geometry-and-trigonometry

[restive void]

dont answer, im stupid

lol

is that 23?

thats supposed to be a 3

wait no

its not

aaaaaaaa

[white swift]

😄

keep calm

[restive void]

so the one right next to it is 23

[white swift]

do the 4 central angles

whats DEC
CEB
AEB
AED

[restive void]

im trying to find dec

aeb = 117

[white swift]

dec is given

[restive void]

not dec

[white swift]

[restive void]

aed

im silly

[white swift]

its okay

[upper sedge]

aed and dec are supplementary angles.

[white swift]

o hoh lol mb

[restive void]

wait are they

yea they are

[white swift]

DEC=AEB=117
AED=BEC=180-117

and then you go from there

[restive void]

63

[white swift]

and fill out the rest of the triangles

[restive void]

is <ECB 80?

[upper sedge]

Yes.

[restive void]

would <BAD be 120?

[upper sedge]

Assuming that the parallel looking lines are parallel...

[restive void]

yea

[white swift]

EAB=180-117+23

im useless lol

so yes

anyway gnight you've pretty much figured it out

[upper sedge]

Sleep well, Blackout.

[restive void]

how would i find <ADC?

i got 137

but apparently thats not right

[upper sedge]

Well, BDC=ABD, and you found ADE, so ADC = BDC+ADE

[restive void]

60?

[upper sedge]

That's what I'm getting, yeah.

[restive void]

got that problem done, now uh

theres this madness

[upper sedge]

Another way to get that is to realize that ADB = 180-BAD

[restive void]

thats what i did lol

now uhhh

how the hell do i do that

[upper sedge]

That's not too bad, assuming MP, QT, RS, and NO are parallel, and that MN, QR, TS, and PO are also parallel.

[restive void]

everything are parallelagrams

[upper sedge]

Oh, that makes life SUPER easy.

[restive void]

not for me :)))))))))

[upper sedge]

大丈夫

I've got your back.

[restive void]

thanks

okay

so where do i start

PN

okay

is there any trig involved in this

[upper sedge]

I'm sure it could be done with trig.

[restive void]

how else can you do this

i really dont want to try to use trig here

[upper sedge]

Note that RN is length 3.

Also note that R is a midpoint.

[restive void]

does that make PT 3

[upper sedge]

Yes.

It also makes TX and RX 3 as well.

[restive void]

so 12

[upper sedge]

Yep!

[restive void]

why cant x be a midpoint

:(

[upper sedge]

Now MQ isn't bad because you're given XS.

X is a midpoint.

[restive void]

it is?

[upper sedge]

It is! =D

[restive void]

how?

thats not what the problem says only q r s t

[upper sedge]

Because it's the intersection of MO and NP in a parallelogram.

[restive void]

oh yeaaaah

so mq is 5

[upper sedge]

Yes.

[restive void]

xo 10?

xo

XO

jesus

how do you find nmq?

[upper sedge]

I got distracted.

Sorry.

[restive void]

its okay

[upper sedge]

Yes, XO is 10.

Now for the angles, which are essentially all given for you in one way or another.

[restive void]

i dont see something for <nmq

teacher didnt show us this btw

hate my teacher

[upper sedge]

NMQ isn't so bad. You know seceral things in advance. Let's just draw all of our connections. NMQ = SOP = 180-(TPO+132), and TPO = 125-XPM, and XPM = 180-(37+MXP), and finally, MXP = 180-132.

Work backwards from this information, and you have your answer.

See? Not so bad. ^^

[restive void]

18

i think i got all of them

im not sure

probably all wrong

but its okay

lol

[upper sedge]

It's alright to double check your work. ^^

Write it all down, keep the expressions simple.

[restive void]

so 23)18

#24 48

#25 30

#26 30

#27 55

right or wrong

lots of that

is guess work

so probably all wrong

[upper sedge]

I didn't crunch the numbers myself. Let me get my whiteboard.

[restive void]

i did random bull crappery

that seemed to make sense

so i did it

[upper sedge]

No! don't do that!

Random bull crappery isn't how to do the math!

Let the math tell you what to do instead.

Strange thought, that, I know.

Still, you have definitions. Those definitions will lead you to the answers one-by-one every time (where the answers exist).

[restive void]

math is hard for me

[white swift]

wat

math is relatively hard

[restive void]

shut up its hard

lol

[upper sedge]

I left the whiteboard in the car. I'll be back in just a moment.

[restive void]

wait

[white swift]

some people do better than others

[restive void]

how would i do this

thanks

are e and f a linear pair

if so f is 135

and the second one, would <f increase while e decreases?

or vice versa

[upper sedge]

I'm back. I'll go verify the previous results, and we'll address those if issues have arisen.

Your numbers for 23-27 were correct.

[restive void]

hell yes

[upper sedge]

You're right that f is 135.

However, as the lift raises, e is the one which will increase, and f is the one which will decrease.

[echo willow]

yo

any bois help me?

[keen aspen]

With?

[echo willow]

polygons

[keen aspen]

Ok

[echo willow]

ok cool

one sec

[keen aspen]

Send the question

[echo willow]

different question

sorry bout that

[keen aspen]

Haha okay

[echo willow]

"The numbers of sides are the ratio 1 : 2|
the interior angles are in the ratio 2 : 3
how many sides do they both have?"

sorry if its a bit basic

[keen aspen]

The number of sides are the ratio 1:2?

