geometry-and-trigonometry

[zinc grail]

sorry im bad at geometry

[astral hornet]

you have parallel lines and transversals

so you can find congruent angles

and use those to prove AA ~

[upper karma]

yeah, and then you can deduce that CE : DA = 8 : 10

Need help

which question?

dm me

i gtg

is 1 a-c

in question 1

[worthy nimbus]

Can someone please help me with 8

I've proved it's a right triangle but I dont know how to prove it's a rectangle

Since I thought all I had to was use slope because rectangle follows the same rules of a parallelogram so I used slope, but the point I got doesnt follow the rules of rectangle because using that point would mean the diagonals aren't congruent and that opposites sides are congruent

[astral hornet]

you would need 2 pairs of opposite sides to be parallel and congruent, as well as one angle to be right

[worthy nimbus]

I have the parallel sides

[celest flax]

How would i calcualte the height of a equilateral triangular pryamid without volume

[young echo]

First, you have to find the distance from the "center" of the base to any given vertices
You will get a SAS problem, which you can use cosine rules, but since it's right angle triangle, you can simply use SOHCAHTOA
By using CAH, cos(30) = (s/2) /x
where x = (s/2) / cos(30)
x = (s/2) / (sqrt(3)/2) = s/sqrt(3)
Then you can find the pyramid's height by using Pythagoras theorem
s^2 = h^2 + x^2
h^2 = s^2 - x^2
h^2 = s^2 - (s/sqrt(3)) ^2
h^2 = s^2 - s^2/3
h^2 = (2/3) s^2
h = sqrt(2/3) s

The side view triangle is slightly "squashed down" as it isn't standing upright (leaned down)

@celest flax

[celest flax]

thanks

❤

i also found out you can do h=(√6 / 3) * a A being the side length, but im going to show both solutions on my answer

[young echo]

√(2/3) = √[(2/3) * (3/3)] = √(6/9) = √6 / 3

[celest flax]

ye gotcha

[young echo]

both are equal, but with no surd denominator (square root), to make it "prettier" and easily calculatable

glad to help

[celest flax]

thx

[hazy granite]

what exactly is this question

i found it in similarity

[young echo]

surface area vs volume i think

[hazy granite]

peel wouldn't be part of volume though

actually probably volume/volume = surface area/surface area

[young echo]

It is surface area vs volume, cause real life peeling will remove part of the volume

imagine you peel the potato into cross section, you will pill the whole potato away

if you have same total volume of large potatoes and small potatoes (say 5 large potatoes = 38 small potatoes), which one will have larger surface area?

[upper karma]

Question 5 and 6 please

I really don't understand

[young echo]

Q5. that is triangular pyramid, where all its faces are made of triangles (including base), use this information to help you.
It probably has angle inside the pictures, but it is now pitch black, so it may or may not be solvable

Q6. It is given the base is a square, use Pythagoras theorem to find a)

@upper karma for b and c, you have to find the plane between what it stated
b - angle between line AB and the base
c - angle between line AM and plane ACD

[upper karma]

From what I see is that (tri)MAC should be figured out first

so it solves everything in there

if EB, BC, CD, DE is each 16 cm, first we find EC.

Then divide that in half, and you found MC.

(this is the longer way of explaining how to get (tri)MAC)

[upper karma]

Need help plesae 😦

just ask lol

You know gemotry?

yes?

ok

Problem B

and C

Not sure how to do

Anyone help me

@upper karma

@young echo @upper karma thank you both

so much

however

is cm just half of ce

huh?

Hey @upper karma

Can you help me with my gemotry problem b and c

for c I think it's cause the angle ratios are the same

For b

It says cos(X) = 3/5

cosine is given by adjacent and hypotenuse

so the adjacent side is 3

And hypotenuse is 5

To find out opposite

do square root of (5^2 - 3^2)

= 4

[glad ocean]

yes

So it's 3,4, and 5

[upper karma]

Now you have adjacent = 3 hypotenuse = 5 opposite = 4

[glad ocean]

just use SOHCAHTOA from there

[upper karma]

Yep

Just plug em into formula

ok

(sohcahtoa)

got it?

why opposte 4

cause Pythagoras

[glad ocean]

because 3^2 + 4^2 = 5^2

[upper karma]

yep

[glad ocean]

since it's a right triangle and the Pythagorean theorem applies

[upper karma]

mhm

so then just work out opposite since u know hypotenuse and another side

wait

where u get 4 square?

from the theroem

[glad ocean]

ok so a^2 + b^2 = c^2 where c is the hypotenuse

[upper karma]

Look what it says at the top of your paper

well its 3/5

there is no 4

lemme write it down

ok

[glad ocean]

so basically there are three sides in a triangle (TRIangle)

3 and 5 are two of the sides

5 is the hypotenuse of this right triangle (c)

3^2 + b^2 = 5^2

b^2 + 9 = 25

b^2 = 16

b = 4

[glad ocean]

nvm

[upper karma]

uh

[glad ocean]

oops

That's my surge protector thing oof

[upper karma]

oof

huh

ooh

is that how u got 4

From this

@upper karma

3^2 + b^2 = 5^2
b^2 + 9 = 25
b^2 = 16
b = 4

yes

ok

do you understand

yes

Great

is that for 1

problem A?

