geometry-and-trigonometry

[candid peak]

my teacher just screams at us because he gives us definitions instead of teaching us how to use formulas

[heavy oar]

My teacher literally has to look up formulas on her computer 💀

[slow moon]

💀

you guys have bad teachers

jeez

[heavy oar]

I mean, she is in her 80s

[slow moon]

!?

she needs to retire asap

[heavy oar]

Even my grandma didn't even live that long 😭

She said she won't get properly paid unless she works certain amount of years and she's gonna meet the "requirement" after this year

[candid peak]

my teacher is a 42 year old man who wears exoskeleton shoes and hes gay he was great first now i have a 15 he refuses to turn in my assignments theres petitions to get him fired ive walked out of his class

[candid peak]

oof

[knotty magnet]

by defintion

[crude gull]

Oooke so im doing fairly well in my math class, but theres one thing that feels super obvious but i dont quite get it

What if theres a triangle behind a right triangle?

[gentle haven]

How many combinations of nothing are there

[gentle haven]

Wdym

[crude gull]

Sorta like the example, do i treat the other one like another right triangle and then solve that with the pythagorean theorum or something?

[upper karma]

What you stuck on

@crude gull

[crude gull]

Im not super sure about finding the area when you have two triangles connected i guess?

[upper karma]

First image

I'll help you

So you know the formula of area?

[crude gull]

Yeah width times height

[upper karma]

...

Well

[crude gull]

Or divided by two for a triangle?

[upper karma]

1/2 * 6 * 6 * 5

[upper karma]

Only works if the triangle is 90 degrees yk I don't know how it's called in english

[upper karma]

S = 1/2 a * b * h

Everything is understandable?

[crude gull]

Yeah ill have try this for a couple problems and see if it makes sense

[upper karma]

Now second image

Pythagoras

Theorem

15^2 + 15^2 = x^2
x = 15sqrt(2)

Sqrt is root btw

Idk if you learn this

Idk what grade you in

Everything understood? This is like 5th grade right? @crude gull

[crude gull]

Yeah it could be, i forgot it though

[upper karma]

no.

that is not precise.

$c=\sqrt{a^2+b^2}$

[somber coyote]

Mr. Macro

[upper karma]

this gives the precise answer

it depends on how much digits your calculator gives you

also it does not solve area

for this problem we use

[somber coyote]

Mr. Macro

[upper karma]

or to make it easier

[opal veldt]

i need help 😭

really really bad

i’m probably over complicating this so hard

but i have no clue how to solve for both variables

[maiden brook]

how to do?

does this count as geo or competition math?

[raven prism]

96 is supplementary to the third angle in the triangle

They are on a line

And 46 and x are alternate interior angles since there are parallel lines

[upper karma]

remember that opp angles are equal and the sum of the angles in a triangle should add to 180 degrees

x=46

as a hint

[upper karma]

Find the locus of midpoints of all chords of the fixed circle O such that these chords subtend right angle at a fixed point A which lies inside the circle and is not the center.

[fast pasture]

Could someone please help me with this?

no. 10 - 17

[upper karma]

10. 180-75=105°

[upper karma]

11. 75°

12. °105°

13.180°

14. 360°-75°-118°-45°=122°

15. 122°

16. 118°+45° = 163°

17. 75°+ 122°= 197°

@fast pasture

[fast pasture]

thanks for your help bro aprreciate it

[upper karma]

how not

[upper karma]

np

[upper karma]

i gave precise answer? 15sqrt(2)

i didint approximate please discuss this if im wrong i want to learn

[upper karma]

also the second image didint ask for area did it?

ofc your helper you know better but i dont see the problem in my work

[fast pasture]

btw, would u mind explaining why i need to subtract?

[upper karma]

well whole circle is 360°

[upper karma]

idk if it makes sense to you?

[fast pasture]

oh, well i thought angle COF would measure 118° cuz it's a verticle angle

[upper karma]

Prove that the distance between the midpoints of opposite sides of a cyclic quadrilateral whose diagonals intersect at right angles(and at a fixed point) is fixed

[frosty rose]

can anyone send me a pic of the range of the arc if the arc E ] π/2 ; π [

[neon folio]

Drink water

Stay Hydrated

[frosty rose]

eat food

stay full

[granite hollow]

i have no clue how to make ABC and DEF similar

[upper karma]

ok so depending on how much digits your calculator gives you rounding it to a certain decimal is not precise or inaccurate, it also depends on if you have the ans function but still could go wrong due to previous calculations on the calculator.

hence why we use 👇

[somber coyote]

Mr. Macro

Mr. Macro

[upper karma]

and for this question

we do not have the base of the triangle to be eligible to find the area

so we use Pythagoras theorem and rearrange it to find one of the smaller side

like so 👇

[somber coyote]

Mr. Macro

[upper karma]

so to find the area of that triangle we construct our own formula

[somber coyote]

Mr. Macro

[upper karma]

and we substitute into the formula 🎉

[somber coyote]

Mr. Macro

[upper karma]

Andddd we have our answer

[somber coyote]

Mr. Macro

[upper karma]

wooooooo 🎉

[somber coyote]

Mr. Macro

[upper karma]

use the rule of were opposite angles are equal on the parallel

also the bad printer makes it hard to see

[upper karma]

a yeh right some people might have issue pretty common

[scarlet siren]

yo can someone help me with my math homework its pretty basic but im just dumb

nvm i got it

[foggy violet]

I have this statement to prove

(ABC is the triangle, and a, b and c are the opposite sides of the angles A, B and C respectively)

Can I say that this expression is the same as the LHS, WLOG?

I think that way I can add the two up, show that their some is 0, and since they are equal, each of them are 0. But I'm not sure if this is a valid method.

[upper karma]

try writing cos square as 1 - sin

then multiply a^2 and all

then see what terms cancel out

[upper karma]

yes

[upper karma]

have you tried expading

[somber coyote]

Mr. Macro

[upper karma]

then simplify

and rearrange

[foggy violet]

I did solve it, like this, but it took a lot of steps, and I felt like the above was more elegant.

[upper karma]

Hello everyone, one question, the side faces are perpendicular to the bases in a regular prism, right?

[red flax]

does anyone understand how too solve this?

