geometry-and-trigonometry

[past mantle]

Since $$L\sqrt{3} = 7$$

[white swift]

wat @past mantle

$$\sqrt3$$ can be treated just like any other number

[upper karma]

Idk how to do this

can someone help me?

<@&286206848099549185>

pls it's easy maths

<@&286206848099549185>

[lavish drift]

What part are you having trouble with?

[upper karma]

Idk how to find the scale factor for anything

Im totally confused

[lavish drift]

Scale facor is simply Big/small

So that if you take small and multiple by scale factor you get big

[upper karma]

81/49??

[cursive nexus]

Observe the exponents

cm^2

in^3

[upper karma]

9/7?

[lavish drift]

@cursive nexus Generally I agree with you but its asking for the scale factor for volume

as separate from the others

So I think it might actually want a straight division

[upper karma]

For volume length and area

[lavish drift]

Can't really tell without context

[upper karma]

For all scale factors

What would the scale factor length be?

[lavish drift]

L_big / L_small

[upper karma]

So

9/7?

[lavish drift]

Well the definition of volume for that shape isn't L^2

it's $$\pi r^{2} h$$

[upper karma]

Right

Ahhh I see

[lavish drift]

Wait sorry 49 is the surface area

so is 81

[upper karma]

But idk what the height is

So how would I find it out?

Would I have to find out SFA first?

scale factor area?

[lavish drift]

Well because they are similar you know that if the small one has length L and radius R then the big one has xL and xR

[upper karma]

Right

So

I have to find the radius then?

Pi r^2 = 49

[lavish drift]

$$2πrh+2πr^{2} = 49$$ and $$2π(xr)(xh)+2π(xr)^{2} = 81$$

[lavish drift]

thats the surface area formula

[upper karma]

So I have to deconstruct

Right, thanks man

I found the radiuses

[lavish drift]

Wait how did you find the radius lol

[upper karma]

2 * Pi * r^2 = 49

oh wait

[lavish drift]

Well thats not quite what 49 is

[upper karma]

remove the 2

[lavish drift]

49 is the equation I have above

which is the caps of the cylinder

plus the sides

[upper karma]

ohhhh

so this is hard

we have 2 variables

so how are we supposed to find out?

[lavish drift]

I'm going to be honest with you

Unless I'm missing something very obvious

I'm not sure what the third equality is that you need

[upper karma]

what do u mean?

[lavish drift]

Well you have 3 variables

and two equations

[upper karma]

but since it's 49 and 81

doesn't it mean something?

[cursive nexus]

Your intuition is right

You don't even need the formulas for volume and surface area

Since the shapes are similar

[upper karma]

So what would I have to do to find the scale factors

This is really blowing my mind lol

<@&286206848099549185>

i really need help pls

[tropic stirrup]

Scale factor of length $$= s$$ \
Of area $$= s^2$$ \
Of volume $$=s^3$$

[upper karma]

I know

but i how do I get s?

[lavish drift]

Does that work for a cylinder @tropic stirrup

[tropic stirrup]

Ugh
I'm talking about when the two shapes are similar

No matter if it's a cylinder
If the two are similar it works

The scale factor of the surface area is 81/49 for (1), right

s^2 = 81/49

[upper karma]

ahh okay

so s would be

9/7

sorry i meant just s

[lavish drift]

Oh wow you're right

My excuse is having not taken geometry in many years

[upper karma]

Lol

Thanks @tropic stirrup sorry for pinging, u saved my LIFE

[tropic stirrup]

It's no problem ;p

[keen aspen]

Can someone prove to me how triangle ACB is congruent to triangleABD

[waxen gorge]

Angle

Side

Angle

[keen aspen]

Side BD?

[waxen gorge]

If those two triangles are right triangles

[keen aspen]

No they arent

[waxen gorge]

Wait

[keen aspen]

Yeah I had to sketch it out and BD is not perpendicular to AC as shown in the problem

[waxen gorge]

Well do u have any more info

[keen aspen]

Nope

[waxen gorge]

Cuz one triangle can be 40,45, 95 and the other can be 45,85,60

So they're not congruent

[keen aspen]

How did you say that?

What led you to that assumption

40,45,95 and 45,85,60

[waxen gorge]

I didn't say they are, I'm just providing a case that opposes the question

Cuz that is possible, right?

[keen aspen]

Yeah

[waxen gorge]

If so they're not congurent

Cuz the angles are different

[keen aspen]

Yes but it could also be 45 85 50 or something else

[waxen gorge]

Ya it could

[keen aspen]

I was thinking something like using alternate interior angles and transversals

To try and prove but idk

[waxen gorge]

Bdc will always be 180-adb

Therefore there r a bunch of cases where they're not congurent

As long as bdc=adb it's congruent

[keen aspen]

So they are only congruent when they are right angles

Ohhh I understand now because for example if BDC was 95 that would make angle CBD 40 which means angle B is 85 and angle A 150

And that isn't congruent and that'll apply to any angle values except for when they are the same which is 90

[waxen gorge]

Mhm

Cuz if it's 90°, it's congruent by ASA

[keen aspen]

Now that I thought about it I think that's incorrect

[waxen gorge]

?

