geometry-and-trigonometry

[upper karma]

if I have 1 point on plane and I want to find tangent point on circle how do I do it?

[silk patio]

You draw them

Draw any two rays from P and it works

[upper karma]

ahh ok I see

thank you

@silk patio Thats ingenious how do people find these solutions to these problems?

Is there any logic behind it?

[silk patio]

Yeah but it involves projective geometry

[upper karma]

@silk patio Do you know what is logic behind that solution? How did someone come up with that?

I was trying to find solution on my own and it was too hard

[silk patio]

Unless you know projective geometry it’s gonna take too long to explain it in an intuitive way

And I’m lazy

Yeah it’s hard

[upper karma]

yea ok sure no problem

will look into it, thanks aniway

[upper karma]

Can sm plzzz help me with this

[torpid hearth]

can you tell me what the problem is saying?

[upper karma]

I’m supposed to find how much ice cream he lost

[torpid hearth]

Whenever you lose something, the total amount you have is reduced. What mathematical operation describes reduction?

@upper karma

[upper karma]

I'm supposed to get 729 for h but it isn't working

[sturdy vortex]

https://cdn.discordapp.com/attachments/269573202018041856/853861074503860224/Screen_Shot_2021-06-11_at_8.57.02_AM.png

can someone help? very beginner trig question but I'm a freshman doing some extra credit trig courses

[silent plank]

@upper karma23° isn't 27°
also based on my calculations, h should be over 1000

[grim rampart]

can someone helpme

how did they get this first step that its 90 degrees?

@dark sparrow

[dark sparrow]

???????????

[grim rampart]

im confuse how they got that first step

i mean the second sentence

angle ACB = angle CDB = 90 degrees

but how do they know its 90 degrees?

[tepid crow]

The question says "Triangle ABC is a right triangle"

[silent plank]

the diagram implies that triangleABC is right triangle with a right angle at C

it is also given that CD perp AB

[grim rampart]

so are comparing two triangles?

[silent plank]

which by definition means that <CDB is 90°

[grim rampart]

oh okay, but how do we prove that the pythogorean thm works?

thats where i get confuse

[silent plank]

consider corresponding sides of similar triangles
and set up a few equations

[grim rampart]

got it ty

[ember ruin]

Hi

I don’t understand it

@here

Can someone help in proving?

[lethal robin]

If ∠ B and ∠ Q are acute angles such that sin B = sin Q, then prove that ∠ B = ∠ Q.

How do i solve this?

[storm portal]

If B and Q are acute angles, then they are in between 0˚ and 90˚

or 0 radians and π/2 radians

In this range for B and Q values, sin B and sin Q never repeat values

If B could be any real number, then there are multiple inputs that yield the same output (the outputs repeat)

[storm portal]

Given this, if you know the value of sin B, then you only have 1 possibility for B

The same applies with sin Q and Q

[nocturne remnant]

For a less trig proof, consider a triangle with b and q as angles

[storm portal]

As a result:

if sin B = sin Q, then B must be equal to Q, since there are no other possibilities for B or Q

[storm portal]

I like this

[nocturne remnant]

Sine law tells us that triangle is isosceles and then done

[twilit mist]

If COS-1=(3/6) cause the hypotenuse is the longest

[cloud stump]

The rectangle with max area is a square

[narrow plinth]

Are you sure about that because i remember solving it and don't remember it being a square

[cloud stump]

I am not xD

It's a square if it's a 45 45 90 triangle but just noticed that's prob not true for other right triangles

The rectangle of max area has half the area of the triangle that's for sure

[astral kettle]

is -ln(x+x) = ln(x-x)?

[cloud stump]

No

ln(x-x)=ln0 there is no ln of 0

And i dont think this is a geo/trig question

[upper karma]

need help

[wise pawn]

have you learned SOH CAH TOA?

[robust crane]

can it be said, that if two arc segments intersect that you can check the line intersections of the radial vectors coming from each end of the arcs? such as like this terrible drawing:

this is after a radius check from both circles aswell

because im surprisingly not finding an easy way to check if two arcs intersect each other

[pure cape]

not really, because the radii when extended will always intersect

no matter if the arcs intersect each other

[wise pawn]

depends on how much you like geometry I guess

[silk patio]

I read it and it’s not worth

Unless you’re interested in constructive geometry

Most of what people call Euclidean geometry isn’t that style anymore

Just do simple Olympiad problems and work up

[unique spire]

yea

[cloud stump]

Is a rectangle-trihadral angle a trihedral angle with a dihedral of 90° or a trihedral with a 90° angle on one of its faces

Different sources give different answers

What is the most appropiate

[sinful niche]

my teacher was writing this

isnt csc pi/6=2

[glacial dawn]

Csc pi/6 = 1 / (sin pi/6)

Sin pi/6 = 1/2

1/ (1/2) = 2

So yes, csc pi/6 = 2

[grim rampart]

can someone help me

i cant use trig for this one

[acoustic jungle]

looks like a special case of stewarts theorem

[silk patio]

It’s even easier than that

It’s Archimedes theorem or whatever. Make it into a parallelogram and do it that way

Consider vectors a b. c=b-a
2m=a+b

Then dot product with itself and add

[grim rampart]

stewarts theorem?

hmm

[nocturne thicket]

So these are the answers for the questions on my homework

but I thought you can't simplify x^2 and x? Like they're two completely different things?

[silent plank]

x^2 is the product of x and x

[nocturne remnant]

This is Apollonius’s Theorem

It and the angle bisector theorem are both special cases of Stewart

[stuck dragon]

I don't quite understand

because wouldn't the difference be 0 because it would be split into 4 congruent pieces?

[dark sparrow]

they're asking you to consider all the possible ways to cut up your wood into 4 congruent rectangular pieces

and all the possible dimensions that can result

here are a few examples of cutting patterns

@stuck dragon

[grim rampart]

question, how do you multiply three proportions ?

10 / 6 = 8 / b = 6 / c

[dark sparrow]

what do you mean

[grim rampart]

can we solve three proportions ?

i found that b = 48/10

[dark sparrow]

yeah, and you can use 10/6 = 6/c to find c in the same fashion

[nocturne thicket]

Can someone remind me the formulas/process for solving these?

