geometry-and-trigonometry

[vague berry]

idk

how

[weak shoal]

So, for number 3, here's a hint

[vague berry]

cosx sinx at the top

[weak shoal]

Divide numerator and denominator by $\cos^2(x)$

[somber coyote]

Abhijeet Vats:

[weak shoal]

Oh you just copied those steps out from somewhere else, then?

[vague berry]

no?

i just got stuck

i dont get how to divide by cos^2x

like how do you do bothn num and denomm

[weak shoal]

$\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)} = \frac{\cos^2(x)(\tan(x))}{\cos^2(x)(1-\tan^2(x))}$

[somber coyote]

Abhijeet Vats:

[weak shoal]

Make sure you see why the above is true

[vague berry]

huh

[upper karma]

@vague berry aight srry i was helping on another channel and forgot about you

Whats the matter

[weak shoal]

i just used the fact that:

$$\cos^2(x)\tan(x) = \sin(x)\cos(x)$$

[somber coyote]

Abhijeet Vats:

[weak shoal]

Like I said, try to look at that and see why it must be true

[vague berry]

wait is that true

i dont know how it is

OH

YEAh

THATS TRUe

OOO

i see it

[weak shoal]

yea

so you can finish it off from there

for 3, I'll give you a hint

$\cos^2(x) = 1-\sin^2(x) = (1-\sin(x))(1+\sin(x))$

[somber coyote]

Abhijeet Vats:

[lucid orbit]

Is the exact area of a circle found by leaving answer in terms of pi?

[weak shoal]

Yeap

[lucid orbit]

👍

[foggy beacon]

hey guys just a very quick question please

the principal is just the angle so like 340 and the acute angles will be 20

is that what principal angles means..?

[upper karma]

@foggy beacon is it a test

[foggy beacon]

bro lmao

i swear it's not xD

it's an assignment

[upper karma]

Its ok just gotta ask

Dw

[foggy beacon]

do u think I'll have discord on a test 😆

❤️

[upper karma]

Lol

[foggy beacon]

but what does principal angle mean

[upper karma]

Some ppl sadly have

Wait a sec

https://youtu.be/XPGjhHt0Z5A

This'll do

[weary drift]

principal angle is the coterminal angle between 0deg & 360deg

[foggy beacon]

coterminal? uhh

so for like 340 how would I get the coterminal angle

[weary drift]

do you know what coterminal means?

[foggy beacon]

they have the same ratio so I think it depends on the Quadrant

I thought that getting it was like the same way as the acute angle

but plz help me

[weary drift]

do you know what coterminal means?

[foggy beacon]

i think getting it is the same way as the acute

it has the same ratio

as the angel

[weary drift]

wdym ratio

[foggy beacon]

like if we take the sin of that angle its the same as the sin of the coterminal

[vague berry]

@weak shoal

[weary drift]

by that definition, 0 & pi are coterminal. and that's wrong @foggy beacon

[vague berry]

.,rotate

[weak shoal]

,rotate

[vague berry]

how did you get the cos^2x(1-tanx)

that's confusign me

[weak shoal]

i didn't

i got $\cos^2(x)(1-\tan^2(x))$

[somber coyote]

Abhijeet Vats:

[weak shoal]

just factorize $\cos^2(x)$ from the denominator

[somber coyote]

Abhijeet Vats:

[vague berry]

wait hwat

how do you know even to put 1-tan^2

where is that from

[weak shoal]

$\cos^2(x)\tan^2(x) = \sin^2(x)$

[somber coyote]

Abhijeet Vats:

[vague berry]

oh its the same thign ?

[weak shoal]

dude, you gotta think lmao

[vague berry]

oh yeah

thats true

[weak shoal]

just think through it

[vague berry]

but then where the 1 coming from

is this right

[weak shoal]

lmao dafuq

do you really not know how to factorize?

[vague berry]

no but what is being factorized

OH

[upper karma]

oh its the same thign ?
@vague berry an equation by definition can be switched or moved and will keep being true

[vague berry]

oh

[weak shoal]

$\cos^2(x)-\sin^2(x) = \cos^2(x)-\cos^2(x)\tan^2(x)$

[vague berry]

oh

oh

[somber coyote]

Abhijeet Vats:

[vague berry]

oh

oh

ok

im dumb as shit

im sorry for being dumb asf

holy shit

thanks

[weak shoal]

Please engage your brain and think.

[vague berry]

i didnt see that

[upper karma]

No you are not lol

[weak shoal]

You're welcome.

[vague berry]

i didnt understand what you meant

thank you so much i get it

[weak shoal]

Then for question 3

[vague berry]

i skipped that and did 4

4 was 2 steps

but 3 was also easy

[weak shoal]

$\cos^2(x) = 1-\sin^2(x) = (1-\sin(x))(1+\sin(x))$

[vague berry]

i can do that in my head

[somber coyote]

Abhijeet Vats:

[weak shoal]

aight, good then

[vague berry]

sin^2x-1/tan(sinx-1)

then factorized sin^2-1

and then sinx+1/tanx

easy

[weak shoal]

use parentheses please

[vague berry]

sorry

[weak shoal]

and no, that implication does not immediately follow

there's an intermediate step that you're forgetting

but the idea is essentially correct

[vague berry]

tysm @weak shoal

i appreciate your help and dealing of me

[weak shoal]

It wasn't an issue. You're welcome.

