geometry-and-trigonometry

[dark sparrow]

what is your native language and the word in it

[opaque gull]

Yeah

Swedish

En cirkel tangerar linjen 3x-4y=0 på punkten (8,6)

Tangerar is the veb

Verb

[dark sparrow]

okay so that sounds like... touch

[opaque gull]

BRUH

Yeah touch

[dark sparrow]

a circle touches the line 3x-4y=0 at the point (8,6)

[opaque gull]

Yes

[dark sparrow]

and you are asked for its center

[opaque gull]

Whats the midpoint yes

I used the midpoint formula

(x1+x2)/2 (y1+y2)/2

I used another point (0,0)

[dark sparrow]

the word midpoint is used for line segments only

usually

[opaque gull]

Which I thought I could do idk

Oh sorry, the center then.

(Was called midpoint on the exam tho)

I think cuz the line seg. goes through the circle

Like a diameter and radius

[trim breach]

Center is the midpoint of a diameter, yes.

[opaque gull]

Oh no

So it's not that easy?

I had to calculate the radius and all that?

[trim breach]

I haven’t actually looked at the problem. Just clarified why it might have been called a midpoint.

I am not really in a place to sit down and help, sorry.

[opaque gull]

That's fine

[opaque gull]

this is how I thought the graph looked like

but it could have been a teeny tiny circle too so I am just dazed

[wanton edge]

You said “tangents”, but the circle in the picture you just show isn't tangent to the line 3x - 4y = 0.

Y'know what, I will try

[nocturne remnant]

Correct me if I’m wrong but isn’t the question asking for something like this?

[wise pawn]

that's what I had in mind too

but the wording was kind of vague from the translation so not 100%

[opaque gull]

yes

SORRY GUYS

THATS WHAT I MEANT

Oops caps

the pic elon musk sent, something like that is what i meant

WAIT

OH YEAHHHHH THE CIRCLE IS TANGENT TO THE LINE

FUCK

I made it so the line goes through shit

anyways can you explain how to find the center?

[nocturne remnant]

well are many things we could do

firstly, we can show that the point of tangency of the circle and the x-axis is (10,0)

[opaque gull]

oh shit

you're right

[nocturne remnant]

this is because the lengths of the tangents are the same

[opaque gull]

so what does it mean that the point of tangency of the circle is (10,0)

means we can go up draw a line

thats the center?

[nocturne remnant]

i would probably do sth like

this

we can solve for the center by finding the equations of the two blue lines and then solve for when they intersect

[opaque gull]

that's all i needed

thanks man

[nocturne remnant]

:)

[opaque gull]

OMFG

NOOOOOOOO

i knew it was too good to be true

and the distance I knew it was 10 cuz of 8^2+6^2

sqrt(100)

[opaque gull]

very nice thanks man

[wanton edge]

You're welcome 🤗

[opaque gull]

:D

Geometry is fun but tough

[vestal siren]

can someone help me

[runic beacon]

,rotate

[opaque gull]

Use pythagorean theorem but now you want b^2

[wanton edge]

This one basically use trigonometry. But we can approximate by using the first few terms from infinite series of sin(x) and/or cos(x) [in radians]

[opaque gull]

Oh

Nevermind, thought that the other side was known

[solemn osprey]

(Unrelated to question above) "Parallel lines are two lines that never meet. Find an example that contradicts this definition. How would you change the definition to make it more accurate?"

for the first part of the question I'm having trouble figuring out an example of a parallel line that contradicts the afore mentioned definition

[wanton edge]

I think this is like "Hey, we want to find right triangle that doesn't follow the Pythagorean theorem"

[runic beacon]

https://www.youtube.com/watch?v=XXzhqStLG-4

[opaque gull]

(Arya can I DM you?)

[runic beacon]

this might be useful?

not sure though

hope that is what you were looking for

[vestal siren]

i’m confused on this one

can someone help me

[runic beacon]

you dont need to ask to ask

[vestal siren]

i need help with this

[opaque gull]

,rotate

[terse breach]

is there a trigonometric rule for finding the height of the right triangle with 20 degrees?

I mean, atan will be (x/3) in this case

[wanton edge]

I'm pretty sure it's yes. But unfortunately, it uses cube roots and the formula is more complicated

[terse breach]

cubic roots in trigonometry? When are cubic roots used in trigonometry?

[runic beacon]

which question are you talking about? i also want to know

[terse breach]

.

17

[wanton edge]

"Is there a trigonometric rule for finding the height of the right triangle with 20°?
I mean, atan will be (x/3) in this case", he/she said

[wanton edge]

Let's take the example for cos(3x). Look,
cos(3x) = 4cos³(x) - cos(x).
Substitute x = t/3 and we get
cos(t) = 4cos³(t/3) - cos(t/3)
Now the real challenge is solve the cubic equation in terms of cos(t/3).
And surprisingly, it's possible.
Unsurprisingly, it's more complicated than quadratic equation
And If you want to see the cubic equation, it uses cube roots and square root inside it.

[terse breach]

I think I should do this on Khan Academy

but ffs my school doesn't even go through trigonometry

[terse breach]

alright then, show it for this case

[wanton edge]

The cubic equation is proven to be solvable. I'm not gonna do that this time

[coarse prism]

<@&286206848099549185>

[upper karma]

<@&286206848099549185>

in r=+-sqrt(x^2+y^2) is +- determined by the quadrants the given coordinates (such as (1,4) are in on the tangent graph?

[upper karma]

can i get help?

im stuck on a question

[rain bay]

<@&286206848099549185>

[silent plank]

what;s your issue with this question?

[upper karma]

a prism is just half the volume of a cube

so bwh/2

love the pfp btw, aot is fire

[nocturne remnant]

This is not a prism

It’s a pyramid, whose volume is [1/3 x base area x height]

[strong mica]

how to solve

[gleaming nova]

that shaped is composed of 6 equilateral triangles

[strong mica]

i figured it out eventually

tks btw

[weak patio]

I have question i have to prove that BO AND EC make 90 degrees

It’s given that EOC =90 and BEO AND BCO = 90 and this shape total angles is 360 so EBC = 90 too

EB =BC and CO = EO

so this shape is dalton (or wt ever it’s name in English ) is that right ??

[wanton edge]

,rotate

[wanton edge]

I make the accurate version of your drawing

[weak patio]

Thx

[weak patio]

it’s given that EOC = 90

And BEO = BCO =90 right

And this shape angles is 360 so EBC also 90 right ?

So EO = OC =r and BE =BC

What this shape called in English xd ?

3 sides r equals and the angles in front of each other’s r also equals

[weak patio]

Why 95 ?

[wanton edge]

It's 45°, not 95°

[weak patio]

I see xd

And yes those r 45

This is square if I look it that way 😅

[weak patio]

But why u write CB and BE = r ?

