geometry-and-trigonometry

[silent plank]

that above comment was intended as a joke,

anyway attempt the substitution again

[strong pilot]

ok this may take a minute if thats alright

[silent plank]

try not to overthink it

[dark sparrow]

intended as a joke,
and explicitly marked as such, mind you

[strong pilot]

8(10) + 10(10) = 180
because 8(10) =80 and 10(10) = 100

im too dumb to understand the joke x3

[silent plank]

i mean those are true statements but it wasn't what i asked for

what i was asking for only required to to replace:
m<A with 8m and
m<K with 10m
resulting in the equation 10m + 8m = 180

the stuff above doesn't tell me how you got the (10)s

and/or whether you achieved it via guessing or whether you actually got it through using what i requested but didn't tell me

[strong pilot]

can we combine the like terms (10m +8m) in the equation and then divide?
i tried to think what i could multiply each number by to achieve 180--i figured that i could multiply the 8 by 10 to get 80 and then the 10 by 10 to get 100

[silent plank]

i figured that i could multiply the 8 by 10 to get 80 and then the 10 by 10 to get 100
that would be considered guess/check mentality

can we combine the like terms (10m +8m) in the equation and then divide

yes, they are like terms and you can combine them

and you can perform some division if you do if properly

[strong pilot]

i got 18m = 180
18m/18 = 180/18
m=10

[silent plank]

ok thats great

determining the requested angles should be relatively simple now

[strong pilot]

thank you so much ^^! i appreciate it c:

ill get on to it rn

[wild eagle]

can someone help me with these questions? ^^ my ex boyfriend was supposed too but he never did ( ≧Д≦)

[vale nimbus]

@wild eagle is the whole angle A 32.4°?

[wild eagle]

no only that part in the middle

[vale nimbus]

or just the part in that triangle with the 106.6° corner

[wild eagle]

the part with 32 is a right angle

overall

[vale nimbus]

oh is that given?

[wild eagle]

yes its square

[vale nimbus]

k

aight so since you know 2 angles in BAD

you can calculate the third, what's the third angle? (B)

[wild eagle]

57.6

[vale nimbus]

no

in triangle BAD

[wild eagle]

can you get da awnser first then explain how you got it

thats how my exbf used to do it

[vale nimbus]

bruh

to calculate angle B u gotta do 180° - A - D

A is 90° and D is 41°

so u get that B is 49°

now this is a question for you again

you have all angles and 1 side

and you're searching a side

what formula/law are you gonna use here?

[wild eagle]

trigonometry

[vale nimbus]

yes that too

what ratio?

[wild eagle]

uhhh isn't that llike 1:2 stuff

[vale nimbus]

wait wdym by trigonometry?

[wild eagle]

thats like sin cos tan

right

[vale nimbus]

yeah

which one of those are you gonna use and in which angle

just realised u dont actually have to know angle B when doing it this way xd

[wild eagle]

im confused

[raw tiger]

Can you resend the question with all the information you have done filled it? (Like all the angles and sides you have found)

[wild eagle]

i did it for like a min then fell asleep

[vale nimbus]

you know that sin(A) = opposite/hypotheneus

cos(A) = adjacent/hypothenuse

tan(A) = opposite/ajdacent

[wild eagle]

yes

[vale nimbus]

yes so look at the figure and look at what you're looking for

@wild eagle if i just tell you will you atleast try remembering how to do it for in the future?

[wild eagle]

yes

can we do 2 aswell

[vale nimbus]

if you take the sin(D) = AB/AD

sure thing

try drwing 2 out

think it'll be obvious then

well yeah for 1a

take sin(D) and form it around to give AB = 20sin(D)

which is 13.12118058

[wild eagle]

is that the awnser?

[vale nimbus]

yeah for 1a

also have 2a but dont know how to type it in coherent english holdon

[wild eagle]

also can you draw me out how you did the quesiton

on like paint

so i can stick it into my notes

just draw it quickly in like the simplest way

[vale nimbus]

work in angle ADC
angle A = 32.4°
angle C = 90°
angle D = 57.6° (180°-A-C)
use law of sines to get 20cm/sin(90°) = CD/sin(32.4°)
then form it around and u get CD : 20sin(32.4°) = 10.70165359

draw question 2?

[wild eagle]

yes please

is this 1 b or question 2

[vale nimbus]

thats 1b

[wild eagle]

alright i can write that downb

so do another one of those for question 1 c

[vale nimbus]

ur looking for angle theta

[wild eagle]

yes

[vale nimbus]

question 1c is just pythagorus

idk how to spell his name xdd

[wild eagle]

i think these are the awnsers

tan(x)=81/63
tan(14) = x/500 => x = 500tan(14)
32 + 32 + x = 180 => x = 116

for 2,3 and 4

can you double check?

[vale nimbus]

can you do 1c aswell

i'll check 2 now but not gonna do the others cuz i have to go soon

[wild eagle]

okay

[vale nimbus]

@wild eagle whats your answer for 2?

[elfin ingot]

Can someone explain this question really quick? I don't know how to approach it.

[wild eagle]

81 / 63 so that would be

1.29

[vale nimbus]

@elfin ingot let me draw it for you real quick holdon

@wild eagle you're looking for an angle

[elfin ingot]

fuck nevermind

[vale nimbus]

u figured it out?

[wild eagle]

um

so is that formula not right?

[vale nimbus]

are you allowed to use calculators?

