geometry-and-trigonometry

[grim musk]

but r,0 is the same as rcos theta and rsintheta because it lies on the circle

so it only becomes r,0 if theta is 0

but here theta is 30 degrees

so how does this work

!help

[signal swallow]

as you may have heard, we still don't know exactly what this means...

[spark stag]

but r,0 is the same as rcos theta and rsintheta because it lies on the circle
i don't follow, what do you mean by "the same as"?

what part is "the same"?

[signal swallow]

you also contradict yourself, "so it only becomes (r,0) if theta is 0 [degrees], but here theta is 30 [degrees]"

[spark stag]

the distance of (r, 0) from (0, 0) is the same as the distance of (r cos theta, r sin theta) from 0

[grim musk]

but r,0 is the same as rcos theta and rsintheta because it lies on the circle

[spark stag]

if you want to describe the point (r, 0) in terms of angles, you would indeed need to use a different variable

like (r, 0) = (r cos theta_2, r sin theta_2) or whatever

[grim musk]

this means if you put theta as 0 you will get rcos0 and r sin 0 cos 0 is 1 and sin 0 is 0 so yu get r,0

[spark stag]

or you could use theta' instead of theta_2

or whatever

just something that shows its a different angle

[grim musk]

o i see

thanks

[spark stag]

could use a separate variable entirely as well

[grim musk]

x?

[spark stag]

sure, though greek letters are most common for angles

[grim musk]

but i didnt get one thing

[grim musk]

isnt the angle just 0

[spark stag]

right

and this should make sense, since the sine of 0 is 0

and the cosine of 0 is 1

[grim musk]

so something like angle =Alpha where alpha is 0 degrees

[spark stag]

so (r cos 0, r sin 0) = (r * 1, r * 0) = (r, 0)

[grim musk]

yes but instead of theta while naming just use alpha even tho its 0 right

[spark stag]

if you want to represent it as a variable, yeah, you could use alpha

[grim musk]

ok

[wintry tundra]

So what do u need help with it

[manic forge]

this

[wild harness]

need help with this, im confused

[wintry tundra]

U should find the length and the slope of that line on the right side

Then translate to top point of it to the left side

And u will have the 4th coord

[wintry tundra]

For this, you need to find when sin(x) + 1 is 0 and when sin(x) is 0

[wild harness]

and what after

[wintry tundra]

And thats it

[wild harness]

i checked the trigonemtry graph and they are irrelevant to it

[wintry tundra]

Ur trying to solve for x right

[wild harness]

yes

[wintry tundra]

So just find when each of the those are 0

[wild harness]

what are the numbers

[wintry tundra]

So basically when sin(x) is 0 or -1

I'm not giving u the answer but I'll help u find it

Look at the unit circle and think about when sin(x) would suffice for those

[manic forge]

so 1,6 -2,1?

[wild harness]

thats what ive been confused about

which part of the circle is sin, tan, cos?

[wintry tundra]

The vertical line

[wintry tundra]

Yep

[dire vector]

If I have three random points that constitute a triangle

What is the best way to calculate the area?

Can't use 1/2(base * height) because I only know the coordinates of the points

And thus I only know the length between the points

[eternal phoenix]

Hmm u can use the Area of Triangle formula for coordinates

[silent plank]

determinant and/or formula mentioned above ^
herons

[eternal phoenix]

see this

[dire vector]

Most interesting

And if the triangle is in 3D space

I assume you just add a third coordinate?

[grizzled kindle]

@astral hull you can try to prove BD is the diameter of the circumcircle of ABCD

[upper merlin]

The straight line AB of triangle ABC is bisected at O and the perpendiculars AX, BY, CZ are drawn to any straight line OP. Prove that if A, B are on the same side of OP, then 2CZ=AX+BY

I am not able to understand how to make a diagram for this problem

[slow igloo]

Ouuuu

[slow igloo]

Wtf is this

I’ve never seen this

Yoooo I’m mad confused

[grizzled kindle]

What are you talking about ? We were forced to memorise this

[slow igloo]

What r u taking about

Memorize what

[grizzled kindle]

This formula

[spare temple]

@Nardo Wick#6946 -1 for x XD

[low wraith]

Can anyone help me with this?

[spare temple]

right traingle good

[vernal aurora]

One message removed from a suspended account.

[dire vector]

Drawing it on paper is one thing. Getting a computer do calculate it via code is another 😉 my restriction was that I could only code it from knowing the three vertices.

[grim musk]

hello guys
Image
for this pic we have an angle theta
and so (x,y) which is at some point P would be rcos theta and rsin theta respectively
now i had this one doubt
if theta here is some angle( it can be anything but lets take it as say 30 degrees)
and radius is 1
we get (x,y) as cos theta, sin theta
now lets say the initial side which is x also lies on the circle like the point P
then we will get the (x,y) of the side x to be r,0
but r,0 is the same as rcos theta and rsintheta because it lies on the circle
so it only becomes r,0 if theta is 0
but here theta is 30 degrees
so how does this work
!help

so for this thing

if somehow the line segment doesnt lie on the circle

then could we say its coordinates are acostheta,asintheta( if a is the length of its side)

[silent plank]

do you have a pic

[grim musk]

yea

gimme a sec

here

[silent plank]

repost it here so i don't have to scroll up

[grim musk]

?

[silent plank]

not sure what you're actually asking

[grim musk]

yea so basically

if the point x,y somehow doesnt lie on te circle

will it have the coordinates acostheta,asintheta if a is the length of its side

[grim musk]

just assume it

[silent plank]

still on the same terminal side?

