geometry-and-trigonometry

[upper karma]

yah

learn it from aops

not wikipedia

[abstract scarab]

Yaaa wiki has a lot lot moreover

More*

[upper karma]

aops teaches problem solving and higher levels of logic so u can battle those types of problems in contests

[gray marten]

Whats AOP?

[upper karma]

art of problem soling

solving*

have u ever heard of AMC or AIME

[gray marten]

I've heard of AMC, the broadcasting company

[upper karma]

......

the math contest

[gray marten]

https://en.wikipedia.org/wiki/AMC_(TV_channel)

Nop

[upper karma]

like AMC 10 or AMC 12

[gray marten]

Never heard of it.

[abstract scarab]

https://artofproblemsolving.com/community

[upper karma]

^

yee

[gray marten]

Thought it was book.

Hmm looks like a messy forum.

[upper karma]

u can buy the aops books if u want......

they teach high level theorey

[gray marten]

Sounds scammy...

[upper karma]

no not at all

[gray marten]

Can I get it for free?

[upper karma]

most students in north america who do contests usually read their books

there are free topics

https://artofproblemsolving.com/wiki/index.php/Geometry/Introduction

https://artofproblemsolving.com/wiki/index.php/Geometry/Intermediate

https://artofproblemsolving.com/wiki/index.php/Geometry/Olympiad

[gray marten]

Soo its for competitive math people?

[upper karma]

yah

but introduction and some parts of intermediate u dont really need to do it for competeive

[gray marten]

Does it have stuff on Metric geometry?

[upper karma]

olympiad is those for doing olympiad math

[gray marten]

Or Topology?

What are the topics based on?

Like other than the competition.

[heavy osprey]

I have one more problem if that’s okay it might be a lot of work so if you don’t want to help it’s okay

[gray marten]

Is it like most useful ones for applications?

@heavy osprey Yeah sure

[heavy osprey]

Is that clear enough

[gray marten]

Dimension & Cost of this Gazebo is given. Floor is an octagon. Which gazebo costs least.

Based on square per inch.

[heavy osprey]

So do I find the area then divide it by the price and find the lowest rate?

[gray marten]

Yeah basically.

Find the area of the octagon based on A and B

and yeah divide and find lowest rate.

[abstract scarab]

Yaaw

[heavy osprey]

Okay only thing is my geo class hasn’t learned area of octagon so should I just split it into two trapezoids

Two trapezoids and a rectangle*

[gray marten]

Yeah I guess you could

They say its a regular octagon

[gray marten]

The thing is you should be able to find area of octagon just by A

[abstract scarab]

Ya

You can

[gray marten]

Formula for octagon: 2(1+sqrt(2))a^2

But idk

[high zephyr]

rare formula

[abstract scarab]

But here it's easy to use b

[gray marten]

You can derive it from triangles.

[abstract scarab]

Yaaa...

[gray marten]

a^2 = (2c^2)
c*3 = b

[abstract scarab]

???

[gray marten]

wait nvm

b = sqrt((a^2)/2)*2+a

Just checking to see if b was correct.

So yeah it is a regular Octagon.

They probably want you to use two trapezoids

But you can verify your answer using formula for octagon:

2(1+sqrt(2))a^2

[heavy osprey]

My teacher said we can’t use are formula for octagon we have to split it up into shapes this is what I got

[abstract scarab]

Yepp

[heavy osprey]

I think I need a triangle to find the height

[gray marten]

Yeah sounds reasonable

[abstract scarab]

Yaaa

[gray marten]

I think it was 36.5 not 36.75

[heavy osprey]

Ooh yeah good catch

[gray marten]

Also I would probably use a triangle formula.

b/2 is the height of the triangle

[abstract scarab]

Ya .. I used that

[gray marten]

so 1/2(base * height)

(a/2 * b/2) * 8

Because its made of 8 triangles.

Oh wait

Might be worng with that

[abstract scarab]

Correct

[gray marten]

Nvm it works.

[heavy osprey]

Okay so is that for the area of the triangle

[gray marten]

1/2 base * height

[abstract scarab]

Yees

[gray marten]

So the base of the triangle is 36.5

And the height is half of b, which is 88/2 = 44

And you get one triangle piece, and then you can multiply it 8 times

[upper karma]

Hi

[high zephyr]

channel looks taken

[heavy osprey]

Okay

[high zephyr]

xi do u have a question?

[upper karma]

No

[heavy osprey]

So let me get this straight we’re making this into 8 triangles and forgetting about the trapezoids

[gray marten]

Yeah, its kinda faster.

[heavy osprey]

Alright

[gray marten]

So you cut it into 8 triangle pizza pieces.

And the base is a

And the height is b/2

[heavy osprey]

Okay

[gray marten]

(half the base) x (the height), for triangle formula.

And once you get one triangle, multiply it 8 times.

And tada!

[abstract scarab]

🤣🤣

[heavy osprey]

Is this right then I just multiply it out and divide by prices

[gray marten]

Yep

[abstract scarab]

That's notability?

[gray marten]

Well the notation is a bit weird

[abstract scarab]

I use goodnotes lol

[gray marten]

But as long as you understand it

36.5*44*1/2 will give you the area of a single triangle.

