geometry-and-trigonometry

[upper karma]

um use the diagram @silent plank drew

since u now agle ATB and angle BTS are both 90

ANd aot is isocleses

yah

[silent plank]

where's the 360 coming from?

you said you understood the diagram right?

[upper karma]

180-(90-x)/2

and 360/2-(90-x)/2 thats where the 360 came from

[silent plank]

um wtf is even going on atm?

[upper karma]

k @upper karma post the diagram again

[silent plank]

nfi what that is.

[upper karma]

no like the picture

@silent plank its angle chasing

no like the first picture

that u posted

yah

K @silent plank so OTB is 90 -x degrees

and then u do 180 - that all over 2 to get angle TOB

[silent plank]

its actually also equilateral

[upper karma]

^

oh........

[silent plank]

there are multiple ways to do this problem

[upper karma]

then just do 90-x=60

thats way more easier

[silent plank]

some are more efficient than others

[upper karma]

im so dumb how did i not see that

[silent plank]

derped, didn't realise it sooner

[upper karma]

k so the easiest way to do this problem

is that triangle OTB is equilateral

and u know angle OTB is 90-x

so 90-x=60

and x=30....................

WAYYYY more efficient than angle chasing

umm post the diagram again

k so angle ATB is 90 degrees right

and angle ATO is x

so angle ATB-ATO=OTB

so OTB=90-x

the reason why ATB is 90 degrees is bc of teh circle theroem and point on a circle that froms a triangle with opposite points of the diamter, that point will always be 90 degrees

k

[silent plank]

there are multiple routes to get what you need.

but you would need to justify how you determined that its equilateral

[upper karma]

i need help with a problem

#competition-math

[rigid siren]

is 81i the same as 81 + 1i

it seems weird

im confused

[devout harbor]

@rigid siren No.

[rigid siren]

@devout harbor are u sure

[devout harbor]

Yes

[rigid siren]

ok wait

i have a question

how do I find the 4th root of a complexnumber

[cedar imp]

https://gyazo.com/5d324bd4193496190371c30ad09e2ab6

can someone help?

[versed river]

Its a rectangle so opposite sides are equal

Thus you can set up an equation where those algebraic expressions are equal and solve for a

(XY=WV)

[cedar imp]

@versed river so far I got 7. I think it’s wrong tho

[acoustic jungle]

verify it yourslef

you can subsitute it back to see if it works

[versed river]

Show work

7 is wrong, show me your work so i can correct you.

@cedar imp

[rigid siren]

a = ||10||

its a simple equation

6a-57 = 3a-27

dont click the spoiler until you solve it urself

https://gyazo.com/e0fe53e2ee0493c8c28d98c0fa4b6778

[upper karma]

Hey is anyone able to help me with 2 trig questions?
im just really dumb and cant get it
if you can help pls dm me

[mellow vector]

u should send it here @upper karma

im sure people will help

[upper karma]

There it is, anyone able to help me understand this and help me get to the solution?

[onyx basin]

when you have sin(-t), we can move that negative sign out of the sin function

so sin(-t) = -sin(t)

iirc

[upper karma]

what about cos, and if the first one is already negative like (b)

[onyx basin]

it applies to all of them

cos(-t) = -cos(t)

if cos(t) is a negative number, and you're asked to evaluate cos(-t), what do you think would happen to that negative number?

[upper karma]

become positive?

[onyx basin]

correct

because let's say that cos(t) = -5

cos(-t) = -cos(t)

substitute -5 for cos(t), because we know what cos(t) is

so then you get -(-5)

which becomes positive 5

so if csc(t) = 4.5, do you know how to find csc(-t)?

[upper karma]

-4.5?

[onyx basin]

yep

cot(t) = -2.5, cot(-t) = ?

[upper karma]

2.5

[onyx basin]

yep

[upper karma]

Thanks dud UwU

[onyx basin]

if you're interested in the proofs, here's a website i just found that explains them well

https://www.mathdoubts.com/negative-angle-identities/

[upper karma]

Thank you!

[languid atlas]

I saw this channel was free, what is the area of the shaded region?

[silent plank]

consider the areas of the triangle AFE and rectangle ABCD

[scenic linden]

how do periodic graphs look like

[plucky marlin]

periodic

[fluid stream]

https://en.wikipedia.org/wiki/Periodic_function

[versed river]

they repeat over an interval

[plucky marlin]

[versed river]

don't, hes on a test lol

deleted message in questions a

[fluid stream]

I'm confused

[quiet mason]

questions deleted

[latent iron]

hi!

[latent iron]

could someone help me

The angle of elevation to the top of a building is 75 degrees from a point on the ground. The angle of elevation to the top of the building from a point 20 m further from the first point is 40 degrees , how tall is the building?

[median atlas]

@latent iron start by drawing a diagram

we know angles in a triangle add to 180° so we can work out the other angles

@latent iron

[latent iron]

THANK U

i appreciate it

[median atlas]

no problem

the most important part of problems like this is getting the diagram right

then normally you can see the relationships between all the values

[knotty barn]

This is not possible correct?

