geometry-and-trigonometry

[main fiber]

i mean graph

[trail tendon]

oh

[main fiber]

or this one

im not sure how to find asymptotes or plot

i only know how to get the 3 points

stuck here

not sure where to go from here

ive been trying for hours and nothing online does the problem like how my teacher said

[idle island]

This can be done using graph transformation

[tardy loom]

Yes, but it was about proving that definition and seeing where it comes from.

[faint pasture]

.... you dont prove definitions?

definitions are just definitions

[tardy loom]

A system which believes that is essentially "because I said so." Now it is faith based. c²=a²+b² is also a definition. Yet there are many proofs for it. Should we have simply accepted the definition?

[faint pasture]

.... the pythogorean theorem isnt a definition

Its a theorem

[tardy loom]

That is also a definition

[faint pasture]

No?

I dont even wanna argue

Just google whats a definition

[tardy loom]

Just did. Does not refute. Essentially says all theorems are definitions but not all definitions are theorems. This sismply tells me that I was looking for someone to prove the Dot Product definition, thus making it a theorem. Which makes sense, since it's a deluxe version of the Pythagorean Theorem.

[faint pasture]

https://users.math.msu.edu/users/duncan42/AxiomNotes.pdf

[tardy loom]

It's like you're arguing squares aren't rectangles.

[faint pasture]

💀 LMFAO

whatever sails your boat

[tardy loom]

Thank you.

[rocky heron]

From Khan academy:

y = Asin(Bx-C) + D
y = Acos(Bx-C) + D

From the web:

y = Asin[B(x-C)] + D
y = Acos[B(x-C)] + D

Which of these are correct?

[obsidian harness]

they're the same thing written in different ways
B(x - C) = Bx - (BC)

but only B(x - C) has the horizontal translation being C units to the right

[rocky heron]

oh

I learnt the first one from khanacdemy, and found the latter from few math websites

so i should stick with khan academy's one?

[obsidian harness]

the web one is more useful I feel

[rocky heron]

oh, I see

[frosty rose]

what does the cartesian equation of a circle represent? (radius, circumference, etc.)

[honest cipher]

Hii can someone explain to me how does pie and rad even connects with trigonometry? I missed the lesson bcs of a competition

Oh and the question basically ask what's the value of p

[thin hornet]

3. 29/36

[austere mortar]

theyre both just different units to measure angles, heres the definition for them:

[honest cipher]

One rad is 180* right

[austere mortar]

and about trig in this, say that you have these 120°, if we'd wonna convert it to radians we could just use the formula to get 2/3π, now if youd want to find the sin(120°) itd be the same as sin(⅔π) which from the tables we know is equal to √3/2

[austere mortar]

one rad is 180°/pi in degrees

when you look at the graphs of trig functions pi is used more so as a unit on the x axis

[trail tendon]

pi radians is 180 degrees

[obtuse stag]

Thanks!! forgot to reply and i did great in my math test today haha

[fluid fern]

Me and my friend are having an argument about if a rectangle can be a square

I said no but he said sometimes

If a rectangle is a square then it’s a square and not a rectangle (what I concluded)

[cunning lion]

a square is both a square and a rectangle

shapes can belong to multiple different categories at once

[fluid fern]

But can a rectangle be a square?

[cunning lion]

yes, because all squares are rectangles

[fluid fern]

But is a rectangle a square?

[cunning lion]

can it be a square? yes. is it always a square? no

[fluid fern]

I’m gonna ask my teacher 😭

[cunning lion]

all squares are rectangles, but only some rectangles are squares

just like all apples are fruit, but only some fruit are apples

[fluid fern]

Alr ty for the help

[trail tendon]

a rectangle is a square in the same way that a cat is a cheetah lol

no wait

there fixed

all squares are rectangles the same way that all cheetahs are cats

:>

[vital rain]

Don't think so a rectangle is not a square as all the sides aren't the same

If a tree falls and no one has witnessed it did the tree really fall

[gray pivot]

Ur friend is right

This can be solved with a 5 second Google search

No need to talk to your teacher

[viscid gate]

Can someone tell me if a triangles area is 3 times bigger than another does that mean the sides are 3 times bigger aswell or no

[upper karma]

no

[viscid gate]

Darn

[buoyant sigil]

lol

@viscid gate what problem are you trying to solve?

maybe we can help

[viscid gate]

Letter c

This is what i have so far

[buoyant sigil]

for which part

oh c

because you know the area of ABC, you know the area of CAD

do you know a formula for triangle area?

[viscid gate]

(1/2)base×height?

[buoyant sigil]

yes!

[regal wedge]

id ratio of the squares of correspond sides is equivalent to the ratio of the areas of similar triangles

[buoyant sigil]

A = (1/2) * b * h

you know two of those numbers

[viscid gate]

🤕

[buoyant sigil]

are u ok

[viscid gate]

I'm trying to understand 😭

[buoyant sigil]

ok welllllllllllllllllllllllllllllllllllllllllllllll

you know A, right? of CAD

A = area

because it is 1/3 the area of the other triangle

[viscid gate]

Yeah

[buoyant sigil]

and do you knwo the base or height?

[viscid gate]

The base yeah

[buoyant sigil]

so if you know A and h

in the equation

can you solve for b?

[viscid gate]

Give me a minute

I got h=6.211

[lyric halo]

i never got the point of giving angles like 85 or 38

[upper karma]

Woah that looks nasty

[lyric halo]

Wanna try it out with me?

I'm gonna go brush my teeth and wash my face

[upper karma]

Way ahead of you

[lyric halo]

give this problem a go

[lyric halo]

oh u already began?

