geometry-and-trigonometry

[spark stag]

so you should be able to find an identity for cotangent based off that

[buoyant idol]

is this the same for the others?

[devout shell]

So the reciprocal of opp/adj will give you?

[buoyant idol]

adj/opp lmao

[spark stag]

yep.

[devout shell]

Yes

[buoyant idol]

so cot is just cos/sin?

[devout shell]

It is the same process for the others as well

[spark stag]

yes, essentially

[devout shell]

Yes

[buoyant idol]

and cos/sin is adj/hyp/opp/hyp

[devout shell]

Yes

[spark stag]

which simplifies to adj/opp, but yes.

[buoyant idol]

so adj*opp/ hyp^2?

wait whot

[devout shell]

No

[buoyant idol]

oh ye

[spark stag]

(i'm assuming some brackets there)

[devout shell]

You messed up with your math there

[buoyant idol]

so cot is basically adj/opp

[spark stag]

yes

the reciprocal of tangent.

[devout shell]

Yes...please we already told you that, what makes you so unsure of that?

[buoyant idol]

sec is hyp/adj

and csc is hyp/opp

right?

[devout shell]

Yes

[buoyant idol]

j wna make sure

[devout shell]

If there’s one thing left to teach you, it’s to be confident in the math. Given the definitions and all, you just apply what you know and what you get is correct. Aside from some mess up on the fraction simplification, all we told you is correct. No need to confirm all the fine points. That’s for you to do so you gain confidence lol.

[buoyant idol]

lmao okok

i make shit ton of careless mistakes tho soooo ykno.....

anyways is there anything else abt trig that i should know abt similar to things like inverse functions

things that can mess up trig

[devout shell]

A website with all trig functions listed would be best to test your math. We’ll give you the concepts here, but we aren’t really ones to check every single answer for you.

[buoyant idol]

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

WIKIPEDIA lmao

[devout shell]

If it has all equivalent forms of the trig functions listed, then might as well use it

Wiki has become a pretty good source of info for general stuff. I haven’t found obviously fake info on it ever.

[buoyant idol]

oh thats reassuring lmao

[flint pelican]

<@&286206848099549185>

[fringe dirge]

What's your question

[flint pelican]

Idk how to do it

[upper karma]

I cna't really read it

[olive solar]

same

[flint pelican]

Here I’ll take another pic sorry

[upper karma]

Type it out fren

or that

[fringe dirge]

It tells you how to do it

[flint pelican]

Uh is that better or still nah

[fringe dirge]

What are you confused about there

[flint pelican]

Ik but I don’t understand I get the 20 by 15 but idk where 18 came from

We are trying to find the area in the white box right but idk what the area of the black part is

.....

[fringe dirge]

After they tile the left and right

you're left with 18

since 20 -2 = 18

[flint pelican]

I’m sorry I’m still confused could you dumb it down for me

[fringe dirge]

Think about it

Draw a picture on some grid paper

Or just make your own grid lines

[tropic turret]

Think about overlaps. We already count the corner squares by doing 15 on the sides. If we did 20 at the top and bottom, wouldn't we stack extra tiles on the corners?

[flint pelican]

So uh was the formula for these bh/3 Or lwh

[weary drift]

You first find the area of the base, which is shaped like a triangle, then multiply the base by the height of the prism

[flint pelican]

Ok thank you

How do they know the radius and height are 6 and 8

[weary drift]

They’re rotating the triangle around side AB

If you can imagine what cone that generates, side AC is the radius of the cone

and AB is the height

[flint pelican]

How do I imagine this like is this triangle standing on top of side ac

[weary drift]

Side AC will draw out the base of the cone, and the cone will sit on top of this base, sure

[flint pelican]

Yes I get it

Thank you

[weary drift]

Np

[finite frost]

Any advice with this never seen it before just came up wondering if anyone knows a good way to go about it

[tropic shard]

I'm guessing like:

Then to say that sin(α)=sin(β) I'd then use the first condition they gave, that cos(α) = tan(β) and do the same step over and notice the connection.

[finite frost]

Thanks appreciate the help

[nimble rain]

how do u read radians

im so bad

[weary drift]

1 revolution (think 1 whole circle) = 2pi radians

pi rad = half circle, pi/2 = quarter circle, pi/4 = eighth of a circle

[dark sparrow]

wdym "read radians"

[spark stag]

in other words, $\pi$ rad $= 180^{\circ}$

[somber coyote]

Namington:

[spark stag]

do you have a more specific question?

[flat geyser]

i need to make sure im not going crazy

can someone add these up for me

add the two waves

this is what i get

[dark sparrow]

yup

checks out

[flat geyser]

so what i sent is right?

[dark sparrow]

yes

otherwise i wouldn't've said "checks out"

[flat geyser]

aight ty

for somereason the website is telling me its wrong

[quiet mason]

i feel it should be independent of time

or the standing wave finale

[flat geyser]

nah what i said was defo right

also this:

i dont understanding what im being asked

[quiet mason]

reposting in diff channels

aight but i guess they just want you to represent c in terms of $\alpha$

[somber coyote]

Lionel:

[quiet mason]

so as $c=\log_a b$

$c=\frac{\log_\alpha b}{\log_\alpha a}$

[somber coyote]

Lionel:

Lionel:

[flat geyser]

Sorry I forgot about the question channel

And didn’t want to delete from here

So I just had to change the base to alpha

Ok ty

[silent plank]

or use another variable so you don't confuse it with a.

[quiet mason]

@silent plank nah dude read the question

[flat geyser]

Turns out what Lionel said isn’t accepted either :/

Doesn’t matter it’s not like I don’t understand

Question just badly worded

[silent plank]

fk me. Who chose to mix a and alpha together.

what's the layout of the answer box?

