geometry-and-trigonometry

[brave kayak]

and since 4 angles are given

if im assuming correctly that 3x and x+25 are angles

[upper karma]

It's irregular looking

Yeah

[brave kayak]

im gonna assume it's a 4-sided polygon

and in a polygon with 4 sides, what should the total angle inside be?

[upper karma]

360

I'm running out of time

[brave kayak]

right

so if we add up all the given angles, it should add up to 360

[upper karma]

The xs are blank numbers that have to be added in

[brave kayak]

yep

[upper karma]

But they must be the same and add to 360

[brave kayak]

if we add up all the angles (leave the x in)

it should be equal to 360

so you should have an equation that has x in it

[upper karma]

x is gay

who is x

[brave kayak]

x gon give it to ya

[storm rover]

I said 25%, it is asking “what percent of the figure is region a”

It’s clearly not 25 percent but we are given no other information

[twin prawn]

Well one triangle is 20% because you can subdivide the entire pentagon in 5 identical triangles

And A is half of what's left after you remove two triangles

So... 30%?

[brave kayak]

but are we sure that it's the center?

[main spire]

Hey what geometry rules do I need to know here to solve something like this?

[supple abyss]

this looks relevant

[main spire]

[marsh crest]

Is there an easier way to do this

I’m so sad that I couldn’t do this on the test

[idle bloom]

Whoa how did those csc turn into sin

[marsh crest]

1/sin = csc

So you can just flip them and change to sin??

[fossil lotus]

Your answer looks right but I’m confused about how you got there

Replacing csc with 1/sin gets you

$\frac{2(\frac{1}{\sin{x}})^2}{(\frac{1}{\sin{x}})^2-1}$

[somber coyote]

GrandPiano:

[fossil lotus]

@marsh crest

Your steps between 2csc^2/(csc^2-1) and 2/(1-sin^2) don’t look right

[marsh crest]

@fossil lotus why can’t u switch it

[fossil lotus]

Because csc isn’t sin, csc is 1/sin

[marsh crest]

YeAh so you can flip it over the fraction and turn it back to sin??

[fossil lotus]

Not in the denominator

If the denominator were just csc^2 then you could

[marsh crest]

How do you solve that problem then smh

[fossil lotus]

So you have

$\frac{2(\frac{1}{\sin{x}})^2}{(\frac{1}{\sin{x}})^2-1}$

[somber coyote]

GrandPiano:

[fossil lotus]

Then this becomes

$\frac{\frac{2}{\sin^2{x}}}{\frac{1}{\sin^2{x}}-1}$

[somber coyote]

GrandPiano:

[fossil lotus]

And then you can turn that into

[marsh crest]

What if I multiplied both top and bottom by sin^4

[fossil lotus]

$\frac{2}{sin^2{x}(\frac{1}{\sin^2{x}}-1)}$

[somber coyote]

GrandPiano:

[fossil lotus]

And solve from there

[marsh crest]

Eh nvm

[fossil lotus]

Just multiplying the top and bottom by sin^2 as I just showed works

[marsh crest]

@fossil lotus is this good now?

[fossil lotus]

Yeah looks good to me

Are you aware you switched the terms in the first step?

Not that it makes a difference

[marsh crest]

Wdym

The pheta sign?

[fossil lotus]

The first term should be multiplied by cscθ-1 and the second by cscθ+1

The numerator and denominator of each term, that is

[vagrant umbra]

Hello, how can I call this 3D shape? Is there a name for it ?

[marsh crest]

Oh yeah whoops

@vagrant umbra maybe a truncated square pyramid?

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

It looks like this

[vagrant umbra]

thanks

[low tusk]

actually, you can just call it "frustum"

[slate coral]

Could someone help me on this? https://gyazo.com/48060cabe0628034bc3828a6fd8ed89d

[dire rampart]

k=1 is a root

so u can factor out (k-1)

[upper karma]

Hi

The book says that 140-3x+x+3x forms a straight line, so you can get the value of x by equating the equation above to 180

[royal crest]

sum of angles = 180

[upper karma]

The thing is, why wouldnt the eq. be 40+140-3x+x+3x+50

Yeah, but...theres 2 whole angles left out?

[royal crest]

no, 50 is left out because it's not in forming the line

imagine yourself doing a rotation from A to E, first you go 140-3x then x then 3x

and that's a whole 180

[upper karma]

Oh, I see

So the 50 is forming EB

Not AE

and the 40 is forming AB

[royal crest]

i'm not sure what you mean by forming but have it your way

you just need to think of a line segment AB as rotating from A to B, and that's a whole 180 degrees

[upper karma]

I see, thanks

😗

[royal crest]

you're welcome :)!

[upper karma]

Wrong channel

[tired terrace]

oh shoot ur right

[upper karma]

Mfw Ive been stuck on a problem and my issue was with summing 3+2+1

[idle bloom]

trust me it keep happening in harder math

there's a common saying in calculus that goes "calculus is easy algebra is hard"

I've gotten end results wrong many times because I can't add or multiply or anything

[upper karma]

any ideas?

[mossy vine]

ok so

the hint in the question is that you have trig functions and something that looks like you can write it as a difference of two squares, because a lot of the basic trig identities involve squares

[upper karma]

How would i do that?

[mossy vine]

write as a single fraction

[upper karma]

i did

[mossy vine]

and you got?