This is my rationale for it, they don't overlap at all and I slapped it into desmos and an integral calculator to make sure I wasn't going mad

https://www.desmos.com/calculator/uk8ckrchz0

Full desmos link^

It's not asking for the values of x where there are solutions either, just a full proof

[drifting parrot]

Can someone help me figure this out?

[quiet mason]

what have you tried

try to express that 13pi/2 as the sum of two fractions

Heres a hint ||13pi=12pi+pi||

[drifting parrot]

I’ve tried subtracting 2pi out of the 13pi

[unique junco]

Why did you do that stuff on the right

Can you just go around the unit circle 4 times and end up with cot (pi/2)

Which would be 0

well, 0/1 = 0

[drifting parrot]

I’m not sure how to do it that way

[unique junco]

sorry, 3 times around the circle

Cora, do u agree that 4pi/2 = 1 lap of a circle?

[drifting parrot]

Yes

[unique junco]

So would 12pi/2 be 3 laps of the circle

[drifting parrot]

Oooh, I see

[unique junco]

Since repeating laps doesnt do anything, we can discard that

leaving us with cot (pi/2) = x/y = 0/1 = 0

[drifting parrot]

What if it was something other than pi/2?

[unique junco]

whatever x is / whatever y is

So if it was 5pi/6

Tell me the x and y of 5pi/6

[drifting parrot]

X = -sqrt(3)/2 y = 1/2

[unique junco]

nice

so cot (5pi/6) = -(sqrt 3)/2 ALL OVER 1/2

do some more math, keep switch flip

-sqrt 3

[drifting parrot]

To get the sine, cosine, etc?

[unique junco]

Well we know what the sine and cosine are

you said waht they were

[drifting parrot]

Oh yeah 🤦🏾‍♀️

[unique junco]

But in order to find the cot, you divide the sin/cos

[drifting parrot]

Ok

[unique junco]

Any more questions?

[drifting parrot]

No that’s it. Thank you! 😁

Soo like this?

I’m not sure if I did this right @unique junco

[unique junco]

well ive never seeon someone start from the left side of the unit circle

Il do a VERY ROUGH drawing

[drifting parrot]

Ok

[unique junco]

https://i.imgur.com/MUztrDX.png

[slate pasture]

i need help with this

can anyone help

[unique junco]

shoot

Im blanking

[knotty barn]

Yea I'll help

Give me a sec

[slate pasture]

thanks

[unique junco]

All i got is that its either 05/13 or -5/12

[drifting parrot]

@unique junco that doesn’t look too rough

How did you get the 5pi/2 and the other values?

[unique junco]

we agreed that 4pi/2 is 1 lap of the circle right?

[drifting parrot]

Yes

[unique junco]

So what would 1pi/2 + 4pi/2 =?

[slate pasture]

when you guys are done with bridge could i get some help next

[drifting parrot]

5pi/2

[unique junco]

@slate pasture This problem is simple, shouldnt take too help you out with

Not trying to be mean when I said simple, I just looked up how to do it 😂

[drifting parrot]

ok. I just want to make sure I fully understand

[slate pasture]

ight send me what u found bridge

[unique junco]

Are you familair with: SohCahToa?

[drifting parrot]

me or @slate pasture

[unique junco]

Dame

[slate pasture]

yea

[unique junco]

So you agree that Sine = Oposite / Hypoteneus

[slate pasture]

yup

but im not familar with the unit circle '

[knotty barn]

Sorry, I'm back

[unique junco]

https://i.imgur.com/1PL4sXo.png

[knotty barn]

You don't need unit circle for this

[unique junco]

You can ignore ALL of the red in this

[slate pasture]

ight

[unique junco]

Now im gonna add 1 lin to help u out

https://i.imgur.com/TEWsUmB.png

Still good?

i guess i should say, still following?

[slate pasture]

yea

[unique junco]

So, since you know about Soh

Sine = Oposite / Hypoteneus

What is the length of the red line, which would be the "oposite" part of Soh

@knotty barn SOrry for stealing ur job, u can take over if uwant

[knotty barn]

Ok sure

[slate pasture]

13right

or 12

[knotty barn]

Hypotenuse is 13

[unique junco]

^

[slate pasture]

alright beacuse 13 is the radius

[knotty barn]

(-5,12) is your coordinate, so 5 is your width, and 12 is your length of the triangle

[slate pasture]

12

[unique junco]

Yes, or you could have done sqrt of (-5^2 + 12^2) = 13

[knotty barn]

Yea

Do you know what the RAA (Related Acute Angle) is

[slate pasture]

so my adjacent is -5

[unique junco]

Nope (To helixion)

[knotty barn]

Ok hold on

[unique junco]

@slate pasture yes Adjacent is -5

but were not looking for that because were find S = o/h

now that we know both o and h, what is the answer?

