geometry-and-trigonometry

[tidal notch]

ummm

first the bottom one

the bottom picture

that looks more confusing

but its probably not

[thorn talon]

Its a trapezium and rhombus right?

[tidal notch]

yeah

well

i see triangle

rectangle

and trapezoid

[dark sparrow]

i'm actually a bit busy right now, sorry

[crude kraken]

If big square is four, radius of each circle is 1

Length of each corner to the edge of the circle is thus sqrt(2)-1

Length across big square is 4sqrt(2) diagonally, and we know that the circles each take up 2+sqrt(2)-1

So length of one side of the small square is

4sqrt(2) - 4+2sqrt(2)-2

I think

4sqrt(2) - 2(2+sqrt(2)-1)

So yea, 4sqrt(2) - 4 + 2sqrt(2) - 2

Or 6sqrt(2) - 6

How?

If one side of big square is 4, then radius is 1

diameter is 2

😛

So the area is

by squaring the above

108-72sqrt(2)

Or roughly 6.177

You're welcome -2

(tfw I'm not -1)

They have an interesting approach actually

drawing the triangle there

Since length of side is 2sqrt(2), the square length is 2sqrt(2)-2

And if you square that you get .6something...

...

Where'd I mess up D:

[dark sparrow]

sqrt(i) is not -1

i^2 = -1, thoguh

[crude kraken]

^

but i^(1/2) is not

[dark sparrow]

and you didn't mess up, dw

[crude kraken]

But I got like 6.something

they got .6something

[dark sparrow]

== (2 * sqrt(2) - 2)^2

[charred spear]

0.6862915

[crude kraken]

Oh

Must've just been me squaring it wrong :p

(-1)^1/4

== (-1)^(1/4)

[charred spear]

0.70710678+0.70710678i

[marsh rover]

Hi

How to solve this

https://imgur.com/a/4dx9K

[dark sparrow]

what have you tried?

[marsh rover]

let BQ = x

let QC= y

x+y / 2 : x I dont know i cant do it

[dark sparrow]

are you allowed trigonometry?

[marsh rover]

Anything will do

[dark sparrow]

then you can do this

[marsh rover]

the ratio is a trig function

[dark sparrow]

indeed

[marsh rover]

Oh when I saw this problem i thought it would be a number just some number

[dark sparrow]

angle BPQ = 180° - 2θ
so φ = 90° - (180° - 2θ) = 2θ - 90°

tan(θ) = 2

is that clear so far?

@marsh rover

[marsh rover]

Im not sure

I havent learn trig identity yet

But is the answer really a trig function? Why cant it be a number

[dark sparrow]

it is a number

and i was about to explain how to find it

=tex \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} = -\frac{4}{3} \ \tan(2\theta - 90^\circ) = -\frac{1}{\tan(2\theta)} = \frac{3}{4}

[marsh rover]

hold on

I remember posting this question on quora

x2+y2=(x+2x−y)2⇒8x2−6xy=0⇒4x=3y⇒xy=34x2+y2=(x+2x−y)2⇒8x2−6xy=0⇒4x=3y⇒xy=3/4

[dark sparrow]

^ for exponents

think you meant x^2, rather than x2

:p

[marsh rover]

I dont understand why the Hypotenuse is x + 2x - y

The answer the guy gave is 3/4

which is quite close if you draw the thing on paper and actually measure

Is there some theorem?

[dark sparrow]

well i mean, the side of the square is 2x

and so QC = 2x - y

which i hope is clear

[marsh rover]

Triangle BPQ's hypotenuse

is 3x-y?

[dark sparrow]

yes

[marsh rover]

why is that

[dark sparrow]

do you understand why QR = 2x - y?

[marsh rover]

QC = 2x-y yes

[dark sparrow]

no, i'm asking you whether you understand why QR = 2x - y

[marsh rover]

oh sorry um i dont

i know why QC is 2x - y but not qr

[dark sparrow]

i take it that you also don't understand why PR = x, then

[marsh rover]

yea

but why

[dark sparrow]

if you draw two tangents to a circle from the same points, the line segments you get will be equal

[marsh rover]

https://www.cut-the-knot.org/Curriculum/Geometry/GeoGebra/Tangents.gif

Like that?

[dark sparrow]

that got cut off

so idk

[inner lantern]

Can anyone create some problem?

And please with the answer, I will trick my friend.

[dark sparrow]

what kind of problem

[inner lantern]

some advanced problem

[dark sparrow]

that's still too vague

[inner lantern]

if you refers “problem” to something like triangle or quadrilateral

[tropic stirrup]

This @inner lantern

:')

[dark sparrow]

@tropic island can you repost those definitions and the problem? i forgot which was which and which one you needed

well, theorems, not definitions

[tropic island]

Wait gtg

Brb is like 10 secs

Theorem (Leg opposite to 30° ∠): In a right triangle with a 30°-angle, the leg opposite to the 30°-angle is half of the hypotenuse.
Proof: Extend AM to point B so that AM = BM. Then, ΔAMC ≅ ΔBMC by SAS and AC = BC by CPCTC. But m∠A = 60° and ΔABC is equilateral (as isosceles triangle with 60°-angle). Thus, if AC = a, AM = a/2.
Theorem (Leg opposite to 30°∠ Converse): If one of the legs in a right triangle is half of the hypotenuse then the angle opposite to that leg is with 30°-measure.

