geometry-and-trigonometry

[eager kraken]

deg

[rich wolf]

what are you doing with that exponent

[eager kraken]

(i would tell by if the units of the numbers are in the degree rather than the pi symbols?)

[rich wolf]

not necessarily

you can have 1.3 radians for instance

[eager kraken]

Well to find sin(t)=0.66862229
I just did the sin^-1 (0.66862229)

[rich wolf]

degrees usually have the degree symbol and radians are naked numbers

[eager kraken]

ok

there are degree symbols for this one

He hasn't really taught us radians yet

[rich wolf]

ok

your algebra is wrong i'm guessing

sec(t)=1.5625809

1/cos(t)=1.5625809

1=1.5625809*cos(t)

1/1.5625809=cos(t)

t=cos^-1(1/1.5625809)

[eager kraken]

t = 50.21065124

YES

thank you

[twin swan]

sinx(2cosx+1)=0

can anyone help me with this????

all i know is to distribute

but after that idk

like its 2cosxsinx+sinx = 0

[thorn talon]

it's almost certainly in your best interest to not expand

it's in fully factored = 0 form

[eager kraken]

I dont understand where I am going wrong with this, when i solve side p using the sin function, the answer is .0049516885

I could be wrong but i don't think it's possible to have this right triangle if side p was that short

[thorn talon]

you are right

[eager kraken]

I found n from doing tan(50.3)=n/129

[thorn talon]

you can't

that's fine

[eager kraken]

wait

I just tried p = 129/cos(36.7) and it resulted in 160.8928893

why didn't the sin work tho?

[silent plank]

what were you doing when using sin

[eager kraken]

sin(50.3)/155.381251

[silent plank]

wrong setup

[eager kraken]

my angle M is also wrong

[silent plank]

sin(50.3) = n/p
→ p = ?

[eager kraken]

n/sin

[silent plank]

also, yeh you used the wrong angle for cos

[eager kraken]

oh i see

How do I know which side of the triangle c is?

[silent plank]

by convention, the
lowercase side is opposite the capital angle

i.e side a is opposite angle A
etc

[eager kraken]

im sure glad my teachers never told me that

[rigid siren]

why does cos (2x) cause a horizontal shrink

Ik it shrinks the period

but why

logically

[thorn talon]

You're doubling the argument

It's essentially going twice as fast to complete a period

[merry rapids]

Can anyone help me with this question. I am tasked with finding the length of the unknown side, x, however I can't seem to answer it. I believe it's 21 x cos47 but I am getting the answer, 14.32, whereas the actual answer is 30.79

Thanks

[idle bloom]

You've done your multiplication and solving for x wrong

Remember Cos(theta) = Adjacent/Hypotenuse

Not Hypotenuse/Adjacent

[merry rapids]

So how would i rectify it?

[idle bloom]

Well

What equation are you starting with

[merry rapids]

What do you mean?

[hardy socket]

can some one help me with these

i know theorems n postulates but my brain just can comorehend how

[idle bloom]

Cos(theta) = Adjacent / hypotenuse. So plug in the values you know

[merry rapids]

cos=21/x

[idle bloom]

cos(47) but yes

Now solve for x

[merry rapids]

Wouldn't that be 21 x cos47?

[idle bloom]

No

[merry rapids]

Then I'm not sure, could you explain please?

[idle bloom]

also don't forget your

Cos(47) = 21/x

so I multiply by x

x * cos(47) = 21

Then how do I get x by itself

[merry rapids]

Divide cos47 = 21?

[idle bloom]

Divide both sides by what?

[merry rapids]

x? I'm having a brain fart

[idle bloom]

You have

x * cos(47) = 21

You want x by itself

So what do you divide both sides by to get x by itself

[merry rapids]

cos47 = 21

[idle bloom]

You can't divide by an equation

[merry rapids]

Sorry my bad, cos47

[idle bloom]

Yes

So you now have

x = 21/cos(47)

Does that make sense how we arrived here?

[merry rapids]

Got the answer, Thanks

Yeah

[idle bloom]

Ok good

[merry rapids]

Thanks for the help!

[hardy socket]

guys does anyone know the reasons for this

[twin prawn]

I think we're missing a problem description of some sorts

or yknow maybe a diagram

[somber coyote]

Pa_u_los:

[twin prawn]

law of cosines

[eager kraken]

Is the answer for
sin(t)=sqrt3/2
60 degrees?

And/or is there only one answer for the problem?

Yea

But an acceptable answer for a trig class would be 60

Neat thanks

[upper karma]

i cant do this one

[eager pendant]

@upper karma can you make your username pingable

also try drawing some lines

[upper karma]

it doesn't let me change my nickane m,

k wait

ther

@eager pendant ?

[eager pendant]

try extending a few lines

[upper karma]

@eager pendant do you know how to solve it yourself?

[eager pendant]

yeah

[upper karma]

im looking at the solution rn but I still cfan't undestand it

what did you get as an answer?

[eager pendant]

well i got 2

so i assume it should be 200

[upper karma]

yea that is the answer

[wind heart]

Hello

Would this be considered by Alt Int. Or by Vertical?

[silent plank]

which angles?

[wind heart]

Triangle ABC and EDC

[silent plank]

what did you actually mean by

considered by Alt Int. Or by Vertical?

[wind heart]

Like

In number 6

It shows 2 perpendicular angles

But it’s also a Vertical angle

And you’re supposed to classify what the triangles are congruent by

Oh wait I’m in idiot

It’s the SAS SSS ASA stuff isn’t it

[silent plank]

yes

[wind heart]

ok thanks lol

[upper karma]

im clueless plz help

[cobalt turtle]

,rotate

[upper karma]

👀

[median crown]

those are all perpendicular bisectors no?

so

they make

90 degree angles

...

