geometry-and-trigonometry

[teal patio]

I think you might have rounded wrong

[cyan scaffold]

yeah

it's all about rounding

[teal patio]

Ye

I got 4416.0

[mossy aspen]

ok i tried again i got that

my delta math crashed guys TOT

its not letting me submit it

the one time i got it right

[teal patio]

💀

rip

[cyan scaffold]

i remember the days of delta math

[mossy aspen]

TOT

sir i most certainly did NOT SUBMIT IT YET PLZ LEMME SUBMIT

thank you guys

🙂

it finally worked and i got it right

[somber coyote]

Nerdyasianguy

[sly urchin]

Very close to a Hagge circle

[dull tusk]

suchitRole icon, he/him — 4:58 PM
can any one say some importent properties of hyperbola

[gilded crescent]

what,s this

[sly urchin]

A normal geo problem?

[slim plinth]

tan(x) = sin(x)/cos(x)

[quaint chasm]

idk if this fits the channel, but does this degenerate shape have a name?
i was trying to see if i could make a sort of extended cube - where each vertex only connects to 3 edges - and accidentally came up with this weird thing

you take a cube, remove half the faces so you only have a corner of the cube, and then merge the 3 vertices that stick out (shown in image 4)

[ocean stream]

help?

[spiral lodge]

!status

[gentle marsh]

Hey guys?

I’m completely new to trig tho I know afew of the prerequisites

[spiral lodge]

Ok

[distant crag]

What grade are you in?

[gentle marsh]

Grade? Nah I’m not in America

And I study above my level anyway so it doesn’t matter

Could u teach me?

[spiral lodge]

It's (obviously) impossible to teach you such a huge topic here, especially from scratch

I suggest you pick a textbook and/or watch some courses online, e.g. on YouTube or on KhanAcademy

[thick compass]

Khan academy is great

[timber kraken]

im learning trig its so easy omg

[timber kraken]

how’s my work

[spiral lodge]

Good

Don't forget the ° symbol though

[winter shadow]

its a geometry group chat for kids at my school

[gusty geyser]

<@&268886789983436800> seems questionable

[lunar hull]

@winter shadow not your personal army

[neon prairie]

This is a good way to get banned from here fwiw.

So, please don't do anything like this again

[sly urchin]

Any ideas? I think the median triangle of ABC will be relevant but i haven't solved this

[viscid gorge]

Any help ?
This is an ellipse with F being a focal point

[patent yacht]

If you understand it’s okay

If you don’t pls let me know

[silver palm]

Why does sin(x)/x=1?

I want directing, not a straight answer please and thank you.

[dark sparrow]

sin(x)/x never EQUALS 1, it approaches 1 as x->0.

really you should be saying lim[x->0] sin(x)/x = 1

look up "small angle approximation" for proofs

[silver palm]

Thanks a lot

[viscid gorge]

Watch a YouTube video about it

[viscid gorge]

?

[silver palm]

Yeah I thought if that

[silver palm]

I think I can help

First

What's the angle inside the triangle

[viscid gorge]

All the information is given

But you can find F being 120

[silver palm]

Well we know that F²+B²=4 by Pythagorean's

[viscid gorge]

Yea

You want to do the cosine law ?

[silver palm]

It's what I was thinking

[viscid gorge]

BN^ 2 = BF^2 + FN^2 - 2BFFN cos(F)

[silver palm]

Wait you want length of F

Right?

[viscid gorge]

How you want to find FN as some parameter of the ellipse

[viscid gorge]

The question wants BN

[silver palm]

Ok

Well

[viscid gorge]

Bump

[silver palm]

F is 120 so BN² = BF²+FN²+2BFFNcos(30)

Which is

BN²= BF²+FN²+2BF(sqrt3/2)

Oh

Im stupid

[low ocean]

||Use BF and the 30 angle to find minor and major axis lengths||, ||let F' be the other focus, then FN+F'N is fixed and you can find it||, ||on the other hand in triangle F'FN with F'FN=150 degree, FF' is known so you can apply cosine rule||,||now you have 2 equations for 2 variables which is sufficient||. ||After we get FN, use cosine rule again to find BN||

[timber kraken]

i got 1.73

but I used simple trif

trig

sin(120)=x/2

but thats not a right triangle

sigh'

[viscid gorge]

I found FN = 2/7

Is it right?

[low ocean]

no idea

I haven't done any calculation

[exotic valley]

https://www.youtube.com/watch?v=zoz-lYuzZzM&list=PLCLfupK6Ie8Vcg6_PSH48WtIpSla5iD9P&index=16

arkadaşlar bu kanalı takip edermisiniz

kendisi benim ocamda

he is my teacher

[long geyser]

How do you find the circumcenter and other centers/stuff of these 3 points? I'm learning analytical geometry right now but learning about the midpoint formula makes me wonder what happens if there's 3 points, how can you find the midpoint of all 3 points in 2 dimensions?

[cunning lion]

the midpoint of two points has the average of their coordinates

[long geyser]

Yep

Ohhh, should I maybe add up all 3 points and divide by 2?

[cunning lion]

what is the "midpoint of three points" though

[long geyser]

Well my thought process is that if you can find the distance that is equidistant from 2 points, then the same must be true for 3 points right? Where like this red point should be in a coordinate that is equidistant in length from all 3 points?

[cunning lion]

well the point equidistant from those points is the center of the circle they all lie on

[long geyser]

Ah so something like this?

[cunning lion]

yeah

[long geyser]

Cool!!!!!!

[cunning lion]

if you think of the 3 points as vertices of the triangle, this is one of the triangle centers (which one?)

[long geyser]

centroid?

hard to tell

[cunning lion]

well the diagram doesn't make it obvious

[long geyser]

Ah so then it's just a circumcenter right?

[cunning lion]

indeed

[long geyser]

i see

if I know the diameter of the circle, and divide it by 2 then I just have to figure out where the radii of 2 of the vertices have to meet to figure out where the circumcenter is?

Like this?

[long geyser]

Fascinating stuff on the wikipedia about circumcircles.

[vast timber]

i am so thrown off here

what am i missing here

the equation is marked as good but then when i actually put it into the calculator i get wrong answer

[keen rose]

perhaps a silly question, but have you tried to answer it in radians or as a positive angle (360-14.2)? I don’t see anything else wrong with it, although shouldn’t it be in de fourth quadrant?

[exotic valley]

you must be wrıte -7.78 X 1.97

[crimson nymph]

did @south leave?