Wait for B

Okay

yep

For Sin is it 4/5 and Tan is 4/3 right?

@upper karma

right?

for B

yep!

does it say to work it out or can u just leave it like

Sin(X)=4/5

Cause if you have to work out the exact value

Just do inverse sin(4/5)

idk

here

idk

ah

it says just ratios

ok

so just leave it like

lAlright

sin(X) = 4/5

also

Do you play league

Why is sin the same for both triangles you drew?

Yes I do play that game.

ahri?

yes

fox lady?

lmaaao

fellow league pkaya

anyways

ye

sin is the same because the hypotenuse will always be the same

depending on the angle, the "adjacent or opposite" side

you see what I mean?

Yes.

like for one angle, the adjacent side ( the one next to it) could be another angles opposite side

sin relies on opposite and hypotenuse

so

that's what I'm thinkin

ok

Problem C?

2 I mean

@upper karma

ok

[pastel bough]

using desmos and plotting sin(x)=cos(2x) is so fun

[upper karma]

@upper karma use Pythagoras

which side looks the longest to you

in that triangle

[pastel bough]

i think its 5

[upper karma]

ecksd

[pastel bough]

6&

[upper karma]

4

nop

6?

Yep

Meaning it's the hypotenuse

That makes it c

Now plug all the values into Pythagoras

ok

so 4 is a square

yep

So 4 squared

(b) squared

There's no b given so we need to find that out

=

6 squared

20

B equals

How'd you find that?

[pastel bough]

\cos\left(x\right)\cdot\sin\left(2x\right)=\tan\left(xy\right)
put that into desmos graphing lol

[upper karma]

it's not the right answer

You need to do

it is

square root(6 squared - 4 squared) = b

4^+b^= 6^

[pastel bough]

also does anyone know how to plot fractals into desmos?

[upper karma]

@pastel bough I'm helping someone here can u not sorry

[pastel bough]

ok

[upper karma]

Ty

4^+b^= 6^
you saids 6 is the hypotenuses

so its C

yep

you need to then find b right?

to get that bottom side

its the ratios

you get the ratios by working out all the sides first

ok

So its not ? 4^+b^=6^. I though you have to do the ratios for it

find the trigonometric ratios

now you have this

ooh

there are your two ratios

do you get it now

yes is the first one sin and second one is cos?

yep

Do you need to reduce the Cos and sin?

no as it only asks for the ratios

ok

that'll be a new skin on league ;)

lmao jk

really lol

What kind of skin?

maybe

kda kaisa

or nightbringer yasuo

🐷

nice

it's so shaky

ok

Let me write it out

The first one is Tan(-60) =

I think you have to use the CAST circle

what that?

I won't be able to explain it properly

uhhh

<@&286206848099549185>

Maybe draw it out

it's like the circle that shows what trig functions are positive where

aaaaaaa

I need to label the side of the right triangle, taking into consideration directions. Find the requested trigonometric ratio.

So?

Uhhh

I'm still kinda confused

let's just wait for helpers

ok

Well can you help mea different problem?

[glad ocean]

hm

You might have to bisect the perpendicularity of the photosynthesis

transversal

[gentle venture]

I have a few questions about a trigonometry ws I have, its on arc lengths and area of a sector

The first one is: a sector has an arc length of 16cm and a central angle measuring 0.95 radians. Find the radius of the circle and the area of the sector.

<@&286206848099549185> does someone know how to solve it? Or at least how to start?

[timber hinge]

wat

r * theta = arc length

solve for r

then use area formula

done

[gentle venture]

the problem im having is its essentially doing the reverse of it

[timber hinge]

[fleet wolf]

thats what colenn did

[gentle venture]

so im confused on what to do with pie

pi*

[timber hinge]

??

0.95 * r = 16

solve for r?

thats literally it

[gentle venture]

well for the 0.95 radians im not sure if it has the pi multiplied into it

[timber hinge]

no

[gentle venture]

so do i need to divide out pi?

[timber hinge]

nope

[gentle venture]

well im over complicating it, thanks

[timber hinge]

np

[upper karma]

Need help.

[drowsy spoke]

Need Question.

[main dust]

+membercount

[hazy granite]

how to find the volume of a larger cylinder when you have the volume of the smaller cylinder and the surface area of both

[dire rampart]

are they related in any way?

[upper karma]

@upper karma no problem lol

if u want I'll explain with detail

[hazy granite]

yeah

they are similar

@dire rampart

[dire rampart]

couldve said that at the start

find the ratio of length between them

then volume of large = volume of small * scale factor^3

[upper karma]

here's a fun easy problem

[neon hornet]

r = 0.4

[tender adder]

@neon stag I would fill in the rest given angles

but idk

Halp

[tender adder]

<@&286206848099549185>

[fallow edge]

Um... I would try setting up a system of equations using what u know

like set x = to the side opposite theta

But I have to go now, so maybe someone else can help you fill in the rest

[tender adder]

$$Cos(θ)=3/x ?$$

[somber coyote]

Outsane:

[glad ocean]

= cos(48°)/15!

?

idk actually

[zinc grail]

hello

how is that x considered a vertical angle

[astral hornet]

Angle x are formed by 2 lines

And share vertex

[umbral snow]

"Vertical angle" isn't a type of angle.

