geometry-and-trigonometry

[remote heart]

that's what she wrote

yea

[upper karma]

ok so z means only whole number

ok

[remote heart]

k is an integer

yes

[upper karma]

ok thank you so much

[remote heart]

sure

[remote heart]

yes

[twilit zenith]

If it ever helps:
ℙ is the set of prime numbers.
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...}
ℕ is the set of natural numbers. It's all whole numbers from either 0 or 1.
{0, 1, 2, 3, 4, 5, 6, ...}
or
{1, 2, 3, 4, 5, 6, 7, ...}
ℤ is the set of integers
{..., -2, -1, 0, 1, 2, ...}
ℚ is the set of rational numbers
ℝ is the set of real numbers
ℂ is the set of complex numbers

[rich wolf]

$\mathbbH$ is the set of quaternions

Also important to know

[somber coyote]

AMD:
Compile Error! Click the reaction for details. (You may edit your message)

[rich wolf]

No blackboard h

Rip

[twilit zenith]

I think you need to surround the letter with curly braces, like so:

$\mathbb{H}$

[somber coyote]

RoiKadmon:

[rich wolf]

$\mathbb{Thanks}$

[somber coyote]

AMD:

[rich wolf]

Um

What

[severe oracle]

I need help deriving this

from this system of equations

im not sure where to start

[twilit zenith]

Seems to me that once you express the sine as a function of the other terms, then you can use trigonometric identities to find tan as an expression of sin, and thus as an expression of the other terms.
Not sure whether this is precisely what you were looking for.

[severe oracle]

im not sure either, this is for physics

[rich wolf]

@severe oracle projectile motion?

Or angular?

[severe oracle]

projectile

@rich wolf

[rich wolf]

What are theta 0 and 1

Initial angle and angle of impact?

[severe oracle]

correct

[rich wolf]

Isolate for sin(theta)_1

[severe oracle]

what about the v

[rich wolf]

?

[severe oracle]

wait isolate for sin(theta)_1 where

im confused

[drowsy marsh]

How do b

Part b

[upper karma]

I can't read it, please take a clearer picture if you can

[drowsy marsh]

Ok hold up

[upper karma]

the greatest angle of elevation will occur when he is closest to the flagpole

[spark summit]

can i get some gelp

[upper karma]

to find the third angle, use supplementary angle rule, that is they add up to 180deg

once you find that, knowing the sum of the angles inside a triangle, you should be able to make an equation to solve for x

and once you have x, you can find y 🙂

let me know if you need more help from there

@drowsy marsh did that make sense? are you still having trouble?

[spark summit]

@upper karma

does that mean its 80??

[upper karma]

Good

yes

Now you can find x 🙂

:p

[spark summit]

im confuesed

soo

[upper karma]

okay

so what do the angles in the triangle add up to?

[spark summit]

180

[upper karma]

yeah

and you know one of them is 80, another is x, and another is x-10

so you know

80+x+x-10 = 180

right?

[spark summit]

so 55

[upper karma]

yes 55 is x

and y?

[spark summit]

45?

[upper karma]

mhm!

do you get it?

[spark summit]

idk

we will find out later

thx u tho

[upper karma]

okay, feel free to ask if you need more help

[spark summit]

need more help

[upper karma]

well a regular pentagon can be broken up into 5 equal triangles

and since you have the distance OX and XD you can easily find the area of half of one of those triangles

[versed river]

Bruh dont you have ED?

[spark summit]

it's 6 no??

[versed river]

Yes, all the sides are 6, ed inclusive

@upper karma what would you use XD for?

[upper karma]

if you have ED, you have XD

[versed river]

yeah, but why would you use that

thats overcomplicating things

area of a triangle is b*h/2

[upper karma]

mm, okay

[versed river]

h=OX, b=ED

[upper karma]

just find the whole triangle then

[spark summit]

it will be 12.39 right

[upper karma]

yes. you're right.

yes

then multiply by 5

[spark summit]

ok

62

[upper karma]

good

[spark summit]

???

[upper karma]

basically same thing as last time

ten triangles this time 🙂

perimeter = 150, each side will be what?

[spark summit]

ohhh

ok

[upper karma]

then the apothem (line to the center) is 23.1

[spark summit]

is 150 the total perimeter

so is each side 15??

so each will be 173.25

then times 10

so 1732.5

??

[upper karma]

sounds right

i didnt punch it into calculator but

[twilit zenith]

By the way, what you did initially actually works, Breylin.

Indeed, if you were to multiply the perimeter, P, by a, the apothem, and divide by two, you could expand P as the sum of the lengths of the sides of the polygon, and deduce that this is equal to the sum of the areas of the triangles, and therefore is equal to the area as a whole.

Here's a visual example, if it helps.
The lengths of the sides here are:
s1, s2, ..., s6
and we get that the area of the first triangle is:
s1 · a / 2
the second one is:
s2 · a / 2
...
and thus, we get the same area, as shown above.

[upper karma]

That's cool, never learnt that myself. Or maybe I did and forgot.

[spark summit]

@upper karma Thank U

[upper karma]

Thanks @twilit zenith

[twilit zenith]

This case is somewhat specific, but I still think it's a nice proof to see.

[upper karma]

No problem, Roi's method is very neat too! Sorry for not catching that.

[spark summit]

Pascal, u got a paypal - i send u over a dollar or two

[upper karma]

ahaha no need for that!

[spark summit]

thanks tho

[upper karma]

my pleasure

[fleet wolf]

yo send me a dollar or two

[upper karma]

no!