[grim musk]

yea

[silent plank]

well then (x,y) would lie on a different circle of different radius (still centred at the origin)

and the same principals apply

[grim musk]

no no

imagine if you have a vector

say v

are its coordinates gonna be vcostheta,vsintheta

even tho it doesnt lie on a circle

like look at the initial side\

it doesnt lie on the circle

the length of the side is x

its coordinates are xcosalpha,xsinalpha?

alpha because theta is different

alpha is another angle=0

so thatll be x,0

[silent plank]

you said earlier that it was on the same terminal side

(i.e. implying that the angle is the same)

[grim musk]

ok

i didnt understand how to ask the q

just go thru the new part

i think that makes more sense

[silent plank]

regardless

well then (x,y) would lie on a different circle of different radius (still centred at the origin)
and the same principals apply

[grim musk]

im talking about the initial side

[silent plank]

consider any random point on the coordinate axis

(excluding(0,0))
you can draw a unique circle centred at the origin the passes through it

[grim musk]

wdym exploding

[silent plank]

*excluding sry

[grim musk]

ok

ok

[silent plank]

(radius * cos(angle), radius * sin(angle))

[grim musk]

look

in this picture a doesnt lie on any circle

but a has coordinates acostheta,asintheta

so if another such length is there

[silent plank]

you can consider basic right angle trig in this case if you want to simplify things

[grim musk]

wait

[silent plank]

and do you agree with

consider any random point on the coordinate axis
(excluding(0,0))
you can draw a unique circle centred at the origin the passes through it

[grim musk]

if you have a side length s

its coordinates are scostheta,ssin theta

even if it doesnt lie on a circle as suc

such

[silent plank]

so even if a circle isn't explicitly drawn, you can draw one yourself if you want

[grim musk]

for the initial side?

because for the terminal side

a circle is already drawn

[silent plank]

wdym

like if you really insist you can draw a crappy circle around it like this

[grim musk]

no no

[silent plank]

sry, i don't get what you actually want...

[silent plank]

might be a case of extreme overthinking

[grim musk]

look at this pic

[silent plank]

uh huh...

[grim musk]

the terminal side lies on the circle

[silent plank]

there is no circle there

[grim musk]

but the initial side might not lie on the circle

imaginary circle

[silent plank]

wdym by might not lie on the circle

[grim musk]

look the terminal side definitely lies on the circle

right?

[silent plank]

a circle

there isn't just only 1 circle

[grim musk]

ok

[silent plank]

there are infinitely many circles

[grim musk]

ignore the circle

if theres a length b

does it have coordinates bcostheta, bsintheta

or no

this is what im asking

[silent plank]

yeh

basic right angle trig

[grim musk]

even if its not on a circle

[silent plank]

why are you so obsessed with circles

[grim musk]

like that vector i showed

[silent plank]

and i'm pretty sure i already adressed that

[grim musk]

just draw an imaginary circle?

ok

[silent plank]

and you blatantly dismissed it

[grim musk]

no no i saw that

[silent plank]

if you feel the need to the point to be part of a circle, then draw one if it eases your mind

[grim musk]

ok

[silent plank]

for Q1, its basic right angle trig

and more generally, its the definitions of sin and cos

[astral hull]

haha, that’s actually what i am struggling with

[serene heath]

i am struggling with parabola could u tell me how to go about learning ti i have completed circles and straight lines

[upper karma]

john 1:1 forms perfect trapezoid

[dark sparrow]

numerology

[dark sparrow]

....

[upper karma]

That's cool

Do you have another one?

[short crow]

How do I go about with this?

[cosmic falcon]

since it is an isosceles triangle mark the base and perpendicular to be each of length a

that gives u the length of hypotenuse to be root(2)times a

the given square has the hypotenuse as one of its sides

so the square of the length of the hypotenuse must be equal to 24

which is 2a^2 = 24

now we were asked to find the area pf the squares drawn on each of the other sides

since the other sides have a side length of 'a'

the area of each square will be a^2

[cosmic falcon]

now use this relation to get the value of a^2 which is clearly 12

so the answer will be 12

[short crow]

Ah, I get it now. I have this habit of overcomplicating things lol

[jolly roost]

Olease i needd some help pelase

[jolly roost]

Just failed

[dark sparrow]

i mean you didn't tell anyone what you need help with

how exactly do you expect anyone to help

[upper karma]

does anyone know anything about Congruent Traingles cause i am struggling to do my homework i would appreciate if someone could help 😅

[humble pulsar]

nah, no one in the entirety of a math server has ever used congruent triangles

[upper karma]

uh alr 😭

[small cipher]

sure what are you struggling with

like the definition?

how to find equal angles and sides?

[upper karma]

😆

[wise pawn]

is every injective cubic function of the form a(x+b)^3+c?

I suppose I just need to consider c an arbitrary constant and check the discriminant 😩

nah that won't work that'll just tell me what I already know, that that form is injective

I think I'm overcomplicating it, maybe I can just differentiate and see what condition forces the parabola to have 1 or no roots

[dark sparrow]

@wise pawn x^3 + x

[wise pawn]

thanks

[ornate coral]

I have trouble with proving the generalized bisector theorem when the point is external

Can you help?

Here's the solution for internal point

Sty it's in spanish

Sry*

I try to do it with sine law

But idk which to find

[serene heath]

how to go about parabola and elipse concepts after i am done with circles and straight lines

[whole vortex]

Need help with 7

<@&286206848099549185>

[silent plank]

what have you tried?

also don't ping helpers before 15min

[whole vortex]

Need help with #7

[silent plank]

you already said that and that doesn't answer my question

[silent plank]

?

[burnt sigil]

Can someone help me with this? I'm not sure where to begin

<@&286206848099549185>

[upper karma]

What are you supposed to do

[burnt sigil]

That's all the instructions I have..Why I'm so clueless

[upper karma]

I think you are supossed to write the angles in the box

[upper karma]

obama

[upper karma]

x=11.25

[icy mortar]

anyone here

i need help real fast

[upper karma]

someone help

<@&286206848099549185>

[mystic oasis]

Can someone help me with a geometry test plz <@&286206848099549185>

[upper karma]

cos(-b) = cos(b), sin(-b) = sin(b): but why does the negative sign become positive? cos(a-b) should still be cos(a)cos(b)-sin(a)sin(b) for the last step

[wise pawn]

sine is an odd function, so sin(-b)=-sin(b)

[royal citrus]

Hey, what would be the value of x here?