[upper karma]

Do you have the answer

[lyric halo]

yes i have the answers

[upper karma]

Is it D

[lyric halo]

nope

[upper karma]

B?

[lyric halo]

partially

[upper karma]

What the hel

l

I was like there's two answers

But D is the most true I thought

It checks out

Just sub it in

r2+r3=12

Then I made r2=d

and r3=2d

x+2x=12

x=4

r1,r2,r3 goes as 2 4 8

In a geometric sequence

It even looks like that in the diagram

[lyric halo]

nah it's BC

[upper karma]

How the hell does C work when we got little to no information

And D doens't check out????

[upper karma]

Tasty question

[upper karma]

There is some formula for the length of QR right ?

Length of direct common tangent

[lyric halo]

Yes

there is

[upper karma]

Could you type out what that is

,rccw

There

You could work out from there I suppose

[upper karma]

And since you know PQ,
And that tan(theta/2) = r1/PQ

You can substitute it

:))

@lyric halo

,rccw

[upper karma]

And for D,it's going to be quite lengthy

[upper karma]

Similar process

Write Pr as PQ+QR

[upper karma]

Sub value of r1 into pq and replace it in this equation

Simplify simplify simpligy

And you will get it

I guess so :p

There is probably a shorter method for this

@lyric halo wake up

Dang felt good

[lyric halo]

im not reading what u sent

im gonna solve it on my own

[upper karma]

🥲

[upper karma]

Ping me if you figure out how to do A

[vital rain]

How do I find the length of a helix

Something like

[obsidian harness]

you need calculus, bud

apply the arc length formula on $(a \cos t, a \sin t, bt)$

[vital rain]

Hm

[somber coyote]

south, just south

[vital rain]

What's the formula tho

[obsidian harness]

$\int_0^T \sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2} \ dt$

[somber coyote]

south, just south

[vital rain]

All I know is that you deffericiate I think

[obsidian harness]

it's based on 3D Pythagoras

[vital rain]

Oh thanks alot

[obsidian harness]

sqrt(dx^2 + dy^2 + dz^2)

no worries!

[vital rain]

What if the helix cuts the cylinder

Is there someway to know the permiter of the shape it makes

[obsidian harness]

it wouldn't be a helix then

oh wait ok

but I still don't get what you mean

[vital rain]

Like let's say the cylinder is hollow right

And you grab scissors and cut the helix

What's the permiter of the shape

[obsidian harness]

ohhhhh okay

[vital rain]

I did this but it's wrong

[obsidian harness]

https://youtu.be/Ce64D_yGDvY

you don't need calculus then

what you said made me realise

if you unroll the cylinder, the helix will be a straight line on a rectangle

so when you cut it, you get two congruent right triangles

[vital rain]

I looked at that but it doesn't make sense

I cut up a cylinder and it makes a parallelogram

[obsidian harness]

this is useful if the number of revolutions is not a whole number

[obsidian harness]

it's not a parallelogram

[vital rain]

It does make it tho

[vital rain]

That video is only 1 revolution

[obsidian harness]

oh okay so you want to do it in general

[vital rain]

Should i show you the questions

The one I'm working on

[obsidian harness]

yeah

[obsidian harness]

ohhh

[vital rain]

I keep getting it wrong for some reason

This is what I know so far

I hope that's right

[proper gale]

About the sides of the quadrilateral 𝐴𝐵𝐶𝐷 it is known that 𝐴𝐵 = 𝐵𝐶 and 𝐶𝐷 = 𝐴𝐷. The diagonal 𝐴𝐶 divides this quadrilateral into two triangles: equilateral and right. The side of the equilateral triangle is 6. Calculate the area of ​​the quadrilateral 𝐴𝐵𝐶𝐷. - pls explain me how to do this ( I should be able to do this knowing that the height of a 90 60 30 triangle is b√3 and the hypotenuse of a 90 45 45 triangle is a√2 )

[lyric halo]

Draw a figure

[proper gale]

thats the main reason I'm here... I can't

[lyric halo]

@cedar quarry so sorry to disturb u but do u know this

[proper gale]

I mean I dont know how to

[lyric halo]

Have you tried to?

[proper gale]

yeah

but I can't make it match the description

[lyric halo]

Alright, gimme a few minutes to get on my computer

Then I'll help you out

[proper gale]

alr

[lyric halo]

@proper gale

I assumed that ABC was the equilateral triangle

and that ACD was the right angled triangle

Now since AD = DC

the only angle which could be 90 degree was ADC

and since AD = DC

the rest two angles are equal to 45 degrees

so yeah, question solved

[proper gale]

is it possible that the ADC angle = 90? idk it just doesnt look like it is

[glass hound]

Hey guys?

[proper gale]

oh wait nvm it is possible

thanks

idk I overcomplicated it lol

I was able to do the abc but I couldnt figure out how to do the right triangle

I didn't notice it can be an isosceles

thanks

[sour mesa]

Yo

[rose sierra]

YO

[sour mesa]

Sup

[pliant monolith]

p = cot(x), q = tan(x)

If, tan(pi* p) = cot(pi* q) then show that,

tan(x) = 1/4 (2n + 1 plus/minus √(4n² + 4n - 15)) ,

Where, n ∈ Z and n < - 2 or, n > 1

[buoyant sigil]

dawg u gotta use latex

[astral onyx]

do you have any idea how to prof this:
THEOREM:
Any pair of non-concentric circles can be inverted into themselves in an infinite number of ways.

[upper karma]

into each other? or themselves?

[honest geyser]

someone help pls

[obsidian harness]

wait

what's angle BAO then?

