geometry-and-trigonometry

[blazing pond]

<@&286206848099549185>

[patent plume]

@blazing pond Do you remember the properties of a rhombus?

[blazing pond]

yes

waiiiiit

so its 16 right @patent plume

[patent plume]

👍

[blazing pond]

LEZZ GETIT

ur sure rught?

[dry ore]

Ok this should be right geometry

So

I need help with number 5 and 8

Can anybody help me

?

[limber osprey]

wth my math exam is friday aswell

[wintry tundra]

i wonder why people have so much trouble with similarity ratios

anyways since SDE and SWT are similar, that means coordinated sides have similar ratios

so $\frac{56}{40} = \frac{5x + 3}{4x - 3}$

[somber coyote]

hiidostuff

[wintry tundra]

@dry ore

[pulsar swan]

Can anyone help me. My exam is next week..

[tired pine]

lots of alternate interior angles there, get two angles and prove that ABE is similar to CDB, then the desired conclusion follows

[stable gull]

idk how to set this up

its trig can someone help

[potent tree]

Guys can someone assist in D4

[versed river]

find |PA| and |PB| in terms of x and y using pythag theorem\distance formula, then apply the hint

[nocturne thicket]

I got x = 200/tan(28)

376.145293069

but the answer sheet says im wrong

[versed river]

How did you get that

[nocturne thicket]

90-Angle of depression=28 degrees

Tan 28 = 200/x

x(tan 28) = 200

x = 200/Tan (28)

[nocturne thicket]

@versed river you still there?

[versed river]

oh sorry that seems right actually, what does the answer sheet say?

[nocturne thicket]

2. 176.6 ft

[versed river]

ah so the intended meaning was actually this then

[nocturne thicket]

wtf

[versed river]

i interpreted it the same way as you but if thats the answer this is the intended interpretatiuon

[nocturne thicket]

200 as the hypotenuse then?

[versed river]

ye

[nocturne thicket]

yo that's fucked

so it'd actually be cos 28 = x/200

[versed river]

ye

[nocturne thicket]

oh my fucking

thank s

i hate this textbook

[dry ore]

@helpers

<@&286206848099549185>

Can anyone help me?

This is the only question I need help with

[tiny snow]

@dry ore In similar triangles corresponding sides are proportional to each other. You can use that to create an equation.

[dry ore]

How do I create I equation with the sides?

[tiny snow]

If BC is proportional to DE, do you know what that means?

[dry ore]

Yeah they equal eachother right?

Or there similar

[tiny snow]

No, it means they differ only in scale. Similar triangles are scaled and possibly mirrored versions of each other.

[dry ore]

So then 2x+3=15?

[tiny snow]

So BC relates to DE as AB relates to AD.

BC/DE = AB/AD

[dry ore]

Yeah

[tiny snow]

Use that to build your equation, by plugging in the numbers and expressions you have for the lengths.

[dry ore]

Oh ok so I was right is it 2x+3=15?

[fast pulsar]

Dude, we discussed this exact thing yesterday...

I think I even explained it in depth

[dry ore]

I know but I'm still confused

[fast pulsar]

And you didn't ask yesterday because...?

[dry ore]

I thought I understood it but when I got to this question I forgot how to solve it

[fast pulsar]

I'll leave you all alone. I do not mean to be rude, I am just a bit frustrated

[fast pulsar]

Go back to my response from last time

And if that confuses you when you look at it, then feel free to keep asking questions

[tiny snow]

No worries, I’ll let you read up on what dackid wrote first.

[hasty flume]

So in this server people can help other people with homework?

but u just cant cheat or give answers right?

[summer relic]

Right, no tests or quizzes ✅

[dim cliff]

can someone help me with this question please?

[lunar sand]

I had trouble with solving/what to do with this problem, so i tried searching it up to get some clues. Apparently, it's supposed to be $15,000 e^-0.05(3). Where does the negative come from and why?

[jaunty sail]

Are these triangles solvable?

[summer relic]

you have all three angles

[jaunty sail]

But only one side

[summer relic]

Question a is weird

[jaunty sail]

Right

You can’t figure it out can you

[summer relic]

No I'd need at least one side value for a and one more side value for b

What unit is this?

[jaunty sail]

This isn’t even my assignment

Someone asked for help lol

[summer relic]

yeah they aren't giving the full problem

[jaunty sail]

Right

Alright thanks

[summer relic]

np

[upper karma]

can anyone help with this please ?

[tired pine]

Get the radius of the circle

Since E is the center of the circle, EF and EG are perpendicular bisectors of AB and CD, respectively

Construct either AE or BE and solve it using pytha

Then just reverse for CG

[empty fulcrum]

I have the lengths of all sides of a triangle, how do I find the height?

I’m having trouble with this one problem

[tired pine]

There are actually multiple heights of a triangle

[empty fulcrum]

if the red point is G

how to I find AG

If I have AC and AB

Also sorry I can’t draw straight but BG is the height

[tired pine]

Use ABsin(α)

Though it seems like you’re finding the area of the base since that looks like a prism, yes?

[empty fulcrum]

Yes triangular prism but I have to first try to solve the area for ABC

[tired pine]

you can do (1/2)(AB)(AC)(sin(α)) instead

Essentially the same thing as bh/2 but in terms of 2 sides and included angle

[summer relic]

I've filled in the equation I need and I put it into desmos, anyone know how I can get the angle from desmos?

and also if I should be in degree/radian mode

at least for the second part

[summer relic]

Still confused on this one <@&286206848099549185>

[fast pulsar]

So what does it look like in desmos?

[summer relic]

This is what it looks like in degree mode

[fast pulsar]

Can you put it in radians please

[summer relic]

Yes

[fast pulsar]

May wanna zoom in

[fast pulsar]

Keep going

[fast pulsar]

Now shift it to the right til you see the maximum

[summer relic]

im hoping this is it

[fast pulsar]

Alright... Maybe put it back in degree mode, not sure why this one looks worse 😆

[summer relic]

haha will do

[fast pulsar]

And that's your maximum?