[wild eagle]

yes

[vale nimbus]

take the tan inverse of 81/63

and put that into an angle

and thats your answer :p

[wild eagle]

whats an inverse

[vale nimbus]

tan^-1

[wild eagle]

my calculator sucks can you do it for me

its not on here

only just tan

[elfin ingot]

ok here is a real question

Would angles OLP and MLP be equal?

it is a rhombu

rhombus

[vale nimbus]

@wild eagle 52°7'30.0569''

[elfin ingot]

press 2nd

then press tan

[vale nimbus]

yes they're equal

[elfin ingot]

aii ok

[vale nimbus]

do u know why?

[elfin ingot]

cpctc

[vale nimbus]

what

not a native english speaker so idk what that means

[elfin ingot]

basically, you can prove that triangles LMN and NOL are congruent

Corresponding Parts of Congruent Triangles are Congruent = CPCTC

[vale nimbus]

yup

👍

[elfin ingot]

wish me luck 😦

[vale nimbus]

you got this bro

@wild eagle do u still need help?

[wild eagle]

yeah can you help me find the awnser for 3 aswell

like is my formula correct?

[vale nimbus]

is the angle i marked the angle of depression?

[acoustic jungle]

I don't understand why teachers add so many decimals to questions

[vale nimbus]

yeah lmfao

[acoustic jungle]

literally waste of time

[vale nimbus]

@wild eagle i got 2005.390467 for the height of the cliff ._.

you got it wrong, its 500/x and not x/500

so x ends up being equal to 500/tan(14°) = 2205.390467

[nocturne musk]

Hi I need some help

[raw tiger]

post your question

[raw tiger]

Use your sin, cos, and tan identities

[nocturne musk]

Im sorry what do you mean exactly

[raw tiger]

sin(x)=opposite/hypotenuse, cos(x)=adjacent/hypotenuse, and tan(x)=opposite/adjacent

[nocturne musk]

So I have to do each one by their corresponding identities?

[silent plank]

no. for this question you should apply the cosine rule

[raw tiger]

They will all give you the same answer for each angle since they are ratio's

[silent plank]

the ratios for sin,cos,tan can't be applied directly here since you don't have a right triangle

[raw tiger]

ahh yeah you are right, i am pretty rusty on geometry lol. Try this formula: a^2 + b^2 − 2ab cos(C) = c^2

[nocturne musk]

Thank you

[late current]

You could also take the highway and just draw a line in the middle and end up with two right angled triangles

[silent plank]

sure, that gets you right triangles, but doesn't really help you solve the problem

[nocturne musk]

Wait

How do I know which one is A, B,

[upper karma]

is this the answer

[arctic apex]

Can somebody help with number 4

[upper karma]

no u fucking tard help me first

[dark sparrow]

k first off don't be a rude fuck

[arctic apex]

I dont fucking know Im in geometry and 8th grade

[dark sparrow]

second @arctic apex you might wanna move to a questions channel just to not clog this up

now with that out of the way: @upper karma i'm afraid all of your answers are wrong

55° is not coterminal with 125°, and yet you ticked it.
485°, 845° and -595° are coterminal with 125°, and yet you didn't tick any of those.

sin is positive in quadrant II, and yet you didn't tick it.

csc(A) < 0 in quadrants III and IV, and among those it is quadrant IV in which cos(A) > 0, so you should have ticked IV and not I.

i'd appreciate an acknowledgment if you read my reply

[arctic apex]

Wait but this says geo/trig so where do i put it?

[upper karma]

@dark sparrow ty so much

[dark sparrow]

the questions channels (#help-1 through #help-0) are not bound to any specific topic. you can just put your thing in any of those that isn't occupied atm

[arctic apex]

ohhh

ok

[dark sparrow]

i mean

otoh

mayoo and i seem to be done here

[upper karma]

wait

[dark sparrow]

or not

sure go ahead

[upper karma]

https://gyazo.com/fd31b77845cb48b4e90782684bbe75ed

[dark sparrow]

oh god that second question ugh

what even is "QSFR" meant to be lol i've never seen this abbreviation before

anyway! you got it right on the first and third questions

and it looks like whatever QSFR is it's the only thing that can be the correct answer: SOHCAHTOA is the definitional mnemonic for right triangle trig, s=rθ is the formula for arc length, and ASTC is a mnemonic about which functions are positive in which quadrant

[upper karma]

https://gyazo.com/12adc8ae104819348ca3805ea8457f96

wat ab these

[dark sparrow]

these are both wrong but now i have to go to sleep

[acoustic jungle]

@upper karma second and third are wrong

sec is 1/cos

if cosine theta = -3/5

what do you think 1/ cos is

and for the second question you just made a mistake you should be able to figure it out

[plain forge]

This is technically a calculus problem but it’s the trigonometry I am not understanding how are they getting those values for theta? If you solve for theta those are not the values I get

Wait nvm I think I get it now you need to subtract tan inverse(4/3) from Pi and 2pi to get those values

[acoustic jungle]

yes and that is because tan is negative in quadrants 2 and 4

[plain forge]

Is there a way to know which r value you use with each theta without plugging it in?

[ancient axle]

http://prntscr.com/rucl7m

Can anyone help me with this?

[rich wolf]

@ancient axle think about it

[ancient axle]

I've been staring at this for like 15 minutes

[rich wolf]

bro just think about it

look at question 3

[ancient axle]

;-;

I mean is W 10?

[rich wolf]

what shape is that

[ancient axle]

Triangle-

[rich wolf]

no it's not

how many sides does the shape inside the circle have

[ancient axle]

four

[rich wolf]

lol

[ancient axle]

;-;

online learning is hard ;-;

[rich wolf]

preschool has failed this generation

[ancient axle]

excuse me ;-;

my 8th grade math teacher is giving out assignments without any directions ;-;

[rich wolf]

bro that doesn't mean you should confuse a triangle with a quadrilateral

[ancient axle]

i mena it looks like a triangle

[rich wolf]

alright enough messing around be serious

[ancient axle]

alr

my fault

[rich wolf]

what do you know about the angle sum of a quadrilateral

any quadrilateral

[ancient axle]

They add up to 360?