This goes with the first one

[dark sparrow]

mhm

and what problem were you doing again?

idr which of these you actually needed there

of these two, the "direct" theorem can be summed up as "if angle = 30°, then leg = hyp/2" and the "converse" can be summed up as "if leg = hyp/2, then angle = 30°"

[tropic island]

In △ABC, m∠CAB = 60° and point D ∈ BC so that AD = 10 in, and the distance from D to AB is 5in. Prove that AD is the angle bisector of ∠A.
Which theorem used to prove problem?
a) Leg opposite to 30°∠
b) Leg opposite to 30°∠ converse
c) Median to hypotenuse th.
d) Median to hypotenuse th. converse

[dark sparrow]

ok gimme a sec

[tropic island]

I'm really sorry about last week...

@dark sparrow You still there?

[dark sparrow]

ok so here's a sketch

i am

i was making that

[tropic island]

Yeah I got that picture

[dark sparrow]

so since you want to prove that AD bisects angle A, and you're given angle A = 60°, that means you want to prove angle DAB = 30°

[tropic island]

And you know DHA is 90°

[dark sparrow]

yeah, you know ADH is a right triangle

in this case, you have leg = hyp/2, and you want angle = 30°, so you'll need the converse

[tropic island]

So it's B

Thanks!

[dark sparrow]

as i said earlier, i feel like the forceful division of a single biconditional ("both-ways", if you will) theorem into a "direct" and a "converse" is idiotic

[tropic island]

Yeah!

Me too

Someone actually agrees with me!

[dark sparrow]

because like, both this labeling and the reverse are equally valid

[tropic island]

Yep

[dark sparrow]

personally, i was taught this fact as one theorem which went both ways

it required a slightly lengthier proof, but it was better in terms of organization

and honestly, that's how all the theorems i've been taught have been presented

[tropic island]

Yep

Wait you're a graduate right?

[dark sparrow]

i'm in my first year of uni

[tropic island]

Oh

Wow Joey's thing was deleted

XD

Somehow that just made my day!

[dark sparrow]

and i'll let you in on a little secret: in uni, none of these biconditional theorems are broken up

if an implication (an "if this, then that" statement) only goes one way, it's presented as such, but otherwise, it's like "these statements are equivalent: 1. [blah] 2. [blah]"

i've had a few theorems stating the equivalence of three statements actually

which would be a nightmare to break up into one-way implications

since you'd have 6 of those

1->2, 2->1, 1->3, 3->1, 2->3, 3->2

[cobalt folio]

Does anyone remember the matrix for symmetry with the yz plane? in R3

Can't find it on my notes and I can't find it on Google

does symmetry mean the same as reflection?

We use "simetria" in Spanish I never heard of Reflections

[dark sparrow]

like, mirror image?

diag(-1, 1, 1) i'm pretty certain

[cobalt folio]

yeah that was it I ended up finding it online

but thanks

[upper karma]

kj,h

[marsh rover]

What is your answer?

[dark sparrow]

what've you tried, and where did you get stuck?

[marsh rover]

Well i tried and my answer was 60 something

i just want to confirm

I used similar triangle

[dark sparrow]

8 cos(θ) = 8 sin(θ) + 6 cos(θ)

8 sin(θ) = 2 cos(θ)

tan(θ) = 1/4, so cos^2(θ) = 1/(tan^2(θ) + 1) = 16/17

area = 64 cos^2(θ) = 64 * 16/17

== 64 * 16/17

[charred spear]

60.23529412

[dark sparrow]

looks correct

[marsh rover]

ok

[dark sparrow]

does my solution need explaining?

[marsh rover]

It's okay I can handle it

Can I just post 1 more Question

I know the answer and saw the solution just dont get it

[dark sparrow]

ok

[marsh rover]

actually make that 2 geometry questions

[fluid hawk]

@marsh rover

tan(ABM) = AM/AB = (AB/2)/AB=1/2

second one is either too hard, or I'm just dumb. 🤷

probably the latter

[tropic stirrup]

B

13

[fluid hawk]

how tho

if i'm proving congruency, then yea, i can see it now :+1:

[tropic stirrup]

draw a foot of perpendicular from F to AD and call it H
then by ASA congruence (AD = FH, FHE = ADG, HFE = DAG)
triangles EFH and GAD are congruent
AG = 13 by pythagorean theorem, and therefore EF = 13

you shoulda seen it earlier tho 😩 👌🏼

[fluid hawk]

which channel do i go if i need to ask people about college math

[tropic stirrup]

#question-anynumber

[fluid hawk]

cool

which lvl math do most people here talk about?

college?

[dark sparrow]

all kinds tbh

[fluid hawk]

oh cool

[marsh rover]

The first question I posted answer is E while the second question answer is 13

[fading thorn]

how do you proof this determinant identity for an nxn matrix?

oh wrong channel i guess

[dark sparrow]

what are you taking as the definition of the determinant?

[fading thorn]

have it here, one moment

these 3 properties, n is element of {1,2,...}, r is e. of R; v1,v2,...,vn,w are e. of R^n; i is e. of {1,...,n}; j is e. of {i+1,...,n}

[fading thorn]

any idea?

[red mango]

the levi civita symbol has the same symmetry properties as the determinant

Maybe you can exploit that somehow

[upper karma]

@fading thorn just decompose in your basis and use the properties there is strictly nothing else

you will get a sum on the n-uplets by the n-linear property

and using alternate you will reduce it to the sum on S_n

[urban sonnet]

Can someone help me?

I suck at maths rip.

[round isle]

Yes

[urban sonnet]

Okay, if it is possible @round isle please walk me througgh it.

It isn't super hard, it should be relatively easy for you but i'm just a bit dumb aha. 😅

Question 4 please.