[pine oar]

someone help here, https://i.imgur.com/TDuE8gh.png how would you know if the point of intersection is a right angle

[weak shoal]

measure the angle using a protractor

[pine oar]

lmao

[upper karma]

Sine's law im pretty sure

@pine oar well you should draw it first

[pine oar]

well i do draw it, but it's hard to get accurate angles without a protractor

[upper karma]

Then post it and ill confirm if its sine's law

No lol doesnt need to be accurate

Just a sketch

Vats was joking

[pine oar]

ok im sketching

thats the best i can do

lol

forgot to label b

Wait I think I just figured it out

Convert the angles to decimals so it's easier to work with

330 27' is around 29.55 away from the start

[upper karma]

Wait I think I just figured it out
@pine oar did you

[pine oar]

+60.45

yeah

29.55 + 60.45 = 90

i think thats it

[upper karma]

uh idk where you get 29.55

[pine oar]

360-(330+27/60)

part in parenthesis is the conversion from degrees and minutes to decimal

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

[upper karma]

do you mean $360-(330+\frac{27}{60})$

[somber coyote]

Al𝟛dium:

[upper karma]

or the 60 is all the way

[pine oar]

yes that

[upper karma]

,calc 360-((330+27)/60)

[somber coyote]

Result:

354.05

[pine oar]

u put the parenthesis wrong on the calc

,calc 360--(330+(27/60))

[somber coyote]

Result:

690.45

[upper karma]

,calc 360-(330+27/60)

[somber coyote]

Result:

29.55

[pine oar]

there u go

yes

[lucid orbit]

What is the area formula of a cone that has semi circle basically an ice cream?

[pine oar]

@upper karma is the way I figured it out wrong or is that okay

[upper karma]

This is my drawing

C is coral thing

[ripe dove]

@pine oar drawing was right

[upper karma]

Does the word "bearing" change anything

[ripe dove]

Yes

Bearings are shit

[upper karma]

Im not native so idk what that is

[pine oar]

malix the drawing may be right but it doesn't look right lol, how am i supposed to put a right angle onn that

[ripe dove]

If a bearing is given strictly as an angle it is measured clockwise from north

[upper karma]

Oh so it does change lmao

[ripe dove]

Yeah

[upper karma]

Ok my drawing is voided

[ripe dove]

And there’s also bearings given as (N or S) angle (E or W)

N 30 degrees E means 30 degrees east of north

[upper karma]

https://cdn.discordapp.com/attachments/326138757474680852/713163551442731118/20200521_175701.jpg

,rccw

[pine oar]

i didn't label b

but

it works

[ripe dove]

It’s where you put the angle

[upper karma]

As "bearing" isnt familiar for me, id trust malix

[ripe dove]

@pine oar you already have 2 right angles up there you can use

[pine oar]

true

well

on the right

how would i use the right angle

[ripe dove]

And a coincidental and convenient third you will have to realize by calculation

[pine oar]

330 27' - 270

and then that gives me the lower part of the angle

[lucid orbit]

If i have a right triangle, are the other two angles 45 degrees?

[pine oar]

and then subtract that from 90

[ripe dove]

Gets one internal angle on the right

[pine oar]

ok perfect

[upper karma]

If i have a right triangle, are the other two angles 45 degrees?
@lucid orbit not at all

[pine oar]

i get it

[lucid orbit]

oof

[ripe dove]

Only if both (non hypotenuse) sides are the same length will both other angles be 45 degrees

[pine oar]

i got another question

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

[upper karma]

Well its basically doing sine definition twice lol @pine oar

[pine oar]

wat do u mean sine definition?

[upper karma]

$\sin(x)=\frac{\text{opposite side}}{\text{hypothenuse}}$

[somber coyote]

Al𝟛dium:

[pine oar]

can you do that without a right triangle?

[upper karma]

No

But its a right triangle

[pine oar]

well we have no values for either side of the right triangle

[upper karma]

Do a system of equations

Two variables

Sine definition twice

[ripe dove]

h is a side for both triangles

[upper karma]

ok

And x forms part as well

Of both eq

@pine oar have you got it or may i write it

[ripe dove]

i was gonna say the other known side of both triangles are functions of x you can easily determine

[pine oar]

write it pls

i don't get it

[upper karma]

Aight wait a sec

[pine oar]

aite

[ripe dove]

base of one triangle (with larger angle) is x

base of other triangle (with smaller angle) is x+155

[upper karma]

$\begin{cases} \tan(11"30')=\frac{h}{x+155} \

\tan(38"40')=\frac{h}{x} \end{cases}$

[somber coyote]

Al𝟛dium:

[upper karma]

" is °

@pine oar

Im srry it was tan not sine

Its kinda late for me lol

Need sleep

,ti

[somber coyote]

The current time for Al𝟛dium is 01:52 AM (CEST) on Fri, 22/05/2020.