[flat geyser]

It just lets you input an answer

[silent plank]

you entered (45,45) for (x1,y1)
similar error for (x2,y2)

the entry layout is horrible

[silent plank]

what have you tried?

[flint pelican]

Wait I looked at the key

I think ik

[silent plank]

what have you tried?

[balmy pelican]

what have you tried?
^

[frozen bluff]

uh idk

nothing

abt this

tbh

i just need an answer ;-;

[fringe dirge]

Nah fam

[frozen bluff]

...

thats racist

jk bnut idk how to do this

;-;

[fringe dirge]

"what have you tried?"

[frozen bluff]

x1,x2 -y1-y2 / 2

^

[fringe dirge]

And why are you doing this?

[weary drift]

thats racist
we do tend to discriminate against those who don't try much and only ask for answers

[frozen bluff]

bruh i said jk

;-;

[weary drift]

:>

[frozen bluff]

what is it tho?

yall boy math nerds

i need help thats why im in this discord

[weary drift]

did you answer "why are you doing this?"

[frozen bluff]

bro lmfao

i just want the answer

lmfao ur a fucking nerd

[silent plank]

what were you trying to achieve in each part of your calculations?

[balmy pelican]

I can't even understand their calculations

[frozen bluff]

IT WAS 45

fck im smart as hell

[balmy pelican]

good and how did you find out?

[weary drift]

[frozen bluff]

my nigga bc i im smart

like fr

yall green

[balmy pelican]

ok...

[frozen bluff]

why the door?

[fringe dirge]

Lmao this guy's a moron

[frozen bluff]

lmao idc

ur probably gay

[fringe dirge]

There's literally nothing wrong with being gay

[frozen bluff]

yes there is

[fringe dirge]

<@&268886789983436800>

[twin heron]

not here squidy boy

[frozen bluff]

actually everyone has there own opinion i dont like gay people and i think something is wrong with them "Fredom Of Speech"

i cant get arrested for saying gays arent ment to be

@twin heron

[twin heron]

this isn't the place buddy

[frozen bluff]

lmfao

[fringe dirge]

@frozen bluff You have no idea what freedom of speech means buddy

[frozen bluff]

bro thats discord

[twin heron]

hes gone

this isn't for discussion about killing gays, so I just banned him

[fringe dirge]

sucks having kids with such horrendous world views

idk how they pick that shit up

[umbral snow]

The worst part is the light theme

[small raptor]

@fringe dirge Usually from their parents.

[frosty steppe]

@umbral snow

[umbral snow]

Yep that's awful

[quiet mason]

😂

[next saffron]

lol i have to put my disc light theme cause if its dark my screen starts to do dim even though brightness is max

[stuck torrent]

Do three non colinear points in 2D uniquely define a circle?

[umbral snow]

Yes, since
(x - a)² + (y - b)² = r²
Has three unknowns

[sly marlin]

^ really nonrigourous argument

just intersect perpendicular bisectors

[stuck torrent]

Well

That doesn't quite show uniqueness

[sly marlin]

Well, since the points are not collinear, there are 2 non-parallel perpendicular bisectors. The centre of the circle lies on the intersection.

[stuck torrent]

yes

That's how you construct the circle using those three points using straight edge and compass, but I still don't see uniqueness. 🤔

maybe this is enough for uniqueness since my geom is shit

[dark sparrow]

what other point could it possibly be

[sly marlin]

if you can construct it, then it is unique

[stuck torrent]

🤔 ok

[sly marlin]

(of course, your construction must only construct one point)

this construction only constructs one point

[buoyant mirage]

I have the equation and radius of a sphere in R3, how do I find where it intersects the yz plane?

[fringe dirge]

What do you know about points on the yz plane?

[buoyant mirage]

The x coord is always 0

Do I set x to 0 in the equation and solve?

[fringe dirge]

correct

[buoyant mirage]

Awesome, thanks

Now there's another question. I have the center of a sphere, and I need to find a radius so that the sphere touches the yz plane. How do I solve that?

[fringe dirge]

By touches I assume you mean just touches at one point?

[buoyant mirage]

I assume, yeah

[fringe dirge]

Well what do you think you'd do?

[sly marlin]

so, where is the tangency point on the plane?

[buoyant mirage]

Set the radius to 0, set x 0 and solve?

[sly marlin]

well, can you figure out what radius you need now?

[buoyant mirage]

Not sure. I don't have the problem in front of me, sorry ^^"

[sly marlin]

well, say the point is (x, y, z), can you find the radius in terms of x, y, z?

[buoyant mirage]

I think so. It would have to be the point where x is 0

And then I could solve that for the radius

[sly marlin]

so, the tangency point is (0, y, z)

so the radius?

[buoyant mirage]

The radius is x, right?

[sly marlin]

no

what if x<0

[buoyant mirage]

|x|

[sly marlin]

yeah

[buoyant mirage]

Alright. Thanks. I never really did geometry so I really appreciate the help

[flint pelican]

I’m lost

Why is the height 4c isn’t that the hypotenuse because it looks like line y=4x goes up diagonally and how do you know what line x=c is

[scenic glade]

(diagram not drawn to scale / no scale provided)

[flint pelican]

Ok

Was that um to me

[spark stag]

yes, the line y = 4x goes up diagonally

where does that intersect the line x = c?

well, if we have y = 4x, and x = c, then it intersects when y = 4c

so the y value where the lines intersect - ie, the height of the triangle - is 4c.

then, the answer key shows you how to find the value of c

[wanton berry]

i know the identities of sin(x)cos(y) but in this case I have sin(x)cos(x)

i have no idea what to do with that

[mellow flicker]