[hollow lily]

He did

[rancid basin]

on another account he said he was unable to access his south acc or smth

[hollow lily]

I don't think so

[crimson nymph]

oh

[hollow lily]

You can still see him in IBO discord server

[neat ravine]

Hey I have a question which bothers me all the time
We know that trigonometry is used in triangles, (name says it)
Well but, if we talkin about sin0, cos0, etc, how can we talk about 0 degree?
Cuz if theta is 0 degree, then it would no longer be a triangle, but coincident lines?
Trigonometry isn’t valid there,
And moreover, we know that hypotenuse is the longest side of the triangle, greater than Perpendicular and the base, but if we are talking about sin0,cos0 etc, we are saying that base=hypotenuse and perpendicular=hypotenuse….. WHAT-?
This really bothers my mind 😭😭

[shut thorn]

unit circle

you use the unit circle to define values such as sin0 cos0 etc etc

wait lemme show how

green is the radius one thetha is the angle between the sides

wait let me give them a name

also made it a bit clearer.

OC is the radius. And AB perpendicular to OC

so from here

since OA = 1

sin(thetha) would just be the side AB

similarly. cos(thetha) would just be the side OB

try to figure out why and if you cant. Ask

[shut thorn]

the radius is one. i.e OA = OC = 1 unit

[shut thorn]

now as thetha reaches zero. The base reaches the radius OC. So when thetha becomes zero. sin(thetha) being the perpendicular. Would also be equal to it. but since theres no perpendicular. sin(thetha) must be zero

similarly. you can say about cos(thetha)

[jade cypress]

@kindred scaffold

[kindred scaffold]

omagah thank you so much

life saver

saved my exam tomorrowww

[kindred scaffold]

please help

guys in the solution, i don't understand how they got the fractions in the second line

!trigonometry

[lime crown]

#geometry-and-trigonometry - euclidean geometry, coordinate geometry, trigonometry

[strange pond]

from the last part of the first line, expand (1 + cos theta)^2

[storm ridge]

is woud be helpful to learn Ceva’s theorem and Menelaus’ theorem? I can't found good resources to learn them

[soft cedar]

@plush copper

oops

<@&268886789983436800>

[onyx lynx]

not really unless you're gonna grind competition math...if you're curious it's really not a lot to learn, it's just two theorems

[neat ravine]

Ohh 😮
it makes this so simpler if we take unit circle
You explained very nicely and clearly, thanks for that!
All understood 😁

[neat ravine]

Wait
One last confusion…

Isn’t sin(theta)= perpendicular/Hypotenuse?
So if we put theta zero, there literally isn’t anymore hypotenuse and no triangle is there!

[dark sparrow]

the right triangle definition works only for angles strictly between 0 and 90°

[neat ravine]

Not including 0 and 90?

[neat ravine]

Oh well
But that means then we aren’t focusing or right angle triangle here?
Like AOB is right angled triangle?
(I mean I know it is right angle triangle, but we aren’t focusing on this fact, right?)

[kindred scaffold]

thank you so muchhh

[plucky crag]

hi all! just need some tips of using identities. i have learnt a bunch of them but struggle with which ones to use when ecspecially when doing induction questions. and example question is above in which i struggled to know when and where to use them. ik to try and find the way to the k+1 term but i still really struggle. anything to keep in mind apart from just practising these types of questions

[shut thorn]

that right triangle is only there so that sin(thetha) and cos(thetha) can be AB/OB and OB/OA respectively.

since then.

because it's a unit circle.

The values just become the sides AB and OB

and then we can see what their values are for every and every angle

[shut thorn]

think of it like uh

in a unit circle. The x and y value of a point on the circumference is given by cos(thetha) and sin(thetha) respectively where thetha is the angle made by the point from the x-axis (if I'm sure)

[dusk notch]

hey

anyone

[zinc bluff]

soh cah toa

[full mauve]

How do i memorize all these formulas for ellipse circle hyperbola and parabola for trig it felt like there was so much info thrown at me that it all just slipped out my head

[shut thorn]

do questions on them?

cant really comment much since i havent been taught that yet

[full mauve]

I did

[shut thorn]

what if you

try to mess around with the formulas

or try to derive it yuorself

[frosty delta]

You are a 10th grader?

Lmk if you need help with 10th maths (CBSE),I have studied it

[long geyser]

When using the distance formula, is it ok to represent delta x and delta y or the change in them like this where it's subscripted? h is just a variable name I assigned for the height of a triangle in the coordinate grid/plane to figure out it's length because I know 2 points of it.

For reference, this is the original diagram, I'm trying to figure out the area of trapezoid COWL.

[frosty delta]

I guess change in x or delta x shouldnt be used as distance formula is just pythagoras theorem for coordinates of two points

[long geyser]

Can you check my work?

I'm not sure how to simplify 4sqrt5+2sqrt45+15

[frosty delta]

Um sqrt 45 is just 3sqrt5

[long geyser]

Ty!

[frosty delta]

Area is just 7sqrt5 + 15

[long geyser]

Then the answer is just 7sqrt5+15

Yea!

thanks

[frosty delta]

Wlcm

Sqrt 5 is approx 2.23 if you need a simplified answer

[long geyser]

Thanks!!

[vast timber]

i figured it out, i was supposed to add 180 to the result

[long geyser]

How do you find out the coordinates of all the points on a circle with a certain radius?

[queen juniper]

there are an uncountably infinite number of points on a circle's circumference, but with the circle with center (a,b) and radius r, the formula for that circle is (x-a)^2+(x-b)^2=r^2

plug in any x value [-r,r], and find ur y

[long geyser]

<@&268886789983436800>

[long geyser]

Tysm!

[shut thorn]

you can also use trigonometry if you know the angle it makes with the x axis right?

x or y axis

[queen juniper]

polar coords

yes

: )

[shut thorn]

:3

no wati wrong face 😭

:b

there

[winged oyster]

how do i practice geo and trig

[silent plank]

do questions

[frosty delta]

Understand the concept and start solving questions

[polar merlin]

wait im new in here do we do complex numbers as well

like anything euler's identity related

or does that classify as precalc

[shut thorn]

i think that should be preaclc

calc

precalc

[polar merlin]

ok ty

[long geyser]

Do you eventually learn an easier/less-tedious method to calculate perimeters/area of shapes in the coordinate plane than having to use the distance formula?

