geometry-and-trigonometry

[celest wave]

broo

[upper karma]

,calc 4*sqrt(26)

[celest wave]

whats the total surface area

[somber coyote]

Result:

20.396078054371

[celest wave]

ok

that

[upper karma]

24.39

per container

[celest wave]

bro

[upper karma]

@silent plank is that right

[celest wave]

u just said it was 4

[void roost]

😦

[upper karma]

no

[celest wave]

yes

[upper karma]

show

[void roost]

😿

[upper karma]

circle o has radius 6

[upper karma]

STOP CRYING

[void roost]

Yes

[celest wave]

so whats the total surface area for year i dont get it

[upper karma]

,calc 20.396078054371*365

[somber coyote]

Result:

7444.5684898454

[upper karma]

answer

[celest wave]

is that my answer

or his answer

[upper karma]

i guess so

its ur

bruh

[celest wave]

ok

thanks

[upper karma]

distance o from p is 4

ab is parallel to op and dist. between it is 2

op = radius of circle + distance

[void roost]

Yep

[upper karma]

so its 10

let m be the midpoint of ab

[void roost]

👍

[upper karma]

then om = radius - distance from ab to op which is 4

now there is a triangle

u use the pythagorrus

u get am^2 = 20

am = root(20)

am = 1/2 of ab (midpoint), then ab = 4*sqrt(5)

if circle is at 0,0 in cartesian plane

then p is at 10,0

wait

no

[void roost]

Hmm?

[upper karma]

my result is diff

from the answer

[void roost]

What is it

[upper karma]

224

its PA^2 right

[void roost]

Ye

[upper karma]

we need someone who specialises in geometry

[void roost]

My answer came 278

[upper karma]

wtf

how

[void roost]

There is error in that ans

Hold up let me send screenshot

[upper karma]

can u send the whole ques

i think the question has smth missing

[void roost]

It is the entire ques

[celest wave]

rachit

[upper karma]

can i use trigonometry in it? @void roost

[void roost]

Yes

[upper karma]

is it in the syllabus

[upper karma]

what?

[celest wave]

/canu help me with last question

[upper karma]

later

[celest wave]

when is later

[upper karma]

nvm i cant find whats wrong

[upper karma]

12hrws

[celest wave]

what

no

r u serious

[upper karma]

yes

bye

[celest wave]

nooooo

[void roost]

This was my approach

The extended A is the problem

[sturdy kayak]

Can someone answer

F

[trail tendon]

undefined

[sacred nimbus]

how do i set up the two similar triangles?

[flint ingot]

i think ADC is similar to DCB

because DCB+DCA=90 DCA+CAD=90 DCB=CAD

so the answer is 1

i think

[bronze flicker]

it's 2

cuz ∆ACD~∆CBD

I think

[harsh crow]

nope

that would be implying ad = bd

which is false

[bronze flicker]

how?

[harsh crow]

cancel out the cds

btw 1 is the correct answer

because triangle ACD is similar to the whole triangle, ABC

AD corresponds to AC, while CD corresponds to BC

[bronze flicker]

the ratio would come out to be
AD/BC=CD/BD=AD/CD
how are you gonna cancel out the CDs?

oh wait

I didn't read the options correctly

My bad

yeah, have to prove that it's similar to the whole triangle

[harsh crow]

or

that its similar to the other

which i think works too

but im not sure

thats wht @flint ingot

did

[bronze flicker]

works

I didn't read the options correctly

[neat plover]

hii

quick question

x must be OP right?

like, the radius

[maiden brook]

i mean u dont necessarily know its a rectangle just from the right angles

[neat plover]

mmm you're right

[upper karma]

Can someone give me the table of trigonometric ratios please?

[royal pawn]

Hi! If I have a triangle with 2 sides equal to 3 and 4, will the 3rd side always be 5?

[neat plover]

Could i write it as x^2=(r-a)^2+(r-b)^2?

[maiden brook]

yes

[maiden brook]

nope

there is a triangle with sides 3 4 4

infinitely many triangles with two sides 3 and 4

if there were only one triangle then all triangles having those sides would be congruent, and i bet u havent heard of SS congruence

[royal pawn]

Ah yes… forgot that it could also be 3 4 4. Thanks!

Will ‘AC’ in arbitrary pentagon inscribed in a circle always be greater than any of the triangle sides? Or is there a way to get AC be equal to one side?

[turbid salmon]

By "the triangle sides" do you mean AB,BC,CD,etc?

[royal pawn]

In that case the triangle ABC with sides AB, AC, CB

[turbid salmon]

Ok so, for any triangle, we can circumscribe a circle, right?

And then once we have that circle, we can then pick two points on that circle to be our last two vertices of the pentagon (D and E)

And clearly we can pick triangles where two sides have equal length

[royal pawn]

So in this case, if I say that AC will always be greater than one of the sides, it will be a wrong statement?

[turbid salmon]

It would be wrong because it would be equivalent to saying that for every triangle ABC, AC is greater than one of the other sides

[royal pawn]

Alright, thanks!

And last question about this task. If I want to proof that AC=CE by using the sine theorem, will I have to make a system like on photo (got result by using sine theorem, then with triangle angle property (the sum is 180) and trigonometric formulas get proof that they are equal?

[golden thicket]

have some questions on how to do this, can anyone help me rq?

[upper karma]

bro i keep getting mid scores in geo

93%

in the last qiz i did i got 96.7%

30 question quiz 20 mins

[upper karma]

Hey there. Could someone help me learning Law of Cosines?

[royal pawn]

Hi, what do you need help with?

[upper karma]

Actually, i would like to know how to know the sin, cos or tan of an angle that is not 30, 45 or 60.

