geometry-and-trigonometry

[keen aspen]

He's on his last one

[waxen gorge]

idk ur time zone tho

[narrow sleet]

Wow

[restive void]

@waxen gorge well

It’s twelve

And last one

And I’m gonna die

[narrow sleet]

So quickly do it

[restive void]

Because I have spanish homework

[keen aspen]

SPanish1?

[waxen gorge]

Then hurry

[restive void]

It’s difficult for me okay lol

[waxen gorge]

Spanish 1 is easy

[keen aspen]

It rly is

[waxen gorge]

I have a 95

[restive void]

Es la clase de español 2

[keen aspen]

I have a 100 but its Spanish2

[waxen gorge]

Lol

[restive void]

Y tengo 102%

[keen aspen]

Find the side lengths por favor

[restive void]

OH YEAH

MATH EXISTS

[waxen gorge]

lol

[narrow sleet]

Do it quick, so u can escape

[restive void]

Is ab sqrt 29

[keen aspen]

uh

[narrow sleet]

I got sqrt 13

[restive void]

I did it in my head

[keen aspen]

no

[restive void]

Late at night

So yeah probably wrong

[keen aspen]

Yeah sqrt(13)

[narrow sleet]

Write it down

Find one more and you're free

[restive void]

Yep, I see what I did

[keen aspen]

except DC

Find BC

[narrow sleet]

Or AD

[keen aspen]

ye

[restive void]

Bc = sqrt 13?

Say yes thanks

[keen aspen]

yes

[narrow sleet]

Yes

[restive void]

YES NEVER DOING THIS SHIT AGAIN

[keen aspen]

gj

[narrow sleet]

@restive void You have exam

[restive void]

Thanks ily guys <3

[narrow sleet]

Better expect that

[restive void]

@narrow sleet don’t remind me

[keen aspen]

Aquiver you still play gd or no?

[narrow sleet]

Just try to practise

[restive void]

Woah woah wish

Woah

Pjs

How do you know I play that

[keen aspen]

Haha I subscribe to you

[restive void]

Haha uh

How do you know I play that

[keen aspen]

I knew you back when we called with Cemplix

[restive void]

Oh

[narrow sleet]

@keen aspen Subscribe?

[keen aspen]

On Youtube

[restive void]

Oh btw bloodlust got verified today

[keen aspen]

Yep 😄

121k atts

[narrow sleet]

Cool

[restive void]

Pjs

Have you had a name change

[keen aspen]

I did not lol

[restive void]

I’m trying so hard to remember you

[keen aspen]

Ik you probably dont know me, I think we spoke on the GDD server

Im a VIP on there

[restive void]

Dw I’ll remember you today, you just helped me with three hours of math

[keen aspen]

Haha alright ill catch you around if you need anything

[restive void]

<3

[glad falcon]

HI

[hasty thistle]

This should be really basic for you.

[glad falcon]

Ok

[hasty thistle]

Lemme see

[glad falcon]

k

[hasty thistle]

for the pirst pic

lol gimme time to draw an illustration

[glad falcon]

ok

bro hurry

its twelve over here

imma about to go to sleep

[hasty thistle]

ohh

its okayy

[glad falcon]

quick

nah tell em

me

[hasty thistle]

welp this is harder than I expected

welp

lets say

o my

lets say line segments are <--->

@glad falcon

reply when you can

[thorn talon]

what do you need help with?

[hasty thistle]

I need an equation in X°

@thorn talon

[thorn talon]

is that the image of the problem?

[hasty thistle]

yes

[thorn talon]

i don't think there's enough information from the image drawn

[hasty thistle]

oh

wait

[past mantle]

Hold on, linear pairs

It's possiblr

Possible*

[hasty thistle]

^

[past mantle]

Just get the number for the side adjacent to x

Then it will be possible

Plus kang is right, u need more info

[hasty thistle]

The problem is a congruent angle

[thorn talon]

i think it's an issue with how you drew the lines

cause here

[hasty thistle]

the angle beside 2x° is a right angle

[thorn talon]

it looks like they intersect, and the go off in different angles

[past mantle]

Yea

[thorn talon]

if they intersect, then continue straight on

[hasty thistle]

oh they are diferrent lines though

[past mantle]

If u drew them better, then it's obvious

[thorn talon]

then you can use angle properties of the intersecting lines

[hasty thistle]

lemme redraw?

[past mantle]

Yes

[hasty thistle]

wait

I tried

@past mantle and @thorn talon

[thorn talon]

what's that supposed to represent?

[hasty thistle]

?

what do you mean?

[past mantle]

Do u not have other angle Measures?

[thorn talon]

cause like

[past mantle]

Omg, I see it

[thorn talon]

?