[upper karma]

1 sec'

Just made a common denom

[mossy vine]

ok, good first step

[upper karma]

Then subtracted

ohhhh because then you would make it csc^2(x)-1

difference of 2 squares

[mossy vine]

yeah

[upper karma]

makes sense

[mossy vine]

and then you should hopefully have an identity for that

[upper karma]

Yep

[mossy vine]

to make it "simpler", which is a bit ambiguous in this case imo but it's probably what they're going for

[upper karma]

csc^2(x)-1=cot(x)

[mossy vine]

cot^2(x), yes

nice

[upper karma]

wait

But what would i do with the numerator?

It becomes (cscx-1)-(cscx+1)

all divided by cotx

[mossy vine]

stare at what you've just written

specifically, the numerator

[upper karma]

huh

[mossy vine]

(also make sure it's cot^2(x), that's what it should be)

[upper karma]

confused on the numerator

[mossy vine]

if you need, multiply out the -(cscx + 1)

it should hopefully become obvious what to do

[upper karma]

oh

isnt this just 0

[mossy vine]

no

not quite

[upper karma]

still confused

[mossy vine]

numerator should be ||-2|| (check after you've redone)

[upper karma]

been working on this for 3 hours

Just dead at this points

point

[mossy vine]

the cscx cancels out, yes

but you should get -2 rather than 0

as the numerator

[upper karma]

ok

makes sense now

ty

[mossy vine]

probably the best way to write it is ||-2 tan^2(x)||

[upper karma]

another question

How would I approach this?

The bottom one

I tried making secx times sinx=tanx

Didnt do much

made tanx= sinx/cosx and cot cosx/sinx

Made common denoms but didnt help either

[mossy vine]

secx sinx = tanx I think is a good idea - or at least htere is a solution using that

|| tanx/(tanx + cotx) = 1/(1+cotx/tanx) = 1/(1+cot^2(x)) = 1/csc^2x = sin^2x||

I think writing everything in terms of sin and cos should work though

[upper karma]

where did you get the second step?

The ||1/(1+cotx/tanx)||

[vagrant elk]

you could multiply both the top and the bottom by the same factor and remain equivalent

in this case by ||1/tan||

which distributes across the denominator to give 1 + ||cot/tan||

[eternal hare]

Can someone help me

[upper karma]

still confused

How do you multiply by ||1/tanx||

[mossy vine]

divide top and bottom by tanx

[upper karma]

i didd

did

so tanx/(1/tanx)=1

[vagrant elk]

which gives ye

and cot/tan = cot^2 since cot = 1/tan

[eternal hare]

Can someone hlep me

help*

[vagrant elk]

@eternal hare with which

[eternal hare]

Do the top one first I got the bottom one

Do (a) first

I don't know what the relationship is

[vagrant elk]

Well, what theorems/rules does the class give you to solve these problems with?

[eternal hare]

That's the thing

I forgot

[vagrant elk]

alright boyes, we got em
i guess i'll look them up since i ofc forgot them too

I believe this might be the one you need

@eternal hare

[eternal hare]

@vagrant elk Thanks

[mossy vine]

you want angle at the centre, angle at the circumference and then just some standard angle facts about straight lines and triangles @vagrant elk @eternal hare

although I guess it gives you the angles at the circumference :^)

so just angle at centre = 2 times angle at circumference

[vagrant elk]

i no circle good : )

[mossy vine]

:^)

[upper karma]

sorry for so many reposts, phone keeps lagging

How would I construct the bearing for this

What angle would I need to find?

anyone know what to do?

[fossil lotus]

65° + 25° = 90°

Since it’s a right triangle, you can use the pythagorean theorem

@upper karma

[upper karma]

@fossil lotus it’s looking for the bearing

[fossil lotus]

Oh I see

So the angle of the path directly from the airport to the final landing point?

Couldn’t you take the arctangent of 1/4 (from 120/480) and subtract 25° from that?

[upper karma]

,ask define bearing

[somber coyote]

Display results online and refine query

Requested by rudy#0446

[fleet lake]

I can't figure out how fast upward or downward the graph of line goes.

I know the 1) graph is positive but can't figure out if it's <=> than 1.

[dark sparrow]

lines with slope ±1 are at 45° angles to the axes

[fleet lake]

I dont know trig

[dark sparrow]

you don't need to know trig you just need to know visually what a 45° angle looks like

you're not supposed to calculate anything

alternatively, if you don't want to think of angles, you can look at the grid lines

[fleet lake]

I think the first graph is less than 1 because it moves rightward much faster then upward. Or $x_2-x_1$ is greater than $y_2-y_1$ this implies that the slope is less than one . Is my argument correct (

[dark sparrow]

if your line crosses more horizontal gridlines than vertical gridlines, then its slope is far away from 0 (i.e. greater than 1 or less than -1)

[somber coyote]

Krishna_The monkey boi:

[dark sparrow]

yes

[fleet lake]

I mean run is greater than rise hence proofed not grater than 1

[dark sparrow]

exactly

rise < run so rise/run < 1

[fleet lake]

I struggled with this yesterday but the concepts of graph are interesting .

I like it

[fleet lake]

Ann how about the graph C @dark sparrow

c)*

I k ow the slope is negative

[dark sparrow]

yes. and it's steep, too.

[fleet lake]

Steep means closer to vertical line

??

