geometry-and-trigonometry

[silent plank]

for the problem below that you didn't write in the sqrt for the sqrt(61) in the first line

to identify whether 10,24,26
is a triple you can apply similarity
divide all sides by 2 to get 5,12,13
which is a common triple

[hallow crystal]

I added (Square Root Method) so I know there’s methods besides square root

And I added subscripts to Leg 1 and Leg 2 like u said uwu

no parentheses around the (2x)
@silent plank
AHHHH I fucked up, it would = 4x^2 right?

[silent plank]

yes

[hallow crystal]

2sqrt(6) wasn't squared properly either
@silent plank
The square doesn’t cancel it out?

[silent plank]

the 2 would also be squared

[hallow crystal]

Ah

X = square root of 19?

@silent plank

[silent plank]

looks like it

lowercase x

which is another issue

[hallow crystal]

I did nut know that

Well, it’s always lowercase in my notes uwu

for the problem below that you didn't write in the sqrt for the sqrt(61) in the first line
@silent plank
Omg

I’m glad that’s the only part I messed up on cuz the answer still remains the same

And about the baseball one it’s kind of a dumb question so ignored it but what would the answer be?

@silent plank

[silent plank]

just need to know one of the sides

since you're told its a square

[hallow crystal]

@silent plank
Fixed?

Oh and I also added the divide by two method like u said

[silent plank]

yeh, mistake looks fixed

[hallow crystal]

What about these two?

[silent plank]

round answer to the satellite problem

[hallow crystal]

Uhh

I have no idea what the square root of it is and we’re not allowed to use a calculator

[silent plank]

factorisation i g

[hallow crystal]

Actually this is old hw and he apologized for not noticing how big the numbers were, so I’ll round it yea

x = 2829?

[silent plank]

nearest 10

so itd be 2830

looks like there's issues with 37

[hallow crystal]

looks like there's issues with 37
@silent plank
I see what I did wrong, but 4 sq root 5 is the b length right?

[silent plank]

you should apply pythag twice

once for the altitude and a second time using that value to get x

[hallow crystal]

I see!

@silent plank fixed?

[silent plank]

couldn't be bothered checking every single one but looks like you got the idea

[hallow crystal]

couldn't be bothered checking every single one but looks like you got the idea
@silent plank
Noo I meant just 37 cuz that’s the only one I changed

Ohh or do u mean u never checked all of em to start with when I first sent it? Cuz that’s understandable

[silent plank]

yeh

[hallow crystal]

yeh
@silent plank
How am I supposed to do this

[silent plank]

determine whether the sum of the squares of the 2 shorter sides is less than, greater than or equal to the square of the longest side

[hallow crystal]

Oh I know that, but how does he expect us to do this without a calculator

[silent plank]

calculating the squares of these values is pretty simple

[quasi dove]

yup

[hallow crystal]

WHOOPS

my dumbass was tryna find 3 sqrt 2 😭

Aight @silent plank last one for the day could u check these 😄

Oh actually I didn’t understand number 2 😔

[silent plank]

parentheses issue again with Q1

[hallow crystal]

Damn it

[silent plank]

you didn't expand Q3 properly

2x*2x is 4x^2

[hallow crystal]

Ahh I always mess up on multiplying x

How do I do number 2?

[silent plank]

rearrange to get all terms on one side

factorise or use alternate method like QF

[hallow crystal]

But then it’ll become three terms on side instead of two since there’s no like terms to combine

[silent plank]

you'll have a quadratic equation

[hallow crystal]

oo I see

[silent plank]

as for Q7, you mixed up your identities

[hallow crystal]

Acute and Obtuse?

@silent plank

[silent plank]

yeh,
first one is obtuse
second is acute

[hallow crystal]

Are my signs right?

[silent plank]

which signs

[hallow crystal]

Greater than less than

[silent plank]

yes

the conclusion was wrong

[hallow crystal]

I thought less than meant acute because smaller angle?

[silent plank]

look up some visualisations online

[hallow crystal]

Ohhh it depends on C not a^2+b^2

Understood

No factors of 1 add up to -2 >:(

For number 2 ^

[silent plank]

you have the factorisation right there

[hallow crystal]

Isn’t it wrong

[silent plank]

(x-1)(x-1) = 0

and the only solution to that is x=1

[hallow crystal]

-1 and -1 don’t = 1 😔

I need to find two factors of 1 that equal -2

OH WAIT

Ahhhhh

x = 1 alright

Everything fixed?

[iron shadow]

Hi, could anyone work through these questions with me?

[quasi dove]

about the first question, you have to express y as 5-3x and put it into the first equation

about the second one, what about using graphs?

[whole carbon]

<@&286206848099549185> Can someone help me with Quantitative reasoning

[dark sparrow]

@whole carbon you should first post your problem and only then ping helpers if 15 minutes went by with nobody responding

[whole carbon]

Oh @dark sparrow

remember me

[dark sparrow]

please tell me you're not coming back with the same problem sheet as last time

[whole carbon]

LOL

no its kinda same but new one

I need help with green only

[dark sparrow]

this formatting is making me cry

i am not willing to do this

[whole carbon]

I need help badly

Please ann you helped me a lot last time

[upper karma]

quality is low

Can you zoom in on the column then post screencap again

[whole carbon]

This is all I got , not sure if correct

[upper karma]

Do you understand boolean algebra well

This is pretty easy

[whole carbon]

nope

Idk what boolean is

[upper karma]

Mathematical logic

[whole carbon]

is the middle part correct?

Its just T or F

I gotta fill em out for each row

[upper karma]

The middle column is correct

[whole carbon]

would left go

F

F

F

T

[upper karma]

Also this doesn't belong in #geometry-and-trigonometry, it should be in #proofs-and-logic

[whole carbon]

Oh ok

Not sure where it went tbh

Is the left correct?

[upper karma]

What is your answer?

[whole carbon]

F

F

F

T

[upper karma]

Yes

[whole carbon]

Ok

Idk about the last one

Is it T F T T @upper karma

?

[upper karma]

The answer to the third column? Or the truth table for 'implies'?

[whole carbon]

3rd

[upper karma]

That would be wrong

[whole carbon]

What is it then

[upper karma]

T
T
T
T

So it is a tautology

[whole carbon]

Its all T?

[upper karma]

Yes

[coral dagger]

How do I solve for the horizontal shift?

Can I assume

that y=0 then solve for c?

