geometry-and-trigonometry

[quick skiff]

this is your solution

[empty yew]

(8th question)How to find the value of k? I got

$k=\sqrt{1-\tan^{2}\theta}$

[somber coyote]

KingDanger

[empty yew]

Anyone help me

<@&286206848099549185>

Anyone help me

[empty yew]

I proved it but Idk how to find the value of k with this

[empty yew]

Sorry It's plus or minus square root

[formal geyser]

Thanks a lot

Now i understand it

[hard moon]

Hint: instead of solving for k, solve for k^2 in “tan^2 (θ) = 1 - k^2”, and plug that into “(2 - k^2)^3/2”.

Can you reduce both the left and right hand side of “secθ + tan^3 (θ)cscθ = (2 - k^2)^(3/2)” to something they are both equal to?

As for the portion involving “find values of k”, I’m not sure yet

Oh wait

Lmfao I should’ve read you already proved it

[median wyvern]

anyone can solve given the gollowing graphs, find the equations of each graph in standard form? i will send the 2 graphs in dms

[hard moon]

Sorry

[empty yew]

Np and sorry

[median wyvern]

anyone good at vertex form to standard form? in graphs

?

[empty yew]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[median wyvern]

oh okay

give me a second i will send the equation

[empty yew]

If you found how to find the value of k then ping me

[median wyvern]

@empty yew

i got it all wrong and i got a chance to redo it so if u guys could help would be nice

[empty yew]

I can't see a single thing could you please send it clearer?

[median wyvern]

i dont have my phone on me i will send it next period when i can use my friends phone

[empty yew]

It's night in my country so please ping helpers for further help @median wyvern

[quick skiff]

if you dont mind

im trying to help someone and i got x = 58

[median wyvern]

ok thansk

[empty yew]

One min @quick skiff I will see your question

[quick skiff]

oh wait, i see where i might have gone wrong

wait no, its fine

im confused why its not on the list of available answers

i get this

same thing for the other angles

(its not x, but getting x is simple because theres an isoceles triangle)

[hard moon]

It seems that the only values of k that are allowed are -1 ≤ k ≤ 1, since tan^2 (θ) = 1 - k^2. Since tan^2 (θ) ≥ 0, we must have 1 - k^2 ≥ 0. Solving this you obtain -1 ≤ k ≤ 1.

I don’t think that there are any restrictions in this range that dont work for the identity we proved

[empty yew]

Yes you're correct the answer key saying so

Thank you so much @hard moon

[hard moon]

yw

[median wyvern]

@empty yew

better now?

[quick skiff]

just that question?

[median wyvern]

ye

its saying "given the following graphs, find the equation of each graph in standard form."

[quick skiff]

hello

$a$

[median wyvern]

bro im not gonna lie i have no idea what your saying

is it possible for you to do it for me?

[quick skiff]

ok i really dont know how to use it but

[median wyvern]

so i can better understand it that way

[quick skiff]

to find the formula

its easiest if you have the vertex of the parabola

because i assume you're doing quadratic formulas

[median wyvern]

yea

quadratic formulas

leon would it be possible to do one of the graphs and send me how you did it so i can better understand it

[dark sparrow]

@quick skiff you should not put normal text inside dollars

dollars are supposed to delimit math formulas, and you can do them in the middle of a message (and in fact SHOULD do so), like this: $y = x^2$

[somber coyote]

|Ann⟩

[quick skiff]

thank you!!

where (x, y) is that

[dark sparrow]

yeah but don't forget to put dollars at the beginning and end

[quick skiff]

i give up

yes i forgot

[median wyvern]

brother i got no idea what your saying

i would understand if you could just do it for me lol

[dark sparrow]

the vertex of a parabola $y=ax^2 + bx + c$ is the point $\left( -\frac{b}{2a}, \frac{4ac - b^2}{2a} \right)$

[somber coyote]

|Ann⟩

[dark sparrow]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[quick skiff]

4a btw not 2a

[median wyvern]

yes i understand

[quick skiff]

:D

[median wyvern]

but it helps me better

[dark sparrow]

ok right typo on my end

2 notes on that tho

1. this formula goes the other way from what M wants to do. it's useful when you have a standard-form eq and you want the vertex.

[quick skiff]

true

but i always find it easier that if you have the vertex point

[dark sparrow]

2. the x-coordinate of the vertex matters much more than the y. once you have the x coord, the y coord is just a matter of plugging the x into the parabola's equation

[quick skiff]

use the vertex form

[dark sparrow]

i mean sure

you can write down the equation of the parabola in vertex form and then expand it, converting to standard

[quick skiff]

thats how i learned

[dark sparrow]

thats how you're meant to do it

and for good reason bc anything more high-caliber is not appropriate here

[quick skiff]

yeah i know

thats how i did it at a competition

wasnt sure if it was /4a or /4ac

so i just spent time doing f(-b/2a)

because it was an equation w/ parameters

it just helps me save time imo

[quick skiff]

sure, but practice makes perfect

i mean i practiced for an exam today and i still had the biggest brainfarts of my life, math is like that

which is why i would advise you to try it yourself

the vertex form is $f(x) = (x + x_0) = y_0$

[somber coyote]

leon

[median wyvern]

this is my test leon i got 30 mins to redo it and i got nun so far

im depending on u guys to help me finish this

cus so far i dont understand shit to be honest

[dark sparrow]

uh what

<@&268886789983436800> timed test cheat attempt

[median wyvern]

he gave us a period to redo it

[dark sparrow]

mangled beyond recognition.