[slate pasture]

12/13

[unique junco]

Great

[slate pasture]

so now i need to find the rotational

[knotty barn]

$sin(\Theta)= \frac{12}{13}$

[somber coyote]

Helixion:

[unique junco]

Cool bot

[drifting parrot]

When your done, could you finish helping me @unique junco

[knotty barn]

@slate pasture You following along?

[slate pasture]

yea

[unique junco]

@drifting parrot Oh sorry

Thought we were done

Yes so

you agree that 1, 5, 9, 13 are all increased by 4?

[drifting parrot]

yes

[knotty barn]

@slate pasture Do you know what Related Acute Angle and Principal Angle are?

[slate pasture]

nah

[unique junco]

Lastly, you also a gree that pi/2 gets you to (0,1) (the top of the circle)

[drifting parrot]

yes

[unique junco]

Awesome

so we do 3 laps of the circle

WHich would be 12pi/2

add 1pi/2 and get to 13pi/2

[knotty barn]

@slate pasture Moving this to #help-1

[unique junco]

All of those points I marked (5pi/2, 9pi/2) were the number of rotations needed to get to (0,1)

I feel like im doing a terrible job at explaining this

[drifting parrot]

uhh

[unique junco]

can you join a call?

[drifting parrot]

no your not

it's me

i might be able to

hold on

[unique junco]

everyone gets stumped by math, dont worry bout it

[drifting parrot]

Ok, I can join a call

[slate pasture]

im guessing that it makes more sense for the eqaution to be the last one for #5

[slate pasture]

if anyone mind checking that the last answer for number 5 is correct let me know

[dark sparrow]

what do you think it equals?

what have you tried so far to find it and where are you stuck?

[acoustic jungle]

@slate pasture you should cover your email adress

[slate pasture]

how

[silent plank]

use win+shift+S
to use the windows snipping tool instead of screening the whole window

or use alternative programs

[ionic bluff]

lmao time to dox you now

[acoustic jungle]

the question is ambiguous unless it says from the starting point to the center of the circle to the wall in 90 degrees

[dark sparrow]

@upper karma why do you think x=2?

[acoustic jungle]

and if it says that the answer is not 90 seconds

[dark sparrow]

what

ok ducksmeat let's move

[dark sparrow]

this graph looks like a sinusoid overlaid on top of a line

[slate pasture]

yea it says that two functions are being added and im confused a bit

[dark sparrow]

what function's graph looks like a line?

a straight line, that is

[slate pasture]

linear

[dark sparrow]

well there you have it

[slate pasture]

oh that simple thanks

another thing

is this equation correct

[dark sparrow]

your answers to questions 6, 7 and 9 are correct, but your answer to question 8 is not.

[slate pasture]

alright so the period

is not 3

its 2 pie then possibly

[dark sparrow]

pi, not pie

and no, the period is not 2pi

[dusky cobalt]

yea

[slate pasture]

ok i get it its 4 pi cause it starts at -1

[dusky cobalt]

:DDD

[drifting parrot]

I need help again

[devout harbor]

, rotate

[devout harbor]

@drifting parrot You can rewrite this as sin(theta) = (91) / (-10*sqrt91)

Sin(theta) is opposite/hyp, so make a right angled triangle with the given sides

[devout harbor]

, rotate

[drifting parrot]

,rotate

[devout harbor]

Okay, forget the negative sign for now

Sin(Theta) = Opposite/Hyp, make a right angled triangle with the given sides

@drifting parrot Can you make it or should I?

[drifting parrot]

i'll try it

,rotate

[devout harbor]

Yes. The side opposite to the angle is of length 91

And the hypotenuse will be of length 10sqrt(91)

put em in and show me

[drifting parrot]

How do you know this

Ok

[devout harbor]

It's because sin(theta) = opposite/hyp

[devout harbor]

Yes

Now use the pythagorus theorem to find the 3rd unknown side

[acoustic jungle]

why

don't you just need to figure out theta

why do you need to know the unkown side

[devout harbor]

So that we can find the other trig functions

Once we find the unknown side, we can find cos(theta), tan(theta)

and in exact form

[drifting parrot]

ok. I finished solving it. I'll show you in just a minute

[acoustic jungle]

I thought the question just asks for theta

ohh

[devout harbor]

ah look above at the actual question

[acoustic jungle]

ignore me

[devout harbor]

all g

[acoustic jungle]

I have only seen the cropped version of the problem

[devout harbor]

Okay now you have b = sqrt(819)

You have all the 3 sides of the triangle, now you can use cos(theta) = adjacent/hyp and tan(theta) = opposite/adjacent

find their values

[silent plank]

simplify the radical

[drifting parrot]

@devout harbor

[hollow lava]

x = 12/5 correct?

[silent plank]

yes

[hollow lava]

Thank you.