Or is there an alternative way to do it

Wait thats to find a sine function to find a cosine, can I assume the highest point on a graph which is 7:12 or 7.2

[dark sparrow]

what do you mean by swap

i don't know what you're talking about.

is there a problem you're doing? this could clear up what you're trying to do.

this does not help

what do you mean by "some calculus"? why are you refusing to post the full problem so i can figure out exactly what you're trying to do and how to do it?

oh boy, sideways

so you want to find x in quadrant 2 such that cot(x) + 2 = -tan(x)?

[coral dagger]

Is my V Window too big if I’m seeing a simple straight line?

I’ve double checked my trig functions

And they are all in radians

h(t)=2cos(pi/372(t+7.2)+9

[dark sparrow]

what's the Y range of your viewing window

[coral dagger]

what's the Y range of your viewing window
@dark sparrow -10 to 10

[dark sparrow]

yeah too big

the range of your function is [9,11] so a window which doesnt go too far beyond that would be better

say, [5, 15]?

[coral dagger]

ok lemme try

[dark sparrow]

hold up

[coral dagger]

it’s somehow still a straight line

[dark sparrow]

oh lmfao

your x window is too short

[silent plank]

period is mega huge

[dark sparrow]

try making it like 0 to 600 or something

[coral dagger]

ok

Ohhh

Ok

Thanks!

I made it

0-2000 (that pic isnt a represensative of 0-2000)

[silent plank]

might be limitations of the calc

[upper karma]

If I rotate a line around its normal line, in a 3d space, do I get a plane?

[paper vale]

Yea

But u can’t say “it’s normal” as there are >1 lines (♾️) which are normal to it

[upper karma]

But if I was given 2 of them which are normal that is a plane right

[paper vale]

Yea

[upper karma]

Great thanks

[supple onyx]

So I came acroos a geometry question where the basic setup is that you have 2 skew lines, both given in vector form, and a point M. You need to find a line that intersects the 2 lines and passes through point M. I struggled to find a simple method (I mean really simple, the method I used is still not particularly hard but I ended up with a peculiar system of equations where my calculator just couldnt figure it out if I didnt first solve for one of the variables and then plug back in so that it was a system of only 2 variables). Is there a simple method for these types of questions, or do I have to end up solving a system of 3 non linear equations (again not particularly hard but I wonder).

Basically my way to solve it was to take a general point on line 1 and line 2, then write the vector equation for the line you want to find as M + t(M-P) where P is vector for a geenral point on line 1. Then equate that with H a general point on line 2 and solve the system of equations for the three parameters (and in the end you only have to care about the solution to one of the parameters)

[upper karma]

I need to review my coordinate geo, but here's what I can think of

Take any general point on the line 1 as $\overline{r_1} = \overline{a_1} + \lambda\overline{b_1}$
And on line 2 as $\overline{r_2} = \overline{a_2} + \lambda\overline{b_2}$
Now, on substracting,
$\overline{r_1} - \overline{r_2} = (\overline{a_1} - \overline{a_2}) + \lambda(\overline{b_1}-\overline{b_2})$

[somber coyote]

livvney:

[upper karma]

This would give you a line intersecting through both the given lines,

And any general point on it would be given by $\overline{r_1} - \overline{r_2}$, which, you can just consider as $\overline{r_3}$

[somber coyote]

livvney:

[upper karma]

Equate r_3 with the vector equation for point M and solve to get lambda

[zinc dragon]

Is this channel free?

[upper karma]

apparently

The person hasn't responded yet so

[mental vessel]

This doesn’t make sense to me, shouldn’t the side that says 189 be the longest side?

[high shell]

not at all

for example, if they traveled in one direction, they could have travelled any number of kilometers and still have 0 distance between them 🙂

you're probably just confused because the triangle you drew makes it look like 189 is the longest side 😛

this is a task about the cosine theorem, basically.

[mental vessel]

oh okay i think im hanging onto the right angle rules too much lol

thanks

[zinc dragon]

https://cdn.discordapp.com/attachments/409596215697866773/760130369675591720/IMG_20200928_185445.jpg

A hint would be good enough.

[dark sparrow]

8 or 9?

[upper karma]

triangular inequality

[dark sparrow]

@zinc dragon the constraint not mentioned explicitly is c < a+b

[zinc dragon]

So you're claiming that the question is incomplete?

[dark sparrow]

no, it's not incomplete

the constraint is just the strict triangle inequality

i'm guessing you're looking for some strategy other than brute-force counting

[upper karma]

it follows that ||c < 10||

[zinc dragon]

So c<10?

[upper karma]

correct

that's what I said in the spoiler thingy

ok this sounds bad

[zinc dragon]

But for a, b: are there any specific constraints that fix a type of values for them?

[upper karma]

a<=b<=c

[zinc dragon]

Ohh yeah, ig i see what you're talking about hobo

[upper karma]

you could even brute force this

c = 9: a = ....

[zinc dragon]

Yepp💯

[mental vessel]

I know I did something wrong

sorry, go first sidharth

[upper karma]

I think he's done (?)

[mental vessel]

Yeah I thought so I waited a couple minutes to make sure lol

[zinc dragon]

I just have one doubt. So (a, b, c) cannot be (9, 1, 10) because c<a+b so if i choose (8, 1, 10), there aren't any other constraints holding the formation of a triangle, are there?

[upper karma]

remember a + b + c = 20

you got 19 there

there are actually very few triangles

[zinc dragon]

Okay got it.

[upper karma]

how many?

[zinc dragon]

I haven't calculated yet. Wait.

8

@mental vessel i am sorry for holding you. This channel is all yours

[mental vessel]

no its okay

thank you

dont mind my triangle cause its a little weird compared to the question

but i know the angle cant be 50.6 degrees its over 100, what did i do wrong?

[upper karma]

Why do you think it's over 100°

@mental vessel

[zinc dragon]

244 is the largest side and the angle opposite to should be the largest(unlike your diagram)

I see no mistake in your solution.