[median wyvern]

like make corrections

he let us take it home n stuff to correct it

[obtuse field]

are you supposed to be doing it by yourself

[median wyvern]

nah hes fine with me asking my friends and stuff in the class

but no one has a clue of how to do it

[quick skiff]

which is: $f(x) = (x + -\frac{b}{2a})^2 + \frac{4ac-b^2}{4a}$

[somber coyote]

leon

[quick skiff]

typo

and another typo

because its + b-2a

[median wyvern]

anyways goodbye i dont find any help in this thanks for your time

[dark sparrow]

leon, i think you are making it all like 40 times worse

[quick skiff]

LMAO

ur right

i just realised wtf im writing

they need an image to text because my god im new to latex and its confusing me

[maiden brook]

How do u do this: Find the second smallest value of a such that a, b, and c are all integers and a triangle with these three side lengths has area (a * root(a))

[dawn sonnet]

pls help

[hard moon]

the choices seem strange: “39 < n < 10” and “10 > n > 39” are equivalent, but no number is less than 10 but greater than 39”

although, it looks like something is blurred in front of the rightmost part of each of the choices

[plain jetty]

Hey I was wondering, Is it easier to factor using x method or quadratic formula? Just preference

[grave pond]

What is "x method"?

[trail tendon]

there actually is

something called the x method

[hoary totem]

it seems like just another middle school trick to make people learn factoring easier

[trail tendon]

sometimes its better to try factoring, with or without using x method, and then if u start having trouble use the quadratic formula

[trail tendon]

yea, i never heard of it until like a month ago 🗿

[grave pond]

As already mentioned, the only option that makes internal sense is 10 > n > 39. However that doesn't match the diagram either.
An upper bound of 39 looks right (it corresponds to angle L approaching 180°), a lower bound of 10 doesn't. n=10 would make angle L be 35° and by the law of cosines it is clear that angle L has to be larger than angle S.

[grave pond]

All my attempts to google what it is just leads to endless lists of videos. Apparently nobody has deigned to write down what it means?

[trail tendon]

it's just factoring but where they draw an x, like this

the numbers on the sides should be originally empty

and you have to figure them out based on the top and bottom

the two sides multiply to the top, add to the bottom

[grave pond]

How do the 4 and 3 arise there?

[trail tendon]

they put them there

with factoring

[grave pond]

But what's the method?

[trail tendon]

the x method isn't really much different from factoring

they just put the x

[trail tendon]

drawing the x 😂

[grave pond]

How does that produce the numbers 4 and 3 in this case?

[trail tendon]

they have to figure out the number on the sides

by simply knowing that the numbers on the sides multiply to 12 and add up to 7

[grave pond]

Yes, but what's the method for that?

[hoary totem]

none

u just guess

lol

its the same as normal factoring

u just arrange it to be an X

[trail tendon]

^

its not really a method at all lol

[grave pond]

What is "normal factoring"?

[hoary totem]

where u dont draw an X

and u guess

lol

nobody has the time to solve
a + b = 7
ab = 12

[trail tendon]

(x+ blank)(x+blank), for binomials

[grave pond]

So "know what the answer is already, and check that it is correct"?

[hoary totem]

just guess 3 an 4

[trail tendon]

basically 😂

[hoary totem]

well if u want to describe trial and error that way then yes

[trail tendon]

thats why i always use quadratic formula unless factoring is obvious

[hoary totem]

^

[trail tendon]

quadratic formula is actually methodical

[grave pond]

I wouldn't call that a "method".

[trail tendon]

exactly

i mean maybe guessing could be referred to as a method... maybe

[clear anvil]

[grave pond]

Well, I suppose if bogosort is an algorithm, then the limits of the acceptable are wide.

[trail tendon]

🗿

i had to search up what bogosort was ☠️

very appropriately named

i don't know what a bogo is but its probably the dude who came up with it

😂

[clear anvil]

or just bogo cuz it’s bogus

[grave pond]

It's derived from "bogus".

[clear anvil]

[trail tendon]

😂

[clear anvil]

damn i was right on the money lol

[trail tendon]

😂

[hoary totem]

a method need not be an algorithm tho

[grave pond]

Right, graphical methods are methods too, but are not algorithms.

But "write the problem in this form, then write the answer here" stretches the concept rather a lot.

[hoary totem]

true

[tranquil wing]

Can someone help me with sequence s

[trail tendon]

i didn't mean every chat 😂

[jagged wyvern]

Can someone check if this is right?
Using area of a sector of a circle i think we just switch the proportionalities around
I did this: n/255 = 115/360
360n = 29325
n = 81.46

[jagged wyvern]

Ignore the n = 2507.81

[deft sail]

The denominator of 115/225 should be 360 instead of 225. I’m not exactly sure

So I think it’s the fourth option

[median garnet]

The angles measures should be on the same side of the proportion i think

[jagged wyvern]

@trail tendon So 81.46 should be correct? since the proportions are switched around?

[jagged wyvern]

'

[trail tendon]

yeh

[twin spindle]

Howcome the csc² and sin² cancel?

[trail tendon]

they...don't ._.

[twin spindle]

Nvm

I shoulda watched the whole thing

He did that on accident 😭

Bruv I was so confused

[quaint sequoia]

Can someone help me ?

[jagged wyvern]

Think of the lines LK and JM lines on a plane

Do you know the interior angles theorem?

[quaint sequoia]

Yes thank you I already answered the question is it ok if you help me with this other question?

[jagged wyvern]

I'll try

[quaint sequoia]

I tried setting them as supplementary

But it didn’t work

Idk wht to do

[jagged wyvern]

Ok

Well you might not see it but you can assume that the triangle inscribed the circle is a right triangle

So with that, you can assume angle H is a 90 degree angle

And we know that all angles in a triangle all add up to 180

You can set up an equation

[jagged wyvern]

Adding it all up to 180?

[quaint sequoia]

Yeah I set 5x - 2 + 2x + 8= 180 because there where on the same line so I assumed they added up to 180

[jagged wyvern]

Can I ask what you got for x?

