geometry-and-trigonometry

[upper karma]

I think I know what the ratio of shaded to non shaded is

@high zephyr yea

I was mistaken

I thought it was circumference

Very sorry

[copper valve]

k i just kicked them

[acoustic jungle]

hmmmmm....

[upper karma]

Kicked who

Oh damn

[high zephyr]

why does he think fishraider is being a smartass

[upper karma]

Fish raider does seem smart ngl

@acoustic jungle Your a good boi

👌

[acoustic jungle]

I am not sure

I was so confused.

[high zephyr]

im confused also

[copper valve]

not a big deal

[upper karma]

He mentioned how The Formula is 1/2 R^2 x Theda

What does this mean

He wanted to do it another way

[high zephyr]

that's another formula

[acoustic jungle]

that's the way with radians I believe

[upper karma]

How I do this

[high zephyr]

which question?

[upper karma]

The 31 5.12

And the 8 11

For the following write the correct equation

Is it

Sin(b) 31/c?

[high zephyr]

nono, 31 degrees is the angle here

you can take angle A for trig, which is 31 degrees

[marble parcel]

Do i just rotate each point by 45 degrees?

[high zephyr]

@marble parcel channel seems taken atm

[upper karma]

@high zephyr so how do I write it

[high zephyr]

first, do u know your trig ratios?

[upper karma]

No

[high zephyr]

an easier way to memorise this is SOHCAHTOA

[upper karma]

Ohh yes that

I know those my teacher taught us sohcahtoa

Idk how to do the correct equation tho

It’s for sin

@upper karma what

Idk

Need help wirh msth

Help with what

The write the correct equation

Write correct equation of what

What equation

Sin

Sin(31) = 5.12/c

Since 5.12 is the opposite side

Ohh ty

And c is the hypotenuse

How abt

Sin = opposite/hypotenuse

Is it

Sin(x) 8/11?

Yea

Ty

You could use arcsin to find what x is

But idk if ur trying to get that rn

Not yet

Alright

Your welcome

🙏

[acoustic jungle]

||you're 💩||

[upper karma]

Lmao

[silent plank]

apply the intersecting secants theorem

then solve for x

example of what?
what's the equation you have after applying that theorem?

https://lmgtfy.com/?q=intersecting+secants+theorem

[upper karma]

Bruh

Look it up

Thats like him doing your homework

At least put some effort

X = 66

If 66 is the central angle

[silent plank]

it's not 16

[upper karma]

Oh wait

[silent plank]

read the theorem carefully

[upper karma]

Actually it’s 2

X is 2

Since it’s inscribed angle

@upper karma x = 2 my friend

[silent plank]

can you show your work?

[upper karma]

https://media.discordapp.net/attachments/326138757474680852/702650411311562873/unknown.png

X = 2

Since inscribed angle

Oh

Am I missing something

What is it then

[silent plank]

the 1 is just a label

[upper karma]

Oh

[silent plank]

ok. type your work.

[upper karma]

66 then?

@silent plank

[silent plank]

yes

[upper karma]

Yay

👏

You can do it

Gotta go fast 💨!

[silent plank]

what equation do you have after applying that theorem

[upper karma]

Don’t procrastinate son

Do it earlier in the day

[silent plank]

^

intersecting chords theorem

(because those are chords that are intersecting each other)

[eternal solar]

are you finding X

[novel flax]

use tangent chord theorem: 53 is half of x

[eternal solar]

also it's a triangle

[silent plank]

and also consider alternate segment theorem

[novel flax]

yes

[silent plank]

no

[upper karma]

It’s 4:00 turn in ur homework

California?

[silent plank]

how are you getting those values?

[upper karma]

Nope it isnt mate

Umm wait i might have spoked too far

I didnt said anything yet lol

[silent plank]

that does sound right

[upper karma]

Grats

[silent plank]

waiti mean diesnt

fkn auto correct

doesn't

[acoustic jungle]

x is not 88-64 over 2

[silent plank]

sru

y

[upper karma]

Lol

F

[acoustic jungle]

(88+x)/2=64

wut?

I didn't say it was 64

[upper karma]

👍

[high zephyr]

u're trying to find x right?

[upper karma]

Its an equation

,w (88+x)/2=64

[silent plank]

,w 9(9+x)=12(12+12)

[somber coyote]

Display results online and refine query

Requested by ramonov#0921

[upper karma]

Yes

[acoustic jungle]

what question are you talking about

the angle one or the side one

the side one is 23 but the angle one is 40

[upper karma]

Lmao

[marble parcel]

@high zephyr did you ping me

[high zephyr]

yes, to tell u that the channel is taken

at that time i pinged u

[gray marten]

Hey, so I was thinking about a cone today, and I was trying to derive the volume of the cone.
My first thought was:

• Get circumference of circle: 2πr
• Get area of triangle: ½bh
• And then multiply the two, so it's like a bunch of triangles appearing on the circumference:
• 2πr * ½rh = πr²h

This is clearly wrong. But I'm not sure why.