[mental vessel]

ohh

so the angle should be up here

[upper karma]

No

Redraw the diagram

[mental vessel]

in the question it says the two airplanes are 189km away from eachother, which is what i have

one airplane goes 168km and the other goes 244km

[silent plank]

244 is the largest side and the angle opposite to should be the largest(unlike your diagram)

i.e. your diagram isn't scaled properly

[upper karma]

Your solution is correct, it is 50.6° but diagram is incorrect

[mental vessel]

So like that then

then it asks the angle between the two flight paths, which would be opposite to 189 right

[upper karma]

Yes

[mental vessel]

okay cool

thanks

[upper karma]

how would i find a quadratic function when given the x intercept of x= -5/8 ? yes its only one x-int

[dark sparrow]

is it known that that's the only x-intercept?

or is there another one at an unknown location

[upper karma]

yes

exactly one x intercept

which im pretty sure means its a perfect square

[dark sparrow]

and you're absolutely right.

the equation will take the form $y = a(x + \tfrac{5}{8})^2$, where $a$ is either arbitrary or to be determined by other info

[somber coyote]

Ann:

[iron shadow]

Can someone explain to me this proof
[18:42]

[hallow crystal]

After someone’s answered the person above me’s question, could someone check my work and tell me if this is right?

[pliant osprey]

@hallow crystal didn't check the entirety (looks okay though) but yes, khun

[iron shadow]

<@&286206848099549185>

[fresh pike]

what would question 2 look like compared to quesiton 1? I'm not sure how to differentiate a-b from a+b or how to show it with a compass and a line

[silent plank]

what did you do for Q1

place your arc in the other direction

[fresh pike]

on b?

[fresh pike]

ohhhh i see

and for this one?

does c go in the other direction too?

[silent plank]

yes because subtraction

[fresh pike]

i get it now

thx

does the dot go on the arc on (a-b)?

can you show me what a+2b-c would look like?

[silent plank]

if you understand 1 and 2, combine the two ideas

[iron shadow]

@silent plank could you answer my question?@pliant osprey

[silent plank]

which part of the proof don't you undertand?

[iron shadow]

the explanation

the part that says these are equal

[pliant osprey]

you've counted the area of the big square in two ways

[silent plank]

the line before that says
we're going to write the area of the larger square in two different ways

1. is one way (using the sides of the large square)
2. is the other way (using the smaller square and the 4 triangles)

those both represent the area of the larger square

and would have the same value

[iron shadow]

how come they are using the ways to equate then simplify to get the theoreom?

2 ways

i know they mean same thing

[silent plank]

because that's the approach they're using

[vernal dawn]

Given the directed line segment PR below, find the coordinates of q on pr such that the ratio of pq to qr is 2:1. P being -2,5 and R being 1,-7

[silent plank]

to prove pythagoras,

that a^2 + b^2 = c^2

[vernal dawn]

whats that

[silent plank]

to get a relation between the variables a,b and c

[iron shadow]

ok but using area to find an equation that only gives you answers to missing sides doesnt make sense

[silent plank]

wdym

why wouldn't it make sense

i mean its all outlined

[iron shadow]

areas of all squares triangles in both ways are used then equated to simplify but the simplified equation is pythagoras thereom however, the theorem when used gives answers to mising sides

[silent plank]

but what's the issue with that exactly?

and why is that an issue

[iron shadow]

how does maths go from areas to sides?

[silent plank]

its not only about area

its about the relationship of your variables

[hallow crystal]

@hallow crystal didn't check the entirety (looks okay though) but yes, khun
@pliant osprey

[silent plank]

1. gives you information about a and b (in relation to Area)
2. gives you more information about a and b as well as c (in relation to Area)
   and together you get a relation between those 3 variables

[vernal dawn]

Given the directed line segment PR below, find the coordinates of q on pr such that the ratio of pq to qr is 2:1. P being -2,5 and R being 1,-7

[hallow crystal]

Jinnnnnnnnnnnn

Can I squish 🥺🥺🤲🏼🤲🏼

[silent plank]

i mean you agree with all the math right...

[hallow crystal]

I’m sorry I love everyone’s pfps-

[silent plank]

a^2 + b^2 = c^2
is the logical conclusion to everything set up before that

[iron shadow]

ahhh is a matter of cancelling the 2ab giving you equation to find all sides only

[hallow crystal]

Is there a symbol for midpoint? Do I rlly gotta write it all out

[iron shadow]

@silent plank

[silent plank]

well yeh...

[iron shadow]

how do i ask questions and get a reply quicker btw@silent plank

[silent plank]

be explicitly clear upfront about which parts you need help with

[hallow crystal]

LOL professional tutoring ❤️

[iron shadow]

ok, just asking im quite new to server 🙂

[silent plank]

M for midpoint

[vernal dawn]

@silent plank can you help me

[silent plank]

subscripts to indicate which segment

[hallow crystal]

M for midpoint
@silent plank
OHHH I was wondering why the formula has Xm and Ym instead of X1 and Y1

[silent plank]

should be capital M

[hallow crystal]

XM and YM?

Ohh

So I would write the answer as

---

HX M

Ahh wait no

[silent plank]

Given the directed line segment PR below, find the coordinates of q on pr such that the ratio of pq to qr is 2:1. P being -2,5 and R being 1,-7
parentheses around points, math is also case sensitive, keep labelling consistent
have you drawn a diagram?

[vernal dawn]

P is (-2,5) and R is (1,-7)

i have to find q on the line pr

The ratio of pq to qr is 2:1

[silent plank]

the change in the x and y coordinates will also be in the same ratio

Q on the line PR,
RQ to QR

[modern sorrel]

How do you factor a five term polynomial

[high shell]

well, you can guess 3 roots and get left with a quadratic

Why do you need to, though?

[modern sorrel]

my teacher doesnt really like us so hes torturing us with 5 term polynomials

i'm just tryna study for the test tmrw that this will most likely be on

[silent plank]

rat root theorem

[high shell]

oh, yeah, I remember that

don't think I ever used that except for the test that was for 😅

[modern sorrel]

thank you boys

[hallow crystal]

Why do u have to subtract point A’s coordinates from the midpoint’s coordinates in order to find point B’s coordinates?

I know you need to subtract because the original midpoint formula is adding, but shouldn’t it be vice versa?

Shouldn’t you subtract the midpoint from Point A instead

In order to find Point B

[hallow crystal]

<@&286206848099549185> :

[upper karma]

Wut?

[quasi dove]

A(a) B(b) M($\frac{a+b}{2}$),
$b = \frac{a+b}{2}*2-a$

[somber coyote]

ThisIsMyName:

[onyx tapir]

Hello I need help with 2- d)

Here's what I've worked on

<@&286206848099549185>

[upper karma]

3 D right

i dont see your work on 3 D

[onyx tapir]

@upper karma

My apologies, 2D *

[upper karma]

sin^2(x) cos^3(x) = cos(x) to be sure

right ?

sin(2x) can never be +/- 2 in the real domain

so only solutions for cosx = 0 count

[coral dagger]

I’m trying to “visualise” this question

I’m unsure if my depiction of it is correct

[coral dagger]

How do I find the mininum value

for question c

Im trying to find the vertical shift for my cos function

Vertical shift = Max+Min/2

Max-Min /2 was equal to = 25.4

Oh nvm

[marble oriole]

Hi!

