geometry-and-trigonometry

[nova prawn]

Your strandars and here's standart different.

Yup, eventually.

[upper karma]

sigh

[nova prawn]

Here, %300 increases to %500. If you go outside protests, you get arrrested

Money is everything. Dollar is important.

[upper karma]

If you go outside to protest, loot buildings, and burn buildings down, no one cares.

Since it's in the name of justice.

I gtg lol.

[nova prawn]

cya

[upper karma]

Gl on your endeavors :3

[nova prawn]

it will be hard, but will worth it.

thank you 🙂

1:30 a.m here...

[upper karma]

Np lol.

[upper karma]

America is going to collapse in the next few years

It’s predictable.

[eternal crag]

@ripe osprey

[upper karma]

@upper karma Lol it's right in everyone's face.

Everyone knows America gonna be fucking wack in a few years, they just think it's either these shit political correctness police that'll cause it.

Or white privilege.

That'll ruin America.

@ripe osprey Nerd.

ah good race and politics in the #geometry-and-trigonometry channel

Race and politics is the fundamentals of geometry.

Or it's the other way around.

[dark sparrow]

fuck off yall this shit in particular does NOT belong in the channel yall chose to post it in

[upper karma]

[fickle mist]

staph pls

[dire sand]

Hey all, hopefully this is the appropriate channel, but I'm wondering if someone could explain this in a bit more of a simplistic manner

I'm not quite understanding the point

This follows the irrationality of root 2 proof

so the way i'm interpretting it is that the greeks were saying, give me two vectors, and the resultant vectors magnitude can be divided into 2 pieces, and both would represent the magnitude of the original vectors

Which i'm kinda confused about

or is it saying give me two line segments and i can a line that isn't the resultant

hm

idk

Yeah but if someone could please dumb this down it would be appreciated 😄

[upper karma]

It is saying that Greeks thought lengths were “constructible” only if the length was rational

But this is not the case. Make a right triangle with leg lengths of 1, then the hypotenuse construction has length sqrt(2)

[dire sand]

oh

whats with all the divides evenly and rational multiple stuff

[upper karma]

“Divides evenly” means hypotenuse construction I suppose (when interpreted geometrically)

The rational multiple part is another way of saying the same thing, but with multiplication instead of division

[dire sand]

I don't get whats even about the hypotenuse construction though

[upper karma]

I think it means “connected”

[dire sand]

wdym by connected? sorry for my unfamiliarity of the terminology and stuff

[upper karma]

Suppose I put two lines of length 1 perpendicular to each other and touching at only their end points

The Greeks would say

There is a line of rational length which connects the other end points on the lines together

[dire sand]

but the straight line from the tip of one line to the tail of the other would be root 2

since it's the shortest path

[upper karma]

Yes, so the Greeks were wrong

[dire sand]

ah ok

Thanks

[upper karma]

If you view multiplication of 2 lines as a rectangle

Then that’s how hypotenuse being related to “divides evenly” might make sense

I don’t exactly know how the book is using that term but that’s my guess

[dire sand]

multiplying 2 lines gives a triangle ?

xd

[dire sand]

rectangle*

lol

sorry

is that like japanese multiplication

[charred zephyr]

is this channel free?

i guess it is

can you give me a hint to prove this proposition:
Let A be a point, (D) a line, H a point such that (AH) is perpendicular to (D). The distance AH is equal to the distance from A to (D).
what i think we are going to use is the following property:
The distance between a point A and a line (D) is the smallest distance between point A and a point H of line (D).
and i think i need to prove that the smallest distance "happens" when (AH) is perpendicular to (D) but i dont know how

are we going to use angles

or maybe pythagoras

nevermind i got it

[raw spear]

When dealing with trig values of special angles and forming a right triangle in a unit circle, how would you figure out the lengths of the adjacent and opposite side if the triangle isn't isosceles?

[royal edge]

I don’t understand the problem. Teacher assigned some optional work

[umbral snow]

If you know the height and hypotenuse of a right triangle, then you know the base.

@royal edge

[royal edge]

I still don’t get it

[umbral snow]

Let's say I attached 5 wires all of known length to the same height on a post

Then all of the wires will attach to the ground at the same distance away from the sign

[royal edge]

Ah I see! Thanks

[wooden current]

h

[charred zephyr]

hi can someone give me a hint idk what to start with with this:
Let (C1) and (C2) be two circles which intersect in points A and B.
(L) is a line which goes through A.
(L') is a line which goes through B.
C and E are respectively the intersections of (L) with (C1) and (C2).
D and F are respectively the intersections of (L') with (C1) and (C2).
Show that (CD) and (EF) are parallel lines.

diagram i made on geogebra

i first though about using thales proportionality theorem but it seems like it doesnt work but idk

apart from that i am stuck

maybe similar triangles

with AA

FGE = CGD is clear i may need to prove CDG = GFE or GCD =GEF

[silent plank]

consider stuff like inscribed angle theorem, cyclic quadrilaterals

[charred zephyr]

the inscribed angle theorem is the one about the center, double half etc?

like AMB = 1/2 AOB

[upper karma]

Got it

Consider cyclic quadrilater CABD

<CAB+<BDC=180, <CAB+<BAG=180

You continue...

