geometry-and-trigonometry

[fickle rose]

start at 0,0 and go from there

[frail temple]

Wdym

[fickle rose]

growth factor is 4 so you multiply y by 4 as x increases by 1

[frail temple]

I need formula

[fickle rose]

what do u mean formula?

like for an exponential equation?

[frail temple]

The process of solving

the x is -3 or what?

[fickle rose]

there is no one value for x

what are u saying

[frail temple]

Idk my prof

[fickle rose]

it’s a function you can put any value for x inside of it

since there is no initial amount start graphing at 0,0

[frail temple]

Is this correct?

[fickle rose]

yes

[frail temple]

The graph?

[fickle rose]

yeah graph is good

[frail temple]

umm.. thanks

[zenith lodge]

🙏❤️

If anyone wants to learn me how to do it then it would make my day 🙈

[zenith lodge]

I figured it out

No need for help anymore 😄

[strange edge]

You have all the school theorems that apply

[heady scaffold]

2e, 0 after first reflection
2e, -6 after second reflection
final answer is 2e, -6

[solemn smelt]

Yo can someone help me with this math?

I’ll do the bonus

[grand scaffold]

I found a solution based on two assumptions:

a) Right-Angled Triangle
b) Isosceles triangle

Please see traingle on right side.

[solemn smelt]

Wait is this for me or another individual?

[grand scaffold]

Anyone who can check my logic.

[solemn smelt]

Interesting.

[grand scaffold]

I understand that angle ABC is not bisected (that is it is not split into two equal parts) based on the fact that the length of line AC is 2 + 3 = 5 Units and 2 ≠ 3 and 2<3)

Based on the iscoseles assumption, length AB = length BC

Finally, I assume that a square is formed within the triangle.

[upper karma]

This is the channel I need

🇾🇴

[grand scaffold]

I did all the mathematics guys...

Please note: I assumed an iscosceles triangle. (I graphically estimated the values of 2cos45 and 3sin45 respectively, but ended up using a calculator for simplicity).

[mossy estuary]

Can you guys help me with this question

1. If Robert has a gyatt at a 45 degree acute angle, what is the length measures?

[grand scaffold]

QUESTION:
Using special triangle:

cos45 = 1/√2

Does anyone know a method to calculate 2cos45 or 10 cos45 by hand (i.e without calculator) ?

(I am simply increasing the amplitude of the graph)

[thick fable]

so you mean that you have $\cos{45^{\circ}}=\frac{1}{\sqrt{2}}$ and you need to calculate $2\cos{45^{\circ}}$?

[somber coyote]

!Yajat!

[grand scaffold]

Yes. Can I do that without a calculator?

[timber cargo]

I though you asking to prove why cos(45) is root 2 over 2

[thick fable]

,, \cos{45^{\circ}}=\frac{1}{\sqrt{2}}\[em] 2\times \cos{45^{\circ}}=\frac{2}{\sqrt{2}}

[somber coyote]

!Yajat!
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[thick fable]

@grand scaffold just multiply by 2 both side?

[honest summit]

sorry for late reply tysm

[woeful glen]

A little late...

Hello @everyone! I am new to the server, so please show me the ropes!

[rocky dome]

can i have help

@woeful glen

[frozen ocean]

do you know what the points are in this given graph?

[rocky dome]

wdym like the figure points?

[frozen ocean]

ye

[rocky dome]

(-2,2), (-1,4), (-1,9), (-6,9), (-7,7)

[frozen ocean]

it asking you to reflect them

[rocky dome]

yeh over that line

[frozen ocean]

switch numbers

[rocky dome]

wdym switch numbers?

[frozen ocean]

you given the points you have to switch them

[rocky dome]

i dont think that is right

yeah?

do u know the answer?

[frozen ocean]

for example, your point gives as (-2,2) what happens if you switch them?

[rocky dome]

2,-2?

[frozen ocean]

!nosols

[lime crown]

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

[frozen ocean]

correct

continue with others

[rocky dome]

hol up i dont think thats gonna be right

[frozen ocean]

[rocky dome]

that wasnt right

now its onto a new question

[frozen ocean]

I'm not done yet

you haven't reflected the other points

[rocky dome]

yes i did

i switched the numbers

[frozen ocean]

look up for Reflecting shapes article on Khan Academy

[rocky dome]

bro u do know that those are set lines

basically reflecting over the x axis and y axis

[frozen ocean]

I know right?

[rocky dome]

those are specific lines that are not "daigonals"

perfect squares

the question doesnt have daigonals

diagonals*

when u do diagonal trick it does work

[frail temple]

(F(x) = 2^x)

[somber coyote]

Janno

[smoky jetty]

can u send your sol? or is there a description givne for it? im just curious

[solemn crescent]

can someone help me with this - surface area of this:

[vast saffron]

Someone's been doing geometry on lord of the rings map

[upper karma]

can anybody please help me with this problem?

[inner vapor]

https://media.discordapp.net/attachments/805086058382491648/1164417867324866581/image.png

how to find domain

[crystal egret]

Whenever you have a root write what is inside

Then >0

Solve to get domain

[upper karma]

😭

[upper karma]

I need help, plzz solve with steps

[lilac otter]

x and y should be of the same sign and x/y should be greater than 1

[hasty folio]

im in honors geometry in eighth grade (bad decision on my part to take that class) and i dont understand angle relationships, like with the correspondant angles and that stuff

[molten oriole]

If I have a triangle with EC+DB=min(DB+EC), ∠BDC=90* and ∠BEC=90*?

Or not?

[upper oracle]

Why don’t they do this for a midpoint symbol, I hate writing b is the midpoint of ac over and over

[desert gust]

The tangent lines that intersect f and g at f(x) and g(x) are parallel lines. f'(x) = g'(x) can be true at a point x for an F and G of different degree.

if f'(x) = g'(x) for every x, then f = g + c, where c is some constant value in R. Meaning F is just a copy of G translates up or down by c

If F and G are first degree, i.e. on the for y = mx + b then yes, f'=g' means they are parallel lines

[silent plank]

common midpoint notation is M for midpoint with the segment in subscript.

$M_{AC}$

[somber coyote]

ℝαμΩℕωⅤ

[grave pond]

Not to mention (A+C)/2.

[proper eagle]

Does anyone, have ever seen this formula pi/(x*sin(pi/x))?

[silent plank]

an expression by itself is not a formula

is there more context?

[proper eagle]

This represents the quotient between the arc delimited by a regular polygon inscribed in a circumference and the length of the side of this polygon.

[small parcel]

let's say there's a right angle, and you're trying to find the cosine of 90 degrees. What do I do if I'm surrounded by two adjcanet angles?

[silent plank]

you mean two adjacent sides?

the sine,cos,tan definitions in a right triangle only really apply to the acute angles

when applying them to non-acute angles, unit circle definitions are more ideal

[vast saffron]

Oh yeah I have this question too, like what exactly the difference between an expression, a formula, an identity, an equation

Especially an identity and a formula

Don't need to answer it it's a bother lol

[rigid otter]

How do I prove that G exists?
Points G1, G2, G3, G4 are the centers of gravity of triangles CBD, ADB, BCA, DAC.