Two angles opposite like that are always the same

[solar shale]

I need helping doing cos and tan

[glad ocean]

SOHCAHTOA

S = O/H; C = A/H; T = O/A

[solar shale]

Ok I think I got cos L 49/41 and I’ll watch a video about it after

[glad ocean]

erm

see L is the angle

cos L = LM/LN

[solar shale]

So 139/90?

Or would I just leave it with letters?

[glad ocean]

letters

you see angle measure and side length are different things

side length isnt degrees

[tender adder]

Is it true that arcsin and inverse sin are the same

and for all the other functions, arccos and arctan

[teal axle]

none of my answers match the letters

[astral hornet]

a calculation mistake was prb made

[glad ocean]

@tender adder yes arcsin = sin^-1 or inverse sin

same with all the other basic trig functions

including cot, sec, and csc

[wild flame]

hello i am here to say trig sucks booty

[glad ocean]

booty cheeks

[upper karma]

@neon hornet wrong

how did you even get to that result for r

[neon hornet]

Not a clue m8

@upper karma 0.3852?

[upper karma]

just write normally

who even writes results with floating points

[neon hornet]

Because I’m on mobile

[upper karma]

im pretty sure you can still write it

[upper karma]

nope

[neon hornet]

Fuk

[upper karma]

i cant even imagine how you got there tho

what's the reasoning

[neon hornet]

Big up pythag

[upper karma]

where

[neon hornet]

(2r)^2 + (2-r)^2 = (5r)^2

Idk why it doesn’t work tbh

[young echo]

what's the question?

[neon hornet]

Just scroll up till you see some circles

[neon hornet]

My method should work

[upper karma]

no

(5r) ^2

really?

just think about it

see if it makes any sense

@neon hornet

but you are close, in a sense

[neon hornet]

The (5r)^2 comes from the hypotenuse

[upper karma]

ik

but it's wrong

[young echo]

r = 1/(sqrt(3)+1)

[plucky marlin]

6r ?

[neon hornet]

Why

[plucky marlin]

[upper karma]

@young echo yeah smth like that

[neon hornet]

I accounted for that

With (2-r)

[young echo]

this is how i do

[neon hornet]

Can you explain why it doesn’t work

Ye nvm

I’m a tard who can’t count

[upper karma]

@young echo yeah same proof

[neon hornet]

(2r)^2 + (2-r)^2 = (4r)^2

[upper karma]

i think i got some calculation wrong

tho

[neon hornet]

I had the right method I just counted one too many radii

[upper karma]

nah

it's fine

i did it right

[neon hornet]

4r^2 + 4 - 4r + r^2 = 16r^2

[neon hornet]

Ew

[upper karma]

now don't spoil this, i did have time to solve it

currently at work

but think about it

[neon hornet]

Doesn’t look too bad

I’ll try it later when I have some paper

[upper karma]

oki

[nova pine]

The top of the trough is 86cm the depth is 46 cm. Find the volume

And the volume if it contains 26cm of water

Rectangular prism

[vapid condor]

is thiss right?

[drowsy spoke]

what's PLS?

[vapid condor]

perpendicular line segments

[drowsy spoke]

and SAS?

[vapid condor]

side angle side

[drowsy spoke]

looks right

what do you think is the answer for 4?

[vapid condor]

congruent angles?

[drowsy spoke]

i was thinking 3 implies 4

[vapid condor]

Yes

[amber raven]

You know when I did these 2 column proofs in 10th grade geometry, I actually wrote whole sentences in the second column. Maybe that's why I don't think 2 column proofs are that bad

[upper karma]

Hey guys, just joined this server cause I honestly am clueless on how to do these questions that got dumped on me. Any help is very appreciated!
https://gyazo.com/44c083cfb6fdadb09f45cad0d0b17276
https://gyazo.com/79c4a1318ed70cbc92988ece1e71639c
https://gyazo.com/d3ba2870599dec9e47cf2a7e3e94d244

[spice ridge]

can someone explain equations of circles to me

and how to find them

?

@copper valve you're smart

ping me back when someone has answered pls

[astral hornet]

Equations of circle is (x-h)^2+(y-k)^2=r^2

Where (h, k) is center and r is radius

@spice ridge

[spice ridge]

ok i have to memorize that

what if i have to find center and radius

given the equation

[astral hornet]

Reverse the procedure of writing the equation, pick out the k h r

[spice ridge]

cat

i dont even know how to do factoring

please just explain

[astral hornet]

Is the equation in center-radius form or another form?

[spice ridge]

lemme find the picture of my test

[astral hornet]

Move to #math on tech server

[young echo]

@upper karma
for the first picture, you have a SAS triangle, what rules you can apply to get another side length? After that, what rules you can apply to get the angle?
for second picture, this is Pythagoras theorem, try to identify the triangle from the length it wanted
for the third picture, it's Pythagoras theorem too, but try to identify the suitable triangles to find the length it wanted

[stone birch]

so i just had a test like couple of hours ago and i was totally lost on this question, so i had to sketch a graph of y=1/2sin(180t) from 0<t<10, i was getting nothing else but zeros

can someone help me please

thanks

[umbral snow]

(1/2)sin(180t)? Or 1/(2sin(180t))

[stone birch]

(1/2)sin(180t)

[umbral snow]

A few things to note:
Amplitude is 1/2

Period of a normal sin function is 360°, but we're compressing this one by a factor of 180°. So, the period is now 2

If you did find the zeroes, you would find there's a zero at t = 0, t = 1, t = 2...