I imagine its because the relation between area/volume and circumference/perimeter of shapes doesn't work the way I think.

[acoustic jungle]

hmm.

That is not making any sense because you are multiplying by the area of the triangle

it would've made more sense if you did area of base * height /2 instead of /3

[gray marten]

Area of b*h/2 instead of b*h/3?

[high zephyr]

it would've made more sense if you did area of base * height /2 instead of /3
@acoustic jungle he did do base*height /2

[acoustic jungle]

oh no I meant like area of base (circle) * height

[high zephyr]

👌

[gray marten]

Right, the other intuitive way that seems to work is by getting the volume of a pyramid: ⅓lwh
And then replacing the base area: lw
With area of a circle: πr²
Getting you: ⅓πr²h

Im still slightly upset as to why the circumference*triangles didn't work, but I imagine something like the left side of the image happened.

This whole ⅓ thing about cone's pisses me off, but turns out when people make cone cutouts, it usually looks like ⅓ or a circle too.
https://www.firstpalette.com/images/printable-mainpic/cone.png

[acoustic jungle]

oh I see what you mean now.

[gray marten]

It feels unintuitive that, it doesn't work.

[acoustic jungle]

However why do you have 2pi * r inside the circle

[gray marten]

Oh, that's the circumference.

Because I initially imagined the triangles appearing out from the rims of the circle.

[acoustic jungle]

You extended the inside of the circle at origin O to 2pi*r

so you are getting a larger value.

[gray marten]

Oh crap yeah.

[acoustic jungle]

it's like saying the area of the circle is 2pi*r^2

[gray marten]

So yeah, it'd be impossible for that to work, since no matter what I do, there's going to be the big gap in the middle of the cone.

[acoustic jungle]

Well the thing is you can't even make a cone with that.

since it's a rectangle.

[gray marten]

Yeah big rip.

I think my misunderstanding is coming from mixing up circles/cones with how squares/rectangles work.

[brazen quarry]

How do i figure this out i need a little help

This is a problem on my homework

[acoustic jungle]

,rotate

[acoustic jungle]

inscribed angle theorem.

that's all you need.

[brazen quarry]

Appreciate it

[upper karma]

please help me undersatnd this problems

[remote heart]

what do you need help on?

[upper karma]

hrm

i need some help in understanding the relationship between 79, then 118

and then separately, 54 and x

errr uhhh. like how to solve it

[humble pebble]

i think its around 30

that sounds right

@upper karma

[upper karma]

okayy

why is it around 30 ?

[tawdry mural]

hi

so

how would one

solve this

from my hw assignment

Determine the height of the hypotenuse if alpha = 34 ° 12'33 '' and b = 9 cm.

[high zephyr]

do u mind sending the question?

which side is b?

[rose sphinx]

Length of hypotenuse will sound more grammatically correct

[high zephyr]

oh he left the server

[acoustic jungle]

@upper karma it's not around 30, it should be 15 off from 30, use the secant secant angle theorem.

[onyx epoch]

how do i get the area of ABC

i know the coordinates for A,B and the angle v

and also calculated the circles equation

[acoustic jungle]

What unit is this ? (I don't know how to do it either)

[onyx epoch]

unit?

[acoustic jungle]

like what unit is this

[onyx epoch]

what is a unit? sry english isn't my main language

[acoustic jungle]

What have you learned during this time when solving this question.

[onyx epoch]

I've calculated the circles equation and the equation for the tangent going through B

if that's what u mean

[acoustic jungle]

What I mean is, what similar questions have you done, what is this question related to?

[onyx epoch]

Well it's just part of some excercises for the summers exams

it's all mixed stuff

[acoustic jungle]

is AB the diameter

[onyx epoch]

Yes

[acoustic jungle]

OH

that should be easy then

ACB is a right triangle

[onyx epoch]

yep

[acoustic jungle]

calculate length of AC and CB

[solid swan]

i thought it was the area including the curved part

[acoustic jungle]

then multiply then divide by 2.

[onyx epoch]

It is

[acoustic jungle]

and add the area of that chord thing (the central angle is 2 times that angle)

[onyx epoch]

how do i get length of CB tho

i don't know coordinates of C

[acoustic jungle]

Is the question supposed to be tedious.

[onyx epoch]

well idk

just supposed to find the simples way of solving it

[acoustic jungle]

You can find the length of CB and AC using trig since ACB is a right triangle

[onyx epoch]

how

[acoustic jungle]

and you know the length of AB using the length formula.

[onyx epoch]

i was thinking of making a straight line through AC and finding C from intersection with circle

[acoustic jungle]

Well you wouldn't need to find the exact coordinates

if yo know the lengths you can find the area of the triangle

[onyx epoch]

but how

[acoustic jungle]

I'm really not sure if it's the easiest way, because after finding the area of the triangle you have to calculate the area of that chord by doing 2v/360 * pi *r^2 - area of small triangle OBC

I'll post your question again, maybe someone will have an easier way

[onyx epoch]

how do i get the area of ABC
i know the coordinates for A,B and the angle v
and also calculated the circles equation

[solid swan]

triangle abc?

or the whole slice of pizza

[acoustic jungle]

You can find the area of ABC by finding AC and BC by doing sin(v) = BC/AB

but this is probably not a good way to do the problem.