I'm having trouble but I dont know how too find the interior angles

this ones a little more puzzling for me

[sterile widget]

Assume that angle BAC is x

We know that all angles of a triangle add up to 180 degrees

X + (4m +3) + (3m -3) = 180 degrees

While (6m+22) + x = 180 degrees

We can rewrite the equations as x + (4m + 3) + (3m-3) = (6m+22) + x

Remove x from both sides

(4m +3) + (3m-3) = (6m+22)

7m = 6m +22

subtract 6m from both sides

m = 22

sorry if my explanation is a bit confusing

[red flax]

no I understand it

[sterile widget]

ok

for the second image

[red flax]

thank you 😁

[red flax]

yes for the second one

[sterile widget]

We know that all angles of a triangle add up to 180 degrees

(5x-7) + (x+16) + (8x+3) = 180

14x - 7 + 16 + 3 =180

14x + 12 = 180

14x + 12 - 12 = 180 -12

14x = 168

x = 12

[red flax]

and I just plug those in for the equations right

[sterile widget]

yep

[red flax]

okay thank you 😁

I appreciate you a lot

[sterile widget]

you're welcome

[red flax]

I need some help with this one

😔

[royal willow]

what is m here

[red flax]

idk if they want me too create a equation for x or y

idk just confused rn

[red flax]

m doesnt have anything too it

it might mean measurement

[royal willow]

so it is saying that angle x is 5 times as great as angle y

[red flax]

yes

and I think angle z is 60

I am not confident in the answer I am getting for this question which is why I need help

[royal willow]

what have u done so far

u know angle z is 60 right

[red flax]

yeah

I suppose x is equal too 60 too

[royal willow]

and x + y + z = 180

and we know z = 60

so x + y + 60 = 180 right

[red flax]

yeah

[royal willow]

ok and we also know that x is 5 times as big as y right

[red flax]

yes

[royal willow]

so lets put x = 5y into this

and solve for y

now we have 1 variable so we can solve

5y + y + 60 = 180

so 6y = 120

so y = 20

so far u with me?

[red flax]

yes I am

[royal willow]

then using that information we know x is 5 times as big as y

so if y is 20 then x is 5x20

so x = 100

[red flax]

ohhh

okay now I understand

I was just overthinking it too hard

[royal willow]

all good

[red flax]

thank you for the help I appreciate it

[royal willow]

np

[red flax]

wait so why would you assume angle BAC as X instead of assuming angle A is X? 😔

because I am rereading and looking at it wondering why A isnt assumed as X since BAC = 180

or am I wrong 😭

[sterile widget]

I mean that you could assume that angle a is X

[red flax]

ohhh okay thank you because I am taking these for notes

since I missed class today

[sterile widget]

damn

ok

[red flax]

😁

[sterile widget]

np

[red flax]

but

before you go

how would I find MN?

[sterile widget]

sorry

had to go

bye

i guess

[red flax]

its fine I figured it out

[sterile widget]

ok

[royal willow]

Those are two ways to write the same thing. Angles in the form of ABC correspond to angle B between A and C

[red flax]

ahh okay okay

[solemn lion]

Do I cancel out the 9/4 sec^2theta. Or do I simplify the radicand first?

[thick fable]

no u can't

[ebon pier]

bro see

dude*

see

if you check the radicand you can multiply 4 and then you can take the 9 common and you get sec^2theta -1 which becomes the sqrt(tan^2 theta) hence it becomes

9/4(sec^2 theta/tantheta)

[upper karma]

4d+3 = 6d - 13

[solemn lion]

I did that after some help and realization

Thanks

[twin crag]

how do i solve this

[stark hinge]

Solved it (atleast one answer)

@twin crag

[thick fable]

!nosols

[lime crown]

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

[stark hinge]

So, tip cot a = 1/ tan a

[stark hinge]

Mb

[opal salmon]

Help, I don’t get this

[royal willow]

hint: common factoring

[opal salmon]

How should I continue

[royal willow]

well (x^2 + x - 6) doesn't equal x(x-6)

as for the (2x^2 + 4x) u can factor out an x yes, but u can also factor out a 2

[opal salmon]

What should I do when I am baout to factor x out of x

kinda confused here

[royal willow]

do u know how to factor simple trinomials?

[opal salmon]

yes

[royal willow]

ok so what do u do when u are trying to factor (x^2 + x - 6)

find two numbers that multiply to (-6) and add to 1 right?

[opal salmon]

yes

[royal willow]

so what are the numbers

[opal salmon]

-2,3

[royal willow]

okay so then it factors to what

[opal salmon]

(x-2)(x+3)

[royal willow]

yes

so now u have what

[royal willow]

continue from this using what u just solved

(x-2)(x+3)(2x^2+4x)

u can still factor out using common factoring

on the (2x^2+4x)

[opal salmon]

I found that the greatest common factor is x

[royal willow]

is it

isn't the greatest common factor 2x?

[opal salmon]

Oh yeah

your right

[royal willow]

lol

so what do u have now

[opal salmon]

(x+2)

[royal willow]

yeah

so put it all together

|| 2x(x-2)(x+3)(x +2) || is what u should get

and that is completely factored

[opal salmon]

Thanks bro

[royal willow]

np

[opal salmon]

rlly helped me there

[royal willow]

u knew how to do it I just guided u

[sick kraken]

Does anyone know if I made mistakes

I'm unsure about Arc FG and Arc AC

[pulsar garden]

when it says "arc AT" it asks for the lenght of the arc AT?

or it asks for the angle to the center?

that faces AT?

[sick kraken]

Length

[pulsar garden]

Oooh so to find the lenght u can just get the angle in the center and then do the conversion with Pi

[pulsar garden]

U can divide the circumference with 360 triangles, each triangle with the angle on the origin of the circumference has 1 degree

Every triangle is congruent to each other because of SAS axiom

So u know that you have an angle of 41 degrees on AOT(O is the center of the circumference)

And the angle insists on AT

So the formula should be (41/360)*2π

Wait are u sure ur looking for the lenght of the arc?

Oh nvm now i saw better what is given

My bad

About FG just do the conversion from degrees to radiants, same thing to AC

And if ur asking if FC is correct it is, you can check that by taking the chord CG and knowing that u get two congruent angles

Not sure about AC

Im taking my time

So i dont have anything in mind other than difference to check on AC, if FG is correct then it is AC

And same thing otherwise

Oh now i saw ur solution to the left, that's not correct

Because it is an intersection point, not a vertex of the circumference

[sick kraken]

Hi im back

[pulsar garden]

Also u can't really say that arc FA is insists of an angle of 148 degrees because FA is secant to the circumference but the other line isn't secant or tangent

So u can't use the theorem ur looking for

[sick kraken]

Hmmm

[pulsar garden]

Where did u take that 2*148?

By which vertex of the circumference?

[sick kraken]

The 2*148=x+(23+x) is a formula for when an angle is inside the circle but not at the center

2a=Arc+arc

[pulsar garden]

Where did u take 24?

[halcyon vector]

i think this question is wrong and if it isnt pls help me with the ans

[sick kraken]

I mean 23

Not 24 mb

[pulsar garden]

Its an intersection point inside the circumference, you can'use that theorem

It has to be a point of the circumference

[pulsar garden]

When a quadrilateral is cyclic?

[sick kraken]

This is the theorem i used

[pulsar garden]

What the, do u have the proof?

