geometry-and-trigonometry

[wintry tundra]

do you see how the 5 is touching angle c?

[slender mist]

yup

[wintry tundra]

the side with length 5 touches angle c

[slender mist]

thats why the call it

tanc

[wintry tundra]

that means they are adjacent

[slender mist]

oh

thank you

[wintry tundra]

so you have to find the side thats opposite of angle c

which would be?

[slender mist]

12

[wintry tundra]

yep

so 12 is on top because its opposite over adjacent

so opposite is on top

[slender mist]

i think

i got the answers

[wintry tundra]

and the adjacent side is 5

so the tangent is 12/5

alright tell me what you get next

so i can see if you did it right

[slender mist]

TanA = 5/12
TanC = 5/13
SinA = 12/5
CosA = 12/13

:pain:

[wintry tundra]

yeah cos a is right

sin a you have to find the opposite side

so which side is opposite of angle a?

by the way the opposite side is the side that the angle doesnt touch

[slender mist]

5

[slender mist]

that would be 5

right

[wintry tundra]

yep

and sine is opposite over hypotenuse

and the hypotenuse is the longest side

[slender mist]

so 5/13

[wintry tundra]

yes!!!

nice

oh also

you got tan c wrong

[slender mist]

then tan a

would be 12/5

[wintry tundra]

no

because 12 is adjacent to a

remember its opposite over adjacent

[slender mist]

it

i do not follow

:confusion:

so tan a

would be 5/12

@wintry tundra right

[wintry tundra]

yes

because 5 is opposite of a

since it doesnt touch it

and 12 is adjacent since its not the hypotenuse and it touches a

[slender mist]

ahhh

and that leaves us

with tan c

being

[wintry tundra]

yep

which is?

[slender mist]

12/5

[wintry tundra]

dude i can tell it clicked

bc u got it right immediately

that is indeed tan c

[slender mist]

oooo

thank youu

[wintry tundra]

np

[quiet bramble]

can anyone help me with this? please @ me

[ivory path]

where's B?

[simple frigate]

where does sine cossine , tangent etc comes from

[wintry tundra]

What do you mean

[simple frigate]

where does the value come from

why is cos30° √3/2

etc

[wintry tundra]

These are what the values happen to be

Sadly there's no actual way to explain why these values are the way they are besides Taylor series

[signal swallow]

or you can just use the properties of a right triangle

the proportions for a 30-60-90 right triangle are always 1, sqrt(3), 2

[valid harbor]

alternate mean is to go from the height of an equilateral triangle

[wintry tundra]

for a random value tho like if the angle was 46.57543543 then you cant really specifically say besides just it is bc it is

[molten falcon]

Aren't the sqrt(3) and 2 swapped?

[silent plank]

yeh, it should be the other way around

[soft bridge]

yo can anyone in alegbra 2 help me on this

for most of the questions you have to write sentences

but some of them require to write an equation, expression, or function

[signal swallow]

yeah i'm tired lmao

[molten falcon]

ah i see

[dim pier]

Hi, this might be a bit advanced but do the addition rules etc (e.g. sin(x + y) hold for the hyperbolic variants of these functions?

Imagine a vertical line through the middle of the rectangle and spin the rectangle about this vertical line as your axis. Can you see that the path swept out is a cylinder? Then which rectangle when swept out gives you the cylinder in the question?

Which part is confusing? We can break it down step by step to help you understand more

Alright, first do you know what an axis of rotation is?

And if not, do you know what an axis is?

Do you know the word rotation? (Sorry if this is rude, not sure if english is your first language)

Ehhh sort of

So have you ever eaten skewers? Like maybe cubes of meat or some veggies on a skewer and they're roasted over a fire of some sort

Or actually if you've seen a kebab machine, my analogy works too

The rod which the food is skewered on, is an axis

And when we flip the food over, or spin it around, we spin this food about or around this axis, and the axis becomes an axis of rotation

The difference between about and around is semantic and people might use them interchangably, so it's important to know what people mean with the word. More commonly you see about

Right, so now you get what an axis of rotation is?