[prime barn]

Anyone know how to solve this? I

I'm getting stuck every move

[acoustic jungle]

I got it

start from the right side

and also note that cos(pi/2 - x) = sin x

@prime barn

[prime barn]

why would you start on the right? doesn't the left look more complicated?

[acoustic jungle]

sorry I am getting my directions mixed up

you are correct

🤦‍♂️

[prime barn]

any chance I could see a picture of your work?

[acoustic jungle]

it's a bit messy

i'll redo it

[prime barn]

its fine

[acoustic jungle]

@prime barn

[prime barn]

on the second arrow, did you multiply both sides by sinx?

*sinx over sinx

[acoustic jungle]

yes

watermelon poo

hmm I should've just multiplied both by sin^4 x on the second arrow would've been simplier

[prime barn]

well thanks anyway

[prime barn]

what about this one?

you need to simplify in a form that can be graphed

[upper karma]

Damn that’s hot

[acoustic jungle]

idk what it means for it to be in a form that can be graph

you should try it yourself first

cot=cos/sin

csc=1/sin

[prime barn]

i did but im lost

when it means in a form to be graphed it means usually only one trig function

like 2 sin x

or 1/2cos x

[prime barn]

nvm figured it out

[acoustic jungle]

@prime barn what is it

I cannot simplify it to 1 term

[prime barn]

you need to convert cot into cscx -1

using the pythagorean identity

and then factor that equation

so that its

[acoustic jungle]

just tell me the answer

[prime barn]

(cscx +1) (cscx-1)

so it becomes 3cscx -3

thats the answer

[acoustic jungle]

oh wait I had that

I had (3(1-sinx))/sinx

bruh.

[pastel anchor]

send help my brain is blowing up

[acoustic jungle]

@pastel anchor have you tried drawing right triangles

[pastel anchor]

huh

[acoustic jungle]

There are 2

[pastel anchor]

how would right triangles help, i don't get it

[acoustic jungle]

just do it

[pastel anchor]

alright let me draw them real qucik

ok what about them

[acoustic jungle]

and after if you don't know what "similar triangles" mean then search it up

[pastel anchor]

i still don't get it

[acoustic jungle]

Do you know what similar triangles are

[pastel anchor]

yea but this is about secant tangent theorem

[acoustic jungle]

ok that works too

AB*PB=CB^2

use Pythagoras to find CB^2

[pastel anchor]

u need PB to find CB^2

and the question is asking me for PB

[acoustic jungle]

No you don't

You need AB to find CB^2

and you are given AB

[upper karma]

Geometry is cool ngl 👍

[pastel anchor]

yo can u give the steps m8

[acoustic jungle]

ACB is a right triangle

[pastel anchor]

ok i found the CB

its 164

but i plugged it in for secant tangent formula

[acoustic jungle]

it doesn't work because it's wrong

AB is the hypotenuse

not the side

you should be subtracting

[upper karma]

6

6 right

Cb is 6

@acoustic jungle 👍

[pastel anchor]

oh

[upper karma]

Ya I’m correct

Woohoo

Don’t know what to do next tho

[pastel anchor]

what is x?

[acoustic jungle]

use the secant formula

find y first

[pastel anchor]

this is secant secant

[acoustic jungle]

yes

use the secant secant formula\

[pastel anchor]

its 32 right

[acoustic jungle]

maybe maybe not

[pastel anchor]

bruh

[acoustic jungle]

tell me what you did

[pastel anchor]

so i found the top secant

which was 48

48=2y+y

y=16

x=32

[acoustic jungle]

https://www.google.com/search?q=secant+secant+formula&rlz=1CARWXF_enCA890&oq=secant+s&aqs=chrome.0.69i59j69i57j69i59j0l5.1452j0j7&sourceid=chrome&ie=UTF-8

[pastel anchor]

wait

[acoustic jungle]

product, you accidentally used addition

[pastel anchor]

ah

but the formula is correct right?

[acoustic jungle]

Yes?

[pastel anchor]

idk the answer

all ik is y = 2squredroot6

[acoustic jungle]

and x = 2y

what is stopping you from getting the answer

[pastel anchor]

can x be a decimal

[acoustic jungle]

a variable can be anything

yes

[pastel anchor]

so i got 9.8

becuse i round

[acoustic jungle]

ok.

[modest spear]

https://cdn.discordapp.com/attachments/686262935554621441/699770358412345384/unknown.png

im so bad at trig lmfaooo

[umbral snow]

Any other way to write sin(2θ)?

[modest spear]

would i use

sin2x = 2(sinx)(cosx)

?