D:

[indigo field]

hi guys

i need help with this problem

[upper karma]

You can try to find the area of big circle and subtract area of little circle 🙂

[indigo field]

i would but i don’t even know how to do that lol 😹

[static hollow]

The area of a circle is πr^2

and π is 3.14

[upper karma]

I got this correct, but the teacher guided me at the time. Was wondering if someone can help me with this question.

K

[twilit zenith]

Are you familiar with the formula for
cos (α + ß)?

[upper karma]

yes, but i think they usually go by U and A

[twilit zenith]

As in cos (U + A)?

[upper karma]

Correct.

I think.

[twilit zenith]

In this case, the formulas are:
cos (U + A) = cos (U) cos (A) - sin (U) sin (A)
cos (U - A) = cos (U) cos (A) + sin (U) sin (A)

From these formulas, one can find that:
sin (U) sin (A) = 0.5 (cos (U - A) - cos (U + A))

And why is that?
Let us try to plot the formulas for the sum and difference of angles under cosine to get to the left expression:
[I'll write it now. Please give me a minute.]

[supple wedge]

theres like a constant at the front tho

[upper karma]

Appreciate it ❤️

Exactly why I'm asking for help.

The constant is ruining me.

[twilit zenith]

There you go.

[upper karma]

I'll try to test it out. Thanks

Another quick question, this would be in quadrant 1 correct?

or 3?

[twilit zenith]

First one, I believe.
Remember that cot Θ = 1 / tg Θ = cos Θ / sin Θ.

In this case, since both cot (Θ) and sin (Θ) are positive, then cos (Θ) is positive, too.

[upper karma]

The 4 quads went, A(ll) S(tudents) T(take) C(alculus) although

Oh

Appreciate it. Thank you.

[fallen jungle]

my teacher taught me all silly tom cat

he was a very silly man >:D

[upper karma]

thats adorable

[twilit zenith]

Here's a more visual way to memorize it.

[upper karma]

Can someone check if this is right?

ill be saving that, thank you very much.

I had a question regarding identities.

tan(30-y)

I just want someone to tell me whether my answer is right or wrong.

I got, ((1/√ 3)-tan y) / ((1+(1/√ 3)tany)

[twilit zenith]

That's correct. 😁

[upper karma]

Thank yous. I feel accomplished, sorta :3

[twilit zenith]

You should feel so. 😎

[upper karma]

❤️

cos (α+ß) would equal to cos α cos ß / sin α sin ß

because they're reciprocals right?

[twilit zenith]

I don't think so, unfortunately.

[upper karma]

Do you know why it would be positive?

[twilit zenith]

Are you referring to cos (α + ß), or to cos (α) cos (ß) / (sin (α) sin (ß))?

[upper karma]

cos (α) cos (ß) / (sin (α) sin (ß))

If it's a long explanation, no need to stress, I could always ask my teacher tomorrow.

[twilit zenith]

For this term to be positive (for simplicity, let's assume that sin (α) and sin (ß) are both non-zero, so the expression is defined)), then either none, two, or four of the terms must be positive.

For example: If we have α = ß = 45°, then all the terms are positive, and we get a positive quotient.

Not sure if that answers your question. 😅

It's not always positive, though. For example, if:
α = 45°, ß = 135°, then:
sin (α), cos (α), sin (ß) > 0
cos (ß) < 0
and so we have 3 terms which are positive, and thus the quotient is negative.

[upper karma]

so these two would be practically identical if it were negative cos (α+ß), (cos α cos ß - sin α sin ß), or would it be positive (cos α cos ß + sin α sin ß)

that's what im stumped on.

[twilit zenith]

Are you referring to when:
cos (α + ß) = cos (α - ß)?

[upper karma]

It's just asking what the equations of sine would be.

[twilit zenith]

As in sin (α + ß) and sin (α - ß)?
Sorry if I'm being slow with this. 😅

[upper karma]

Youre great help, I'm just bad at wording things.

cos (α + ß) = cosα cosß (+/-) sinα sinß

Was wondering whether it was positive or negative, my assumption was positive although I'm having second thoughts

[twilit zenith]

Ooh, I think I see what you mean now.
cos (α + ß) = cos (α) cos (ß) - sin (α) sin (ß)
cos (α - ß) = cos (α) cos (ß) + sin (α) sin (ß)

If you're unsure, I would recommend you take a numerical example.
Say: α = ß = 45°.
In this case, we get:
cos (α + ß) = cos (90°) = 0
on one hand and:
cos (α) cos (ß) - sin (α) sin (ß) =
0.5√2 · 0.5√2 - 0.5√2 · 0.5√2 = 0.5 - 0.5 = 0
on the other.

If you're interested regarding where the formula comes from, it's originally derived from the formula:
sin (α + ß) = sin (α) cos (ß) + cos (α) sin (ß)
as proven here.

[supple wedge]

,w expand cos(a+b)

[somber coyote]

Display results online and refine query

Requested by Hmm#2554

[supple wedge]

yes

i remember it by cos(a+b)=cos cos - sin sin

[twilit zenith]

That works too.
The formula for sin (α + ß) can be derived from cos (α + ß) and vice versa, assuming you remember at least one of them, so I'd focus on trying to remember at least one for any future exam you have, and if you get stuck, then try working it out with the other, if you don't remember.

[upper karma]

I'm pretty sure I'll remember.

Hopefully

On second thought

Let me write it down.