[manic glen]

sooo... i'm currently looking at a NASA document for work (as it's the only source i've found explaining euler-to-quaternion conversion for all interpretations of euler angles), and they're using a lot of formulas of this form: "q1 = "sin½θ1 sin½θ2 sin½θ3+ cos½θ1 cos½θ2 cos½θ3"

what does sin½ and cos½ mean?

[dark sparrow]

it's likely that they meant sin(1/2 θ_1), etc

[manic glen]

actually it's not...

sin½θ1 actually ends up meaning ±√ [(1 - cos (θ1))/ 2]

[dark sparrow]

well, thing is

=tex \sin^2(x/2) = \frac{1 - \cos(x)}{2}

[dark sparrow]

that equation i wrote there is true for all x, so what i wrote still stands 😛

[manic glen]

but then what is sin^2

like... i currently have 3 explanations, each of which is in terms of one of the other explanations

[dark sparrow]

$$\sin^2(x)$$ is just $$(\sin(x))^2$$

[dark sparrow]

i assumed you were familiar with trig and the slightly clunky but commonplace notation therein

sorry

[manic glen]

i have $$\sin^2(x) = \frac{1-cos(x)}{2}$$ and $$\sin\frac{1}{2}(x) = \frac{1-cos(x)}{2}$$ and $$\cos2(x) = \cos^2(x)-\sin^2(x)$$

[dark sparrow]

okay

first off

that last one

you're not putting the parentheses correctly

=tex \cos(2x) = \cos^2(x) - \sin^2(x)

[hidden oak]

cos(2x)

ye

also sin(x/2) in the second one

[dark sparrow]

the second, even with the parentheses corrected to $$\sin(\tfrac{1}{2}x)$$, is demonstrably wrong

[manic glen]

actually, yes, it's this: https://puu.sh/z3whu/c63a35c0a4.png

[hidden oak]

so it's the same as the first one

[dark sparrow]

honestly

that ±sqrt() thing is really really cringey to me

[hidden oak]

lmao

[manic glen]

welp... turns out i spent the entire morning trying to figure out a problem that was due to notation - yay

[umbral rivet]

aww

Glad you got it fixed tho :3

[restive void]

okay so lets say this is a square

like

awful drawing

but yes whatever

would <bac be 45?

[white swift]

that's a square

[restive void]

shhh

[runic jackal]

that's a square

[white swift]

it's a square btw

[restive void]

well

okay

yes

[runic jackal]

im aware

[white swift]

we need a new emote, squareBTW

[restive void]

i dont know how to do mathematics

[runic jackal]

yes and that seems like 45°

[white swift]

how'd you get the small o

45o

I don't have that...

[restive void]

the degree?

[white swift]

nono

[runic jackal]

samsung keyboard lel

[restive void]

alt something

[white swift]

alt+o does nothing form e

[restive void]

idk

[white swift]

=tex 45^o

[runic jackal]

SAMSUNG KEYBOARD

[white swift]

got em.

[runic jackal]

LOL hacky

=tex 45^{\circ}

[restive void]

whats the area of a squareeEEEE

[restive void]

lxh

?

[runic jackal]

side x side

lel

side2

[white swift]

side, SQUAREd

:^)))))))

[restive void]

also

is five root three over three just 5

[white swift]

wat

[runic jackal]

no

[restive void]

or would i have to rationalize

the thingy

[runic jackal]

no

[restive void]

then what the fuck do i do

[runic jackal]

=tex \frac{5\sqrt{3}}{3}

[runic jackal]

leave it like that

[restive void]

no i mean

[runic jackal]

dont do anything

[restive void]

five root three over root three

[runic jackal]

OHHH

[restive void]

not just three

[runic jackal]

yeah its five

u said three just now

[restive void]

i see

[eager pendant]

Can someone help me with this?

I tried making a topdown view of the cube and did some weird stuff with pythagoras

but couldn't really get anywhere

[waxen gorge]

Well we want to minimize the distance

Treat one rate of change as r

And another as 2r

The shortest distance is when they're both closest to the center of both top and bottom

So 3r = 10root110 so 2r = 20root110/3 and r = 10root110/3.

Mm

Wait lemme think about it

I have a solution

But it's kinda weird

So

We can imagine this as a 3d graph

Let one be z = (a, b, c) and the other j = (A, B, C)

(where the two beetles are att

We can use the distance formula to get the difference between the points

(DE)^2 = (A-a)^2 + (B-b)^2 + S^2

S is the side length of the cube

(try looking at it in the 3d plane

But we only have 2 differenenxes squared now

So that looks like it would be in the 2d plane

And we have two diagonals

AC and FH

These are equal

And they're also equal to the side length of the cube times sqrt(2)

One bug is always twice as fast as the other

[runic jackal]

by looking at it

hmm

[waxen gorge]

One is (2d - s/sqrt2, 0) and other at (0, ((s/root2) -d)

As for the distances in comparison

So now that we've found the points

We put them into the equation for (PQ)^2

And we gettt

(2d- s/sqrt)^2 + (s/root2 - d)^2

= 5d^2 -3sqrt2ds + 2s^2

Hmmm

How do without calc..