[quaint sequoia]

Um I set it up like that and I got 12 it’s says it’s wrong

[jagged wyvern]

Ah

[quaint sequoia]

I got a super long decimal so i knew it was wrong

[quaint sequoia]

Oh wait sorry I forgot to plug in lol

[jagged wyvern]

I mean

Its gonna be a decimal either way trying that

Im assuming they want a whole integer

Let me know if you got it

[median garnet]

Isnt it always 90 degree?

[jagged wyvern]

If the diagram is up to scale yes

[median garnet]

Yep

[median garnet]

If its in the same line it doesnt mean the total is 180 degrees

But h is always going to be 90

And the degree of a triangle adds up to 180

Try making an equation with that

[paper token]

i need help from someone

badly

this is due soon

[supple gulch]

Help

[hallow dust]

Hello

[supple gulch]

Hi

[hallow dust]

Sakes

[supple gulch]

Show me photo

[hallow dust]

I need help with this

It does not ask me to solve

It asked for only the answer

That’s why I asked for only the answer then the man got mad and kept stalking me

Thanks btw

[supple gulch]

Do you at least know what alternative exterior/interior angles

Corresponding angles

[hallow dust]

Yes

They tought us that

I know interior angles is always 180 right

[supple gulch]

Yes Consective interior angles

Not alternative interior angles because they’re the same

And are congruent

Do you know how to do the first one or need help

[hallow dust]

Uh

For the first one I subtract 180 by 91 right

[supple gulch]

There you go

[hallow dust]

We can just go question by question

[supple gulch]

Yep

[hallow dust]

For #1 X=89

[supple gulch]

Yes

Now 2 do you know how to solve

[hallow dust]

Time for number 2

Number 2 is exterior right

[supple gulch]

Yes but what specifically

Consecutive or alternative

[hallow dust]

Alternative

[supple gulch]

Good

[hallow dust]

Do we do 98-45

[supple gulch]

So that’s means it’s the same

[hallow dust]

Or 180

Oh 180

[supple gulch]

No because remember it’s congruent

[hallow dust]

Ok

So 180 -98

[supple gulch]

No that’s only for consecutive interior angles and exterior angles

They are the same so it’s just 98

[hallow dust]

Oh

So number 3 is interior

[supple gulch]

No

Remember interior is inside

Exterior is outside

[hallow dust]

I looked at wrong problem my bad

It’s just same

So 118

[supple gulch]

No

It’s the same as number 1

What do you subtract

[hallow dust]

Breh

I said that already

I told u 180-118

Answer is 62 right

[supple gulch]

Yea

[supple gulch]

Mb

[hallow dust]

Why u say I was wrong at first lol

[supple gulch]

No I didn’t say u we’re wrong at the first question

[hallow dust]

Oh

Number 4 just like 2

So it’s 53

[supple gulch]

Very good

Wait brb

[hallow dust]

Number 5 equals 58

180-122=58

[hallow dust]

Ok

[supple gulch]

Srry im back

[hallow dust]

Ok

Number 5 is 58 right

[supple gulch]

Number 5 yes ur correct

[hallow dust]

6 is same as 2

So it’s 88

Right?

[supple gulch]

Yes and what type of angles are those

[hallow dust]

It’s exterior

[supple gulch]

Good

[hallow dust]

Interior would be inside

[supple gulch]

Yes

[hallow dust]

I get it now

Ima speed run and give u answers rq

[supple gulch]

Yea just do 7 and 8

[hallow dust]

7 =62

8=107

[supple gulch]

W

Let me give you this guide

[hallow dust]

Ok

[hallow dust]

Thanks so much

Ima take picture of my homework to make as notes to

Add me back

[supple gulch]

Do you want one challenge question before u go

It’s requires a bit algebra

[hallow dust]

Yes

Let me try

U think u can be my homework helper from now on until I graduate

Which is in 3 weeks

Where like u help me better understand to I’m able to do on my own

[supple gulch]

What math do u take

Geometry right

[hallow dust]

Weve been doing stuff like this lately

[hallow dust]

I think we’re just doing this to learn high school stuff early

[supple gulch]

Im ngl I finished that like a year ago

I have no idea what type of class ur in

[supple gulch]

Ok here it is

First tell me what angles these are

[hallow dust]

Ok

Hmm

Exterior

Am I wrong

[supple gulch]

Yea

[hallow dust]

NOOOOOOOOOOOOOOOOOOOO

NOOOOOOOOOOOOOOOOO

[supple gulch]

Bro dw like I said it’s inside

[hallow dust]

Wait so it’s not exterior right

It’s interior then

[supple gulch]

Yes cus it would be outside

Yes it’s interior

[hallow dust]

Ohhh

[supple gulch]

Nice

[hallow dust]

My notes tricked me😂😂

[supple gulch]

They are alternative interior angles

[hallow dust]

That’s why I shouldn’t have cheated

Kk

Ima not that down

[supple gulch]

So what does that mean

[hallow dust]

*note

That means 180 is apart of it

[supple gulch]

Not quite

That’s the consecutive interior and exterior

[hallow dust]

Ohh

[supple gulch]

If they are alternative interior angles, then they are what

[hallow dust]

If it’s alternative interior angles

Would it be the same

[supple gulch]

Yes

[hallow dust]

Or am I wrong again

[supple gulch]

Very good

[hallow dust]

Ok

[supple gulch]

Now the tricky part is just the algebra

[hallow dust]

So only alternative angles are the same

[supple gulch]

Yes and corresponding angles

[hallow dust]

Consecutive are 180

Right

[supple gulch]

Yes

[hallow dust]

Ima put these in my notes holon

[supple gulch]

Btw same side means consecutive as well

180

[hallow dust]

K

Thanks so much

Thanks for all the notes I have took them down!