[rich wolf]

yep

[ancient axle]

right

[rich wolf]

so make an equation

[ancient axle]

5m + 10m + 8m + x(m) = 360?

[rich wolf]

yes

[ancient axle]

23m + x(m) = 360

[rich wolf]

wait i'm sorry

there are two unknowns you won't be able to figure it out this way

[ancient axle]

yeah thats where i kinda got stuck

[rich wolf]

you'll have to use the other theorem

[ancient axle]

huh-

[rich wolf]

if an angle has its vertex on the outside of the circle

then the measure of its arc is double the measure of its angle

[ancient axle]

yeah so...

10-

[rich wolf]

let me send a pic

[ancient axle]

yeh an inscribed angle is 2x the intercepted arc

[rich wolf]

measure of the arc is double the measure of the angle on the outside of the circle

ok

[ancient axle]

right.

[rich wolf]

do you see any inscribed angles in that problem

[ancient axle]

H

http://prntscr.com/rucl7m

[rich wolf]

an inscribed angle is an angle with its vertex on the outside of the circle

[ancient axle]

yeah...

[rich wolf]

make an equation

you know that the whole circle is 360 degrees

so the sum of all the arcs must be 360 degrees

[ancient axle]

33m = 360

[acoustic jungle]

or use the theorem that says if a quaderalteral is inscribed in a circrle that the opposite sides sum to 180

[ancient axle]

I dont see opposite sides

[rich wolf]

you did it wrong teddo

[ancient axle]

they can't be supplementary ;-;

i figured

[rich wolf]

the arc is double the angle

[acoustic jungle]

A and K sum up to 180

[rich wolf]

10m + 16m + 20m = 360

[ancient axle]

How did u get that?

[rich wolf]

omg

the measure of each arc is double the measure of the inscribed angle

arc AK is two times angle H

[ancient axle]

oh alr

[acoustic jungle]

46 m = 360?

[rich wolf]

arc HK is two times angle A

[acoustic jungle]

what??

[rich wolf]

and arc HA is two times angle K

[acoustic jungle]

That isn't right?

[rich wolf]

and arc AK + arc HK + arc HA = 360

[acoustic jungle]

HW is two times angle K

[ancient axle]

I used the opposite angles are supplementary thingy

[rich wolf]

shoot my bad

[ancient axle]

And I got m=18

[acoustic jungle]

M should be 10???

[ancient axle]

yeah 10

misstyped

[acoustic jungle]

The reason the theorem works is because angle K subtends arc WAH and angle A subtends WKH

and WAH + WKH = 360

so k+a is half of that

[ancient axle]

http://prntscr.com/rucl7m

right

so

How do we find angle W?

[acoustic jungle]

Way 1: use the fact that the sum of angles in quaderlateral is 360 and you know m =10

way 2. use the theorem angle and get 5m+W=180

[ancient axle]

oh

I like the first way

[acoustic jungle]

ok.

[ancient axle]

i got 180-

wait no

[acoustic jungle]

ok.

[ancient axle]

130?

[acoustic jungle]

ok.

[ancient axle]

um

is it right?

[acoustic jungle]

I am not telling you 🙂

[ancient axle]

;-;

fair enough

can u help me with the second problem

[acoustic jungle]

ok

sum of triangle = 180

and use the theorem again

[ancient axle]

i got 60

[acoustic jungle]

wait brb let me try it first so I don't mislead you

[ancient axle]

ok

[acoustic jungle]

ah you would use the inscribed angle thoerem

[ancient axle]

how do u see an inscribed angle ;-;

[acoustic jungle]

and 60 degrees is not correct

[ancient axle]

damb

[acoustic jungle]

wait one sec

yes not 60

[ancient axle]

i dont understand

[acoustic jungle]

look at the arc SAE is subtending

that arc would be double the degree of SAE

so that arc would be 160

ARC SE = 160

do the same thing with angle SRA

[ancient axle]

;-;

[acoustic jungle]

maybe this will help you https://www.youtube.com/watch?v=GppOSNTi5OA

and also if you manage to prove every single theorem then you will be able to remember them

[ancient axle]

okie

ty

[digital gulch]

Hey are u a 138° angle?

Because I can be your angle bisector 😳

[simple slate]

how do I find the total amount of sides with the interior angle sum?

[digital gulch]

The measure of each interior angle of a regular n gon is (180(n-2))/n

So if you have the interior angle sum, then you can solve for n

[simple slate]

how

[digital gulch]

The sum of all the angles of an n gon is 180(n-2)

What’s the interior angle sum? @simple slate

[simple slate]

2340

[digital gulch]

Ok

2340=180(n-2)

Solve for n

That’s the answer

[simple slate]

how

[digital gulch]

Divide by 180 both sides

[simple slate]

oh'

[digital gulch]

You get 13=n-2

Add 2 both sides

15=n

That’s the answer

15 sides

[simple slate]

ok

[lilac gull]

what is this problem asking for?
32 ° 15' 66" is the same as ______

[acoustic jungle]

measurements?