[stone birch]

oooo

i understand

thankyou

[upper karma]

How does he know that the angles all add up to pi?

@ me if u are able to give an explanation

[limpid basin]

angles add up to 180 degrees in a triangle .-.

[neon stag]

Need some help on this question

[paper glacier]

since it's a rectangle, the diagonals are of equal length

[neon stag]

So I would find x by doing: 4x-54 = 3 +1x?

[paper glacier]

yep

as for b, you know E is the midpoint of A&C

[neon stag]

Aright so i made it to letter C and now I’m having problems trying to even understand it

[paper glacier]

Haven't been taught that notation sadly... You know what the m< thing means? The angle at A on lines BA&AC?

[hard gale]

means measure of whatever angle

[paper glacier]

huh?

[hard gale]

m<BAC : the 'number of degrees' of angle BAC

[paper glacier]

$\widehat{BAC}$

[somber coyote]

Froguchan:

[paper glacier]

I guess this? why don't ppl use the same notations universally smh

[hard gale]

yeah this also

smh our horse

[paper glacier]

horse?

but yeah @neon stag for c, you know that m<BAD is a right angle so you can find the correlation between the two angles

[upper karma]

How can we do Q43 without calculator?

[twin prawn]

Hmm

t!wiki Heron's formula

[loud cedar]

📖 | ** https://en.wikipedia.org/wiki/Heron's_formula **

[upper karma]

Oh, yea we can use heron's formula

Thank you!

[twin prawn]

Np :3

🍮

[upper karma]

But isn't it difficult and lengthy to find area by heron's formula?

Cuz calculator is not allowed

And have to do 1 question in less than 2 min

[swift thorn]

half times base times height

two of the vertices have the same y-coord so they're the base and the third gives you the height

@upper karma

[upper karma]

But the height is not given

Ot is not a right angled triangle

@swift thorn

[swift thorn]

half times base times height applies to every triangle, as long as height is measured perpendicularly to the base

if the base is (-sqrt 10, sqrt 8)(sqrt 10, sqrt 8) then that's parallel to the x-axis, so the height needs to be measured parallel to the y-axis

the base is at y = sqrt 8 and the tip is at y = 8 so the height is the difference

@upper karma

[upper karma]

Oh,yea
I thought you are saying that (-sqrt32,8) is the height 😄

Thank you!

[neon void]

sies this look right

does*

[upper karma]

looks right

[neon void]

thank you

[rustic cairn]

How do i figure this out

[vast girder]

how do i go about answering this?

<@&286206848099549185>

[dire rampart]

please wait 15 minutes before pinging helpers, read #info

work out angle CAD

[vast girder]

thank you

😃

[mild cargo]

The ratio of the sides of a 30-60-90 triangle is $$1 : \sqrt{3} : 2$$

[somber coyote]

Almax:

[mild cargo]

@vast girder

[vapid mica]

How do I prove GI and AM parrallel, the only logical way is with Thales Theorem I have the measure of AB and IO also AM and BM. Help?

aslo G center of Gravity if that helps???

and AMB is right ofcourse

[vapid mica]

<@&286206848099549185>

[rustic vine]

Hey so I'm trying to solve an SoE and I've gotten to about here but I'm not sure if I messed something up because I can't figure out how to get past this point algebraically

I substituted in the denominator for x

<@&286206848099549185>

[upper karma]

#info

please only ping helpers after 15 minutes has passed and how did you post this 1 minute in the future?

[rustic vine]

Ah, sorry, also did i?

[hard gale]

you are in the past

[upper karma]

I am for some reason 1 minute in the past, it is what my laptop is saying at leasr

[rustic vine]

If it helps this is what I have overall, trying to solve for x and y

[upper karma]

I need help really bad

[naive walrus]

Help

[main dust]

I forgotten most of this

[marsh spruce]

Hi could anyone dm me? I'm trying to finish some assignments and I don't want to get side tracked

[weak zenith]

Hi, does someone know if this is possible to resume a 3d space into a 2d one without loosing information? Like a 3d point would be resumed into a 2d point with a formula, and then there is a revert formula that would revert it into a 3d point

[twin prawn]

This reminds me of space filling curves

[weak zenith]

I thought it could be possible if we add a Z axis on a 2d plan and work around with that but then I realized that there are 8 different parts on a 3d space (+++;-++;...) And there would be only six parts here, so we should add two z axes

Oh, will look about it

Hmm not sure space filling curves would do the work here, at least I don't see how

[twin prawn]

I'm doubting it would be a practical solution, or if it's a solution at all

Otherwise I'm inclined to say that you'll always lose info if you're trying to go from 3D to 2D

[weak zenith]

I'm thinking that we could be able to, because since both are infinite space, we could just organizate them in another way

[twin prawn]

That's why I thought of space filling curves. You have them in 2D and 3D, so you could map from 2D to 1D to 1D to 3D and backwards

[weak zenith]

Oh didn't know about that lmao

[robust socket]

$$ 3sinx = 4k $$

[somber coyote]

Umma.Gumma:

[robust socket]

I should determine k here, so that the equality is verified

not sure how to proceed tho

[sudden lance]

It is the way to do it. Since sin(x) is between -1 and 1, you should try to search for k when it is between those values

[robust socket]

you were absolutely right thanks mate

solved it

[oblique forge]

I have a C in trig and that will drop me out of senior top 10, so i need mega help for my next test 😦

[cold plaza]

[trail minnow]

how would you guys solve the following problem:

"Two flag poles are placed 100 m apart. One flag pole is 70 m tall and the other is 10 m tall. Draw a line from the top of each pole to the bottom of the other pole, and find the point at which the lines intersect, Find the height from the point of intersection to the ground."

while using only properties of triangle similarity and basic trig functions (literally sin, cos, tan, and arcsin arccos and arctan and nothing else)

@copper valve plz help

[copper valve]

no

not now

[trail minnow]

oof

[copper valve]

read #info, if you are new to math server ^^

[trail minnow]

o

oof sorry

[astral hornet]

@trail minnow a hint (maybe ) : tan is the slope of the hyps

[trail minnow]

lskjflsjdfiodjfoiwe fuck how did i not think of that

glllllldjgl

luckily everyone else failed too

so

he probably will discount that problem maybe??

@astral hornet i'm unironically reeing rn this was a short answer worth 6 points out of 48 on my geo test

literally why he never even mentioned cartesian geometry so i didn't think to use it

brb petitioning my math teacher to curve test by (x*y)

[desert ruin]

can i get some help with some geo questions on the computer

[winged saddle]

SHORE

i mean sure

wot doo u need help with @desert ruin

[glad falcon]

How do i find out the point if i know the distance and one point

[winged saddle]

wot

use trig functions

lol

[idle bloom]

Ehhhh

seems over kill

[winged saddle]

what are u talking about anywae

y

[idle bloom]

like if you have a point and a distance

you find the other point

[winged saddle]

oh

yeah ur right

[idle bloom]

tho you would need more info

[winged saddle]

im overthinking it

[idle bloom]

cause it could be in any direction

[winged saddle]

yeah it would draw a circle

but if it precised the direction

[idle bloom]

(that's the definition of a circle lol)

[winged saddle]

bruh

im pointing out the obvious im bad

[idle bloom]

I know

i'm just agreeing with you chill

[winged saddle]

ok

sorry

[clear haven]

lol

[strange zealot]

any germans here?

that know trigonometry

[dark sparrow]

why do you need Germans specifically?

[strange zealot]

I have a german textbook with tasks

Idk if anyone else would understand:P

I could post a pic if you want

[dark sparrow]

sure

i mean i don't speak much German but i can try

[strange zealot]

😛

[upper karma]

Seems like a simple trigonometry

Picture says it all

[dark sparrow]

there's a picture yeah

i take it that you are asked to find α and β?

[strange zealot]

yeah

[dark sparrow]

uh huh

[strange zealot]

that was just an example, I have a lot more to show lol but understanding this first would be great

[dark sparrow]

yeah ok so obviously you'll need a calculator for this

[strange zealot]

👌 got that on me

[dark sparrow]

so you're basically given cos(α) right away

it's 1468/1622

,calc acos(1468/1622)

[somber coyote]

Result:

0.43928615589185

[dark sparrow]

^ that's α in radians

,calc 0.43928615589185*180/pi

[somber coyote]

Result:

25.169242731129

[dark sparrow]

in degrees

@strange zealot do you follow so far?

[strange zealot]

sec

cos a is adjacent/hyp

nvm

yes

[dark sparrow]

yes ok so

DE can be found with the Pythagorean theorem (as sqrt(1622^2 - 1468^2)), and it is equal to FC

this lets you find BF and hence sin(β)

[strange zealot]

sec let me write all of that down and try BF + sinB myself

[dark sparrow]

ok

[strange zealot]

not sure how to get BF tbh

getting DE is clear

wouldnt you need CE to get BF?

@dark sparrow

[dark sparrow]

you are given BC

it's 2132

BF = BC - FC = BC - DE

[strange zealot]

right

getting DF is pythagorean again, not needed but should be right? and with this I can use sin cos and tan to get β?

and β is 45,01°?

sinβ 1442,18/2039 = 0,707*180/pi = 45,01°

@dark sparrow

[dark sparrow]

assuming you didn't mess up the arithmetic, that sounds about right

[strange zealot]

Thanks! that's it for today though lol but I actually learned more in these 30mins than in half a month of a math teacher trying to teach maths

I appreciate it, might come back tomorrow or later today

hope you have a great weekend!

[dire rampart]

yw

[warped torrent]

@frosty flame i did that formula :0

but im doing the process wrong I think @_@

[frosty flame]

sin^4(x) = (sin^2(x))^2

so its 15[1/2(1-cos(2x))]^2

[warped torrent]

i got that part

but then there was more that I had to do yeah? @-@

@frosty flame

see @_@

[frosty flame]

i am in game

sry

[warped torrent]

oh sry

[frosty flame]

i think ia m right hto

sin^4(x) = (sin^2(x))^2
so its 15[1/2(1-cos(2x))]^2\

[upper karma]

Need some help calculating the shaded area of the Yin Yang symbol (without the two circles that float in the middle of the white and black area).. isn't the shaded and non shaded area equal to half the circle area?