I’ll just text u tomorrow then

[supple gulch]

Kk

We can solve next time

Cus it’s midnight for me rn

[hallow dust]

Kk

Cya have a nice night

[hallow dust]

11:13 for me lmaoo

[north kindle]

I’m having a bit of trouble understanding triangle congruences

Mostly on which angles and sides exactly must be the same for stuff like SAS, ASA and AAS

[lime dune]

the configurations need to line up

for example if i had triangles ABC and DEF with AB=DE and BC=EF but angle B = angle F

[dark sparrow]

SAS = two sides with an angle between them

[lime dune]

then we can't conclude anything

[dark sparrow]

ASA = two angles with a side between them

SSS = three sides

[lime dune]

^ yea that's probably the simpler way to put it

[dark sparrow]

it might make sense to try to do it visually

to put hatch marks on a diagram to represent which sides are equal

then you should get two configurations in your two triangles which "look the same" by the markings

[north kindle]

Like are these two triangles congruent?

I can’t tell

[dark sparrow]

yes, DEF = HGI

note you have to "flip" one of the triangles to make the equal-element configurations "look the same"

[north kindle]

How are the first and fifth here congruent

wait nvm I see

[dreamy obsidian]

hi

[obsidian harness]

hi, just ask your question if you have one

you will get help faster this way

[dreamy obsidian]

ok

[dreamy obsidian]

Did I do it right

@quaint basalt

[obsidian harness]

Yes this is correct

[dreamy obsidian]

are you positive

[obsidian harness]

Yep

x = 3 is not differentiable cause of the sharp point but it is continuous there

[dreamy obsidian]

Okay this one was a little hard for me but lets see

These both

I tried

I doubt I got it right tho

[obsidian harness]

Yep so I can see you factored 8 right

[dreamy obsidian]

yea I did

[obsidian harness]

Yep so x = 5 is a removable discontinuity

Yeah that's correct

[dreamy obsidian]

8 is right

[obsidian harness]

mhm

[dreamy obsidian]

what about 9

[obsidian harness]

Yeah k = -1 is possible

Basically any k < 0 would work

Cause by the intermediate value theorem, there would be at least one zero between x = -3 and 0, and there would be at least another zero between 0 and 3

[dreamy obsidian]

thats what I thought, but then I was like hmm

I guess it worked out for me tho

[obsidian harness]

Yeah

[dreamy obsidian]

alright thanks man I appreciate it

[obsidian harness]

No worries

[dreamy obsidian]

1 MORe

sorry

actually nvm I got it

[obsidian harness]

all good

[ruby wigeon]

In a quadrangular cone $P-ABCD$,the base $ABCD$ is square, $AB=4,PC=PD=3,\angle PCA=45^\circ$,the area of $\triangle PBC$ is

[somber coyote]

e_waste

[ruby wigeon]

rly dunno

[winter apex]

hi

new here

i need a very difficult question on tignometry

pls

?

[spiral lodge]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[spiral lodge]

Btw what do you mean with difficult?

It could be difficult for me and not for you, or viceversa

[fathom grail]

AAS

[shadow hedge]

Is this correct ?

[dark sparrow]

show your work @shadow hedge

this looks sus to me

[shadow hedge]

I used desmos but idk whether i did correct or not

o did arcsin 8/9

but it doesnt mention if its rad or not

I assume it is degree?

[shadow hedge]

it may look a test for you but its not. Its our schools online system and this is an optional work we should do. im doing this to imporve myself but i got some points where i dont understand it.

need to learn these before our tests

actual tests*

[dark sparrow]

no i mean your answer looked sus

[shadow hedge]

Oh ok

[dark sparrow]

1. it actually does not matter
2. you do not need x itself. not in degrees, not in radians, not in any other angle unit.

3. you do not need desmos

you should know double angle identities

do you?

[shadow hedge]

tbh idk im very behind class

this is wrong right?

[dark sparrow]

yes your answer is wrong.

[dark sparrow]

ok then i have given you a keyword to look stuff up. look up "trigonometry double angle identities" on ochem tutor, khanacademy, etc.

[proven summit]

cos2x= 1-2sin^2 x

1-(8/9)^2

[dark sparrow]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dark sparrow]

(preemptive)

[old finch]

https://tenor.com/view/matematica-gif-7322246

[north kindle]

I have no idea where to start with this problem

[dark sparrow]

O is the center of the circle. what are OB and OC?

[north kindle]

Radii of the circle

[dark sparrow]

what can be deduced from this?

[north kindle]

That it’s an isosceles triangle

[dark sparrow]

that what is an isosceles triangle?

call it by its name

[north kindle]

The triangle OBC

[dark sparrow]

right

what can be deduced from that?

[north kindle]

That the angle OCB = OBC = OCD

[left ibex]

Y'all how do I do this

The awnswer gonna be smth like 3π but probably not 3

[dark sparrow]

indeed. you are two steps away.

[rugged shuttle]

central and inscribed angles

what do you know about them

[left ibex]

Not enough

[rugged shuttle]

and how can they be used to find answer

[left ibex]

We just got these ixls and in trying to get them done before we even learn the unit

[rugged shuttle]

lel

ok so a central angle

is an angle stemming from center and touching 2 points on the circle

[left ibex]

This is also the last IXL I gotta do

[rugged shuttle]

the measure of central angle = measure of arc

an inscribed angle goes across the entire circle and the arc it corresponds to is 2 times the actual angle measure

[north kindle]

I know that BOC must be 180-2*OBC, so I feel like I should try to find OBC

[left ibex]

Well @rugged shuttle how would I go about solving this

[rugged shuttle]

ill give u hint, find arc length of PQ and then try to use central or inscribed angle properties

[left ibex]

Arc for PQ is 126

[rugged shuttle]

yes

and angle PRQ seems to be an inscribed angle

[north kindle]

But I’m not sure how I would go about that

[left ibex]

So it would be 252?