I need help with intersection of bounding boxes

Does someone happen to know how to calculate the intersection of rotated bounding boxes?

[meager pendant]

find angle aoc

[upper karma]

what have you tried

hint: ||consider inscribed angle theorem||

@meager pendant

[meager pendant]

im trying to figure out what the shared arc is

i think i can use opposite angles of quadrilateral

if i construct cb

[upper karma]

do you know inscribed angle theorem?

[meager pendant]

this?

[upper karma]

i think i can use opposite angles of quadrilateral
i highly doubt you'll get anywhere with that if you don't use the inscribed angle theorem

yes

[meager pendant]

I got 150

[upper karma]

hold up

[meager pendant]

using the opposite angles of quad and the inscribed angle theorm

[upper karma]

yes 150 is correct

good job

[meager pendant]

nice 😎

thanks for the help

[upper karma]

yw

[narrow crag]

is this shit trig or geo

cuz im taking geo

and this is scary

[meager pendant]

It's circle geometry

trigonometry is about triangles but i think it falls under geometry ( triangle geometry?)

[narrow crag]

oh ok

im starting geo this year and im kinda worried

looks hard

[meager pendant]

the proofs aren't too hard to learn/ understand

[upper karma]

just need to practice and get used to it

[meager pendant]

its just sometimes its diffficult to see how they can be applied in the question

its pretty fun tho

[upper karma]

Yeah

[meager pendant]

like brain riddle

[narrow crag]

r proofs the thing with postulates

[meager pendant]

creative thinking or whatever

yeah they are

[tardy nest]

Sina=-5/13

How can one side of the triangle be 12 is it correct?

For 4,1

Never mind I figured it out

[muted hollow]

Hi i have a geometry question
Can someone help please?
Based on this

https://cdn.discordapp.com/attachments/698664724426129418/760534132001734716/unknown.png

I have 10 questions

1. What numbers are corresponding to 3?
2. What two numbers are alternate interior to angle 4?
3. Which two numbers are alternate exterior to 10?
4. What number is vertical to 14?
5. Angle 11 is alternate exterior to what number, and alternate interior to what number?
6. Angle 13 is alternate exterior to what number, and alternate interior to what number?
7. Angle 5 is alternate exterior to what number, and alternate interior to what number?
8. Angle 1 is alternate exterior to what 2 numbers?
9. Angle 8 is alternate exterior to what 2 numbers?
10. Angle 3 is alternate interior to what 2 numbers?
    I need help with the alternate interior and exterior angle questions
    I did all of them but i still dont get it and dont know if they are right

[upper karma]

@muted hollow can i see your answers to them in order to check if they are correct

[muted hollow]

Ok hold on

A little sloppy

[upper karma]

On 1 if you think that 7 and 2 are corresponding to 3, what stops you from 6

[muted hollow]

You cant do diagonally

Its only downa nd across

Down and across*

That was my teachers answer

[upper karma]

Wait what? I'm assuming the 4 lines are parallel right?

[muted hollow]

Yes but he said its only downa nd across

Idk

Down and across*

[upper karma]

Oh wait

Hold up

[muted hollow]

Just its right xd

[upper karma]

I may have missread it

[muted hollow]

Oh ok

I mainly need help with the alternate interior and exterior

I have no idea how to do it on a 2 by 2 grid thing

I know how to do it on a 2 by 1 though

Just not a 2 by 2

[upper karma]

2 seems correct

3 is too correct

4 as well

5 as well

6 as well

is the 7) a 1 on exterior?

[muted hollow]

Yes its a 1

[upper karma]

and the rest is okay

good job

[muted hollow]

Oh really?

[upper karma]

yeah i couldn't find any mistake at all

[muted hollow]

I still dont understand fully when the number srae interior and exterior on a 2 by 2 like that

Could you explain?

[upper karma]

wdym by 2 by 2

[muted hollow]

Here hold on

This is a 2 by 1

This is a 2 by 2

[upper karma]

oh okay so where are you confused on 2 by 2

you did all of them right

[muted hollow]

Yeah but like

Wait give me a minute

[upper karma]

sure

[muted hollow]

See look

Wait

For exaplme on number 8

Example*

OH WAIT

I think i get it

Its asking angle 8 is alternate exterior to what 2 numbers

But 5 is on the interior

But 8 is the alternate to 5

And 8 is on the exterior

I think i get it

[upper karma]

what i usually do when alternate exterior or interior angles is to like forget the other line we are not gonna use, and focus on the 2 by 1

and then change lines

to get the other one

that might be useful

[muted hollow]

Yeah i get what you mean but like

I feel like the exterior and interior changes when it gets to a 2 by 2

But it r3eally doesnt

Its confusing

Really*

[upper karma]

practice and practice and the more problems you do the less it'll appear confusing

especially in maths

[muted hollow]

Yeah but i am just nervous that the teacher will mark them wrong as a homework assignment

[upper karma]

the more situations you've come upon before the exam, the more experience you'll have when dealing with the test

[muted hollow]

Thats why i have to make sure they are all right

Yeah exactly

[upper karma]

Yeah but i am just nervous that the teacher will mark them wrong as a homework assignment
they are all correct lol

believe in yourself

[muted hollow]

Wait what grade are you in?

College?

Or?

[upper karma]

is that really necessary to ask?

[muted hollow]

Im just curious...

Thats all

[upper karma]

last year before uni

idk about the grade or whatever

[muted hollow]

Im only 14 turning 15 on october 13th

[upper karma]

nice!

[muted hollow]

What about you? Or are you comfortable telling?

Prob around 18?

Or higher?

[upper karma]

around 18 is a good guess

[muted hollow]

Ah ok

[upper karma]

hey uh

does anyone know the answer to this

[muted hollow]

Well nice talking to you thanks a lot~

!

[upper karma]

same, yw!

@upper karma is that an online test

[bitter jetty]

for triangle abc prove that (a+-b)/c = sina =- sinb/sinc

how do i prove that

ive been at it for like an hr with nothing

[mental vessel]

I got a=5.7cm
im not sure how to get the other lengths for the triangle BCD tho

[eager hatch]

Hi so I’m in geometry and me and my friend got different answers on a problem, with his being right. Can anyone explain where I went wrong/why it’s wrong?