[silent plank]

what have you tried

[royal citrus]

finding srq

which is 70

also know that pqa is psr cause exterior angle of cyclic quadrilateral is = opposite interior angle

other than those things

nth come to mind

[small cipher]

Well by alternating segment theorem, since AQ is a tangent, AQS=QRS for any point R on the circle between q and s

[small cipher]

this also isnt true because the exterior angle is PQA + RQB, not just pq

a

[royal citrus]

Ye

realized it

[stoic mural]

is someone able to help me out with my math cct

im struggling lmao

[silent plank]

cct?

[stoic mural]

canada sorry, its like a end of the semester assigment

thats worth alot of your mark\

[gleaming nova]

@stoic mural you would get help quicker if you immediately posted your question instead of waiting for someone to respond

[gray stone]

Need help figuring out the area of this

[stoic mural]

@gleaming nova i was told to google it

didnt really help

[stoic mural]

the thing is telling me to explain primary trig, and when it is used

[gleaming nova]

the shape is symmetrical

[gray stone]

not to scale

[gleaming nova]

trash question then

i mean

it obviously is

those two 40cm are the same length

[gray stone]

yeah but the 20 doesn't look like it's twice the 10

[gleaming nova]

basically if it’s not accurate like that then this question is impossible

[gray stone]

I'd think so to

just need to get a second opinion

[gleaming nova]

i mean they want an answer

so just assume it’s symmetrical

[proven veldt]

900m²

[proven veldt]

Answer is 900m²

[dark sparrow]

how are you getting an answer at all @proven veldt

can you show your work?

[proven veldt]

Ok

[dark sparrow]

how are you getting 400 m^2 for the bottom-left part?

(also it should be cm^2 not m^2 but that's kinda beside the point)

[proven veldt]

Look i tried

Whats ur answer

[dark sparrow]

why get so defensive all of a sudden

my answer is NEI

[turbid star]

ye i dont think the answer is determinable, if it is not symmetric

[neon bobcat]

is there any other info given over here

[upper karma]

is there any good websites for geometry help

[tepid folio]

can someone help with sin law and cosine law

ping me if you can

[iron spoke]

what do you need help with exactly?

@tepid folio

[tepid folio]

when to use each

[iron spoke]

both can be used in a lot of situations

for example the sine law is often used to find out sides of non-right triangles and the cosine law for figuring out the hypotenuse of non-right triangles

[tepid folio]

and for right angle triangles you would use sah coh toa?

[iron spoke]

yeah

so with these laws instead of doing funky stuff with the triangle you can just directly use them

[turbid star]

is there a general formula for the equation of an arc of a circle in the coordinate plane?

or do we just define it using inequalities

[broken anvil]

Well it’s easier to make a bound on the angle of the arc….becz of that It forms an inequality for the possible values of point coordinates….

[turbid star]

hm

[crimson perch]

Formula for segment when given is radius and arc

[rare sierra]

The arc length or section area?

[turbid star]

arc length

[turbid star]

wdym?

[zealous pivot]

what is known about generalizations of Hilbert's third problem? Like, cutting n-dimensional "polyhedra" (don't know how are they called) and assembling the pieces into new n-dimensional "polyhedra".

[grizzled kindle]

Dehn the guy who solved the problem hypothesis that there exist certain invariants like Dehn's invariant and area of a face ... But nothing more is known about these invariants

[zealous pivot]

Btw

An important restriction of the hypothesis is that only finitely many cuts are allowed

What if infinitely many cuts are allowed? I suppose then the claim would hold for any number of dimensions, but you would be making choices for infinitely many elements, so it's weird

[grizzled kindle]

@zealous pivot yes that is called calculas

Integral calculus to be more specific

[zealous pivot]

Don't think you call that integral calculus, but you can attack it with it probably

But there would still be some subtleties that would need be adressed

Or you say it's a part of it

[grizzled kindle]

Like?

[zealous pivot]

You probably right

[golden widget]

How would I be able to combine these 2 terms into 1

= 2+tan^2x

[rare sierra]

you need the central angle and the radius

Find the circumference of the circle then multiply that by the arc's angle over 360

I think this is an accurate example

[ebon pulsar]

how would i be able to find what x equal to?

im working on Exterior Angles and its a bit confusing

[wintry tundra]

alright ill help with this

so first thing

what do the angles of a triangle add up to?

[ebon pulsar]

180

[wintry tundra]

yep

but also notice how theres an angle on the other side of the 127 degree angle

[ebon pulsar]

yeah i see

[wintry tundra]

what angle is a line that is completely flat

its either 0 degrees or its what?

[ebon pulsar]

180?

[wintry tundra]

nice nice

so that means the flat angle with the 127 has to add up to 180\

so 180 = 127 + ?

where ? is the other angle

and finding ? will help to find x

so now u just need to solve for the ?

[ebon pulsar]

it would be 53 but we have another angel which is 23

[wintry tundra]

yep

[ebon pulsar]

so would it be 30?

[wintry tundra]

oh no

ur getting that part confused

we just need the 127 and ? to add to 180 so it can account for the full rotation

the 23 angle isnt a part of that

[ebon pulsar]

ohh

[wintry tundra]

so now inside of the triangle we have a 23 angle and a 53 angle

and remember triangles add up to 180

so we can show that as 180 = 23 + 53 + x

and now you can find x

[ebon pulsar]

104?

[wintry tundra]

yep

and there we have it

you solved that problem

but the most important part is

do you understand how all that worked?