And multiplying it by 8 will give you the area of 8 triangles (the whole octagon)

Also @abstract scarab since they're using approximations for b and a, the answer will "slightly" be different from the real area of the Octagon. But I don't think they mind.

Weird question tho.

[abstract scarab]

Lol

Anything else??

[gray marten]

There are technically 3 answers?

[abstract scarab]

🤣🤣🤣🤣

[heavy osprey]

Does this look right, gazebo 2

Oh crap nvm it says least per square inch

So gazebo 1

[gray marten]

Area of Gazebo 2 looks pretty correct.

Same with 1

[heavy osprey]

So would the final answer be 1 cause it’s the lowest

[abstract scarab]

Yaaa

[heavy osprey]

Thanks a lot!!

[gray marten]

Yeah! Np!

[digital gulch]

I’m confused on this

[abstract scarab]

Ya .. one side of the other

[digital gulch]

Wdym

[digital gulch]

Bruh

Each cylinder has to be tangent to the other 6

In that picture, each cylinder is only tangent to either one or two cylinders

Also tangent in here means touches only in one point

[upper karma]

Yes

[digital gulch]

In that picture, each cylinder is tangent via a line

[upper karma]

it doesnt specify on size or formation so yes

[abstract scarab]

Yaa

[upper karma]

one big cylinder

[digital gulch]

Yeah it doesn’t specify

[upper karma]

and many tiny ones touching it at one point in the air

like its floating lol

[abstract scarab]

Good imagination

Lpl

[digital gulch]

But the smaller cylinders also need to touch each other

Each cylinder needs to touch the other 6

[upper karma]

they all need to touch other 6?????

[digital gulch]

Yeah

[upper karma]

thought it just said 1

[digital gulch]

Nope

[abstract scarab]

Use my soln guys .it's the easiest

[upper karma]

dont think so

[digital gulch]

But that solution is wrong

[upper karma]

bc

[digital gulch]

Yeah I also don’t think so

[upper karma]

lets say we have a cylinder

A

[digital gulch]

Yeah

[upper karma]

cylinder A can touch each of the other 6 cylinder s

[digital gulch]

Yeah

[upper karma]

but the maximum amount of cylinders next is 5..4..3

and so on

[digital gulch]

I don’t get what you mean

[upper karma]

would help if i had paper on me but

the outer most cylinders

[digital gulch]

Draw it

[upper karma]

can only touch a maximum of 3 cylinders

dont have a paper with me

[digital gulch]

No

On your device

[drowsy marsh]

How to do 15

Wait i mean 16

[abstract scarab]

One sec

Got it??

[drowsy marsh]

Ohhh i see

Thanks :)

[upper karma]

I know Area = 1/2 * sin A * B *C

I find that A = 70, B= 16cos70, C = 8

When I sub it in

It's not in the multiple choice

where did I go wrong?

[abstract scarab]

Here I go

[upper karma]

woah

That was quick

Thx

[abstract scarab]

Yepp

Let me know if u have any other qs

[digital gulch]

I’m still stuck

I’m just going to say that it is not possible

[abstract scarab]

🤣🤣🤣

Leave that question

Bro

[digital gulch]

Ok

[dull remnant]

do these addd up to 180

[silent plank]

wdym by "these"

[dull remnant]

a and b

im p sure they

do because of the ambiguous case

[silent plank]

assuming the diagrams are reasonably drawn to scale yes

[dark sparrow]

there are hatch marks, ramonov

[dull remnant]

i think they are

[silent plank]

as in the triangles are actualy obtuse and acute respectively

[dull remnant]

cant u do like the law of cosines

[silent plank]

some people can draw some really fked up diagrams

[dull remnant]

lol

[upper karma]

I see an ASS

Angle side side

[silent plank]

but yes, your reasoning is fine

[upper karma]

If you overlap the triangles, dont you get an isosceles triangle on the right?

Which would also imply that a and b add to 180

[tawdry pivot]

yeah

[visual mist]

can someone help please

[acoustic jungle]

I don't know how to do this but If you can prove triangle PMB is isosceles then you can prove AM = MB

Oh

[inner sandal]

APC is a right triangle that is similar to ABC @visual mist

[acoustic jungle]

Ok I have. $\frac{PB}{AB}=\frac{PA}{CA}=\frac{AB}{CB}$

and $\frac{PA}{AB}=\frac{CP}{CA}=\frac{CA}{CB}$

[somber coyote]

Fishraider:

Fishraider:

[acoustic jungle]

@inner sandal what do you do next

[gray marten]

I was gonna do this entirely on a cartesian plane.

[upper karma]

Did you just made semejant triangles

[gray marten]

Okay, I proved it with pure Cartesian coordinates.

[acoustic jungle]

gay

[gray marten]

i agree

Here are the three functions.
$$y=\sqrt{\left(\frac{CA}{2}\right)^{2}-x^{2}}+\frac{CA}{2}$$
$$y=\frac{-CA}{B}x\ +\ CA$$
$$y=\frac{2CAbx-CAb^{2}}{\left(CA+b\right)\left(CA-b\right)}$$

[somber coyote]

Lang:

[gray marten]

Circle, line, tangent line.

M is found by making the tangent formula = 0.

And you'll get B/2.