[hazy yew]

what equal to? sinˆ2(x)-1

[rigid otter]

sin^2(x)+cos^2(x)=1 so sin^2(x)-1=-cos^2(x)

[hazy yew]

thanks, as i think

[grave pond]

The center of gravity of the whole quadrilateral lies on both G1G2 and G3G4.

[rigid otter]

Thanks!

[atomic saddle]

guys

i need some help

given the triangle ABC,where AB=c , BC=a and CA= b. Prove that when c=(the square root of) a(a+b) then mC=2mA

[grave pond]

I would start by the law of cosines applied to C and A, and see if I can get derive something that looks like the double-angle formula for cosine.

[manic rain]

anyone know how to solve this using proof?

[restive spruce]

I confused

[timber cargo]

How do you add 1/2 with 1/3

[restive spruce]

with equaling 2 and 3

[rain leaf]

Basically you need a common denominator

so multiply the first fraction by (1-sinx)/(1-sinx)

And the second by (cosx/cosx)

that'll probably lead to some sort of cancellation

the other thing that might work is this is cot(x/2) + tan(x/2)

[heady scaffold]

r u fr?

r u absolutely sure ur old enough to be using discord?

if you are then im really sorry

but thats wild man

[timber cargo]

I'm providing him with an analogy

Lol

[heady scaffold]

oh okay

i got scared for a second there lol

[fickle rose]

what are u solving

that’s not a question

[wise karma]

I have this question and I solved it this way. Please someone check my solution and evaluate them if they are correct or not.

[wise karma]

Solution for questions B. And the TSA formula for A (no need checking on this).

[wise karma]

This is the solution for question number C

[foggy parcel]

Do a normal sum
It will guide u to a more simple form

[hoary panther]

,rotate

[royal bear]

,rotate

[royal bear]

,delete

[royal bear]

,flip page

???

[gentle arch]

someone can help me calculating the area of the shaded figure? I don't really know how to start

[long bone]

anyone know how to solve this pls

[late monolith]

first off, do you know how to find the area of a hexagon?

[gentle arch]

I know the formula but I'm quite lost as regards finding the areas of the semicircles

[late monolith]

Area of a semicirlce is half the area of the circle

well that is quite obvious

can you make a guess on what the radius for each of the semicircles would be

[wise karma]

you guys are just rotaing and filping. is not this math interesting? are these correct?

[late monolith]

how they do this

[hollow vortex]

Guys Which youtube channel is better in teaching affine geometry?

[upper karma]

Have you marked the wrong lines as equal?

Ah I get it

[long bone]

no its ce and de that are equal

[upper karma]

Where are you getting stuck

[long bone]

yeye

[long bone]

so i drew a line parallel to ab

and part of alpha is 50 degrees ofc

but i cant find the rest

[devout raft]

ABC is a triangle, m(ABC)=90°, [AD] Bisector, |AE|=|EC|, AD is perpendicular to DE in degrees. According to these data: |BD|/|DC| what is the rate?

[devout raft]

Anyone please?

[stone wyvern]

Rate?

[timid bone]

what does > mena with lines?

[devout raft]

|BD| and |DC| The question asks the ratio of these lengths to each other.

Like 3k and k

[sour matrix]

pls help me with this math problem

[autumn thistle]

dropping here just in case anyone else can help/is interested in helping

[sour matrix]

someone help me with the middle two problems pls

[wicked jetty]

Why don't they averave

[icy cobalt]

Can anyone explain how/when using double and half angle identities would be useful? I see how they are derived yet I just don't see any practical use for it or really understand the point of doubling an angle

[graceful bane]

can someone help me in geometry

PLEASE

[upper karma]

Sorry for the late reply.
Find angle DEB
Then use exterior angle property on alpha with triangle ADB

What is the answer given

[dark sparrow]

!da2a

[lime crown]

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

[strong vault]

Hi

SOES ABYOJE KNOW IF THIS IS CORRECT

[dark sparrow]

@strong vault YOU SHOULD REALLY STOP SHOUTING!!!

,rccw

[dark sparrow]

anyway, no, your drawing is incorrect.

[long bone]

i honestly couldnt find it. what did u get for the answer?

[upper karma]

I got 85

I think I m wrong, I will recheck

[upper karma]

Is it 100?

[royal bear]

why is Euclidean geometry so hard😭

[long bone]

i got 100 but im pretty sure that was wron g

[long bone]

can you shpw your working for 85 please?

[trail apex]

anyone here tutor?

[maiden spade]

i am having trouble with some college level trig homework, is anyone willing to help atm?

[hazy lava]

helo

[maiden spade]

helloo

I am completely stuck, math aint my strong suit at all

[hazy lava]

show me the problem

[maiden spade]

alr one sec

[hazy lava]

ok

[hazy lava]

ok

theta is 45

[maiden spade]

yeah the only issue is i gotta show my work lol

i just need to figure out how to get the answer, my professor gives credit as long as we try

[hazy lava]

first sin2theta/cos2theta=1

sin2theta=cos2theta

cos2theta-sin2theta=0

cos(2theta)=o

[maiden spade]

thanks : D

[hazy lava]

cos(2theta)=cos{(2n+1)xPie/2}

2xtheta=(2n+1)xpie/2

theta=(2n+1)xpie/4

thats it

lol

its really hard to type all these

[heady tinsel]

Decide angel x so that the incline on AB is 15 deg

[grave pond]

If you decide on a scale such that the height of the slope is 1, you can compute the lengths of the two fat lines that form the tops of the 15° and 20° angles using trigonometry. Once you have them, you have two of the sides in the right-angled triangle where x is one of them, so a bit more trig will give you the angle.

[hazy lava]

x is 40.822 degrees

[grave pond]

!nosols

[lime crown]

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

[hazy lava]

what does that do

[burnt arch]

@patent root okay so i think i have the concept down for what i was doing yesterday

so i think a lot of confusion came from the fact that i was essentially solving for a new equation, not a specific value

and i didnt realize that

and also the reason why we dont just slap +2pi n at the end also became clear after a bit of thinking

(referring to sinusoidal equations)

since they're periodic there can be a bunch of solutions since the function will repeat it's output every so often (period) and so we need to generalize our answer to work for any period

and 2pi n doesnt get added onto the end because our angle needs to be represented by some multiple of x, and so since x is being multiplied, we need to algebraically malnipulate 2pi n around to keep it proportional

^^^
does this all sound right?

[wicked jetty]

bump

[bronze wind]

Can someone walk me through this

[frozen ocean]

,rccw

[bronze wind]

<@&286206848099549185>

[opal storm]

Have u heard of the abc rule for triangles

ABC being sides of a triangle

Look up that formula

[bronze wind]

I dont really understand it

can you walk me through it

and how I would use it

[balmy lagoon]

are you talking about the pythagorean theorem? a^2+b^2=c^2?