I'm not good at math lol

and also i didn't understand a lot about law of cosines

[upper karma]

that's what i am studying on school rn

[royal pawn]

I’ve learned several methods in school. Bradis table (there will be angles and their sin and cos and you can find tg (tan) and ctg by using them), some trigonometrical conversions (like a formula for sin of sum of angles) can help you with getting e.g. cos(120 deg) also sometimes it is useful to build a rectangular triangle and find sin by using the cathetus and hypotenuse (I think sometimes it might be too hard to build it) and of course calculators. I think you should also understand the trigonometric circle, at least how to work with it and understand which quarter gives e.g. negative cos and positive sin

[upper karma]

thank you so much.

[royal pawn]

As far as I know, usually it is enough to know the main formula. Then you can get a formula of an angle (it’s cos)

I hope I didn’t say any misinformation. Please correct me if I was wrong somewhere. Good luck with trigonometry!

[upper karma]

Thank you so much, Triste! Btw, are you brazilian? because "triste" is a portuguese word.

[royal pawn]

No. Word was translated to French or Italian and became my nickname, don’t really remember which of them, but not to Portuguese

[upper karma]

oh, but cool username anyway

[royal pawn]

Ty :D

I assume it will be angle-angle (4) because there are two pairs (actually 3 pairs, but 2 is enough) of equal angles given in the conditions. I hope didn’t miss anything.

[maiden brook]

the answer for the first image is 3), 4) for the second, 3) for 3rd, 1) for 4th and 2) for the 5th

[rocky lintel]

4, the other 3 are postulates for congruence, but since its looking for similarity best answer is angle angle since we can tell at least 2 of the angles are equal meaning so by definition, the second triangle is some scaled version of the first.

basically, aa (4) shows similarity since as long as 2 triangles are proportional, theyll have equal angles

[paper nimbus]

Hey I'm in this class next school year!

[tranquil musk]

<@&286206848099549185> can tell me the the diphenthine equation try to explain like you are explaining a ninth class students and also can anyone tell me BPT or Thales theorem

[digital kindle]

@frail hawk is the value of 'x' 105 degree ? u can use basic mathematics rule to find it out

[tranquil musk]

@digital kindle can you tel me the bpt and diphenthine equation and explain like you are explaining 9 class student

If you can

I am a 9 grader

[frail hawk]

Yes i forgot about the equilateral triangle
They are equals, i forgot about it in that time

[tranquil musk]

Pls

Any one

@frail hawk

[digital kindle]

no problem

[tranquil musk]

Hello @digital kindle

Can you tell

Me

The bpt

And diphenthine equation

Explain like you explaining 9 grader

<@&286206848099549185>

@frail hawk

Hello @frail hawk

[frail hawk]

Well i cant even spell English well and i must using Google translate to know the English of equilateral triangle

[digital kindle]

When you add a non-volatile solute to a solvent, the boiling point of the solution increases. This is BOILING POINT ELEVATION @tranquil musk

[tranquil musk]

In maths

I don't know the the word exactly

Can you think

Diphenthine equation

It has some mistake

Sorry

@digital kindle

[digital kindle]

Actually, "Diphenthine equation" doesn't seem to be a standard term in chemistry

[digital kindle]

no prob

[tranquil musk]

Wait lemme check

[digital kindle]

i think u are referring to basic proportionality theorem

[tranquil musk]

Ya

Thales theorem

And also can you tell me the fermats theorem

Explain like you're explaining to class 9 student

I am 9 grader

[digital kindle]

Definition:
If a line is drawn parallel to one side of a triangle, intersecting the other two sides, it divides those two sides proportionally.

In Simple Terms:
In a triangle, if a line parallel to one side cuts the other two sides, the segments created on those sides are proportional.

for eg. in a triangle ABC , if DE is parallel to BC then AD/DB=AE/EC

[tranquil musk]

Ok

Thanks

And fermats theorem

Are you Indian? @digital kindle

Hello

Is there anyone here

[digital kindle]

Fermat's theorem states that there are no positive integers with value a,b,c which satisfies the equation - a^n + b^n = c^n . Provided that n>2 @tranquil musk

[digital kindle]

yeah brother

[tranquil musk]

Thanks

What is your grade

Or you are in college

@digital kindle

[digital kindle]

i am presently in 10.

[tranquil musk]

What?

[digital kindle]

why?

[tranquil musk]

So how do you know fermats theorem

In which school and where

You study

[digital kindle]

im from kolkata n i study in NATIONAL GEMS HIGHER SECONDARY SCHOOL (ICSE)

[tranquil musk]

Ok

[digital kindle]

@tranquil musk I read about it.

[tranquil musk]

I am in cbsc

[digital kindle]

from ?

[tranquil musk]

Ok

Chandigarh and school is kV

[digital kindle]

Fermat's Theorem is nothing tough .

[digital kindle]

oo. nice to meet you my friend

[tranquil musk]

Me too

Nice to meet you

Can I send you friend request

Can you be my friend in Discord

[digital kindle]

yeah ok

[tranquil musk]

Ok

[formal geyser]

Hello

[formal geyser]

All the radiuses equal 1 cm

What is the area of the rectangle

I understand that the bottom side of the rectangle is 6

[faint pasture]

whats being asked?

[nocturne remnant]

the area of the rectangle

[weary remnant]

^

[formal geyser]

Does anyone have any ideas?

[grave pond]

The all-yellow circle is the inscribled circle of the right triangle. There's a formula for the inradius of a right triangle that should let you write an equation to find the vertical leg when you know the other leg is 6 and the inradius is 1.

[formal geyser]

Thanks!