[past mantle]

Think of 2x+that angle

If u have that

Than u get same side interiors

[hasty thistle]

yes

wait what

which angle?

[past mantle]

Plus if there was another angle, u could do vertical angles

[hasty thistle]

the right one?

[past mantle]

The one u narked

Marked*

And the one to the right

[hasty thistle]

oh

its measured 90°

[thorn talon]

ok

that changes everything

[hasty thistle]

then solve for x

[thorn talon]

now you have enough information

[past mantle]

It's simple, it's just that u really need to see it

[hasty thistle]

I need a solution

I'm really bad at this

[thorn talon]

that marked angle is 90 right?

[hasty thistle]

yep

[thorn talon]

well

can you see

x, 2x and 90

are on the same line?

[hasty thistle]

which means?

[thorn talon]

they add up to 180 right?

[hasty thistle]

uhm

X° and 2x°?

[thorn talon]

x + 2x + 90 = 180

[hasty thistle]

so all three measurements are 60°?

[thorn talon]

no

you just need to solve that equation for x

and you can then determine each measurement

[hasty thistle]

oh?

probably x° = 45°

I need a proper equation @thorn talon

[thorn talon]

just making sure

this is one line right?

[hasty thistle]

yes

[thorn talon]

then

that equation above works

x + 2x + 90 = 180

[hasty thistle]

so to solve for x

180 - 2x + X?

[thorn talon]

collect like terms

3x + 90 = 180

subtract 90 from both sides

3x = 90

divide both sides by 3

x = 30

[hasty thistle]

oh

wait, "3x"?

[thorn talon]

x + 2x

[hasty thistle]

ohhh

soo x = 30

thanks! 😃

[thorn talon]

yes

np

[hasty thistle]

It should be really basic for you then?

[thorn talon]

idk

[hasty thistle]

honestly, I'm still having a hard time.

[thorn talon]

geometry is a weak point for me still

[hasty thistle]

I mean, like I said earlier

this is like for 7th graders

A little advanced I guess.

[thorn talon]

i sometimes miss the obvious solution

and circle around

going a long unnecessarily hard solution

[hasty thistle]

wow

[thorn talon]

i think this one

[hasty thistle]

but still the same answer

[thorn talon]

had the optimal solution though

yeah

[hasty thistle]

optimal?

[thorn talon]

well

not doing unnecessary stuff

[hasty thistle]

oh

lemme study the solution

brb

[thorn talon]

ok

[hasty thistle]

I don't think I'll be able to finish this assignment

lol

it took me an hour just to solve 1

[thorn talon]

well, should still try i guess

see what you can learn from it

and it should get easier in the future

[hasty thistle]

ahh

I'll try and show you my sol.

[thorn talon]

ok

[hasty thistle]

in this problem

I mean its not informative

[thorn talon]

do we know the angle in between?

[hasty thistle]

it doesn't say anything

All I know is that this assignment is about Congruent lines.

[thorn talon]

do you have an image of the task?

[hasty thistle]

uhm

I just obtained my phone

I guess I can take a pic

My handwriting on the other hand

is really bad

kek

[thorn talon]

yes please, of the task

[hasty thistle]

uh sure. give me a minute

[thorn talon]

cool thanks

[hasty thistle]

I'm sorry, my phone is all out. Amma try charging

[thorn talon]

it's ok

[hasty thistle]

okay im sending

is the image clear btw

@thorn talon

[thorn talon]

yeah

unless we assume that angle to be 90 degrees

then there isn't enough information

[hasty thistle]

let's assume

only option

[thorn talon]

if it was 90 degrees

then it's another angle on a straight line question

x + 3x + 90 = 180

[hasty thistle]

the same from before I assume

[thorn talon]

setting up to the same concept

different equation

[hasty thistle]

oh

so

4x + 90 = 180

[thorn talon]

yep

[hasty thistle]

then

subtract 90 from both sides

[thorn talon]

yeah

[hasty thistle]

what do you mean by that btw

[thorn talon]

?

[hasty thistle]

I mean where do I subtract?

[thorn talon]

from both sides

this eliminates the 90 on the left side

and the 180 on the right becomes 90

[hasty thistle]

oh so

4x = 90

[thorn talon]

yeah

and then one step left

[hasty thistle]

try to remove the 4

divide both sides by 4?