[opal storm]

There’s also rules

For when the sum of two is less then or equal. To the third leg or

Or something

Like certain situations are impossible

And based off those situations

U can find all possible@values

[bronze wind]

Im not looking for hypotheticals or whatever

[opal storm]

No

[bronze wind]

I just need an example to work off of

[opal storm]

All possible third sides

Here hang on

Lemme@pull up the formulas I’m referring to

https://byjus.com/maths/triangle-inequality-theorem/

@bronze wind

^^

There’s an example worked out

[bronze wind]

Ok

[opal storm]

Hope this helps 🥂

[bronze wind]

thanks

[upper karma]

I erased it, but the procedure I followed was wrong. Like accidentally took two variables as a sum rather than their difference...why do you think 100 is wrong?

[patent root]

Yes, you got the idea

[burnt arch]

and if im right, whatever is in sin ---> (whatever number) <--- is the angle

no matter if it's being multipled by x, n, pi, gravitational constant

whatever is in that sin(x) thingy is the angle right?

[patent root]

yes, the argument of a trig function is an angle

[burnt arch]

oh wait

i have a question

in a function

where i have some f or sin, or whatever(x)

what is the ( ) called?

[patent root]

argument

[burnt arch]

oh okay

[patent root]

f(x), x is the argument

[burnt arch]

oh okay

i just called it the input

[patent root]

f(82x), 82x is the argument

input, argument, same idea

[burnt arch]

mhm

i prefer input because when i talk about inverse functions i can use the word output as well

the input of f(x) is the output of f inverse(x)

[patent root]

that's fine

[burnt arch]

but argument makes sense as well

[patent root]

argument is just alternative vocab for the same thing

[burnt arch]

mhm

synonym

[patent root]

sure

[burnt arch]

and so since the argument of a sinusoidal function can be multiplied by x, like i said (just to be sure), we cant just slap +2pi n at the end but rather have to divide that by whatever x is being multiplied by to keep things proportional right?

in regards to the solution set i mean lol

[patent root]

Yes

[burnt arch]

okay, that actually makes sense

[patent root]

I wouldn't necessarily put it like that, but yes

[burnt arch]

lol ill come up with a better explanation down the road, i'm just trying to make sure im absorbing it correctly and understand the logic

i like to put things simply so that the phrasing works with my intuition

[patent root]

You will get a better way of thinking about it when you get to things like periodic motion in physics

[burnt arch]

yeah for sure

btw i do have plans on teaching all of this math too, i just gotta learn it first lmao

i wanna teach hs algebra (hobbyist thing, like an online educator deal) but i obv need to learn how to do the math so i'm really big on clear explanations

so please be patient if i repeat the same question a thousand times, it's just cuz i'm trying to make it make as much sense to me as possible lol

[patent root]

I used to tutor math for extra money

[burnt arch]

lol i just wanna do it for fun

[patent root]

tbh, the help section here is pretty similar to what I used to do

[burnt arch]

it's very stimulating and rewarding

[burnt arch]

mhm, i dont understand anything in those help channels so i cant help as much as i would love to

if i was good at math, i would sit in each help channel helping everyone like every day but yk i gotta get good at math first lol

[patent root]

We probably shouldn't talk about this in this channel btw

[burnt arch]

yeah, imma go to math discussy

[dreamy barn]

Hi

[sage bison]

how do i proof that cos(x)^2 + cos(2x)^2 + cos(3x)^2 -1 = 2cos(x)^2(2cos(2x) - 1)cos(2x)

[wicked jetty]

Because exponents just go to the left of the number like logs do @sage bison

[sage bison]

what

[wicked jetty]

Exponents become the coefficient

[sage bison]

...???

[wicked jetty]

Oh I read it wrong

[long bone]

yep i did the same thing and got 100 💀

but its the wrong method ofc

[true trellis]

Is there anyone who can help me with 18

[hazy lava]

helo

[hazy lava]

tan2x+tan2x.tanx and then expand tan2x

tan2x=2tanx/{(1-tanx)(1+tanx)}

[true trellis]

Thank you

[placid coyote]

I somehow got 34 help me out

[wicked jetty]

68

bc 34*2

[placid coyote]

Thank you, just found that out not too long ago lol.

[fickle rose]

there are vertical angles and alternate interior angles

which should help you solve

[upper karma]

Can someone explain why sine was used for N_1 and N_2? The angle is w.r.t y-axis.

[patent root]

Try finding the y component of N1 and N2 in terms of that angle

It might help to draw a really big wedge and make some triangles

[coarse fossil]

yall how do i do like y = -x for graphing polygons and reflections an stuff

[upper karma]

I still cant see it. Why don't you draw it for me so I can see what you mean.

[patent root]

You should try to draw it and then show me what you don't see

[upper karma]

We can decome N1 into x and y components

decompose

I still don't get where they are getting sine from. We take cosine when angle is with respect to y-axis

and looking for y component

[upper karma]

So?

[twilit inlet]

https://cdn.discordapp.com/attachments/908078124405755914/1166466394045567106/image.png?ex=654a977c&is=6538227c&hm=3b0840f6ea9c5a31da6f64314c31e379d7ff3688b76ea140709bf35ac0582e5f&

[wicked jetty]

SAS I think

[upper karma]

someone??

[wise ridge]

i helped you with this

<@&268886789983436800> this guy is spamming

[upper karma]

bruhhh

it wasnt help full

[thin sentinel]

hi, please get an open channel from the list of open channels and ask there

[upper karma]

k

isnt that geometry

[thin sentinel]

the ones that are labeled help-### without a name

[wise ridge]

bruh if my explenation didnt help nothing will

[upper karma]

u didnt

[thin sentinel]

yeah but for specific homework help like this it's best to have a dedicated channel for it

[upper karma]

ok tq

[wise ridge]

I did💀 bruh if you dont take math seriously leave this server please

[upper karma]

where can i take my help from, could you redirect me with some channel

[upper karma]

dm

[upper karma]

i did not come to this server to listen to your advises

[wise ridge]

then why did you join this server

you are just trolling and asking the same question to everyone

[thin sentinel]

alright dude you need to stop being so weirdly mean, let us handle the moderation issue and you just focus on being helpful and kind

[cosmic locust]

How to solve a triangle in which you know its 3 angles and its area

[thin sentinel]

pick a side and call its length x, then find an expression for the area of the triangle in terms of x

[thorn crow]

what do you mean by solving a triangle

[cosmic locust]

Guys, I have a triangle and I have to find the height, I know that the area is 50m2, and I have the 3 angles of the triangle, a triangle cannot be solved with only 3 angles, I have already determined that I have to find the base to get the height only isolating the equation of the area of a triangle

[cosmic locust]

know all his lenghts and height

[thorn crow]

which type of triangle is it

?

acute, obtuse, isosceles or right or some arbituary triangle

[thorn crow]

or construct another triangle upon yours and make a parallelogram, take a side as base, divide the length of base by 2 times the area (100) and you get the height

[ripe lance]

if i restrict my domain, can i model my hairline using a sinusoidal function?

