doesn't mean anything until you explicitly define it to mean some limit of finite calculations.
The details about how you choose to cut the infinite thing down to finite approximations you can calculate turn out to completely dominate which result (if any) you'll get.

[tepid hatch]

https://imgur.com/a/KMiGhHX
what witchcraft is this? does this come under similarity of triangles

[nocturne remnant]

look up angle bisector theorem

[tepid hatch]

thanx

[zinc agate]

The second is one satisfied for cos x = cos π

But both of them are just

The numerators and denominators cancelling each other out

[grave pond]

The second one is satisfied by everything.

[zinc agate]

Well yes

But he didn't seem satisfied

By the answers

[grave pond]

For all we know he hasn't even been back since he asked.

[zinc agate]

Umm could u help with a problem

I opened a help channel but

Can't get anyone to help

#help-22

@grave pond

[grave pond]

I'd file it under the law of sines. $$\frac{BL}{BA} = \frac{\sin\angle BAL}{\sin\angle BLA} = \frac{\sin\angle LAC}{\sin\angle ALC} = \frac{CL}{CA}$$ (where the denominators are equal because supplementary angles have the same sine).

[somber coyote]

Troposphere

[grave pond]

Triangles BAL and CAL are not similar, though .

[grave pond]

Don't ping specific people for help.

[upper karma]

help pls (traductor?)

[ember stratus]

how to prove

[eager ivy]

are circles congruent?

I just wanted to know if that was necessarily true

[ember stratus]

yeah they're congruent

[eager ivy]

thanks, let me look through it

[hollow moth]

GEOMETRY PROS I need help

[eager ivy]

Hello, so you know that RS and RW are perpendicular bisectors of the chords

[hollow moth]

yes

what's next

[eager ivy]

So then, you can use the fact that radii that intersect congruent chords are congruent

or not really radii, mor like segments where one endpoint is the center

basically, RS and RW are congruent

So then, you have to prove the triangles congruent

[hollow moth]

do you mean SW

[eager ivy]

Yes

Sorry

[hollow moth]

alright

its ok

[eager ivy]

So we already have an angle(the right angles) and a side for both triangles

so we need to prove another pair of angles to prove the triangles congruent through SAA critetia

[hollow moth]

oh

[eager ivy]

So connect S and T

Then you get an isoceles triangle

And with base angles congruent

You can do the same for the other chord

And CPCTC proves STU and STW congruent angles

So we have another congruent angle pair

[hollow moth]

I see

[eager ivy]

So, with that, CPCTC proves that SW and SR are congruent and you can solve for x

Use x to find SR and pythag to find ST

SU*

(UW is 26 because it is half of the chord)

DOes that make sense, i tried to type fast and so it may be confusing

And on to the second problem

[hollow moth]

yeah no worries

[eager ivy]

Whenever a radii intersects a chord and it is perpendicular to it, the chord is bisected

That is a proven theorem

So PR = RQ = 16

And MN = 40, so LN = 20

So LP is 20 because all radii are equal in length

[hollow moth]

yes

[eager ivy]

And you can use pythag to solve for LR

We already have PR

And angle PLN is 80 degrees because central angles are always equal to the intercepted arc

So you have all 4 parts

Glad I could help

arcs*

[hollow moth]

Thank you bro

[eager ivy]

np

Good luck

[viral patrol]

howdo u get the surface area of a sphere ;-;

[near harness]

Are you asking for the derivation?

The surface area is 4πr^2

[dark sparrow]

do you just want us to act as your personal google or did you mean to ask "where did the sphere surface area formula come from"

[viral patrol]

whatever nonsense

[dark sparrow]

??

[ember stratus]

Hi dajiahao, how did they use the sine rule to calculate for the circumradius in:
https://artofproblemsolving.com/community/c6h1951086p13467971

[upper karma]

https://brilliant.org/wiki/sine-rule/#:~:text=The extended sine rule is,radius of the circumscribed circle.

See Extended Sine Rule

[oak mural]

how is the answer this? i dont get it

[outer sinew]

$$\sqrt{2x^2} = 12$$

[sick hemlock]

Dude how did you get 45√2 💀

[somber coyote]

Kiameimon | Welt Rene

[oak mural]

bra ion kno this geomotry stuff lmaoo

[sick hemlock]

sin(45°) = x/12 = 1/√2

[oak mural]

what is sin

[sick hemlock]

since sine is opposite/hypotenuse

[outer sinew]

U don't even need that

[sick hemlock]

uh

[outer sinew]

U just need pythag theorem

[golden marten]

oh this is one of the identities no? or am i wrong

[oak mural]

ion kno i was going off of the last answer solutions

[sick hemlock]

I mean, clearly the question was asking for trigonometry to be used since it specified the angle

[golden marten]

i dont know any of the identities but there was something with trianles oe something i forgor

[outer sinew]

Seeing how they don't know what sine is?

[oak mural]

ight bet im finna watch a yt vid on dis rn

[outer sinew]

No need to

Just apply pythagorean theorem

[sick hemlock]

and yet, what's the point to include the angle if it's not required

[outer sinew]

Ask the setter

[outer sinew]

Not me

[sick hemlock]

the question is ambiguous on what method it requires

[oak mural]

i dont kno how to

[outer sinew]

... just go for the simpler method no?

[outer sinew]

What does the pythagorean theorem tell us

[sick hemlock]

also for the record, I mentioned sine because you already mentioned Pythagorean theorem

[outer sinew]

In your case

[outer sinew]

@oak mural

Do you know what it is?

[golden marten]

||a^2 + b^2 = c^2||, no?

