geometry-and-trigonometry

[pure void]

oh ok

[unique flower]

... thats annoying, ig you have to create an equation with law of sines like 9^2 = 5^2 + x^2 - 10xcos(y) and then solve for y :/

do you need to show your work?

[pure void]

nope

its just practice work

[whole granite]

cant you split in two right triangles?

[unique flower]

ok then don't even use law of cosines for y

[north heart]

you dont have enough info

[unique flower]

use law of cosines for x then law of sines for y

[pure void]

kk

[whole granite]

why not?

[unique flower]

if you want to go hardcore mode you can use this for as cos(27) but just use decimals, its probably the only thing necessary

[unique flower]

idk how helpful that would be

[north heart]

that's not two right triangles

and tbh that's not that useful

actually

hm

i think you can solve for y from there

looks like a pain though

i think law of cosines/sines is just easier

[whole granite]

oh, I see

[pure void]

i think a=5, b= x, C= y

[unique flower]

you have 2 unknowns

that won't work

I recommend looking at the 27 angle

and see how you can apply law of cosines with it

annoyingly the law of cosines is always drawn with an obtuse angle being the "angle C" which can be misleading when solving a problem

it works for smaller angles

[pure void]

ye

and the triangle

is upside down

[opaque gull]

system of equations?

[unique flower]

ig you can when attempting to incorporate the 27 angle

but its harder

and more annoying

[opaque gull]

yeah i guess

but it's a method =)))

tricky, what would be the easiest

[unique flower]

thats how to find x

and then you can use LoS for y

[opaque gull]

ahaa

ok

@pure void

here

[pure void]

oh

[vagrant ore]

Does anybody know how to solve this? I already found the ratios

[wise bear]

find the formula for area of a pentagon idk what it is, multiply that by the length and you get the volume
the given volume is 300 so solve for the length
the red prism's length should have a ratio of 6/5 to the original length and you can use that value to fint the total area

correct me if im wrong guys

or maybe there is an easier way to do it

[tulip vector]

Hey guys, can someone help with this question? My friends wouldn't helpp me out..

I know the other questions, but the B question is kinda hard for me.

[wise bear]

for b) you know that the whole straight line angle is equal to 180 degrees and that the two angles other than the one youre looking for are 52 and 38

that gives you an equation: 38 + QOR + 52 = 180

from here i think you can figure it out

[tulip vector]

tysm!

[wise bear]

np!

[tulip vector]

tysm for the help, i got 6/8 :>

[wise bear]

nice, congrats!

[coral axle]

i need some help with trig
so i have a polygon with am area of 225 square meters
if the area is doubled, how does each side length change

[whole granite]

maybe you can develop by:

# we have (x for original area)
x² = 225 -> x = sqrt(225) -> 25

# then (x for doubled area)
x² = 225 -> x² = 225² -> x = sqrt(225²)
x = sqrt(225²) -> x = sqrt(50625) -> x = 225 

diff = 25 - 225 -> 200

[coral axle]

whats sqrt

[whole granite]

square root, same for √9

[coral axle]

bro im dum

[whole granite]

hmm, but its any polygon

I was considering your polygon a square

[coral axle]

lets give a more clear thing

area of a regular pentagon has an area of 30 cm

whats the area of it if we multiply the side lengths by 4

im not sure how to get the side lengths

sorry for being so dumb

[whole granite]

no bro, dont worry about it

[coral axle]

ok

[whole granite]

just let me check some things

[coral axle]

can you show me how to get the side lengths?

[whole granite]

you can use the formula to find the side perimeter from the area, but I strongly recommend you to understand the structure of a regular polygon

[coral axle]

whats the formula?

[whole granite]

I found something interesting:

[coral axle]

so i divide by 5

to get the area of one triangle

[whole granite]

exacly

[coral axle]

and where would i go from there?

im not really sure

wait its isoseles

but i would have too guess the side length or something

no the top angle would be 72

ok ik kinda where to go from here

[whole granite]

not really you can split the isosceles triangle in two right triangles

[coral axle]

and ik the angles

[whole granite]

and then divide the area by 2

[whole granite]

I guess you know

a regular pentagon have 108 degress

[coral axle]

ye

but i would need to figure out one

and im not sure how to get that one

if i can i would be able to use cos,sin,tan

[coral axle]

how would i find the sides tho

alright i need to go sleep

im gonna have to figure it out ig

idk

[whole granite]

https://www.youtube.com/watch?v=CG8ipDUeyRs&ab_channel=LameeStorage

seems helpful, I'll watch

[dreamy quest]

hi could anyone answer this quick question for me?

is a circumscribed circle of a triangle constructed inside or around the triangle?

<@&286206848099549185>

[dreamy quest]

thank you!!

[fluid jetty]

<@&286206848099549185> how do i know when its decreasing or increasing

[whole granite]

maybe:

if d/dx > 0 # -> increasing
if d/dx < 0 # -> decreasing

[fluid jetty]

what is d

[whole granite]

no, its not a var, its a symbol

d/dx means a derivative of a function

[fluid jetty]

whats that mean i forgot

[whole granite]

if the derivative is positive then the function is increasing otherwise decreasing

geometrically its a slope of the graph of a function

this is a good example, we can see the m = -0.34, which means the function (green) is decreasing

[unique flower]

@fluid jetty are you in calculus or no

[fluid jetty]

im confused

[unique flower]

because usually increasing /decreasing problems are high school math content

[fluid jetty]

im in adv algebra w trig bc

[unique flower]

ok then

as x increases, does the y value increase or decrease?

ex: if the x value is 1 in stead of 0, would the y value be higher or lower

and "increasing or decreasing" should apply to all points, not just 0 and 1

[fluid jetty]

so

-1/3 to the 1st power would be what

[unique flower]

-1/3

[whole granite]

This is the graph of your function

[fluid jetty]

at -1/3 to the second

[unique flower]

without testing values, judge by eyeballing if it goes "up" or "down"(you have to be careful in some cases)

[fluid jetty]

i understand

so like

-1/3 to the negative 2nd power would be -9

so as x gets lower

y gets lower

[spare moss]

Could you please explain, the equation beneath this one(1 - cos2x = 2sinx^2)

Why has it done this way and how has it done (1 - cost = ... )

[silk wren]

is this an AAS/ASA Postulate? because there's ∠N, ∠Z, the vertical angle M, and the ZM = NM

[tepid wave]

R
|\
|_\__N
Z M\ |
    \|
     G

[silk wren]

waz dis OwO

[tepid wave]

its just angles are the same

they're similar triangles

thats waht ur diagram actually looks like

[silk wren]

But but

the RM and GM

it didnt say theyre congruent

[tepid wave]

interior angles to show RMZ = GMN

then all angles are the same

they share same length on sides with same angles

[silk wren]

sosososo

uhh

if there's a vertical angle

uh like

all the sides are congruent

so RM = GM?