[ebon pulsar]

Yes i did

thank you for the help

[tawdry condor]

Im kinda confused about something, if we have a triangle like this

how would i use trig to get the sides so that i can find the area?

im completely lost i have some idea of how i would go about it but at the same time i dont

i used to know how though, ping me if you can help

[dark sparrow]

giving names to every relevant point will help

writing out right triangles by name will also help

[tawdry condor]

wdym

ohh

[dark sparrow]

you know how in geometry we usually give each point a capital letter name?

[tawdry condor]

yes

[dark sparrow]

and how polygons and the like are described by listing their vertices

[tawdry condor]

im just having trouble with the basic concept

[dark sparrow]

i'm saying that writing out something like "XYZ is a right triangle" can and will help

[tawdry condor]

like how would i find the bottom side length

[tawdry condor]

ok

[dark sparrow]

it'll take a few steps

[tawdry condor]

well I gtg now but Im willing to learn how to take steps

[dark sparrow]

i have given you some pointers

look at right triangles and lengths and angles therein

[kind reef]

Hello, my geometry is very weak, but I was wondering if anyone can aid me in finding a mapping between alpha and alpha'?

(for now I haven't tried analytical approach)

[manic shell]

well

cant you just do 180-alpha

and you get this angle

then you can use the law of sines i guess maybe

hm

actually

idk

[kind reef]

actually using law of sines didn't occur to me directly, however that would still require me finding this length

[manic shell]

im assuming this is the midpoint of the smaller circle

right?

[kind reef]

anyhow i spent some time using analytical approach and i found the mapping in terms of d. i get the following for d=0 and d=5. the trend is that the max value of y lowers and it is trying to get flattened. i don't really know if this is any trivial or popular function. (ignore this message for present conversation if you want)

[kind reef]

yeah, you can assume the radius of smaller circle to be 1

[manic shell]

idk tbh

my geometry was never good

probably because i have no imagination visualizing things

[kind reef]

i think this problem might actually be very complex for a "pure" (synthetic) approach

with the analytical i found the formula in terms of tangents

[frosty remnant]

I hope this is okay to ask here.

I want to model a kind of stopper here to 3d print, but I don't know how to match the curvature. How can I do this?

[cosmic smelt]

do you guys know how can i do this?

the question 1

i dont know how start

[turbid star]

I didn't ask for how to find the arc length or the formula for the arc length
my question was, that is there a general equation for a given arc, without bounding/inequalities
like the general equation for a circle is x^2 + y^2 = r^2

from what I could find, bounding the angle theta seems like the only proper way

like someone else said earlier

[gleaming nova]

only other thing i can think of is an integral

[turbid star]

hmm how so?

[gleaming nova]

because an integral can calculate arc length

[turbid star]

hmmm

that just comes back to the formula ig

[gleaming nova]

i know

but a circle was an example

[turbid star]

wdym?

[gleaming nova]

this seems to imply that you said the circle was just an example

[turbid star]

ye I want to plot only a part of a circle using an equation

[gleaming nova]

hmm

i can’t think of anything that can do that

i mean you can get a full circle and half a circle

but i don’t know of anything else

[turbid star]

yeah y = sqrt(r^2 - x^2) gives semicircle

hmm I wonder if we could make a quarter circle

quarter circle is the intersection of y = sqrt(r^2 - x^2) and x = sqrt(r^2 - y^2)

[gleaming nova]

,w plot y = sqrt(1-0.5x^2

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by quantum#6320

[clever fable]

Does anyone know any website/app/ online graphing calculator where I can graph subsets of the complex plane

Like this one for example:

[nova galleon]

is it just the argz = pi/3 ray rotated pi/2 anticlockwise, then shifted right by 1?

[upper karma]

given any triangle ABC, let M be the midpoint of AC, P be a point on BC and O be the intersection of AP and BM. If OB = PB, show that OM/PC = 1/2 using auxiliary lines.

help

[wispy nexus]

hello

I am making a browser animation where I move an object to xz coordinates and I want to make it arrive with a small offset so it dose not collide. is there a formula to somehow add an offset to the target xz coordinates so that no matter from what side it comes it will always have offset? hope it makes sense

[median dirge]

<@&286206848099549185>

[grave pond]

If you ask an actual question of the helpers, they might be more inclined to answer it.

[gleaming nova]

i don’t even know what you want

that’s not a joke about you not actually asking a question either

[median dirge]

I just need help finding the perimeter of the shaded area- the triangle

I found the are

a

[upbeat helm]

also you need to wait 15 minutes before pinging helpers

not 7

also wonder

[grave pond]

It looks like you have found the length of the circular part of of the perimeter. What's preventing you from adding the length of the straight line?

[upbeat helm]

this is a 45 45 90 triangle right

[median dirge]

yes

[upbeat helm]

so then we know that

right this is the side lengths in a 45 45 90 right triangle

[median dirge]

its an isosceles triangle, correct?

[upbeat helm]

yes

because the two angles are both 45 degrees

[median dirge]

yes

[upbeat helm]

and you know that x = 12

so we know the longest side is just 12sqrt(2) right?

[median dirge]

yep

substuting if x=12, thatd be it

[upbeat helm]

but you understand why right?

[median dirge]

yea

[upbeat helm]

okay so whats the perimeteR?

[median dirge]

would it be 6 pi-12 times sqrt 2

[upbeat helm]

why would it be subtracting?

[median dirge]

thats what I did for the area, I subtracted the shaded area from white triangle

would it be diffrent for perimeter

[upbeat helm]

yes

[median dirge]

ok so 6 pi+ 12 times sqrt of 2

[upbeat helm]

because we want to add the arclength + the side length

yep

[median dirge]

Ok

[upbeat helm]

and that makes sense right

cuz we don't want to remove a portion from the arclength

we want to add the other side of our shape

[median dirge]

Ohhh

that makes sense

tysm

[median dirge]

I need help finding the area and perimeter of the shaded region... any help would be much appriciated!