@lapis pewter

[lapis pewter]

If the measure of an angle is 38 find the measure of its complement

[mellow flicker]

Ok

[lapis pewter]

I keep getting different answers

[mellow flicker]

So an angle and it’s compliment add to 90

[lapis pewter]

Yes

[mellow flicker]

You know that right

Ol

[lapis pewter]

I do I have the notes

[mellow flicker]

So what to add to 38 to equal 90

[upper karma]

lol

[lapis pewter]

Idk it’s self

Blitz your definitely a virgin stfu

“lol”

Stfu

[upper karma]

lol

ok, a "hint"

what is (90-38)+38

[mellow flicker]

XD

[lapis pewter]

Sicko mode😔

It’s 90 that isn’t the answer

[upper karma]

90-38 though

[lapis pewter]

Ur telling me it’s 52

[upper karma]

yes

👏🏿

[lapis pewter]

Alright coach

Why this so confusing

I’m about to fail my class and there’s a test Tuesday

About to read a whole book on geometry

[sly marlin]

It says "Be careful of the first one though..."

@coral solar

because they rounded the values to get the classes

[coral solar]

Ohhhh I finally see it thank you

[sly marlin]

sec theta

[dark sparrow]

everything except sec(θ)

[sly marlin]

how does sec relate to sin and cos?

don't think of the triangle anymore for this

triangle is useful when you know the angle is from 0 to pi/2

when it's outside, the sign would be a problem

[dark sparrow]

yes

[dark sparrow]

what do you think

[dark sparrow]

no

both, actually

now they are both even more wrong

[devout shell]

Don’t be aggressive like that please, and never say that word here again, you know which one I’m referring to @mossy marlin

[upper karma]

God bless the unit circle in pins

[upper karma]

can someone help me with this

It's time to put it all together! Now that you know how to construct, it's your turn to create your own construction. Using your compass and straightedge, complete the steps below.

Draw a line segment and copy it to the right of the original segment. Explain your steps and justify each step used.
Bisect the original line segment from problem one.
Draw an angle and copy it to the right of the original angle. Explain your steps and justify each step used.
Bisect the original angle from problem three.

[coral solar]

How do I do 10a? Because it only has one angle and one value stated for the hypotenuse

[silent plank]

what information can you get from
D is the midpoint of AC?

[coral solar]

That AC is 2x

[silent plank]

and what information can you get from that?

[coral solar]

Is Ac equal to BC

I don’t know

[silent plank]

how sure are you about whether they're equal?

[coral solar]

It says Bc is 2x that’s why

[silent plank]

but do you know why its 2x?

[coral solar]

But I still don’t know how to work out x

Because d is the midpoint and ac and bc are equal

[silent plank]

why are ac and bc equal?

[coral solar]

Ohhh

Because they’re both the radius of the circle

[silent plank]

ok.

you have 2 sides in terms of x
a known angle
and a known side opposite that angle

what do you think can be applied here to find x?

[coral solar]

Sine rule?

Wait no because it’s not asking for the angle

Ohh

I have to do sine

Twice?

[silent plank]

what rule involves 3 sides and an angle?

[coral solar]

First to get the angle then I take it away from 180 then I do the sine rule again to get the side?

Cosine

But I don’t know what angle B is so I need to do the sine rule?

[silent plank]

the known angle and side are opposite each other,
you already have all the information you need to find x.

[coral solar]

But if I have those then I don’t need to work it out

[silent plank]

you're trying to find the value of x

x and 2x will be your other 2 sides

what is the cosine rule?

[silent plank]

sqrt(14) and 120 deg are opposite each other, so which variables are they?

[coral solar]

a

So I have to rearrange the cosine rule?

[silent plank]

after substituting in the values
you'll have an equation in x
then just solve for x

[coral solar]

Thank you

[ocean junco]

Hi! I had to do statements for my geometry homework and I’m afraid that it just doesn’t make sense to me. If someone could explain, that would be ever so helpful.

[umbral snow]

That's logic, not geometry lol

[upper karma]

if x implies y and y implies z then x implies z

[umbral snow]

If A, then B
If B, then C

Naturally, you can also conclude that if A, then C

That make sense?

[upper karma]

dumb thing without context for learning geometry if you ask me

[umbral snow]

Maybe for learning proof basics

[upper karma]

could at least make something that correlates

[weary drift]

special triangles may help

[tropic shard]

$$4 \sec{\Big( \frac{π}{4} \Big)} + 7 \cot{\Big( \frac{π}{6} \Big) } = \frac{4}{\cos{( 45^{\circ} )}} + 7 \frac{\cos{( 30 ^{\circ} )}}{\sin{( 30 ^{\circ} )}}$$

[silent plank]

the calculations seem to have been done for \cos and \tan
instead of \sec and \cot

[tropic shard]

?

[weary drift]

^{\circ}

[somber coyote]

HenryCastle:

[tropic shard]

Neat.

[dark sparrow]

@upper karma if you're able to evaluate sec(π/4) and cot(π/6) individually, then you're able to evaluate 4sec(π/4) + 7cot(π/6).

[upper karma]

How do I define [tanx] ?

[tropic shard]

Pardon?

[upper karma]

Greatest integer function

But it has tan x

[tropic shard]

So like this?

[upper karma]

Holy moly

[tropic shard]

where ceil(tan(x)) ≡ ⌈tan(x)⌉

[upper karma]

Yep

[tropic shard]

Looks cool.

[upper karma]

How to write it down?

Mathematically

In terms of range?

-infinity to +infinity?

[tropic shard]

Range is from negative infinity to infinity, domain is-

yeah

wait

good question, what'd be the range of a floor/ceiling function..