[twilit zenith]

Don't worry.
Even if you don't remember it today, I'm sure it'll sink in soon enough.
Writing things down generally is a good idea, though.

[supple wedge]

just bash through angle sum questions

it will stick

[twilit zenith]

Practice is also a good way to remember, as Hmm suggests. 👍

[upper karma]

Practice makes perfect.

[supple wedge]

no

only if u practice properly

if u practice w misconceptions you will be reinforcing them

make sure u got the right concept first

ok

ima head ouyu

out

[upper karma]

Bye, have a great night.

one more question I need checked

hopefully i got it right

(-secθ / 1- cosθ) = (-1 - secθ / sin^2θ)

i calculated it out, and got

1-secθ / sin^2θ

[twilit zenith]

Just to clarify: Are you trying to work out the expression, or solve the equation?

[upper karma]

im verifying identities

for

(-secθ / 1- cosθ) = (-1 - secθ / sin^2θ)

I'll try asking a friend, have a great night.♡

[twilit zenith]

Oh, don't worry, I'm working it out. I'll try to have it ready in a minute.

[upper karma]

Take your time, no pressure it's just review. :>

[twilit zenith]

This is what I got for the two of them.
I don't see a direct argument to go from one expression to the other, but I suppose that if you were to try to equate the two expressions in the bottom right of each segment, you'd get an equality that always holds true.

[upper karma]

The squared threw me off a bit.

Thank you

Other than that, I can practically complete the rest of the review I just had some issues with the previous problems.

I hope you have a wonderful night.

or day, depending on your location

[twilit zenith]

Thank you very much. Same to you. 🛏️ 🌙

[dark sparrow]

@upper karma

[upper karma]

https://cdn.discordapp.com/attachments/699655642306052107/701394374633717820/unknown.png

Soooooo

For part a)

[dark sparrow]

uh

okay before we start

[upper karma]

Yeah

[dark sparrow]

this is a low res pic and i am having trouble reading it

[upper karma]

wait wot

Can you try opening original maybe

[dark sparrow]

okay i guess i can work with just zooming in on it

whatever

alright, so back to your point of confusion

[upper karma]

Okay, well

My point of confusion is that

I have never, ever in my life

Did trigonometry or geometry

Because I didn't do maths from year 8 to year 10

Now I am in Year 11, doing IB Maths HL and this is messing me up

So idk how my THOUGHT PROCESS

should be

[dark sparrow]

uh

wow

i mean

[upper karma]

Yes, I went in not even knowing algebraic manipulation.

[dark sparrow]

okay so like yeah this is gonna be tough

[upper karma]

I'd study for like 8 hours daily but moving on xd

Yeah it is

I don't know how to think

Not systemized at all

K

[dark sparrow]

they give you a diagram

this diagram on the right

[upper karma]

Yeah

[dark sparrow]

in which the yellow lampshade is expressed as a difference of two circular sectors

[upper karma]

I don't know what that means

The expression

"Difference of two circual sectors"

circular

[dark sparrow]

i'm talking about this from a purely geometric standpoint

do you know what a circular sector is

[upper karma]

I know what a sector is I guess

Wdym by circual sector

like, the region between two radii

of a circle

is what I know as a sector

[dark sparrow]

a circular sector is a piece of a circle bound by two radii

like a pizza slice

[upper karma]

yea

[dark sparrow]

i guess i am calling circular sectors what you are calling just sectors

[upper karma]

I see

Yeah I get you

[dark sparrow]

so... ok, so do you understand what i meant when i said

the yellow lampshade is expressed as a difference of two circular sectors

[upper karma]

yeah

[dark sparrow]

ok great

so

[upper karma]

Aaaaaaaaaaaand the arcs subtend the same center

[dark sparrow]

the circular sectors in question have the same angle, θ degrees

and their radii are R and r respectively

[upper karma]

Yeah

r+28 should be R

I am guessing

[dark sparrow]

not quite

[upper karma]

oh

[dark sparrow]

it's best to do everything in one unit

in this case meters

(and hence square meters for area)

[upper karma]

r+0.28

[dark sparrow]

so R = r + 0.28, yes

now

we've gotten that. we'll set it aside for a moment.

[upper karma]

Yea

[dark sparrow]

our goal for part a

will be to express the area of the lampshade in terms of theta and r

that's our long term goal and we will not jump to it immediately

[upper karma]

I see

[dark sparrow]

so since the lampshade is what's left of the big sector when the small sector is cut out

we know that the area of the lampshade is equal to the area of the big sector minus the area of the small sector

does that make sense

[upper karma]

Why did I not think of that

.

Yes it does

[dark sparrow]

alright great

so now

do you know how to find the area of a sector given its radius and angle (in degrees)

[upper karma]

theta/360 * pir^2

my bad

[dark sparrow]

superb

that's going to be the area of the small sector, since you used small r

[upper karma]

and

theta/360 * (pi*(0.28+r)^2)

[dark sparrow]

$A = \frac{\theta}{360} \pi R^2 - \frac{\theta}{360} \pi r^2$

[somber coyote]

Ann:

[upper karma]

should be the other one?

oh

[dark sparrow]

yes

that is in fact what i was going to do next

$A = \frac{\theta}{360} \pi [(r + 0.28)^2 - r^2]$

[somber coyote]

Ann:

[dark sparrow]

does this make sense to you

[upper karma]

Not quite

R^2 became what

[dark sparrow]

R^2 became (r+0.28)^2

[upper karma]

(r + 0.28)^2

[dark sparrow]

as you wrote

yes

[upper karma]

but then

theta/360*pir^2

[dark sparrow]

and then i factored πθ/360 out.