[winged saddle]

yeah

@upper karma

u calculate half the big circle, then add half a semicircle

oh wait

nvm jus half the big circle

[upper karma]

The weird part is that in the answer sheet the area of the shaded area (the one with all the symbols) is $4\pi r$

[somber coyote]

观察者:

[winged saddle]

formula for area of circle is pi r^2

r is 10

so 100pi

but divided by 2

so 50 pi

[upper karma]

Actually, r is unknown

I just forgot to remove the numbers on the graph

xS

[winged saddle]

bruh

[upper karma]

Ikr.. just noticed

[winged saddle]

ok

wait

so they give u this thing but with no number at all?

[upper karma]

Nope

Only unknowns

[winged saddle]

...

ok well its half the circle

[upper karma]

With this info, the colored area is limited to three half circle arcs

[winged saddle]

yeah are u calculating the area or circumfernece?

[upper karma]

Circumference

[winged saddle]

ohh

yes indeed 3 circle arcs

1 is the big semi circle and 2 small semi cricle

sorry if im not a big help

i cant do much with no numbers only unkwons

[upper karma]

I don't know why I bumped into a wall suddenly..

I'll post the answer when I find it

[winged saddle]

ok

[upper karma]

Thanks for the help though

[winged saddle]

;-;

yw i guess

[hard gale]

Half circle circumference $= 2\pi (2)r \Rightarrow$ Half circle circumference $= 4\pi r$

[somber coyote]

emeric75:

[upper karma]

Sorry for the latex.. wrote it messed up

[somber coyote]

观察者:

[upper karma]

Question.. why is your font different than mine? @hard gale

[hard gale]

i have a different preamble lel

[upper karma]

Look at the LaTeX

Lol

[hard gale]

,tex --preamble

[somber coyote]

emeric75:

[hard gale]

oops

,preamble

[somber coyote]

Personal LaTeX Configuration

Config Option Values:
​ ​ ​ keepmsg: Keeping your message after compilation
​ ​ ​ ​ colour: Using colourscheme grey
alwaysmath: ,tex renders in textmode
allowother: Other users may not view your source and errors
​ ​ showname: Your name shows on the compiled output

Current preamble

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{fancycom}
\usepackage{color}
\usepackage{tikz-cd}
\usepackage{physics}
\usepackage{chemfig}

\renewcommand{\familydefault}{\sfdefault}

\renewcommand{\ot}{\leftarrow}

\newcommand{\bb}[1]{\mathbb{#1}}

\newcommand{\fct}[5]{\begin{equation} 
\begin{split}
#1 :\hspace{0.5em} &  #2  \to  #3 \
 \hspace{0.5em}& #4  \mapsto  #5
\end{split}
\end{equation}}

[hard gale]

"\renewcommand{\familydefault}{\sfdefault}" this

[upper karma]

Hmm

[hard gale]

well you gotta add it

[upper karma]

,tex \renewcommand{\familydefault}{\sfdefault}

[hard gale]

lemme remember how we do it

[somber coyote]

观察者:

[upper karma]

Like this

Half circle circumference $= 2\pi (2)r \Rightarrow$ Half circle circumference $= 4\pi r$

[somber coyote]

观察者:

[upper karma]

Nah

[hard gale]

[[ ,preamble \renewcommand{\familydefault}{\sfdefault}

[upper karma]

,preamble \renewcommand{\familydefault}{\sfdefault

[somber coyote]

Your new preamble has been sent to the bot managers for review!

[upper karma]

Thanks!

[hard gale]

gotta wait a bit cause it's manually checked for the moment

should be p quick tho

[upper karma]

Well it sure looks better with this font, thanks for the help!

(someone should purge this chat until my last message about the geometry problem)

[hard gale]

iirc you can also change the font of the math but i'm lazy

[upper karma]

Lol

[hard gale]

i could purge the messages one by one

[upper karma]

Nah, faster to do it if you are a mod

[hard gale]

well i'm not a mod lel

[warped torrent]

ive been stuck on this assignment all day and have only been able to do 10 questions ;-;

can someone help me with these problems?

[hard gale]

really looks like a quadratic doesn't it?

that's what we're gonna use @warped torrent

[warped torrent]

i started with that but then I think i messed up along the way @_@

[hard gale]

so what did you do?

[warped torrent]

oh wait, I factored it

i factored and got (2sin theta-1)(sintheta+1)

then I got sin theta=1

and sin theta= 1/2

[hard gale]

"sin theta=1" should be -1

2sin(t) - 1 = 0 or sin(t)+1 = 0

[warped torrent]

oh wow...

im dumb

thank you

[hard gale]

[supple haven]

does anyone here know what sec2theta is equivalent to?
I know that the sin2theta is equal to 2sin^2cos^2

i wasnt sure which chat this belonged in since both involve trig

[old juniper]

tan^2(x)+1

@supple haven

[supple haven]

can you tell me how you got that please?