Wait no

My bad

I misunderstood what you said

[north kindle]

Or I could try to prove the triangle is equilateral

[rugged shuttle]

so the measure of an inscribed angle is half of what its corresponding arc is @left ibex

sorry i worded it weird

[left ibex]

Oh

So 63?

[north kindle]

I feel like I’m missing something trivial

[rugged shuttle]

you got it

[left ibex]

So is that the final awnser?

[rugged shuttle]

yes

[left ibex]

That was right

But now there's a triangle

[rugged shuttle]

angle UWV is an inscribed angle

[left ibex]

So half of 82?

[rugged shuttle]

what arc does it correspond to?

[left ibex]

Wait

I'm so confused

[rugged shuttle]

look at UWV and see what arc the 2 ends of it make

gtg but the answers 49 try to figure out why

[left ibex]

Will do

[north kindle]

Is it 48?

Wait no I don’t think so

It really seems like the path is proving that the triangle is equilateral

And I think I’m missing a fact that makes that trivial

[sharp gyro]

its correct guys?

...

[hard moon]

disregard the comment where I said “that is very close” (i deleted it); I thought you were referring to the most recently posted problem in this channel and I did not read carefully

As for a hint, mark down the segments and angles which you know are congruent

Also, notice your goal is to figure out one of the angles of the triangle; since you know the sum of the angles in a triangle add to 180°, we know we can solve the problem if we can figure out what the marked angle is

[north kindle]

I know that

I’m not stupid

[hard moon]

ok, so then what have you done

[north kindle]

I know the triangle is isocoles, I know OCB = OBC

[hard moon]

hint: parallel lines, same side interior angles

that’s all I’m going to say

[sharp gyro]

can u answer my hard question (lol)

[grave pond]

Did you swap 6 and 10?

[sharp gyro]

Are the places wrong?

[grave pond]

What exactly is it you concluded has length 6?

[sharp gyro]

Pisagor rules

3,4,5

[grave pond]

No, which distance in the diagram is it you say is 6?

[sharp gyro]

side of right triangle?

[merry pollen]

im in geometry right now, learning about radians. if somebody could dm me and call me and take me through it id really appreciate it

[grave pond]

Sigh. Which side?

[sharp gyro]

idk 🥸

a long side

[grave pond]

Then the solution is not correct, no.

[sharp gyro]

how to find the right solution?

if u want show me

)))

[hard moon]

what did you find x to be in your original work

[sharp gyro]

yep but the answer key says the correct answer is 10...

[trail tendon]

the correct answer is 10

how do you find the side that i colored red

[obsidian hornet]

OB-bisector of angle AOC, it means that half of your picture is symmetrical, AO=OA”=15, CA”=sqrt(BA’^2+CA’^2)=sqrt(8^2+6^2)=sqrt(64+36)=sqrt(100)=10(PYTHAGORIAN THEOREM FOR TRIANGLE BA’C), BC=10.

[trail tendon]

CA?

i know the length you mean but that aint no CA

and also you did not explain how you got 10 ._.

[obsidian hornet]

CA””””””””

[trail tendon]

i know u did pythagorean theorem but idk how he supposed to nkow that from that

[trail tendon]

oh

ok ig that works

lol

[obsidian hornet]

He said that he know PT

[obsidian hornet]

.

[trail tendon]

but you gotta show the work bre-

XD

[obsidian hornet]

Now fixed

[trail tendon]

👍

XD

[grave pond]

We don't need to find the length of OB, though.

[obsidian hornet]

Of course

this is in order to confuse with unnecessary information

[fickle heart]

How do I solve for x?

(Looking for help, not answers. Otherwise I learn nothing)

[forest blade]

when working with half angles, how do you know where the half angles quadrant is? all the examples I can find involve angles in radian form and then they divide that radian, making it easy

what if all I know is something like cos -4/5

which doesn't have pi in it but is in quadrant 2

[trail tendon]

can you set up an equation using the pythagorean theorem?

[fickle heart]

I have no clue where I'd start

Rn my brains in

[trail tendon]

do you know the pythagorean theorem?

[fickle heart]

"O crap, I don't know what's happening, I usually know what's happening, PANIC MODE!!!"

[trail tendon]

💀

[trail tendon]

is the answer to this question no?

[fickle heart]

A^2+b^2=c^2

[trail tendon]

yes

[fickle heart]

Yeah

[trail tendon]

do you know what a is?

[fickle heart]

X

[trail tendon]

sure

do you know what b is?

[fickle heart]

X+10

[trail tendon]

not quite, x + 10 is the hypotenuse, right?

[fickle heart]

Oh

Well I feel awfully silly

[forest blade]

Nvm I figured it out!

[fickle heart]

So now I have

X^2+20^2=(x+20)^2

[trail tendon]

yeh

do you know how to solve that using algebra?

wa

[fickle heart]

If there's no exponents yah

[trail tendon]

its actually x^2 + 20^2 = (x+10)^2

[fickle heart]

I fatfingered the 1

[trail tendon]

try expanding (x+10)^2

[trail tendon]

o alr XD

[fickle heart]

x^2 + 20^2 = x^2 + 10^2

[trail tendon]

not quite

(x+10)^2 is (x+10)(x+10) right?

[fickle heart]

I guess

I'm getting bullied 😭

[trail tendon]

why ☠️

[fickle heart]

I kid I kid

[trail tendon]

🗿

[trail tendon]

(x+10)^2 does not equal x^2 + 10^2
(x+10)^2 = (x+10)(x+10), and then you have to factor it out (or use foil)

do you know foil?