[verbal beacon]

for the 19th question , join the center of the circle to the point of contact of the tangent and then as you know that the tangent to a circle is perpendicular to the radius , you'll have a right triangle ,now just use pythagoras theorem to get x

[silent plank]

tan-sec
sec-sec
or more generally power of a point

[wheat helm]

does anyone know how to do this and can teach me

[humble pulsar]

Intersecting Chords Theorem

product of the 2 segments on the chord is constant

[wheat helm]

so mk is 17 and x= 0?

[sage jay]

Dill

will u offer me help cuz math exam soon

[wheat helm]

nah i was just asking for help

[sage jay]

;-;

[wheat helm]

@humble pulsar did i solve that right ?

[sage jay]

9th grade u sure cant help?

[humble pulsar]

x=0, yes

[wheat helm]

how would i solve for mk then?

[humble pulsar]

plug x=0 in...

[stuck dragon]

oh wow ok

I have not thought of that

thank you

[junior light]

Don't ask for help on tests.

[sage jay]

Explain cuz i confusion

Why 1=2

[stuck dragon]

my imagination got stuck on the last one so it was confusing

[junior light]

This is #chill -tier content. Don't meme in help channels.

[wheat helm]

@humble pulsar do you know how to solve these

[sage jay]

[graceful patrol]

Hey guys I got a question

[dry jungle]

does anyone know how to do this and send work?

Exit Slip Question

[silent plank]

what have you tried?

[agile ether]

somebody help

<@&286206848099549185>

[pure cape]

please ping after a minimum of 15min

and notice all of those three angles add up to 180

or those two unknown angles add up to 90

[agile ether]

uh

[raven gulch]

and using this fact, u can get a linear equation

which u can solve to get x

now try to do it

[agile ether]

ok

[narrow plinth]

If AB and CD are parallel diamaters of 2 bases of a cylinder with height H and the point E splits AB into a ratio of 1:2.Find the volume of the cylinder if line through E and D makes a 45° angle with the base of the cylinder

I did it like this and hgot the wrong answer

It should be H^3*pi/3

But idk how to get that answer

I tried doing it a xouple of times and keept getting this

<@&286206848099549185>

[nocturne thicket]

ooohhh, thanks

[verbal beacon]

np

[upper karma]

Can sm help, I’m supposed to find the area of this

<@&286206848099549185>

[trim breach]

Separate the shape into a right triangle and a rectangle. Find the areas of each, then add them together to get the total area of the shape.

[upper karma]

What would the l and w be for the triangle

I mean b and h

[trim breach]

Well, to make the rectangle, opposite sides need to be congruent. So, the rectangle should have a dimensions of 4.8 and 13.5.

That means the triangle’s base is 5.7 - 4.8.

And because the triangle shares a side with the rectangle, the height is 13.8.

[upper karma]

Okay thank you, when I do the work can I show you to make sure it right?

@trim breach

[trim breach]

Not quite. I think I might have confused you.

The rectangle’s area is correct, but the triangle is not.

[upper karma]

Can you draw on paper how it’s supposed to look

[trim breach]

You sketched it correctly.

But the base of the triangle is not 5.7.

[upper karma]

Oh okay so what did I do wrong

Oh

What is it?

[trim breach]

5.7 is that total segment. 4.8 of that makes up the rectangle.

So the expression to calculate the base of the triangle is:

,,5.7 - 4.8

[somber coyote]

Lidoh

[upper karma]

Ah okay so is the square right?

[upper karma]

^

[upper karma]

@trim breach is that right

[trim breach]

Yes!

[upper karma]

Thank you

[old fable]

At what values does cos x = sinx

?

[acoustic jungle]

at tanx = 1

[edgy juniper]

in the example it said find b
then it just said c = 90 with no explaination.
is there any rule on this?
my guess is the rule is if line a and b pass through the centre in a straight line, then c will always be right angled?

[silent plank]

Thales/inscribed angle theorem

[alpine wadi]

can someone help me with these

[acoustic jungle]

@alpine wadi are there more than 1 possible value for 1?

or are we assuming -90<=sin-1(),cos-1()<=90

[alpine wadi]

@acoustic jungle im not sure, thats all the teacher gave me

[acoustic jungle]

then it's probably between -90 degrees and 90 degrees.

you will need to know sin(a+b)=sinacosb + sinbcosa

for sin(arccos)) you'll need to draw out a triangle.

for example let theta = arccos(7/25)

the height of this triangle with sides 7 and 25 is sqrt(25^2-7^2) = 24

so sin(theta) = 24/25

and so sin(arccos(7/25))=24/25

do the same thing with the others.

[near fiber]

Solve pls

[gusty breach]

angle c=90 degree so A+B+C=180 B=180-90-37 B=53 DEGREE

[acoustic jungle]

use arctanx - arctany = artan((x-y)/(1+xy)) 4 times

or change 4tan-1(1/5) to arctan(120/119) and apply tan to both sides

[near fiber]

Thanks

[acoustic jungle]

question is bad though

looks like a lot of fractions

which is dumb.

[nocturne thicket]

I forgot how to do this, the answer is D but how do you do this again?

[earnest basin]

Simplify

[nocturne thicket]

[edgy juniper]

yeah i know but, how did you get angle c ?

[vital perch]

From her eye, which stands 1.61 meters above the ground, Savannah measures the angle of elevation to the top of a prominent skyscraper to be 24^{\circ}
∘
. If she is standing at a horizontal distance of 340 meters from the base of the skyscraper, what is the height of the skyscraper? Round your answer to the nearest hundredth of a meter if necessary.

what could I do for this?

[silent plank]

have you drawn a diagram?

[vital perch]

not yet no

[silent plank]

start by drawing one

[vital perch]

what do i draw for it sorry for asking?

[silent plank]

draw some lines/triangle/rectangle to represent the problem

[vital perch]

ok so i drew a triangle

would there be a specific side I put down the 340 and 1.61?