But yeah idk how to do this properly, other than cartesian hellscape:

[acoustic jungle]

There definitely is an easier way

it feels like you are using calculus to calculate the area of a square

[gray marten]

I used calculus to find the tangent of the circle.

But yeah the whole things overkill.

But my god Cartesian coordinates makes such simple questions gross.

[silent plank]

construct AP

angle in alt segment

vertical angles / angle sum of triangle (and/or similar triangles)

to show triPMB is isosceles

[gray marten]

Okay. I need to figure out how to actually do this.
As you can see I lack knowledge required, since I only know how to do this cartesian.

[rose sphinx]

Hey

Anyone with a good hold over triangles&circles ?🙄

[gray marten]

Well thats what we're kinda doing rn.

[silent plank]

i was deliberately semi-vague above.

how many of those terms do you understand

[rose sphinx]

I hv a Q if u people wanna try

[gray marten]

This isn't my question, its from some rando guy, above. I just solved it in an overkill-cartesian way and realized I didn't know how to solve it properly.

I understand most but I have no idea how to go and do it.

@rose sphinx whats your question?

[gray marten]

Mmm time to snap my neck.

[rose sphinx]

Jst rotate it or ur phone

[acoustic jungle]

wtf I can read vertically

[silent plank]

justify with appropriate theorems

[gray marten]

@rose sphinx not on phone

on PC

[rose sphinx]

Ops

[gray marten]

Im reading it now tho

[silent plank]

post it again

so we can get the bot to rotate

[gray marten]

I am bot.

[rose sphinx]

Gg

[gray marten]

Let ABC be triangle, and let O be its circumference.
The internal bisectors of angles A, B and C intersect O at A1 B1 and C1 respectively

And internal bisectors of angles A1, B1 and C1 of the triangle A1, B1 C1, intersect O at A2 B2 C2 respectively
If the smallest angle of triangle ABC is 40*, what is the magnitude of the smalelst angle of triangle A2 B2 C2 in degrees.

[rose sphinx]

Go fr it like it bolded me so I assume it will be hard fr anyone else

[gray marten]

Im not a fan of wordy problems.

And im not good at geometry clearly.

But i'll try understand it

[rose sphinx]

I challenge whoever thinks he's good enough

[gray marten]

I don't think im good enough m8

I can solve it, using gross cartesian stuff probably

[quiet mason]

he challenges u

[gray marten]

But thats never what you want to do.

[rose sphinx]

@silent plank what about u friend

[quiet mason]

lmao do you just want your homework question solved

in disguise as a challenge

[gray marten]

Get me to do it lol.

[rose sphinx]

Ha ha its not one

[gray marten]

I'll do it using purely cartesian

And then ask you to tell me how to do it geometrically.

[quiet mason]

that looks like comp math

isnt it?

and its comp geometry=death

[gray marten]

What the hell is an internal bisector lol

Oh huh

A big weird line coming out of the middle of the line

[rose sphinx]

They don't give engineering lev Qs as homework to school kids

[acoustic jungle]

engineering?

cosx=1?

[gray marten]

Trying to understand this thing? Feels like im summoning satan.

[rose sphinx]

And I know the geometric sol to it

Its one of the hardest I hv solved

[gray marten]

Hey

On line 2

it says

"respectively, and the internal bisectors of the angles"

A1, is that B2 or B1?

[rose sphinx]

Didn't understood what u say

[gray marten]

Line 2? Of the question?

I can't read the photo since its kinda pixely.

[rose sphinx]

U wrote it bro

[gray marten]

Wut

I invented this problem?

[rose sphinx]

Naah

I meant sceoll up

[gray marten]

I was asking you to confirm?

From the image?

Since I assume you have the original text?

[rose sphinx]

Let ABC be triangle, and let O be its circumference.
The internal bisectors of angles A, B and C intersect O at A1 B1 and C1 respectively

[gray marten]

:/

[rose sphinx]

And internal bisectors of angles A1, B1 and C1 of the triangle A1, B1 C1, intersect O at A2 B2 C2 respectively
If the smallest angle of triangle ABC is 40*, what is the magnitude of the smalelst angle of triangle A2 B2 C2 in degrees.

[gray marten]

Bruh.

U just copied my text.

I eyeballed it so its got mistakes, and its probably wrong?

So I was asking you to check the second line?

[rose sphinx]

No its correct

[gray marten]

:/

Well it's not.

Are you sure its correct?

[rose sphinx]

Yaah

[gray marten]

Well it's not, I wrote "Circumference", but the question says "Circumcircle"

:/

[rose sphinx]

Op I missed that

Sry

But now u got that right

[gray marten]

Look bruh, the question is confusing, idk what its even saying, I swear it repeats itself in the middle sentence too.

If you can translate what ever this is into more understandable form maybe.

But hell I ain't had no Geometry classes other than basic trig.

[rose sphinx]

Hoops

The Q is very clear

[gray marten]

Well rip me

me smooth brained

[rich wolf]

@gray marten sounds like you ain't had no english classes neither

[tulip heron]

Can someone help me solve this?

[acoustic jungle]

ABC=(3/2)^2 * BFG

BFG = (2/1)^2 * BDE

[gray marten]

Ok gonna try solve it properly this time.

Oooh is CAP, similar to PMB?

Or not.. idk.

What theorems, ideas, concepts are you even supposed to use here?