[tepid wave]

thats what it actually looks like

[silk wren]

oh why did u remove the vertical angle mark?

[tepid wave]

looked gross

[silk wren]

lmao

HAHAHA

soooooooooooooooooooooooo

it is an ASA Postulate?

[tepid wave]

i mean u can call it that

[silk wren]

okayyy thanks ❤️

[tepid wave]

but its literally just the two triangles look the same and are the same

[silk wren]

mmmmmmmmmm

tyty

[north heart]

notice that it's symmetric about bd

[grave pond]

And the little white triangles outside the given shape are all congruent.

So the sum of the hypotenuse and the short leg is 4.

And so the hypotenuse is the solution to x² = (4-x)² + 3²

[austere pendant]

did you know the volume of your mom is

wait nevermind

[wise bear]

a calculator only has so many input spaces

[vague token]

,rotate

[vague token]

How do i solve this I don't know where to start as every approach I know of doesn't suit this question

[clear crypt]

Any you study Euclid’s bro

[nocturne remnant]

Why would I study Euclid’s bro

[upper karma]

I would calculate area of bunch of triangles and find solution there, but its not the elegant way i guess

[tepid wave]

This is how I'd solve it

And then generalize it for the other case

where CD is scaled by some factor

[upper karma]

Have you slved this?

[tepid wave]

I think its like 9.375, since BC = 3.125

[upper karma]

how did you found BC

[whole granite]

how sine is calculated since its not a linear interpolation between (-1, 1) and (0, 2PI)? So what kind of interpolation is it?

[dark sparrow]

it is not any kind of interpolation at all

if you want to look for ways to calculate sin(x) using a computer you might want to look into things such as taylor series or maybe the CORDIC algorithm

[burnt sigil]

Can anyone here assist me with this? I know to multiply by the conjugate and use pythag ID but I'm not able to simplify enough to get credit.

[storm portal]

@burnt sigil What's the conjugate of the numerator?

[tepid wave]

See what tropo said, I meant the reddish line not BC.
The red line = 4 - x

Looking at the poking out white triangle,

it has hypotenuse (4-x), adjacent x, opposite 3
So (4-x)^2 = x^2 - 3^2, which solves to 3.125

[north heart]

if you only need to rationalize the numerator i think there's a funny trick here

wait, nvm

im a dum

sorry abt the ping lmao

[tepid wave]

yeah u just multiple the whole thing by:
$$1=\frac{\sqrt{\left(3-3\sin y\right)}}{\sqrt{\left(3-3\sin y\right)}}$$
And the numerator is rationalized

[somber coyote]

ryаn

[terse bane]

thought the operation sign stayed the same when it's under the radical sign

oh wait the question asked to rationalize the numerator

[terse bane]

that's the conjugate of the numerator right?

[tepid wave]

yee

[grave pond]

I don't think "conjugate" makes much sense for "reverse the sign of the sine".

If we have something like $a+b\sqrt c$ we can rationalize it by multiplying by $a-b\sqrt c$ because that ends up with $a^2-b^2(\sqrt c)^2$ and $(\sqrt c)^2$ \emph{is rational}.

[somber coyote]

Troposphere

[grave pond]

But $(a+b\sin x)(a-b\sin x)$ is not especially nice compared to $a+b\sin x$. We just get $a^2-b^2\sin^2 x$, and $\sin^2 x$ is not a priori an improvement over $\sin x$.

[somber coyote]

Troposphere

[digital agate]

suppose there is a set of smaller squares that form a larger square. There are circles inscribed in each of the smaller squares. is there proof that the sum of the areas of the smaller circles is equal to the area of the circle inscribed in the larger square?

[drowsy bobcat]

no because that isn’t true

here’s a counterexample:

wait

hmm

i drew up a quick proof

it could be wrong tho

idts >.>

say there’s a square made of squares that arent all exactly the same

you can apply this process to them since i think the #n = x^2 portion is generalizable

and then make sure each of them have the same size

and boom

i’m like 99% sure it’s correct tho and i cant be bothered to write out the generalization since it’s 1:10

yeah i know my handwriting sucks dont tell me ¯_(ツ)_/¯

there are squares like this one where my argument sorta falls apart

but this should again be generalizable so it shouldn’t matter

too tired to actually show it tho, if someone wants to do it for me to wake up to in the morning that would be cool

nvm i lied lmfao

i was too curious

i did it

here ya go @digital agate

i’m like 99% sure I’m correct

[drowsy bobcat]

that should say “number of x = y^2”

oops

[drowsy bobcat]

actually fuck me now that i think of it i think this is sufficient proof

since all the squares are equal

awwww fuck this

the last one is still better

but both work

damnit

yeah the first one works because you can break down every square into the 1x1 case and then the 1x1 argument applies to the big square too

[drifting pollen]

@drowsy bobcat

May I ask you a dumb question?

[drifting pollen]

Find CE

[limber shale]

I think that i saw a very similar problem on the Mind Your decision YouTube channel, but i forgot how to solve it😂

[nocturne remnant]

Power of a point

[drifting pollen]

Lmao

I was doing optics

Then a question struck in my mind

Whats the relation between thickness/ apprature of the lens with the radius of curvature

But logic is pretty simple

[upper karma]

[drifting pollen]

You can solve this by similarly

Or there is an another complex way to solve

[drifting pollen]

Not actually*

First find the radius of the circle

[upper karma]

following a question i asked in competition math channel, is there any analytic expression for θ in terms of R1, R2, R3 and the angle α?

i can find an expression for α in terms of θ, but not vice versa

[dark sparrow]

what's your expression for alpha in terms of theta?