[upper karma]

where did that go

Ohh

my bad I didn't even see the theta/360pi

at the start

Been awake for 3 days or so so I am missing a lot of details

[dark sparrow]

uh

wow

that is like

bruh you're seriously sleep-dep

catch some Z's

[upper karma]

I am

suffering from a very bad case of insomnia

sooooooooo yeah

[dark sparrow]

alright... moving on

$(r + 0.28)^2 - r^2 = r^2 + 2 \cdot 0.28 \cdot r + 0.28^2 - r^2 \ = 0.28 \cdot 2r + 0.28^2$

[somber coyote]

Ann:

[upper karma]

hm

you just expanded it right

[dark sparrow]

yes

[upper karma]

what was the point of expanding it

simplifying?

[dark sparrow]

yes, the purpose is simplification

[upper karma]

i see

so now it's

theta/360pi(that)

[dark sparrow]

$A = \frac{\theta}{360} \pi [0.28 \cdot 2r + 0.28 \cdot 0.28]$

[somber coyote]

Ann:

[dark sparrow]

and now you can factor out 0.28 and you will arrive at exactly the formula they give

[upper karma]

so

oh

i see

okay I would have never thought of that

[dark sparrow]

this is a matter of algebra really

[upper karma]

It's not the algebra that's my problem

it's knowing what to do

to begin with

[dark sparrow]

the more algebra you do, the easier it becomes for you to see these sorts of things

[upper karma]

I see..

tysm

[dry phoenix]

Hey guys, is this true

disregard that |

on the right

[dark sparrow]

no

[high zephyr]

btw, nice pfp ann

[dark sparrow]

thank you

[dry phoenix]

Isn't it true without ^2 ?

[storm lily]

No. arccos(cos(720°)) = 0° as an example

No matter what you put into arccos, it only spits out angles from 0° to 180°

[high zephyr]

spits

i like it

[storm lily]

Oh oops thats probably not a very mathy term XD. When we were doing functions my teacher used vomit a lot. Like f eats up x and vomits up f(x) and g eats up f(x) and vomits out g(f(x)) a bit of a weird and kinda gross thought

[high zephyr]

[hasty ocean]

Can anyone help me with 10-20 question at 12:50 pm est today in geometry surface area and volume we could do call if you want thx.

[wheat river]

u tryin' to cheat at an exam fam

[hasty ocean]

nah

its not an exam

its hw

were given a random amount of questions and we have to do them starting at 12:50 but im confused on this chapter

so I came here maybe for some help

[wheat river]

fair enough; don't have the time tho unfortunately, maybe someone else feels up to the task!

[hasty ocean]

Ok I will be waiting for a reply from someone hopefully

[idle bloom]

Just send the questions here

[hasty ocean]

hellow spamkin

[idle bloom]

And what's confusing you

[hasty ocean]

surface area and volume

like the formulas

sometimes it will ask find lateral area

of a ceratin thing

im in 10th grade by the way

im doing like the basic geometry

[twilit zenith]

Do you have some example questions you've been struggling with?
Perhaps I'll be able to clear up a thing or two.

[hasty ocean]

lateral area i have to find with 4yd height and base of 3

[twilit zenith]

A base of 3 meaning that the base is a square whose dimensions are 3 yards x 3 yards?

[hasty ocean]

i just need the formulas for lateral area, surface area, and volume so I can hopefully pass my exams in the futre

yes 3 and 3

its a traingle

figure

[twilit zenith]

To start, drawing the shape will usually give you a good intuition of what you're looking for. In this case, you need a square prism where the sides of the base are 3 yards and the height is 4 yards.

[hasty ocean]

I see

[twilit zenith]

On, wait, sorry, do you mean that the base is an equilateral triangle, that all the sides of which are 3 yards?

[hasty ocean]

yea

Base of the triangle is 3 and the height is 4

yards

[twilit zenith]

Here is the drawing.
Do you know what faces you need to calculate the area of in order to find the lateral area?

[hasty ocean]

not sure

would it be 4 and 3

[twilit zenith]

Yes! 😁 👍
You need to calculate the areas of all of the faces that "connect" the two bases. These are rectangles whose dimensions are 3 yards x 4 yards.

[hasty ocean]

lateral area is 12?

[twilit zenith]

Not just yet.
Keep in mind that this is the area of one face.

[hasty ocean]

oh

12 times 5

[twilit zenith]

Not precisely.
Remember 2 things:

1. The area of the triangles is different from that of the rectangles.
2. You don't have to calculate the areas of the triangle bases, because they don't count towards the lateral surface area.

[hasty ocean]

ok

thanks

[twilit zenith]

Sorry to ask the same question, but do you understand now which faces you need to use to calculate the lateral surface area?

[hasty ocean]

ssort of

im just not to strong in geometry

[twilit zenith]

May I color them in, just to be sure?

[hasty ocean]

sure

[twilit zenith]

There they are.
Those are the faces that connect the two triangular bases.

[hasty ocean]

oh ok the ones taht connect you multiply

ok thanks for the help

[twilit zenith]

Exactly.
I'll rephrase:
You need to:

1. Find all of the faces that connect the two bases.
2. Calculate their areas.
3. Sum them up together to calculate the lateral surface area.

And you're very welcome.

By the way, may I add one more thing before I leave you alone?