[onyx basin]

i would say that it's safe to assume that triangle CPA and triangle APB are both right triangles

line BA is tangent to circle

AC is diameter

[acoustic jungle]

Bro Lang Ramovv guy already told you how to do it

did you not understand it

[gray marten]

lol no, i aint no geometry person

https://media.discordapp.net/attachments/326138757474680852/698547477632253962/SPOILER_unknown.png?width=339&height=219

i dont see how this shows AM = BM

[onyx basin]

AM and BM clearly both have one congruency mark on each line

therefore they're congruent, therefore they're equal

[gray marten]

I think its by showing MPB = MBP

which somehow shows its an isoceles

[gray marten]

Welp this sux, idk whats going on, so ill do it anotehr time.

I can see APM and PMB are iscoeles or scalene.
Since, the only right angles are CAM, CPA.
So triAPM and triPMB clearly don't have any right angles.

[half rover]

Can I get some help with number 4,5,and 6

[acoustic jungle]

@gray marten no you have to use the theorem rommoovvoe said

I forgot what it was called but the tangent line angle is the same as the other angle

So angle CP(something) = CAP

and angle PAB= ACP

[visual mist]

@inner sandal yep, what next?

[acoustic jungle]

Sushi you are 12 hours behind lol

[inner sandal]

similar triangles have congruent angles

thus you can show the two triangles AMP and BMP are isosceles

[gray marten]

lol mebe i should get a compass and straightedge, so i get better unnerstanding of geometey

[visual mist]

i was sleeping lol

How is APM isosceles

[silent plank]

tangent theorems

whoops I forgot to indicate the right angle (which applies another theorem)

[visual mist]

is the theorem like two tangents start from the same point r equal in length

[silent plank]

for a circle yes

[eager pendant]

what you can do is just show MP tangent to (APC) through alternate segment (defining M to be the midpoint)

[visual mist]

how is PM and MB the same length?

[silent plank]

tri PMB is isosceles
which can be proven by showing that those red-dot angles are equal through theorems

which i have left as a task to you

(that comment was left out when the image was reposted)

@visual mist

[visual mist]

i dont see any theorems that can be applied to prove that PMB is an isosceles triangle

[silent plank]

angle in alternate segment, vertical angles, thales/angle sum of triangle

can be applied to show that the red dots are equal

and see if you can spot how those could be applied

(i.e if you can prove angles MPB and MBP are equal, you've essentially proven that triangle BMP is isosceles with MP =MB)

@visual mist

mostly a repeat but reworded a bit

[visual mist]

hmmm

[silent plank]

i can break it down further if needed

[visual mist]

i still dont see anything lol

[silent plank]

lets start with that, can you see how the angles in alternate segment theorem could be applied here

[visual mist]

yea

then PAM can be 90-theta?

[silent plank]

yes

[visual mist]

and angle M is too because isosceles

wait no

p

[silent plank]

also angle M is ambiguous

[visual mist]

wot does that mean

[silent plank]

it means if you jst say angleM, i have no idea which angle you're actually referring to

[visual mist]

oh ok

sorry

[silent plank]

can you identify the size of the angle APC?

[visual mist]

uhh no

[silent plank]

hint: CA is the diameter of the circle

[visual mist]

idk

[silent plank]

you mentioned 90°

but not quite because its the point of contact

its 90° because it is subtended by the diameter

[visual mist]

yeah

thats why i deleted it

[silent plank]

which is an application of thale's theorem if you were taught the specific name

[visual mist]

nope

[silent plank]

ok. so what would be the size of ACB in terms of theta?

[visual mist]

90-theta

[silent plank]

and angleABC?

[visual mist]

idk theta?

[silent plank]

how sure are you?

don't guess. justify your answer

[visual mist]

wait

yea its theta

[silent plank]

ok. good

now are you able to determine angle BPM?

[visual mist]

90-2theta

[silent plank]

how are you getting that?

[visual mist]

180-angle PAB-angle PBA

oh wait

wrong

oops

[silent plank]

try not to overthink it

[visual mist]

im not sure

[silent plank]

like really try not to overthink it

hint: v a

[visual mist]

v?

[visual mist]

oh x

[silent plank]

and similarly for your problem?

[visual mist]

ohh theta

[silent plank]

so now after all that we have

and have proven that angleMPB = angleMBP (=thetha)

are you alright with all of that?

[visual mist]

yeah

[silent plank]

which proves that triangleBMP is isosceles and that MB and MP are equal

[visual mist]

yes

[silent plank]

AM and MP are also equal from tangent theorems (and/or RHS congruency)

[visual mist]

yea

[silent plank]

thus AM = MB and the point M bisects AB

[visual mist]

ohhhh

thanks

i get it

[rose sphinx]

Hey

Any updates on what I posted yesterday

🙄 🙄

[upper karma]

@rose sphinx repost it

[dark sparrow]

,rccw

[rose sphinx]

@upper karma

[upper karma]

Have you drawn it?

[rose sphinx]

Yeah

[upper karma]

Post it please :)

@rose sphinx

[warm mountain]

im blanking

completely forgot how to do it

[upper karma]

@warm mountain okay so

First off we want to name the base of the first little triangle

If we have 18 as the big one, and the little second base is x, the other little base is ... @warm mountain

[warm mountain]

18-x?