[storm rover]

I have a problem, "ABCD is a parallelogram, express vector BD in terms of vectors AD and BC"

Not picture or anything

[quiet mason]

do you know the triangle law of vector addition

[storm rover]

no we havent learned that yet

nvm i think i got it

would it be AD-BC)

and that quanitity is divided by two

[somber coyote]

Pa_u_los:

[marble mesa]

https://discordapp.com/channels/268882317391429632/326138757474680852/671358313929244695

[silent plank]

cos(C) > 1...

did you forget to divide by 2 or something?

[marble mesa]

@silent plank Why should I divide by two? The law of cosines is $c^2 = b^2 + c^2 - b \cdot c \cos C$

[somber coyote]

Pa_u_los:

[marble mesa]

Afaik, although I don't understand it.

I don't even understand what a cosine is

[silent plank]

wrong variables

and formula

$c^2 = a^2 + b^2 - {\color{red}{2}} ab \cos(C)$

[marble mesa]

oh my god... Thank s

[somber coyote]

ramonov:

[silent plank]

wtf

[marble mesa]

yeah

wtf

[silent plank]

erase that source

what is it so i know what to avoid it

[marble mesa]

how can I do that?

is the first thing it appears on the internet

dumb google AI search

[silent plank]

i seriously doubt that

also don't round your sqrt(8656) to 93

it will make your calculations for the angle less accurate

[marble mesa]

oh, ok

ok

[silent plank]

maths is fun shows the correct formula

[marble mesa]

yeah, I will click in the first website instead of seeing the first image

[smoky plover]

I’m trying to do angle proofs, can anyone help me on this question?

[warped olive]

,rotate

[brave gust]

Hi

Question.

How do I know the coordinates of pi/4

okay

yes

Yep

[somber coyote]

WhipStreak23:

[brave gust]

yea

tangent 45 degree is 1

[somber coyote]

WhipStreak23:

[brave gust]

ok got it

[somber coyote]

WhipStreak23:

WhipStreak23:

[brave gust]

yes

wait root 1/2

then we get root 2/2 ?

lol

[brave gust]

yea i got it 🙂

[brave gust]

@upper karma Thanks!

[marble mesa]

how can I say that de part that is not in black is 20 cm^3

the total volume is \approx 160.146

cm^3

[upper karma]

Theorem 8.6 states that if a quadrilateral is a parallelogram, then its diagonals bisect each other. Write a two-column proof of Theorem 8.6

[upper karma]

What's the difference between cos and cos^-1? When would I need to use cos^-1 instead of cos?

[wind heart]

Did I do number 2 right

[wind heart]

<@&286206848099549185>

[upper karma]

where is number 2 on this sheet

i see 6-12

[river forge]

presumably the top line

if so yes this is correct

[upper karma]

yes

[wind heart]

Wait I mean like

The Line B one

[river forge]

yes

[wind heart]

Oh ok

[balmy pelican]

@upper karma cos takes an angle and gives you a ratio, cos^-1 takes a ratio and gives you an angle

[marble mesa]

@upper karma $\text{you would need to use cos^-1 to know what's the value of the angle given. i.e.}$ $ 1.55684829 = \cos C$$\text{; then you would have to do} $\cos^{-1} 1.55684829$ \text{which will give you a value meaning your angle}

the thing is that I don't know why that is like that. Ping a helper for solving your question

[eager kraken]

How do I find the lengths of side h and x with this data?

The two formula i have been told are
tan(29)=h/x+10
tan(52)=h/x

[thorn talon]

Sure

[eager kraken]

and then to just plug it in, but when i do that and solve the variables cancel out

[thorn talon]

What's your working out after?

[eager kraken]

so i plug em in like
tan(29)=xtan(52)/x+10

maybe im just doing the math wrong from there

[thorn talon]

How'd you get that?

You can multiply through the first equation by x + 10

And the second by x right?

[eager kraken]

plugging xtan(52) into tan(29)=h/x+10 as the h

[thorn talon]

Ok

Sure

[eager kraken]

10tan(29)=-xtan(29)+xtan(52)

if this is the right direction, i am not sure where to go from here

Can i solve this with similar triangles?

[tacit karma]

dunno

try solving for the hypotenuse using x and h

[eager kraken]

but i dont have x or h

[tacit karma]

no

just

(10+x)^2+h^2=(c1)^2

and plug it in to one of the trigonometric functions

it won't cancel out

[eager kraken]

is there a way if i can check whether or not a side length is correct?

I kind worked it out a different way and got 5.543090515

10tan(29)+ (h/tan(52))tan(29)=h

[sleek vine]

@eager kraken have you learned the sine law?

[eager kraken]

barely

[sleek vine]

what do you mean barely?

like, has it been covered in class and you're not sure how it works, or you've never heard of it?

[eager kraken]

i've heard it mentioned, but looking through my noted we didn't learn it at all

so could i just do (sin(23)/10)=(sin(128)/c1)

to solve for c1?

[sleek vine]

yes that's how i would approach it

and the same works for c2

i'm trying to figure out how they expect you to solve it without the sine law though

[eager kraken]

This is an example from the book

must not be using law of sines because the question is just looking for the lengths of x and h

[eager kraken]

This one is seemingly easier but I cant figure out the length of x

c=18.27804155 y=14.59749304

(ping me)

[high bough]

@eager kraken what have you tried

[upper karma]

if only we knew the length of h

[summer condor]

Hi guys. Can you help me with some tips which can help me with it?

no, i'm not

ok

[somber coyote]

WhipStreak23:

[summer condor]

Oh, thank you

@upper karma , do you know some books or website about geometry?

[summer condor]

no, i don't

[summer condor]

Thank you

[high bough]

@upper karma yes thats me xD

Nice helper role

[upper karma]

how do i calculate the S and P of a triangle

[eager kraken]

@high bough I tried the law of sines as well as the other method of just plugging things in to get that c value, from which I used the pythagorean theorem to get y

[summer spire]

@upper karma what do those letters mean?

[upper karma]

here comes the problem

i dont know

[summer spire]

can't help if we don't know what the question means ¯_(ツ)_/¯

[upper karma]

LIKE THE SIZE OF IT

i dont

ehhhhhh

the rotation i guess?

[summer spire]

rotation of what?

