geometry-and-trigonometry

[gentle bear]

big time lost

[soft dagger]

okay

it has a magnitude of 98

and forms an angle with the x axis of 331

can you try and draw that

[pulsar merlin]

@upper karma my teacher just told me to skip the assignment after i just did 10 questions....

[devout kraken]

So the equation would be. C = arcsin(x/40)

?

[upper karma]

@pulsar merlin damn

[soft dagger]

@gentle bear the angle is supposed to 331°

[upper karma]

@dim gyro what do you want

[soft dagger]

331 is the angle not the magnitude @gentle bear

[dim gyro]

@upper karmai need help with picture number 1

idk what to do

[upper karma]

Ok

Are you here

@dim gyro

[dim gyro]

yes

[upper karma]

I gtg to eat so quick, use the complementary angles in 70° to get the inner angle of the triangle. Use the cosine law using 2 side lengths and one angle (the one you get by doing what its on the beginning of the message).

Ill be back in one hour to confirm your results, good luck.

[noble token]

Hey, I'm having trouble visualizing this problem and I was wondering if somebody could help me work through it.

[upper karma]

Can someone help me in #help-4

@noble token well, you could google "right prism"

and see what it looks like

you could also google "what is the slant height from a right prism"

from that drawing, you need "s"

you already know "h"

and you know "r" (which is half of "a")

the perimeter of the square base is 24

that means "a" = ... ?

then once you have "r" and "h", Pythagoras' theorem

and you get "s"

[noble token]

ah

[upper karma]

you with me so far, @noble token ?

good

[noble token]

yea so h = 8

r = 4

no

r = 3

[upper karma]

3, yes

[silent plank]

uh. prism and pyramid are different.

[noble token]

ah

[upper karma]

slant height?

[noble token]

that's where i was getting tripped up

because that ain't a prism

that's a pyramid

[silent plank]

a square prism doesn't have a "slant height".

[noble token]

so the question is flawed

[silent plank]

it is unclear what the question bwants

[noble token]

that's what i suspected

i'm gonna assume pyramid for the sake of the question

then it's just a 3 4 5 triangle

so the slant height would be 5, yes?

[upper karma]

maybe it looks like this?

"slant height"

[acoustic jungle]

your end faces are not equal

nor parallel

and the sides are not parallelograms

[dusky surge]

,w frustum

[somber coyote]

Display results online and refine query

Requested by Biscuit#5208

[dusky surge]

Opps

[upper karma]

in a right prism, all angles are 90 degrees

right?

so there's no "slant height"

[dim gyro]

@upper karma i think i did it correctly

[silent plank]

not quite. how are you getting 21.6?

[dim gyro]

i put16.66−14.7×cos(110) in the calculator

[silent plank]

which gets you:

[dim gyro]

21.6

[upper karma]

that's not the law of cosines
edit: it is

[silent plank]

Is the calculator saying anything else?

did it's just put 21.6 and stop there?

[dim gyro]

21.6876961069

[silent plank]

which doesn't round to 21.6

also you should not be rounding yet

the stored value above is what you should be square rooting

it is also a good idea to write out the full equation instead of just one side of the cos law

[upper karma]

it rounds off to 4.7, not 4.6

Hey im back

I told @dim gyro what he had to do

It worked good?

heh, I was off a bit

fricken' rounding off

but it still gets you to 4.657, which is still closer to 4.7

Good it worked

[upper karma]

SSS

[rapid siren]

thx

i got few more

[upper karma]

AAA

[silent plank]

do you know the theorems related to similarity and congruency?

[upper karma]

@upper karma stop giving the answers, its totally what you aren't suppose to do in this server

You might give him a clue, method or anything to help but an straight solution

ight then

Good

[humble pebble]

rent free i need answers too

[upper karma]

i'll give you hints since i can't give answers

@humble pebble be like

(kidding)

[humble pebble]

@upper karma be like 👨🏿

fun at parties

[upper karma]

Not really

Lmao

[humble pebble]

oh ok thanks for the answer

[upper karma]

But if you could read my previous messages and rules, which is what you have to do in a server, you'll understand the matter

[humble pebble]

oops wrong dm

[upper karma]

if you're only given sides... what did you think the answer was gonna be?

and if you're only given angles...

lmao ikr

I mean, come on

Yeah lol

[silent plank]

however if you're only given angles it doesn't mean that the triangles are ||necessarily ... ||

[devout kraken]

what’s the equation to find opposite from a hyp and angle? Is there one?

[upper karma]

sin(theta) * hyp

[versed river]

$\sin(\theta)=\frac{O}{H}$

[somber coyote]

Sneaky:

[versed river]

$H\sin(\theta)=O$

[somber coyote]

Sneaky:

[upper karma]

Yes! 👍

soh cah toa

Yes!

[devout kraken]

What abt adjacent?

given angle and hyp

[silent plank]

refer to you trig ratios and which do you think would be appropriate

[devout kraken]

cos(theta)=adj/hyp ?

[silent plank]

yep

and then isolate adj if that is what you want to find

[devout kraken]

$\cos(\theta)=\frac{A}{H}$

[somber coyote]

leaf:

[devout kraken]

okayokay thank you

[acoustic jungle]

Just a thought, proving triangle congruency all comes down to SSS and the fact only 1 triangle can be constructed with 3 given side lengths

[devout kraken]

There’s also SAS , ASA, and AAS?