[upper karma]

Yup

There's a lot of ways of solving this problem lol

@warm mountain what are those red lines at the top? Are they the same angle? I can't see really

[warm mountain]

Same Angle, yes

[upper karma]

Oh ok then

Do you know trigonometry?

[warm mountain]

A bit. I'm just kind of blanking out on this, haven't done this kind of thing in a while.

[upper karma]

Ok no prob, it kinda happened to me a few weeks ago lol

So

Lets call that red angle alpha

We want a system of equations with alpha and x to solve it

@warm mountain how can you do the system

(If you dont know still whats the system tell me)

[warm mountain]

Still don't know, drawing a blank

[upper karma]

Ok

So the trigonometric identities

We'll take sin

Sin (alpha) = opposite side/hypothenuse

Remember?

[warm mountain]

Yes

[upper karma]

Okay so

As we know the opposite sides of both little triangles and the hypothenuse of both

We can do the system...

[warm mountain]

but are the two angles combined 90 degrees?

[upper karma]

Umm unless the question tell you, no

[warm mountain]

hmm

I'm still drawing blanks, I havent done this stuff in a while

[upper karma]

We know that both 2 lines red are both alpha, the same angle

Right?

[warm mountain]

would that mean that 18-x = x? Since the angles are the same?

[upper karma]

Wowoo what did you do there

[warm mountain]

or did i just botch that

[upper karma]

I think so

[warm mountain]

oof

[upper karma]

Wait a sec

[warm mountain]

But if sin(alpha) = x, because it's on the opposite side, wouldn't that be the same with the other angle since they're the same?

[upper karma]

When did i said that sin (alpha) = x

[warm mountain]

You said that sin(alpha) = the opposing side, right?

[upper karma]

No

[warm mountain]

isn't x the length of the opposing side from the 2nd angle?

[upper karma]

Sin (alpha) = opposite side/hypothenuse
@upper karma look

isn't x the length of the opposing side from the 2nd angle?
@warm mountain yes

But you are missing hypothenuse

Got it now?

[warm mountain]

Don't you only use hypotenuse in a right triangle?

[upper karma]

Nope

The hypothenuse is always the largest side

Wait

Now you are confusing what lol

Uhh ask another you just got me lol im srry i gtg

@warm mountain

[warm mountain]

... ok then

[upper karma]

<@&286206848099549185> help him

[silent plank]

i'll take over

[warm mountain]

I really don't get this, I'm drawing a blank, my apologies

[silent plank]

are you familiar with the trig formula for area?

(of a triangle)

[warm mountain]

Formula for area of a triangle?

[silent plank]

the trig formula specifically

[warm mountain]

No, I'm very drawn out of it

[silent plank]

Area = 1/2 ab * sin(C)

[warm mountain]

Alright

[silent plank]

have you seen that before?

[warm mountain]

I believe so

[silent plank]

if you apply that to the left triangle and right triangle on the right,
the areas would be in a ratio of 10:14

[warm mountain]

Alright

[silent plank]

do you understand how i'm getting that?

[warm mountain]

A bit... not quite sure

[silent plank]

let the common side be s and those red angles be u

https://cdn.discordapp.com/attachments/326138757474680852/698928876272746597/fa2b59a08a90d008ad257a7441652ba8.png

Area_left would be 1/2 * s* 10 * sin(u) right?

[warm mountain]

Yes

I see that

[silent plank]

similarly, what would be the area of the triangle on the right?

(in the same format)

[warm mountain]

1/2 * s * 14 * sin(u) correct?

[silent plank]

yes

and the ratio of their areas,
Area_left / Area_right = ( 1/2 * s* 10 * sin(u)) / (1/2 * s * 14 * sin(u))

those 1/2,s and sin(u) cancel

which leaves you with 10/14

[warm mountain]

I see that

[silent plank]

which can be further simplfied to 5/7

ok good

[warm mountain]

so question... are we applying that ratio for the base then?

[silent plank]

can you determine the ratio of the areas of the triangle on the right and the biggest triangle?

getting there

[warm mountain]

Ah

Uh... I don't know. Do you combine the left triangle and the right triangle and put that against the right triangle? like 12/7?

[silent plank]

sort of. wasn't exactly what i asked for but that's the ratio
of big/right (instead of right/big)

[warm mountain]

Oh alright

so it'd be 7/12 in Right/big

so would you apply that as x/18 = 7/12

[silent plank]

yes

[warm mountain]

so X is 10.5?

[silent plank]

lowercase x but yes

everything make sense?

[warm mountain]

Yes, thank you very much

[upper karma]

@warm mountain we could also have done it by semejant triangles

If you are curious tag me and ill show you, im srry i had to leave before tho @warm mountain

[warm mountain]

It's all good... though I do feel a bit stupid now

[upper karma]

Why tho?

[warm mountain]

I was lost like 90% of the time lol

[upper karma]

Man no worries, i was also a lil blancked out

With the hypothenuse thingy lol

So you dont want to know the semejant triangles method? (Its easier than ramonov's method imo)

[silent plank]

takes 3-4 lines if you know your theorems and are efficient

is semejant similar?