[wanton edge]

Because it forms a rectangle OCBE. But since those two adjacent sides have the same length, OCBE is a square

[weak patio]

Might be

Hmm

Kk thx u

[wanton edge]

I think the answer is greatest common factor of 1186 and 4151
(593)

[magic lichen]

why will it be the greatest common factor

shouldnt it be like 2

cuz at max u can have 1 edge touching 1 edge of the other polygon

[wanton edge]

we'll talk about vertex, not edge

[magic lichen]

ye i know

in 1 edge u have 2 verticies

so at max u can have 2 vertices from each polygon touching each other

but why?

[magic lichen]

at max only 1 edge can touch which is 2 vertices touching

[opaque gull]

How do you make a circle in GeoGebra freely?

[magic lichen]

wuts geogebra?

[opaque gull]

A graphing calculator

[smoky cave]

can someone enlighten me

[opaque gull]

Also a 3D calculator and such

Search it, Alpha.

[magic lichen]

ye ye

[magic lichen]

with what

problems?

[smoky cave]

I forgot how to answer this step by step

[magic lichen]

which one?

[smoky cave]

this

[magic lichen]

what are u supposed to find?

[smoky cave]

sin 30

[opaque gull]

Sin(30)=1/2

[magic lichen]

ye

its a 1: root 3 : 2 triangle

root 3 is ur 2 root 3

[smoky cave]

is c= square root of 3

answer

[magic lichen]

no

its 2

.-.

1: root 3 : 2

[opaque gull]

,rotate

[somber coyote]

Couldn't find an attached image in the last 10 messages.

Apr 27, 2022, 9:34 AM

[magic lichen]

.-.

wait guys sin 60 is 1/root 2 right?

dude geo

here

applies for all 30 : 60 : 90 triangles

[wanton edge]

I use mobile for using Geogebra. I'm pretty sure that the same procedure can happen on computer.

[opaque gull]

Oh thank you a lot.

[magic lichen]

is this guy a bot or a real person cuz he just does not not know an answer to a question

[opaque gull]

A person.

What are you talking about?

[magic lichen]

jeezus christ man learn to not know stuff

its usefull.

trust me

[wanton edge]

A person

[opaque gull]

What?

[magic lichen]

i k n o w

[opaque gull]

Arya is smart.

[magic lichen]

no hes not smart

hes google itself

[magic lichen]

imma dip

[wanton edge]

Nah I'm not that smart, but why can't I be smarter than myself in the past?

[opaque gull]

Knowledge is great

[smoky cave]

I use the hand trick

[opaque gull]

I know them by heart

[magic lichen]

how

[opaque gull]

Something like that

[wise pawn]

this seems silly and more difficult than just thinking about the literal 45-45-90 and 30-60-90 triangles

[tidal patrol]

can someone plz help me with this

[dark sparrow]

have you done problems like this before but with only one angle?

such as "given sin(x) = [number] and x in such-and-such quadrant, find tan(x)" or whatever

[upper karma]

sin(A+B) = sinA cosB + cosA sinB, you're given sinA which is -(6/7), you can use tanB to find sinB. cosA = sqrt of 1-sin^2A, so you just need to find sinB

If I'm not mistaken

[nocturne remnant]

You messed up the angle sum formula

[upper karma]

sin(A+B) = sinAcosB + cosAsinB no?

oh yeah my bad I put another + by mistake

[nocturne remnant]

It’s good now

[rain bay]

<@&286206848099549185>

[nocturne remnant]

Didnt you post this a while ago

Also don’t tag helpers immediately

[rain bay]

Don’t think so, it’s different.

[rain bay]

Okay

[silent plank]

what have you tried?

the formula for volume of any pyramid is pretty much the same

[tidal patrol]

ooh ok got it thx!

Explain briefly, in words, how you would determine which equivalent trigonometric identities can be used in a
scenario modelled by the following diagram.

could someone help me with this?

[azure crown]

sinx = (45/53)
then sin ^-1(45/53)

[frigid hinge]

i feel bad for the 8th graders in my school who taking geometry and trigonometry this looks scary

[robust dune]

What are you exactly lookin for?

The area of the triangle created?

In that case it is sin x × cos x × 0.5 i think

[terse breach]

you must first find the base area of the triangle, the height which you need to find using pythagoras theorem

so 6^2 yd - 5.5^2 yds = height of the triangle

then when you have calculated the area of the triangle, you multiply the area by the height, then divide the product by three

that way, you should get the volume of the triangular pyramid

I hope this helped

[upper karma]

school overcomplicates maths imo. The teachers should explain why you are learning trigonometry and calculus etc, it would engage the students a lot more

[frigid hinge]

omg they take math as something like if u dont memorize a formula youll be stupid

[robust dune]

My teacher wrote his master of education how there is 2 types of doing maths

One type is doing it through understanding the other through memorizing

Of course the understanding one has better chances but it's not easy to make someone understand maths especially in higher maths since it gets more abstract

[terse breach]

Easiest question ever

Use the inverse trigonometric formulas to find the angle

[slim anvil]

very confused on what im supposed to be trying to find the area of

[azure crown]

means entire trapezoid i think

[wanton edge]

This is what I made using Geogebra

[opaque gull]

^

[opaque gull]

The trapezoid
(or find the area of the rectangle then add it together with the area of the triangle however that doesn't really work here)

[still hatch]

A polygon of n sides has 𝑛(𝑛−3)/2
diagonals. How many sides a polygon has with 54
diagonals?

i got the answer but would appreciate if anyone would give suitable explaination

[dark sparrow]

do you want an explanation of how the formula came about, or an explanation of how to use the formula to solve the problem?

@still hatch

[still hatch]

I would like both pls@dark sparrow

cuz I didn't used any formula

i want to see how it's done so yeah

[dark sparrow]

which one first?

to derive the formula: fix one of the vertices, and observe that it has n-3 diagonals coming out of it: one each to each vertex other than the one we started with & its two neighbors. adding n copies of this together (one for each choice of starting vertex) gives n(n-3), but doing this counts each diagonal exactly twice, so we divide by 2 to get the true number of diagonals.

to solve the problem: solve the equation n(n-3)/2 = 54 for n.

[still hatch]

🙏 thanks

[obsidian urchin]

I have a quick question: Is the formula for a mathematical pendulum's oscillation time (the unprecise one) the same as taking the area of the triangle that's formed in the circle

mg*sin(v)

[vagrant ore]

I need help with this problem

[hasty sun]

Have you tried anything?

[vagrant ore]

No I'm not sure what to put for a and b

[hasty sun]

Say you roll a die. What is the probability of you getting a 1?

[vagrant ore]

1/6?

[vagrant ore]

So for the red die is it 3/6 or 1/2? And for the green die it's 2/6?

[hasty sun]

justify your claims : }

[vagrant ore]

Because there are 3 numbers on a die bigger than 3 and 2 that smaller

[hasty sun]

could you list them for me please?

which numbers on a die are greater than 3?

and which ones are smaller than 3?

[vagrant ore]

Bigger: 4 5 6
Smaller: 2 1

[hasty sun]

ah, i'm sorry.. i was thinking otherwise

[hasty sun]

so your claims are true!

now

what's P(A) and P(B) then?