[shut thorn]

a less tedious method would be to write "similarly" on everything which is similar to something else 😭 🙏

[long geyser]

I don't understand, if you meant this example then it's pretty clear imo that even if you didn't get the stripes on the lines which say that they're congruent in length to each other that the coordinates make it clear that they'd have the same length

I did the latter before realizing that it was already given that they're congruent

[frosty delta]

You can use ladder method for obtaining area

Idk about any other formula for getting perimeters

[long geyser]

Ty

[slim plinth]

Use distance formula

[cursive oracle]

what does tangent give information for

because cosine and sine give the x and y intercepts

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

also on a related note, the angle between two lines formula

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

obviously there's more but my fingers are cold so

[cursive oracle]

hm

so wait

it finds the inner angle?

and what is the arc in arctan

[queen juniper]

arctan is the inverse of tan

[cursive oracle]

ah

[queen juniper]

ie, if tan(a)=b, arctan(b)=a

[cursive oracle]

does that carry for all the functions

arcsin arccos

[queen juniper]

arcsin, arccos, arctan, etc

yes

hyperbolics too

[cursive oracle]

ok good to know

i dont know those yet lol

or the terminology at least

[queen juniper]

you'll learn them eventually

very useful

[cursive oracle]

ohh

ya no non-euclidian for me just yet

[queen juniper]

not hyperbolic geo

[long geyser]

Arcsin etc can be used to find the angles

[queen juniper]

hyperbolic trig

like cosh or sinh

[cursive oracle]

well hyperbolic trig applies to hyperbolic geo doesnt t

[long geyser]

https://www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-solve-for-an-angle/a/inverse-trig-functions-intro

[queen juniper]

yes

[cursive oracle]

thankfully i know that lol

[long geyser]

Nice

[queen juniper]

but sinh and cosh can be written as (e^x+-e^-x)/2

so just fancy name basically

[cursive oracle]

wait im doing a problem here

so sin and cos only give the intercept points right

or is it length

it would be the same thing wouldnt it?

for a unit circle

[queen juniper]

hm?

(sin,cos) defines a specific point on the unit circle

[cursive oracle]

how would i derive side length

its been a while since i took geo

[queen juniper]

for a line?

just use distance formula

[cursive oracle]

for a triangle in a unit circle sorry

[queen juniper]

ohh

which side?

[cursive oracle]

opp and adj

hypo is 1 bc unit circle

[queen juniper]

oh that triangle 😅

ye, sinx gives height

[cursive oracle]

early trig

[queen juniper]

cosx gives width

only bc h=1 tho

[cursive oracle]

what is x

theta?

[queen juniper]

no theta keyboard shortcut, so i have to use x 😢

[cursive oracle]

alr good good

no new operations

[cursive oracle]

so whats the point of arctan finding the inner angle if arcsin and arccos both do that

[queen juniper]

hm?

arctan works iff tan(a)=b

sin and cos work similarly

[cursive oracle]

so they all output angles

but i was under the assumption that they output the angle of the hypotenuse to x

[queen juniper]

no, it's the inverse

not the reciprocal

[cursive oracle]

hm

so arcsine gives

the angle of y and the hypo?

[queen juniper]

arcsin(sina)=a

arccos(cosb)=b

arctan(tanc)=c

[cursive oracle]

oh wait im dumb

inverse operation

[queen juniper]

same for arcsec, arccsc, and arccot

[cursive oracle]

so it finds what you put in

ok

[cursive oracle]

are sec csc and cot the hyperbolic functions

[queen juniper]

no

sec=1/cos

csc=1/sin

cot=1/tan

[cursive oracle]

what are they unabbreviated

[queen juniper]

hm?

secant, cosecant, and cotangent

sine, cosine, and tangent

[cursive oracle]

ok so those are ratios

[queen juniper]

mhm

[cursive oracle]

ok nevermind i get what reciprocals are now

[cursive oracle]

does this all look good

[queen juniper]

yes, gj

[cursive oracle]

oh i have errors in there

[queen juniper]

areas r right

[cursive oracle]

i have x written over the hypotenuse in the first one

[queen juniper]

didn't check the calculations tho 😅

[cursive oracle]

lol

[fluid jackal]

Need help

..

[feral holly]

CAB = 180-(45+30)
CAB = 105

CAO = CAB - OAB
CAO = 105 - (180-90-30)
CAO = 45

I mistook 30deg as a measurement for the whole ABO

[fluid jackal]

Yeah..

The final answer is 60 °

[feral holly]

Yeah...

[fluid jackal]

How to find OAB?

[feral holly]

OAB is an isosceles triangle

AOB = 90deg

So there is 90deg left

Divide it by two

You get 45deg (OAB)

[frosty delta]

Um in this Q, ACB and AOB will be equal that is 90, because angles subtended by same arc are equal so CAB will be 60 (by angle sum property). Now OAB = OBA = 45, so CAO will be 60 - 45 = 15, but you said answer is 60, how?

[wraith pumice]

how are you taking isoscles if its not given that o is the centre of the circle?

[frosty delta]

I guess its pretty common to assume that O is the centre

[wheat iron]

and o is for origin

[wraith pumice]

i get it but stll

[wraith pumice]

the answer given is 1930.97 cm^3

nvm i got it

[simple vigil]

Which class ?

10th ?

[wraith pumice]

ya

[frosty delta]

That is if we are talkin about points whose coordinates are given
This is a 1 mark Q of 9th or 10th Grade, think like that for once

[hard crater]

Imma need some help [precalc] if n= tan A - sin A and (m²-n²)²=16mn prove that m=tan A + sinA

[drowsy knoll]

what's the relation between the direction ratios when 2 lines are intersecting?

this is what google says, just wanted to confirm

[frosty delta]

Um as far as i remember, if the below thing is true (a1/a2 is not equal to..), then the two lines are said to be parallel or have no solution that means they dont intersect

[drowsy knoll]

for parallel lines a1/a2 = b1/b2 = c1/c2 (in 3d)

[frosty delta]

I am talkin about 2d, just x and y

[drowsy knoll]

i'm talking abt 3d

[frosty delta]

Um idk about 3d, sorry

[drowsy knoll]

😂

[drowsy knoll]

oh ohk

no worries

[frosty delta]

Man i just watched dhurandhar and your pronouns made me remember it again

[drowsy knoll]

😈

i'm listening to fa9la rn lol

nice

[frosty delta]

I did a same Q like this, but m and n were given like that, it was just to prove that m² - n² = 4(mn)^1/2

I dont think there's another way rather than just assume that m = TanA + SinA
Then add m and n and subtract them also, now multiply both results and you will get m² - n²
Then keep this result aside and multiply m and n, after getting the value, check it by putting in the given equation (m² - n²)² = 16mn or m² - n² = 4(mn)^1/2
If LHS = RHS then this means our assumption was correct
I could not find any other way to prove m = TanA + SinA
@hard crater

If m and n were given then it is possible to prove m² - n² = 4(mn)^1/2
But to find m, its difficult or i guess not possible without assumption

[wraith pumice]

ig i have a similar question, see if you get any ideas from here

[wraith pumice]

yeah

i cant see any ends trying to find it any other way

[spiral lodge]

<@&268886789983436800>

[long geyser]

I just derived the equation of a parabola!

