geometry-and-trigonometry

[twilit girder]

wow

I mean so far I just started on it

[tiny burrow]

for us in grade 9, we can choose to do math 10

[twilit girder]

oh actaully?

[tiny burrow]

yep

I need help lol

[twilit girder]

I dont know if ill be much help though

you might need to work with an expert since I recently just started this

good luck though

[tiny burrow]

same

the fraction and the sin part is kinda hard

<@&286206848099549185> help us

[twilit girder]

oh opposition/hypotnuse?

[topaz oar]

i don't see any negative signs

[twilit girder]

like are you trying to find the angles or the side lengths of the triangle?

[indigo wadi]

Right, but COT can also be in Q4

Which would make both X and 1 negative

Tan*

or at least

1 is negative

And x as a value is negative

Which would make sin of that also negative

If you draw out the triangle

[topaz oar]

if it is in Q4. yes it is negative. If it is in Q1 then it is positive.

you made a valid point there

[tiny burrow]

both

[jade sierra]

how do i do this .-.

i kind of remember you multiply everything by the demonimator

but after that i forgot

[obsidian harness]

$\frac{4 \cdot 2 + 4i \cdot 2 + 4 \cdot 9i + 4i \cdot 9i}{2^2 - (9i)^2}$

then remember that $i^2 = 1$

[somber coyote]

south

south

[jade sierra]

My teacher taught i to the power with some wheel

Which is mad confusing imo

[obsidian harness]

yeah I mean it's not that bad

$i^1 = i$

$i^2 = -1$

$i^3 = -i$

$i^4 = 1$

[somber coyote]

south

[obsidian harness]

so the cycle repeats every 4 powers

[jade sierra]

Oh sick

Thanks

[obsidian harness]

so if I had like $i^{11}$, well that's just $i^4 i^4 i^3 = 1 \cdot 1 \cdot -i$

[somber coyote]

south

[obsidian harness]

that's how it works, so just divide your number say 11 by 4, and take the remainder

[obsidian harness]

np!

[jade sierra]

My brother Taught me a method that if you divide any number by 4, and if it ends on .25, it’s square root -1, 0.5 -1 , 0.75 -i , 0.0 1

I think it works

But not sure

[cunning lion]

that comes out to be the same as taking the remainder (since .25 = 1/4 means a remainder of 1, .5 = 2/4 means a remainder of 2, etc)

[jade sierra]

Ig it does work

[indigo wadi]

okay cool, thanks

[round blade]

My current progress with this question:
I figured PA×PB=PC×PD and then put the values which resulted in:
|z0-a|×|z0-b|=|z0-c|×|z0-d| and squared both sides and expanded a lot
Final result got a bi-quadratic where I let z0=x
Found that the sum of the roots of the equation divided by 2 is the exact same as the RHS to be proved in the question
What should I do now?

[soft bolt]

need help with this

[velvet burrow]

Hello, calmest. I think I can help you to solve this problem.

[obsidian harness]

there is one unique point where CD and AB intersect (assuming that CD and AB aren't parallel)

hence the roots of the biquadratic should be the same

and $2z_0 = \text{sum of roots}$

[somber coyote]

south

[round blade]

Wait so all of them to be same?

Oh

Nvm

[round blade]

Hm

[round blade]

Can you please elaborate because I don't fully understand

[obsidian harness]

two non-parallel lines always intersect at one unique point

hence there can only be one value of $z_0$ that satisfies the biquadratic

[somber coyote]

south

[round blade]

Ohhhhhhhh

That makes a lot of sense

So 2 roots of each line are there to make it possible

Thank you so much omg

I've been stuck in this question for days

But so if all roots are same shouldnt it be 4Z0?

[obsidian harness]

no it's just $z_0 + z_0$ = sum of roots

[somber coyote]

south

[round blade]

Wait how

Is it 1 z0 per line?

And because there is 2, there should be 2 z0?

[obsidian harness]

no, z0 is just the intersection of the lines

hmmmm

[round blade]

Hmm

;-;

1 root per line then?

Common root

[round blade]

Oh wait nvm

I took Z0 As x

So only z0 is possible

But then it still doesn't make sense to me

[upper karma]

complex numbers and geometry

a match made in heaven

[upper echo]

This one is not too difficult if you know the equation of the line in complex numbers.

[round blade]

Yeah no I solved it

I reviewed my notes and found that writing the 2 complex slopes of the lines and subtracting them gave the answer

[chrome token]

On the surface of the planet lives one inhabitant, that can move with the speed not greater than u. A spaceship approaches to the planet with its speed v.
Prove that if v/u > 10 , the spaceship can find the inhabitant, even it is trying to hide.

This is geometry right?

[silk rock]

Could anyone try to explain gr 12 trig identity’s? I’m very lost

[hazy blaze]

Be more specific

[round void]

Could anyone help me with angles of triangles

[steel hinge]

Hmmm?

[round void]

My teach said if I dont lock in Im cooked

💔

[steel hinge]

Wdym?