[fickle heart]

Yeh

[fickle heart]

Well, know, yes. Good at it? No

[trail tendon]

try it

on
(x+10)(x+10)

[fickle heart]

x^2+20x+100?

[trail tendon]

yep!

so back to the original equation

x^2 + 20^2 = (x + 10)^2

what do you have now

[fickle heart]

x^2+20^2=x^2+20x+100

[trail tendon]

yeah

and whats 20^2

[fickle heart]

400

[trail tendon]

yeh

do you know how to continue?

[fickle heart]

300=20x

X=15

[trail tendon]

yep

[fickle heart]

Siccc

[trail tendon]

noic

e

[fickle heart]

I'm now beeg brain

[trail tendon]

yeez XD

[jagged wyvern]

How do i do this?
Am i suppose to set up the revolutions like this: C = x(pi)(r)
As x implies the number of revolutions because I got a decimal

[hoary totem]

(2x + 10) * 10 = 20^2 = 400
2x + 10 = 40
2x = 30
x = 15

[hoary totem]

u get a decimal

the question asks for complete revolutions

[jagged wyvern]

Yep

[hoary totem]

so if say ur answer was 6.4293 then it only made 6 complete revolutions

[jagged wyvern]

So I just round my answer?

[hoary totem]

not round

floor

u just take the part before the decimal point

cuz even if u had 5.99 rotations

its only 5 full rotations

[jagged wyvern]

Well

If the bicycle travled 40 meters, would C = in terms of pi?

Like 40pi

[trail tendon]

how (2x + 10) * 10 = 20^2 ?

[jagged wyvern]

This is what I did
40pi = x(pi)(32.5)
40pi = 32pi(x)/32pi
x = 40/32.5
x = 12

[jagged wyvern]

If this'

Uhh I got x = 12 incorrect

[trail tendon]

why 40pi?

also 40/32.5 is like 1.2 not 12

[jagged wyvern]

1 is also incorrect

[trail tendon]

yeah because why 40pi?

[jagged wyvern]

Uhh i dont know

[trail tendon]

xD

[jagged wyvern]

Is it suppose to be 40pi??

[trail tendon]

the bicycle traveled 40 meters

pi is only used to get the length of the wheel

so u can count the rotations

[jagged wyvern]

Ohh

Alright, so my equations should be set up like 40/32.5pi instead

So approx 3 revolutions?

[trail tendon]

i mean i would use 32.5, and also keep in mind this is the radius

do u know circumference of circle formula in terms of radius?

[jagged wyvern]

C = 2(pi)(r)

[trail tendon]

yeah

so whats the circumference

[jagged wyvern]

I just thought the "2" was how many revolutions so I changed it to x

40

[trail tendon]

i mean the length of the circumference is one full revolution right

[trail tendon]

no, wat?

[trail tendon]

C = 2(pi)(r)
r = 32.5 cm
what is C?

[hoary totem]

this thing called the secant tangent theorem
dont worry about it

your method works perfectly fine

[jagged wyvern]

Oh what lol i thought 40 was Circumference

[trail tendon]

o mk

its one of those weird geometry tricks prob XD

[trail tendon]

no, 40 is total distance

[jagged wyvern]

Why am i given 40??

[trail tendon]

so the wheel is moving 40 meters

the wheel's circumference is the distance it takes for the wheel to make one rotation

[jagged wyvern]

204.2

[trail tendon]

i would keep it in terms of pi for accuracy

now also keep in mind

[jagged wyvern]

65pi

Yes

[trail tendon]

yeh

the radius is in centimeters

the distance is in meters

[hoary totem]

BP * AP = TP^2

the secant tangent theorem

the solution to the problem posted doesnt even require a quadratic if u use this

[trail tendon]

😭

why is this true

i hate geometry mahn

[hoary totem]

actually ive never proved this

[trail tendon]

its just a thing

that exists

because someone proved it, once

💀

[jagged wyvern]

I dont understnad

Am I suppose to convert cm to m?

[trail tendon]

you can...

[jagged wyvern]

Or do I have to divide 65pi with 40

[trail tendon]

i would convert meters to cm

[trail tendon]

i mean you have to convert them into the same units definitely first

[jagged wyvern]

65pi/4000cm

So if i were to use this and find the "total" revolutions,

[tidal canyon]

can someone help me get the answer for these and the formula so ik how to do it?

[jagged wyvern]

It would be 0..?

[trail tendon]

why are you dividing the total distance by the distance of the bike?

i mean

why aren't you

lol

[jagged wyvern]

I dont know man 😮‍💨

Guide me through pls

[trail tendon]

i just wanna try to add a visual

it may not help

but in order to make a complete revolution, the wheel has to have gone the same distance as the length of the circumference of the wheel

[jagged wyvern]

Ye

[trail tendon]

so if the blue line is 4000 cm

[grave pond]

We probably need to assume that the author meant to say that the "random" hits are uniformly distributed within the outer square (to the left) or outer circle (to the right). That doesn't really make a lot of sense (especially for the archer), but otherwise the problems are hopelessly underspecified.

In that case, what the problem really asks for is just how large a fraction of the outer area the inner area is.

[trail tendon]

and the red distance is 65pi cm

how many red distances make up the blue distance

not makin sense?

[jagged wyvern]

4000/65pi

[tidal canyon]

dont really know thats all the info given

[trail tendon]

yeh

[jagged wyvern]

🤷‍♂️

[trail tendon]

so what is 4000/65pi

[jagged wyvern]

193

[trail tendon]

...no

[jagged wyvern]

WHat

[trail tendon]

calculator error?