[silent plank]

in your diagram, there should be an indication of where the ground is

[vital perch]

so my adjacent

[silent plank]

no

[vital perch]

oh

what should I do for that if you don't mind me asking

[silent plank]

first start by drawing the ground

and vertical lines to represent a human and a skyscraper

[vital perch]

did that

[young bloom]

Give 2 sphere S1 has radius of 4 S2 is 6 they share both center point of (4:0:0)
Give point A (-4:0:0) and line d tangent with S1 and intersect S2 at B and C

In what condition does the area of the triangle ABC is the largest

And how to calculate it

[soft arrow]

Hello
I'm in need of calculating a "Look at" rotation/vector between 2 points in space. It is 2d vector operation
So far I have this in pyhton:

import math
# X = ------ x
# Y = |||||| Y
vecA = [100, 0] # look right 90 deg
vecB = [0, 100] # look up 90 deg
eye = [0,0] # centre of world - ignoring for now lets assume were at 0 anyway

dot = vecA[0] * vecB[0] + vecA[1] * vecB[1]
det = vecA[0] * vecB[0] - vecA[1] * vecB[1]

print(math.atan2(det, dot)) ## should give me 90 deg? if its in deg...

This should give me... some value, but all I got is 0 :- (

[fervent valley]

guys I need help

[tepid crow]

Well they have given the volume as 1000, so start from there

pi r^2 h = 1000

[fervent valley]

yep

[tepid crow]

And they have given the TOTAL surface area as T

TOTAL surface area will be the curved surface area + the area of the top and bottom circles

[fervent valley]

so that would be 2pirh + 2pir^2 right?

[tepid crow]

exactly

we basically got the answer already

see the RHS of what they're asking us to prove

we already got 2pir^2

[fervent valley]

so then I go back to this formula

[tepid crow]

yup

[fervent valley]

then I find the height?

[tepid crow]

and write 2pirh in terms of v

[tepid crow]

ye

[fervent valley]

Then I put that formula in the SA formula roght?

[tepid crow]

yup

you'll get 2V/r

[fervent valley]

then thats how u get the T ig

[tepid crow]

you got it dude

[fervent valley]

nicee

thanks man

[tepid crow]

anytime

[fervent valley]

but the next q asks abt a min SA

Why is a graph required for that?

[tepid crow]

you know differentiation?

[fervent valley]

hmm no not really

[tepid crow]

kk

im trying to think

nothing's coming up lol

I really think we need calculus for this

[tepid crow]

<@&286206848099549185> is it possible to find the minimum surface area here without differentiation?

[granite forge]

Can anyone help me solve my math problem about trigonometric ?

[tepid crow]

[granite forge]

Make The D and F to product

i'm stuck with that

i tried many times

many formulas

[tepid crow]

the question itself asks you to "convert into product"?

[granite forge]

yes man

[tepid crow]

do you know the formulae for sin a - sin b and cos a - cos b ?

[granite forge]

i know all of them

sum to product

product to sum

[tepid crow]

great

did you apply it in the sums?

[granite forge]

yes now im stuck

i use sum to product

[tepid crow]

🤔 yeah im getting stuck too, ill get back to you if i find anything

[granite forge]

okay sure thanks man

[frail nebula]

can someone help me?

[tepid crow]

yeah im not getting anywhere @granite forge

[granite forge]

Angle A = 180 - 90 - 37 , Side b = 15cm/ Sin(37), Side a = Cot(37) x 15 @frail nebula

@tepid crow thanks for your time anyway

[frail nebula]

thank you i have two more question

[granite forge]

@frail nebula do you know law of sines & cosines?

the formula?

or you're already pass that level

[frail nebula]

im pretty sure i learned it but ive forgotten it

[granite forge]

what about now

which formula are you in

[frail nebula]

wdym

[granite forge]

i mean there are many formula

like this

are you archive all of this?

these*

[frail nebula]

man i dont even know online messed me up

[granite forge]

but idk if i answer your problems it would too high for you or not

[granite forge]

like this

[granite forge]

in your problems they asking about what formula will ur use

they will talk about law of sines & cosines

[frail nebula]

its the law of sins i think

[granite forge]

i think i should give you an answer with that laws

[frail nebula]

oh wait i figured something out

thank u for ur help

[granite forge]

this must help you

[frail nebula]

and sorry for not making sense😅

[granite forge]

keep concentrate on study online

[old fable]

When drawing a normal line to PQ does the angle become θ/2 or does it stay θ?

<@&286206848099549185>

[shadow reef]

dont spam!

[old fable]

bro

[shadow reef]

also normal line throught where?

[old fable]

to R

[shadow reef]

through R?

[old fable]

yes

i want to compute PQ

[shadow reef]

i mean draw a normal line on PW through R

what does that look like?

[old fable]

its a right triangle

with angle theta/2 i suppose

is that right?

[shadow reef]

jup

[old fable]

so when multiplying it by 2 for PQ

it still stays theta/2

?

[shadow reef]

multiply by 2 for PQ for what?

for the lenght PQ ?

[old fable]

yes

[shadow reef]

just calculate 1/2 of PQ with the angle you know have

thats just a length

but its only half of what you need

[old fable]

yes

[shadow reef]

so take it by 2 in the end

[old fable]

yes

[shadow reef]

no need to mess about the angles

[old fable]

ohh ok

seems kinda counterintuitive lmao

[shadow reef]

lol you just dont have the right intuition yet

[old fable]

nvm im dumb as fuck

yeah i meant i was dumb that i didnt see it cuz I saw it right away after you said it lmao

what class is that?

[shadow reef]

maths

[old fable]

yes i know that lmao

what

calc 2

[shadow reef]

maths 2 i guess

[old fable]

huh

[shadow reef]

we dont have it split up like that

[old fable]

dont you have calculus?

like as a class

[shadow reef]

integrals, matrices, differential equations, fourier expansion is one class here. its called maths 2

[old fable]

ohhh i saw it was german

[shadow reef]

jup

[granite notch]

Any idea of how I should do this

[shadow reef]

#prealg-and-algebra

[old fable]

What did I do wrong? Idk how to proceed

im getting a imaginary term and idk if that's what supposed to happen

A(theta) and B(theta) is the area function

[old fable]

<@&286206848099549185>

[acoustic jungle]

@old fable what is the question

[slate pendant]

how do you find ab

[acoustic jungle]

you can either use AB = CD or prove that using the pythogorean theorem

[old fable]

find the lim as theta -> 0+ of A(theta)/B(theta)

A(theta) and B(theta) are the area functions

[acoustic jungle]

of what

[upper karma]

Amateur math

pff

[old fable]

A is semi circle

B is triangle

[old fable]

what

[acoustic jungle]

couldn't you find the answer online

[old fable]

no

[acoustic jungle]

pretty sure that proof or smth is in every calculus intro

[old fable]

what proof

[acoustic jungle]

for A do you mean segment

not semi circle

semi circle is half a circle

[old fable]

yes

area of a semicircle

[acoustic jungle]

wut

ok.