[round void]

Like

Idk how to explain it but thats what the assignment is named

Its finding missing angles and the triangles are complex

and like triangles in real world stuff

[steel hinge]

Okay, show me an example problem

[round void]

alr holdup

mind if I send u a few

[steel hinge]

Sure

Dms or here are fine

[round void]

I used ai to solve them ignore that I got it “correct” but I wanna go over them to actually understand it

[steel hinge]

Okay 👍

Small chance I may not know how to do it though. I'm in algebra 1 in terms of school, and my knowledge is pretty scattered past that

[round void]

its alr if u dont know it

I’ll post it here js incase anyone knows it if u dont know it

[steel hinge]

Gimme a moment. I know I can solve the first one

[round void]

the one with a lot of boxes?

[steel hinge]

The first image

The triangle

[steel hinge]

Oh, yeah

Nvm I see what you meant

Yeah, that one

For angle 2, it would be 130°+30°+x°=180, since it's a straight line below, and any full turn around that equals 180°, so angle 2 =20°

Same with angle 7; 110°+x°=180°

So angle 7 is 70°

And you can find angle 5 with either the known angle opposite to it or angle 7, either way, you get 110 (opposite angles like 110° and angle 5 are always the same.)

Now, with the triangle formed by the 30° angle, angle 4, and angle 5, you can find angle 4, since we now know angle 5 to be 110°

Triangles interior angles always add up to 180°

So 110°+30°+x°=180°

X=40

Do you understand how to do it so far? @round void

Oh, and question 15. As said earlier, triangles interior angles always add up to 180°

So you add up all of the x's

You get 2x+3x+4x=9x

So 9x=180°

And x=20°

So you just plug in x as 20°, and there's your answer

And question 26. As I've said several times already, triangles interior angles always add up to 180°

So 72+77+(angle 2)=180°

So, it seems it's actually not drawn to scale, and angle 2 is 41°. As for angle 1, I need more info

Assuming the cars roof is ignored, it would be 41°+(angle 1)=180°

So in that case, it'd be 139°

But I don't know if that's what it's asking.

@round void

As for question 12, it's just time consuming, but basically, here's the 3 rules you NEED to know. Opposite angles in intersecting lines are always the same. Adjacent angles in intersecting lines always add up to 180°. A triangles interior angles always add up to 180°.

You can find the answer to all of those questions if you know those 3 rules.

I'm too lazy to finish the one with all of the boxes, it's super time consuming, and I get bored kind of easily. Not to mention I do math for 88+ minutes a day.

Sorry

[round void]

ty for the help🙏🙏

im reading this as we speak

[steel hinge]

Np, just memorize these rules

[round void]

No matter the shape or how weird the triangle is it will always add up to 180?

[steel hinge]

Yup 👍

[round void]

Wait what is adjacent angles again

[steel hinge]

One sec, I'll send a pic

1 and 2 are adjacent, 1 and 4 are adjacent, 1 and 3 are opposite.

So like on opposite sides of the line

(Sorry it's sideways, didn't notice till it was too late)

[round void]

so if they make a straight line

im confused

[steel hinge]

Okay, gimme a moment..

Just to reinforce this, it doesn't matter the angle. 1 and 2, and 1 and 4 are still adjacent.

1 and 3 are still opposite

So long as the lines are straight through each other, and there are no turns, those rules will always apply

If you still don't understand, tell me. I'm not mad or anything, I promise. I enjoy being able to succeed in helping people

@round void

[round void]

So if they like cross eachother

kinda

I kinda see it and I kinda don’t see it

also when ur looking at stuff like that which side of the line do u know the number correlates to

like how the 1 is in between 2 lines

which line would it be

[steel hinge]

Oh... you thought I was numbering the lines?

[round void]

nono like

[steel hinge]

The numbers represent angles

[round void]

oh

so 1 represents that v

[steel hinge]

Yes

[round void]

I always saw it as like the line being numbered

[steel hinge]

Not the actual angle, but like a variable

[round void]

my teacher never rlly went over any of this just flew through it

ohh

[steel hinge]

So all of those numbers in the screenshots you sent are essentially variables

Unless it had the degree symbol

Or the (2x) ones

Do you understand it now?

@round void ?

[round void]

Yea

[steel hinge]

Yay

[hot latch]

help

i got everything but d

im just not getting the right answer

what am i doing wrong pls help

[upper karma]

They said beta is between -pi/2 and 0

I.e the fourth quadrant

Where tan is negative

So your tan beta would be -1/2 and not 1/2

Also you shouldn't write it like this

[hot latch]

wait how should i then

[upper karma]

Tan^-1 (3/4) and tan^-1(-1/2)

[hot latch]

oh no wonder cuz i drew my traingle in quadrant 2

wait let me do it and show u so u can tell me if its right

[upper karma]

Wow

This explains why cos and sec are positive in fourth quadrant

Dang

I thought they took it as a convention

[hot latch]

idek whats going on

[upper karma]

Lol nvm

[hot latch]

i do fine on all of them besides tangent

[upper karma]

It's correct

Go on

[hot latch]

the 1 in negative right

is*

omg

thank you im so slow bro

i didnt realize the 1 had to be negative

okay wait so what abt thisone

[upper karma]

why is the image 12 mb 💀

[hot latch]

ask my cell phone

[hot latch]

u literally can just tilt ur head

[upper karma]

lol

whats the question agai m?