[dark sparrow]

yes

[fleet lake]

Ok so it's less than -1 right?

d) the slope is undefined .

[dark sparrow]

yes to both

[fleet lake]

e)the slope is negative.

But I can't figure out how steep it is

[dark sparrow]

it's exactly -1, actually.

[fleet lake]

How you figured that out

[dark sparrow]

the line goes diagonally through grid squares

[fleet lake]

So this implies that slope is-1

But why?

Ann?

[dark sparrow]

why what

[fleet lake]

How you figured out that the slope is -1

[dark sparrow]

well ok consider two points on it that are the opposite corners of a grid square
run = 1 and rise = -1, so slope = -1/1 = -1

[fleet lake]

Let's come back to it later.

f) the slope is positive

And less than 0 because it's not sttep

???

Isn't it @dark sparrow ?

[dark sparrow]

positive
less than 0

(f) is steep, actually. very much so

[fleet lake]

Ok so it's more than 1 as it's clswe the y axis

[fleet lake]

Just asking if I am right

So

Point Q is 8 units left and 3 units above the P

So the T must also be 8 units left and 3 units above the Q . If yes coordinates of T are (-11,9).

I checked my guess.

The slope formula and the distance formula show me a green flag so I am right

Also the midpoint formula showed me a green flag.

[fleet lake]

I am just clueless

[fleet lake]

<@&286206848099549185>

[fleet lake]

@dark sparrow can u help me this problem

[dark sparrow]

uhhhh one moment i'm a tad busy rn

[upper karma]

@fleet lake for part a use
|| midpoint formula or (x1+x2)/2, (y1+y2)\2||

Forget how to do part b

I think you have to ||put variables in the midpoint formula and solve for a given point||

Like ||(x1-3)/2,(y1+6)/2=5,3||

[fleet lake]

What ? @upper karma

8.34

[upper karma]

No for 8.33

@fleet lake

[fleet lake]

I want help with 8.34 @upper karma

[wet ice]

Um I need help on geometry

10-4 11,12,16 and 10-5 3,4,5,6

<@&286206848099549185>

[spiral dirge]

can someone help me with these two problems

[wet ice]

Ig everyone’s busy

[spiral dirge]

on 10-4 problem 3 would be 42 i'm pretty sure @wet ice

on ones like that the angle is half the arc

so for #4 u would do 360-70-120

[wet ice]

I need help on bottoms

Oh

[spiral dirge]

and then youd get 170 and divide by two

oh idk how to do those

[wet ice]

Wait what

On 4? Too one ?

Top *

[spiral dirge]

no in 10-4

[wet ice]

Oh ok ty

[fossil lotus]

There doesn’t seem to be

If you search “trigonometric identities” you’ll find some lists

[vagrant elk]

I don't know of any in general, unless A and B happen to be cos(b) and cos(a) respectively

[thin slate]

Lengths of the consecutive sides of quadrilateral ABCD are 12cm, 22cm, and 15cm. Circle O is inscribed in ABCD. Find the length of the fourth side.

[thin slate]

<@&286206848099549185>

[dark sparrow]

there's a certain property that a quadrilateral has if and only if it admits an inscribed circle

[thin slate]

in this case the quadrilateral is inscribed

what do i do

[dark sparrow]

no, it is not. read the problem.

Circle O is inscribed in ABCD.

[thin slate]

oh yeah, im so dumb lol

ok wait

ill try now

im still stuck

if anyones here

@dark sparrow can you help

[dark sparrow]

there's a certain property that a quadrilateral has if and only if it admits an inscribed circle

what is that property?

[thin slate]

hmm

[dark sparrow]

if you're having trouble recalling it, maybe try setting the problem aside for a moment and just draw a quadrilateral with an inscribed circle. see if that makes it click

[thin slate]

okay

[rotund swan]

I think diameter is equal to 35.5 , is it?

[thin slate]

hmm

i dont know

[fleet lake]

@dark sparrow Ann I figured out that problem

[rotund swan]

There is a theorem between circles’ chords

[fleet lake]

https://cdn.discordapp.com/attachments/326138757474680852/581067526675365898/2019_05_23_16.04.19_edit.jpg

[dark sparrow]

@rotund swan the diameter of the incircle is of no importance here

[fleet lake]

Yes nima

[dark sparrow]

nor can it be obtained from the information that is given

[fleet lake]

?

To me?

[dark sparrow]

no that's to nima

[fleet lake]

Oh

[thin slate]

oh I think, Ann thought you were talking about my problem

[dark sparrow]

yes

sorry

[rotund swan]

Nope 😂

@thin slate
Value of x??

[thin slate]

oh i remembered the property

22+x=12+15

[rotund swan]

Yes that’s it

[thin slate]

I have another problem tho

:dd

Quadrilateral ABCD is inscribed in circle O. <C=160. Find degree measure of acute angle included between the bisectors of angles B and D.

[rotund swan]

Can you write it in peace of paper??

[thin slate]

wdym? do you want me to draw it? or

[rotund swan]

Yes draw it please 👍🏻

[thin slate]

👌👌

its a rly bad drawing but maybe it will help

[rotund swan]

Good

Value of à?

[thin slate]

yes

if you figure it out ping me

or dm me

[broken wyvern]

This looks circle theorem related help

[rotund swan]

@thin slate ok 👌🏻

[thin slate]

thank you

[jade lotus]

@thin slate I cant understand your drawing. The quadrilateral is not inscribed in the circle.