[upper karma]

Guys, I want a place to practice trig equations, formulas and inequalities

[blazing veldt]

https://www.khanacademy.org/math/trigonometry

[errant igloo]

Could someone help me solve this?

I’m sorry if I’m being a burden, but I’m truly stuck

[upper karma]

do you remember how midpoint is found

afaik

its

[errant igloo]

You are given the midpoints in this case

[upper karma]

midpoint = (x1+x2)/2 + (y1+y2)/2

[errant igloo]

Yeah

[upper karma]

so let's consider GH part
||Bro I'm gonna try I haven't done these things in like years||

[errant igloo]

Honestly I tried but my answer was a fraction and was wrong

[upper karma]

lemme see what you tried so far

[errant igloo]

Uhhh let me rewrite it

I have chicken scratch handwriting

Also, I’m not looking for the midpoint, it gives me the midpoint, but I need the two points that allow that to be the midpoint

[upper karma]

don't you want the G

[errant igloo]

Yes

But G is not the midpoint

K is the midpoint of segment GH

[upper karma]

3, 9 = (x+x0)/2, (y+y0)/2
3 = (x+x0)/2
Cross multiply maybe

and then maybe after you use substitute equations

[errant igloo]

This is getting very confusing

I’m trying to solve for point G H and J

While point K is the midpoint of segment GH and point L is the midpoint of segment HJ and M is the midpoint of segment GJ

and it give me the places for all the midpoints

So instead of using the midpoint formula, I’d use the slope formula, right?

[upper karma]

how's slope involved in this

[errant igloo]

Because you are finding the points that make the midpoint

[upper karma]

I also don't get what's really confusing you rn, sorry 😦

[errant igloo]

I’m a freshmen trying to figure this out

I didn’t have a teacher today to teach this topic because we had a sub

And I was trying to ask someone to kindly help me solve it

[ripe lance]

id checkout a video on the midpoint formula

[upper karma]

can someone help me with this

[leaden radish]

im stuck

[upper karma]

trigo

[flat willow]

hi

[upper karma]

trigo is fun and geometry is madness

[flat willow]

so sin2x = 6/13

and then you can just substitue some theta = 2x

[upper karma]

wouldnt you just do percentages

[flat willow]

and then divide by 2

[upper karma]

its ab but divided by 4

just add

idk

[flat willow]

oh i'm talking about eclipseryu's problem

[burnt arch]

13sin(2x)-3=3 has at least two solutions.
Let the first solution be x_1 and the second be x_1. Also let n be some integer.
13sin(2x)=6
sin(2x)=6/13
2x=arcsin(6/13)
arcsin(6/13) is approximately 0.4797 radians.
By the sin(x)=sin(x+2pi) identity:
2x=0.4797+2pi(n)
x_1=0.240+pi(n)
For x_2:
By the sin(x)=sin(pi-x) identity:
2x=(pi-0.4797)+2pi(n)
x=1.331+pi(n)

^^^
This is just one example

[upper karma]

Explain in Fortnite terms

[leaden radish]

i have not worked with percentages this whole semester

[burnt arch]

solve for x

theres two x's

the first is 0.240+pi(n)

and the second is 1.331+pi(n)

[prisma estuary]

so f and g are supposed to both be 65 degrees, but 54+64 dont equal 180 and I thought same side exterior angles were supplementary?

[sharp cloud]

huh

[heavy oar]

Did you already ask something similar in a help channel?

[heavy oar]

I haven’t gotten far in geometry yet so I don’t know how to explain it properly but here’s a visual to help. The reason why the same side exterior angles don’t count here is because there are three angles instead of two, but they still all add up to 180. The two red lines form a vertical angle. Two angles across from each other both equal 61° and to find the rest you subtract that from 180, giving you 119. You have a missing angle but you were given the other angle, which measures 54. Subtract 54 from 119 and you found the measurement of angle f. Another way you can do it is add all the 3 angles up and make them equal to 180 because they lay on the same line (blue line.)

[prisma estuary]

ohh thank you I just was confused why g couldnt be 126 but I see it now tysm

[heavy oar]

Np

[mint garden]

I got this

Someone please check my work

[frozen ocean]

,rccw

[frozen ocean]

@mint garden if you still need help, would you mind showing us the original question is?

[mint garden]

This, it's a question I submitted but I need to have the right answer for it to be accepted

[celest wave]

hi

help

[astral stratus]

Pls help, its asking to write 2 column proofs.

[wicked jetty]

-7 -4

[foggy parcel]

Assuming two points A(x1,y1), B (x2,y2)
The coordinate of the midpoint(M)of the straight line AB would be
((x1+x2)/2), ((y1+y2)/2)

[thick fable]

@astral stratus use congruence rules

[vernal pilot]

are there any good vector proof worksheets? and if possible, can it include a few 3D vector questions.

[kindred laurel]

how do I simplify a radical with a decimal in it?

more specifically the square root of 179.84

[silent plank]

you could start with multiplying the inside by 100/100

applying sqrt laws leads you to
=sqrt(17984)/10
then simplify the numerator
also not geo

[white dome]

since it’s an isosceles and you found angle 2, 122, you would then subtract 122 from 180. Resulting in 58, after just divide by 2.

[last ocean]

Need help

[crimson pawn]

what does the greater than signs means with a line under

[thick fable]

it is a sign of 'greater than equal to', that simply means that the thing that is being pointed with it should be greater or equal to the value there is, suppose its written x ≥ 5, this means that x can be greater to 5 and also equal to 5 at the same time @crimson pawn

[crimson pawn]

i’ve just never seen it written like that lol

appreciate it

[smoky jetty]

for no.4 (the octagon problem) B, remember that the sum of an octagon's interior angles is equal to 1080 deg, so each interior angle has 135 deg. Now if QT is formed such that PQ is perpendicular to UT, then triangle PQT is a right triangle. So, <PQT +,TQR = 135
after this, you'd get the base angles oft he isosceles trapezoid QRST. then by refering to 45-45-90 special triangle, u should get the needed measures to find both polygons' areas

[upper karma]

Can I have help in geometry anyone

[smoky jetty]

dont ask for help, share it here and let anyone give it a try!

[hazy lava]

hm

[upper karma]

[lime crown]

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

[signal zephyr]

I know I got the minimum wrong when I did my submission but how is my maximum wrong?

[sick kraken]

Can someone review my notes please. Some geometry and trig fundamentals

[frozen ocean]

neat handwriting

[sick kraken]

Thank you I'm still adjusting to the Apple pen

[sick kraken]

Some of it…

[frozen ocean]

nah they all seems neat

im impressed

[sick kraken]

Haha Ty, do you know if the notes are correct?