[fleet wolf]

draw the centre of the circle. Call it O, say and observe that it is on AB. Draw line OC. This will split your region into two areas, triangle AOC and sector BOC. Find the areas of both and add them.

[solid swan]

yeah and the angle for BOC would be 2v because of angle at center

[onyx epoch]

wouldn't i need the coordinates for C for all of these solutions?

[solid swan]

no not really

you have the equation for the circle right?

then you also have radius

[onyx epoch]

yea

oOOOH

OC is also radius

[solid swan]

area of sector BOC is r²(2v)/2
area of triangle AOC is r²sin(π-2v)/2

i think

[acoustic jungle]

That's a much better way than what I had.

[onyx epoch]

area of sector BOC is r²(2v)/2
area of triangle AOC is r²sin(π-2v)/2
@solid swan shouldn't it be BOC is r^2*sin(2v)?

[solid swan]

area of a sector is r²θ/2

and area of triangle is ab(sinθ)/2

[onyx epoch]

aaah

i thought sector was just an area

mb

i split it like this

[solid swan]

A_2 + A_3 still gives the same thing so works too

[acoustic jungle]

If you are given the angle in degrees you should use degrees to find that sector PBC

40/360 * pi *r^2

and to find triangle APC use the 1/2 ab*sin(c)

I just realized something, APC and PBC have the same area? (The triangles)

since they are both r^2 * sin(40)/2

[solid swan]

wait no

angle APC is 180° - v_0

which is 140°

oh

nevermind i forgot about identities

[onyx epoch]

they have same area yea 😛

[solid swan]

cool

but is that generally true

like if i slide c across

no that wouldn't

[acoustic jungle]

I think it will.

[solid swan]

but at 90 degrees though

let PC be perpendicular to AB

then area of triangle APC should be equal to that of triangle BPC

but what about the remaining part

[acoustic jungle]

angle APC is always 180 - angle BPC

I believe they are always equal.

since the area of APC is always r^2 sin(180 - BPC)/2

or just r^2 sin(BPC)/2

[solid swan]

wait no no no no

sinθ = θ only applies for small angles of θ only

the formulas are different

area of sector is r²θ/2
area of triangle is ab(sinθ)/2

there's no sine in area of sector

[acoustic jungle]

I was talking about the triangle.

[solid swan]

yeah i guess

because the area that gets left out increases as the angle gets larger

[stark snow]

Probably I'm trying to be too smart for my own good by using De Moivre's theorem

Why did this way not work?

(the reason why this way immediately jumped out to me is because I thought it would give me an easy way to evaluate out the sum)

[fleet wolf]

what did you do on line 2?

looks like you said $s_n=-e^{i\theta}s_n$

[somber coyote]

rockpaperscissors:

[upper karma]

Need a bit of help—

When given an intersection on the unit circle, I need to determine an angle measure.

And I have no idea how to do that

“Finding angle measures (in degrees) in standard position”

How do I do that?

[stark snow]

@fleet wolf I tried to multiply through so that I could get a term with k+1 so that I could cancel stuf fout

Like

You know

The standard trick for evaluating a geometric series

[upper karma]

;-;

<@&286206848099549185> my question hasn’t been answered in 30 minutes...

[stark snow]

For your question

Draw a diagram

Take the x coordinate of the point of intersection

You then have how far along it is on the x axis and you know the radius of the unit circle

You can then find the angle using the standard trigonometric functions

Or the y coordinate

Depends on what you want to do

[upper karma]

ok...

Thanks.

I think I got it.

[stark snow]

Good

[stark snow]

<@&286206848099549185> It's been an hour and a half please help me

[upper karma]

hi

Can anyone dm me

[gray marten]

o yea btw, most people ehre dont like doign questions in dm

[acoustic jungle]

Why not?

oh there is no bot I guess.

[upper karma]

Yeah

Im included in that ppl

[upper karma]

@acoustic jungle thank u

[upper karma]

hi

[agile tree]

hello

[upper karma]

is anyone willing to hellp me with a lot of problems

law of sines

[humble quail]

Send problems here

[upper karma]

determine the remaining sides and angles of the triangle ABC A=140 40', C=20 20', AB=9

140 and 20 are degrees

does anyone know how to do this u need to find the area of the shaded area

[humble quail]

@upper karma
I don't have functions of degrees with minutes on my calculator, so calculate it yourself pls :<

[novel flax]

@upper karma find the radius of the small circle and big circle, then subtract the area of the small circle from the area of the big circle

[upper karma]

do i subtract 21-3.5 to find the diameter?