[rotund swan]

Why not??🤔

[jade lotus]

Because the circle is inside the quadrilateral not the other way around

[rotund swan]

But you can draw it in paper, therefore it’s possible...

[jade lotus]

This drawing doesnt help, in my opinion. When a quadrilateral is inside a circle (inscribed in the circle), the opposite angles have an interesting property

[rotund swan]

🤷🏻‍♂️🤔

[thin slate]

ye ye its completely wrong, i figured it out

I was mistaken because, when I sent the first problem I thought it was this one and when Ann corrected me that the circle was inscribed in the quadrilateral I thought it was about this problem

[rotund swan]

So you mean that this is totally wrong?

[thin slate]

only the drawing

:dd

[rotund swan]

But why?

[thin slate]

the quadrilateral was inscribed in the circle

just like that guy said

do you get it?

[rotund swan]

Vice-versa ???

[thin slate]

wait ill send a picture

[rotund swan]

No

I got it

[thin slate]

ok

👍

[rotund swan]

✋🏻👍🏻

[thin slate]

thanks for your help btw

[rotund swan]

😉

[thorn kraken]

anyone able to help me with some trigGraphing and equations, im quite stuck

[cinder portal]

oh

just post your question

[thorn kraken]

its a multiparter, should i post my attempts or just straight q

i feel like im super overthinking and making it hard for myself but i really dk

[cinder portal]

did u do part a

[thorn kraken]

i plotted x1, found the amplitude, period and phase shift for X2

but then i wasnt sure what to do with the phase shift

its 0.5 but my x axis is ranging from 1/100 pi to 1/25pi

[cinder portal]

X2 is really just X1 shifted to the left by 0.5, with different amplitude

[thorn kraken]

or 0 to 1/25pi

"X2 is really just X1 shifted to the left by 0.5, with different amplitude" ik that but i dont know what im doing to convert that so its in terms of pi

if that makes sense

[trail minnow]

what do you mean you don't know how to convert it in terms of pi?

like how to write 0.5 in terms of pi?

[thorn kraken]

give me a second to clarify my thoughts, im being a bit dumb rn

[trail minnow]

hmm

try just graphing out $X_1 and X_2$

[somber coyote]

hegel:

[trail minnow]

for me when i was learning sin waves it was helpful to think of breaking them down into a bunch of composite parts or "modifications" to the original sin wave

and then moving from there

original sin wave meaning sin(x)

[thorn kraken]

okay so im moving it by pi/2 right, but that seems like a super high amount in comparison to the scale on my x axis

does that make sense

[trail minnow]

hmm

what is the scale on your x axis?

[thorn kraken]

my period is 1/25pi

[trail minnow]

ah

[thorn kraken]

and im plotting for 1 period

[trail minnow]

oh i see

so are you confused with how to graph X_2?

[thorn kraken]

no, im just not sure if im right because of the scale difference

[trail minnow]

hmm

[thorn kraken]

which is why i want someone else to confirm im doing the right thing

[trail minnow]

wait, are your functions supposed to be 300 sin(50 pi x)?

[thorn kraken]

300Sin(50 pi t)

[trail minnow]

ah

um yeah this is an incredibly odd thing to ask a student to graph

[thorn kraken]

right?!

[trail minnow]

because that's... a really thin really tall wave

[thorn kraken]

i thought i was brain damaged

[trail minnow]

you aren't doing anything wrong mathematically

it's just fucking weird

i just checked on desmos its literally like that

[thorn kraken]

especially considering this is way beyond what we normally do

[trail minnow]

is there a typo or something lol?

[thorn kraken]

i have no idea

[trail minnow]

for reference at a decent large scale

the graph of $300\sin(50\pi x)$

[somber coyote]

hegel:

[trail minnow]

is htis

my scale here is on the level of ones as well

[dark sparrow]

uh

[trail minnow]

goes from about -4 to 8 horizontally

[cinder portal]

what in tarnation

[dark sparrow]

yikes

[thorn kraken]

yeah ik, its like the crackhead of graphs

[trail minnow]

lmfao

[cinder portal]

what am i looking at HAHAHAHAHA

[trail minnow]

like

[cinder portal]

b r U H

[trail minnow]

the period is so tiny that it literally approaches a flat out just

[thorn kraken]

@cinder portal thats the graph i have to plot

[trail minnow]

square????

wtf is this??

[thorn kraken]

and rn we are trying to figure out if im brain damaged or my teachers are

[trail minnow]

you're fine lmao

this is just moronic

it makes sense that you are getting what you are

300 is an incredibly high amplitude

so your waves are going to be extremely tall

[cinder portal]

the important parts is where it intercepts or touches the x axis

[trail minnow]

50pi x is an extremely small period, so your wave is going to be really thin

[thorn kraken]

well the function was 300Sin(50 Pi t)

[cinder portal]

which points

does the function

interesect the x axis or touch the x axis

[thorn kraken]

and then i have to plot 200Sin(50 Pi t +0.5)

@cinder portal 0,0

[cinder portal]

here

give me a second

[thorn kraken]

x

im on drugs

hod on

[trail minnow]

lmfao

holy shit

[thorn kraken]