[thorn talon]

yep

exactly

[hasty thistle]

90 is not visible to 4 tho

22.4

hm

[thorn talon]

22.5

[hasty thistle]

22.5&

22.5*

[thorn talon]

which is fine

[hasty thistle]

yeah I'm getting it

[thorn talon]

degrees, can go into minutes and seconds if necessary

though i just generally leave exact

[hasty thistle]

thanks

[thorn talon]

np

[hasty thistle]

why didn't we used transposition btw

[thorn talon]

what's that?

i don't remember terminology from geometry

[hasty thistle]

nvm

kek

what about the 3rd problem its diferrent

[thorn talon]

they all kind of have

the same concept

just with different values

[hasty thistle]

oh

can you help me with the solution T.T

[thorn talon]

what are the values?

[hasty thistle]

5x, 4x and x

[thorn talon]

and they all lie on the same line right?

[hasty thistle]

yes

[thorn talon]

so

x + 4x + 5x = 180

[hasty thistle]

the same concept like the same?

lol

[thorn talon]

well, same concept

setting up different equations

[hasty thistle]

combine like terms

[thorn talon]

yep

[hasty thistle]

11x = 180

?

[thorn talon]

should be 10x

[hasty thistle]

hm?

oh ye

[thorn talon]

1x + 4x + 5x

[hasty thistle]

sorry

[thorn talon]

don't need to be

it's not a problem

[hasty thistle]

10x = 180?

[thorn talon]

yep

[hasty thistle]

each side is 18

I'm starting to think, 180 is not the total measurement

I lack information

e-e

[thorn talon]

it's the sum of angles on a line

stuff like that

[hasty thistle]

good point

i mean

any angles on a straight line?

[thorn talon]

yes

[hasty thistle]

so the last one, I guess\

has the same concept from the problems from before?

[thorn talon]

yep

[hasty thistle]

okay thanks

I finished

the assignnment

[thorn talon]

nice

[hasty thistle]

and found out this assignment is a piece of cake

kek

[thorn talon]

nice

[hasty thistle]

with your help ofc

I do hope my answers are correct

if not, I guess I can tell that tomorrow

[thorn talon]

yeah

[hasty thistle]

I mean can I talk to you more about this

if I can clarify the answers?

I really need more knowledge

I'm good with basic statistics though

[thorn talon]

ok

[vital oxide]

help?

[umbral snow]

How far have you got? Have you drawn the triangle? Have you applied cosine law?

[glad falcon]

Anyone wanting to help me

[echo willow]

E=mc vagina

[white harness]

Question ?

[vital oxide]

help pls

pls <@&286206848099549185> 😦

[upper sedge]

Oh man.

[crude kraken]

Oh my

[upper sedge]

Let me think about how I want to approach that. There's a really helpful theorem.

Has to do with the circumcenter.

[cyan sonnet]

^

[abstract arch]

Law of cosines?

Nah that wouldn't work.

Unless there's some way to find one of the angles

Within a triangle with x as a vertex, that is

[vital oxide]

i try to use it but idk how

[upper sedge]

Alright. I have my ideas together. Do you know how to find the circumcenter of a triangle, Nam?

[vital oxide]

idk 😦

im only a freshmen 😦

[waxen gorge]

Perpendicular bisectors!

[upper sedge]

Yes!

[vital oxide]

oh!

[upper sedge]

Since the point we're dealing with is equidistant from all of the verticies, we're dealing with the circumcenter, which we can find by locating the intersection of the perpendicular bisectors of every side.

[vital oxide]

is there anyway of solving it using law of cosines

cuz my teacher is so strict

:((

[upper sedge]

Is that what your teacher is expecting?

[vital oxide]

yes 😦

using law of cos

[upper sedge]

Draw in all of the right triangles that are used in the construction of the circumcenter first.

Before we go nuts doing trig everywhere, let's just get everything we know about the triangles involved together.

[vital oxide]

oh ok

oh shit i solved it

HOYL

[upper sedge]

=)

Well done!

[vital oxide]

1 more tho xD

😦

[upper sedge]

How interesting.

Alright, this one is also not too bad.

Think about what it's telling you.

You have all of your angles, so you know which sides are bigger than which sides, and you know what all of the sides add up to.

[vital oxide]

yes

and then?

[upper sedge]

If you have all of the angles, and you know how the sides are related, you can get the values expressed in terms of the law of sines, can't you?

[vital oxide]

OH

[past mantle]

The last problem

You can you cosine law applied with SSS or SAS

[main spire]

How does one find the equation of a circle given three points?

In this case, I have the origin, (-3, 6,), and (2,1)

[white harness]

Three points on the circle ?

[copper valve]

i think theres a way using the triangle formed from the 3 points

but i don't know how to do that :(

[upper karma]

doesn't that give u radius

(x+3)^2+(y-6)^2=50

you're given (h,k)

@Big Bad Bug#1448

smh why can't I tag him

[cursive nexus]

Just substitute lol

Equation of a circle is

[upper karma]

did I do it right?