Now imagine this rectangle skewered on your axis of rotation

If the rectangle were to represent a part of the x-y plane, the axis of rotation could be the y-axis

So your "skewer" goes from one edge to the opposite edge, instead of through the rectangle directly

So it's not like a normal vector to the plane, the axis lies in the plane

You following?

So now we need a bit of mental visualization, and maybe if you have paper around it would help

Can you see that if you fix this axis, and you spin the rectangle around, the path you trace is a cylinder?

Right!

Now the question is simply asking you, which rectangle do I spin, to get this cylinder?

Ah nope

More like flipping the rectangle over and over

Like imagine if you had a piece of paper on your table, and you flipped it over and over

Yeah

Yep!

So the axis of rotation would go from left to right now

Can you see that?

That is a axis of rotation, but the question asks for the other axis of rotation

Which is perpendicular to this one

Maybe I just misunderstood your drawing here

Imagine if you had a flat piece of paper, and you flipped it from right to left instead of up down

So now your axis of rotation would be up down

Can you see that?

Right, so now the question is asking which rectangle you have to spin (flip) about this up-down axis of rotation, to get the cylinder with dimensions described in the question

So you look at each rectangle, let me ask you - how does the height of the rectangle influence the height of the cylinder?

Perfect

Just took a bit of visualization

And so how does the radius relate to the width of the rectangle?

Hahahha

Okay so for simplicity's sake, we'll describe the rectangle as a width and a height, can you see that

Now, we spin the rectangle about our up-down axis of rotation. Where is this axis of rotation, relative to the width of the rectangle?

Hahahha

Alright, let me just say that the axis of rotation is in the middle of the width of the paper

Can you see that?

So you see how when you rotate this rectangle about this axis, you get a cylinder?

Right, so now knowing that the axis is in the middle of the rectangle, how does the width of the rectangle relate to the radius of the cylinder?

Nope, not quite

Hmmm, let me see if I can find a video to help visualize

https://www.khanacademy.org/math/geometry/hs-geo-solids/hs-geo-2d-vs-3d/v/rotating-2d-shapes-in-3d Would this help?

It's not exactly your problem, but it's close to what I'm asking you to visualize

NP, let me know if you still need help

[unreal flax]

Hi community help please!
I have a overdetermined system with full column rank of linear quations of the form Ax=b ( A is long matrix)
I have A_1x = b1 which is composed of rows of Ax=b but height is smaller than A.( A_1 is Wide matrix)
I want a least squares solution of Ax=b but such that A_1x = b1 strictly holds.

[verbal swallow]

If tanA+cosA=2, what's tanA-cosA=?

[dark sparrow]

@verbal swallow are you sure you copied the problem correctly

it looks to me like there would be two possible answers

[verbal swallow]

@dark sparrow yes. I checked and it's cosA

[dark sparrow]

and you're not told anything else?

[verbal swallow]

Nope. "If tanA+cosA=2, what's tanA-cosA=?" it's the ques 😦

[wanton prawn]

Angle ACM = 25° I have to find the other green marked angles. I have no Idea where I can find my 2nd angle. Anyone that can help me out? 🙂

[upper merlin]

is M the center?

of the circle

[wanton prawn]

yes

[upper merlin]

MA=MC cuz its the radius

can u find the others now?

and to find angle ABC you need to use a property of circles which you would have learnt

[wanton prawn]

the engle acm is the one that is already marked green on the top right?

angle*

thx for your help

[buoyant bison]

Just a quick check, is writing this correct ?

[wanton prawn]

if you want to write l = length in cm and w = width in cm

then yeah thats correct 😄

[marble lark]

Can someone help me calculate the area of this compound solid please

[olive cove]

Just calculate the 3 volumes of the 3 main shapes or 2 if you see how

[marble lark]

Yeye I was able to achieve it, thank you very much :DDDDD

[past jungle]

I have a question on asymptotes. Why is x=-1 not an asymptote in this case? I used the quadratic formula to get x=9 and -1 but only x=9 was the correct answer.