[silent plank]

Your y^2 turned into y

and your 4 and y look too similar

[long geyser]

Yep my bad

[long geyser]

Are your y's curved? I'm not sure how to change that, other than writing x curved for one of the diagonals or making the bottom of the y more slanted to the left.

[silent plank]

i loop the tail of my y
and the top part of my 4 looks more triangular (like how it appears here)

[long geyser]

Ty!!

Like this?

[winged copper]

Yea

[long geyser]

Thanks

[winged copper]

Thank @silent plank not me

[queen juniper]

does this not look like a person sitting down?

4°

[long geyser]

It doesssss!!!!

[shadow imp]

that are grade 10th questions

[tall dove]

I don't see CBSE questions of grade 10 are very elegant, I mean standard and basic is just of name, both are just simple.

[shadow imp]

yes that's true

very simple

[tall dove]

Proving, and proving converse are different.

[tall dove]

You just need to know how to solve - ax⁴ + bx² + cx + t type equations.

[frosty delta]

This is whats asked generally, the one which he gave is not possible ig without assumption

[frosty delta]

Elaborate a bit, whats t here, where did this come from

[tall dove]

Verification is easier way, but ig, it is not always good to know m. truths.

t=√m

[frosty delta]

The sign is wrong
It should be t⁴ - n² - 4t(n)^1/2

How do you solve this?
There are two variables, one is t that is root m and other is n = Tanx - Sinx

[opal grove]

I need help with 4 please

[frosty delta]

Um a tennis ball is just a sphere

[opal grove]

I crammed my economics hw so my brain is fried

[tall dove]

Oh yes, solve that, @ab

[frosty delta]

It requires area of material required for 100 such balls so calculate total surface area of 1 sphere that is 4 pi r² and multiply it by 100

[opal grove]

Oh okay thanks

[tall dove]

It's easier to solve power 4 e. of above type

[frosty delta]

How tho? Idk how to solve 4 degree equation

[tall dove]

Put, and happy time😂

[frosty delta]

Wth

You dont have any logical explanation

[tall dove]

No solve, its going to cut down

[frosty delta]

What grade are you in?

[tall dove]

I'm in 10.

[frosty delta]

Who is teaching you all this?

That is the most time taking method

[tall dove]

Of course you can solve this, if you know abelian theory, this have radicals, no matter how long.

[frosty delta]

That was just a proof of a linear equation, you turned it into a 4 degree shit and now this formula

[tall dove]

Ahhh! Verification, is correct but still not suitable.

Anyways one may assume m = s/c + sk, and prove k = 1,by comparing too.

[simple vigil]

What's this formula

[sharp rover]

Oh Maths Gurus hlo

[slim plinth]

I have this one

Inverse trig

[frosty delta]

Pi/3

[winged oyster]

woah the derivative of the volume/area of geometric shapes is the circumfrence/surface area

but what about squares

because the area is x² but the circumference is 4x

and cubes too

x³ but the surface area is 6x²

[winged oyster]

4pi(3.5)²*100

[frosty delta]

I guess circumference is used for curved surfaces

[wraith pumice]

maybe hes asking about perimeter?

[modest sluice]

any clue how to get obtuse ABC?

[winged oyster]

oh ye i meant perimeter

[frosty delta]

BAD will be 30° (360-330)
Join B to AD to form right triangle then the remaining angle in that right triangle will be 60° (remember)
Join B to CD at point E, another right triangle will be there, now CBE will be 59° (360-301)
Now as you joined B to E and B to AD at a point, a square is formed there, so Obtuse ABC will be 90+59+60 = 209° (because obtuse angle is greater than 180°)
Hope it helps and check the answer as Idk if this is correct or not, just my solution

[modest sluice]

Thanks

[long geyser]

I tried doing it like that, and now my writing looks really pretty! Thank you!

[slim plinth]

What’s directrix

[queen juniper]

direction tractrix

trust

[slim plinth]

Is it the length or the distance

[queen juniper]

it's actually just a straight line

search it up

[slim plinth]

Ok

[queen juniper]

too lazy to define rn

[slim plinth]

Ok

[solar sky]

How do u this?

[queen juniper]

triangle sum theorem and rhombus diagonal rule

[solar sky]

Alr thank u

[idle coyote]

Finally constructed a pentagon!

[queen juniper]

good job, but it scares me

[idle coyote]

It isn't the most elegant, but at least it's mine...

[shut thorn]

Did you also summon a demon in the process by chance?
😭🙏

[idle coyote]

I can assure you I didn't.

[shut thorn]

🤨

[shut thorn]

"Q.) The base of an isoceles triangle is 16cm. if its area is 48cm^2. Find its perimeter"
im stuck on this question

is there a better way to do it than to just use herons formula and square it?

[shut thorn]

ohh thanks

[exotic yarrow]

[shut thorn]

i forgot that the height is a perpendicular bisector

😭

[slim plinth]

(6,8,10) triple warning

[exotic yarrow]

mhm yes it’s a scalar multiple of the (3,4,5) triple

But writing out words takes more effort than just writing the square root expression

Hence why I just chose to “use” the Pythagorean theorem

Even if I used the triple to shortcut the actual result

[idle coyote]

Update: it turns out that my pentagon was very slightly irregular. My virtual compass wasn't representing the angle with enough precision.

[dark sparrow]

so was it an error in the construction or was it merely a precision defect in your software?

[iron geyser]

yo

[long geyser]

Why can't you just add the vertical angles that correspond to APD and DPC to get 174=major-arc-ADC?

Is it because AP is not parallel to PC and DP is not parallel to PB?

They're not necessarily parallel (and can't be otherwise this wouldn't be a circle with 360 degrees) because AC and DB aren't given as diameters, right?

[cunning lion]

is there any more information given in the problem?

[long geyser]

No

These are the official steps though

[cunning lion]

but yes you don't know what APD and DPC are, so idk how you would work with them

certainly AP doesn't look parallel to PC and similarly with DP and PB

[long geyser]

I thought that you could say DPC=70 because of vertical angles and that APD=104 because of vertical angles

but you can't

i can't

[cunning lion]

that would wok if APC and DPB were straight lines, but they don't appear to be on the diagram

[long geyser]

Yeah, I got confused, I thought for a second that they were straight lines and so i was really confused when my answer got flagged as incorrect.

tysm!!! ❤️

[cunning lion]

one clue that you can also get is that if AP and PC were colinear, then angle APC would equal 180 degrees, which it does not

[dark sparrow]

wok

[long geyser]

That makes sense! Tysm

[upbeat lava]

Is it possible to calculate any exact trig value without a calculator?