[cursive nexus]

$$(x-h)^2 + (y-k)^2 = r^2$$

[cursive nexus]

Plug in all 3 points

You get a system of 3 linear equations

[upper karma]

dex did I do it right

(x+3)^2+(y-6)^2=50

[cursive nexus]

Let me check

How did you get the 50

[upper karma]

distance from origin to one of the points squared

that's the radius

[cursive nexus]

The origin is on the circle

It's not the center

[upper karma]

oh

[cursive nexus]

@main spire

[upper karma]

my bad

@cursive nexus

[cursive nexus]

Origin is (0,0)

[upper karma]

but he says origin is -3,6

[cursive nexus]

No he says he has the origin, (-3,6), (2,1)

Those are 3 points on the circle

[upper karma]

oh

ok

yeah I agree there is a system of equations

3 equations

[main spire]

Lol

Sorry for the confusion

I have to find the center point of the circle

Given that the circle crosses the origin

And also crosses the intersections of 2 polynomial equations

But I solved for those points already so it’s irrelevant

So, is there a method without substitution?

[white harness]

$$\left(\frac{-3}{2}, 3\right)$$

[cursive nexus]

Yeah there's a method with perpendicular bisectors

[main spire]

Ooh go on

[cursive nexus]

So you draw a triangle with the three points

Use two of the edges

Find the midpoints of those edges

And then find the equation of the perpendicular bisectors

It's just -1/slope of original

For slope of perpendicular

Find where the perpendicular bisectors intersect

That point is the center of the circle

Then find the distance between the center and a point

To get the radius

[main spire]

Ah okay

And I don’t even need the radius for this problem

Since it’s just asking for the coordinates of the center

[brittle surge]

Can someone help

i need to construct a square with d-a given

[chrome fiber]

what's d-a?

[brittle surge]

the diagonal minus the side

[main spire]

Nice profile hahaha

[brittle surge]

soo

can anyone solve this ?

[main spire]

Give us the d - a

[brittle surge]

what do you mean

[cursive nexus]

$$d-a = (\sqrt{2}-1)a$$

[brittle surge]

oh

ty

[chrome fiber]

how does that help though?

without approximation, i mean.

[brittle surge]

i mean

i found an answer online

but thanks for the reply

[silver niche]

"Find the area of an iscocles triangle with a base of 3x and sides of 2x"

[upper sedge]

Neato. Have you drawn it out?

[silver niche]

Yes

[upper sedge]

Did you perhaps notice that the line which bisects the unique angle gives two congruent right triangle's whose unknown side is the height of your isosceles triangle?

Additionally, do you know the formula for the area of a triangle?

[silver niche]

Yes I did
1/2b*perpendicular height

[upper sedge]

Good. Now solve for the missing side of a right triangle using the Pythagorean theorem.

It will be your perpendicular height.

[silver niche]

(2x)^2 + (1.5x)^2

[upper sedge]

Not quite.

[silver niche]

sqrt

of

[upper sedge]

1.5x would be either a or b. Solve for which ever letter (b or a) that you don't have.

[silver niche]

(2x)^2 - (1.5x)^2 = b^2

Yes?

[upper sedge]

And take the square root, yes.

That will be your height at the end of the day.

[silver niche]

Ok, I can take it from here
Thank you 😄

[solar bane]

How to solve?

[eager pendant]

@solar bane have you heard of brahmagupta?

[solar bane]

no

[zealous patrol]

use brahmagupta's formula

google it

[chrome fiber]

it's a cyclic quad.

[zealous patrol]

yeah

[solar bane]

Right

[chrome fiber]

s is the semiperimeter.

[solar bane]

2s=a+b+c+d

Awesome

[zealous patrol]

yep

[hasty thistle]

all the answers were correct

@thorn talon

Thank you!!

[robust gulch]

okay

so im working on a program

and i was wondering if i could make a program that takes 3 values

[upper karma]

Yeah....

[robust gulch]

and returns whether you can make a triangle with the 3 integers, and the numbers are the size of the triangles length

let me finish first 😛

[white harness]

AC + AB > BC in Euclid's geometry.

[robust gulch]

alright

[celest swan]

Is the following statement true?
(1) if you give me a relation that defines a certain shape, there’ll always be a complexer but somewhat similar shape defined by an equation pf superior degree.

Per exemple, if you give me an n-sphere, there’ll always provably be a (n+1)-sphere

(2) a flat plane that expends infinitely is a sphere.