[umbral snow]

Got it

[modest spear]

awesome thank you :-)

[pastel anchor]

lmfao whats the chances of me finding a tarzaned fan

[acoustic jungle]

baby puto

use pythagoras

[pastel anchor]

@acoustic jungle do i make my own line on top or the bottom

[acoustic jungle]

if you draw the question on paper

the perpendicular bisector of that line passes through the center of the circle

and you would use pythagoras

if you think about it, the center to B would be r-4

[pastel anchor]

ah

u see its those small things i can't remember

[modest spear]

https://cdn.discordapp.com/attachments/695856813303922798/699785570905489519/unknown.png

sin A is 15/17
need help with cos(2A)
it says the identity is cos(2x) = cos^2(x) – sin^2(x)

[devout harbor]

Yes.

cos^2 (x) is same as [cos(x)]^2

sin^2 (x) is same as [sin(x)]^2

you have cosx and sinx

[modest spear]

ah okay thank you

<3

helped a lot

[clever flame]

Can anyone help me out with this question? "A person is looking at a boat out of a lighthouse window that is 32 m above the water at an
angle of depression of 29°. How far away is the boat from the base of lighthouse?"

[supple wedge]

diagram pls

[clever flame]

there's none ;-;

[supple wedge]

draw 1

visualise

[clever flame]

this is the thing i was drawing but idk how to solve it if i dont know where to put the values at

plus not sure if this is even right

[supple wedge]

er

flip the triangle around

cuz u put the 90 degree as 29

xd

@clever flame

[clever flame]

like this? lol

[supple wedge]

29 is on the left side

[clever flame]

o ok

[supple wedge]

ur triangle is wrong

,rotate

[supple wedge]

,rotate

[supple wedge]

ok this shape is correct

use sine rule

btw

@clever flame

[clever flame]

oh

my internet died LOL

o and ya thank you

=12

[junior thunder]

I'm confused how they got the final answer

[devout harbor]

what's confusing u

[junior thunder]

the radical 3 over 3. how do you get that from 1/3

[versed river]

what do you get if you multiply sqrt3 by sqrt3

[junior thunder]

3

[versed river]

so maybe they multiplied the numerator and denominator of the fraction by root3?

[junior thunder]

Ok

Thanks

[compact spire]

Need some help figuring this out

[devout harbor]

What's confusing u

[compact spire]

All of it is confusing me at the moment, I don't quite know where to start with part a.

[devout harbor]

Well V=<1,-5> means V= 1i - 5j

You can do the same for W now

V - W means you have to subtract their respective components( i with i, j with j, etc)

[compact spire]

W=11i+5j

[devout harbor]

Yes.

What about V-W?

[compact spire]

(1-11)(-5+5)

?

[devout harbor]

(1-11)i + (-5-5)j

[compact spire]

Okay

(1-11)i+(-5-5)j

[devout harbor]

Yes.

And that's equal to -10i -10j

[compact spire]

Okay

[devout harbor]

Or -10i + (-10j)

For part b, you can try making a diagram if you want

[compact spire]

So that is the answer for it?

[devout harbor]

Yes.

[compact spire]

not too bad

Okay so part a done

[devout harbor]

Can you make a diagram for part b?

[compact spire]

diagram?

like a graph

[devout harbor]

yeah

[compact spire]

sure

give me one sec

for part B it'd just be 1i-5j

[devout harbor]

?

no

[compact spire]

Oh

[devout harbor]

It's asking for the angle

[compact spire]

okay

so just draw 1,-5

[devout harbor]

Go 1 unit to the right(x-axis direction)

[compact spire]

and down 5

[devout harbor]

Then 5 units down

Yes

[compact spire]

okay got it

[devout harbor]

and draw the resultant vector from the origin

[compact spire]

okay

drawn

[devout harbor]

how do we find the angle

[compact spire]

well its like a triangle

[devout harbor]

The angle this vector makes with the positive x-axis.

Yes.

tan(theta)...?

[compact spire]

tan = sin/cos

of theta

[devout harbor]

tan(theta) = opposite/adjacent

[compact spire]

sorry

opp/adj

[devout harbor]

opposite = 5, adjacent = 1

find theta

[compact spire]

5/1

sorry

one sec

[devout harbor]

find arctan(5)

radians/degrees, both work

[compact spire]

1.37

[devout harbor]

okay, that's in radian

actually the question asks in degrees

[compact spire]

78.7

[devout harbor]

do it again with degrees

okay

so the angle this vector makes with the positive x-axis is -78.7 degrees

[compact spire]

Okay

[devout harbor]

for part c, make the diagram if you want

use the pythagorus theorem to find the length of this vector

i g2g sorry

[compact spire]

Thank you for the help

[devout harbor]

np

[versed river]

c and d draw a diagram and use pythag theorem,

you dont need a diagram if you feel you dont need it

[compact spire]

so for c I got 12 sqrt 22

[versed river]

im in a league game rn if noones helped you when im back i will check your answer

[compact spire]

thank you sneaky

I appreciate y'all

[upper karma]

Anyone can help me with this? how I can solve this trig problem?

[versed river]

@compact spire im done

ping me when you're finished your warzone game

$12*\sqrt{22}$ is wrong

[somber coyote]

Sneaky:

[little osprey]

@upper karma Are you still here?