[versed river]

probably either radians or decimal form .is there a context as to where this question is from

by decimal i mean like 32.5 degrees

[lilac gull]

its in degrees so decimal form I guess

idk ill just stick the measurements on a triangle and use inverse trig functions

[verbal summit]

may i ask how to find a perimiter using coordinates

perimeter

[versed river]

could you give an example of what you mean? like given 4 points making a quadrilateral, find the perimeter of that? if so, you would use distance formula(pythagoras) and find the length of each side

[verbal summit]

A=(-4,0)

B=(-1,6)

C=(3,-1)

[versed river]

ill help you in a sec, gotta have some lunch first, unless someone comes before i can come back

ill ping you when im back

[acoustic jungle]

either use the distance formula or the shoelace formula @verbal summit

[verbal summit]

whats the distance formula

or shoelace whichever is easier

[versed river]

@!hypegore

so

first lets calculate that right side

idk what shoelace formula is

so ima say distance formula

which is

now, how might we apply this formula to that side?

[acoustic jungle]

sorry my bad shoelace is for finding the area

facepalm

[ebon pelican]

pls help

[dusky surge]

Which part of the question do you need help with?

[latent sapphire]

Hello? Thank you for allowing me in! Im really struggling with this sin cos problem can anyone help

[silent plank]

our ability to help depends on whether or not we can see the problem

[latent sapphire]

Could you possibly break it down as well?

[silent plank]

what trig identities do you know?

and/or what are you allowed to use?

i'll give semi-vague descriptions
option 1: use inverse trig
option 2: apply something that is essentially pythagoras' theorem

[latent sapphire]

Im semi familiar with the inverse trig and i know for this problem you’d use the pothag identity theorem

My notes^ but this stuff still kinda confuses me

[silent plank]

there's quite a bit of bad notation in there

$\sin^2(\theta) + \cos^2(\theta) = 1$

[somber coyote]

ramonov:

[silent plank]

$(\sin(\theta))^2 + (\cos(\theta))^2 = 1$

[somber coyote]

ramonov:

[silent plank]

is also fine

$\sin\theta^2 + \cos\theta^2 \umwhat 1$

[somber coyote]

ramonov:

[silent plank]

is not

anyway, use one of the acceptable forms of this identity and solve for cos(theta)

$\sin(\theta)^2 + \cos(\theta)^2 = 1$ is also a bit questionable

[somber coyote]

ramonov:

[latent sapphire]

Thats how far ive gotten... maybe? Im terrible at math sorry guys

[silent plank]

one of the basics in algebraic manipulation: do the same thing to both sides

you indicated an intention to subtract 0.2209 from the left side, and you would also need to subtract that same amount from the right side

[latent sapphire]

1-.2209?

Which would give me -1.2209

[silent plank]

its good practice to include the 0 before the decimal point
and no, 1 - 0.2209 isn't the same as -(1 +0.2209)

[latent sapphire]

So .7791

[silent plank]

its good practice to include the 0 before the decimal point

[latent sapphire]

0.7791

[silent plank]

that's better

now you have

$\cos^2(\theta) = 0.7791$

[somber coyote]

ramonov:

[latent sapphire]

Right

So would i sub in 0.7791 for theta?

[silent plank]

no

the **square **of cos(theta) gives 0.7791

what could you do here to help find theta?

[latent sapphire]

Divide on both sides

[silent plank]

note that i bolded the word "square"
what did you plan to divide by and how would it help?

[latent sapphire]

Dividing by 2 to remove the square but i dont think thats right

[silent plank]

dividing by 2 doesn't remove the square

lets try something simpler.
how would you solve something like:

$x^2 = 5$

[somber coyote]

ramonov:

[latent sapphire]

Square rooting?

[silent plank]

yes

and what would that give you?

[latent sapphire]

2 square roots of 5

[silent plank]

what do you mean by "2 square roots"?

[latent sapphire]

2 outside the radical and 5 inside

[silent plank]

$2\sqrt{5}$ or $\sqrt[2]{5}$

[somber coyote]

ramonov:

[latent sapphire]

The one on yhe right

[silent plank]

$\sqrt{\ \ }$ already indicates that it is the square root. the 2 there is uncessary.

[somber coyote]

ramonov:

[silent plank]

however, there is more than 1 solution to that equation

[latent sapphire]

As a decimal 2.2

[silent plank]

not what i was looking for

consider negative solutions

[latent sapphire]

I actually have no idea what i could get negative out of that

[silent plank]

lets use even simpler numbers.

$a^2 = 1$

[somber coyote]

ramonov:

[silent plank]

a = 1 would be a solution here
what other number when squared would give 1?

generally: $\sqrt{x^2} \neq x$

[somber coyote]

ramonov:

[silent plank]

$\sqrt{x^2} = |x|$

[somber coyote]

ramonov:

[silent plank]

for something like: $x^2 = 5$, \
applying the square root to bot sides gets you: $|x| = \sqrt{5}$,
which has the solution: $x = \pm\sqrt{5}$

[somber coyote]

ramonov:

[silent plank]

depending on how you are taught, the concept of absolute values may be glossed over

it isn't completely necessary to write the intermediate step and for that reason it is occasionally skipped at this stage.

alternatively you can attempt to solve it by applying the factorisation of a difference of 2 squares.

and achieve the same result

note: for your current trig question, you should identify that there are 2 solutions to cos(theta) from that equation and then choose the correct one based on your given interval

[olive scarab]

is there any way to do problem 1?
I have only worked with $sin^{2} and sin^{3} $ so this is too new for me :( dont even have any idea to solve this
find x that $cos^{2020}x - sin^{2020}x =1$

[somber coyote]

.,.:

[dark sparrow]

note that $\cos^{2020}(x)$ and $\sin^{2020}(x)$ are both guaranteed to be in $[0,1]$

[somber coyote]

Ann:

[dark sparrow]

and there's only one way to have $a - b = 1$ with $a, b \in [0,1]$

[somber coyote]

Ann:

[olive scarab]

oh oh

thanksss

[silent plank]

what seems to be the issue?