@charred zephyr

[charred zephyr]

ok thank you ill try!

i think i got it thank you

[upper karma]

What did you do?

Don't write the whole proof just tell me what angles you used

[upper karma]

Rip

👻

[upper karma]

Okay so what? @arctic vortex

[arctic vortex]

i dont undestand this

do i find the scale factor?

@upper karma

[upper karma]

So first idk why the guy decided to put b instead of putting literally 20 but ok

[arctic vortex]

lol

[upper karma]

And do you know the general approach to these type of problems or you are new to them?

[arctic vortex]

new to them

i have a lot of homework to do which is why im asking for help here, and studying really hard to understand them once im done with everything

[upper karma]

Okay so

https://youtu.be/fA2EZoYgMgA

For the basic understanding

[arctic vortex]

oh ok

wait but i only have 30 mins to finish these questions lol

i already did 2

i have 2 left

both like this

[upper karma]

Watch it

Trust me

[arctic vortex]

k

[upper karma]

We'll work it out after together

[arctic vortex]

20 / 7 = 2.9

so i multiply 12 * 2.9

[dark sparrow]

20/7 is not equal to 29/10.

[upper karma]

Oh ok nvm

[arctic vortex]

its 2.85

[dark sparrow]

no

[arctic vortex]

20 / 7

[dark sparrow]

20/7 is not 285/100 either

[arctic vortex]

oh

[dark sparrow]

do not round intermediate calculations at all

[arctic vortex]

oh

[dark sparrow]

only round the final answer and ONLY if you're asked to do so.

[arctic vortex]

i did 20/7 on google

and it said 2.85

[dark sparrow]

do not use a calculator

[arctic vortex]

ok

[dark sparrow]

are you allergic to fractions or something?

[upper karma]

And you aren't appropiately applying it

[arctic vortex]

yeah lol

[dark sparrow]

aledium what contrals is doing is fine in principle

it's the inappropriate rounding that i am pointing out

[arctic vortex]

ohh

so it just do 2.8

just be*

[upper karma]

[arctic vortex]

20 / 7 ?

[dark sparrow]

no

no! no!!! 20/7 is not equal to 28/10 !!!

20/7 is 20/7 !!!

[arctic vortex]

ohh

so the answer is like 34

if u multiply 12 * 20/7

wait

1 second

i undestand it

bc its scale factor

ok im stupid

[dark sparrow]

12 * 20/7 is correct.

you just insisted on inappropriately rounding 20/7.

[upper karma]

Formally $\frac{y}{12}=\frac{20}{7}\implies y=\frac{12\cdot 20}{7}$

^

[arctic vortex]

oh wow haha i wouldve done that

[upper karma]

Welp

[dark sparrow]

,calc 12*20/7

[somber coyote]

Result:

34.285714285714

[arctic vortex]

thats a nice bot

[upper karma]

Latex died again

[dark sparrow]

34.3 is the answer you will get after rounding to one decimal place.

[arctic vortex]

oh ok thanks !

[dark sparrow]

note that rounding prior to multiplication by 12 would have resulted in an incorrect answer.

[upper karma]

aledium what contrals is doing is fine in principle
Yeah i confused the y with the b

[somber coyote]

Al𝟛dium:

[upper karma]

Now @arctic vortex

[arctic vortex]

yeah i got it right

thanks !

[charred zephyr]

if two triangles are similar can we directly apply thales theorem to prove parallelism

[gloomy slate]

ok

[upper karma]

Post it

Again lol

[gloomy slate]

i need to find exact value of sin and cos

[upper karma]

So you are given literally $\theta=\frac{23π}{6}$

Wow

It died again

[somber coyote]

Al𝟛dium:

[upper karma]

Okay not theta but its to name it

[gloomy slate]

k

[upper karma]

Can you post the original question

[upper karma]

Okay better

[charred zephyr]

Again lol
@upper karma what?

if two triangles are similar can we directly apply thales theorem to prove parallelism
@charred zephyr can someone answer me

if two triangles have same angles does it mean their bases are parallel when they're put in the same summit form or butterfly configuration

[upper karma]

@gloomy slate okay so 23π/6

Draw it on the axis first

Hello?

[gloomy slate]

hey

[charred zephyr]

hello

[gloomy slate]

@upper karma im not sure how id draw it

[upper karma]

So

Think of unit circle

23π/6 seems VERY big, compared to what we will use to graph it

Bc remember its 0, 2π

And 23π/6 is WAY bigger than 2π, do you agree? @gloomy slate

[gloomy slate]

yes

its almost 4pi right

?

[charred zephyr]

i'll just wait

[upper karma]

Yep

So first lets just remove 2π so its on our 0, 2π $$\frac{23π}{6}-2π$$

[somber coyote]

Al𝟛dium:

[gloomy slate]

What do you mean by its on our 0?

[upper karma]

Our 0, 2π region

[gloomy slate]

k

11pi/6

[upper karma]

Yep

So now let's move to: on which quadrant is 11π/6 on?

[gloomy slate]

4th

[upper karma]

Now draw it, it'll help you visualise it

We are almost done

Just wanna build the base strongly

[upper karma]

Okay fine

Now let's recall finally a known thing

$\cos(\theta)=\cos(2π-\theta)$

[somber coyote]

Al𝟛dium:

[upper karma]

Apply it

And we'll get a very known angle to check on the unit circle for cos

[gloomy slate]

Howd you get that equation

[upper karma]

Trigonometry

[gloomy slate]

hm

pi/6?