<@&286206848099549185>

I know the 2nd one must be right, but like... at the same time...:

[tropic stirrup]

How dere

[celest swan]

$$\lim_{Circumference\to\infty}\text{Circumference}=“\infty “$$

[celest swan]

$$\text{Circumference}=2\pi r$$\
$$”\infty”=2\pi r$$\
$$”\infty”=r$$

[celest swan]

No???

But then

Its volume is its own circumference and its own radius and thats very disturbing

@tropic stirrup

[upper karma]

🤔

what's the question

[celest swan]

True or false: a flat plane that expends infinitely is a sphere.

[upper karma]

if it's flat how is it a sphere

[celest swan]

Well

[upper karma]

jkjk

that was a joke

[celest swan]

Okok

But do you why im hesitant about that one?

[upper karma]

why

[celest swan]

Because:

(1) if it had an infinitely small inclination, like: 1/infty. Then since it expends infinitely, it would be sphere no?

Also

[upper karma]

[celest swan]

We agree the thickness of the plane isn’t 0 roght?

[upper karma]

yeah

[celest swan]

So

Say we use pythagore to find the thickness, point A being on the bottom face and B on the top

There’s a crazily small difference between face A and face B

Now we place the point on the extremities of the figure:

We get

$$(\infty)^2+(thickness)^2=(distance AB)^2$$

[celest swan]

Whelp

$$\infty+thickness^2=AB^2$$

[upper karma]

you can't add infinity tho

🤔

[cursive nexus]

please don't troll

[upper karma]

infinity isn't a number

[celest swan]

I AINT

[upper karma]

you can't do that

you shouldn't even think about things like that

[celest swan]

Im asking a question that’s all

Is a flat plane expending towards infty a sphere?

[cursive nexus]

you can take the one-point compactification of a plane

and that's a sphere

[upper karma]

yeah like C

you add a point called infinity

[celest swan]

...smiles...

[upper karma]

but it's for reasons wholly unrelated to pythagorean theorem

and we just call the point infinity

[celest swan]

Oh ok

[upper karma]

we could call it whatever we liked lol

like "the happy point"

[celest swan]

But is the inclination of the flat plane infinite?

[upper karma]

no

it's just 0

[celest swan]

How’s that?

[upper karma]

because it's flat...

[celest swan]

Could prove it algebraically for me...?

[cursive nexus]

we can define a homeomorphism between the sphere and the plane with infinity

it's called the stereographic projection

[upper karma]

look up Riemann sphere

[celest swan]

Yes?

[upper karma]

can you clarify your question

I think we are talking about different things now

[celest swan]

Of course! Gimme a minute tho

[cursive nexus]

🙃 the prime has trolled before so watch out

[upper karma]

Did you know

every plane contains a point

[celest swan]

Yes i did

[cursive nexus]

😱 no

how preposterous... planes contain points...

[upper karma]

I almost got trolled

[celest swan]

you have a flat surface, a plane.
that surface doesn’t end: you could walk in a straight line as long as you want, you wouldn’t get back to your initial point.
is that plane a sphere?

[upper karma]

it's one point away from a sphere

add a point to the plane and it's the same as a sphere

[celest swan]

Oh but of course!

Because then inclination wouldn’t be 0!

[upper karma]

it is homeomorphic to a sphere

[celest swan]

But aren’t points infinitely small...?

[upper karma]

it's not like the plane magically turns itself to a sphere

we just say they are the same thing

[celest swan]

Ok

[upper karma]

and yes

[celest swan]

Then what difference between a number n of infinitely smoll and n+1???

Oh

Infinity aint a number sorry

[upper karma]

wat

[celest swan]

Like you said if you add one point it becomes homeo~thing to a sphere

[upper karma]

yeah

[celest swan]

But adding infinitely small isn’t adding nothing?

thinks
No nvm

It’s not that simple laughs

[upper karma]

so do you think (0, 1) and (0, 1] are the same?

because we just added the point {1}

lol

idk lol

he was like

is a flat plane really flat?

no he thinks a plane is a sphere because if you walk on a plane in a direction you never get back to your starting point

a sphere is a one point compactification of a plane

like the Riemann Sphere is for C

idk what he's talking about like

"infinity^2 + 1/infinity^2 = AB^2"

[celest swan]

Yeah kind of

And like

Im so mixed up haha

[solar bane]

How to solve?

[celest swan]

$$\shared$$

[charred spear]

Rendering failed. Check your code. You can edit your existing message if needed.