[acoustic jungle]

SAS - SSS congruence by cosine law

ASA - SSS congruence by sign law

AAS is literally the same thing as ASA

[silent plank]

don't forget rhs

*sine

[upper karma]

I mean, doesnt cosine and sine's rule/laws have more uses than those

[upper karma]

Yes

[oak pawn]

Could someone help me find ACO

[devout harbor]

@oak pawn There's a property, the tangent to a circle always forms a 90 degree angle with the radius at that point

[upper karma]

isn't it a right angle?

[devout harbor]

yes

[upper karma]

ok

[oak pawn]

Oh I wasn’t sure usually they put the right angle sign then in this book, thanks

[vague berry]

i got a trig test tomorrow 😮

[tight tide]

heya is anyone else here doing High School Khan Academy Geometry ?

[quick schooner]

does anybody know how to do number 3

[vague pagoda]

yeah

[upper karma]

@quick schooner do you even SOH CAH TOA, bro?

[vague pagoda]

what have u tried

[quick schooner]

no what do i do here sin36=hypothensuse/opposite

sin36=4cm/b

[upper karma]

what does sin(36) evaluate to?

(set your calculator to degrees first, not radians)

,w sin(36)

[somber coyote]

Display results online and refine query

Requested by DSpider#6917

[quick schooner]

our teacher said not to use a calculator

[upper karma]

ok, then just keep it as "sin(36)"

[quick schooner]

k

thanks

[upper karma]

so you have:
sin(36) = 4 / b

what does "b" equal to?

solve for "b"

[quick schooner]

sorry if i sound dumb here but i have no idea how to find b

all i know is that b is the hypothenuse

[upper karma]

or...

you could think of it like:

3 = 6 / 2

[quick schooner]

ohhhh

thanks

[upper karma]

and move the "2" over to the other side

(so that the equation evaluates to true)

"b" is the "2" in that example

[rich wolf]

Omg

Do you really need that many steps to solve a division problem

That is painful to look at

Lol mb actually i scrolled up and saw

[woeful nova]

can someone help me on this? how would I solve it?

[dark sparrow]

let me reproduce your diagram in a more easily readable format

[woeful nova]

alright

[woeful nova]

so would a ratio solve since similar?

3 X

_ _

•  -
  

4 5

and then cross multipliy?

x=15/4

[dark sparrow]

15/4 is the correct answer yes

[woeful nova]

thanks!!!!!!!

[woeful nova]

when you add radicals, is it just like how you'd think you add them 7 |2| + 7|2| = 16 ?

[dark sparrow]

uh

are you using |2| for the square root of 2 or something

[woeful nova]

yeah

[dark sparrow]

do not

that's not what | ... | means

$|x|$ and $\sqrt{x}$ are two entirely different things

[somber coyote]

Ann:

[woeful nova]

`|----

[dark sparrow]

that is even worse.

just write sqrt().

sqrt(2)

like this

[woeful nova]

okay

[dark sparrow]

anyway, no, $7\sqrt{2} + 7\sqrt{2} \neq 16$.

[somber coyote]

Ann:

[hollow tiger]

I think he means to ask if you can add radicals

in which case yeah you can

7 * sqrt(2) + 7 * sqrt(2) = 14*sqrt(2)

[dark sparrow]

but that isn't what he said.

he asked if 7sqrt(2) + 7sqrt(2) was equal to 16

and the answer to that is an unambiguous no

[hollow tiger]

ok I guess I'm just confused now lmao

[woeful nova]

7sqrt(2) + 7sqrt(2) = 14sqrt(2)

?

[hollow tiger]

yes

[dark sparrow]

see that's actually correct this time

[woeful nova]

okay thanks

[gentle bear]

https://gyazo.com/d11463f4ab086994a22b7b5c5eb236d4

this is the graph I came up with

here is the work I did

Maybe I don’t understand I should be doing fully but I was under the impression I take the middle values and multiply it against 2

[upper karma]

How will x^4 - 16

Be factored?

[silent plank]

consider x^4 = (x^2)^2 and 16 = 4^2

[vale nimbus]

sin(x) + sin(3x) = 2cos(x)

i combined sin(x) and sin(3x) and then divided both sides by 2cos(x) so im left with sin(2x) = 1

but then i only get 1 of the 2 solution sets

i guess the other way to solve is to move 2cos(x) over to the left, combine sin(x) and sin(3x) again and then factor out 2cos(x)

but i dont see why what i did doesnt give me the full solution

[silent plank]

dividing both sides by cos(x) erases the solution you would get from cos(x) = 0

[vale nimbus]

thanks man

wait what even when i factor that stuff out i still only get 1 of the sets

[silent plank]

can you write out your work?

[upper karma]

What type of factoring will be used for 20x^2 - 4x?

[silent plank]

in use

(also not geo/trig)

[vale nimbus]

sin(x) + sin(3x) = 2cos(x)
sin(x) + sin(3x) - 2cos(x) = 0
2sin(2x)cos(x) - 2cos(x) = 0
2cos(x)[sin(2x) - 1] = 0

[silent plank]

did the give a restriction on x?