[old juniper]

sin^2+cos^2=1 right

so divide the whole thing by cosine

divide by cosine^2

@supple haven

[supple haven]

uhh gimme a sec lemme write that donw

[old juniper]

let me know if you need any help

[supple haven]

hmm

hmm, so like I know how to find sec^2(theta)
but, I dont know how to find sec(2theta)

@old juniper

[old juniper]

oh sorry, you could use the double angle formulas

[supple haven]

im not sure how that works for sec csc and cot though

[old juniper]

since sec is 1/cos(theta) just do 1/(cos^2theta - sin^2theta)

[supple haven]

oh

thanks

that makes a lot of sense

I wish I had thought of that

[old juniper]

one really nice thing about trig is that it revolves around these theorems s you can just plug and play 😃

[supple haven]

tyyyy

okay one question

for some reason this always slips my mind but

I now have 1/7/25 as an answer

how do I make this fraction "normal"

[old juniper]

25/7

[supple haven]

multiply by 25/25 right

[old juniper]

yep or what I think is just flip it up

[supple haven]

does that work for every question like that?
like 1/x/y

it always becomes y/x?

[hard gale]

1/(y/x) would be x/y ye

[supple haven]

(1/x/y) :/

[hard gale]

what if you take (1/y)/x ?

[supple haven]

uhhh ]

[old juniper]

1/xy

[supple haven]

^

[hard gale]

i was asking the question lul

so yeah just be careful

[supple haven]

thanks y'all

[old juniper]

your welcome anytime 😃

[supple haven]

Im really hoping I pass the retake test on monday

[hard gale]

good lucks bro 🍻

[supple haven]

I already failed the first one and cant afford to fail again, so im hoping if I study hard I'll get a good grade on it

what about something like

(7/16)/(3/2)

[hard gale]

it's the same thing as $$\frac{7}{16}\times \frac{1}{\frac32}$$

[somber coyote]

emeric75:

[hard gale]

ie $\frac7{16} \times \frac23$

[somber coyote]

emeric75:

[supple haven]

thanks

can you tell me what makes that possible?

[hard gale]

which part?

1st or 2nd ? ~~or both 😊 ~~

[supple haven]

or like why (7/16)/(3/2) is the same thing as (7/16) * (2/3)

[hard gale]

for the first step, it just comes from definition of multiplying two fractions

$$\frac{a}{b} = a\times\frac{1}{b}$$

[somber coyote]

emeric75:

[hard gale]

(except a and b are themselves fractions here)

and then 1/(3/2) = 2/3 you already ok with it i suppose

[supple haven]

ohhhh

I get it

jeez so much of math is really really interconnected

i'd do well to start paying more attention to that

[hard gale]

hell ye

[upper karma]

if you know normal fraction division, it helps to put 1 as 1/1, i e 1/1 % 3/2

[hard gale]

then directly do 7/16 % 3/2 lel

😋

[upper karma]

that as well

[upper karma]

https://i.imgur.com/GYcJcRT.jpg

Does anyone know how I can calculate the ratio of the colored area vs non colored area?

[limpid basin]

gonna need some more information here ._.

[upper karma]

The question says, find the relationship between the whole triangle and the colored area. If desired use a ruler

That's it

Any ideas?

[fleet wolf]

are the altitudes of the colored triangles half of the whole triangle

?

[swift thorn]

it's the same as the ratio between the whole height and the coloured height

[upper karma]

Just came back after a long journey in the kitchen xD

Height of the whole triangle using my ruler shows 1,6cm, colored area is 8mm

Meaning that the height of the colored area is the same as the height of the non colored area @swift thorn

@fleet wolf

[swift thorn]

the ratio between the areas is equal to the ratio between the heights

[fleet wolf]

^

[swift thorn]

should not be too much trouble to prove

[upper karma]

Mind proving it?

[fleet wolf]

the sum of the bases of the colored triangles is equal to the base of the whole triangle

use the area formulas

[upper karma]

Another question which I proposed earlier, which I claimed to have solved.. I was mistaken

I cannot seem to remove the numbers in geogebra.. nvm the numbers (the only given we have is r, f, c, g

Wanted, find the circumference of the colored area which is made out of three half circle arcs.

[wild hamlet]

that's why u use desmos

I'm busy rn anyway

:\

[upper karma]

Second, if a line were to be drawn between F and A to divide the colored area into two equal pieces

I just got stuck on this question hard.. don't know why I cannot seem to understand it

[swift thorn]

"Mind proving it?" yeah i kinda do, given it's a homework problem for you

it even says you can use a ruler so they probably expect you to measure and calculate the ratio instead of proving it

[upper karma]

I already fixed it nvm the old question

Thanks for helping though

And I just solved the other.

[supple haven]

Can you guys help me out with this by checking my work?

Gah damn discord lowers picture quality

Hol up

HNNNNNNG discord needs to suck a potato

why does the quality get so low?

[drowsy spoke]

,rotate -90

[drowsy spoke]

you've got the right idea

but i'm not going to calculate everything to see if it's right

means the working's right

[dark sparrow]

which ones do you want checked

[supple haven]

Just the first 3

[dark sparrow]

the photo quality is too low for me to read your work for 3, and your answer looks rather strange too

#1 and #2 are good though

can you do #3 once more but on another piece of paper and in pen?