[upper karma]

Yes

@little osprey I’m here

[little osprey]

Ok well do you know how to find the period

[versed river]

am i missing something or is that not ashby

[little osprey]

I mean he responded immediately

Yeah maybe not

Plus Ashby wasnt online until I pinged

[upper karma]

i pm u

but yeah, any help with that problem would be great

how do u go about solving those?

[little osprey]

Well do you know what the period is

[upper karma]

no, please explain

[little osprey]

A period is when the graph completes

okok

Do you know what the formula is

usually

lets start withthat

Welp if youre busy

https://www.youtube.com/watch?v=Y8XlieOhyHs

This can help

[upper karma]

i dont get how he says the period is 1

btw im in the middle of something rn, but im juggling this with the other thing im doing

so im sorry for the delayed responses

[little osprey]

Ahh understandable dw about it

Well a period is the number of x axis units needed to complete one cycle

So how many x units are there before the graph repeats itself or completes a cycle

[upper karma]

i cant tell on this one

what is the period?

6?

[little osprey]

Ok can you see the point (0, -4)

how long is it on the x axis before you see another point that has the same y co-ordinate

[upper karma]

-5 u mean?

[little osprey]

oh my b

yeah

you can tell me

Like dont be scared to be wrong

[upper karma]

2.5?

so 2pi/2.5?

[little osprey]

no like

on the x axis

You see (0,-5)

When is the next time you see

(x,-5)

Like what is the x coordinate

[upper karma]

I love geometry ❤️

[little osprey]

4pi

Can you see it on the x axis?

[compact spire]

@versed river damn I thought I had it

so sqrt (11^2)+(5^2) isnt what im doing then

[versed river]

yes

it is

are you sure thats 2 root 22

or 12 root 22

whatever

should be sqrt(146)

thats irreducible

12root 22 is massive

thatd be more than 48, which is obviouslyt more than root146

@compact spire sqrt(11^2+5^2)=sqrt(146), which can only be factored into sqrt(2*73) which is obviously irreducble

[compact spire]

Okay

I guess I was doing it wrong somewhere

[versed river]

how did you get 12root22

im curious

[compact spire]

11^2 = 131

thats how

lmfao

I wrote a 3 instead of 2

[versed river]

would still be wrong lol

[compact spire]

yeah

[versed river]

do you know how to do d

https://cdn.discordapp.com/attachments/326138757474680852/699823520003129425/325558b58a648eae2638701322a1b234.png

[compact spire]

Is it kind of like doing A but I get the |w| part of it

of both

[versed river]

well what did you and sup? get for V-W in standard vector form?

[compact spire]

78.7 degrees

-78.7

[versed river]

thatd be polar form

im talking i and j

[compact spire]

Oh

-10i-10j

[versed river]

yup

so do the same process you did for W

[compact spire]

Okay

[versed river]

you shouldnt need to draw a diagram if you understand how c worked

[compact spire]

Im confused on where to start on this now

[versed river]

well, itd be a triangle

so you can use pythag theorem

the minus sing only changes the direction of the vector

the magnitude is still the same

can you derive a general formula for the norm of a vector xi+yj using the pythagorean theorem?

norm being magnitude, length whatever

the double lines

[compact spire]

im not feeling too well right now. going to eat some food real quick

be right back

Okay so I need to get the double ling thing that I did with c but for v and then just do part a again?

[versed river]

you dont need to do anything from part a again, you already have your result from part a

[compact spire]

So I just use my result and do the part c on it?

[versed river]

yes

[compact spire]

Perfect, eating now and then I'll post my answer in a little

[versed river]

sure thing

[rich wolf]

Damn it really be like that

[compact spire]

Okay Im slightly confused on this now

[versed river]

do you have a general idea of what the vector would look like?

[compact spire]

so I have $\sqrt{-10i^2-10j^2}$

[versed river]

\sqrt

[somber coyote]

Pony Boy:

[versed river]

i mean its probably more appropriate to write $\sqrt{(-10)^2+(-10)^2}$

[somber coyote]

Sneaky:

[compact spire]

$\sqrt{(100)+(100)}$

[somber coyote]

Pony Boy:

[versed river]

yup

[compact spire]

$\sqrt{200}$

lmfao

[somber coyote]

Pony Boy:

[versed river]

yes. you can simplify that if you want

[compact spire]

20

?

oh wait

[versed river]

nope

that would be sqrt(400)

its not an integer

[compact spire]

$10*\sqrt{2}$

[versed river]

$

[somber coyote]

Pony Boy:

[versed river]

yes

[upper karma]

Any helpers on?

[versed river]

are you still struggling with the trig graph ashby?

[upper karma]

i was busy, so i wasnt able to work on it

but now im free

[versed river]

repost the graph

[upper karma]

im so confused on it bro

[compact spire]

thanks for the help @versed river

[versed river]

Did you watch the video the other dude sent?

thats not me

[upper karma]

yeah i did

but he didnt explain how he got the period, and JC explained it, but i still dont see it

[versed river]

Well repost the graph and we can have a look

[versed river]

So period is, in layman's terms, the time it takes for a period graph to repeat

can you see where the pattern starts to repeat on that graph?