[vagrant ore]

P(A) would be 3/6 and P(B) would be 2/6

[hasty sun]

correct! now.. what about P(A and B)?

[vagrant ore]

P(3/6 and 2/6)

[hasty sun]

not really, no

[vagrant ore]

Do you add them?

[hasty sun]

event A is that the outcome of the red die is greater than 3. event B is that the outcome of the green die is smaller than 3. are the events dependent or independent?

For independent events A and B, P(A and B) = P(A)*P(B). So you multiply the probabilities.

So P(A or B) = P(A) + P(B) - P(A and B) = P(A) + P(B) - P(A)*P(B). Don't forget to reduce your answer to lowest terms as your question asks you to.

[vagrant ore]

Independent

[hasty sun]

indeed

[hasty sun]

so you can use this

wait

err.. thinks

[hasty sun]

yea right, you can use this

[vagrant ore]

So P(1/2) + P(1/3) - P(1/2) × P(1/3)?

[hasty sun]

nooo

P(A) = 1/2 and P(B) = 1/3

[hasty sun]

now correct what you wrote here

[vagrant ore]

1/2 + 1/3 - 1/2 × 1/3

[hasty sun]

and that simplifies to?

[vagrant ore]

2/5 - 1/6

[hasty sun]

what does that simplify to?

[vagrant ore]

1?

[hasty sun]

wait

1/2 + 1/3 is NOT 2/5 : |

[vagrant ore]

Is it 2/6?

[hasty sun]

no : |

[dark sparrow]

do you know how to add fractions

[vagrant ore]

Wait it's 5/6

[hasty sun]

Tell me. Did you just crunch them in a calculator? stares intimidatingly

[vagrant ore]

Yes

[dark sparrow]

you really should review fraction arithmetic

you will get nowhere in probability if you do not know how to add fractions

[vagrant ore]

I'll do it when I have spare time

[hasty sun]

apparently my spare time's over so.. good luck!

[vagrant ore]

Ok

[dark sparrow]

also tbh

why is this in geometry anyway

[vagrant ore]

Ok I got a little refresher on adding fractions and it's 1/6

Idk where else to put it

[dark sparrow]

what's "it"

1/2 + 1/3 definitely is not 1/6

1/2 + 1/3 - 1/6 definitely is not 1/6 either

[vagrant ore]

I thought you were supposed to multiply the numerators and denominators

[dark sparrow]

i thought you were supposed to not be cryptic in what you mean when you say "it"

it is not clear what you were trying to do and why you got 1/6 and what you were supposed to do instead and where exactly you messed up

[vagrant ore]

I'm trying to do 1/2 + 1/3

[dark sparrow]

okay

and if you tried to claim that this is (1 * 1)/(2 * 3), then congratulations, you correctly multiplied those fractions together.

you multiplied, but you were supposed to add.

[vagrant ore]

I thought that's how you add

[dark sparrow]

no, that's not how you add fractions. that's how you multiply them.

by your logic, two halves added together make a quarter and not a whole

which should strike you as nonsense

[vagrant ore]

Wait it is 5/6

I just added them and it is 5/6

[dark sparrow]

1/2 + 1/3 is 5/6, yes.

[vagrant ore]

I said that before and you guys made it seem like it was wrong

[hasty sun]

It wasn't wrong. But you shouldn't have used a calculator, is what I'd meant.

[vagrant ore]

Oh ok

[vagrant ore]

Alright so to continue the problem it simplifies to 5/6 - 1/6

[hasty sun]

This simplifies further, of course. Do it.

[vagrant ore]

Simplified it would be 2/3

[hasty sun]

Did you use a calculator again? stares intimidatingly

[vagrant ore]

No I did 5-1 and got 4/6 then simplified that

[hasty sun]

Hm. Tell me what 5/6 - 1/5 would simplify to.

[vagrant ore]

Don't you need common denominators to subtract?

[hasty sun]

it makes life easier, but it's not necessary

what's 5/6 - 1/5?

[vagrant ore]

Alright one sec let me do the maths

[hasty sun]

DO NOT use a calculator

[vagrant ore]

I won't

[hasty sun]

i'm watching you

watches

[vagrant ore]

19/30

[hasty sun]

very good!

[hasty sun]

this is your answer

[vagrant ore]

Cool thanks

[hasty sun]

Want another similar problem to check if you actually got it?

[vagrant ore]

No I have more like this one on my computer

[hasty sun]

Ah. Best of luck with them!

[vagrant ore]

Thanks

[rancid sequoia]

Can someone help me with this

[static jolt]

choose a diagonal and divide it in half

[rancid sequoia]

Bro idk what a diagonal is man I’m dumb as shjt

[bright pivot]

guys

can someone help me with an easy problem pls

its urgent

[static jolt]

so so one diagonal would be (2,4) to (4,-4)

so the ((2 + 4)/2),(4+-4)/2)

which is (3,0)

[rancid sequoia]

Ok thanks man

[static jolt]

you could also choose the other diagonal

[bright pivot]

help pls

[static jolt]

to double check

they both should be (3,0)

DB = sqrt(6^2 + 2^2)

DB = sqrt(40)

[rancid sequoia]

Dude I’m screwed I have a geomtry assignment by 8:25 and I don’t know like any of it I have to go to tutoring I think

[static jolt]

then AC = sqrt(6^2 + sqrt(40^2))

AC = sqrt(36 + sqrt(40))

which is the answer but i would use a calculator

[rancid sequoia]

Contrapostive is like all opposite right

[static jolt]

idunno what contrapositive is

[rancid sequoia]

I have this like 20 question practice I have to do for a grade I can repeat it but this stuff is hard

[static jolt]

the line of reflectoin is the origin i think

but idunno which one that is from the multiple choice

i guess its y=x and not origin

just because the other 3 arnt right

[rancid sequoia]

Thanks

[static jolt]

idunno

[rancid sequoia]

This one I think is more advanced im sorry

[static jolt]

i wouoldnt do y=x actually

[rancid sequoia]

It was right

[static jolt]

lol nice 😜

(8,-6)

becaus its symmaetric across the x-axis

[rancid sequoia]

Damn

You on something man damn

This one is less confusing but I don’t really understand it to much

[static jolt]

G

because alternate exterior angles are congruent not suplementary

[rancid sequoia]

I only have like 2 questions left do you think you could help me with them

[static jolt]

mebbe

[rancid sequoia]

Ok I got that one write I just need 2 more and than I’m done

[static jolt]

was it 130?