[pine oar]

hmm

[grizzled rose]

Can someone help me telling me if it’s good? I have to calculate the dimensions of a rectangule if the perimeter has a length of 44cm and his area is 105cm2

[pine oar]

@upper karma ok i worked it out

ty

sorry for late response

was writing down all the info

[upper karma]

Np

[foggy beacon]

isn't -53 already an acute angle

how can i find the acute angle of -53 can someone please guide me

[slate bay]

I'm not sure about the terminology, but I think the principal angle is the angle from 0 to 360 degrees that coincides with your angle.

And the related acute angle is the angle you form with the x-axis.

As in, if you had -30 degrees, you can see it forms an angle of 30 degrees, ignoring the sign, so the related acute angle of -30 degrees is 30 degrees.

Not sure how to phrase that.

[loud glade]

hi! im almost done with school but i have two assignments left that i cant figure out at all. One is over finding the arc or central angle. The other is Inscribed angles and sector area. This is geometry stuff, if anyone is able to help id love for you to dm me! im desperate

[surreal bolt]

er you can ask for global help in the channels a,b,c, ...

[upper karma]

Or ask here

@loud glade just send here

Many people would help you

[loud glade]

okay one moment

[upper karma]

@loud glade just use the fact that a circle is 360 degrees and a diameter splits the circle into 180 degrees

Watch some videos online

Inscribed angles and area of circle or sector

@loud glade Have you tried them yourself first atleast?

[loud glade]

yeah a few times

ive even looked at her reviews

i think i understand the super simple ones like number 3 on the second page

but everything else is difficult

[soft gulch]

There's BPT

No wait I'm dumb

[upper karma]

what strategies do you reccomend I do to organize my time bc I want to take geomtry over the summer as an online class. They expect you to do 7-8 hours of work alone with 90min online classes and and 3 exams that are proctured virtually. Any resources, textbooks, youtuber you reccomend since im a slow learner and we have a month and a half to learn everything with a quiz every week

bc this is stress full since its from june 22nd to june 30th

th up side its only pass or fail

[upper karma]

MindYourDecisions

wat?@upper karma

Its an awesome yt channel dedicated to geometry

oh sorry I mistunderstood, thanks for the reccomendation

help #help-2 👉 👈 😳

[foggy beacon]

hey guys can someone please please guide me through this one question

here's the equation

idk what to do next

its 2

can someone please tell me what I should do after this step

[white cradle]

ok

in the interval [0,360°] which angles give sin(y) =1/2?

hint: there's two of them

[foggy beacon]

sin 30 , 150

is that correct

[white cradle]

indeed

so can you proceed now?

[foggy beacon]

yes but like i dont know what to do in the bracket

do i just like plut it in the calulator or what

[white cradle]

here y=x/2 - 90°

and you just said that y is either 30° or 150°

[foggy beacon]

yes

[white cradle]

that will give you two possible solutions for x then

just plug in and solve

for example,
one solution is y=30°=x/2-90

so x/2=120

x=240

get it?

so find the other solution, and you're done

oh shit wait

x is in [0,360] not y

[foggy beacon]

bro i swear you're too smart

[white cradle]

if x is in [0,360°] then y is in [-90,90]

so then, only 30° is needed

150° is outside the range they're asking

sorry for the confusion

so only one solution

[foggy beacon]

okay thank u lots lots bro ❤️

[upper karma]

What are dependent events in probability?

[upper karma]

@upper karma events that depend on the other for example, if A occurs it will change the result of B but if A doesn't occurr it won't change B. Numerically, gotta prove that $P(A)*P(B) \neq P(A \cap B)$

[somber coyote]

Al𝟛dium:

[upper karma]

There are more twins to this, let me know if you need more

Ok ty

Also this isn't the channel to ask about probability but oh well

Np

[tame ore]

HEY

ILL PAY SOMEONE 20 DOLLARSSSS TO DO MY HOMEWORK

LIKE

[crystal pike]

for what

[tame ore]

FR ITS ONLY 20 QUETIPOSN

chords

[crystal pike]

send me screenshot of 1 problem

[upper karma]

@tame ore ay deal

[upper karma]

I’ll doo all of it

[tame ore]

ugh

period

[upper karma]

How will I get the 20$ tho

@tame ore

[tame ore]

paypal

[upper karma]

@crystal pike Law Of Cosines Big Boy

@tame ore What else can u give

[tame ore]

wdym

[upper karma]

In exchange for the 20$

I’m just a kid

[tame ore]

u want more?

[upper karma]

No

I don’t have paypal

What can u give in exchange of giving the 20$

[tame ore]

hmmm

cash app

maybe

[upper karma]

U play any games

@tame ore

[tame ore]

👁️ 👄 👁️

[upper karma]

Do u

Nah esmerelda, keep your money

I’ll help

@tame ore Aight fuck it, I’ll do it for free

@tame ore lets go dms

Bruh

Hf

Bitch

😆

[tame ore]

oh youll help

aye

bet

[upper karma]

You see how it resembles a right triangle

[tame ore]

i guess

[upper karma]

Ok, so for this, we can apply the Pythagorean-

Ok man, all yours

[tame ore]

mkay

so

[upper karma]

Man

I dmed the dude

The same thing

Plz my turn

[tame ore]

but i dont wanna solve it

[upper karma]

@upper karma see what I’m talking about

He wants someone to do all his work

And I’m the man for that

[tame ore]

aye

[upper karma]

@tame ore lets go dms plz

I thought that was against the rules or smth

😔

It is

But I’m

A

Idk the word

Haha you’re fine

@royal dew Wait is this against the rules?