[hot latch]

i just need to solve d

i sent the stuff the text gave me

[upper karma]

you dont have to worry about alpha since they already gave it in tan

as for beta

which quadrant is it present in ?

[hot latch]

well yea but i still need to plug it into the thing

i have to solve for tan(a-b)

[upper karma]

.

[hot latch]

2??

bro

[upper karma]

🔥

[hot latch]

pi and pi/2

is q2

[upper karma]

im asking for beta

not alpha

[hot latch]

then 1

[upper karma]

.

[hot latch]

can u stop being mean

[serene hedge]

damn

[upper karma]

im trying to be funny 🥲

mb

sorry

[hot latch]

its ok im laughing too

im just kidding\

[upper karma]

yes the first quadrant so tan would be ?

[hot latch]

positive

[upper karma]

you have cosbeta as 1/2 so your tan beta would be ?

[hot latch]

its sqrt 3\

[upper karma]

yes

substitute

[hot latch]

i put it in my triangel

i did

wait

let me set it up

[upper karma]

why -1 ?

[hot latch]

bc cos 1/2

i assumed the negative

[upper karma]

if it was cos\theta = -1/2 then you would be right but that is not the case

[hot latch]

well i got all the other ones right so why is this one not working

[upper karma]

where?

[hot latch]

i boxed them

[upper karma]

you took cos beta as 1/2 not -1/2

you did no mistake there

[hot latch]

haha

wtf ami doing

im so confused so do i assume the negative or noit

[upper karma]

why assume ?

[hot latch]

cuz it goes in q4

[upper karma]

which one ?

[hot latch]

the 1

in beta

[upper karma]

ah

i see

its negative when the vertical line goes to the 4th quadrant

[hot latch]

oh wait but do i not do that since cos isnt negative there

[upper karma]

but the 1 here is horizontal

[hot latch]

what vertical

oh

i see

so how do i write it

[upper karma]

?

[hot latch]

is that right

[upper karma]

yes

[hot latch]

yay let me try it

ok i give up

[upper karma]

lol

why

you already got the answer

you just have to simplify ti

[hot latch]

i am doing something wrong

💀💀💀

this is the answer btw

oh should i have done the 3 denominator first

[upper karma]

this is you rn

[hot latch]

DUDE i am losing my min d

[upper karma]

you are on the right path

go on

[hot latch]

i erased it

[obsidian harness]

99% of crypto bros dip before the peak

be the 1%

[hot latch]

i cant do algebra i surely cant do crypto

can you please write it and show me how to do it

[upper karma]

@hot latch

[hot latch]

so where did that 9 come from

[upper karma]

i multiplied and divided the whole fraction with -9

[hot latch]

but why

[upper karma]

i am not sure how to answer that

south

[hot latch]

did you just think "oh i like this number" and did it

[upper karma]

lol no

[upper karma]

@obsidian harness

[obsidian harness]

ah the 9 comes from the denominator, 16sqrt(3) / 9

generally you want to multiply top and bottom by the LCM of all the denominators

[hot latch]

so i took the 9 out of the -16sqrt3/9 and just distributed it to everything

take*

yall r too smart for me i cant\

[vagrant fiber]

Help

[vernal pilot]

Circumference is minor arc + major arc, and then you can solve for r by using C=2pi r.

then i would suggest drawing a triangle to find the angle.

[cold crag]

Is this a geometric series question?

[upper karma]

No

Trigonometry

[upper karma]

How to find the equation of the ellipse which is formed by the intersection of the cylinder of x^2 + y^2 = 1 and the plane x+y+z=1 ?

i moved z from left to right side and sqaured it on both sides

and substituted x^2+y^2 =1

leaving me with 2xy=z^2 - 2z

is that right /

?

what to do adfter tha t?

[upper karma]

Is this related to volume and area?

1/3 x pi r² for cylinder ig

@upper karma

i just want to find the equation of the ellipse

what shape is the ellipse

[trail tendon]

its in 3d ._.

[trail tendon]

what about
<cos(t), sin(t), 1-cos(t)-sin(t)>?

0<=t<=2pi

:>

[upper karma]

ah

i didnt think about paramatrizing it

thanks

[trail tendon]

the equation of a line in 3d is usually parameterization 😔

i dont know if you can like do it without parameterization at all lol

[upper karma]

i was expecting an equation in just x-y

💀

[trail tendon]

[upper karma]

.

[noble zinc]

you parameterize.

[upper karma]

got it ty

[upper karma]

Did you use elimination method to solve it?

Linear inequalities in 2 variables

[gloomy osprey]

is this correct?

[upper karma]

Isn't a circle sum up to 360⁰?

We know that EF and BF both have the same angle

(180⁰-48⁰)/32⁰ = 66⁰

@gloomy osprey

[gloomy osprey]

E and f aren’t the base angles right?