[sick kraken]

So i used the 148 degrees and used 23+x cause that's the arc of it

[halcyon vector]

like a cyclic quadrilateral is a quadrilateral whose opposite angles are supplementary

[pulsar garden]

If its a theorem and it works, everything is perfect, never saw that theorem

Its weird that u get the theorems but not the proof

[sick kraken]

Not really but my advanced geometry just gave me the formulas

Lemme show u

[pulsar garden]

I agree about the first two, never done ther two

I should see the proof

Also the last one seems useful to optics

[pulsar garden]

If ACD is hysoscele then its easy lol

But u dont know if it is

[sick kraken]

Same

But if it's true did I do anything wrong?

Because I'm not sure if

[pulsar garden]

But we know it is because of the SAS

[pulsar garden]

If its true then its ok

[sick kraken]

2 x 148 = X + ( 23 + X ) works

Wouldn't work?

[pulsar garden]

Yeah it works if the formula given is correct

[halcyon vector]

umm what is hysoscele

[sick kraken]

Idk if I can consider both the Arcs X tho

That's my only doubt

Cause maybe I could've done

[pulsar garden]

A triangle with the two obliquous side congruent

[sick kraken]

2 x 148 = X + ( 23 + Y ) ?

[pulsar garden]

Wait lemme write

[pulsar garden]

Hold on, it doesnt say if BD is the bisector

It isnt a parallelogram so we can't say that the two diagonals has the intersection point on the medium segments

I have an idea

Wait

ACB is 30 degrees because AB = BC;
180 - 70 - 2*30 = 50

[halcyon vector]

yeah that much is easy

[pulsar garden]

It means that E isnt the medium point

[halcyon vector]

u can even find AEB and BEC but that will also be pointless

[pulsar garden]

Yeah it seems wrong but lemme check

ADC = 60 degrees

[halcyon vector]

yeah got it

[pulsar garden]

Very weird, we have infinite solutions about the angles

Seems impossible but idk, im taking more time

[halcyon vector]

😂😂😂

[pulsar garden]

Because we have that (30 + x) + (30 + y) = 180

[halcyon vector]

i got this in my test for 3 marks and if this is wrong im definitely getting full marks

[pulsar garden]

It means that x + y = 120

[halcyon vector]

yeah i thought of that too but it just leads to mmore pointlessness

[sick kraken]

I'm so bad at circles…

I don't think it's right 😭

[pulsar garden]

It isnt in fact

[sick kraken]

😔

[pulsar garden]

30 + 30 + 140 > 180

[sick kraken]

But are the bottom triangles irght

[halcyon vector]

yeah cause like we dont know if EAD = DCE

[sick kraken]

Wait

Then it isn't a triangle

Is the question itself wrong?

[pulsar garden]

If x = y then that triangle is equilateral but we have to understand more, there are infinite solution, maybe we need more time

[halcyon vector]

THATS WAT IM SAYING

[halcyon vector]

no

[sick kraken]

Yeah now I see not

[halcyon vector]

cause BEC =AED

same AEB = DEC

[pulsar garden]

Actually it seems impossible to me

Infinite solutions...

[halcyon vector]

thank god

[pulsar garden]

But to me

Im a student as you

[halcyon vector]

i got so scared when i was not able to do this

[pulsar garden]

Ur going to fail a lot, no worries

[sick kraken]

I feel like it has 0 solutions as you said 30+30+140>180 meaning a triangle doesn't exist

[pulsar garden]

That's how you learn i guess

[pulsar garden]

Yeah but that's because u taken the diagonal as a bisector when it isnt

[halcyon vector]

yeah but like i was so overconfident that todays paper is gonna be a piece of cake

[pulsar garden]

That angle next to the 70 degrees is 50

[halcyon vector]

and then i wasted atleast 5 to 10 mins on this

[pulsar garden]

Ur in highschool, wait for pure mathematics

[halcyon vector]

🥲

[sick kraken]

Scary

I'm probably only gonna do calculus in uni

Because I wanna be a Vet

I don't think that deepdives in to a lot of Math

[halcyon vector]

Thank you for your time pochineko and holi

[sick kraken]

Yup

[pulsar garden]

Im sure that its wrong

[sick kraken]

So this is correct? @pulsar garden

Using these formulas

[pulsar garden]

And we can prove it by contraddiction

[pulsar garden]

Lemme say this to him and then we check

[sick kraken]

Why would the angle next to 70 be 50 if AB=BC

[pulsar garden]

It isnt bisector

Its just the image

That makes u think that

[sick kraken]

If AB=BC wouldn't 70=70

And not 50

[pulsar garden]

180 - 30 - 30 - 70 = 50

[sick kraken]

In geometry you can't rely on images even tho it's kind of all visual too 😭

[pulsar garden]

The other angle is 50

Not 70

[halcyon vector]

ACB = 30, BEC = 80, BEA = 100, ABE = 50

[sick kraken]

Ohhh

[halcyon vector]

BY amngle sum prop

of triangles

[sick kraken]

Ok wait

[pulsar garden]

So lemme try by proving that it isnt possible

Wee are gonna use contraddiction that it can't be 60 degrees

Nah i forgot

Sry

[halcyon vector]

😂😂

[sick kraken]

How did you guys find 50* 😭

[halcyon vector]

no worries

[pulsar garden]

The sum of every angle of a triangle is 180, there's a theorem

[sick kraken]

Ik

30+80+70=180

[halcyon vector]

AC is a straight line

[sick kraken]

AB=BC

[halcyon vector]

means 180

[pulsar garden]

Do u see that 70 u added

[halcyon vector]

80+80 = 160

[pulsar garden]

Substitute that with an x

[sick kraken]

Oh true

I overlooked it mb

Oh I see now

[halcyon vector]

yesh

[pulsar garden]

And we even know that ADC is 60 degrees

[sick kraken]

How can we even find other angles

[halcyon vector]

how?