[echo willow]

like if one had 3 sides the other would have double

[keen aspen]

Okay sorry

I had to do something for a bit

[echo willow]

its all good

[keen aspen]

Now we have the number of sides being the ratio 1:2 so, you know the formula to find angle measures in a polygon?

[echo willow]

interior or exterior?

[keen aspen]

Interior

[echo willow]

180(n-2)

[keen aspen]

Yes

[echo willow]

but how do i use both the ratios together?

[keen aspen]

So if we distribute the 180 we get 180n-360

[echo willow]

yea

[keen aspen]

So now once we have that we know the ratio of the sides is 1:2

So we can say that 180n-360:2(180n-360)

[echo willow]

ah

where does the interior angle ratio factor in?

[keen aspen]

Yes now, we have the ratio 2:3 for the ratio of interior angles

So we can set thay equal to 2/3

[echo willow]

2/3 = 180n-360/2(180n-360)?

[keen aspen]

Yes

Now cross mulitply

[echo willow]

oh

I didnt realize it was that simple

lol

[keen aspen]

Yes although I might have messed up maybe

[echo willow]

oh

one more thing:

what would 360/360n be?

i might've messed up before or something

[keen aspen]

What do you mean

I'm working it out on paper and I'll show you

[echo willow]

Ight

You've been very helpful, thanks a lot

[keen aspen]

Yep nvm its 4 sides

[echo willow]

Thanks man

Night

[keen aspen]

Haha gn that's no easy problem by any means

[lapis night]

if i have a right triange on the complex plane with sides 1 and i, is the hypotenuse length 0?

[mint sandal]

No

Its sqrt2

Or sqrt2 e^3pi/4 as a complex vector

[celest swan]

Anybody knows the equation defining this body?

[copper valve]

t!wiki Kummer surface

[loud cedar]

📖 | ** https://en.wikipedia.org/wiki/Kummer_surface **

[copper valve]

@celest swan

[celest swan]

Thanks!

[copper valve]

np

it's hard to say what equation defines it, cause it sits in projective space, not euclidean space, and also I don't know enough algebraic geometry :p

[celest swan]

I caught that after reading a bit

[simple geyser]

http://prntscr.com/ibouan

[upper karma]

==15*1.2

[charred spear]

18

[upper karma]

@simple geyser they are

B

[simple geyser]

http://prntscr.com/ibpb84

http://prntscr.com/ibpbds

<@&286206848099549185>

these are my last two problems i made alot of progress but i dont understand these

[halcyon quest]

In the first problem, you are essentially being told to find the projected height. You're given the projected width and the original width. I'd start by finding a ratio from those two values.

That ratio is the new projection divided by the original. Again, since you're given the width for both the new projection and original.

Then multiply the old height by the new scaling constant (new/original).

[simple geyser]

done

thanks for your help

and everyone who else helped me

[upper karma]

dont give me an answer just tell me how to do it plz

[zealous patrol]

lel

[copper valve]

8/(2x-2) = (6+8)/(3x)

[brazen roost]

First and foremost

Google and wolfram alpha exists on the internet

[copper valve]

solve for x

[brazen roost]

It's free

[zealous patrol]

no

[brazen roost]

And it's pretty fucking useful

[zealous patrol]

bad

[copper valve]

discord is also free

[zealous patrol]

^^

[copper valve]

and pretty useful

[zealous patrol]

^^

[upper karma]

thats why im here

[zealous patrol]

^

[copper valve]

😄

[brazen roost]

Yeah

If that problem is from khanacademy I need to stab someone

[upper karma]

no its not

[copper valve]

@upper karma similar triangles have equal ratio of matching side lengths:
(2x-2)/8 =(3x)/(8+6)
solve for x

[copper valve]

hi

[wild vapor]

@copper valve

soo i do not know where to put this.. so we have to make a model of something which is composed of atleast three solid shapes. could you guys suggest something? the one on the pic is a model of a fountain. our teacher want it to be easy so no half sphere no half anything, and no holes on everywhere.

it can also be anything from, structures to regular everyday objects

[umbral snow]

Snowman?

[celest swan]

If I have a point $$A =(0, m+1)$$ and a point $$B =(n+1, 0)$$, and a point $$C =(\frac{n+1}2, \frac{m+1}2)$$, am I right to say that $$mAC=mBC$$ ?

[celest swan]

<@&286206848099549185>

[slender shore]

@celest swan yes but you will have to prove that, also m is positive since you're dealing with distances

[neon fossil]

wow nice pic

anime girl with math book

very original

[upbeat sparrow]

welp i have a test in geo tomorrow

[copper valve]

what kind

[upbeat sparrow]

and i think ive got the gist i had to stay after to review

uhhh

the word popped out of my head

[copper valve]

algebra

euclid

there was a third one 🤔

differential

projective

[upbeat sparrow]

similarity triangles, and proportions

[copper valve]

Ok

idk if i can do that ;o

[upbeat sparrow]

i legit forgot the new/old thing

and screwed up my whole entire homework

lol

i was so confused

OH

scale factors too

and dilations etc etc

[celest swan]

Ok huuumm

<@&286206848099549185> If I have a point $$A =(0, a)$$ and a point $$B =(b, 0)$$, and a point $$C =(\frac{b}2, \frac{a}2)$$, am I right to affirm that $$mAC=mBC$$?