[short fable]

Because for x=-1 it's undefined but it's not an asymptote

[past jungle]

but why is it not?

[short fable]

if you were to graph this function, it'd look like this

[past jungle]

without the graph how would i know its not an asymptote?

[short fable]

you'd just find all x for which denominator is equal to 0

but

numerator has to be nonzero

which in case of x=-1 doesn't happen $x^2-1 = (-1)^2-1 = 0$

[somber coyote]

Michał

[past jungle]

ah i see

well thank you for the answer!

[cerulean nest]

1/x

[dreamy quest]

can someone pls help me with this?

<@&286206848099549185>

[cerulean nest]

Ah, now no one is gonna help you because you did not wait 15 minutes

[short fable]

So like, in 1. these statements are true cuz they r given. In 2. you can use corresponding angles (cuz HT || SF)

and for 3. you can use the fact that the sum of degrees in a triangle is 180°

[dreamy quest]

i didn’t notice it in the rules, mb

@short fable thank you sm :)

[cerulean nest]

haha all good

[buoyant bison]

Thanks

[clever drum]

Proofs were a nightmare with me, I struggled with them back in 6th grade

[dreamy quest]

ESPECIALLY TWO COLUMN PROOFS OMG

[clever drum]

1. Is always given no matter what

[clever drum]

I once had 8- at one point

8 column proofs were pure nightmares

[dreamy quest]

8????

[upper karma]

@dreamy quest did u solve it

I can help

[dreamy quest]

oh yea, someone helped me w it. but thank you though :)

[winged hound]

Can anyone tell if this is right

[dark sparrow]

yes

[winged hound]

when i am trying to find the orthocenter, does it matter which two coordinates i pick to find the slope?

[leaden atlas]

i need some help
calcule the radius of the circle

[upper karma]

What is the value of the angle between the line that passes through points L1 A (-6,1) and B (2, -1) and another line L2 that passes through points C (-8, -4) and D (6.2)?

[olive cove]

Make vectors and use dotproduct

[upper merlin]

Really hard to give hints in this cuz it requires constructions so take my beautiful diagram

[olive cove]

Beautiful it is but line paper for mathematics is questionable imo.

[leaden atlas]

dont worry,i did, but thanks anyway

[upper merlin]

😅

[wintry tundra]

I use line paper constantly

[formal tartan]

yo that's a good one...

[warm totem]

How do you get that answer?

[mighty rune]

set up a proportion

you get 60/40 = x/240 (I converted to inches)

so 40x=14400

x=360

and converting it back to feet gives you 30 feet

[warm totem]

what does / mean

[mighty rune]

a fraction or divison

[warm totem]

So you just do 60/40 =1.5 then to get rid of the 240 you just multiply so 360=x

[mighty rune]

$$\frac{60}{40}=\frac{x}{240}$$

[somber coyote]

Omega Warrior

[mighty rune]

here

like that

[mighty rune]

you could also do it in the cross method where you do 60 times 240 equals to 40x

and gives you the same answer

[warm totem]

Oh okay

That one?

[mighty rune]

for this one you get: $$/frac{x}{6}=/frac{x+1.8}{8.4}$$

[somber coyote]

Omega Warrior

[mighty rune]

it didnt format right rip

$$\frac{x}{6}=\frac{x+1.8}{8.4}$$

[somber coyote]

Omega Warrior

[mighty rune]

here

@warm totem

[warm totem]

okay

[warm totem]

I keep getting 4.5

[mighty rune]

hmm

oh wait

@warm totem x is the value of CF, to find CD, you need to add 1.8 to x

so then you will get 6.3

[warm totem]

ohhh okay

[mighty rune]

yeah

[daring bolt]

