geometry-and-trigonometry

[outer lagoon]

Can someone explain double and half angle formulas and any good way to memorize them

[near timber]

why cant i just use the negative angle?

[dark sparrow]

you can

it'll be offset by 2pi from their solution

but if you're solving an equation of the form cos(x)=c that doesn't matter

[near timber]

ohh ive been so confused because when i graphed it one of the awnsers was correct but the (negative) angle one was off by alot

[terse breach]

I'll translate this

In an ABCD romb, the diagonal AC = 12, the AB side is equals to 3 square roots out of 5. Find the tangent of the angle BAC

Is this the correct solution?

[dark sparrow]

@terse breach is it ok if i respond in russian?

[terse breach]

yes

[dark sparrow]

что это у тебя вообще за расчеты?

откуда взялись числа 45 и 36?

и откуда взялось, что 9 = 3?

[terse breach]

3 из корня 5 будет 5 если мы взять формулу в квадраты

[dark sparrow]

??

не поняла

[terse breach]

вот пример

[dark sparrow]

то есть ты вычисляешь BO по теореме Пифагора

[terse breach]

да

[dark sparrow]

почему бы об этом сразу не написать?

[terse breach]

было бы и лучше так

[dark sparrow]

короче, значения BO и tan(BAC) у тебя правильные. проблема - с оформлением

[terse breach]

Да, у меня получается большой кошмар на доске в школе, хотя решаю правильно

Спасибо за обзор зато

[dark sparrow]

в работе было указано нецелые ответы писать десятичными дробями? если нет, то я бы записала ответ как 1/2

[terse breach]

Видно, что не указывают, в какой форме записывать ответ, так что я написал в десятичных

[dark sparrow]

лучше так не делать

простые дроби предпочтительнее

особенно если что-нибудь типа 1/3...

[terse breach]

понятно

[terse breach]

теперь формулировка лучше?

[dark sparrow]

в четвертой строчке в пункте 1 вместо плюса должен стоять минус

[terse breach]

исправлено

[dark sparrow]

вот так нормально

[terse breach]

понял

[near timber]

to determine the highest value of 2sin(3x)+cos(3x), do i have to take the derivative and then look for max min values or is there a more efficient way?

[dark sparrow]

sure is

$a \sin(x) + b \cos(x) = \sqrt{a^2+b^2} \cos(x - \varphi)$

[somber coyote]

Ann

[near timber]

thanks!

[upper karma]

why was angle 60 chosen here?

[magic lichen]

cuz theta' ig

[terse bane]

Line h goes through point (0, -k) and has a gradient of 1. If that line is tangent to the circle which equation I already sent above, then the value of eight times all possible values of k is?

the radius is an imaginary number which means the circle doesn't even exist, but can you get the answer anyway?

here's the slope intercept form of the line

then just substitute it to the equation

and since the line is tangent to the circle, that means the discriminant is equal to 0

and the value of all possible ks multiplied is

since they ask for the value of 8 times all possible values of k

is this a valid answer?

[vestal siren]

#3-12

[whole granite]

try to show some effort, or at least split the questions, it'll help you to get help

[vestal siren]

okay

[vestal siren]

wait wdym split the questions

like put it in sections

?

[whole granite]

post only a single question, solve it, then post the next one, not all in one like you did in the image

[long parcel]

Could somebody pls explain this to me ?

[vestal siren]

okay

[soft marten]

help?

[desert shore]

this is kind of silly but does anyone understand what this question is asking me

[formal hatch]

Hello, I am just a bit confused, in the textbook (french) it says that the value of the angle between vector u and vector v is 7pi/12; But I don't see how

It says that the angle between two vectors is the smalles angle theta when they have the same origin

[honest thistle]

take hight as Y and t as X and do the quadratic formula

[desert shore]

No i meant c in specific

[honest thistle]

just do t=0 lol

should be just before the time of launch

[desert shore]

thats what i assumed but needed to make sure im understanding this correctly

ty

[honest thistle]

do you have the answers?

[desert shore]

no

[honest thistle]

well i would do t=0 idk about what else we could do

[onyx wasp]

Anyone who knows the name of this figure? It’s supposed to be a topped off cone with a curved side

[signal swallow]

frustum

oh wait curved side

:(

[onyx wasp]

The side is a x^2 curve

And I can’t find it omg

[honest thistle]

lets say sinx = a so the question becomes a quadratic formula:
6a^2 - 5a + 1 = 0
if we solve for a we get 3a = 1 and 2a = 1 so a= 1/3 or 1/2

[signal swallow]

don't do people's work for them

[honest thistle]

for 12
from the B point draw a line to |AC| so that it makes a 90° angle lets call the point that it contacts as D.
DBC angle is 45° and you can try to find |DA| from 30 60 90 triangle and get |DC| along with |DB| and find tan45

[honest thistle]

alright, im sorry

[signal swallow]

again, don't do people's work for them, not to mention this is not correct

[vestal siren]

dang 💀

[north heart]

lmfao

[crisp turret]

How would one solve a problem like this

[tall coral]

It's between pi16 and 324 if that helps at all
Idk how to solve it though

[tall coral]

you can break the shape like this

shouldn't be that bad from there

[tall coral]

wait nvm

[low jungle]

hey why cant arcsin have its range restricted to [π/2,3π/2] instead of [−π/2,π/2] it would still be injective right?

[dark sparrow]

do you mean [+pi/2, 3pi/2]?

[low jungle]

yeah my bad

[dark sparrow]

well you could take the inverse of the restriction of sin to [pi/2, 3pi/2] sure

that exists

it's just different from the canonical restriction

and is somewhat less elegant

[low jungle]

i see.. so its just a convenient convention of sorts right?

[dark sparrow]

essentially yes

[low jungle]

alright thanks!

[compact jasper]

that is identity rule ? sin(a+b)=sinacosb+cosasinb
but why ? why just not sin (a+b) = sin(a) + sin(b) 🙂

[nocturne remnant]

Why not?

[dark sparrow]

sin(a+b) = sin(a) + sin(b) is simply not true lol

sin(90°+90°) ≠ sin(90°) + sin(90°)

[nocturne remnant]

“Hey god why didn’t you make sin(a+b)=sin(a)+sin(b) that would make things so much easier bruh”

[mild briar]

Just like (a+b)² = a²+b²

It would be pretty, but it's not reality

Not to mention, that would imply sin(x) = λx for some constant λ

Which would bode ill for the shape of circles as we know them

[limpid spruce]

Higher than snoop dogg

[simple rain]

are supplementary congruent angles right

[slate phoenix]

https://en.m.wikipedia.org/wiki/Pitot_theorem
@aakhilv#0001

[lethal robin]

does anyone know how to approach geometrical questions that want you to proove something

i look at the question, i think about how i can solve it

no methods help

then i look at the solution and it seems so obvious

[simple rain]

send me one of the questions

[lethal robin]

Aight

[lethal robin]

Prove that the centre of a circle touching 2 intersecting lines, lies on the angle bisector of the lines.