[summer relic]

I believe so

[fast pulsar]

Did desmos point it out?

[summer relic]

The problem before had no decimals in the maximum

[fast pulsar]

It tends to show maximum and minimums

[summer relic]

yes desmos did

[fast pulsar]

Okay, that's your maximum

[summer relic]

the y is? or just in general the coordinate

[fast pulsar]

Round it to 24 ft if they do not want decimals

[summer relic]

okay will do

you sure its not the y they want?

27

[fast pulsar]

"At what distance should you stand from the wall..."

[summer relic]

so we need to know theta

from the wall

ahhhh good point

24

[fast pulsar]

24 is our x value

[summer relic]

we cant levitate

so 24

[fast pulsar]

And theta is whatever you found on there

[summer relic]

oh how would I find theta?

on desmos?

[fast pulsar]

Well.... you can't 😆

Well, ya found it already

[summer relic]

Solid point Bob Ross

[fast pulsar]

That maximum has an x and $\theta$ coordinate

[somber coyote]

dackid

[summer relic]

oh wow im silly

[fast pulsar]

So the ordered pair we have is $(x, \theta)$

[somber coyote]

dackid

[summer relic]

plug the x into the tan equation

[fast pulsar]

Or... Just look at the graph because we already know the value

[summer relic]

26.387

[fast pulsar]

Yep! So roughly 26°

[summer relic]

what do they mean by "do not use calculus"

do they want pi?

[fast pulsar]

You can calculate this with derivatives

[summer relic]

sounds fun

way above my pay grade

[fast pulsar]

Nah, you're angle has been in degrees forever at this point

[summer relic]

also true

[fast pulsar]

Do you know basic derivatives?

[summer relic]

My major hasnt required calculus, I think I'll take it when I go for BS

I am going into stat

[fast pulsar]

Ah okay. I highly recommend Calc 1 at least

[summer relic]

Yep UW will make me take it

But I'm living the good life not taking it rn

[fast pulsar]

Nah, Calc is actually pretty interesting :)

[summer relic]

Well I guess I'll know in 2 years haha

Thanks again

[fast pulsar]

You bet

[summer relic]

I'm thinking this is a vector question. I am given a lot of information and I think it's making me stumble a bit.

[upper karma]

can someon explain how i would get the phase shift? when i tried this for my hw i got the phase shift of pi/12 yet that is not an option

phase shift is -c/|b| so that means if it is pi/6 the c is pi and the b is 6

and for the sin graph i inputed the same thing except the c is pi/2 and it appears the same

yet pi/12 isnt an option for phase shift, so what did i do wrong?/

[upper karma]

okay the answer is 2p/3

not sure how

[somber coyote]

web
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[strange wigeon]

Like why is it 180/pi to convert the radians into degrees.

[summer relic]

because 180 degrees = pi

its a ratio

[hot ice]

hey besties can someone help

[copper marten]

do u know SohCahToa?

[hot ice]

no

[copper marten]

ok so:
sine: opposite/hypotenuse
cosine: adjacent/hypotenuse
tangent: opposite/adjacent

did u learn any of those in your class?

[copper marten]

can u send a clearer picture, the angle measures are kinda blurry

[spring mural]

<@&286206848099549185> can someone help me please with 15?

[spring forum]

hey uh does anyone know how to find the reference angle of an angle

i dont really get it

[olive cove]

I guess they want you to change degrees to radians and radians to degrees

you know one degree is pi/180 radians so 678⁰ = 678*(pi/180) radians

[tiny snow]

@spring forum Yes, the reference angle is the angle modulo 360 degrees, or 2 pi radians.

[spring forum]

so the first one would be 42 degrees right?

[tiny snow]

42?

[mint dune]

yes it would

[mint dune]

reference angle is basically shortest angle from x axis

[tiny snow]

@mint dune You are right. I am wrong.

[mint dune]

lol

[jolly cargo]

hey, can i get some help with geometry?

im not gonna lie, im honestly hopeless when it comes to math so i don't know pretty much anything

[summer relic]

can't help with tests king

[summer relic]

Sorry my guy mods can look at message history and deleted messages, the rules say no tests

[fast pulsar]

@summer relic was there any evidence that says this is a test? 🤔

I don't see it tbh

[iron shoal]

How do I prove that this is a parallelogram

[quick plank]

slopes of the lines that are parallel

[summer relic]

Yeah before he deleted the message, didn't want to be rude about it but I can't be doing rule violations

[fast pulsar]

No no, you're absolutely right man. Academic honesty is very important

[undone field]

Hi can I get some help on a feo question ?

[quick plank]

post it as long as it's not a test

[nocturne thicket]

Can someone verify my answer?

[quick plank]

is it a test/quiz

[nocturne thicket]

its homework dude

why would i post a test here that's not allowed

[quick plank]

people still do it

anyway I think it's pythagorean twice

[nocturne thicket]

what

Pythagorean theorem?? here???

[humble pulsar]

3 times

[humble pulsar]

Yeah.. why is that a surprise?

[nocturne thicket]

I can't find the left triangle with pythagorean theorem

[nocturne thicket]

this is geometric mean...

[humble pulsar]

It's a bunch of right triangles

[nocturne thicket]

Right, but the only triangle i can find is the right one

[humble pulsar]

Yeah, good place to start

[nocturne thicket]

OK but i won't be able to find the left triangle

@quick plank @humble pulsar how exactly do i solve this equation with pythagorean theorem here

[quick plank]

I'm doing it rn

almost done

@nocturne thicket

[nocturne thicket]

holy shit how does that work

wait you can't cancel the x^2 by x^2 can you?