[halcyon quest]

I mean, you can use the midpoint formulas perhaps.

xmid = (x2+x1)/2, ymid = (y2+y1)/2.

between points (x1,y1) and (x2,y2) respectively.

[upper sedge]

Since C is the midpoint on a line between A and B, the slope from A to C is the same as the slope from C to B.

You can prove it using the two points of each line if you'd like.

[celest swan]

I don’t know if AB is a straight line

[upper sedge]

Is this euclidean?

[celest swan]

That isn’t specified.

[upper sedge]

First axiom of Euclidean Geometry: There exists a straight line between any two points.

[celest swan]

But AB might not be straight.

I mean

The curve AB

[upper sedge]

What class is this for?

[celest swan]

I’m a self learning person

[upper sedge]

Alright, what are you studying specifically?

[celest swan]

Logical operators, naïve set theory

I must prove that and formulate it with proof symbols (forall, if, exists, and, etc)

[upper sedge]

xD

I was about there.

[celest swan]

I dou

Ok

Is there a subdivision of that in particular?

[upper sedge]

Sure, there are non-euclidean geometries which exclude or replace euclidean axioms.

[celest swan]

I mean i could prove it through mere algebra easily but that ain’t the point

Vectors?

[upper sedge]

If you're asking the questions in terms of geometric relationships, you need the axioms of your system of geometry before you can construct a deductive proof. Without them, what are your premises?

[celest swan]

Is that question rhetorical?

[upper sedge]

Somewhat.

[celest swan]

Mmhf

Well

We always do

So basically i want to prove the curve AB is linear right?

@upper sedge ?

Ok ima do some more research on these axioms tho

Ill prolly be back Tomorrow

[upper sedge]

Very good plan.

[celest swan]

Mmmmh

I was expecting more than 5 axioms to learn lol

[upper sedge]

Nope!

It's a pretty nice system.

[celest swan]

Yeah but like

It ain’t like the imaginary numbers 1001 axioms

It’s figurative

Lol

[upper sedge]

They have definitions. =P

[celest swan]

Well ok so

IfAB curve is straight, a circle passinge by A B and C has an infinite radius mmh?

[upper sedge]

My dude, what are you doing?

[celest swan]

Im trying to understand the consequences of the axioms

And

One of the consequences is that if you have 3 points that aren’t aligned, you can draw one and only one circle from them no?

Where the 3 points touches the circumference

[upper sedge]

perimeter

But I see what you're doing, now.

[celest swan]

Perimeter

So

[upper sedge]

Look at what you've said, though.

[celest swan]

Yes

[upper sedge]

if you have 3 points that aren’t aligned,

The 3 points you have are aligned.

[celest swan]

We don’t know it they are

[upper sedge]

By the points you've given, we do.

[celest swan]

Can you elaborate?

Oh! No we misunderstood

[upper sedge]

He's not accepting that they're collinear, yet.

[celest swan]

I don’t know if they’re collinear

[upper sedge]

$$A=(0,a)$$, $$B=(b,0)$$, and $$C=(\frac{b}{2},\frac{a}{2})$$ are the points he's given.

[upper sedge]

First axiom.

From any two points, there eixsts a straight line.

[celest swan]

Yeah

[upper sedge]

So we now have $$\overline{AB}, \overline{AC}, \overline{BC}$$

[celest swan]

Indeed

Gtg

?

[upper sedge]

Let's see what those are. $$m_{\overline{AB} }=\frac{\Delta y}{\Delta x} = \frac{a-0}{0-b} = -\frac{a}{b}$$. Since A is a y-intercept, we get $$y = mx+b$$ for the line as $$y = -\frac{a}{b}x+a$$. Now, consider it a function: $$f(x) = -\frac{a}{b}x+a$$. $$f(0) = a$$, $$f(b) = 0$$, and $$f(\frac{b}{2}) = \frac{a}{2}$$. Since all points exist on the line generated by f(x), they're collinear.

[upper sedge]

Since they're collinear, no circle.

I mean, if we're in projective geometry, are we globally or locally Euclidean?

Maybe I misunderstand.

Doesn't projective space have more points than Euclidean space? Are we talking about a Euclidean subspace of a projective space?

I mean, I'm not an expert.

I just figured that's what you were intending to do.

Why go through the trouble of forcing Euclidean axioms to ignore them on a projective plane if we don't intend to see that they apply to a euclidean subspace of that plane?

[reef cave]

Could somebody help with this

[neon fossil]

it's usually just chasing the numbers around

try to make triangles out of the weird parts

[eager pendant]

@reef cave are you allowed to use trig?

[ruby swallow]

Help me

[thorn talon]

For the parellelogram question

DA and CB should have the same slope right?

And it looks like positive slope

Right?

[ruby swallow]

I don't know

[thorn talon]

Well intuitively, when you go from left to right, these line segments seem to be going uphill

2 and 4, have both lines with same slope

And 2 has both lines with positive slope

So 2 right?

[ruby swallow]

Ya I think our teacher puts stuff we haven't learned yet on here

So I don't fully know

[thorn talon]

Have you done stuff related to linear functions or linear equations?