[novel flax]

the diameter of the big circle is 21

and the diameter of the smaller circle is 21-2(3.5)

[upper karma]

which is 7

radius

i got the ans

there is another diffuclt one

idk how to do

[novel flax]

ok

[upper karma]

i found the parallel 11.2

i think that is correct

[novel flax]

draw a line connecting the vertices at the top @upper karma

you will find that the shaded area is the area of parallelogram minus right triangle

[upper karma]

how do i find the other side for the rectangle

[novel flax]

what rectangle

draw the line first

[upper karma]

where

where is the top

[novel flax]

the two points

[upper karma]

the parallel points?

[novel flax]

yes

[upper karma]

ok

likes this

@novel flax hello likes this?

[novel flax]

,rotate

[novel flax]

no

the line should be horizontal @upper karma

[upper karma]

the on in the top?

it is

[novel flax]

draw this blue line @upper karma

[upper karma]

i did it just didn't show

[novel flax]

oh

[upper karma]

whts a phase shift

i did the ploy thm

and is it 25.81?

for the base

[novel flax]

which side @upper karma

you move the basic sinusoidal function vertically or horizontally @upper karma

[upper karma]

the bottom

so

is the answer a graph

or a number

[novel flax]

The Phase Shift is how far the function is shifted horizontally from the usual position

the bottom side is 23.8

[novel flax]

@upper karma

oh the phase shift for that particular equation

[upper karma]

ya

[novel flax]

is pi/2

[upper karma]

so 2PI

[novel flax]

pi over 2

or one half pi

[upper karma]

how do i do this then

[novel flax]

find amplitude, period, and midline

also it helps to identify whether it's better to use a cosine or sine graph

[upper karma]

can u help me wit this @novel flax

[novel flax]

the diagonal of the square is the diameter of the circle

[upper karma]

its 8?

[novel flax]

the diagonal is 8root 2 which is also the length of the diameter

the radius is half of the diameter so radius is 4root2

the area of the circle is the radius squared so it's 32 pi

the area of the square is 8*8 = 64 so subtract it from the area of the circle

and u get 32pi-64

[upper karma]

isn't 4 squared 16?

how did u get 32

[novel flax]

(4root2)

squared

[upper karma]

ok ty

for the help 😄

[jolly bear]

I don’t understand why 2b is wrong

[acoustic jungle]

Do you mean 2c.

[jolly bear]

No, I know 2c is wrong

2b

[acoustic jungle]

which part is wrong on 2 b

the negative sign?

[jolly bear]

Yes

I don’t understand

[acoustic jungle]

cos u is -4/5

[jolly bear]

Oh

I see why now

Wait

[acoustic jungle]

also your teacher is very smart to put two variables that look like the same thing

[jolly bear]

What variable

[acoustic jungle]

lol the u and the v

[jolly bear]

Lol

[acoustic jungle]

sin v should be negative as well

[jolly bear]

Hm

My teacher is so lenient lol

[acoustic jungle]

I agree.

[jolly bear]

Make sense now

[granite cypress]

Hello

I just need help with an exponents question

[upper karma]

Algebra

[rich wolf]

Algebra

[granite cypress]

[rich wolf]

@granite cypress what is it

Dont hesitate to ask

[granite cypress]

Sorry i just needed to answer this

[silent plank]

is that 70 or 7°?

[rich wolf]

Whats that thingy next to the seven

[granite cypress]

0

[rich wolf]

70

?

[granite cypress]

no its 7^0

[rich wolf]

Oh

[granite cypress]

its the exponent

[rich wolf]

0s are ovals

Not circles

[granite cypress]

hehe sory

[rich wolf]

I think romanoff got this

Cya

[silent plank]

I guess that's a good place to start.
what's 7^0?

[granite cypress]

Would it be 1?

[silent plank]

yes

[granite cypress]

ok

[silent plank]

do you know how negative exponents work?

[granite cypress]

They are not negative they just only are being like divided which makes them positive

like

a^2 = 1/a^-2

like that?

[silent plank]

negative exponents is a term used to refer to exponents expressed with a negative power

yes.

[granite cypress]

oh okay.

[silent plank]

so what would 3^(-2) be?

[granite cypress]

It would be 5?

sorry i dont know.

[silent plank]

apply what you just posted moments ago.
maybe rearrange it a bit

[granite cypress]

Hm.

oh 9.

right?

[silent plank]

3^(-2) isn't 9

[granite cypress]

So I need to use the negative exponent?

[silent plank]

I mean the exponent is negative here...

[granite cypress]

so..

1/3^2

[silent plank]

yes

[granite cypress]

oh wow

I even thought about it

before

[silent plank]

and overall what will the entire expression simplify to?

[granite cypress]

1/3² * 5/7⁰

[silent plank]

simplify everything

[granite cypress]

5/9?

[silent plank]

yep

[granite cypress]

Wow

tysm

[silent plank]

also what's this doing in the geotrig channel

[granite cypress]

lol idk

[rich wolf]

Its not that out of place lol

I learned factoring quadratics in Geometry class

Even though polynomials are more algebra

(x+a)(x+b)=x^2+(a+b)x+ab

Isnt that cool

[frail rock]

Is this channel open? I was kinda hoping to ask about the Hinge Theorem, I don't think I quite understand it.