[dark sparrow]

@median dirge have you made any progress so far?

[median dirge]

Yes

[dark sparrow]

okay, show your progress

[median dirge]

I believe I have but im not sure

so the empty half circle I removed by adding the full half circle to create a square ABCD

I have not gotten the change to write it down... Im just thinking for now...

do you think that could work

I found the area, its 576

I just need the perimeter now

[gleaming nova]

i guess AB = BC?

[grave pond]

I guess the curved parts of the perimeter are semicircles? 😛

[gleaming nova]

[upbeat helm]

somehing like this right

[dark sparrow]

that is what wonder did for the area.

anyway

the perimeter is composed of four parts whose lengths should be easy to calculate

(two lines and two half-circle arcs)

[upbeat helm]

or he could find the area of the square, subtract one side, and then add the half circle

either way works

[median dirge]

so that would be

96-24+12 pi?

[silent plank]

wheres 96-24 coming from

[median dirge]

one side is 94 correct

[silent plank]

no

[median dirge]

2

4

24

[silent plank]

yes, 1 side of the square is 24cm

[median dirge]

so 96-24

• semi circle

=12 pi

[silent plank]

wheres 96 coming from

[median dirge]

perimeter of square

[silent plank]

why are you only subtracting the length of one side

[median dirge]

is it 2 sides

[grave pond]

Um, can you explain in your own words what "perimeter" means?

[median dirge]

l+w+l+w

[silent plank]

is not the definition of perimeter

[grave pond]

No, an English sentence explaining the word "perimeter" to someone who has never heard it.

[median dirge]

the outlines of a close-sided figure maybe

[grave pond]

In this context, perhaps the total length of the outline would be better.

[median dirge]

agreed

yea

[silent plank]

(two lines and two half-circle arcs)

[grave pond]

Since you seem to think there should be 3 times 24 included in the result, can you mark up what those three twentyfours correspond to in the figure?

[silent plank]

are you able to identify those components in the outline?

[median dirge]

72

right?

[grave pond]

Do you have an image editor available so you can post a diagram where you have highlighted those three times 24 in red?

There should only be two, and I'm having trouble guessing which third one it is you want to include too.

[median dirge]

im confused

[grave pond]

So am I. If you don't have an image editor, can you explain in words which three sides of the square you want to include in your number for the perimeter?

[median dirge]

I added the full semi circle and empty one to form a square

for area I did l times w

which is 576

[silent plank]

that's for area

[median dirge]

for perimeter idk what to do

[silent plank]

perimeter is different

erase those red markings that you have

[grave pond]

Sorry for the silly question, but do you think that 24+24=72?

[silent plank]

and applying the definition of perimeter

[median dirge]

no thats 48

someone typed three times 24

[grave pond]

Okay, good.

[silent plank]

mark each piece of the outline (of the original) shaded shape in a different colour

[grave pond]

However then I'm still confused about where you got 72 from?

[median dirge]

Ok so area is 576, one side is 24 and a half circle is what

12 pi correct

[grave pond]

(The image with the contours of the two half-disks marked was Yottachad's).

Yes.

[upbeat helm]

yeah

so you have the perimeter of a rectangle

minus one side of it plus the arclength

[median dirge]

which is 576-24+12 pi correct

[upbeat helm]

not 576

[silent plank]

no

[upbeat helm]

you are looking for the perimeter

not the area

[grave pond]

576 was an area.

[upbeat helm]

perimeter of a rectangle is the sum of its 4 sides

and each side is 24

[median dirge]

ok so the perimeter would be

96

[upbeat helm]

so you have 24 + 24 + 24 + 24 = 4*24

yes

[median dirge]

I tried that

its wrong

[silent plank]

no

[upbeat helm]

96 - 24 + 12pi

[silent plank]

mark each piece of the outline (of the original) shaded shape in a different colour

is wrong

[grave pond]

Um, no.

[upbeat helm]

really?

[silent plank]

stop obsessing over 96-24

[upbeat helm]

ur right there are 2 arcs

[grave pond]

Since you have an image editor, can you post an image with that proposal marked?

[silent plank]

don't try to be fancy and go back to basic definitions

[upbeat helm]

u on windows?

[silent plank]

mark the lengths you need to find (by definition)

[grave pond]

No.

[upbeat helm]

okay

[grave pond]

(But why on earth is that relevant?)

[median dirge]

ok

so the formula is

[median dirge]

Im so confused

[upbeat helm]

i was gonna tell you how i've been doing these images

[silent plank]

mark each piece of the outline (of the original) shaded shape in a different colour

[upbeat helm]

its this thing called snipping tool (windows + shift + s)

[grave pond]

How do you get that to be 96 - 24 + 12pi?

[silent plank]

start with a clear idea of the pieces that make up the outline of the shape

[upbeat helm]

it would be 96 - 24 - 24 + 24pi***

cuz there are 2 arcs and only 2 sides

[upbeat helm]

anyways wonder

just look at this picture and then find all the sides

lengths*

then just add them all up to get the whole perimeter

[median dirge]

got it

that makes more sense

[upbeat helm]

yeah i was overcomplicating it

[silent plank]

there no point in being fancy here for the perimeter

[upbeat helm]

yep

[median dirge]

I need help witth one more thing

[silent plank]

simply add up the lengths instead of unecessarily involving stuff like subtraction

[grave pond]

Same procedure as last problem, miss Sophie?

[median dirge]

.

perimeter

[grave pond]

Same procedure as last problem!