I can't think of a way to find the minimum of a two variable function without calc :(

Welp

Make it look horrific

[waxen gorge]

Orrr

[keen aspen]

@bleh

[waxen gorge]

O that

[keen aspen]

Regarding back to this it is congruent

ABD and ABC

[runic jackal]

is that congruent?

[keen aspen]

I found that angle ABD and ACB are the same and A maps onto it self

[runic jackal]

it still doesnt satisfy

they are similar but not congruent

[keen aspen]

So that means angle ADB and ABC are the same

[waxen gorge]

@eager pendant from that equation ignore d and s and do (-b/2a) on it

[runic jackal]

wait they are not even similar

[keen aspen]

Yeah

[waxen gorge]

Wut

That's flawed tho

[keen aspen]

ADB and ABC?

How are they not similar

[eager pendant]

@waxen gorge that sounds horrific

[waxen gorge]

Well u have to do it that way without calc

[keen aspen]

Angle ABD and Angle ACB are both 45°

[waxen gorge]

(3root2s/10, something)

[keen aspen]

Angle A is the same angle

[waxen gorge]

Plug in (3root2)(s)/10 into it @eager pendant

[keen aspen]

So by deduction Angle ADB and angle ABC have to be the same

[waxen gorge]

Angle a can be anything

[keen aspen]

Yeah but it will always be the same as itself

[runic jackal]

omg somewhere else pls why are there two topics going on

[keen aspen]

Bc it is geometry

Unless you want me to move

[waxen gorge]

Look at that

Waittt

It right ☺

I'm so dumb lol

[upper karma]

yes

[keen aspen]

Yesterday we went off track and was talking about ADC triangle

And I thought about it laying down in my head and realized

[waxen gorge]

But just because that's true doesn't mean they're Congruent or similar?

[keen aspen]

I forgot which one means same length for side

For congruent do they have to be same side lengths?

[waxen gorge]

They're r multiple ways

Search up triangle congruence theorems

[eager pendant]

@waxen gorge given this is basically a standard yr 8 test id doubt that's the only solution w/o calc

[keen aspen]

So for it to be congruent it has to have a congruent side

[waxen gorge]

what's yr 8

Is that like

Alg 2?

/geometry b?

@keen aspen if it has one congruent side it has two have two other congruent angles

[keen aspen]

We already proved that all three angles are congruent

[eager pendant]

idk about America but I think our yr8 is equivalent to the final year of your middle school

[waxen gorge]

Where do u live?...

In 8th grade here, they just learn how to sine cos and tan of right triangles at the end of the year

[keen aspen]

You are trying to prove triangle ABC is congruent to BCD

Like we did yesterday and its the wrong triangle you are looking at

[waxen gorge]

Ohhh

I thought it was the the bottom to the top

[keen aspen]

Yeah lol

So can I say that both triangles are congruent

[waxen gorge]

Ya

[keen aspen]

Sweey

Sweet

Geometry can be a bit of a brain teaser at times thanks for the help though :D

[waxen gorge]

Yw xD

[tawny pewter]

a parallelograms diagonals

one diagonal is 8+x and the otherside of it is 2y+5

the other diagonol is 2x and the other side of that is 3y

how do i evne do this

[waxen gorge]

What do u mean by other side

Can u send a pic?

[tawny pewter]

its bisected

[upper karma]

hm

[tawny pewter]

by

ok

[waxen gorge]

Kk

[waxen gorge]

Ok

Well

Aren't the top and bottom triangles the same

So 2y+5 = 3y

And 2x = x+8

Cuz the angles r bisected

[tawny pewter]

the two diagonols dont equal eachoterh i think

[waxen gorge]

Ik

[tawny pewter]

oh

i tried doing that

i got 8 and 5

and that doesnt work

[waxen gorge]

Well

Is 3y/2x = 2y+5/x+8

Ohhh wait

Just realized it was a rectangle lol

x+y = 13

Sooo

[tawny pewter]

o

thats still 8 and 5

[waxen gorge]

Mmm

Well

U can make a system of equations

Using angle bisection theorem

U see a bunch of triangles with the same lengths

[tawny pewter]

i dont know

[waxen gorge]

https://en.m.wikipedia.org/wiki/Angle_bisector_theorem

Find two equations that equal the same thing

Set them equal

Find another

Solve

Probably using substitution

[tawny pewter]

how do i do that if they have differnet variables

[waxen gorge]

Make the shorter side a and longer b

x+8/2y+5 = a/b

2x/3y=b/a

2y+5/x+8=b/a

[tropic stirrup]

Err

Is that a parallelogram

[tawny pewter]

ye

[tropic stirrup]

If that is, in a parallelogram two diagonals bisect each other

[waxen gorge]