[grave pond]

E is the center of the circle through A, B, D. The angle marked in red is from the central angle theorem: alpha is half of angle AEB. Now you can compute the coordinates of point E. Since |ED|=|EB| you then know all the side lengths in triangle EDC, which you can solve to find angle ECD. Add or subtract this from angle ECA.

[zinc mural]

Is it possible to calculate lower and upper sum without seeing a graph? For example like f(x) = lnx^2 [1,7] with 30 subintervals.

[grave pond]

Sure -- especially since you know ln x² is increasing on the entire interval, so the maximum in each interval is always at its right end.

[zinc mural]

Yea, but wouldnt that take a lot of time? If each interval is 0.2

[grave pond]

Perhaps -- but I don't see how having a graph to look at would make it faster anyway.

[dark sparrow]

are you forced to do it by hand

[grave pond]

Nobody does that by hand anyway, except a few times in "make sure you've understood the definition" exercises.

[zinc mural]

Well its always good to be prepared to do them by hand. But I think the usual lowersum(lnx^2, 1, 7, 30) works 😛

[dry basalt]

According to the lenght, a_1 and _2 is equal to what here ?

[dry basalt]

<@&286206848099549185>

[fickle abyss]

https://cdn.discordapp.com/attachments/268882317391429632/960376590926245928/unknown.png

HELP

[upper karma]

helps with solution

[dark sparrow]

@upper karma do you want someone to give you the answers to these?

[upper karma]

i tried answering it

[dark sparrow]

so you have tried these problems but got stuck, or got the wrong answers?

[upper karma]

yes

[dark sparrow]

okay, then show your work.

[upper karma]

d cc

d,c,c

[dark sparrow]

your work, not your answers

there's no telling where (or whether) you made a mistake if you don't show how you got these answers

[drifting pollen]

4x+2+3x+14+(180-(9x-14))=180

For first

For 2nd it's 180-52 = 128

For 3rd Heptagon

[drifting pollen]

4x+3x+2+14+180-9x+14 =180
7x+16+180-9x+14=180
-2x=180-180-16-14
-2x=-30
so x = 15

Nvm u already solved

[dark sparrow]

@drifting pollen do not give out answers or work.

[drifting pollen]

What?

[dark sparrow]

you did mingmei's work for them.

don't do that in the future.

[drifting pollen]

@Ann#0413

@dark sparrow

Assume that there is an equal chance of being born on an odd or even numbered date in a month. A family has two children. If one child was born on an odd date, the probability that the other was born on an odd date is

Oops wrong section

[dark sparrow]

what

why ping me

[drifting pollen]

Find the probability

Pls

[dark sparrow]

no, why ping me specifically?

[drifting pollen]

Cuz ur genius

[dark sparrow]

😒

get lost

[drifting pollen]

Srry

Btw there is no harm in solving a problem

[worn glacier]

Is there anyone interested in making 10-20 dollars helping with geometry

Dm me ASAP

[nocturne remnant]

#rules

Oh he’s muted

[hoary grove]

Hello, I'm trying to prove corresponding angles are congruent. Can you check it out?

Lines T and L formed an intersection and so does lines T and M. Because lines L and M are parallel, the angles formed between x & y and x & z must be congruent. Therefore <gamma is congruent to <beta, proving corresponding angles are congruent.

[drifting pollen]

Wait

I have proved this too

If you consider line as collection of points then the rate of change will be same for all points

@hoary grove

It's little bit similar like why Vertically opposite angles are same

[upper karma]

I only needed to find the area of the shaded parts, basically those grey spots in the picture, i already got an answer but i wanted to check if im right, i got the answer of 36.28cm^2, did i get it right or not?

Pretty sure its area of square

[silent plank]

can you show the calculations your performed that led you to 36.28

[signal swallow]

don't do people's work for them

also ??

[drifting pollen]

@signal swallow tell me rules

Idk

[signal swallow]

#rules ?

named appropriately

#❓how-to-get-help as well...

[drifting pollen]

Alright didn't know that

[young vector]

can someone help me plot this in geogebra?

[hoary grove]

Then do I say: "Because the rate of change are the same for lines L and M, they form the same angle with line T"?

[drifting pollen]

Yes

[drifting pollen]

Since rate of change of T with respect to rate of change of L is equal to rate of change of T with respect to rate of change of M cuz L and M both are parallel so rate of change of both are also equal

Wtf

[north heart]

isnt that just a pic of like

stokes theorem

lmao

[drifting pollen]

I accidentally

Sent it

And can't delete now

What a bs

Oh so where was I

[upper karma]

tysm

[drifting pollen]

A thought came to my mind to prove it easily

[neat horizon]

anyone?

[dark sparrow]

@neat horizon have you made any progress on this or are you stuck not knowing how to begin?

[neat horizon]

stuck

[dark sparrow]

mkay

heres a rough sketch of the scene

given this sketch, would you be able to show where the angles of elevation and depression appear here?

[neat horizon]

yes, from the base of the building is 18 deg. 50 min = 18.83 deg elevation then the angle of depression if from the top of the building to the base of the pole 48 deg. 10 min = 48.167 deg.

is that right?

[dark sparrow]

sorry, let me be more clear.

are you able to draw these angles on the sketch that i gave you?

just to ensure you understand where they are geometrically

if you need to label any relevant points then just say which ones you label and how

[neat horizon]

is this right?

[dark sparrow]

okay, so you got the 18°50' angle correct

but not the angle of depression

this is where the angle of depression is supposed to go

angles of elevation and depression are always measured from the horizontal, after all

[neat horizon]

Okay, I see. I dont know the next part of it.

[dark sparrow]

there are some right triangles here that you can make use of

would you like me to give names to relevant points or would you rather give the names yourself?

@neat horizon

[azure kelp]

So I'm currently doing trig Identities and my teacher gave a answer key but I don't understand one of the steps. In this problem why are you allowed to multiply the equation by (1-cosx)/(1-cosx) right in the beginning of the problem?

[upper karma]

It's a fraction mate,if you multiply a value with the numerator and denominator the fraction's overall value will remain the same. Your teacher did that to make the denominator into sin^2x

[azure kelp]

So with any type of problem like this when simplifying for one side I can multiply it by anything as long as it's done to the top and the bottom?