Because they got E being 60 degrees with x being 4 and y being 10

[silent plank]

,w 10x+3y +4y -7x +8= 90, 5x+4y = 3(4y-7x+8)

[eager hatch]

?

[silent plank]

mistake the start

your very first equation at top left

[eager hatch]

Oh wait it’s 7y right?

[silent plank]

yes

[eager hatch]

Lol dumb mistake thanks for the help

And that’s it right?

@silent plank it’s just that right? From there is it all correct

[silent plank]

what?

if the mistake is at the very first line

[eager hatch]

Like if I change that and proceed with said changes it’s correct right

[silent plank]

it pretty much renders everything after irrelevant

[eager hatch]

Like adjust based on first equation changes

[silent plank]

redo the whole thing

[eager hatch]

Ok but like for the systems that’s right?

Like original 2 equations

Sorry I’m kinda tired today

So I sound really dumb lol

[silent plank]

the other eqn looks ok

[eager hatch]

I’m sorry I sound so dumb

[vernal dawn]

guys

can someone help me with something really easy

i dont know how to do it though

[upper karma]

!r1

[silent plank]

we won't be able to either

[upper karma]

rule 1, don't ask if you can get help, just post it. #❓how-to-get-help

[bitter jetty]

for triangle abc prove that (a+-b)/c = sina =- sinb/sinc
how do i prove that
ive been at it for like an hr with nothing

[strange beacon]

Hey, can anyone give me any tips with the following problem: You have a DIN A4 sheet of paper from which you want to make the largest possible sugar sachet. Make a suggestion on how to make such a sugar cube - also give the dimensions of the sugar cube

[upper karma]

Would this just be 24/(sqrt3/2)

[silent plank]

(a+-b)/c = sina =- sinb/sinc
something looks off about that.

no, where are you getting sqrt3/2

[upper karma]

That side of the is sqrt3/2 and the hypotenuse is 1

[silent plank]

that doesn't tell me how you're getting that specific value though

[upper karma]

Ok so sin58=x/24

Wait no no

24/x

[silent plank]

°, but that would be ok

[upper karma]

But how would I isolate the variable

[silent plank]

algebra

same operations to both sides etc

[upper karma]

So sin58=24/x

[silent plank]

where are you suddently getting cos

[upper karma]

Oh sorry

[silent plank]

note that my main question was specifically where the sqrt(3)/2 is coming from

[upper karma]

In a 30/60/90 triangle isn't the leg sqrt3/2

[silent plank]

ratio of the leg opp 60° and the hyp would be sqrt(3)/2

but in what way is this a 30°-60°-90° triangle

[upper karma]

Yes thats what I meant

Oh

Its not i just realized

[twilit steeple]

can someone link me a review page or packet for geometry i need to brush up

[upper karma]

I think I need a brain

[upper karma]

Ok wouldn't this be 19/sin(63)

[silent plank]

nope

check how you're setting up your trig ratio

you're saying that the ladder is touch the building at a place higher than the ladder itself

[upper karma]

Oh oh

Its 19sin(63)

[bitter jetty]

for triangle abc prove that (a+-b)/c = (sinA +-sinB)/sinC
how do i prove that
ive been at it for like an hr with nothing

[silent plank]

@bitter jetty something looks very off about that statement

[bitter jetty]

o sry

@silent plank i fixed it lol

[silent plank]

still looks off

[bitter jetty]

how

thats the problem i got

[silent plank]

math is case sensitive

[bitter jetty]

o ic

ok

now is it gud

for triangle abc prove that (a+-b)/c = (sinA +-sinB)/sinC

can u help

@silent plank

[silent plank]

start with the sine rule

$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$

[somber coyote]

ramonov:

[bitter jetty]

ye

ik that

[strange beacon]

one question

Does it mean that the volume of a pyramid for infinity n is infinity=

[silent plank]

using that can get you the equations: \
$\frac{a}{c} = \frac{\sin(A)}{\sin(C)} \
\frac{b}{c} = \frac{\sin(B)}{\sin(C)}$

[strange beacon]

Because I thought it would become a cone

[somber coyote]

ramonov:

[bitter jetty]

yes

[silent plank]

and then add/subtract them

[strange beacon]

how can anything have an infinity volume

[bitter jetty]

wdym

o add them together?

[silent plank]

yes

[strange beacon]

this for n to infinity

[silent plank]

you'd be adding or subtracting the same value to/from each side

[strange beacon]

yea but for infinite n the volume of a pyramide is infinity isnt it

but in this formula I don't get how a changes for more n's

because a has to become shorter if we add more sides to a pyramid

[hallow crystal]

For number 16 which one is right?

<@&286206848099549185>

[bitter jetty]

i got to sinA = sin(180-B+C)

and that a/c = sinA/sinC

but im stuck again

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[silent plank]

sinA = sin(180-B+C)
not quite

[bitter jetty]

ya

@silent plank thats y i was trippin

and i feel thats the issue in my proof

[silent plank]

insufficient parentheses

[bitter jetty]

@silent plank cud u expand

on that

[silent plank]

what's the relationship between A,B,C in a triangle

[bitter jetty]

o its sin(180-(B+C))

they all add up to 180

[silent plank]

yes,better

and then apply properties of the sine function

[bitter jetty]

🤦‍♂️ ur a genius

@silent plank tysm

[modern sorrel]

What is a discontinuity im so confused

[vague crypt]

$2y + (4x - 100) = 3y + 30 + (x + 5)$

[somber coyote]

LifeSource:

Sidharth:

[zinc dragon]

@modern sorrel

[vague crypt]

$-3y - 6x = -420 \qquad 3y + x = 145$

[somber coyote]

LifeSource:

[vague crypt]

$-5x = -275$

[somber coyote]

LifeSource:

[vague crypt]

$5x = 275$

[somber coyote]

LifeSource:

[vague crypt]

$x = 55$

[somber coyote]

LifeSource:

[vague crypt]

$55 + 3y = 145 \implies 3y = 90 \implies y = 30$

[somber coyote]

LifeSource:

[upper karma]

Kinda need help ;-;

[vague crypt]

$(5y + 38) = (8x + 26) \qquad 3x + (5y + 38) = 180$

[somber coyote]

LifeSource:

[junior light]

Uh where's $\vec{AR}$?