[noble zinc]

nah, just the standard method of parametrization

[upper karma]

No

[noble zinc]

don't see anything wrong with it

[gloomy osprey]

Not trying to sound dumb jsut trying to process it I thought cus the alternate angles equal the same and then the base equal the same so the remaining angle is the difference

[gloomy osprey]

Is he jsut trolling or smth

[noble zinc]

me?

[gloomy osprey]

Nooo

[noble zinc]

oh sorry

[gloomy osprey]

The other guy who said it’s 66

[gloomy osprey]

.

This xd

[upper karma]

B amd F are

[gloomy osprey]

They not?

[upper karma]

Mb i misunderstood the concept

[noble zinc]

could you elaborate further on this?

[gloomy osprey]

Xd

U jsut fried my brain

[upper karma]

I applied lines and angles formulae on this

[gloomy osprey]

I just want x 😭

[upper karma]

180⁰-interior angle

[noble zinc]

x = 84

[gloomy osprey]

❤️

[upper karma]

(n-2)x180⁰ = x?

[noble zinc]

48 + 48 + x = 180

[upper karma]

Let x remain on the left side

180-48-48=84

I understand now

Thx

[upper karma]

In the graphics, A, B and C are tangency points. calculate 𝜃

Ik the triangle with theta in it has 45 as one of the angles I just need the other

[loud void]

See why can’t we do stuff like this in my geometry class?

[faint hollow]

thoughts anyone?

[stray junco]

How do you solve this

[obsidian harness]

Pythagoras gives you the lengths of DC and DA

or distance formula, same thing

[stray junco]

i did that i used it

for some reason they dont take 4 root 10 times 40

nvm i fumbled it

[random ridge]

use distance formula

[fading cradle]

guys if we have a right triangle ABC in A
and we draw E the orthogonal projection of point A on (BC)
does this prove that the triangle AEB is right-angled at E?????????

[faint pasture]

uh yes?

[fading cradle]

really?

[faint pasture]

idk why would you ever need that though

but yea

[faint pasture]

by orthogonal projection you mean the altitude right

well yea

in this case it does mean that according to a quick google search

[fading cradle]

ye the altitude of E

[faint pasture]

tho doesnt the concept of triangles in itself break when you start to get to vectors stuff

[faint pasture]

yea then obviously by definition of altitude

[fading cradle]

no there is no vectors stuff

in this case

[faint pasture]

then yea

[pseudo abyss]

anyone can help?

[fading cradle]

thx

[rustic plover]

Start by joining bc

[arctic grove]

after joining BC,
α=2<ACB
β=2<DBC
<ACB=180-90-<DBC=90-<DBC
α+β=2<DBC+2<ACB=2<DBC+2(90-<DBC)=180

according to cyclic angle theorem

[shrewd pecan]

I think 72 but not sure

[shrewd pecan]

why is there a square root before the 12^2 + 4^2

shouldn't it just be 12^2 + 4^2 which is 160

and then the square root of 160 is 12.65

then you multiply that by 6.32

and you get 79.948

which is basically 80

cuz there is no need to do 1/2 (80)

cuz this is rectangle

are for rectangle is simply l x w

you calculated as if the shape is a triange

[obsidian harness]

agreed that there shouldn't be a 1/2 being multiplied by 80

but why are you rounding

why are you rounding early when you can get an exact answer

[shrewd pecan]

well

I guess not rounding would have been better

but even though I rounded I still get 79.948 and then I can just round again

rounding early isn't ideal but it still worked

[faint pasture]

I got 49/48 but the ans key says 64/15 can someone verify the answer

[gloomy osprey]

b = 7 a = 83 right?

[shrewd pecan]

yes

that is correct

angles on line = 180

and then angles in triange = 180

[upper echo]

49/48 is correct

[ocean saffron]

ay guys any idea on how to do this

i need to make a diagram and its trigonometry

so is this like a 3D shape

[coarse wing]

Can y'all help me with this? Honestly I was enrolled into school really late, hardly know anything about geometry and am trying to get extra credits. Any help would be great 👍🏾

All I know is that the first shape's supposed to be a kite, the third one's a rectangle and yeah

[faint pasture]

Thanks

[stray junco]

Can someone explain this to me and how to make it a decimal

[wise pawn]

looks like a rectangle with the coordinates listed, past that there's not really anything to say without more info

[stray junco]

Oh yeah sorry

Im figuring out the area with decimals

[wise pawn]

ahh

assuming it's a rectangle, you only need to find the length of the sides AB and BC. The way I think of them is as the hypotenuses of right triangles

like this, so then I'd find the length of these legs here

[stray junco]

I finished all those stages and i was left with two root numbers with different radicals

I was wondering how you make that into a decimal

[chrome token]

Prove that You can always pose a circle of radius S/P inside a convex polygon with the perimeter P and area S.

[wanton sinew]

How do I start learning trigonometry as a 13 y.o with no experience?

[chrome token]

use area = base*height for a rectangle

[obsidian harness]

Khan Academy

and org chem tutor on YouTube

[chrome token]

why org chem?