[vale nimbus]

nope

[silent plank]

can you show me the rest of your work? that looks ok so far.

[vale nimbus]

im stupid

my handwriting is so ugly i wrote my 1/4 so poorly it looked like a 1/2

i got it now dw

[upper karma]

No you are not

Lol

[vale nimbus]

should i try to avoid dividing by stuff when solving these sorts of equations?

so i dont eliminate possible sets

[silent plank]

try not to don't divide by stuff that could be 0

[upper karma]

@silent plank Do you know bout my question?

[vale nimbus]

oki, thanks

@upper karma factor out 4x

4x(5x-1)

[upper karma]

But what type of factoring?

[vale nimbus]

uh

idk? factoring?

[silent plank]

common factors?

[upper karma]

Alright

For what values of k can 3x2 + kx - 8 be factored?

[silent plank]

(also not geo/trig)

[night summit]

can someone help me?

i dont know what i did wrong

[upper karma]

you have to find the value of x

when x = cos ^-1 (-8/9)

[night summit]

where did cos come from?

the video has nothing to do with it

thats just an exampe

example*

[silent plank]

show your work

how are you getting the 170.4 and 189.6

[night summit]

i substitute x for sinx and y for tanx and solved

then found reference angle because when solving for y or tanx you get a reference angle of 9.6

i added that to 180 and subtracted from 180

[vale nimbus]

cant u factor out tan(x)

[silent plank]

substitute x for sinx
using the same varaible bad

why did you add it to 180?

[night summit]

cus thats what the video did

the guy in the video*

[silent plank]

how you use your reference angle depends on the sign and the trig function

where is sin positive?

[night summit]

quad 1 and quad 2 only

all strippers take cash

i know

[silent plank]

9.6° is already in quadrant 1

[night summit]

yea

[silent plank]

subtracting that from 180 gets you 170.4° (the solution in quadrant 2)

[night summit]

yea

[silent plank]

and that's it

[night summit]

wat?

theres no

3rd answer?

i have a question a tho

a question*

[silent plank]

well your 0° and 180° for tan(x)=0 are fine

[night summit]

how do i know

if its 0 and 180

is it always 0 and 180?

or does it depend on the sign?

[silent plank]

if what's 0 and 180?

[night summit]

the first 2 options

so if cos was 0

[silent plank]

what sign?
0 is neither positive nor negative.
and x= 0, 180° are the only solutions to tan(x)=0 in the given interval

[night summit]

yea i get that

but if cos was 0

what would the first 2 options be

and if sin was 0

[silent plank]

refer to the unit circle if you are unfamiliar with these

[night summit]

ok ty

wait

what do I do if its a whole number

[silent plank]

if what's a whole number?

[night summit]

when i solve for sin

it gives me 7

and when i solve for tan it gives me 0

[upper karma]

how did u solve it

[night summit]

i substituted s for sin and t for tan

so i did st=7t

[silent plank]

you shouldn't need to resort to subtitutions for these questions

[upper karma]

^

what you need to do is bring 7tan x to the left side

[vale nimbus]

uh i have a question can i ask now or should i post it in one of the question channels

[upper karma]

question channels

[silent plank]

anyway sin(x) =7 has no real solutions

so ig leave 2 boxes blank

and only input the solutions you get from tan

[night summit]

tan is 0

so

i only put

0 and 180?

yes i got it ty

@silent plank sry but if a function comes out with a whole number like 1 and increasing, does that mean it has no solution?

[silent plank]

can you be more specific?

[night summit]

so if sin(x) = 1

it has no real solution?

so sin(x) cos(x) and tan(x). if any of these 3 functions equal to a whole number

they have no solution?

[silent plank]

sin(x) = 1 definitely has real solutions

so does cos(x) = 1

sin(x) = c and
cos(x) = c have no real solutions if
|c| > 1

[vale nimbus]

tan(x) + 4sin²(x) + 2 = 4sin²[(pi/4) + x]

how to solve

asked in #help-1 but we couldnt figure it out and now the channel's been robbed

[night summit]

ok ty ramanov

ramonov*

[silent plank]

and tan(x)=c has solutions for all real c

[vale nimbus]

after 5 hours i still cant solve trigonometric equations

and im supposed to be able to do inequalities and inverse trigonometry by monday

[upper karma]

Inequalities are ez

Just post questions here

[vague berry]

Yo

[vale nimbus]

i just cant do equations

im hoping inequalities are similar to equations

but i cant even do those so it aint looking too good

[vague berry]

hey guys how would you write the domain of 2sec(x)?

like what is the correct notation?

given that cos58˚ = a/c and sin 58˚ = b/c, calculate the expression tan302˚ + csc212˚ + sec 122˚. Answer as a fraction in simplest form

please help me

ping me maybe

[novel mango]

what have you tried het

evaluate tan(2pi-x)

tell hows it related to tan(x)

Hint here :||360-58=302||

[vague berry]

ooh ok

i kinda get what u mean

[novel mango]

just try to get these in terms of the standard angles

[vague berry]

wait can you give me an example

[novel mango]

also like 212=270-58=3*90-58

[vague berry]

like tan(2π-x) like how do you evaluate that

[novel mango]

there

[vague berry]

oh

[novel mango]

do you know how to evaluate the trigonometric ratios of compound of angles

i.e like a sum of angles

[vague berry]

no we haven't learned it such a lazy teacher

[hollow tiger]

I have a sheet of formulas if yall need one

[novel mango]

its useful and helps you relate
tan(A+B) with tan(A) and tan(B) individullaly

[vague berry]

Hmmm

how can i apply it for each one

[novel mango]

have you learnt complex numbers?