I hope you wouldn't mind.
Then I'd like to add:

1. Note that sometimes, the faces aren't rectangular. In this case, you'd need to use a different formula to calculate them, or try to "unroll" them in your head in order to get geometric shapes that you know how to calculate the area of.

2. Sometimes, there aren't 2 bases, but only 1. In this case, the lateral area will be anything except the base, much like last time, it was anything except the two bases.

Another example.
In this case, to calculate the lateral surface area of the cylinder, you can "unroll" the surrounding face of the cylinder into a rectangle whose length is the circumference of the bases - 2πr - and whose height is the height of the cylinder - h.

[drifting parrot]

I'm having trouble with this problem

The restriction is [0,2pi)

[twilit zenith]

I think you're on the right track.
Keep in mind there could be multiple possible values for tan (x).
Do you know how to progress from here?

[acoustic jungle]

^

factor tan x

[drifting parrot]

No, that’s where I’m stuck @twilit zenith

Could you explain it to me in detail?

[upper karma]

Do you know how to factor it?

You're on the right track, just to try to simply the left side so that it's written as two factors

[drifting parrot]

I don't

[upper karma]

Okay

Well you have tan^3(x) - 3tan(x) = 0

[drifting parrot]

ok

[upper karma]

then, what is the greatest common factor between tan^3(x) and 3tan(x)

[drifting parrot]

x?

[upper karma]

it would be tan(x) right?

does that make sense?

[drifting parrot]

why would it be tan(x) when there's a tan^3?

[upper karma]

one sec ill write it out

...

we could let them do it

Thats what you would write anyways tho?

factor the apple

🙂

[drifting parrot]

ok

[upper karma]

perfect

now switch back, what is apple equal to?

wait

no

inside the bracket

you should only have 3

because 3x 🍎 = 3 🍎 , what you have is 🍎X 3 🍎 = 3 🍎 ^2

[drifting parrot]

oooh

ok

[upper karma]

Hope the apples didn't confuse you haha 😅

[drifting parrot]

They don’t 😂

[upper karma]

yes that's great

so now we have tan(x)(tan^2(x) -3) = 0 right?

[drifting parrot]

yes

[upper karma]

so tan(x) = 0 or (tan^2(x) -3) = 0

[drifting parrot]

👍🏿

[upper karma]

and then can you solve for x from there?

[drifting parrot]

I think so

[upper karma]

there will be more than one possible x value

okay, let me know if you get stuck

[drifting parrot]

ok

[upper karma]

Okay so you can't make the tan disappear there at the bottom :p

then square root of 🍎 ^2 = 🍎

so square root of tan^2(x) ?

[drifting parrot]

tan(x)=sqrt(3)

[dark sparrow]

then square root of 🍎 ^2 = 🍎

$\sqrt{x^2} \neq x$ in general

[somber coyote]

Ann:

[upper karma]

I have never written ±tan(x), but sure. Yes.

eaiser to just write ±3, I was getting there

[foggy beacon]

hey guys just a very quick question please

will the ratio of that triangle be the same as the one for that angle?

[dark sparrow]

what triangle

[foggy beacon]

the one that i put a red mark on

[upper karma]

how to know which inverse trigonometric functions are even and odd?

[foggy beacon]

you google it.

[drifting parrot]

@upper karma

Here’s what I have so far

[upper karma]

looks good to me, so which of all the values you got fit into your domain?

[drifting parrot]

The positive ones

The restriction is [0,2pi) so I assume that excludes any negative values

[upper karma]

Yep

I also don't quite understand what you wrote there with x + pi/3 + pi(n)?

ah, you are trying to write it periodically

[drifting parrot]

Yes

[upper karma]

Okay so what are all the value of x? write them out

[drifting parrot]

Ok

[upper karma]

and 0

[drifting parrot]

Got it

[upper karma]

great!

[upper karma]

looks good

[vocal hound]

what would a cross section formed by a plane intersecting a cube where the plane cuts off a corner of the cube look like

[latent iron]

help pls

[upper karma]

did you learn sine law?

[upper karma]

@latent iron you need to use Law of Cosines

look it up

once you know 2 angles, it's easy to find the 3rd (since all angles from a triangle need to add up to 180 degrees)

^

Also can be done with trigonometry

But yeah whatever you choose to

[rapid temple]

wait law of sine and cosine isn't trig?

[upper karma]

it is

:p

it's the

eye of the tiger

its the thrill of the fight

[rapid temple]

:/

[upper karma]

🙂

[upper karma]

wait law of sine and cosine isn't trig?
@rapid temple i meant trig definitons mb

[rigid siren]

How do I find all possible parabolas given a focus and a containing point (also need to find directrix)

[tame berry]

I am unable to find a solution for this

[upper karma]

@tame berry this seems a fun problem

Lemme try it first

[tame berry]

Well it was for the first 2 hours I spent it on

[upper karma]

Lmao

[rigid siren]

thats a tricky one

[upper karma]

Yeah indeed

I have seen a VERY similar one on MindYourDecisions

Almost equal

[latent iron]

oh

[upper karma]

Almost got it

Im not sure of the last step

In the last step you might want helpers

Hi

I solved that geometry problem

@latent iron you have to do this step a LOT of times on different triangles

Do you know the angle theorem which says that all the angles on a triangle sum up to 180? This is the theorem i used here

@upper karma post it to my dms or here

@tame berry these angles are identical:

all the angles for a triangle need to add up to 180

a line is 180 degrees

are you still with us, @tame berry ?

well?