[old fable]

so 1/2pi r^2

[acoustic jungle]

nevermind then, it's not the proof I was thinking of.

[old fable]

ok

[old fable]

where did i make a mistake?

[acoustic jungle]

I'll take a look at it

you should probably post this in calculus

[old fable]

alright

[acoustic jungle]

I haven't taken calculus, but here's what I have. wlog the side lengths be 1
area of triangle is sinx/2. Area of semicircle is d^2pi/8. (d is diameter) d^2 = 1^2 + 1^2 -2cos(x) (cosine law). Substitute in, the area of semicircle is (1-cosx)/4 @old fable

you'll need to figure out the limit yourself

[old fable]

wlog the side lengths be 1
area of triangle is sinx/2.

what do you mean by this

[old fable]

Nvm

I already figured out the answer

[acoustic jungle]

@old fable what did you get?

[north ether]

anyone able to help me with some questions on Interior and Exterior Triangle Angles

[distant ravine]

sure! what's the problem?

[alpine wadi]

can someone help me with these questions

[acoustic jungle]

sin(a+b) = sinacosb + sinbcosa

have you done that

[cloud stump]

Use the inverse trigonometry complex cool method

Convert into arctan and then do arguments

$\arctan{\left(\frac{a}{b}\right)}+\arctan{\left(\frac{c}{d}\right)}=\arg{(b+ai)}+\arg{(d+ci)}=\arg{((b+ai)(d+ci))}$

[somber coyote]

September22nd

[covert swallow]

Use the addition formula. Then for the first part, draw a right triangle with sides 7, 24, and 25, and mark the angle theta where cos(theta) = 7/25, and find sin(theta). Repeat with the other angles and add everything up

[old fable]

0 I didn’t use you method

My first attempt I did Pythagorean theorem but that came out to complicated

But it was way easier

[alpine wadi]

no im not quite sure how to do that

[acoustic jungle]

then you'll need to use that and do what mchen10 said after.

[acoustic jungle]

nice, I also got 0.

[old fable]

It was way easier than you method haha

We over thought it

[acoustic jungle]

what's your method then?

if the radius was 1, the area of the triangle was sin x/2. Using the cosine law, the area of the semicircle is (1-cosx)/4

[old fable]

PM was 10sin(θ/2)

PQ was 20sin(θ/2)

MR was 10cos(θ/2)

And that was just by looking at the triangle

So calculating the limit it was like π/2 • lim tan(θ/2) as θ ->0+

[acoustic jungle]

nice, that works too.

[old fable]

Yeah so no cosine law haha

[tender palm]

can u guys help me

[acoustic jungle]

If a triangle has a right angle, then a^2 + b^2 = c^2. If a^2 + b^2 = c^2, then the triangle has a right angle. In other words, a triangle has a right angle if and only if a^2+b^2=c^2.

[tender palm]

thankkk youu so much

[signal phoenix]

Help?

[acoustic jungle]

let ? = x. Use similar triangles.

@signal phoenix

[upper karma]

How to find if it is cube instead of square

?

[covert swallow]

try cubing the whole equation, and plugging in what you know

[upper karma]

Ok thanks

[old fable]

Cosec θ 🤢

[alpine wadi]

like this?

[runic magnet]

Yeah ok

You need to use the formula for the slope first

[somber coyote]

ΔToni=Toni-Toni_0

[runic magnet]

y_2=-9, y_1=-7

x_2=-6 x_1=1

and then, when you have the slope, m, you need to use the formula point-slope

[somber coyote]

ΔToni=Toni-Toni_0

[runic magnet]

then replace y_0 and x_0 with any of the two points given, (1,-7) or (-6,-9)

and there you have it, your function equation

Hope that this helped a bit

[rigid acorn]

Anyone here?

[humble pulsar]

no, clearly no one is here

[runic magnet]

people these days

[humble pulsar]

k

[runic magnet]

I was agreeing with you

[humble pulsar]

no reason to ping me

[runic magnet]

sorry ig pfft

[pallid plover]

Can somebody help me with this question?

<@&286206848099549185>

[acoustic jungle]

@pallid plover is h the midpoint of the lines

[pallid plover]

I don't believe so, I'm pretty sure it's arbitrary?

[acoustic jungle]

what is the answer supposed to be in terms of.

[pallid plover]

im not exactly sure

[rotund scaffold]

help?

[silent plank]

what have you tried?

[rotund scaffold]

Nothing

[signal swallow]

well try somethin

what do you know that can relate the side of length 13, side x, and the angle 40 degrees?

[rotund scaffold]

Can you give me the equation?

I don’t know it

[signal swallow]

if you don't know the basic trig equations you should probably watch a video on it

before attempting to solve problems using trig

https://www.youtube.com/watch?v=I3jyBUyjg48

[lavish geyser]

why if i use theorem of marlen or euclides theorem gives me different results?

[simple rune]

What are the rotational symmetries of the rational plane Q^2?

[dark sparrow]

those would be the rotations by angles whose sine and cosine are both rational i suppose

[spring tusk]

can sum1 help me on 2

[waxen flume]

@spring tusk exam?

[spring tusk]

nah jus homework

[hollow coral]

sin(60)/cos(60) = tan(60)= sqrt(3). Dunno if this is what they want tho.

[rigid acorn]

Do I use this formula to solve b?

[rotund scaffold]

2 rad 6/rad 3=? in simplest radical form?

[humble pulsar]

what's rad?

radians?

[rotund scaffold]

radical

[humble pulsar]

specify which radical

[rotund scaffold]

2

[humble pulsar]

so sqrt()

[rotund scaffold]

can someone explain steps to solving this?

[silent plank]

consider the special properties of the type of triangle you are given

[earnest basin]

Hi i need help with trigo

[signal swallow]

have you attempted any of them?

[upper blaze]

This one looks interesting to try, do you have any ideas?

[high bramble]

Hm, good question

Mhm

[marble copper]

POOP

[dark sparrow]

wrong server?

[upper karma]

Damn I thought you were talking about the rectum

that chamber at the end of the large intestines

[pallid plover]

Nope, I'm stuck

[spring tusk]

Given tan 2theta = 0.8234 solve for 0 degrees <= theta <= 720 degrees to the nearest degree (Hint: there are several angles that satisfy )

how would i solve this?

do i jus do the recipricle to find the answer?