[upper karma]

Yep

[azure kelp]

alright thanks

[upper karma]

Is the probelm to prove it equals to (cotx - cscx)^2

[azure kelp]

yeah

[upper karma]

Then it's good

It's a correct way

[azure kelp]

I wasn't sure because since you can't change what's on the right side I thought it would limit what you could do on the left

[hollow vapor]

Ok i have a question: Can tan45°=1 be proven like tanh(x) =1 where x=∞?
And the same question but framed differently is that
Can we prove mathematically tan45=1 without any geometrical help?

[broken anvil]

Ultimately I think ull have to use geometry cz it’s itself a geometric ratio

[hollow vapor]

Maybe that's why Euclidean geometry is limited.

[dark sparrow]

pretty sure sin(45°) = cos(45°) = sqrt(2)/2 can be derived by non-geometric means

it might be painful but it prob can be done

[upper karma]

A circle has a 12cm radius, is the answer 226.08cm^2 always the answer or other solutions can have different answers aswell?

[dark sparrow]

wym by "the answer"?

what are you looking for exactly?

are you looking for the area of the (entire) circle?

[silent plank]

the answer to a question depends on what's being asked

[upper karma]

Yes ig?

[dark sparrow]

"ig"?

so you do not know your own goal?

[vagrant ore]

I need some help with this problem

[hollow vapor]

Does anyone know the process of finding it? Cause that is quite important.

[dark sparrow]

why is it important?

do you have a problem that absolutely positively requires abstaining from anything and everything geometric?

[hollow vapor]

It is important for my understanding.
No i don't have a problem but I'm looking for an alternative.

The main reason being we can't use Euclidean geometry to map space-time but instead we use hyperbolic trigonometry to find rapidity.

[dark sparrow]

...??

[hollow vapor]

Do you have the proof?

[dark sparrow]

are you ok with:

• defining cos and sin via their taylor series
• defining pi to be 2 times the smallest positive solution of cos(x)=0
  and having the proof be based on those

[north heart]

you could probably frame the geometric argument in the complex plane

and then solve using like vectors and such

which is technically non-geometric

[hollow vapor]

Yes I'm ok with it.

[dark sparrow]

okay

it can be shown just from the taylor series alone that cos(2x) = 2cos^2(x) - 1

substituting x = pi/4 into this we get 2cos^2(pi/4) - 1 = 0

noting that cos(0) = 1 (plug x=0 into the taylor series) and that cos is infinitely differentiable and hence continuous we get that cos(x) is positive for x ∈ [0, pi/2)

therefore cos^2(pi/4) = 1/2 implies cos(pi/4) = 1/sqrt(2)

[north heart]

wait

how do you get the cos^2(pi/4) = 1/2 part

[dark sparrow]

elementary algebra

[north heart]

oh

[dark sparrow]

from 2cos^2(pi/4) - 1 = 0

[north heart]

ohhhh

yeah i misread it

that makes sense

nice proof

[dark sparrow]

i suppose you could also prove sin(2x) = 2cos(x)sin(x) (from taylor series fuckery) and cos^2(x)+sin^2(x) = 1

use the latter to conclude sin(pi/2) = 1

@hollow vapor TL;DR taylor series fuckery will give you everything you want.

[hollow vapor]

Why just π/4? If I substitute only π as well it would still give me 1/√2.

[dark sparrow]

??

what are you talking about?

[hollow vapor]

The substitution

Substitution of x =π/4

[dark sparrow]

are you trying to claim that substituting x = pi, and not x = pi/4, into cos(2x) = 2cos^2(x) - 1 would give you that cos(2pi) = 1/sqrt(2)?

y/n

[hollow vapor]

Y

[dark sparrow]

i mean

you're wrong

you would get cos(2pi) = 2cos^2(pi) - 1

and cos(pi) = -1 so this leads to cos(2pi) = 2*(-1)^2 - 1 = 1

so i don't know what part of your ass you pulled this out of

do you still insist that substituting x = pi into cos(2x) = 2cos^2(x) - 1 would give you that cos(2pi) = 1/sqrt(2)?

@hollow vapor

[hollow vapor]

I made a mistake considering cos π/2 = 0.
See you don't understand what I'm trying to know. Lemme put it clearly.
We know for a fact that cos 45 = 1/√2
Sure, but how do we know that? And how can it be found? Whatever be the value. I'm talking about the very fundamentals if you know what I mean

And i don't want the geometrical answer for it.

[dark sparrow]

there are numerous ways to see why cos(45°) = 1/sqrt(2)

well

i guess most of them are geometric

you asked for something non-geometric

[north heart]

what's wrong with the geometric answer

[dark sparrow]

you got something non-geometric

and now you are unhappy with the non-geometric proof that you received

[hollow vapor]

Nothing wrong, I'm just trying to find something which is bothering me bad.

[dark sparrow]

what is bothering you?

[hollow vapor]

Because you already considered cos π/2 to be 0. I mean if I know how cos π/2 =0 without the geometrical proof. I'm good.

[dark sparrow]

i asked if you were ok with:

• defining cos and sin via their taylor series
• defining pi to be 2 times the smallest positive solution of cos(x)=0
  and you said yes.

that second point is the definition of pi that does not rely on any geometry

just as you wanted

and from this definition it follows in the most direct way possible that cos(pi/2) = 0

does this resolve your confusion?

[hollow vapor]

Yes it does. Thank you.

[worthy gust]

Does quadratics fit this channel?

[wide pumice]

Thats algebra

[opaque rampart]

Hello I need some help on my homework

It’s about law of cosines

[outer lagoon]

cos 45 is root2/2

[trim breach]

@tulip vector There are 180 degrees in a triangle.

The sum of all three angles must equal 180 degrees.

[tulip vector]

i know that, but I'm kinda getting confused on how to use the x's

from the 2x-12 and x+16

[north heart]

add em

[trim breach]

If all the angles add to 180, then your equation is:

,,x + (2x - 12) + (x + 16) = 180

[somber coyote]

Lidoh

[trim breach]

And then you solve for x.

[tulip vector]

Ok, thank you so much for the help! 🔥

[upper karma]

Can someone help me with proportions dm me

[stuck flax]

can someone help me with this problem, it says solve for x and assume all tangent lines are tangent

[north heart]

is (2x+35) the angle qrs?

