geometry-and-trigonometry

[pallid cloud]

It's in 4)

It is to be shown later

[acoustic jungle]

oh sorry.

[pallid cloud]

I have R=a/2sinA

But they want me to express it in terms of let's say a b and sinC

And S=abc/4R comes after I show that S=1/2.bcsinA and R=a/2sinA, it's not shown yet and I don't think it gonna help

[acoustic jungle]

I don't think it does

that's why i deleted the message

[pallid cloud]

I think there's something wrong in the question idk why

S can be expressed the way they asked

And R the simplest form of it is a/2sinA

I tried to use vector product

I got ab=R^2.|1-|sinA|-|sinB|+|sinC|| or something like that

I even said R=a/2sinA=b/2sinB

So R^2=ab/4sinAsinB

Used some trigonometric formulas

But didn't get anywhere
Maybe I missed something or the question is wrong

I never encountered a simple expressing of R in the form they're asking for

[white dew]

Is there someone I can dm about proving trig identities.

I do not want to cut in on the problem being worked on here atm

[glacial haven]

rewrite tanx and secx in terms of sinx and cosx

[white dew]

So focus on the left side, not right

[glacial haven]

you can do either but left side looks easier

[deep trail]

https://gyazo.com/1adb1fd3ce458bf2ac67a0e3465d4953 so if I have this then cos(63) = 23/x and i forgot how to find x pls help

[silent plank]

do the same stuff to both sides of the equation to work towards isolating x

[deep trail]

divide by 23?

[glacial haven]

use trig

[silent plank]

that may help depending on what you do afterwards

[glacial haven]

oh wait you did

[deep trail]

cos(63)/23 = 1/x

[silent plank]

(if you meant dividing both sides by 23)

[deep trail]

now im stuck

[glacial haven]

dont divide by 23

[rich wolf]

Now take the reciprocal

[silent plank]

you could take the reciprocal of both sides

[deep trail]

i forgot wat that means

[rich wolf]

You can do this because the reciprocal function is injective

[deep trail]

wait so just cos(63)*23 = x

[silent plank]

no

probs better to approach it another way

from cos(63°) = 23/x
consider first multiplying both sides by x

[deep trail]

well honestly i can probably just guess from the asnwer choices i have

[silent plank]

don't guess

[deep trail]

oh i got it

ty

[white dew]

This is correct I hope

[upper karma]

It is.

[gritty sail]

can someone draw this for me

i drew it incorrectly

and got wrong answer

i cant imagine it

[rich wolf]

@gritty sail use geogebra

[gritty sail]

how do i make it 3d @rich wolf

nvm i found it

uh nothing shows up

idk how to use

it doesnt work

[rich wolf]

@gritty sail search up "geogebra 3d"

On google

[gritty sail]

i did

[upper karma]

?

[dark sparrow]

!

[upper karma]

@upper karma use the section of a circle formula or derive it from the circle one

[pallid warren]

Hi, how do i calculate x and y axis of a triangle if i know that the hypotenuse is 10 meters and the x and y is in a different coordinate system. How do i convert x and y to be in the same coordinate system as the hypotenuse.

[dark sparrow]

you mean in a different measurement system?

i.e. that the legs are 0.3 and 0.2 respectively of an unknown length unit?

[pallid warren]

ye

[dark sparrow]

well

letting x be the length of this unknown length unit in meters

you have $(0.3x)^2 + (0.2x)^2 = 10^2$

[somber coyote]

Ann:

[pallid warren]

ye

[dark sparrow]

well

[pallid warren]

how do i get x, long time ago i did this kind of stuff 😄

[dark sparrow]

do some algebra to it

0.09x^2 + 0.04x^2 = 100

0.13x^2 = 100

[supple wedge]

Find, in radians, the acute angle for which $\frac{1}{(1+cosec\theta)(sec\theta-tan\theta)}=3cot\theta$

[somber coyote]

Hmm:

[supple wedge]

fucking shit i overcomplicated it for myself now im stuck 🙂

ping me if u wanna help

[upper karma]

Jeez

Try putting them in terms of cos and sine

[supple wedge]

u mean

make it into non fraction?

[dark sparrow]

"put things in terms of cos and sin" means "put things in terms of cos and sin"

and then perhaps do some algebra to get rid of the skyscraper fractions

[supple wedge]

lets see

i got rid of the fraction

cuz

1/sec

all that

thats what i meant

@dark sparrow

idk what im doing

[dark sparrow]

hhf wow you

overcomplicated it way too much

gonna replace $\theta$ with $x$ for simplicity:

$\frac{1}{(1 + \frac{1}{\sin(x)})(\frac{1}{\cos(x)} - \frac{\sin(x)}{\cos(x)})} = \frac{3\cos(x)}{\sin(x)} \ \frac{\sin(x)\cos(x)}{(\sin(x) + 1)(1 - \sin(x))} = \frac{3\cos(x)}{\sin(x)} \ \frac{\sin(x)\cos(x)}{1 - \sin^2(x)} = \frac{3\cos(x)}{\sin(x)} \ \frac{\sin(x)\cos(x)}{\cos^2(x)} = \frac{3\cos(x)}{\sin(x)} \ \tan(x) = 3\cot(x) \ \tan^2(x) = 3 \ \tan(x) = \sqrt{3} \ x = \pi/3$