I thought about this inscribled circle but i dont know any formula related to it

[grave pond]

I didn't either, but then I googled inradius in right triangle :-p

[formal geyser]

But this formula uses two legs, hypotenuse and inradius. We know one leg and inradius, so since we dont know the hypotenuse, we cannot find the second leg

And if we knew the hypotenuse there would be no need to use this formula, since we can apply pythagoras

Maybe i should solve system of equations

[formal geyser]

Thanks

I understand it now

It looks so easy now

I struggle with geometry

[sharp salmon]

Try using Khan Academy, it’s free and it has excellent lessons on Geometry and Trigonometry.

[formal geyser]

Thanks, i have tried to learn limits using khan academy

But from what level of geometry should i start? What stage should i begin with?

[sharp salmon]

The preference is all up to you, and your struggle points. Personally when I have lots of free-time I go on there and do lessons of both Geometry and Trigonometry, just to keep my mind refreshed on the fundamentals and all of the identities.

[grave pond]

I was imagining saying $$\frac{6+x-\sqrt{6^2+x^2}}{2} = 1$$ and then rearranging algebraically to get a quadratic equation.

[somber coyote]

Troposphere

[grave pond]

(Oh, that is what Able scribbled on the diagram too).

[plain cedar]

Im pretty bad at math but I need to know this for programming purposes.

How the fuck do vectors work if there are two different units lmfao

There are two connected bodies, the right circle does a 10m/s^2 force down.

What is the angular velocity of the left circle? (the distance is 5m)

[grave pond]

Can you show the full problem statement, please?

It sounds like you have some unstated assumptions along the lines of "the left circle is fixed in space", but I'd rather not have to resort to guessing.

[plain cedar]

I might as well just say the full problem

The two circles are connected, theyre a bit useless actually, just look at their center.

If the circle on the right does a force of 10m/s^2 down, what would be the equivelant for the vector in the left circle?

Heres the problem: I don't know actual vectors

so i cant word my problem well

[grave pond]

To be honest, that sounds like gibberish. I don't think "a circle does a force of 10 m/s² down" means anything.
First off m/s² is not a unit one can measure force in.
But even "a circles does a force of 10 N down" would be nonsense.

[plain cedar]

yes i know

i havent slept and i feel like im stuck at progress because im working in 10% efficiency im sorry lmao

[grave pond]

Are you sure you don't have an original statement you can show an actual photo/screenshot/scan of?

[plain cedar]

uhh let me think of an example

No

What number makes these two forces identical

if I push on the wing

just ignore.. everything like drag and mass lmao

[faint pasture]

well the circle with no acceleration doesnt change its position basically? and initially their centres are at a distance of 5
then you wanna calculate the angular velocity if i understood what you mean

[faint pasture]

None

[plain cedar]

Dont think of the rotational velocity in the roll axis though

[faint pasture]

what is the "roll axis"

[faint pasture]

oh do you mean as in the plane is rotating about its centre or some shit?

[plain cedar]

No, I mean everything except that

Ok im sorry im gonna go sleep and fix this when im awake

[grave pond]

Before you spend to many words on saying what not to think about, you first need to explain in an understandable way what it is we should think about.

[faint pasture]

well idk what you mean but
tan(theta) = 5t^2/5 = t^2
where theta is the angle made by the initial 5 meters with lines joining the final position of the moving body and the stationary circle
you could differentiate this to get the angular velocity

i got the 5t^2 from the second equation of motion

ie s = ut + 1/2 at^2

[stuck sleet]

The answer is the center of the sphere, Q and the concentric spheres with center Q, right?

[obsidian harness]

yes

[plain scarab]

Could anybody help me out?

[harsh crow]

it reflected over the y acis

axis

meaning that it goes to the left of the y axis now

and it must be the same distance from rhe x and y axises as it originally was

do you still need help?

[plain scarab]

yes please

[upper karma]

I'm back guys

Who needs help 🙈

[upper karma]

I think c

[kind yoke]

yea (2, 4)

[kind yoke]

left ?

[upper karma]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[plain scarab]

thank you

[plain scarab]

thank you

also I have one more

I missed one question

[kind yoke]

sorry I am unfamiliar with this format "r(90, O)"

[upper karma]

theres nothing to explain

its just common sense

[plain scarab]

a rotation of 90 degrees around the origin

[upper karma]

there's many things to explain

[plain scarab]

Does a composite transformation have to be 2 different types of transformations?

[upper karma]

you could explain that reflecting a point (a,b) along the y axis will result in (-a,b)

[upper karma]

No

[golden fern]

I couldnt solve this and i looked at the answers and there was no steps

how do u do it?

[plain scarab]

my teacher is tweaking then

Hold on

I’ll show you my answer

in the comments she says that a composite transformation need to be two different transformations. Not 2 of the same type

[upper karma]

really weird

notational issue?

[plain scarab]

nah

word for word what she said

😭

[upper karma]

issues in definition thne ig

[plain scarab]

ok i got it

would this work

[upper karma]

no

[plain scarab]

damn it

[upper karma]

you didn't transform it to (-4,7) at the end result

[plain scarab]

how do i do that again?

[vernal pilot]

All sides are equal. And you can deduce the interior angles. Split it into triangles and rectangles.

[upper karma]

reflect it again

[plain scarab]

so just reflect (4,7) again?

just making sure

[upper karma]

yeah across the y-axis

[plain scarab]

Kk

[upper karma]

so it'll be (-4,7)

[plain scarab]

This seem about right?

Thanks for all the help

[upper karma]

the last line should be (4,7) over the y-axis to obtain (-4,7)

yea

real

you're translating it to (4,7), reflect it two times to get the same thing

[plain scarab]

Got it, thanks

final answer?

sorry im a fucking idiot

[upper karma]

yes

[plain scarab]

aight ty

[upper karma]

your teacher should be fine with this

[spiral lodge]

Since it's not specified, one of the transformations could be the identity

At least I would have used it, after translating

[languid forum]

That would be technically correct and "objectively" the best way.