[stuck flax]

Im guessing yes

[north heart]

what's the 70 degrees then

[stuck flax]

man I dont know Im so bad at math thats why Im on this server 😭

[north heart]

yeah im not gonna lie that's just confusing lol

i have no idea how to help u lmfao

[azure crown]

i think theres that one theorem that says the angle qrs is half of 70

and if <qrs is 1/2 70 then uhh

[north heart]

that's the inscribed angle theorem

[azure crown]

do algebra

to find x

[north heart]

but how are you getting the half of 70 thing

[azure crown]

wait yea

the inscribed angle is half the intercepted arc

[stuck flax]

I have the answer sheet It says the answer is 0

[north heart]

wait, no it isn't.

inscribed angle theorem says something else

although according to the answer sheet it's correct??

i think this is just a shit question tbh

[azure crown]

well then x is 0

so it's just 35

[north heart]

70/2 = 35 + 2x, 2x = 0, x = 0

is the logic

[azure crown]

and also where is the tangent line

[north heart]

it's dumb tho, since we don't actually know what the 70 represents

^^

[stuck flax]

the 70 is the arc

[azure crown]

it says assume tangent lines are tangent

[north heart]

it's in degrees

[azure crown]

i think 70 is arc qs

[north heart]

.

[stuck flax]

yea

[north heart]

nobody labels arcs in degrees

[azure crown]

well yea u can tho

[stuck flax]

so how whould i show the work for it

[azure crown]

2x + 35 = 1/2(70)

[north heart]

say that 2x + 35 = 35 by the inscribed angle theorem

and solve algebraically

[stuck flax]

How about this one

[north heart]

gods, you might have the dumbest labeling scheme of all time on your hands here

im not even gonna spend brainpower on this

[azure crown]

wait

so uhh

the angle is half 8x+10

[north heart]

nope

you cant assume that

[azure crown]

so 49 = 1/2(8x+10)

yea u can because

it said

[north heart]

since it's not an inscribed angle

no you cant

it isn't an inscribed angle

[stuck flax]

The answer for this one is 11 I just need to know how to do it

[azure crown]

noo it's a theorem bc fe is tangent line

[stuck flax]

oh yea so thats right

[azure crown]

see he said it's 11 so if u plug 11 into this equation i put it works

[north heart]

? prove it then

[azure crown]

it says it

[azure crown]

up

here

wait no

[north heart]

that's not the statement of the inscribed angle theorem

[azure crown]

wrong image

[stuck flax]

it says assume all lines that appear tangent are tangent

[north heart]

not that

that's fine

[azure crown]

no thats another theorem

[north heart]

bruh which theorem is this then

[azure crown]

ima get out my fucking notes just to type it out ill brb

[stuck flax]

I kinda remember the therom is says that one of the angles is twice as big as the others, something along thoes lines

[azure crown]

"the measure of an angle formed when a secant and a tangent intercept at a point on the circle is a 1/2 the measure of the intercepted arc."

[north heart]

huh

ooh i should try to prove that

[azure crown]

alright

[stuck flax]

yea that

[azure crown]

i also think our school is using the same textbook as urs cuz we just finished that same chapter and had the test

[stuck flax]

yea probably but we dont use the book we just get notes in class and the teacher gives out these woksheets

its just tricky questions like that, that are bothersome

[azure crown]

yea

[stuck flax]

Last one I promise 😅 I just don’t know how to do these type of problems

idk why the picture got compressed like that

[azure crown]

can u make it bigger

image

[stuck flax]

fuck hold up its still bad

[azure crown]

ok so

is that vertex the center of the circle

there is like no given

information

[stuck flax]

not really

[azure crown]

cuz if so it would just be 5 because theyre both radii

[stuck flax]

the answer is 5 it says

[azure crown]

yea so

that one point in the middle is the center im assuming

idk if i can

but since that radius is 5

and the line x is also radius

it is 5

[stuck flax]

oh ok that makes sense

ok but how about this one, its more tricky

[azure crown]

ok

[azure crown]

4

[stuck flax]

thats correct how did you get that

[azure crown]

so basically the radius diameter whatever is perpendicular to the chord

and if that happens then the radius bisects the chord

so if it bisects and one side is 4

the other is 4

[stuck flax]

ohh ok ok i see

[azure crown]

idk check ur notes to see if u learned that theorem

[stuck flax]

i was absent that day thats why i was so confused about it

[azure crown]

ok

[stuck flax]

this one shouldnt be as obvius right?

[azure crown]

yes

this one harder

wait no

just use

a2 b2 c2

[stuck flax]

ohhhhhh

[azure crown]

11.7^2 + x^2 = 14.2^2

[stuck flax]

yea that whould be 8

[azure crown]

yea it would round to that

[stuck flax]

I done with it now, thank you Mr.LazyDuck you and valley were the only ones that could help out 👍

[azure crown]

np

[stuck flax]

can I add you just in case if I ever need help?

[azure crown]

yea

[stuck flax]

thank you : )

[whole granite]

lazyduck solution is simple and better but you can also use it to find x

[azure crown]

for first one i know 23x-2 = 1/2(247- arc qs)

and u can find arc qs by subtract 247 from 360

then that will give u x so u can get angle measure and arc

[uncut palm]

Thank you so much bro

[azure crown]

ngl no clue about second one

or third

[uncut palm]

You got help with this one?

[azure crown]

what am i finding

[azure crown]

just 147 i guess

[uncut palm]

holy shit

i didn't see the question mark

[azure crown]

because central angle equal to intercepted arc

[uncut palm]

Bro i need glasses 😂

[stuck flax]

Hello I’m back with a quick one, can I assume that the chord is the same length as the radius and then use pathagreom theorem?

wait nvm it gives the chord lenght

[whole granite]

same here

[stuck flax]

so whould x be C or B

[azure crown]

wait u need help cuz i can think of a way @stuck flax but it seems kinda complicated

and idk if theres a simpler way

[stuck flax]

sure go at it

[azure crown]

so u make a right triangle using radius length 13.2

one side is 10.9

other is x

no i mean y

then solve for y

then subtract y from 13.2

and that is x

ur answer

[stuck flax]

wait so how whould i make a right trangle

[azure crown]

u see the 10.9 side

yea

the hypotenuse

is like on top of that

or opposite the given right angle

[stuck flax]

like that?