[frozen ocean]

im too lazy rn

[cunning abyss]

wrong channel oops

[harsh garden]

can someon help me with this

[foggy parcel]

Great work buddy!
handwriting's fine af like wtf

anyways, there's some tips for the notes

1. U could have added the simillar triangle theory to the trig part. It could put extra realism to the fact of the ratios of sides being same for same angles
2. It'll be appreciated if u put proof of the distance formula and other associated formulas in coordinate geometry to have a better glance at those things. Cuz it's no fun understanding or learning the fundamentals of this without using graphs uk!

[gentle haven]

is sas congruence an axiom or a theorem

in some textbooks it says axiom and in others it says theorem

and the proof is really weird

so im not sure

[autumn gorge]

hello guys

im struggling with trig and mostly inverse functions of cos, sin, tan, sec, csc, cot

do you g uys know any tops

tips*

[proven sun]

Someone please help I need to find the area of multiple shapes put together and I can figure out how to do it. Dm please

[upper karma]

1. Find θ

θ =

2. Using the unit circle, find the following values:

cos 0o =
sin 0o =
cos -90o =
sin -90o =
cos 135o =
sin 135o =
cos 315o =
sin 315o =
cos 60o =
sin 60o =
cos 150o =
sin 150o =
cos 240o =
sin 240o =
cos -30o =
sin -30o =

Find these values and write them in radians:
arccos 0 =
arcsin 0 =
arccos 1 =
arcsin 1 =
arccos (-22) =
arcsin 22 =
arccos (-12) =
arcsin (-32) =
arccos(-32) =
arcsin 12 =

can somone help me

[sick kraken]

Okay thank you I'll be sure to add them!

[frozen ocean]

Are you allowed to use a calculator?

[naive stream]

what did you use to write these notes? It looks nice

[vale oxide]

a

[thick fable]

how is the angle in red circle equal to angle in green circle

[frozen ocean]

they look same angle

[thick fable]

?

how? that's my question

i just cant see how they r equal

[frozen ocean]

can you recognize that they're same size angle?

idk how to explain it mathematically tbh

[thick fable]

ikk they are same but how

like in this they are though

vertical interior angles

but what about the above one

[frozen ocean]

I mean in the blue circle

[thick fable]

what?

[frozen ocean]

like this

yeah sorry i suck at explaining

[thick fable]

yea i can see lol

[thick fable]

okay youre back again

[river sapphire]

This purple one would be 90-dtheta/2

Since yellow line is tangent

[river sapphire]

Lol

[river sapphire]

And this orange thing is also 90°

[thick fable]

and this one too?

[river sapphire]

Yes

[river sapphire]

Since these 2 are radius and tangent

[thick fable]

ohh okay it doesnt look like a right angle, thats why i was having trouble

okay got it

thanks both

[upper karma]

yo

does physics apply in this

[frozen ocean]

discord.gg/physics

[upper karma]

thx

[sick kraken]

Goodnotes

[naive stream]

thanks

[wooden wharf]

guys i just got my report card ive got all A's in math

[frozen ocean]

good job

[hasty hare]

hi

ppl

i think im new here

i dunno

[small parcel]

SSA criterion works for right and obtuse triangles right?

[fickle rose]

no

it just doesnt work'

[smoky jetty]

ugh after 5 days of trying to comprehend how this formula is dervied (searched everywhere on the net) I gave up and instead looked at it from a different approach. So upon doing some graphs, I realized I can simply create a triangle using the lines' x&y intercepts then say line 1's x&y intercepts would be its base, and the d (which perpedicularly intersects l_1 & l_2) would be the altitude. On line 2, I can use either the x or y intercept as a point of reference. Then I can refer to the formula to find shortest distance of a line to a point, which I was able to understand (simply Height = 2Area/Base on a coordinate plane). Tadaa same result, but if anyone knows how the formula in the pic came up as it is, feel free to share!!

using green line's x-intercept as the point of ref

or using the green lines' y-intercept as the point of reference

then yeah pretty much finding altitude of triangle on a coordinate plane

where intercepts create the sides

[foggy parcel]

That's actually more deep
I'd prefer doing problem solving in this way

@smoky jetty

I don't know that formula as well lol

[smoky jetty]

i didnt get the procedures online so I thought maybe triangles may do the job and yeah came up with another formula that I understood well enough

which was this

but if anyone can guide me to the other formula, then id love to

[grave pond]

If you're happy with the distance-from-a-line formula, I'd suggest this:

Convince yourself that if you remove the absolute-value signs, you get a signed distance-from-a-line formula that gives you a positive distance on one side of the line and a negative distance on the other side.

Now, apply it to a fixed line through the origin with equation ax+by=0 (that is, no c).

Then you can use it to find an equation for all the points whose distance to your fixed line is some number d:

(ax+by)/sqrt(a²+b²) = d <==> ax + by - d·sqrt(a²+b²) = 0

And if you do that once again for different d, say d', we have

ax + by - d'·sqrt(a²+b²) = 0

Thus, the c's for lines at various distance from our fixed line are just their distances times -sqrt(a²+b²).
If we have ax+by+c1 = 0and ax+by+c2= 0, then their distances to the fixed line are -c1/sqrt(a²+b²) and -c2/sqrt(a²+b²).
The differences between those is the distance between the lines, in other words |c1/sqrt(a²+b²) - c2/sqrt(a²+b²)| and you can now lift the division by sqrt(a²/b²) out of the absolute value.

[smoky jetty]

i didnt get what u meant, could u rephrase this?

[grave pond]

I meant:
$$\frac{ax_1+bx_1+c}{\sqrt{a^2+b^2}}$$
is the "signed distance" from $(x_1,y_1)$ to the line $ax+by+x=0$, in the sense that its \emph{magnitude} is the actual distance and its \emph{sign} tells with side of $ax+by+c=0$ the point is.

[somber coyote]

Troposphere

[smoky jetty]

what do u mean by "signed distance"? is it sort of like the raw value of distance even if it's negative?

could u elaborate this?

im not that familiar with "magnitude" in this sense

[grave pond]

"Magnitude" = "absolute value".

[smoky jetty]

oh

how about this part? Im not sure what it meant either

[grave pond]

I can't immediately think of a different way to say that, other than just repeating myself.

The sign will be + on one side of ax+by+c=0 and - on the other side.

[smoky jetty]

it's alright, but can u like graph/draw it?

[smoky jetty]

coz im tryna connect that to this (your prior response)

[smoky jetty]

hmm how would the result look like?

[grave pond]

Umm....

[smoky jetty]

i mean in the equation (mb if im confusing atm)

[grave pond]

I'm sorry, I confused about what it is that's confusing you.

[smoky jetty]

ig to make it less confusing, do u know how this formula was derived? If u could, can u show your process and ig some bits of explanation, then i'll just ask/clarify parts of it

i couldnt comprehend to your response a few hours ago, sorry😔

[ocean shale]

can someone help me with this

[grave pond]

That's what I was trying to explain a derivation of, starting from the distance formula that it sounded like you had already proved to your own satisfaction!