[jolly bear]

How do u do 4

[rich wolf]

Ill break this down into four steps

[jolly bear]

Yes

[rich wolf]

Find the midline

[jolly bear]

What’s the midline

[rich wolf]

Omg rly

The middle line

Of a wave

[jolly bear]

U mean like the symmetry of y

Then shouldn’t it be 6

[rich wolf]

Middle line

Midline

[jolly bear]

That’s what I thought

But I never heard of midline

What do we do now

[rich wolf]

Find the midline of the function

[jolly bear]

Y=6

[rich wolf]

It is in the middle between the maxes and mins

Yes

You take the avg of 10 and 2

[jolly bear]

Yes yes, continue

[rich wolf]

Now you must find the amplitude

[humble quail]

Is this channel open? I was kinda hoping to ask about the Hinge Theorem, I don't think I quite understand it.
@frail rock
Why don't you understand it? It looks intuitively easy to understand, for example with this picture I found in the internet

[rich wolf]

Bro can you go to a diff channel

Wtf

[jolly bear]

What dis

[rich wolf]

Not u donnie but deltaphi guy

[frail rock]

Oh, thank you. The course I'm doing rn didn't really explain it well.

Sorry if I caused some trouble.

[humble quail]

Ofc u didn't

[rich wolf]

@jolly bear find the amplitude

[jolly bear]

Yes

Is it 4

[rich wolf]

Yes

Now find the period

[jolly bear]

The formula for it is 2pi/b

But I don’t know b

[rich wolf]

Ur thinking backwards

You use the period to find b

Not the other way around

How often does the function repeat itself?

[jolly bear]

Depend, for trig function alway

[rich wolf]

Let me just tell u

The horizontal distance between a min and a max is half of the period

2pi/period = b

[jolly bear]

Really

[rich wolf]

Yeah

[jolly bear]

Shouldn’t it be full period

[rich wolf]

No

A period is how long it takes to get back to the same position

Like from a minimum to the next minimum

But you have a minimum and a maximum

[jolly bear]

I got it now

So the period is pi/4?

[rich wolf]

B=pi/4

The period is 8

[jolly bear]

Oh my bad, messed up

[rich wolf]

One last step

Find the shift

[jolly bear]

Vertical shift?

[rich wolf]

Horizontal

[jolly bear]

Lemme think

[rich wolf]

It is "c" on that paper you sent

[jolly bear]

Yes

[rich wolf]

Yes

[jolly bear]

Is it three right

Right 3

@rich wolf

[rich wolf]

Where is your "starting point"?

[jolly bear]

8,2

[rich wolf]

You cant use that

This is a sine curve

A sine curve starts at the midline

[jolly bear]

I keep thinking it a cos function

My bad

So 4

[rich wolf]

What did you choose as your starting point?

[jolly bear]

Uhm,I choose 0,4

[rich wolf]

0,4 isnt even on the curve

This is a sine curve

[jolly bear]

0,5

[rich wolf]

So your starting point must be on the midline

And the midline is at y=6

[jolly bear]

I’m so bad at calculating lol

[languid swan]

Can someone please help

PLEASE

with trig

[rich wolf]

Go to #help-1

[jolly bear]

What do we do with the midline

[rich wolf]

The function is equal to its midline halfway between the min and the max

So that must be your starting point

Because sine curve starts on the midline

[jolly bear]

So that halfway distance is the horizontal distance?

[rich wolf]

Yes

Half the horizontal distance between min and max

[jolly bear]

I never learned this

So it is 2

@rich wolf

After this, only d is left

[rich wolf]

It is not 2

Im sorry i phrased that weird

The starting point for a sine curve is halfway between the min and the max

So in this case it is as x=6

And y=6

So 6,6

[jolly bear]

Wat

[rich wolf]

That is your starting point

Which is how you determine the horizontal shift

It is 6 to the right of a normal sine curve

[long estuary]

I have figured out a) 52

But what is b) ?

How do bearings work?

[jolly bear]

So Basiclly to find the starting point. U take the average of the max and min coordinate?

Which is 6,6 right

[north basin]

@long estuary come to #help-2

[long estuary]

alright

[jolly bear]

@rich wolf

[upbeat bone]

Hello all i have a qeustion
I need to find 24 points evenly spaced around a circle where the ceneter of the circle is 0,0 and the radius is 33
I have no idea how to do this im no math man

[twilit zenith]

Do you mean that you need to place 24 points on the circumference of the circle such that the arcs created by these points are of equal lengths?

[limber patio]

Doing Euclidean geometry

whats the second line mean

For context, the problem is

[frail rock]

I'm pretty sure it means parallel.

[limber patio]

oh wow im dumb

[twilit zenith]

Rat is correct.
A bit strange, since generally there are only 2 end points picked for the line. 🤔

[limber patio]

i read that as srtu

my bad

[frail rock]

Nah, it's fine. I don't think I've ever seen the parallel sign written like that til now.