[honest geyser]

1 rad

[obsidian harness]

yeah sector OBD - (sector ABC - triangle AOB)

[obsidian harness]

nope

do you recall the circle theorem about the angle on the circumference being....

[honest geyser]

twice angle at the centre

[obsidian harness]

other way around

angle at the centre is twice angle on the circumference

[honest geyser]

yeh i mean

[obsidian harness]

cool so then you need 1/2 ab sin C for triangle AOB

ah you also need the cosine rule (or two congruent right triangles) to find side AB, which is the radius of sector BAC

that should be enough info for you to try

[honest geyser]

get it let me try

[honest geyser]

10.518 cm^2

OCB = 8.7354 cm^2 ig

final answer 3.76

[obsidian harness]

,calc 1/2 * 5 * 5 * sin(pi - 1)

[obsidian harness]

no that's not correct

also please show your working

[honest geyser]

?

[somber coyote]

Result:

10.518387310099

[obsidian harness]

ah shit sorry

right

,calc 2 * 5 * sin((pi-1)/2)

[somber coyote]

Result:

8.7758256189037

[obsidian harness]

yes that's correct

nice work!

[alpine mason]

,calc 4 * (pi(2^3))

[somber coyote]

The following error occured while calculating:
Error: 'pi' is not a function; its value is:

[alpine mason]

uh

how is mine any different that it breaks

,calc 4 * (3.14159(2^3))

[somber coyote]

Result:

100.53088

[alpine mason]

correvt

[cunning lion]

,calc 4 * (pi*(2^3))

[somber coyote]

Result:

100.53096491487

[cunning lion]

just add a *

[alpine mason]

ohh

[cunning lion]

otherwise it thinks you are trying to use pi as a function name

[alpine mason]

can we do like

,calc B*3
B=2

,calc B*6
B=2

[somber coyote]

Results:

6 B
2

[cunning lion]

,calc is not advanced enough for that. you can replace ,calc with ,w and it will be able to process that

[somber coyote]

The following error occured while calculating:
Error: Unexpected operator not (char 4)

[alpine mason]

,w B*6
B=2

[somber coyote]

Results provided by WolframAlpha

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

Requested by whitest_of_whiteout

[alpine mason]

how does that make any sense?

2x6 ≠ 2

[cunning lion]

look at the "input" section to see how it interpreted it

i would edit the command to have a comma, that would make it realize they are separate equations

[weary plume]

I don't really see how, and why, we jump from the familiar definitions of trigonometric functions in the context of unit circle (sine of angle theta corresponds to the y coordinate of the point theta goes to on the unit circle) to these in the picture. Suddenly r joins the party. Can someone clarify things a little bit for me?

[obsidian harness]

oh you just scale the entire unit circle by a factor of r in all directions

so if the hypotenuse increases by r, the opposite side must also increase by r

instead of just y = sin theta, we now have y = r sin theta

similarly for everything else

[weary plume]

I was working on a problem yesterday

tan(theta) was given as -2/3

I assumed the point was on the unit circle and made all the calculations on top of that

I got the answer wrong

[obsidian harness]

correct

then you must have went wrong somewhere else

[obsidian harness]

did you check your theta was in the right quadrant?

[weary plume]

Yep

[obsidian harness]

wait which quadrant is specified then

[weary plume]

Gimme a sec

When I got beyond the unit circle in the second try, I got it right

Considering the r as well, which turns out to be √13

[obsidian harness]

okay so the x-coordinate is positive, so that would be theta in the 4th quadrant

[obsidian harness]

yeah you drew a right triangle with opp = 2 and adj = 3 right

then yes the hypotenuse is √13

that's the most natural method I feel

[weary plume]

Given that tan(theta) reveals the x and y coordinate of the point, yes I just applied Pythagorean theorem

[obsidian harness]

yes

but then the question you did is not really related to $x = r \cos \theta, y = r \sin \theta$ I feel

[somber coyote]

south, just south

[obsidian harness]

you're just using the fact that the angle is the same as long as opp/adj is constant

so the triangle can be scaled up to have hypotenuse = whatever

[weary plume]

Don't the broader definitions of TFs stand upon that anyway

[obsidian harness]

yeah

[weary plume]

So the upshot is that the data obtained from natural phenomena aren't necessarily compatible with a unit circle

That's why we extend our definitions beyond it

Am I right

[obsidian harness]

I don't know what you mean by that

it's just that the unit circle is a special case of (x, y) = (r cos theta, r sin theta), when r = 1

(also ignore the i, that has to do with the real-complex plane, which is the same as the xy-plane but you have complex numbers)

this gives us a coordinate system that is called polar coordinates

and polar coordinates are super useful

[weary plume]

We use those functions to model things

[obsidian harness]

oh yes correct

[weary plume]

And that must require us to extend our definitions as much as we can. I don't know which came before: unit circle or the broader approach, but it's sure that the latter is what makes trigonometry useful

Maybe the unit circle is just devised as an introductory concept to make the following material easier to consume

Because it surely did so for me

[obsidian harness]

yes I think this is true also

most people start with SOHCAHTOA

and then they realise through practice that cos and sin can be defined through the unit circle, by making the WLOG assumption that hypotenuse = 1

[obsidian harness]

then if you scale everything by r you get this definition

so basically full circle I guess, we went from hypotenuse could be anything to hypotenuse could be anything

[weary plume]

The answer was kinda obvious indeed

But it doesn't hurt to double check with someone who knows better

Thanks

[vital rain]

Oh thx it was right i also found a simpler way of doing it

[vital rain]

I need to find the permiter of the blue, the sidelengths are 1 any hints to go on about answering the question

[pulsar hatch]

You could try thinking about what happens when you put all the individual blue pieces together

[vital rain]

Yh it makes a cross

But the questions is asking for 4sf and I'm a bit confused

[pulsar hatch]

4 what?