[shut oxide]

help me with proofs #help-17

[wicked jetty]

Z is 33 @upper karma

X is 32

Y is 40

[smoky jetty]

i see, must've been my comprehension to new math concepts

im gonna try to understand it again later

[cerulean star]

As someone who is doing algebra 1 that looks super impossible to answer

[smoky jetty]

still having a hard time understanding, but do u think that the distance from a line to a point formula is sufficient enough to use for distance betweeb parllel lines?

[rain plaza]

neeeeed hrlppppppppp

[frozen ocean]

#❓how-to-get-help

[floral shale]

You peaked with this one

Top tier comedy

[pastel barn]

Yo anyone good with trig ratios? A quick and simple question

[fringe crater]

how can i get ADE = EDI

[paper violet]

is ED a bisector?

[thick fable]

#help-43

[smoky jetty]

@grave pond (sorry for the ping)

[thick fable]

@smoky jetty what exactly is your question

[smoky jetty]

tl;dr above: I don't know how distance between parallel lines was derived so I instead used the concept of distance from a point to a line (which I learned how it was derived) and got the same distance anyway. Troposhere showed the derivation which I unfortunately had a hard time comprehending. So Im asking if the distance from a point to a line should do fine for parallel lines.

to not bother him more on making it more understandable, and I think I've already spent 6days trying

[thick fable]

from what i studied, i had 2 different formulaes to find distance b/w 2 lines and distance between a line and a point

[somber coyote]

!Yajat!

[thick fable]

this for parallel lines

[somber coyote]

!Yajat!

[thick fable]

this for point and line

@smoky jetty

or perhaps i did'nt understand your question

[smoky jetty]

yeah there are 2 formulas, but I realized, the intercepts can serve as reference points for the second formula u shared

[thick fable]

tho i have thee derivation of these formulaes

[smoky jetty]

i learned the derivation of distance-from-a-line-to-a-point, but I couldnt understand the one for parallel lines

might be slacking in comprehension and algebra

i'd love to see how u derived it

[thick fable]

these are my notes

tell me if you understand it

x=0 and y=0 in this case

lines are ax+by+c_1 and ax+by+c_2

you can notice how a and b are same for both the lines

its because they are parallel lines

[somber coyote]

!Yajat!

[thick fable]

like we have already derived the formula to calculate distance between a point and line, we are taking help of this formula to derive this result

[smoky jetty]

oh wait omg

WOWW

holy crap

no fcking way lmaoo

so these lines or pretty much any line as long as they have the same slope, would have the same A and B constants? When transformed to general form

[thick fable]

i mean those conditions are for parallel lines so yea we can use this formuka

also

not necessariliy the coefficients have to be equal

like consider 2 equations

[somber coyote]

!Yajat!

[thick fable]

they do not have same coefficients rn

but you have to diviide the equation by 5

then it would have same coefficients

[smoky jetty]

right!

[thick fable]

yea

[smoky jetty]

so basically, A's and B's of each line must be proportional?

for parallel lines

[thick fable]

just this

[smoky jetty]

yeah

[thick fable]

if it follows this you can use this formula

you can pick random equaitons, plot them on desmos and you can see that they are parallel

[thick fable]

equations that follows this would be parallel

[smoky jetty]

so basically if two lines are parallel, their A and B constants must be proportional, then we can just exclude it in the formula to find distance from a point to a line, resulting in |C1-C2| numerator?

[thick fable]

to find distance from a point to a line

is this a typo?

u mean from a point to a line or between 2 parallel line

[smoky jetty]

yeah from point on the parallel line to the parallel line, but since they're parallel, we can reduce Ax1 and By1 in the expression?

[thick fable]

yes as x and y are zero here

[smoky jetty]

whats the length of d_2 in this case?

[somber coyote]

!Yajat!

[smoky jetty]

i mean is the point of ref also the origin?

[thick fable]

yes

[smoky jetty]

OHHH

[thick fable]

it is origin

[smoky jetty]

now everything makes sense!

[thick fable]

yea

[smoky jetty]

basically subracting the distances of two lines from origin, right? but since parallel lines have proportional A and B constants, we can just cancel them out in the expression?

[somber coyote]

!Yajat!

[thick fable]

@smoky jetty

[smoky jetty]

i see!

Here was my work on deriving it, tell me if there are mistakes in reasonings and stuff

[thick fable]

we cannot ssay a=b

but we can say that a_1=a_2 and b_1=b_2

[smoky jetty]

yeah, this got me confusing as well

[thick fable]

lol

[smoky jetty]

oh wait nvm

[thick fable]

okay for better clarifiaction lest just solve for the distance between lines 3x+4y=10 and 12x+16y=7

when you divide the second equation by 4 it givess you 3x+4y=7/4

now you have 2 equations 3x+4y=7/4 and 3x+4y=10

can you see now how a_1=a_2=a?

and b_1=b_2=b

[smoky jetty]

can we set it to a_1/a_2 = a?

and b_1/b_2=b?

[thick fable]

no

i dont see why r u so confused

[smoky jetty]

lmao hold on lemme think more

[smoky jetty]

yeah they are proportional, but if a_1 = 3, and a_2 = 12, how can a_1 = a_2?

[thick fable]

.

[smoky jetty]

i still dont see how a_1 = a_2 😭

but if those gen. form equations were transformed to slope-intercept form, I could see that a_1 = a_2

[smoky jetty]

@thick fable

[thick fable]

sorry but its literally the same thing i was telling you to do

what happens when you divide second equation by 4

tell me

[smoky jetty]

yeah you'd get the same

[thick fable]

then?

lmao

[smoky jetty]

since they're just the same when we transform standard form to slope intercept form, we can already say that in parallel lines a1=a2=a?

[thick fable]

not initially however you can say that after simplifying it a bit

okay igtg now

[smoky jetty]

yeah but I just couldnt grasp around the idea that a_1 is already equal to a_2 and is equal to a, when u actually must first divide a_2 by smth (e.g. coefficient of y > 1 in a slope intercept euqation) to make a_1=a_2=a

[smoky jetty]

exactly! ig a_1 = a_2 = a only after transforming the equations to slope intercept form

[jagged venture]

The equilateral triangles ADF and BCE are constructed outside the square ABCD. If
DE ∩ CF = {G} show that △ABG is equilateral.

need some help with this one...

[smoky jetty]

from (so, a=b) I changed the reasoning @thick fable . Is this more accurate?

[weary haven]

If a single point is removed from a line, do the remaining points form a convex set?

[nocturne remnant]

No because of the point removed is P and you take two points Q and R on opposite sides of P, then the line segment QR contains P but P is not in the set.

[weary haven]

Thank you for the response. I was quite confused because someone claimed on the internet otherwise.