[rancid sequoia]

I believe so

I don’t really understand the other one

[static jolt]

i think its H

its basically asking how to draw a perindicular line using a special tool

but dont do h yet

caause i dunno

[rancid sequoia]

It wasn’t right and I’m kinda fed I need to get 2 more right

[static jolt]

😦

[rancid sequoia]

It’s ok I need 80 percent I just need 2 more

[static jolt]

this one idunno

i would say F

because i dont think its rotation

[rancid sequoia]

Good job man

One last one

[static jolt]

that was luck lol

[rancid sequoia]

This is the last one and then I’m good

[static jolt]

bisector is the middle of the line

maybe use a ruler?

thats a weird question being on a computer screen

[rancid sequoia]

I went with x and was correct

[static jolt]

ok nice good job

[rancid sequoia]

3 total left I don’t really understand this one

[static jolt]

H

[static jolt]

H

[rancid sequoia]

Somehow wasn’t right I just need one last one though

[static jolt]

H

[rancid sequoia]

I can’t thank you enough you saved me dude

[static jolt]

okay np

[vague bison]

missed class when learning this- how would i solve sinθ=-.7 from 0 to 2pi. ive figured out how to graph it with desmos but i dont know how to actually find the answers algebraically

[nocturne remnant]

You need to solve for theta?

[sly vale]

Sin^-1 (-0.7) is -45 or -pi/4 (approx.)
More accurately, it would be 44.5 ish

[river marten]

I have this geometry problem to show that the side of a rectangle with an inscribed 'snowflake'(beginning of fractal) is 1 by 2/3 sqrt3. It seems quite straightforward but using 30 60 90 triangles i can't seem to solve it..

[terse flume]

geometry question , what is sinus ? (sin)

[silent plank]

sine is a trigonometric function

[terse flume]

ye ye i know but like what is it?

[silent plank]

there are multiple levels of answers to that

[terse flume]

but like what should i imagine if we re talking bout sinus , like wtf is that alien shit

[silent plank]

think of it as a function that can do certain things

in a right triangle
the sine of an acute angle gives the ratio of the side opposite that angle and the hypotenuse

and you can also consider the unit circle definition

[river marten]

what is the calculation behind trigonometric functions

what does your calculator do when you take the sine of a number

[sly vale]

Perpendicular divided by Hypotenuse Of a right angled triangle in accordance with an angle.

[sly vale]

^^

[terse flume]

yea yea okey okey i exactly understand what u mean

its so easy

[sly vale]

It is quite easy but it's application ranges from easy to extremely hard.

It's more like a tool

[sly vale]

Bro no one is that free to tutor you every day. Can surely help you on a concept or two but asking for a tutor is probably too much.

[vagrant ore]

can you slice a tetrahedron and make a square cross section?

[sly vale]

Yes

[sly vale]

It can be a square

[vagrant ore]

Ok thanks

[buoyant garden]

I'm not sure if this is considered Algebra or Geometry but I've been put of school all week because I've been sick and I'm extremely confused. Can someone help?

[sly vale]

Number is a perfect square of 17

@buoyant garden

[buoyant garden]

Ok thank you!

Can someone explain how to do these?

[sly vale]

Just cube root it

As volume is mentioned in unit ^3

[buoyant garden]

How do I cube root it on a online calculator?

[sly vale]

So basically it says, side x side x side = something

What's side?

[buoyant garden]

Do you divide the number by 3?

[sly vale]

6 would be the answer

No

What's the formula for volume of cube?

[sly vale]

Do you know it?

[buoyant garden]

Not really.

[sly vale]

Ok

So volume of a cube is Side ^ 3

Or side x side x side

So you're given Side ^3 = 216

What's Side?

Side will be root cube of 216

You know what's a root cube?

[buoyant garden]

No

Sorry, im just very confused.

[sly vale]

You know what's a square root?

[buoyant garden]

Yes

[sly vale]

Man you will have to study about them from your school textbook before doing these questions.

[buoyant garden]

I dont have a textbook for this year 😅

[sly vale]

Then you can look up online what are square , square roots, cubes and cube roots.

There you'll find a detailed explanation

You can try Khan Academy or YouTube

[buoyant garden]

Alright thanks.

[vagrant ore]

I have a question. Two cards are selected from a well-shuffled deck of 52 playing cards. What is the probability that neither of them is an ace? My question is would I do 48/52 times 47/51?

[dark sparrow]

that is one possible way to phrase the solution

yes, 48/52 * 47/51 is correct

[vagrant ore]

ok I thought so

[dark sparrow]

-> #probability-statistics tho

[vagrant ore]

ok

[sly vale]

48C2 / 52C2

~~ 0.85 . So yep your answer is correct

[upper karma]

Have been struggling with this question lately and would greatly appreciate it if someone could give me some tips or advice on how to solve it. Thank You.

[umbral snow]

Beni is not 0.8 km away from the pad

Let Alessio be "a" km away from the pad

[grave pond]

I wonder if the author of that problem has ever seen any video of a rocket launch. They generally pitch over (i.e., stop following a vertical line through the launch pad) long before they "disappear from view". That makes the task woefully underspecified, even if we assume that "disappearing from view" is somehow an intrinsic event that happens at the same time no matter where you're viewing the launch from.

[upper karma]

Yes, you're correct, however, I'm not sure how i could use it to find the altitude tho, along with the angles of elevation

[ember stratus]

I have a coordinate system, shown in black below, in which a point is situated along the 𝑥-axis. There is a different coordinate system rotated along the 𝑧-axis by 𝑎 degrees, shown in red in the figure. I can switch between these two systems by a rotation matrix 𝑅.

Two lines (in the black coordinate system) have constant 𝑦-values. These two lines intersect the projections of the 𝑥-point onto the red coordinate system.

I would like to find these two points 𝑃1 and 𝑃2, but it is not clear to me how I can do that. Can I get a hint?

enter image description here

[grave pond]

Is the point x supposed to lie on the red line between p1 and p2?

[strange umbra]

"Let ABC be a triangle with AC < BC. The circle with center C going through B intersects the line AC in D. The line DI, with I being the incenter of ABC, intersects the circle around C in another point (E).

Prove that CE is tangent to the circumcircle of ABC."

or something like that, I'm not really good at translating

[upper karma]

whats F for

oh

[strange umbra]

ignore it

[sly vale]

Altitude of rocket would be 3.11 km and the distance of Alessio would be 5 km from launch pad.

[upper karma]

sorry i couldnt solve it, imma keep it and lyk if i solve it

[ember stratus]

yes

[upper karma]

@sly vale your correct! I just did the calculations, and the total distance from the launching pad was 5km and the altitude (h) of the rocket was 3.1 km. However my text book says it is 4.2 km...

I'm not sure why, but 4.2 km was the distance between beni and the launching pad, though.

[sly vale]

That's for Beni

4.2 km for Beni and 5km for Alessio

[upper karma]

Yes, I figured that

And the altitude of the rocket is 3.1km right?

[sly vale]

Yes

[upper karma]

Okay, than I have no idea why my textbook says it is 4.2km for some reason

[sly vale]

It's wrong

[upper karma]

I'll have to consult my teacher about this, this textbook has had numerous mistakes from the past chapter...

[sly vale]

Most probably a printing mistake

[upper karma]

Definitely, but thank you for the support 💖

[sly vale]

Welcome

[grave pond]

Then you know the angle between the red line and the black x-axis, so you can draw some right triangles and use trigonometry to find the diffence in x-coordinate betwen x and p1 or p2.