[ruby swallow]

Probably just do remember what a linear equation is

[thorn talon]

Its got concepts related to that

You know how to do the next question?

[ruby swallow]

No

[thorn talon]

actually alternative approach

you have expressions for all the exterior angles

the sum of these must not be over 360

[ruby swallow]

Okay

[thorn talon]

there's enough information to straight out solve for x

not sure why the question doesn't ask for that though

[ruby swallow]

Okay

?

Nvm

[thorn talon]

you can determine DEC using angle sum of a triangle

then FEC by angles on a line

[ruby swallow]

Okay

[waxen gorge]

@reef cave set them equal to line equations

[smoky arch]

I’m allowed to use mid-segment, angle bisector, altitude,median, and perpendicular bisector theorem

[neon fossil]

similarity of triangles

the right one and the top one

[smoky arch]

I’m guessing you’re referring to question 10

[neon fossil]

9

[smoky arch]

Oh yeah sorry

But on what bases is it similar

By which postulate?

[neon fossil]

angles

[smoky arch]

I know but angle-angle-side?

[neon fossil]

call one of them alpha, the other beta, and try to match them

3 angles

well they're not exactly the same triangle, I'd say proportional more than similar maybe

like one is the other times a constant

[smoky arch]

Ok, so I’m going to just hide in a hole

I’m just doing basic geometry

Not yet looking at cos,tan and sin

So it isn’t quite appropriate to use your method

I think I’m just going to ask my teacher

[waxen gorge]

Use similarity

No need for trig

[echo willow]

Herro

@smoky arch eyo I'm doing that shiz

Need help?

[smoky arch]

Well

I got help

I misunderstood the Angle Bisector Theorem

[echo willow]

Ight

[west dune]

@upper karma

Here, I think this is empty

[lucid dirge]

Hi there, so I'm facing real difficulties with these ! I would be glad if someone would help, by help I mean either help me to find formulae to do this question or send me youtube tutorial. Thank you, even for the time you spent reading this 😄

If for any reason you will not be able to read my terrific writing then A = (4.00 , 1.50, 3.00) C = (5.13, 2.58, 5.00)

[neon fossil]

you need the formula of the distance between two points

it's fairly simple

sqrt((x1-x0)^2 + (y1-y0)^2 + (z1-z0)^2)

[lucid dirge]

So, I did this

Before that I just subtracted them for each other, i.e. 4.00 - 5.13

1.50 -2.58

3.00 - 5.00

[neon fossil]

you're taking square roots of negative numbers

you need to take the square of the whole expression

including the negative sign

[lucid dirge]

Um, I don't understand sorry 😭

[neon fossil]

=pup sqrt((4 - 5.13)^2 + (1.5-2.58)^2 + (3-5)^2)

[charred spear]

Query made by @neon fossil
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=sqrt((4+-+5.13)^2+%2B+(1.5-2.58)^2+%2B+(3-5)^2)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[neon fossil]

like so

[lucid dirge]

and that will be the final result?

[neon fossil]

yeah

[lucid dirge]

Thank you very much, may you be blessed !

[neon fossil]

your mistake is that for example in the first one

you're doing

4-5.13

= -1.13

[lucid dirge]

Other guys spend 30mins with me

[neon fossil]

and instead of taking (-1.13)^2

you're taking -(1.13^2)

which is wrong

[lucid dirge]

I supposed to to them all at one go

Am I correct?

[neon fossil]

not sure what you mean

you're leaving the minus sign out of the square

and it should be inside

[lucid dirge]

=tex sqrt(-1.13^2 + -1.37^2 + -2.00^2)

[lucid dirge]

that's what I did for google, and google put minus outside of the brackets

But thanks for your help, I appreciate it !

Have a virtual cookie 😄

🍪 and a hug 🤗

[neon fossil]

👍

[lucid dirge]

If you don't mind, can you just send me to a tutorial which would clarify this for me?

[neon fossil]

https://www.vitutor.com/geometry/distance/line_plane.html

[lucid dirge]

Thank you very much !

Ahrw, I almost don't understand anything from there but will try my best 😭

[neon fossil]

try to find another one then

just google for "angle between line and plane"

you can probably find videos and such

[lucid dirge]

I'm even trying to look through my lecture notes and it;s not there 😭 Bad luck upon me

[lucid dirge]

Nope, I still need help sadly 😦

[lucid dirge]

@neon fossil Sorry for mentioning you 😭

[neon fossil]

?

[lucid dirge]

I still need help, can you go through it with me?

[neon fossil]

just google for "angle between line and plane"
you can probably find videos and such

[lucid dirge]

I watched 3 and still don't get it 😭

[neon fossil]

https://www.youtube.com/watch?v=iPfjrNtYFg0

try this one

in your case, the line is A + tC

and the plane is the xy plane, aka z = 0

[lucid dirge]

So in my case I use : 2.54

or whole equation ?

xy plane is 1,1,0 apparently

[neon fossil]

xy plane is the equation z = 0

it has normal vector (0,0,1)

[lucid dirge]

his R is my A, his t is my C

but what do I do with the CAB angle ?

[neon fossil]

you want the angle between the line A + tC and the plane z = 0\

[lucid dirge]

um.

I don't get it..

[neon fossil]

the line that passes through A and C

you can write it as A + tC

[glacial bear]

AB=BC=CD, AC=BD=AD, what is angle ABC?