I dont think so lmao

I hate when this happens

I hate it when ... yeah

Rude af

you beat me to it

?

Oh yeah

Lol

you typed it before I could hit enter

Lmao

Well

it's easy to find it

you just need to visualize the triangles

Yeah

[tame berry]

I am sorry, I had to go eat dinner

@upper karma yea, I was able go till there

[upper karma]

@tame berry ok, so these two angles are 50, right?

any doubt why they're 50 degrees?

[tame berry]

Nope

[upper karma]

you see where I'm going with this?

Vertical angles right

That’s why it’s 50

And those angles are supplementary

Ya I’m with u my man

And then 3 angles of a triangle = 180

So that’s 40

@upper karma I understand

Since that’s 40

Oh wait

Mhmm

70 + 10 = 80

60 + 20 = 80

Isosceles triangle

Top angle is 20

@upper karma right?

yep

Yay

I’m currently taking geometry right now too

So I’m proud of myself

Trig is next year for me

Next school year

[tame berry]

Got it

This is where I got stuck

[upper karma]

@tame berry Your welcome son

[upper karma]

.

[drifting parrot]

Ok, I’m here

[upper karma]

@drifting parrot what do u need son?

[silent plank]

that problem is evil

[drifting parrot]

I’m waiting for @upper karma

[upper karma]

do you know where i got the parts under then yellow and green?

those are the double angle identities

[drifting parrot]

Ok

[upper karma]

since we have sin(2x) and cos(2x)

Yes that’s very hard

I do not know that

So then the constants (1) cancel out

and we can factor out sinx

[drifting parrot]

So sine and cosine both have a constant of 2 and that gets canceled out?

[upper karma]

No that's the double angle, a constant is something like + 1, - 5, etc

these are our double angle formulas

[drifting parrot]

got it

I learned about these two weeks ago and still forgot

[upper karma]

and for the cos, it can be sometimes tricky to choose one since there are three

I choose the second one because then we have 1 - 2sin^2x -1 and 1-1 =0 so we get rid of those ones

and it is more simple

do you get that?

[drifting parrot]

I’m rereading this

[upper karma]

Okay, take your time ^-^

@surreal flume do what

https://tenor.com/view/alice-facepalm-gif-5094236

K

[drifting parrot]

Ok. I still don’t get it 😣

[upper karma]

hmm which part?

[drifting parrot]

I choose the second one because then we have 1 - 2sin^2x -1 and 1-1 =0 so we get rid of those ones
@upper karma

[upper karma]

See if you can follow this more

[drifting parrot]

ok

This makes sense!

I see how you substituted the cos(2x) for 1-2sin^2(x)

[upper karma]

yep!

Do you get the rest?

[drifting parrot]

I think so. I'm going to try to solve the problem and show you what I have once I'm done

[upper karma]

https://media.discordapp.net/attachments/599013597703831554/701867584525566072/image0.jpg The 3rd choice?

Is it da third choice?

is this a test

Nah

This is someone else’s homework

\

Just wanna see for myself

If I’m right

mhm

That’s why I sent it as a link

And not my own picture

So is it,

Full circumference of a circle is 2 Pi r

yes

Half of a circle is Pi R

So Pi r divided by 10

Since it was cut into 20 segments

mhm

And half is 10 segments

So Pi r over 10

Right

@upper karma answer me

Ok

Thanks dudes

[acoustic jungle]

.

[upper karma]

no problem

This is proof it wasn’t my homework

I just wanted to see if I was correct

For myself

no worries haha

👍

[drifting parrot]

I think I was able to get it @upper karma

[upper karma]

Yeah?

[drifting parrot]

I went through the problem the same way you did and I was able to get it

[upper karma]

Good!

Now can you solve for x?

[drifting parrot]

Sine(x) = 0

[upper karma]

And

Tan(x) = sqrt3/2

[drifting parrot]

Is that for the second problem?

[upper karma]

The second term, yes

[drifting parrot]

Umm

Are you using another identity?

[upper karma]

Yes, tanx= sinx/cosx

Move -2sinx to the other side

Then divide by sqrt 3 and divide by 2

You get

Sqrt3/2 = Sinx/cosx

Sqrt3/2 =tanx

And then solve for when tanx if sqrt3/2

[silent plank]

there seems to be a mistake

$\sin(2x) \neq \sin(x)\cos(x)$

[somber coyote]

ramonov:

[silent plank]

@drifting parrot

[drifting parrot]

yes?

[upper karma]

2sinxcosx

My bad

So then you just get rid of the other 2

[drifting parrot]

Like this?

[silent plank]

,rotate

[silent plank]

no, 1 sec

[upper karma]

@drifting parrot

[drifting parrot]

oooh

ok

Why is it done this way?

[upper karma]

Its like manipulating the problem

Actually a clever way of doing it

[drifting parrot]

This is how I had done it

[upper karma]

@drifting parrot look, if its an equation it has to be true. And the only way that the equation...

$2sin(x)\cdot(\sqrt{3}cosx - sinx) =
0$

...is true is if either 2sin(x) is equal to 0, bc it'd be 0*(√3cosx - sinx) which is 0 and then its true

Or the fact that √3cos-sinx is equal to 0, bc it'd be 2sin(x)*0 which is 0 and then its true

[somber coyote]

Al3dium:

@CoraXSkull look, if its an equation it has to be true. And the only way that the equation...