[vital rain]

Significant figures

[pulsar hatch]

oh

[vital rain]

Mabye it isn't 12

[pulsar hatch]

It's not 12

You have to add the hypotenuses

[vital rain]

Oh the sides

[pulsar hatch]

Ye

[vital rain]

Wait but how is that possible

[pulsar hatch]

wdym

[vital rain]

Tidk the other lengths to find it

Idk*

[pulsar hatch]

You can figure it out

[vital rain]

I'll try

[pulsar hatch]

ok

see how it goes

[vital rain]

K thx

[alpine mason]

,calc 47*((pi-2)^3)

[somber coyote]

Result:

69.924819491536

[stray junco]

Easier way to understand dividing line segments?

[stray junco]

Uh.. wut?

[upper karma]

Nvm
It was a question of my own

[junior mesa]

Hey is anyone on to help by chance?

I'm very lost

[trail tendon]

just post your question

[junior mesa]

Alright

[trail tendon]

:>

[trail tendon]

you know the half angle identity for sin?

[junior mesa]

I don't understand

like half angle identities

I know theta

and cos and sin

and tan

but

I don't really understand the half stuff

[trail tendon]

does this look familiar?

[junior mesa]

kind of

It looks familiar but like

with different variables and a little bit different

[trail tendon]

well

if you know cos(theta) you can plug it into this formula right?

[junior mesa]

uhhh

so

U mean like

replace the sin(theta/2)

and plug it in

sorry I'm lost

[trail tendon]

you want to find sin(theta/2)

you know cos(theta) = -7/25 right?

[junior mesa]

wait

I thought theta was just like

a variable

similar to X

but not X

[trail tendon]

it is

[junior mesa]

more like for angles

How do I solve for theta in these problems

[trail tendon]

this is for a specific theta

but you don't need/want to solve for theta here

[junior mesa]

oh ok

[trail tendon]

so ur gonna wanna plug cos(theta) = -7/25 into +- sqrt( (1-cos(theta))/2 )

basically replace cos(theta) with -7/25

[junior mesa]

ohh ok

so

I would do like for exmplae

+-sqrt( -7/25))/2 )?

sorry If I'm wrong

I'm use to like the

+-sqrt(b^2 - 4ac/2)

[trail tendon]

close

[junior mesa]

or the quadratic formula

oh

[trail tendon]

1 sec

should be this my mouse glitching lol

but thats not the answer yet btw

thats just the possible values of sin(theta/2)

does it make sense so far tho?

[junior mesa]

hmmm ok

I have a question

[trail tendon]

yuh?

[junior mesa]

inside the square root how come it changed from

1 -

to 1+

[trail tendon]

i replaced cos(theta) with -7/25

so it became double negative

[junior mesa]

ohhh

i see

[trail tendon]

1 sec

yea

like

1 - (cos(theta)) = 1 - (-7/25) = 1 + 7/25

1 + 7/25 = 32/25

[junior mesa]

let me write this down to follow along easier

[trail tendon]

ok

[junior mesa]

At the very end is that

+- 4/5

[trail tendon]

yes, but thats not the final answer

[junior mesa]

ohh ok

[trail tendon]

the two possible values are -4/5 and 4/5

but

theta has to be between pi and 3pi/2 right?

[junior mesa]

uhhh

between pi and 3pi/2

[trail tendon]

yes

[junior mesa]

I'm confused on that part

[junior mesa]

OHHHH

yeah

Wow

you just clearred

the problem up

so much more for me

[trail tendon]

💀

[junior mesa]

Like I can finally read it

[trail tendon]

XD

do u got now or...

[junior mesa]

Since Cos(theta) = -7/25 and its greater than pi and less than 3pi/2

Let me check

if i can do it

[trail tendon]

alr

[junior mesa]

wait so

If Cos is

-7/25

and cos is the cah part of soh cah toa

does that mean Adjacent and Hypothenus are

-7 and 25

[trail tendon]

the ratio between them is -7/25 yes

[junior mesa]

ohh ok

[trail tendon]

the sides could be -7 and 25, or -14 and 50, ect.

[junior mesa]

yea

ok so

and since sin is

opposite over hypothenus

would that mean the denominator by default

has to be like

25

or am i confusing myself

[trail tendon]

actually

naw

that would be for sin(theta) nvm

[junior mesa]

ohh yeah

It's theta/2

[trail tendon]

ima delete since technically not accurate lol

[junior mesa]

alright

[trail tendon]

yeah

but basically

you know that since theta varies from pi to 3pi/2

[junior mesa]

yea

[trail tendon]

when theta is pi, sin(theta/2) is sin(pi/2) = 1

when theta is 3pi/2, sin(theta/2) is sin(3pi/4) = sqrt(2)/2

[junior mesa]

I'm gonna be honest

I'm lost

[trail tendon]

see i wanna graph it in polar 💀 😭

[junior mesa]

😭

Is there like

a formula for this

[idle island]

is the final answer +4/5?

[junior mesa]

Nah

It means like

It could've been either or

[trail tendon]

yeah he said is it the positive one

and it is

[junior mesa]

Oh

[trail tendon]

im tryna draw it tho

lol

[junior mesa]

ohh

[trail tendon]

im thinking how

like basically

when you go from 1 to sqrt(2)/2 with the sine function

[junior mesa]

yeah

[trail tendon]

sqrt(2)/2 is about 0.7 something

[idle island]

ik but since its theta/2 so it is defined from pi/2 to 3pi/4 where sin is positive

[trail tendon]

what?

anyway

sin(pi/2) is bounded between 0.7 and 1

4/5 =0.8 is bounded there

rather than its negative

[idle island]

i was telling this to kazruki i think

[trail tendon]

i mean idk what u said

[idle island]

ah nothing much really

[junior mesa]

oh

wait so

[trail tendon]

theta/2 is defined from pi/2 to 3pi/4 yeah

[junior mesa]

the theta / 2 means its positive

[idle island]

yes

[junior mesa]

automatically

?