[supple wedge]

xddddd

o

i am big sumb

[somber coyote]

Ann:

[supple wedge]

dumb

oooooooooo

ok

thanks <3

[supple wedge]

how do i find area of equilateral triangle with side x meters and 60 dergrees

do i use $\frac{1}{2}ab sin c$

[somber coyote]

Hmm:

[acoustic jungle]

Sure

[supple wedge]

i did it anyway

[rich wolf]

\sin

[heavy crow]

how can i use the sum to product formula on $a\sin(\alpha)-b\sin(\beta)$?

[somber coyote]

nix:

[thorn grotto]

can you give more context on your question?

[deep trail]

https://gyazo.com/1613123cd359dc385a9888285fede9dc im confused on which angle it is asking for

[upper karma]

The right one if im not wrong

[deep trail]

ok

[upper karma]

The one in the right i mean eh

Not the right angle lol

[deep trail]

of course

[upper karma]

Ok good

[gritty sail]

i know i asked this but i still cant get it

[heavy crow]

@thorn grotto I just want to know if its possible to use a trig identity to write that sum as a product

[languid ocean]

can anyone help me with "Determine the exact value of the trigonometric functions. cosin210"

do i literally just do cosin 210?

(in deg)

[glacial haven]

no

use the unit circle

[languid ocean]

ok i didnt think so

[glacial haven]

210 is also 7pi/6 radians

[languid ocean]

i watch tutorial videos and it always brings that up

but how does that help me? they always explain it in a way i have a hard time understanding in

let me see if i can explain which part i dont understand

[glacial haven]

use this

[gritty sail]

can someone help with my q i still cant get it

[languid ocean]

thank you @glacial haven

[dusky surge]

Jbao I'd like to help, but it's hard to explain without proper drawing tools

Does this help?

[surreal bolt]

Random Tip: trig formulas based on the unit circle for complex numbers. (less memorization or a way to check your memorized formulas)

z1 = cos(A) + isin(A)
z2 = cos(B) + isin(B)
z1 * z2 = (cos(A)cos(B) - sin(A)sin(B)) + i(sin(A)cos(B) + sin(B)cos(A)) = cos(A+B) + isin(A+B)

A = B
z1 * z2 = (cos^2(A) - sin^2(A)) + i(2sin(A)cos(A)) = cos(2A) + isin(2A)

C = 2A
cos(2A) = 1 - 2sin^2(A)
cos(C) = 1 - 2sin^2(C/2)
sin(C/2) = +/- sqrt((1 - cos(C))/2)

... and others.

[gritty sail]

oh wait that said floor function! I thought that was absolute value symbol! @dusky surge thanks!

[acoustic jungle]

@surreal bolt yeah I saw that in 3blue1brown quaranteen video

[teal frigate]

can someone help me pls

idk what to do?

[gritty sail]

i would assume isosceles triangle

[teal frigate]

what

[acoustic jungle]

then use pythagorean theorem, then cosine law

[teal frigate]

i srsly have no clue what im doing

[acoustic jungle]

assuming that's a isoceles triangle, you can split it into two right triangles, sides with 1.3 and 1.8

[teal frigate]

okay

[acoustic jungle]

then you can find the third side with pythagorean theorem

wait.

you don't need to do that

[teal frigate]

hmm

[acoustic jungle]

then just use tan(x)=1.3/1.8

then find the value of 2x

[teal frigate]

tan^1(1.3/1.8) = 35.838

thats doing something

idk what to do next

[acoustic jungle]

do you understand what you are doing

basically we are splitting the isoceles into 2 right triangles

does that part make sense

[teal frigate]

ye i got it now

[tender iron]

How do I write the equation for this?

[upper karma]

@tender iron do you know the equation of a circle?

[tender iron]

no

[upper karma]

Look it up on google

[tender iron]

I have the equation formula - I don't understand how to fill it in

[acoustic jungle]

try messing with desmos.com until you get it right, and also understand why it is right.

(x-b)^2+(y-c)^2=r^2

[upper karma]

Oh yes

b and c are the points at the center

Also to clarify, they want you to express that formula in x and y terms, so:

(x-"number")^2+(y-"number")^2=number²

[tender iron]

alright ty

[upper karma]

Np

[upper karma]

@spark lintel here

[spark lintel]

ok

[upper karma]

Oh ok

Use the double angle formulas that i gave you

For tan 2x and cos 2x

[spark lintel]

tan(2x)=2tan(x)/(1-tan²(x))

[upper karma]

Yeah

[spark lintel]

what about sec2x?

[upper karma]

$tan(2x)=\frac{2tan(x)}{1-tan²(x)}$

Now

[somber coyote]

Al3dium:

[upper karma]

@spark lintel as i said, put it in cos terms

And do cos doble angle formula

[spark lintel]

so we would have two fractions then?