0 1/50 Pi 1/25 Pi

im plotting for one period

[trail minnow]

hahahahaha

[cinder portal]

hope this helps

omg i tried plotting it

[cinder portal]

you were right

[trail minnow]

look how thin it is

[cinder portal]

it was like a freakin brick

HAHAHAHAHA

[trail minnow]

THIS IS SO USELESS LMFAOOO

fslfjsdf

[thorn kraken]

i thought i broke my monitor when i put it into desmos

[trail minnow]

this is literally ridiculous lmao

[thorn kraken]

this in only have of pat a of the question boys and girls

[trail minnow]

why is your teacher giving you this

unironically i bet it was a typo or smth

[thorn kraken]

its a homework assignment

idk why

[trail minnow]

a true galaxy brain

[thorn kraken]

im just gonna double check everything i do step by step

my next job

[trail minnow]

lmao day in the life of helpers on the mathematics server 😩

[thorn kraken]

is to move X2 to the left by 0.5

which in my opinion is by a quarter period

so in this case

1/100 pi

[trail minnow]

are you in 10th lmao???

are they giving this to 15 year olds to spite them?

or 16 idk how the fuck age works

[thorn kraken]

im 18

[trail minnow]

still this is so dumb

what a pointless problem >.>

@clear haven check out this fucking problem lmao

this is what happens when you plot it on desmos

[thorn kraken]

the full question if anyone cares

[dusky pike]

You sure thats not a fill function on ms paint @trail minnow

[trail minnow]

unfortunately lmao

[thorn kraken]

you guys dont know how relieved i am to know im not just stupid

[trail minnow]

this problem is too galaxy brain for us to comprehend lmao

[thorn kraken]

i drew the graph

huzzag

huzzah

@trail minnow do you know how to go about part iii.

[upper karma]

grrr struggling c:

I know the acute of Z + X = Y

Wait

Z = 180-97?

Is Y = 39? lol

[fossil lotus]

I got 40° for Y

180 - ((180-97) + (180-123))

[upper karma]

180 - 57 - 83

,calc 180-57-83

[somber coyote]

Result:

40

[upper karma]

Oh, yes, my bad.

Mental maths be bad

wow k

I was going a way...worse route

I was doing system of equations

lol

[fossil lotus]

Oh

Yeah it’s simpler than that lol

[upper karma]

123 = Y + Z ; 97 = X + Y

but its so much simpler smh lol

[eager pendant]

assuming Q is the centroid works

[tepid rapids]

https://imgur.com/a/MWpv81Q

hey guys how do I show that angle RPS = theta?

actually i think i got it

oo thanks dude

the say I did it was QPR + SPR = 90 deg, and theta is the complement angle of QPR

*the way

is that incorrect?

[dark sparrow]

wrong channel @upper karma

[tepid rapids]

@timid imp thanks man

but I need help with another question

same triangle https://imgur.com/a/MWpv81Q

how to prove ab = h^2

actually i think this ties back to the other question

tan theta is h/a and also b/h

h/a = b/h

and cross multiply. huh

THANKS AGAIN! 😛

[keen glen]

hey

[keen glen]

pl3ase

how is this done

[proper citrus]

Are the faces of a pyramid, with a rectangular base, all isoceles triangle?

Other than the base itself, of course, that's not a triangle.

quick question

Anyone?

<@&286206848099549185> ?

okay fuck this

[fringe dirge]

No they don't have to be

[heady juniper]

feels bad man discords dinged me like 50 times in the past 10 mins

but still no reply SadPepe

<@&286206848099549185>

[rotund swan]

Does AB cross through the center??🤔

[heady juniper]

not necessarily

[fringe dirge]

Think power of a point

[iron saddle]

https://gyazo.com/75cc0e33685910d6ed2c7e25abc2de25

anyone able to help ^^

[rotund swan]

@iron saddle

And you’ll get 34 for AB

[heavy crow]

why is arccos(10/7^(3/2))/3=arccos(5/(2sqrt7)) and how can i prove that

[dark sparrow]

it might be easier instead to prove that $\arccos\left( \frac{10}{7^{3/2}} \right) = 3 \arccos\left( \frac{5}{2\sqrt{7}} \right)$

[somber coyote]

Ann:

[dark sparrow]

and that can be done by letting $\theta := \arccos\left( \frac{5}{2\sqrt{7}} \right)$ and then showing $\cos(3\theta) = \frac{10}{7^{3/2}}$

[somber coyote]

Ann:

[river vale]

Could anyone explain to me how this is a line? I'm very confused atm

I need to show that the lines a and b form a plane, and then find the equation of that plane

Which appears to be this:

But seriously, how are a and b even lines to begin with?

Also, at the very start of the solution they start by saying this:

How could one come to that conclusion? I'm beyond confused

[heady acorn]

what's r?

This is a kinda problemthat is solved using linear algebra tools, ont sure if you studied them

they are lines, since a is described by 2 equations with 3 variables, meaning dimension of a i 3-2 = 1 which means its a line

its rather linear algebra channel question overall, you would get more response if you posted it there

[river vale]

Well, I'm only in high school (last year), I don't think I've dealt with linear algebra yet

Why is a not a plane?