[obsidian harness]

jacked at maths

he doesn't only do org chem

[rose karma]

khan academy better tho

[lone nova]

Could someone help me with this pleaseee it's stereometry

At least the first problem 😮‍💨 it's asking to calculate the volume

[lone nova]

I've got it now 😭

[grim haven]

guys A F and C algebraically should be aligned but even in the 3d visualisation it doesn't look aligned got any idea?

note that no measures were given for the ABC triangle

<@&286206848099549185> \

[gloomy osprey]

The attached diagram shows a regular pentagon and a regular hexagon. It is not drawn accurately.
Work out the size of the angle marked x.
4 marks\

[blissful spoke]

Can someone help explain how to choose what x values to use for function plotting tables?

[gloomy osprey]

someone pls draw on this and work it out

[lunar seal]

can i progressively do this and someone help check my answers along the way

for number 1 i got the perimiter of the playground to be 273 yards=819 feet

so i divided 819 by 50 to get 16.38 = 17 rolls

and that 17 x 90 = 1530 is the cost of the fencing

[lunar seal]

final answer i found for 1 after finding grass stuff is 2682

[drifting zealot]

Are there any textbooks or free sources that help with proofs

[pale vault]

how do i do dis

[somber wolf]

oh i hate planes

[pale vault]

you don't know how to do it?

[somber wolf]

i slightly remember it was a rough year when i did algebra

[pale vault]

oh

this is algebra

[somber wolf]

geometry

not algebra

[pale vault]

do you have any friends that know how to do it?

[somber wolf]

uh let me think

c or d

[pale vault]

no i don't want the answer

i just wanna know how to do it

[somber wolf]

ok ok

so

think of a plane like the walls celing and floor of a room

[pale vault]

is a plane always 4 points?

[somber wolf]

no

[pale vault]

how many points does it have to be

[somber wolf]

more so each plane is like a wall

its a infinte surface that extends in all directions

[pale vault]

ok

[somber wolf]

and you want to find the planes that the line ab lands on

[pale vault]

so

[somber wolf]

so you have the left wall

[pale vault]

could it be ABE

?

[somber wolf]

and the celing as planes that contain line ab

[pale vault]

im kind of confused

[somber wolf]

ok

um

oh ew

ok

each side of the cube has a plane

the plane is the light bluee area in this picture

you want the planes where your line is on both

[round blade]

Can someone please help me with question number 2?

[upper echo]

Represent z_m=z0+w*z^m, where w is some nonzero complex number and z is a primitive root of degree n. Then expand binomials and change the summation order. Use the fact that z^j+z^(2j)+....+z^(nj)=n, if j=0 and 0 otherwise, for j<n.

[flint saffron]

could someone please help me showing that ABC~A'B'C'

[round blade]

Wow that is something I didn't expect to be using. After using that method I managed to reach the answer. Thanks!

[vernal pilot]

do you know the internal angles of regular polygons?

[gloomy osprey]

Nope

[vernal pilot]

well the angles add up to (n-2)*180

then you can further divide by how many corners there are

[sturdy oak]

How am i supposed to calculate the surface of this shape??

[merry knot]

count the squares

[frosty kraken]

A regular polygon P has n sides and lies in 2d cartesian plane. Find all possible value of n if all vertices of P are a lattice point

Any hint?

I suspect that n must be even

[maiden brook]

does a triangle work? A square? A pentagon? A hexagon

[frosty kraken]

Triangle no

Squares work

[upper echo]

This is a classical fact. Only squares fit. https://hrcak.srce.hr/file/2836

[royal charm]

hi! Is there anyone who can help me on this

[floral sun]

is there a specific question you need help with?

[royal charm]

well iim very stressed bc its my last year and they gave me geometry and pre calculus at the same time ad they gav eme pcakets to do these math and i jst need soemone to show me how to do tuis

or if i did smth wrong

[floral sun]

alr let's go one by one

[royal charm]

thank you :(

i have to do 8 packets

i did like 5

but i have to check fi

[floral sun]

oh noo :((

[royal charm]

if i did it rigth

[floral sun]

let's just take it slowly and make sure you understand

[royal charm]

so i jsut need someone who knows geometry :(

[floral sun]

ok so for the first question is it asking you to find all the angles?

[royal charm]

okay!

its something like that

[floral sun]

ok

[floral sun]

32 is correct for that one gj

ok so

do you see the angle directly opposite of that one?

do you know what that kind of angle is called

[royal charm]

for the first one?

[floral sun]

yes

[royal charm]

i do know that with the square it's rigth angle

[floral sun]

the one drawn in purple

[royal charm]

ohh

i think acute angle

[floral sun]

it's called a 'vertical angle'

[royal charm]

ohh

[floral sun]

yes gj!!

more specifically it's a vertical angle

[royal charm]

:3

[floral sun]

if you have two intersecting lines, the opposite angles will be equal

so based on that, what is the purple angle equal to?

[royal charm]

what do u mean by that?

[floral sun]

the dark blue is the two straight lines that are intersecting

the two light blue angles are equal

[royal charm]

ohh

[floral sun]

these are called vertical angles

so what would the purple angle be equal to?

[royal charm]

intersecting

intersection

[floral sun]

like the angle measure?