[outer sinew]

Yes

Wait lmao I pinged the wrong person

XD

In any case

So our c is 12

Wait

I pinged the right person

[sick hemlock]

uh, didn't you ping the right person

[outer sinew]

My brain

[sick hemlock]

lol

[golden marten]

lol

[golden marten]

||wait so how do we solve for x tho||

[sick hemlock]

just apply Pythagorean theorem

in this case what do you think sides a and b are?

[golden marten]

i should know this i went thru precal last year

[outer sinew]

||\sqrt{2x^2} = 12||

Or if u wanna do it the other way, ||2x^2 = 144||

So ||x^2=72 and sqrt that to solve for x||

[golden marten]

wait its like ||5 12 13 right?||

[sick hemlock]

Just stop looking at special triangles, simply think about the Pythagorean theorem

[golden marten]

like a ||5 12 13 triangle or something i forgor but there were things like 3 4 5 and all these traingle combinations||

[golden marten]

oh ok

[sick hemlock]

a² + b² = c²

what do you think sides a and b are in the triangle?

[golden marten]

how do we know both sides can be solved with 2x tho what if they are different lengths

[golden marten]

leg

[outer sinew]

They are not different lengths tho

[golden marten]

opp and adj

[golden marten]

oh ok. how would we solve it if it was tho

[outer sinew]

We use sine/cosine

[golden marten]

ah i see. could you give an example n how to do dat?

[outer sinew]

Depending on which angle is given

[oak mural]

no

[outer sinew]

You shouldn't be doing this question then...

[golden marten]

you dont know pythagroem therom? how does your tracher want you to solve it?

[outer sinew]

You need to learn the fundamental concepts of triangles

[oak mural]

idk i barely go to school

[golden marten]

oh

[golden marten]

khan academy or youtube dawg just search somethig up you can learn anything if you want to

[outer sinew]

Ok so let's just say we have a triangle which hypotenuse is 4 and the angle between the hypotenuse and length we wanna find is 30 degrees

So we just take 4×cos(30) since cos is adjacent over hypotenuse and the angle (30) is adjacent to the side we wanna find

[golden marten]

ohhh i remeber dis

thank brodie!

[sick hemlock]

sry my battery died

It seems his question is already answered

[oak mural]

i was asking how to get the soluation

cuz this geomotry shi to me jus looks like u get random letters nd numbers nd multiply them and get a random ah solution this shi dont make sense

[sick hemlock]

Alright you need to learn Pythagorean theorem first

and by learn I mean actually dedicate time to it

If you don't go to school watch some videos online

khan academy vids or any other math ytuber

that question can not be solved without some basic understanding of geometry

[supple robin]

guys, how can I prove that the central angle of a circumference has the same measure of its arc?

[oak mural]

YESSIR

I divided 18 by 2 then put the checkmark thingy

[sick hemlock]

why...?

[oak mural]

cuz its the right way to do it?

[sick hemlock]

it's really not

[raven sable]

by "checkmark thingy" do you mean the square root symbol?

[oak mural]

ya i didnt know the name of it

[sick hemlock]

what you're doing is memorising a pattern between the solutions and question data, then applying the same pattern

If I gave you any other angle, your answer would be wrong, regardless of whatever you divide the hypotenuse with

that's why I'm suggesting you to learn Pythagorean theorem and then learn some basic trigonometry

[oak mural]

give me one i need practice

[outer sinew]

You need to build on your fundamentals first...

[oak mural]

is there a shortcut

[sick hemlock]

it's a big jump to go from that to trigonometry, but if your only focus is solving those questions then that's my only suggestion

[outer sinew]

Not at all.

There is no such shortcut in maths if you wanna get good at it

[sick hemlock]

You must have time right? you don't have your school finals or anything rn probably

[upper karma]

Need help with this question

[oak mural]

naa

its like in 3 weeks

[sick hemlock]

what do you think the amplitude of y = sin(3x - π) is?

[upper karma]

4? @sick hemlock

[sick hemlock]

Then dedicate time to maths, build on your fundamentals like kiameimon mentioned

[sick hemlock]

Why do you think it's 4?

[oak mural]

im gettin money doe

[sick hemlock]

I'm simply asking for the amplitude of sin(3x-π)

Not 4 × sin(3x - π)

[upper karma]

Wouldn’t it be 3

3pi/2

[sick hemlock]

no, it wouldn't

[upper karma]

Sorry 2pi/3

[sick hemlock]

Alright, the trick here is remembering that 3x - π represent a number themselves

what do you think the amplitude of y = sin(t) is?

[upper karma]

Oh ok

4?

[sick hemlock]

where t represents angles

[upper karma]

4?

[sick hemlock]

do you know what an amplitude of a function is?

[upper karma]

No

[sick hemlock]

alright, have you seen the graph of sine function?

[upper karma]

Yeah

It’s a bunch of swiggly lines

[sick hemlock]

It's something like this is it not?

[upper karma]

Yes !

[sick hemlock]

Do you see that the maximum and minimum points on the graph are basically at equal distance from the x-axis?

@upper karma

Can you see that the lengths a and b are equal?