[upper karma]

hold oN

SURFACE AND SCOPE

im just gonna google it

or in other words browse the net

[summer spire]

@upper karma you could always make flashcards

or maybe just do a lot of trig problems

but I don't know of a "least boring" way to memorize that kind of stuff directly

[copper moss]

Hola

Could a french person explain this sentence (in french or english) to me please? Ignore the red

"s'ils dirigent des droites des droites orthogonales"?

It it saying that two vectors are orthogonal if lines parallel to them are orthogonal?

[autumn sky]

heya guys

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

I need to make a line between those points

and also extend that line infinitely

the thing is that I can't figure out how to

Wait, nvm I just did

[surreal roost]

Huh?

[thorn talon]

🤔

[limpid tinsel]

😐

https://tenor.com/view/ight-imma-head-out-spongebob-imma-head-out-ima-head-out-gif-14902967

[turbid lily]

[upper karma]

how many right triangles are there @surreal roost

[surreal roost]

2

[upper karma]

take a deeper look

[knotty merlin]

lemme do this in my head a sec

ok

done i think

[surreal roost]

so there two similar triangles

i think

uh

yeah thats all I got

[knotty merlin]

....

[surreal roost]

🥄

70?

[rich wolf]

They are similar triangles

Which means their sides are in proportion to each other

Which means you must use segment BD to solve

Give it a variable name like y

[gritty sail]

Yes it's 70

[surreal roost]

Well I did the other one, and I started this one. I got that the answer is the square root of 320, but I need to simplify it which should be the easy part, but apparently its not.

[knotty merlin]

Just use surds

[surreal roost]

?

[upper karma]

it looks like you might be able to make a system of equations

[surreal roost]

🙃

[tidal ridge]

this is prob reallhy simple

but

how to find alpha angle?

we're given mic1 at [0,2] and mic2 at [0,0]

so i use sin(alpha)=d/2 and cos(alpha)=d/adj

also sin^2()+cos^() of alpha result in d^2+adj^2 = 4

which doesnt help too much either..

am i missing an identity?

[tidal ridge]

how to find d value?

was thinking could split mic2 angle to 45deg

but that seems wrong?

[cosmic juniper]

https://gyazo.com/01cab3eb1e662d526f0ca505c170e86b

need help

[white silo]

@cosmic juniper still need help?

[cosmic juniper]

Nah

[white silo]

Ok

[rigid saffron]

I need to compare the distance between two points in 3D-space, p_1 and p_2 to some origin in the most efficient way possible. When calculating the distance to a specific origin I obviously don't have to take the square root, but what about squaring each value. If i simply take the absolute value and omit the squaring and square-root in the distance formula will I get a correct comparison?

|p_1.x-origin_x| + |p_1.y-origin_y| + |p_1.z-origin_z|

[queen python]

Second part

Can’t figure out how to make the triangle equilateral

[upper karma]

Alright please help

This is due in like 10 minutes

[idle bloom]

When the bridges are closed, it spans the whole distance across

So find the length of each bridge using trig (45-45-90 triangles)

[upper karma]

Ohhh thanks

Here’s breakfast

Pls help

[silent plank]

since it's asking for exact answer,
you shouldn't be putting cos(45°) into your calculator

(generally you shouldn't be putting that in unless they're specifically asking for a decimal approximation)

[twin prawn]

@queen python you basically get your answer for free if you know how to construct sqrt(6) given a unit line segment

[upper karma]

Ok I found a different way

It will be 80sqrt2 I think

[upper karma]

can someone help me with this problem....... If sin(θ) =-3/5, and θ is in quadrant IV then find, cos(θ), tan(θ), csc(θ), sec(θ), cot(θ)

because i dont get this at all 💀

[silent plank]

draw a (right) triangle in quadrant 4 where sin(θ) =-3/5

[upper karma]

ok

then what lol

[silent plank]

pythag

[upper karma]

so its x^2 + (-3)^2 = 5^2

x^2 + 9 = 25

x^2 = 16

idk how to find the cos or tan or the others tho...

[silent plank]

what's your x?

and what are the definitions of those trig functions

[upper karma]

x would = 4 right ?

give me a second I’ll screenshot

This is what it's asking

[silent plank]

can you show the triangle you drew?

[silent plank]

nope, that's not quite right

[rich wolf]

Very nice drawing

[upper karma]

lmao

how would it look like then ???

[silent plank]

well first you need to label your angle

what's your definition of sin?

[upper karma]

isn't opposite / hypotenuse

[rich wolf]

Sure

Wait

You should actually know this to be a special sort of right triangle

If you use pythagorean theorem you will see why

[silent plank]

sides need to be labelled correctly first

[rich wolf]

True lol

[eager kraken]

Do I pretty much have to memorize the pythagorean identities?

[thorn talon]

I mean

You need to know one

Which follows from well

The Pythagorean theorem

The rest you can derive easily

[eager kraken]

If I am finding the 6 trig functions derived from a line when the only values given are an equation for a line, should I just pick a point on the line to find the trig values?

Otherwise I am not sure how to find the 6 trig values from an equation of a line

[thorn talon]

🤔

what's the context

[zealous sierra]

hi! does someone know what’s the best way to solve this? i only have to use tan, and you don’t have to show me calculations because that would take a lot of your time, but just a push in the right direction would be great! :)

[summer spire]

just start finding the values of missing parts

eventually you'll come up with enough info to find D

you can start with finding AB and AF

every little bit will help

[gloomy frigate]

I have a question related to the transformation of position and rotation of objects on a 3d grid. Would this be a good place to ask?

Well, the problem is this: I have a bookshelf with items contained on the shelves. If the bookshelf is moved and rotated, what formula can I use to transform the position and rotation of the items on the shelves?