[simple rain]

do you have a diagram

[lethal robin]

The question doesn't have a diagram.

[fleet idol]

Can anyone explain me how to calculate side BA please?

[trim breach]

BAC and BAD are both right triangles, if I am reading this correctly. AB is a common side. You should be able to use a system of equations to represent the situation.

[vagrant ore]

Is the answer 1 line?

[trim breach]

Yes.

[pastel pewter]

can some one help me to find the measure of the angle between the diagonal of trapezoid inscribed in circle?

[light fable]

what did you try?

you were on the right track using cos

what's the equation with cos from soh-cah-toa?

right, and what is it in your case, with those labels

I mean write the equation about cosine

with your given diagram

[wheat walrus]

similar concept

write out the cos equation you would use to solve for x

are you still on this question?

you’re just solving for x right?

you’re not going to use inverse cosine (cos -1) when trying to figure out a side length

should look something like this

you only need to do inverse sin, cosine, or tangent when figuring out angle measures

side lengths are simpler

[light fable]

^ that's the way

[wheat walrus]

that’s good, the more practice, the better

yea what you have in looks right

the horrors of deltamath

[outer bison]

i gotchu

[north heart]

the angles of a triangle add up to 180

that's all ya need

[outer bison]

im in algebra 2 so im not 100% sure but add all those degrees up and is all equal to 180

and solve for x

[wheat walrus]

YOOOOOOO

[north heart]

motherfucker this a test?

<@&268886789983436800> not sure if it's a practice exam or smthn?

but ig i should ping

[silk charm]

banned

[ember stratus]

To locate a whale, two microphones are placed 6000 feet apart in the ocean. One microphone picks up a whale's sound 0.5 second after the other microphone picks up the same sound. The speed of sound in water is about 5000 feet per second.

a. Find the equation of the hyperbola that describes the possible locations of the whale
b. What is the shortest distance that the whale could be to either microphone?

[wheat walrus]

it was a practice website called deltamath…

[daring wind]

can there be an algebraic expression for

cos(20°)

i don’t care how complicated it is, as long as it uses all real numbers and functions

like how cos(45°) is sqrt(2)/2

something like that, but for cos(20°)

[wise pawn]

I'd probably look at trying to work something out of the real part of the corresponding cyclotomic polynomial

[grave pond]

cos(20°) is the real part of an 18th root of unity, so it is certainly algebraic.

[grave pond]

However, the minimal polynomial for cos(20°) is 8x³-6x-1. It has 3 real roots (namely the real parts of the other primitive 18th roots of unity), so it is a casus irreducibilis where the roots of the polynomial cannot be expressed using real radicals only.

[stiff tulip]

I’m quite unsure tackling how to solve for X when it’s in the exponent

In example, “Solve for X” 10^3x=25

[grave pond]

That's not even an equation.

[stiff tulip]

Oop

Brb

It says round nearest hundredth

Do I begin by cube rooting ?

I got log(2.92)=X

X=0.47

I’m not sure if it’s correct or not

[grave pond]

Punch it into a calculator and see if you get 25 as you should.

[stiff tulip]

Yeah sort of

[grave pond]

But closer than either 0.46 or 0.48 gets you.

[silent plank]

ideally you'd avoid rounding in the intermediate steps

and use
x = log(cbrt(25))

[grave pond]

You get a slightly nicer expression if you start by taking the logarithm: log(25)/3.

(Which of course has the same value).

[stiff tulip]

Oh ok

[pallid pilot]

chat any idea how i can get AE if i have AC equals 25? (ignore the writing down there)

chats dead

[wheat walrus]

oh what’s up

[simple holly]

i fucking hate geometry dash

awful game

ruined my life

[wheat walrus]

AE is proportionate to AB as AD is proportionate to AC

that’s about the only way i see it

[timid escarp]

Yo I’m in accelerated algebra 1 and I might take geometry next year is there any thing I should know before hand like things that are gonna be annoying

[unborn heath]

just pay attention in class and you should be fine

depending on the teacher u might have alot of memorization as theres a decent amount of theorems and angle relationships

[lean quarry]

Guys do any one of you have, the formulas of triangle
Like

Areas

Similarities

Medians
Bisectors

And plenty more, all in one file ?

[glacial otter]

<@&268886789983436800>

[slender wasp]

tagging 8 times

[silk charm]

why are you pinging random people?

[weary drift]

man

[gaunt lion]

that was quick

[daring wind]

is it possible to solve binets formula for x

it’s like

coukd you solve this for n

[daring wind]

this but multiplied by cos(pi•n)

[grave pond]

What do you mean by solve here? If you have a large Fibonacci number and want to know which it is, in practice you can just ignore the ((1-sqrt(5))/2)^n term -- it goes to zero for large n anyway.

[grave pond]

Hint: It's a red herring that one of the lines goes through the center of the circle.

[rocky tide]

I can't understand those two different question can someone help me plz?

[odd dust]

What it supposed to be 148 degrees?

[heady forge]

What don't you understand ?

Let's start with the first one @rocky tide

Ask anything.

[vagrant ore]

I need help with this

[heady forge]

Try to form two triangles

That could help

[vagrant ore]

How would that help? I tried it

[heady forge]

X= 60°

[vagrant ore]

Oh I was just going to say that lol

I noticed that it did make two 30 60 90 triangles

[heady forge]

Tangents that extend from the same point are identical

[vagrant ore]

I didn't see that at first

[heady forge]

They form identical triangles

And from this we extract the following

x+β=180°

β is 120° in our case

[rocky tide]

Is the first one good
cos(30°) = MN / 10
10 cos(30°) = MN
10 (0.866) = MN
8.66 = MN

[heady forge]

Yes

[rocky tide]

@heady forge can u explain me this one pls if I am not borthing you .

[heady forge]

What do you want me to explain?

You are welconed

Welcomed

[rocky tide]

How to work this

[heady forge]

You try to find X?

[rocky tide]

I think so

[heady forge]

Because there is no question. I'm assuming you need to find value of x

[rocky tide]

Yeah it should be it

[heady forge]

Ok

Cos40°=X/10

0,766=X/10

X=7,66

[rocky tide]

Thx man but if I get the question in sin=but in fraction will it work with the formula u make.