[upper karma]

starts on (-pi, -2)

[versed river]

sure, lets use that

when does it repeat again from there?

[upper karma]

what do u mean by repeat

Sorry im really slow

[versed river]

don't worry

complete one cycle, you could say

[upper karma]

it ends on 5pi

[versed river]

yeah, but its already repeated since then

actually

[little osprey]

do you know what amplitude is

[upper karma]

yea it is the max height of it

[little osprey]

ok good

[upper karma]

for that graph to get period, we need to divide 2pi/B

how do u get B?

[versed river]

Denton do you wanna take over or do you want me to keep help them? and you find B by setting the period = to 2pi/B and solving for B

[little osprey]

Idk I have online class soon

Do you have time?

[versed river]

I have time, school holiday for me rn

[little osprey]

Ahh ok, can you please take over then :)

[versed river]

yeah

[little osprey]

Thank you

[versed river]

soso @upper karma you can also think of period, in the same vein, as the distance between two maximums or two minimums in a graph

[upper karma]

@versed river can we PM?

[versed river]

sure

[teal frigate]

hey can someone help meout

[little osprey]

just ask the question

[teal frigate]

i have a screen shot

[versed river]

sohcahtoa

[teal frigate]

yea

kinda get it but not like this

[versed river]

you have sin(Θ)=O/H

can you identify which sides are O and H in this case?

[teal frigate]

hold on

8 and 10?

no 6 and 10

[versed river]

yes

6 being the opposite and 10 the hypotenuse?

[teal frigate]

ye

[dark sparrow]

well then, unless you forgot how to divide, you now know the value of sin(θ)

[teal frigate]

o

ok

thx

[surreal bolt]

Is it possible that the parallel postulate is equivalent to "all triangles have 180 degrees as the sum of their angles"?

[granite mortar]

yes

it's not only possible, it's also true

[modest vine]

What are the units for the measure of volume?

[acoustic jungle]

^3 @modest vine

[dark sparrow]

(length unit)^3

[drifting parrot]

I need someone to check my work. I feel like there's something I did wrong

[silent plank]

,rotate

[silent plank]

be careful of signs when taking square roots

and don't mix lowercase with uppercase

simplifying $\sqrt{\sin^2(B)} = \sqrt{\frac{25}{169}}$ gets you \
$\sin(B) = \pm\frac{5}{13}$ \
and you should choose the appropriate sign based on the location of B

[somber coyote]

ramonov:

[drifting parrot]

you're right

I forgot about the signs

🤦🏾‍♀️

[deep trail]

so there is sin cos and tan for things like opposite over hypotenuse and stuff but there is nothing for adajcent over opposite. Or is there one and what is it

[drifting parrot]

opposite over adjacent is tangent

[deep trail]

i know

[drifting parrot]

soo the opposite of that is cotangent

[deep trail]

oo

[drifting parrot]

which is adjacent over opposite

[deep trail]

ok that makes sense

[silent plank]

you seem to want the reciprocal functions for sin,cos,tan which are
cosec, sec, cot respectively

[deep trail]

ok

https://gyazo.com/774c7b4fe795c9ccb6f1b49d118b123e help ):

[little osprey]

have you drawn it

[deep trail]

i dont know what to draw. what does it mean by elevation to the top of the pole is 22 degrees. and elevation to bottom is 20 degrees

[little osprey]

well

you know the pole is on top of a building

right?

[deep trail]

yeah i dont need to draw that

[little osprey]

no but you do

you know the height of the pole from the ground is the height of the pole + the building

[deep trail]

https://gyazo.com/23eaf20951dda3aa10fcb60a87729ade

perfect

[little osprey]

put the pole at a side

so that it makes more sense

[deep trail]

https://gyazo.com/44e1b4d310beb0d065eb2d6fd8359d33

[little osprey]

Bahahhaha no I meant at the top of the building but the side

[deep trail]

ok

[little osprey]

youre trying though

[acoustic jungle]

lmao

[deep trail]

i did it

[little osprey]

So can you see that the height of the pole of the ground is 400 + height of pole

[deep trail]

yes

[little osprey]

so what angle does the building and the ground make

[deep trail]

90

[little osprey]

so its a right angled triangle

[deep trail]

its a building

[little osprey]

yup youre right

coming back to the question

the angle of elevation to the top of the pole is 22degrees

[deep trail]

what does that mean...

[deep trail]

is it like https://gyazo.com/93d53ad285b3b050225fa6bcce34115b

[little osprey]

Fishraider's diagram

[deep trail]

oh

[little osprey]

Gameson, do you understand Fishraider's diagram?

[deep trail]

i think so

the answer would be 20 right

wait no 40

[little osprey]

the answer to the question?