I know this out out of topic but y’all think y’all can help me 😓

[weary drift]

@daring bolt pls keep offtopic posts out of here

move this to smth like #chill

[daring bolt]

Oh my bad I didn’t any thing to offended you or anyone else

[weary drift]

dw feel free to repost in #chill

[daring bolt]

Yea sorry

[grizzled ether]

Hi! How do you do this type of question? KInda confused with it

[upper merlin]

triangle AFD is similar to triangle EBA

sometimes marking the angles can solve questions which involve only side lengths

[vagrant bolt]

why does this correspond to a tilt on the axis?

[upper karma]

<@&286206848099549185>

[sick sable]

you should not do this

[ember heart]

Well a horizontal line would've 180° in total
So on one part you've got 90° so can get the other and find the value of p

[jagged ferry]

Hey why would it be 2r in the parentheses instead of just r this is my math teachers work.

[upper karma]

yoo this channel is lit

i’m gonna stay here

[upper karma]

is anyone here good at proofs

<@&286206848099549185>

[jagged ferry]

D=90 e =80 and f=100 they are supplementary angles

[west burrow]

how would we find the back bearing

im really confused cuz i cant just minus 55 from 360

[jagged ferry]

What angle are you looking for?

[mighty rune]

what about 180-55?

[west burrow]

my prooblems been solved

no more help neededc

ty tho

[mighty rune]

oh

[wild lion]

Hi can somone help me solving this one

[dark sparrow]

@wild lion do you still need help with this?

[wild lion]

Yes

[dark sparrow]

have you made any progress on this or are you stuck not knowing where to begin?

[wild lion]

I'm stuck with the center of the circle haven't encounter a center that is monomial in form

[dark sparrow]

"a center that is monomial in form"

anyway, may i suggest graphing the line y = 4x and the points (0,0) and (2,8)?

something might jump out at you.

[wild lion]

I get it

[quiet bramble]

'

can someone please help me with this?

[median perch]

Hope this helps

[dark sparrow]

do not give out answers

[median perch]

Oh oops

So we have qr=wz/2 as qr is a midsegmemt

That means 2x-2.5 = (2x+7)/2

Just solve for x

And find wz and qr from the X we found

Hope this make sense idk

[austere jewel]

yo guys is sin a . 2 cos a = 2 sin a . cos a

[graceful halo]

assuming '.' is implying multiplication here yes

[austere jewel]

yes

alright thanks

so then 2 cos a . sin a = 2 sin a . cos a

[upper karma]

Hi

[dark sparrow]

yes, this is just the commutative law of multiplication

[river marten]

Pls solve Q-21

I dont understand

[kind bone]

Hell

[austere jewel]

Ye ofc i know that but with this new trigonometry that im learning I just wanted to make sure

[kind bone]

Someone pls tell me how to get good at geometry and not fail and good resources and do good on tests

Im a freshman taking honors geo and need at least a B+ to not fail bc i got a d- first quarter

[clear turtle]

can someone help

[fallow kiln]

try drawing the triangle

A'B'C'

then draw lines from A to A' to C' to B' to B to C, since thats exactly what its asking about

then look at the shape, think and try firgire out how u could find the area of that shape

reflect means mirror

[hearty lantern]

You can split the forces into their horizontal and vertical components

[daring bolt]

Oh my bad

I just came back to delete it

[hearty lantern]