[dawn dome]

yi @tropic stirrup can you help me out with geometry >//<

=tex \text{the equations of perpendicular bisectors from sides AB and AC of a triangle}\ \text{ABC are }x-y+5=0\text{ and }x+2y=0 \ \text{ respectively. If the point A is}(1,-2)\text{, find the equation of the line BC}

[dawn dome]

i have a solution in mind but its cancerously long

[tropic stirrup]

Hi

Ok

So the trick is this

the solution to that simultaneous equations is the circumcenter, right?

now draw two lines from A, one perpendicular to x - y + 5 = 0 and the other to x + 2y = 0

You can easily find the equation of the circumcircle by its center and A(1, -2)

find the intersections of the circle and the two lines you've found to get the coordinates of B and C

to finish it all, find the equation of BC

I guess this is the "cancerously long" solution you talk about

[dawn dome]

nah

my thing was

find the equations of AB and AC

then solve those with the respective bisectors

you can obtain the remaining 2 vertices

then you have those and finding a line through 2 points is ez

that would work right?

[tropic stirrup]

Yeah that's it basically

[dawn dome]

no better way?

because it gets irritatingly long

[tropic stirrup]

don't think so

uh no

[dawn dome]

and the calculations suck

[tropic stirrup]

it's not long

[dawn dome]

long as in those calculations

yeah i just sovled it

but the calcs suck lol

fractions and stuff

[tropic stirrup]

y = x + 5; y = -x/2

perpendicular lines that pass through A:
y = -x - 1, y = 2x - 4

Intersections:
(-3, 2), (8/5, -4/5)

Reflect:
(-7, 6), (11/5, 2/5)

not too dirty

[dawn dome]

ye but then

you left out the line part lol

i mean yeah its 100% doable but annoying

=calc ((-4/5)-(2))/((8/5)-(-3))

[charred spear]

-0.60869565

[dawn dome]

yup

gurl did you hear about mathbeta

it can factorial 10000

[tropic stirrup]

=pup 10000!

[charred spear]

Query made by @tropic stirrup
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=10000!
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[tropic stirrup]

No, but I've heard of Wolfram|Alpha

[dawn dome]

no

[tropic stirrup]

[dawn dome]

like in the calc

[charred spear]

Error: Cannot perform factorial on a number greater than 300

[dawn dome]

=calc 10000!

an update is coming out

[tropic stirrup]

Oh ok

[dawn dome]

it can even do 3^(10^1000)

[tropic stirrup]

K ek

That's pretty big

Can it do integration stuffs?

[dawn dome]

not in the calc, no

but it now has sympy

so if you input sin(π/3) it'll give sqrt(3)/2

oh yeah @tropic stirrup wanna see something interesting

ill just dm it because idt its safe to send here

[tropic stirrup]

Ok

[cursive nexus]

[cosmic skiff]

Could use some help on a formula

I have function, the square root of a paraboloid with two variables. What would make the function open ended, closed, limited, unlimited, coherent or incoherent?

[misty knot]

I asked a question geometry related in #help-3 but just realised this would have been a better place. We can talk about it in either room.

[upper karma]

help

I think we'd need the equations of the parabolas?

Maybe not but it seems we could draw a lot of different parabolas that pass through those points

[crude kraken]

*with those tangent lines

[upper karma]

oh

I did not read that

[crude kraken]

fun problem it seems, interested in answer when someone saves ye

[upper karma]

just gonna make a guess that it's either 4 or 8

lol

[crude kraken]

Maybe if we reorganize as two parabolas that cross the x-axis at points distance 5 apart with the perpindicular tangent liens...

Since it's a rectangle it'd have to be symmetrical

so just multiply that by 2

Constructing such a parabola is beyond me

but that seems reasonable, no?

[upper karma]

oh wait

so rotate the rectangle so it rests on a side of length 3

it's 8

uh

why

I just guessed

ok wtf

Beizer curve equation

I did 2 / 3 * area of triangle

since that's some random formula I remembered for the area of an arc

yeah that's right too

but that's sketchy lol

[past mantle]

im confused

[pulsar bay]

Whaf do you want to find?

The vol or surf area?

[past mantle]

let me check

volume

[thorn talon]

don't you just work out each part individually then add?

[pulsar bay]

Yeah

So work out the volume of a sphere and divide by two and add the volume of the circular base cone

2/3pir^3 +1/3pir^2h

[ruby swallow]

Help me

[thin hound]

In degrees list all solutions between 360 deg and 720 deg

3sin(2x) + 5 = 4

<@&286206848099549185>

[upper karma]

Well, we can start by rearranging the equation.

[thin hound]

Oo lemme do that

Hold on

3sin (2x) = -1

[upper karma]

Okie

So, sin(2x) = -1/3, right?