[quick plank]

i don't understand your question

I just subtracted x^2 from both sides

[nocturne thicket]

wait how did you find A

[quick plank]

I never found a

but it would have two different values

in terms of x

so the left triangle has a as the hypotenuse and x and 6sqrt13 as the sides

so I plugged that into pythagorean for the top equation

[nocturne thicket]

right, so how do you go from 637+a^2=(x+13)^2 to 637+(468+x^2)=x^2+26x+169

[quick plank]

okay do you understand how a^2 has two values

that are equivalent to each other

[nocturne thicket]

yes

[quick plank]

so I substituted one of the values (the top one) into the bottom equation

and I expanded (x+13)^2 using sum of a binomial squared pattern

[nocturne thicket]

i still don't understand what you substituted

wait

[quick plank]

i'll explain it on paper starting from the top left 2 equations

[nocturne thicket]

did you simplify a^2=(6Rad13)^2+x^2

and get a=(6Rad13)+x?

[quick plank]

okay wait

the value of a is never needed

although you could do it that way if it helps

[quick plank]

in both equations, I only used a^2

[nocturne thicket]

but then you have two values

a^2 and x^2

[quick plank]

yes

and a^2 is in terms of x^2

what happens when you substitute a^2 is that there are no longer a^2 terms

it becomes a quadratic, then a linear equation

[nocturne thicket]

sure, but how did you substitute a^2 here?

did you use a^2=(6Rad13)^2+x^2?

[quick plank]

yeah

then I used that in the other equation in the system

[nocturne thicket]

ahhh i see

i honestly didn't know that was possible

system of equations + pythagorean theorem

thank you so much, seems like I was right

[quick plank]

goes a bit more in depth for each step

[royal citrus]

Hey how do i do this?
i am stuck with a^2-b^2 when I do the LHS
and Idk what to do after that

[fast pulsar]

$(\sin(A)\cos(B))^2-(\cos(A)\sin(B))^2=\sin^2(A)\cos^2(B)-\cos^2(A)\sin^2(B)$

I know I did not write anything revolutionary, but see if $\sin^2(\theta)+\cos^2(\theta)=1$ may have a role here

[somber coyote]

dackid

[royal citrus]

leme check

once again

[fast pulsar]

Also, what is the goal of this problem?

[royal citrus]

to prove

[somber coyote]

dackid

[fast pulsar]

Ohhh, yea pythagoras' Identity will be useful here :)

I'll give you a hint for the first one. Let $\cos^2(A)=1-\sin^2(A)$

[somber coyote]

dackid

[royal citrus]

aight, thankyou

[fast pulsar]

You bet. If you're stumped, ping me so I can help out

[tiny snow]

Animated geometry problem: Let A be the center of a circle and B a point on its perimeter. Two tangents CD and CB meet in C. Between the tangents and the circle around A another circle is inscribed. If r is the radius of the circle around A, then C is moving away from B at constant velocity r. Find an expression for the relationship between the area of the inscribed circle (orange) and the area of the original circle (blue) over time.

Fiddle: https://www.geogebra.org/geometry/z2yfscn8

[tiny snow]

I need one other condition, since equations (1) to (4) here are not independent (the final statement will reduce to 1=1). Any ideas?

[somber coyote]

PROnoob

[supple echo]

@tiny snow

I derived it using trig mostly

[young inlet]

does anyone have a tip on memorising pi/180 and 180/pi for radians?

i keep forgetting which is for which

[supple echo]

from degrees?

[silent plank]

$\pi = 180\deg$

[somber coyote]

ℝamonov

[silent plank]

that's pretty much all you need to remember for that

(and multiplication by 1)

units are important

[tiny snow]

@supple echo Did you use angles from other triangles than the ones I've already drawn?

[young inlet]

both ways

i cant remember if degree -> radians is pi/180 or 180/pi i get mixed up

[supple echo]

think of pi/180 with units as pi rad/180 degrees then u need to multiply it so as to get the correct units

[supple echo]

no only one angle

angle ACB

[young inlet]

i'll try to remember that thank you!!

[tiny snow]

@supple echo Thanks, I'll try the trig way 🙂

[supple echo]

ya i think its pretty straightforward that way

[mint dune]

@wooden pelican did you unsend again

[wooden pelican]

Yes because I am so stupid and looked at the wrong lesson.

[mint dune]

👍

[wooden pelican]

I sat there for 10 minutes trying to use my limited physics knowledge to translate the logic

I'm the perfect example of trying hard in the wrong places

[wooden pelican]

https://puu.sh/HoYcs/40285b09ab.png

@mint dune Can you help me with this one?

https://puu.sh/HoYcC/0da86188ba.png

So it's just sqr 288?

https://puu.sh/HoYcs/40285b09ab.png

[mint dune]

sorry bruh im busy helping like 2-3 others

if your thing is still unsolved, ill help

[wooden pelican]

okay ty

[versed river]

it is sqrt(288). depending on whatever the software you're doing this on is it might want you to simplify the radical

[mint dune]

this is incorrect

@wooden pelican here, a = 12 and b=12i, can you tell me what a^2 + b^2 would be

[versed river]

...

mate a and b are the real and imaginary parts

of the complex number

its clearly looking for the magnitude

i mean tell them to submit 0 if you want lmao

but you answer to them when they get it wrong

[mint dune]

oh oops i wasnt paying attention

sorry this is me being extremely stupid

im extremely non-knowledgeable about anything complex

[wooden pelican]

I'm having a lot of trouble with this can somebody help

heres notes

https://puu.sh/HoYBe/60eb1229f1.png

https://puu.sh/HoYBs/1352f2eef4.png

[versed river]

you suggested an answer and i told you it was right.

[versed river]

.