[pulsar garden]

Fixed the message

I meant ADC

[halcyon vector]

yeah

got it

[pulsar garden]

Because it is cyclic

Lool we have even more variables

Because we dont know how it divides

So we have x, y, z and w

Because we dont know which angle is ADE and CDE

And the not all the triangles with x angles are congruent with each other

[halcyon vector]

yeah and we cant make another eq to solve for z and w

[pulsar garden]

Alr i know how to prove that it is possible

[sick kraken]

I think I solved it but idk

[pulsar garden]

Dont let me forget

[halcyon vector]

YAYYY

[pulsar garden]

Since it is cyclic it means that x = y

[halcyon vector]

ADC IS 60

[pulsar garden]

Yep

[halcyon vector]

50 +40 is 90

[pulsar garden]

But there is a problem

[halcyon vector]

which angle are u taking x and which y

[sick kraken]

We’d need to solve everything via equations right

[pulsar garden]

Nah it isnt

[pulsar garden]

Lemme get to my computer

[sick kraken]

We can't just interchange numbers visually using the angle triangle theorem

[pulsar garden]

it means that x is 60 and y is 60 degrees

ADC is equilateral

[halcyon vector]

yeah got it

[pulsar garden]

we were only thinking a way about congruence theorems

so we were thinking that it was impossible

and thinking that a problem is impossible makes us impossible to solve it lol

[sick kraken]

Guys

[halcyon vector]

bruh u dont need to correct ur grammer we understand u😂😂

[sick kraken]

Sorry to bother but would the arc that idk be X or Y

I wrote an equation for it

[halcyon vector]

sorry i am not at that level yet😭😭😭

[halcyon vector]

i dont understand anything in this fig

[sick kraken]

Its ok dw I'm just using formulas to solve I'm also bad at circles

[pulsar garden]

we trusted too much the draw

[pulsar garden]

using the formulas works, the problem is that i hate to use something that isn't mine

that's because i always look for proofs before using them

[halcyon vector]

wat do we have to find in this?

[pulsar garden]

the formula works, and also u can use the difference of angle to see if even AC is correct, so it works

[pulsar garden]

dont use variables name that are points

u might get confused

[halcyon vector]

oki guys i have to go bye

[sick kraken]

I did

[pulsar garden]

cya

[sick kraken]

2x=80+(Y+76)

There's no proof that the Angle(X) = The Missing Arc (Y)

Also if I use x a the only variable it gives me 156

And as you can see there's a diameter so that whole arc should equal 180 and 158+76>180

Using Y gave me 78 as a value

[sick kraken]

Ok

[pulsar garden]

akr

alr

the first angle u find it easily

well u already wrote everything

ur just looking for x?

y = 80 + x;
x = (360 - 2y)/2

x = (360 - 160 - 2x)/2 => x = (200 - 2x)/2
2x = 200 - 2x => 4x = 200 => x = 50

@sick kraken

damn, this is a 26, not an x

sorry

x = (360 - (2*106))/2

x = 74

[sick kraken]

Sorry I had to go do something

Look what demos gives me hmmm

[pulsar garden]

u can use geogebra for euclidean geometry

[sick kraken]

What's geogebra

[pulsar garden]

a tool

[sick kraken]

Ohh ok

[versed mulch]

Im learning geometry😓

[wide oyster]

I need a hexagon with a width of 18in what would the side lengths be?

[misty tusk]

so you've got your hexagon right

and its composed of 6 equilateral triangles and you have a width of 18 inches

so if we split that into 2 and break out one equilateral triangle we get this

here

now we know that since these are all equilateral triangles (and we're splitting this one in half) we can use similar triangles because this will be a 30-60-90 right triangle

so we're able to see that these two triangles are similar by angles and now we can find the side length using ratios

so since our side length is b we should be able to solve for it

$\frac{9}{\sqrt{3}} = \frac{b}{2}$

[somber coyote]

architecture2

[misty tusk]

and that's how you'd find the side length

[wide oyster]

thanks

[wicked arrow]

if A,B two physicial or geometrical objetcs defined as a set of points,atoms ,proove that :d(A,B)=inf({T(x,y) | for all (x,y) belongs to A ,B respectively }) and T(x,y) is the set of standarts of trajectories beetween x and y .
the space is R^3
and the distance is usuale

[heavy crow]

,w simplify sin^3(x)

[somber coyote]

Results provided by WolframAlpha

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Requested by taylornix

[heavy crow]

I can do this with eulers formula but what's the intended way to do it

like... power reducing + product to sum?
I know it's quick with the triple angle formula but who knows that off the top of their head?
(this came up in an ODEs class, solving y''+4y=sin^3(x))

idek anyone who remembers the product to sum formula

[upper karma]

④ A circle with a diameter of 24 cm is cut in half. A cone-shaped cap is made from one part. Calculate the height of this hat.

[royal lotus]

Sum-of-angles formula. The power reduction will bump down the coefficient of the angles by powers of 2, so 3x in the argument won't help, yet.
I never remember product-to-sum / sum-to-product. I don't find them worthwhile, yet.

[royal lotus]

What measurements from the semi-circle are preserved, and which change (and into what)?
Where does the angle of the chunk cut out / chunk remaining go?

[upper karma]

idk

[upper karma]

doesnt say, maybe just thrown out

[subtle girder]

Are all u guys mathematicians?

[royal lotus]

Picture the cone you make (or literally make one). There is a triangle you can create between the vertex, drawing an altitude to your table, and going out to the edge.
So if you have an almost-circle (like 359-degrees), the angle of the "triangle" made with the vertical under the cone is almost 90-degrees (it's almost flat), and at only 1-degree, it's super tall, and the top angle is practically zero.
We can conclude the cut of the central angle is probably going to have a linear relationship with the angle of the triangle. We can check to make sure, as we go through.

@subtle girder I teach math at a 2-year college

[subtle girder]

Wow

[royal lotus]

The arclength of the remaining parts of your circle become what?

And where does the radius of the original circle go?

[outer marsh]

so are you turning a semi-circle into a cone?

[royal lotus]

Yes

I would suggest actually making a cap to explore the situation.

Label the parts, and see what they become.

[floral cairn]

Are angles JLK and LJM congruent alternate interior angles?

[royal lotus]

Since JK & LM are parallel, yes, LJM & JLK are alternate interior angles, and congruent.

[floral cairn]

Ty bro

[river abyss]

geometric proofs are making me crazy

im acrually about to fail geometry because of it

[thick fable]

[royal lotus]

@river abyss How comfortable are you with some of the more basic proofs?

[river abyss]

I don’t understand any of it

[royal lotus]

and/or have you worked in or read any of Euclid's Elements?

Most of the time, you look for "what can I guarantee to be true" at every step in the proof. Look at your options, and then choose something that gets you a step closer to your target statement.

[floral cairn]

Review the thereoms,postulates, and properties

[river abyss]

Okay 👍

[floral cairn]

If u have any questions and what the properties mean you can ask me

[knotty magnet]

lol u here

[smoky jetty]

u olympiad nerds lol

[knotty magnet]

xd

[river abyss]

Do you use proofs in any other math y besides geometry

[thick fable]

yea

[river abyss]

Bro

[thick fable]

if im not wrong, i think its a whole branch of math

could be wrong tho

[gritty mural]

delta epsilon proof my beloved

[river abyss]

This is making me sad

[gritty mural]

its ok delta epsilon proofs are fun

u do them in calc

theyre way more satisfying

geometry proofs are 🤮

[floral cairn]

Geo proofs aren’t that hard

What do you need help with

[floral cairn]

Geo proofs are satisfying

[gritty mural]

...

who is this guy

[floral cairn]

Highschool Freshie

[heavy crow]

the majority of math is proofs after a certain point.
you generally forget about proofs until lower div linear algebra, and then upper division maths is mostly proofs.
grad math is almost all proofs. many classes have absolutely no computation at all. it's just all theory.
upper levle math proofs a lot more interesting than geometry proofs though.