[harsh crow]

ma bad right

i cant distinguish left from right if my life depended on it

[sturdy stone]

Fr

[plain cedar]

dyslexia?

the rigid body is going 10m/s at around -120 degrees

[grave pond]

You keep not explaining what the heck that means.

What does it mean to "push down on someting" with a "force" measured in units of acceleration?

Which other forces act on your object?

Is it in free fall?

[plain cedar]

Lmao it weighs 1

The units really do not matter

[plain cedar]

Only these

[grave pond]

Perhaps it does not matter to you that you're asking a question that does not make any sense, but then you should not expect any answer.

[plain cedar]

I want to know how the vector of the body's velocity changes depending on the distance of the force from its center

[hearty raven]

Do you draw the velocity vectors from the center of the object?

[plain cedar]

Yes

[hearty raven]

Then it gets longer with higher velocity and shorter with slower velocity

It's pointing at the position where the object will be in the next "blink of an eye"

that could be a second, or a year whatever unit you use for time

[grave pond]

Then you need to explain a whole lot more about what your situation is. At the moment you insist on giving just that one data point and refusing point blank to give any other information other than the nonsensical statement that there's some "force" that has units of acceleration.

[plain cedar]

You can totally measure force in units of acceleration if you assume mass =1, I already said this

[hearty raven]

Bruh

Just draw a vector on the vectors top

[grave pond]

You're still actively refusing to explain what the heck is up with your situation.

[hearty raven]

that point sin same direction and is (original vectors length)² or something

² = acceleration factor

[grave pond]

Each of the numerous requests for you to clarify the situation has been answered with you repeating the nonsensical question without adding any of the explanation that was asked for.

[plain cedar]

A body is made of several parts, if I wanted to do a force on one of those parts, how would the net angular velocity of the body change depending on this part's position

[hearty raven]

idk what exactly is going on but I know speed(t) = acceleration * t

How are the parts connected?

[grave pond]

That was repetition number about 42 of the question you have been asked to clarify.

[hearty raven]

If the connection is rigid, you can connect all mid points of the body parts and in the middle of all these connecting lines find a mutual middle point

If the connection is flexible or elastic it gets harder

[plain cedar]

Yes its rigid

[hearty raven]

Then you want to look into "center of mass"

[plain cedar]

What do you mean by this?

[grave pond]

When you are asked to clarify what it is you're asking do not just repeat the same question that you have been told does not have enough information.

[plain cedar]

Wait a minute someone useful is answering my question

[hearty raven]

Gotta ninja into the weird thoughts that are hard to describe as a math guy

[grave pond]

Tzaot is also asking you for clarification.

[hearty raven]

yeah but I do it without blaming on my own lack of understanding, the way I do it is add to where I know something and see if it helps.

I'm ready to be told its useless so I don'Ät need a perfect picture to start trying stuff

[plain cedar]

You helped a bit with this

[hearty raven]

and that, in the end, is all I can hope for todelo

https://en.wikipedia.org/wiki/Centroid

The "edges" of this triangle become individual shapes centroid

There is something I don't know:

The starting points of these vectors with individual speeds are not ON the center of mass

so there could be a missing step in my solution

PS: Don't call people who try to help you not-useful

Trying at all is a comendable attempt

convey hard to phrase things as sketch or something idk, don't make ppl loose patience

[plain cedar]

They werent

They were just making snarky remarks at the fact that I couldn't convey my question correctly

Its the sort of inflated ego behaviour youd expect at this server

[plain cedar]

Youd be surprised, this helped me tons in a bit of a different way already.

👍

thanks

[plain scarab]

@upper karma Got a 29/30 on that thanks to you 🙏

thanks

[upper karma]

Np!

[formal geyser]

Hello

The area of right square is 8 cm^2

There are 3 squares

The orange triangle is equilateral

We need to find the area of the pink square with question mark

I'am completely lost

I think the diagonal of pink square is the side of the orange triangle, so if we find the side of triangle we can find the side of the square

[lime dune]

is the equilateral triangle's side length the same as that of the square?

or can we not assume that here?

[formal geyser]

It's not said they are equal

[lime dune]

hmm aight

[formal geyser]

I think we can prove it using this triangle that seems to be osoceles

[lime dune]

looks like you might need some sort of trig

[formal geyser]

.

[lime dune]

hmm

ok yea that seems to be the case

just tried it in geogebra

[formal geyser]

Yeah man i proved it, look...

[lime dune]

in that case it seems the answer would be ||4||

[formal geyser]

Sorry for terrible picture

[formal geyser]

That 30° part is 90 - 60 = 30°. And left top angle in the isoceles triangle is 90 - 15°. The other angle 75° too

[lime dune]

i'd be hesitant to call that a "proof" but in this case it does work out lol

if we're under the assumption that the equil triangle and big square have equal side lengths

[formal geyser]

Why isn't it proof that the side of equilateral triangle's side equals the square's side. No offense, i am just curious

[lime dune]

you're assuming what you're trying to prove

which is circular reasoning

"suppose X, therefore X"

[formal geyser]

So is it proof when we don't now exactly if a statement is true or not, so we need to prove it right?

[lime dune]

if we move the leftmost point around we can get stuff like this

[lime dune]

i would call this more of a "by inspection"/"guess" that happens to turn out right

[formal geyser]

Okay

[lime dune]

same way as if i told you to solve sqrt(x+8)=x+2

and you said x=1 immediately

[formal geyser]

By the way this geometry problem seems interesting to me

[lime dune]

this was a pretty cool problem yea thanks for sharing

[faint pasture]

well a challenge for those who want it, try to find tan(21degrees) ||consider tan(37degrees)= 3/4||

[stuck sleet]

6 is the answer, right? here is my solution

[weary remnant]

I got 6 too

[stuck sleet]

thanks for confirming the answer

[weary remnant]

👍

[stuck sleet]

at #43 is my solution correct?

angle FOA is congruent to angle FOB

angle FOB is supplementary to angle BOC

therefore

angle FOA and angle BOC is supplementary

or Am I wrong?