[nocturne remnant]

We can see that DAIB is cyclic because angle CDI=angle CBI(by symmetry)=angle ABI
Then angle BAC=2angle BAI=2angle BDI=angle BCE(by inscribed angle thm)
Then done by tangent chord thm or sth

[strange umbra]

oohhh

[strange umbra]

thx

[median matrix]

Can someone help me with this exponential regression problem that i've been having trouble with

i tried a help room but the person didn't know how to solve these types of problems

[terse flume]

@Rock123222#6687 just put some ice in the soup it ll be quicker

[lapis moon]

put y = 80 and solve for x in the equation in part (a)

when you've finshed you may use wolfram alpha to check your answer

,w solve 80 = 180(0.9)^x

[somber coyote]

Results provided by WolframAlpha

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

Requested by vin100#3191

[grave pond]

Note that for temperatures,, a function of the form 180·0.9^x doesn't really make sense -- that would mean that if left for long enough, the temperature of the soup would approach 0 °F -- but the zero point of the Fahrenheit scale is arbitrary, not something soup will spontaneously converge towards in the real world.

The exercise is almost certainly looking for a best-fit solution in the form f(t) = b·a^t+c rather than just b·a^t.

(Hmm, the asker already left. Oh well.)

[scenic gust]

Could somebody help me?
Al I got is a triangle with 1 know side but don't know hot to continue.
Thanks in advance

That's what i got so far.

[cursive fossil]

Let d equal distance of the person from the wall.
tan = (altitude of painting-altitude of eyes)/d

Instead of solving for theta, the question asks for a general expression of tan of given a distance.

[scenic gust]

Thanks let me try to work it out

So tan theta= (2,25-1,75)/d

@cursive fossil ?

[grave pond]

Why do you subtract 1.75 from 2.25?

You already got 2.25 by subtracting 1.75 from 3+1...

[scenic gust]

Yeah sorry my mistake

Ah so it is tan theta = 2,25/d

[grave pond]

I find it unclear in the problem whether the angle it is talking about is the angle to the top of the painting (as in your sketch) or the middle of the painting -- as well as whether the 3 meters are measured between the floor and the top, middle, or center of the painting. The best you can do is probably to explicitly state what your assumptions are in the solution.

[scenic gust]

Yeah thanks

You got some more time @grave pond struggeling with another question as well trigonometry really fucks with me

[grave pond]

In truth I don't really. Trying to pry myself away from the laptop ...

[scenic gust]

No problem have a good day

[willow vessel]

Why a line in hyperbolic geometry must contain two distinct ideal points?

[unborn gulch]

Anyone able to help me with this? I don't really understand the question and got it wrong the last time I did this same practice exam (not for a grade)

[keen furnace]

Please help

[rare wasp]

is 91 degrees trisectable?

[lapis moon]

that's impossible by compass and straightedge construction

consider

,,\cos(3x) = 4\cos^3(x) - 3\cos(x)

[somber coyote]

vin100

[lapis moon]

$z = \cos(x)$ is the root of the cubic polynomial

[somber coyote]

vin100

[lapis moon]

,,4z^3-3z-\cos(3x) = 0

[somber coyote]

vin100

[lapis moon]

for a fixed $x$

[somber coyote]

vin100

[lapis moon]

so the extension degree $[\Bbb{F}(\cos(x)):\Bbb{F}(\cos(3x))] = 3$ if $\cos(3x)$ isn't constructible

[somber coyote]

vin100

[lapis moon]

here $\Bbb{F}$ is a field of constructible numbers

[somber coyote]

vin100

[lapis moon]

classical rule & compass construction can only construct numbers whose extension degree with respect to $\Bbb{F}$ is a power of 2.

[somber coyote]

vin100

[cyan glacier]

fuck trig

just kidding

where can i get help again

for trig

[lapis moon]

to trisect 91°, you're constructing (91/3)°

[lapis moon]

math help channel

[gleaming nova]

are you aware of what the channel you’re in is called

[cyan glacier]

yes

[lapis moon]

once we know that an angle of 91° isn't constructible by rule & compass, then we may apply the above arguments with x = (91/3)° to conclude that that's not construtible.

looking back the condition that $\cos(3x)$ isn't constructible is too strong. i've quit U for a few yrs. i can only recall the rough ideas.

[somber coyote]

vin100

[lapis moon]

but once we can that $\cos(91\pi/180)$ isn't constructible, then we would see that the desired trisected angle is neither constructible.

[somber coyote]

vin100

[lapis moon]

by a rotation of right angle, we see that that's equivalent to asking whether $\cos(\pi/180)$ is constructible

[somber coyote]

vin100

[inner sandal]

for this question i assumed for all x in S² that f(x) is in S² too and tried to reach a contradiction. if T is a translation T(x)=x+b b≠0 then ||T(b/||b||)|| = ||b/||b||+b|| = ||b||+1 > 1, which means f can't contain any translations (reflections in parallel planes), so its decomposition of 4 reflections which preserves orientation must be 2 rotations, 1 rotation and 2 reflections, or 4 reflections. Since the pairs of reflections aren't translations, they're rotations around the intersection of the planes, so we just have 2 rotations either way. But that can cancel to just 1 rotation, which contradicts the fact that f is no fewer than 4 reflections. Is this correct

[vagrant ore]

I need help with these problems

[rare wasp]

thx so much! the last point is a good point

but if we suppose 91 degrees to be constructible, would it be trisectable?

[magic lichen]

guys where do i learn intermediate trignometry?

[lilac whale]

It would be C. Basically you use the top number and multiply it to what t is. Then you use that number for t and use algebra from there. Add each sum.

Is it okay to give the answer or nah?

[rare wasp]

I believe 3pi/7 is trisectable but not constructible for example. I'm having trouble finding a definitive way to determine if some angle is trisectable

[magic lichen]

wuts trisectable?

[lilac whale]

To divide into three equal parts

[magic lichen]

check if its divisible by 3?

[lilac whale]

Yea

[magic lichen]

how is it hard then

check using divisibility by 3 rule

[lilac whale]

It has to be an integer answer

Thats probably why

[magic lichen]

ah ok

[lapis moon]

the premise "91° is constructible by rule & compass" is false, so whatever the following conclusion is, the implication is logically true, but meaningless.

from the half-angle formula for cosines

,,\cos(\frac{x}{2}) = \pm\sqrt{\frac{1+\cos(x)}{2}},

[somber coyote]

vin100

[lapis moon]

so we consider $\cos(\pi/45)$. apply the tripe angle formula to see that it's of extension degree 3 from the field $\Bbb{F}$ equiped with $\cos(\pi/15)$, which is not constructible since that again of extension degree 3 from $\Bbb{F}(\cos(pi/5))$.

[somber coyote]

vin100

[austere hull]

I have a doubt

Plz anybody can come voice chat

Anybody ??