[umbral snow]

@glacial bear
That is half a regular hexagon. So, you just need to know the angle of that

[cursive nexus]

Can't assume thtmat

Look for congruent triangles

[glacial bear]

Say the middle point is E. Then only ABE and CDE are congruent triangles, right?

[cursive nexus]

ABC is congruent to DCB

[glacial bear]

i see i see

[cursive nexus]

ABD congruent to DCA

[glacial bear]

yes yes

But still don't know any angle measurement 😦

[cursive nexus]

Sum angles of a triangle and sum angles of a quadrilateral

Also angles that are congruent from congruent triangles

[mossy vine]

and from isosceles triangles

[glacial bear]

180 for triangle and 360 for quadrilateral

Am I not seeing something? I still don't get it.... 🤦

[cursive nexus]

what do you know about isosceles triangles?

[glacial bear]

Both sides are equal

Both bottom angles are equal

[cursive nexus]

2 of the sides are equal

and 2 of the angles are equal

[mossy vine]

write in an angle, call it x or whatever, do some angle chasing

you should be able to write an angle in multiple ways, allowing you to find x and hence the angle you want

[glacial bear]

It doesn't work....

[mossy vine]

it does

because I've done it

I would go through I got it more but I am hella tiref

[cursive nexus]

Let x be the angle opposite of a red side in a red green red triangle

Let y be the angle opposite of a green side in a green red green triangle

write some equations about the angles

solve for x and y

[cursive nexus]

no

don't write any angles in the center

I'm talking about the big triangles

forget there is a center for now

[glacial bear]

🤔

OK, do I start with this?

[cursive nexus]

no don't use any of those angles!

do you see the red line segments?

they're equal when used all across

AC = BC = AD

you don't know anything about the relationships of the lengths from the intersection

what I mean is let y = <CBD

x = <BAD

then now use every geometry fact you know to make a bunch of equations

then you can solve for x and y

which allows you to solve for <ABC

[cursive nexus]

you also know <ABD = x

<BDC = y

and so on

[cursive nexus]

also use that the sum of angles of a triangle is 180

[cursive nexus]

yes

[glacial bear]

OMG it really is half a regular hexagon 🤦

[cursive nexus]

you should really put 3 points when you write an angle

um no

the angles of a regular hexagon are 120 degrees

the angles of a regular pentagon are 108 degrees though!

[glacial bear]

oh right

i thought i remember seeing 108 somewhere

[cursive nexus]

great

[upper karma]

I have no idea

[reef cave]

@ #4045 which one do you need help on

[errant talon]

i can help

soooo

1. It has four right angles

[weak edge]

hey there

can i ask something to do with the thinking behind geometry? or is it too philosophical

[neon fossil]

sure

ask

[weak edge]

I'll start with the pythagorean theorem: The general notion is that joining the end of two perpendicular lines has a line with a length given by the 'law/rule' of the pythagorean theorem. Generall people see this as if this is some property of triangles, separate from the triangles.

I think the better way to look at it is: The pythagorean theorem is the fact you defined to lines to be perpendicular. Look at this picture i just drew: https://i.imgur.com/lZvxQVR.jpg
The moment you called it true on two finite lines being perpendicular, you also called true/implied the pythagorean relationship. In a way, it builds on itself, in analogy it's like: A=B, B=C, THUS C=A, type of logic.

It is the moment you drew the lines you implied/created that, as opposed to the general notion of laws.
from this, i like to also think the rest of the universe is that. This 'logic', without any mystery or magic, just things building on one another. i can give some more examples eg light/shadow relationship and geometry. We can explain things like light relationships with simple lines and geometry, and the stuff in physics in which we can't, we call it 'properties' of the universe. I believe it is the same geometry - just much more complex and not theoretically explained to a full extent yet

[restive void]

so i have to do two things

classify the quadrilateral

and then find x and y

anyone wanna show me where to start

[neon fossil]

4 equal sides

that should tell you what quadrilateral it is

[restive void]

couldnt that be a square or rhombus?

its a rhombus right

since <T

[neon fossil]

squares are rhombuses too

but yeah

[restive void]

but a rhombus isnt a square

[neon fossil]

it's a rhombus that's not a square

so now you can find all the angles yeah?

you know the properties of the rhombus?

[restive void]

uhh

[neon fossil]

opposite angles measure the same

[restive void]

all the parallelogram properites

[neon fossil]

and angles add up to 360

[restive void]

all sides congruent

[neon fossil]

yeah

[restive void]

diagonales are perp right

[neon fossil]

yeah

you dont need diagonals here though

[restive void]

oh

[neon fossil]

just the fact that opposite angles are the same

and that all sides are the same

that will give you an equation for x and an equation for y you can solve

[restive void]

so is 2x-1=127 what i do

wait

no

[neon fossil]

no it's the other angle

[restive void]

hmm?

[neon fossil]

there's four angles, right?

[restive void]

yea

[neon fossil]

two of them measure 127

[restive void]

so v

[neon fossil]

two of them measure 2x-1

and they all add up to 360

that's the equation

[restive void]

would x be 53

[neon fossil]

no

write the equation

[restive void]

set it = to 360?

im not seeing an equation to write, 2x-1=360? i dont do anything with 127?

should i do 2x-128=360?