[autumn wind]

how can the squares of a tesseract move?

[half oasis]

Could someone explain to me why AB/sinC = 2R ? Where R is the radius of the inscribed circle and the angle C measures pi/4 (45 degrees)

[dark sparrow]

uhhh

i don't think it does

AB/sin(C) would be greater than AB

which even 2R pretty clearly isn't

[half oasis]

the answer is R = 1 , i got it from the answers of an exam

i drew the diagram maybe that's why it looks wrong

[dark sparrow]

what was the question?

[half oasis]

Find the radius of the inscribed circle of the triangle ABC. where AB = sqrt(2) and C = pi/4

it's from the Romanian baccalaureate

[dark sparrow]

wat

what you're saying makes zero sense

what triangle are you applying the pythagorean theorem to?

what point are you calling D?

ahem?

C = 45°

the problem says so

ABC can't be equilateral

[spiral lintel]

Question : how do you solve for the big triangle created with the diagonals of an isos. trap.

[spiral lintel]

The angles

[turbid bear]

How would you solve this problem?

I think splitting it into triangles is how you do it but I’m not 100% confident

[white swift]

what are you trying to find?

what is this drawing

== 90+51+12

[charred spear]

153

[white swift]

😄 ?

[round isle]

What do you even need to find in this question?

[turbid bear]

Length of A to D using trigonometry

[white swift]

what is b

like is this a triangle?

[turbid bear]

It’s a non-straight triangle
@white swift

[white swift]

that's a very interesting definition

is it curved?

[turbid bear]

Can’t say it’s curved but the hypotenuse is just split into two bits

[white swift]

draw AB intersecting DC

and do math

with angles

[turbid bear]

Drawing a line from point b intersecting DC?

[white swift]

continue AB to intersect DC

[turbid bear]

Ok I managed to solve by splitting it into two triangles and working out their adjacent and opposite lengths thanks for the help tho @white swift

[white swift]

good job 😄

[whole pumice]

done with my hw, lmk if you need help

[marsh rover]

repost

[topaz valley]

What have you tried?

[marsh rover]

nothing I cant think of any related problem

George Polya XD

[marsh rover]

is that bad

i dont know where to start

[marsh rover]

-hi-

First I let the square side be 3. Hence, DN= 1, NC = 2 (stated)

Next I realise that triangle MDN and triangle NCG are similar

DN/NC = 1/2 so MN/NG = 1/2 etc

my problem then turns into find the length of MD Why? So that I can take 3 - MD to find AM

I let MD = n

Since The black triangle is an isosceles

BC + CG = MN + NG = 3 + something

that something is 2k because of the 1/2

so 2k+3 = 3(sqrt(k^2 + 1))

solving for k gives me 2.4

ratio is hence (3-2.4) / 3

= 1/5

the problem was hard really a lot of insights had to be made

[surreal bolt]

would ya like a separate solution?

I'll take that as a Yes. heh.

We'll use coordinates. Cartesian plane style.

[marsh rover]

Okay sure

[surreal bolt]

N is the origin.
N (0, 0), D (0, 1), C (0, -2), M (a, 1), B (-3,-2)

Let K be the midpoint of BM ((a-3)/2,-1/2). Note since BMG its isosceles, GK perpendicular to BM.

MN is a segment on a line.
Call it line L.
L has slope 1/a through (0, 0)
or y = (1/a)x.

BG has the equation y = -2.
G is (-2a, -2)

Again:
Let K be the midpoint of BM ((a-3)/2,-1/2). Note since BMG its isosceles, GK perpendicular to BM.

BK has slope 3/(a + 3).
GK has slope (-1/2 + 2)/((a-3)/2+2a)

negative inverses:
3/(a + 3) = -((a-3)/2+2a)/(-1/2 + 2)

3(-1/2 + 2) = -(a + 3)((a-3)/2+2a)
9 = -(a + 3)(a-3+4a) [multiply by 2]
9 = -(a + 3)(5a-3)
9 = -5a^2 - 12a + 9
a = 0, -12/5

M (-12/5, 1); A (-3, 1)
MA = 3/5
AB = 3
tan(theta) = 1/5

I admit to leaving out (forgetting) for a long time that M had a y-value of 1 in this setup. I was trying to have M be a generic (a, b) and terrible expressions happened with nothing falling out at the the end.

But the problem turned out okay after finally noticing b = 1. Hope you got something out of this solution. 😃

[turbid bear]

1=1

[surreal bolt]

... I feel mocked somehow.

[marsh rover]

sorry i was afk for some bit

[surreal bolt]

Br b

[ruby swallow]

Impossible

[dark sparrow]

uhh

yeah you can't

CIM and HAM are similar but that's it

[spiral lintel]

wait the last problem, I got 1/5

I found ABM = 26.56...

1/2**

@ruby swallow can't, you need you know that CI = AH for M to be the midpoint

to&

@marsh rover I got 1/2 but I might be. First I assume that each side of the square is 2, so AB=2 and AM=1 just for proportions.

root of x^2 + 2x^2 = root of 5 times x

tan(cos-1(2/root of 5)) = .5

or tan(cos-1(2x/x(root of 5))= .5

maybe im wrong tho

[marsh rover]

You are assuming AM = 1 as M is the mid point which isnt stated

[upper karma]

Does anyone mind explaining why the sum of the exterior angles for regular polygons will always be 360 degrees?