[upper karma]

One is higher and one is the lowest

[sick hemlock]

but the distance is the same right? In a way the x-axis divides the function into 2 parts which are equidistant from it

so we call the x-axis the midline of the function

Now the distance between the midline and the function's highest/lowest point is called the amplitude

so here the amplitude is a

[oak mural]

do i use patheg thereom for this

[upper karma]

No

[sick hemlock]

You can't use Pythagorean theorem for that

[raven sable]

you'll need sohcahtoa

[upper karma]

^

Well actually you can use logic to find the other angle

[oak mural]

how do u even pronounce it lmaoo

[upper karma]

90+60

[sick hemlock]

anyways what are the minimum and maximum values of the sine function? @upper karma

[upper karma]

The highest point

[sick hemlock]

yes, the highest and the lowest point

[upper karma]

Ye

[raven sable]

pronounce it how it looks basically. "soh" rhymes with "low", "cah" rhymes with "the", and "toa" has the "same" vowel sounds as those... if that makes sense

[dark sparrow]

/ˌsoʊ.kə.ˈtoʊə/

[worthy matrix]

you'll want to memorize your special right triangles

45-45-90 and 30-60-90

[grave pond]

You can perfectly well use Pythagoras for that problem. Mirror the ramp triangle in the ground; together with the original triangle it forms an equilateral triangle, so we know that the vertical side of the triangle must be 3 ft. That's enough to plug into the Pythagorean theorem to find the horizontal side.

[novel canopy]

hi need help

[grave pond]

You need dollar signs around that to trigger the bot.

[novel canopy]

$\left[\tan \left(x\right)+\cot \left(x\right)\right]^2$

[somber coyote]

SLOWWWWWWWWWWW

[silent plank]

soccer tower

[grave pond]

And what do you want to do with that expression?

[errant lake]

Any help? This is what I’ve done so far

[grave pond]

Interesting -- my simplification would have given 1/(sin²(x)cos²(x)), but that is the same as one of your options.

Write the tangent and cotangent in terms of sine and cosine. Them multiply out the square-of-a-sum. That gives you two copies of 1; add each of them to one of the remaining terms.

[novel canopy]

thx

[grave pond]

You can also just metagame it: you want something that explodes when x approaches 0° as well as when x approaches 90°, and only one of your four options do that.

[dense canyon]

I need help in 4

[pine silo]

i can solve it

what is the value for pi in this?

[grave pond]

Don't just solve people's homework for them please.

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[pine silo]

oh ok

[burnt basalt]

How do I memorize formulas easier

[wide valley]

"soh-ka-toe-uh"

[wide valley]

1. It is much easier to memorize a formula if you understand what it means and how it works
2. Practice, Practice, Practice
3. Use acronyms, mnemonics, or other memory aids to help you remember formulas. E.g. "Please Excuse My Dear Aunt Sally" = PEMDAS.
4. Teach it to someone else.

[wide valley]

Do you have an answer key for this?

[dense canyon]

I solved it but it just took a while

thank you

[wide valley]

Can you share the solution?

[dense canyon]

sure

[wide valley]

It was an interesting problem

[dense canyon]

1899.45

[wide valley]

Thanks, do you have the solution written down too?

[dense canyon]

yea

[wide valley]

May I have a look?

[dense canyon]

sure

my handwritings messy sorry

[wide valley]

So did you multiply π5^2 by 5 to get the total area of those circles with radii 5?

No sorry, there was only 1 circle with radius 5

But 5 circles with radius 2.5

Right?

[dense canyon]

yes

[wide valley]

,calc 5*19.625

[somber coyote]

Result:

98.125

[wide valley]

So shouldn't it be 98.125 here?

[dense canyon]

hold on

there is 6 circles on the cheese

with a radii of 2.5

[wide valley]

Yes my bad

[dense canyon]

so i think 117.75 is right

[wide valley]

,calc 253.14 + (5/2)^2 * 3.14 + 6(2.5^2)3.14 + 4^2 * 3.14 + (2.5/2)^2 * 3.14 + 13.14

[somber coyote]

Result:

274.16125

[wide valley]

Okay so that should cover all the circles

Now what about the greater quadrilateral polygon?

[dense canyon]

wait like the whole trapezoid?

[wide valley]

What was your approach to solving that

I'm talking about the quadrilateral polygon, so the figure with 4 sides

So the trapezoid

[dense canyon]

oh i just did the formula for that so 1/2(50+41)48

I got 2184

[wide valley]

What formula is this, heron's?

Does (50+41)/2 calculate the average of the lengths of the two parallel sides?

[dense canyon]

yes

[dense canyon]

I'll be honest I have no idea what that is

[wide valley]

Okay but could you tell me what formula this is 1/2(50+41)48

[dense canyon]

area of a trapezoid

area = (1/2) · (p + q) · h

[wide valley]

Where h is 48, I assume

[dense canyon]

yes

[wide valley]

,calc 1/2 ( 50+41) 48

[somber coyote]

Result:

2184

[wide valley]

Okay so you added that to 274.16125?

Or did you subtract 274.16125 from 2184?

,calc 2184-274.16125

[somber coyote]

Result:

1909.83875

[wide valley]

You wrote 1899.45

I suppose you got a different answer than me for 274.16125

At any rate, good job solving the question, I learnt something new in the process 🙂

[dense canyon]

cool

thanks

[grave zealot]

Does anyone know what does "|||" symbol means when talking about triangles?

[opaque thunder]

how

[upper karma]

similar?

i had a very old book that had that notation

[lucid breach]

somebody help plsss

😰

[outer sinew]

Are there any pythagorean triplets where a²+b²=c² and a+b=c?

[vivid plinth]

No, because then $(a+b)^2=a^2+b^2$, which leads to $2ab=0$, which means that either a or b would have to equal 0 which is an impossible side length for a triangle.

[somber coyote]

Northstar

[outer sinew]

Ah

[outer sinew]

How'd u come to the conclusion that this would imply that tho?

[vivid plinth]

Wdym?