[summer spire]

I'm guessing you'd have to find each item's offset from whatever the point of rotation of the shelf is and rotate around that, if it's a simple rotate or translate, at least

but if you're talking a bookshelf tumbling down stairs and keeping track of each item, you'd likely need a physics engine at that point

[jagged holly]

https://media.discordapp.net/attachments/411010725956550657/672834955356667915/Screenshot_3.png?width=960&height=224

can someone help me solve this problem i figured out what sin was but i cant figure out cos

[marble topaz]

Draw yourself a lil unit circle and a diagram

You've already got two sides out of three, you should be able to find the other stuff just fine

[olive solstice]

Can anyone help me with math?

[quiet mason]

@upper karma is a god at geo and can help you

[olive solstice]

It's highschool level of math 🙂

Thanks ill ask him

[mint bronze]

I dont understand how to convert degrees to radians, my math book has an explanation but their explanation is extremely generic, can I get some help with this?

[devout harbor]

You can just use ratio method, or that's what I call it.

I always use this instead of just memorizing

[mint bronze]

@devout harbor I dont understand it.

This is what they are showing and explaining and I don't understand or see the pattern of steps that they took. there is nothing else to explain it either

[dire rampart]

there are 2pi radians in 360 degrees

so each degree is pi/180 radians

[devout harbor]

Just look at what I did.

That's basically it.

Or what lemon said.

[mint bronze]

I'm gonna try to read this back, just let me know if I'm wrong or something. 360degrees = 2pi rad, 45degrees = pi rad, 360 * pi = 2pi (45), pi = 90pi/360. At this point you would simplifiy it?

[vague pagoda]

how is 45 degrees = pi rad

[mint bronze]

the 45degrees is what the problem is, to convert to pi rad

[vague pagoda]

multiply 45 by pi/180 radians

and u get your answer

[mint bronze]

Okay Thank you.

[vague pagoda]

@mint bronze also the photo had 45 degrees = x rad i think

not pi rad

[mint bronze]

It was hard for me to see lol

[vague pagoda]

yeah was for me too

[mint bronze]

x makes more sense though

[devout harbor]

Yeah because we don't know it, so we keep it x

[mint bronze]

Thanks @devout harbor for the image, it helped

[devout harbor]

np

[upper karma]

find the value of height h

the uppermost quadrilateral is a square btw

[median crown]

tf is that drawing

[upper karma]

Lol

[summer spire]

is quadrilateral connected to that square also a square?

[median crown]

it has 4 equal sides

so it must be a rhombus

but one of those angles is 90 degrees

so the middle quadrilateral must also be a square too

[rich wolf]

This shape is badly drawn

[formal hamlet]

Which scientific calculator do you guys prefer for Trig?

Casio 115-es plus, Casio 991-ex, or TI-36X pro ?

can you guys help me with B) please. I dont understand the relationships correclty and on the right Im trying to find a correct way to list relationships

Which is correct? top or bottom?

[silent plank]

your tables are a bit unclear

are those supposed to be the ratio of sides opposite those angles?

[winged zinc]

@formal hamlet

You are attempting to solve for x and y

Because you have two unknowns, you require two equations

What two equations do you think you could use?

[formal hamlet]

yes the tables are ratios. The bottom is correct I was told

So therefore 16/2=8

90 degrees=2x

so therefore x=8

plug in 8 square root 3=Y for 60 degrees

[winged zinc]

Erh, unless I made a mistake, I think y=8 and x = 8sqrt3.

[formal hamlet]

Yeah youre right

[winged zinc]

And that logically makes sense as the angle opposite side y is les than the angle opposite side x

hence y should be less than x

[formal hamlet]

Right right. That's correct. Thank you

[winged zinc]

All good

[formal hamlet]

the way the angle is oriented is tripping me up a little. gotta be more attentive

[winged zinc]

I solved it by creating two equations. $$x^2+y^2=16^2$$ and then the sine rule with $$\frac{x}{sin(60)}=\frac{y}{sin(30)}$$.

[somber coyote]

Lachlan:

[winged zinc]

From there it is just a matter of substituting in values

[formal hamlet]

nice thank you.

[winged zinc]

There are a number of ways you can solve it

But do what ever works and makes sense for you.

[formal hamlet]

well its nice to see other ways to solve it and pick the best one. it also helps to understand the relationships better even if I go with a particular method

thanks again

[winged zinc]

Granted you can just solve it with trig identities

so sin(theta)=O/H

That can get you there

Likely the easiest

[formal hamlet]

what is O and H in this problem?

[winged zinc]

Well, choose some angle in the triangle (preferably not the 90 degrees) and O is the side opposite this angle, and H is the hypotenuse.

So for example, sin(30)=y/16

so y=16 sin(30)

[formal hamlet]

pretty sick bro thank you

[winged zinc]

You ever heard of the saying SOHCAHTOA?

[formal hamlet]

definitely not

[winged zinc]

Haha

It is an acronym for trig ratios

Sin(theta)=opposite/hypotenuse
Cos(theta)=adjacent/hypotenuse
Tan(theta)=opposite/adjacent

[formal hamlet]

perfect. Im saving that.

[winged zinc]

Adjacent is the side which is next to the angle but not the hypotenuse.

[formal hamlet]

noted

[winged zinc]

I you don't mind me asking, what year level are you in? (feel free to not reply)

[formal hamlet]

Im a second semester sophomore at university. I was a business major, but changed to computer science which requires a higher level math accomplishment

[winged zinc]

Ohh nice

[formal hamlet]

yeah definitely more challenging but Im up for it

this is my first time in Trig. I'm gunning for an A.

[winged zinc]

Nice

If you ever need some help, feel free to dm me.

[formal hamlet]

thank you I will. Ill friend you rn

[somber coyote]

DeadEye:

[formal hamlet]

Ignore This^, I cant delete it

[green roost]

Can anyone quickly clear somthing up for me? Is sin4X=2sin2X? That seems logical to me but the question seems to be worth too many Mark's for it to be as simple as that (I calculated sin2X in the previous question)

[devout harbor]

Is the question asking you to solve for x where sin4x=2sin2x?

Because sin4x=2sin2xcos2x in general.