[royal charm]

ohh

[royal charm]

62?

wait no

[floral sun]

it would be equal to the opposite angle, so 32

[royal charm]

OH

😭

[floral sun]

it's ok lol

[royal charm]

so i write 32 in thathat angle?

[floral sun]

yes

[royal charm]

now im rmemeber

so tyen 180 -3-32?

oops

[floral sun]

these two orange angles are equal

[royal charm]

180-3180-32

[floral sun]

they are also vertical angles

[royal charm]

OMG

HUBABYUIA

[floral sun]

so it would be 180-32

[royal charm]

yeah

[floral sun]

because 180 in a line right?

[royal charm]

my tablet double it

yeah

[floral sun]

yeah and you should be done for this one!!

do you have any questions on question #1?

[royal charm]

No!

[floral sun]

ok let's move on to question 2

your answer looks correct to me!!

[royal charm]

really?

[floral sun]

yes!!

if you dont have any questions we can do question 3

[floral sun]

what would the orange angle be equal to?

considering the 125 degree angle

[royal charm]

would it be 180?

[floral sun]

no the entire angle including 125 would be 180

the orange angle would be 180-125

[royal charm]

ohh

[floral sun]

yes

[royal charm]

so wait to check every angel is to negative w the nukber

number

[floral sun]

wdym by that

[royal charm]

like

ahh lemme. Thjnk how to say it

so everynumber we see i just have to put "180-#" with the number

[floral sun]

this is only the case for a straight line

[royal charm]

ohh

[floral sun]

since the total angle measure would be 180

[royal charm]

i got 55

[floral sun]

yes gj!!!

in the triangle above, what is the total angle sum?

for the entire triangle

[royal charm]

how do you do that again

[floral sun]

ok so for any triangle, the total angle sum would be 180

the sum of these yellow angles would be 180

for ANY triangle

[royal charm]

ohhh

so i put the 55 next to the 125

[floral sun]

yes gj

can you figure out the bottom left angle?

[royal charm]

yes,!

[floral sun]

you know the total sum of the three angles is 180, and the other two are 91 and 55

[royal charm]

89?

[floral sun]

so you would calculate 180-91-55

[royal charm]

ah shoot no

i forgot to put 55

lrmme try again

okay 34!

[floral sun]

good job!!

[royal charm]

yay!

do i do the same w 25

[floral sun]

using the same logic with the straight line,

can you find the purple angle?

[royal charm]

yes!

[floral sun]

ok!!

[royal charm]

hmm

is itt

9?

wait

noo seems to small

[floral sun]

no, the total angle sum for the line is 180

if you know one angle is 91,

the other would be 180-91

[royal charm]

so if i were in the yellow angle i would use all of the numbers?

i got 66!

[floral sun]

close, the purple angle is 89

since 180-91 = 89

[royal charm]

reallyy

awww

[floral sun]

so in this diagram, the yellow angles sum to 180

[royal charm]

i see what i did srong

i did 180 - 25

[floral sun]

good job!!

the next angle we calcualte may use that!

[royal charm]

64?

[floral sun]

since you knwo that the total angle sum in a triangle is 180, can you find the light purple angle

[royal charm]

oh wait

i put the 89 wrong side

[floral sun]

really close

[royal charm]

25?

waitr

NO

[floral sun]

the light purple one

[royal charm]

it be 180-25-89

o,g

[floral sun]

it should be 180-89-25

[royal charm]

yeahhh

[floral sun]

just subtracting the other two angles from the total

[royal charm]

i got 66

[floral sun]

good job!!!

[royal charm]

:))

[royal charm]

now we got this

[floral sun]

can you find the magenta/pink angle?

[royal charm]

OH WE GAVE MORE?

[floral sun]

loll

[royal charm]

oh i get it

bc it shows a whole circle

[floral sun]

yes!

i think you should be able to solve the rest with this knowledge

ping me if you need any help tho!!

[royal charm]

okkk

wait i add eveeverything igth?

[floral sun]

yes

[royal charm]

Okay

I got 180

Wait

Do I add of all of the things we put the numbers at

@floral sun

[floral sun]

no

no need

[royal charm]

Only the ones that are in the corner?

[floral sun]

you dont' need to add any angles together

[royal charm]

Then, so what do I do?

I added 89+89+91

I got 269

[floral sun]

no you don't need to add any angles

[royal charm]

Then how do I know what I put in it

[floral sun]

you dont' need anything else for that question

[royal charm]

Ohh

@floral sun

Sorry to bother :(

[floral sun]

it's alr

is it question 4?

the 86 looks correct to me

[worthy nova]

can somoeone be my trigo tutor

for like 2 days

pretty please

[floral sun]

i will ;DD

[steel hinge]

If you're fine with the fact that I am currently learning trig in my own time, then sure.

[floral sun]

i took trig 2 yrs ago but i still remember

hehe

[worthy nova]

what

grade

[steel hinge]

I know a pretty large amnt so far. I use kahn academy

[worthy nova]

are you?