[static pasture]

It may be useful to some of you to know there is a rational parametrization of the unit circle
It is e(t)= [(1–t^2)/(1+t^2), 2t/(1+t^2)]
This gives you all rational points on the unit circle except for [–1, 0] but we can modify our previous parametrization to get
e(u : t)= [(u^2–t^2)/(u^2+t^2), 2ut/(u^2+t^2)] (know that a : b is the proportion of a to b and a : b= c : d if and only if ad–bc= 0) now all rational points on the unit circle can be reached through this modified parametrization e(0 : t)= [–1, 0] and e(1 : t)= e(t)

[hard coral]

Quick question, for a cut hollow cylinder, how do I find the surface area of that segment/chrod if I have its radius

Its arc length is 24.74 cm

but the segment is 6 cm

[strong mica]

geometry: how to find the missing value

[silent plank]

apply formula for volume

[lilac whale]

oh thats actually cool

[pliant roost]

it is pretty simple
First, we know r = 8
We also remember that circle's area is πr²
Area of the sector is just α/360° * circle area, so, in final we get 80/360 * 3.14 * 8², and you can calculate that

[pliant roost]

Circle circumference is 2πr, knowing r = 5 cm, it is 10π.
Remembering that arc is a part of the circle again, the length is just (substituting for π) 72/360 * 10 cm * 3.14, which you can calculate too

[tulip vector]

how to solve for x and y?

[dark sparrow]

this question is impossible to answer until you've posted a problem

[dark sparrow]

if you had a similar question but with only one variable, would you be able to do it?

[magic lichen]

u can take 17x-70 = X
3y + 5 = Y
5y+15 = Z
2x+5 = A

u dont need to :U

actually its simple

17x-70 = 3y+5

thats a hint

[magic lichen]

i dont know how to calculate the area of a sector bruh i forgot to learn

[pliant roost]

isn't it just logic

[magic lichen]

i know

wuts the formula

im lazy to use logic

._.

[pliant roost]

you don't even have to learn it tho

[magic lichen]

tell me formula >:(

oh wait

[pliant roost]

just α/360° * πr^2, and if you use radians it' just α/2 * r^2

[magic lichen]

dude

πr^2 is what u did

._.

[pliant roost]

f

[magic lichen]

he screwed up his exam XD

[pliant roost]

i don't even have trigonometry

[magic lichen]

i just started learning so i basically dont

[pliant roost]

though it's not trigonometry

[magic lichen]

ye

[pliant roost]

though I somewhat know it (trig) since it's pretty simple

[magic lichen]

ye its a part of a circle

so u calculate full area of circle

and find out how much of the circle

[pliant roost]

and multiply

[magic lichen]

which is ur angle/360

ye

[pliant roost]

easy 5th grade math

[magic lichen]

like they even know what pi is

XD

not like i didnt learn in 5th

[pliant roost]

well we did know

[magic lichen]

oh ok then indian math is bad

._.

-._-.

[pliant roost]

though we didn't pass circles on geometry yet

[magic lichen]

XD

[pliant roost]

I mean I am in like 7th grade

and even though my class is math one it's still slow

so I just cover math by myself

[magic lichen]

ye its what i hate about my school math

[pliant roost]

I mean how can you live without trigonometry

[magic lichen]

it doesnt teach what im learning at home

just boring at school

:U

we havent started trig and im in 9th.

they start in 10th

how bad is this school

[pliant roost]

we start in 8th

[magic lichen]

.

name of the school?

[pliant roost]

it is not indian

[magic lichen]

dc

tell name

[pliant roost]

school №1540

[magic lichen]

im transfering rn

yo what is that

[pliant roost]

I mean it's just a school

[magic lichen]

.

aight

imma go help some duds in alg

[pliant roost]

ok

[dark sparrow]

[magic lichen]

got it got it

[steep axle]

Has anyone here taken the Geometry SOL

[echo siren]

Hi could anyone kindly go through how the ans (2nd pic) solves the qn (1st pic)? Been looking at the ans for a while but not really sure what's the approach used

[lapis robin]

You can find sin c by making a triangle with the tan c value

Looks like you just expand sin(x+c) into sinxcosc+cosxsinc

Find cosc as well

Also you didn't give all the information provided for this question

[echo siren]

oh ok thanks! Makes sense to use the sin expansion identity

[echo siren]

That's the whole qn actually haha

[outer lagoon]

Can anyone help me with unit vectors

[lapis robin]

No shot that's all the information

I don't believe that

[echo siren]

really lol

[lapis robin]

There's information for the start of 1

?

[echo siren]

Nopee, it just starts with 1A-1 and so on

the information you thinking of is trigonometry identities?

[lapis robin]

no

[low karma]

Can someone fill this completely out with the correct answers and message me when complete please and thank you

[upper karma]

Can anyone help me understand unit circles?

I cannot do them fast enough and always get it wrong

[trim breach]

This server isn’t meant for you to solicit completed assignments. #❓how-to-get-help

[wanton edge]

Is this true?

I mean proving
$$\alpha = 2\beta \Leftarrow a = b$$
is easy for me. But how about the other way?

[somber coyote]

Si Arya

[nocturne remnant]

Suppose on the contrary that a ≠ b. Then there must exist some other point on side a that is b units away from the uppermost point

I think you can fork out some sort of contradiction from there

Well.. I don’t think we have to use a “contradiction” proof; a phantom point construction will work as well

[magic lichen]

unit circles are circles with radius 1

[sudden anchor]

you can prove it like this

[wanton edge]

Thanks. And now we want to prove the converse of that.\
$\alpha = 2\beta \implies a = b$

[somber coyote]

Si Arya

[sudden anchor]

Well since $\alpha=2\beta, AD=AC \implies B\in k(A,AB)\implies AB=AC=AD=a=b$

[somber coyote]

Лукар

[nocturne remnant]

I feel like you’re skimming over the most important part here… how did you get “B ∈ k(A,AB)”?

[wanton edge]

It might be better to prove the contrapositive first.\
$a\neq b \implies \alpha \neq 2\beta$

[somber coyote]

Si Arya

[royal portal]

heyo folks

im having trouble solvin a geometry problem

id need just an idea'

[sudden anchor]

Are BB1 and AA1 angle bisectors(just making sure)

if AA1 and BB1 are bisectors then this is good

[sudden anchor]

well B is element of k(A,AB) cause the angle A (which is subtending BC)= 2 * angle D(which is subtending BC) so B must be a part of a circle centered at A

[pliant roost]

How can I prove this identity?