1/cos2x

[upper karma]

Yeah

[spark lintel]

and the tan

[upper karma]

Yeah

[spark lintel]

are they subtracting each other?

[upper karma]

Yes

[spark lintel]

One sec

[upper karma]

Just tell me how it'd look after substituting the formulas

Ok

[spark lintel]

Got this ^

[upper karma]

Good

Now apply the cos double formula

[spark lintel]

cos²x + sin²x

[upper karma]

cosine one is cos²x + sin²x

Yes

[tawny beacon]

Can anyone help me with a question that has been stumping people

[spark lintel]

Like this ^

[upper karma]

@tawny beacon occupied channel. Move

@spark lintel yes

[tawny beacon]

oof

[spark lintel]

ok what are my next steps

subtracting the fractions

[upper karma]

Yeah try

[spark lintel]

Not sure how

[upper karma]

Look

Tan²x=sin²x/cos²x

Put it tan²x

After done that, do common denominator with 1

Lmk when you are done

With that

@spark lintel

👌

[spark lintel]

one sec

[upper karma]

Sure

[spark lintel]

Not this is it?

[upper karma]

Yes

But do common denominator with 1

[spark lintel]

oh so this

[upper karma]

Not sure what you did

Take them away from the paper it'll look easier

Like this

Do common denominator on:

$1-\frac{sin²x}{cos²x}$

[somber coyote]

Al3dium:

[upper karma]

Imagine this a separate exercise

If you cant still get it, do it simpler, common denominator on:

$1-\frac{a}{b}$

[somber coyote]

Al3dium:

[upper karma]

So you can visualise it better

[spark lintel]

is it 1-a / 1-b ?

[upper karma]

No, eh do it with numbers

$1 - \frac{2}{3}$

[somber coyote]

Al3dium:

[upper karma]

Hint: ||common denominator is 3||

[spark lintel]

1/3

[upper karma]

Yes

[spark lintel]

ah so you just subtract the numerator

[rich wolf]

Yes

[upper karma]

The process is, look for common denominator(a factor that is present in both denominators), which in this case is 3 (bc 1 doesnt matter). And multiply by the common denominator with the fraction that you want to have as common denominator, which in this case we want "1" to be in common denominator. So 1/1, with the new common denominator, 3, 1/3 and multiply it with the numerator 1, so we get 3/3. So its:

$1=\frac{1}{1}=\frac{3}{3}=\frac{3}{3}-\frac{2}{3}=\frac{1}{3}$

@spark lintel

[spark lintel]

Ah okay

[somber coyote]

Al3dium:

[upper karma]

Good

Now try again with the example with letters a and b

[spark lintel]

b-a/b ?

[upper karma]

YES

Good good

Now try with the cos² and sin² example

The one we need

[spark lintel]

uhm

cos^2x-sin^2x/cos^2x ?

[upper karma]

Yes

:)

[spark lintel]

Perfect

[upper karma]

Now notice we have

[upper karma]

$\frac{1}{cos²x+sin²x}-\frac{2tanx}{\frac{cos²x-sin²x}{cos²x}}$

Yeah lol

[somber coyote]

Al3dium:

[upper karma]

Now

We'll do division in the second term

Do:

$\frac{2tanx}{1}:\frac{cos²x-sin²x}{cos²x}$

[somber coyote]

Al3dium:

[upper karma]

Now

This is basically the second fraction we have in the original problem

Do it

[spark lintel]

does the : represent division?

[upper karma]

Yes

[spark lintel]

ok so I think we can do

2tanxsec^2x? as numerator

[upper karma]

Why sec?

And not cos²?

[spark lintel]

1/sinx = secx?

[upper karma]

Dont bring back sec, we solved it before

1/sinx = secx?
@spark lintel no, secx=1/cosx

cscx=1/sinx

But we dont want that here

Just do the division

[spark lintel]

2sinx cosx / cos^2x-sin^2x ?

[upper karma]

Why 2sinx?

Where do you get that from

[spark lintel]

hold on I screwed up one sec

[upper karma]

The denominator is ok

Okay

[spark lintel]

Im not sure

[upper karma]

Ok just try

The denominator is ok

[spark lintel]

cos^2x + sin^2x / cos ^2x - sin^2x ?

[upper karma]

Id love to see from where did you got that sin^2x in the numerator

Its simpler than what you are may thinking

[spark lintel]

is 2sinx and sin^2x the same thing?

[upper karma]

Its cross multiplying

is 2sinx and sin^2x the same thing?
@spark lintel no

Its not the same a^2 than 2a

[spark lintel]

ya true

[upper karma]

Do the cross multiplying thing

Quick i have to go sleep very soon

[spark lintel]

is the denominator: 2sinxcosx?