[warm mountain]

Semejante? That's similar in spanish

I was originally going to do 18-x/x = 10/14 as the entire thing... that works

[upper karma]

Oh uh maybe that is not the translation

I was originally going to do 18-x/x = 10/14 as the entire thing... that works
@warm mountain yeah basically that

takes 3-4 lines if you know your theorems and are efficient
@silent plank ikr but this one takes a line xD

[silent plank]

well 2

[upper karma]

Yeah idk

[silent plank]

that's quite nice

[upper karma]

Yup, quite simple if you dont make an error rotating the triangles

[silent plank]

that formula can be derived from using sine rule

also parentheses

[upper karma]

Oh interesting

[silent plank]

in fact this question is very sine heavy

[acoustic jungle]

@upper karma show me what triangles are similar

also you can use the angle bisector theroem

14/x = 10/(18-x)

[silent plank]

for some reason i keep forgetting that

and i end up deriving something similar every time

[acoustic jungle]

@upper karma look
@upper karma

Sin (alpha) = opposite side/hypothenuse
Where is the hypotenuse

[upper karma]

There i said i made a dumb mistake

[acoustic jungle]

alright what about the similar triangles

I do not see any

[upper karma]

@acoustic jungle that's the one im referring to bc in my language its called triangles "semejantes" which i translated to semejant

14/x = 10/(18-x)
@acoustic jungle

Sorry for double tag

(They both have one angle in common) idk how its called in English lol

[acoustic jungle]

They only have 1 angle in common

you need to know at least 2 angles to prove they are similar

@upper karma

[upper karma]

@acoustic jungle

@acoustic jungle that's the one im referring to bc in my language its called triangles "semejantes" which i translated to semejant

Do you know what i mean

[acoustic jungle]

No.

Similar triangles means all angles are similar

you can't prove they are similar with 1 angle

I don't know what similar triangles you are referring to

@upper karma

Where are the similar triangles

[upper karma]

Uhh

Let me explain

14/x = 10/(18-x)
@acoustic jungle So in Spanish we call semejant to this theorem

[acoustic jungle]

you said semejant is similar triangles

[upper karma]

But in English is not called Semejant

When

Did i?

I didn't

[acoustic jungle]

oh? well okay I just misunderstood

[upper karma]

No prob

Out of curiosity

[acoustic jungle]

or you might've used the word wrong

[upper karma]

What does similar triangles mean in english

[silent plank]

so what would similar triangles actually be in spanish

[upper karma]

Triangles "semejantes"

Which i translated to semejants

[acoustic jungle]

similar triangles means triangles have the same angles

[upper karma]

Lol

[silent plank]

similar triangles, have the same angles, corresponding sides in same ratio

[upper karma]

Oh nice

Triangles "semejantes"
@upper karma this solves your question

[silent plank]

semejantes → semejants
that just removed an e, and still isnt english

[upper karma]

Ikr thats the mistake i made

Lol i said that look above

That it was a mistake

[acoustic jungle]

Well at least now you know semejant doesn't refer to the angle bisector theorem

[upper karma]

Yeah i guess xd

[versed river]

what have you tried @crystal pike ? try drawing a diagram of the vector

[crystal pike]

SOH-CAH-TOA

[versed river]

yeah, so lets start with the i component

get a diagram, and figure out whether you wanna use sine cos or tan

[crystal pike]

Tan

[versed river]

this is for the i component

[crystal pike]

right?

[versed river]

not tan

can you show me your diagram?

[crystal pike]

Dont think i could send it

[versed river]

ill draw one then and you tell me if you agree

[crystal pike]

okok

[versed river]

so do you agree that the magnitude is on the hypotenuse?

[crystal pike]

yes

i do agree

[versed river]

and whats the i component on?

is it opposite or adjacent?

[crystal pike]

hmmm

im not sure and dont want to guess

but my best guess would be adjacent lol

[versed river]

well you can see the angle 80

and the j component is opposite thnat

so it must be adjacent

so then you have sin80=what?

sorry

not sin

cos80=what?

sine you have adjacent and hypotenuse you're using cos of course

[crystal pike]

So i find Cos(80)

[versed river]

mm its best to leave it as cos80 for now

just sub into the formula first

cosΘ=A/H

[crystal pike]

ah alright

[versed river]

so if you sub in your numbers what do you get, and can you rearrange that to get an expression for the i component?

[crystal pike]

so all i need to sub in is Cos(80) into the cosΘ=A/H rn?

[versed river]

yes

H is the magnitude, A is your i component

[crystal pike]

how do I find my H and A tho

[versed river]

look at the diagram

H is the magnitude, 19, right?

and A is the i component, its what you're trying to find

[crystal pike]

so i plug in Cos80 = X/19?

[versed river]

yes

then what do you do to get X on its own

[crystal pike]

subtract 19 from both sides?

[versed river]

what no you're dividing by 19

[crystal pike]

whoops

[versed river]

you have to multiply both sides by 19 to get X

so then you have X=19cos80

and then you can do the same for the j component, with sine

i have to go now though. ping helpers or whatever if you still need help with that in a bit

[crystal pike]

@versed river

before you go

[tame ore]

so it says its a figure but the area is wrong. for figures isnt the formula just width x height? i did 22 x 11 to get 242

[versed river]

ye

[crystal pike]

did you get 80?

just real quick

[upper karma]

Isnt the angle 65 @versed river

[crystal pike]

so o know

[versed river]

bruh

why did i do 80

when did that come into my head

[upper karma]

Lol xD

[versed river]

but same process

[crystal pike]

its 65?