[jagged wyvern]

Bruhh

[trail tendon]

oh make sure

[jagged wyvern]

This is rigged breh

[trail tendon]

pi is on bottom

[jagged wyvern]

Yes teh pi is on the bottom

[trail tendon]

if you write 4000/65pi technically to a calculator it looksl ike its on top

[jagged wyvern]

Is it suppose to be on degree or radian mode

[trail tendon]

you have to write

4000/(65pi)

[jagged wyvern]

Alr

[trail tendon]

shouldnt make different

[jagged wyvern]

Should be 19 right?

[trail tendon]

its 19.58, which means it made 19 full rotations yeh

[jagged wyvern]

😮‍💨

[jagged wyvern]

I humbly appreciate it ty

[hoary totem]

from the law of cosines:
OP^2 = p^2 =
= AP^2 + r^2 - 2r AP cos(a)
= CP^2 + r^2 - 2r CP cos(c)
thus
AP^2 - 2r AP cos(a) = CP^2 - 2r CP cos(c)
AP(AP - 2r cos(a)) = CP(CP - 2r cos(c))
and
2r cos(a) = 2 * AH = AB
2r cos(c) = 2 * CT = CD
thus
AP(AP - AB) = CP(CP - CD)
now
AP - AB = BP
CP - CD = DP
thus
AP BP = CP DP
this is the secant - secant theorem, move one of the secants to a tangent point and the secant - tangent theorem follows.

pretty cool proof ¯_(ツ)_/¯

[tepid burrow]

does anyone know what this projection is called?

I found it at http://www.quadibloc.com/maps/maz0206.htm and they call it the airy projection but i think there's another name for it because i can't find any details on it. i have a latitude and longitude and i am trying to map it out using a projection like this, thanks

[hoary totem]

The Airy projection is an azimuthal minimum error projection for the region within the small or great circle defined by an angular distance, , from the tangency point of the plane ( λ 0 , ϕ 0 ) .

This azimuthal projection is an analytic solution to the problem of minimizing shape and area error in a map projection: It was described by the British Astronomer Royal, George Biddell Airy, in an 1861 paper.

[grave pond]

BTW, it can be done without trigonometry:
By algebra we have CP·DP = (TP+CT)(TP-TD) = (TP+CT)(TP-CT) = TP²-CT².
Since CT² = r² - OT², this gives CP·DP = TP² - (r² - OT²) = (TP² + OT²) - r² = OP² - r².
A completely similar calculation gives AP·BP = OP² - r² too.

[hoary totem]

thats smarter

[tepid burrow]

it only accounts for latitude in the paper though, not longitude

[hoary totem]

http://www.quadibloc.com/maps/maz0206.htm

here are some formulae

https://mathworld.wolfram.com/AiryProjection.html

[tepid burrow]

omg thank you for this resource

[grave pond]

For an azimuthal projection, the only formula you need for specifying the projection is a function from (spherical distance between your point and a fixed center point) to (straight-line distance between the image point and the center of the map).
The direction from the center of the map to the image point should always be the compass bearing from the fixed center point on Earth towards the point of interest (along a great circle).

[tepid burrow]

would direction be longitude in this case?

[grave pond]

Only if your center point is the north or south pole.

In the example above it looks like the center point is somewhere in India, so there you need to figure out the compass bearing of a great-circle line from somewhere-in-India to the point in question. That takes some spherical geometry (which I can't immediately rattle off a formula for), but it's the same spherical geometry for all azimuthal projections, which is why it's not stated explicitly in the article about Airy.

[tepid burrow]

alright thank you very much, this had me very lost

[grave pond]

(Or in other words: An azimuthal projection means that great circles through the center point are mapped to straight lines through the center of the map, and the angles between those lines are true. So the only thing one needs to specify for a particular such projection is how distances along one of those great circles map to distances along the straight line that represents it).

[tepid burrow]

what could i search online to find out more about this?

im having a hard time visualizing it in my mind

[grave pond]

I'm not sure what a good source would be, but I'm working from https://en.wikipedia.org/wiki/Map_projection#Azimuthal_.28projections_onto_a_plane.29 myself.

[tepid burrow]

you think this is a good source 😂
https://apps.dtic.mil/sti/tr/pdf/ADA037381.pdf
1977

[grave pond]

No idea. Looks thorough, though.

[jagged wyvern]

Can someone help me with this?

[twin spindle]

If I factor out the sin, can it cancel out the cos that's in the () with the cos in the denominator?

[hard moon]

no

[maiden brook]

no lol

for the sake of the argument, let sin theta = 1/2 and cos theta = sqrt3/2

[obsidian harness]

No sin theta also needs to be divided

If you try expanding the brackets in the numerator maybe it will be easier

[jagged wyvern]

@trail tendon I tried setting up a proportionality then simplifying it like 45/30 --> 3/2 (3:2) it didnt work wondering if you could help

[trail tendon]

uh

i mean

arclength formula is s = r(theta)

where s is the arc length, r is the radius, theta is the angle

and note that s is the same, and you already know theta

for both

right?

[jagged wyvern]

Uh I was given a different eqt

AB/2nr = mAB/360

[jagged wyvern]

Yes

[jagged wyvern]

But how do i set it up here

[trail tendon]

uh

whats AB

is it the radius?

nah wait

that wouldnt make sense

whats AB

lol

[jagged wyvern]

Im not sure if im given

[trail tendon]

how do you use a formula if you dont know what the variables mean ._.

[trail tendon]

whats n?

[jagged wyvern]

pi

[trail tendon]

o alr

[jagged wyvern]

mAB is the central angle

[trail tendon]

r is radius im assuming

[jagged wyvern]

AB is the arc

[trail tendon]

ohh alr

[jagged wyvern]

Yes

[trail tendon]

so yeah that works

[jagged wyvern]

ratio between arc and circumference

[trail tendon]

yeah

[jagged wyvern]

But how do i plug it in

[trail tendon]

try multiplying both sides by 2 * pi * r

[jagged wyvern]

AB = mAB/360 * 2nr

Hm ok

[trail tendon]

AB is the same for both circles, right?

so you can set them equal and plug in your values to find the ratio between the radiuses

radii?

radii

💀

[trail tendon]

does this make sense though?