Yeah, the latter line confused me a bit.

[limber patio]

My teacher's handwriting just gives me a headache lmao

[frail rock]

Wonder why they did PRTU instead of PU or something kek.

[limber patio]

Yeah that's what tripped me up

I misread PRTU as SRTU so I was like... tf is the relationship between these 2 segments and that segment

And then I was thinking if he meant angle and jesus christ almighty the sudden realization that that was a parallel symbol just makes me feel like such an idiot

Thanks a lot!

[upper karma]

Your welcome

[frail rock]

Happy to help. :D

[limber patio]

:)

[twilit zenith]

Cameron, if you're still there, is it still relevant?

[rich wolf]

sorry for late reply

@jolly bear but yes

the starting point is at (6,6)

[jolly bear]

But what does it tell us about the shift

[rich wolf]

where does the normal sine curve start

[jolly bear]

At 0,0

[rich wolf]

and we are trying to find the horizontal shift

[jolly bear]

So from 0 to the x of the starting point which is 6

So it is -6

[rich wolf]

yes

t-6

because it's 6 to the right

[jolly bear]

Yes

[rich wolf]

or whatever the variable is

[jolly bear]

Now how do we find the d

[rich wolf]

https://cdn.discordapp.com/attachments/326138757474680852/703003538770493531/image0.jpg

the d is the midline

the vertical shift

which we already found

[jolly bear]

What

[rich wolf]

d is the "midline" of the function

[jolly bear]

Oh damn

[rich wolf]

which is the amount the function is shifted vertically

[jolly bear]

Ok, this make sense now

[rich wolf]

so write the equation we have

one more thing i forgot to account for

[jolly bear]

What else

[rich wolf]

just write the eq and i'll show you

[jolly bear]

Gimme a sec

[rich wolf]

first of all

it's a shift to the right

so it should be t-6

[jolly bear]

Ok

[rich wolf]

second of all

look at the problem

it gave us a max and a min

[jolly bear]

Yes

[rich wolf]

but the max came before the min

so the way our sine curve goes, according to our starting point, is down first, then up

[jolly bear]

Yes

[rich wolf]

so if you have a sine curve that goes down first, what does that mean?

[jolly bear]

It mean they are not showing u what is before what go down first

[rich wolf]

sure.

you can calculate that point though

because we know the period

but there's an easier way

[jolly bear]

Wait

What point are u referring to

Oh the starting point, keep going

[rich wolf]

the red curve starts at (0,0) and goes up

[jolly bear]

Yes

[rich wolf]

the red curve is sin(x)

the blue curve starts at (0,0) and goes down

what is the blue curve?

[jolly bear]

Yes

Yes

The negative sin

[rich wolf]

yes. it is negative sin(x)

so if a sine curve goes down first, its amplitude must be negative

so take that eq you have there

[jolly bear]

Ok

[rich wolf]

$f(x)=4\sin(\frac{\pi}{4}(t-6))+6$

[somber coyote]

AMD:

[rich wolf]

and make the amplitude negative

[jolly bear]

Ok

Wait

Then it just be the same function with the 4 being negative

[rich wolf]

yes

$f(x)=-4\sin(\frac{\pi}{4}(t-6))+6$

[somber coyote]

AMD:

[rich wolf]

that is your answer

[acoustic jungle]

@limber patio did you get it? could you tell me what you did because I used the angle bisector theorem to prove it which probably isn't that good of a way.

[jolly bear]

My head hurt, I’m gonna finish my English hw

Thanks

[rich wolf]

@jolly bear

midline, amplitude, period, shift

MAPS

[acoustic jungle]

Amplitude, Shift, peRiod, Midline

ASMR

[rich wolf]

[limber patio]

Oh I'm not assigned to prove it

I'm just asked to look at the problem

[acoustic jungle]

oh..

lol ok.

[limber patio]

I'll try and see how to prove it though

[acoustic jungle]

I did it by the angle bisector theorem and similar triangles.

[limber patio]

similar triangles

how did you use the angle bisector theorem @acoustic jungle

[acoustic jungle]

Could you take a picture of you've drawn.

I can't find my phone

[limber patio]

there's the problem again

my diagram is pretty much exactly that

Alright i got it

I was being the big dumb

So first you know that v is the midpoint of arc pvu via the given

As such, arc pv and vu are congruent

and because angles spv and uqv intercept congruent arcs, theyre congruent

then you know that sq and prtu are parallel

so because the arcs between two parallel chords are congruent, sp and qu are congruent

then finally because sq is congruent to sq via reflexive property

we can use arc addition and get arc psq is congruent to arc squ

and because angles spr and qut intercept congruent arcs

then are congruent

so we have ASA congruence

and as such, pr is equal to ut

[acoustic jungle]

What triangle is ASA congruence?

ah I see.

that's a good way.

[tame ore]

hiii

does anyone know how to convert to radicals?

or to pi

i mean

[acoustic jungle]

just "ignore" the pi and add it later

[tame ore]

wait so

3(pi symbol)2

[acoustic jungle]

You would do it in terms as pi meaning don't evaluate pi

it should look something like (some number) * pi

[tame ore]

so 226(pi symbol)

[acoustic jungle]

no.