[idle island]

thats what i meant

[junior mesa]

ohh ok

[idle island]

its sin so yes

[junior mesa]

ohh

sin is always positive

[idle island]

its positive from 0 to pi

[junior mesa]

oh

Is that just sin

or does it apply to

tan and cos

[idle island]

tan is positive from pi to 3pi/2
cos is positive from 3pi/2 to 2pi

[junior mesa]

ah ok

I see

wait so

taking that information

If I'm doing this one

would I just do

Ok

@idle island

@trail tendon

I don't know if I did this correct

but I got the exact same answer

this time it was

negative -4/5

[trail tendon]

you have to use a slightly different formula

[junior mesa]

oh

[trail tendon]

basically just changed the - to a +

when its cos(theta/2)

[junior mesa]

oh ok

wait so

it would endu p being

inside of the parenthesis

1 + -7/25

[trail tendon]

yeah

divided by 2

and then sqrt of that

and then +-

:>

[junior mesa]

so

I have

Sqrt(-16/25)

but

I can't SQRT a negative

Is it just undefined?

[gray pivot]

Undefined for real numbers

[junior mesa]

what do you mean by that

[gray pivot]

Have you learned imaginary numbers yet

[trail tendon]

you shouldn't get imaginary

you forgot the 1

[junior mesa]

I don't think we have

[trail tendon]

maybe

[junior mesa]

where did i forget it at

[trail tendon]

actually idk what u did

[junior mesa]

Let me send my work

[gray pivot]

Well then it's undefined 🤷 if u got the correct sqrt

[junior mesa]

So I did

[trail tendon]

1 + cos(theta) = 1 - 7/25 = 18/25

divide by 2 and you get 9/25

take sqrt and you get 3/5

[junior mesa]

wait

[trail tendon]

so it should be +- 3/5

[junior mesa]

I ended up doing

1+-7/25

and getting

-32/25

[trail tendon]

💀

[junior mesa]

is that why

I thought it looked strange

[gray pivot]

What

[trail tendon]

maybe he means -1 - 7/25

idk 💀

[gray pivot]

How do you get -32/25 from 1+7/5 or 1-7/5

[junior mesa]

because since it's

a fraction 7/25

woulnd't

wouldn't a full number 1

be equal to the 25

so I added +25

or

[gray pivot]

Not if it was negative

[junior mesa]

in this case -25

oh

[gray pivot]

Only for the positive

25/25 - 7/25 is 18/25

[junior mesa]

but

its a -7/25

[gray pivot]

Not -32/25

[gray pivot]

Thats what I said...

[trail tendon]

1 - 7/25 = 25/25 - 7/25 = (25-7)/25 = 18/25

[junior mesa]

wait

holdon let me re-check my work

ok I see what I did wrong

I was supposed to do

+25

instead of

-25

[gray pivot]

Yes because 1 does not equal -25/25

≠ oh cool I can do the sign from mobile

[junior mesa]

ok I'm starting to understand a little

so

even if it's a -1

[gray pivot]

1≠-25/25 1≠-(25/25)

[junior mesa]

regardless its going to equal -25/25

-25/-25*

[gray pivot]

No

[junior mesa]

which the negatives cancel out and make

25/25

[gray pivot]

-1≠-25/-25

[junior mesa]

oh ok

[gray pivot]

You cant just make up a new negative sign

[junior mesa]

wait so

is -1=25/25?

[gray pivot]

Nope

[junior mesa]

oh

[gray pivot]

One of them needs to be neagtive

Or the fraction needs to be multiplied by -1

Basically negative signs anywhere in the fraction apply to the whole fraction

[junior mesa]

so would it be the denominator?

[gray pivot]

Can be either

[junior mesa]

oh ok

[gray pivot]

Normally written in the numerator

[junior mesa]

But

The thing I'm really confused on though

[gray pivot]

And if it's your final answer you probably would want to simplify to no negatives in denominator

[junior mesa]

is how did 1 + cos(theta) = 1 - 7/25 = 18/25

wait

ok

I see

I feel dumb now ngl

I finally realized

It's a +1

[gray pivot]

Is cos(theta)=-7/25?

[junior mesa]

yea

[gray pivot]

Then

[junior mesa]

I got my final answer is 3/5 @trail tendon

[gray pivot]

1+cos(theta)
1+(-7/25)
(25/25)+(-7/25)
18/25

[junior mesa]

yea

[gray pivot]

.

[junior mesa]

oh

[gray pivot]

Correct

[junior mesa]

the thing I'm wondering is

since it's a COS

would it make it a Pos

3/5

[gray pivot]

I dont think so. Are you solving for theta?

[junior mesa]

Uhhh'

[gray pivot]

Because a -3/5 in cosine should be fine

,w cos(-3/5)

[somber coyote]

Results provided by WolframAlpha

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

Requested by warbled1

[junior mesa]

This is what I was solving for on it

[gray pivot]

Is there context to that or is it a standalone question

[junior mesa]

yea there is

[gray pivot]

Negative should be fine here

[junior mesa]

got it

Hey

Are u able to help me with another problem?