[upper karma]

It's defined like X-1<[x]<=X

@tropic shardHow did you plot it?

[tropic shard]

I used Desmos

[upper karma]

But how do I plot it without a software?

[tropic shard]

Are you really being asked to do that? This doesn't seem to be something someone could do without software. Atleast no where with accuracy.

Idk how to do set builders, but a guess would be {y | y ∈ (-∞, ∞) ∧ y ∈ ℤ, 0}

[upper karma]

Yeah we are asked to do that

Basically it's a calculus question

A part of it is [tanx]

[tropic shard]

What'd I'd do is plot the vertical asymptopes of y=tan(x), graph y=tan(x), then tick mark the y-axis. After all that, I'd start plotting the graph y=⌈tan(x)⌉ by eye.

[upper karma]

But the steps are getting smaller as you go up...

How did that happen?

[tropic shard]

Because the function is increasing at a faster rate as you get closer to a vertical asymptope

[upper karma]

So I have to keep in mind that too?

[tropic shard]

Yep. The steeper the slope of the graph, the smaller the step should be.

[upper karma]

Ooof

[tropic shard]

It isn't too important the size, but I'd make sure to not make something where y=5 have a bigger step than where y=0.

[upper karma]

Should I dm you the question?

[tropic shard]

Y'know, cause the steps should be getting more and more smaller the higher/lower it is from the x-axis.

Sure.

[upper karma]

I sent

[hot iris]

Would anyone know how the concept of surface area is extended in linear algebra? All I can find online is formulas for highschool stuff

[narrow storm]

hello

someone explain how 6/sqrt(3) is 2*sqrt(3)

@violet nest

[violet nest]

wat the heck just do @ helper next time

anyways

[dark sparrow]

$\frac{3}{\sqrt{3}} = \sqrt{3}$

[somber coyote]

Ann:

[narrow storm]

wait

nvm

i got it

[violet nest]

=tex \frac{6}{3^{\frac{1}{2}}} = \frac{6}{3^{\frac{1}{2}}} \cdot \frac{3^{\frac{1}{2}}}{3^{\frac{1}{2}}}

oh

$\frac{6}{3^{\frac{1}{2}}} = \frac{6}{3^{\frac{1}{2}}} \cdot \frac{3^{\frac{1}{2}}}{3^{\frac{1}{2}}}$

[dark sparrow]

there's a diff bot here

[somber coyote]

AbdullahTrees:

[narrow storm]

changing the denominator to a 3 so as not to include a rational number i the denominator

[violet nest]

@narrow storm honestly next time ping people when you are really stuck

[narrow storm]

what did i just do?

i pinged you

people

[violet nest]

you figured it out before i even finished typing

ie you could do it on your own

[narrow storm]

lol i thought i was stuck

i was on it for like 8 mins then i decided to search up the exact question

didnt think it would yeild anything but it did

sorry bruh bruh

[grim condor]

Write an abosolute value equation with solution x = 3.5 or -2.5.

[umbral snow]

|x - a| gives the distance between x and a

So think in terms of distances

[grim condor]

okay. got it

[weary drift]

50in is the distance it rolled, not its radius

sure

[wild flame]

I took trig last year but I forgot how to do this: "If cos(theta) = 5/13, what is tan(theta)?"

How would I approach this?

[mortal mason]

x=r*cos(theta)

polar coordinates

[umbral snow]

Make a triangle where cosθ = adjacent/hypotenuse = 5/13

[wild flame]

OHHHH

yea now i remember

thanks 😄

[umbral snow]

Lol, glad it was that easy

[silent plank]

are you given the domain of theta?

[flint pelican]

Idek

What to do

[silent plank]

Q56?

[flint pelican]

Yes questions 56 and the answer is J

[silent plank]

what's the formula for the area of a triangle?

[flint pelican]

Bh/2

[silent plank]

do you know any other formulas for area that involve
2 sides and the angle in between?

[flint pelican]

I don’t think so

[silent plank]

are you familiar with
Area = (1/2) ab sinC

[flint pelican]

No

I’ve never done this in school yet

But if you could give me a link to learn it on khan I could get it maybe so ab is the sidelength?

[silent plank]

C is the angle between the sides a and b

[flint pelican]

Ok so 70 and 110

But how do I know what x and why are

Y*

[silent plank]

you don't need to know what they are

[flint pelican]

Ok

Ah because they give me the area

Maybe

[silent plank]

using the formula what would be the area of triangle1 in terms of x,y

(don't bother calculating sin70)

[flint pelican]

Why

Wouldn’t it be 30=1/2 *sin(70) xy

[silent plank]

yes

now what would be the Area of triangle 2

[flint pelican]

1/2 ab sin(110)

[silent plank]

*xy

[flint pelican]

Sorry yes xy instead of ab

[silent plank]

what's the relation between sin(70) and sin(110)?

[flint pelican]

Am I allowed to use a calculator lol

[silent plank]

if you know your trig you don't need to

[flint pelican]

I haven’t taken trig yet so I don’t know it well

Oh shoot they are the same

So the area is the same too since we are multiplying by the same things

Thank you

[silent plank]

yeh

hmmm, that stuff should've been covered in the book before these problems

[south junco]

How do I even start

[weary drift]

did you try anything?

[south junco]

no cause the teacher didnt even talk about it

[weary drift]

you sure?

what's the relationship between angles formed by intersecting lines?