[rain canyon]

i'm not sure if this is the right explanation but if you divide 360 by n (360/n) , n being the number of sides of a polygon, you'll know what one of the exterior angles of the polygon is
for example, a hexagon has 6 sides and you want to find out what one of the exterior angles is. so you divide 360/6 and get 60. so 60 is one of the exterior angles of a hexagon

[spiral lintel]

@marsh rover is M the midpoint of AD?

[spiral lintel]

nvm

[dark sparrow]

@upper karma imagine that you're on a motorcycle going around the polygon

at every corner, you will have to turn

and each individual turn is equal to the corresponding exterior angle

but if you come full circle that means you've turned a total of 360°

[ruby swallow]

Is this the most effective way to prove this statement

[dark sparrow]

two column proofs 🤢

[ruby swallow]

Is it at least correct

[dark sparrow]

idk what step 3 is meant to be

[pallid ledge]

I can't read it tbh

maybe he means the 2 triangles are similiar ?

[whole forge]

I also need help with that but like when they give u a number then u have find the other pair...eg...AB=5 find CH idk how to do them

[dark sparrow]

?

[whole forge]

What she sent @ruby swallow

[whole forge]

Help me plz

@here i need help

[umbral snow]

The interior angles of a triangle add up to 180 degrees

[whole forge]

Yes

[umbral snow]

So
x + 48 + 76 + 65 = 180

[whole forge]

Okay thx what about the second picture?

[umbral snow]

Similar triangles. The triangle doubles in size for every side. So 10

[whole forge]

That one i didnt get it

:(

The red things what does it mean?

[dark sparrow]

the red marks?

these denote equal segments

[gritty flare]

JR = RK

LQ = QK

thats the red marks

[whole forge]

Thanks i got it

Please explain me one of these

I dont get them

Is the same solving for x?

@here plz

Im sorr

Sorry

[dark sparrow]

don't ping @here

it's disabled, but still rude

[whole forge]

Ok

Help.me plz

[dark sparrow]

okay so which one of these do you want explained?

[whole forge]

1 or 2

[dark sparrow]

ok

i take it that SV is meant to be a median?

[whole forge]

What

[latent turtle]

You just set both of the equations equal to eachother and subtract to find X

The go from there

[dark sparrow]

no.

you can't just blindly set the sides you're given equal to each other

you don't immediately know if they're equal

[whole forge]

@latent turtle i was doing it but i dont think is correct

[latent turtle]

Oh didn't see that my b

[dark sparrow]

do you know what a median is? @whole forge

[whole forge]

Yeah ig

[dark sparrow]

so

[whole forge]

The half

[dark sparrow]

no.

it's the line from a vertex of a triangle to the opposite side which divides said side in half

now, is SV supposed to be a median in that triangle?

[whole forge]

Yes....

[dark sparrow]

okay

so you know RV = VQ

[whole forge]

Ol

Wait

Yes i see

Continue

[dark sparrow]

and so RQ = RV + VQ = 2RV

[whole forge]

Im confused..sorry csn u put the equation in a piece of paper its gonna be easier for me

[dark sparrow]

...how will me writing that on a piece of paper and sending you the pic make any difference?

[whole forge]

Yes it will i will understand the process

[dark sparrow]

RV = VQ

and RQ = RV + VQ

do you understand these two?

[whole forge]

Ohh yes i think i do

[dark sparrow]

RV + VQ = RV + RV = 2RV

do you understand this?

[whole forge]

Kinda but idk what it has to do with the numbers :(

[dark sparrow]

i was about to get to that.

[whole forge]

Okkay continue

[dark sparrow]

so is it clear why RQ = 2RV?

[whole forge]

Yeah

[dark sparrow]

so now we can replace RQ and RV with the expressions we're given

x + 5 = 2(x - 2)

do you understand what i just did?

and do you need help solving that for x?

[whole forge]

Yeah cuz he is smaller he need to be doble?

Ik how to solve for x

[dark sparrow]

what?

[whole forge]

Nothing

[dark sparrow]

ok

[whole forge]

Thanks

I feel bad are u making any profit from this? :(

[dark sparrow]

what, like money?

no

[whole forge]

Ouch...u should make videos on youtube ill watch ur videos and u can get them from there

Lol ur good man

[dark sparrow]

psh, i would, but i sound like a dude and until i fix that i'd rather minimize the amount i need to use my voice

[whole forge]

U can use a program for ur voice tho

[dark sparrow]

that's unlikely to make me sound the way i want to sound, honestly

[whole forge]

But people care more about the content of what you teaching than ur voice...Well im just saying cya i gotta finish this assigment 50 questions

[dark sparrow]

pff

first and foremost i'd be trashed in the comments for being trans, that's for certain

because youtube

[gritty flare]

@dark sparrow btw i think in america they call Median the altitude?

[whole forge]

But u wont tell ur sex gender...u decide tho i just u should be taking advantage from ur knowledge

U also can do like c-khan academy they just shiw their work very clearly and only using a voice

[gritty flare]

@dark sparrow 😗 but dont be that way... look at the opposite... people who actuallywill enjoy your vids 😄 i mean you explain like a beast in math

[whole forge]

Ikr^^

And the vids u put on youtibe ppl on this server will go to watch then if they really need help just create the playlists for the topics and make the kinda short

May be u wont earn millions but is better than not making anything here

[tropic yarrow]

my dumbass cant go fukin geomety

im in calc III lol

anyway how do i scale things around a center poitn

like this but with a square

like that

lol nvm my dumass fogot vectors existed

i got it

[muted sequoia]

Can someone help me with this please?

I'm in desperate help

[muted lichen]

I need someone to go to mathematics in vc to help me on my 6th grade test corrections

[muted sequoia]

This is how much I understand...