[floral cairn]

I’m gonna. Take linear algebra in my senior year

[heavy crow]

linear algebra is a lot more interesting than geometry in my very biased opinion

[upper karma]

Well geometry is fun because it has a lot of intuition unlike linear algebra, where things get abstract

[heavy crow]

linear algebra is far more intuitive to me than geometry lol

[upper karma]

Are you sure about that?

[heavy crow]

yeah, actually.

[upper karma]

You think vector spaces are more intuitive than for example Thales theorem

[heavy crow]

oh 100%

[upper karma]

Hmm

[heavy crow]

people have different inclinations in math. there is no one subject that's ultimately more intuitive or easier than another.

[upper karma]

Not ultimately, but you can see a big difference between geometry and linear algebra, especially the proofs

[heavy crow]

vector spaces are intuitive for me because it's just generalizing the concept of adding and scaling. and the whole point of linear algebra imo is showing that scary abstract vector spaces are really just adding and scaling vectors in F^n under the hood.
linear algebra proofs are way more intuitive for me than geometry/trig proofs. that's why i just use euler's formula for all my trig proofs

[upper karma]

But that’s not intuitive at all it you‘re defining sine and cosine with complex numbers instead of understand it’s geometrical meaning

It’s very good for proofs

[heavy crow]

eulers formula has a very strong geometric meaning

and the analysis definitions of sine and cosine involve plenty of geometry

[upper karma]

True but I still think it’s better to begin easy and then go abstract

[heavy crow]

indeed. but once i did that, then i found that starting abstract made everything easier

my point is not that my way is better. it's that your way isn't. people will find different approaches/subjects easier for them.

[upper karma]

Yes abstract makes a lot of things easier but one can’t ignore the history before, because absactness was build from that

[upper karma]

Sure

Everyone has different approaches

Maybe my university is just bad, idk, it’s just not fun to get thrown by abstract definitions and theorems without showing some good applications or intuition

[heavy crow]

most people don't like abstraction of the sake of abstraction, you aren't alone. that's why i'm getting a phd in math and most people don't. but intuition is really important, yes. that said, i generally want the more abstract version of something before i care about using it to solve problems.

[upper karma]

I see

[royal lotus]

Not necessarily, but your teachers may be less interested in teaching than their own research.
This is fairly common.
I teach at a two-year college, and we are expected to become better teachers, trying out new approaches every semester or every couple of years, depending on how it's going with the other one(s).

Personally, I prefer learning via abstraction, like ET, but prefer teaching through historical context and highlighting motivation via the problems that existed, then focus on the solutions with the math I'm teaching that day.

[graceful torrent]

For -cos^2(x)can we use identity to make it into -(1-sin^2(x))

[silent plank]

yes

[misty tusk]

my personal ranking of proof difficulty (easiest to hardest) is probably geometry, epsilon-delta, trigonometry, and linear algebra

they all have a lot of similarities but its NOT worth it at all to call them the same

[runic mica]

who here is taking the sat on dec 2

[fathom beacon]

hello, how do i find the area of this trapezium

[round lagoon]

Find the hyp then find angle BDC , hence find angle BDA and BAD then solve the second triangle with sine rule.

[tiny citrus]

Hello, could somebody pleaase help me find the value of Xg and Yg? I have no idea how to find them

G is at the center point of the rectangle

One last thing, the height of the rectangle is given, even if i didnt write it

[proper otter]

for some reason i pretty dislike geometry, especially 3d geometry

with calculus theres actually stuff going on and going through step by step solving stuff is engaging and fun

on the other hand staring at a bagua for hours isnt as fun, this is somewhat more tolerable with 2d geometry, but for 3d i just die

[worthy latch]

can somebody help me simplify this equation using double angle identities:

\frac{sin2\theta \left(6tan\theta \right)}{\left(2-2tan^2\theta \right)\left(2cos^2\theta -1\right)}

if the bot doesnt automatically just simplify the problem so it is easier to see, enter this in symbolab and it will give you the equation

i used chatgpt and all the other math solvers online but couldnt get the same answer as the one i had on my answer key, which was (3/2)(tan^2(2(theta)))

[grave pond]

You need to surround the TeX with dollar signs to get the bot to render it.

[thick fable]

,, \frac{sin2\theta \left(6tan\theta \right)}{\left(2-2tan^2\theta \right)\left(2cos^2\theta -1\right)}

[somber coyote]

यजतलमाओ

[thick fable]

!status

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[thick fable]

@worthy latch

[worthy latch]

no 2

[thick fable]

show ur work

[frozen ocean]

do you know how does double angle work in general?

[worthy latch]

yes

i got the answer nvm

[grave pond]

Please don't crosspost a question to several channels. #math-discussion is probably the better one here.

[smoky jetty]

a more geometry-inclined approach

[upper karma]

$A=\frac{h}{2} \times (a+b)$

[somber coyote]

Mr. Macro

[upper karma]

h is the Hight

a is BC

b is AD

but first you are required to find the missing values using pythag

[hard coral]

Is there a way to find the angle between two vectors without the angle being incorrect? I say this because the way many people say to do, keep getting me the wrong answer. I am trying to find the angle from the x-axis to the red vector which is anchored at the origin but no matter what I do, it says the angle is less than 180 when its actually greater

the angle should be 360-θxy but that isn't possible when θxz is less than 180 which makes it difficult

this is why

[junior flame]

help

[sullen quiver]

Okay so you need to interpret the slope

A slope of -2 means for every x your y decreases by 2

So you find your difference between -9 and 3

And then multiply that by -2 to get the amount your line goes down

And then since you start at 7 you then do 7-amount line goes down

[junior flame]

ohh

[vocal bison]

idk 🤷‍♀️

[pastel kite]

send the complete pic

i cant see the right part

[vocal bison]

oh

i think F is the midpt of DB but 🤷‍♀️

[upper karma]

it wouldnt be since its logical

wait you might be right

ABCDEF only point left is in the middle

forming GF and cf

[upper karma]

Yes the mid point is F

[hard coral]

Can I get some help? I have a vector that starts at the origin and goes to some randomg point. How can I find the angle between this vector and say the X-axis defined as the vector <1,0,0>. Before you Google the answer, Google's answer doesn't work. I need to get the angle that gets the correct angle in all quadrants. For example, arctan cannot tell the difference between quadrant 1 and 3 which other trig functions have similar issues but this is very important that I get the exact angle

Hell I even tried law of cosines

I need this angle because I can then use it in two functions that came up the x and y components of a vector. However when x or y is 0 using the law of cosines method, the angle is entirely wrong when I alter the x or y that isn't 0. Which then throws out the law of cosines method. Is there anything I can do? I feel like I'm at a loss but then again, there is no way this is impossible I know this for sure

[vernal pilot]

Have you learnt about the scalar product? $|a|\cdot|b|\cos(\theta)=a_{1}b_{1}+a_{2}b_{2}$?