[heady forge]

Since we know Hyp and an angle and we need to find adj,we take cos40°

What do you mean?

If you get sin then you have
Sin40°=BA/10

[rocky tide]

Like the question is similar but just numbers are in fraction

[heady forge]

Oh

[rocky tide]

Can I use this format to get the answer

Wait

[heady forge]

You will multiply fraction with 180°

And then replace the answer

[rocky tide]

Like this

[heady forge]

So

(15/17) multiply by 180°

And then same process

[rocky tide]

Ohhhh thx man

[heady forge]

Np

[rocky tide]

I got 15/17 X180
=2700/17x1
=158(14/17)

[heady forge]

Then we take the formula

[grave pond]

What? 15/17 is the cosine, not a fraction of a circle (or of a straight angle).

You don't even need to know the angle in degrees because the problem is telling you the values of all the trig functions.

[heady forge]

Oh yes

[grave pond]

The only thing you need to do is know which of them to use.

[heady forge]

Oythogoren theorem

[grave pond]

Pythagorean? Not even that.

[heady forge]

Cos =adj/Hyp=15/17

And then replace Hyp with 34 and adj with y

If I remeber correctly

[grave pond]

y is the side opposite to angle A, not the adjacent one.

[heady forge]

Then take sin

[grave pond]

@rocky tide, are you still here?

[rocky tide]

Yeah

[heady forge]

Sin=opp/Hyp=y / 34=8/17

Solve for y

[rocky tide]

My level is grade 9 so by far in trigonometry I had learn Pythagoras theorem

[heady forge]

Sorry,I thought that the angle was 15/17 rad. My apologies

[grave pond]

You don't need Pythagoras for this, but you do need to know what sine and cosine are.

[rocky tide]

How to solve then

[heady forge]

SinA=opp/hyp

[vagrant ore]

Is the answer circle?

[grave pond]

Yes.

[vagrant ore]

Thanks

[rocky tide]

Then x 180

[heady forge]

SinA=y/34

[grave pond]

No, there's no 180.

[heady forge]

No forget that

8/17=y/34

solve for y.

@grave pond my bad!

[north heart]

just isolate y

[heady forge]

So y=16cm

That's your answer.

[rocky tide]

The 34 need to be divide by a number right

[heady forge]

Just gave you the solution.

[grave pond]

It's not good to go all the way to a complete answer just because the asker isn't getting it.

[north heart]

^

[heady forge]

@rocky tide I'll write it down again :

sinA=opp/Hyp(1)

sinA=8/17
opp=y
hyp=34

So we put all of these at (1) and solve for y
So y=16

[rocky tide]

Yes 8x2

Is 16

[heady forge]

No it's 8/17=y/34

[rocky tide]

34 dividend by 17

2x8

[grave pond]

Yes.

[heady forge]

Yes,you got it

[rocky tide]

[vagrant ore]

I need help

[north heart]

look up the inscribed angle theorem and use properties of triangles for the rest

[vagrant ore]

Is the answer 30?

Wait I mean 20

[north heart]

idk im not gonna do it lmao

if it checks out then probably

[vagrant ore]

Wdym checks out?

[north heart]

oh

yeah

it's 20

lmao

i think you're supposed to use something much simpler than inscribed angle lmao

[vagrant ore]

Lol

[silent plank]

inscribed angle is one of the simplest things to apply

[north heart]

im 99% sure they're meant to use basic trig lmao

[grave pond]

Could you show the solution you think is simpler than the inscribed angle theorem?

[north heart]

by simpler i mean uses just trig and properties of triangles

[grave pond]

I don't understand what it is you're proposing.

What could possibly be simpler than "divide this number from the figure by two and you're done"?

[north heart]

im not proposing anything, i'm just saying that i think they were expected to use just trig and triangle properties to solve

[north heart]

bc ngl im not sure if this guy is taking ms geo, but when i was, i didn't know wbt the inscribed angle theorem at all

[grave pond]

You're saying that you think they are supposed to use a method that you refuse to reveal what is?

[north heart]

?

fine i'll try and find it lmao

gimme a bit

[gritty plinth]

Can someone please check a problem for me quickly?

[north heart]

ok so

i probably should've known better than to go about disagreeing with tropo

i genuinely thought there had to be a simpler way >.>

but i dont think it's possible to do w/out inscribed angle

my b

[vagrant ore]

I need some help

[grave pond]

draw another secant line (connecting two of the points on the circle that don't have a line between them yet), derive more angles using the incribed angle theorem (I think a pattern is showing by now), and you will be able to find x using the angle sum of a triangle.

[vagrant ore]

Is it 15?

[grave pond]

Hmm, that's not what I get, but I might be missing something. What's your reasoning?

[vagrant ore]

Because it looks like two 30 60 90 triangles and 30 ÷ 2 = 15

[dark sparrow]

looks can deceive

also there are no triangles here

[vagrant ore]

I remembered what to do and got 26. Is that right?

[dark sparrow]

seems ok now

[vagrant ore]

Ok thanks

[celest radish]

Help

[hidden crypt]

Try using sinx+sin3x=2sin2x*cosx and cos2x=2cos^2x-1, this must help

[celest radish]

Did help somewhat but still couldn't solve it

[hidden crypt]

What did you get?

[silent token]

help mee

if O= 26,656° find (x+y)/z approximately

[celest radish]

4sinxcosx + 2sinx = 2cosx + 1

[acoustic jungle]

I would wlog z = 1 and trig bash everything.

not sure why they chose the value 26.656, seems kinda stupid.

[hidden crypt]

you should've got sin2x+2sin2x*cosx=2cos^2x+cosx

[celest radish]

Yes then I divided both sides with cosx

[hidden crypt]

you can get something better

sin2x+2sin2x*cosx=sin2x(1+2cosx)

[azure crown]

x=1/2(66-14)

[celest radish]

How is that better?

[hidden crypt]

because 2cos^2 x+cosx=cosx(2cos+1)

[celest radish]

Oh damm

[silent token]

I don't get the result, I found the others and it shows that (x+y)/z is 1

[acoustic jungle]

Show your work

[hidden crypt]

firstly, find angles that are equal to theta, there are plenty of them

[silent token]

I did it again, okay? @acoustic jungle @hidden crypt

So it's option C? 1,9?