[dark sparrow]

it's the intersection of two planes pretty much

[river vale]

hmm I see

now, what I did to make a parametric equation of a, was take 2 random points that work in a

I took (0,0,4) and (0,-8,0), but then the vector would be (0,-2,1)

meaning the x value never changes

but the point (-16,0,0) seems to be perfectly fine

so I'm doing so much wrong

also Wolfram is saying it's a plane, did I type it wrong?

[heady acorn]

you wrote it wrong

these equations are not equal

its first equation AND second one

[river vale]

but both are equal to zero?

[heady acorn]

yeah but for example the first equation is 0 for vector (1,0,0) and the second equations is zero for (1,0,0) and (2,0,0). If you set these equations equal you won't get the (2,0,0) solution (becuas it doesnt satisfy first equation)

basically 'a' is intersection of two planes, so think about it, if you have two planes that are at different angles, their intersection will be a line

[dark sparrow]

@river vale but you didn't put =0 there lol

just because two things are equal to each other doesn't mean they're both equal to zero

[river vale]

ah true

so how could I get the equation of this line?

instead of having to use those two planes

[fleet lake]

Midpoint of segment BC is (-1,-2).

The slope of line passing through A and B is 9+2/5+1=11/6

Sorry I just realized that I made an arthimetical mistake

Ignore my question

[elfin ibex]

I've been given a task to find out the maximum volume of a prism given a perimeter of 360cm and have no clue where to start. I've spent several hours on it so far but I'm not getting anywhere. Can someone please provide some pointers?

[elfin ibex]

Our class is covering algebra II

It seems like an optimization problem but our class hasn't covered calculus so I'm stumped.

[elfin ibex]

Yes

It's the perimeter

The task seems really stupid.

[hot badger]

@elfin ibex what type of prism

if its a rectangular prism

actually leme guide through the solution

basically

4x + 4y + 4z = 360

x+y+z = 90

maximize xyz

now let's just let z = z for now

x+y=90-z

and we need to maximize xyz, but we have an arbitrary value already established for z

so x + y = a, where a is the corresponding arbitrary value, 90-z

and we need to maximize

xy

basically let's say x+y = 40, which is not the actual optimal value, but anyways

just think about that simplified question

[elfin ibex]

We got told it could be any priam

*prism

[hot badger]

any prism

hmm

[elfin ibex]

I'm assuming its a triangular prism

[hot badger]

but only straight edges?

[elfin ibex]

Yes

[hot badger]

ok

leme think for a minute

[elfin ibex]

Thanks

[hot badger]

obviously the two bases have to be regular

so let's say the perimeter of one base is x

and the number of bases is n

side length is s

sn=x

[elfin ibex]

(360-2sn)/n for height?

[hot badger]

we'll get there later

[elfin ibex]

Ok

[hot badger]

does the teacher want a rigorous proof

or just the answer

[elfin ibex]

They said to provide "justification". Our class hasn't covered any proofs yet so I'm pretty sure it just needs to be supported.

[hot badger]

oh ok

one sec

so

if x is the area of a base

then shouldn't the volume be increased by adding a side, albeit slightly shrinking each side length

[elfin ibex]

The more sides added, the more edges though

[hot badger]

yeah ik

but

the area of the circle is larger than the square

which is in turn larger than the area of the triangle

[elfin ibex]

Ah, that helps!

[hot badger]

but also a cylinder isn't a prism

did it ask for integer side lengths?

ohhh wait

nvm

this wouldn't work ic

yeah i think u were right about approaching the whole solid

ok so if n is number of sides per base, a equals base side length, and b is height

letting a = a and b = b, two arbitrary value

values

the volume is [a/(2tan(180/n)) * a/2 ] * n * b

brackets are each triangle in the base

so we need to maximize n/tan(180/n)

[elfin ibex]

Thank you

[hot badger]

if the teacher lets you graph it, just use that

to find n

[elfin ibex]

Yeah, that's gonna be the easiest method. Thank you for your help?

Fuck did not mean the ?

I meant !

[hot badger]

yw

[elfin ibex]

!rep hogman

[hot badger]

oh there's a rep system?

[elfin ibex]

Yeah, I think I did it wrong though

t!rep hogman

Maybe they removed it

[hot badger]

prolly

i haven't seen it recently

[elfin ibex]

Without you I probably would have failed

Thanks

[whole swift]

Let u=1√3(1,−1,1) and v=1√6(1,2,1)

Find the two points of intersection on S2 of the spherical line L_u and spherical circle C(v,π3).

Any idea how to start this?

[dry glen]

Can anyone solve this:

In quadilateral ABCD the measure of angle a = 60 the measure of angle c = 120, AB = 41m, DC = 22m. The length of AD is 2m longer than BC, find the perimeter

I get negative values for my answers and it's driving me crazy

[fleet lake]

I figured out the a) part . I need help with b)

[spark stag]

Consider the line y=x

[fleet lake]

Mmm?

y=x?

0=0

[fleet lake]

<@&286206848099549185>

[halcyon coyote]

x intercept is where y = 0

y intercept is where x = 0

can this happen in the same point?

[fleet lake]

No I figured that out.

[zealous trail]

Looking for some help with my trig homework if anyone is around

here is the first question, if you feel like helping a dood out!

[devout shell]

do you remember the conversion from rectangular to Cartesian?