[neon fossil]

2(127) + 2(2x-1) = 360

[scenic timber]

how do you simplify cos^2(x)+cos(x) to find x? is there an identity that will not loop it

[scenic timber]

omg thanks

[sudden helm]

Hello,
I have to prove that every  quadrangle: S=pr
S - area
p - semiperimeter
r - radius
(circle is drawn in the quadrangle, (don't know how to say that correctly))

[upper karma]

<@&286206848099549185>

[waxen gorge]

Ohhhh this

Havent done it in a while one sec

[sudden helm]

oh

[runic jackal]

what is quadrangle

[sudden helm]

figure with 4 sides. i forgot the name

because square is regular

not in this case

[waxen gorge]

https://proofwiki.org/wiki/Area_of_Triangle_in_Terms_of_Inradius

Meh works fineeee

[sudden helm]

if it's for triangle, it also works for other figures?

[runic jackal]

ohhh

[waxen gorge]

Ye

The figure u want can be split into triangles

[restive void]

how do you find <K? My answer sheet says 125, but i want to know why

given this is a kite

[thorn talon]

Well

The total angle sum should be 360 right?

And M and K are equal

[restive void]

yeaah i figured it out, but uh

wanna help me with another

[thorn talon]

Sure

[restive void]

i have to find everything

[thorn talon]

What do we need to find

[restive void]

side lengths anyway

[thorn talon]

Pythagoras

[restive void]

yeah, but uh

i have to write it as a simplified radical and i kinda forgot how to do that

so say for ST

[thorn talon]

Ok

[restive void]

its root 50

right?

[thorn talon]

Yes

[restive void]

so how do you simplify it

[thorn talon]

What's the largest square factor of 50?

[restive void]

5?

[thorn talon]

Square

[restive void]

wait

no

[thorn talon]

25 right?

[restive void]

yeah

and then same thing for 25?

which would be 5

[thorn talon]

So we can write this as sqrt(25) * sqrt(2) right?

Yeah

Which is 5sqrt(2)

[restive void]

so is it 2root5

your gonna have to help me with this a lot

[thorn talon]

Ok

[restive void]

would sv be the same thing as st?

[thorn talon]

Yeah

Also 5sqrt(2)

[restive void]

wait why

aaaaaa

[thorn talon]

Sqrt(5)

[restive void]

so starting from root 50

[thorn talon]

50

Sqrt(50)

[restive void]

you go root 25 root 2

[thorn talon]

Yep

And root 25 is just 5

[restive void]

then just 5 outside of a root, * root 2

[thorn talon]

Exactly

Yep

[restive void]

and 5 is outside the root why?

oh

because it doesnt have a square factor

for TU its root169 right?

[thorn talon]

Because sqrt(25) gives a nice intergers answer

Root 2 stays as is, nothing we can do about it

Yep

[restive void]

is there an easy way for finding square factors

of a number

[thorn talon]

Technically you don't need to straight away

But its not too difficult, so anyway

Just know the first few square numbers

And see if it divides the number nicely

[restive void]

so thats like

4, 9, 16 etc

[thorn talon]

Yep

First 12 are nice to know

[restive void]

so basically guess and check

[thorn talon]

I guess kinda

There are more systematic ways to do this

[restive void]

explain?

[thorn talon]

But I think just brute forcing it is fastest in general

[restive void]

uh

is root169 already simplified

[thorn talon]

13

[restive void]

didnt get there

lol

[thorn talon]

5 12 13 is a Pythagorean triple

[restive void]

yeah it seemed like it

i was like hey they're all whole numbers

[thorn talon]

Knowing some Pythagorean triples can sometimes be useful

But not necessary

[restive void]

is 8, 9 in a pythag triple

[thorn talon]

No

[restive void]

damn

woulda made life really easy

so for this one

i kinda know what to do

[runic jackal]

@tropic stirrup give me korean geom problems pls im bored

[restive void]

i know the midsegment is half of the sum of both bases

but i dont know the equation to write out

[runic jackal]

2x-1 = (10+44)/2

[restive void]

wait could i do 2x-1 = 1/2(10+44) and double both sides

[runic jackal]

thats what i wrote love

[reef cave]

😂

[runic jackal]

cuter approach

[covert osprey]

http://prntscr.com/ieheex

Any1 know how to solve this?

[tidal grove]

@covert osprey

[covert osprey]

@tidal grove ty

[tidal grove]

👍

[upper karma]

H1L1NG T4TS 33SY

BUT 1T L00K3D H4RD 4T F1R5T GL4NC3

[final prairie]

Don't shitpost

[covert osprey]

http://prntscr.com/ieubs0

[waxen gorge]

well y=-5x

Is basically

like

[umbral snow]

You want the point where the line intersects the unit circle

Or, you can set up a triangle

That might be easier

[waxen gorge]

No circle intersection is easier

:/

so plug in y=-5x

In ur x^2 + y^2 = 1

solve

Then bleh

[umbral snow]

y = -5x
x² + y² = 1

x² + (-5x)² = 1
26x² = 1
x = ±sqrt(1/26)

[waxen gorge]

but it's negative

Since it's 3rd whadrant

[umbral snow]

y² = 1 - x²
y = ±5/sqrt(26)