Already did that

Supposedly it didn't work

🤷

[tawny pewter]

the asnwer is x=9 and y=6 but idk

how to get there

[tropic stirrup]

x + 8 = 2y + 5
2x = 3y

3y + 16 = 4y + 10
y = 6
x = 9

[waxen gorge]

M

[tawny pewter]

wat

how u get 3y+16

[waxen gorge]

Multiply both sides by 2

[tawny pewter]

wat

taht doesnt give me 3

im eve4n more confused now

zzz

[waxen gorge]

Ya it does

x =(3/2)y

Plug that in

[tawny pewter]

how do i get that

[waxen gorge]

Multiply both sides by 2

[tawny pewter]

x + 8 = 2y + 5
2x = 3y
3y + 16 = 4y + 10
y = 6
x = 9 how do i even get to the second part

muiltiplay what by 2

where do i get x=(3/2)y

[waxen gorge]

From 2x=3y

Isolate x

Use substitution

[tawny pewter]

what do i plug in for x

or

y

[waxen gorge]

so we have x = (3/2)y

[tawny pewter]

where

where

where

[waxen gorge]

From 2x=3y

Divide both sides by 2

[tawny pewter]

ok

then

i mulitply by 2 and get 2x=3y

[waxen gorge]

What

so you know we have x=(3/2)y

So plug that into x+8 = 2y+5

[tawny pewter]

oh

do i do it the same way for x

how do i find the x

theres a 2(2/3)x

what do i do

do i multiplyu both sides by 3

i fouind it

ty

i got it

tyty

[eager pendant]

no idea how to start with this, though im guessing it has something to do with similar triangles

i know the answer is 80 if that helps, but i dont know how to get there

[hardy meadow]

Well, the acknowledgement of twelve triangles in total, divided by the area of PQR would result in eighty. But evidently that wouldn't be the approach, based upon the granted information. If there isn't anyone else, I'll attempt to help you at a later time. Quite a bit of physics work to complete.

[crude kraken]

How does one represent angles in 3d?

[waxen gorge]

Wha u mean

They're just angles lol

[crude kraken]

I mean like

On a sphere

Or rather

on a circle

you have theta

what would that be on a sphere? Theta + Some rotation factor?

Or are angles 3d as well, in which case they're an angle theta, something sideways, something else sideways making triangular segment?

[waxen gorge]

Dihedral angles?

[crude kraken]

?

[waxen gorge]

Search it up on Google

[crude kraken]

New question, if you take two intersecting planes and cut a third between them that can move around, if you take the third and find the angle, will the angle stay the same as it moves side to side?

nvm this is horribly worded

if you move the green one along its 'z axis' will the angle between red, cross, and blue always stay the same?

[waxen gorge]

I guess

[crude kraken]

I guess the question still doesn't make sense

:/

[waxen gorge]

So

U see those r semicircles

The semicirxle going out fits into the indented portion

So the shaded region is half the circle

The radius is 5/2

[fierce tide]

2.5

[waxen gorge]

So the area is 25pi/4

[fierce tide]

2.5*3.14

[waxen gorge]

So the area of the shaded region is 25pi/8

[fierce tide]

where do you get the 25 from?

[chrome fiber]

5 squared.

[waxen gorge]

The area of the circle is 25p/4

[chrome fiber]

it's uh.

[fierce tide]

o

[waxen gorge]

(5/2)^2 * pi

Is the area

Since the radius is 5/2

[fierce tide]

so

25*3.14/4

?

[waxen gorge]

/8

Cuz it's half the circle

So 25 * 3.14/8

[fierce tide]

9.8125

[waxen gorge]

Ya

[fierce tide]

9.8 is the answer

thx

[waxen gorge]

Yw

[fierce tide]

o no

[waxen gorge]

No

There r a bunch of quarter circles

They add to one circle

[fierce tide]

but they want the shaded area

[waxen gorge]

So the empty spots r a circle with radius 9/2

The area of that is 81pi/4

[chrome fiber]

=text 9^2 - pi * (9/2)^2

[waxen gorge]

The area of the square is 81

So 81 - 81pi/4

[chrome fiber]

=tex 9^2 - pi * (9/2)^2

[chrome fiber]

i can't.

[chrome fiber]

is there a documentation for the bot?

[fierce tide]

you guys are making this way to complicated

[chrome fiber]

oh, latex.

never been good at that.

okay, look.

you have four quarters each with radius 9/2, right?

[fierce tide]

yez

[chrome fiber]

area of a quarter is 1/4th of a circle.

[waxen gorge]

Whenever u see a crazy geomtry problem

[chrome fiber]

so 4 quarters would make a complete circle.

[waxen gorge]

Try to make it into things u know

[fierce tide]

yep

[chrome fiber]

so you get the area of the square.

[fierce tide]

huh

[chrome fiber]

and subtract the area of that complete circle with radius 9/2.

[fierce tide]

oke

[chrome fiber]

and that's your shaded region.

there.