[zealous trail]

To find the polar coordinates you do the distance formula I think

for the x value

[devout shell]

the x coordinate is already given

and the y one as well

[zealous trail]

so polar coordinates are (r, Θ)

[devout shell]

oh yea

you can find what Θ is though

you know that tan(Θ)=y/x

[zealous trail]

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

[devout shell]

that's a fair way to do it

[zealous trail]

yeah tan(Θ)= y/x, but to find the angle we need to find Θ = arctan(y/x)

[devout shell]

correct

[zealous trail]

or Θ=arctan(2/-5)

[devout shell]

yes

[zealous trail]

so then what's the answer

[devout shell]

notice that x coordinate is negative

what quadrant are you in then?

[zealous trail]

II

[devout shell]

correct, we are in quadrant II

it says you can use approximations as long as you denote properly so no idea what they want

[zealous trail]

yeah that's what stumps me, like should I put it in a calculator

[devout shell]

they want numbers or the numerical value of arctan(2/-5)?

[zealous trail]

it comes back as approx 338°

[devout shell]

lol

that's a bit much

it should be between 90 and 180

,calc 338-270

[somber coyote]

Result:

68

[devout shell]

oops, overshoot

[zealous trail]

wait nevermind I did that math wrong

[devout shell]

oh lol

[zealous trail]

arctan(2/-5) = -21.8°

ty for helping by the way, I'm so lost in the sauce rn

[fleet lake]

Is the answer 1?

[dark sparrow]

yes

[fleet lake]

Ok then I am right

[vivid yew]

YIKES

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[fleet lake]

@dark sparrow ?

[dark sparrow]

what do you expect from me rn

[fleet lake]

I think a line always hit with x or y intercept for even both .

There no such linear equation of form $Ax+By+c=0$ such that it does not intersect with x or y intercept.

[somber coyote]

Krishna_who likes geometry:

[tacit turret]

can someone help me on how to find a vector perpendicular to a= [1,2,4] and b= [0,3,-2]

[dire rampart]

cross product easiest

[tacit turret]

i did the cross product and it's -16, 2, 3 which are all supposedly the numerators of the answer but i don't get how to get the denominator sqrt269

[devout shell]

They wanted unit vector then

Take all the components, square them, add them together, then take its square root

Then divide all components of the vector by that number to get a unit vector

[storm rover]

What’s a sinh

[devout shell]

hyperbolic sine

$\frac{e^{x}-e^{-x}}{2}$

[somber coyote]

⚡Amphy⚡:

[storm rover]

Lol

[devout shell]

that's what it is though

[storm rover]

I know

It’s just weird

What about csch

[devout shell]

that's just 1/sinh

[storm rover]

What about sec

[devout shell]

1/cos(x)

you meant sech?

[storm rover]

No just sec

[devout shell]

secant is 1 over cosine

[storm rover]

Okay

Now I think I understand

[tacit turret]

how do i solve the tension of the shorter rope using cartesian components

[devout shell]

you use vector algebra

is the tension in one of the ropes given?

[tacit turret]

no just the weight

[devout shell]

and no distances either, so seems like we have to assume tension in ropes is the same? I'm trying to solve this quickly as well lol

[tacit turret]

the question asked for the shorter rope tension so i assume that they're different

[devout shell]

using equations of equilibrium maybe we can get the force

ok, so we can definitely do this then

[tacit turret]

i know the force acting down is 98N

[devout shell]

we need to establish a coordinate system

[tacit turret]

and that the vector that connects to the 40 degree angle is the one i'm looking for

[devout shell]

up is +y and right is +x, agree?

let's make that the convention

[tacit turret]

yes i assume that where everything intersects is 0,0

and that the force acting down is -98N

[devout shell]

to let's examine the force in the x direction

the tension in the left rope call that T1, tension in right rope, call that T2

so T1 and T2 are vectors

[tacit turret]

yeah

[devout shell]

we can decompose the vectors

using the angles

gather the x components and solve for T1 in terms of T2

then write the equation for all forces in the y direction, the total force is 0 because it's not moving

same thing for the x components, they sum to 0 since it is not moving

after you solve for T1 in terms of T2 then you can solve for T2 in the sum of forces in y equation

[keen aspen]

Find x

Neat little problem

[cinder portal]

don

t you find the diagonal

and ez klap

gimme a sec

[devout shell]

not as easy as you would suspect

[dire rampart]

x is right there lol

found it

[devout shell]

gottem

[cinder portal]

o this question is genius

holy shit

[devout shell]

lol

as good as stokes question

[cinder portal]

cant argue with that

that stokes one was genius too

[devout shell]

you'll never stop being in awe of it huh?

[restive fern]

i joined this server for help because i have the eoc in two days and flvs is skipping over some stuff it should have covered. I am in the middle of a practice exam and I am not sure what to do with this problem. I understand that BC is 5 due to the 30 60 90 triangle thing and that DB is 10. Aside from that I have no idea what to do

[cinder portal]

my brain is forever traumatized from that

[cinder portal]

horizontal length of helicopter to library subtract horizontal length of helicopter to zoo

[restive fern]

that was the wrong picture

[cinder portal]

o nvm

[restive fern]

sorry

[cinder portal]

find BC and find AC

AC - BC = AB

do you know how to find AC and BC?

if not I can show you how

[restive fern]

I found BC by the 30 60 90 right triangle thing

[cinder portal]

what is BC

[restive fern]

Im not sure how to find AC

I think its 5

[cinder portal]

correct

ok to find AC

what iwant you to do

is ignore BD, and ignore the 30 degree angle

redraw the picture if you have to

what you drew, you should be familiar with

[restive fern]

Im not sure what you mean

[cinder portal]

redraw the picture

but this time

dont draw BD

dont draw the 30 degree angle

[restive fern]

so that angle D is 75?