[upper karma]

Yep.

Now

https://images.app.goo.gl/wzkbYY1aaFdPgdLR7

For cos

[gloomy slate]

sqrt(3)/2?

[upper karma]

Yep

That is for cosine

BUT

$\cos(\theta)=\cos(2π-\theta)$ IS TRUE BUT NOT $\sin(\theta)=\sin(2π-\theta)$

[somber coyote]

Al𝟛dium:

[upper karma]

Ok?

[gloomy slate]

ok

i see that chart has 11/6 as well

but when they ask something -3pi/4

[upper karma]

i see that chart has 11π/6 as well
@gloomy slate yeah for the sine you have to check on 11π/6

But for cosine check for π/6 which is more comfortable bc of the identity we said above

[charred zephyr]

🙋‍♀️

[upper karma]

but when they ask something -3pi/4
Yes?

[gloomy slate]

I can refer to the chart and see 3pi/4 is this

but when they ask for the negative version

sin turns into -sqrt(2)/2 as well

but how would I know that based off that circle chart or whatever

[upper karma]

Well

$\sin(-\theta)=-\sin(\theta)$ and $\cos(-\theta)=\cos(\theta)$

[somber coyote]

Al𝟛dium:

[upper karma]

And THIS is what you should use

[charred zephyr]

it's like functions parity

[upper karma]

$\sin(-\frac{3π}{4})=-\sin(\frac{3π}{4})$ which of you can check on the unit circle

Yes

[somber coyote]

Al𝟛dium:

[upper karma]

@gloomy slate

[gloomy slate]

hmm okay thanks

[upper karma]

Np!

[charred zephyr]

if ABC and ADE are similar we do agree that angle a should be equal to the red angle and angle b should be equal to the blue one isnt it

[upper karma]

If BC and DE are parallel, yes.

Well you proved they are similar so yeah

[charred zephyr]

two triangles can be similar AND not have parallel bases?

;

[upper karma]

Uh

[charred zephyr]

ok

but thanks anyways

but just if you know would angle a be equal to purple or green angle? same question for angle b?

forgot to say the little triangle is similar to the bigger

[silent plank]

it could be either

[charred zephyr]

it depends on some specific thing?

[silent plank]

depends if the top and bottom lines are parallel

[charred zephyr]

so similar triangles dont necessarily have parallel bases?

[silent plank]

depends

on their relative position

[charred zephyr]

on what?

[upper karma]

Their Relative position

If you can gave an example, it'd made it easier for us to explain

Like not an invented one

Lmao

[charred zephyr]

ok wait

ABCD is a square of lenght 4
Diagnols AC and BD intersect in point G
The cocyclic points A;E;B;F form quadrilateral AEBF
Angle BOD is equal to 170° with O center of the circle
Show that AD and BC are same line or parallel

corrected something...

[silent plank]

as always, draw a diagram

[charred zephyr]

uh you know what just cancel my question im too tired to follow the lines

[upper karma]

F

[wary wind]

We can even cancel you if we want...

[tawdry pivot]

can we cancel their name

[upper karma]

What's some tips to find the family of lines tangent to a circle, if we know its radius and origin?

[arctic vortex]

can anyone help me with like 15 geometry questions about triangle congruency?

[dark sparrow]

oh boy

[upper karma]

Post them

[arctic vortex]

wait tho

i will solve them

u just check if they are good

[upper karma]

Aight

[arctic vortex]

can i get 10 minutes pls

or like 5

[dark sparrow]

sure

ping me when you're done and send me the questions and your answers

[upper karma]

Take as much as you need

[arctic vortex]

ok tysm

[dark sparrow]

actually you can send the questions now so i have time to look at them

[arctic vortex]

o 1 sec pls

@dark sparrow @upper karma

[upper karma]

What's the difference between a and c

[dark sparrow]

is there gonna be fourteen more of these?

@upper karma the difference is in which vertices are paired

[arctic vortex]

no there are like 15 questions on my homework packet but i dont undestand like 5

and the rest i just ask u to check them

so can u help me with question 1 pls ??

@dark sparrow srry for ping

[dark sparrow]

ok so what's troubling you here

[arctic vortex]

i dont understand q#1

at all

i see the triangles

so i find which of them are similar/

the lines

[dark sparrow]

you are given four similarity statements
you are asked to choose which one of them is actually true for the diagram

[arctic vortex]

oh ok

wait i will solve it can u check it?

[dark sparrow]

i thought that was what we were gonna do anyway

[arctic vortex]

oh yeah

and wait what is the "B"

in the question

does it belong to the big small or both

[dark sparrow]

it's (2,2) as far as i can tell

anyway, you solve your 15 questions, give me the questions and your answers, and i check them and tell you which ones you got wrong

[arctic vortex]

ok

wait whats the answer to #1

or what does b mean

what triangle does b belong to

[dark sparrow]

wdym

B is a vertex in three triangles

[arctic vortex]

ohh

[dark sparrow]

it's the intersection of AC and DE

[arctic vortex]

is the answer B??