@gray pivot

[gray pivot]

Depending on the problem

I haven't done geometry in a while

[junior mesa]

This is it

[gray pivot]

Yeah I dont know that

I'm a sophomore in high school

[junior mesa]

Ah okay

Dang your smart for one

[gray pivot]

If you need help with algebra I can help tho

[junior mesa]

Do you know this?

[gray pivot]

Nah sorry

I'm learning trig next semester

[junior mesa]

Ah okay

Ah

I see

thank you though

Is anyone online able to help right now?

[obsidian harness]

Look up the formula for sin(a + b)

[obsidian harness]

Simplifies to (2 tan t) / (1 - tan^2 t)

[faint pasture]

Its just Sin(A+B)

[lyric halo]

what do i do next

[kindred narwhal]

someone pls help

[copper oak]

in what way

[obsidian harness]

literally tells you to draw it

then apply the sine rule

also for part c

[kindred narwhal]

ya i got the answer

[scenic quest]

what is the answer?

[kindred narwhal]

h=6.3 i think

and the third side was like 5.5 or something

[scenic quest]

oh

i did

my answer is 6,1

but your answer is correct

[harsh pond]

In this picture can someone explain how did he get an angle by adding 30 and 45 together. I'm trying to figture it out but i can't

What I did know is somehow 3 angle have to be added to get 180 degree

[upper karma]

Hey

Guys

[upper echo]

use
70=90-20
50=90-40
sin(90-t)=cos(t)
sin(2t)=2sin(t)cos(t)
and so on

[upper karma]

Can someone help me in this question
Idk I am not able to prove

Show that the perpendicular drawn from (4,1) on the line segment joining (6,-5) and
(2,-1) divides it internally in the ratio 8: 5.

[upper echo]

that's not true. Just draw that.

[upper karma]

Yea q is incorrect

Right?

It should be (6,5) not (6,-5)

[silk dawn]

that would take a reallly long time

[upper echo]

nope. it took me 6 lines with one equality sign in each 🙂

[silk dawn]

how do u continue from there

[upper echo]

what have you got?

[silk dawn]

1/4 sin^2 10 cos 40 cos 80

= alpha- 1/16 sin 10

[upper echo]

you did it wrong

[silk dawn]

how

[upper echo]

2sin40 cos(40) = sin(80)

[silk dawn]

my bad

[upper echo]

not cos(80)

[silk dawn]

let me do it again

still stuch

[silk dawn]

sin 40sin 80

[upper echo]

If A(4,1), B(6,5), C(2,-1) then you can use the dot product for vectors CA, and CB. As |CA| * cos(ACB) is a projection of CA on CB it equals (CA * CB)/|CB|. So, you just need to prove (CA * CB)/|CB|^2=5/13. Which is kind of obvious: CA=(2,2), CB=(4,6), so (2 * 4+2 * 6)/(16+36)=5/13.

[upper echo]

use the same thing for sin10*sin80

[silk dawn]

then

1/8 sin 10 sin 40 sin 20

[upper echo]

use some formula for sin(a)*sin(b)

[silk dawn]

there isnt any such formula

[upper echo]

there is

[silk dawn]

let me see

i just derived it

[silk dawn]

nice

80 is the ans

cool method

[upper karma]

Gotcha

Thanks

[burnt ice]

Can someone please help me with this?

[manic tusk]

Is that supposed to be a regular trapezoid

[burnt ice]

nothing is mentioned

[manic tusk]

Ok it is a regular trap

Probably should've mentioned that lmao

Anyway you can solve for the other side of that triangle, x, since 2x+4.5=9.5

[burnt ice]

That was what I was looking for, THANKS

also can I just assume every trap to be reguar in case nothing is mentioned?

[manic tusk]

I guess so... I solved for the angle quickly and it's 16.3 smth rounding to 16.4

They should mention that it's regular though

[burnt ice]

nice, that's what I got too

thanks alot for the help

[manic tusk]

Essentially if it's multiple choice and they do smth like that I'd assume it's regular and solve for the angle, if your answer is nowhere close to 1 of the choices try smth else

But in this case it is just 16.4

[burnt ice]

hmmm okay, got it

[manic tusk]

Come to think of it maybe you don't have to assume it's regular

But I have badminton rn so I can't work it out, but if you divide the lengths as x and y where x+y+4.5=9.5 maybe you can solve for area of trapezoid, which is sum of rectangle + 2 triangles, and maybe that would show x=y

[burnt ice]

I thought of that too but since they told us that these multiple choice quesstions should take 2 mins each, I don't think they need that much work

[tight goblet]

hello, I need help understanding the behavior of hyperbolas, specifically their asymptotes

[austere mortar]

how could I find the max and minimal value of this function without using derivatives?

[obsidian harness]

take tan x = +infinity or x = pi/2 for instance and the max value is clearly +infinity

on the other hand, for the minimum, sub u = tan x

min value of u^2 + u = (u + 1/2)^2 - 1/4 is -1/4

and this value is attained at x = arctan(-1/2) so this is not a false minimum

[austere mortar]

so the min value would just be y = -1/4?

[obsidian harness]

yes

[austere mortar]

alright, thank you very much ^^

[manic tusk]

Yeah and I thought about it and that doesn't work either

[weary plume]

If we're using the reciprocal properties to graph cosecant and secant functions, then how come we're using 2sin(2x) to graph 2csc(2x) and not 1/[2sin(2x)]?

[regal sonnet]

Hi folks! I'm trying to place n angles X = x_1, ..., x_n on the unit circle, such that they all live at multiples of some fixed angle k, and such that no matter how much I rotate them, no two rotated angles ever coincide with any two non-rotated angles. That is, if we call theta + X the set of rotated points in X, the intersection of X and theta + X has size at most 1, for all theta except multiples of 2pi (of course).