[tropic yarrow]

rotte question pls

[muted sequoia]

This

[blazing eagle]

Which part do you think is the problem

[muted sequoia]

I don't understand most of this, I know the basic thing of vectors like beginners stuff

I know that OBA makes a triangle

Q is somewhere between AB near B

OP= 2/5 OQ

I don't really get the "AP produced meets OB at R"

But I think it means that if I draw a line straight through the point AP it'll touch a point in OB

And that's where R is

I'm sorry if I'm asking stupid questions but I really need to learn this

Woops I think I'm asking in the wrong channel

[tawny drum]

Still in highschool, so nothing impressive, but something caught my interest. According to my book this rule follow: https://gyazo.com/4b0da1795fa479d139a11e2d2b946633

[tropic stirrup]

uh

[dark sparrow]

you don't need the circle there 😛

[tropic stirrup]

that's an "of course" thing

[tawny drum]

Yeah, but I'll get to that

[dark sparrow]

perpendicular lines have the property that their slopes multiply to -1

[tawny drum]

That isn't my question

My question is

if the slope is a straight line vertically

that means the slope is infinite

and if the other slope is 0

0 x infinite = -1?

[tropic stirrup]

To be precise, the slope does not exist

[dark sparrow]

the m1 * m2 = -1 thing doesn't apply in that case

[tawny drum]

that doesnt make sense to me

[dark sparrow]

yeah that's the edge case

so no wonder it doesn't make any sense

m1 * m2 = -1 for perpendicular lines if both slopes exist and are finite and nonzero

[tawny drum]

then they should imply that in my school book 👌

😛

thanks for the answer

[tropic stirrup]

They must have said the conditions somewhere in the corner of that book

[tawny drum]

They actually haven't

[tropic stirrup]

:disgust:

[tawny drum]

I can send you a picture, but it's dutch

so you might not like it 😄

[tropic stirrup]

nope

nah nope never I don't speak dutch

[tawny drum]

is it true that lim a -> infinite and b -> 0 with (|a| x -1 x |b|) = -1?

the first -1 is there, because 1 line always approaches from a negative side

[dark sparrow]

...no, if your a and b are independently approaching infinity and 0 then that limit simply doesn't exist

[gritty flare]

Slope is rise/run

right?

[dark sparrow]

it's defined like that, yes

[gritty flare]

lets move to algebra

wanna show you something amazing

[mighty slate]

Hi I have a question about Diamond Hierarchies. I'm trying to understand this paper: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.540.4575&rep=rep1&type=pdf

I can't figure out how to find the parents of a diamond. Section 3.2 talks about parent diamonds but doesn't really tell you how to find them. Figure 3 shows visually where these parent diamonds are located. But I don't know how to explicitly find the offset of these parent diamonds from the split vertex of a diamond

[upper karma]

wtf?

[gritty flare]

@dark sparrow one thing i dont understand

Arent I correct?

[dark sparrow]

i can't read anything you've written

[gritty flare]

shit

ok

so you see that triangle?

I need Angle B

[dark sparrow]

yes

[gritty flare]

so i just used Law of Sine

[dark sparrow]

i understand that

[gritty flare]

sin(51)/11= sin(b)/10

[dark sparrow]

uh huh

[gritty flare]

so i just multiplying everything by 10

[dark sparrow]

uh huh

[gritty flare]

10sin(51)/110=sin(b)

[dark sparrow]

ahem?

[gritty flare]

so

[dark sparrow]

10sin(51°)/11 = sin(b), surely?

[gritty flare]

OH

THATS WHAT I DID INCORRECTLY

shit

thanks

=calc arcsin(10sin(51)/11)

[charred spear]

Failed to parse equation: Invalid syntax at position 8

arcsin(10sin(51)/11)
        ^

[dark sparrow]

mathbot calls it asin

asind if you want degrees

[gritty flare]

=calc asind(10*sind(51)/11)

[charred spear]

44.95055729

[gritty flare]

finally

thank you

[dark sparrow]

but ok

what have you tried so far?

[dusty gazelle]

Umm

[dark sparrow]

much better

[dusty gazelle]

Well, something like this:

The part where the radius touches the tangent would be 90

[dark sparrow]

you've got plenty of named points here.

refer to angles by their names.

[dusty gazelle]

The angle TSR=90

[dark sparrow]

👍

[dusty gazelle]

And the only other part I got is <y=2<x

[dark sparrow]

alright

well, consider the triangle RST

[dusty gazelle]

Okay

[dark sparrow]

you know and TSR, and you know angle TRS

[dusty gazelle]

180?

Sum.

[dark sparrow]

and the three angles in triangle RST have to add up to 180°, yes

this'll let you find x

then y, since you stated correctly that y = 2x

[dusty gazelle]

Yes

[dark sparrow]

and then z, similar to how you found x

[dusty gazelle]

Got it.

[upper karma]

There should be a functions theme for this category because grade 11 is functions

[upper karma]

How can I have the variation of P_b ?

S, P_a and P_b are momentum, and F a force

[umbral snow]

? @upper karma

[dark sparrow]

1. why are you writing in allcaps?

2. care to post the problem that's confusing you?

[upper karma]

Anyone know if I did this right

[tropic island]

In a right triangle the leg opposite to the acute angle of 30° is 7 in. Find the hypotenuse and other leg.

Hypotenuse= [] in.
Leg= [] in.

[dark sparrow]

okay what have you tried so far?

[tropic island]

Hypotenuse is 14 in.