[thin hound]

Oh yea

You right

[upper karma]

2x = asin(-1/3)

[thin hound]

What does the a mean

[pulsar bay]

is the 360 and 720 deg for the solutions x has to lie between?

arc

its another way of saying inverse

[upper karma]

asin, just means inverse of sin

[pulsar bay]

if 360<x<720, remember to base your results for 720<2x<1440

[upper karma]

^

[thin hound]

So whats after that

After we get the inverse

@upper karma

[upper karma]

@pulsar bay Helf, I g2g

[pulsar bay]

oh right

sure thing @thin hound

2x=arcsin(-1/3) right?

so the principal value you get is -19.5

and so we can either sketch the sinx curve

and see where the principal value turns up at -1/3

and see the rest of the corresponding values

or you can do it via rules

a rule is that sin resets after every 360 degrees and for negative principals, -180+19.5 is another solutions and 180+19.5 is also another solution

continue doing that until you get into the range of 720 and 1440

and then divide all answers by 2

and you get your result

[thin hound]

I got it dw

[whole forge]

How do i round to the nearest tenth?

Example:
25.7142871

And 51.42857143

[pulsar bay]

look at first and second decimal point

if second decimal point more or equal to 5 round the 1st decimal up

and if it is below 5 round first decimal down

[upper karma]

What does it mean by “fully labelled” is there anything more I can add for each?

[pulsar bay]

dont see anything else you can add apart from labelling the axis

[upper karma]

K thanks

[crimson wing]

@upper karma desmos is a great utility to use to visually verify a trig function's graph. Way better than my ti-84, imo.

can I divide: cos(2x)=-1
to cos(x)=(-1/2)

or wait, should cos(2x)=-1 turn into this formula

[waxen gorge]

@crimson wing u can't do cos(2x)/2 = cos(x)

[crimson wing]

how did you get cosx*sinx @waxen gorge

[waxen gorge]

cos(2x) = cos(x+x)

Oh cos

Thought that said sine

I'm dumb

Ya so cosx = 2pix

[heavy kernel]

would anyone be able to help me with proof

is u able to halp

well imma post the question anyway

Pls assist on number 1

[abstract arch]

If ST is the perpendicular bisector of RQ at the point P, then that implies RP = PQ

Since triangles SPR and SPQ have two congruent sides and one congruent angle, they are congruent by SAS

This implies angles SQP and SRP are congruent

Similarly, we can say that since triangles PQT and SPT share two congruent sides and an angle, they are congruent by SAS

This implies angles TQP and TRP are congruent

angle SRT = angle SRP + angle TRP by angle addition
angle SQT = angle SQP + angle TQP by angle addition

Since angle SRP is congruent to angle SQP and angle TRP is congruent to TQP, you can say by substitution,

angle SRT = angle SRP + angle TRP
angle SQT = angle SRP + angle TRP

Therefore, angles SRT and SQT are congruent

@heavy kernel

[heavy kernel]

this hurts my head

like a lot

thanks man

ill try to understand

[abstract arch]

pfft. It's more difficult to understand without a diagram, tbf

I would recommend labeling the angles

[heavy kernel]

TSP is a line tho

so how would that works

[abstract arch]

Also, I technically skipped a step in saying that since ST is a perpendicular bisector of RQ at point P, angles SPQ, QPT, RPT, and SPR are all 90 degrees

[heavy kernel]

well thats given

[abstract arch]

Should've been TRP. I was reading a picture on discord, it's hard to see without clicking on it

[heavy kernel]

ah

[abstract arch]

I was just mentioning I skipped it, idk how hard your teach/prof is about it

[heavy kernel]

she just want a 2 column thing

where i state the proof and the reasoning on each side

unless the question specifically tell me to do something else of course

thanks again

[abstract arch]

ye

[sterile mason]

Isnt the volume of any 3d figure its surface area * height?

If so why is the volume of a cone 1/3pi*r^2 * height?

[thorn talon]

i don't think that's true

in general

[sterile mason]

hmm

[thorn talon]

what figure does that apply to anyway?

[sterile mason]

cube

[pulsar bay]

Works for general prisms i think

[thorn talon]

surface area of cube with edge length 3

is 54 units^2

but volume is 27 units^3

though?

unless i'm misunderstanding something

[sterile mason]

the area of the bottom surface is what I mean

which is 9

[thorn talon]

ah right

so the base

[sterile mason]

yeah

[thorn talon]

yeah, that applies to prisms in general

base area * height

pretty sure it falls apart for cones and pyramids

[sterile mason]

it works for cyllinders too

[thorn talon]

yeah

that's basically a prism with circular base

[pulsar bay]

^

[sterile mason]

for pyramids and cones it seems to be the same but divided by 3?

[thorn talon]

yeah

[sterile mason]

ok thats easy to remember 👍

[glacial grail]

Hello Guys can somebody help. How to find angle a here ?