Among all possible solutions, I'd like to maximize the minimum distance between two points. That is, to have as much minimum spacing as possible.

For n = 2, k = 5, the solution is [0, 175°] or [0, 185°], assuming wlog one of the points lives at zero. For n = 3 we get all of these: [[0,115,235],[0,115,240],[0,120,235],[0,120,245],[0,125,240],[0,125,245]].

What does the solution look like in general?

[spark sparrow]

Does any one here know why the point on a quarter ellipse where y = b/2 is the formula for the height of an equilateral triangle? (a * sqrt(3))/2 ? lmfao

[obsidian harness]

not sure exactly what you mean

but isn't it just because $(1/2)^2 + (\sqrt 3/2)^2 = 1$

[somber coyote]

south, just south

[obsidian harness]

yeah you mean a quarter-circle right

[spark sparrow]

Quarter circle or quarter ellipse. https://www.desmos.com/calculator/wnc8kvgokt

[obsidian harness]

wow, nice graph

[spark sparrow]

Its wild because Im smart dumb xD. I had to find this point for work and instead of deriving it properly, I just bruit forced it by taking the minimum difference between array slices of the delta of the ellipse. xD

Almost a year later I derived it properly.

What job would require this? Financial astrology of course hahah (f)

[spark sparrow]

Thank you!

You can verify that the change in y on either side is b/2 by integrating the quarter ellipse from 0 to n and from n to a

Im sure there is a real name for a point like that, but Im just self educated for better or for worse :/

[upper karma]

Does (n-2)180 work for any shape to find what the angles add up to?

[dusty canopy]

Hi, may i ask you where you get the question from?

[silver cliff]

How would you find the area of the tilted square?

[upper echo]

the slope is 1/(4-1) and the height is 4+1.

so the area of the tilted square is 5^2+15^2.

[silver cliff]

So 250?

[upper echo]

yes

[silver cliff]

thank you

[austere mortar]

why isnt this correct? Im supposed to find all of the values of m for which the function has solutions

[elder halo]

How do I draw this? I can't think how to draw this at all

AOD is the diameter of a circle, centre O. The cyclic quadrilateral ABCD is such that COB is 42º and OD is parallel to BC.

[honest oxide]

Anyone need help?

[obsidian harness]

Draw A and D on opposite ends first

B and C lie on the circumference, definition of cyclic

OD parallel to BC means that BC horizontal if you drew AD horizontal

Finally angle BOC is 42

A B C D should be in clockwise order

In other questions they also can be in anticlockwise order

You just can't go out of order

[obsidian harness]

Should be -1 not 1

In the 1st line

[stray junco]

In this situation where youre dividing line segments how do you draw the triangle

[obsidian harness]

what triangle?

oh do you mean two right-angled triangles?

[stray junco]

I think youre supposed to draw a triangle and put the points that follow to the ratio

If the ratio is 1:3

[obsidian harness]

yeah I suspect it's this then

[stray junco]

This is what i saw

[obsidian harness]

yeah you get the idea

[stray junco]

How do you draw it up

Up or down

Frick my spelling

How do you know its drawn that way? Is there a way to know?

[obsidian harness]

both up and down work

[stray junco]

Oh ok

[stray junco]

If ab is 5 times as long as bc is it 5:1 or 1:5 im dumb

[obsidian harness]

then AB:BC = 5:1

[elfin trail]

Guys, what does geometry typically go over (other than shapes)

[meager thistle]

Hey guys could you check my work, is there anything else I can do to simplify this fraction?

[trail tendon]

not really unless you want to rationalize the denominator

[meager thistle]

From what I remember it was mostly dealing with triangles and circles, I think, how to find one or more sides of the triangle using the Pythagorean theorem or sine, cosine, and tangent, finding the angle using sine, cosine, and tangent. Learning about corresponding angles, alternate interior and exterior angles, and lots of other stuff. You can find videos on YouTube that are hours long, if you’re interested, that go over all basic geometry stuff and more.(Ima be honest I’m not sure if you guys will use sine cosine and tangent)

[meager thistle]

Alrighty, I don’t want to rationalize it, thank you!!

[elfin trail]

Thanks

[maiden brook]

geometry is about shapes so i dont know what u mean. but at its core its just triangles and circles

[novel shadow]

Guys pls help 🥲

[ripe breach]

anybody good w geometric proofs??

[pallid smelt]

Angle 1 is 35 degrees, Angle 2 is 65 degrees, Angle 3 is 29 degrees, Angle 4 is 115 degrees, Angle 5 is 65 degrees, Angle 6 is 29 degrees, and Angle 7 is 144 degrees.

Keep in mind the figure to the right is a quadrilateral, not a triangle. Also, angle 7 is a separate angle from the 90 degree one

[novel shadow]

Oh got confused on 7 lol

[pallid smelt]

depends on the problem, but in general im decent

[ripe breach]

this is what ive got

[pallid smelt]

do you just want the straight up entire proof

[novel shadow]

Hmm some of what you said were my original answers but the teachers been confusing me with the area around 7

[pallid smelt]

Oh what did they say?