[outer sinew]

Like how'd u know that a²+b²=c² and a+b=c implies a²+b²=(a+b)²

Oh

...mb (Square the 2nd equation)

LMFAO

Tyty

[vivid plinth]

$a+b=c$ so $(a+b)^2=c^2$ so $(a+b)^2=a^2+b^2$

[vivid plinth]

Np

[outer sinew]

Yeah... got that thanks man

[vivid plinth]

Lol I just noticed arrows don't work with latex well

I'm gonna edit the message

[somber coyote]

Northstar

[vivid plinth]

Ok that's better

[outer sinew]

Lmao did u try \implies or \to?

[vivid plinth]

No I didn't lol

I haven't used latex before

[outer sinew]

They look better

$a+b=c \to (a+b)^2=c^2 \to (a+b)^2=a^2+b^2$

[vivid plinth]

Ya that's definitely better

[somber coyote]

Kiameimon | Welt Rene

[outer sinew]

Usually it's used for stuff like summation notation but hey it works here fine too

[vivid plinth]

I'm pretty active on the help forum but every now and then I find a situation where the latex bot would be helpful in explaining something or another.

[outer sinew]

Wait whaaaaaaaaaa

[somber coyote]

Northstar

[vivid plinth]

Made a slight adjustment

[somber coyote]

Northstar

[vivid plinth]

It works because when you rotate a line 90 degrees, its slope $x$ becomes $-\frac{1}{x}$, and $tan^{-1}(x)$ gives you the line's angular displacement off of the x axis, so $tan(tan^{-1}(x)\pm90)$ makes the two expressions equal (with a few conditions).

[somber coyote]

Northstar

[vivid plinth]

Oops I made a small mistake earlier

I meant $x(tan(tan^{-1}(x)\pm90))=-1$ if x is not the tangent value of a multiple of $\pm90$ that is not a multiple of $\pm180$

[somber coyote]

Northstar

[vivid plinth]

Ok ykw using latex is too much work I'll just let you figure out what I mean

[dark sparrow]

(currently being resolved in #help-14 )

[vivid plinth]

The easiest way for me to describe this is with this graph:

The function equals -1 for most values of x...but not all. You can try it out and see for yourself.

The strange thing that I do not understand is that when you create an equality to -1, the graph of the intersection is unresolved...

[coarse lance]

Given that a car is traveling at 30 miles per hour, and each wheel is spinning at 2000 rotations per minute, can you figure out the circumference of the wheel in inches?

[dark sparrow]

@coarse lance do you still need help w this

[coarse lance]

yeah

[dark sparrow]

any progress?

[vivid plinth]

This looks like dimensional analysis to me. Do you know how to solve such problems?

[coarse lance]

not sure, I have not heard of dimensional analysis

[vivid plinth]

It's just a fancy term. Try to make the information given to you into two expressions, and then find a common denominator to equate them and solve.

[dark sparrow]

idt you need dimensional analysis as such

it's just unit conversion

within the same quantities

ish

@coarse lance you will need to know how many inches are in a mile and how many minutes are in an hour.

[coarse lance]

63360 inches in a mile, 120000 rotations per hour

[dark sparrow]

okay

so how many inches does your car travel in 1 hour

[vivid plinth]

Mb there's one more step

[coarse lance]

1,900,800

[dark sparrow]

ok cool

that is 120,000 times the circumference of the wheels

[coarse lance]

Whoa

so the circumference of a wheel in inches equals inches per hour over rotations per hour

[vivid plinth]

Yes

So what do you get if you use that ratio?

[coarse lance]

15.84 inches the the circumference

[vivid plinth]

To put this into perspective, $30$ miles per hour $\to 30\cdot5280$ feet per hour $\to 30\cdot5280\cdot12$ inches per hour, so the car is traveling $1,900,800$ inches per hour. With the rotational information, each wheel is spinning at $2000$ rotations per minute $\to 2000\cdot60 \to 120,000$ rotations per hour. Now we can equate the two expressions since they have a common denominator. $1,900,800$ inches = $120,000$ rotations, so the circumference of the wheel is 15.84 inches.

[somber coyote]

Northstar

[coarse lance]

i am a bit confused on how distance traveled in one hour over rph gives the circumference

how do we know thats not giving the diamater or another measurement of the wheel?

[vivid plinth]

It's just in this case

[vivid plinth]

The circumference of the wheel is what's touching the ground when it rotates

[coarse lance]

ohh right

well pretty cool math to know tbh

i wanna verify the accuracy with a test drive

[dark sparrow]

this is not math this is geometry

also if youre gonna do a test drive make sure you maintain a constant velocity throughout and measure exactly how long you maintain it for and your tachymeter reading

[coarse lance]

shouldn't it give the same result regardless? Assuming my tire doesn't pop and the circumference stays the same

[dark sparrow]

this scenario assumes you have a constant tachymeter reading and velocity for some length of time

[sour cedar]

I need help -.-*

Oh #help

My bad

[rancid heart]

someone help, its 46/2 right?

[torn gazelle]

Could someone help with this?

[obtuse patio]

!show

[lime crown]

Show your work, and if possible, explain where you are stuck.

[obtuse patio]

draw it out large

or use colored pen on top

[queen trellis]

Does anyone know the equation to this?

[outer sinew]

complete the square

[edgy cradle]

hey

Can someone give me some trigonometry tips?

[torn gazelle]

For what?