[vagrant ore]

I need help solving this

Not really

I know you're supposed to take half of 6y

[vagrant ore]

I solved it

[mystic shadow]

$$\cos(x)-\cos(y)=\frac{5-\pi}{10}$$

[somber coyote]

MiracleMan

[mystic shadow]

$$\cos(5x)-\cos(5y)=\frac{1}{2}$$

[somber coyote]

MiracleMan

[mystic shadow]

This is easily solvable by plotting

I was wondering if this was analytically solvable

I know 0 < x,y < pi/2, and y>x

[dire hull]

Is it okay if I ask for help here and trig 😅

[upper karma]

I tried using maclaurin series but nope

[somber coyote]

Holo

[rare sierra]

can someone give me the equation I can use to solve this because their explanation is bogus?

I know in this scenario
A= 18/30
B=15/30

[tulip vector]

hello, I'm having a bit of a problem on this one. If there are two equal sides/angles the "N", what should I do? My answer is supposed to be largest to smallest order

[valid harbor]

what

can you clearly state your problem

[polar jewel]

Please help! I dont know this topic. Need this asap!!

Also the part of what i sent

[trim breach]

This is just knowing definitions. I could help, but I really think going back and ensuring you understood the lesson fully is probably a better starting place.

[spare moss]

I want to calculate this integral, but could you help me with the visualization?

How could it be drawn?

it is the shape which is restricted by axis

[merry dew]

help

[silent token]

9x-4=4x+2
9x-4x=2+3
5x=5
x=1

LO= 9 (1)-3 = 6

64=8+7x
56=7x
x= 8

m <NPI = 8+7x = 8+7 (8)= 64

I think

[molten marsh]

can anyone help

nvm i got it

[terse breach]

Why is tan multiplied by three and powered up to 2?

[worthy siren]

hey, I need help on some math homework thats due in an hour. Im currently in the unit of adding vectors, and we recently learned sine cosine and tangent, which we're supposed to use for this example. How can I do it?

[hollow vortex]

you can use pythagoream theorem to find the magnitude of the movment

[orchid gate]

in geometry of triangle what this value mean?

[somber coyote]

Ausländer

[proud plover]

Guys is my solutions correct?

Can anyone help me?

[upbeat helm]

Integrals go in #calculus . Also post the question before your solution

[nova aurora]

can someone help

[sterile orchid]

can someone pls help me with this question, i've drawn a diagram but i do not think i have the right equations:

[dark sparrow]

so far so good

what equations did you get for AB and AC

[upper karma]

can someone help me pls ?

[clear lark]

Hi, I agree with your answer. I wonder if there is a typo and the coordinates of D are meant to be (-2,-2)

[upper karma]

they are all incorrect

[errant flare]

How?

[past charm]

im pretty sure i've seen something similar on youtube

like mind your decisions probably

[errant flare]

what do you mean?

[past charm]

thats a youtube channel that makes videos on math problems

[errant flare]

thanks

[wanton jetty]

Can someone please help me with these 2 questions

[vestal siren]

can someone check my work

[trim breach]

The first triangle was labeled incorrectly, so you used the wrong trigonometric ratio.

Actually, looking at it, all of them are.

The opposite side is one the angle does not touch.

Though the second triangle seems right, but you still labeled the sides incorrectly. And I think your final answer is incorrect as well.

[vestal siren]

i'm still confused

[vestal siren]

oh alright thank you

[trim breach]

“Adjacent” means “next to”.

Many of the steps you did were right, but you just flipped opposite and adjacent when you labeled them, so you ended up using the wrong ratio.

[vestal siren]

oh alright bc

wait can you show me an example how to do one

[trim breach]

@vestal siren

[vestal siren]

okay thank you

[sterile orchid]

like y=am+c because i don't know the y values

[dark sparrow]

........

okay so that took me some time to decipher.

seeing as that's nearly a full day later you're saying this

but whatever

do i understand correctly that you have actually written down The equation of line AB is y = am + c as part of your work, or did you just blurt that out

[sterile orchid]

yeah i wrote it like that

sorry it's y=-am+c because the x value is -a

also line AC would have the same equation

[dark sparrow]

what is m?

[sterile orchid]

i don't know i don't have any definite coordinate just variables

[dark sparrow]

.....

so you're writing things whose purpose you're not even aware of.

strange!

[sterile orchid]

no like that is the equation of the lines with the information that is currently known to me

[dark sparrow]

anyway, i think you're confusing the equation of the line itself with what happens when you plug one of your points into it.

[sterile orchid]

hmmmm, what do you recommend that i do then?

[dark sparrow]

consider point A and its coordinates first and foremost

and then the angles made by AB and AC with the x axis

[sterile orchid]

right right, i am sure that i have no idea what the coordinates of point A are apart from the fact that the x value is greater than a

[dark sparrow]

the medians of your triangle intersect at the origin

and the intersection point of the three medians of a triangle divides each of them in a 2:1 ratio

[sterile orchid]

sorry i dont think i understand lmao my b

medians?

[dark sparrow]

the lines in a triangle that connect each vertex to the opposite side's midpoint...

[nocturne remnant]

What “further depth” do you mean exactly

Are you struggling with something in particular?

What equations?

Well
Best I can do is to link you to some resource
https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func

The unit circle definition of sine and cosine is much more general and powerful than the “ratio of sides in right triangle” definition that students are usually introduced to

So just think of using sine and cosine to solve for triangle sides/angles as an application of trig functions, not the definition of trig functions itself

[nocturne remnant]

true

Whereas if you look at something like uniform circular motion it’s much more sensible that the derivative of sin would be cos

[muted sail]

Still confused by this.

I dont remember the formula on how to find surface area of cone. ;-;

[upper karma]

Surface area is = 𝝅rl + 𝝅r²

Where l is the side of it (e.g the 11cm on the 1st one)

The 𝝅r² is the surface area of the circle face so if you cone is open at the end you don't need to use it

Hope it helped a little :)

[muted sail]

thanks :D

[upper karma]

By any chance they don't give you the 'l' side just use Pythagoras assuming they would atleast give you the height for it

[muted sail]

I dont really get what it means by show your awnser in terms of pi

[upper karma]

Oh usually your answers on your calculator will have 𝝅 with it since it was multiplied in

Just don't hit the S->D button (the button that converts fractions to decimals, still don't know the name of it lol)

Basically they dont want decimal answers

[muted sail]

oh, okay.

[upper karma]

I could go through one with you if you still feel lost

[upper karma]

Can someone help me out with this?

I just don't know how to set up the equation

[trim breach]

The diagonals in a rhombus bisect one another.

So BD = 2BE

[upper karma]

So I just double the equation of BE?