[azure crown]

no

[stuck flax]

where

[stuck flax]

ohhhh i see

[azure crown]

yes what he did

[stuck flax]

wow

[azure crown]

@stuck flax make sure u use a different variable for the side length because u have 2 x side lengths

so u see he used y

[stuck flax]

ahh i see

damm thats some big brain I whould be stumped other wise big thanks

Also for this one I keep getting weird answers, the answer is supposed to be 9 according to the answer sheet, but I can’t seem to get the answer 9

[whole granite]

is it complete? the question is just it?

[stuck flax]

well for these types of problem thats how you usally set it up

i can show an example of one i did right

sorry for the sloppy hand writing but you can see that the number alone which is 40 in this case you put it to the power of 2, then you get the other outside number (32) and multiply it by that whole line (32+x)

[stuck flax]

so following that logic for these type of problems I still couldnt get the answer 9 for this question

I dont know if the answer key is wrong or if I am wrong

[whole granite]

why did you multiply (32) by (32 + x)?

[stuck flax]

thats just the formula for these types of tangent questions

theres probably a legit answer to that but i dont know

[whole granite]

I see, I need to understand where this formula came from

[stuck flax]

ummm

theres proabaly a name for it that I dont know

but it has to do with tangents

[whole granite]

hmmm

[stuck flax]

Here’s another example of it working

[whole granite]

they create a rect from the 9 times 9+x

but why 40² = x*y??

[stuck flax]

this problem whould be 40²= 32 (32+x)

and from there its just a algebra equation

[stuck flax]

so for this it would be x-3^2=x-5(x-5+5)

[whole granite]

https://geometryhelp.net/tangent-line-to-circle/

lets try to understand this formula to re-apply it to any kind of problem

[stuck flax]

could you try and solve x-3^2=x-5(x-5+5) as an algebra equation

[whole granite]

the algebra dont matter so much right now, the geometric logic is what I'm looking for

[stuck flax]

ahh ok

Its that I was just wondering if I was doing the algebra part wrong or not

[whole granite]

wtf

[whole granite]

I guess you missed some algebra because the formula works, but I can't understand it yet, the why it works

[stuck flax]

huh so it was the algebra

that's weird because i tried to put it in mathway and that didnt work either

[whole granite]

someone knows how can we prove this relation? or name of it?

I mean, it's obvious D = B and then D² = B² when both lines are tangents to the circle

But how about when the lines are not equal (image 2)?
Why we can assume the same rule as image 1 applies for the image 2?

[nocturne remnant]

Go look up power of a point or sth

But the proof is simple: just check that the purple triangle and the orange triangle is similar, and derive the length relation from there

[analog valve]

Can someone help me with trigonometry

[fair cloud]

what is the phase shift of this function

im so confused

wait i got it its solving like an equation

[cold totem]

ctg x <= -sqrt(3)/3
1/tg x<= -sqrt(3)/3

tg x <= -3/sqrt(3)

tg x <= -3/sqrt(3)
tg x <= -3sqrt(3)/3
tg x <= -sqrt(3)

tg pi/3 = sqrt(3)

tg -pi/3 = -sqrt(3)

q.e.d.?

now this

sqrt(3) = tg pi/3

so
(sin pi/3)/(cos pi/3) * cos(2x - pi/3) = 1.5

idk what now

[cold totem]

<@&286206848099549185>

[hoary grove]

The test is over, but still, what is the thought?

[drifting pollen]

Dm @hoary grove

[hoary grove]

sure

[broken anvil]

Hey I’m trying to find out what’s wrong in this working, as my answer for sin18 is coming different from what it should be…can someone pls tell what’s wrong?

[silent plank]

cos^2(2t) isn't 2cos^2(t) - 1

[upper karma]

can i have help with this pls ?

[icy lake]

what does co-terminal angle mean/

[broken anvil]

Ohhhh ya

Thanks…

[zinc mural]

How can you find the intersection points of y1 = -x^2 + 4ax + 5a^2 and y2 = x^2 - a^2?

[drifting pollen]

By using brain

[desert galleon]

this is review but i completly forgot can ppl jus help me

[north heart]

seriously?

look up the formulas for area and perimeter of these shapes and just plug it into a calculator

[desert galleon]

i came here bc im not aloud to use a calc

[opal blaze]

You don't need to use a calculator. Just solve the formulas by hand using the formula.

[desert galleon]

idk the formulas y would i b here

circumfernce and area of circles

[upper karma]

i need help plssss

[timber ridge]

Then post your question and clarify what you tried and where you're stuck

Depends on context. Where are you getting the angle from?

[mystic coral]

What am I missing?

[timber ridge]

Well, how'd you get that solution, firstly?

[timber ridge]

Oh man I'm silly I'm overlooking the obvious

[timber ridge]

Your work is right but [0,2π) does not include 2π

The only solution you care about is 0, then

[mystic coral]

oh, duh, thanks lol

[upper karma]

Guys can 3.14 be 3.1416 in geometry?

[grave pond]

3.1416 is one of the numbers that would round to 3.14 if you are to state them to two decimal places, yes.

That's as true in geometry as it is everywhere else.

(Both of 3.14 and 3.1416 are also decimal approximations to pi, which I suspect may have something to do with your question).

[upper karma]

For SAS rules and AAS rules for law of sines, do you apply SAS when the triangle is right side up and for AAS when the triangle is upside down?

Pls?

[trim breach]

The orientation and labeling of the sides and angles is arbitrary. Do you have an example of what you mean?

[upper karma]

This is AAS (upside down)

This is right side up (ASA)

@trim breach

[trim breach]

Yeah, the orientation really does not matter here. What you are looking at is what parts of the triangle are given to you. AAS means you have two consecutive angles then a side. ASA means you have two angles and the length of the side between them.

You apply the Law of Sines the same way regardless.

[upper karma]

I think I have it now thnx

Appreciate the help

[crimson junco]

would appreciate some help on this

[crimson junco]

i ended up getting the vector ON as 3/5 a + lambda b

i have no clue what to do from there

<@&286206848099549185>

[crimson junco]

never mind i got it

[placid parrot]

Can i have help with these problems

[trim breach]

Have you drawn them?

@placid parrot

[placid parrot]

i dont really know where to start for 199 but ill show what ive got for 200

[trim breach]

Okay. I do not really understand your work, but we can start.