[spring tusk]

<@&286206848099549185>

[hazy valve]

so the inverse of tan (and then dividing by 2) will give you 1 answer
but from 0 to 720, tan(2x), you'll get the same answer multiple times, so you need to think about the period of tan(2x)
(also reply to my messages so i get ping'd)

[spring tusk]

so do i just get the recipricle answer and subtract it by 720?

and devide by 2?

[hazy valve]

i think we should do a few steps before doing this question if that's alright

can you solve tan(x) = 1 for me, and how you get there?

[spring tusk]

tan^-1 1 =45

[hazy valve]

be careful with brackets but yes, that's right
what you're using here is "tan inverse" (sometimes called "arctan") - not to be confused with the "reciprocal", which means 1/tan (sometimes called "cot"). i know they use very similar notation but unfortunately it's something to keep in mind

[hazy valve]

now, 45 degrees is one correct answer. can you give me another answer that solves tan(x) = 1?

[spring tusk]

uh i cant think of one

root 2025?

[hazy valve]

here's a bit of a clue

[spring tusk]

225

[hazy valve]

yes exactly
now, tan is periodic by 180 degrees. that means that you if you add (or subtract) 180 to any degree, you get the same value of tan (notice that 45 + 180 = 225)
with that, can you give me another answer to tan(x) = 1?

[spring tusk]

405

[hazy valve]

perfect

now let's try tan(2x) = -1
can you give me a few answers for that?

[spring tusk]

135,315,495 and so on

[hazy valve]

so tan of (those values) gives me -1
but i was very sneaky and asked for tan(2x) = -1
can you modify your answer? (also reply to my messages to ping me)

[spring tusk]

67.5,157.5,247.5

[hazy valve]

perfect 🙂 now you can try your original question

[spring tusk]

ok i think i got the answer for it

[spring tusk]

i got 379.73402064

[hazy valve]

i think you're slightly off by a few digits, just double check your calculation (also don't forget to give your answer to the nearest degree)

[spring tusk]

ok thanks a lot man

[hazy valve]

want me to check your final answers?

[spring tusk]

ya if u can

[spring tusk]

i double checked it n got the same answer

[hazy valve]

great, if you're happy 🙂

[earnest basin]

@hazy valve hi can u help me with this

[tame pumice]

I’m just looking and I am wondering

What are we solving for?

[earnest basin]

Huh

[hazy valve]

i don't understand the question?

also in general you're better off asking in the Math Help category, and if no one responds in 15 mins, using the @helpers ping

[acoustic jungle]

p = QR (by opposite side convention)

then use the sin law.

[upper karma]

yeah that makes sense but they should specify to avoid confusion

Like the same triangle, not the same type

I mean you could have the same type of triangle that being an equilateral but not congruent to the initial

So you could have been given a unit equilateral

hm you're probably not getting me

idk how to explain it verbally

What i'm asking is this

so lets say you've been given a 5 cm equilateral triangle

are you only allowed to use that

so we can't change the size of it

yeah but angles won't change

[upper karma]

so take in this image for example

all interior angles are 60

apart from the exterior angles

[spring tusk]

quick question both of these can be found in quadrant 2

[lyric bridge]

yes i believe so

[versed river]

no lol

cos theta cant be less than -1

the latter can occur in 2 and 4

[toxic ravine]

could someone help me with f

[silent plank]

What's the question asking for?

[unique condor]

İs there any proper method other than solving for annoying equations and substuting

[earnest basin]

Help meeeee

<@&286206848099549185>

[silent plank]

what have you tried

[earnest basin]

Help me

[humble pulsar]

someone did try and help, you ignored it

[earnest basin]

Where?

[humble pulsar]

literally read after you posted the picture..

[earnest basin]

Idk who she's talking about i didn't know

[humble pulsar]

I mean you also pinged helpers right away, so clearly you havent read the rules

clearly they were talking to you...

who else would they have been talking to?

[earnest basin]

A lot were asking

Not so obvious should've pinged me

[humble pulsar]

it was very clearly directed at you, given it was 7 minutes after you posted, and the previous question was from an hour prior to yours

[earnest basin]

I didn't read the hours after the girl posted her question.

U should've say, " oh someone will help me, look up"

I didn't IGNORED it

That's different

[humble pulsar]

then why didnt you respond to it

you also have yet to answer the question Ramonov asked you

[earnest basin]

What respond? Why would i respond if i think i am not the one ramonov talking to

Didn't i told u multiple times i didn't know ramonov is talking to me bobalicious kengenemers

[silent plank]

regardless, now you know

[earnest basin]

Yeah i know now

[silent plank]

now since you want a ping to make sure:
@earnest basin
what have you tried?

[earnest basin]

I answered it already, ramonov. Thanks

Thanks mosh huh for being the bobaers

[acoustic jungle]

what is a bobalicious kengenemers

[toxic ravine]

sorry about the confusion, i shouldve added detail and not sounded so vague but essentially i needed to determine an equation for F

[silent plank]

did you still need help with the question?

[toxic ravine]

yeah

[silent plank]

are you familiar with the standard equation of a circle?

[toxic ravine]

not really

[silent plank]

the equation of a circle with centre $(h,k)$ and radius $r$ is:
$$(x-h)^2 + (y - k)^2 = r^2$$

[somber coyote]

ℝamonov

[toxic ravine]

yes, i remember noe

now*

[earnest basin]

Beautiful😀

[south igloo]

Can someone explain this

[restive nexus]

<@&286206848099549185>

[cursive jewel]

what part do you need help with?

[south igloo]

All

[cursive jewel]

alright so a line is composed of infinitely many points

but a few points are defined in the graph

what are 2 points that form a line where one of them is C

[south igloo]

yes

[wheat dome]

I want to put a bullet in my head

the law of cosine/sine are so annoying

[cursive jewel]

yeah but they are pretty helpful

[dark sparrow]

@wheat dome if you are suicidal then you should contact a suicide hotline and/or get a therapist, preferably both.

[wheat dome]

oh I'm not lol

I prob should have said that 😳

I've never been more happy to be fair

[wheat dome]

yea I guess

and too the calculator does make it much easier

[upper karma]

Okay, I've been studying trig for about 2 years now and I still don't know what sine ACTUALLY means, when you say sin(30) what are we doing to that 30? why do we have trigonometry at all? I can't comprehend the sine function or the curves in my head, just what??