[tiny halo]

im confused because im answering the question and it says its worng

[silent plank]

what's your answer?

[tiny halo]

1,074.7

I have to round to the nearest tenth also

[silent plank]

round everything at the end

not for each component

it feels like you did something like
18.5^2 * 3.14

[tiny halo]

yeah

[silent plank]

which you shouldn't

[tiny halo]

thats what they are gettin

its an example but im confused on how to do it

[silent plank]

it doesn't seem to evaluate to ~4300.7 either. what's the full question?

,calc 18.5^2 * pi

[somber coyote]

Result:

1075.2100856911

[tiny halo]

thats the whole question

[silent plank]

where are those 4300.7
7853.8
15393.4 from?

[tiny halo]

those are the same quesiton just different numbers

[silent plank]

who's "they"?

[tiny halo]

?

[silent plank]

because those values don't make any sense

[tiny halo]

ik but its saying its wrong

[acoustic jungle]

those values are wrong because they did 37^2 instead of 18.5

[tiny halo]

really

so they are doing 37 instead of 18

18

[acoustic jungle]

Yes

[tiny halo]

18.5*

wow

[acoustic jungle]

which is wrong

[silent plank]

they did pi*d^2 which is wrong

anyway, to get the proper answer, you should only round at the very end

[supple wedge]

rounding at every step wtf

[silent plank]

note that your calculator should have "pi"

[tiny halo]

yeah it does

[silent plank]

and you should enter the value as that

[tiny halo]

ok

Thank you guys

[vale nimbus]

(x-h)² + (y-k)² = r²
Center (h, k)
Radius r

idk if this helps but it might idk

[dire rampart]

@upper karma what have u tried

[rich wolf]

@upper karma you shouldnt be getting decimals if you keep everything as fractions lol

Find the exact solutions

[acoustic jungle]

https://www.desmos.com/calculator/yifcuokg4l

[gritty osprey]

how to find side of a triangle of non right angle given area

[silent plank]

depends on what other information you are given

[gritty osprey]

@silent plank

[silent plank]

trig formula for area (of a triangle)

[gritty osprey]

1/2 ab * sin(angle)

but how to find

am rlly confused

[silent plank]

Area = 1/2 ab * sin(C)
by convention the angle C is the angle between the sides a and b

[gritty osprey]

yes

so far i get

0.5 * 17 * sin(68) / 150

what is wrong

[silent plank]

that isn't an equation

what's that supposed to be

and how does it relate to x

[gritty osprey]

x = 0.5 * 17 * sin(68) / 150

[silent plank]

show me what you intitially started with

[gritty osprey]

150 = 0.5 * 17 * sin(68)

[silent plank]

where's the x?

[gritty osprey]

150 = 0.5 * 17 * x * sin(68)

now what

[silent plank]

now do the appropriate manipulations to both sides of the equation to isolate x

[gritty osprey]

what

my equation was correct

[silent plank]

which equation are you referring to?

[gritty osprey]

x = 0.5 * 17 * sin(68) / 150

[silent plank]

justify why you think that is correct

[gritty osprey]

you are just wasting time

just be to the point

pld

*pls

this isn't literature

[silent plank]

i want you to describe clearly what you did and hopefully realise the mistake yourself

i.e. does what you did to get from eqn1 to eqn2 make sense mathematically

ignoring the degrees, starting with:
150 = 0.5 * 17 * x * sin(68)
is fine

from there, what would you need to do to both sides of the equation to get x by itself?

[gritty osprey]

divide

[silent plank]

by:

[gritty osprey]

150 and x

[silent plank]

divide both sides by 150 and x?

[gritty osprey]

divide both sides by x

[silent plank]

that would get you:
$\frac{150}{x} = 0.5 \cdot 17 \cdot \sin(68\deg)$

[somber coyote]

ramonov:

[gritty osprey]

yes

[fiery steppe]

wait wait wait

how this bot works?

lol

[silent plank]

is that what you want?

i mean its mathematically valid...but doesn't really work towards isolating x

lets try a simpler example.
how would you solve something like
3a = 7

[gritty osprey]

a = 2.3

2.3 reccuring

[silent plank]

and how did you arrive at that?

[gritty osprey]

divide

[silent plank]

by:

[gritty osprey]

3

[silent plank]

divide both sides by 3 because 3 is the coefficient of a right?

[gritty osprey]

yes

[silent plank]

and then apply the same idea to your question

currently you have:
$$150 = 0.5 \cdot 17 \cdot x \cdot \sin(68\deg)$$
which can be rewritten as:
$$150 = 0.5 \cdot 17 \cdot \sin(68\deg) \cdot x$$
to make it more convenient for you

[somber coyote]

ramonov:

[silent plank]

what is the coefficient of "x"?

[gritty osprey]

the answer is 19

i just wolframalpha

you take too much time

be more efficient

[silent plank]

if you only want the answer you've come to the wrong place

does putting that into wolfram teach you how to solve equations?

[gritty osprey]

I figured out how to do it in just 2 seconds after wolframalpha

[silent plank]

will you continue to struggle with subsequent problems if you have no idea what you're doing wrong?

[gritty osprey]

I never struggle

[silent plank]

ok then i'm done here.

[gritty osprey]

ok bye

pls be more efficient

learn to think like a machine

[silent plank]

i just have to interject here

my questions are relatively simple and should take you no more than 30 seconds to have answered each one

and were aimed to break down the problem and identify the source of the problem

[gritty osprey]

they are taking too much time

be fast

and efficient

[rich wolf]

Lmao my man couldnt even isolate for x and he says he never struggles

[acoustic jungle]

hi.