[dark sparrow]

nope

it's A

[arctic vortex]

what

how??

[dark sparrow]

the legs of ABD are in a ratio of 3:2, as opposed to those of EBA which are in 2:1

EB/BA = 2 = DB/BC

[arctic vortex]

ohhh

@dark sparrow are these correcT?

[dark sparrow]

first one only

[upper karma]

^

[dark sparrow]

just because it looks like a parallelogram doesn't mean it is one

[upper karma]

Sad truth

[arctic vortex]

would it be C then?

for the second one

[upper karma]

Yes

[arctic vortex]

oh ok

@dark sparrow @upper karma

[odd hedge]

pinging HoboSas pings me for some reason

[arctic vortex]

o sorry hobo

wrong person

[odd hedge]

lol

np good luck

[arctic vortex]

i clicked hobo and i thought it would direct to hobosas

[warm basin]

why did u select C

did u just guess

look at the coordinates of points K, L and M

its clearly d

k is located at (1,1)

for L y = 4 and x = 5

for M y = 1 and x = 5

[arctic vortex]

no i went to desmos.com and i put all the points in

[dark sparrow]

why are the minus signs weirdly offset

[warm basin]

^

[dark sparrow]

anyway

(-1, 1), (-1, 4) and (2,1) do not form a 3-4-5 triangle

[arctic vortex]

oh

[dark sparrow]

while KLM here is one

[upper karma]

Try drawing every triangle

[arctic vortex]

yeah thats what i did

and c looked like the best option

wait

is it d then

@dark sparrow

[dark sparrow]

yes

[arctic vortex]

really?

no joke like really

how is it d that was my last option

@dark sparrow whats the second one ??

[warm basin]

you should stop guessing "Oh ummmm is it C?"

[arctic vortex]

wait no its not a

its c

[warm basin]

you should actually know why your choosing an option

[dark sparrow]

it's one of the angle congruences

[warm basin]

multichoice homework sucks

[arctic vortex]

yeah

i think its either c and e | or f and d

no not d

a

[dark sparrow]

honestly

this is a poorly worded question since there are multiple correct answers

or wait hold on

[arctic vortex]

?

[dark sparrow]

no wait there actually is only one my bad

[arctic vortex]

?

[warm basin]

directions says you must select BOTH

and its written BOTH in caps

[arctic vortex]

yeah

[dark sparrow]

i am not spoiling the answer.

[arctic vortex]

oh

[warm basin]

now if there arent more than 1 correct answers, i would be pissed.

and nah dont spoil it

[arctic vortex]

can u give me a hint

i know one is either c or f

[warm basin]

@arctic vortex why do u think its c or f?

stop guessing just like that its stupid

[arctic vortex]

because it makes sense unlike a where bc =/ qr

[dark sparrow]

okay so let's just establish this: it's gotta be one of the ratios and one of the angle congruencies

[arctic vortex]

yeah

[warm basin]

$\Delta ABC \sim \Delta QRS$

[somber coyote]

trece:

[warm basin]

do you know what this means Contrals?

in words

just wanna make sure

[arctic vortex]

triangle abc is similar to triangle qrs ?

[warm basin]

ok

[arctic vortex]

wait

are we still on this one

@dark sparrow @warm basin

[warm basin]

of course

ann gave u a hint already

okay so let's just establish this: it's gotta be one of the ratios and one of the angle congruencies

[dark sparrow]

option a would require angle C = angle Q to work

but we don't have that in the options

wait

uh

no, option a doesn't work at all

BC would correspond to RS, not QR, under the similarity statement they give you

option a is out entirely

[arctic vortex]

yeah

so the ratio thing is either c or f

and the congruence is either d or e

[warm basin]

look at option c

do u notice anything

or why it would be correct/incorrect

[arctic vortex]

its correct bc it matches up

with everything

[warm basin]

okay

what about the angles

[arctic vortex]

its not d

either a or e

b or e

@warm basin

[warm basin]

what do u think about f

[arctic vortex]

its like the same

[warm basin]

so were between b),c) and e),f)

[arctic vortex]

yeah

@warm basin

[warm basin]

do u know trigonometry

[arctic vortex]

no

[warm basin]

like basic functions sine, cosine

ah

[arctic vortex]

im in geometry

not trigonometry

im taking that next year i think

here is the question again by the way

so u dont have to scroll up

[silent plank]

highlight the sides and angles of your choice on the diagram

and see if it satisfies SAS

[arctic vortex]

yeah

because b is the angle side

angle **

[upper karma]

So what are the answers?

[warm basin]

SAS stands for Side angle side

[arctic vortex]

its either f and e or c and b

[silent plank]

specifically the congruent angle needs to be between the 2 corresponding sides

[upper karma]

That's the key

[arctic vortex]

yeah i know sss and sas and asa and stuff

[warm basin]

the ratio between two sides is the same as the ratio between another two sides
also the included angles are equal

thats how u know

[arctic vortex]

oh

yay only 4 more

and then i am done with homew ork for the week

and i can do anything iwant

yay

[silent plank]

you should check again whether e,f is actually valid

[arctic vortex]

how do i check that

[upper karma]

You think

[arctic vortex]

oh

yeah i think that e and f are correct

or is it b and e

@silent plank i think its b and e

[upper karma]

Are you sure <C is between AB and BC

[arctic vortex]

actually

[silent plank]

highlight the sides and angles of your choice on the diagram
and see if it satisfies SAS

[warm basin]

angle C is not between AB and BC

[arctic vortex]

no

[warm basin]

therefore its not that

[arctic vortex]

oh so its not b and e

[upper karma]

So it's?