[vague berry]

Yes I have

[novel mango]

nice

e^ix=cos(x)+isin(x) do you know this VERY famous relation?

[vague berry]

well

No I don't

[novel mango]

ok then you know it now

its known as eulers identity

so observe how i use this

to get the trig ratios for sin(A+B)

e^(i(A+B))=e^(iA+iB)=e^(iA)*e^(iB)=(cos(A)+isin(A))(cos(B)+isin(B))

read this take time

[rich wolf]

let x=a+b

[novel mango]

do you understand everything?

@vague berry

[vague berry]

um

yeh im still here

[novel mango]

do you underatand the line?

[vague berry]

im trying to

[novel mango]

yea sure

[vague berry]

but its kinda challenging for me since we have never once gone over it

[novel mango]

its very useful and you can skip all work

its just algebra

see i used the fact that a^(b+c)=a^b.a^c

you can multiply it out for getting the formula

[vague berry]

Ok I think I got it

how would one attempt

Īt is known that k is an x-intercept for the function f(x) = sin (x-m), where m is a positive integer. Find ONE of the x-intercepts fro the function g(x) = cos (x-m) in terms of k.

[novel mango]

Ok you can see that later anyways
$\tan{(A+B)}=\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$

is the formula You would need

it would follow after some work with the e^ix thig

[somber coyote]

God V2:

[digital gulch]

Can someone help me with this?

[upper karma]

well if K is perpendicular to L, then its slope it the negative inverse of L's slope

[digital gulch]

Yeah

So k’s slope is 1/2

[upper karma]

yah

[digital gulch]

Yeah i got that part

[upper karma]

divide the number in the point that passes L with the slope of L

[digital gulch]

Why

[upper karma]

cuz thats how u find r in the first point

[digital gulch]

Oh ok

So you divide it by -2?

[upper karma]

yes

[digital gulch]

R/-2,3/2

[upper karma]

now do the same thing with your second point and the slope of K

[digital gulch]

12,2r

[upper karma]

yes

[digital gulch]

What do i do with that?

[upper karma]

oh hang on

6 was the x-coordinate, not the y-coordinate. my bad

what you should do is multiply 6 with the slope of K, not divide

my apologizes

[digital gulch]

Oh ok

[upper karma]

so your 2nd r should be 3

[digital gulch]

3,r/2

[upper karma]

so your points should be (3/2 , 3) and (6 , 3)

[digital gulch]

Oh

[upper karma]

you know what? the 2nd r should also be 1.5 actually. i read the question incorrectly

and ignore what i said about the slope of K.

so your first point on line L should be (1.5 , -3) and your second point on K should be (6, 1.5)

[digital gulch]

Oh

[upper karma]

yah i fucked up again. my bad i don't mean to confuse you

[digital gulch]

Its ok

[upper karma]

alright

[digital gulch]

Yeah the rest was easy

[upper karma]

ok good

did u solve it

[digital gulch]

Yeah

[upper karma]

do u mind sharing your answer?

i got .6

[digital gulch]

-1.2?

[upper karma]

yah

[digital gulch]

Ok

Thanks

[upper karma]

yah np

i graphed it and i got (.6 and -1.2)

[digital gulch]

Oh wait its asking for a

My bad

Its 0.6

[inland bluff]

Yo

Anyone down to help

[rich wolf]

Bro

Posting the same question twice

🤡

[plush burrow]

Given that the terminal arm passes through the point (-3, -4), find the exact values of all trigonometric ratios

can someone help me on this please

i need 3

[merry abyss]

plot on Cartesian plane

and find hypotenuse

which is also radius

then find the trig ratios and use CAST

not a full answer sry

[plush burrow]

thanks dude

[crimson olive]

can someone help me with this please?

[upper karma]

What's 36 degrees

[crimson olive]

i'm not even sure lmao thats what was so confusing

maybe the arc but im not really sure?

[rich wolf]

This is simple

Angle DCA is half of arc DA

And opposite angles of a cyclic quadrilateral add up to 180

[upper karma]

@crimson olive

what is the number of degrees of arc AB?

remember, the whole thing is 360 deg

arc AD + arc AB = ?

maybe this will help:

https://www.youtube.com/results?search_query=inscribed+angle+theorem

I guess you could also think of it like this:

[crimson olive]

i got it thank you so much!

[digital gulch]

Im bad at coordinate geometry

[upper karma]

https://www.mathopenref.com/images/coord/coordplane.gif

plot it and see what you get

or rather, if you didn't have to plot it, what would AC be?

hint: ||law of cosines||

this also sounds like it could be a 30-60-90 triangle

[digital gulch]

How do i plot it?