@copper valve

[copper valve]

hm?

don't do work for other people lol

that's not helping

[silent plank]

no free lunch

[tame ore]

help is smelly

work ethic is

beneficial

🙂

[ocean axle]

Is it true that those triangles are similar to each other? There is a bisector.

[upper karma]

angle bisector

[ripe dove]

It is not necessarily true

@ocean axle

[ocean axle]

Ok, so there must be some additional information.

[upper karma]

Yup

You should look up the triangle similarity theorems

Like AA, for example, where two angles have to be congruent

[ocean axle]

What can we say about areas of those triangles? Are they the same?

[upper karma]

what

[keen forge]

I am trying to understand this problem. If anyone could help that would be great!

https://www.homeworklib.com/questions/914745/this-problem-gives-you-a-preview-of-something-you

[upper karma]

We can’t assume anything about the area of those triangles, niko ^

[ripe dove]

What we know about those triangles: they each have an angle and a side that have the same measurements.

[upper karma]

^

But that’s not SAS or ASA

[next dew]

hey guys, I was wondering if in this problem i had to subtract the area of the base of the cone from this when solving for total surface area of this shape. Do you have to remove the base of the cones surface area in order to solve?

for surface area of the total problem

[gritty sail]

yea

only the stuff u can see from the outside

[rigid peak]

Question asks me to
Eliminate the parameter to find a Cartesian equation of the curve. of x=sin^2theta, y=cos2theta, -pi/2<=theta<=pi/2
By adding the equations I get x + y = cos^2theta
Not sure where to go from there, any advice?

$\begin{cases}
x=sin^2{\theta}
y=cos2\theta
\end{cases}$

[somber coyote]

Agent:

[umbral snow]

θ = arcsin(√x)

y = 2cosθsinθ
y = 2cos(arcsin(√x))sin(arcsin(√x))
y = 2 √[1 - x] √x

[river stream]

What is the length of the radius of the radius of a circle whose area is 100pie

[umbral snow]

Oof, please see the rules on helper pings

[fallow edge]

Wait 15 mins before pinging

[river stream]

k sorry i deleted it

[umbral snow]

The radius of a circle with 100π area:
A = πr²
A/π = r²
√(A/π) = r

See how that rearrangement works?

[rigid peak]

Hey Kaynex was the previous response for me? Not sure how cos2theta goes to 2cosθsinθ

[umbral snow]

Oh shoot haha

I got that confused with sin(2θ) mb

Use cos² - sin² instead

Actually, use 1 - 2sin²θ

[rigid peak]

yea thats what i did, and tried combining them and was left with x + y = cos^2θ

either way you are left with cos^2θ

[umbral snow]

See what I did

[rigid peak]

ah

oh its just 1-2x lol

thanks

[river stream]

100 divided by pi gives me 31.3

83

i get the square root for that and its just 5

[umbral snow]

Yeah I guess it is haha

@rigid peak

@river stream
A is not 100, but is 100π

[rigid peak]

@river stream you have 100π = πr^2

the π would cancel out

[boreal narwhal]

Does anyone know how to answer this?

[upper karma]

@boreal narwhal Do you know the properties of a parallelogram?

[boreal narwhal]

I believe so, Why?

[upper karma]

Then apply your knowledge in solving them

What do you know about the sides of a parallelogram

?

[boreal narwhal]

That GH is equal to CB

?

[upper karma]

Yes

Just set them equal

And solve for x/y

Opposite sides in a parallelogram are equal

[boreal narwhal]

When I try to set them equal I get a fraction

[upper karma]

Ya so what

[boreal narwhal]

so would the answer be a decimal/fraction?

[upper karma]

Depends if your teacher wants it a specific way

Otherwise anyway is fine

Fractions are good tho

Since decimal formation can be enormously long

@boreal narwhal Do you understand the 2nd problem?

[boreal narwhal]

Yeah, i got 39/2 for A

[upper karma]

Alright

Do you know anything about the angles?

Did you get that part?

[boreal narwhal]

no not yet

[upper karma]

Have you not learned anything about the angles in parralelogram?

Well, opposite angles in a parallelogram are equal.

And consecutive angles are supplementary.

@boreal narwhal Do you understand?

[boreal narwhal]

so would I set it equal to each other or180?

[upper karma]

Explain more

What would you do

@boreal narwhal

Didn’t quite understand what you are implying

Sorry

[boreal narwhal]

do I do 2b+41 = 180

[upper karma]

No it’s adding the 2 angles together

Another tip is that all angles in a parallelogram added up = 360

If you had that angle = to 180 then it wouldn’t work

@boreal narwhal Do you want a video explaining this?

[boreal narwhal]

sure

[upper karma]

Hold on

Theres this one https://youtu.be/Kj1p8j-aoN0

Which specifically explains the angles

And this one

https://youtu.be/t42sLuns4Qs

Goes more in depth about all the properties of a parallelogram

@boreal narwhal Good luck!