[ripe breach]

yeah that would be great lol

[novel shadow]

Dude she just sucks at teaching tbh 🥲 and when she does she barely explains thats why i got confused with this
Cause to me 2=5 and 115=4

[ripe breach]

i have like 4 more though; everyone tells me to go through the theorems i could apply i just dont know how to do them in general

maybe i just dont understand the jist of it lol

[pallid smelt]

Statement: <1 is congruent to <2, Reasoning: Base Angles Theorem
Statement: <1 is congruent to <3, Reasoning: Corresponding Angles
Statement: <2 is congruent to <3, Reasoning: Transitive Property of Congruence
Statement <3 is congruent to <4, Reasoning: I think you can just put "Previous steps" because logically it makes sense
Statement: Triangle NEB is isoceles, Reasoning: Converse Base Angles Theorem

should be right

[ripe breach]

thank you

[pallid smelt]

no problem

[novel shadow]

Im not sure about yall but i keep getting 11.6 when trying to solve for x

I need like 4 more problems to finish this jsjsbd

[odd herald]

help plz 🙏

[odd herald]

x might be 15

[novel shadow]

Is this multiple choice? Orr?

[odd herald]

no

its single

[novel shadow]

Cause from what im seeing all are true but C

Its worded weirdly imo

[odd herald]

you have to match the choices with the question

[odd herald]

fr

my teachin buggin

[novel shadow]

20 should probably be A because its the answer choice thay relates to the line

[odd herald]

ya i think so too

[novel shadow]

Im confused with the rest tho

[odd herald]

well... do you need help or som

[novel shadow]

This ig +3 other problems

[odd herald]

aight ima see if i got you

this 10th grade math right?

[novel shadow]

Yeah

[odd herald]

k i just learned that

[novel shadow]

Dawg she aint teach us none over here

Im confusedd isjdbfjf

[odd herald]

a) AM, b) KP, c) CM, d) C, e) CM, f) L

[novel shadow]

Thxx

[odd herald]

hol up

i messed up on one

[novel shadow]

She straight up never told us this 💀

[odd herald]

e) A

[novel shadow]

Oh ok xd

[odd herald]

ya i have no idea lol im failing that

[novel shadow]

I have a 2 hour exam friday 😭 kill meee

[odd herald]

good luck

remember chatgpt is your friend

[novel shadow]

Yeah!!!!

[ripe breach]

is this SSS SSA and stuff?

[novel shadow]

What

Idk what that means 🥲🥲

[ripe breach]

side side side

side side angle

it is im pretty sure

[novel shadow]

Uh i dont think so?

[ripe breach]

your teacher didnt teach you that?

[novel shadow]

Yeah

She doesnt teach anything

[ripe breach]

a) ASA
b) SSS
c) SSS
d) HL

im pretty sure thats what the answer is

[sonic coral]

Guys does anybody know what p=x/2 mean

I'm learning bout circles

Radius

That stuff

[trail tendon]

we're gonna need more context...

[stray junco]

AB : BC = 2:3 and AC = AB + BC = AB =2/5AC
How the heck does this work

[trail tendon]

what?

i think you miswrote something

also more context would be nice...

💀

[weary plume]

Can someone explain why those two angles are the same

[obsidian harness]

angles on a straight line add up to 180 degrees

so in the quadrilateral the angle sum is 90 + 90 + (180 - theta) + theta = 360

as expected

[plain halo]

can someone explain like the alternate interior angles alternate exterior angles remote angles etc?

im so connfused

[stable harbor]

Pls help thank you

[upper echo]

BDC=PCB and AEDB is cyclic

[stable harbor]

So how much is x

Is it 110?

[upper echo]

why?

[stable harbor]

Every opposing angle has the same angle

I think

[upper echo]

no just read about cyclic quadrilaterals.

and angles between a tangent and a chord

[kindred agate]

plis help

[regal wedge]

draw in OC, it’s a radii of the circle, then use pythagorean theorem

[kindred agate]

tysm

[regal wedge]

yep

gotchu

[kindred agate]

srry i need more help
i hate this

[kindred agate]

😟

[drifting zealot]

The HL Theorem is only with a leg and Hypotenuse right?

[earnest musk]

I'm new here can I ask for any help here? im in secondary school rn

[drifting zealot]

Yes!

[earnest musk]

Probably very simple but I'm dumb so need help solving this

[trail tendon]

it doesn't mean prove right?

[earnest musk]

I think it might have been show 😅

[trail tendon]

how about 'solve' 💀

usually you 'prove' or 'show' an identity, but this isn't an identity

its only true for some theta values

[earnest musk]

Something like that those terms are used interchangeably in my book

[trail tendon]

'prove' or 'show' are interchangeable, but i think this one would be more of a 'solve' since you'd want specific theta values

you're looking for theta right?

[earnest musk]

No not looking for theta It says if cosA+sinA=√2 sinA the prove that secA=2√2 SinA

I cant solve it I'm kinda feeling like it's unsolvable

[trail tendon]

oh

[trail tendon]

yea

cos(theta) + sin(theta) = sqrt(2)sin(theta)
cos(theta) = [sqrt(2) - 1] sin(theta)
sec(theta) = 1/[sqrt(2) - 1] csc(theta)

hmm

[earnest musk]

the result I'm getting is sec(theta)=√2+1/sin(theta)

[trail tendon]

you mean sqrt(2) + (1/sin(theta)) = sqrt(2) + csc(theta)?

idk how u got that result anyway

[earnest musk]

Let me do it in pen and paper and I'll give a pic

[trail tendon]

ok

[earnest musk]

This is what I got correct me if I made any mistake pls

[trail tendon]

yea looks fine

[earnest musk]

I wrote sin(theta) twice ignore it lol 💀

[faint pasture]

Viper check alg oncee

[trail tendon]

i don't think what you are trying to 'prove' makes too much sense though

[trail tendon]

?

[faint pasture]

Theres a que in there

Looks kinda good but idk if it solvable

[earnest musk]

I'm wondering is it possible to get 2√2sin(theta) from √2+1/sin(theta)