[deep trail]

yeah

[little osprey]

Wait do you know your trigonometric ratios

[deep trail]

maybe

[little osprey]

Its ok if you dont :)

[deep trail]

is that like tan, cos, sin

[brittle knoll]

any one want to help me with a test

[little osprey]

wait a test?

Gameson

YUP

So do you know like the abreviations

[deep trail]

yes

[little osprey]

could you name them so I could be sure

[brittle knoll]

yes

[deep trail]

i jsut DID

[brittle knoll]

\

[deep trail]

sin cos tan

[brittle knoll]

i got a 40

[little osprey]

I meant like opposite

[brittle knoll]

i need help sir

[little osprey]

hypotenuse

and adjacent

[deep trail]

oh

its just sohcahtoa

i know them

[little osprey]

@brittle knoll Have you drawn a diagram?

Ok ok good

[brittle knoll]

wdym

[little osprey]

Like have you drawn how the question would look like

@deep trail So.... you have the height for this right

[brittle knoll]

no sir

[deep trail]

@brittle knoll sqrt(412.25)

[brittle knoll]

thats the answer?

[deep trail]

yes

jc denton i think i have the answer already

[little osprey]

ok whats your answer

[deep trail]

40

[little osprey]

and give me like 3 minutes to check it

[acoustic jungle]

did you do tan 23 * 400 - tan 20 * 400

[deep trail]

3 minutes

um no

i saw that the 2 degrees was 1/10th of the building so i thought the pole would be 1/10th of the buildings height

[brittle knoll]

if i go live can u give me answerrs

[deep trail]

20 degrees / 10 = 2 degrees soooo 400 / 10 = 40

[little osprey]

yeah no

its not 400

[deep trail]

it says its 400 feet from the gruond

[little osprey]

its asking for a height of a flag

[brittle knoll]

whats the formula for this

[little osprey]

do you think a flag is 400 feet high

trig

[brittle knoll]

if i go live who can help me

[deep trail]

No

[acoustic jungle]

soupy this channel is occupied

[deep trail]

I said it's 40 feet high

[brittle knoll]

sorry

[little osprey]

Still wrong

[brittle knoll]

fish raider can u help me

[deep trail]

Way is it then

[little osprey]

Gameson, can you show your steps

[acoustic jungle]

No, I have math jams meeting thing

[little osprey]

soupy can you draw a diagram

[deep trail]

Just a sec

Building is 400 feet tall. And the angle thingy from the bottom is 20. But when it is 22 you can see the top of the pole. So when the angle was 10% more u could see the top. So I just multiplied 400 by 10%

[acoustic jungle]

No that is not how it works

and the building is not 400 feet tall

you are 400 feel from the base

[little osprey]

^

they are not similar triangles

you cant do that

[deep trail]

Uhhhhhh okayty

[little osprey]

you have to use trig man

[brittle knoll]

where does the 44 go

[deep trail]

But it's not a triangle it's a BUILDING

[brittle knoll]

hold on my bad

[little osprey]

what

no

its a kite

youre almost right soupy

[brittle knoll]

Item 2
You are holding a kite string in your hand. The angle of elevation from your hand to the kite is 44degrees and the distance to the kite is 250 feet. Your hand is 5 feet above the ground. How high is the kite? Round your answer to the nearest tenth of a foot.

[little osprey]

Gameson was referring to his one question

Ok imagine a person flying a kite

or not flying one

5 feet is pretty short ngl

[deep trail]

RIDING a kite

[brittle knoll]

i was thinking the bottom?

is this right?

[little osprey]

wait did you say the hypotenuse is 5 feet

no

[brittle knoll]

yes

[little osprey]

could you draw a human with a string attached to the kite

to the 44 degrees

[brittle knoll]

where is the kite?

[deep trail]

https://gyazo.com/7186cdc00426f0402c9ce6af54ef9408

whys it not SHOWING UP

[little osprey]

soupy imagine your hand is 5 feet above the ground

[upper karma]

Hi idk how to do this. 😳 anybody knows? formula?

[little osprey]

@upper karma Use exact trig ratios

[brittle knoll]

im very stupid sir

[little osprey]

@deep trail Could you send again I cant see

youre really not

[deep trail]

https://gyazo.com/7186cdc00426f0402c9ce6af54ef9408

[little osprey]

it takes time

[deep trail]

idk it snot working

it doesnt work

d

there it worked

[acoustic jungle]

bro

wtf is that

[deep trail]

lol wut?

thats my school picutre bro

my yearbook photo

[brittle knoll]

is the red right?

makes more sense

[deep trail]

is that a cracked version of adobe products i see

[little osprey]

ok the 250 is right

that's a start

Ramanov feel free to step in

[silent plank]

the 5s are in the wrong place.

include the human and the ground in the diagram

[little osprey]

I think thats the part where he is confused

[little osprey]

OOOO CLOSE

[brittle knoll]

im confused where the human goes plus where thee 44 goes

i havent done this in 3 months

would the 5 be inside?

[little osprey]

Ok imagine