Okay

Got it

[upper karma]

hello

does anyone know about circles

[fallen sleet]

given an arc, is it possible to construct with compass and unmarked straightedge the circle upon which it lies?

this works

[fallen sleet]

i understand now

[wicked fossil]

Do you have a specific question about them?

[near sand]

Hi, so I'm working on the first one (finished the other two) and I'm a bit confused as to what to do, even with the hints. If someone could help, that would be much appreciated

,rotate

[warm plinth]

I've been trying to solve this and i found a contradiction when i tried 2 different approaches, i cant manage to find the mistake

This is all the information i got, heres the original exercise, (its in italian)

,rotate

[warm plinth]

"A right pyramid has as its base a square with a diagonal of 18(sqrt2)cm. The hight of the pyramid is 1/2 of the hight of its triangles, find the volume of the pyramid in exact values."

I found only 2 approaches and one of them involves trigonometry, which we havnt learnt yet.

Exact values is like 10(sqrt2) not ≈14.14

[warm plinth]

I just realised that i might be in the wrong channel

[dark sparrow]

no, this is ok

your question is very clearly geometric

[runic beacon]

ok, so i am not sure if my approach is correct, but, the diagonal of the square should give you the side (by the pythagorean theorem) So, now let the height of triangle be y and then height of the pyramid will be 1/2 * y.
Now the centre of the square, the height of the triangle and the height of the pyramid also form a right angled triangle in a different plane.
So, height of pyramid by the pythagorean theorem, as in (1/2 * y)^2 + (1/2 * 18 )^2=y^2
solve for 1/2 y , you will get, height of pyramid
Then you can apply the formula for volume

[warm plinth]

Oh i tought i had to ask in #help

[dark sparrow]

no

[warm plinth]

Oh ok 👍

[warm plinth]

(1/2 * y)^2 + (1/2 * 18 *sqrt2)^2=y^2
im strugling over here, where did you get (1/2 * 18 *sqrt2)^2

[runic beacon]

it is half of the base

[warm plinth]

oh and by hight of the triangle i mean the "apotheosis" i didnt know the word for it

[runic beacon]

what is the apotheosis?

[warm plinth]

wait google translated f-ed up lmao

i meant the apothem

[runic beacon]

then i think i need to think what i have done

[warm plinth]

Heres everything i tried so far (all giving me contradicting results) i have in 'a' the apothem

(Btw the dot (•) is multiplication)

The hight of the triangle is the apothem

Thats what i meant by hight

[runic beacon]

i was talking about the actual height now

[warm plinth]

The hight of the pyramid is 1/2 of the apothem of its triangles

[runic beacon]

wait i misunderstood the whole question

[warm plinth]

Oh xD

So the diagonal of the base of a quadriangular pyramid (not sure thats a word) is 18 * sqrt2 cm, the hight of said pyramid is half of the apothem of the pyramid's triangles, find the volume in exact values.

What i found is that the side of the base is 18cm, and that the triangle that forms using the apothem as a hypothenuse and the hight and half of the length of the base (9cm) as the cathetus' is a 30 60 90 triangle

Anyways i gtg

[upper karma]

had

[warm plinth]

im going for option 3 and 4 as this is the only case where a = 2h, im strugling to understand the mistake in 1 and 2 tho

[magic lichen]

guys how do i start trignometry?

whats the best place

to start learning trignometry

[magic lichen]

thx

damn bro 8th?

me too but im learning in 9th

i just came to 9th so im still 8th?

tries to make sure that i learn hard stuff in the earlier grades

yes yes

yeah just the basics not too deep into it

[trim breach]

Your mistake in method 1 is only applying the exponent to h, not 2h.

[warm plinth]

ohhh thx

[candid flume]

how dis

Write the general form for all the solutions to cos0= -1/2;
based on the smaller angle.

I swear my prof didn't say anything about this s***t

[simple rain]

is anyone here

[simple rain]

how to do 10

[wise pawn]

I'd imagine it's a complete cone with the top chopped off, so you can find the volume of the whole cone and the volume of the piece cut out, and subtract it

[simple rain]

you have to use similar triangles but idk which

[simple rain]

yea but idk height

of the full cone

[coarse saddle]

that's correct

[simple rain]

can u tell me the sides for it

[coarse saddle]

I don't want to give you the answer

I want you to figure it out yourself

[simple rain]

how to set it up

[coarse saddle]

try to complete the cone

draw the lengths you need, etc.

[simple rain]

that’s what i said

i don’t tho

the teacher showed in class but i forgot

so like this

[coarse saddle]

that's good

you're very close

[coarse saddle]

remember

you aren't looking for x

you're looking for the height

[simple rain]

oh height is 12

3-4-5

i think

ok thx i got it

[coarse saddle]

[simple rain]

noooooooo

[coarse saddle]

what's wrong?

[simple rain]

i got it wrong

it’s B somehow

[coarse saddle]

you see

the answer doesn't matter as much as the work

[simple rain]

what i do wrong tho

there might be mistake on the key

[coarse saddle]

probably just a calculation mistake

[simple rain]

can you try it and see what you get

[coarse saddle]

emm

sure

[simple rain]

ping me cuz ima close discord

[coarse saddle]

@simple rain did you forget to delete substract the volume of the smaller cone?

[simple rain]

what did you get as answer

ooohohh wait yea

oop

[upper karma]

hiii can anyone help me with this? it’s geometry snf it’s surface area of composite figures

[lapis moon]

divide and conquer: outer curved surface area is $2 \pi r h$

[somber coyote]

vin100

[lapis moon]

upper and lower surface area is circle's area minus that of the square

inner surface area is four times each rectangle $sh$, where $s$ is the side length of the square

[somber coyote]

vin100

[lapis moon]

it suffices to establish a relationship between $s$ and $r$.

[somber coyote]

vin100

[lapis sequoia]

i need help with something in my geomotry

[deep niche]

idk how to do this

[vestal siren]

can someone explain to me how to do this

[wheat walrus]

you comfortable with sine, cosine, and tangent?

[vestal siren]

yes

I just don't understand the angle 0 part

[wheat walrus]

the “0” is called theta and it’s just a variable

same for x, y, z, or any other alphabetical letter

[upper karma]

can someone help me with 1 hw question?

yes? or no?

[wheat walrus]

depends

what is it

[upper karma]

geometry

circle

[wheat walrus]

send a picture !

oh circle

[upper karma]

circle

[wheat walrus]

ew

nah i don’t know this

and if i learned it, i’ve expelled it from my memory

[upper karma]

so no help?