[green roost]

No I was asking if my assumption that sin4X=2sin2X, I've just to find the exact value of sin4X given that cosX=4/5 but that's fine I know what I'm doing now

[tardy comet]

Does a diagonal split an angle in 2 identical pieces inside a parallelogram

[upper karma]

can someone explain how to find the measure of angle B?

[marble topaz]

Law of Sines

Do you know what that is?

[upper karma]

mm

well

kinda

but its confusing me

[marble topaz]

What part of it is confusing?

[upper karma]

idk honestly I think if I just go back and practice more with sin and sin^-1 I'll understand it

so I guess I'll just have to do that to be ready for law of sines

[marble topaz]

Alrighty

Lmk if you'd like some more help ^^

[upper karma]

ok thanks

[upper karma]

Yo

If you have a cosine equation and the number in front of x is pi, then would the period be 2pi/pi = 2?

[median crown]

yes

[upper karma]

Alr thx

@small mauve hey 👋

[small mauve]

Hi

[upper karma]

What math class do you have

[small mauve]

calc and lin. alg

but

why

[upper karma]

Oh your in university cool what’s ur major

[small mauve]

math and cs

[upper karma]

Cool

I was just

Curious

[upper karma]

I just started trig this semester. My professor isn't doing the best job explaining trig identities to me, is there a resource that anyone knows of with videos or tutorials that could help me?

Right now, just having trouble verifying identities.

I'm looking for a good book to learn proof-based Geometry/Trigonometry, anyone would recommend any?

[weak shoal]

@upper karma Well, the proofs are available online if you search hard enough

You could just see the proof of one or two of those identities. Then, use those proofs as the basis for proving all the others.

@upper karma I’d recommend Trigonometry by Gelfand and Coordinate Geometry by Eisenhart

[formal hamlet]

Off to a slow start in Trig

But im gonna get this A.

[thorn talon]

[tall flower]

Hey all can someone help me with a trig problem?

[thorn talon]

🤔

[tall flower]

okay here: if "tan a = 0.1327" find "cot a"

[fallen cedar]

1/tan(a) = cot(a)

[tall flower]

oml

thanks

[fallen cedar]

np

[upper karma]

how do I calculate the projection of a leg on a right triangle

i already calculated that the shorter leg is 30 units

[topaz hearth]

Can anyone explain what is going on here? i dont understand how to rearrange like this.

[upper karma]

@topaz hearth They're just doing abc = (a)bc = a(b)c = a(bc) = ab(c) = (abc)

Parentheses don't matter in multiplication at all

[daring python]

Can someone explain this like I'm 5

[topaz hearth]

Thanks!

[vague pagoda]

@daring python use sin and cos to find out them

[daring python]

Yeah, only brushed over trig last semester and now I'm expected to know its ins and outs. I just need a simple explanation of how to do so.

[upper karma]

What does your book or notes say

[upper karma]

@daring python are you aware of the unit circle?

[daring python]

@upper karma Loosely, but not really.

[upper karma]

https://www.mathsisfun.com/algebra/sohcahtoa.html

this should help you get to the answer

focus on the sine and the cosine formulas

[lament brook]

I finished the question and the answer is cos = - sqrt of 37/37 and the csc is 37sqrt37/6. It simply looks extremely wrong. I can show my work if needed. Not sure if im overthinking things or not

[devout harbor]

Okay first of all.

You can think of tan(theta) = -6 / 1

Yes?

[lament brook]

yes

[devout harbor]

Draw a triangle.

Think of -6 as 6 for now.

Find the remaining side using pythg theorem.

[lament brook]

root 37

[devout harbor]

Sin(theta) = 6/sqrt(37), so cosec(theta) = sqrt(37)/6. Yes?

@lament brook

[lament brook]

yes

[devout harbor]

Now cos(theta) = 1/sqrt(37), so sec(theta) = sqrt(37).

sec(theta) = sqrt(37) can also be written as 37/sqrt(37)

[lament brook]

ohh

[devout harbor]

I messed up my coding. Anyways.

$\sqrt(a) = \sqrt(a) \cdot \frac{\sqrt(a)}{\sqrt(a)} = \frac{a}{\sqrt(a)}$

[somber coyote]

Sup?:

[devout harbor]

Given that a is not equal to 0 and in the example above, sqrt(37) is definitely not equal to 0.

So we can do this and we have sec(theta) = 37/sqrt(37).

Anyways now.

Do you know in which quadrants sin/cos/tan functions are positive or negative?

[lament brook]

yes i do

since it's in q2 cos is -

[devout harbor]

Yes, so since cos is negative, sec is also negative.

[lament brook]

yeah

[devout harbor]

So we have sec(theta) = -sqrt(37) = -37/sqrt(37).

Sin is positive so cosec is also positive so that remains the same.

And that's it.

[lament brook]

ah

thank you so much

[devout harbor]

np

[pliant karma]

so uh

could i get some help with a geometry question

:^)

In triangles GHI and RST, ∠G ≅ ∠R, ∠H ≅ ∠S, and segment GI ≅ segment RT. Is this information sufficient to prove triangles GHI and RST congruent through SSS? Explain your answer.

[drowsy walrus]

no

You’re given angles congruent

and one segment

and you can’t do the reflexive property because none of the triangles share an endpoint

it helps to draw out and mark what’s congruent

[daring python]

https://prnt.sc/qx3lfb https://prnt.sc/qx3mf1

I need help man, I got no idea what I'm doing

[tribal urchin]

hm

damn

well u know it's in quadrant 4

quad 1 = ALL +
quad 2 = SIN +
quad 3 = TAN +
quad 4 = COS +
*including their variants like sin and csc

u gotta draw a triangle i think and sin = opp/hyp and u know it's -1/2

[native summit]

for the first problem, where on the unit circle is $sin(\alpha)=-\frac{1}{2}$

[somber coyote]

Arc1llusion:

[native summit]

In the 4th quadrant

[daring python]

???