[boreal narwhal]

Thanks

[boreal narwhal]

@upper karma

Would this be A or D?

[azure reef]

look

do you know properties of angles and parallel lines?

[boreal narwhal]

A little

[azure reef]

so angles a and angle of 36 deg are opposite angles, yeah?

is the same true for angles b and 113 deg?

[boreal narwhal]

I thought angle B was 36 degrees

@azure reef

[azure reef]

no

i cannot draw it now, but try to draw two parallel lines by youself

and any line crossing them both

and look at properties of inner angles

[boreal narwhal]

Wait but angle b and 36 are identicle

[azure reef]

why?

[boreal narwhal]

what do you mean?

https://cdn.discordapp.com/attachments/326138757474680852/713600573504618568/unknown.png

If you look closely B and 36 look the same

[azure reef]

we don't care how do they look

maybe painter was drunk, huh

what we care is numbers and properties

look, move upper base of trapezoid down that way it overlaps lower base

what can you see?

i mean angles a and 36 will yield some property

[boreal narwhal]

hmm ok

and how do I solve for X and Y for this?

[azure reef]

opposite sides are equal

[regal shell]

Use system of equations after that

[azure reef]

so u'll get two linear equations

[boreal narwhal]

Would it be 7x+6=12y+7?

[regal shell]

Move all the variables to the same side of equation and then use system of equations

Make 2 separate equations

[azure reef]

Would it be 7x+6=12y+7?
well, yes but it is senseless

make two eqs

[boreal narwhal]

wym?

[azure reef]

2 eqs - 2 unknowns

1 equation for sides

1 equation for bases

[regal shell]

7y+7-5x=0

[azure reef]

2x+6-5y =0

[boreal narwhal]

Why do you subtract 5x?

[azure reef]

Why do you subtract 5x?
opposite sides are equal

[regal shell]

Because 7y+7=5x

[azure reef]

if you subtract equal from equal...

[regal shell]

And then I moved all variables to one side

[boreal narwhal]

So how would you solve 7y+7=5x?

[regal shell]

You also have the other pair of sides

[azure reef]

by the help of second equation

[boreal narwhal]

5y=2x+6?

[regal shell]

Yes

Do you know how to use linear combination?

[boreal narwhal]

Not really

[regal shell]

Substitution?

[boreal narwhal]

How would I do it

[upper karma]

@boreal narwhal what u need help on

Sorry I was offline

[azure reef]

How would I do it
@boreal narwhal express x in terms of one of the eqs

and substitute it into the second

[night summit]

https://gyazo.com/35b118df4179a0a5b23df3d0d975695c

can someone help me pls

I know this is geometry sort of which is why I dont know what im missing

[azure reef]

do you know sines etc

[night summit]

i used sin to find 1 other side

yes

sin 30 is 1.5/3

so the opposite to sin is 1.5

and I tried to find sin of 60 because we know its a 30 60 90 triangle

[azure reef]

just use pythagor theorem then lol

[night summit]

it give s me the same result

it just gives me a decimal

and I dont know how to accurately calculate perimeter and area

[azure reef]

and to find area btw you even do not need third side

Area=ACsin(A^C)1/2

[silent plank]

um wtf is that...

[night summit]

im confused

[upper karma]

Just use 30 60 90

To find all 3 sides

[night summit]

I did

[upper karma]

Then what’s the problem

[night summit]

I know one side is 1.5 and the other is a decimal

[silent plank]

you should know the exact ratios for special angles

and not be entering this stuff into a calculator

[azure reef]

sin60=sqrt(3)/2 sin30=1/2

[night summit]

where did u get that

im slow af unfortunately

[silent plank]

exact trig ratios for special angles

[night summit]

oh we r doing that triangle??

[upper karma]

That’s the triangle u have

So 3m=2a

[night summit]

wtf why am i so dumb

[upper karma]

So do u get it now?

[night summit]

yea 3m=2a so a is 1.5

[upper karma]

Yes

[night summit]

and that means c is 1.5rad3

?

[upper karma]

That’s not a radian

[night summit]

radical

[upper karma]

It’s 1.5 root 3

[silent plank]

use sqrt() for square root

[night summit]

ok

so c is 1.5sqrt(3)?

[upper karma]

Yes

Can you solve the rest?

[silent plank]

if c is the variable you are using to denote your base then yes

[night summit]

but it wants me to find the exact perimeter and area of the triangle

and when I add everything it gives me a decimal

how do I know when to stop

[silent plank]

don't evaluate the square root

leave it as is

and not be entering this stuff into a calculator

[night summit]

I dont know how to solve it normally if I dont use a calculator

[silent plank]

your sides are: 3m, 1.5m and 1.5sqrt(3)m right?
add/combine the terms you can

[night summit]

oh wait

nvm

i thought it was 4.5m and you could add the 1.5 from the sqrt

[silent plank]

you can combine the 1.5m and 3m
and your exact perimeter would just be:
(4.5 + 1.5sqrt(3)) m

[night summit]

Oh I didnt know they wanted me to leave it in that form

so for area its the same just (4.5 x 1.5sqrt(3)) m

?