I found out the area of the white circles, which is just pi, I’m stuck on what I do next…

[karmic path]

HELP ME BRO

HELP

[obtuse patio]

To draw lengths

To figure out what is going on

[torn gazelle]

I already figured it out

Can you help me real quick

[obtuse patio]

yeah but you didn’t understand why I said something

[torn gazelle]

It’s another problem

[obtuse patio]

justask

If I can’t someone else can

[torn gazelle]

Nobody is willing

[obtuse patio]

cause you didn’t show any effort

and you’re showing 3 problems

[torn gazelle]

I didn’t show any effor?

Cause I didn’t understand it…

[obtuse patio]

Show your attempt

And you didn’t explain what you don’t understand

We cant read your mind

[torn gazelle]

Ok I’ll show this

Is problem c correct?

In a 30-60-90 triangle this hypotenuse is double the short side, so is 1/2R double that?

[snow crystal]

yeah that's right

[torn gazelle]

Then what would the area of the triangle be?

It would be 1/2b•h

So 1/2(r)•1/2r?

[snow crystal]

but what is the lenght of the base

[torn gazelle]

r

[snow crystal]

no

base can't be as long as the hypotenuse

use the fact that it's a 30-60-90 triangle

[torn gazelle]

So AC is the base?

[snow crystal]

it could be AC or OC

it's up to you

[torn gazelle]

AC

So then 1/2(1/2r)•r

?

[snow crystal]

no, you're not multiplying by hypotenuse

what is the height of the triangle if AC is the base

[torn gazelle]

OC

Oh so it’s

Radical 3

[snow crystal]

what would the length be in terms of r

[torn gazelle]

Radical 3r

(Radical 3)r

[snow crystal]

no

you said that AC is 1/2r

[torn gazelle]

Yeah

[snow crystal]

and now you have to multiply it by sqrt(3)

[torn gazelle]

And the longer leg is radical 3 times the short leg

[torn gazelle]

So then 1/2rsqrt3

[snow crystal]

yeah

your 2a is r

[torn gazelle]

Ok

So then the area is 1/2(R)• radical3?

Cause the base is R

And the height is radical 3

[snow crystal]

the height is 1/2 multiplied by sqrt(3)

rad3/2

the base is 1/2r as you said

[torn gazelle]

That’s the area?

[snow crystal]

no

now you have to use it to evaluate the area

[torn gazelle]

And the height is r right?

[snow crystal]

1/2 * base * height

[snow crystal]

no, r is the hypotenuse

[torn gazelle]

So what’s the height?

1/2sqrt3?

[snow crystal]

yup

[torn gazelle]

So this is it?

Then I need to do that right?

[snow crystal]

yeah

it should be r*rad3/2 actually

bc it's r/2 * rad3

[snow crystal]

use this

[torn gazelle]

SOo

@snow crystal it would be rad3/4?

As the area

[snow crystal]

rad3/8

[torn gazelle]

As the area

[snow crystal]

yeah

222

wait wtf

its 2 * 2 * 2

[torn gazelle]

Ok

Thank you so much

[snow crystal]

np

[torn gazelle]

Then it says

Find the area of the dodecagon in terms of r

[obtuse patio]

try it out on your own first

you literally just finished part 1

[torn gazelle]

It would be rad3/8•12

Right

[snow crystal]

@torn gazelle the area was actually supposed to be r^2*rad3/8

I made a mistake

[torn gazelle]

Oh

Do I leave it like that @snow crystal

[snow crystal]

yeah that's right

[torn gazelle]

Find the area of dodecagon in terms of r

So then what?

That’s one triangle and there are 12

[snow crystal]

but those triangles are different from the one which we just calculated

use the formula area = 1/2 * sin(of the angle between two sides) * the two sides that create the angle

I gotta go

[torn gazelle]

Oh no

Can we finish this rq

[snow crystal]

you know something about trigonometry?

[torn gazelle]

A little

[snow crystal]

you use this formula and multiply everything by 12

done.

[torn gazelle]

Area= what over 2?

[snow crystal]

1

[torn gazelle]

And the angle would be 360 divided by 12 right

[snow crystal]

bro you said it yourself that the angle is 30°

[torn gazelle]

/

So that’s what the formula is

Sin30 is .5

So 1/2 and .5 cancel

So then the area is equal to 12r^2?

@snow crystal

[snow crystal]

why would 1/2 * 1/2 cancel out

@torn gazelle

maybe at least I'll get the active role lol

[torn gazelle]

Oh whoops😂

In your picture it looks like it says it’s subtracting

[snow crystal]

ik it's kinda hard to make clear drawings in paint lol

[torn gazelle]

It would be 1r^2

Times 12

Then it’s 12r^2

So the same thing?

[snow crystal]

whered you get that from

[torn gazelle]

.25*

[snow crystal]

so what is your final answer

[torn gazelle]

3r^2

[snow crystal]

LMAO

hell nah

[torn gazelle]

.25 times 12?

[snow crystal]

what is the area of a single triangle

[torn gazelle]

Idk

Would that be the .25r

[snow crystal]

no

figure it out

[torn gazelle]

I’m confused

[snow crystal]

you've got the formula

put some effort in it

[torn gazelle]

I already used the formula

[snow crystal]

you used it wrong

[torn gazelle]

See

Area=1/2•1/2•r•r

So then we do 1/2•1/2

.25•r•r

.25r^2

[snow crystal]

yup you got it right

[torn gazelle]

Then that times 12?

Cause that’s just one triangle

[snow crystal]

yeah!