[versed river]

and deleted user, youre wrong but i dont ahve time to explain why

its a composite shape

break it down

[upper karma]

@tame ore what

Could you explain better

[tame ore]

its asking for the area of the figure

so i did the width times the height

11 x 22

and it says its incorrect

is there a different formula?

[upper karma]

Yeah bc that is the formula of a rectangle

And that is not a rectangle

[tame ore]

what should i use

[upper karma]

Has a rectangle cut off

A of rectangle - A cut off

[tame ore]

what does that mean

[upper karma]

A is area

The cut off means the white thing where i attributed to "cut"

[tame ore]

oh wait so

do you want me to cut it into two parts

and find the area and conjoin them

[upper karma]

You could do that

[upper karma]

Or get the ( A of the total rectangle which is b*h which is 22 multiplied by 11 ) - ( A of the cut off rectanlge)

Where did that 2 come from?

Yeah you can also do that

But that 2?

And 1?

Oh you numbered them llol

@tame ore so you got it? Or need more further explanation?

[tame ore]

i got it

thank you

[upper karma]

Np :)

[crystal pike]

@upper karma

For this how would i find x?

tan=19Cos(65)/x

[tame ore]

so for this

it says the formula is

1/2 the height x (b1 + b2)

so since its a square the height must be a 10

so i did 1/2 of 10 x (10 +15)

which is 125

and i did many different ways with it being 125 and it says its incorrect

[viscid ginkgo]

if a point is on the corner of plane R, is it said that the point is on the edge of plane R?

[onyx basin]

@tame ore 10 is only the height of the rectangle, not the whole trapezoid

[tame ore]

but isnt it the base

on one of the sides

since its 10 then 15 as the bases

[onyx basin]

1/2 the height x (b1 + b2)
so i did 1/2 of 10 x (10 +15)
but here you used 10 as your height

yes, 10 and 15 are both bases

[tame ore]

yeah

so 10 x 10 is 100

then + 150

so 250

/2

is 125

ive done it so many different ways

[onyx basin]

we can't assume that that is a square

[tame ore]

but it says its wrong

ohhh

then what else could I do

[onyx basin]

use pythagorean theorem to find the other side of the right triangle at the top

then that's your height

[tame ore]

whats the theorem again 💀

[onyx basin]

$a^2 + b^2 = c^2$, where $c$ is the hypotenuse

[somber coyote]

andrwxlin:

[upper karma]

that shape is ridiculously out of scale

[upper karma]

A vector is an object that has both a magnitude and a direction. Geometrically, we can picture a vector as a directed line segment, whose length is the magnitude of the vector and with an arrow indicating the direction. ... Two examples of vectors are those that represent force and velocity.

Vectors are used to represent a quantity that has both a magnitude and a direction. The vector is normally visualized in a graph. A vector between A and B is written as
—>
AB

I got that from google

[vale nimbus]

lmfao that picture

[upper karma]

Ya that’s vector from despicable me

[sly marlin]

ok victor

[tame ore]

HEY

BFFS

[upper karma]

Well the equation for the area of a rhombus is 1/2 • d1 •d2

1/2 times the length of the first diagonal and times the length of the second diagonal

@tame ore so just plug it in

[sly marlin]

looks like a deleter

@grim stratus you can define a vector as an element of a vector space

vector space as a set over a field with scalar multiplication, and adds like an abelian group

[viscid ginkgo]

if a point is on the corner of plane R, is it said that the point is on the edge of plane R?

[full ruin]

hello can someone help me

if anyone is awake

ok

but is anyone there?

[versed river]

there are plenty of people here, whether they can help you depends on if they can see your problem

[dark sparrow]

just ask your question

[full ruin]

If A n B has 2 elementsA has 6 elements and B has 5 elements, then how many elements does A U B have?

this is like venn diagram stuff

[versed river]

n(A U B) = n(A) + n(B) - n(A ∩ B)

[versed river]

n(A) being the number of elements in set A

use the formula

n(A U B) = n(A) + n(B) - n(A ∩ B)

[full ruin]

please give me that answer then i’ll learn i promise

[versed river]

lol

thats the worst way to learn

also thats

looking like a test

[full ruin]

wait the answe r

[dark sparrow]

are you not able to do some addition and subtraction on your own???

[full ruin]

is 9

LMAO

[dark sparrow]

sneaky ALREADY basically gave you the answer

[full ruin]

i’m not dumb i’m just lazy

or is that wrong

[dark sparrow]

i don't know is it

[full ruin]

hmm no it’s not

woah you did basically give me the answer

but where did you get the formula

i need more

[versed river]

do you not have access to google lmao i googled aub formula

[full ruin]

oh lmao

[dark sparrow]

principle of inclusion-exclusion

but i mean

it is pretty straightforward to explain in a case as simple as this

[versed river]

if you have an intuitive understanding of what these kinds of sets are the formula follows pretty logically

[dark sparrow]

if you know what union and intersection are

while n(A) + n(B) accounts for all the elements in A ∪ B, it overcounts those in A ∩ B, counting each one twice.