[wooden pelican]

Pasted the wrong question sorry

https://puu.sh/HoYC7/e380348a61.png

r = 16.97 or sqr 288 as we know

angle in degrees is 45 i believe

[versed river]

that's correct, but i'd guess it wants radians if ti doesn't explcitly say

so ig it wants you to write r(cos(theta)+isin(theta)) then

theta being 45 degrees in radians

[wooden pelican]

Thank you I just figured it out

I had to think simpler

I was trying to solve and derive everything lol

I realized the question is just asking for form

Yeah same answer ase you

thanks for verigiyng

z = 16.97(cos theta + i sin theta)

im working on the inverse atm

z = 16.97(cos 45 degrees+ i sin 45 degrees)

from tan = b/a and referencing unit circle

[versed river]

I'll leave whether it will accept an approximate answer and whether it wants degrees or radians up to you since idk anything about what the software you're submitting your answer to, but that's correct and reasoning's all good.

[wooden pelican]

I know it is but i am just being transparent in case others need help

That way maybe it can lead to a discussion but idk i am new here

I genuinely enjoy math i am just rushed for time

I'm giving up for now, gonna think about it tomorrow

My hobby is math I literally take it for fun and wish to master my community college

I mean the fact i can use it one day for a degree is cool too

I enjoy it due to physics though

My patience is shit for theory

Maybe I should just learn on my own and give up school, it's too fast paced

nah its fine ill save the energy when it gets hard

[fast pulsar]

I absolutely encourage for you to learn on your own, and I strongly discourage for you to give up school.

[tired pine]

^

If anything, most high schools discuss math pretty slowly

To make sure students get the topic

but then again, what is “fast” and what is “slow”

[reef parrot]

can someone help me analyze if this is whether true or false
if two angles are not vertical, then they are not congruent

[pure cape]

take in the case of parallel lines

and think about the corresponding angles

they are not vertical angles, but are still congruent

[reef parrot]

what is congruent basically

[pure cape]

congruent is equal

or well, technically it means identical in shape/form

basically equal

[reef parrot]

its just two angles that share a common vertex?

[pure cape]

what is?

congruent? or vertical angles?

[reef parrot]

congruent angles

[pure cape]

no

congruent angles are angles that share an identical shape/form

they are identical to each other

which makes them equal to each other

for example in this picture

these are not vertical angles

they are corresponding angles of two parallel lines

which makes the equal

you can prove this by using similiar triangles/shapes

or by lots of other methods like parallelograms and stuff

[reef parrot]

i kinda get the congruent angles

what about vertical angles

is the letter Z for example, are vertical angles?

[pure cape]

no

vertical angles is just a special case where two angles area congruent when they share the same vertex

[reef parrot]

is letter Y one?

[pure cape]

what is the Z and Y letter you are talking about?

[reef parrot]

the shape

[pure cape]

huh?

[reef parrot]

the literal letter Y

its form

[pure cape]

i really dont know what you are talking about....could you draw it out?

[reef parrot]

is it or nahh

frick, im college yet i dont remember these kind of stuff

[pure cape]

no...

see how those two angles lie on the same line

[reef parrot]

here, what do you call this

[pure cape]

hence making them supplementary (adding up to 180deg/pi rad) to each other (i think? i dont know if thats exactly how its called in english)

[pure cape]

two angles made by two rays?

[reef parrot]

umm, yes?

[pure cape]

so....

[reef parrot]

sooo?

[pure cape]

i dont know what you would expect tbh

it's just two angles

[reef parrot]

is that a vertical angle?

[pure cape]

like i said, and you said

[reef parrot]

im losing my mind, i dont know what to think

[pure cape]

vertical angles are two angles that share the same vertex

im not sure what you are confused at here...

[reef parrot]

ok so i guess this is a vertical angle

omg i think i need to re-learn thin

[pure cape]

these are pairs of vertical angles

[reef parrot]

ill read math is fun

[pure cape]

it shares the same vertex

which is the point in the middle

[reef parrot]

oh yeahh

ok ok ok

[pure cape]

alright then

[reef parrot]

if two angles are not vertical, then they are not congruent

wait

ill relearn this

[pure cape]

no...if two angles are not vertical, they are not vertical

[reef parrot]

no, thats the statement we need to determine if they are true or false

this is false then haha

its alright, i already knew the answer a while ago

what about this
if two angles are not cogruent, then they are not vertical

this is false right?

because not all verticals need to be congruent

[pure cape]

then that is true

[pure cape]

uhh all vertical angles implies a congruence

but a congruence DOES NOT implies a vertical

look, this is what you are getting it wrong at

the statement "if two angles are vertical, they are congruent" statement is one-way, not sure how to call it in english

[reef parrot]

yeah i already said it

[pure cape]

but it means that if the hypothesis is true, the result is true

but does not work the way around

[pure cape]

no, this is wrong

ALL vertical angles are congruent

because of the statement i have said

[reef parrot]

ohhh mb mb

[pure cape]

but NOT ALL CONGRUENT angles are vertical

[reef parrot]

i need to restate it

[pure cape]

showing that the statement is only true one way

[reef parrot]

ok ok

[pure cape]

what i meant by this is that there is nothing special that we can conclude about two non-vertical angles except for them not being vertical

[reef parrot]

if two angles are not cogruent, then they are not vertical
from that, can i use this as an example?

[pure cape]

im not sure because thats some really bad representation

you should just draw two different angles

on a line

for example

[reef parrot]

that is an example of what

congruent angles?

am i right

[pure cape]

of two non-congruent angle...

your statement

[reef parrot]

omg

i am now questionning why am i in college

[peak mesa]

How can i get the blue vector between two orange vectors but references to a green vector

ik that will be by a triangle but idk how

maybe i need to find the distance between the green dot and the brown line and with that calculate something

[peak mesa]

my goal is to find this distance

[fast pulsar]

I am very confused on your question. Can you try rewording it please?

[peak mesa]

i have a line and i have an external point, i want to get the intersection(90º) of the line and get the distance between one point or another

[fast pulsar]

Is there any other information we know? At the moment, I don't think we can conclude anything with just that info.

[peak mesa]

i want to get the height

between the intersection of a point in a line

but a line made by 2 vectors

[fast pulsar]

We don't know any angles, or lengths of the other sides, so we can't conclude anything

[peak mesa]

we know that lengths

[fast pulsar]

We know lengths exist, do we actually know what they are?