[somber coyote]

TedNowKaczynski:

[junior light]

Could you provide the complete image?

[coral dagger]

I’m trying to find the horizontal phase shift and I attempted to do this

By plugging in my Y as 36 and x as 1

And I have the other variables and what they mean

Though I hit a dead road

Because I got “syntax error” when I try doing

-1^-sin-1 to get rid of the sin

[dark sparrow]

-1^-sin-1
what.

your calculator is going "what the fuck are you trying to say" at you

[coral dagger]

Like I want to remove sin

but I got that response

[dark sparrow]

that would be sin^-1(-1**)** = sin^-1(sin(pi/6 + c))

you APPLY sin^-1 to both sides

not raise it to the power of a function

[coral dagger]

oh

When I mean by

sin^-1 I meant like

I applied -1 first then sin^-1

like this

i thought that’s how we do it lol I guess I was wrong

ok now I got

2.094=C as my horizontal shift

[onyx tapir]

Hello.

Is there anything wrong with this intuition: to find the angle a vector makes with a particular axis, find the dot product between the vector without that axes component, with the unit vector in that given axis.

The reason I'm asking this is since my book gives these formulas above which was not what I was thinking.

[upper karma]

$\bar{r} \cdot \bar{i} = \lvert r \rvert \lvert i \rvert cos(\alpha)$

$cos(\alpha) = (x\bar{i}+y\bar{j}+z\bar{k})\cdot\bar{i}/\lvert r \rvert$

So the angle $\alpha = cos^{-1}(\lvert x \rvert)/\lvert r \rvert)$

[somber coyote]

livvney:

[upper karma]

Similarly you can find the other angles

[jolly bear]

Help with number 6???

[left storm]

I need help with some geomotry

Those 2

[upper karma]

@jolly bear make the vector representation of the situation and solve accordingly

@left storm what have you attempted?

@left storm for 11, you only need to know the distance formulae, for 12 you just have to solve the linear eqns

[coral dagger]

Without solving for C using algebra, is there any ways I can find the horizontal shift

by simply looking at the graph

Im just wondering

https://gyazo.com/6579bd4bdefccfd28f9127b71e6e0b5e?token=ae52e44e957c60e7aae485c8f9abe99c

by simply observing the sine function's picture

[coral dagger]

Wait can anyone here help check my calculations

something went off

Because Im supposed to have a horizontal shift of -2.17 according to my graphing calculator

though when I plugged in a month and temperature (the first month and temperature) to solve for C

I get -4

Moved to #help-9

[upper karma]

hi

<@&286206848099549185>

[paper vale]

it is trivial

[paper vale]

bruh why did u take a picture of your laptop from your phone

[upper karma]

1. distributive prop, addition prop, substitution prop
2. addition prop, addition prop, divison prop

shush gimma a break

LOL

[paper vale]

what are u asking

[upper karma]

if my answers were right

[paper vale]

isnt that just trivial though

[upper karma]

wdym

its proofs

and im stuck

😔✊

i think im right but idk forsure

[shrewd bay]

Can two points be collinear?

Or does it have to be 3 or more?

[remote mortar]

does a unit on a coordinate plane mean 1 box on it or like 1 of the things that the number on the sides represent because like 1 box doesn’t mean one when ur graphing but if the question says 1 unit = 1 inch are they reffering to one box on the plane or not

[upper karma]

Can two points be collinear?
@shrewd bay what do you think

[mental vessel]

Just checking if I did this correct ?

[opal forge]

Can someone ease explain where 1600 is coming from out this equation. I'm considering my options here, I'm doing 9-1 and 9+1 which are equaling 8 and 10. But 8^2 is 64 and 10^ is 100

What should I be doing differently

[upper karma]

They're simplifying 1 - (9/41)^2 right? And you might know 1 - (9/41)^2 = 1 - 9^2/41^2. Turn 1 into something so the expression can be written as a single fraction and you'll have answered your own question :)

[opal forge]

@upper karma I think I might be a little slow. So 1 - 9^2/41^2 = 80/1681?

You have to write all numerators above the common denominator

I get it now!

thank you

[upper karma]

Err 1 - 9^2/41^2 = 1600/1681 but glad it helped

[opal forge]

Lol yeah so it's 1681-81 / 1681 = 1600 / 1681

[upper karma]

Great

[shrewd bay]

Can two points be collinear?

The definition I was giving said collinear has to be three or more points.

But isn’t collinear points on the same line?

So wouldn’t two points be on the same line?

[hollow raven]

Considering there needs to be a set of points on the same line for them to be colinear, yes, your observation would be correct

[shrewd bay]

I’m not sure why the definition from my teacher says three points.

[flint crypt]

can anyone help me with my hmw

[upper karma]

So we just talk here for geo?

@upper karma yeah, there are other things too like pre calc and pre algerbra

Yeah I saw that

oh mb

It’s alr💀

so your new here?

you're*

Yeah I joined like a little while ago

that's cool, i joined a few days ago so, im basically new too.

what grade are you in

https://tenor.com/view/umm-confused-blinking-okay-white-guy-blinking-gif-7513882

sorry that sounded kinda creepy. im just wondering bc i wanna know what grade ur taking this class in. yk what i
mean

Lmao yeah

I’m a freshman.

oh me too

i took the algebra regents in 8th. Thats my nyc curriculum though.

That’s lit. I’m from ny too

thats crazy

[high geode]

anybody know the exact value of sin 16pi/3?

[upper karma]

It can be written as sin(5pi + pi/3)

you get it now ig

[high geode]

i found the answer is squar3/2 I just don't know why

[upper karma]

It's -√3/2

[high geode]

yup that's what i ment

[upper karma]

sine is negative in third quadrant

[high geode]

right

[upper karma]

5pi is just 4pi + pi i.e 2 full rotations then a pi

Add pi/3 to it , it brings it to third quadrant

[high geode]

thanks

[onyx tapir]

can someone please explain the reasoning behind why this is the case

if an angle is in the second quadrant.

I don't get these relationships at all

all i know is arcsin(X) + arccos(x) = pi/2

[upper karma]

Do you want a proof of why arcsin(x)+arccos(x)=pi/2? @onyx tapir

[onyx tapir]

@upper karma Yes please

Also, the reason why and how you can create a relationship between arccos( - value ) and arcsin (+ value)

[upper karma]

okay

hold up

@onyx tapir have you learnt differentiation?