That will give you pi larger than the value you want.

[rich wolf]

Surface area of the cylinder

Is two circle

Plus a rectangle

[tame ore]

im confused i dont really understand all these symbols

[acoustic jungle]

you should have something that looks like 6pi*12+9pi * 2

and you factor out the pi.

[rich wolf]

@tame ore bro think about it this way

What is a cylinder

It is a "net" (folding) of some 2D objects

[acoustic jungle]

He's asking what in terms of pi means.

[rich wolf]

Like the "net" of a cube

Is just 6 squares

Folded together

[tame ore]

so 6 pi

[rich wolf]

Omg

Where did u get that 6 from

[tame ore]

you said its 6 squares

omg im stupid

please

🤪

[acoustic jungle]

Treat pi as a variable and put your answer in terms of that variable.

[rich wolf]

A cube is 6 squares

What is a cylinder?

A cylinder is a combination of some 2d shapes

[tame ore]

so 226 divided by pi and thats the answer?

[rich wolf]

Ok just listen to fish

[tame ore]

i dont understand what fish is saying

👀

[rich wolf]

"In terms of pi"

Do you know what that means

[tame ore]

so my answer is 3.14 pi symbol

[rich wolf]

Omg

How are you getting this

[tame ore]

yall said terms of pi

idfk

[rich wolf]

Let me make this one thing clear:

Pi is a transcendental number

Which means its exact value is not known

[acoustic jungle]

try y-5=2x

what is y in terms of x.

[rich wolf]

Which means that we can only approximate it to a certain extent

If we say "in terms of pi" that means do not approximate

Leave pi as pi

[tame ore]

so

the number 226 turns into

[surreal ingot]

When would a function not have a limit?

[rich wolf]

If it doesnt approach the same value from both sides

[surreal ingot]

By chapter of Limits using properties of limits

[acoustic jungle]

45 degrees.

[tender coyote]

opps

can someone help me with this

[acoustic jungle]

the circumference is 2pi*r

and that shaded circumference is 200/360 of total length

[tender coyote]

thx

[rich wolf]

Theta =s/r

[tender coyote]

how do i find s

[twilit zenith]

The area of the sector corresponds to the portion that the angle is out of the entire 360° turn that is the circle.

Meaning, for example, if you had a sector with 90°, which is a "quarter turn," then its area would be a quarter that of the circle's.

[glad gorge]

Can someone please help tell me what I need to do to find r

[teal frigate]

can someone help me find the angle of a right angle triangle

[acoustic jungle]

inscribed angle theorem @glad gorge

[glad gorge]

Oh

Thank you ^^

[twilit zenith]

Obama, could you please show the question, for context?

[teal frigate]

ok

[acoustic jungle]

use arcsin to find x.

[twilit zenith]

Are you comfortable with sin, cos, tan and what they mean?

[teal frigate]

whats that?

ye i am comfy

but whats arcsin?

[acoustic jungle]

sin^(-1)

[twilit zenith]

The inverse function of sin.
It's a function that takes the value of sin and returns the angle that creates that sin value.

[teal frigate]

hmm

how do i do this with my calculator?

[twilit zenith]

Could you please take a picture of your calculator?
Alternatively, can I show you how to do it with Google?

[teal frigate]

i have the casio fx-82 plus 2

[teal frigate]

oh

ok

[twilit zenith]

By the way, do you know how to tell whether you're using radians or degrees?

[teal frigate]

im using degress

degrees*

[twilit zenith]

I see.
It's just that sometimes, people don't know they're using the wrong units on the calculator and get a wrong answer.

[teal frigate]

ye

so how should i lay out the equation?

just like if im trying to find a hypoteneus or adjacent?

[twilit zenith]

You know two sides in the triangle:
The opposite side, and the hypotenuse.

In this case, you must figure out the trigonometric ratio with:

[teal frigate]

ok

thanks

[tame ore]

BRO

HELP ME

I JUST NEED THE ANSWER

[novel flax]

surface area: 2 pi r h + 2 pi r^2

[high zephyr]

^ we're not gonna straight off give u an answer

[solid swan]

ok think of the cylinder as 2 circle plates and a piece of paper wrapped around it

the area of each of the plates is πr² = 9π so if you multiply it by 2 you get 2πr² = 18π

and then the piece of paper wrapped around

the width of it is 12cm given

but then the length would be equal to the circumference of the circle plate

which is 2(3π) = 6π in this case

so the area of the piece of paper would be 12(6π) = 72π

[upper karma]

@teal frigate you can also use the Law of Sines

17/sin(90) = 15/sin(x)
17/1 = 15/sin(x)
17 = 15/sin(x)
15/sin(x) = 17 (same as above, only reversed)
1/sin(x) = 17/15 (move the 15 to the other side)
sin(x)/1 = 15/17 (reciprocal gives the same result but makes it easier to get sin(x))
sin(x) = 15/17
x = sin^(-1)(0.882) (also called arcsin)
x = ~62 degrees