[fierce tide]

so

all i need to do

is

add up the squares

[chrome fiber]

?

[waxen gorge]

Quarter circles

There r 4 of them

4 quarter circles = 1 circle

[fierce tide]

so what then

how do i find the area

with out the d

[waxen gorge]

U know the radius

Which is (9/2)

[chrome fiber]

think about how you can get the area of the shaded region.

[waxen gorge]

Cuz it's half the length of the square

[chrome fiber]

you'd need to get the area of the square, obviously.

[fierce tide]

radius 9?

or 4.5

[chrome fiber]

nope.

[waxen gorge]

9/2

4.5

[fierce tide]

ok

so

14.13

[chrome fiber]

look carefully. the radius of two quarters is 9.

[fierce tide]

4.5*3.14

[chrome fiber]

so that of one would 4.5.

[fierce tide]

14.13

is that the answer?

[chrome fiber]

nope.

[waxen gorge]

No

[chrome fiber]

you have the radius.

[waxen gorge]

pi * r^2 = area

[chrome fiber]

but area is pi * r^2.

[fierce tide]

right

4.5*4.5

[waxen gorge]

Yap

[fierce tide]

20.25*3.14

[chrome fiber]

and this would give you the area of the portion that's not shaded.

[fierce tide]

68.85

[chrome fiber]

yeah.

now you have the area of the square.

and the area of the region that's not shaded.

so you subtract the two to get the area of the shadded region.

[fierce tide]

subtract what?

[chrome fiber]

area of square - area of the circle that you just got.

[fierce tide]

k

so the area of the square is 68.85?

[chrome fiber]

nope.

[fierce tide]

or is that the circle

[chrome fiber]

it's the area of the circle you get when you combine all the quarters.

81 - 68.85 should be your answer.

[fierce tide]

o

ok

its not

the answer

[waxen gorge]

Did u round

[fierce tide]

yep

[chrome fiber]

pi*4.5^2 is not right.

[fierce tide]

i did 12.2

than what is/

[chrome fiber]

it's 63.61.

17.4 should be your answer.

[fierce tide]

17.4

k

lemme try it

finally

its right

[chrome fiber]

great.

[fierce tide]

took long enough

[chrome fiber]

do you understand how you got the answer though?

[fierce tide]

kinda

[chrome fiber]

feel free to ask if there's something you don't understand.

[fierce tide]

ok

ty

[chrome fiber]

no problem.

[eager pendant]

In , points D and E lie on BC and AB respectively such that ∠DAE=10° and ∠ADE=20° . If ∠CAB=∠CBA=50°, then what is the value of ∠CED?
Given that ∠DEB=30°, I did some angle chasing and got to here:

but im not sure what to do next

[chrome fiber]

can you give a better figure?

[eager pendant]

sure

[chrome fiber]

with all the points labeled?

[waxen gorge]

Well so u want to prove the top right triangle and the one below r similar

At this point

Soo

[eager pendant]

well, this was the original figure

[waxen gorge]

20 + x = 180 - (150-x+10)

Nvm

Not useful :p

haha

[eager pendant]

for what its worth, in the diagram ad and ce look like they intersect at right angles

and so i guessed 70, which turned out to be correct

@chrome fiber is this better?

[chrome fiber]

i think you mean ED and BC.

yes, definitely.

lemme think.

[waxen gorge]

Yes so ud just have to prove theyre perpendicular

And how we'd do that ~

Hm

Mm Everytime I try something I always get x=x :/

[chrome fiber]

oh, i got it.

it's kinda messy though.

lemme rewrite.

[waxen gorge]

Kk

[chrome fiber]

waaait, nvm.

i got an x - x.

damn.

[violet nest]

ABC=ADB=90, Ar=t and cs=2t, Find Angle acb

[tropic stirrup]

Hm

I have a not-so-nice solution

at least in my mind, dunno if it works though lemme write it out

Lemme denote O and O' as the center of the circles, small and large respectively

And without losing generality AR = 1 and CS = 2

the foot of perpendicular from O and O' to BD is H and H'

[tropic stirrup]

Eeehhh welp nevermind x'D

[tropic stirrup]

@violet nest you allowed to use tangent addition/subtraction formula?

[violet nest]

Yep do anything just get the answer

[eager pendant]

do you want me to just give you an answer without explanation?

because i probably could do that

@violet nest

[final prairie]

Comic Sans for vertices

Fucking damn it

[tropic stirrup]

$$\tan \angle {\rm OAR} = x,$$ such that $$\overline{\rm OR}=x.$$ \
Then $$\overline{\rm O'S}=2\tan(45^{\circ}-\angle {\rm OAR}) = 2\cdot \frac{1-x}{1+x}.$$ \
So $$\overline{\rm CD} = 2\cdot\frac{1-x}{1+x}=\frac{4}{1+x}$$ \
Now we have $$\overline{\rm BD}^2=\overline{\rm AD}\cdot\overline{\rm CD}=4,$$ and therefore $$\overline{\rm BD} = 2.$$ \
$$\overline{\rm HR} = 2-x,$$ and so $$\overline{\rm AB}=\overline{\rm AR} + \overline{\rm HB} = 3-x.$$ \
From triangle $$\rm ADB,$$ we get $$(1+x)^2+2^2=(3-x)^2,$$ which simplifies to $$x=\frac{1}{2}.$$ \
To conclude, $$\angle {\rm ACB}=\angle {\rm ABD} = \arctan \frac{\overline{\rm AD}}{\overline{\rm BD}}=\arctan \frac{3}{4}.$$

[tropic stirrup]

@violet nest

[eternal orchid]

I posted the solution to this problem in general-1 yesterday when he posted the problem but he ignored it lol.