[peak mesa]

yes

the problem is that the triangle can have another variation

[fast pulsar]

I agree. Please label the information we know.

If we know a side, label it

If we know an angle, label it

[peak mesa]

all distance from the 3 vecotrs

[fast pulsar]

Otherwise, there are infinitely many possibilities and the question is meaningless

[peak mesa]

we know the lengths from red/brown lines

its vectors (x,y) so we can know the distance between them

[fast pulsar]

Please label what we know

[peak mesa]

label?

i already said the lengths that we have

the values of the lengths is abstract in this case ofc

[fast pulsar]

As in label them in the picture

[peak mesa]

this?

[fast pulsar]

Do you know trigonometry?

Also, if you have $\vec{DE}$ the height is explicitly given to you

[somber coyote]

dackid

[peak mesa]

if i dont know GE and i want to calculate GD how i can have DE

but can apply for different cases

if DG in that example above is inside the triangle or outside

i want to get the height to calculate the vector by the intersection from 90º

idk if its a better way

[fast pulsar]

My brain's having a hard time right now. Not sure if I can help you at the moment.

[upper karma]

Someone please help

[narrow plinth]

I think it will just be (PQ+SR)/2

And that's 48

[fast pulsar]

I don't know exactly, but I am not comfortabke saying that is it

[tired pine]

since this is an american trapezoid (the one with one pair of parallels)

Then yes that is it

It wouldn’t be the case for a british trapezoid, which has no parallel sides

btw why do americans and british still not agree with each other on what is a trapezoid and what is a trapezium
Every single international problem with either, I have to think which trapezoid is which

[honest jolt]

who knows lol

[fast pulsar]

Wait... It says midsegment

[upper karma]

Pls help again

😔

[summer relic]

180-112 I believe

[fast pulsar]

It helps to draw an identical parallelogram right next to that one :)

[upper karma]

I got BD is congruent to DE

what's after

I forgot how to do

<@&286206848099549185>

[toxic trench]

use SSS

[upper karma]

how do I get AD to be congruent to CE

that's the only side left

or BE is congruent to BE

I'm trying to clutch 14 assignments in 3 days in math

and 5 in history

I will be here alot

and some reference is 1 assignment usually takes a night so I gotta be efficient

[toxic trench]

uff

use parallelograms as proof

[unborn python]

Find the quadrants in which the points ( other than the origin ) lie whose abscissa is
equal to the additive inverse of ordinate.

Find the quadrants in which the points ( other than the origin ) lie whose abscissa is
equal to the additive inverse of ordinate.

pls help

<@&286206848099549185>

[fast pulsar]

Please do not ping right away

[unborn python]

apologies

but i need help

[candid tartan]

could I get bearings questions for the following topics
Right angle triangle
Cosine rule
Sine rule
area of the triangle.

[last perch]

https://cdn.discordapp.com/attachments/798391517923770368/820872383690440715/unknown.png

https://cdn.discordapp.com/attachments/798391517923770368/820872464157638656/unknown.png

https://cdn.discordapp.com/attachments/798391517923770368/820875156447821864/unknown.png

https://cdn.discordapp.com/attachments/798391517923770368/820877085241376828/unknown.png

[fast pulsar]

One question at a time mate. We aren't just gonna do your entire homework for you

[molten rivet]

,w 4^6

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ShatteredSunlight#7267

[molten rivet]

You can try all the possibilities to be exhaustive

[fast pulsar]

That is a bit inefficient 😆

[molten rivet]

Wait it's only 3 questions

[last perch]

yea the first question was

just a few options

in 2 screenshots

sorry

[molten rivet]

If you really need help, you should ask directly what you don't understand rather than copy pasting screenshots and just leaving it

[last perch]

its for my friend who is just overall kinda struggling with this thats all and cant really figure it out

and shes pretty shy so she wanted me 2 go and ask

[molten rivet]

Ask your friend exactly what is confusing.

There is a lot of elimination that can happen from the lack of certain information for some of the questions above

That would be a good step forward

[cold siren]

hi can someone explain this to me?

[maiden coyote]

can anyone please help me with my test? its on inverse of trig functions and compositions of trig functions

[cold siren]

<@&286206848099549185>

[subtle jacinth]

@cold siren I will try

[cold siren]

ion think u can ask for help on tests

[cold siren]

ok ty, if you get it explain it to me because i have a lot of problems like that

[subtle jacinth]

Ok

Repost picture

[maiden coyote]

can someone please help me with this

[subtle jacinth]

So that's not a midpoint @cold siren correct?

[cold siren]

i dont think its a midpoint

or if it is its not given so idk

[subtle jacinth]

You need to be clear on that as if it is a midpoint then the triangles would be isosceles

[cold siren]

ya i dont think its a midpoint

[subtle jacinth]

Ok

Both have same height y

[cold siren]

uhh

[subtle jacinth]

hard

[vapid stag]

uhh

SOH CAH TOA!

yay

[subtle jacinth]

no angles

[sacred matrix]

Hello, anyone can link me up to some documentation or something that explains this? I can't wrap my head around the fact that D is correct answer. Because isn't D and E period the same in terms of value? which means that their periods are the same? Anyway thank you for any help!

[toxic trench]

no 2pi and pi are half a period apart

[sacred matrix]

I am confused now a bit, but how do you get to the fact that it's the smallest period? Yea they are half a period apart but isn't the expression 2*x? I am definitely missing something but can't figure out what

[lethal mist]

Let the length of third side of big triangle be l then x²=l²-21²-14²
And l²+14²=(21+x)²

[silent plank]

the period would be 2pi/2 = pi
it is true that the function would also be 2pi periodic, but pi is smaller than 2pi.