$2sin(x)\cdot(\sqrt{3}cosx - sinx) = 
0$

...is true is if either 2sin(x) is equal to 0, bc it'd be 0*(√3cosx - sinx) which is 0 and then its true

Or the fact that √3cos-sinx is equal to 0, bc it'd be 2sin(x)*0 which is 0 and then its true

[upper karma]

There's no 2

Who you are referring to

Both of you

Idk i only took it to explain him the reason

You just have tanx=sqrt3

So x=pi/3

Exactly

Or x=60

But that is more elegant

[drifting parrot]

I get that if it’s an equation, it has to be true but I need to know why in order to understand what I’m doing

[upper karma]

But i thought what you needed to understand it was that

What do you need then to understand?

[drifting parrot]

What I did wrong. I sent a picture of how I solved it, but I didn’t do it right

[silent plank]

it looks like you didn't fix the mistake that was pointed out

$\sin(2x) \neq \sin(x)\cos(x)$

[somber coyote]

ramonov:

[drifting parrot]

@upper karma

What about this?

[upper karma]

@silent plank but he didnt evaluate the 2sin(x)

[drifting parrot]

This is what @upper karma showed me.

[upper karma]

Yeah Pascal seems to be right

[drifting parrot]

Can you explain what she did?

[silent plank]

correctly manipulated:
sin(2x) to 2sin(x)cos(x)

instead if incorrectly replacing sin(2x) with sin(x)cos(x)

compared to what you did earlier

[drifting parrot]

That’s how she wrote it

With the 2x

[acoustic jungle]

Then there would be a mistake

[silent plank]

that was a mistake

that was acknowledged ~40 minutes ago but glossed over

[drifting parrot]

I didn’t know it was a mistake

[silent plank]

did you actually read:

$\sin(2x) \neq \sin(x)\cos(x)$

[somber coyote]

ramonov:

[drifting parrot]

Yes

[silent plank]

when i first typed it?

[drifting parrot]

Yes

Look, I’ve never done a problem like this before. I’m already frustrated with this problem as it is.

[silent plank]

can you attempt it again with this updated information?

[drifting parrot]

I’ll try

Give me a minute

[drifting parrot]

I found the correct identities

[upper karma]

👍

[acoustic jungle]

it's good to memorize the more general identities

Sin(x+y) and cos (x+y)

since sin(2x) derived from that

[drifting parrot]

But I still couldn’t get it

[silent plank]

you didn't sub properly

sin(2x) = 2sin(x)cos(x) right?

what should sqrt(3) * sin(2x) be?

[drifting parrot]

2sin(x)cos(x)

[acoustic jungle]

*sqrt(3)

[drifting parrot]

Sorry

2sqrt(3)sinxcosx

[acoustic jungle]

do you see the mistake now.

[drifting parrot]

Yes

Let me try again

[south inlet]

i have a simple question with trig..

um

[novel flax]

what have you tried so far

[south inlet]

wait do i just pick from either hyp or adjacent?

i havent really tried anything ive been confused

[novel flax]

is the distance between the tee and the hole the hypotenuse or a leg of the right triangle

[south inlet]

a leg

the distance hes lookin upwards is the hypotenuse

[novel flax]

the leg would be just the horizontal distance

the full distance from the tee to the hole is the hypotenuse

do you see why @south inlet

the hole is on the surface of the ground

[south inlet]

ohh yeah that makes sense

i can see that

[drifting parrot]

@acoustic jungle

This is what I have so far

not sure

[storm lily]

sin 4a = 2 sin 2a cos 2a

Technically, you could write sin 3a = 2 sin 1.5a cos 1.5a but no one really does that much, but there is a triple angle formula

[upper karma]

I need help with an online test

I have no clue how to do any of it I thought someone here could help

@upper karma me

Can I see

yep just a sec

@upper karma

[high zephyr]

I need help with an online test
@upper karma pretty sure you're not allowed to ask for help during an exam

[upper karma]

ok

my bad

[acoustic jungle]

tbh

online tests are very dumb

[upper karma]

K

I agree

[drifting parrot]

@acoustic jungle

[acoustic jungle]

what

[drifting parrot]

Here's what I got so far

[upper karma]

Nice handwriting

👌

[acoustic jungle]

ok you got x=0 and pi, plug them back in to see if it's right.

[drifting parrot]

thanks! @upper karma

Do I have to factor?

[acoustic jungle]

also where did the 1 go.

[hollow finch]

$\arcsin(0)=0$, $\pi$ and $2\pi$

[drifting parrot]

the one is gone because when I replaced cos(2x) with 1-2sin^2x, it canceled out the 1 that was part of the original equation.

[somber coyote]

Lσνιηg✧Sσνєяєιgη:

[hollow finch]

Yes

Now the second part

$\sqrt{3}\cos(x)-\sin(x)=0$

[somber coyote]

Lσνιηg✧Sσνєяєιgη:

[drifting parrot]

Do I factor anything out?

[hollow finch]

No

[drifting parrot]

Ok

[twilit zenith]

May I give a small "hint"?

[hollow finch]

Any identities you can think of?

[drifting parrot]

Sure

[acoustic jungle]

square them.

[drifting parrot]

Tanx?

[acoustic jungle]

move them first though.

or you can do that.

[hollow finch]

Tanx?
@drifting parrot Ya

I was thinking of this

[twilit zenith]

You got it, Cora. 👍

[drifting parrot]

😁

[acoustic jungle]

quadrants 1 and 3. 30 60 90 triangles.

if that makes sense

I gtg.