[tropic stirrup]

Oh ya?

[eternal orchid]

yeah just look up my comment history.

[tropic stirrup]

Yep, just did

[eager pendant]

is there a name for this theorem?

[tropic stirrup]

I don't think there's a "fancy" name for that tbh

[eager pendant]

can i just call it a well-known fact

[tropic stirrup]

Sure

[eager pendant]

done

ty

[tropic stirrup]

You can call it a property of circle, or any kinda shit

np

[eager pendant]

oh apparently its called the inscribed angle theorem

[tropic stirrup]

....why....

I guess x'D

[eager pendant]

okay, as a follow up question (not really)

whats the central angle of a regular polygon?

[tropic stirrup]

I have no idea, I've never heard a "central angle" of a "polygon"

[eager pendant]

oh

apparently its this value of theta

so the central angle of an n-sided regular polygon is just 360/n

[tropic stirrup]

Oh lol

[dark sparrow]

REGULAR

[eager pendant]

oh, forgot to edit my message above

but to my defense, in my book they didn't mention 'regular'

well

do you know how i would prove/disprove this claim?

wait, gimme a moment

For the angle between two diagonals in a regular n-gon to be x, there must be two vertices of this polygon such that they divide the preimeter of the polygon in ratio x : 180-x

hm

i guess i'll ask stackexchange later

[glass zealot]

hey guys

I am looking to learn some geometry

And I need help With this task

[dark sparrow]

[gritty flare]

and no task was poted

:/

[crude kraken]

shh it could still happen

just gotta wait 'till 12:46 today

[dark sparrow]

what have you tried so far and where did you get stuck?

@vast pasture

[vast pasture]

I don't understand how to set it up

[dark sparrow]

okay

so... let's call that point in the very middle Q

as in, the one where the five lines meet

[vast pasture]

k

[dark sparrow]

you're told here that AQ and CQ are bisectors

which i hope is clear from the diagram

[vast pasture]

yea

[dark sparrow]

this means that Q is the meeting point of all three bisectors, even if the third one isn't drawn here explicitly

[vast pasture]

ok

[dark sparrow]

the meeting point of all three bisectors is also the incenter of your triangle

does that need explaining?

[vast pasture]

so

Do I do 5x-3=7x-11?

[dark sparrow]

yup

i hope you can solve that equation yourself

[vast pasture]

4=x?

[dark sparrow]

indeed

[vast pasture]

ty

[gritty flare]

Question: Find 2 numbers, such that it creates a rectangle whose Perimeter is half its Area

and we indeed got result: 6 and 12

but i wondered, if there was a simpler way to do it

@copper valve heyy!! could u help

😄

<@&286206848099549185>

[gritty flare]

ye so there is actually no simpler way

=tex \frac {ab}{a+b} = 4

[gritty flare]

it is a rectangle

no like one side doesnt equal the other

[violet nest]

@tropic stirrup I just saw that and you're so awesome lol you're the best geometry expert in this entire server I hope you stay Alive for long

@eternal orchid really? I didn't see can you post your version again

[gritty flare]

and then i got reciprocal

and turned it

[gritty flare]

yes and 12 and 6 match

but it was too late

q-q

ye

we were on a competition

time had already ended when i realized the answer

[violet nest]

Why is this in geometry

[gritty flare]

LOL

YES A

it is geometry

thats why

i was doing this alone

AND THERE WERE 5 PPL IN MY TEAM FOR CRYING OUT LOUD

😐 so

we failed

couldnt get to second level

:/ but oh well

you wn some you lose more

[eternal orchid]

I wrote it as several comments @violet nest , just look in my post history if you want to see it.

[eager pendant]

Three circles of radius 1, 2, and 3 centimetres just touch each other as shown. A smaller circle lies in the space between them, just touching each one. What is the radius of the smaller circle?

ok

so i marked the centres of the circles

and we get a 3-4-5 triangle

i let the radius be r

and started applying a coordinate system like this

now i could let the coordinate of the centre of the smaller circle be x,y

but that didnt lead anywhere

<@&286206848099549185>

[hazy field]

use descarte's kissing circles theorem

[tropic stirrup]

Hm

That should be fairly easy

[hazy field]

yep

[tropic stirrup]

The three points even make a right triangle so

[eager pendant]

thanks!