So the acute angle between F1 and x axis is 180-124=56 degrees. Hence you. Have a triangle with the horizontal component adjacent to the 56 degree angle and 55N is the hypotenuse. Hence using trig the horizontal component for F1 = 55cos(56). Using a similar method the vertical component of F1 is 55sin(56). The same method can be used to resolve F2 into vertical and horizontal forces. A triangle can be constructed with angle 4 degrees, hypotenuse 90, horizontal component opposite 4 degree angle and vertical component adjacent to 4 degree angle. Thus, horizontal component of F2 = 90sin(4) and vertical component of F2 = 90cos(4). So the resultant force components are:
Horizontal: 55cos(56)+90sin(4)=37.034N (180 degrees to positive x axis) Vertical: 90cos(4)-55sin(56)=44.184N (270 degrees to positive x axis. Create a right angled triangle with hypotenuse being the resultant force, and other sides being horizontal and vertical components.By using pythagoras, the resultant force is sprt(37.034^2+44.184^2)=57.654N By using trig on the triangle, the acute angle between negative y axis and resultant force = arctan(37.034/44.184)=39.969 degrees. Hence the angle of the resultant force from the positive x axis is 270-39.969=230.031 degrees

[daring bolt]

I’m so sorry

[hearty lantern]

Np, enjoy the answer anyway

It’s good revision for me 😉

[daring bolt]

Yea my bad tho

[hearty lantern]

Dw abt it

[urban bolt]

sin2x < cos2x

[upper karma]

someone plz tell this.....

[upper karma]

can u tell me with a rough diagram plz

[upper merlin]

also s = (a + b + c) / 2

that formula for r would be very helpful as one the angles in your case is 90 degrees

[upper karma]

what is the derivation of this?

[upper merlin]

i will have to google it up ig

[upper merlin]

are you comfortable with hindi?

[upper karma]

yes

[upper merlin]

https://www.doubtnut.com/question-answer/derive-the-formula-of-inradius-i-delta-s-ii-s-a-tan-a-2-iii-4r-sin-a-2-sin-b-2-sin-c-2-1339329

[upper karma]

wow

have u qualified prmo ?

[upper merlin]

no i ll be giving my first try this time

[upper karma]

me too

[upper karma]

how tan(A/2) = sqrt((s-b)(s-c)/s(s-a))

[upper merlin]

https://www.quora.com/How-do-you-prove-Tan-A-2-s-b-s-c-s-S-a

no this is better ig: https://math.stackexchange.com/questions/1691711/how-to-prove-sin-fraca2-sqrt-fracs-bs-cbc-cos-fraca2-sqrt

[upper karma]

thanks

tangent is such an inconvienient topic

function

sry

I meant function

[spring mauve]

Problem 6. Assign to each side b of a convex polygon P the maximum area of a triangle
that has b as a side and is contained in P. Show that the sum of the areas assigned to
the sides of P is at least twice the area of P.

pls help with this 🙂

[clear turtle]

ok

can we find what c is since m and l are parrellel?

[clear turtle]

in my opinion i think b is pretty much like 75 degrees

[upper merlin]

There is very little chance that people are able to help you with IMO problems

[spring mauve]

our teacher gave us these stuff.....not understanding some parts of it

[upper merlin]

👀 your teacher gave you the 6th problem of IMO 2006?

[spring mauve]

yep
not all of em only one

extra credit

are u able to help?

[upper merlin]

no I can just read the solution online which is also pretty tough to understand

[spring mauve]

i can also read it online but then there is no fun in doing that way ig

[upper merlin]

yea you can just keep trying

[upper karma]

find area

i have to use circle concept

[dark sparrow]

is anything else known?

this looks like it might be insufficient data

[upper karma]

no

[dark sparrow]

also what's "circle concept" supposed to mean

[upper karma]

means theorems related to circle

[dark sparrow]

circle types?

are you sure the problem doesn't say anything else other than what you have marked? are you absolutely 100% certain there's nothing else?

can i see the text of the problem

[upper karma]

my teacher only gave this to me

r u trying it?

[dark sparrow]

your teacher only gave you that diagram? nothing else?

honestly, i can't come up with a way to attack this problem right now

maybe it's possible to do it by coordinate-bashing but i'd like to avoid that if possible

[upper karma]

ok but if u will get any so. please tell

without using circles

[dark sparrow]

we are allowed to use trigonometry, right

[upper karma]

definitely

[upper karma]

ok i got the solution

i used angle bisector theorem

[dark sparrow]

hm?

now i'm interested. can you share your solution?