For the people who can't work with series notation, the left-hand side is sin(α) + sin(α + β) + sin(α + 2β) + ... + sin(α + (n - 1)β)

An example is sin(2°) + sin(4°) + sin(6°) + sin(8°) + ... + sin(176°) + sin(178°) (α = 2°, β = 2°, n = 90)

[pliant roost]

Basically the step in solving a problem I have stumbled upon and can't get over

[wanton edge]

My first thought is\
$\sum_{k = 0}^{n - 1}\sin(\alpha + k\beta)=\sum_{k = 0}^{n - 1} Im\left(e^{i(\alpha + k\beta)}\right) = Im\left(\sum_{k = 0}^{n - 1}e^{i(\alpha + k\beta)}\right)$

[somber coyote]

Si Arya

[pliant roost]

I would like the proof not to use the exp(x) function properties, also this is a sum anyway, when I need that exact formula

Maybe it is somehow connected to series like geometric or arithmetic

But I didn't study series yet so I am kind of stumbled. I could study them tomorrow though, but I don't think it will help that much

[lapis moon]

When I was in high school, I was fascinated with exponential function with complex indices, but with improved comprehension, I'm more comfortable explaining with trigo identies that you wanted

[pliant roost]

I mean I am not even in high school yet, just a year before it

exp(x) is cool but boring

you can use it sometimes because of its nice properties but I of course would like a more... trigonometric proof?

[lapis moon]

Let's consider a simpler problem: forget about the sine, just take sum on some consecutive terms in an arithmetic progression. The most difficult part is $$\sum_{k=1}^n k.$$. Gauss has a nice idea, but we would take a "differential approche", rather than his "global way". We observe that each term is "equidistant" with its two neighboring terms.

[somber coyote]

vin100

[lapis moon]

we write k to be a difference of two things, then do telescopic sums

,,k=(k+1/2)^2-(k-1/2)^2

[somber coyote]

vin100

[lapis moon]

This can be easily seen using difference of squares identity $a^2-b^2=(a-b)(a+b)$.

[somber coyote]

vin100

[pliant roost]

Yeah sure, I remember it

no need to explain simple math

[lapis moon]

The choice of 1/2 is no magic

[pliant roost]

though wait

isn't that 2k?

[lapis moon]

That's the mid-point of our "1D-grid system"

Oops sorry

[pliant roost]

arithmetic))

[lapis moon]

,,k=\frac{(k+1/2)^2-(k-1/2)^2}{2}

[somber coyote]

vin100

[pliant roost]

is 1D-grid system just a line?

sorry to interrupt

[lapis moon]

I got that fixed.
It's like marks on a ruler. Sorry idk what's the right word in our daily lives.
But that doesn't affect the main idea.

[pliant roost]

yeah seems about right

the point from which we build up on

[lapis moon]

,,\sum_{n=1}^N a_n = \sum_{n=1}^N (b_{n+1}-b_n) = b_{N+1}-b_1

[somber coyote]

vin100

[lapis moon]

I'm on 📱 excuse me for speed

[pliant roost]

no problem

[lapis moon]

The second sum is called telescopic sum.

[pliant roost]

ok, I will derive this sum by hand tomorrow (too late today)

[lapis moon]

That allows us to get the sum for n-th term

No worries it's also late on my side.

I'll just type so that you can read tmr

[pliant roost]

yeah of course

is a_n that k term but with index to clarify its... index?

well yes

why wouldn't it be

[lapis moon]

In fact in prefer writing

,,a_k=b_{k+0.5}-b_{k-0.5}

[pliant roost]

so you just have a set and a_n is it's nth element

[somber coyote]

vin100

[pliant roost]

oh ok

[lapis moon]

A set with order, or a sequence

[pliant roost]

well yes

and what is b?

fractional indices seem scaryish

[lapis moon]

$b$ is a hidden sequence to be found out

[somber coyote]

vin100

[pliant roost]

ohh ok continue

and fractional indices?

[lapis moon]

There're some rules to make an educated guess about this sequence

[lapis moon]

That might sound weird at the first time. Unluckily idk how to draw on mobile

Like Ms paint

[pliant roost]

there can be some apps you may download

[lapis moon]

wwwwww

[pliant roost]

my mobile has image editor for example with a marker

[lapis moon]

imagine the 'v' connected like vvvvvv

below are a_n's

above is b_?-b_?

[pliant roost]

so you have a_n correspondence to b_? - b_?

uhh I am bad on terms, it has a better name

[lapis moon]

Yes, and I want the b_?'s in the middle cancel out.

[pliant roost]

hm

and what about fractional indices

is it like set length is 2 but there are 3 elements?

so you have {a,b,c} 's length 2
1 1.5 2

[lapis moon]

Sorry I'm not thinking about counting here. Actually I don't really care what's inside. Instead I'll focus on the boundaries

[plush portal]

hello i need help with trigonometry at #help-11

[pliant roost]

ok

[pliant roost]

so it gets shifted? and bigger by 1 in length?

[lapis moon]

It's just my personal habit that's not seen elsewhere.

[lapis moon]

The indices of b are shifted by 1/2

[pliant roost]

continue so I understand

[lapis moon]

Yeah nice catch

[pliant roost]

oh

[plush portal]

hello i need help with trigonometry at #help-11

[pliant roost]

just on the pic it looks like some items in a set are inbetween the items

[lapis moon]

My strange habit would turn out relevant to your desired identity coz in the denominator your see beta over two

[pliant roost]

lol

cool

anyways we shall continue on

[lapis moon]

Yes I give up on online drawing pad. I'm gonna download drawing app afterwards

The problem with splitting a_n with b_? using integer indices is that you can split that in at least two ways

        a_n
        |    \
        |       \
      b_n    b_{n+1}

and

a similar picture for

,,a_n=b_n-b_{n-1}

[somber coyote]

vin100

[pliant roost]

so you can do both + and -?

[lapis moon]

That's in our construction of telescopic sums

[pliant roost]

yes, like you do it in one way or another

[lapis moon]

Represented in pictures, if we use integer-valued indices for b_n, you'd need to face "|\" vs "/|"

[royal portal]

Could you explain how you find out the first phase like ą+B=64?

[lapis moon]

Let's get back to your problem. You have $(\sin(a+kb))_k$

[somber coyote]

vin100

[lapis moon]

It's like "arithmetic progression" wrapped with sines

[sudden anchor]

Well you should notice triange ABO

and do sum of angles

[lapis moon]

Sorry I'm using latin alphabets to save time

[pliant roost]

yeah np

[lapis moon]

It's like a 1D grid system with points ...,a-b,a,a+b, a+2b,...

wrapped with sines

So we are writing each grid point as a difference of two terms

,,a_n = b_{n+1/2}-b_{n-1/2}

[somber coyote]

vin100

[lapis moon]

with

,,a_k = \sin(a+kb)

[somber coyote]

vin100

[lapis moon]

I wonder if you know sum-to-product and product to sum formulae

[pliant roost]

yeah I do

It's just sin(a + b) = sin(a)cos(b) + sin(b)cos(a)

[lapis moon]

Nice observation of the principal behind

[pliant roost]

I don't think you will care about needing to do smth with sin or cos(kb)

since it's just sum in a sum, which is overly complicated and useless I think

[lapis moon]

triple angle formula would too complicated
Before I go out and take metro, I will type this:

[pliant roost]

And sin(a) + sin(b) = 2sin((a+b)/2)cos((a-b)/2) if I am not lying

[lapis moon]

observe that in our "splitting of a_n" into b_n, the n is the midpoint/average of the two indices in b_?

You've found the right formula

in face the (a+b)/2 captures this idea of "midpoint"

[pliant roost]

Ohhhh!!!