And yeah you'd take the negative x and positive y

[covert osprey]

http://prntscr.com/ieuoyp I dont know what im doing wrong here

[celest heart]

maybe write the hypotenuse as the square root of a^2+b^2

so in terms of a and b since those are givens

[covert osprey]

@celest heart its not working :/

[celest heart]

a isn't opposite

a is sin

and b is cos

not adjacent

so cos is b

a is sin

1 over a is csc

and 1 over cos is sec

try that

[timber stone]

Why can't a = 2sin theta and b = 2cos theta. Doubt the OP is still here but that's not the right approach

[covert osprey]

@celest heart didnt work

http://prntscr.com/iewm93

[runic jackal]

henlo

draw a triangle to aid u

[orchid sphinx]

Use the Pythagorean theorem

[waxen gorge]

Here

sinx/cosx = a/b

acosx = bsinx

should be obvious from here

[runic jackal]

how is it obvious

hmm

[waxen gorge]

How is not tho

[runic jackal]

how u gon find sinθ from the equation u gave

[waxen gorge]

Easily

[runic jackal]

u just draw a triangle and u can deduce straight away

[waxen gorge]

No

U would get the same answers he has

sintheta = a/b/costheta = a*costheta/b

costheta do the opposite

Then flip them

[runic jackal]

wym

[modest adder]

is the center of an equilateral triangle the location where all three lines perpendicular to the midpoint of each side meet?

[runic jackal]

i guess

[chrome fiber]

orthocenter, yes.

[tidal grove]

7th one pls

Tell me if it's not clear enough

[waxen gorge]

It's simple

U want to make it so the area is a minimum

bh/2

U know it passes through a circle

So if u just draw it out ull easily see iy

[tidal grove]

@waxen gorge wdym area should be min?

N why bh/2? Is it necessary that CA should be perpendicular to AB?

[smoky shuttle]

yo, I have a question

with my GCSE maths knowledge, is it possible to solve this problem I made? or is this polar calculus or some shit

(what is the white area in terms of 'a' and 'A')

[mossy vine]

how is the smaller circle in that being defined?

[smoky shuttle]

both equations are on the left, lemme rewrite them to make them easier to read rq

[cobalt pine]

probably has origin at (-5,-5) and radius 5

or well -a,-a and a

cause a just happened to be 5 in this example

[smoky shuttle]

big one on the top, small on the bottom

[cobalt pine]

or is it a-A for the coordinates?

yeah it is

but yeah can't remember geometry for shit

[smoky shuttle]

imagine there's a square around the big circle, both circles have to be touching that circle at the bottom left corner

[copper valve]

😮

[smoky shuttle]

so like this, no matter the radii

[mossy oriole]

@smoky shuttle Is the problem to find the area of the intersection, because if you yeah its easily solvable

[smoky shuttle]

it's the white area, which is the small circle minus the intersection

so yeah, I guess

but I have no calc or advacned geometry knowledge at all

[mossy oriole]

Ah

ok

so

Yeah this is very doable

Do you know the equation of the white circle?

the equation of the purple is of course (x)^2+(y)^2=25

and the area of the purple is therefore:

[smoky shuttle]

I did it like this so both circles can be adjusted

[mossy oriole]

oh alright

so then, just to let you know, you wont get a number for this problem

[smoky shuttle]

yeah, one rarely does for generalised one

s

[mossy oriole]

So two circles, one with rad A and another with rad a?

[smoky shuttle]

yeah

[mossy oriole]

ok

So the area of the white circle is just:

=tex \pi a^{2}

That's what you're looking for correct?

[smoky shuttle]

excluding the red part though

[mossy oriole]

ah

[smoky shuttle]

just the very white part

[mossy oriole]

ok

Still doable

So the red area is the same as area of circleA - Area of circlea

[smoky shuttle]

what? no it isn't

[mossy oriole]

Oh wait no

Im an idiot

[smoky shuttle]

😄

[mossy oriole]

LMAO

[white harness]

$$\int_{{(x, y) \in \mathbb{R}^2 | x^2+y^2 < R^2, (x-R+r)^2+(y-R+r)^2 < r^2 }}dxdy$$

[mossy oriole]

My brain breaks sometimes

I was thinking about this problem: https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_14

[smoky shuttle]

a guy from a chess server did it without calc btw, so that's what I'm aiming for

mainly because I know no calc

[mossy oriole]

similar problem but with different shapes

[upper karma]

@smoky shuttle do you still want to know how?

[smoky shuttle]

Sure

[upper karma]

basically key is to find the red area

which one are looking for?

white part?

[smoky shuttle]

Yeah, the white

[neon fossil]

you can do it with trig

being clever with angles and stuff

not worth it honestly

[past mantle]

6 and 10

Here's what I got

[chrome fiber]

i don't understand what you've done for #6.

#10 is correct.

[keen aspen]

Essentially for finding the area of a hexagon, its basically the area of 6 equaliteral triangles

So you have the side length of one triangle, so what you need is the vertical height of the triangle which you can find by the Pythagorean Theorem

Since when you construct the triangle, the 10 is the hypotenuse, 100=(1/2)(10)^2+b^2

So can say that the vertical height is √75, now you can find the area of a triangle, 1/2ab

Which is 5(√75)

Now since there are 6 of these triangles, you mulitply it by 6, giving you 30√75 which is roughly 259.8 in. ^2

[waxen gorge]

It's 3s^2sqrt3/2

[past mantle]

I had to use a different way

Using $$\frac{1}{2} aP$$

[covert osprey]

http://prntscr.com/ig8x4w

[umbral snow]

Make a right angled triangle where cotθ = x, can you find sinθ of this same triangle?