[somber coyote]

TheLord26

[vernal pilot]

Is this what you are after?https://www.desmos.com/calculator/amaqkz4gj0

[upper karma]

ABC ia a triangle that has the side lengths of a, b and c.
a, b and c have an equation like this: a^2 + b^2 + c^2 -6a -10b - 14c + 83 = 0
What is tan(ACB)?

[hard coral]

That's essentially what I did and it gave lesser results

I did that and the law of cosines

Law of cosines showed potential though I have two inputs and it all falls apart when one input is zero and I try to move around another input. That is, the law of cosines is good but it is applied to two functions which have different inputs for each. Those two functions make up two components of a vector and when one input is zero, the other one just isn't on the right side of the graph when I move around it's input

[lament tree]

anyone knows where this formula for a conic section comes from?

I suppose I could derive it with patience, but I'm kinda lazy rn

[mint nexus]

can anyone help me do the statement and reasonings for proofs? im extremely confused on how to do it and i failed my test because of it. feel free to dm me if you can help me :)

[obsidian harness]

you can show that this quadratic form is invariant under translation, rotation, reflection, and dilation

it's just some algebra

[lament tree]

what does this have to do with it being a conic section

[obsidian harness]

the main reason why we need the cross term xy is cause of rotation

a circle already has a, b, d, e, and f

so does an ellipse

[obsidian harness]

that you don't need any more terms to represent all conic sections

you're not missing a term somehow

[lament tree]

ok but like

what does this have to do with a cone

and a section

of such cone

knowing the formulas it's easy to know this represents all conic sections I want to know if you can go from cone and plane to this formula

[obsidian harness]

oh right, I assumed you were working from the standard forms of a circle, ellipse, parabola and so on

[lament tree]

no

[obsidian harness]

mb

[lament tree]

well I already know proofs for the ellipse and circle conic sections, which would lead to the formulas, but I'm curious on the others, if the formula pops out or something

I suppose I could just set the formula of a plane equal to the formula of a cone

I mean, the equation

formula is weird in this context

[lament tree]

I can also find a formula for a rotated cone and set z = 0

[lament tree]

I think I got this actually

I made the equation of an inclined cone and after setting y = 0, it's giving me all the conics

[lament tree]

apparently at least

[stuck sleet]

Hello I cannot understand the third property of the pyramid:

How did they derived it???

The ratio said:

base on this figure:

[lament tree]

basically, the area grows linearly as it goes down and down

[stuck sleet]

how this is derived? Thanks!

[lament tree]

I'm not sure how to prove that without unecesseraly fancy stuff

[lament tree]

actually, it grows quadratically

just realised I read the thing wrong

the perimeter of the section grows linearly and as a consequence the area grows quadratically with the altitude

[lament tree]

but basically, if you focus on each side of the slice, you can spot pairs of similar triangles, and you can also construct other similar triangles by drawing both altitudes from the tip to the bottom

[stuck sleet]

yeah

[lament tree]

that's the idea that makes the segment grows linearly, and as a consequence, the area quadratically

(just a property of areas)

omg I gained the ative role 🥹

[stuck sleet]

does this property have a proper name so I can search how is the derivation?

I was learning solid geometry

[lament tree]

uhh idk, and the only common derivation I've seen of it is kinda handwavey, it's a not very rigorous limiting process, there's probably something better tho

the idea I've seen is that if you divide the area in a bunch of very small squares to aproximate it well, if the ration between two similar figures is k, than after multiplying "the figure" by k, in all directions, the little squares will have multiplied by k both the height and the base, so the new area of each square is the previous one * k * k or * k²
I think that's a good intuition tho

oh actually, I think if you prove that for triangles, you can prove that for any polygon since they can be divided by triangles

[stuck sleet]

thanks!

[lament tree]

np

[stuck sleet]

Though I did derive the formula in similar triangles...

[lament tree]

nice

so that's it

[smoky jetty]

anyone wanna give it a try?

[stuck sleet]

for two similar triangles

So if I said all the triangle faces of the two pyramids are similar, the bases are also similar?

[minor arch]

I'm having trouble with this proof:

https://i.imgur.com/Xz05qt3.png

I get lost at the "Then apply the distributive property twice on the left side of (6):

[minor arch]

So Khan Academy explained the proof more clearly. But it seemed to have arrived at the proof through other means. Anyone knows wtf this book is saying?

Rofl no, I'm dumb. I got it now fr. The way they phrased it, I didn't realize that step 7 was only working on the left side and foiling it out. STUPID. I should be sleeping.

[weary torrent]

They factored out the u-v twice

[minor arch]

Yeah, TY. It makes sense. I was thinking about it too hard.

For that step, I thought that the righthand side was a simplification of the right handside for the version at (6). When they were just focusing on the leftside.

it says so plainly as well, but my mind didn't register that.

[lucid jasper]

Find m,k in R s.t.
(y+mx)²=ky(3x+4y-9) is a circle.

[wide field]

Why does the axial section of a cylinder, which is parallel to the cylinder axis, look like this?

Why is it parallel to the radius?

[bronze wind]

Can someone tell me how to solve for X?

[wide field]

I think I understand, I can move the virtual line to the radius as much as I like. And obviously, no matter how I draw the section, I can move the virtual line to the radius as much as I like. And it will always be parallel 😅

[upper karma]

hellow their

so using the cosine rule

we can solve the angle of tirangle HLJ

and since angle H is equal to angle F

once we find the angle

we can use it to solve x by substitution into the sinerule

e is opp to andgle E

g is opp to x+2

$\frac{e}{sin(E)} = \frac{g}{sin(G)}$

[somber coyote]

Mr. Macro

[upper karma]

oh sh**

we need angle G

[smoky jetty]

see if they're similar. If so, set ratios of corresponding sides in proportion

[shrewd prawn]

Someone help please

[quiet sand]

im learning that as well ;-;

uhh

no clue

[fickle rose]

honestly just use pythagorean theorem to find the height. normally u should use geometric mean especially if u have ugly numbers like that on the bottom but you don’t have enough lengths to do it

[shrewd prawn]

Can u tell me the answer at least?