[upper karma]

ok wait

[upper karma]

this is the sol..

[dark sparrow]

AD^2 = AE * AC - ED * DC
where did this come from?

[upper karma]

take a random triangle

draw angle bisector

[amber wraith]

stewart theorem?

[upper karma]

now apply stewarts theorem and angle bisector theorem

[upper karma]

there are 3 ways to prove this

u can also prove it by applying cosine rule in both triangle made by angle bisector

[dark sparrow]

hm

[upper karma]

and one by using a circumcircle

[dark sparrow]

i personally have never heard of stewart's theorem in particular

[upper karma]

ohh

[dark sparrow]

everything else seems like just algebra

[upper karma]

it is stewart's theorem

[gaunt iris]

Help

[prisma briar]

Hello? anyone need some help?

[dapper marlin]

could you help with 8 9 10

[shy wedge]

Im kinda new to trigonometry but I think I figured out another solution

@dark sparrow Soryy if you didnt want to be tagged but I figured you might want to since you were involved too

Thing is I'm not sure how to follow now, I'm not even sure if I did it right

[worldly lintel]

hi, i wondering if someone can help me to know how to do a problem from the "Equation topic of the line that passes through 2 points"

[shy wedge]

Cuadratic functions?

Send the problem I can try

[silent plank]

do you have a specific question that you're stick on

[upper karma]

How do I do this?

[upper merlin]

first try to calculate its maximum height in terms of l and r, then try to see how it would change with respect to the angle

[upper karma]

I have projected it onto a flat surface and found the max height to be 2rcos(alpha)

Where alpha is the angle between the circle and the surface

But I need to find the link between alpha and theta

@upper merlin

[upper merlin]

umm yea i ll help later i am busy rn

[tawdry merlin]

So the problem is this

Sry it's vertical

[wintry tundra]

hmm

i might not be able to find it but i might

[tawdry merlin]

Ok

[wintry tundra]

cot is expressed as cos/sin

i think you have to find sin theta then

[tawdry merlin]

You mean first I have to find sin

Hm

Ok

[wintry tundra]

any rules you think you could use here? i forgot a fair sum of intricate rules in trig lmao

[tawdry merlin]

I also don't have any idea 😅

[wintry tundra]

ahh

[tawdry merlin]

😅

[wintry tundra]

sorry i dont think i can help

[tawdry merlin]

Ok

Np

[wintry tundra]

i know some trig stuff but not a lot

[tawdry merlin]

I have the formula but idk how do I implement it

Let me find sin theta first

Then

I'll find cot with the sin

[wintry tundra]

ok

[rich wolf]

@tawdry merlin problem?

[tawdry merlin]

Ye

So

How can I solve this

[rich wolf]

cos(t) = x/(x+1)

find cot(t)

i'm using t because i don't want to type out theta haha

[tawdry merlin]

B = x
P = x + 1

H = ?

[rich wolf]

draw a triangle

a right triangle

[tawdry merlin]

So I'm using formula
h² = p² + b²

And I got 5 idk who

*how

[rich wolf]

what is p

[tawdry merlin]

Perpendicular

[rich wolf]

and h is hypotenuse?

[tawdry merlin]

Yeah

And

[rich wolf]

then you're already wrong. recall the definition of cosine

cosine = adjacent/hypotenuse

[tawdry merlin]

Oh k

[rich wolf]

sorry made a typo there

cos = adj/hyp

@tawdry merlin

this is what we know

[tawdry merlin]

Hm

[rich wolf]

did you figure it out?

[tawdry merlin]

A big NO

[rich wolf]

do you know how to solve for the third side of a right triangle?

what is O?

[upper merlin]

recall there are 2 altitutdes on 2 different bases

so what would be the value of h in terms of l and r?