[obsidian harness]

No, this reasoning is not correct

BFO and AFO are not necessarily congruent

what is true however is that AOF and AOE are congruent

because AF = AE (tangents to a circle from the same point are equal) and AO is common, and OF = OE as they are both radii, so you have SSS congruence

(RHS congruence is of course possible, or SAS etc)

so you have three pairs of congruent triangles in total, just like AFO and AEO

this should allow you to see how if you label all 6 angles around point O

[faint pasture]

well a good proof will be
||AOF = 90 - A/2, BOD = 90-B/2, COD = 90-C/2 its pretty straightforward from there||

[stuck sleet]

RHS you mean Right Triangle-Hypotenuse-Side?

[obsidian harness]

yes

[faint pasture]

i got these because O is the incentre which is the meeting point of angle bisectors and by the fact that radii are perpendicular to tangents

[stuck sleet]

I'm in my fifth year in engineering and still I am really weak at basic geometry...

[obsidian harness]

geometry is a different skillset you know

[stuck sleet]

Kinematics and Dynamics of Machines and Mechanisms have a lot of geometry, I really need to master the basics.

[obsidian harness]

wait are you still in high school or something?

[stuck sleet]

Right now I discovered the angle measurements of angles inside, on, and outside the circle.

[undone quest]

Anyone has some sort of quesiton for grade 9

[stuck sleet]

I am in college I stopped recently from enrolling because my basic math foundation is really bad. I learned calculus and differential equations but I want to fill the gaps missing.

[undone quest]

I wanna do some geometry practice

Class 9 10 11 geometry

[faint pasture]

trig too?

[undone quest]

Ya ya

Anything

Trigo log inverse

[stuck sleet]

this is my 3rd book I am completing on geometry

[faint pasture]

find tan(21 degrees) ||consider tan(37 degrees) = 3/4||

[undone quest]

Pie is eternal

[stuck sleet]

the first two was from

Moise, Geometry
Kern and Bland, Solid Mensuration

[undone quest]

Yo what

Bro how do i find 21

-0-

[faint pasture]

think about it

you asked for a challenge

[undone quest]

Ok ok

I accept it

I will give u the answer in my own way

No copy no internet

[obsidian harness]

you mean approximate tan(21 deg)

tan(37 deg) is not exactly 3/4

[faint pasture]

😭 yea thats why i said consider it to be

[faint pasture]

hmmmm

[flint ingot]

is the circumcentre of a right angled triangle always the midpoint of the hypotenuse?

[faint pasture]

yea

[flint ingot]

k

how do i use circumcircles to do triangle problems?

[obsidian harness]

it comes from the fact that the midpoint of the hypotenuse has the same distance to all 3 point by Thales' theorem

[flint ingot]

ye

thanks

[undone quest]

Solve this

[obsidian harness]

you should specify

[faint pasture]

not that bad

[undone quest]

Jee mains question

Solution of three pages

[flint ingot]

i dont have one rn but i know that you should always draw the circumcircle when given a triangle problem

idk how to use the circle though

[faint pasture]

lemme try

[obsidian harness]

https://math.stackexchange.com/questions/2191058/prove-that-sin-frac-pi7-sin-frac2-pi7-sin-frac3-pi7-frac-sqrt7

there's a neat solution with roots of unity

[undone quest]

Yup

[obsidian harness]

https://socratic.org/questions/how-do-you-evaluate-sin-pi-7-sin-2pi-7-sin-3pi-7 - this one goes on for way too long

[undone quest]

Actually

Anyone heard of sophie germain identity?

[obsidian harness]

yeah

[undone quest]

The number theory one

To prove composites

[flint ingot]

what theorems should i learn to solve geometry problems for bmo questions?

https://ukmt.org.uk/competition-papers/jsf/jet-engine:free-past-papers/tax/challenge-type:79/

these types of difficulty

[stuck sleet]

Thanks I get it now!

[stuck sleet]

thanks too

[obsidian harness]

npnp

[limpid ginkgo]

Hey u guys got any reccomendations for dynamic math software? For algebra and functions and such. I currently use geogebra and hate it sm

[upper karma]

what u utilise them for?

[flint ingot]

what does this mean

[upper karma]

x and y arent equal to a

[flint ingot]

oh ok

if x y and a are points on paper does it mean x and y arent on a

[upper karma]

yes

they have different coordinates

[flint ingot]

k thanks

[limpid ginkgo]

Solving equations, plotting functions etc

[upper karma]

can't u do the first part yourself

[faint pasture]

that goes for everything

[formal geyser]

Hello

I need to find the length of the tangent line

I have no idea what to begin with

[dusky locust]

consider the radii of the circles that touch the line

[formal geyser]

13

[dusky locust]

do you know why the answer's 13

[formal geyser]

It is the sum of radius 5 and radius 8

But i don't know why the tangent line also equals 13

[sturdy stone]

Cyclic quads, alternate segment theorem, isosceles triangles, inscribed angle theorem, angle sum of triangle, similar triangles, area formulae, parallel lines, trigonometry. Thinking of it as "learning theorems to solve problems" is wrong though, there aren't really that many theorems and it's mostly about recognising stuff and being creative rather than having memorized every theorem.