[cinder portal]

look carefully

angle D isnt 75

its 45

don't worry, i thought it was 75 at first too

[restive fern]

Uhm

ok

OH

I see

thank you sm

[cinder portal]

you got it?

[restive fern]

Yes

[cinder portal]

cool

[restive fern]

thanks man

[cinder portal]

ye anytime

[fleet lake]

I am just clueless

Ann can u help me?

[dark sparrow]

make a picture

i'm too busy being pissed off at god knows what to provide any more assistance

[sharp fern]

are u doing ok?

[dark sparrow]

obviously not

[sharp fern]

did something happen/do u want to talk about it?

@fleet lake draw the two squares on a graph and then find the steepest line and get find the slope of that

[cinder portal]

$slope = \frac{\deltay}{\deltax}$

[somber coyote]

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

[cinder portal]

whats latex delta

[dark sparrow]

v]b23hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh+

[cinder portal]

If you want to maximize slope, you want to find a HUGE y, and a VERY SMALL x

what the

[sharp fern]

uh

[cinder portal]

Ann broke

any1 know how to fix an honorable

[sharp fern]

what happened ?

[fleet lake]

$slope = \frac{\Delta y}{\Delta x}$

[somber coyote]

π is Krishna's irrational friend:

[fleet lake]

@cinder portal I can't understand the problem can you draw me a picture or something

[cinder portal]

give me a second

here you have two squares

now

choose any random point on any of these two squares

choose a combination such that

the change in y is the greatest, while the change in x is the smallest

we can take this slowly if you want

lets say we choose two points, which two points, would give you the smallest x value?

[fleet lake]

I will think about it

Later

[cinder portal]

i gave you a drawing already

you can see it

just think about the x distance

[sharp fern]

[fleet lake]

Yes

I will but later after 40 mins

[cinder portal]

.-.

[sharp fern]

are u gonna watch a show

[cinder portal]

r00d

[fleet lake]

Learning inequalities

On kc

Ka*

[west moat]

I got arctan[2/sqrt(5)]

I drew the base first then found the length from the centre of the pyramid to the corner

and then I drew the triangle from the slant edge to the perpendicular height

Anyone know what I did wrong?

[twin prawn]

Can you state those lengths you've found in the process?

[west moat]

so for the base and perpendicular height I made the length of it 2x

and from the corner to the centre the length i got was x*sqrt(5)

so my final triangle had base x*sqrt(5) and perpendicular height 2x

@twin prawn

[twin prawn]

I don't think you calculated the corner-center length correctly

Remember that, with your choice, the perpendicular distance from a base edge to the base center is x

[west moat]

oh yeah

i see what i did wrong 🤦

[twin prawn]

As well as the distance from the corner to the midpoint of a base edge

So Pythagoras tells you it should be x*sqrt(2)

[west moat]

But it still isn't right?

i got arctan[sqrt(2)]

[twin prawn]

You're right, it's not; that's because the corner-center length is completely irrelevant lol

[west moat]

Wait what

[twin prawn]

You wanna have the apex-base midpoint and apex-corner distances

[west moat]

Mid point?

[twin prawn]

You have now calculated the angle between the slanted edge and the imaginary line connecting the base center and the corner

Ah I meant midpoint of a base edge

[west moat]

isnt that what im supposed to find?

[west moat]

oh, fuck that question.

thanks mate

[tepid rapids]

i have done a) but cant do b)

[west moat]

aha i had the question last time

good luck with 20 @tepid rapids

[tepid rapids]

@west moat ay u in Year 11?

[west moat]

Yes sir

[tepid rapids]

sydney?

[west moat]

Yes sir

[tepid rapids]

what a fuckin beast

[west moat]

ayy

4 unit next year?

[tepid rapids]

i'm doubting myself frfr

but maybe

never give up hope 🙏

[west moat]

same lol

[tepid rapids]

bruh i legit can't do 20) either. forget 'Enrichment' xo

[west moat]

i have 20 ill show u

[tepid rapids]

yes please

[west moat]

i had to ask same question last time

ahha 19b and 20

basically u draw a line from C to the other line

to make a rectangle

and then u solve the triangle CHB that u just made

and since u formed a rectangle CH = PQ

Im assuming u know how to find AP

[tepid rapids]

ye ye i got it

i had no idea you're supposed to construct a line like that

[west moat]

Neither did I

[tepid rapids]

how do i figure this out by myself lol

[west moat]

Ikr Ann told me hwo to do

I would never think of drawing a line like that

[tepid rapids]

@west moat ok what about 19b

[west moat]

Yeah thats afucking joke

didnt do it lol

I legit copied the answers for 19a

hahaha

[tepid rapids]

nicely done

[remote radish]

Can somebody help me with this I don't understand it

[honest bay]

what do you notice when you look at these two triangles

@remote radish

[remote radish]

They're parallel?

@honest bay

[honest bay]

parallel triangles?

DE is Parallel to CB

we see angle A is a common angle between the two triangles

any idea?

[remote radish]

Ac is 20