[upper karma]

start from the origin

make AB a horizontal line

[digital gulch]

Ok

How do i plot point C

[upper karma]

BC is a diagonal line

(the hypothenuse)

[digital gulch]

It cant be a 30-60-90 triangle

BC would be 6

But that would make AB 6 times sqrt(3)

And 8 is not 6 times sqrt(3)

So it cant be a 30-60-90 triangle

We also dont know if its a right triangle

[upper karma]

I plotted it wrong 🙂 ABC = 30 deg

not 90

[digital gulch]

Im just gonna use law of cosines

[silent plank]

it doesn't really make sense to use coordinate geo to do this question

[upper karma]

it would look something like this:

maybe if you used a ruler

and measured it "manually"

and a protractor for measuring the angle

[digital gulch]

Yeah i got that

Thanks

[potent crane]

Anyone know if I need to label something for the -1 on f(x) axis?

since x != pi/2

[upper karma]

@digital gulch @silent plank I think there's a way:

30-60-90 triangles have a certain property

that the hypotenuse is equal to ||twice the length of the shorter leg||

so the dotted line (i.e. the shorter leg) would be...?

then you can find the side with the question mark

and subsequently this side:

you can Pythagoras' theorem your way through this without any protractors and stuff

@digital gulch you understand what I'm saying?

using the law of cosines is a shortcut, but that's not what the question asked you to do 🙂

they chose 30 degrees for precisely this reason

@digital gulch

@potent crane what is the midline of that graph?

@potent crane

don't forget the ">" arrows on your axis

the vertical line is the y axis

and it doesn't have anything with "pi" on it

what is the amplitude of that function?

the midline?

and the period

[upper karma]

@potent crane
UPDATE: this is wrong. Should be cos(x)(1/(x-(pi/2)))

how about now? do you know what the amplitude and midline are now?

from "cos(x)(something)"

hint: ||1cos(x)(something) + 0||

aaaand you're offline

nice

[dark sparrow]

they're not just offline, they straight up left

[old holly]

hello can i ask something

[upper karma]

if it's related to trig, sure

[old holly]

this is really dumb

but

since when is the volume of this not $\frac{1}{3} \cdot 100\pi \cdot 13cm^3$

[somber coyote]

Im confused; get over it:

[old holly]

it's kinda geometry

[dark sparrow]

10 cm is the diameter

and not the radius

[old holly]

ah

ok got it

fucc im dumb

[acoustic jungle]

@upper karma

this is not right.

[upper karma]

yep

should be cos(x)(1/(x-(pi/2)))

[upper karma]

True i was gonna say that

[dark sparrow]

f r e s h m a n s d r e a m

[vale nimbus]

could i have some help with goniometric inequalities?

dont really have a specific problem in mind but i think a good place to start is with tan(3x) <= sqrt(3)/3

we luckly only have to do basic inequalities but they're still too hard for my pea-sized brain

[acoustic jungle]

Well, if you let 3x = u then tan (u) = 1/sqrt3 and u=30 or 180+30

so 0<u<=30 and you will have to do it for the third quadrant as well

[vale nimbus]

https://cdn.discordapp.com/attachments/706156873744908318/706160636769337394/JPEG_20200502_171003.jpg

dont really know what i did wrong

[vale nimbus]

<@&286206848099549185> where did i go wrong?

[dusky surge]

,w graph tan(x/3)

[somber coyote]

Display results online and refine query

Requested by Biscuit#5208

[hallow rose]

The general solution for $ \tan(x)= \tan(y)$ is $x= n\cdot \pi +y$

[somber coyote]

FinalBoss:

[vale nimbus]

i dont get it

i mean i see why its that

[vale nimbus]

but i dont even understand how to solve these inequalities

in the book they draw a circle which is an understandable thing to do, but then they get pi/2 involved too

which i can see why but not how

they do that cuz tan isnt defined when the angle is pi/2

but idk how you're supposed to take that into account in your answer set

[dusky surge]

As you can see from the graph, for tangent(x/3)≤√3/3
The values are clear

[vale nimbus]

bro im not supposed to do it graphically lol

[dusky surge]

We can do it with the help of reading graph and understand the pattern

For the 'first'curve , that inequality holds for -3π/2 <x ≤π/2

So you can see for these questions about tan(x) smaller than a number, it will have a pause on every π/2+nπ

Like the second' curve, the inequality holds for 3π/2<x≤7π/2

Do you see the pattern?

[vale nimbus]

no

[dusky surge]

Ah... Okay, lemme use a question for example

[vale nimbus]

in class we always did these using the unit circle

but its been so long that ive forgotten

[dusky surge]

Okay... But using unit circle on tangent values is harder than sine and cosine

[vale nimbus]

i mean like i understand the drawing now but i dont see how they got their part with π/2 from it

[dusky surge]

Very well. Let's continue with your question then.

I'll type it out

[vale nimbus]

the solution should be (-3π/2)+3nx < x < π/2 + 3nx

[dusky surge]

$\tan{(\frac{x}3)}\leq\frac{\sqrt3}3$

[somber coyote]

Biscuit:

[dusky surge]

I'm handicapped with Latex...

[vale nimbus]

if you feel more comfortable writing it down on paper go ahead :p

[dusky surge]

Okay, gimme some moment

For this first range I picked, everything under the line y=√3/3 and on the tan(x/3), it will suits the inequality

And since it happens for every repeated copies of this first range, we can easily generalize the inequality

[vale nimbus]

i can understnad that

but on the drawing in my notes there's some weird stuff

holdon let me send a picture

the top part is also marked

[dusky surge]

Hmm... You mean the pi/2 part?