[onyx cloud]

@gritty osprey i hope you realize you were the inefficient one in this case

[gritty osprey]

sure

[onyx cloud]

as you couldn't answer his simple questions effectively

[gritty osprey]

ok

I don't rlly care abt maths

I am more of a programmer

so yea...

[onyx cloud]

that's not the topic of discussion

[silent plank]

1. identify problem
2. debug

[gritty osprey]

programming > maths

too slow man

[onyx cloud]

you were complaining about an inefficient helper where 1. they were offering their help for no compensation, and 2. you clearly couldn't grasp the concept

[gritty osprey]

you gotta finish things fast

[onyx cloud]

you gotta understand them first

which you clearly didn't

[gritty osprey]

wolframalpha did

and I learnt the thing in 2 secs

[onyx cloud]

great

[silent plank]

great

[onyx cloud]

now stop complaining about free help

[gritty osprey]

he took like 10 mins

[acoustic jungle]

hello.

[onyx cloud]

hello

[inland ice]

there's a difference between being "inefficient" and explaining something in such a way that it helps you understand the material lol

[acoustic jungle]

hello.

[unique blaze]

given point 1 and point 2 on a 3d grid and a length how would i get the coordinates of point2+length perpendicular to point 1 -> point2 and where y is constant?

[onyx cloud]

if y is constant then we have a 2d grid correct?

[unique blaze]

for point 2 and the length, not point 1

hope that makes sense xd

[onyx cloud]

first issue with the question

there are infinite perpendiculars to a line in a 3d grid

[unique blaze]

but only 1 where y is the same for both the length and point 1, right?

[onyx cloud]

isn't the length just a scalar

like a number

it doesn't have coordinates

[inland ice]

may you send a screenshot of the question

would probably help more

[onyx cloud]

^

[unique blaze]

nvm, i think ive worked it out by calculating the gradient between point 1 and point 2 taking the reciprocal and then using the gradient to find the coords. Its not a like textbook question, its for game design and i couldn't wrap my head around it. thanks tho, apologize for the confusingness of the question.

[onyx cloud]

you can use vectors and dot products as well i'd imagine

[round isle]

Hello everyone

Is there any algorithm to find the shared volume of two intersecting cuboids?

It is easy to calculate it if both cuboids are axis aligned but what if they can have any arbitrary rotation?

Sorry, one more clarification, when I say they are rotated, I mean they are rotated about one axis only. So that means, the cuboids are rotated about the y-axis only

My previous method was to just treat it as a 2D problem first and find all the possible intersecting points (from a top down perspective) but that introduced a lot of case statements (intersecting shape could be a triangle, square or rectangle)

Is there any clever algorithm to deal with this?

[hasty spoke]

Can someone please help me with this question

I just don't know where to start

[fiery scroll]

Hope this is the right channel.. hard to find the right one when im a little lost in here lol

hey guys.. i dont mean to interrupt....new member with the need for a simple formula based on some constants.. any one wanna help me tackle it? its for a script similar to c# BUT i can do all of the scripting part, i just need help coming up with the formula.. we know that

If yaw=0 then x= -1.5 & z= -1.5
If yaw=90 then x= -1.5 & z= 1.5
If yaw=180 then x= 1.5 & z= 1.5
If yaw=270 then x= 1.5 & z= -1.5

so we need to find the formula for a floating scale for the following

while yaw floats between 0 and 360 then:
X floats between -1.5 & 1.5 
&
Z floats between -1.5 & 1.5```

[round nacelle]

@clear comet plz help these peasant

s

[summer spire]

@round isle sounds like a multivariable calculus problem

[fiery scroll]

one guys says calc, one guys says trig, ive tried questions.. this discord is a gold mine of information...🙄 🙄 🙄

[summer spire]

@fiery scroll no, your question looks like trig to me. I'm just a bit lost on what everything represents

I suppose my first question is what is the radius of your circle?

[fiery scroll]

ok.. i can explain.. i had an object on the map that is 2m wide and 5meters long. the model can rotate on the center point 360d, when facing forward, yaw is 0 (we will say south) X and Z are a point on the object that is offset from the center point by they value.

when the object is at yaw of 90, then x and Z need to change. i have the 4 known points when the object is at any given yaw.

i need to be able to grab the X and Z IF i know the yaw AND we know the max the x and Z can be..

not sure if im explaining it properly.

[gleaming nova]

Are sin, arcsin, sinh, and arcsinh the only versions of the sin function?

[summer spire]

@fiery scroll from what I can work out:
$$f(x)=\frac{3\sin(x-135)}{\sqrt2}$$
$$f(z)=\frac{3\sin(z-45)}{\sqrt2}$$

[somber coyote]

Googol30:

[plucky jacinth]

can someone eplxain this or have like a notes sheet about this

[rich wolf]

@plucky jacinth think about it

what 2D shapes make up a 3D triangular prism

[unborn jacinth]

I had a question about graphs and equations

when you have y = (x+3)(x-1)^2

logically speaking, how come the even multiplicity causes it to bounce off the x axis but odd multiplicity crosses the x axis

[rich wolf]

@unborn jacinth

Let's look at a really simple example

[unborn jacinth]

ok

[rich wolf]

y = x^2

That is a second degree polynomial with one root

Of even multiplicity

[unborn jacinth]

yes

oh

so because its squared its never going to be negative

[rich wolf]

Yes pretty much

[unborn jacinth]

but in my example

there are other factors that affect it too

like for example if I had (x-3)(x+4)^3(x+2)^2

[rich wolf]

This is a fun thing to play around with if you're interested

[unborn jacinth]

sure!