[arctic vortex]

e and f

[upper karma]

angle C is not between AB and BC
^

[arctic vortex]

oh 1 second then

f and b?

[upper karma]

Stop guessing

[arctic vortex]

ok

[upper karma]

Sit down and think about it

[arctic vortex]

wait

its sas right

idk

[silent plank]

highlight the sides and angles of your choice on the diagram
and see if it satisfies SAS

[arctic vortex]

ok

[silent plank]

like use paint or highlight it on paper or something

[silent plank]

which options are you using to get those?

[arctic vortex]

b and e

a =q and c =s?

[silent plank]

where are you getting <B = <R though?

[arctic vortex]

im not getting that

[silent plank]

then why is it indicated on the diagram

[arctic vortex]

bc its the angle

[silent plank]

that's not what i'm telling you to do

[arctic vortex]

oh

[silent plank]

indicate what the options (b) and (c) are saying

if that is what you're checking

[warm basin]

b c

[silent plank]

typo sry

and do the same for (e) and (f)
since you were concerned about that too

[arctic vortex]

wait its b c

[warm basin]

look contrals, the option c is basically saying that the ratio between AB and QR is the same as ratio between AC and QS

the angle between AB and AC is A right?

and the angle between QR and QS is Q

[arctic vortex]

yeha

yeah

[warm basin]

makes sense now?

[arctic vortex]

oh

[warm basin]

we can move on to the next question now, i dont wanna leave u all day long on the same one xD

[arctic vortex]

so a is the center of ab ac

and q is the center of qr q

qs

so the answer is b c

[warm basin]

i wouldnt say center

[arctic vortex]

the middle

midpoint

[warm basin]

A is the angle between AB and AC

[arctic vortex]

ok now i finished everything else

can u just check

@warm basin can u check these

@dark sparrow

[dark sparrow]

first one is correct, second one isn't

[arctic vortex]

whats the second one?

also is this correct?

[dark sparrow]

you posted two problems

[warm basin]

on the second one i want u to look on option c and d

[arctic vortex]

ok

they both look equal and good

[warm basin]

do they?

[arctic vortex]

wait no its d

because

one of c

does left

[warm basin]

okay good

[arctic vortex]

and the other one goes to hte point

R

so its d

[warm basin]

so d it is

[arctic vortex]

OH MY GOD

I GOT A 90% ON MY HOMEWORK

[warm basin]

just remember, a correct answer is a answer that makes sense

[arctic vortex]

AND I ONLY ASKED HELP FOR LIKE 5

IM SO HAPPYT

DAD MOM

GDT HTE CMAREA

looks like

i dont need to do

the problem sets

because i am just too smart

yesssss

[warm basin]

[arctic vortex]

i am just too smart

[warm basin]

im sorry son but u shouldve gotten 95% or better

[arctic vortex]

oh

[warm basin]

how many problems were there in total

[arctic vortex]

24

[warm basin]

ok ok

[arctic vortex]

i actually got 1 half wrong

because i misread it

wait a minute

i shouldve gotten a 100

wtf i misread #24 and #5

but i make 1 mistake on hte half wrong question

[upper karma]

Good job

[arctic vortex]

thanks

[warm basin]

21,6 questions right

[arctic vortex]

now i am einstein jr

no 24 questions

got 2 wrong 1 half wrong

[warm basin]

21.6 are right out of 24

[arctic vortex]

so 2.5

[warm basin]

90% of 24 dude

do u know how to calculate percentages ?

[arctic vortex]

yeha

yeah

look

1 half wrong

[warm basin]

24 times 0.9

[arctic vortex]

2 wrong

[warm basin]

xD

[arctic vortex]

lol

[silent plank]

,calc 34.04/38

[somber coyote]

Result:

0.89578947368421

[silent plank]

who the fk gives 0.04 points

[warm basin]

^

[arctic vortex]

idk

i have weird teachers

[warm basin]

probably non engineers

[arctic vortex]

and srry if i was annoying

this is my first time doing something like geometry

im more of an algebra person

[soft gulch]

Someone please send a hard geometry problem

[graceful narwhal]

🤔

[upper karma]

Someone please send a hard geometry problem
I agree

[upper karma]

Using a ruler, what's the circumference of this circle?

[umbral snow]

7

[warm basin]

umm is it c)?

[upper karma]

Measure the diameter then multiply by pi

[warm basin]

i mean

u can come up with a formula

including the monitor size, resolution

lmao

[upper karma]

Also depends on the browser zoom and whether you clicked on the image or not.

[arctic vortex]

can anyone help me with this test correction

[eternal crag]

CBE + ABE = CBA

[upper karma]

x + CBE = 180

ABE = 90

[arctic vortex]

yeah

what

so x = 90?

wait no

[upper karma]

x=139

[arctic vortex]

wait how

[upper karma]

What

[arctic vortex]

also is this d?

c**

[upper karma]

a

wait

[arctic vortex]

oh

is it because of the consecutive int angles

[upper karma]

I have no idea how to say that in english

[arctic vortex]

oh

[upper karma]

Alternates angles?