[torn gazelle]

.25 times 12

Is 3

[snow crystal]

you forgot about something

[torn gazelle]

3r^2

[snow crystal]

yes thats the right answer

[torn gazelle]

Bruh

Look

[torn gazelle]

^

Had that before

[snow crystal]

thought you were talking about a single triangle

[torn gazelle]

Oh ok

[snow crystal]

either way you got the right answer

[torn gazelle]

Yes thank you

The final question says, Why is the formula for the area of a regular dodecagon so much like the formula for the area of a circle?

[snow crystal]

what's the formula for circle area

[torn gazelle]

Pi r^2

And the formula for the dodecagon is Mr^2

[snow crystal]

uhh, M?

[torn gazelle]

Bro I’m dumb

That’s what it says right here

Mr^2

[snow crystal]

bro you just computed the area of that dodecagon

wtf are you saying rn

[torn gazelle]

I’m lost

😂

Low key tired

[snow crystal]

you said it's 3r^2

[torn gazelle]

PiR^2 and 3r^2

[snow crystal]

see any similarities yet?

[torn gazelle]

So it’s just .14159 off

[snow crystal]

yuh

[torn gazelle]

Awesome

For the true or false questions, do I have the correct answers?

For D I need to drawn it out

[snow crystal]

nah bro my head is falling off

[trail pebble]

Idk what to do all of these look right

[upper karma]

can someone help me w my homework plsss

@everyone

[worthy matrix]

dont ask to ask

just do it

[proven zodiac]

I need help with Cavalieri's Principle

When do I use V=w * h * l and V=3.14r^2h

also is there any yt vids for studying the EOC for Geometry?

[short echo]

The first one V = whl is for the volume of a rectangular prism

And the second one is the equation for volume of a cylinder

[wheat crest]

I need help on these questions.

[snow crystal]

maybe draw them out first

[torn gazelle]

Ayy could you help me real quick.

For this, the central angle is 40deg

So how would I find arc ab?

[snow crystal]

in terms of radius?

or like an actual number

[torn gazelle]

The measure of the arc

So the arc angle

[snow crystal]

divide the nonagon into equal triangles

and try to figure out what to do next

[torn gazelle]

If the central angle is 40deg, wouldn’t the arc be double so 80deg?

[dry current]

Can someone give me the answer to this PELASE and explain I’m very confused

[snow crystal]

what's the formula for the arc

[torn gazelle]

Idk

[mental bobcat]

AE = AD because tangents from the same external point are equal. You can do the same with CD. Then add AD and CD

[snow crystal]

$$L\ =\ \frac{\theta}{360°}\cdot2\pi r$$

[somber coyote]

IamMax420

[upper karma]

aaaaa what is this

[outer sinew]

Just take area of the circle minus the rectangle

[inner stag]

[violet raft]

do you know sohcahtoa?

[inner stag]

Nope not at all-

[violet raft]

well i would look at that first.

sin cos tan

[dapper lodge]

i got you

[dapper lodge]

shouldnt they teach u that if its a question 💀

[inner stag]

My math teacher is the type of guy to give you the homework and not give any explaination

He gave us all our work for the quarter at the start and its our job to finish it all

[dapper lodge]

i took a ton of notes on this lol, i didnt understand it at all either

[inner stag]

Ty dawg

[inner stag]

Its kinda wack trynna understand this w no context of any work-

[dapper lodge]

my math grade dropped to a 29% once 💀 but i got it back up

[dapper lodge]

hold on im not done

[inner stag]

My teacher hasn't taught us anything throughout the year

[inner stag]

Oh not you dw-

Mb if you sounded aimed towards you

[dapper lodge]

here's examples of sine, tangent, and cosine

[inner stag]

Tysm

[dapper lodge]

do u have a scientific calculator?

[inner stag]

Uhh i dont think so, i can try use one online

[dapper lodge]

oh yea you'll need one for sin, cos, and tan

look into buying one, or if ur teacher has some in class

or look it up but theres different ones idk how reliable itll be

but

[inner stag]

Ill look tmr

Ill try use one online today

[dapper lodge]

alr

im sure u can get it

do u have to learn about finding the angles of a right triangle with only side lengths too?

[inner stag]

I think so?

[dapper lodge]

thats where the sin^-1 comes in

idk im sorry, i hope my notes helped, they helped me a lot but i cant teach you good over online

oh yea dont mind the mexican restaurant at the top of the page 😭

just remember

hypotenuse= across from 90 degrees
opposite= across from the given angle
adjacent= line between given angle and 90 degrees

the notes are a little messed up, where it says sin(ø)=opposite/hypotnenuse/adjacent

the adjacent isnt supposed to be there, its a label for the triangle

[raw forum]

BCDE is a square.

What theorem proves that triangle ABC is similar to triangle CEF

[dapper lodge]

sorry mate ik theyre similar but idk what can officially prove it

havent learned that

[upper karma]

just make up ur own name for it

il798li theorum

too late i already named it

exeterishere theorum

too bad now i get the credit

[dapper lodge]

whats the theorem tho 💀

[upper karma]

if a square is placed inside a right triangle, where the corners of the square are touching the hypotenus, opposite, and adjacent side of the triangle, the two triangle that are separated from said square are similar

best i could do

actually wait

ok one of the corners has to touch another corner of the triangle

[raw forum]

but whats the name of the theorem

[snow crystal]

They're similar bc their angles are the same

I don't think that it has its specific name

[undone moat]

The naming of the sides are part of the Pythagorean theorem.

[brazen veldt]

it is called the corresponding angle theorem. sorry if im too late i just joined the server and had to wait 10 mins to type. basically because BC and DF are parallel the left F corner on the side between the two parallels is just as big as the left C corner on the outside of the two parallels.