[rich wolf]

Just curious, have you ever heard of the concept of a limit?

[unborn jacinth]

yeah

as x reaches infinity / - infinity

[rich wolf]

It doesnt have to be infinity

[unborn jacinth]

ive done end behavior and asymptotes and all that stuff

now I'm just trying to analyze

[rich wolf]

It can be as x approaches any value

[unborn jacinth]

why those actually cause the changes in the graphs that they do

oh ok

[rich wolf]

Let's take that polynomial you mentioned there

$(x-3)(x+4)^3(x+2)^2$

[somber coyote]

AMD:

[rich wolf]

Let's look at the even multiplicity root

[unborn jacinth]

ok

[rich wolf]

(x+2)^2

Obviously if x = -2, then f(x) is 0

[unborn jacinth]

yeah

[rich wolf]

But what happens immediately around x = -2 ?

[unborn jacinth]

everything is 0

[rich wolf]

Try it from both sides, getting closer and closer to -2

[unborn jacinth]

oh

ok

around

[rich wolf]

Like try plugging in -1.9 and -2.1

[unborn jacinth]

ok

-1.9 is about -0.5

and -2.1 is about -0.34

so they "bounce"

[rich wolf]

Can you start to see the pattern?

Try it again with the odd root

Take the limit as x->-4 from the left and right

[unborn jacinth]

ok

yeah if its less than -4 its + but greater is negative

[rich wolf]

Now obviously there's a pattern here

The challenge is can you put it into words

And explain why

[unborn jacinth]

ok

well the pattern is that even multiplicty causes it to bounce off the x axis right

and odd causes it to cross

im not rlly sure logically why that happens

if I plug in numbers it works

but like idk

[rich wolf]

let's think about it this way

any real number to an even power is always positive, isn't it?

[unborn jacinth]

yeah

[rich wolf]

$(x-3)(x+4)^3(x+2)^2$

[somber coyote]

AMD:

[rich wolf]

let this expression be f(x)

[unborn jacinth]

ok

[rich wolf]

think about what the other roots do

what is the sign of (x-3)(x+4)^3 immediately around x = -2 ?

[unborn jacinth]

negative

[rich wolf]

it is always negative around x = -2

and (x+2)^2 is always positive

because it is a real number to an even power

[unborn jacinth]

yeah

[rich wolf]

so you've got a negative number

multiplied by a positive number

[unborn jacinth]

so thats negative

o

but what if instead of x+4

it was x-4

then it would be positive right

[rich wolf]

idk off the top of my head

[unborn jacinth]

ok well x-3 is negative

the x+2 part is positive

meaning x+4 has to be negative

because the value we get at -1.9 is negative

so if it was x-4 im pretty sure it would be positive around -2

lemme graph on desmos

ye ok

I think I get it now

there's isn't much of an explanation besides "thats how math works"

its basically just plugging in numbers

[rich wolf]

but the point is that the behavior of a function around the root of even multiplicity always has the same sign

there's more formal ways to describe it but yeah pretty much

[unborn jacinth]

ok

wait same sign?

same sign as what

[rich wolf]

f(-1.9) has the same sign as f(-2.1)

[unborn jacinth]

oh oh

yeah

ye

ok tysm!

that helped a lot

[upper karma]

@rich wolf Why isn't this a 45-45-90 triangle?

[unborn jacinth]

thats 30 60 90

[upper karma]

but why is it 30 60 90 if the cos and sin are the same?

[rich wolf]

they aren't

[unborn jacinth]

are they?

I think thats just a question

its not saying they're the same

right?

[upper karma]

they're not

[rich wolf]

it's just asking you to find the cos and sin of that angle

[upper karma]

nvm i see what it is pi/3 is 1/3 which is 60 degrees

[latent iron]

is anyone able to help me ? i dont really get what it's specifically asking for

[rich wolf]

@latent iron draw a pic

[latent iron]

@rich wolf yeah thats what

i need help with

im not sure how to draw it bc i dont get what

"54m from the foot of the building" intends at

[rich wolf]

you are standing 54 meters from the foot of the building

so that is a line from you to the foot of the building

[latent iron]

so would it be the btm

like from bottom of building to point they're measuring

[rich wolf]

yeah

[latent iron]

oh i see

i put it as the hypotenuse whoops

[ionic bluff]

hey quick trig question

why is wolfram alpha dumb

[rich wolf]

@ionic bluff wolfram is only as smart as you are

[ionic bluff]

i lodged a formal complaint with wolfram alpha

i can't believe they would lie like this

[tame ore]

so

this is 60 right?

cause....

it all adds up to 60

[silent plank]

nope

what are you adding up and how are you getting 60?

[tame ore]

the sides

22 - 8 = 14

[silent plank]

and the rest of your calculations?

[tame ore]

they add to 60

and perimeter is when u add them all

right?

[silent plank]

"they add to 60" is a claim and not work.
i want you to show me specifically what you did, and the numbers you added to reach that 'incorrect' value

to identify what the issue is

[tame ore]

i dont understand-

[silent plank]

fill in the blank
"i got 60 by adding the numbers : ___ together"

?