[sacred matrix]

@silent plank do you mean that D has smaller period than E just because its 2pi + pi/4 compared to 4pi + pi/4? Because I imagine it being the same as sine begins at 2 pi the same way it begins in 4 pi so + pi/4 is same period for both of them, that's what I am thinking. I can't figure out where am I thinking wrong :/

[silent plank]

where's 2pi and 4pi coming from

from transformations, the period is
2pi/b (where b is the coefficient of x)
in this case, b=2, hence period = 2pi/2 = pi

[maiden coyote]

help please

[sacred matrix]

@silent plank okay I get that part that from 2*x in the argument we can calculate period which is like 2pi/2 or pi/2 or (pi/4)/2 but wouldn't that result in getting that A is the smallest period? like (pi/4)/(2/1) = pi/8 which is the smallest?

[silent plank]

why are you dividing pi/4 by 2/1

2pi isn't a random value being used.

[sacred matrix]

in A we have pi/4 hence I would do (pi/4) / 2 to get the period as with the example you did. I think I am getting it all wrong.

[silent plank]

yeh, you're getting it wrong

from transformations, the period is
2pi/b (where b is the coefficient of x)
in this case, b=2, hence period = 2pi/2 = pi

that has nothing to do with the options in your multiple choice

for a trig sine function:
$$f(x) =A\sin(Bx-C)+D$$ \
the period is: $\frac{2\pi}{|B|}$

[somber coyote]

ℝamonov

[bleak stag]

A sin lmao

[sacred matrix]

I think I have some kind of weird idea of what's going on which is why I don't understand how to do it. I will look into this more and what you wrote! Thank you for your time and help much apprecited

[bleak stag]

this is christian mathematics

[feral idol]

can anyone help

[silent plank]

look up transformations of a trig function

do you know your trig ratios / soh cah toa?

[feral idol]

kinda

[silent plank]

wdym by kinda?

[feral idol]

sin cos and tos?

[silent plank]

what's tos

[feral idol]

i mean tan

[silent plank]

yeh, do you know that those do in a right triangle?

[feral idol]

not really

[silent plank]

have you ever been introduced to a mnemonic like SOH CAH TOA?

[feral idol]

ye

[silent plank]

do you know what it represents and
can you apply that here?

[feral idol]

my assingment is that i gotta do the radios

for

sin cos tan

on the diagram

she showed

[silent plank]

yes..that part is clear...

also doesn't answer the question i posed

[feral idol]

shi

i got 20 mins left eeeeighhhhhhhh

i got

B AND A

54

and

12

623

is it right

[silent plank]

wdym by B and A

and 54 and 12

623 would be correct for Q6 though

[feral idol]

forget the 54

i meant 12

[silent plank]

no, that doesn't seem right

don't multipost. the answer posted in the other channel was fine

[agile tree]

okay sorry sir.

[sacred matrix]

@silent plank is this the right answer as to why the D is the smallest? or completely not the point?

[silent plank]

your colouring is unclear

[sacred matrix]

I mean the from Pi its the shortest to complete the period

[silent plank]

what are A,B,C,D,E supposed to represent

[sacred matrix]

A,B,C,D,E are the value of options in my example up

[silent plank]

also that isn't even the trig function in your question

[sacred matrix]

A,B,C,D,E are represented here

[silent plank]

also that isn't even the trig function in your question

[sacred matrix]

Ohh wait, am I in the wrong chancel? If so I am sorry

[silent plank]

ur in the right channel

you've graphed y=sin(x)

instead of y=sin(2x + pi/4)

[sacred matrix]

Yeap I see the problem now my bad.. I got caught in the moment and thought I had it :/ Thank you gonna think about it again

[feral idol]

hi

could anyone help rq

[olive cove]

Is (b / cos(a)) * sin(a) = b * tan(a) true?

[honest jolt]

ye

[olive cove]

Ok

[grizzled talon]

<@&286206848099549185>

[mint dune]

what are the options

[unreal abyss]

This is for practice does anyone know how to find Angle 4 ,9, and BA I'm a bit stumped

[heavy flume]

I ve been stuck on this for an hour

Need help ASAP

[humble pulsar]

draw a diagram, then apply trig rules/laws

[chilly narwhal]

Hello I've been stuck on these for hours and I need help

1Find the value of y if the measures of the interior angles of a convex decagon are 5yº, 5yº, 8yº, 8yº, 9yº, 9yº, 9yº, 9yº, 9yº, and 9yº.​

2.For a regular pentagon, the area is 336 mm^2 and the apothem is 9.6 mm. Find the side length, in millimeters of the pentagon. Round your answer to the nearest whole number​

3.A quadrilateral was graphed on a coordinate grid and then was rotated 180° counterclockwise with the origin as the center of rotation. If one of the vertices of the original quadrilateral was located at (1,-1), what is the ordered pair of this vertex of the new quadrilateral after the transformation?​

Thank you in regards

[somber coyote]

Dimartiz

[vapid stag]

how to tell what quadrant cos theta can be in?

[onyx moth]

https://www.youtube.com/watch?v=aZzTFiY444o&t=0s

[humble pulsar]

if cos is positive, then quadrants 1, 4
negative is quadrants 2,3
(Look up CAST rule)

[vapid stag]

yesssirr

already got it

[vocal lodge]

for this question

i got

4a times the square root of 2

is that right?

[tiny snow]

Looks correct @vocal lodge

[vocal lodge]

cool thanks

[feral idol]

can anyone help me with khan academy

<@&286206848099549185>

[next flower]

hey can anyone help me with a quick question on my practice

[lethal sigil]

@next flower sure

[summer relic]

2 to the right

and should be 5 down

just count the boxes for that one

[celest tangle]

https://cdn.discordapp.com/attachments/373217754666500097/821458712451219506/20210316_150334.jpg

is there a way to calculate B's coordinate relative to the Cartesian plane and not relative to the line?

[tired pine]

given the fact that you have a distance and you can calculate a slope

i'd say yes

based on the figure there's one answer, though I believe there would initially be two

[inland rock]

Can anyone help me with this question? I have no idea how to do it: https://i.imgur.com/ZMsRGzI.png

[trim breach]

What formula have you considered using? Do you know what those two lines are?