[long geyser]

Would you say that point P is collinear with line segment AC if you wanted to describe the collinearity of that line segment?

[long geyser]

Maybe this? https://math.stackexchange.com/questions/501660/is-there-a-way-to-get-trig-functions-without-a-calculator

[fading geyser]

Not exactly, but with Taylor approximation, you can get pretty close

(Its not easy to find the approximation with Taylor by hand though)

[upbeat lava]

What is Taylor approximation?

[fading geyser]

its a way to approximate functions using polynomials

Its not pre-university (doesnt mean that you shouldn’t learn more about it)

But just so you know

[rapid path]

Dont understand anything from geometry

[fading geyser]

Make specific questions and we will try to help you!

[fading geyser]

https://youtube.com/shorts/lYzChTh5YVo?si=y5FB3ETP84Rhwq4w

[rapid path]

Literally anything, I just can't remember what to do or use.

[upbeat lava]

Ty

[upbeat lava]

What is sin(x)?

[fading geyser]

Have you taken trigonometry?

I guess not

[long geyser]

It's the sine of an arbitrary angle, I think.

[upbeat lava]

I’m at high school math and we haven’t covered it I guess

[fading geyser]

But is basicly a periodic function that has a relation between degrees of an angle of a right triangle and its sides

[fading geyser]

You will, its fun

[upbeat lava]

We’ve covered some trigonometry but not this yet

Also it definitely is fun!

[long geyser]

Sin gives you the length of a segment, right?

[fading geyser]

Mmm yeah, its the vertical segment of the triangles from the circumference of radius 1

[long geyser]

How about for an arbitrarily right triangle? it just gives the ratio then? Like with a right triangle with hyp=5

[fading geyser]

But in general is the division between the opposide side of the angle and the hypothenus

[fading geyser]

Yeah, then its a ratio

[long geyser]

So is it easier in the context of a unit circle?

[fading geyser]

Yes , because the radius is 1, so there’s no denominator

[long geyser]

Right

[fading geyser]

And you can generalize sin and cos for angles bigger than pi/2

[long geyser]

Because Sine=Opposite/Hypotenuse and anything over 1 is itself.

[fading geyser]

Yep

[long geyser]

Nice!

[ruby lake]

Can anyone answer my question on MSE? I’ve tried to solve it many times, but I’m stuck. I need the solution to learn these kinds of math problems (which are very common) and to continue my 2D physics engine.
https://math.stackexchange.com/questions/5122141/line-segment-sliding-acros-2-surfaceslines?noredirect=1#comment11040579_5

[upbeat lava]

I watched a video on Taylor series but it has a lot of prerequisite so I’m not yet able to apply it

Also I realise that I do in fact know what sin x is but didn’t realise because I’ve been doing trigonometry with a formula triangle instead of algebraically

But yeah, I’ll try and make a start on calculus and hopefully I’ll be able to understand it

[shut thorn]

i dont think thats possible is it?

[slim plinth]

The green one can use Pythagorean theorem

[shut thorn]

then nothing else

literally nothing else 😭

[slim plinth]

Like this

[slim plinth]

And laws of cosine

[kindred scaffold]

wsg yall

[fleet bramble]

Hey guys

I need a little help with vectors

Anyone available to explain

[sly urchin]

How to prove that the Gergonne point of a triangle is the Lemoine point of its intouch triangle?

[obsidian hornet]

If I understood correctly, this is should be just by definition of symmedian for intouch triangle

[pearl oxide]

it is (not certain anymore, my eyes deceived me and I saw two whole right triangles)

[bitter hinge]

anyone do gcse

[strange pond]

!da2a

[lime crown]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[upbeat lava]

I do

[low rune]

can someone help me understand how to find the values to this problem?

[exotic yarrow]

reposted in help channels ^

[sly urchin]

Oh yeah

[green yew]

igcse yeah

[visual glacier]

chat i am in grade 10 studying trignometry rn and cant seem to understand what sin cos and tan are, sm1 mind explaining, will help me greatly!

[green yew]

they are basically the ratios of sides of a right triangle

in respect to an angle

[visual glacier]

i already know for right triangle, i dont understand when we start using circles

soh cah toa is easy, but circles i dont really know

[green yew]

hmm alr

so in a circle

in a unit circle

all the radii are

1

which we consider the hypotenuse

[rustic lynx]

its just quadrants

[upper karma]

https://youtu.be/57VrEiEPD1I?si=39Ky-nUs55dHzbpm

,tex .unit circle

where are factoids

[nimble hearth]

Heavily recommend watching both of these, especially TR-15Z as it shows where the common values are derived from:
https://youtu.be/oJgBJfstOOU?si=MF0KFGhovLHSqQEw
https://youtu.be/i9ahDcV-bVg?si=7vrKzUtus5wUx_8T

[tribal rose]

I forgot the identity to find the second time. Isn't it something along the lines of 0.927... + 2pi?

[exotic yarrow]

$\cos(x)=y \implies x=\arccos(y)+2\pi n$ or $x=\left(2\pi-\arccos(y)\right)+2\pi n$, where $n$ is an integer

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

you can justify this to yourself with the unit circle + periodicity

[tribal rose]

Thank you!

[orchid zinc]

How to prove heron's formula

[queen hemlock]

Every point on the circle is from an angle + radius of the origin, you can create a triangle to map the origin going to the point on the circle, and trigonometry can be used to find that point

[exotic yarrow]

you can google this

https://proofwiki.org/wiki/Heron's_Formula

[orchid zinc]

Ty

[sacred nova]

do you guys know how to make the dotted/dashed lines graphing out the movement of the points on these circles appear as a trail left behind by them, instead of simply being line equations?
I know I should use some kind of time slider, but how do I incorporate that into the graph?
https://www.desmos.com/calculator/6tejaeooel

dont question the name btw

[sacred nova]

I'm talking about the parametrics
\left(30\cos a+10\cos-5a,30\sin a+10\sin-5a\right)\operatorname{for}0\le a\le8\pi

Thanks!

[visual glacier]

aight ty

[edgy shell]

Beautiful geometry.

https://www.tiktok.com/t/ZThPkDm82/

[sacred nova]

its...interesting

dare i say peculiar?

[molten swan]

guys gen how do I get better at geometry?? My english abilites are much superb, but my math not so much

[dark sparrow]

make diagrams for every question, and in these diagrams give names to everything relevant

[molten swan]

I just like don’t get how to set up an equation, especially with all of the theorems and such

Like im supposed to be smart im in the smart boi math class

[dark sparrow]

kinda vague tbh

theres a lot of different equations that could crop up

do you have any specific examples of questions you have struggled with

[molten swan]

Ignore the other tabs..