<@&286206848099549185>

[humble pulsar]

it's the y value of the co-ordinate on the unit circle when the arm makes a 30 degree angle

[upper karma]

I still don't quite understand, can you elaborate?

so is there a specific "expression" we have to follow? like f(x) = 4x+3 so sin(x)= ?

what are we doing to the x?

@humble pulsar

[humble pulsar]

I mean, yeah there's taylor expansions for sine and cosine, but what I said is sufficient

a point (x,y) on the unit circle can be written as (cos(t),sin(t)) for the corresponding angle t

[upper karma]

So what lead to that said (x,y) = (cos(t),sin(t))?

@humble pulsar

[humble pulsar]

definition of cosine and sine

you also dont need to ping me every message

[upper karma]

nono, why do we have sine, cosine or anything really? how was it invented and how was it defined as originally?

[upper karma]

yeah, sorry

[humble pulsar]

"sorry for pinging, let me ping you again"

[upper karma]

yeahh, no pings, got it lol

[humble pulsar]

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

modern math just stuck with sine cosine tangent secant cosecant and cotangent

[upper karma]

I'm much more confused now, What is chord theta? How was it derived? Why did they use sine to define sine? I've had the same problem with other explanations, What are we doing to that 30 in sin(30)??

[humble pulsar]

"why did we use sine to define sine"
this floor is made of floor

and idk im not a historian

[flint nova]

Sin(x) can be defined in a lot of ways. The most common one is the one defined for right angled triangles where sin(x) = opposite side / hypotenuse.

[flint nova]

This can be used in the real world to find angles given sides or sides given angles.

[low estuary]

can someone help 🥲

all of them are asking the degrees dont care about the lang

<@&286206848099549185>

[grim rampart]

can someone help me with a proof

i got stuck

i need to prove

d => e

this is what ive tried so far

if someone can please help

im not sure if im on the right path

[upper karma]

I was trying to show in $\mathbb{H}$ -Poincaré half-plane model- that hypercycles defined for a given line $l$ and a given $a$ in $\mathbb{R}$ as $$H(l)={z \in \mathbb{H} : d_{\mathbb{H}}(z,l)=a } $$
are not lines.

To prove something I tryed to argue that given that $Mob(\mathbb{H})$ acts transitvely on the lines of $\mathbb{H}$ I can find the hypercycles of the imaginary axis and via the $\gamma \in Mob(\mathbb{H})$ that maps my line in the imaginary axis i can state $$H(l)=\gamma^{-1}(H(\gamma(l))$$

Where I'm using the fact that a Möbius transformations preserves distances. At this point I defined [already here i have doubts] $d_{\mathbb{H}}(z,l)$ for a given $z \in \mathbb{H}$ and a line $l$ as the inf of the distances $d_{\mathbb{H}}(z,w)$ with $w \in l$.

Given this and the general assumption $z=a+bi$ and $w=ki$ i'm left with trying to calculate the inf of $$arccosh(1+\frac{|a+(b-k)i|^{2}}{2bk})$$.

Is any of this correct? There is any "distance from a given point to a given line" standard formula in hyperbolic geometry? Thanks for any help and hint.

[somber coyote]

Stephen

[grave marsh]

hi

we get, 180-(96+x)=180-(96-x+56) implies x=28 and angle BAC=96-x=68°

[grave marsh]

btw can you share the question booklet with me?

[low estuary]

oohh

danke sehr schön

[low estuary]

ofc but its in turkish

[grave marsh]

nvm can you share? they look awesome

[low estuary]

it is this and if i find the pdf i will send ti

it*

[grave marsh]

i think i got the pdf. thanks

[low estuary]

nice 👍👍👍

the yellow ones are easy, blue ones are medium and pink ones are the hardest ones

[grave marsh]

okayy

[grim rampart]

what book pdf is that

is there someone who know geometry

[cursive jewel]

yeah i took geometry

[uneven dust]

Geometry

[upper karma]

use sum to product formulae

club the 1st and 3rd terms

@uneven dust

[uneven dust]

I converted -cos2pi/7 to +cos5pi/7 and then we get all 3 in the n*pi/7 where n is odd..but iam stuck there

[upper karma]

try what I suggested

[uneven dust]

Ok sure I'll get back with u

[upper karma]

club the 1/7 and 3/7 terms

[uneven dust]

Yea I'll do it

[naive storm]

find the sector bounded by an arc whose measure is 2/3 pi radians and whose diameter is 4 ft to the nearest whole number

[dark sparrow]

what do you mean by 'find the sector'?

@naive storm

[naive storm]

area of sector*

[dark sparrow]

and 4 ft is meant to be the diameter of the circle that the arc is taken from, right?

[naive storm]

yeah

[dark sparrow]

okay, so what is troubling you?

[dark sparrow]

it seems that @naive storm chose not to respond to me even two and a half hours later

who, me? or them?

[dark sparrow]

where's D?

do you find the 2D equivalent of this bizarre too?

that the two tangents to a circle from the same point have the same length?

[upper karma]

Can sm help with thiss

[dark sparrow]

do you know how to find the sum of all interior angles of a polygon?

[upper karma]

No sorry

[dark sparrow]

(hint: it depends only on the number of sides)

[upper karma]

Yeahhh... I don’t know what that means

So do you mind helping me solve it?

[dark sparrow]

to find the sum of all interior angles in a polygon, you don't need to know its exact shape, only how many sides it has

and the sum of all interior angles in a polygon with n sides is 180*(n-2) degrees.

[upper karma]

It has 10 sides

[upper karma]

So what do I do

What’s the formula

Would is be (180) 10-2?

@dark sparrow

Since it has 10 sides

[dark sparrow]

me: [literally gives you the formula]
you: what's the formula???

[dark sparrow]

no. be careful about your placement of parentheses

[upper karma]

I got this is this right?

[dark sparrow]

you've applied the formula i gave you correctly.

congratulations.