[thick fable]

yea it is correct but i dont know why are you saying slope intercept, like its necessary that you have to simpify it in slope intercept

[smoky jetty]

oh I just elaborated why a_1 = a_2 = a

since it might be confusing to assume that initially, a_1 = a_2 = a already

[thick fable]

hm yea

[smoky jetty]

alright man, I really appreciate your help & clarifications. Forgive me if I have been pretty confusing/lost earlier, but yeah thanks for easing things up for me ❤️

[thick fable]

np!

[dark helm]

Hi, is there any proof/explanation for why this works? Ive tried it on some examples and it seems to work.

[astral stratus]

How do i find x, ive already found y

[astral stratus]

pls help i got the class in 1 hour

[smoky jetty]

u're given two isosceles triangles

[leaden hearth]

90 - (2y+64) - x/2 = 0

[smoky jetty]

what do u know about an isosceles triangle's property

[astral stratus]

2 sides r equal other isnt

[smoky jetty]

yeah

how about their angles?

[leaden hearth]

90 - (2y + 64) = x/2
x = 180 - (2y + 64)

[smoky jetty]

dont give out answers prematurely, !

[noble heath]

Isn’t the text in the picture the explanation?

Notation errors notwithstanding

[leaden hearth]

you know y so other side of that is 180 - (2y+64)

and another triangle is Saquin parallel triangle so
(45 - x/4) * 2 + (180-(2y+64)) = 180

[leaden hearth]

easier answer

[astral stratus]

thank you so much

[smoky jetty]

So apparently it's called the normal form of a line or the shortest distance from a line to origin. I tried deriving its formula and got this. Is it correct?

theres also this sort of other ver. of the formula: x cos a + y sin a = p
where angle a is the angle created by the perdendicular line d with the positive x axis

[thick fable]

@smoky jetty what is the second equation sorry

[smoky jetty]

in the picture?

[thick fable]

yes

[smoky jetty]

this?

[thick fable]

ye

i that theta

[smoky jetty]

theta? those are zeroes in parentheses

[dark helm]

I mean why is the cos value equated with 2pi. Why does it work? Thats my question. More of a logical question so I can make sense of it in my head instead of just memorizing.

[wet grotto]

can someone pls help me with this question?

do i need to use the angle bisector theorem?

[forest prairie]

Because it's periodic with 2π, which means that we'll get the same value everytime we move 2π

I hope that I understood your question and that I answered it

[noble heath]

Ok, maybe this helps:

Fact 1: cos(2 * n * pi) = 1 for every integer n

Now when is 2pi/3 * m of the form 2 * n * pi?

Answer: when m is a multiple of 3

[cerulean marten]

can someone help me

[smoky jetty]

is this correct @thick fable ?

[smoky jetty]

ight this one is pretty understandable

[dark helm]

Yup, you did. Thanks!

[dark helm]

Thanks 🙂

[strong yoke]

When do you flip the sign of an inequaltiy? (I've looked online, can't find anything clear...)

[safe tiger]

it says that- two right circular cones x and y are made. x having three times the radius of y and y having half the volume of x. calculate ratio between the heights of x and y.

[thick fable]

its inside the mod so you have to take 2 cases when c>0 and when c<0

[smoky jetty]

mod?

[thick fable]

modulus

absoulute value

[smoky jetty]

oh

the arrow tells that the numerator C must be >0 after getting the absolute value

[thick fable]

yea

now make one more case with -c and show this when c<0

thats the basic definition of moldulus

[upper karma]

How do I find AB

[forest prairie]

Could you send a picture of the problem ?

I guess you work with complex numbers? I’m Not at that level yet, but afaik, we do it because the degrees would be above 270 in the 4th quadrant. Cf. The picture

[smoky jetty]

afaik 360 is subtracted to an angle in 4rd quadrant since the result is the positive coterminal angle of that angle, which is an acute one

and a reference angle for the trig functions & ratios

[wicked jetty]

y is 10

[zealous meadow]

but how to find x

[wicked jetty]

idk

[merry pecan]

You can find the intersection of line AP and BP

And for the equations of line AP and BP, you have a cordinate for both so you just need it slope

It's*

[steep shuttle]

AP=BP are equidistant

[zealous meadow]

its fine
i found it

using

dist formula

[merry pecan]

Cool

[wet grotto]

do i have to use law of sines or similar triangle for this question

pls help me

[upper oracle]

Does this count as an octogon?

[leaden hearth]

I think so

[wicked jetty]

No it has like 11 sides

[leaden hearth]

how?

it has 8 line and 8 point

(8 angle)

[wicked jetty]

They bend a bit

[leaden hearth]

this is how I see the picture

[wicked jetty]

Well from where it bends it has 11 lines

[leaden hearth]

eye lvl 2000

[wet grotto]

can someone help me with this pls?

[leaden hearth]

to make it easier

<B = <ABD = <ABC
<C =  <ACD = <ACB

<DAC = 2<DAB

and as in shape we know that

<DAC + <DAB = <BAC
3<DAB = <BAC
2<BAC = 3<B
2 *(3<DAB>) = 3<B

remove 3

2<DAB = <B
<DAC = <B

and we know that trianlge have 180 deg inner angle

180 = <C + <BAC + <B
180 = <C + <DAC + <ADC = <C + <B + <ADC 
<C + <B + <ADC = <C + <B + <BAC

remove <C + <B
<ADC = <BAC

we found that triangles CAD and CBA have same angles
so there is a x that

BC * x= AC
AC * x = CD
AB * x = AD
ax = b
bx = e
cx = d

and about x you can use first one

ax = b
x = a/b

sorry for my bad english if you don't get something ask me

[leaden hearth]

you don't need sin

[wet grotto]

wait ive got another sol

[leaden hearth]

sin lead you to wrong answer trust me

look

A is 3x

and <CAD is also 3x

so all angles on CAD and BAC is same

so that sides multipy by same number

we call it a

[leaden hearth]

let me check

[wet grotto]

k

[leaden hearth]

(this is another x)
x is b/a
so
(b/a)* b = b^2/a
cx = bc/a

yep

our answer is same

but I made a small mistake in last

(come on I calculate all of it in just chatbox not paper)

[leaden hearth]

i wrote ax =b so x = a/b
but it's x = b/a

[wet grotto]

ye

[wet grotto]

it happens

anyways TYSM

[leaden hearth]

yw

[upper karma]

already did the assignment but is it possible to reflect the pre image to the image in under 3 translations (other than reflections cannot go through the grey)

[wicked jetty]

reflect over y axis and then rotate it twice

[upper karma]

wouldn't work

and i said in under 3

[surreal escarp]

https://cdn.discordapp.com/attachments/894762638553260043/1169434262697680926/IMG_6334.jpg?ex=65556387&is=6542ee87&hm=2a3daf3992cd41e0156c57c5177b53f058429cfcd89a1ec593ed92d3384b1eda&

how do i do this??? im so lost

i just wrote whatever i thought it was but i actually have no clue how to solve it

[bronze wind]

<@&286206848099549185> Can someone walk me through this?