[arctic vortex]

ok thats fine

[upper karma]

Something like that

[arctic vortex]

yeah

opposite of alternate

[upper karma]

Yes

[arctic vortex]

https://cdn.discordapp.com/attachments/326138757474680852/729303302948519946/Annotation_2020-07-05_075039.png also how did u get 139?

here

[upper karma]

The problem gave you 149

139

[arctic vortex]

yeah

[upper karma]

I called BCF x

[arctic vortex]

o

wait

angle b = 90

right

[upper karma]

What's b

that is not an angle

[arctic vortex]

i still understand this

dnt*

dont*

[upper karma]

What don't you understand

[arctic vortex]

the question

[upper karma]

Given that <BCF = 139, find <CBA

[arctic vortex]

how do i do that

also @upper karma is this C?

[upper karma]

No, it's a

how do i do that
By noticing that x + CBE =180, CBE + ABE = CBA, with ABE=90

[steep temple]

Why is a line's gradient defined as rise/run and not run/rise?

Given point $A = (0, -3)$ and Point $B = (5,7)$, and Point P such that $AP:PB = 2:3$, why does P not lie on the coordinates $P = (\frac{2(5+0)}{5}), (\frac{2(7-3)}{5})$?

[somber coyote]

dlp ω:

[steep temple]

And lies on $P = (\frac{2(5-0)}{5}, \frac{2(7--3)}{5}-3)$ instead?

[somber coyote]

dlp ω:

[steep temple]

I understand the -3 in the y point comes from $\bar{AB} \equiv y = 2x-3$, but my question is why is subtraction performed on the 2 x, y coordinates as opposed to addition?

[somber coyote]

dlp ω:

[steep temple]

Why is addition performed when finding the midpoint of a segment, but not in this case?

In fact, upon trying a variety of divisors, I have found 2 is the only non-1 divisor which allows a point to lie on the original line.

Actually, why is the y-intercept factored into the former equation, but not the latter? It seems like something intuitive but the inconsistencies may be putting me off.

damnit where is @earnest pagoda when you need him

[steep temple]

How do I prove that a curve with the equation $x^2 + y^2 = n, {x, y>0}$ does not pass through the origin of the plane?

[dark sparrow]

plugging in (x,y) = (0,0) results in n = 0, which obviously goes against your assumption that n > 0.

@steep temple

[earnest pagoda]

look for section formula for what you asked previously

[steep temple]

corrected to what I meant

[somber coyote]

dlp ω:

[dark sparrow]

the origin is not even in the region {x,y > 0}

[steep temple]

haha amazing proof!

thank you ann.

[dark sparrow]

i'm just pointing out the obvious

[steep temple]

(that was not sarcasm)

look for section formula for what you asked previously
@earnest pagoda my question was more like why does the section formula differ from the midpoint formula?

[earnest pagoda]

midpoint formula is section formula with ratio 1:1

[steep temple]

got it.

[arctic vortex]

how do i get the answer

@dark sparrow

[dark sparrow]

well given that they basically tell you $m \angle CBE = 41\dg$, and that $ABDE$ is a rectangle, it shouldn't be too hard to calculate

[somber coyote]

Ann:

[dark sparrow]

unless you're allergic to addition or the number ninety

[arctic vortex]

oh

@dark sparrow wait so a=90 and b=90 right

[dark sparrow]

what are lowercase a and lowercase b?

neither option a nor option b lists 90 as its value

[arctic vortex]

the angle a

and the angle b

because there is a right angle and same side interior angles are a thing so does that mean e is 90

and since its a rectangle everything has to equal 90

so A=90 and B=90?

wait no that doesnt make sense

@dark sparrow im confused

[dark sparrow]

don't refer to angles by single letters unless it is unambiguous which angle is referred to

which in the case of angle B it is not

here's the diagram stripped of everything except the four relevant points

[arctic vortex]

131

[dark sparrow]

WELL THERE YOU GO DON'T YA?

was that rly so hard

[arctic vortex]

ok im stupid

i didnt see that

i know the answer to this but idk how to prove the answer

is correct

@dark sparrow

[dark sparrow]

what?

[arctic vortex]

i know the answer to this but idk how to prove the answer but idk how to prove this is correct

is it A?

i think its A

because it is supplementary

w and 118

and 118 + w must equal 180

[dark sparrow]

you answered your own question

[arctic vortex]

wait

really

oh

then im stupod sorry for pinging toy

you

and thanks

[tidal river]

can someone help me with this question

[dark sparrow]

whats giving you trouble

@tidal river

also don't post across multiple channels

[tidal river]

my bad I went to a channel to find the answer but realised there is a speific one for geometry

[charred zephyr]

use trigonometry

[tidal river]

the question is supposed to be geometry based

[dark sparrow]

no sucette

@tidal river thank you for completely ignoring my actual question.

let me repeat myself.

what is giving you trouble for this problem?

[charred zephyr]

you know people dont pay attention to thing sometimes and its better to react in a nice way if its the first time

[tidal river]

i don't know how you make a list for descending or ascending order

[dark sparrow]

do you know what it means for a collection of things to be listed in descending order?

do you know what "descending order" means?