[upper karma]

Look, please just do cross multiplying, idk what you did there

Also the denominator you said before was right

Ig ill expand it

Do:

$\frac{2tanx}{1}:\frac{cos²x-sin²x}{cos²x}$

$\frac{2tanx\cdot{cos²x}}{1\cdot({cos²x-sin²x})}$

@spark lintel

[somber coyote]

Al3dium:

[upper karma]

I only crossed multiply

This is the solution

Of the common denominator obv

And we'll remove the 1

$\frac{2tanx\cdot{cos²x}}{cos²x-sin²x}$

[somber coyote]

Al3dium:

[spark lintel]

if you cross multiplied how does 2tanx end up in the numerator ?

[upper karma]

Up down up

Down up down

[spark lintel]

oh

[upper karma]

Here

Kinda unclear but you get it

So understood everything above?

[spark lintel]

Ya cross multiplying makes it so much easier to solve now that you've explained it

[upper karma]

Good

Now put this back onto our original thing

[upper karma]

$\frac{1}{cos²x+sin²x}-\frac{2tanx\cdot{cos²x}}{cos²x-sin²x}$

Yes

[somber coyote]

Al3dium:

[upper karma]

Ok

Do you know the (a-b)(a+b) identity

[spark lintel]

Not really no

[upper karma]

Put it on yt

It'll save some time

[spark lintel]

ok

https://www.youtube.com/watch?v=75DXDXjLiIg

good ^ ?

[upper karma]

Yeah lol

Although seems a comic video

Lol

[spark lintel]

he explains it -> (a+b)(a-b) it that fine?

[upper karma]

Yes

[spark lintel]

Oh is it just foiling?

[upper karma]

?

[spark lintel]

nvm I understand it though

so the answer to that is

a^2-b^2

[upper karma]

yeah

@rich wolf can you take over me?

I gtg sleep lol

The original question is

https://cdn.discordapp.com/attachments/269573202018041856/706976832796491796/image0.jpg

[rich wolf]

Sure looks fun

[upper karma]

We are trying on RHS

It is

I think i can quickly resume all the steps we did

$\frac{1}{cos²x+sin²x}-\frac{2tanx}{\frac{cos²x-sin²x}{cos²x}}$

[rich wolf]

Are you allowed to do things to both sides

[somber coyote]

Al3dium:

[upper karma]

@rich wolf idk but its easier, true

Its easier that way

Anyway, my steps were

https://cdn.discordapp.com/attachments/326138757474680852/707014662259998790/Capture.PNG

After original question

Go with double formula ^

Tan x for sinx/cosx

Common denominator in the second fraction below

In the 1-sin²x/cos²x

Which is

$\frac{1}{cos²x+sin²x}-\frac{2tanx}{\frac{cos²x-sin²x}{cos²x}}$

[somber coyote]

Al3dium:

[upper karma]

Now i made the division

And we are here

$\frac{1}{cos²x+sin²x}-\frac{2tanx\cdot{cos²x}}{cos²x-sin²x}$

[somber coyote]

Al3dium:

[rich wolf]

Equals what?

[upper karma]

Oh, but i took only the RHS

[rich wolf]

Oh i see

[upper karma]

You can do whatever you want

Either continue this

Or do things on both sides

[rich wolf]

How are you cross multiplying if theres no equality

[upper karma]

?

Oh

I call it like that lol

This thing

Idk what to call it to explain him lol

So yeah gl @rich wolf @spark lintel

[spark lintel]

Thanks for the help

[upper karma]

Np , AMD will do an amazing work as well

[spark lintel]

@rich wolf are you ready?

[rich wolf]

Sure

Tbh i dont see exactly what al3dium is going for here

[spark lintel]

there is where we are at

[rich wolf]

Just from manipulating the right side?

So that is equal to sec2x-tan2x ?

[spark lintel]

trying to make it look like the left side

[rich wolf]

Yes but youve only done things to the right hand side

Right?

[spark lintel]

Exactly we are trying to take the right side and make it look like the left

you only work with one part of the equation not both

[rich wolf]

Just making sure lol

[spark lintel]

He left me off with (a-b)(a+b) identity

@rich wolf do you know what the next step is?

[rich wolf]

Give me a sec im solving it

On my own before i walk you through it

[spark lintel]

No problem

[rich wolf]

Ok

Youre gonna have to cross multiply for this problem lol

[spark lintel]

cross multiply this ?

[rich wolf]

So right now we have $\frac{cosx-sinx}{cosx+sinx}=\frac{1-sin(2x)}{cos(2x)}$

[somber coyote]

AMD:

[rich wolf]

So

[spark lintel]

Wait what happened to the tan fraction?

[rich wolf]

Oh i just started from the beginning because tbh i dont really get what al3dium was going for

Pretty simple that sec2x-tan2x=(1-sin2x)/cos2x

[spark lintel]

You're not able to take a call are you?