[silent plank]

no

you're using the wrong values

[azure reef]

area is half of product of cathetes

[night summit]

what is cathetes

not knowing what cathetes is and using area is half the product cant I just make it (4.5 x 1.5sqrt(3))/2 m^2

[silent plank]

fancy term for legs of a right triangle

A = 1/2 bh (where h is the altitude relative to b)

[night summit]

ahh

[azure reef]

or generally, A = 1/2bc times sine of an angle between sides b and c

[night summit]

https://gyazo.com/9ea2fb069f374cbf9c0478508752063e

?

[azure reef]

yep

but are cathetes alright?

[silent plank]

that can be simplified further

[night summit]

idk this site is very simple and maybe wont recognize it

[silent plank]

i would also suggest using fractions

makes simplification easier

[dark sparrow]

@azure reef the sides of a right triangle adjacent to the right angle are called legs in english.

[night summit]

how would I use fractions here?

[azure reef]

@dark sparrow and cathetes in Greek

[silent plank]

1.5 = 3/2

[night summit]

oh

[silent plank]

continue to simplify: \
$= \frac12 \times \frac{3\sqrt{3}}{2} \times \frac32$

[somber coyote]

ramonov:

[azure reef]

$\frac{9\sqrt{3}}{8}$

[dark sparrow]

\sqrt{3}

{}

[somber coyote]

Commander Vimes:

[silent plank]

also wasn't directed at you

[night summit]

ohh

I really need to practice math again im a disgrace

so this would be the final answer right?

https://gyazo.com/c5d7e7dad8cff9bff0509d92094a0c1a

[azure reef]

yep

[night summit]

thanks

[ocean axle]

I have tetrahedron with sides all equal to 6. There is a sphere inscribed in that tetrahedron. What is height from base of the figure to the mid of this sphere.

[upper karma]

No idea

[upper karma]

quick question --> might be dumb, but it's 3:00 AM so here goes:

is it valid to say that if all of the angles in a polygon are equal, so are all of the sides?

[dark sparrow]

no

take a regular hexagon and stretch out one pair of parallel edges

[regal shell]

Or just any non-square rectangle

[civic jolt]

Can anyone explain?

[silent plank]

elementary limits

$\lim_{a\to0} \frac{\sin(a)}{a} = 1$

[dark sparrow]

_

[somber coyote]

ramonov:

[civic jolt]

?

0/0 ?

0/0=1
Or m I getting it wrong?

[dark sparrow]

"0/0" does not mean "limit does not exist"

[civic jolt]

So it's 1?

[dark sparrow]

ramonov just told you this particular limit is 1, yes

[civic jolt]

Ahh

So it's fixed?

[dark sparrow]

what's "it"

[civic jolt]

sin (x)/x = 1

[dark sparrow]

sin(x)/x never EQUALS 1.

sin(x)/x APPROACHES 1 as x goes to 0

[civic jolt]

It approaches 1 but is written =1 ?

[versed river]

the limit is equal to 1. when a limit is equal to a value, that means it approaches that value

[dark sparrow]

no, ramonov did not write $\frac{\sin(x)}{x} = 1$

[somber coyote]

Ann:

[dark sparrow]

he wrote $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$

[somber coyote]

Ann:

[civic jolt]

Wait I hv to process this

[civic jolt]

Aight I firgured out y

Y is theta approaching 0?

[dark sparrow]

2x -> π/3

2x - π/3 -> π/3 - π/3

[civic jolt]

?

I don't get the 2nd part

-> π/3 - π/3
How?

[dark sparrow]

...

[civic jolt]

-> 0
To compensate for the 0?

[dark sparrow]

you're overthinking it

like do you really not understand why $\lim_{x \to c} (x+5) = c+5$ for example

[somber coyote]

Ann:

[civic jolt]

x-c=0

[dark sparrow]

...

[civic jolt]

That's how I've always looked at it

[plucky marlin]

[civic jolt]

2x - π/3 -> π/3 - π/3
How?

[dark sparrow]

2x approaches pi/3

what do you think 2x - (constant) would approach, if not pi/3 minus that same constant

[civic jolt]

0

[dark sparrow]

sigh

[civic jolt]

Ono

pi/3 - pi/3 = 0
Right?
pi/3 is there to compensate for the 0

[dark sparrow]

idk what the hell you mean by "compensate for the 0"

[civic jolt]

pi/3 - pi/3 = 0

[sick isle]

Guys why does this work I cant get it

[rich wolf]

@sick isle cos^2+sin^2=1

So the other segment of the hypotenuse is sin^2

[sick isle]

yes but i dont get how to get the cos^2 there

[rich wolf]

And then the vertical part of the triangle is sin

Pretty sure its something with geometric mean

[silent plank]

that smaller triangle is also a right triangle

cos(theta) = a/cos(theta)
a= cos^2(theta)

[rich wolf]

What is a

Altitude?