[steel hinge]

8th grade

[worthy nova]

bro im so far behind on math in general

dude im in 11th

im so cooke

[floral sun]

you'll be fine

[worthy nova]

i might just cry

[floral sun]

i can tutor u

[steel hinge]

I learned a tiny bit of calculus too (why I'm learning trig)

[worthy nova]

dude i know its not always the teachers fault but this teacher is geniuny terrible

[steel hinge]

Damn

[worthy nova]

she just explains one lesson and immediately moves to te next one with no practice or no info in how it actually works

[floral sun]

damn that's bad

[worthy nova]

i have a test on thursday

i shouldve locked in earlier but i had other grades too

now or never i guess

[floral sun]

wait dm me topics or soemthing

[steel hinge]

Look. If I were you in that situation, I highly recommend you use kahn academy. It's amazing for math.

[worthy nova]

I just need to pass and ill try to get abetter grade on the next semeseter

[floral sun]

aops is rly rly good

[steel hinge]

What is aops

[floral sun]

art of problem solving

[steel hinge]

Ah

I know them from yt

[hot latch]

can someone check my answers

[forest ivy]

Can someone help me with this please

[carmine nexus]

which ones

[coral yew]

somebody help

[obsidian harness]

try dilating each point on the triangle

so (-8, -8) goes to (-2, -2) for example

[round blade]

Can someone please tell me question 1?

[dim narwhal]

The number 1 is 50 degrees

[steel hinge]

Nvm, for a sec I though I was wrong (why i deleted the messages.). I was right though. 302°

[steel hinge]

For number 2

[steel hinge]

Oh.. wait.. we're supposed to find the arc length, not the angle

[dim narwhal]

It says measurement

[steel hinge]

You sure?

[junior trench]

yeah the measure of the arc is the degree measure of the central angle

if they wanted arc length they would've stated it directly

[round blade]

Can someone please tell me question 1?

[obsidian harness]

consider $z_1^2 + z_2^2 = r^2$ which implies $z_2 = z_1 e^{i \theta}$

[somber coyote]

south

[round blade]

Hm

Yea

But what after that

Now we have theta in the mix

I attempted the question again and again and found that (z1-z2)^3=40

[obsidian harness]

$z_1 (z_1^2 - 3z_1^2 e^{2 i \theta}) = 10$ and $z_1 e^{i \theta} (3z_1^2 - z_1^2 e^{2 i \theta}) = 30$

hmmm

[somber coyote]

south

[obsidian harness]

interesting

[round blade]

Literally just added those 2 equations

[obsidian harness]

ah

idk maybe try dividing?

[round blade]

Hm

[obsidian harness]

$\frac{1 - 3e^{2 i \theta}}{3 e^{i \theta} - e^{3 i \theta}} = \frac{1}{3}$

[round blade]

Hmm

[obsidian harness]

cause everything has (z1)^3

[round blade]

Yea

[obsidian harness]

,w (1 - 3x^2)/(3x - x^3) = 1/3

bruh

[round blade]

That is not great...

But enough to work with

[somber coyote]

south

Results provided by WolframAlpha

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

Requested by sooouth.

[obsidian harness]

wait I fixed it and it got wrose

[round blade]

💀

Lemme try it on paper

No wait it's a cubic 💀

,w (x^3-9x^2-3x+3)=(0)

[somber coyote]

Results provided by WolframAlpha

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

Requested by nemdt

[round blade]

,w ((1-(e^(2ix)))/((e^(ix))-(e^(3ix))))=(1)/(3)

[obsidian harness]

well we have

$z_1^2 + z_2^2 = \frac{10}{z_1} + 3z_2^2 + 3z_1^2 - \frac{30}{z_2}$

[somber coyote]

south

[obsidian harness]

by rearranging the original equations

[somber coyote]

Results provided by WolframAlpha

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

Requested by nemdt

[round blade]

Hm

What to do from here tho

[obsidian harness]

idk

[round blade]

There is the fact that Im(z1^2+z2^2)=0

So either both of them are real numbers or their imaginary parts are same

I'm guessing the later

[obsidian harness]

$z_1^2 + z_2^2 = \frac{1}{3} (z_1^2 - \frac{10}{z_1}) + \frac{1}{3} (\frac{30}{z_2} + z_2^2)$ also

[somber coyote]

south

[round blade]

Hmmm

Those gonna have graphs

[obsidian harness]

3a^2 + 3b^2 = a^2 - 10/a + 30/b + b^2
a^2 + b^2 = 1/2 (-10/a + 30/b)

yeah you get the same equation either way

[round blade]

I think there must be something to do with the Im(z1^2+z2^2)= 0 thing

[obsidian harness]

https://www.wolframalpha.com/input?i=x\left(x^{2}-3y^{2}\right)%3D10+and+y\left(3x^{2}-y^{2}\right)%3D30

yeah I think z1, z2 are real or something

https://www.desmos.com/calculator/fv4xc19ysv

,w a(a^2-3b^2)-b(3a^2-b^2) simplify

[round blade]

Hm

[somber coyote]

Results provided by WolframAlpha

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

Requested by sooouth.