[thorn talon]

🤔

[oblique tiger]

damn is that webwork

ptsd

[daring python]

Bruh

I only get 2 attempts at it

[rich wolf]

Does the altitude of an isosceles triangle bisect the angle

[daring python]

https://prnt.sc/qx5laz

[thorn talon]

🤔

I mean

The logic is valid

But

Might want to look closer at the circle

I think it's a reasonable assumption that each point is equally spaced apart in terms of angles

So instead of going forwards by two points

Go backwards by two points

[upper karma]

So I’m reviewing my test answers and I want to know how tf does this work?

Every answer I came up with doesn’t make sense

[devout harbor]

Should be part b since (2,6) gives us (3,9)?

@upper karma

[upper karma]

Oh ok

I’m new to these dialations

[potent steppe]

Help

[stark snow]

a) The real projective space $\mathbb{R}P^n$ is the space of lines through the origin in $\mathbb{R}^n+1$. What does $\mathbb{R}P^1$ look like?

b) The canonical line bundle over the real projective space is constructed by attaching, to each point of $\mathbb{R}P^n$, the line that it represents. What does the bundle over $\mathbb{R}P^1$ look like?

So, by drawing a diagram, it was obvious to me that $\mathbb{R}P^1$ looked like a circle, $S^1$, touching the origin once and extending however far I liked. However I am having a slight degree of trouble understanding part b intuitively. I thought that since a line looked like $\mathbb{R}$, I could understand the line bundle as looking like $S^1 \times \mathbb{R}^1$, which would just be a cylinder. However, this wasn't right

[somber coyote]

Daniel Cann:

[devout harbor]

@potent steppe You can find the distance BD using formula of distance between two points, draw a triangle and use cosine rule. I think..

[fringe dirge]

@stark snow This is more suited for #point-set-topology

[stark snow]

@fringe dirge Oh, okay I'll put it there

[noble coral]

hello

can someone help me

at the

voice channel

@fringe dirge

hello

[upper karma]

Sorry if the image is sideways

can someone explain why BC and AD are congruent?

[cedar yew]

help

[upper karma]

you could use cosine

to find the angle of MFP

then you would know the angle of MPF

then since you know 12 and 10, you can find MP

but there might be an easier way I dont know of

[sturdy bay]

I've got a doubt

consider a pair of lines ax^2+2hxy+by^2+2gx+2fy+c=0

then, to find their point of intersection, you take the partial derivative of the equation wrt to x, set it to zero and do the same for y

which gives you two linear equations in x and y which can be solved to obtain the intersection points

How is this working?? Where is the partial derivative even coming into picture here??

[south junco]

Can someone help me in questions a

[winged zinc]

@south junco, what do you need?

[sturdy bay]

Could someone help me with my doubt earlier?

[twin prawn]

Why would you write a pair of lines like that

[sturdy bay]

huh

A homogenous equation. wont it represent a pair of lines?

by pair of lines.. I meant you take two line equations and you multiply them.

as in (a1x+b1y+c1)(a2x+b2y+c2)=0

[twin prawn]

I'm lost as to why you would want to solve the intersection point of two lines using partial derivatives in the first place

What's the motive behind it?

[sturdy bay]

I only know that doing that weird ritual of partial derivatives will give us the intersection point

and thats why I wanted to know how things are working out

[sturdy bay]

the weird thing I observed about this is.. the resulting two linear equations that we get after taking the partial derivatives aren't even the actual pair of lines

as far as I noticed.. they are just two lines having the same intersection point

[twin prawn]

Can you show a worked example?

[slim needle]

Does someone know how to do this?

(find perimeter and area of the colored part)

I see there is an "internal" arc and "external" arc

[slim needle]

No one?

[cinder portal]

find the area of the big ice cream cone, and subtract small ice cream cone done

ur answer will be a bit nasty

[silent plank]

consider drawing in the centre and constructing radii to those points of intersection

[slim needle]

@cinder portal hahahah nice answer. But what do you mean when you say "small ice cream"?

@silent plank I don't get it. A drawing?

[cinder portal]

damn i dont know how to explain this without drawing it

fuck me

[slim needle]

Wait. This is an equilateral triangle, which can be "divided" into an isosceles triangle and two right triangles

But how would I apply the formula? 1/2×alpha×r^2

[silent plank]

uh whats that formula meant to be for?

also what's alpha

[devout harbor]

That's the formula for area of a sector and alpha is supposed to be the angle in the sector.

[slim needle]

Exactly @devout harbor

[silent plank]

you should know you angles and the radii and boom

[slim needle]

But what's the angle of the two right triangles? Should I use pi/2?

[silent plank]

right triangles were used to find the radii of the big circle
they aren't used when finding the area of your sectors

[slim needle]

I'm not understanding... I'll have to ask the teacher

[silent plank]

i'll guide you through a few steps

were you able to find the radius of the circle?

[slim needle]

The radius? Why?

[silent plank]

because the radius is essential for finding area's of sectors and arc length

and as you just stated you need to apply
A_sector = 1/2×alpha×r^2

[slim needle]

The radius is 6

Isn't it?

[silent plank]

there are 2 radii involved

one is for the the arc which is 6
and there other is for the circle as indicated by my 3 red lines

[slim needle]

Uhm, ok

[slim needle]

Ok

[silent plank]

the blue sector + 2 isosceles triangles would be your "big ice cream cone"

the green sector would be your "small ice cream cone"

[slim needle]

2 isosceles triangles???

Where?

[slim needle]

Ah ok

And now?

[silent plank]

were you able to find the radius?

[slim needle]

Is it L/sqrt3?

[silent plank]

whats L?

6?

[slim needle]

Oh sorry, it's the side of the triangle

Yes, 6

[silent plank]

simplify it

[slim needle]

2*sqrt3

[silent plank]

ok good

what would be your angle(s) at the centre?

[slim needle]

🤔

[silent plank]

several ways to get it:
circle theorems, properties of equilateral triangles, other

[slim needle]

Ok, tell me

[silent plank]

angle at the centre is twice the angle at the circumference

[slim needle]

So, 2*60?

120 deg

[silent plank]

yes

[slim needle]

Ok, I found the perimeter

[silent plank]

also those 2 triangles are congruent and would also have angles of 120°

[slim needle]

By using the arc formula radius * angle (in radiants)

But now, the area?