[south junco]

what

wait let me try something i have an idea

[weary drift]

[south junco]

e

How would you write down a linear pair for example

[weary drift]

@south junco you got 14 right

[south junco]

ok cool

[coral solar]

how do i find the intersections of lines for (y=4x-9 and y=2x+3) because the x isnt on it's own

[south junco]

i dont know how to find it but i think i have the answer

probably right

but its worth a shot so here

[coral solar]

the answer is 6,15

[south junco]

oh

then there you go

lol

[weary drift]

@south junco nothing special about writing down linear pairs, you just gotta know what they are

[coral solar]

i just need to know how to work it out for that type of question

[south junco]

i know what they are but dont know how to right them on paper

[weary drift]

like if for example angle A and angle B are linear pairs

you literally write this

$\angle A \text{ and }\angle B \text{ are a linear pair}$

[somber coyote]

RokettoJanpu:

[coral solar]

i know how to work it out if the x has 1 in front of it but this doesnt 😦

[south junco]

sorry for the wrong angle of the picture

[spark stag]

,rotate

[weary drift]

@coral solar do you know anything about the point where those two lines intersect?

[coral solar]

that the point is 6,15

[weary drift]

no, nothing specific

in general if two lines intersect, what do you know about the point where they meet?

[coral solar]

perpendicular

[weary drift]

are the lines always perpendicular if they intersect?

[coral solar]

parallel*

[weary drift]

if they're parallel do they intersect?

[coral solar]

yes

2 different parallel lines itersect

[weary drift]

parallel lines do not

no idea where you got that from

[coral solar]

okay thank you

i worked out the answer its fine thank you

[radiant nimbus]

how do I get the cross product of 3 vectors?

distributive property?

[weary drift]

cross product takes in 2 vectors and outputs 1 vector O.o

[radiant nimbus]

Maybe I'm not understanding

[weary drift]

unless you got something like A x (B x C)

[radiant nimbus]

So I know cross product gives a vector

but you need cross product to get area right?

[weary drift]

hold on a sec, what does your hw look like?

[radiant nimbus]

the length of the cross product gives length of a parralelogram

er

area of parralelogram

buuut

it wants the area of a parralelopiped

lol

er

volume ugh i suck at typing

[weary drift]

oh, parallelepiped

[weary drift]

you're looking for the triple scalar product then

[radiant nimbus]

that sounds familiar but I don't have it in my notes 😦

[weary drift]

http://mathworld.wolfram.com/ScalarTripleProduct.html

[radiant nimbus]

so it's set up like this ||AxB||*||C||?

||AxB||*||C||

[weary drift]

$A \cdot (B \times C)$

[somber coyote]

RokettoJanpu:

[radiant nimbus]

so cross product of B and C and dot product of A and resultant vector?

[weary drift]

mhmm

you can do the cross and dot product or take the determinant of this matrix

https://i.gyazo.com/c03e07592109eac3c6dfed63a47358cb.png

[radiant nimbus]

I've been doing matrix for cross product lol, but I couldn't figure out how to do a matrix for 3 vectors

@weary drift do I stop at A*(BXC) or do I get magnitude of the whole thing?

[weary drift]

wdym

[radiant nimbus]

well for area its |UxV|

like the area of a parralelogram

[weary drift]

base area is indeed |BxC|

[radiant nimbus]

https://tenor.com/view/yes-gif-9118517

for dot product if I have a vector U = 2i+3j and V= -9i-4j+3k.... K would be 0 x 3 right?

[neat vigil]

Hi guys, im trying to solve this one for a while, but i don't know what to do.
"For the right triangle ABC, a = 5 and r = 1. Find sin and cos values of each angle."

[silent plank]

was there a diagram?
where is the right angle located?
what is r?

[upper karma]

i think this is what he said

[neat vigil]

@upper karma yes, that's it

[south junco]

I think I’m wrong

What would be the answer to this problem be?

I got x=7 8/9 and y=5/17

[spiral path]

x = 21 and y = 8

[south junco]

How

What did I do wrong

[spiral path]

on a straight line, the angle is 180 deg
so,
( 10x - 61 ) + ( x + 10 ) = 180
and
( 18y + 5 ) + ( x + 10 ) = 180
Solve them, and you will get x and y

[south junco]

Ok thx

[fleet knot]

@spiral path when do you take trigonometry

[spiral path]

I didn't get the question

[fleet knot]

What year

[spiral path]

you mean in which grade?

[fleet knot]

Yea my bad

Cuz it looks like you already took it

[spiral path]

For Indians, it's at 8th grade(IIT). Normally, it is included in 10th grade

[fleet knot]

@spiral path no

Geo is 10th

Cuz im in 10th

[spiral path]

you don't have trig in 10th?

[south junco]

im in 8th and have geo

[fleet knot]

@south junco hold up

In 8th I was taking pre

[south junco]

yeha

im advanced

i go to a high school

but in middle school

i have to take a bus

The question is name a linear pair

I don’t know exactly what to write

[junior harbor]

EKF and HKG

[spiral path]

where's example 1?

In a unit circle, sin thetha = the value of y component

so for (a), sin (0) = 0 as y component/co-ordinate is 0

similarly the rest

for (b) , the angle is 90 deg / (pi/2) rad , which is on y-axis, so sin(pi/2) = 1 , similarly the rest

[junior harbor]

Yup

[quiet mason]

thetha

did you just write theta wr to your pronunciation

[limber mango]

Someone help me

[dark sparrow]

can you not zoom out

so that the entire polygon can be seen at once

maybe this will make the problem easier to do too

[limber mango]

I can’t zoom out. Sorry.

But the polygon is a nonagon

And the last answer choice is 56 degrees

[dark sparrow]

it's a nonagon

so it has how many sides?

[limber mango]

9

Ah

If I’m correct, the nonagon would rotate 360 degrees

[dark sparrow]

a 360 degree rotation will leave ANY shape unchanged.