[vale nimbus]

yeah

[dusky surge]

But tan(pi/2) is undefined...

[vale nimbus]

yes

holdon quick question inbetween

[dusky surge]

Yea?

[vale nimbus]

does this one look right?

[dusky surge]

Let's see

[vale nimbus]

the order of my stuff isnt right

i just realised

just imagine the 3pi/4 part being on the left

[dusky surge]

I think it's the other way round lol

[vale nimbus]

what do you mean?

[dusky surge]

just imagine the 3pi/4 part being on the left
@vale nimbus
This

[vale nimbus]

i was told to put the smallest angle to the left of x

im dumb nvm

yeah it should be the way it is

[dusky surge]

You should

But....

[vale nimbus]

is the solution correct?

[dusky surge]

The right side should be pi/4 +2pi+2kpi

[vale nimbus]

why?

[dusky surge]

Then 3pi/4 will be the smaller one :D

Lemme draw it out

[vale nimbus]

is there a reason for that?

[dusky surge]

It should be below this straight line

If it's you're solution, you are describing the things above the straight line

[vale nimbus]

huuuuh

[dusky surge]

I'm sorry for the ambiguity, I am just not used to using unit circles to solve these kinds of problems

[vale nimbus]

i understand the drawing just not what you mean by mine is above the straight line

nor do i see how adding 2pi to the right side would fix that

[dusky surge]

From what I understand from your last question, the 2nd line is describig something like this

[vale nimbus]

but why add 2pi

thats whats throwing me off the most

[dusky surge]

But when we actually need this

Since the angle is moving anti clockwise, we add 2pi to make the inequality easier to be written out

[vale nimbus]

i'm sorry but i still dont see how that solves the problem

is that just to make it easier? or does that fix the issue of the shaded area being in the wrong place

[dusky surge]

Yea, I fix the issue of the shaded area being in the wrong place, it's about angle moving anti clockwise

Because we cannot describe clockwise from π/4 to 3π/4 the angles between them

So we add 2pi to fix the problem

[vale nimbus]

ooooh i think i get it

so you always describe the angles clockwise?

[dusky surge]

By adding 2pi, turning it back to an anticlockwise situation

Piece, you have given me a new insight in solving problems using unit circles, thank you!

[vale nimbus]

[dusky surge]

So does that make sense now?

[vale nimbus]

abit yeah!

let me try another

[dusky surge]

Sure!

[vale nimbus]

you dont have to add 2pi here since you can describe clockwise from -2pi/3 to -pi/3

i think

[dusky surge]

Hmmm

Remember, we always describe them in the anticlockwise direction

[vale nimbus]

so the angles have to be positive?

[dusky surge]

Nah, you are already describing anticlockwise direction😀

Try to draw the 2 angles on a unit circle and you'll see it

[dusky surge]

Ahhh, it's the other way round XD

I kinda know what you don't understand

You know sine it's the y in the unit circle right?

[vale nimbus]

yes

[dusky surge]

So, for sin(something) < N for some constant N

On the graph, it's just everything that is under the line y=N and on the unit circle

So in this case, it's anticlockwise from -2π/3 to -π/3

[vale nimbus]

[dusky surge]

Sorry....

[vale nimbus]

no problem man its not your fault

[dusky surge]

Don't mean to make your mind explode

[vale nimbus]

but wait why does -pi/3 have to be on the left

isnt that -60°

and -2pi/3 = -120°

[dusky surge]

Yea, -2pi/3 is on the left, what you wrote was right.

[vale nimbus]

oh ok

sorry lol

so how do i know in which direction i should be going

like counter-clockwise or just clockwise

[dusky surge]

For me, I just read the graph...

[vale nimbus]

can you give me another problem?

[dusky surge]

This is what I read which range should it be.

can you give me another problem?
@vale nimbus
Wanna try cosine?

[vale nimbus]

uh 1 more sine, i think it just clicked

not sure tho

[dusky surge]

Okay

[vale nimbus]

brb 2minutes

[dusky surge]

$\sin{(x)}\leq\frac12$

[somber coyote]

Biscuit:

[dusky surge]

Okay

[vale nimbus]

back

k let me see

i feel like i had to add 2pi

[dusky surge]

Yea

[vale nimbus]

i cant say why again tho fuck sake

[vale nimbus]

mhm

so when it has to be below i have to describe counter clockwise and when i have to say when its above i have to describe clockwise?

or is that a bad generlisation

[dusky surge]

Nah, it's a bit more complicated than that

[vale nimbus]

[dusky surge]

when it's below,
Value negative, then it's cool
Value is positive, need to add 2pi

And the situation is different for 'above' case

[vale nimbus]

but wait

https://cdn.discordapp.com/attachments/326138757474680852/706186484679573564/JPEG_20200502_185244.jpg

here the value was negative and it wasnt cool

[dusky surge]

This is correct

[vale nimbus]

wait waa

so that one's correct?

that solution

[dusky surge]

Yea

Has always been

[vale nimbus]

omg i think

i really get it now

i dont do it with a graph but i try to interpert the (something)<x<(something)

and i think i get it now

[dusky surge]

Oh okay

[vale nimbus]

the smallest my angle can be is the thing on the left and the biggest my angle can be is the thing on the right

idk i cant really explain it

can you give me another problem with sin?