[tidal river]

from greatest to least

[dark sparrow]

ok great

so the problem asks you

"if the angles are ordered from greatest to least by measure, then will the result be A, C, B?"

[tidal river]

so is it true of false

i'm not sure

[dark sparrow]

thats what the problem is asking you

ok let me dumb it even more down

[tidal river]

can you give me the answer and explain how you got to it

[dark sparrow]

"is it true that A is the largest angle, C is the second-largest, and B is the smallest"?

[tidal river]

i need to learn

[dark sparrow]

here

i dumbed it down for you as far as i possibly could

i'm not giving you the answer to a true/false question

[storm gale]

Christ man, getting an answer handed over to you is not a good way to learn, you gotta struggle.

[dark sparrow]

@tidal river if you want to give up then please let me know so that i know not to expect a response from you

[tidal river]

I trying to learn this out of my own interest. Not to just get an answer handed over to me. If tell me the answer and explain how you got there then maybe I'll understand

[dark sparrow]

the question asks you
"is it true that A is the largest angle, C is the second-largest, and B is the smallest"?

[charred zephyr]

but theyre trying to lead you to the answer step by step

[dark sparrow]

there arent even that many steps
im just trying my best to reword the question in a way that is more understandable to you

[storm gale]

@tidal river take a break for a couple of mins and then reread a "triangles" section of your favorite basic geometry book. This is actually a VERY simple question once you remember the basics, I'm sure it will click for you when you refresh that part.

[sinful anvil]

I have an algebra book lol, called algebra demystified

And plus, isn’t Sine law or cosine law acceptable to find the angles of A, B and C

[steep temple]

@earnest pagoda i have read the brilliant article on the section formula and i still do not see where subtraction comes into play

this is the closest thing that comes to subtraction and even then i think it is irrelevant to my case

going by their logic, https://cdn.discordapp.com/attachments/326138757474680852/729353592754012210/180738241824030720.png should not hold up

the section formula holds up (as expected), so this leads me to believe the formula the book provides is either a fluke (which I doubt), or some weird formula based off it

[earnest pagoda]

@steep temple
They are rewriting the formula

the thing you posted is relevant to your problem

[steep temple]

i see it now

[zinc pulsar]

brushing up on trig and wondering how he just changed inverse tangent into inverse sin?

[arctic vortex]

can anyone help me with some 30-60-90 and 45-45-90 triangle questions

[umbral snow]

@zinc pulsar
That's what the triangle in the top right is for.

Both tan^-1(3/5) and sin^-1(3/√34) represent the angle at the base of the triangle, and so must be equal

This is a common method that can turn any inverse trig into any other inverse trig

[livid sequoia]

is angle 3 and 1 the same?

[sly belfry]

yes

since those two lines are parallel

you can take angle 1 to be equal to the complementary angle of 2

which is the opposite angle of 3 (thus, equal to 3)

[livid sequoia]

thank you

[livid sequoia]

For part A do I add 9x+6x so it can become (15x+30)?

[dark sparrow]

for part A, first and foremost you write down the equation that says "these two angles add up to 180°"

only after that should you do any algebra to it

[livid sequoia]

So ill put 6x+(9x+30)=180 for A?

[livid sequoia]

How do I find the measures for part B?

[dark sparrow]

well a good first step would be finding the value of x wouldn't it

[upper karma]

[upper karma]

@arctic vortex Yea I can

[arctic vortex]

o nbm

nvm

im good

thanks anyway!

[upper karma]

Was gonna refer you to myself

but seems as thought you’ve got a handle on things

[finite vigil]

Hello

Is anyone available to help me with some geometry?

pls @ me if so

[cold tangle]

Just ask

[upper karma]

Just send question.

[sleek thistle]

can i get help with this?

how do i find ABE?

wait is it 34?

[upper karma]

,rotate

[sleek thistle]

oh yeah sorry about that

[upper karma]

Lemme give you a hint.

Wait.

[sleek thistle]

okay

[upper karma]

Shit idk how to give a hint without giving the answer.

Okay.

AEB is similar to ECD.

[sleek thistle]

🙂

[dense leaf]

guys, how is tan 1 > tan 2
And what would be the apporach to compare in such questions

[silent plank]

know your unit circle

0 < 1 < pi/2 < 2 < pi

[dense leaf]

I still don't understand why tan 1 > tan 2, sorry if this is a dumb question to ask

[silent plank]

you are working in radians
consider the inequality above

2 (radians) is in quadrant 2 (where tan is negative)
1 (radians) is in quadrant 1 (where tan is positive)

[dense leaf]

Oh! Thanks..that made it clear

[silent plank]

which is enough to conclude that tan(1) > tan(2)

[dense leaf]

Thanks!

[royal edge]

Can someone draw this out for me or explain it? I don’t understand it

[arctic vortex]

my sine cosine and tan thing on the calculator wont work someone help

[dark sparrow]

try putting in acos(10/12).

[arctic vortex]

33.6

wow u a genius

u are a genius*

thanks

u solved all my problems

[royal edge]

I got it nvm

[arctic vortex]

@dark sparrow why wont this work on my calculator

4=x(tan30)

tan50*

[dark sparrow]

4 = x * tan(50°), so x = ?