[lost elm]

heyy

anypne here

[copper valve]

can u not spam every channel

[slow charm]

:-(

[glass zealot]

hey

u there?

[nocturne estuary]

yes

[glass zealot]

I will send a pic

k?

[nocturne estuary]

yes

[glass zealot]

nex

@nocturne estuary

[nocturne estuary]

yes?

which part is 7.2?

[glass zealot]

ab

[nocturne estuary]

ok

[glass zealot]

it just feels weird that BC is 3.3

[nocturne estuary]

== sqrt(7.2^2 - 6.4^2)

[charred spear]

3.2984845

[nocturne estuary]

you didnt draw it to scale

[glass zealot]

yes I did

I drew it right

[nocturne estuary]

hmm

[glass zealot]

ab 7.2

6.4 ac

[nocturne estuary]

3.3 is right

[glass zealot]

ill draw it again

[nocturne estuary]

draw 6.4 at the bottom

makes it easier

for me anyways

This is roughly what it should look like

[glass zealot]

ill show u how my product will end up like

k???

[nocturne estuary]

sure

i used an online triangle constructor

[glass zealot]

ab is 7.2

ac 6.4

[nocturne estuary]

you said ab is 6.4 and ac is 7.2

thats what it says on your diagram

[glass zealot]

soz¨

im doing similarities

[nocturne estuary]

its not a right angle then

[glass zealot]

ik

im almost done

@nocturne estuary

that is right???+

[nocturne estuary]

what is right?

[glass zealot]

pic

i sent

[nocturne estuary]

which one? when? where?

[glass zealot]

I just drew that

that seems right

[nocturne estuary]

you have no lengths, im confused

[glass zealot]

ab 6.4

ac*

ab 7.2

[nocturne estuary]

i really dont understand why you're drawing it instead of just calculating it

are you told to draw them?

[glass zealot]

ye the task 😛

frist draw, then get BC then AD and CD

lol 😛

so b) pythagoras theorem to solve

c) similarities

[nocturne estuary]

looks right to me

hard to judge it by eye tho

[glass zealot]

Nex

how do I find AB

sdsssssssss __

CD

AB I know, CD is x

@dark sparrow

[dark sparrow]

i'm kinda busy right now

:x

[glass zealot]

okay

anyone else???

[charred spear]

3.2984845

[marsh crest]

Is a trapezoid a parallelogram? I do not think that trapezoids are parallelograms but parallelograms are trapezoids. It’s like comparing apples to fruits, trapezoids can be parallelograms but parallelograms are trapezoids. The definition of a trapezoid according to google’s dictionary is a quadrilateral with only one pair of parallel sides. That means that a trapezoid does not have multiple parallel sides. Additionally the definition of a parallelogram is a four sided figure with opposite sides parallel. A trapezoid is not normally a parallelogram but a parallelogram is always a trapezoid.

how's this lol?

[eager pendant]

if you define a trapezoid as having exactly one pair of parallel sides

then no

but if you define it as a shape with at least one pair of parallel sides

then parallelograms are trapezoids

and trapezoids are parallelograms if they have two pairs of parallel sides

[marsh crest]

    I do not think that trapezoids are parallelograms nor that parallelograms are trapezoids. Trapezoids cannot be be a parallelogram or vice versa. The definition of a trapezoid according to google’s dictionary is a quadrilateral with only one pair of parallel sides. That means that a trapezoid does not have multiple parallel sides. Additionally the definition of a parallelogram is a four sided figure with opposite sides parallel which is more than one pair of parallel sides. Trapezoids with two pairs of parallel sides are no longer trapezoids but parallelograms. Imagine you were looking for a new dog and you see that a bulldog and a french bulldog are for sale. Sure they are both bulldogs, but a bulldog is not a french bulldog. Furthermore a french bulldog is nothing like a bulldog, frenchies are maybe 30 pounds while the bulldogs are over 50!

[eager pendant]

hm

are you trying to write a paragraph on why a trapezoid isn't a parallelogram?

[marsh crest]

yeah

i know the teacher thinks trapezoids are parallelograms so im going against her

[eager pendant]

well you could make your paragraph much more concise

[marsh crest]

i need the filler

[eager pendant]

why

[marsh crest]

Answer the question with support in one (large) or two paragraphs...infuse geometry and comedy in the essay...make me laugh. I'm not sure how many points it will be worth...

[eager pendant]

.>

[marsh crest]

it needs to be funny!!

[neon fossil]

ew

[marsh crest]

lol

its for extra credit lol

i need this

oof

i currently have like a B in the class, if i get like 2 more points I think i'll have a b+!!

[eager pendant]

as stupid as this question might be, what subject is this assignment for?

[marsh crest]

Hn geometry

[eager pendant]

i mean, all we can probably help you with is Answer the question

as for ...with support in one (large) or two paragraphs...infuse geometry and comedy in the essay...make me laugh., idk

[marsh crest]

I have so much hmwk still to do

sigh..