[hollow finch]

$\sqrt{3}=\frac{\sin(x)}{\cos(x)}=\tan(x) \implies $ basic$\angle=\arctan(\sqrt{3})=60^{o} \implies 1st: x=60^{o}$ or $180^{o}+60^{o}=240^{o}$

I think

[somber coyote]

Lσνιηg✧Sσνєяєιgη:

[acoustic jungle]

That's right.

[drifting parrot]

Thanks for helping me @acoustic jungle

[upper karma]

@drifting parrot thats what Pascal got lmao

[drifting parrot]

lol

[hollow finch]

Hmm

[upper karma]

In the very beginning

[hollow finch]

Can you think of any other methods to do this?

[drifting parrot]

To find the tangent?

[upper karma]

No i mean the answer

He got too 60

Or π/3

[drifting parrot]

The other one is 4pi/3

[hollow finch]

To solve $\sqrt{3}\cos(x)-\sin(x)=0$

[somber coyote]

Lσνιηg✧Sσνєяєιgη:

[upper karma]

He only got π/3

Thats what i saying

Ikr these things

Just saying he had the same thing as you

[drifting parrot]

I found a chart that has all the tangent, cotangent.... values

That’s what I used

[hollow finch]

$\sqrt{3}\cos(x)-\sin(x)=2\cos(x+\frac{\pi}{6})=0 \implies x+\frac{\pi}{6}=\frac{\pi}{2}$ or $\frac{3\pi}{2} \iff x=\frac{\pi}{3}$ or $\frac{4\pi}{3}$

[somber coyote]

Lσνιηg✧Sσνєяєιgη:

[hollow finch]

Oooooooooo

[upper karma]

Can anyone help me out? I’m stuck on a question

[hollow finch]

Sure

I'll try

[upper karma]

I figured out how to get the angle but I don’t know how to get the sides of the triangle

[hollow finch]

,rotate

[upper karma]

A and b

[hollow finch]

Do you know about sine and cosine rule?

[upper karma]

Yeah a bit

[hollow finch]

Try to apply that

$\frac{\sin(a)}{A}=\frac{\sin(b)}{B}=\frac{\sin(c)}{C}$ in a triangle

[upper karma]

Ok I’m lost

[somber coyote]

Lσνιηg✧Sσνєяєιgη:

[upper karma]

Use trig functions

[hollow finch]

You dk how to apply the formula I assume

[upper karma]

It’s easier

Yeah I don’t know

I’m new to trig

[hollow finch]

Oh ya trig functions can

[upper karma]

It’s a right triangle anyway

Just use trig functions

And calculator

[hollow finch]

Omg

[upper karma]

How do I do that?

[hollow finch]

sine=?

[upper karma]

Cos

Tan

[hollow finch]

soh cah toa

[upper karma]

Yes

I’m lost

[hollow finch]

$\sin(x)=\frac{opp}{hyp}$

[somber coyote]

Lσνιηg✧Sσνєяєιgη:

[upper karma]

Watch video

I did I wanted half of youtube

@upper karma look video online

And couldn’t find a YouTube video

[hollow finch]

khan academy

[upper karma]

Plenty of YouTube videos

[hollow finch]

There's alot

[upper karma]

What did u type

Ya khan academy

That’s good too

But how do I find the adjacent and opposite if I only have the hypotenuse

[glad oak]

use letters

[upper karma]

Get x by itself

Use letters 👌

Cos = x/ hypotenuse

Multiply hypotenuse both sides

X = hypotenuse times Cos

Idk

[glad oak]

I need some help knowing what SAS triangles are and how I'm supposed to use trig functios to solve them

[upper karma]

Just look on YouTube

@glad oak Sin And cos rule

Sas triangles are when

[glad oak]

that's where I'm lost

[upper karma]

You know 2 sides and an angle between them

[glad oak]

yes

side angle side triangles

[upper karma]

Sin(a)/Side A

[glad oak]

but incorporating the sin and cos rules are where I get extra lost

[upper karma]

= Sin(B)/Side B

[glad oak]

and knowing whatthey are

[upper karma]

Look online watch videos

Learning them by reading stuff isn’t the best choice

Watch some videos

@glad oak Or shall I just explain it to u

[twilit zenith]

Do you mean proving that triangles are congruent, Elfthan?

[glad oak]

nah

I need to figure out the sin and cos rule

[upper karma]

Sin(a)/Side a = sin(b)/Side b

$\frac{\sin(a)}{A}=\frac{\sin(b)}{B}=\frac{\sin(c)}{C}$ in a triangle

[somber coyote]

𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫:

[upper karma]

@glad oak

[glad oak]

I think I'm like I'm missing several steps so imma watch a youtube vid thanks for the help

[upper karma]

Your welcome

👌

[acoustic jungle]

@upper karma your angles are wrong.

Why did you write 53+17=90

and b is not isosceles

[upper karma]

law of sines

[vital bough]

am i overthinking this

[high zephyr]

seems correct

[vital bough]

okay thanks

[acoustic jungle]

that is wrong

ASS doesn't 100% prove congruency

that is why it's called ASS - disclaimer (not the actual reason)

@vital bough

[high zephyr]

oh sheet i mixed up ASS and AAS

holup is that an exam?

[humble pebble]

yea that is

[high zephyr]

fml, how did i mess that up

@vital bough

[glad gorge]

Can someone help me pls ;-;

Idk how to do this

[acoustic jungle]

inscribed angle theorem

thales theorem - essentially inscribed angle theorem

[glad gorge]

.......