If you read the first chapter of the book Euclidean Geometry in Mathematical Olympiads then you probably know enough to solve 90%+ of BMO1 geo

[flint ingot]

thanks so much

https://ukmt.org.uk/textbooks/2-a-mathematical-olympiad-primer-2nd-edition

do u think this book covers everything u need for bmo?

[sturdy stone]

yeah it's written to cover it

https://archive.ukmt.org.uk/docs/BMO Preparation Sheet.pdf also try reading this

[flint ingot]

k nice

[formal geyser]

Can anyone help me with it pls?

[upbeat pond]

Hey can anyone suggest a video from which I can study trigonometry (grade 11)

These r the formulas I am studying this year

[formal geyser]

This looks scary

I mean, the amount of formulas needed to be remembered

[sturdy stone]

most of them follow from cosine and sine angle addition formulae

there is honestly no reason to memorize all of those

[upbeat pond]

Where I live these should be on ur finger tips otherwise you will lose all the marks in trigonometry

[flint ingot]

if the square of one lenght of a triangle equal to the sum of the squares of the othwr two is the triangle always right angled?

[upbeat pond]

the triangle is a right triangle with the angle opposite the first sides as right angle

[sturdy stone]

what country?

[upbeat pond]

India

[sturdy stone]

sounds plausible

[upbeat pond]

It is

[flint ingot]

k

What quadrilaterals have diagonals of same length?

[undone quest]

Rectangle and square

[flint ingot]

not even trapezium?

k thanksss

Are u sure only square and rectangle?

[sturdy stone]

Isosceles trapeziums too

https://en.wikipedia.org/wiki/Equidiagonal_quadrilateral

[flint ingot]

do the diagonals intersect at 90 degrees?

[undone quest]

Yup

100 percent sure

Just prove the triangles congruent and check it

For any 4s polygon

Bro wtf

Only square and rectangle

Bro is iso trapezium has diagnols equal how u ginna prove that respective triangles are congruent

In square yes but not in recti

I cna give u chart

Of every property of quadrilaterals

[flint ingot]

100percent sure if it has equal diagonals the diagonals intersect at 90 degrees?

[undone quest]

Bro

I told only sqr

Not a rectangle

U need this frnd

[sturdy stone]

isosceles trapezium also has equal diagonals

[undone quest]

Prove it then

Using congruency

Isosceles have equal diagnols but not normal trapezium

I would rectify my mistake

[sturdy stone]

yes

[undone quest]

Yes 🤓

In this scenario the triangles r congruent as i see

[sturdy stone]

Let ABCD be the isosceles trapezium and let E be the intersection of the diagonals. All isoscles trapeziums are cyclic, so ABE is congruent to DEC. It is similarly easy to show that ADE is isosceles and the result follows.

[undone quest]

Yes yes

Hehe

Hehe

[loud hatch]

Hello everyone

I have a question

Consider a rectangle ABCD with sides AB = 4 and BC = 3. A point P is randomly chosen within this rectangle. What is the probability that triangles APB and BPD, if they exist, have an area less than or equal to 2 at the same time?

[flint ingot]

1/9

thanks @sturdy stone and @undone quest for da help

[sturdy stone]

yw

[loud hatch]

Almost there. Here it says the correct answer is 2/9. I don't get it

Maybe the template is wrong

How did you do it?

[flint ingot]

oh i read the question wrong

wait no I didn’t

[flint ingot]

heres how i did it

if P is 1 or less from AB then APB has area less than or equal to 2

if P is 4/3 or less from AD then APD then APD has area less than or equal to 2

the area of the points where P can be is 4/3*1 which is 1/9 of the area of ABCD

[loud hatch]

you read the question wrong actually. the triangles are APB and BPD

[flint ingot]

Oh lol

[vernal nexus]

brother

use pythogoras

to find y

1.6 square + 10.2 square

is y square

then with the value found for y

do 1.6 square + x square is = 1.6 square

to find x

then do 0.5 times 1.6 times 10.2

plus 0.5 times 1.6 times 3.4

add both for area

times 22$

Y IS 10.342

x is 3

[wide slate]

sorry im not good at math, is this a or b?

[vernal nexus]

i gave ans

for a

lenght of x and y

[wide slate]

how to do b?

[vernal nexus]

sent above

just use pyhogoras

a^2 + b^2 = c^2

and 1/2 x base x height

[wide slate]

is this b?

[vernal nexus]

ye

[wide slate]

go it thanks @vernal nexus

[vernal nexus]

np

[wide slate]

is 6 - 205.37 and 7 is 3.57m

[vernal nexus]

cant see it

[harsh crow]

no one can read that bruh

[trail tendon]

How'd u get ur answers

[sterile patrol]

can someone tell me if i get full credit for dis proof

cuz like answer key is showing diffrent method but idk if im right

[tall coral]

I think it would work

[grim lichen]

yo quick question im confused on the formula for period is it 2pie/b for sin and cosine and pie/2 for tan grapths

[plain scarab]

Would this be RP?

Nvm

I think it’s MQ

[dusky locust]

yoo mq

[plain scarab]

Is that right?

[sour cosmos]

it's RP

[plain scarab]

Is this RQ

[sour cosmos]

PS

[plain scarab]

I get it

Do they have to be along the same line?

[frosty swift]

Can anybody tell me what is meant by that sign

[frosty swift]

Ya it means parallel line to that

[sour cosmos]

because a line extends infinitely in both directions then if the points are on the same line they will extend infinitely in the same direction

[plain scarab]

Ah ok

[sour cosmos]

bad explanation but whatever

[plain scarab]

Thank you for your help

[sour cosmos]

yes

[sour cosmos]

the arrow above means its a line in this case

[frosty swift]

Ohh thanks bro

[sour cosmos]

np

[gentle haven]

Can someone please explain what exactly this theorem means?