[full ruin]

ok thanks for the help!

great teachers

fire mine

[viscid ginkgo]

if a point is on the corner of plane R, is it said that the point is on the edge of plane R?

and if JQ = ND is 2JQ = 2ND, what property is this?

Multiplication or distributive?

[upper karma]

30

actually im not 100% sure but if a line is tagent to a circle it forms a 90 degree angle

since |AO|=|OB|=|BS|

that 90 degree is split into 3 equal parts

90/3=30

and if we form a a radius |OT|

u have an isoceles triangle with angle TAO = angle ATO

so therfore x=30

but im not sure if u trsicted a line if that angle gets trisceted to?

actually nvm

i wrong

that line isnt tagent

that touches the circle

i mean |AT| isnt perpendicular to it

@viscid ginkgo distributive

[silent plank]

have you learned circle geo theorems yet?

[upper karma]

@upper karma

its 45?

[silent plank]

and/or what are you allowed to use

[upper karma]

i dont know alot of circle theorems though

[silent plank]

aimed at renar

[upper karma]

@silent plank is the answer 45?

i did assume angle ots is 90 degrees

actually it is

the lines tagent

k u only need to know 2

if a line is externally tagent to the circle the radius is perpendicular to it

and if u have a point on the circle and form a triangle with that point nd the opposite ends of the diamater its 90 degrees

since u now angle ots is 90 degrees

and it splits the hypotnuse in half

since ob=bs

that means the splited angle is 45 degrees

yah

and u know angle ATB is 90 degrees

bc of the 2nd theorem

so 90-45=45

and make a radius from O to T

and u get an isoceles triangle

AOT

and since angle TAO equals to angle ATO

and angle ATO is 45 degrees therfore TAO is 45 dgrees

so x=45

[silent plank]

no

its not 45

[upper karma]

idk then

[silent plank]

drawing a line like that doesn't necessarily bisect the angle

yeh, its 30

[upper karma]

well if its 30

that means angle ATB is 90 degrees

which i thought at first

[silent plank]

pls stop guessing if you are unfamiliar with circle geo and how to approach the problem

consider constructing the tangent to B

[upper karma]

im only familar with certain circle theorems

[silent plank]

you're allowed to add certain things to help solve problems

[upper karma]

@silent plank if u trisect a line does that mean that angle that got triscted equals 30?

[silent plank]

depends where the vertices are

[upper karma]

https://discordapp.com/channels/268882317391429632/326138757474680852/699293136278782013

is angle ATO equals OTB equals BTS

[silent plank]

1 sec, i'll see their approach. i got it through another method

[upper karma]

k my first thought was 30 but i wasnt able to proove it with the methods i did

so i thought ATS was 90 degrees

and since the line was trisected i assumed the angle will get trisected to

so aot=30 degrees

ato i meant

and since its an isoceles triangle

tao is the same

so there both 30

[silent plank]

gimme a min, drawing some stuff

[upper karma]

yah at first i thought 30 and then i thought there was no way angle ATS equals 90 degrees

i forgot how tagents work

[silent plank]

sry for the bad quality

[upper karma]

yah its 90

[silent plank]

angles ATB and OTS are both right angles

[upper karma]

since its tagent

OHHHH I SEE WHAT I DID WRONG

for angle OTS

i thought it got bisected

[silent plank]

which can be justified by stuff like tangent and/or thale's theorem

anyway, do you see how all those angles were obtained?

[upper karma]

yah i mean i saw that originally but i did alot of assumtion after

OTS is jus tagent to the line

and ATb

is bc

the endpoints of the diameter

and a point on a circle

will be 90 degrees

but after what i did was that TO =ST

so i thought it bisceted

it was just a median

[silent plank]

that doesn't answer my question

confirming that you understand before we move on

yes

[upper karma]

i understand it now

so (180-(90-x))/2

and 180 minus that

• 2x equals to 180

[silent plank]

yes. OTB is isosceles

[upper karma]

(360-90-x)/2+2x=180

so u solve for x

[silent plank]

do you agree with the values of those angle presented in the diagram?

and then apply the angle sum of a triangle

[upper karma]

(270+3x)/3=180

yah

meant 450

thats to tedious

do what ramonov did

[silent plank]

is this a different problem?

oh 1 sec

do you have a better pic?

can't see their alpha and beta on the diagram

and that 108 is a typo

$\beta$

[somber coyote]

ramonov:

[silent plank]

greek letters are commonly used to denote angles

but they should be present / labelled in his diagram

yeh thats trash

[upper karma]

^

[silent plank]

improperly labelled work and no justification for what he did

[upper karma]

@upper karma solve it the way ramonov did

[silent plank]

from that

[upper karma]

like once u get the 90-x

[silent plank]

he's pretty much saying that

[upper karma]

ust do angle chasing

[silent plank]

like i said earlier, apply angle sum of a triangle

you have sufficient angles involved

[upper karma]

like180- (180-(90-x))/2 +2x =180

[silent plank]

no like seriously wtf, that work isn't even mathematically consistent

[upper karma]

triangle TOB is isoceles

and sicne the base angle is 90-x

u minus that from 180

divide by 2

to get one of that angle

yah

and then minus that from 180

and theres a 2nd isocles triangele

with the 2 angles being x

so u add that to 2x

and u get 180