[silent plank]

(where a is the length of the indicated segment in question)

[sick isle]

Ok so i still dont get it here is a picture

[wooden current]

@sick isle Nice bucket

[sick isle]

thanks man its my home

[wooden current]

ok so what's bugging you

[sick isle]

I cant figure out why that length is cos²x

Where x in the angle from x axis to line drawn

[wooden current]

i mean Ramonov explained it

You apply cos x in that little triangle

cos(x) = cos^2(x) / cos(x) <=> cos^2(x) = cos^2(x) and it checks out

[signal perch]

Yo

[ripe dove]

Yo

[signal perch]

Yo

[tawdry pivot]

@​yoyoyoyo

[upper karma]

help

how do I find a surface area of a triangular prism with height and a base?

@upper karma base times height

Oh wait

Nah that’s area

yea was aboutta say

It’s

You just calculate the area of all the sides

And add them all together

https://youtu.be/7YInNiIzK9o

ohh okay

[dark sparrow]

You just calculate the area of all the sides
And add them all together

ngl i find it kinda weird that people seem to not know or not remember this simple idea

[upper karma]

@dark sparrow yes

[regal shell]

Lmao

[rich wolf]

same

[upper karma]

I came here for math help not to be insulted by smart asses

[dark sparrow]

this wasn't intended as a jab at you specifically

[upper karma]

It was implied

[dark sparrow]

[long briar]

I don't think it's their fault specifically.

If someone is confused about something like that, it's more likely that their teachers failed to explain to them what surface area actually IS, which is a bit sad : /

[upper karma]

reminds me of a sad story that my professor told me

he said his friend was teaching at a middle school, and the principal was spectating the class one day

one of the students asked where the surface area of a cylinder formula comes from, and she gave the intuitive and memorable explanation

then afterwards, the principal told her "that's not part of the curriculum, you can't teach that"

[dark sparrow]

wat

[graceful narwhal]

the feels

😦

[upper karma]

https://tenor.com/view/wait-thats-illegal-halo-meme-gif-14048618

[royal otter]

Can someone help

[dark sparrow]

there are 6 questions here

which one do you need help with

if multiple, which one do you want to start with

@royal otter

[royal otter]

2nd and 3rd

@dark sparrow

[dark sparrow]

alright, so what is giving you trouble for #2?

[royal otter]

That sin alpha is that stuf there and i would say everything can u do at least the 2nd question

I have a test so id ont have time for explaining

[dark sparrow]

uh

what

are you having a test right now????

[royal otter]

Sorry if this is against the rules i dont have time

Yes

[dark sparrow]

this IS against the rules

[royal otter]

I just joined

[dark sparrow]

this IS against the rules.

[royal otter]

Okay

[dark sparrow]

<@&268886789983436800> get their ass

[royal otter]

I wont ask for help then

Thanks anyway

But still good server for getting help thi

;-;

Pls dont

Hey hey be undearstanding a bit

[long briar]

Nah @royal otter, screw that. You are cheating over your classmates who did honest, fair work up to this test : (

Oh wait, shoot why did I ping you you are in a test

[copper valve]

their butt has got got

[dark sparrow]

thamk u

[tawdry pivot]

this is against the rules of not only this server

[graceful narwhal]

maybe there a specialized cheating servers

I feel there is a demand for those

[weary drift]

google found me chegg servers

[upper karma]

Chegg study :O

[clear forum]

😮

[upper karma]

😳

[clear forum]

idk if this is the right place to ask how to calculate the surface normal on a sphere 😳

[ripe dove]

It’s almost like the 10 min time to read the rules is ignored...

[upper karma]

@clear forum circumference

Of a circle

Multiplied

By

[clear forum]

wot

[hot ember]

I have this problem where "cos b = -(9/13)" and b is an angle greater than 90deg but less than 180

Using arcos b I got that angle b = 133deg

But I am so lost as to how to get the rest of the trig functions now, was wondering if someone knew what steps to take now

[silent plank]

did you mean arccos(-9/13)?

also that doesn't round to 133°

[hot ember]

oh yeah i used (-9/13)

[silent plank]

use pythagorean identities to help find the other trig ratios

[hot ember]

Although I'm not sure if I should have or not, since the text is this:

use pythagorean identities to help find the other trig ratios
@silent plank ah okok will do

thnx

[vale herald]

A line segment has one end point A at (-3, 5) and a slope of -2/3. State Two locations (x, y) of the other endpoint B.

[weak shoal]

Okay so what do you propose might be a good way of attacking the problem?

[upper karma]

BRUH, dont multipost. @vale herald

[weak shoal]

? He posted in linear algebra and I told him to come here

[vale herald]

-2x-y-1

is that right/

[upper karma]

@weak shoal he posted at #prealg-and-algebra too smh

and we answered him already

[weak shoal]

oh

okay

[vale herald]

What do you mean you answered me?

[upper karma]

you may wanna check #prealg-and-algebra again maybe

[vale herald]

u can't use point slope formula to graph it..

[upper karma]

who told you it was used to graph

[ripe dove]

Also, you could use point slope form to generate a graph...though that won’t necessarily help with your specific question

[viscid ginkgo]

https://i.imgur.com/kpY0JsA.png****

can I use alternate interior angles here?

[slate bay]

Why wouldn't you be able to?

Wait, sorry, what's > mean?