[gritty apex]

In the book I'm studying I'm suppose to prove the law of sin(A+B)=sin A cos B + cos A sin B, using Ptolemy's theorem. However, I have only managed to prove it in case of A + B = 90º. Does anyone can give some suggestion or insight?

[dark sparrow]

ptolemy's theorem?

what's that

...nvm

[gritty apex]

The product of the diagonals of a inscribed quadrilateral are equal to the sum of the products of the opposite sides of this quadrilateral

[chrome fiber]

use a unit circle.

and the angle subtended by two chords at the circumference is half of that subtended by the radii at the center, given the two chords and radii meet.

unit circle, O is the center. ABCD is the cyclic quadrilateral.

i've used the fact that sine of an angle is half of the chord that subtends twice of that angle.

so, sin(α) = 1/2 * chord that subtends 2α (which is BC).

same goes for sin(α + β) (which is angle BAD in the figure).

sin(α + β) = 1/2 * chord that subtends 2(α + β) (which is BD).

you can figure out the rest on your own.

actually, you need this definition only for sin(α + β), since you can get sines and cosines from triangles ABC and ADC, since they are right angled at B and D respectively (diameter subtends a right angle at the circumference).

[gritty apex]

I looked at the book but I couldn't find so far this theorem "fact that sine of an angle is half of the chord that subtends twice of that angle.". Is there any chance that you could direct me to where I could find proof of it?
And thank you for providing a solution to my problem.

Never mind, I managed to get this part by myself. Once again, thank you.

[junior scaffold]

how do u do this

[dark sparrow]

what have you tried so far and where did you get stuck?

[hidden oak]

how many servers are you asking this in lmao

[dark sparrow]

@junior scaffold?

[hidden oak]

he got the answer

[dark sparrow]

ಠ_ಠ

[long harbor]

is this the right place to ask ab trig

[neon fossil]

probably either here or precalculus or questions

[long harbor]

ok so i have a questrion ab ambiguous cases

with trig

[thorn talon]

?

what's your question about it

[primal tide]

In this problem:

The first thing they do the solution is say this:

how do we know that m∠AFE + m∠AEF = m∠DFC + m∠AFE

This isn't immediately clear to me

[thorn talon]

A and D are right angles

[primal tide]

yeah

[thorn talon]

AFE + AEF = 90

cause it's a triangle

[primal tide]

oh

[thorn talon]

so triangle sum

[primal tide]

I'm dumb

I see it now

thnx

[restive void]

how do you find tan a, tan b?

tan is opposite over adjacent right?

[crude kraken]

Yep

[restive void]

so a

24/7?

[crude kraken]

Yep

As long as your teacher wants it in radians

[restive void]

unsure as to what he wants

[crude kraken]

As it's a triangle I would assume degrees

So multiply that by 180/pi

[restive void]

okay, we havent learned that

im assuming he wants radians

wait, just write the answer as 24/7

[crude kraken]

^

And B is 7/24 :0

:0

[restive void]

adjacent can never be the hypoteneuse right?

[crude kraken]

^

[restive void]

so 25/24?

i see

[crude kraken]

== tan(pi/2)

[charred spear]

1.63312394e+16

[restive void]

why though?

[crude kraken]

🤷

tan(c)?

It's basically asking for the slope of a vertical line

[restive void]

oh, so the hypotenuse cant be in the equation at all?

is that only tan?

what about cos

[crude kraken]

Cah

[restive void]

what if the adjacent angle is the hypotenuse

[crude kraken]

adjascent/hypotenuse

[restive void]

so i can do hypotenuse over hypotenuse?

[crude kraken]

no

[restive void]

for cos

[crude kraken]

Pick a random angle

Adjascent is the side next to that angle that is touching the 90 degree angle

the hypotenuse is the side next to it that isn't touching the 90 degrees

[restive void]

so the only thing that cant have hypotenuse

is tan

i see

teacher didnt explain that welll

my teacher is such a nice guy, not a teacher lol

hes smart

but

can't teach very well

[crude kraken]

Also worth noting that sin, cos, and tan have hidden beauty that sohcahtoa doesn't go over :0

[restive void]

uh

is it like

good to know

[crude kraken]

Sotto doesn't want me to cover unit circle D:

and x coordinates and y coordinates and slope D:

[restive void]

welp

complicated stuff sounds like

[crude kraken]

it's really not

[restive void]

ill stick to the simplier lol

[crude kraken]

in fact, I'm gonna go over it anyways

:0

but it's so coool

and things will make so much more sense when you get to later points

especially if the teacher is bad as per se

[restive void]

but if its too complicated, will it get in the way of future stuff i might learn about sin tan cos

or is it like

getting the degree in the triangle

[crude kraken]

It's not complicated

and it will make it awesomer

[restive void]

degrees

i see

[crude kraken]

Degrees and radians is easy stuff tho

Degrees = Radians * 180/pi

Radians = Degrees * pi/180

really all there is to it

[restive void]

tan(a) 35/12?

[crude kraken]

Yup

[restive void]

k good

[upper karma]

==atan(25/12)

[charred spear]

Error: Unknown name: arctan