[limber mango]

Ah

[dark sparrow]

this nonagon can be rotated by a much smaller angle.

[limber mango]

40?

Since it has 9 sides

And since 360 degrees would leave any shape unchanged

[dark sparrow]

well there you have it

you were able to do this w/o me

[limber mango]

Thanks

[dark sparrow]

i didn't supply any information whatsoever

[dark sparrow]

why don't you try proving it yourself

idk, is it?

[somber coyote]

euler/fermat:

[dark sparrow]

are [these] supposed to denote area

can you show your diagram

i don't have anything to draw with atm

i'll get back to you in like 20 mins hopefully

[upper karma]

@upper karma as for why the altitude to the hypotenuse is 2.4 in a 3-4-5 triangle, it's just inverse pythagoras theorem

[coral solar]

hi how do i solve 6b

i found x = 4/3

when i did 6-4/3 i got the answer but then when i divided it by 3 i got a wrong answer

the answer is 14/3 but i dont know how to get it

[silent plank]

where is 10b...?

[coral solar]

6b*** sorry

[silent plank]

why did you want to divide it by 3?

[coral solar]

because i rearranged 3y+x=6

and y=(6-x)/3

[silent plank]

well you found the x-intercepts to be
(4/3, 0) and (6, 0)

6 - 4/3 is already the distance between the 2 intercepts

and you need to do anything else

[coral solar]

it's asking for the x intercepts

[silent plank]

which you found by plugging y=0

after you got those values, you don't need the equations/lines anymore

find the distance between the x-intercepts*

[coral solar]

nvm i got it

it says this

[silent plank]

where's the calculation of the x intercept for the 2nd line?

[coral solar]

6/3=2

[silent plank]

no, that would be the y intercept

i may have misunderstood you earlier, i thought you already found the value of the 2nd intercept by yourself

what's the value of y for the x-intercept?

[coral solar]

i asked my friend

she sent me this

she told me i had to plot the lines

[silent plank]

and then what will your x be in
3y + x = 6

[coral solar]

2

[silent plank]

what's the value of y for the x-intercept?

(where it intercepts with the x axis)

what value did you set y to for the first line?

hmm,
it seems that you havent done a) yet?

[coral solar]

y is 0

[silent plank]

ok, and what will x be when y=0 in the second equation?

[coral solar]

4/3

[silent plank]

6a): sketch on the same diagram, the lines
if that was done you should already have this info

in your second equation

[coral solar]

i already did that

[silent plank]

3y + x = 6

when y=0, x is definitely NOT 4/3 for this equation

[silent plank]

that's your first equation, i'm asking about the 2ND equation

[coral solar]

both of the equations are on that graph

it asked me to sketch it on the same diagram

0,-4 is y=3x-4 and -4 is the y-intercept

[silent plank]

that equation matches to lower line on the graph

but the sketch for the 2nd equation is wrong

and i'm going to ask again,
what is x when y=0 for 3y + x = 6?

[coral solar]

2

oh

3

[silent plank]

still no

[coral solar]

-2

[silent plank]

3(0) + 3 = 6
0 + 3 = 6?

[coral solar]

3

[silent plank]

no

3(0) + x = 6
x = ?

[coral solar]

6#

6

[silent plank]

that's better

[coral solar]

omg thank you

now i get it

[silent plank]

now that you have the correct x-intercept does that clarify things

[coral solar]

yes omg now i get it thank you

[silent plank]

btw, on your graph. all intercepts should be labelled

as well as the axis

[coral solar]

i'll label it now thank you

[steel hawk]

Can anyone help me with this please

[kindred dock]

whats the maximum arc length a circle can have?

[weary drift]

have you given this any thought yourself?

[kindred dock]

it cant be 360

isnt it just infinitely big then?

because the radius can be any size so

[weary drift]

360 what

i assume you're talking about 360 degrees, but since when were degrees used to calculate arc length?

[kindred dock]

no?

[weary drift]

not. at. all

[kindred dock]

i didnt even say that but ight

[snow socket]

https://gyazo.com/e8d8f4a8e82f5837014c5967465922a6

can somone tell me why i got this wrong ?

or what i did wrong

[silent plank]

the radius term was calculated incorrectly

[snow socket]

okay so i did this

[silent plank]

why is
(2 - (-5))^2 = -23?
that's way off

firstly, it came out negative
did you do
2 - 5^2 = 2 - 25 = -23?

[snow socket]

so it would be 49 right

[silent plank]

yes

[snow socket]

oh okay

i see now

so the radius would be 85

[silent plank]

not quite

radius would be sqrt(85)
but that equation uses r^2 which is 85

[snow socket]

so like it says put it in an equation it would sqrt of 85

[silent plank]

the question above has the form
(x-a)^2 + (y-b)^2 = r^2

[snow socket]

yeah it says put it in equation form

[silent plank]

so you would be using r^2 here

[queen river]

I have Geometry class tomorrow and I'm nervous someone help

[tranquil pilot]

with what?

[snow socket]

ohh okay got it

[tawny merlin]

OK so

i have a problem

so arc lenght and radius should be in what unit exactly

to find the angle

[silent plank]

the same unit

[grave fulcrum]

So what do you recommend I do?

Ig, I'll repost problem

@dark sparrow

Are you still there?

Just tryin not to lose ya

[dark sparrow]

repost the problem

[grave fulcrum]

how do I find 3rd side/angle?

[grave fulcrum]

thank you so much for drawing that

would I use soh cah toa here?

[dark sparrow]

would you?

[grave fulcrum]

that's my guess

[dark sparrow]

why don't you act on it

there's nothing in the problem telling you explicitly to do this or that

[grave fulcrum]

alrighty

but doesn't that require 2 sides still?