Anywho, others find it very simple but it just hasn't clicked with me yet.

[dark sparrow]

so it's specifically with angles, ok

what's the most fundamental thing you can say about angles in a certain very simple kind of shape

[molten swan]

Most share congruent diagonals and many paralells??

[frosty delta]

In Q11, join W to Y
In Triangle ZWY, ZW=ZY
So Angle ZWY=ZYW, then use angle sum property
Now do this in triangle WXY
And find the two angles
This is my way of solving it at first
Hope this helps

[molten swan]

This is a good way of explaining it thank you!!

[tribal rose]

Does anyone have a list of identities for these kinds of problems? The ones I'm finding online are missing some of them.

[spring lion]

Legit can't do triangles 💀

Geometry is so hard man

I can do all of math but not geometry

I was scared to ask such a simple question here but yeah

I hope you guys don't judge

[marsh wing]

use coord bash

[exotic yarrow]

Reposted here: #help-20 message

[exotic yarrow]

We don't care about this. However, when asking for help, please show what you've done so far - it gives us more context and saves time from explaining things unnecessarily. \ \

Since we had the equal lengths $AD$ and $BC$ that are not part of the same triangle, we are motivated to construct a \textit{new} triangle. Consider constructing $\triangle ADE$ on $\overline{AD}$ that is congruent to $\triangle CBD$ and angle chase.

[somber coyote]

Civil Service Pigeon

[spring lion]

Okay but I legit couldn't do anything here

[exotic yarrow]

triangle BCD is fairly simple to angle chase tbf

[spring lion]

I can't do these kind of questions at all

I started from the beginning but still don't understand it

Like this?

No clue

Ugh is there a tip for this stuff man

The whole tests are full of questions like this and I can't do any of them unfortunately

[exotic yarrow]

"on AD" means that one of the sides should be AD

[spring lion]

Oh

[exotic yarrow]

eh these angle puzzle questions before trig are basically all constructing things

try to look for the odd thing out and force it into being useful

like BD=CD is easy to use because they're both part of the given triangle BCD

so you can say that CBD=BCD from that directly

but AD=BC, not so much

so you should make a construction that you can use it (this is what I meant by "force it into being useful")

[spring lion]

So I draw this

Man what do I do next

[spring lion]

Okay

[exotic yarrow]

Again, there's some easy angles you can fill in

like BD=CD is easy to use because they're both part of the given triangle BCD
so you can say that CBD=BCD from that directly

start with this

[spring lion]

This?

Is CBD 10

[exotic yarrow]

mhm

[spring lion]

Okay

Then

Um

BDC is 160?

Would that help somewhere

[exotic yarrow]

mhm

[spring lion]

Okay wait

AD=BC

I need to use that

[exotic yarrow]

mhm

that's why I constructed triangle ADE

[spring lion]

How do you just construct a triangle like that man like how am I supposed to think of that

😭 what do I doooo

I need to figure out geometryyyyy

[exotic yarrow]

You have a triangle with side BC, so try to construct a congruent triangle with side AD

[spring lion]

This?

[exotic yarrow]

mhm

[spring lion]

But how do I use that

[exotic yarrow]

did you fill in all of the angles that you possibly can

[spring lion]

Hmmm

Idk man I don't see anything

I'm trying hard

But

I just don't see anything

It's just 70

ADB is 70 as the questions says

[exotic yarrow]

Note that we constructed congruent (aka identical triangles)

this means that the corresponding angles in these triangles are all the same

[spring lion]

AED 160?

[exotic yarrow]

mhm

[spring lion]

Okay

EDA

10

But ADB is 70

[exotic yarrow]

mhm

[spring lion]

Sooo...

[exotic yarrow]

angle ADB is split into angle ADE and angle BDE

[exotic yarrow]

so $\angle ADE+\angle BDE=\angle ADB$

[somber coyote]

Civil Service Pigeon

[spring lion]

60?

[exotic yarrow]

mhm

[spring lion]

Okay how do I use that 60

Hmm

Ahhhh I'm so god damn dumb man

What do I do

With 60

No clue

[exotic yarrow]

try making another construction

[spring lion]

2 sides are congruent so the third should be too?

Or is this completely wrong

[exotic yarrow]

$\triangle EBD$ is equilateral, yes.

[somber coyote]

Civil Service Pigeon

[spring lion]

Alright

60+70+10+x

Wait nvm

So ABD is 60?

[exotic yarrow]

$\angle EBD=60^{\circ}$, not $\angle ABD$.

[somber coyote]

Civil Service Pigeon

[spring lion]

Oh

Okay

So I need to find one more thing

So how do I find ABD?

[exotic yarrow]

go back to angle chasing

[spring lion]

Sooo

BAD is x

ADB is 70

EBD is 60

EAD is 10

😔

Mannnnnnn

[exotic yarrow]

||look at the angles around E||

[spring lion]

Um

Bro you might kill me for this but I don't get it

Like

[spring lion]

.

I know this

It's an equilateral

[spring lion]

Should I use the 160 degrees?

[exotic yarrow]

yeah that would be relevant

I'm referring to these three angles

literally the angles with vertex E

[spring lion]

OHHHH

I KNEW IT

160+60+y

y=140

[exotic yarrow]

mhm

[spring lion]

So then what?

[exotic yarrow]

keep going

you don't need me to spoonfeed you every step

[spring lion]

I know but I don't know what to do mannn

If I can't even do this I am just dead

[exotic yarrow]

you need x

work backwards

what angle(s) would suffice to find x

etc

[spring lion]

Well

I would need to find out

Only ABD

EBD is 60 and we need to add something to ABD

To find x

[exotic yarrow]

but you drew BE

alright nvm

angle chase the remaining two triangles

EBD and AEB

[spring lion]

For AEB I gotta use 140 right?

[exotic yarrow]

yes

[spring lion]

140+x+EB=180?

[exotic yarrow]

what is EB

angles are named with three letters

[spring lion]

Oh

Right

Wait

EBA?

[exotic yarrow]

Not the entirety of x

$140^{\circ}+\angle EBA+\angle EAB=180^{\circ}$

[somber coyote]

Civil Service Pigeon

[spring lion]

Oh

Wait one of them is 30

Since

Hmm

That should be x-10?

[exotic yarrow]

sure

[spring lion]

Man

EAB is x-10

😭😭😭

I should just start over the whole topic

Angle chasing is killing me

Do you know a place I could learn this type of stuff?

[exotic yarrow]

sum of angles in a triangle...