[dusky surge]

It's okay, as long as you get it xD

$sin(2x)<-\frac12$

[somber coyote]

Biscuit:

[dusky surge]

Hmmm...-5pi/6 is smaller than -pi/6

[vale nimbus]

... right

brainfart sorry

been at it for a while now :p

[dusky surge]

Yea, I know it'll be real tried for ya

[vale nimbus]

if u switch the left and right around tho, is it correct?

[dusky surge]

Anyway, if those two interchange, it's correct

[vale nimbus]

YES

[dusky surge]

Yea

[vale nimbus]

ok 1 last one

and then its dinner time for me

[dusky surge]

Let's try the 'above' case

[rich wolf]

Bro its only 1:30

[vale nimbus]

im from belgium

7:30pm for me

[dusky surge]

We're from different timezones XD

[rich wolf]

Lmao

[dusky surge]

01:30 for me

[vale nimbus]

AM?

[dusky surge]

Yep

[rich wolf]

Japan?

[vale nimbus]

that doesnt sound very healthy

[dusky surge]

Hong Kong

[rich wolf]

LOL

I was close

[dusky surge]

Indeed... It's unhealthy, staying up doing maths

[rich wolf]

Im pretty sure thats the same time zone though

HK and Japan

[vale nimbus]

could u give me 1 last problem? then its bedtime for you 😤

[dusky surge]

Doubt that> Im pretty sure thats the same time zone though
@rich wolf

Oooo

Sure

Lemme type

$\sin(x+\frac{π}4)>\frac{\sqrt2}2$

[somber coyote]

Biscuit:

[vale nimbus]

kk

[vale nimbus]

[dusky surge]

Correct!

[vale nimbus]

thanks so much dude

[dusky surge]

You're welcome!

[vale nimbus]

i think i understand now

hoping its kind of the same thing for cos

[dusky surge]

Graphically it's similar XD

[vale nimbus]

xD

[dusky surge]

But I'm really not sure about it

[vale nimbus]

could you give me a cos problem? or are you going to sleep now

[dusky surge]

For written ones

I can, lemme think

$\cos(x)>\frac{\sqrt3}2$

[somber coyote]

Biscuit:

[dusky surge]

Correct!

[vale nimbus]

and if it was < instead of >

would the thing above be correct?

the one that i kidna scribbled out

[dusky surge]

Yea!

[vale nimbus]

i get it dude

you're a life saver

[dusky surge]

You're good!

[vale nimbus]

thanks so much man

[dusky surge]

No problem man

[vale nimbus]

have a good nights rest now

[dusky surge]

Have a good dinner XD

[vale nimbus]

anyone else willing to help me get the hang of this?

think ive got sin and cos down

tan and cot are next

[devout harbor]

@vale nimbus https://www.math-exercises.com/equations-and-inequalities/trigonometric-equations-and-inequalities

good luck

[acoustic jungle]

Ok

I'm getting an impossible triangle

A triangle that doest obey the sin law

Could you guys check if the base triangle is not possible

[upper karma]

re-draw it

the angles of depression should be right beside the 315 m line

[acoustic jungle]

I'm just verifying whether or not the bottom triangle is impossible, I did the angles of depression correctly

Also don't refer to my drawing, I had to include it to fit the question in

[acoustic jungle]

also that 0 looks like a 2, it is 70 not 72 although this doesn't matter since it's not given in the question anyways.

[upper karma]

Lmao how is that a 0

then write 0 over it

so it's clear

yah how is that a 0

if it's an obtuse angle, you need 180-angle

[acoustic jungle]

I said before the drawing is in the picture because I had to fit it in

You know I'm going to crop it then

Actually doesn't even matter

The bottom triangle is impossible. The question has an error

[white dew]

can someone explain to me how we get from the second part to the third?

i tried multiplying but I believe I did it wrong because im getting something different

[acoustic jungle]

cross multiplying

QS * sin 2 = 100 * sin 31

wait.

something doesn't seem right...

are you sure that answer is right

[white dew]

my professor sent this problem as an example

so I am just going of the example so I can do the problem he gave

[acoustic jungle]

alright the answer 1475 is correct

[white dew]

he just did not breakdown his steps completely

[acoustic jungle]

but that third step is wrong

[white dew]

wrong how? the process?

[surreal flume]

How do I find the length of an arc in a circle when I am given the diameter and the measure of the angle of the arc

[upper karma]

do u have a diagram u can show us

[surreal flume]

its on the last page on the you try

[upper karma]

what question

[glacial haven]

arc length is given by the equation r theta

[surreal flume]

2 and 4

[upper karma]

what does the degrees outside the circle represent

[surreal flume]

the angle measure inside

[upper karma]

ok

Angle over 360

Multiply by circumference

Easy

for question 2) you have to find the area of the circle then multiply it by 47 then divide it by 180

and for question 4) u have to find the circumference then multiply it by 160 then divide by 360

[acoustic jungle]

[lofty dust]

what about the circumference of this?