[silent plank]

the blue sector + 2 isosceles triangles would be your "big ice cream cone"
the green sector would be your "small ice cream cone"

blue - green would give you the desired area

[slim needle]

Can you guide me?

We can find the first sector by doing 1/2 * 120 * (2*sqrt3)^2???

[silent plank]

you need to convert the angle to radians

[slim needle]

Yes of coursr

[silent plank]

which you should've also done for the perimeter otherwise you wouldn't reach the answer provided

[slim needle]

2pi/3

That would be the blue sector

But the green cone?

[silent plank]

the angle and radius for that are given

angle of 60° = pi/3
radius of 6

(which you should've also used when calculating the perimeter)

[slim needle]

@silent plank so, do I have to do for both 1/2 × pi/3 × 36?

[silent plank]

wdym by do both?

[slim needle]

?

[silent plank]

whoops mistread that
wdym by

do I have to do for both

1/2 × pi/3 × 36 gives the area of the green sector
as mentioned earlier, you also need to find the areas of those 2 triangles

[slim needle]

Oh

How do I do?

[silent plank]

(trig formula for) area of a triangle

[slim needle]

Side × height /2?

With side I mean the base

[silent plank]

i mean you could split your isosceles into right triangles

or you can apply

(trig formula for) area of a triangle
direclty

i.e. Area = 1/2 ab * sin(C)

[slim needle]

Oh true, the two triangles are isosceles, not right

Well, how can I find the height?

[silent plank]

the whole point of the trig formula is that you don't need the height/altitude

only 2 sides and the angle between them

which you have

[slim needle]

We haven't done the area with sine and cosine yet

[silent plank]

strange

ok, then split it into right triangles and find the height using basic trig

and/or pythag (w/ properties of special triangles)

[slim needle]

But do I have to do what?

30 degrees triangle?

Right?

[silent plank]

i mean you should've made some of these calculations already when you found that radius to be 2sqrt(3)

[slim needle]

?

I don't get

[silent plank]

trig, pythag or w/e to find the height

[slim needle]

Sqrt [(2*sqrt3)^2 - 3^2] @silent plank

[silent plank]

simplfiy it

[slim needle]

Sqrt3

[silent plank]

sure

now you should be able to find the areas of the triangles

[slim needle]

3*sqrt3 is the area of the triangle

But if the height is in common...

I got 3*sqrt3 doing (6 * sqrt3) / 2

@silent plank

[silent plank]

sure.

and you have 2 of these triangles

[slim needle]

Now, to find the area of the sector I do 1/2 * pi/3 * 6

[silent plank]

6^2

[slim needle]

And I get 6pi as area

[silent plank]

that's the area of the green sector

what did you get for the blue one?

[slim needle]

The blue one is 1/2 * 2pi/3 * (2*sqrt3)^2?

[silent plank]

which simplifies to:

[slim needle]

4pi

[silent plank]

yes

and as mentioned earlier,
desired area = blue sector + area of triangles - green sector

[slim needle]

Lemme try

3*sqrt3 - 2pi

[silent plank]

3*sqrt3 is the area of 1 triangle

there are 2 triangles

[slim needle]

6*sqrt3

[silent plank]

yes, you'll get
6sqrt(3) - 2pi
which is equivalent to the answer provided

[slim needle]

Thanks a lot!

Beer paid

[odd cairn]

how do i make a parabola out of an equation

in this example it turns the equation directly into a graph, but how do you do that

[olive glacier]

Given tanΘ = 3/2 ; and Θ is a reflex . Work out the value of sinΘ #can someone give me a solution. Thanks.

[wind heart]

I do not comply

How to do this

[odd cairn]

<@&286206848099549185>

[wind heart]

<@&286206848099549185>

[summer spire]

@odd cairn I'm not sure what you mean. Are you asking how to graph an equation?

[odd cairn]

yes

[summer spire]

plot f(0), f(1), f(-1), f(2), etc?
so for the example, to find the y value at x=0, you substitute 0 for x:
$$-1/160^2 + 1$$
$$-1/160 + 1$$
$$0 + 1$$
$$1$$
then you fill in the point (0,1) and move on to the next x value

[somber coyote]

Googol30:

[odd cairn]

oh so you just fill in the x values until the graph is complete

by substituting the x

[summer spire]

that's the most reliable way, yes

[odd cairn]

ok got it

[fallow edge]

Am I crazy or should the plane line up with the vertices of the square?

[dire rampart]

which square

video doesnt wanna work

[wind heart]

Uh

I guess I’ll go for a second and last attempt

https://media.discordapp.net/attachments/326138757474680852/674742973099147264/image0.jpg

CN anybody explain me this

Can*

I’m kind of confused about it

[fallow edge]

@dire rampart I meant cube

[dire rampart]

what

by line up u mean the vertices lie on the plane?

[fallow edge]

Shouldn't the plane intersect the vertice of the square at the origin and the vertice opposite it

[dire rampart]

oh I thought u meant every single vertex lmao

what are the coordinates of G?

doesnt seem like it intersects it to me

[fallow edge]

G is a point on a cube

(1,1,1)

Wait

[dire rampart]

which opposite vertex do you mean

G clearly doesnt lie on the plane

[fallow edge]

<@&268886789983436800> ban @placid basin

[stuck torrent]

Why the aggression?

[twin heron]

[fallow edge]

Thanks

[stuck torrent]

Guess we’ll never know

[trail minnow]

we're attracting so many smoothbrains today

[rich wolf]

smoothbrain is a nice insult

but i don't see why you have to ban that individual

what did he do

[copper valve]

a raid occured

relatively small one

[fallow edge]

@dire rampart isn't the vector normal to the plane ax+by+cz=d (a,b,c)?

[dire rampart]

yes

if you want to know whether it lies on the plane or not just sub it into the plane equation

[fallow edge]

Sure, but isn't the vector normal to the plane intersecting the two points in question the vector (-1,-1,sqrt(3))?