[dark sparrow]

you can and are expected to find sin(25°), cos(25°) or tan(25°), whichever one you may end up needing, using a calculator.

[grave fulcrum]

I'm still confused on how

[dark sparrow]

how what

[grave fulcrum]

how do I use angle+side to find hyp

nvm

I forgot it's 90 degree

[somber coyote]

euler/fermat:

[sly marlin]

hmm

oh yeah

@upper karma what did you try?

Any results you found?

(hint: diagram totally not to scale)

[dark sparrow]

well i DID try to make the right angle at least appear right

[sly marlin]

yeah, but AB is shorter than BC in your diagram

@upper karma Hint: Draw a diagram to scale

[somber coyote]

euler/fermat:

[sly marlin]

Cool

The next step would have something to do with that right angle, since you haven't used it yet

@upper karma So, any ideas?

hmm

remember that right triangles can be inscribed in semicircles

@upper karma

well, you might need to construct more points

to relate the area of ABC with ACD

Well, they do not

necessarily

The easiest way to relate areas is to try to subtract copies of the first shape from the second shape

Looks good

you constructed the important point

done some trigonometry?

What's O?

yeah

oh you guessed that the 4 triangles have the same area

can you prove it?

idk what you managed to find

and how did you define the constructed point?

yeah, the pink extension

yeah it is, but can you prove it?

haha you just stated why z=x earlier

well
O is the circumcenter of triangle ACD
and when I connected OC, I saw that it was parallel to AB
and that told me that z = x

It's just inside here

Okay, but you want to show z=x

you know why

but you just haven't put 2 and 2 together

okay, and then you want to show that OC is parallel to AB

that follows from z=x

Well, notice what shape AOCB is

you have z=x

what shape is AOCB?

(based on the properties of the shape you found)

I thought you already had it?

O is circumcenter -> z=x, because they are both radii of the same circle

directly

ah, I see what you mean

So, notice AOC and ABC

what can you say about them

yeah, and?

and?

so...?

yeah

erm

you mean AOC=ABC?

wait, ACO and ACB

cool, yeah also works out

didn't think of that

nah, I just don't see how they motivate this lol

yours is perfectly motivatable

plotting midpoints opposite right angles is well known

[dark sparrow]

such solutions are always polished

[sly marlin]

yeah, maybe they didn't want to quote Thales

https://en.wikipedia.org/wiki/Thales's_theorem

[silent plank]

heh , i don't remember being told the name of that theorem

[lavish umbra]

So i have triangle

A right triangle

And I plot a point into a space insid the triangle

Anywhere in it

Then what you do is you add up all the distances from the dot to each side

And this sum is always constant

How can I prove this mathmaticly

[upper karma]

Suppose 5 cos x + 12 cos y = 13. The maximum possible value of 5 sin x + 12 sin y is

How do i approach this problem

I know the solution but it feels vague

[sly marlin]

@lavish umbra False, are you sure you don't have an equilateral triangle?

@upper karma R-formula?

[upper karma]

@sly marlin What R formula?

[sly marlin]

https://brilliant.org/wiki/trigonometric-r-method/

[upper karma]

Yeah so it is gonna be R formula, but how is it to be applied here?

On the sum of sines?

Or derive a relationship between the two

And if we are doing that then why?

@sly marlin

[sly marlin]

because it is easier to analyse the a trigonometric function than the sum of two trigonometric functions

[tranquil drum]

why doesnt inverse tan approach infinity at 0?

inverse tan is cos/sin right?

[gritty siren]

huh no

You're talking about arctan, right?

[tranquil drum]

arctan is inverse tangent and cotangent right?

[gritty siren]

arctan is the inverse of the restriction of tan on ] -π/2, π/2 [

the cotangent is something different

for any real x, arctan(x) is the unique real number strictly between -π/2 and π/2 such that tan(arctan(x))=x

for any real x such that sin(x)≠0, cot(x) = cos(x)/sin(x)

[tranquil drum]

kk

[wide ferry]

pls help

[silent plank]

what have you tried?

[sly marlin]

@wide ferry what equations can you form out of the problem?

[wide ferry]

is it 6x+44 + -10+65 or 6x+44 = -10+65

or are both of those wrong

[upper karma]

Both of those are wrong.

For that you want to find what x is.

[wide ferry]

ohhhh

[upper karma]

And you have a known entity, 117.

[wide ferry]

thanks so its 6x+44 + -10+65 = 117

[upper karma]

👍

[sly marlin]

nope

[wide ferry]

wait

yes or no?

[upper karma]

It is smh.

[wide ferry]

ok thanks

[upper karma]

@sly marlin Wait how is it not.

[sly marlin]

you are missing the x in -10x

[wide ferry]

oh

[upper karma]

Bruh moment.

[dim python]

hey there, anyone know if theres a way to solve for surface area of overlapping circles?
I'm familiar with http://mathworld.wolfram.com/Circle-CircleIntersection.html but for the case where the radius is larger than the centroid distance of the other circle i don't think it works.

(solved scroll down to video)

[primal hull]

r2 is half d

hmm thats interesting

[dim python]

thank you, i don't see anywhere that claims the wolfram solution doesn't work, but i feel likt it would't the approach i'm using right now is more arc sine than arc cos which maybe is why the above fails?

i'm thinking maybe something like the below would solve it for the case where the smaller radius triangle around the angles needs to be added? not sure if this makes sense entirely?

[upper karma]

I think x=900

Is dat right?

[snow socket]

https://gyazo.com/996f2901f4905c4121a4a9706fa449e9

why is this wrong ?