[dark sparrow]

mhm

[tropic island]

I think it's like this...

[dark sparrow]

uh huh

you have enough information to find x from that

[tropic island]

It's not sqrt of 147

I don't get what I'm doing wrong

[dark sparrow]

147?!

[tropic island]

==14*14

[charred spear]

196

[tropic island]

==196-49

[charred spear]

147

[dark sparrow]

okay phew you didn't just concatenate 14 and 7

it is sqrt(147) though

a.k.a. 7 sqrt(3)

😛

[tropic island]

No its not

[dark sparrow]

what do you mean "no it's not"

try putting in 7sqrt(3)

[tropic island]

?

the heck is that

[dark sparrow]

=tex 7\sqrt{3}

[tropic island]

oh 7* sqrt of 3

oh

[dark sparrow]

...what did you put in for an answer that your system rejected?

[tropic island]

Thanks

wait I don't get that tho

[dark sparrow]

don't get what

[tropic island]

sqrt of 147

How you got 7*sqrt of 3

[dark sparrow]

okay you know $$\sqrt{ab} = \sqrt{a}\sqrt{b}$$, right?

[tropic island]

Yeah

[dark sparrow]

=tex \sqrt{147} = \sqrt{49 \cdot 3} = \sqrt{49} \cdot \sqrt{3}

[tropic island]

==49*3

[charred spear]

147

[tropic island]

oh

Smackes head

But why not just do

=tex /sqrt{147}

[dark sparrow]

\

not /

[tropic island]

=tex \sqrt{147}

[dark sparrow]

but i mean, sqrt(147) = 7sqrt(3) so it doesn't really make a difference other than aesthetics

[tropic island]

whats that?

Oh

I see what I was doing wrong

I HAD A SPACE!

[dark sparrow]

🤦

[tropic island]

Smackes head a little too hard

Given: ∆AFD, m ∠F = 90°
AD = 14, m ∠D = 30°
Find: Area of ∆AFD

For this

Do you change it to?

=tex \ (7*7sqrt(3))/2

[tropic island]

=tex \ 7*7sqrt{3}/2

[tropic island]

Damn it

[dark sparrow]

=tex 7 \cdot 7\sqrt{3}/2

[tropic island]

I give up!

[tropic island]

No

It's all over 2

=tex 49\sqrt{3}/2

[dark sparrow]

lmfao it doesn't matter tho

it's a product

[tropic island]

Is that it?

[dark sparrow]

yes

[final prairie]

=tex a\frac{b}{c}=\frac{ab}{c}

[tropic island]

Wait

=tex 8\sqrt{3} \cdot 2 = 16\sqrt{6}?

[tropic island]

=tex (8\sqrt{3} \cdot 2 = 16\sqrt{6}?

[white swift]

🤔

what is happning

[tropic island]

Life

[white swift]

8*2 = 16

[tropic island]

Yeah

[white swift]

how does your $$ \sqrt(3)$$ turn into$$ \sqrt(6)$$

[tropic island]

IDK

[white swift]

WHAT DO YOU MEAN you don't know lol

[tropic island]

What is $ \sqrt(3)$$ \cdot 2

[white swift]

okay so $$8a2=16*a$$

[tropic island]

no

[white swift]

$$a=\sqrt3$$

[tropic island]

wait

[white swift]

what do you mean no

[tropic island]

Oh

Wait

I meant squared

so

• becomes 64

8*

and

[white swift]

=tex (8*\sqrt3)^2 ?

[tropic island]

sqrt of 3 squared is?

tex= sqrt{9}

[white swift]

=tex (\sqrt3)^2 =?

[white swift]

are you legit 😄

[tropic island]

Thats it

IDK

Cyberbully

Jk

[white swift]

https://www.khanacademy.org/math/algebra/rational-exponents-and-radicals/alg1-radicals/v/introduction-to-square-roots

[tropic island]

Done that

[white swift]

have you?

[tropic island]

Yep

I go to Khan Academy first, before comin' here

[white swift]

=tex 3^2=9

[white swift]

=tex \sqrt9 = 3

[tropic island]

OH so it's just nine

So it cancels each other out!

[white swift]

yes so its just 3

[tropic island]

YES

[white swift]

great

[tropic island]

So

If I had to multiply them by two, what whoul I do

=tex 16 \cdot 2sqrt{3}

[tropic island]

=tex 16 * \2sqrt{3}

[charred spear]

Rendering failed. Check your code. You can edit your existing message if needed.

[white swift]

=tex 2*(16*\sqrt3) ?

[white swift]

@tropic island

[tropic island]

Um

I guess

No

=tex 2(8\sqrt3)

[white swift]

oh

well whats $$a*(b*c)$$

[tropic island]

tex= abc

[white swift]

$$2*(8*\sqrt3) \to 28\sqrt3 = 16*\sqrt3$$

[tropic island]

tex= abc

lol

sorry

[tropic island]

No

[white swift]

????

[tropic island]

because

$$8sqrt3$$

[tropic island]

WTH

[white swift]

you need a \sqrt

backwards slash

because wat

[tropic island]

$$8\sqrt3$$

[tropic island]

Thats the number

You need to multiply the whole thing by 2

[white swift]

????????????????/

[tropic island]

$$2(8\sqrt3)$$

[white swift]

$$a(bc)=?$$

[tropic island]

$$32 \cdot 2\sqrt3$$

[shy shale]

hey

does anyone know how to do this

I need help with all of those problems

[thick coyote]

the first one is 45

[shy shale]

send me the work with it on how to get to the answer

what about the rest of the question number

[upper karma]

what

[shy shale]

I need help with all the questions on that screenshot

[upper karma]

each acute angle 45?