[rich wolf]

Sorry dude i dont have a mic rn

At my pc

[spark lintel]

ok

so if we are proving the right side

we shouldn't be touching the left side what so ever then

[rich wolf]

Eh it depends

What does your teacher say

[spark lintel]

I can show you an example

[rich wolf]

No i mean what does your teacher say about proving trig identities

What do they tell you to do

[spark lintel]

@rich wolf did you see what I sent

[acoustic jungle]

you can only prove from both sides and them make them equal

you are not allowed to do something to both sides

this is just a note I don't know if you did that or not

but just a reminder

[spark lintel]

@acoustic jungle are you able to hop in a vc room

[acoustic jungle]

No because I'm like 2 years old

I can show you what I did though

let me try to find my phone

@Fern#1010

@spark lintel

it's a bit messy because I only had a pen

the x is a multiplication symbol

Fern my bread is going to mold before you come back

[spark lintel]

I got it thanks guys

[merry abyss]

@acoustic jungle mold I want to see it mold

[acoustic jungle]

@merry abyss is my father

[hollow finch]

Yo

Can I try to help

[kindred linden]

guys i got a few questions i need help with

anyone up for vc?

[floral elbow]

Hello how do you "find the value of the trigonometric ratio."

[upper karma]

Show us the question

[floral elbow]

We just started this in geometry

[upper karma]

So tan = opposite / adjacent

Do u know what opposite and adjacent are

@floral elbow

[floral elbow]

I got it. Thanks for offering to help.

[upper karma]

Ok. What did u get as the answer

Thanks

[floral elbow]

You're welcome

[upper karma]

Mhm

[floral elbow]

I got 35/12 @upper karma

[upper karma]

Yah correct

Let’s gooo got it rite

[ruby widget]

https://cdn.discordapp.com/attachments/297275767568465921/707115748421664818/unknown.png

i need some help

been at it for awhile

can't wrap my head around this

[upper karma]

Pls help solve it 😦

[ruby widget]

its fine guys

figured out somehow

[upper karma]

thx

Your welcome

[misty plume]

@upper karma Man of culture I see, ain't gonna leave anyone hanging

[upper karma]

calculate da area of dis shaded region.

[silent plank]

where are you stuck?

[upper karma]

calculatin da area of dis shaded region.

[dark sparrow]

if you were asked to calculate the area of the whole semicircle, would you be able to do it?

[upper karma]

yes

[silent plank]

if you were asked to calculate the area of the triangle, would you be able to do it?

[upper karma]

maybe

[dark sparrow]

there's no maybe, either you can or you can't

[upper karma]

yes

[dark sparrow]

okay

great

so

does it make sense to you that your shaded region is the semicircle but with the triangle cut out

[upper karma]

yes

but I don't know what the radius of the semicircle is

and I think I need that

[silent plank]

that contradicts how you said that you could calculate the area of the semicircle

[upper karma]

yeah

I kinda lied

[silent plank]

consider thales's theorem.
i.e. that is a right triangle

which you would have needed to apply to find the area of the triangle

[upper karma]

I know the area of the triangle

but not the semicircle

[silent plank]

and if you know 2 sides of a right triangle, what can you use to find the 3rd side?

(which in this case would also be the diameter)

[upper karma]

my eyes

[dark sparrow]

why thales theorem ramonov

[silent plank]

inscribed angle works fine i guess.

[upper karma]

pain

[silent plank]

?

[upper karma]

it never ends

[silent plank]

what never ends?

were you able to complete the problem or do you still need help

[upper karma]

I guess I still need help

it's multiple choice btw

I just don't pay attention to math but this is a test and I just wanna get a halfway decent grade

[silent plank]

i.e. that is a right triangle
and if you know 2 sides of a right triangle, what can you use to find the 3rd side?

[upper karma]

yeah I don't actually remember, usually for simpler problems I literally look at it and compare the length to the known sides and often it actually works out for me

[silent plank]

perhaps a famous theorem starting with Py

[upper karma]

right

so it's 10

that's what I guessed in my head

but now what

[silent plank]

you found that the diameter is 10
apply certain formula to find the area of the semicircle

[upper karma]

I get 50 x pi

but that doesn't fit any of the answers

[silent plank]

how are you getting 50*pi?

what formula are you applying?

Area of semicircle =

[upper karma]

I DID IT

it's A

[silent plank]

A isn't correct

[upper karma]

bruuh

🤯 🔫

[silent plank]

what formula are you applying?
Area of semicircle =

[upper karma]

Area of semi circle is cool

r^2/2 x pi

[silent plank]

ok and what is r in your figure?

[upper karma]

10

[silent plank]

no

[upper karma]

Aha

[silent plank]

10 is the diameter

not the radius

[upper karma]

oh right

so we never figured out the radius

or I didn't

You didn’t

Son

@upper karma So do you know the radius?

no

What is a diameter of a circle son?

And what is a radius?

Do you see the relationship?

Between them

why are you calling me son

Ok Trey

Now answer my question Trey

Explain to me what a diameter and a radius is

random lines inside shapes that will hardly be found useful to me in the future

Very true but specifically random lines in circles

Now tell me

Do you know what a radius is?

The actual definition

Take your time

I’ll be here all day

🙊

So you do not know?

half of the diameter

🙊🙊🙊

Son, tell me... did you search that up?