[wheat walrus]

unfortunately

perhaps consult khan academy

[upper karma]

is there someone here who specializes on geom?

[upper karma]

?

[wheat walrus]

khan academy

[upper karma]

ive never used it

[wheat walrus]

search up the concept on google or youtube idk

[upper karma]

do i upload picture?

[wheat walrus]

no it just teaches you the basic concept

[upper karma]

dang ok

ill be back

[wheat walrus]

good luck

[upper karma]

ty

[vestal siren]

so the problem i had would be 55 right

or wrong

[wheat walrus]

what question are you referring to in the worksheet?

[azure crown]

i think it's 53

[upper karma]

i also got 53

but im not sure if the angle is supposed to equal the bigger arc

i was absent a day and the teacher didnt teach me just gave me notes

[azure crown]

equation is angle = 1/2(151-45)

[upper karma]

why 1/2?

[azure crown]

thats the theorem

[upper karma]

i did —> angle= 151-45/2

[azure crown]

well that doesnt equal 53

unless u mean (151-45)/2

which is the same thing as mine

[vestal siren]

the first question

[upper karma]

oh yeah

[plucky mantle]

Can someone give me some tips for memorizing the unit circle? I've got an exam next Tuesday (pls ping me)

[upper karma]

whats the unit circle @plucky mantle

[azure crown]

write down what u know then memorize what u dont

then repeat until u can write the whole thing

[upper karma]

whats the unit circle duck?

is it in geometry?

[azure crown]

idk im doing geometry rn i just looked it up

i didnt learn it yet

[upper karma]

u need to memorize this whole thing?

i would just print out a tiny version of it and place it in a pocket at that point ngl

[azure crown]

when do u learn that

[upper karma]

idk

[wheat walrus]

no

[upper karma]

i havent

[azure crown]

is that precalc or geometry

[wheat walrus]

@vestal siren when you’re finding the angle measures, use reverse sine, cosine, or tangent

don’t base the answer off of another angle measure you figured out in case your solving is inaccurate, stick with the information you’re given

[vestal siren]

ohhh

[upper karma]

Can someone help me with the vocab ?

5. A (blank) is the set of all points in a plane, equidistant from a given point called the center.

6. A (blank) is a segment whose endpoints are the center of the circle and a point on the circle.

7. An angle whose vertex is at the center of the circle and whose sides intersect the circle is called is a (blank).

8. A chord that contains the center is called a (blank).

[warped atlas]

circle, radius, central angle, diameter?

[upper karma]

ok ty lad

[warped atlas]

👍

[plucky mantle]

Trig

It's used with radians

[vestal siren]

can someone check something for me

[storm portal]

[north heart]

https://dontasktoask.com

[vestal siren]

okay so

i'm doing this problem and I'm confused

and i'm very confused for 3

[wicked fossil]

For x, what type of triangle is it?

[vestal siren]

nvm i got it

[wicked fossil]

ok

[opaque gull]

Yes, includes trigonometry too.

[opaque gull]

For next time use Pythagorean theorem then use arcsin, arccos or arctan (sin^-1, cos^-1, tan^-1)

To measure the angle

[plucky mantle]

Or the law of sines

[opaque gull]

That too

[upper karma]

when do i learn it

[opaque gull]

Like 10th 11th grade

Somewhere there

At least we do here

[upper karma]

ok

[opaque gull]

u can read about it if ya want

[tepid vault]

help

is this valid for all integer p

$\forall p \in \mathbb{Z}$, $cos(2\pi p) = 1$ and $sin(2 \pi p) = 0$

yeah right?

just wanted to confirm cos me bad at trig

[somber coyote]

texaspb

[gleaming nova]

first one yes

[wise pawn]

second one yes

[gleaming nova]

second one could instead be sin(pi*p) = 0

[tepid vault]

okay

then

$cos(2\pi p + \theta) = sin(\theta)$?

[somber coyote]

texaspb

[tepid vault]

[wise pawn]

the fundamental thing to know is they both have a period of 2pi

so you can always add or subtract an integer multiple of 2pi from the argument

[tepid vault]

I wanted to check if $cos(2\pi p + \theta) = cos(\theta)$

[somber coyote]

texaspb

[tepid vault]

this is vewy important for me to understand a thing here

[wise pawn]

cos(x)=cos(x+2pi)

[tepid vault]

perfect

thanks

[wise pawn]

yw

[tepid vault]

because like

the guy did this

now i gotcha

[mystic coral]

What did i do wrong?

[pure cape]

what you did wrong is that you assumed the point was in the first quadrant

there isn't just one angle that satisfies tan(theta)=sqrt(3)

[mystic coral]

How do I find the other angles that works?

[pure cape]

try to solve the equation tan(theta)=sqrt(3) generally

and then choose the angle that would satisfy the following condition

[mystic coral]

Im not entirely sure how to go about that. I assume it would have to do with the range of trig functions?

So if my assumption is correct, then my other coordinate would be (5,-pi/3)?

[pure cape]

not really

[pure cape]

and no

that's in the fourth quadrant, which is also not where the point lies

you can search further into this since i dont have a lot of time to explain, the equation tanx=a has the general solution x=arctan(a) + kpi where k is in Z

and then just restrict this to the interval [0, 2pi], and then choose the appropriate angle

[mystic coral]

Alright, ill look into it, thanks

[ancient spoke]

Can y’all help me on this

[dark sparrow]

do not ping everyone.

[ancient spoke]

My bad

[lapis moon]

sorry we're here to do your homework

[dark sparrow]

we're not here to do your homework.

[ancient spoke]

I said help

[lapis moon]

my bad +1

[ancient spoke]

Not do

[dark sparrow]

in any case

there are two parts to this problem

one about the length of BC, the other about the area of the whole thing

[lapis moon]

but we can give you some hints

i suppose

[dark sparrow]

which of these would you like help with

if both, which one first

[ancient spoke]

I’ve been stuck on this for an hour

[dark sparrow]

which of these would you like help with?
if both, which one first?

[ancient spoke]

I just need some tips

[lapis moon]

what? the question writer has spoonfed students with answer? sosad 😦

you're supposed to find the length, but he/she has just stated that

and ask for an explanation for that

[dark sparrow]

there are two parts to this question. would you like to get help with part a first or part b first?

ok

so look at the diagram

[lapis moon]

just use symetry

[ancient spoke]

Thanks much appreciated

[dark sparrow]

what are you thanking us for

i for one havent even started

i was gonna take you through it

[ancient spoke]

For youre time

[lapis moon]

upper & lower half symmetric

[dark sparrow]

do you want to continue?