I don't get what it means

[sturdy stone]

It's not possible to read all of it

[lime dune]

draw a diagram

this is the most basic cyclic quadrilateral criterion

[gentle haven]

"If a line segment joining two points, subtends eqial angles at two other points lying on the same side of the line then these, the four points lie on a circle (i.e. they are concyclic)."

[sturdy stone]

this is such a bad formulation

If ABCD are four points such that <BAC=<BDC then ABCD is cyclic

[gentle haven]

ohk that makes a lot more sense

[sturdy stone]

(given that B,C lie on the same side of AD)

[gentle haven]

ic

ty

[sturdy stone]

np

[silent knoll]

Hey guys

anyone online to help with stuff on proofs

?

[lime dune]

!da2a

[lime crown]

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

[upper karma]

for some reason i'm enjoying geo a lot

i mean i'm rushing through the course that i'm doing

but i'm enjoying it

[upper karma]

gotta get transcript as soon as possible

then i'll learn the material in depth

i'm basically just skimming material then i'm finna learn everything

(specially proofs)

cause they're too many

[lime dune]

if you want to work through more proofs

may i suggest combinatorics 😃

geometry class "proofs" usually are of the "let's prove this blatantly obvious conclusion" flavor, which isn't always the case in general

[forest anvil]

I must be missing something trivial, but how should I get this formula?

[faint pasture]

Directipn of x* is the same as that of x

[forest anvil]

Yeah

[faint pasture]

And its magnitude = a²/|x|

[forest anvil]

But why

[faint pasture]

From |x||x*| = a²

[forest anvil]

Yeah that’s what I don’t understand lol

[forest anvil]

Why this is true

[faint pasture]

|x|=|x*|= radius

[forest anvil]

Wha

I don’t think that’s true

[sturdy stone]

this is clearly false

the text explicitly says |x|<radius

[sturdy stone]

i'm pretty sure it's a definition

as in the reflected point is defined to fulfill those conditions

but i've never heard of this terminology before so idk

[forest anvil]

I don’t think so? The “reflected” just mean it’s the same distance to the sphere as the original point

Oh sorry

I forgot to say

[faint pasture]

Oh shit

[forest anvil]

It is a sphere

[sturdy stone]

"the reflected point is defined by two properties. [...] Its distance from the origin is determined by the formula |x||x*|=a^2."

[faint pasture]

Thats just the definition

[sturdy stone]

I read it as the two properties defining this point are

1. it is collinear with the origin and x
2. it fulfills that formula

[forest anvil]

Oh, shit

I’m stupid😭

Thanks folks

I keep thinking this is some geometric property that I forgot

[plain scarab]

Is this 72?

[obsidian harness]

,w 2x+5+x+1 = 90

[somber coyote]

Results provided by WolframAlpha

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

Requested by sooouth.

[obsidian harness]

,calc 2 * 28 + 5

[somber coyote]

Result:

61

[obsidian harness]

no

how did you get 72 btw?

[plain scarab]

so first i did 2x + 5 and x + 1

got 3x + 6

[obsidian harness]

yep

[plain scarab]

and this is complementary?

i think

i forget

[obsidian harness]

yeah so they add to 90, correct

[plain scarab]

3x + 6 = 90

then i get 28

and that was wrong so i just took a guess lol

[obsidian harness]

oh ah I see

so you've found that x = 28

[plain scarab]

Yea

[obsidian harness]

but then angle ABC is 2x + 5

[plain scarab]

oh so plug in

28 for x

61

[obsidian harness]

yes exactly

[plain scarab]

ah ok

i see

oh so would this be 60

solve for x and I get 31

Then i plug it back in

to 2x - 2

and get 60

[obsidian harness]

yep

there's actually an easier way, because 4x - 4 = 2(2x - 2)

so you have 2(2x - 2) + 2x - 2 = 180

3(2x - 2) = 180
so 2x - 2 = 60 directly

[plain scarab]

Also could someone explain this?

[obsidian harness]

so you know that LON is a right angle

because the diagram says thaat LOK is a right angle

so LOK + LON = 180, 90 + LON = 180

that's why you can do 90 - 60 = 30

[plain scarab]

How do I even do this?

[trail tendon]

you don't. give up.

nah jk they're the same angle xD

because the lines are parallel you know that those angles are the same, so you can set them equal and solve for x

[viral sun]

The answer is

||5||

Isn’t it?

[plain scarab]

uhh what does parallel mean again

90 or 180 degrees

i forgor

[trail tendon]

parallel lines will never intersect

they go in the same direction/have the same slope

[plain scarab]

Ah ok

I’m dumb

I just do 21x + 5 = 23x -5

Bruh

[plain scarab]

Yeah that’s what i got

@trail tendon is this parallel as well?

[trail tendon]

yes

like the lines with the arrows

the red arrows

are parallel

thats what the red arrows mean

[plain scarab]

Alr

[viral sun]

Yeah the two horizontal lines are parallel

[plain scarab]

Do I just set them equal to each other again?

[trail tendon]

yes

[plain scarab]

Okay

Alr i got 7

[trail tendon]

yes

[plain scarab]

guessing i just do the same here

[steel mountain]

yes the two angles are equal to each other. the answer is 8

[plain scarab]

Wow this is a piece of cake now

[plain scarab]

What’s one red arrow mean?

Perpendicular?

[trail tendon]

it still means parallel

[trail tendon]

parallel lines are where the lines will never cross each other, perpendicular is when they intersect at a 90 degree angle

the + sign is created by two lines perpendicular from each other

the = sign is two lines parallel to each other

[plain scarab]

Ohh

[trail tendon]

just as an example

[plain scarab]

should I just do 5x + 10 = 5x

but then I would just get 10 right

[trail tendon]

not quite

[plain scarab]

oh

[trail tendon]

you would get... nothing

[plain scarab]

Yikes

😭

[trail tendon]

💀

um

but

the angles add up to 180 degrees, right?