[drowsy spoke]

it says tana + tana/1 - tana tana = 2tana /1-tan^2 a
his tana = 12/5

did you change 144 to 114?

nvm

[dark sparrow]

i can read $\frac{\frac{24}{5}}{1 - \frac{144}{25}},$ but i cannot read what is written afterwards

[somber coyote]

Ann:

[drowsy spoke]

ah alright, he continued above it, hence the arrow
1 - 114/25 = -119/25

so he wrote it finally as 24/5 * 25/119

and he forgot the negative sign

[supple haven]

osdaifjsdaklfj

I forgot a negative

careless mistake on my part

son of a gun

okay thanks for checking my work

[strange zealot]

@dark sparrow Hi! do you have time for me again please? x

[dark sparrow]

sure, a little bit

[strange zealot]

How to get A2 🤔

[dark sparrow]

the area?

[strange zealot]

yup

[dark sparrow]

hmmm let me take a look

[strange zealot]

🙏

[dark sparrow]

oh yes, i see it now

here. both the triangles i added are supposed to be right, with the right angles at B and E

the idea is that you can calculate the areas of both of these triangles, and the area of the trapezoid ABEF

and then A2 = area(ABEF) - area(ABC) - area(DEF)

it's actually a common strategy: add something to your shape to make the calculation easier, and then take off what you added later

[strange zealot]

hmm Im gonna try that and see if I can get it right

[dark sparrow]

ok

[strange zealot]

ok, can you check

A2=133,42?

@dark sparrow

[dark sparrow]

oof idk

i don't have a calculator with me rn

[strange zealot]

rip

[dark sparrow]

if you show your work i can look through that

[tropic shard]

Eh, I got ≈ 208.397 m² by using another method (breaking A₂ into rectangles and right triangles) rip.

[dusty axle]

The circles are all the same size and they fit exactly into this rectangular box
What is the area not taken by the circles
https://cdn.discordapp.com/attachments/496719668929626122/551822202710917120/IMG_20190303_174414.jpg

[tropic shard]

Hm.

[dusty axle]

I got the answear to it but the answear im provided doesnt agree with the answear i got
so i want someone to tell me what they get

[wild hamlet]

20cm across

is 2 diameters

r = 5

5 * (rrpi) = circle area

tho now we need the diagonal and height

starting from the top left circle's center and going diagonal right and down yo the bottom right circle is a distance of 20 cm

the circles are in a triangle of 1x,2x,sqrt(3)x where the hypotenuse is 5

so 2x = 5

then add up hieght * 4

x = 5/2

4 * 5/2 * sqrt(3)+ 5 + 5 = hieght total

right?

please correct me If i made any mistakes

[tropic shard]

Height, my dude. Did you get 180.5059991 in the end?

Hm lemme read that

[wild hamlet]

,w 45/2sqrt(3) + 10

[somber coyote]

Display results online and refine query

Requested by Yeat#9716

[wild hamlet]

ooh

[tropic shard]

That's the height?

[wild hamlet]

ye

[tropic shard]

I got around 28.66025, maybe bad rounding error on my part.

Hmm I worked it again, I still have 28.66025, I'll show my work so you can see.

@wild hamlet Agreed?

Drawn poorly(not an art major), but the horizontal lines should be where the circles touch.

[hallow smelt]

how do you notate a subscript? I dont see a downward arrow

[idle bloom]

what just in text?

I think you can only do that with LaTeX

$1_2$

[somber coyote]

Spamakin🎷:

[idle bloom]

like that ^

but you can't do it inline in a normal message

[tropic shard]

Or you can copy and paste, 1₂.

[idle bloom]

w0t

from where

[tropic shard]

Copy and paste the image.

Do another one

[idle bloom]

$sub_script$

[somber coyote]

Spamakin🎷:

[tropic shard]

ok you got me there

[idle bloom]

lmao

[tropic shard]

Only works for numbers

;)

[idle bloom]

$6_9$

[somber coyote]

Spamakin🎷:

[tropic shard]

6₉

[idle bloom]

classic

ok w0t

[tropic shard]

Indeed

[idle bloom]

cause on my computer that's an image

not text

[tropic shard]

Lol, it is, I'm just trolling

[idle bloom]

ok I figured

[tropic shard]

Unicode has some subscripts, superscripts.

I have text replace, so I don't have to copy and paste.

[idle bloom]

neat

[tropic shard]

eˣ, log₁₀(eⁱᶿ) etc.

Looks bad sometimes, which is why it's good to have LaTex to be doing all the work >:D

[bitter crater]

how do you do this? feel like I've learned it before but forgot https://cdn.discordapp.com/attachments/277682465814872065/551725188903862314/unknown.png

[idle bloom]

@bitter crater what kind of triangle is that?

triangle EBC

[bitter crater]

forgot is it right?

[idle bloom]

yea

cause it's a square

so that corner is a right angle

cause all corners of a square/rectangle are 90 degrees right?

[bitter crater]

yea

[idle bloom]

so

[bitter crater]

yea so I kinda forgot all this stuff lol can you walk me through it again

[idle bloom]

ok but we're on the same page that we got a right triangle right?

[bitter crater]

yea

[idle bloom]

ok so do you know Pythagoras's theorem

[bitter crater]

yea I remmeber htat

[idle bloom]

sweet so let's take advantage of that

[bitter crater]

so 7 sq = 3 sq * x sq

[idle bloom]

perfect

so solve for x and you get that missing side right?

[bitter crater]

yea did that but it was irrational

square root 40