[lapis moon]

left & right half also symmetric

so ...

[ancient spoke]

Yes

[dark sparrow]

ok

ok so you see how the shape is rectangular all throughout, yes?

so if you were to connect C and F with a straight line, it would be aligned with BC and FG

hence the line segments BC, CF and FG add up to the height of your shape, which is 20 as the diagram says

do you understand this

therefore

no

x is just x

it is just a number, and all your work will involve expressions in terms of x

the line segments BC, CF and FG add up to the height of your shape, which is 20 as the diagram says
do you understand this

what did i say?

did i say they were all 20 or did i say they added up to 20?

well that answers your own question now doesn't it

divide what by what?

...

ok you know what nevermind

[ancient spoke]

Thanks

I got it

[mystic salmon]

Hi, I asked this ques in a help channel, but I'll put it here as well since it is to do with trig

Question: I'm trying to simplify it and get it into terms of one trigonometric function

I know tan x = sin x / cos x
I also know that it is the same as: tan x * (1-sin^-2)^1/2
so I get to the stage tan x * (1-sin^-1 x)
but every step after this I get confused

could someone offer some guidance, thanks, ps please ping @ me if you reply 📐

[slate kraken]

Well...let me give u hint...sin^-2x means 1/sin^2 x....now we get within square root...1-1/sin^2 x ...again...there's the other formula of trigonometry...sin^2x+cos^2x =1 ..try and see if u can utilize these infos

[wispy iris]

hi, I was wondering if "sec = sin^-1" or if they're different. Thanks in advance.

[grave pond]

The notation sin^-1 is almost always used to denote the inverse function of sine, not for talking about 1/sin(x).

This is inconsistent with sin²(x) meaning (sin(x))², but that the convention we have, like it or not.

[wispy iris]

So, if I have the Adjacent and the θ of a triangle, but my calculator only has sin^-1, then "(sin^-1 θ) * Adjacent ≠ Hypotenuse"

[grave pond]

Yeah, definitely not the same thing.

And not just because sec is the recprocal of cosine rather than of sine.

Your calculator ought to have a way to type adjacent / cos(theta), though.

[wispy iris]

huge thanks

[umbral radish]

can anyone solve x for me?

[dire hull]

idk much but i do know theres the right angle which is 90

add 90 + 42

then substract by 180 to get your third angle

damn it says trig ratios

mb

[umbral radish]

all good

[reef oxide]

Use a tangent with 42°

[reef oxide]

Then remember TOA

[umbral radish]

I really don't understand, im sorry

[wicked fossil]

Tan(angle)=opposite/adjacent

What side is opposite from 42?

What side is adjacent?

[reef oxide]

That's okay! We're solving for x, and given an angle of 42° and a length of 5. If we can put these 3 terms in one equation, we can find their values relative to each other.

Since this is a right triangle, we can use the trig ratios to set that equation up.

From the angle of 42, we have the hypotenuse above it and the adjacent side, 5 below it. To it's left, we have the opposite side, x.

The tangent of this angle is equal to the ratio of the opposite side over the adjacent side.
tan(x) = opposite/adjacent
tan(42) = x/5

Multiply both sides by 5.
(Move the 5 to the other side while observing sign rules, that's how I think about it)
5*tan(42) = x

Punch the left side of the equation into a calculator, round to the nearest tenth as the question asks, and that's your answer.

Phew

[opaque gull]

@umbral radish

[opaque gull]

Great explanation, man.

[upper karma]

Hello

[opaque gull]

hey

[upper karma]

I have normalized coordinates from -1.0 to 1.0 for X and -1.0 to 1.0 for Y
-1.0 X is left, 1.0 X is right, -1.0 Y is down, 1.0 Y is up. I want the center of an euqilateral triangle to be on 0.0, 0.0, the lower left point to be on -0.5, 0.0 and the lower right point to be on 0.5, 0.0
Which one of these methods best represent the center of the triangle in this case?

My idea was to get the height of the triangle whichs is sqrt(0.75), to divide it by 2 for the Y position of the upper point and have the negative value (-sqrt(0.75)/2) for the lower points

this is the result of that

and these are the coordinates I chose

Another person used these coordinates

which had this result

And says that his triangle in the center

Does anyone know which one is actually in the center?

Btw the triangle is equilateral

[magic lichen]

but dont confuse between arcsin and csc

[magic lichen]

another way to check the center is by taking 30 degrees from each corner and draw a line. whereever all ur lines meet, that is ur center

[wispy iris]

Yeah, thanks!

[upper karma]

A = 3

Segment AC = 3 SQRT 2?

D

AB = AC multiplied by SQRT 2

AB = 3 SQRT 6

[hazy holly]

Hey. I'm trying to solve this problem. I drew the radii and prepared the triangle shown. I found that $r = \dfrac{3a}{8}$. I've tried using trigonometry in different ways and I keep finding that $x = \dfrac{18a}{25} = 0.72a$, but the answer says $0.6a$. This is not homework, this is problem 206 from Solving Problems in Geometry by V. Gusev, V. Litvinenko and A. Mordkovich.

[somber coyote]

EvilSonidow

[terse breach]

I need some help

I need to figure out the height of the tripets

[sleek mirage]

can someone show me how to get the answer for this or point me in the right way?

[hazy holly]

Nevermind, found the answer. Using similar triangles you can show that $$\dfrac{x}{a} = \dfrac{r}{a-r},$$ and since $r = \dfrac{3a}{8}$ it follows that $x = \dfrac{3a}{5} = 0.6a$.

[somber coyote]

EvilSonidow

[hazy holly]

Hard to understand what is asked

[hazy holly]

Angles E and B are inscribed in the circle and "see" the same arc CD. Use inscribed angles https://en.wikipedia.org/wiki/Inscribed_angle.

[terse breach]

the area of the trapezium

I've used pythagoras theorem to calculate the side and the area of the first two triangles, but the third one is missing it's height

[sleek mirage]

I get that but whenever I set them equal to each other and solve i dont get the answer given on the answer key

[hazy holly]

Do you mean x or did you substitute back to find the angles?

It asks for the measure of the angles

[sleek mirage]

Both

I didnt get the right x value or the angle measures

[hazy holly]

What x did you find?

[sleek mirage]

like 7.07...

the answer is x=5 and m<dec =48

[hazy holly]

The equation is $2x^2 -2 = 10x-2$, where did you go from here?