[onyx tapir]

@upper karma Yes I have.

apologies for taking some time to reply as i am in a class

when you can respond, i'll check it out and let you know

[upper karma]

okay, i'll maybe go through diffentiation method then

[onyx tapir]

@upper karma yes 🙂

[upper karma]

I've been fighting the bot to make it work this time

Gimme some more time

[onyx tapir]

Take your time

[upper karma]

@onyx tapir done

You ready?

[onyx tapir]

Yes I am

[upper karma]

Let's start by drawing a triangle as such $\$ $\begin{tikzpicture}
\draw (-4,0) node[right=50pt, below=5pt] {b} -- (0,0);
\draw (0,0) node[above=40pt, right=5pt] {h} -- (0,3);
\draw (0,3) -- (-4,0) node[above right=50pt] {1};
\draw (0,0) -- (-0.5, 0);
\draw (-0.5,0) -- (-0.5, 0.5);
\draw (-0.5,0.5) -- (0, 0.5);
\draw (0,0.5) -- (0, 0);
\filldraw[fill=green!20!white!, draw=green!50!black] (-4,0) -- (-3.4,0) arc [start angle=0,end angle=38, radius=0.6];
\filldraw[fill=green!20!white!, draw=green!50!black] (0,3) -- (0,2.4) arc [start angle=270,end angle=218, radius=0.6];
\node at (-0.4,2.2) {$\alpha$};
\node at (-3, 0.3) {$\beta$};
\end{tikzpicture}$
$\$ Now let's do a little bit of trigonometry, $$\sin(\alpha)=\frac{b}{1}=h$$ $$\sin(\beta)=\frac{h}{1}=h$$ $$\cos(\alpha)=h$$ $$\cos(\beta)=b$$ now applying pythagoras $$b²+h²=1²$$ $$b=\sqrt{1-h²}$$ so now we will change the trig statements we did before to make all of them in terms of h. $$\sin(\alpha)=\frac{b}{1}=b=\sqrt{1-h²}$$ $$\sin(\beta)=\frac{h}{1}=h$$ $$\cos(\alpha)=h$$ $$\cos(\beta)=b=\sqrt{1-h²}$$ and now pulling out knowledge of $\sin(\alpha+\beta)$ $$\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$$ and from what we got before $$\sqrt{1-h²}\sqrt{1-h²}+h²=(1-h²)+h²=1$$ hence $$\sin(\alpha+\beta)=1$$ now the final move, we know by the unit circle that $$\sin({\color{green}{\frac{π}{2}}})=1$$ $$\sin({\color{green}{\alpha+\beta}})=1$$ hence, $$\alpha+\beta=\frac{π}{2}$$ and finally by the trig rations we did at the very beginning $$\cos(\alpha)=h\implies \alpha=\arccos(h)$$ and $$\beta=\arcsin(h)$$ and subbing that back into $\alpha+\beta=\frac{π}{2}$ we get the desired proof $$\arccos(h)+\arcsin(h)=\frac{π}{2}$$ which works for either letter.

The drawing took me 90% of the time but there we go

Typo

[somber coyote]

Al𝟛dium:

[onyx tapir]

@upper karma Thank you so much!

Very clear explanation

I appreciate it heaps

[upper karma]

Yw!

[onyx tapir]

With regards to the example i sent above

why is that the case logically?

I know arccos(4/5) = pi/2 - arcsin(4/5)

From the proof you showcased

but not sure about this?

[upper karma]

@onyx tapir i will add the reason to that to the text so that you have it all jampacked

Let's start by drawing a triangle as such $\$ $\begin{tikzpicture}
\draw (-4,0) node[right=50pt, below=5pt] {b} -- (0,0);
\draw (0,0) node[above=40pt, right=5pt] {h} -- (0,3);
\draw (0,3) -- (-4,0) node[above right=50pt] {1};
\draw (0,0) -- (-0.5, 0);
\draw (-0.5,0) -- (-0.5, 0.5);
\draw (-0.5,0.5) -- (0, 0.5);
\draw (0,0.5) -- (0, 0);
\filldraw[fill=green!20!white!, draw=green!50!black] (-4,0) -- (-3.4,0) arc [start angle=0,end angle=38, radius=0.6];
\filldraw[fill=green!20!white!, draw=green!50!black] (0,3) -- (0,2.4) arc [start angle=270,end angle=218, radius=0.6];
\node at (-0.4,2.2) {$\alpha$};
\node at (-3, 0.3) {$\beta$};
\end{tikzpicture}$
$\$ Now let's do a little bit of trigonometry, $$\sin(\alpha)=\frac{b}{1}=h$$ $$\sin(\beta)=\frac{h}{1}=h$$ $$\cos(\alpha)=h$$ $$\cos(\beta)=b$$ now applying pythagoras $$b²+h²=1²$$ $$b=\sqrt{1-h²}$$ so now we will change the trig statements we did before to make all of them in terms of h. $$\sin(\alpha)=\frac{b}{1}=b=\sqrt{1-h²}$$ $$\sin(\beta)=\frac{h}{1}=h$$ $$\cos(\alpha)=h$$ $$\cos(\beta)=b=\sqrt{1-h²}$$ and now pulling out knowledge of $\sin(\alpha+\beta)$ $$\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$$ and from what we got before $$\sqrt{1-h²}\sqrt{1-h²}+h²=(1-h²)+h²=1$$ hence $$\sin(\alpha+\beta)=1$$ now the final move, we know by the unit circle that $$\sin({\color{green}{\frac{π}{2}}})=1$$ $$\sin({\color{green}{\alpha+\beta}})=1$$ hence, $$\alpha+\beta=\frac{π}{2}$$ and finally by the trig rations we did at the very beginning $$\cos(\alpha)=h\implies \alpha=\arccos(h)$$ and $$\beta=\arcsin(h)$$ and subbing that back into $\alpha+\beta=\frac{π}{2}$ we get the desired proof $$\arccos(h)+\arcsin(h)=\frac{π}{2}$$ which works for either letter.

And for your case, $$\arccos(-\frac45)=\frac{π}{2}+\arcsin(\frac45)$$ notice that arcsin is odd, which means that $f(-x)=-f(x)$, and using common sense $-(-\frac45)=\frac45$ : $$\arccos(-\frac45)=\frac{π}{2}+\arcsin(-(-\frac45))$$ using arcsin's parity $$\arccos(-\frac45)=\frac{π}{2}+(-\arcsin(-\frac45))$$ which ends up with the original identity $$\arccos(-\frac45)=\frac{π}{2}-\arcsin(-\frac45)$$ $$ \arccos(-\frac45)+\arcsin(-\frac45)=\frac{π}{2}$$

[somber coyote]

Al𝟛dium:

Al𝟛dium:

[upper karma]

@onyx tapir does all this clarify it?