[lapis moon]

You might worry about the second factor containing (a-b)/2

but that depends on the grid length

note that our 1D grid has evenly-spaced grid points

I'll let you think about these observations and take a look in the metro compartment on my way home

[rancid sequoia]

Yo I’m kinda dumb as shit can someone help me with this

[pliant roost]

ok, thank you very much! have a good day/night/whatever, I will work around this tomorrow since no time today (00:08 AM)

[slender mist]

can anyone help.

[pliant roost]

√(24²-10²)

[pliant roost]

Pythagoras again

[slender mist]

i tried

that

[pliant roost]

What's the problem?

oh no msg edited

[slender mist]

i got it wrong when i used pythag

[pliant roost]

?

[pliant roost]

it's Pythagorean theorem, don't you see

[slender mist]

i did the pythagrous theorem and i got it wrong 🗿

[rancid sequoia]

I thought it was

[pliant roost]

well in the wrong place prob

or bad rounding

[slender mist]

can you try..

[pliant roost]

Wait lol

I see now

[slender mist]

oh sorry.

[pliant roost]

well it's still easy since it's just √(24²-5²)

Since if you connect center and chord ends you get an isosceles triangle with a height in it

[slender mist]

oh.. i did do that 🗿 my answer was off by 0.3 😭

i put it in the calculator wrong

[pliant roost]

23.5 should be it

if rounding to the nearest tenth

[pliant roost]

For you it's a right triangle with legs 15 and 24

[rancid sequoia]

Yeah I got it i just wanted to make sure this one I have no clue how to do

[pliant roost]

Angle R is the least

There is an interesting property of triangles

to the bigger side corresponds a bigger opposite angle

and in reverse

[rancid sequoia]

This is a bit confusing to me

[lapis moon]

Oops sorry I've mistaken. Nice try but there's no difference of two terms on the LHS of this identity that you've quoted. That's important coz we're gonna do a telescopic sum to condense the entire sum into two terms

The one that we're searching should look like ?(a)-?(b) = sin((a+b)/2) * ?

The problem with sin(a) - sin(b) is that it "wraps the average (a+b)/2 with cosine"

[upper karma]

A triangle ABC is intersected by a line I at point F on AB, point E on AC, and point D on
the extension of BC. If triangle ABD has a right angle at A, BC=CD=13, AD=24, and DE
bisects angle ADB, what is the length of AF?

I got 12/5 for this question however the answer is 24/5, can anyone tell me what i'm doing wrong? thanks

[rancid sequoia]

I don’t really know how to do this one

Yeah but I don’t know which one

To choose the answer

None I just know they bisect each other

Reflexive thing

Sas

How do I know which one is the side and angles

I know some but not a whole lot this is my first year

I just forgot a lot of this stuff

So it is just Side side side

It is asking which Theron can be used to prove it is congruent so and you said their was 2 sides so is it h side-angle-side

I’m still confused

[slender mist]

can any1 here, thats free, provide some assistance

[rancid sequoia]

Oh no second one

[lapis moon]

i hope my lengthy introduction would convince you the choice of this formula

,,\cos \theta -\cos \varphi =-2\sin \left({\frac {\theta +\varphi }{2}}\right)\sin \left({\frac {\theta -\varphi }{2}}\right)

[somber coyote]

vin100

[lapis moon]

use $\theta = a + (k+1/2)b$, $\varphi = a + (k-1/2)b$.

[somber coyote]

vin100

[lapis moon]

their ``average'' is $a + kb$.

[somber coyote]

vin100

[lapis moon]

their ``semi-difference'' is $b/2$. observe that in the formula that i quoted, it's "wrapped by sine".

[somber coyote]

vin100

[lapis moon]

rearrange to match the term in the original sum
sum "average" = sum "difference"

,, \sin (a+kb) = -\frac12 \frac{\cos[a+(k+1/2)b] - \cos[a+(k-1/2)b]}{\sin(b/2)}

[somber coyote]

vin100

[lapis moon]

,, \begin{aligned}
& \sum_{k=0}^n \sin (a+kb) \
&= \sum_{k=0}^n -\frac12 \frac{\cos[a+(k+1/2)b] - \cos[a+(k-1/2)b]}{\sin(b/2)} \
&= -\frac{\cos[a+(n+1/2)b] - \cos[a-(1/2)b]}{2\sin(b/2)} \
&= \frac{\sin[a+(n/2)b]\sin[(n+1)b/2]}{\sin(b/2)}
\end{aligned}

[somber coyote]

vin100

[lapis moon]

,tex % https://q.uiver.app/?q=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
[\begin{tikzcd}
& {a_1} && {a_2} && {a_3} \
{b_{0.5}} && {b_{1.5}} && {b_{2.5}} && {b_{3.5}}
\arrow["{+}"{description}, color={rgb,255:red,92;green,92;blue,214}, from=1-2, to=2-3]
\arrow["{-}"{description}, color={rgb,255:red,214;green,92;blue,92}, from=1-2, to=2-1]
\arrow["{-}"{description}, color={rgb,255:red,214;green,92;blue,92}, from=1-4, to=2-3]
\arrow["{+}"{description}, color={rgb,255:red,92;green,92;blue,214}, from=1-4, to=2-5]
\arrow["{-}"{description}, color={rgb,255:red,214;green,92;blue,92}, from=1-6, to=2-5]
\arrow["{+}"{description}, color={rgb,255:red,92;green,92;blue,214}, from=1-6, to=2-7]
\end{tikzcd}]

[somber coyote]

vin100

[lapis moon]

\begin{align}
a_k &= \sin (a+kb) \
b_{k+1/2} &= -\frac12 , \frac{\cos[a+(k+1/2)b]}{\sin(b/2)}
\end{align}

[somber coyote]

vin100

[lapis moon]

in the picture, $$a_1 + a_2 + \dots + a_n = b_{n+1/2} - b_{1/2}.$$

[somber coyote]

vin100

[lapis moon]

there's a deep math meaning behind this trigo identity. this gives us a way to find us the definite integral of the sine function.

https://cdn.discordapp.com/attachments/326138757474680852/971856327695544371/unknown.png

[slender mist]

is their anyone here... who can explain this

[lapis moon]

use the formula $\pi r^2 \cdot \frac{\theta}{360^\circ}$

[somber coyote]

vin100

[lapis moon]

where $\theta$ is in degrees

[somber coyote]

vin100

[lapis moon]

🔗 : https://study.com/cimages/multimages/16/sinx_0_to_2_pi_start9218414468356518937.png

🔗 : https://study.com/cimages/multimages/16/sinx_0_to_2_pi_finish5308746109252968357.png

say we have a closed interval [a,b] divided into n equal subintervals.
we write the partition P = {a = x₀, x₁, ..., xₙ = b}
xₖ = a + k Δx with k = 0, 1, ..., n.

the sum of the area of "these approximating rectangles" is called the Riemann sum. The height of each strip depends on both the function and the choice of point in each subinterval.