Oh. it isnt

Nvm

[dire rampart]

not quite sure what u mean

[unborn jacinth]

why is that sin^2(x)/cos^2(x) = tan^2(x)

and not just tan(x)

ik sin/cos is tan

[dire rampart]

there isnt just one vector normal to the plane

why would it be tanx

[unborn jacinth]

so @dire rampart its always the same power as the sin and cos?

oh yeah that makes sense

basic algebra

otherwise the fraction wouldn't equal the same thing right

@dire rampart

[somber coyote]

IAmJon:

[unborn jacinth]

oh yeah that makes sense

but then @fallow edge what if the sin and cos powers were different

you wouldn't be able to say its tan right

unless they're the same

[fallow edge]

If there's a sine over a cosine, you can "factor out" a tangent, but a sine or cosine might remain as well

[unborn jacinth]

yeah

ok

[fallow edge]

@dire rampart if (a,b,c) were orthogonal to the plane, you could still write the equation of ax+by+cz=d with it

[spark stag]

$\frac{\sin^{3}(x)}{\cos^7(x)} = \frac{\sin^3(x)}{\cos^3(x)} \cdot \frac{1}{\cos^4(x)} = \frac{\tan^3(x)}{\cos^4(x)}$

[fallow edge]

So if (-1,-1,sqrt(3)) were orthogonal to the plane, you could use those coefficients

[somber coyote]

Namington:

[spark stag]

just as an example @unborn jacinth

this example is pretty contrived admittedly, but those are the sort of algebraic things you can look for

[unborn jacinth]

ohk

thx!

[upper karma]

guys I need help
https://cdn.discordapp.com/attachments/387447700234174464/674817172337590309/image.png
I have no idea where to start. I think I shoudl start by finding the two big triangles and using sas to prove they are similar

the triangles cad and bca

actually I can prove they are congruent but I don't know where to go from there

anyone have an idea?

I have no clue how to get to triangle tes and ver

[tacit karma]

TEA~VEC

[surreal ingot]

Does anyone know how to find x graphing point for -3sin(pie/3 ×-3pie)

I got beginning point 9 and ending point 15

[twin prawn]

Did you just write pie instead of pi

🥧

Also I'm very confused about your question, first of all what do you mean by "x graphing point"?

[upper karma]

Hello

[carmine sundial]

do the diagonals of a rectangle bisect opposite angles?

[thorn talon]

🤔

[lament brook]

Just curious, if I get the value for -2/sqrt 13 for a sin value let’s say, and since the bottom tells me the radius do I still have to simplify?

I’m not sure if it will mess up like what’s the y value or whatnot

[thorn talon]

I mean

Simplify to what?

[lament brook]

Uh I didn’t do it yet but basically move the radical up

But not sure if it will mess up my answer or not

[thorn talon]

I mean

Sure

If you want to

Rationalise if you want

[lament brook]

It’s optional?

[thorn talon]

I mean

It's the same number if you did it right

[lament brook]

Ig but not sure if teacher will take points off for having a radical in the dome

Dom

[thorn talon]

Then do it

Nobody really cares other than picky teachers

[lament brook]

Kk thanks!

[tall flower]

I need help with 57 I keep getting 135 but the book says 140

[lament brook]

I think this is what I’m suppose to do. Just wanna make sure

[thorn talon]

Seems fine I guess

[upper karma]

Can someone explain the whole cos(x) = cos(-x) and sin(-x) = -sin(x) thing? I am retarded

[umbral snow]

Not much to explain, you just said it yourself

[upper karma]

Thanks...

[umbral snow]

These are two facts:
sin(-x) = -sin(x)
cos(-x) = cos(x)

[upper karma]

Yeah but why

[umbral snow]

Remember that EVERYTHING sin and cos is determined from the unit circle

[upper karma]

So

[umbral snow]

Make sure you're comfortable with the unit circle to determine trig stuff

[upper karma]

So can you explain it

What I am missing

[umbral snow]

So let's say you wanted to compute sin(-π/6). You'd start at the x-axis, then go π/6 radians clockwise

[upper karma]

Yes

[umbral snow]

Whatever the y-value of your terminal arm is, that's sin(-π/6)

If you're smart about it though, you'll realize that you could instead go counterclockwise, then you'll have the right value, but wrong sign

That is, if you can calculate sin(π/6), there's a quick step to finding sin(-π/6)

[upper karma]

Ok but why does it hold for all x

[umbral snow]

Here's a bit more of a proof. You agree with the summation formula, right? That is:
sin(x + y) = sin(x)cos(y) + cos(x)sin(y)

[upper karma]

No??? What is that

I just started trig

[umbral snow]

Welp you'll learn that one soon lol

Then forget that one for now you'll learn it in school

[upper karma]

Ok

But what about my question tho

[umbral snow]

Basically, if you rotate by an angle θ, flip the circle over the x-axis

That would be the same as rotating backwards by an angle θ

[upper karma]

Ok

So?

[umbral snow]

sin(θ) is the rotation by an angle θ

-sin(θ) is that rotation, but then a flip over the x-axis

sin(-θ) is a backwards rotation by θ

By my comment just before, -sin(θ) = sin(-θ)

[upper karma]

And cos(x)?

@umbral snow ?

[umbral snow]

Okay, similar idea. Rotate an angle by θ. Flip the circle over the x-axis. That's the exact same as rotating backwards by θ

Now, when you flip over the x-axis, this turns sin(θ) into -sin(θ). That's why sin(-θ) = -sin(θ)

In the cos case, when you flip over the x-axis, this doesn't change cos(θ). So, doing this whole thing again, shows cos(-θ) = cos(θ)

[upper karma]

In the cos case, when you flip over the x-axis, this doesn't change cos(θ).

Why

[umbral snow]

Remember that cos(θ) represents the x-value of the terminal arm after rotation by θ

Flipping the circle over the x-axis doesn't change the x-value of the arm

It's very hard to get this across without pictures lol

[upper karma]

Can you get pictures or no