[upper karma]

globama

Same thing C = 2 π r

Just find the radius

[lofty dust]

Thank you so much

[upper karma]

And plug it in

Your welcome

[rigid pawn]

-3cos(t)/-5sin(t)

[glacial haven]

what about it

[rigid pawn]

is that csc^2 or cot

[glacial haven]

3/5 cot(t)

[rigid pawn]

bc i put in the calculator it says csc^2

[glacial haven]

its not

[versed river]

,w -3cosx/(-5sinx)

[somber coyote]

Display results online and refine query

Requested by Sneaky#0928

[versed river]

See, its cotangent

[glacial haven]

isnt that what i said

wow how do you even do that indefinite integral

[versed river]

Yes. I was confirming it to the person who doubted you

[glacial haven]

how do you find the integral of cot(x) @versed river

wait nvm

you just use u sub

[desert pond]

dunno if this belongs here

but not quite understanding how they get the circle equation in part b

i understand the dot product, the vector on the right of the dot product is just the vector a

but how is the vector equation of the circle (r-a) ?

[upper karma]

UwU

[tired dagger]

@eager kraken

[glacial haven]

70

[winged hill]

can someone help me with this? i figured out the others but im still stuck on this

(sorry for the low quality pic my teacher uploaded it that way)

[dark sparrow]

inscribed angle theorem

[winged hill]

the sides dont have to be equal right?

equal in length*

[dark sparrow]

you don't give a shit about lengths here at all

[winged hill]

oo alright thankss

so its gonna be 113 x 2 which is 226

[trail whale]

Can someone help me w these 2

[trail whale]

I think the first one is 32

Can anyone help??

[sly grove]

@trail whale for the first one if you work out the length of all the sides of the triangle ACG then you could use the cosine rule

[trail whale]

Is the answer 29

[sly grove]

no idea lol, i havnt put it in a calculator

although looking at the diagram that seems about right

for the record the diagram isnt to scale but its close enough that you know that you are looking for an acute angle probably below 60 degree

[trail whale]

Ok thanks

[sly grove]

reckon you can handle the other one?

[trail whale]

No lol my teacher set me this and we haven't studied it once so it's hard

[sly grove]

ok so i just put the first one in the calculator and i also got 29

[trail whale]

Yep

[sly grove]

for the next part the process is exactly but the slightly harder part is choosing the triangle

[trail whale]

Ok

[sly grove]

well what triangle would you pick?

[trail whale]

Idk lol

[sly grove]

the first thing is do you get what the mean by the angle between the planes

[trail whale]

I suppose it means these angles

[sly grove]

not for this one

[trail whale]

So what angles

[sly grove]

AVB is the triangle and ABCD is the square

so its asking for the angle at where the two shapes meet

now a reasonable guess might then be to find the angle VAD, but that triangle is at angle so it would be larger than the real value so it wont work

so youll want to find a triangle that is "upright"

[trail whale]

Ok

[sly grove]

i dont want to fully give it away so ill let you know that two of the points have letters and the other one is the midpoint of a line

does everything i said make sense?

[trail whale]

So ve

I think

[sly grove]

VE are two of the points but you need one more for a triangle

the last one doesnt have a letter on it

[trail whale]

Ok

[sly grove]

figured it out?

[trail whale]

I think the answer is 67

[sly grove]

what did you take as the third point?

[trail whale]

That thing that don't have a ketter

[sly grove]

just ran it through my calculator and got 67 aswell, nice job

[lime storm]

I’m having difficulty with this problem. Originally i got 54 but the multiple choice only has. 32, 44, 52, and 76

the last parts says what’s m<jkm

[silent plank]

there's a theorem for it. i forgot the name but you can derive it from inscribed angles and external angle theorem (by constructing PJ or LM)

[upper karma]

@lime storm what is the value of angle JPM?

hint: ||inscribed angle theorem||

well... it's been 5 hours, and you're offline, so I'll just go ahead and post it

[rich wolf]

@upper karma bro dont just give out the answer

[upper karma]

yeah, you're right

he's not gonna learn anything like that

but I am gonna leave this here as a hint:

and this: ||remember that all angles from a triangle add up to 180 degrees. And once you have angle PKL, JKM is what's known as a vertical angle (look it up)||

hopefully it's enough to get him on the right path

[pallid cloud]

I couldn't find what they asked for in 3)

In 1) I showed that using Chasles
In 2) I showed the law of sines using 1)
In 4) I showed the law of cosines using the law of sines already shown in 2)

In 3) I showed that 2R=a/sinA=b/sinB=c/sinC using the law of sines and some geometry in the triangle and its circumcircle

But I couldn't express R in terms of 2 sides and the angle between them

Any help please?

Ping me when you answer please

[pallid cloud]

<@&286206848099549185>

[ancient axle]

um

is anyone alive

nvm

[acoustic jungle]

@pallid cloud deleted message

wait..

You have the answer don't you?

In 3) I showed that 2R=a/sinA=b/sinB=c/sinC using the law of sines and some geometry in the triangle and its circumcircle
c is just sqrt(a^2+b^2-2abcosC)

maybe I am misunderstanding the question?

[pallid cloud]

Fishraider I'm not allowed to use the law of cosines