[zealous lotus]

Anyone know?

[twin crag]

how much of geometry requires a protractor?

like is it really required?

[timber cargo]

no

[worthy matrix]

power of a point

i've also heard that theorem called chord-chord/chord-secant/tangent-secant, etc

[ember stratus]

Points $A, B, C, D, E$ match the following.
$$AC = AE, \angle CAD = 2 \angle CBD, \angle BAE = 2 \angle BDE$$Find all values of $\frac{AB}{AD}$.

[somber coyote]

aSome1gussy

[twin jungle]

yay i love trigonometry

[ember stratus]

then solve the problem

jiangster

[foggy parcel]

<@&286206848099549185>

we have to get the area of the half circle

thought it in a multiple ways but still got stuck

getting the value of AB would help

[lucid jasper]

If sin(6x)/sin(2x) = 5/2

The value of

cos(6x)/cos(2x) is ?

[prisma raft]

seems right no?

[foggy parcel]

yea seems fine to me thanks dude

[prisma raft]

you are welcome

[frozen ocean]

[shell vale]

i am 14 can someone dedicate their time in helping me with geometry and trigonometry

[runic beacon]

if you have any problems you are having trouble with, you can ask them here in the server, maybe in one of the available help channels. that way, if anyone is able and willing to help you out, they can do so

[neat python]

Hi can someone tell me the area of axles in conus

Ty btw

[grave pond]

"The area of axles in conus" doesn't make much immediate sense, even if "conus" is a typo for "cones". Can you try to state your question in different words?

[neat python]

I don't know that word in English ah

What about shaft of S

Maybe it's correct word for it XD

[grave pond]

Sorry, still doesn't make sense. Are you willing to show your problem in the original language -- someone might know it?

[neat python]

I don't think so if anyone there would know that language XD

[neat python]

What about pivot of S

[grave pond]

What is S?

[neat python]

Area

[grave pond]

Huh. I'm completely lost, sorry.

[neat python]

I will tell evey area formulas and one won't be there XD

S base= πR^2

S lateral = πRl

S complete = πR(l+R)

And one left that I don't know

😀

@grave pond

Now?

Or still lost XD

[grave pond]

No, I have no idea what you're talking about.

[neat python]

K

<@&286206848099549185>

[main grove]

how to convert rect coord (0,-pi) to polar coord... always choose angle theta to be in interval (-pi,pi]

i got r but having problems with the angle part..

<@&286206848099549185>

would it be undefined?

[main grove]

omg nvm i am so dumb I got it

[somber bay]

yo

need some help

You slide a 1 m wide table along the straight wall of a 2 m wide corridor, as in the figure below. How far from B, in meters, can you bring corner C of the table, so that you can still fully open the door [AB]?

[young violet]

The diameter of a circle has endpoints P(-6,-11 ) and Q(4, 7).

Write an equation for the circle. Be sure to show and explain all work.

To write the equation of the circle, we need to find the center and the radius.

First, we can find the coordinates of the center of the circle by using the midpoint formula:

x-coordinate of center = (x-coordinate of P + x-coordinate of Q)/2 = (-6 + 4)/2 = -1
y-coordinate of center = (y-coordinate of P + y-coordinate of Q)/2 = (-11 + 7)/2 = -2

So, the center of the circle is (-1, -2).

Next, we can find the radius of the circle using the distance formula between the center and one of the endpoints of the diameter:

r = sqrt[(x2 - x1)^2 + (y2 - y1)^2] = sqrt[(-1 - (-6))^2 + (-2 - (-11))^2] = sqrt[49 + 81] = sqrt(130)

Therefore, the equation of the circle is:

(x + 1)^2 + (y + 2)^2 = 130

This is the standard form of the equation of a circle with center (-1, -2) and radius sqrt(130).

is this right?

[gilded rampart]

What is the formal definition we need to prove?

Two vectors are parallel? How is that defined in vectors?

[near harness]

You have to show that two opposite sides of the quadrilateral formed by joining all the midpoints are equal and parallel. Because if that is true, the shape formed will be a parallelogram

[fringe ridge]

Someone please explain a segment to me

[umbral nymph]

why u swearing at me

SAay it to my

[digital knoll]

hi

[digital knoll]

a segment is just the area of a sector minus the area of the triangle that can be drawn from the three points of the sector

[tawny skiff]

Hi, what are the zero's of cos

or what is cos(x) = 0

i know sin(x) = 0 is n * pie

[flat pawn]

(1/2)pi + (n*pi)

Isn't it better to make the connection using graphs?

Talking about segments. I still don't get why vertices aren't called edges. Also, why aren't edges called line segments?

[wary ember]

someone can help me here

[slow merlin]

hi, can someone please explain to me what dilations, transformations, and translations are? im p confused on it and i have state testing tomm so

thanks

[frozen ocean]

what have you tried so far?

hint: use reciprocal trig identities formula

[wary ember]

hi, ive done another problem but im struggled with this one

[wary ember]

well, that's nothing

[frozen ocean]

wdym that nothing lol

,tex .recip trig

[somber coyote]

Akira

[frozen ocean]

this is what I meant

[wary ember]

yeah of course i mean that´s too obvious

but forget it man

[frozen ocean]

what is the question asking you

[obtuse patio]

Google it

[wary ember]

someone to help me to solve it, that´s it, is just a problem

no 2 or 3 or 4
just one, and im not gonna bother here anymore

[frozen ocean]

do you know how to prove the trig identities?