[obsidian harness]

yeah the signs are off

you don't have (z1 - z2)^3 = 40

[round blade]

What

Oh

L

I think the question is wrong then

Well

I have a question related to P&C

I'm not sure if my answer is correct

Question 3

I did summation of n^2+r^2-2nr with r=1 to n which gave a cubic of n and got exactly n=9

As a solution*

[frank ermine]

Q1 is 10

ig

[round blade]

@obsidian harness Had my friend ask chatgpt paid edition

[round blade]

Oh

[dense fractal]

hi

[winged mountain]

@obsidian harness Alright I got my math test back

the linear programming problems were correct

I actually ended up with a 76/71

[trail tendon]

now look i don't pretend to know math but uhm

hmm

if you have 71 points available

and you got 76

the math aint mathin

💀

[winged mountain]

Extra ptss

[bright python]

I need help with trig identities

[coarse wing]

Guys how do you visually tell the difference between opposite rays and lines?

I know that opposite rays have the same end point while lines go on forever, but they look the same to me.

[hot latch]

WHAT AM I DOINF WRONG HERE ITS SOMETHINF WITH MY NEGATIVE SIGNS BUT IDK!!!!

for b,d,and e i literally got the right answer besides the fact its + or -

[native jetty]

what was (b) asking for

[hot latch]

cos(a+b)

sorry just saw

[hazy sedge]

ik the answer jst how do i solve it

[obsidian harness]

Wow good on u

[obsidian harness]

Out of anyone.... GPT???

Ah you were so close

Should have known to investigate that idea

[junior wigeon]

yo where did the 5pi over 4 and others come from?

these are notes from my teacher i just dk how he got them

[obsidian harness]

Tan is periodic every pi

[trail tendon]

thats 2theta when tan(2theta) = 1

[obsidian harness]

So tan(x/2) is periodic every pi/2

But you do not need this fact if you let 2 theta = ....

[junior wigeon]

why is it done 4 times if its asking for intervals between 0 and 2pi

wait i understand it now

im goated

why is q1 and q3 mentioned though

[obsidian harness]

A consequence of 2 theta being in q1 and q3

You meant to ask why are q2 and q4 there right

[junior wigeon]

no i understand that

its just that is that necessary to solving the question

also when would I solve for the quadrants only in comparision to each period like in the example I showed

[junior wigeon]

for ex, why did we solve for all the periods in (a) and just focused on quadrants everywhere else

[knotty cape]

soh cah tuah

[obsidian harness]

ah cause you have $\cos x = \pm k$ for some real number $k$

[somber coyote]

south

[obsidian harness]

so that must cover all 4 quadrants

automatically

same for $\sin x = \pm k$ or $\tan x = \pm k$

[somber coyote]

south

[idle saffron]

may someone prove it why x = 35*?

[silent plank]

more context?

[idle saffron]

only this information

[silent plank]

then there isn't enough to determine/justify what x is

[idle saffron]

i think so

i tried to prove it by describing a circle around a triangle

but couldn't solve till the end

[silent plank]

there's nothing fixing the position of R,
the triangle can be rotated about point Q
giving infinitely many possible values for x

[idle saffron]

oh ok

[idle saffron]

in original problem asked "find bearing of R from Q", i drew a vertical line from Q and then couldn't prove that x=35

Apparently the problem statement is incomplete

[heavy leaf]

Hey guys, do you know where can i find hard high school math question online

Any recommendations??

[merry knot]

Art of problem solving contest collections

[obsidian harness]

yes it's incomplete

you can keep Q fixed and rotate R anywhere you want

[heavy leaf]

Thx for help bro

[frank blaze]

Calculate the field of the figure painted red

[junior trench]

are they right angled? and are the two vertical lines parallel?

if yes u can use similarity

[frank blaze]

Yes

[junior trench]

okay so the smaller right angle triangle and has 2 sides, 3 and 5 as the hypotenuse
so the remaining side is 4

now u can use similarity to figure out the height of the larger triangle

,w 3/(5+3) = 4/x

[somber coyote]

Results provided by WolframAlpha

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

Requested by milooo09.

[frank blaze]

Ok, thanks

[junior trench]

alg

[boreal olive]

Hello
What s the best geometry book you ve read(looking for an advanced technique book)

[round blade]

It solved the question better than wolfram alpha 💀

[upper karma]

idk if yall see that but this is so difficult

[faint pasture]

I guess 4.8z 5.6 z and 7.6z

[upper karma]

yes hes riht

[faint pasture]

Makes sense too cuz sum gives 18z

So pretty answer

[faint pasture]

You found out the value of z?

[upper karma]

hn

it dont say

[faint pasture]

What

[upper karma]

i dont know what z is

[faint pasture]

Its a variable/unknown if thats what you are looking for

[upper karma]

tbh

i have no clue

im done for

[faint pasture]

Cooked

[upper karma]

ok

so

lemme cook

180

wait

no

180

180/5.6?

im fried arent i?

ima say

there all 180

?

im dorry bro

this is so confusing

😦

im done for

4.8/5.6/7.9?

no

those are the angles

right?

ohh\

do

[upper karma]

i add the 180 add these?

[faint pasture]

It is 7.6 anyways

Not 7.9

[upper karma]

yees>

?

[gray pivot]

4.8z+5.6z+7.6z=180