[spring lion]

It's 180

[exotic yarrow]

eh there's just a few things you need to know (for this question at least):

• Angles in a triangle add to 180 degrees
• Angles around a point add to 360 degrees
• Angles opposite congruent sides in a triangle are also congruent
  The rest is having problem solving skills

[exotic yarrow]

now use that on triangle ABE

$\angle EAB=(x-10)^{\circ}$, $\angle AEB=140^{\circ}$, $\angle ABE=\dots$ add to $180^{\circ}$

[somber coyote]

Civil Service Pigeon

[spring lion]

x-10+140+y?

[exotic yarrow]

note that AE=EB.

[spring lion]

Wait

2x-20+140?

30???

[exotic yarrow]

yes

[spring lion]

😭

[exotic yarrow]

good job

[spring lion]

Thank you bro

[long geyser]

How do you notate whether you're referring to the convex or concave part of an angle?

[long geyser]

Outer/inner angle

[slate pumice]

i think the notation -60 degree means that 60 degree is the outer angle

cause u generally measure angle from positive x axis

anti clockwise

[tender shoal]

could be talking about in case with triangles and stuff

where there isnt a clear coordinate plane

could just use greek alphabet

[slate pumice]

i mean u can assume any axis to be x or y

as far as i understand genearlly i use -ve sign to show its the outer angle

like angle ABC -160* means that its the outer angle

not really sure if thats the correct way to denote it

[tender shoal]

yeah but like

lets assume you'd have a triangle

you would have a side of a triangle

would u use that side as your x axis?

[slate pumice]

ye i understood what u mean

that approach wont make sense there

i mean for a triangle if any angle is greater then 180 then its obvious that its the outer

[tender shoal]

are we talking about supplementary angles?

or like

idk lol

u could have

theta

and then denote the outer angle as theta'

[slate pumice]

theta' generally means complementary

which is 90 - theta

[tender shoal]

really?

i didnt know that

[dark sparrow]

...since when lmao

[slate pumice]

I mean my teachers and me generally use that

also it makes more sense i would say

mainly in phy

even math.stackexchange says this

its not standard

but generally used

[spiral lodge]

I have never seen this notation, neither in my country, nor in many other geometry exercises here in this server 🤷‍♂️

[winged oyster]

whats a good way to remember the 30-60-90 triangle

[silent plank]

derive it yourself
by bisecting an equilateral triangle with side length 2

and applying pythagoras

you're more likely to remember if you go through the process of where the values come from
and in the event that you do forget, you'll know how to obtain them again

[winged oyster]

alr

wdym by sidelength 2

do you mean the bottom or the diagonal

[silent plank]

i mean the sides

of the triangle

each side of an equilateral triangle has the same length
set them as 2 for convenience

[winged oyster]

ok

ty

[fierce karma]

I don't get whats wrong with this

[dark sparrow]

do you know that angle AEB is pi/7...?

actually, hold on.

what exactly is x? is it the distance from any corner to the center of the coin? then how do you know AB = x * 2pi/7?

[fierce karma]

x is the equidistant point of all vertices

I thought I was allowed because it's almost like it was an regular polygon with 7 sides

[fierce karma]

Also I just noticed it was supposed to say segment ab

[fierce karma]

what i did is divide 2pi by the number of sides, that should give me the angle of each side, i then just use arc length formula

[dark sparrow]

yeah but that isnt gonna be the center of arc AB then

if you wanted to say arc AB = x * 2pi/7 then the arc would need to be centered on the center of the coin. but it is not

[lunar rain]

https://www.desmos.com/geometry/jjh9354lz8

[spiral copper]

I subtracted the area of the isosceles triangles from the whole sector

[dusk wren]

Help me solve this question (Q from JEE main sample)

[exotic yarrow]

Angle chasing yields that the chord subtends a right angle. Further hint: ||distance from the centre to this chord||.

[dusk wren]

Ok fine I check

[fierce karma]

Ye, that's exactly how the mark scheme has done it

Idk why I didn't think of it 😭

[spiral copper]

what is mark scheme

[spiral copper]

na bro, u inspired me, u did the major part

[dusk wren]

Who will solve this tricky one ??

[real sentinel]

Use concept, length of focal chord= 4a cosec²ø

[pliant mural]

[cos(pi+j)] in the form of [a+bj]

[real sentinel]

?

[pliant mural]

this is unrelated

[dusk wren]

Correct

[sudden folio]

[
\sin \theta = \frac{3}{5}, \quad 0^\circ < \theta < 90^\circ
]

[
\text{Find the exact value of } \cos \theta.
]

[somber coyote]

wonhaukau

[sudden folio]

help

3/5 of 90 is 67

3 4 5 triangle

oh

so cos is 4/5

big brain

[
\cos \theta = \frac{5}{13}, \quad 0^\circ < \theta < 90^\circ
]

[
\text{Find the exact value of } \sin \theta.
]

[somber coyote]

wonhaukau

[sudden folio]

oh so 5 and 13

ohh ok i see how it is

[faint pier]

5 12 13 triangle

sintheta=12/13

[sudden folio]

oh so

sin-1

(12/13)

[faint pier]

yes

but why are you learning it like this

you can use pyth

[sudden folio]

how should i learn it

[sudden folio]

what’s that

[faint pier]

pythagoras

theorem

[sudden folio]

i know but i’m lazy

[faint pier]

🙏

[sudden folio]

like i don’t wanna do sqrt of anything negative anything

i already know pyth

as u call it

[faint pier]

you wont get negative

[sudden folio]

i mean something minus

something

[faint pier]

practice it bro youll need it

[sudden folio]

[
\tan \theta = -1
]

[
\text{Find all values of } \theta \text{ for } 0^\circ \le \theta < 360^\circ.
]

[
\text{In a right-angled triangle, } \tan \theta = \frac{5}{12}
]

[
\text{Find the value of } \theta \text{ to the nearest degree.}
]

[somber coyote]

wonhaukau

[sudden folio]

tan-1(5/12)

ez

[real sentinel]

yea

ik

[tribal rose]

Okay, I found a list of trigonometric identities, and I can almost reliably find the first solution now.

... so how do I get the second solution? I know 180 + -56.44 is 123.56, but I don't know if A is the second solution.

[winged oyster]

is there a way to actually solve tan(theta) = 5/12 w/o calculator

[exotic yarrow]

Recall that
$$\sin x=y \implies x=\arcsin(y)+2n\pi, (\pi-\arcsin y)+2n\pi$$
where $n$ is an integer. This can be justified by considering the unit circle and periodicity.