[heady scaffold]

no

we arent allowed to

we have to have you work through it

[shrewd prawn]

Oh

[cosmic ferry]

can someone help me with this

[misty tusk]

not sure of the exact notation they want you to use but essentially its asking how you would map the points of one onto the other

so for an example lets take a point in Figure W and see where it is in Figure X

these two points here for example

we've got one at (-1, 5) and another at (4, -5)

now if we take a look at what's actually happening to the shape we can see that its reflected a few times across the x and the y axes

so in order to get (-1,5) to (4,-5) we can do a few transformations right

so the first thing I would do is flip it along the y axis and move it over

you can show that as

$$(x,y) \rightarrow (-x+3, y)$$

and if we apply this to (-1, 5) we can get the points (4, 5) which matches up with the blue outline shown above

[somber coyote]

architecture2

[misty tusk]

now our next step is flipping the blue shape along the x axis, which can be done with a transformation of $$(x, -y)$$

[somber coyote]

architecture2

[misty tusk]

and if we put these together we get a transformation of

$$(x,y) \rightarrow (-x+3, -y)$$

[somber coyote]

architecture2

[cosmic ferry]

So reflect once over the y and once over the x?

[misty tusk]

yep
you also have to shift it in the x axis though

that's why the +3 is there

[cosmic ferry]

If only lets you do two series of transformations

No more than 2 no less than 2

[misty tusk]

really? thats. strange

[cosmic ferry]

It’s some practice website and that’s how they want us to do it

And the teacher wasn’t even here today to explain

So I’m lost

[misty tusk]

i don't see a faster way to do it than that ngl

[ornate lava]

hey i need a confirmation for me answer:
find arc length of the sector where central angle is 3pi/8 and the radius is 18.2 cm
would the answer be 67.5/360 * 36.4pi

[misty tusk]

isnt arc length just the radius times the angle
so wouldn't it be 18.2*3(180)/8 cm

[ornate lava]

but that results in a number higher than 36.4pi

the circumference

i thought the formula was angle * circumference (2pi r)

so i tried doing the 3pi/8 which was 67.5

so is 67.6/360 x 36.4pi correct? or am i missing something here

[misty tusk]

ohhh i see what the problem is
its the conversion between degrees and radians

what unit does it want the answer in

[ornate lava]

uh my friend sent me the problem it doesnt really say

[misty tusk]

hmmm okay

[ornate lava]

but the idea is right correct?

find the angle measure and multiply it to the circumference

[misty tusk]

yeah that's right
just make sure the angle is a fraction out of 360

yeah sorry my mistake
you were right the first time
i just didn't really read it all that clearly

[ornate lava]

so what are the possible units to answer in?

wai tnvm

[misty tusk]

your answer above is in degrees but there's another answer in radians which is in terms of pi

[ornate lava]

6.825 pi cm is an answer right ?

[misty tusk]

yep

[ornate lava]

thank you

[gritty mulch]

Why bother… when u can use this?

[vernal pilot]

have you by chance learnt about rotations?

[bronze wind]

How would I solve it out if there is those 3 corresponding sides

[fickle rose]

setup a proportion

set 2 fractions equivilant to each other

simplify each fraction as much as possbile to make it easier

then cross multiply and isolate for x

[bronze wind]

what if its 3

[fickle rose]

what if what is 3

based on the congruent angle u can use that to figure out which sides should be proportional to each other

u may have to do a bit of testing i guess

[mint pebble]

2. (a) Using Euler's Graph Theory theorem, state why the following graph contains an Euler circuit. [5 marks]
   (b) Use Fleury's algorithm to find an Euler circuit in the following graph. Note that full marks will only be given if the steps for Fleury's algorithm are given. Clearly list the order of the vertices visited in your circuit - e.g., abcdef.
   marks]

[bronze wind]

theres 3 sides with numbers

[upper karma]

Can Somebody roughly list Main topics/Things to know in vectors pre College

[upper karma]

Can someone help me in these problems?

[maiden brook]

Find BF

Then take cross section ABFE

[toxic scarab]

could you help me

[upper karma]

The first one you can to pythagoras to get OP first then use it again to solve for PN

OP^2+MP^2 = MO^2

Rearrange to find OP then use it to PN

[toxic scarab]

thx

[toxic scarab]

so its 25^2+7^2=25^2?

[upper karma]

No

OP^2+7^2=25^2

[midnight current]

Can someone please help me with this

[toxic scarab]

what are some of the choices for the answer

nvm i think 25/24

[upper karma]

tangent is opposite over adjacent

[toxic scarab]

oh

[stiff otter]

Tangent = Opposite/Addjacent

You're referencing from the perspective of angle A

So your triangle would be labeled like this.

Opposite is the opposite of angle A

Hypotenuse is always your longest side or the angle across from the right angle formed

And adjacent is whatever is left

Therefore tan(a) = 24/7

[tawdry pike]

Who can help me

[fickle rose]

find the length of the diagnosis of the parallelogram

ur able to easily find them

and Part B’s answer is already on the diagram

[grave nova]

Is the largest angle of a triangle always opposite the largest side?

[sick kraken]

Im lost on this

[atomic merlin]

how do i find the radius of a circle that touches the sides of the triangle (the one i just drew in)

[hard coral]

I need some help regarding angles in 3 Dimensions. I have the correct angle for everything though the issue now is that, the x-component of the reflected vector is seemingly dependent on two angles for its correct x-component. The reflected vector is in Red and as you can see, the angle is nearly 270 yet the reflected vector looks as though its 225. This is because the x-component is following the angle of the normal vector in green, along the xy plane as shown in the second image of the top down view.

Though the x-component is in along the 180 line or the negative side of the x-axis which is correct, the angle from the x-axis on the XZ plane, should be around 630 degrees. How do I incorporate that into Rx so that way the reflection vector looks like its at 270 and not 225? I feel like I need two cos for one function but that obviously isn't correct at all. Could I get some help?

here is the paper work I did

[hard coral]

nvm I solved it

[upper karma]

The detail consists of a cone and a hemisphere. The surface area of ​​this part is 320 cm². The volume of the hemisphere is 2000pi/3 cm³. Calculate: a) the length of the radius of the hemisphere; b) the area of ​​the lateral surface of the cone: c) the height of the cone to the nearest 0.1 cm.

I am a little stuck I get R = 10
b) 120pi
And then c) H = 0

I am doing something wrong I need help

[rugged hearth]

theres a lil soemthing called the incenter i think

and the incenter of a triangle is the midpoint of the largest circle that can fit in the triangle

and the diameter of the circle is 2/3 the length of the line from the incenter or 1/3

u gotta google it

[median vapor]

hey so do we represent coordinate-plane curvature with a Capital Omega?

[fickle rose]

yes

that’s a theorem

[grave nova]

Ok nice

Thanks

[maiden brook]

for b, erect an altitude from C to AB. Does this necessarily have the same length as AD? What can u say about this quadrilateral, and use this to find BC. then find square simply

bruh "complex problem" smh

[coarse lance]

is this a troll geometry problem?

[urban ivy]

Yeah

[vagrant schooner]

Do you think anyone can solve this?