[arctic vortex]

yeah im looking for x

a = 50 and b = 4

[dark sparrow]

are you able to solve this linear equation for x

or do i need to divide both sides by tan(50°) for you

[arctic vortex]

wait what

im confused

this is what i plugged it in as

[dark sparrow]

ok first off what the fuck are you even doing raising the tangent function to the fiftieth power

[arctic vortex]

what

but angle a = 50

[dark sparrow]

i mean exactly what i said. why is tan not only being raised to the fiftieth power but also given no input?

[arctic vortex]

so tan^50=

how do i put the equation then

[dark sparrow]

and second, $4 = x \tan(50\dg)$ is an \textbf{equation} in $x$. why are you refusing to \textbf{solve} this equation for $x$ and only \textbf{then} plug the result into the calculator?

[somber coyote]

Ann:

[arctic vortex]

wait what

idk what u mean

[dark sparrow]

i mean exactly what i said.

$4 = x \tan(50\dg)$ is an \textbf{equation} in $x$. why are you refusing to \textbf{solve} this equation for $x$ and only \textbf{then} plug the result into the calculator?

[somber coyote]

Ann:

[arctic vortex]

how do i solve that equation

[dark sparrow]

are you able to solve this linear equation for x
or do i need to divide both sides by tan(50°) for you

[arctic vortex]

what

so i do 4/tan(50) = x

[dark sparrow]

congratulations, you've done what i (indirectly) asked you to.

[arctic vortex]

wait really

its undefinde

fined

[dark sparrow]

are you sure you're familiar with this calculator's syntax enough to be certain you're not entering the calculation incorrectly?

[arctic vortex]

idk how to use this

idk how to do sin cos tan

how do i plug it in

[dark sparrow]

x = 4/tan(50).

enter 4/tan(50) into the calculator.

[arctic vortex]

undefined

oh nvm

3.4

its 3.4

3.35

@dark sparrow

[dark sparrow]

,w 4/tan(50°)

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ann#0413

[dark sparrow]

you rounded incorrectly.

you were told to round to the nearest tenth, not the nearest hundredth.

[arctic vortex]

so 3.4

or 3.3

o 3.3

[dark sparrow]

well which is it

[arctic vortex]

3.3

[dark sparrow]

which way do you round on a five?

[arctic vortex]

oh right right

3.4

so the answer is 3.4?

[dark sparrow]

yes

[arctic vortex]

yay thanks

u are a genius

yesss i got another one right

@dark sparrow how do i do sin(53)=x/5

this problem

[dark sparrow]

again

this is an equation in x

solve the equation for x

[arctic vortex]

oh

i solved it

i got only 1 wrong

out of 10 questions

yess my homework is complete

actually now ait

i have 1 more packet

and then i am done yay

[arctic vortex]

@dark sparrow how to do this?

[dark sparrow]

make a diagram with only the relevant geometry and give some names to things

[arctic vortex]

@dark sparrow how do i put tan^-1(203/1058) on calculator

[dark sparrow]

your calculator calls inverse trig functions asin, acos and atan.

is this an "i don't know how to use the calculator" again?

[arctic vortex]

ohhhhh

so its asin

ok thanks

i meant atan

@dark sparrow idk how to use the calculator again

tan45=253/x

[dark sparrow]

for that particular equation, you don't even need one

unless you don't know what tan(45°) is

which you absofuckinglutely should

[arctic vortex]

ok

so the answer is 253

idk how to do this

[dark sparrow]

it sounds like you're psyching yourself up too much honestly

[arctic vortex]

oh

wait so how do i do this

@dark sparrow tell me pls

[dark sparrow]

do what

[arctic vortex]

tan45=253/x

what is x

[upper karma]

$a=\frac{b}{c}$ if it were to isolate c, would you be able to do it?

[somber coyote]

Al𝟛dium:

[arctic vortex]

oh yeah multplay a * c

and remove the c

x(tan45)=253?

[upper karma]

Doesn't it look extremely familiar to tan(45)=(253)/x

Yeah

[arctic vortex]

but it says undefinted whe nput nto calculate

[dark sparrow]

for that particular equation, you don't even need one
unless you don't know what tan(45°) is
which you absofuckinglutely should

[arctic vortex]

or

oh is it 253

[dark sparrow]

idk why you're turning a blind eye to tan(45°) = 1.

[upper karma]

Ill just dip

[arctic vortex]

oh it is 253

@dark sparrow how do i do tan57=h/1826

[dark sparrow]

exactly the same way as all the other equations you've been throwing at me.

maybe just for a moment you should read AND UNDERSTAND what i'm trying to tell you.

[arctic vortex]

but i dont understnad

[dark sparrow]

then go back and review linear equations

[arctic vortex]

ok

nvm i get this one

but i dont understand this height of elevation and stuf

[upper karma]

if you posted a problem of it, we could help, we aren't in your mind

[weak shoal]

This question isn't too hard tbh.

Well, can you see that 2 is the radius of the semicircle?

So, what length does XZ have?

Yes, so it's an equilateral triangle

So what does that say about the angles?

Because the sum of all angles in the triangle is pi. So, the three angles are equal. 3x = pi, where x is just one of those angles

Hence, x = pi/3

No, the sum of the angles in a triangle is 180

or pi radians