[wary ember]

well

i already did this problem

this one

but this was 10x easier

[frozen ocean]

nice

[wary ember]

xd

forget it see you

dafaq with this serer

[short echo]

https://www.desmos.com/calculator/th4jvdbsrn

alright

i made sphere

[foggy parcel]

Given Equilateral triangle and stacks of infinite number of semicircles inside. Solve for the total area of the semi circles.

Any hint would be appreciated

<@&286206848099549185>

[foggy parcel]

in the 2nd pic is the triangle's upper two sides equal?

[spare ore]

i assume the secnod one is the top right one? dont really know what you're asking, but I assume ur talking about the two upper tangent lines? cause yes

[spare ore]

@woven shoal

[woven shoal]

What is asked

[spare ore]

i need to find x y and z

[woven shoal]

What else is given?

[spare ore]

nothing

[woven shoal]

Okay what is x?

[spare ore]

idk

[woven shoal]

Is x, 42?

[spare ore]

the angles are 69

no

[woven shoal]

Eh

?

What is x then, if the top angle is 42.

[spare ore]

That's what im trying to figure out

the arc

si

but I can't find it

[woven shoal]

This data isnt enough.

Well wait

It is ig

I didn't see that the two angles with the horizontal were equal

Well, its enough for finding angles.

I just took x as 42 degrees

Otherwise there's no way you can find it

Arc length

??

[woven shoal]

...?

You can't find arc length using this data

Its insufficient

...

Radius!

Let's become radius people

[pliant tapir]

yooo

who's the teacher here?

or anyone who knows trigonometry more?

ohh greatt

thankss

[dark sparrow]

no such position

the helpers on this server are volunteers

[alpine pagoda]

you need solution now?

if so I can help.

@tiny oriole

[ancient ether]

Please help

[snow crystal]

first, evaluate the third angle

[upper karma]

use the law of sines to calculate the side lengths. Your second angle is 90 degrees and to evaluate the third angle, you have to add the corresponding angles and subtract them from 180.

[ancient ether]

Get a calculator?

[upper karma]

a calculator isnt necessary for most of it aside from evaluating the sides with the sine function

Law of sines: ((a)/(sin(A)))= ((b)/(sin(B)))= ((c)/(sin(C)))

a,b,c are the sides; A,B,C are the angles

[rugged veldt]

Can anyone help?

The diameter of the golf ball is 4.267

[snow crystal]

That monitor tho 💀

[grave pond]

I'm very skeptical that "stack as cubes" is literally what the author of that task imagined.

[rugged veldt]

My teacher said we can assume that

[grave pond]

Then it would be a lot more efficient to make the box rectangular with room for 2×3×2 of those cubes.

Note that two layers each of the arrangement

O O O
 O O
  O

would be too densely packed for cubes, so there would need to be a lot of rattle room if you expand your equilateral-triangle footprint enough to have room for cubes.

[rugged veldt]

Ok so let’s assume that they are spheres

I think the two layers in height means the distance between bases

[steady geyser]

I need to understand the trigonometry part can someone help me?

I have no clue to be honest

Like this one from the middle how can i show that

[undone moat]

Translate please?

I'm so sorry.

[cyan wren]

bottom one i get but top 2 idk

[digital knoll]

a = 6/2

formula for apothem: a = Rcos(pi/n) where n is number of sides (in this case hexagon so 6)

3 = Rcos(pi/6)

rearrange: 3/R = (cos(pi/6))
3/R = (sqrt3/2)
R = 3/(sqrt3/2)
R = 6/(sqrt3)

then we have two sides of a right angled triangle, a is the apothem and R is side c

R^2 - a^2 = h^2 where h is the height of the triangle

(6/(sqrt3))^2 - 3^2 = 3
h = sqrt3

area of the triangle is 1/2 bh
1/2 * asqrt3 = (asqrt3)/2

height of the hexagon is 6 times a
6*3 = 18

therefore the volume of the hexagon is: 18 * ((asqrt3)/2)
= 18asqrt3/2
= 9asqrt3

idk if i did it right or if I even reached the right answer

[digital knoll]

that's my response to question 10

[light sleet]

Is there a name of this colored shape?

Part of sector form?

[snow crystal]

segment?

[digital knoll]

annular sector?

[light sleet]

Oo

Thanks!!

[digital knoll]

npp

[cursive sluice]

does anyone know a good vid covering this topic

[iron pike]

does anyone know how to find the square/rectangle inscribed in a parabola and (limited by the parabula and y=0) knowing the equation of the parabula

pls help

[willow patrol]

not sure if that's what you mean

using this formula you can find the area inside the parabola

[last quiver]

do you mean something like this (area of square/rectangle "inscribed" in a parabola and "limited" by y = 0) , in such a case , ig knowing the coordinates of vertices can help (if the equation of parabola is known, just plug in the value of x in the eqn to get the height of your rectangle/square ) ,and it doesn't require definite integrals to compute the area.

[shadow plaza]

hey there, im in 8th grade and just started geoamitry any tips

[blissful spire]

can someone pls explain why the formula for the volume of a sphere is V=4/3 pi r^3

[cunning acorn]

please help:

[worthy matrix]

just generally in math you shouldnt memorize formulas blindly and instead understand why they work (unless the proof requires some higher math to prove)
also proofs are important so learn them

[worthy matrix]

you can find OP and from there OE since OEP is a right triangle, and with that angle EPO which is equal to angle BPC

[tight lion]

oh

im bad at latex

hold up

if i have $cos\theta_{0} = \frac{v_1sin\theta}{V + v_1cos\theta$ and need to plug it into another function that has an argument of $sin\theta_{0}$ which identity should i use to make this easiest on myself