[upper karma]

can someone explain the basics of trig to me

very simple

[frozen ocean]

!da2a

[lime crown]

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

[nocturne remnant]

what about this was worth da2a'ing about

[nocturne remnant]

trigonometry relates ratios of lengths to angles using trigonometric functions

[pure ether]

use sss congruence by using the distance formula

[warped apex]

Is this answer just 2?

I don’t think it’s that simple💀

[nocturne remnant]

consider the range of sin^{-1} carefully

[pure ether]

well depends on if its degrees or rad

if rad then like 2 is in 2 quadrant so

make it first

[warped apex]

The range is -90 to 90, right?

[warped apex]

Yeah as the assignment doesn’t mention whether it is in radian or degree.

It should be in radian

[thick fable]

you take it as radians if it does'nt clarify it

[warped apex]

Ok

[balmy spruce]

wait is the answer for this guys

What is measurement of the indicated angle assuming the figure is a square?

[bronze wind]

do I have to use it for every single side until I get the answer or do I only need to use it for any one side

[misty elk]

can someone explain to me what is  the "equation of a line", and how to derive it ?

[bitter panther]

Hello i have a problem

[forest osprey]

Hello! How can I assist you today?

[bitter panther]

Let ABC a non-isoscelis triangle with orthocenter H and the feet of altitudes A1, B1, C1. Let A2, B2, C2 be the projections of H on B1C1, C1A1, A1B1. a) Show that the circumscribed circles of HA1A2, HB1B2, HC1C2 have 2 points in common. b) Show that the circumscribed circles of HAA2, HBB2, HCC2 have 2 points in common.

[strange dew]

How can I find the OO' segment?

B is an intersecion of the circunferences, AC // OO'

[tawdry condor]

Can I do this?

Corresponding angles is a postulate but like

My teacher said you can’t prove it but surely you can right

Is this the right direction ?

If I wrote a,b,c as lines in slope intercept form

And their slope would be tan(their angle off of the x axis)

and I say a circle is the angle of a off of the x axis and same for b circle and c circle

Then can I do that

I really wanna know where I went wrong because it is a postulate so it can’t be proved but this looks ok to me

[pure ether]

no every side

unless u trig ash to find 2 similar angles

but thats harder than just finiding the dstance

its calleld sss for a reason

[thick fable]

what do you mean by deriving it

[thick fable]

translate the question please

[thick fable]

wait what do u mean

[tawdry condor]

Like

The line that has a y intercept of 0 and is 45 degrees off of the x axis

Is y = tan(pi/4)x

So u write all the lines in this form and show c and d have the same angle

And thus the same slope

So they are parallel

[thick fable]

d?

[tawdry condor]

b*

[tawdry condor]

Typo

So why am I wrong

Is it not a postulate ?

How can you prove a postulate

I added more to the proof

I also showed linear functions can be represented that way

[thick fable]

how can u say that slope of line a would be equal to slope of line b and c?

@tawdry condor also i quite dont get what you main objective here is, like are u trying to prove how corresponding angles are equal?

[tawdry condor]

I didn’t

[tawdry condor]

I’m trying to prove if two corresponding angles are equal, then the lines are parallel

Converse

Since they share that angle

I wrote them in relation to the transversal and that angle

They have the same slope parallel

So uh

[thick fable]

they share equal angles off the x axis so thier slopes are equal that means they are parallel?

dude u there?

@tawdry condor

[misty elk]

how do you discover it? how do you come up with it ? how does it work ?

[thick fable]

that is just a general form of a straight line

[misty elk]

Why is this form used, in which the slope is not apparent:
ax+by+c=0
?

f(x)=mx+b is intuitive , m is the slope

ax+by+c=0 is not intuitive, and I am trying to find some intuition

[thick fable]

that is just a general form of a straight line, f(x)=mx+b is derived from this form only

[tawdry condor]

Yes

So is my proof valid then

The setup

[thick fable]

i mean we can make this whole transversal like the b and c line parallel to x-axis

[tawdry condor]

No

I just drew them like that

But I didn’t make any such assumptions in the math

[thick fable]

sorry but your handwriting is just not redable

[thick fable]

oh ok

[tawdry condor]

lol it’s fine

Does my logic work

Is this proof ok

[thick fable]

yea those angles can only be equal if those lines are parallel but im not sure about it, i mean the logic seems correct

[tawdry condor]

Fire

[wraith summit]

i need some help with geometry

ill send a picture of the homework I have to do

[slow path]

Can someone please help me through this question? I know the steps I think but I've made an error somewhere as my answe doesn't seem right

[thick fable]

you know its a right angled triangle, you can use pythagoras relation to find the vvalue of k, and then use 2 point form to find the eqaution of the line @slow path

[slow path]

I don't know the length of 2 lines though

[thick fable]

use distance formula for that

[slow path]

And I did it a relatively more complicated way but I've ended up at 2 points

Could you double check that k=9?

I have an answer that works woohoo

[thick fable]

i aint solving anything showme your work tho

[bronze wind]

So which sides for example would I need to find

[pure ether]

i literally said all of thekm

[gusty glacier]

hello my friends

[proud socket]

???

[heady scaffold]

context?

[proud socket]

wdym, like the problem? cause its showing how to round 9 pi

[heady scaffold]

yes

[heady scaffold]

pi is 3.14159

or even 3.141

keep that in mind

now multiply

[proud socket]

ok

28.27. . .. .

[heady scaffold]

yes

now round to nearest tenth

[proud socket]

Oh

28.3

right?

[heady scaffold]

yes

it's that easy

[proud socket]

thx

[hard coral]

Can someone help me understand how to get F? I have a slight intuition but this is finding F(θ) where θ is a defined angle between the X-component of the normal vector, and the normal vector. Where the addition of that angle, and the angle between the negative X-component of the normal, to the normal, equals 180°.

Two things I've noticed:

F(θxz) = 3π/2 when θr = θxz
F(θxz) = π/2 when θr = θxz + π/2

[snow rampart]

Anyone has the xbox gamepass 3 months code and doesn't need it

[bronze wind]

can you do out an example for me

[bitter summit]

i had a friend who had it but didnt neeed it, but it expired

[upper oracle]

10th grade trying to solve a 12th grade (according to the one who posted it) problem, I’m stuck pls help

Haven’t learned trig yet if it helps

[merry pecan]

you made a mistake in solving the angles of the triangle ABE, their sum is 68, they aren't 68 each

[upper oracle]

Oh ur right

[merry pecan]

Welp

Couldn't think of a way to solve without using trigonometry

[smoky jetty]

what value did u get?

I tried to find it by angle-chasing, still got nothing after 30 mins

[merry pecan]

Lmao I'm not gonna solve that, just sharing what I got

Though Its probably not that difficult to solve it using trigonometric identities

[smoky jetty]

although I think I'd get the answer rn if I plug in each given choice to angle ABF and if it matches with other angles that I found. But I think it'd ruin the fun out of it XD

[drifting parrot]

could someone help me touch up on some geometry for 8th grade?

[burnt arch]

is your teacher like not doing their job or is this like a personal thing