[plain scarab]

yea

10x + 10 = 180?

[trail tendon]

yep

[plain scarab]

idk why i was thinking 10 x 17 was 1700 for a sec

[trail tendon]

😂

sometimes if you multiply 2 two digit numbers you get a 4 digit number so ig i can see :l

[plain scarab]

Yikes

im thinking this is 6

[trail tendon]

they are actually the same angle

i'm not good at explaining why and i can't draw :l

but do you see how the ? angle is acute (less than 90 degrees) so it couldn't be 96 or 106?

[plain scarab]

yeah i can see that

also how do i tell if a line is parallel again?

Nvm

I can scroll up

[plain scarab]

How do I even do these?

ik we do some type of setting up

[trail tendon]

for the first one, draw/envision a line segment IA that makes a triangle IJA

[plain scarab]

with the last 2

[trail tendon]

wdym

[plain scarab]

like set up a type of equation or something

Idk

[trail tendon]

...yes

[plain scarab]

car

ok what the heck

that just auto wrote car

[plain scarab]

How would I get to doing that?

[trail tendon]

for which one?

[plain scarab]

Any one

cause I could use that to help me with the other

Both problems seem relatively similar

[trail tendon]

dude its so tiny lol

[plain scarab]

Mb

[trail tendon]

alr for that one

two angles make up the big angle, right?

[plain scarab]

Yeah

[trail tendon]

like if you add up EGF and HGE

you get HGF

ri

[plain scarab]

yea

[trail tendon]

EGF + HGE = HGF

hehe

can you make an equation with that

in terms of x

[plain scarab]

2x + 8 = 120

Dude I was doing 22x + 10 = 120

😭

alr I got 46

56

^

[trail tendon]

hold up

HGE + EGF = HGF
HGE = 2x + 8
EGF = 120
HGF = 20x + 2

no?

[plain scarab]

Yes

[trail tendon]

so HGE + EGF = HGF -> 2x+8 + 120 = 20x + 2

right?

[plain scarab]

yeah

I got 7

[trail tendon]

yeah

[plain scarab]

For 6

I got 6

12x - 4 + 42 = 17x + 8

[plain scarab]

yep

[trail tendon]

yeah

[trail tendon]

actually that doesnt apply for the first one i think, i read it wrong the first time

tiny 😭

[plain scarab]

Sorry

[trail tendon]

oh

do you see two angles that add up to big angle?

[trail tendon]

no prob man i just have tiny screen XD

[plain scarab]

K and A?

Wait

[plain scarab]

yes I do

[trail tendon]

XD

which ones XD

[plain scarab]

The line with a and i

shhhh

[trail tendon]

💀

the angles are IJA, AJK, and IJK

[plain scarab]

dang i was pretty far off

[trail tendon]

like if you follow from A to J to K, the angle is at J on that side

[plain scarab]

ah ok

[trail tendon]

i wish i could draw 😔

anyway

angle AJK and angle IJA make up angle IJK

so AJK + IJA = IJK

AJK = 56, IJK = 140, IJA is what ur tryna find

[plain scarab]

alr

84?

[trail tendon]

yeh

[plain scarab]

Thank you for your help

And explanations

[plain scarab]

could someone go over my proof

[harsh crow]

yeah its good

[plain scarab]

Thanks

[topaz glacier]

Can somone help me my assessment

[errant remnant]

Where would the shift button be on this calculator

my old one got lost so I don't have my casio and I have an exam tommorow 😭

[lime dune]

“2nd”

[errant remnant]

2nd F?

[lofty spear]

probably

[errant remnant]

oh it actually was

ty

[vernal nexus]

https://tenor.com/3YP4.gif

[upper karma]

i have not used a calcultaor in the last 5 years

[faint pasture]

ahem our ed system doesnt involve the tan(42 degrees)

[lime dune]

AP calculus more like AP calculator use amirite 😁

[upper karma]

it does

[fair arrow]

Can someone help with a geometry question?

[subtle cape]

Which type of questions ⁉️

[long raven]

I can't cut it out so I have no other way (that I know of) to solve this

[faint pasture]

No?

[drowsy finch]

How can i prove that the triangle FXE is an isosceles triangle

[upper karma]

Yes

Private publishers book

[faint pasture]

They expect you to use a calculator then

Tho 42 is not a good example

As i had someone~~ calculate~~ approximate it here yesterday xS but yea

[subtle crest]

Geometrically prove that sin(180+x) and sin(90+x) are -sinx and cosx respectively

Guys pls help me Ik how to do it algebraic way but idk geometrically

[faint pasture]

Unit circle

[subtle crest]

Wdym

Wait, I'll show you my progress

[faint pasture]

Ydk the unit circle definition?

[subtle crest]

pls head to #help-22

[upper karma]

hmm, ill send example

[covert cape]

what should i do if i legit understand like half of the concepts of geometry and i have a regents in two days

[upper karma]

must be 84 since you know that the overall angle is 140 degrees

so you know how IJA+AJK is equal to the over all angle which would be IJK

so you can do the opposite

if you know ajk is= to 56 an ijk is equal to 140

you can use IJA=IJK-AJK

which is the subtraction property of equality i think?

then you can use substitution with IJK-AJK

basically 140-56

which gives u 84

(sorry if i did it wrong ppl i'm barely starting geo but those r my thoughts)

[upper karma]

ngl i thought that was a transverso i would have said 5 is congruent to 8 because they look like alternate exterior angles

again sorry if it is wrong

but yes i think transitive property can work since you already proved the other angles to be congruent