You did not know that before

nope

I randomly remembered it

Oh I see

So just divide that diameter in 2

man I'm out of it

Ok

can't believe it took that long to realize

Alright

@silent plank Aight it’s your turn again

so the answer would be C

Ya idk the question just wait for Ramanov

do we... do we help with tests?

yup

@upper karma Ya but what’s the question

see #❓how-to-get-help number 6

[silent plank]

yeh its C

[upper karma]

kakoolate da urrea of da shaded reejun

yeah

thank u guys for the help

6. Do not ask for help on tests. Any violation of this will lead to appropriate action being taken at mod discretion.

#❓how-to-get-help number 6

it's less never learning it and more forgetting everything cuz I couldn't give less of an f

What about open note quizzes

Can I ask

For those

but this was a solid refresher

[dark sparrow]

kakoolate da urrea of da shaded reejun

[silent plank]

i'd say no. don't help on things you aren't comfrotable with

[dark sparrow]

are you purposefully misspelling this

[upper karma]

Calculate

I just don't pay attention to math but this is a test and I just wanna get a halfway decent grade
@upper karma

no that's how they actually spelted it

The

Ok

[dark sparrow]

KAWKOOWAYT DAH URREUH UV DUH SCHAYDEED REEJUHN

[upper karma]

👌

@upper karma neat

it's all pointless anyway

[tawdry pivot]

REEEEEEEE

[upper karma]

@upper karma use similar triangles theorem

[upper karma]

What about 11

For 13 i got x = 7 btw

Choose to do similarity or Thales

[upper karma]

How do i prove

[upper karma]

S i m i l a r i t y

How many times i'll say it?

[silent plank]

consider properties of parallel lines

[upper karma]

no idea how to find

the angles

[silent plank]

do you know your circle theorems?

[upper karma]

uh the thing is

i know some

but not all

[silent plank]

the inscribed angle theorems

specifically angles subtended by the same arc

[upper karma]

a=y?

these are the same

?

since theyre on the same arc

or am i wrong

[silent plank]

alpha and gamma would be equal

[upper karma]

yeah

i get it now

thanks

!!

[zealous ermine]

guysss

what is arctan(tan^3(theta)) simplified to?

[upper karma]

how to i find beta on this?

@upper karma inscribed angle theorem

i dont know how to imply that on beta

The angle of a center = measure of the arc

So beta = measure of arc

And the arc you get it by inscribed angle thoerem

i still dont really get it :/

Its simple, notice that gamma and beta have the same arc

yes

because the triangle is rectangular

With the given angle gamma

Use inscribed angle with gamma and the arc

Understood this last step

?

oh

so beta is 66

?

Yup.

ohhhh

yeah that makes perfect sense

Glad to hear

[royal cipher]

can u use area=0.5b*h on a non-right angled triangle?

my book does it

i thought u have to do area= 0.5 * b * h * sin(a)

[silent plank]

area = 1/2 bh where h is the altitude relative to the base

[royal cipher]

even if there is no right angle in the triangle?

[glacial haven]

doesn't have to be a right angle

[royal cipher]

oh ok

thanks

[silent plank]

area= 0.5 * b * h * sin(a)
not a great choice of variables

[royal cipher]

yeah lol i see why

[silent plank]

= 1/2 ab sin(C)
indicates that you're using two sides and the angle between them

[royal cipher]

yup

[granite elk]

How can i solve (ii) (a)

(i) was ok

Im trying to use Tan=sin/cos in this

[rich wolf]

move all terms to the LHS

[granite elk]

Yeah i already did that

[rich wolf]

now common denominator

[upper karma]

How do I find the perimeter for this shape?

i have the length from the middle to a point

and the radius of the circles taken away

what is the length from the middle to a point

[granite elk]

@rich wolf Im still pretty confused on what im supposed to do

[acoustic jungle]

show us what you are given

[rich wolf]

@upper karma find the circumference of each semicircle

There are 6 semicircles

=3 circles

[pallid cloud]

They don't look like semi-circles

They look like 6 inverted portions of a circle

If we turn them to the other side, they form a complete circle

I'm not sure about that, but I'm pretty sure each portion is not a semi-circle

[silent plank]

tangent theorems, angle around a point and angle sum of a quad will allow you to find the degree of each arc

[upper karma]

@upper karma length from middle to point is 5

[granite elk]

https://cdn.discordapp.com/attachments/326138757474680852/707274600928509983/unknown.png

Im fully confused on how to solve ii a)

[upper karma]

bring everything to the left side

[upper karma]

^

big brain

[granite elk]

and...

[silent plank]

express tan in terms of sin and cos
multiply both sides of the equation by something to get rid of the fraction
apply a Pythagorean trig identity

[upper karma]

^

[spare plaza]

https://i.gyazo.com/thumb/1200/233a96032271ec65843a35e696316d1b-png.jpg how do i do this?

ik the simplified equation is

-16t^2+62x-52=0

but idk where to go from there

[upper karma]

Seems hot

[spare plaza]

what does that mean

[upper karma]

Seems cool

no u

[versed river]

You can use the quadratic formula or complete the square

[upper karma]

or use product sum factor

[versed river]

That looks very difficult to product sum

[upper karma]

oh he has -52 as his c value

i was using 4

cuz thats whats in the question

yah use the quadratic formula

[versed river]

Do you know how to use the quadratic formula, or how to complete the square, sam?