[shy shale]

I need answers and the work with it on how to get to the answer

[upper karma]

no way

[thick coyote]

its a scam

surely

[upper karma]

@thick coyote if 2 angles are 45 then 3rd is 90

idk if isocles triangles can have a 90 deg angle

[thick coyote]

no, that has to be 45 the scam is if he really needs these questions to be answered with full working

[upper karma]

yeh

idk how to show work on that

[thick coyote]

well both sides are equal so $$tan{(\theta)}=1, (\theta)=45$$

[upper karma]

yeah

to solve #3 you need trig right

[thick coyote]

no you can use simultaneous equations

but also probs trig lol

[upper karma]

all angles are given and a side

[thick coyote]

for 3 across?

[upper karma]

ye

you can use basic soh cah toa

[shy shale]

for real need help

yeah

[upper karma]

with wot

[thick coyote]

what is this for?

[upper karma]

school

says class copy

i assume ofc

[thick coyote]

yeah i thought so

[upper karma]

what's ur profile pic

[thick coyote]

uhhh cherry slab cake? lol

idek its been that for so long

[upper karma]

but bruh you name is tasty bagel

[thick coyote]

what a shocking betrayal

[upper karma]

🤔

how do i prove something liek this

1 sc

ideas?

[thick coyote]

ummm

just thinking

what is this question for?

[upper karma]

Anyone available?

wait nvm, this question is too stupid lol.

[final prairie]

This is an interesting question

[thick coyote]

anyone made any progress here?

[upper karma]

I didn't

[floral crest]

I think you have to prove a before b.

[shy shale]

Me neither

I worked on a few problems

And then I gave up

[floral crest]

You can prove that n-1/n < n/n+1 but to put the a/b in the middle, you gotta shorten the distance between a/b and n/n+1.

[upper karma]

=tex \frac{a+1}{b+1}-\frac{a}{b}= \frac{ab+b-ab-a}{b(b+1)}=\frac{b-a}{b(b+1)}>0

[upper karma]

the process is increasing so that answers b)

[thick coyote]

the process is increasing?

[floral crest]

This is really interesting.

I think if you start off below 1/2 you always hit one half

the only other numbers that you can hit above 1/2 and not be at n/n+1 are numbers between n/n+1 numbers. Once you hit the next n/n+1 number, the number n-1/n is lower than your a/b

and one half holds, because the next number down that is n/n+1 is 0

[thick coyote]

can you prove that it always hits 1/2

[floral crest]

No I cannot. Or I haven't yet.

[thick coyote]

thats the catch

i cant lol

[floral crest]

I thought maybe thinking about parity would help. Like you first number can't be even/even. But I didn't get very far.

[thick coyote]

was part (a) solved?

[floral crest]

not here.

[thick coyote]

🤔

[floral crest]

@thick coyote well hey. So say you start below one half and you end up with 3/4

subtract one from that in both the numerator and the denominator

you've already hit 1/2

er

subtract one from denominator and one from the numerator

so to be able to hit any n/n+1 number at all, you have to hit the one before it unless you start between the one before and the one that you hit.

[thick coyote]

thats right

[floral crest]

The first one seems to be somewhat related. like b-a is the distance between the numbers on the number line. b has to be greater than a because a/b < 1 . so you're adding the whole distance between them to both the top and the bottom

[thick coyote]

yeah theres a link missing there

[dark sparrow]

k it seems like i'm the one in need of a hand now

since i had to skip last week's geometry class due to an unexpected illness and now the homework seems to contain a thing covered in the class i skipped

so i've got this equation describing what the problem book calls a "projective transformation of the projective line"

and i'm asked to determine when this transformation has 0, 1, 2 and infinitely many fixed points

so far i've rewritten the whole thing a bit more compactly in matrix form

=tex \lambda \mathbf{x'} = A\mathbf{x}

[dark sparrow]

but i'm not sure if this even makes sense

or what lambda is

<@&286206848099549185>

[hard atlas]

Ok, I'm really unsure what I'm doing here, but I'll give it a shot

[dark sparrow]

i mean

if what i've just done makes sense then i think i can take it from there

ish

[hard atlas]

It does look like it makes sense to me

[upper karma]

Can someone please let me know if I did these two problems correctly?:

If you can't see #23, my answer is (-25, -6).

[tropic stirrup]

second one, yes

first one, no

[upper karma]

Thank you. I'm going to redo that problem and send you my answer.

[tropic stirrup]

You've got a good manner on solving math

I like it 👍🏼

[upper karma]

Thank you 😛

[tropic stirrup]

^^

[sullen flint]

area is 20 cm^2, and the perimeter is 18 cm
find a, b

is system

ab = 20
2 * (a + b) = 18

correct ?

[tropic stirrup]

You've set up the equations all good

[sullen flint]

but when i make a = 20/b and plug it into second equation i get mess

[tropic stirrup]

instead,

[upper karma]

Is this correct now, @tropic stirrup?

[tropic stirrup]

simplify the second one and plug it into the first equation

oh

still nope

-18 = -1 + x_2

you did an algebra mistake x'D

[upper karma]

Let me see....

[sullen flint]

2a + 2b = 18
a = -b + 9 ?

[tropic stirrup]

Yes

now put that in ab = 20

I'm assuming you know how to solve quadratic equations

[upper karma]

Is this correct? @tropic stirrup

[tropic stirrup]

Yep it is! Good job~

[upper karma]

Thank you so much 😃

[sullen flint]

i got that b = 5/2, now i plug b = 5/2 into ab = 20 ?