[jagged wyvern]

Uh

Yes

[trail tendon]

alr

[jagged wyvern]

Uh is the arc length and arc measure the same

[trail tendon]

uh

i think arc measure is like

the angle from the center

or smthn

wait lemme google

its "the measure of an angle that the arc creates in the center of a circle"

so yeh

[jagged wyvern]

Alr

[trail tendon]

so like in this case the 45 degree arc is the arc measure

[jagged wyvern]

So I setted it up like this: 45 = 30/360 * 2n(r)

[trail tendon]

uhhh

[jagged wyvern]

Im not sure when they say "mAB" is the "minor" measure

[trail tendon]

i kind of meant you'd have the same formula on both sides but with different values

wait lemme think

idk what you did

lol

anyway what i meant was

AB = mAB/360 * 2nr

you have two circles, with different angles that you know

you want to find the ratio of their radii

[jagged wyvern]

Yep

[trail tendon]

so lets just say the angle of the bigger circle is x, and the angle of the smaller circle is y. (these are like the mAB in the formula)
lets say that the radius of the bigger circle is a, and the radius of the smaller circle is b

AB is the same for both, so you can take the expression mAB/360 * 2nr for each

and set them equal

but they gonn ahave a different mAB (that you know)

[jagged wyvern]

Ye

[trail tendon]

am i making sense to you or nah

o kewl

[jagged wyvern]

But isnt the radiis gonna be different

[trail tendon]

yes, but you're just finding the ratio between them anyway

[jagged wyvern]

Alright 1 second im gonna set this up

[jagged wyvern]

Not sure if im doing this correctly..

[trail tendon]

i think you are, its a lil hard to read but i think so

your r looks like a minus sign XD

[jagged wyvern]

Sorry

[trail tendon]

naw its good xD

what u get

[jagged wyvern]

Im stuck here

If im dividing a fraction by 2pi

Should it be 1/8/2pi??

[trail tendon]

1/8 * 2pi/2pi * r = 1/8 * r = r/8

you're dividing the 2pi on top by the 2pi on bottom

on both sides of the equation

[jagged wyvern]

So both the 2pi's on both sides cancel?

[trail tendon]

yeh

and by the way, keep in mind the r are different

[jagged wyvern]

Ye

So do i write the ratio as 1/8:1/12?

[trail tendon]

you can simplify it

😔

i think you probably just put it in wrong ngl

😂

[jagged wyvern]

I dont get it

Theres 1 in the numerators

[trail tendon]

multiply both sides by 24

cuz you can

and it makes it simpler

[jagged wyvern]

1/3:1/2

Oh

Wait somethings wrong

[trail tendon]

nope, you multiplied by 4, now multiply by 6

its the same answer you got before, and i don't know how you got the answer before for sure, maybe i shoulda asked 💀, and also you probably just put it in wrong

they wanted the smaller:bigger

if you put in 3:2

it woulda been wrong

cuz it opposite

:l

[jagged wyvern]

Isnt it when a bigger arc implies it's a bigger radius

[trail tendon]

no

[frozen sail]

what’s up guys, nobody’s helping me on this. Just wondering if you know how to do this?

[jagged wyvern]

I gtg could we do this tmmrw?

[trail tendon]

lookit this

[trail tendon]

you literally have the answer lol

its 2:3

im like 99% sure

cuz our method is correct and you can just plug it in

[jagged wyvern]

So youre telling me all i had to do was switch the thing around..

[trail tendon]

into arc length formula

yeh

i don't know how you got it the way you did though

[jagged wyvern]

Ah

[trail tendon]

;-;

[jagged wyvern]

Welp ok

Its correct

Ty have a good night/day

[trail tendon]

u 2

[shut gulch]

hello guys

how do u do this

ive been trying to solve it for a while

cant figure it out

[obsidian harness]

You need the angle bisector theorem

[shut gulch]

yea

but like

i only know how to solve given three sides

im only given 1

[obsidian harness]

Yes so AB:AC = 8:6

And also AB + AC = 42 - 6 - 8 = 28

So you know AB and AC

Then using the angle bisector theorem again you can find what AN and CN must be

[shut gulch]

im sorry if i sound dumb but wont i need BC to solve for the side AC?

[obsidian harness]

Yes but you know what CM is and you know what MB is
So you know what CB is

[shut gulch]

oh right my bad

wait no how about AB?

wont i need that also

[obsidian harness]

Yes ofc

[obsidian harness]

Read this part here for AB and AC

[shut gulch]

tysm m8

[obsidian harness]

No worries

[prisma hornet]

Guys I need help with CAST rule in trig

It’s like super simple appearantly

But I don’t get it

why cant i also do 360-15 or use the original 15 as one of the "correct" angles

[obsidian harness]

According to the CAST rule, everything (including tan) is positive in quadrant 1

Tan is also positive in quadrant 3

(note that 0.2679 is positive, that's why I'm talking about when tan is positive)

So 360 - 15 would be in the 4th quadrant

That's the wrong quadrant to go looking

[prisma hornet]

ohh i see

wait but what about the first example

cant i do 360-51 there

cuz Cos is positive there

[obsidian harness]

When cos is negative, so that's when cos is not positive

So cos can't be in quadrants 4 and 1

So it has to be in quadrants 2 and 3

[obsidian harness]

If this confuses you, then the unit circle definition tells you that:
cos theta = x-coordinate
sin theta = y-coordinate

So if cos theta is negative, then the x-coordinate is negative

Bam, that's quadrants 2 and 3