[upper karma]

@spare plaza

[spare plaza]

Yup i just finished it :D

[fading dew]

Hey, can someone help me real quick with a refresher geometry problem? I completely forgot how to do this lol

[versed river]

oh you're being answered in a questions channel

[fading dew]

@versed river

I dont know trig

the other guy couldn't help me

[versed river]

you could estimate it without trig, but i dont see a way to do it exactly without using trig...

[fading dew]

huh

[versed river]

is this from a geo class?

[fading dew]

yeah

I used to be able to do it like heckin good. but now i cant bc its been so long lmao

im in 10th grade

[upper karma]

Use a sum or difference formula to write a formula for cos^2x. Express the result in terms of sinx and cos2x.

<@&286206848099549185>

please

help

[fading dew]

what

[upper karma]

cos72

[versed river]

you sure you aren't meant to use any trig?

[fading dew]

Bro I have NEVER used trig

I dont understand

[upper karma]

its cos72

its a cofunction identity

trust

now

[fading dew]

What is the emote reaction

[upper karma]

someone fuckin help me

😂

[fading dew]

what is this

[upper karma]

Idk

[fading dew]

thanks gang

[upper karma]

I’ve seen a bot like this before

[fading dew]

twas correct

[upper karma]

It auto reacts

[fading dew]

it stopped

tf

[upper karma]

I created one using python

Cool

[fading dew]

OH MY GOD

i just realized something

[upper karma]

Yes?

[fading dew]

this is a p r e t e s t

thats why

Ive never used trig

b r u h

[upper karma]

Ah nice

Pre test are awesome

[fading dew]

i was so confused, I have never seen like any of this stuff b4 lmao

[upper karma]

Yep nice

Good luck

[fading dew]

thanks man

[upper karma]

someone help

please

Use a sum or difference formula to write a formula for cos^2x. Express the result in terms of sinx and cos2x

@here

@here

@here

@here

@here

@here

@here

@here

@here

@here

@here

@here

@here

@here

@here

@here

[fluid stream]

what have you tried

[upper karma]

i have no clue how to approach thjis

because

making cos^2x=-sin^2x+1 wont do anything

[fluid stream]

Well you are given the hint of using a sum or difference formula

[upper karma]

and expanding it to cosxcosx

wont do anything

yeah

thats great

but im gonna have to make cos^2x to become something

before I use those formulas

Im gonna need two angles to be added

or subtracted

I was thinking of doing

cos(x+0)cos(x-0)

and that does

absolutely nothing

[fluid stream]

you said that you can write your answer in terms of sinx and cos(2x)

can you expand cos(2x)?

[upper karma]

I was thinking

of doing a double angle formula?

you can expand cos2x = 2cos^x-1

or -2sin^2x+1

[fluid stream]

yes, that's the same thing as sum formula of cos(x+x)

[upper karma]

yea

where do i go from there

[fluid stream]

from cos(x+x) = cos(x)cos(x) - sin(x)sin(x) = cos^2(x) - sin^2(x)

[upper karma]

yeah

i realize

but I simplified it by substituing one of the pythagoroas identities

so I got

2cos^2x-1

can u just tell me the answer

i got 5 more min

before the hw closes

and i cant turn it in

[fluid stream]

well if you rearrange that, you get cos^2(x) = sin^2(x) + cos(2x)

[upper karma]

yeah

[fluid stream]

which should work, if not, you can use pythagorean on sin^2(x) (by adding cos^2(x) to both sides)

then you'll get the normal power-reducing formula for cos^2(x)

that is 2cos^2(x) = sin^2(x) + cos^2(x) + cos(2x) = 1 + cos(2x), so that cos^2(x) = (1+ cos(2x)) / 2

[upper karma]

ty

[viscid ginkgo]

strruggling to prove an identity

(sec t + 1)/tan t = tan t / (sec t - 1)

[dark sparrow]

you might consider multiplying the left-hand side by cos(t)/cos(t)

and perhaps also doing the same to the right-hand side

[viscid ginkgo]

but

I thought you had to only focus on one side

and find the proof from there?

so confusing./

maybe you mean, move on

from the left side

and try right

omfg

I got 0

[dark sparrow]

I thought you had to only focus on one side
there is nothing wrong with doing some simplification to the left-hand side, then independently of that doing some simplification to the right-hand side, and then proving that the simplified versions are equal

[upper karma]

ann

[dark sparrow]

what

[upper karma]

hi

[dark sparrow]

was there something else you wanted to say to me

beyond a greeting

[upper karma]

no

just wanted to say hi

[glacial haven]

hi

[upper karma]

hi

how has your day been

[glacial haven]

tiring

[dark sparrow]

y'all take the chatter to #chill

[viscid ginkgo]

If tanA = 5/12 with A in Q3 find sec2A

I don't get it

sec2A = 1/2sinAcosA

actually

1/1-2sin^2(5/13)

[dark sparrow]

you're missing some parentheses here

[granite elk]

How do i simplify this?

its supposed to give me

[upper karma]

How can an expression transform into an equation

[pallid cloud]

Akward you have an "=0" missing in the 1st pic you sent

[upper karma]

@granite elk

[granite elk]

Yeah the =0 is missing

but that dosent really help with the question

[upper karma]

No, but it helps us to understand the question.