[tropic stirrup]

No problem! ^^

And yes, ambulas

However

Your answer to b doesn't seem correct

[sullen flint]

5/2 or 2.5

[tropic stirrup]

(-b + 9) b = 20

b^2 - 9b + 20 = 0

[upper karma]

@tropic stirrup, last one...is the last answer correct?

[tropic stirrup]

Oh uh yep it is

[upper karma]

Awesome 😃

[tropic stirrup]

^^

[shy shale]

How do I do number 6

?

[versed summit]

not alot of stuff to go off here

[shy shale]

@upper karma

[versed summit]

probably GH

[shy shale]

Idk what it is asking

How do I do it

[dark sparrow]

uhhh

¯_(ツ)_/¯

doesn't seem clear at all what the question is meant to ask

[crude kraken]

AB/GH maybe?

[versed summit]

if its similar maybe its asking you to find each side's corresponding side in the other shape

intuitively

[crude kraken]

AB/AE

?

[versed summit]

yeah i'd go with GH

what is each side paired with in the similar shape

[shy shale]

Im stuck

[versed summit]

Dromas find each side's scaled down version

on the similar shape

like AB -> GH

and AE -> GF

BC->HJ

and so on

does help?

[shy shale]

Can you explain how

I dont understand

[versed summit]

when one shape is similar to another, it means simply that one shape is a mini version of the other

[shy shale]

Explain from the beginning on how I get to the answer

[versed summit]

each side on one shape is bigger than its corresopnding side on the other shape by some fixed ratio

as there are no distances on this shape you have to eyeball it

notice how all the angles are the same on both the shapes

anyway

if you pick a part on the big shape, the side similar to it will be in the same part on the smaller shape

for instance

side AB kinda makes the right side of the roof on the bigger shape

so the side it is paired with is the side that makes the right side of the roof on the mini shape

which is GH

also the angles on each vertex on one shape will be the same in the mini shape

thats how id do it

gtg hope that helps

[shy shale]

Ok thanks dude. That helped.

Appreciated for helping me

Since you gtg is their any other tutorser that will help me @versed summit

[versed summit]

probably

go ask someone in general or something

good luck

[crude kraken]

Did I find the area right?

[shy shale]

Ok

[crude kraken]

Also, I might be able to, but I forgot proofs if it has to be two-column

[shy shale]

If anyone is here can someone help me with 7

[crude kraken]

Two polygons are similar if there is a way to make it so the ratio of one side to another is always the same.

[shy shale]

I neeed clarification understand how to do it

[crude kraken]

For instance, take a line that we cut in half, to get from the first line to the second we divide by two, and that means the scale factor is .5

Or, if we look at it from the smaller line, the scale factor is 2

[shy shale]

Hold up need to take notes on this but explain please

[crude kraken]

Expanding that to a square, obviously squares are similar (their angles are all 90 degrees), but you can find the scale factor by dividing one side by the other

I dunno if you know this but given an angle and two sides you can figure out everything about the triangle but that isn't necessary. You have two sides and an angle, so the third side must follow the pattern, so the two are similar if the scale factors of each line are the same.

One line is 18/12, the other is 15/10, does 18/12=15/10?

[shy shale]

Yeah im not really too sure about that or my geometry teacher taught it.

But barely understand what my teacher said when I took notes in class in a textbook pages

[crude kraken]

You don't need to be concerned with finding the third side and stuff yet ;p

[shy shale]

Ok

[crude kraken]

But did my point get across for the scale factor?

[shy shale]

Yes

?

[crude kraken]

A'ight

So are the triangles similar?

[shy shale]

They do look similar I think?

Idk

[crude kraken]

It's not how they look, the drawing isn't necessarily to scale

Here, let me put it this way

Two shapes are similar if you can multiply all the sides of one by some number to make it the same as the other. That number is the scale factor.

[shy shale]

Oh so its not ok. I thought it looked similar because of the picture how its scaled.

[crude kraken]

:p

[shy shale]

But says nit drawn to scale wow im dumb 🤦‍♂️

[crude kraken]

oh wait I'm doing this wrong

[shy shale]

Bruh read it carefully

Here

[crude kraken]

You're supposed to divide across both triangles, not across the same triangle

all I really did was prove 32 degrees = 32 degrees 😛

[shy shale]

Oh

For 7

[crude kraken]

Yep

So, you want to know if when you multiply both sides of triangle VUW by any value, it's possible to have it becomes triangle SRT

To do that, you see what each side's length is relative to their counterpart

so 12/10 and 18/15

[shy shale]

Wait so the beginning you said was wrong

The one uo top?

Im confuse

[crude kraken]

When I did 18/12 it was wrong

:p

[upper karma]

@shy shale hey, for #7 you say it is similar by SAS

[crude kraken]

oh yeah you have to say it's similar by some rule or another that I never understand

[shy shale]

Wait what yeah I think I said it’s similar but he said its not similar when he said it was wrong

[upper karma]

state 18/15=WU/TR=12/10=VU/SR

32=32

[crude kraken]

well yea, there's the simple answer

[upper karma]

therefore similar

VUW ~ SRT

tada

[crude kraken]

== 18/15

[charred spear]

1.2

[crude kraken]

== 12/10

[charred spear]

1.2

[upper karma]

mee gud why do you need a calc for those 2 fractions fam

[crude kraken]

I don't

But they need to find the scale factor too

[shy shale]

So basically it’s similar and why?