One of them should be really obvious with no math involved.

Can you identify which one?

[placid parrot]

the altitude going from b

[trim breach]

Yes. It is just 4, and you can just count along the y-axis.

To go further, the strategy I had in mind was getting the other sides of the triangle.

[neat pagoda]

HEY

I’m in need of

Assistance

😭

I had COVID, missed a lot of the beginning of trig

Trig is easy though

But before we learned trig

[trim breach]

We can use the altitude going down from B to get the length of AB with Pythagorean Theorem.

[neat pagoda]

We learned Uhm metric relations

And I cannot comprehend how to answer my question

[placid parrot]

so its just 4?

[trim breach]

Sorry I meant AB.

[placid parrot]

i dont undersand

[placid parrot]

bro

[neat pagoda]

The writing under is the answer but idk why it’s the answer

[trim breach]

Let me draw it real quick.

I just moved the altitude coming from B outside to be able to see better.

But from (0, 0) to (1, 4) is four units up and one unit right.

Those are two legs of a right triangle.

[placid parrot]

what about the lengths for the otther altitudes'

like aare the points i have right?

[trim breach]

What are the points used for?

I’m asking because I can’t follow.

[placid parrot]

to use the distance formula

[trim breach]

Oh.

Okay. That is a pretty roundabout way to do it.

[placid parrot]

ok

[trim breach]

Were you told to use distance formula or is that the strategy you chose?

[placid parrot]

thats how i was taught'

but i didnt know there was another way

[trim breach]

There is something called the Law of Sines, which uses basic trigonometry and would have the problem done much quicker, but I think you can use distance formula.

[placid parrot]

i havent learned trig yet

[trim breach]

I have to be somewhere in a few minutes. Someone else might be able to help because the application of distance formula might take a bit longer. Sorry. I just don’t want to start and then have to leave halfway through.

[placid parrot]

ok

[hushed viper]

Hello I’m looking for some help. So I need to solve for X but I’m kinda lost on what to do

Edit: I’m sorry if this is in the wrong chat it just showed up in geometry homework and I didn’t know where to put it

Does anyone have a video that could point me in the right direction?

[wise pawn]

you can't solve for x with this kind of information because it's scaling all sides

furthermore, this is false, no triangle can be made with these sides

you can check this by seeing that it doesn't satisfy the pythagorean theorem

[hot grove]

Very stuck on this, id really appreciate some help

[timber ridge]

Hint: consider angles CAO & OBD

[stuck dragon]

^

[kindred turtle]

I have found that x = -2y - 101

I don't know where to go from here.

I have to solve for x and y.

I set all the angles added up to equal 360 and solved.

But after that... unclear.

[silent plank]

note that you have a cyclic quadrilateral
not just any random quadrilateral

[stuck flax]

It says “if the radius is 5 and JL is 1 unit longer than KL, what is KL? (Hint: Use x for KL and x+1 for JL.)

Any ideas ?

[silent plank]

you have a right triangle

[stuck flax]

Yea

[trim breach]

You defined all three sides.

[stuck flax]

The erased marking are from other problems that use this same figure

[trim breach]

Use the hint given in the problem.

[stuck flax]

I tried but I kept getting wrong answers

Whould I do 5^2+x^2=x+1^2 ?

[trim breach]

Yes!

But it is (x+1)^2 to clarify.

[silent plank]

missing parentheses

[stuck flax]

Oh that’s why

Thanks

Also how would you get z here it whould be z^2=2(2+12) right? Because my class mate got another answer

[kindred turtle]

Lol, thanks. I forgot about that.

[vagrant ore]

How do you find the altitude of a square pyramid with the slant height?

[simple pewter]

Hi, is anyone familiar with Riemann-Zeta?

[north heart]

a few things

1. https://dontasktoask.com

2. even if someone did, a pre-university class is not where to ask

3. please read #❓how-to-get-help

[slim saddle]

Hey, why is number 4 not a triangle?

[trim breach]

If you try using Law of Sines, you find out it doesn’t work. (sin(C)/c) * b > 1, so you cannot take arcsin of it because anything >1 isn’t in the domain if arcsin.

[slim saddle]

Alright, thank you.

[grave pond]

Be sure you also understand the geometry of it. If you have the side AC with the right length, and draw the line though A that B ought to lie on according to the given angle, it turns out no points on that line are within 21 km of C.

[odd knoll]

Find the area of the darkened figure if A,B,C and D are the centers of the circles. AB=1 and ABCD is square.

[north heart]

note that the square vertices on the bottom and the top point in the cuboid create an equilateral triangle

that's all you really need

[dark sparrow]

cuboid?

[north heart]

eh, not sure of the proper term

squircle?

ignore my terrible drawing skills

that thing

that right there is an equilateral triangle

and this makes solving really easy

[dark sparrow]

where do you see a squircle

[north heart]

uhhhh

good question

idk the term

the hyperbolic square

that thing

OH

[dark sparrow]

it's not hyperbolic either

[north heart]

the shaded thing

that's what i'll call it

[dark sparrow]

if anything it's bulging out so it would be spherical

[north heart]

the shaded portion

[dark sparrow]

the shaded region

[north heart]

say, is there a proper term for that shape?

[dark sparrow]

not really

[north heart]

shame

[dark sparrow]

curved quadrilateral maybe but thats the only thing i can think of rn

[crude inlet]

i'm guessing it's something to do with subtracting areas from one of the 1/4ths of the circle or the square

to find the cuboid-curved area

[crude inlet]

yeah that works too

[crude inlet]

ok nvm i think i figured it out

[grave pond]

The same problem came up here: #geometry-and-trigonometry message

[drifting pollen]

Make a square , find its area and add area of lens( idk what to call it exactly)

[drifting pollen]

Yeah that's better

[tepid wave]

have you guys tried the one with parabolas

its more fun

[drowsy bobcat]

what even is this lmaoo

also you aren’t supposed to give answers otto

oooh but ryan you should share

[tepid wave]

Yeah, it's just find area covered by 3 parabolas rotated around the focus, 120 degrees

[tepid wave]

Given a parabola of the form $a(x-h)^2+k$, the area of the rounded triangle inbetween the parabola rotated $120^\circ$ 3 times around the focus should be:
$$\frac{20}{48\sqrt{3}a^2}$$

[somber coyote]

ryаn

[tepid wave]

The shaded region is 20/(48√3a^2)

[wise pawn]

fun, what I did was ||translate it so that the focus went through the origin, then I just had to worry about finding the area of the bottom triangular "sector". But that can be found by integrating the region wrt x to where it intersects the lines making 120 degree angles and subtracting off the right triangles from the sides.||

||here's a pic of the green region I'm talking about where I cut the triangles off from: ||

[wise pawn]

Similar problem for you, you can make a smooth closed shape by joining parabolas together where their tangents are the same, what's the area of this shape? https://www.desmos.com/calculator/zdlghzapgm

[coarse raptor]

Hi Guys

When we have for example $\abs{z-z_A} = \abs{z-z_B} \equiv AM = BM$ , we can establish the equation of Δ, mediatrice of AB like $y = f(x)$?