[silent plank]

start with a clear idea of the pieces that make up the outline of the shape

[median dirge]

the whole shape is a squarw

square

[silent plank]

wdym by whole shape

[median dirge]

ADBC

[silent plank]

but you're not being asked about the perimeter of the square ABCD

[median dirge]

idk im confused on perimeter

[silent plank]

you're interested in the perimeter of the shaded figure

[median dirge]

Oh

I got it

nvm

solved

[silent plank]

start with a clear idea of the pieces that make up the outline of the shape
mark each piece of the outline (of the original) shaded shape in a different colour

[upper karma]

Someone who can help me please

Proof next equality

Sinacos⁴a= (1/16)(sin5a+3sin3a+2sina)

[dark sparrow]

you're not allowed to use complex numbers are you

this would have been very easy if you were

[upper karma]

no i am no allowed to use them

[dark sparrow]

are you allowed to use sum to product identities @upper karma

[upper karma]

Hello. Can someone tell me how does the nine-point circle looks in a Triangle rectangle ?. I've been looking for it in google, but dont get nothing

[silent plank]

wdym by
nine-point circle
wdym by
Triangle rectangle

[upper karma]

And besides. How would you prove that the angle CBR = PBA (You cant use the concurrent point from cevians)

[upper karma]

Hmmm, nine-point circle: https://en.wikipedia.org/wiki/Nine-point_circle#:~:text=In geometry%2C the nine-point,constructed for any given triangle.&text=The midpoint of the line,lie on their respective altitudes).

And the triangle rectagle: a triangle with an angle of 90

[upper karma]

I know all the formulas but i don't know how you start

[dark sparrow]

well i would suggest doing something to the right hand side, namely that part of the right hand side can be written as

sin(5a) + sin(3a) + 2(sin(3a) + sin(a))

[upper karma]

you can do this, except its not as good as you are thinking algebraically not geometrically

the bisector theorem works with ratios, so it probably can come from similar triangles

[stiff bough]

Person wants to swim over a river, thats flowing 2 km/h, Person swims with a speed of 3.5 km/h.
Solve:
a) What angle should the person swim to get to the coast perpedicular to the starting coast
b) Swimmers final speed.

In the middle of a log are tied up 2 ropes, which are pulling upwards with 2 cranes. 1 rope is being pulled up from 1 crane with a power of 65 kN, the other one is being pulled up by the other crane with a power of 75 kN. The power creates a 46 degrees and 44 degrees angle difference on a vertical.
What is the combined power of both cranes?

[plain ingot]

do you mean what angle he should take to take the black path, whilst the river moves 2km/h?

and he moves 3.5km

/h

[stiff bough]

yeah

[plain ingot]

ok so

imagine the velocity/h of your person's direction to be a vector

[stiff bough]

sorry if you dont understand the question, i had to translate it myself

[plain ingot]

i understand

for your person to move perfectly horizontally, it has to counter the river's direction. Therefore your persons vector's y component is -2

the length of the persons vector is 3.5 as we know

now were missing the x component. We can juste use that we know the x^2 + -2^2 = 3.5^2, therefor x = sqrt(3.5^2 - 4 or (-2)^2)

we just need to convert this to an angle using trih

trig

we can just use the atan2 function to do dis

and i think his final speed would be your vectors x, component

[stiff bough]

ok

[autumn oyster]

Would appreciate a hand to help me create an equation since I'm quite lost and don't know where to start

[grave pond]

That looks very confusing indeed.

Is "quadratic trig equation in standard form" a technical term in the course you're following? If so, what exactly does it mean?

Even so, it seems quite opaque what the task of creating such a thing has to do with the assumption that we have been given an equation of the form ax²+bx+c=0. Does the quadratic-trig-equation-in-standard-form need to have a particular relation to the quadratic polynomial? Since we're not given particular values for a, b, and c but just told to assume that we have them, are we supposed to describe a general algorithm for producing a q.t.e.i.s. given a, b, and c? (But again, satisfying which correctness criteria?)

[autumn oyster]

The course ironically never mentioned that until I got this worksheet since we only done just writing trig equations from graphs and never did much trig quadratic formula equations. I emailed my teacher about that actually and they only really gave me a vague answer (they just said use the form of how trig equations are usually written). So I assumed it would have to written as like the pic I attached below using the quadratic formula to solve it and one of the conditions in the question are met and express it as a cos ratio (At this point, I'm just going off whatever I can do to get it done 😭) and come up with some equation

[grave pond]

I'm afraid I'm as confused as you, sorry.

It's not even clear to me whether a "quadratic trig equation" would mean something like 2cos²(x)+2cos(x)-2=0 or something like cos(2x)+2cos(x)-1=0.

[autumn oyster]

Yeah at that point I'm assuming it'd be written in that form because I don't know how else they'd want it unless they wanted it in quadratic form also

[silent plank]

have you tried drawing a diagram

[upper karma]

I have

I drew two but since we've been given it's on the minor arc I'm confused

I can't make any secants

and among the tangents I have, only 2 are known

but I need more info to find the perimeter

I suppose.

[silent plank]

can you show your diagram

@upper karma

[silent plank]

do you know anything special about the two tangents to a circle from an external point?

[gaunt heath]

they are equal in length

[silent plank]

that question wasn't for you

[spare temple]

First put it in quardratic formula and set it equal to cosx then solve for possible x’s on the interval (0,2PI)

[smoky palm]

Hey, so

I'm teaching my friend about how to find the perpendicular slope of a line

given the original line

So, something like y=2x+3

he already knows about inverse functions and how to calculate them

So, what I did, I told him

Get rid of the constants, then switch x and y

then solve for y

and then make the x side negative

This is a horrible explanation, but we've got a test tomorrow

Is this a good enough way to calculate it

[valid osprey]

theres an easier method

if u have a line y = 3x + 5 or something

3 is the gradient right?

[smoky palm]

the slope is what you mean?

[valid osprey]

yes

[smoky palm]

yes, it is

[valid osprey]

gradient = slope

the coefficient of x is the gradient

so

just multiply the coefficient of x by -1

no wait

i mis worded that

i meant

[smoky palm]

then do the reciprocal?