[rose rover]

How does one solve this? Getting a few ideas but im not rlly sure

[upper karma]

@rose rover

Start off from doing Pythagoream

yeah, there are 90 degree triangles around the edges

Ofc its a square

find the area for those (which is really only one right triangle x 4)

then find the area for the large square

then subtract the two areas

and you get the one in the middle

[rose rover]

there is overlapping tho

[upper karma]

What

You mean with the lines

hmmm

@rose rover pls do what i said

[rose rover]

each large right triangle overlaps each other

yea

[upper karma]

Lmk when you got the hypothenuse

[rose rover]

assuming hypotenuse c, c^2 = 250

[upper karma]

,w c^2=√(15^2+5^2)

,w c^2=15^2+5^2

[somber coyote]

Display results online and refine query

Requested by Al3dium#4084

[upper karma]

Okay

Now

[rose rover]

for that smaller right triangle

i assume we can use pythagorean triplets?

3,4,5

[upper karma]

https://en.wikipedia.org/wiki/Pythagorean_triple

sounds good

once you find the area for this section everything falls into place

it's basically the area of the big 90 degree triangle (at the top) minus the area for the two small 90 degree triangles

you have 4 small triangles and 4 "middle" sections

once you have the area for those, you add them all up and you get the "outer" ring

then you find the area for the entire square

and subtract the area for the "outer ring"

[acoustic jungle]

?

Where did you prove the triangle is a 3,4, 5 right triangle @upper karma

i assume we can use pythagorean triplets?
no??

[upper karma]

Yeah i was wondering the same

[twilit zenith]

If I may give a hint:
||You can note that the triangles around the square are right-angled, and that the vertices connecting the inner and outer square are perpendicular to the hypotenuses. How can you go about calculating the areas of these triangles?||

[acoustic jungle]

I don't know how you would solve it that way. You would be over counting.

I would use similar triangles.

[twilit zenith]

Notice that you are over-counting if you calculate the area of the 15 - by - 5 right-angled triangles.
However, if you were to calculate the triangles generated by the height and the greater portion of the hypotenuse, then you wouldn't be.

I'm referring to these.

Happens to the best of us. Don't let it get to you.

[upper karma]

what's the height of these?

[acoustic jungle]

why do you want to find the height of that?

you can find it by doing 5 *15/2=15 *x/2

[upper karma]

where's the 5 coming from?

[upper karma]

15 is the base, right?
but 5 is not the height. Look closer

[quick schooner]

can someone help

[acoustic jungle]

oh I see what you mean.

[upper karma]

@quick schooner Law of Sines

look it up

[quick schooner]

thank you so much

[acoustic jungle]

you could easily solve the square question using trig, but that's a small brain move.

but I would use similar triangles.

[silent plank]

its pretty simple with similar triangles

^

[twilit zenith]

Hint:
||What can you say about the relation between the sides, heights and areas of right-angled triangles?||

[upper karma]

the image is wrong, probably on purpose (to throw you off)

the sides are almost identical

the blue square looks rotated to 30 degrees, but it's not

lol

[silent plank]

not the law of sines, just basic trig ratios.
@quick schooner what don't you understand about them?

area should be: ||an integer||

[upper karma]

@quick schooner

[quick schooner]

is it law or basic?

[upper karma]

you can use whatever you want 🙂

this is just an example

[quick schooner]

thanks alot

[silent plank]

wow that is OVERKILL

applying the sine law to a right triangle

[upper karma]

it works with any kind of triangle

not just right triangles

but yeah, you can also use SOH-CAH-TOA

[silent plank]

which is the purpose of this exercise

[upper karma]

true

[quick schooner]

our teacher didn't specify but i would assume basic since we just learned basic

but yeah we are using sohcahtah

[silent plank]

is there a particular question you are struggling with atm?

[upper karma]

@quick schooner

as long as you keep in mind that all angles from a triangle need to add up to 180 degrees, you should be fine

the hypotenuse is always the longest side from a triangle

now that I think about it, they probably want you to use use the 70 degree angle

in that case, you'll have to use cosine

SOH-CAH-TOA (the bold letters stand for: sine, cosine, tangent)

so CAH is cos(angle) = adjacent / hypotenuse (A and H)

cos(70) = 36 / x
x = 36 / 0.342
x = ~105.3

[latent iron]

how do you do this?

[acoustic jungle]

are you allowed calculators.

[latent iron]

@acoustic jungle yeah

[acoustic jungle]

then just find the value of theta+60.

using arcsin.

[latent iron]

i got

sin^-1(-0.348)-60

which gives me = -80.36

[acoustic jungle]

nice!

add 360.

because that'll give the same angle.

[latent iron]

oh okay

THANK U

[upper karma]

@latent iron @acoustic jungle

+1.0 so we end up in the same spot on the unit circle with the sine function

then +360 so we end up in the same spot again

but this time it fits into the range they give you