Oh, just realized it’s a late reply by twelve hours.

😬

[agile owl]

Does some one know how to do this question

[pure cape]

Use trigonometry

To find BD

And then from BD use trig to find h

Hint: ||tangent of given angle||

[agile owl]

The answer I got was 43 ( 3 Sig fig )

[humble pulsar]

not sure how 43 is 3sf

[agile owl]

43.0

You get what I meant

[gleaming shard]

anyone know how to do this

i am dead inside rn

i am in need of assistance

idk i suck at it so i need help

oof

[humble pulsar]

what have you tried?

[upper karma]

Any advice for how to optimize mathematics practice?

Have aced chemistry and physics so far, but not sure what the best way to approach self-study for math is (highest level pre-University, then going on to Calculus).

[limpid jetty]

anyone know how i would solve a problem like this? tag me if u know

[trim breach]

BD is an altitude of triangle ABC, if that helps.

I am not sure what theorems you have learned, but there is one where knowing that fact lets you set up a nice proportion to solve for AB.

[mental rapids]

can anybody show me how to do this through factoring

[silent plank]

get terms on one side
identify common factors

[mental rapids]

Yes i did that I got

[silent plank]

what do you have atm

[mental rapids]

0= tan(x) 0=-sqrt(2) csc(x)

[silent plank]

what did you have before that?
you havent factored correctly

[mental rapids]

-tan(x)=tan(x) -sqrt(2) tan(x) csc(x) I added the - tan(x) to the other side and then factored so it was 0= tan(x)(-sqrt(2) csc(x))

[silent plank]

I added the - tan(x) to the other side
what did you have after this

actually thats poor wording too

and bad

[mental rapids]

so after that it was 0= tan(x) - sqrt(2) tan(x) csc(x) + tan(x)

[silent plank]

you should be adding tan(x) to both sides

[mental rapids]

yeah i meant to both sides

[silent plank]

what do you have after simplifying that?

[mental rapids]

I factored it and got 0=tanx(-sqrt(2)csc(x))

[silent plank]

yeh, u stil aren't factoring correctly

where'd the 2 go

$ab - ac \neq a(-c)$

[somber coyote]

ℝamonov

[mental rapids]

so it would be 2tanx(-sqrt(2)csc(x) ?

[silent plank]

no

ab-ac = a(b-c)

[mental rapids]

im lost what should it be when i factor it then

[silent plank]

ab-ac = a(b-c)
apply that

[mental rapids]

I don't understand what would the a b and c in this equation be

when there is multiple tan(x)

[silent plank]

a would be the common factor in both terms

as evident in ab-ac

b and c are what its being multiplied to

[mental rapids]

so A would be tan(x) right

[silent plank]

lowercase a but yes

[mental rapids]

b would be -sqrt(2) ?

[silent plank]

no

$\underbrace{\blue{2}}{b}\underbrace{\red{\tan(x)}}{a} - \underbrace{\red{\tan(x)}}{a} \cdot \underbrace{\green{\sqrt{2}\sec(x)}}{c}$

[somber coyote]

ℝamonov

[mental rapids]

im sorry i don't understand where the 2 is coming from

[silent plank]

you added tan(x) to both sides of the equation

tan(x)+tan(x) = ?

[mental rapids]

ah ok

i see now i was being dumb

[silent plank]

0= tan(x) - sqrt(2) tan(x) csc(x) + tan(x)
what happens when you simplify the right side

fml pls don't tell me you erased those tan(x) from exsistence

and were looking at
0 = - sqrt(2) tan(x) csc(x) the whole time

[mental rapids]

so the factoring would look like

[silent plank]

which i guess may have had something to do with

I added the - tan(x) to the other side

[mental rapids]

0= tan(x)(tan(x)-sqrt(2)csc(x))

[silent plank]

no

[mental rapids]

fuss

[silent plank]

2tan(x) represents 2 multiples of tan(x)

[mental rapids]

2(-sqrt(2)csc(x))

[silent plank]

NOT tan(x) multiples of tan(x)

now your tan is gone completely

lets consider a simpler example

5🍎 - 3🍎 = ?

[mental rapids]

no i mistyped

tan(x)(2-sqrt(2)csc(x))=-

[silent plank]

= stray - sign?

[mental rapids]

0

sry

[silent plank]

ok. that's better

and then zero product property as usual

[mental rapids]

so tan(x)=0 would be x= undefined

[silent plank]

no

there would be infinitely many values of x where tan(x) = 0
trivially x=0 will be one of them

[mental rapids]

-sqrt(2) csc(x)=0 subtract 2 from both sides

-sqrt(2)csc(x)= -2

the 1st one should've been 2-sqrt(2) csc(x)=0

and then divide -sqrt(2) by both sides

csc(x) = -2/-sqrt(2)

cos(x)= sqrt(2)/2

[zinc scaffold]

just a general question- did anybody had a rough time with verifying trigonometric identities when learning it for the first time?

[humble pulsar]

trig identities are always a tricky concept cause it's (typically) first introduction to proof writing/logic

[sturdy stream]

can someone please help me with this question

[trim breach]

Think about the pyramid in the smaller triangles you can make.

Specifically, start with triangle PFT knowing that one leg is 35.

You should be able to solve that triangle completely with the given information.

Then, see what else you can solve with that newly solved triangle.

[upper karma]

Is this correct

[nocturne thicket]

Using pythagorean theorem I got x = 2

but using sines i got x= 4Rad2???

[upper karma]

show your work

[nocturne thicket]

x^2+x^2=8,
2x^2 = 8
x^2 = 4
x = 2

Sin 45 = x/8
8(Sin 45) = x
x = 5.65685

OH my god

@upper karma I did 8 instead of 8^2 I see now

thanks

[upper karma]

orz

[nocturne thicket]

orz?