[somber coyote]

Iηcθmιng

[upper karma]

The task is to find the angles and prove that the lines which are parallel are parallel, I’ve been able to find most of the angles but I’m not sure how to set up the working and the proof for the side lengths

It’s a regular pentagon

[upper karma]

I got the angles

Wait I think I got the proof as well but not 100% sure

I did it using Angle EAF and BAF but I’m not sure how to set it out

[fast pulsar]

Try a slightly smaller claim. Does a trapezoid have any parallel lines?

In particular, we care about the isosceles trapezoid

[tepid wave]

||Did basically the same stuff, after finding that triangle 'a', I found the largest inscribed equilateral then used Archimedes method (quadrature of parabola) of finding area of segment of parabola (which is 4/3 of the inscribed triangle), which is basically integrating. And added them up||

[fast pulsar]

Oh, this was 5-ish hours ago...

[short mason]

hey

anyone can help me on this?

[grave wasp]

SSS

bc the third side is congruent to itself

Does anyone know if there is a relationship between the tangent and secant lines in a circle and the tangent and secant trig functions? And if there is is it something i can understand without having taken calc

[nocturne remnant]

I think there’s a video on numberphile but I need to find it

https://youtu.be/snHKEpCv0Hk

[tepid wave]

,texsp ||your monomial one can be framed in the same way as the intersecting parabola one. not sure if I did it correctly, but assuming $a > 0$, for parabolas $a(x-h)^2+k$, rotating it about $\left(h,4a+k\right)$ by 120 degrees, the area of the intersecting shape should be:

$$\frac{13\sqrt{13}-19}{6\sqrt{3}a^{2}}$$||

[somber coyote]

ryаn

[silent plank]

you have a tangent and some secants
consider the tangent secant theorem

[vagrant ore]

Oh wait I took the wrong picture

[uneven glen]

Just use the fact that 55 = (100+35-x)/2

[lean garden]

need help trying to find weight using height and volume to find weight/mass. (Of a cone / conical pile)

[distant pebble]

nvm i think i figured it out

and to think 16y/o and below me thought taking notes was pointless

guess i sure showed him by turning into a procrastinating idiot who struggles with math

[tulip vector]

Hi, having a bit of a hard time in this question. Do i only need to add?

Thanks!

[dark sparrow]

@tulip vector yes, there is a theorem that says an exterior angle of a triangle (QRX in your case) equals the sum of the two interior angles not adjacent to it

[tulip vector]

m<1 - m<A + m<b do i use this?

[dark sparrow]

what

??

[tulip vector]

nvm nvm

[dark sparrow]

did you... read my message?

in case it was not clear, i affirmed that yes, all it takes to do this problem is to add the known interior angles to get the angle you're asked for.

[tulip vector]

It's all good, I'm just a bit tired from studying all night.

[upper karma]

Is this question related to geometry?
Q: A cylindrical container measures 8.3cm in diameter and is 16.2cm tall. How much water can it hold?

[dark sparrow]

yes, this question is related to geometry.

[upper karma]

Can u give me the formula on how to solve it?

The book never specified how

But i solved it using V=πr^2h method

[wise pawn]

looks like the right thing

[upper karma]

Oh wait

[wise pawn]

show your steps if you like, maybe someone will check

[upper karma]

Ok

[dark sparrow]

so you found the volume of the can

and are we to understand that the value you got disagrees with that of the book?

[upper karma]

This is how i solved it

Sorry if I have bad handwriting

Update: i found out i was right, no need for help anymore lol

[dark sparrow]

there are multiple issues with your work actually

first you're taking the radius as 8.3 when in fact it was stated to be the diameter and not the radius

second you appear to be claiming that 8.3^2 = 68.89**^2** for no good reason

and then the sudden ^3 out of nowhere

[upper karma]

I just followed the instructions of the book tbh

[dark sparrow]

so you did not actually think about what you were writing, then.

[upper karma]

I mean, what the book says is law i think?

[dark sparrow]

no

bad attitude

[upper karma]

An example of the book:

[dark sparrow]

(8 cm)^2 = 64 cm^2

but 8^2 is not 64^2

[upper karma]

This 6th grade math is wrong?

[dark sparrow]

no

i didn't say the book was wrong

the book is actually right this time round

what you are doing wrong is failing to understand that the exponents ^2 and ^3 were being applied to the units, so if you decided to omit the units then the exponents should go too

(to say nothing of the fact that you confused diameter with radius anyway!)

[upper karma]

They did it with other examples too you know ;-;

[dark sparrow]

did what?

[upper karma]

Ye nevermind, i should continue with other activities

Thanks btw

[dark sparrow]

don't thank me for having done nothing.

[upper karma]

Im losing brain cells at a rapid rate

[tulip vector]

bro same, what grade are you in?

[upper karma]

7th

[tulip vector]

nice I'm in 8th

[upper karma]

Though im still applying for 7th grade

My 6th grade about to end

[tulip vector]

oh ok

[upper karma]

Omg you play valorant, all star tower defense and in the mathematics server at the same time

Its like i met a mirror version of myself

And your filipino too

[tulip vector]

noice

[upper karma]

just to be sure, 1 pi is 3.14 radians right? so 1 radian is just a normal number

[upper karma]

Need help

[daring bramble]

Just a quick question, do the diagonals of trapazoids tell you anything?

[soft marten]

?

[lapis moon]

another pythagorean triple!