[valid osprey]

actually yeah u could say that

i was boutta say that the slope of the original line(y = 3x + 5 is the example i gave) and the perpendicular lines slope, should multiply to give u -1

so if u have 3

as the slope

it has to be -1/3

[smoky palm]

because

reciprocal times -1

thank you so much

[valid osprey]

npnp

[silent plank]

it felt like you mixed inverse functions with perpendicular lines or something

[valid osprey]

yeah

¯_(ツ)_/¯

[warm eagle]

I need help

[vernal aurora]

One message removed from a suspended account.

[upper karma]

yes that they are equal in length

[silent plank]

apply that your question multiple times

[upper karma]

AP and BP are equal that way but I can't figure out what to do next

@silent plank

I will have proof that these tangents will be equal(since drawn from same external point) but I do not have the necessary value for finding the perimeter:
AQ = QT
BR = RT

on second thoughts

wait

I can suppose AQ and BR as x and y and then do the perimeter

so by whole part axiom PQ would be 20-x and PR would be 20-y

then perimeter would be (20-x)+(20-y)+x+y

[upper karma]

oh yeah the variables then get cancelled out

thanks @silent plank

I got it

:>

[upper merlin]

just try to represent angle CBD in terms of the angles of the triangle

[amber quartz]

if $x+y+z=180$ , find the range of $$(sin^2x+sin^2y+sin^2z)(cos^2x+cos^2y+cos^2z)$$

[somber coyote]

In and Out

[amber quartz]

<@&286206848099549185>

[south moth]

if you have a question, please read #❓how-to-get-help and use the help channels

[ebon pulsar]

What dose it mean by naming the figure in 2 different ways?

[sour jacinth]

I think that you could a different nomenclature using the points that define the segment

[ebon pulsar]

As a not so smart individual that is new to the topic may you rephrase to simpler terms?

[sour jacinth]

I'm gonna try, but the problem is simple enough that I would be practically giving you answers

How do you denominate a segment?

For example one that goes from A to B

[ebon pulsar]

What do you mean?

the line?

[sour jacinth]

How do you symbolize this?

As a variable/unknown, I mean

...

[sour jacinth]

You still there?

[ebon pulsar]

I am

sorry my wifi went out

[sour jacinth]

Any guess to answer my question?

[ebon pulsar]

No, this is my first time working with this

I was going to lear it

but my teacher called in sick

[mild birch]

is it in degrees ?

[sour jacinth]

@warm oxide, a segment that has endpoints A and B can be noted as

$\overline{AB}$

[somber coyote]

Max Hetfield

[sour jacinth]

As segments do not have directions, you could also note it as $\overline{BA}$

[somber coyote]

Max Hetfield

[sour jacinth]

So knowing this, how do you think we could note the segment in the image you posted?

[ebon pulsar]

So what happend to the letter in the middle

[sour jacinth]

What letter in the middle?

[ebon pulsar]

Here

the letter Q

[sour jacinth]

I know... But let's go step by step

[sour jacinth]

Let's start here

Use only the endpoints...

[ebon pulsar]

P and E?

[sour jacinth]

Using the notation I explained to you

[ebon pulsar]

$\overline{PE}$

[somber coyote]

Some guy

[ebon pulsar]

that ?

[sour jacinth]

Yeah

And something equivalent?

[ebon pulsar]

What do you mean by equivalent?

Sorry if im making it hard

[sour jacinth]

Think of this

[sour jacinth]

You're learning, no problem

Equivalent means something that "is the same"

For example

$\overline{PE}$

And

[somber coyote]

Max Hetfield

[sour jacinth]

$\overline{EP}$

[somber coyote]

Max Hetfield

[sour jacinth]

Are equivalent: They refer to the same segment between points P and E

Got it?

[ebon pulsar]

So your saying that they would ne "the same" using the example you showed me

[sour jacinth]

*ne ?

[ebon pulsar]

ne?

wdym

[sour jacinth]

What do you mean by ne?

Look at what you wrote

[ebon pulsar]

Oh

I meant be

my bad

[sour jacinth]

Yeah, they mean the same, the are different names for the same thing/segment

Now, one can build a segment using smaller segments

For example

You could say

$\overline{EP} = \overline{EQ} + \overline{QP}$

[somber coyote]

Max Hetfield

[sour jacinth]

Yeah?

[ebon pulsar]

sorry im reading

to fully understand

[ebon pulsar]

@sour jacinth would that mean the the "Q" in this problem would be considered the "middle letter"

[sour jacinth]

Yeah, the middle point

It's like you're gluing together 2 segments

One that goes from E to Q, and another that goes from Q to P

Together they form the big segment that goes from E to P

@ebon pulsar And an equivalent way to note that segment $\overline{EQ} + \overline{QP}$

[somber coyote]

Max Hetfield

[sour jacinth]

is $\overline{EQP}$

[somber coyote]

Max Hetfield

[clever aurora]

Hey guys... https://byjus.com/maths/angle-between-two-planes/

I am following this guide to find the angle between two planes... When the angle between the two planes is >180 applying the expression straight up gives me the wrong angle (it seems to subtract the 180, although that is just based on visual inspection)

So, do I need to essentially know when the two planes are infact >180 degrees apart?

and if so, how can I tell that, if my means of determining the angle doesn't work from the get go?

[clever aurora]

here is an example of what I mean, normals in black, plane in pink

I get an angle of ~12 degrees using this method... It seems to work fine for other cases, except something like this

it seems off by a lot more than just 180 here though

[bitter wedge]

Any guidance?

[silent plank]

draw a triangle

and apply pythagoras

[bitter wedge]

Ah brain fart

Makes sense, thank you @silent plank

[iron walrus]

bruh i have a test tomorrow and i dont know shit

[bitter wedge]

Ok now I'm just lost. I'm not very familiar with trig outside of the basic identities.

[grave pond]

That doesn't immediately look wrong. What's the problem?