[white harness]

Dot product & arccos.

There is simpler: a pentagon inside a pentagon and Vertical Angles Theorem. $$\frac{2\pi}{5}$$

[junior scaffold]

It says its in the third quadrant, how do u know tht

sisnce costheta is -3/5

it cld be in the second quadrant too rite

[chrome fiber]

what's the question?

the statement mentions an interval, so it must be given in the question.

[junior scaffold]

ohh

ty

[chrome fiber]

😄

[orchid hill]

it says simplify

[neon fossil]

yes

[orchid hill]

how do i do it

[neon fossil]

what have you tried?

[orchid hill]

removing the sin square

[neon fossil]

yeah, then what do you get?

[orchid hill]

but do i have to take the 3 out

[neon fossil]

nope

[orchid hill]

out in a bracket

[neon fossil]

sin^2 (3x) ( 1 - cos^2(3x)) is what you get

[orchid hill]

but 3x is still inside for cos how

[neon fossil]

it was inside before

the expression is $$sin^2(3x) - sin^2(3x) cos^2(3x)$$

[orchid hill]

but its not squared

[neon fossil]

?

[orchid hill]

what do i do then

[pulsar bay]

use a trig identity after that

[orchid hill]

1-sin'2'

[pulsar bay]

close

[orchid hill]

oh cos then

[pulsar bay]

sin^2(1-cos^2)

and the trig identity is 1=sin^2+cos^2 right?

[orchid hill]

ye

[pulsar bay]

so whats 1-cos^2

[orchid hill]

sin'2

[pulsar bay]

right

[orchid hill]

the i multiply the sin'2 to get sin'4 does that work

[pulsar bay]

yup

[orchid hill]

suppose the 3 was in front of the sin cos or tan

can it be treated as 3(x) x= sin cos or tan

[pulsar bay]

i dont know what you mean

[orchid hill]

3sin

can it be 3(sin)

[neon fossil]

yeah

[pulsar bay]

yeah

[orchid hill]

but what if its in front can it still be the same eg sin3(thieter)

[pulsar bay]

if you have sin3x=sin3(x) but the 3x is still within the function of sin

goes for the rest of the trig stuff

[orchid hill]

so your saying sin3(x)=3(sinx)

[pulsar bay]

no

never

its inside the sinx

you cant take it out

sin3x=/=3sinx

[orchid hill]

thanks man

could you help me out with one more thing

[pulsar bay]

sure

[orchid hill]

so i changed cos^4 to 1-sin^4

[neon fossil]

that's not right

[pulsar bay]

you cant do that

[neon fossil]

cos^4 = cos^2 cos^2 = (1-sin^2) cos^2

[orchid hill]

could you explain why

[pulsar bay]

spec has

the easiest way is to notice that you can factorise that

[orchid hill]

but i can only take the ^2 out

[pulsar bay]

so that =(cos^2+sin^2)^2

[orchid hill]

how

[pulsar bay]

it follows the coefficients of 1 2 1

and treat cos^4 as x

and sin^4 as y

[orchid hill]

you lost me

[pulsar bay]

uhh ok

maybe ill do it more sistematically

[orchid hill]

i got 2^(1+2sin^2cos^2)

[pulsar bay]

hmm

the answer is 1 btw

[orchid hill]

no squares inside

sorry

(1+2sincos)^2

[pulsar bay]

so sustitute x=cos^2 and y=sin^2

[orchid hill]

ok

[pulsar bay]

you can rewrite the expression as x^2+2xy+y^2

right?

since (cos^2)^2=cos^4=x^2

and try and factorise that you get (x+y)^2

[orchid hill]

i got 1+2 (x^2)( y^2)

[pulsar bay]

uhhh

[chrome fiber]

#precalculus

[pulsar bay]

i dont know if thats right @orchid hill

[orchid hill]

can you show me your working out

[pulsar bay]

mine is one line

LOL

[orchid hill]

of all the steps

[pulsar bay]

Im trying to find

a better way of show you

[orchid hill]

then type in all your thought proscesses

of how you got 1

[pulsar bay]

so cos^4+2cos^2sin^2+sin^4

you can factorise it into (cos^2+sin^2)(cos^2+sin^2)

which is 1*1=1

[orchid hill]

how do you factorise

[chrome fiber]

can you guys move to #precalculus? this is not the right channel.

[orchid hill]

this is trig

[chrome fiber]

which is precal.

[orchid hill]

sorry

[celest swan]

(At rank 4)

Define algebraically the function representing the area of the white triangles divided the blacks’

[waxen gorge]

37/64

Yw

Ive memorized this

Appears on every math competition

@ebon vapor no ur supposed to memorize it

Oops