[upper karma]

Thank lmao

[silver fog]

I tried expanding and stuff but nothing seems to help

[flint nova]

do you have to prove the sin identity given that the cot identity is true?

or the other way around xD

[dark sparrow]

both

[silver fog]

Both ig

[dark sparrow]

this looks like it may be very painful to do unless there's some kind of trick for it

hm. i have an idea.

but i'll need some time to write it out in full

all i've managed to do is to reduce the right half of the iff to $$(\cot(\theta) - \cot(A))(\cot(\theta) - \cot(B))(\cot(\theta) - \cot(C)) = \frac{1}{\sin(A)\sin(B)\sin(C)}$$

[somber coyote]

Ann

[dark sparrow]

dunno how helpful that is

[flint nova]

you can prove one of it if you assume the other to be truse

*true

[dark sparrow]

i know how to prove iff-statements, thank you very much

[flint nova]

https://www.toppr.com/ask/question/consider-a-triangle-abc-such-that-cot-acot-bcot-ccot-theta-thensin-athetasin-bthetasin-ctheta/ this is helpful?

[dark sparrow]

man the typesetting on that site is just painful

[viral kestrel]

Yo how does this work?

I don’t understand

[fossil yacht]

what are you trying to accomplish?

[spice plank]

How do I start Bc I have no idea how to start this in a 2 column proof

[rotund garnet]

First of all, you learn that D=E

And you learn that ABC = ACB

So: EBC = DCB (E=D)
DBC = DCB (ABC = ACB)
DBC = DBC

[covert swallow]

$\angle ABC \cong \angle ACB, and \angle BDC \cong \angle BEC$, because angles with the same measure are congruent

[somber coyote]

mchen10

[covert swallow]

you can use the third angles theorem to determine that $\angle EBC \cong \angle DCB$, and because the angles are congruent, then their measures are equal

[somber coyote]

mchen10

[silver fog]

Can someone pls help out ....I tried squaring the original eqn and writing and using multiple angle formulae but ain’t working

[silver fog]

I did this too but to no avail...Could u be a bit more specific

[somber coyote]

Cooler Euler

[rotund linden]

@silver fog

[somber coyote]

Cooler Euler

[lime shuttle]

Hello,
is that someone would have the proof of Simpson's formulas with the complexes

[dark sparrow]

...what formulas?

[obtuse patio]

this is pre uni

not after uni

[lime shuttle]

Tous formulas cos(p)+cos(q)=2cos(p+q)/2*cos(p-q/2

I think we should use Euler's formula

[dark sparrow]

??

oh you mean the sum to product formulas?

[lime shuttle]

Yes this is Simpsons formula

[dark sparrow]

first time i've heard it called that

in any case, their proofs rely on the angle sum/difference formulas, which can be proved in the complex case without much trouble

[lime shuttle]

But i would like to solve it with the complex

[dark sparrow]

what do you mean

you want to spend hours and hours painstakingly verifying cos(p) + cos(q) = 2 cos( (p+q)/2 ) cos( (p-q)/2 ) with careful manipulation of complex exponentials?

[lime shuttle]

I m in upper sixth and not in a university. It s a little bite complicated for me but i found the solution.

[dark sparrow]

$\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$ is not uni-level

[somber coyote]

Ann

[lime shuttle]

Simpson formula IS uni level

[small prawn]

no it's not i have it in my textbook too

upper sixth student as well

and i thought simpson's formula is for integration?

[earnest echo]

Why is it called Simpson formula
I'm pretty sure greeks knew it

[ivory steeple]

This may be an algebra problem. Let me know if anyone can help. Translating y=2^x right 3 units and up 5.

[bitter spoke]

simpson's rule for approximating the definite integrals?

[earnest echo]

No, sum to product formulas in trig

[covert swallow]

translating a function to the right by n units is the same as subtracting n from x first, then applying the function. translating up is just adding to the function

[lusty pollen]

Hey guys I kinda need help I’m stuck on trying to figure out what delta math is trying to tell me

[lime shuttle]

In France, we learn this formula in the first year of university called "classe préparatoire". With my professor we proove it with the complex.

[robust crane]

does any one know if there is an easy way to test whether a point is interior to an arc as in facing the arc from the inside vs from the outside like this:

note that blue point could easily be outside of the radius too on the other side of the circle

so radius check won't cut it

[maiden path]

You can check the concavity of the arc around that point

[maiden path]

https://www.cs.uleth.ca/~fitzpat/apex-calculus/sec_concavity.html

Might help

[robust crane]

hm that looks advanced 😄

will try to decipher it

[rotund garnet]

You could draw a line between the point and the arc so that the line has a 90° with the arc

[robust crane]

not sure i understand

can you draw what you mean

@maiden pathi tried convexity test but it doesn't work for me

the way i tested it might be flawed tho but i created a polygon of 4 points but the problem is the polygon is complex with intersecting edges so it makes it very tricky

[rotund garnet]

Can you show it?

[robust crane]

@rotund garnet

i was hoping to get a nice quadrilateral then use cross products to test convexity

but since the edges some times intersect it didn't work as i hoped

maybe i need to use a different set of 4 points

[rotund garnet]

Why you don't you a straight line?

[robust crane]

what do you mean

[rotund garnet]

P -> Arc so that the dot product is 0

[robust crane]

im not following ?

how would a dot product of an arc to a point work

[rotund garnet]

https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/differentiating-vector-valued-functions/a/derivatives-of-vector-valued-functions

[robust crane]

im not seeing why derivatives help me here though i can't visual it

[rotund garnet]

When you got a straight line from the x-axis through the arc and the point, what could length of O -> A and O -> P say?

[robust crane]

oh you mean check the directions of P-O and ArcP - O ?

[rotund garnet]

The direction ist the same because it's 90° to the arc

[robust crane]

well my original thought was to check the direction to the radial vector like this

so if the two lines point the same way i must on the underside of the curve

but it didnt work

[rotund garnet]

I don't understand why you don't you only a straight line and the angle

[robust crane]

as you see here the dot is negative but im still on the outside of the curve

[robust crane]

because i don't know what you mean by that

can you draw what you mean ?

[robust crane]

but you've got 2 arcs going on there im confused by what is going on there

[rotund garnet]

That's the sign for a 90°

[robust crane]

yeh but your curve is multiple arcs

[rotund garnet]

And?

[robust crane]

im just trying to check this nothing more

im not seeing how your diagram solves that

for convexity

[robust crane]

nevermind i think i got it

[somber coyote]

ryam

[lavish torrent]

quadrature?