[onyx tapir]

That makes so much sense

Thank you so much for your time

You said notice that sin (4/5) is odd

What does that mean?

[upper karma]

f is odd if f(-x) = - f(x)

so for example sin(x) is odd, in fact sin(-x)=-sin(x)

[onyx tapir]

I see

Thank you

[upper karma]

You said notice that sin (4/5) is odd
i did not say such, i said arcsin was odd

[pastel anchor]

does anyone know how to do sqrt(-2) *sqrt(-2)?

[junior light]

Uh you're looking for $(\sqrt{-2})^2=(i\sqrt{2})^2=2i^2=-2$

[somber coyote]

TedNowKaczynski:

[silent plank]

by definition of the square root, sqrt(x)*sqrt(x) = x

[pastel anchor]

where did the negative sign go?

it went from sqrt-2 to sqrt2i

[junior light]

Uh isn't that how you define i though?

sqrt(-1) is defined to be i

So sqrt(-2) is just sqrt(2)i

[upper karma]

you should have put i in front of the root

[junior light]

Umm it doesn't seem clear? I'll change it then.

[upper karma]

it looks like i is still in the root

it looks like i is still in the root
if you dont look closely

[somber coyote]

TedNowKaczynski:

[junior light]

Aight then

[pastel anchor]

ok ty bros does (sqrt-4)^3 equal to -8i or just 8i?

[upper karma]

-8i

[pliant dawn]

Square root and the square takes out eachother

No need to convert to imaginary figures

[upper karma]

You mean (√{-4} )^3 is a real number ?

[pliant dawn]

Nope but if you take the third root instead it will take out ^3 and leave you with - 4 🙂

TedNowKaczynski:
@somber coyote I meant this one when I made my statement before

[paper vale]

it is trivial

[pliant dawn]

Well maths is about to be as efficient as possible while sparing as much time as possible

[celest horizon]

How would I calculate the area of a spherical square?

so it's like 2 equal triangles, one angle is 2 times larger than other 2, and 2 sides are known

is that right?

hmm looks complicated https://math.stackexchange.com/questions/3847850/what-is-the-surface-area-of-a-plane-on-a-sphere-like-earth

[upper karma]

can someone tell me what to do

[silent plank]

apply properties of straight lines to set up some equations

[upper karma]

i did

then?

how shld i set up the system of equations

@silent plank

[silent plank]

wdym

what do you have now?

[upper karma]

what shld my equations be

8x=2x+4y?

or

4x+7y+2x+4y=180

idk i hvavent done this in ages

[silent plank]

both

[upper karma]

oh

[silent plank]

that will be your system

and then solve using sub or elimination

[upper karma]

omg ok

tysm

[upper karma]

help what do I do

What is like of symmetry?

and what are angles

I know one is E, but what do angles mean

[silent plank]

knowing what angles are is kinda manadatory for geo

look it up

[upper karma]

I think I get it now

thanks

[paper vale]

lol why are they using integers to denote a variable

that is silly

but yea just form an equation and then u can solve it

[hallow crystal]

LOL I have no clue what I’m doing

@paper vale

[paper vale]

rip do u know what an angle is lol

[hallow crystal]

I’m in geometry and my friend who Idk what class she’s in asked me to help her with this problem 😭 I don’t think we’ve done this unit yet in my class

rip do u know what an angle is lol
@paper vale
Did I get the 180° part right at least 😔

<@&286206848099549185>

[upper karma]

i cannot figure this out

is this a test

no it’s a warmup

uh ok

what have you tried

if that counts as a test idk

does it count as grade?

the warmup

like 5 points but only as participation

ig that's kind of in the limit of okay

what have you tried

@upper karma ...?

[pliant dawn]

ABC and CDE are two equilateral triangles. The two triangles have A as a common corner, see the figure below. Prove that AD=BE

[twin prawn]

Hint: compare triangles ADC and BEC with each other

[upper karma]

@pliant dawn

since it is equilateral, then each angle inside the triangle equals to 60°;
as they are equilateral, their sides follow a proportion. Then, it's a product between AC and a constant (k > 1, since CD > AC);
the angle opposite to AD is a + 60° and the angle opposite to BE is a+60°. By proportion (as you see in the image), they are the same triangle and BE = AD, as desired

maybe i jumped some steps, but you can do it by yourself just seeing the image

[vapid slate]

Use f(x) = -x + 3 and g(x) = x2 - 2x - 10 to answer:

g(-2) - 5

pls help

[upper karma]

@vapid slate do you mind to send a pic?

of the problem

[paper vale]

it it trivial by spiral similarity

[karmic spruce]

https://www.geogebra.org/calculator/jftndxxz
I have a problem regarding a circle and two points:
If I know:
The coordinates of point B (2,0)
The length of AB = square root of 8
A and B are both on the same circle
The radius and middlepoint of that circle is 4

How would I calculate point A?

I figured you would probably have to use the distance formula between two points:

and I know that point A would have to meet x^2 + y^2 = 4.

But from that point I got stuck

(Disclaimer: This isn't for a test or anything. It's for an animation controller program I'm writing)

[pastel anchor]

yo can anyone explain the steps to me for this question?

(7 - √-9) - (4 - √-36)

[karmic spruce]

Is that a minus sign?
√-9
?
@pastel anchor

[pastel anchor]

yea

[karmic spruce]

You can't do the square root of a negative number without using complex number

[pastel anchor]

it becomes 3i

[karmic spruce]

also this is #geometry-and-trigonometry , you might find more people that can answer that question in #prealg-and-algebra

[pastel anchor]

im learning complex numbers in trig

[karmic spruce]

ooh shit

[pastel anchor]

so are there steps?

[karmic spruce]

give me a sec

Write out the negative square roots in i

[pastel anchor]

so it becomes 3i and 69?

6i*

[karmic spruce]

I meant the whole equation

It would become this

[pastel anchor]

yea thats what i meant

[karmic spruce]

since √-9 = 3i
and √-36 = 6i

remove the brackets by doing the usual method

[pastel anchor]

so is it -3i+6i?

isnt there a suppose to be a subtraction sign between the parantheses