[terse breach]

Basically, I've found the area of the first triangle, but now I need to find the area of the last angle in order to sum it up and get the area

[somber coyote]

EvilSonidow

[sleek mirage]

i got all the variables on one side and then i used the quadratic formula

[hazy holly]

Since it's a trapezium the bases are parallel. Extend the smaller side and draw the height from the lower left point to create a right triangle with hypotenuse 10 and one angle of 30°.

[sleek mirage]

no wait first i factored out the 2

[hazy holly]

No need for that

$$\begin{aligned} 2x^2 -2 & = 10x - 2 \ 2x^2 & = 10 x \ x^2 & = 5x \ x^2 -5x & = 0 \ x(x-5) & = 0 \ x & = 0 \ x & = 5 \end{aligned}$$

[somber coyote]

EvilSonidow

[hazy holly]

Naturally $x=0$ doesn't make sense.

[somber coyote]

EvilSonidow

[hazy holly]

Can't see anything

[sleek mirage]

ohhh

that makes a lot of sense actually i see what i did wrong

thank you!

[hazy holly]

This is wrong, you can't be sure that what's left on the largest base after dropping heights is 6

[hazy holly]

This isn't what I told you to do

[terse breach]

yes, this is what I did initally

[terse breach]

Like this?

[hazy holly]

No, like this:

Crude drawing but you get it. Use right triangle trigonometry to find height AE and therefore find the area

[terse breach]

which program do you use for that?

[hazy holly]

Geogebra

Geogebra.org

Online and free

[terse breach]

the AE height will be 5, as the angle is 30 degrees, therefore the opposite side must be the half of the hypotenuse

[hazy holly]

Exactly

[terse breach]

but for the leftover triangle, what width will it have if it's 12 in total?

[hazy holly]

Doesn't matter, you don't need it for the area

[terse breach]

Really? So it's not included in the area?

[hazy holly]

No. You would need it for perimeter, but since it's area you don't need the length

Do you know the area formula for a trapezium?

[terse breach]

then it's all simple, base times height

[hazy holly]

That's for a rectangle, not a trapezium

[terse breach]

If we find out the area for the right triangle, we need multiply it by four, since there are four of the right triangles

[hazy holly]

No, there are not

At least there are not 4 congruent right triangles

You are very close but you're using the wrong formula

[terse breach]

the area will be the sum of the two pararells, then divided by two and multiplied by the height

[hazy holly]

Yes

[terse breach]

Alright, thank yoy for the info

It's my first time calculating the area for a trapezium

our school doesn't go through it, I have to get maths from the internet

[untold tundra]

so i just post my question here

?

[dire hull]

Yes

[cerulean shell]

https://media.discordapp.net/attachments/387449305885048852/967927079884714064/Screen_Shot_2022-04-24_at_4.16.24_PM.png

how do i do this?? what does this even mean

[grave pond]

Knowing "cot theta is undefined" and "theta is between pi/2 and 3pi/2" allows you to deduce exactly which number theta is.

[trim breach]

Is this an exam?

[lofty depot]

Naw it’s not

[trim breach]

Do you know the formula for arc length?

[lofty depot]

No what is it

[trim breach]

,,2πr(θ/360)

[somber coyote]

Lidoh

[trim breach]

There we go.

[lofty depot]

Aii what’s next

[trim breach]

You have a radius, r, and an angle measure, θ.

You can just plug them into the formula.

[sleek mirage]

Idk where to start with this problem

[dark sparrow]

do you know what standard form is for equations of circles?

[gray hull]

Here's the question, let me get my work.

At first, I thought the areas of these were related but I'm pretty sure they're not. I would say something about trying to relate the lines which bisect these angles, but yeah I don't know how to prove that they go through the centroid of WXYZ.

[runic beacon]

@sweet whale go to an available math help

[sweet whale]

what?

[runic beacon]

#❓how-to-get-help

[wicked dagger]

8^2 + 6^2 = c^2

[sweet whale]

thanks mate, but how did u solve it?

oh nvm

[wicked dagger]

its a right triangle, so you can use the pythagorean theorem to find the third length

[sweet whale]

🫂

[tulip vector]

help me pls

[upper karma]

180 - (19x+3) + 6x + 15 + 9x + 16 = 180

[upper karma]

guys

help me

prove pytagoras' law

[trim breach]

There are a ton of resources covering the proof. You might want to start with a Google search unless there is something specific you don’t understand.

[lost sable]

Anybody know of any cheat sheets for all conic sections?

[steady jetty]

Can someone help me with the top equation, I’m kind of confused what identities to use to deduct sin(x+y)/cosxsiny to 1+cotxtany

[wheat steeple]

I mean, you've already completely 90% of the solution

if it helps, split the numerator and distribute and denominator

side note, the denominator must be given as sinxcosy

or else the identity is wrong

[steady jetty]

Hmm

I’m still kind of confused is it possible if you can write it out

Because after I cross out the denominator I’m stuck and can’t relate back the the other side of the equation

[opaque gull]

Whats the goal here?

Solve for y? For x?

[steady jetty]

Prove that both sides of the equation r equal

Using trig identities

[opaque gull]

Ohhhh ok lemme just eat dinner

[opaque gull]

Look up trig identities and keep trying meanwhile

[steady jetty]

Never mind I got it

[lapis moon]

hi, is there an English name for this formula:
https://zh.wikipedia.org/wiki/定比分点公式

$${\overrightarrow {AD}}={\frac {|{\overrightarrow {CD}}|}{|{\overrightarrow {BC}}|}}{\overrightarrow {AB}}+{\frac {|{\overrightarrow {BD}}|}{|{\overrightarrow {BC}}|}}{\overrightarrow {AC}}$$

[somber coyote]

vin100

[grave pond]

I'd call it something like convex interpolation, but that's not an already established name.

[tepid cave]

Can someone help in #help-6

[opaque gull]

A circle tangents the line 3x-4y=0 at the point (8,6) what's the midpoint of the circle? It also intersects the positive x-axis

I solved it to y=(3/4)x and chose origo (0,0) then found the midpoint between the points I got (4;3) is that right?

I feel like it's too good to be true and a much harder process

[nocturne remnant]

You mean the centre of the circle?

[dark sparrow]

"tangent" is not a verb by the way

[wise pawn]

a circle constructed with its center at (4,3) will be cut in half by the line, so it can't be tangent to it

also it's not gonna be unique if it only intersects the positive x-axis, since there are many circles that are tangent to the line and intersect the positive x-axis, but it would be unique if it was required to be tangent to the x-axis

[opaque gull]

Idk the verb in English

I guess intersects?