[upper karma]

nice

[nocturne thicket]

👍

[steel hull]

hi there any body knows analytical solution of this equation

,w A sin(x)+B cos(x) = C

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by alirad#9692

[dark sparrow]

do you want all the edge cases that can arise here

[steel hull]

no i want know how can i solve it

A sin(x)+B cos(x) = C

A,B,C=Const

[dark sparrow]

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

are you familiar w/ this?

[steel hull]

wow tanx a lot ann

[dark sparrow]

oh, so i guess this alone was enough

alright

[steel hull]

ya tank u

[upper karma]

anyone where i can learn about section formula,locus,equation of straight line, area of triangle and quadrileteral or just coordinate geometry

i am in grade 9

[neon hamlet]

How would you start solving this?

[modern estuary]

if you are giving the area of a sector in terms of pi would you give the answer as a fraction or decimal

also is this correct <@&286206848099549185>

[prime linden]

decimals are cringe, fractions ftw

[humble pulsar]

fractions

[modern estuary]

is it proper to write as a decimal

ok

and is my answer correct

i just need it checked

[humble pulsar]

yes, cause it's a 3rd of the circle

$\frac{\pi r^2}{3}$

[somber coyote]

moshill1

[round prism]

How would I do this problem?

[modern estuary]

also is this correct

<@&286206848099549185>

[prime linden]

,w (4*1/2)^2pi * 120/360
Have I gone insane?

[humble pulsar]

120/360 = 1/3

[prime linden]

,w (4*1/2)^2pi * 120/360

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by rts#2928

[prime linden]

he wrote 6.75pi

[humble pulsar]

4(1/2) isnt 4*1/2

4(1/2)=4.5

[modern estuary]

wait so was the answer correct?

[prime linden]

ye

[modern estuary]

ok

[prime linden]

i was dumb dumb

[modern estuary]

lol

[prime linden]

You can use wolfram you know
,w 65 degrees to radians

stupid bot

,w 65 degrees to radians

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by rts#2928

[modern estuary]

is it correct

[prime linden]

is this a test?

[modern estuary]

No its HW

but i have a B- in geometry rn and i need to get it up and i usually go over my answers with my sister or dad but they are both gone

[prime linden]

Ok, so in the unit circle

what does the arc length represent?

When the radius is 1 unit length

and I say that the arc length is 2 radians

what does 2 radians mean?

[prime linden]

Ok nvm perhaps radians is too early. I just wanted to say that I got a different answer

So i am trying to understand how you reason

[nocturne thicket]

How exactly do I solve this?

[silent plank]

mutliple applications of tan

[nocturne thicket]

Ohh i forgot about Tan

Tan 15 = 200/x
x+250
tan x = 200/250+x?

[silent plank]

umm...

don't use the same variable to represent different things

x is already defined to be an unknwon angle

don't use it again to represent and unknown length

[nocturne thicket]

oh i see thanks

[vague orbit]

What is the tangent of (-√3/2, -1/2)?

[vague orbit]

See

nobody here good at math

all frauds

[humble pulsar]

so sorry you lack patience

what's the quotient identity for tan?

[mental rapids]

where would tangent equal 0 on the unit circle?

[mint dune]

when the opposite side O is zero, tan = O/A will also be zero
there are two points on the unit circle when the opposite side becomes zero

[mental rapids]

would the raidans of those sides be pi/2 and 3pi/2?

[mint dune]

you seem to be confusing adjacent and opposite.
at pi/2 and 3pi/2, the adjacent side will be zero, causing tan to become infinity

[humble pulsar]

tanx=0 when sinx=0

since tanx=sinx/cosx

[mental rapids]

Oh ok so it would be pi and 0 as radians right?

[mint dune]

yes it would

[mental rapids]

ok ty finding the funcions on unit circle always confuse me

[mint dune]

lol np

[upper karma]

what would Given g(x)=−x+3, find g(1). be?

-1+3 is 2 right?

[mint dune]

same thing as earlier. replace x with 1

yes

[humble pulsar]

yes

[upper karma]

oops

LOL

this one different

Given f(x)=3x−5, solve for x when f(x) = 1

[mint dune]

instead of replacing x with 1, replace f(x) with 1 and find x

[upper karma]

so 3 times 1?

[mint dune]

if f(x) = 3x - 5, and f(x) = 1

then
1 = 3x - 5

[upper karma]

so what would the answer be?

[mint dune]

you can work this out yourself. its a simple linear equation

[mental rapids]

add 5 to both sides

[upper karma]

OH

so thats 6

[mental rapids]

it would be 6=3x

[upper karma]

6 divide by 3?

[mental rapids]

yup

then x is by itself

[upper karma]

is 2?

[mental rapids]

yup x=2

[upper karma]

sorry im doing 2 different subject at the same time

[mental rapids]

the goal of those equations is to try and isolate the X on one side

so always try to get x by itself

[upper karma]

We're moving too fast on math

Given h(x)==−3x+2, solve for x when h(x) = -4

subtract 2 on both side?

[mint dune]

yes

[upper karma]

=-6 then divide

oops did that wrong

[copper marten]

how do i find the area? i figured out the rest of the angles but i’m not sure what to do from there. (sorry my trapezoid is very crooked)

[silent plank]

apply symmetry and use trig to determine the altitude

[upper karma]

can someone walk me through this?

Given h(x)==−3x+2, solve for x when h(x) = -4

[copper marten]

wouldn’t that be -4 = -3x+2 because u plug it in and then isolate x

[upper karma]

uh

[copper marten]

tyty

[upper karma]

confusion

[copper marten]

ok um so h(x) equals -4 right?

[upper karma]

yes

[copper marten]

so then replace h(x) with -4

[upper karma]

which I did

[copper marten]

that gives u -4 = -3x+2

then u want to isolate x

[upper karma]

-3(-4) would be a positive

[silent plank]

h(x) = -4 doesn't mean x=-4

[upper karma]

oh

[silent plank]

x is what you're trying to find

[copper marten]

yeah ^

[upper karma]

mb