[subtle gate]

For all shaded points this must apply: |y|<3 AND |x|<5 @snow socket

See the problem?

[snow socket]

no im confused

@subtle gate

would only the middle part be shaded

[dim python]

For anyone curious. the wolfram answer works but you have to retain the sign when subtracting out the triangle area for the smaller one. ie. when the smaller one's center starts behind x (white line), its triangle area you subtract from the circle area must be negative, to add it in instead of subtracting.

[upper karma]

does anyone else take geometry?

[spark stag]

nope, you are the only person to ever take geometry in your life

[sly marlin]

@pearl geode Just ask

[minor goblet]

Hi.
I have this question
Enter the expression asin(x)+15 , where asin(x) is the inverse sine function. Alternatively, you may enter the inverse sine function in either of the following forms, which are also accepted: arcsin(x) or sin−1(x).

[sly marlin]

hmm, so far so good

what's the question?

[minor goblet]

I dont get what it is asking? But that is the question. Is it to write the expression of asinx + 15 in inverse sine form?

Oh i think i get it, I believe it equals arcsin(x) + 15

Or asin^-1 + 15

Well apparently the answer is Sin^-1 + 15 lol

[sly marlin]

...

Is there more context?

[minor goblet]

Lol no sorry

[sly marlin]

screenshot/picture?

[minor goblet]

https://prnt.sc/p2ye1j

It's a simple question I just read it incorrectly

[silent plank]

its a horrible question

a by itself should never be used to represent an inverse function

[obsidian moon]

yeah i dont know many people who use asin lol

[silent plank]

if it IS written like that,
a's just treated as a variable

[quiet mason]

lol a sine function

[flint pelican]

Guys what is the difference between e and a isn’t the square root of x^2= x since that is x^2/2 =x^1

[sly marlin]

e and a?

oh

you did not take a picture of the option labels

@flint pelican what are you asking?

[flint pelican]

Ok sorry one second

The difference between option a and option e

[sly marlin]

and if it is an argument about $\sqrt{a^2+b^2}=a+b$, plug in some nonzero numbers

[somber coyote]

Element118:

[flint pelican]

Wait wait I don’t even need to I can see why it doesn’t work because you need to square them both first then add then square root can’t square square root and then add thank you I got got got got it

[flint pelican]

Guys for this one ik the 2 angles around 120= 60 but is the reason they are both congruent is because the angle that hits a light=same as reflected

The light hits the point where the angle is 120 and then reflects back up

Idk how to explain it

[keen aspen]

Depending on the surface, the light will bounce off of the surface at the same angle it came in

So assuming that this is the case for this surface, your understanding is correct

[flint pelican]

Alright

Also for this one I still struggle with it I’ve done it at least 3 times before could you explain which step I messed up on

[silent plank]

you divided by two twice

[flint pelican]

Wait but I thought I was supposed to divide by 2 twice

Like first b/2 to get radius and the.n square it

And do b^2/4 *pi all over 2

[silent plank]

and you should only divide by that 2 once

[flint pelican]

Which 2 in which step

So I do b^2pi/2 - (b/2)^2 pi/2 right

Lol

[silent plank]

you don't divide each term by 2

[flint pelican]

Oh you divide the answer by 2? Like first you multiply the top then divide by 2?

[silent plank]

whichever way you look at it, you only divide once

[flint pelican]

Alright thank you

[silent plank]

otherwise for some ridiculousness,
(1 * 1 * 1 * 1 .... * 1)/2 would be ...

[flint pelican]

1/2

Ah and you can’t do 1/2 *1/2 because that’s 1/4

[gritty siren]

congrats

[flint pelican]

Alright

[weary drift]

Draw some triangles to find out

[weary drift]

In that example you’ve constructed a triangle using the reference angle instead of the actual angle. You use this triangle as normal, but then remember to adjust the result of cos and sin according to which quadrant you’re in

[dark sparrow]

convention

same reason the y axis goes bottom to top rather than top to bottom

except that there do exist contexts where it makes sense to have the y axis run the other way

in image processing, the y axis is inverted, so as to make the top left corner the origin

[unkempt magnet]

hi, can someone help w this problem

a right triangle has two legs with one leg measuring x and another leg measuring x+8. if the perimeter of the triangle is 4x, find the length of each side

[upper karma]

Okay.

Now the total perimiter is 4x.

Can you find the 3rd side?

Using that information?

[unkempt magnet]

4x-2x+8 right

[upper karma]

Slightly incorrect.

$4x-(2x+8)$

[somber coyote]

leviosa:

[upper karma]

Makes sense?

[unkempt magnet]

yup

[upper karma]

Simple enough?

[unkempt magnet]

yeah

[upper karma]

K.

[upper karma]

I'm trying to visualize trigonometry by watching this video: https://youtu.be/LGBNYjC5jdU

At 14:21, why is the height increasing? Does theta mean height? I'm so confused. I understood everything until I got to that part.

[spark stag]

do you mean the physical height of the angle/lines above the plane?

they're doing that to visualize what a sine/cosine wave looks like

yes, the height corresponds to theta (the angle)

[upper karma]

ugh I'm trying to make sense of this

at 11:38, why isn't the height increasing then?

I'm thinking of the circle as being a flat thing, but is it supposed to be standing up?

[spark stag]

the thing they do with increasing height at 14:21

is just a visualization tool

to demonstrate how sine/cosine waves look

11:38 visualizes a changing angle as a change on the unit circle

this is the "common" visualization

but at 14 minutes

they want to show you graphs of how sine/cosine change as the angle does

the thing is, if we just use the unit circle

we only get 1 value of theta at a given point in time

in order to graph changing values of theta

we need to, well, change theta

they choose to make theta's axis the height axis