Parallel, right?

[viscid ginkgo]

yes

I just don't know the rules that well...

I'm looking at an example in my book

[upper karma]

yeah you could

[viscid ginkgo]

but it doesn't look like this

ty

I'm having trouble feeling like it's "correct" since I have no way to "Verify" it

[upper karma]

https://complete-concrete-concise.com/mathematics/proving-alternate-interior-angles-are-congruent-the-same/ Proving Alternate Interior Angles are congruent

Yes derive theorems and equations to fully understand things

[upper karma]

^^^

I try to derive it before looking at it in the book

Kinda fun actually

👍 👏

🧕🏼🤲

[lavish glacier]

Hello, can anyone help me on my quiziz hw?

[upper karma]

@lavish glacier Sure

Sure

Wait is that a quiz

[lavish glacier]

So if i have am apothem on a hexagon of 11

What is area

So is it 22 times 22?

@upper karma

[regal shell]

Apothem?

[little osprey]

a line from the centre of a regular polygon at right angles to any of its sides (dictionary).

[lavish glacier]

None of them work

I fond the radius

[little osprey]

ok

whats radius

@lavish glacier

[lavish glacier]

31.4

@little osprey

[little osprey]

no

[lavish glacier]

Why?

[little osprey]

What is circumference formula

wait we will get there

[lavish glacier]

Circumference is 20pi

[little osprey]

yeah but the formula

what is it

[lavish glacier]

R times 2

[little osprey]

nope

close

[lavish glacier]

If its 8

[little osprey]

no

[lavish glacier]

Then its 16

[little osprey]

look

circumference formula is

dπ

diameter x π

[lavish glacier]

Its double the radius

Oh waot im dumb

[little osprey]

Search it up

if you would like

[lavish glacier]

Thats diameter

Ok

[little osprey]

I think yo ucan solve it from tehre

[lavish glacier]

So

[little osprey]

Yeah?

So?

[lavish glacier]

100

?

@little osprey

[little osprey]

A or D

No wait

Dont rush

[lavish glacier]

Yeah

[little osprey]

First opiton

or last option

They're both 100

[lavish glacier]

No nc its not aquared

Squared silly

[little osprey]

what?

Area always has units^2

[lavish glacier]

One was in and in²

Yes

[little osprey]

Yeah

[lavish glacier]

Thats why d was correct

[little osprey]

So which option

Ok

Good

You got it right

[lavish glacier]

Haha

[little osprey]

recall the formula for the area of a parallelogram

[lavish glacier]

Ok i finisbed that one

[little osprey]

Why do you post them here if you know how to do them

[lavish glacier]

Idk, when i post it here and you guide me, i somehow know what to do.

It like comes to me

[upper karma]

Awesome!

👏

[lavish glacier]

Not awesome

[upper karma]

Oh

[little osprey]

lol

ok have you learned exact trig ratios

[lavish glacier]

Yes

[little osprey]

well imagine a line from the top two points of the trapezium to the bottom line

[lavish glacier]

Ok

[little osprey]

Im just going to assume its a trapezium

[lavish glacier]

Yes height

[little osprey]

Yes

[lavish glacier]

Whats trapezium?

[little osprey]

I think you can handle it from there

[upper karma]

Yes

You can handle it

I believe in you

[lavish glacier]

Whats trapezium?

[little osprey]

Well if its not a trapezium

actually

[supple sundial]

I have a question regarding the graphing of reciprocal functions, where should I go?

[little osprey]

it doesnt matter in this case

@supple sundial What type

[upper karma]

@supple sundial Perpendicular

[little osprey]

@lavish glacier Forget the trapezium

[lavish glacier]

Ok

[little osprey]

Ok, so you created a rectangle and 2 triangles correct?

By doing that

[upper karma]

Yessir

[supple sundial]

perpendicular?

[lavish glacier]

No

[little osprey]

@supple sundial Answer my question

[upper karma]

@supple sundial Perpendicular slope

[lavish glacier]

Wait what?

So i have 2 teiangles amd a rectangle

Ok

[little osprey]

Do you get it or not

[lavish glacier]

Triangles

No

[little osprey]

ok thank u for telling me

[upper karma]

Ya I don’t get it either

[lavish glacier]

What trig function do i use?

[supple sundial]

@little osprey How does one use a system of equations to manually find the graph of a function in the form y=a/x-h+k

[lavish glacier]

Tan?

@little osprey

[lavish glacier]

Yes

So i use tan to find

[upper karma]

90 60 30 triangle

[little osprey]

@supple sundial Could you just go to one of the #❓how-to-get-help channel

Use exact trig ratios Var

[lavish glacier]

Idk

Wdtm?

Its either sin, cos, or tan right?

[upper karma]

You can also just simply use the 90, 60, 30 triangle rule

[lavish glacier]

Ok

But what am i finding?

[upper karma]

The height

[lavish glacier]

Height?

Yeah

[upper karma]

Because the area of a trapezoid consists of finding the height

[little osprey]

It doesnt say its a trapezoid

[lavish glacier]

So but i dpmt have any other variables

[little osprey]

xi

[upper karma]

Ok

[little osprey]

whats tan 60