[bitter wedge]

I just don't understand it. This is an instructor's work.

[grave pond]

Do you know Euler's formulas?

[sullen lark]

im here! I can't help you though

[gleaming nova]

this person has already had their question answered

don’t help them

this is why you stick to one channel

[sullen lark]

I think if you ping everyone, more people will help you

[gleaming nova]

@upper karma

nah

ping mods

that’s how to do it

preferably ping them a few times actually

[left fractal]

I'd normally ask for help so i can understand it but i'm very close to a deadline so i need answers fast so can please if anyone can help me that'll be so cool

nvm i just read the rules 🤦‍♂️

thanks you guys anyway have a great day!

[iron walrus]

Can

Someone tell me the answer

[pine quartz]

where should i start?

[dark sparrow]

@pine quartz have you done similar problems before where you're given the sin, cos or tan of some angles?

you could begin by finding sin(A), cos(A), sin(B) and cos(B).

not necessarily in that order.

[pine quartz]

aight. do i need to use pythagorean theorem?

[dark sparrow]

you will need to use it a few times yes

[pine quartz]

okay. thank you

[pine quartz]

these are what i get, sin(A)=15/7, cos(A)=8/17, sin(B)=12/13 and cos(B)=5/13

should i apply next the sine sum identity formula and so on?

[dark sparrow]

sin(A) = 15/7?

[pine quartz]

i mean, 15/17. i'm sorry

[dark sparrow]

right

okay, yes, this sounds correct

and yes, you now have everything you need to apply the angle sum identities

[pine quartz]

oh god,,,,, is that all? thank you for guiding me :"))

[upper karma]

hiii i need helppp

[robust cave]

I need some assistance please

I can start off with working out the range and the 1st angles but I'm lacking confidence in my steps

yo @upper karma

you're in the same vc

can you help 💀

wait @verbal flint can you help pls

anyone 🥲

I tried so much but I think I just lack the basic steps

[verbal flint]

yeah sure

[robust cave]

yo

I am here

you can type it gyani

mr knowledgeable ultimate

[verbal flint]

x = 1/3 (-2 π n + 10 - cos^(-1)(-2/5)), n can be any integer.

i will write the steps

wait

[robust cave]

I haven't been introduced to the whole pi aspect in these type of questions

but I'll use it nonetheless if it helps

[verbal flint]

ok no problem. i shall tell you the basic approach. you need to write your R.H.S in terms of cos

can you do it?

[robust cave]

I shall try

I can hear you gyani

[verbal flint]

cos(cos^(-1) (-2/5)) = -0.4

[robust cave]

haeh

why is cos multiplying by cos inverse

[verbal flint]

cos inverse returns angle, and when you feed it to cos, you get -0.4, my goal is to make both sides same types of entities

so that i can remove the cos

and write in terms of angles

do you get my point?

[robust cave]

too complicated 💀

we do it a simpler way

[verbal flint]

umm

[robust cave]

like this look

[verbal flint]

which grade are you in?

[robust cave]

12

[verbal flint]

well another way to do is take inverse both sides.

[robust cave]

e.g.

[verbal flint]

ok then cos(3x-10)= -0.4 , 3x-10 = 113.578 degrees

x = 41.192 degrees

[robust cave]

246.4

[verbal flint]

is one solution

[robust cave]

so x = 85.5

and 473.6

so x = 161.2

[verbal flint]

in degrees?

[robust cave]

yeah

[verbal flint]

that is also one of the solutions

cos(113.578) = -0.3999

[robust cave]

3 solutions in total I'm pretty sure

[verbal flint]

and cos(246.4)= -0.4004

yeah 3 solutions

[robust cave]

thanks for the help

I was asking 2 other people at the same time

you got the same as my friend so it must be the correct answer too

[short crow]

Help please.

[grave pond]

If you draw two additional lines in the diagram, you can get theta to be one of the angles in a right triangle that you can use trigonometry on.

(One of the additional lines is just continuing the one that's already shown as a dashed line).

[short crow]

Oh I see

Thanks man.

Used tan and got 4

Appreciate it.

[grave pond]

4° agrees with my calculator :-)

[bitter wedge]

No

Sorry for the late response

[bitter wedge]

I figured out the answer, but I only got it because I saw where to change my answer. I still don't understand how the formula works. When I search it up, all I get is stuff regarding complex numbers.@grave pond

[warped canopy]

Yo anyone here

[sly scroll]

Guys I have a doubt in 3D Geometry

Can anyone help me with equation of a cylinder? with cross section as x^2+y^2=9
And hight 5?

[hoary mango]

cross section?

bro i thought cylinders were just rh rule

[iron epoch]

U guys know ambiguous case???

[dire moth]

Can someone help me solve this correctly

Like teach me how

Would be appreciated

<@&286206848099549185>

[forest drift]

tan 15 = y/7 = -0.855

solve for y

[forest drift]

I'll try it out

[dire moth]

Ok

Can you tell me the steps after if possible

So o can learn it

I*

[woven gate]

@dire moth

Sry for ping

But

If you can

I need help

[dire moth]

On what

[woven gate]

What me to give you a screenshot?

[dire moth]

Sure

[woven gate]

Okay

I’m just confused how to get x

[hoary mango]

how do i remember law of sines and cosines

[woven gate]

Like what = x

[dire moth]

I honestly have no idea if being honest

Sorry

[woven gate]

Okay

Ty

[dire moth]

Np gl on that tho

[hoary mango]

question

how do i remember the formula for law of sines and cosines

[dire moth]

Law of sines is just cross cross right

If I’m correct

[hoary mango]

no like

the actual equation formula

the non-right triangle's coordinates being diagonal is easy to remember

but the equation feels like hell

[dire moth]

A=b sinA/Sin B

Is that what you’re asking

[hoary mango]

hmm

[forest drift]

that is sin rule