[tame ore]

on the shape

11 8 14 5 22

[silent plank]

ok there we go

you ignored one of the sides

[tame ore]

ohhh

[silent plank]

note that that shape has the same perimeter as a rectangle with dimensions 11m * 22m
and you can simply do ||P = 2(11+22)m||

[rich wolf]

a + b + c + d + (a-c) + (b-d) = 2(a+b)

Omg

That is pretty cool ngl

Didnt know that

Thank you romanoff for your wisdom

[silent plank]

you can consider it as shifting those sides outwards to form a rectangle

and would still apply if you had some weird staircase shape

[rich wolf]

Yeah it makes a lot of sense actually

Like if you rearrange the segments you can make it into a rectangle

Kind of obvious in hindsight

But yeah thanks

[night summit]

can someone please help me

im a junior in highschool and I have no clue how im suppost to do this because our teacher never even told us to study this

https://gyazo.com/951969336e810f7b55541558bd7eef9c

[onyx cloud]

crim

im not exactly sure that vocabulary

but i think you need to make use of the unit circle

and find the cos(pi/6) and sin(pi/6)

[rich wolf]

Oh its asking you the slope of the line

[night summit]

how do i find that

i dont understand this at all

as in

pretend im in elementary school

[rich wolf]

Ok

Think about it

What kind of equation is y=kx

[night summit]

equation of a line?

[rich wolf]

Yes

[night summit]

with no intercept?

[rich wolf]

Eh

Not no intercept

[night summit]

k is the slope?

[rich wolf]

But 0 intercept

Passing through the origin

And yes k is the slope

[night summit]

so

wat does it want from me

it says constant of proportionality

[onyx cloud]

have you ever seen this before?

[night summit]

yea a unit circle

all strippers take cash

ive tought it to myself

[onyx cloud]

ok

look at the line for theta=pi/6

[rich wolf]

Thats a good one lol

[night summit]

so

[onyx cloud]

what they want is for you to find the slope

[night summit]

x is rad3/2 and y is 1/2

they want me to find the slops of rad3/2 for x

and 1/2 for y?

[onyx cloud]

they want you to find the slope of the line

you can do that using those points yes

[night summit]

how

[onyx cloud]

what is the formula for slope

[night summit]

the equation to find slope is y2-y1

over x2-x1

[onyx cloud]

yep\

[night summit]

what do i put in for the other xs

and other y

[onyx cloud]

look at the unit circle

what other points are on that line

[night summit]

it only gives me 1 set of coordinates

rad3/2

[onyx cloud]

look at the line pi/6

what other points are on that line

[night summit]

only rad3/2?

i dont see any other points?

[onyx cloud]

there is 2 other points

[night summit]

ohh shittt

[raw tiger]

The center point is on every line. What are the coordinates of the origin?

[night summit]

ohh

0,0 is origin

oh

[flint bloom]

u could do this

[night summit]

so it wants me to find the slope of those 2 points

[raw tiger]

That might be a really convenient point to use (the origin)

[night summit]

well

when it comes to math i have the comprehension of a squirrel

[wind heart]

Hi

I need late night help

How to get answer

[versed river]

sum of exterior angles of a polygon=360 degrees

so you can set up an equation with that fact and solve for y

[raw tiger]

You could also add up all the interior angles and that should be equal to the total angle of a hexagon (720 degrees)

Then just solve for y

[wind heart]

Oh

That’s easier than I thought

Ok then

I should sleep

This epidemic is really f’ing up my sleep schedule

Even though I don’t have to leave the house

It’s ironic

thanks bye

[rose sphinx]

Hey

Is there a way to calculate sin37 without calculator

[dark sparrow]

no

[quiet mason]

ive used the approximation sin37=3/5 before

,w sin37-3/5

[somber coyote]

Display results online and refine query

Requested by Lionel#5497

[quiet mason]

close

[rose sphinx]

What about sin54 & sin18

[quiet mason]

those are achievable

and sin54=sin(3*18)

[versed river]

i think black pen red pen has a pretty good video on sin18

[rich wolf]

Sin(pi/180)

,w Sin(pi/180)

[somber coyote]

Display results online and refine query

Requested by AMD#4434

[rich wolf]

Idk

Is there a 37th angle multiple formula

Sin(pi/180) has a closed form)

[dark sparrow]

well yeah sure you can expand sin(37x) all the way down to a polynomial of the 37th degree in sin(x) to your heart's content

[rich wolf]

Nice

So any rational multiple of pi is technically computable

[dark sparrow]

that word doesn't mean what you think it means

but yes

you can get a closed form

more or less for any rational multiple of pi

[rich wolf]

Which word

Computable?

[dark sparrow]

yes

[fiery steppe]

guysss help

plz

how much is the tangent of a

i'm stucked here since this morning ç_ç

[silent plank]

what do you know about "tan"

[fiery steppe]

nothing

that's the fact lmao

if you relate to the exercise

i'm tryna learn

[upper karma]

Opposite over adjacent

Side

[silent plank]

read up some notes, that tell you the definitions of trig functions and their applications

[upper karma]

Soh Cah Toa

[silent plank]

and come back if there's anything you don't understand about them

[cobalt bear]

How

I need the equation

(a) and (b)

[silent plank]

which equation?

there are many equations and methods that can be used to solve this problem

and it may also be more efficient to do part b) before a)

[cobalt bear]

um

I mean I need the steps

[silent plank]

did you attempt anything yourself?

[cobalt bear]

no

[silent plank]

how much trig have you learned so far?
put a little more effort first.

[cobalt bear]

Without BC is hard :(

[silent plank]

you could consider using
tan(37°) and tan(69°)

which can be a bit tedious

alternatively, you can determine angleCAD first and then apply the sine rule to get AC (which completes part B first)

though there is a formula which you may or not be allowed to use

which can be derived from tan

[upper karma]

okay so im given a parabola x²+y²+2x-6y-2xy+3=0

i need to find the focus

and the apex