I'm just curious, idk what a quadrature of a parabola is

[upper karma]

hi

i joined this place because i found a proof that i want to share but i am bad at proofs but i discovered this myself

so it starts with these triangles, and the angle between these two triangles (assuming that they are right triangles) is supposed to be 45

idk how to prove it formally, but i made i diagram proving it is 45

[acoustic jungle]

or use the cosine law because the middle triangle is a sqrt5 sqrt10 sqrt5 triangle

cos(theta) = 1/sqrt2 => theta = 45

[upper karma]

i never heard of a sqrt5 sqrt10 sqrt5 triangle in geometry idk how

i discovered it on accident while working on a ball

[acoustic jungle]

I should be more clear @upper karma

[upper karma]

hmm

this triangle?

[acoustic jungle]

Yes. The side lengths are from the pythogorean's theorem.

[upper karma]

and since all triangles are "upright", you can tell that there is a right angle there by calculating the slopes (-2, 1/2) and seeing that they are negative reciprocals of each other. knowing that it is a right triangle and isosceles, you can now tell that that angle must be 45

[acoustic jungle]

yes, that works as well.

[grave marsh]

hey which application do you use to edit these figures?

[upper karma]

i use gravit designer, its an in browser based vector graphics editor free to use

[grave marsh]

okay nice

[upper karma]

its not for graphing equations though, you cant really make things like parabolas and waves

[pure bronze]

Do you still need help with that?

I might have an idea of maybe how to do this, it involves trigonometry though, if that’s fine with you

[untold cosmos]

man i really need help on this

[covert swallow]

all of them?

[upper karma]

gimme a sec

for 33, angle XBC has an angle of 102 degrees i think

that is because of the angle properties in parallel lines and stuff, this is only possible to solve because it is an isosceles trapezoid

[covert swallow]

since the trapezoid is isosceles, the base angles are congruent, so the left lower angle has a measure of 78 degrees. BC is parallel to AD, and by the same side interior angle theorem, the angle is supplementary to angle A.

[untold cosmos]

sadly yea

[upper karma]

idk how to do 34, i think you do 57 - 21 then add 57 but im not sure

[arctic glade]

anyone here good at proofs? i really need 2 proofs done by tomorrow

[dark sparrow]

we won't do your homework for you here

[arctic glade]

never said i need you to do my hw but ok

[humble pulsar]

have you tried asking the question...?

[dark sparrow]

sure did sound like you wanted someone to do those proofs for you

if that wasnt the case then you gotta word your questions better

[untold cosmos]

anyone know how to do this

[subtle palm]

hiii i might can

for 34 use similarities between KLAB and KLMJ

since 2*LA = LM

and 2*KB = KJ

so it should be smooth sailing from here; diff between KL and AB are half as KL and MJ

that's the only one i can do, others seem jibberish

#❓how-to-get-help for the rest

[subtle palm]

that is correct

lol i just realized mchen10 did 34

DAGNABIT

[limber vapor]

i love geometry

[subtle palm]

me too

[limber vapor]

i rlly love it more than anyone else

[subtle palm]

oh is that so

[limber vapor]

wait a sec

u have 11 at the end of ur name too

[subtle palm]

i guess cos i now moved a bit more towards alg 1

linear*****

[limber vapor]

and wait a sec i rlly love kar98 and scope x6 too

[subtle palm]

if you wanna chat go to #math-discussion

or #chill for non math

[subtle palm]

sorry your hints are beyond comprehension

have you ever TRIED cod ww2?

it's the best interpetration

[limber vapor]

no

im talking abt ur bio dude

[subtle palm]

then when the hell did you ever use a kar98

yea in what game

[acoustic jungle]

Ohh he asked a question lol

I thought he just said "I need help" and posted nothing

[limber vapor]

alot i think

[subtle palm]

also move to #chill pls \

[limber vapor]

i know every sniper ever existed

without playing a game

[acoustic jungle]

2*57=JM+21. It's a theorem

you can prove it with similar triangles and area.

[subtle palm]

lol everyone got 34

you seem qualified to do the exercises part

we already did 33 34

[subtle palm]

name all kar98k zooms then

[limber vapor]

i mean it depends

on what game u r playing with kar98

[subtle palm]

not the game

[limber vapor]

ok lemma see

[subtle palm]

i meant the alltime hit Real-Life 2 Alyx

[acoustic jungle]

oh nevermind, he did solve it

I did not see.

[subtle palm]

bruh cheating

[limber vapor]

x4,x6 and x8

no

i didnt google nothing

[subtle palm]

"ok lemma see" do you have a list of scopes from the 20th century lol

[limber vapor]

i meant "ok lemma think"\

[subtle palm]

ah yes the all-new innovative mind-seeker

where you can see minds

[limber vapor]

bruh dude

[subtle palm]

nothin, just sussin u out

btw it is correct, there are x4 x6 and x8

ok now can we move to #chill

[acoustic jungle]

For exercise 1. Use congruent triangles. 2x+20=5x-4. y^2+26 =90 because that is a right angle and AC is double of AF

for exercise 2. Use congruent triangles again

for exercise 3, it seems like trig bash to me.

[dark sparrow]

you mean if a light ray lands exactly on a corner?

it gets scattered.

i suppose so

though a single ray would just bounce off randomly

i think you're overthinking it

in the real world, no corner is perfectly sharp. if you zoom in far enough you'll find rough spots, especially at corners like what you've depicted

¯_(ツ)_/¯

go ahead ridicule me

[patent plume]

Maybe some sort of quantum shenanigans

[grave marsh]

how do you define a tangent

i mean i only know that "tangents are curve that touch a curve at exactly one point"

idk anything else

so you can make this curve yes?

x axis

[dense bough]

it's undefined at a sharp point

[grave marsh]

isnt x axis the tangent line??

[dense bough]

the proof is that from the left, the limit is -x but from the right, the limit is x]

thus we have two different tangents which can't happen

[grave marsh]

no it wont?

a line passing through (0,0) not x axis intersects the parabola at two points

[grave marsh]

isnt that circular reasoning? cuz you define the derivative at the point to be the slope

[dense bough]

well my point is that it's non-differentiable at x = 0

[grave marsh]

is this true? if you zoom in it intersects twice

[dense bough]

so the tangent doesn't exist

[grave marsh]

but it's kinda circular reasoning ig becuz tangents come before calculus

i mean slopes are primitive than differentiation

[dense bough]

you need calculus to rigorously define tangents

except when the gradient is constant, such as for a straight line

[grave marsh]

okay