[celest heart]

(1-sin^2(x))^(1/2)/sin(x) = cot(x)

[river dune]

Random question, but how do I find the formula for the sum of the degrees of n-polygon, and I think it is related to the formula of number of diagnols, so how to find the number of diagnols in n-polygon.
I remember something as making the polygon of triangles created from the diagnols then summing them up (the 180 degrees).
Can anyone explain that to me and the proof for both formulas?

[waxen gorge]

U di

Diagonals is nC2 -n

So n(n-3)/2

@river dune

Sum of angles is (n-2) * 180

[river dune]

What's nC2, I'm not very good with combinatorics

But how did you get to the sum of angles

[waxen gorge]

bc

For nc2

That means

N*n-1/2

But use n(n-3)/2 instead

Because that's the simplified version

Since u have to subtract n cuz of overcounting

Also for sum of angles

U can split a figure into a bunch of triangles

So 180*n

But the center

The center forms a circle

With an angle measure of 360°

So ur overcounting

So 180*n - 360° = (n-2)(180°)

[proven vine]

Yoyo

I need help

So my teacher started the year of pre calc 11 with radical unit. We have to know trig to some extent and I suck at it

Plz help me

The first question

[upper sedge]

I'm having a little difficulty reading it, but I'll do what I can.

The first thing to know is that if ABCD is a square with perimeter 4, then any sidelength of that square is length 1.

Same goes for the mysterious equilateral triangle.

Right.

Don't mind me.

Fix'd.

I'll proof-read more from now.

Alright.

Now, we need a little more information which our happy triangle and square provide. There's a lot of stuff we don't care about, though.

In this case, we know the side length AD is 1, we know that angle FAD is 45 degrees (half of 90 from the square), angle ADF is 30 degrees (90-60 from the equilateral triangle), and angle AFD is 105 degrees (180-(30+45) because the internal angles of a triangle add to 180).

We have enough information to calculate AF because of the law of sines. We can set it up using anfgle AFD, side AD, angle ADF, and side AF. With this much information, do you know what to do next?

[proven vine]

I got to the point where I found the angles

But I don't know what to do after

To solve AF

[upper sedge]

Do you know the law of sines?

[proven vine]

No because my teacher started with Radical unit and didn't tell us one thing... But now I know. I searched it and have the formula but don't know how to use it.

[upper sedge]

Essentially, if you have two angles and a side, you can figure out any other side because the ratio of the length of a side and the angle across from it is the same for all of the sides and their opposing angles.

That is, in our case the following equality holds:

[proven vine]

Oh yeh I did get the answer but I don't know how to make it into an exact value

[upper sedge]

Oh.

Alright, what did you get?

[proven vine]

Lemme check

0.483

Like I need it in radical form

[upper sedge]

And what did you plug into the calculator to get that?

[proven vine]

Sin of 105 multiplied by 0.5 (sine of 45 degrees)

[upper sedge]

First, do you know the unit circle?

[proven vine]

Nope

No clue

[upper sedge]

Do you know the exact values of sine and cosine of the special angles 30, 45, 60, 90, ...etc?

[proven vine]

Yep for 30, 45, 60

We only learned for those 3

[upper sedge]

Alright, that's good.

Do you know the trignometric identities for the sums and differences of angles?

[proven vine]

If I know it, I know it. I'm not familiar with the math names.

[upper sedge]

I see.

[proven vine]

K

I just searched it

And nope

[upper sedge]

Oh, good.

Well, you need those here.

[proven vine]

We only learned the basics

Of trig cause we only have like 2 questions on it

[upper sedge]

The reason we need that is sin(105) = sin (45+60)

From there, apply the sum identity to split it into sines and cosines of 45 and 60, which we can actually get the values for.

Do you know what I mean by that?

[proven vine]

Could you elaborate?

[upper sedge]

Sure. $$\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$$

[upper sedge]

Take $$\alpha = 45$$ and $$\beta = 60$$

[proven vine]

Yeh ok I think I get it now. Now I know why I don't know this. This is far beyond what we have learned.

Kind of pissed at my teacher.

[upper sedge]

If you have more questions, please feel free to ask.

[past mantle]

What is $$\frac{1}{2} aP$$ used for?

[copper valve]

idk

those letters could be anything

[upper karma]

Area of regular polygon

since this is the geometry channel

[past mantle]

Ik P=perimeter

But whats a

[upper karma]

apothem

[past mantle]

Isn't that a line from the center to the midpoint of a line?

If it is, is there a way to prove for it?

[umbral snow]

Let b, c, d... be the sides of the polygon. You can split the polygon into n triangles, where n is the number of sides of the polygon. Then, the area of the polygon is the sum of the areas of the triangles

A = 1/2 ab + 1/2 ac + 1/2 ad +...
A = 1/2 a(b + c + d...
A = 1/2 aP

[past mantle]

Would I use tan and arctan for angle and side?

For apothem

[umbral snow]

Given what?