@granite elk so are you here?

[granite elk]

Yeah

Original question was

[upper karma]

So, do common denominator

[upper karma]

From what you got, do common denominator

[granite elk]

i simplified what i got

[upper karma]

https://cdn.discordapp.com/attachments/326138757474680852/707531340412878908/unknown.png

From this ^, do common denominator

[granite elk]

Thats what i got

[upper karma]

What? Can you follow my instructions

[granite elk]

?

[upper karma]

Do common denominator on the full expression

You should get up to $\frac{1-\cos²{x}}{\cos{x}}-(\frac{3\cos²{x}-2\cos{2}}{\cos{x}})$

@granite elk

[royal cipher]

the fact he pinged akward

nice

hahaha

[upper karma]

Lmao

[granite elk]

rip

[upper karma]

Mmm

You should get up to $\frac{1-\cos²{x}}{\cos{x}}-(\frac{3\cos²{x}-2\cos{x}}{\cos{x}})$

[granite elk]

ohh

thanks

[royal cipher]

also the frog pfp really suits the the reply hahahaa

[upper karma]

Lol

[royal cipher]

akward u havea great name and pfp

[granite elk]

lol thx

[somber coyote]

Al3dium:

[upper karma]

Now

Follow this? @granite elk i only did common denominator which was what i told you to do for 5 mins

[granite elk]

ok

[upper karma]

Understood this step?

[granite elk]

yeah

[upper karma]

Ok isnt a proper answer lol

Good

Now

[granite elk]

is that minus meant to be there

next to the brackets

because its going to affect the - inthe 2 cos

[upper karma]

Substracting

$1-\cos²{x}-3\cos²{x}+2\cos{x}$

[somber coyote]

Al3dium:

[upper karma]

is that minus meant to be there
@granite elk yes it is

[granite elk]

ok

[upper karma]

And thats about it

Substracting

$1-4\cos²{x}+2\cos{x}$

[somber coyote]

Al3dium:

[upper karma]

Arranging it

$4\cos²{x}+2\cos{x}+1=0$

Which is what you wanted to get

[somber coyote]

Al3dium:

[granite elk]

Thanks

it finally makes sense now

[upper karma]

Np

[upper karma]

@strong pilot state the formula for an hexagonal pyramid

Here on this chat

[strong pilot]

SA = hp + b?

[upper karma]

SA?

[strong pilot]

surface area

[upper karma]

No its $A_{total}=(\frac{b*h}{2})*6 + A_{base}}$

[strong pilot]

what does the a represent

[upper karma]

A, area

p perimeter

Ap, apothem

Wait

My bad

[somber coyote]

Al3dium:
Compile Error! Click the reaction for details. (You may edit your message)

[upper karma]

Now lol

[strong pilot]

ok that makes more sense kinda

[upper karma]

Ok now

[strong pilot]

what would atotal represent

[upper karma]

Uhh wait lemme read it again

@strong pilot the area of the whole thing

[strong pilot]

oh ok

alrighty lemme try this

tysm

[upper karma]

Okok np

[upper karma]

Gonna ask this here again, how to find the perimeter? Didn't quite get the answer I was looking for last time.

I have the length from the middle to the point (5cm) and the length of the radius of the circles that are taken away from the sides (also 5cm)

I need to find the perimeter, but I don't know how.

visually, its scuffed

cause if the circles werent taken away then the original circle would be the same size as the circles taken away

[brisk ginkgo]

Can someone dm I’m so confuseeed

<@&286206848099549185>

[acoustic jungle]

@upper karma assuming that shape is a hexagon, the perimter is just the circumference of the circle

if you construct another circle connecting the the radius of the star, it'll be a rhombus

with 60 degrees

and there are 6 of them

[brisk ginkgo]

I NEED HELP PLSSSSs

[upper karma]

@brisk ginkgo just post it bruh

[upper karma]

@brisk ginkgo "help, help, I need help", 5 minutes later he's offline

Lmao

what I understand from that is that the test is over and he's no longer interested in getting help

Prob ^

Which is pretty sad and dissapointing

[brisk ginkgo]

No it’s a review!

Not a test

[upper karma]

Post it!

post the problem that you need help with

[brisk ginkgo]

HELP HELP HRLP PLS

[upper karma]

haven't you learned to figure this out in class or something

[brisk ginkgo]

I’m saying no and our teacher doesn’t teach in class

I mean zoom class

We don’t have like classes for math

[upper karma]

you dont have zoom classes for math?

damn