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

Exact form in terms of $\arctan(5/12)$. The calculator is just giving you a decimal \textit{approximation} (read: not exact form) of this. If you wanted to approximate by hand, you theoretically could use something like the series expansion
$$\tan x=x+\frac{1}{3} x^3+\frac{2}{15} x^5+\dots$$
and use enough terms to get a sufficiently accurate answer. (Use something like the Lagrange error bound to determine how many terms you need.) This is what the calculator is doing under the hood anyway.

[somber coyote]

Civil Service Pigeon

[winged oyster]

oh ok

[exotic yarrow]

[winged oyster]

thats gonna be a really long process

[queen juniper]

ooh maclaurin series

for tanx i never bothered 😅

[cerulean lintel]

5x+7y

/textit 5x+7t

/Texit 75x+72

[cerulean lintel]

How do i make one of these

[faint pier]

latex

[cerulean lintel]

I found out already thx

$Y=-\frac{a}{b}x+\frac{c}{b}$

[somber coyote]

emiya

[solid gate]

hey all,
would anyone please be able to prove that the angle at the centre of this image is 2x

[exotic yarrow]

https://www.khanacademy.org/math/geometry/hs-geo-circles/hs-geo-inscribed-angles/a/inscribed-and-central-angles-proof

[solid gate]

the proofs usually go like this but i dont see how to apply it to the first diagram

[solid gate]

thank you !! yep all makes sense !

[solid gate]

😭 erm actually so i understand the examples on khan academy but im trying to apply it to the question i shared and idk its not working 🙁

[exotic yarrow]

those cases should be exhaustive

is this not just case C

[solid gate]

i believed it was im still trying to make it make sense...i will look at it for a few more mins

[solid gate]

yes it is c ... i understand it now 🙂 thank you 🙂

[exotic yarrow]

[tribal rose]

Thank you so much! I finally figured out how to use this!

I finally got a question right, and not by pattern-guessing the answers.

[exotic yarrow]

[long geyser]

"In circle P there exists a quadrilateral that has all 4 vertices on the circle, and each angle is also an inscribed angle. Prove that any two opposite inscribed angles must sum to 180 degrees." Is this a correct way to frame a question for a proof that I'm trying to do? I'm trying to generalize that for any quadrilateral that has all 4 of its vertices on a circle that the opposite angles inside of it sum to 180 degrees.

I'm aware that there already exists a proof for this, but at the end of this video I'm watching the instructor encourages to try and generalize this for any case; when the angles aren't given.

[queen juniper]

||well, we can draw diagonals for any inscribed quadrilateral making 8 angles, and arrange them into the form 2(angle1+angle2+angle3+angle4) using the fact that angles that intercept the same arc are equivalent. Then, it's easy from there||

[long geyser]

Sorry, to reiterate: I'm just wondering if my question is correctly said for what I'm trying to do "In circle P there exists a quadrilateral that has all 4 vertices on the circle, and each angle is also an inscribed angle. Prove that any two opposite inscribed angles must sum to 180 degrees." I don't want to get spoiled the steps.

[queen juniper]

oh, i see 😅 . well, ur question is mathematically sound, but u could just say "proof of cyclic quadrilateral theorem", as that's the name for it.

[long geyser]

Thank you very much!!

[cunning nebula]

Is this working out and conclusion all correct?

[winged oyster]

why is the volume of a sphere (4/3)pir³

id assume it has something to do with its surface area being 4pir² but uh

why specifically

[exotic yarrow]

https://letmegooglethat.com/?q=volume+of+sphere+derivation

[winged oyster]

ok so it is what i thought it was

derivative of the volume of a sphere

Ok!

[visual jackal]

hello+

[muted cove]

what a good server

[tired nymph]

I just want to know, Im doing mathcounts soon is there any important formulas that i need to know?

@everyone

[long geyser]

Can I say that the cyclic quadrilateral theorem holds because x and y add up to 360 degrees and so half of them must add up to 180 degrees?

[queen juniper]

that works, nice!

[long geyser]

Tysm!!!!!! ❤️

[worn swift]

Desperately need help understanding this

[dark sparrow]

can you show what the dropdown gives you as options

[wispy bluff]

the sine is an odd function, using that fact you should be able to find which of those options equals -3 sin (-2x), try for a bit if you fail to get it come back here

[shut thorn]

The hell does it mean to have a negative angle

[long geyser]

Is it impressive that I solved this in my head?

[drifting cipher]

yes!

[long geyser]

Thanks! I’ve done a lot of practice, so I remember what to use and when for what I’m trying to solve at a given portion of the problem, and it’s a bit tedious for me to use a calculator other than when it’s difficult otherwise or time consuming so I try hard to rely on my brain, and I happened to get to the solution for this specific problem before I needed to use any calculator!

[drifting cipher]

math is so fun when you start to understand and do things in your brain

[long geyser]

Yayyyyy

[dark sparrow]

@long geyser excellent emoji right there

[long geyser]

Where?

[dark sparrow]

in the react to k's message

[long geyser]

Yayyyy ty

[drifting cipher]

i agree!

[drifting cipher]

[raw nacelle]

Tuff schedule for sophomore year?

[drifting cipher]

Hardest class is phys ed right there

[nocturne breach]

Fr

[dapper hinge]

my russian 4d theory. ts terrible

[solid gate]

can you please explain how you solved it? Im trying to but cant yet

[long geyser]

Of course, this is how I solved it step by step:
The problem was on Khan Academy https://www.khanacademy.org/math/geometry/hs-geo-circles/hs-geo-inscribed-shapes/a/challenge-problems-inscribed-shapes

The explanation is on the website too.

[long geyser]

The images are from Khan Academy.

[long geyser]

This can seem quite daunting to try to do every step at once, but when you do it step by step it becomes way simpler. First you are given a diagram/figure with information about it, and you're trying to find the length of a major arc in the circle. From there you can see that the inscribed angle that intercepts arc BCD is BAD, so if you know the measure of it then you can double that measure to get the angle of the arc that it intercepts because of the angle at the center theorem. Knowing the arc measure of its angle lets you solve for known proportions of the circle which will let you find the length of the arc that you're trying to solve for, which completes the solution to the problem. You also have to know algebra and geometry terminology, if you'd like I'd love to explain some more if that interests you.

[tough token]

How do I do this?

[frosty delta]

angle 3 = angle 4 (By CPCT)
So AB = AC
As Angle 1 = Angle 2
So AD = AE
Now in AB = AC
Divide by AD
AB/AD = AE/AD
As AD = AE, we can write it as
AB/AD = AC/AE
By Converse of Basic Proportionality Theorem
DE is Parallel to BC
So Angle 1 = Angle 3 and Angle 2 = Angle 4
By Angle Angle Similarity,
Triangle ADE is Similar to Triangle ABC

@tough token

Hope this helps!