[fast pasture]

can someone pls help?

1 - 5

[obsidian harness]

in that circle, there are two right angles (CD is the diameter)

one 45 degree angle

and the other two angles are equal

the two angles sum to 360 - 180 - 45 degrees

[fast pasture]

so the two angles measure 67.5?

@obsidian harness

[obsidian harness]

yes

[fast pasture]

aight thx

[ornate igloo]

two concentric circles their radii is 12 , 7 respectively, AD is a chord of the bigger circle, A and D are points on the bigger cirlce, AD intersecting the smaller circle at point B, C respectively , prove that AB*BD = 95

Couldnt solve this in exam im in Grade 10

[obsidian harness]

let the midpoint of AD (and of BC) be point M

then AM^2 + MO^2 = AO^2 = 12^2

and BM^2 + MO^2 = BO^2 = 7^2

the way to see this is that 12^2 - 7^2 = 95

so......

[ornate igloo]

isnt there any forced way to find AB or BD to prove AB*BD equals 95
because i dont understand how your solving proved it

[obsidian harness]

use my hint that 12^2 - 7^2 = 95 to see that

this is pretty close already

[ornate igloo]

thank you from your hint i knew it after i factorize i conclude that AB*BD = 95

👍

[obsidian harness]

no worries!

[hard coral]

I need some help here. I have need the x-component of the red vector, to be at around 0. Or rather, the angle should be almost 90 as the Tau_r says it should. But is there any way I can rewrite R_x such that R_x is nearly 0 when θ_xz is nearly 90? I found that the way I have it, works but it requires me to manually switch sin(θ_xz) to cos(θ_xz) at certain angles that I don't know of. That disqualifies me using a piecewise function and that isn't the goal. I need some way to get the R_x function to be correct with two angles it seems. Any advice?

The Tau_r is just a way to display the θ_xz angle dw about it

Here are my x and y inputs. For context, this is for computing the reflection angle on a surface

This is the surface

thanks if anyone could help

[gentle haven]

I made a rotation function in which it gives you the point after rotating any point any angle around any coordinate.

https://www.desmos.com/calculator/kmf7cb0hei

[slow spruce]

Is vertical angle congruence axiomatic and if not how do you prove it from axioms?

[upper karma]

It is not axiomatic and provable

[slow spruce]

@upper karma thanks for the reply

I'll see if I can do it from Euclid's when I feel up to it

[upper karma]

[opaque sluice]

Hello

Can someone explain me how to know the projection of the cos in the sun according to the circle ?

For example, I'll take the equation : cos(x) = 1/2

What is the projection of 1/2 in the sinus axis according to the circle of trigonometry?

[vernal pilot]

are you just asking what $\cos(x)=\frac{1}{2}$ means?

[somber coyote]

TheLord26

[vernal pilot]

or in general, what each of the trig functions are actually doing?

[opaque sluice]

No, like $\cos(x) =\frac{1}{2}\ = cos(x) = cos(\frac{π}{3}$

[somber coyote]

Meyland
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[opaque sluice]

Bruh

But you get it

[vernal pilot]

$\cos(x) =\frac{1}{2}\ = cos(x) = cos(\frac{\pi}{3})$

[somber coyote]

TheLord26

[vernal pilot]

Uhh, this is just trigonometry.

Basically think of x as the angle (usually its represented as theta). and the "1/2" as the adjacent side of a triangle divided by the hypotenouse.

then you solve using trig

what that pi/3 means is just 60 degrees.

but in radians.

$cos(\frac{\pi}{3})=\frac{1}{2}$

[somber coyote]

TheLord26
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[vernal pilot]

yes.

[opaque sluice]

Wanna know how project the cos point into the sun axis

For example the square root of 3 divided by 2 is pi/6

[vernal pilot]

think of it as the xy plane. where any point is can pe represented with (sin(y),cos(x)).

and, in fact, if you graph $(\sin(\theta),\cos(\theta))$, you can find any point on the unit circle.

[somber coyote]

TheLord26

[opaque sluice]

Got you

Thank you

[worldly scroll]

How can I deduce $\cos2a = 1 - \sin^2a using other identities?

,, \cos2a = 1 - \sin^2a

[somber coyote]

sb

[worldly scroll]

By using the trigonometric identity  cos2x=cos^2(x)−sin^2(x) , and other identities, find the positive expression for  sin(A/2)  in terms of  cos(A) .

[twin crag]

try putting theta for both x and y in the cosine double angle formula, and see whta you get

[twin crag]

OH SORRY @worldly scroll i meant angle addition formula for cosine cos(x+y)

[supple abyss]

you don't because it isn't true

[twin crag]

its cos^2x right

[limber kraken]

Here I gota very cool trick for you guys: cos((180-a+b)/2) = sqrt( (sin(a)-sin(b))^2 + (cos(b) - cos(a))^2)/2 for all positive values of cos

the order is from l;eft to right like how you read english

[limber kraken]

Oh yeah the fromula that I rediscovered... haha... anyway its cos(a)cos(b)-sin(a)sin(b)

[limber kraken]

Ignoring compex values and using the formula for cos we we cos(a)*cos(a)-sin(a)*sin(a)=1-sin^2(a) we can calccel out the sin^2(a) and we get cos^2(a) = 1 which means a is 0,180 any anything after that added y 360 such as 360,540,etc.

[plain marten]

Thats not true. Its $\cos 2a = 1 - 2\sin ^{2} a$

[somber coyote]

Cherry Man

Cherry Man
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[bold sandal]

isnt this a trig theorem for the 30 60 90 triangle?

[limber kraken]

Its nota formula but a equation that is supposed to be solve if I am correct? its just find the value of a

[obsidian harness]

it's much more general than that

it works for any value of a

[twin crag]

i don'tt hink so

unless you're talking about the side ratios

[worthy python]

Just aced my trig test guys

[lethal moss]

nice

[celest wave]

hi

someone help me please

i already got a few answers but i still need help plz i dont get it

[lean tartan]

Boolean algebra

[lethal moss]

Hi exloria.
From what I see you have found out that the triangle is equilateral. The thing you are missing is that every angle in such triangle has the exact same measure. Try to find it then.

[celest wave]

how

[vocal mauve]

Equilateral triangles have 60 degrees for all angles

[celest wave]

4(5)-8=12

yes

[lethal moss]

... and bare in mind that BAC and ACB are both inner angles of this triangle (the 60 deg rule applies)

[celest wave]

so how do i find BAC? Do I have to add everything?

[lethal moss]

refer to what Salaam said earlier.

[lethal moss]

Is there something to clarify?

[celest wave]

180

?

[vocal mauve]

What?

[celest wave]

because 60 + 60 +60 is 180

[vocal mauve]

All angles in an equilateral triangle are 60

Yes

180 is the sum of the angles