[upper karma]

I don't know how to do this

It's been way too long since I last did anything like this

I got to B, C and that was it

[upper karma]

If I had a piece of paper on me I can figure it out. But we want a function to find the height at any given theta, where theta is the angle when I roll the cone.

Maybe there was even an altitude theorem for a faster result I can't remember

[upper merlin]

finding the max height is essential to find the final answer

[upper karma]

Once I find the max height, what do I do with it?

[upper merlin]

try to find how it would change with respect to theta

[upper karma]

Well ok

[nocturne remnant]

D lies on segment AB
Do you know what that implies for vectors OD, OA and OB?

[upper karma]

Nope

I want to learn this again

Where should I look?

[nocturne remnant]

Well
I implies that OD is some multiple x,y of OA and OB such that x+y = 1
So OD = (x)OA + (1-x)OB for some x

Try to think about it and see if it makes sense

[upper karma]

Huh

Thanks

[nocturne remnant]

Yeh
So you can use that to fork out x in the question

[upper karma]

I think I'll try to make sense of that first

since I've seen it popping up here and there from time to time

but no idea where this x<something> + (1-x)<something> comes from

[nocturne remnant]

Yeh
Try picturing the case where OA is 1 to the right and OB is 1 to the up

yk what, here’s a short proof

Let M be the midpoint of AB
Then
xOA + (1-x)OB
= x(OM+MA)+(1-x)(OM+MB)
=OM + x(MA)+(1-x)(MB)

Hopefully it’s more obvious now that the above expression covers all the points on segment AB

[upper karma]

Oh okay

Thank you so much

[tawdry merlin]

Can anyone help me ?

[nocturne remnant]

[tawdry merlin]

Huh OK

So

The question goes like this

If sin(t) = x/sqrt(x +1 )find the value of cot(t)

[nocturne remnant]

for starters you can try finding cos(t)

Also, are you given that t is an acute angle or sth?

[tawdry merlin]

No

[nocturne remnant]

Huh weird

[tawdry merlin]

T stands for theta

Yk

So

I used

[upper karma]

nvm i was wrong

it's x+1 Xd

[tawdry merlin]

A pythagors for smth theorem to find that opposite

Value

[nocturne remnant]

But do you know that theta is acute

Otherwise cot(t) can be negative

[tawdry merlin]

Ohk I didn't knew that I just landed in trigonometry

So

Yk

[nocturne remnant]

Um have you learnt trig functions for angles beyond 0 to 90 degrees?

If not then we can probably assume t is acute

[tawdry merlin]

Yeah

I haven't

[nocturne remnant]

Ok good

so cos(t) should be positive

What’s cos(t) in terms of x?

[tawdry merlin]

Um

Idk what are you talking about so I'll just show the question

[nocturne remnant]

Sure

Interesting, you were given cosine
And not sine

[tawdry merlin]

Oh oh

Sry

Sry

My bad

I just

[nocturne remnant]

In this case you would want to find sin(t) instead

Still remember that sin^2(t)+cos^2(t)=1

[tawdry merlin]

Ok

[tawdry merlin]

If we just convert into a figure then ig we should get

And I used that pythagoras theorem to get OPP of the triangle

[nocturne remnant]

Cool

[haughty flame]

could someone explain 16b to me please?

please ping me if you end up answering

[upper karma]

ok i will try by this method also

[livid moss]

Bye

[shy wedge]

Bye :(

I figure out the question said aproximate to the closest non-decimal angle

Tge answe was 117.8 and I wrote 117 instead of 118

[livid moss]

Ah okay

Gj

[tawdry merlin]

if tan(t) = 3/4 then find the value of cos(t) and cosec(t)

How can I solve it

[shy wedge]

What's wrong here?

[olive cove]

,rotate

[shy wedge]

Oh nevermind

I misstyped on the calculator

[upper merlin]

Did you try drawing the hexagon?

then try all you have to do is divide that hexagon into regions and calculate the area for each of them

[ionic bluff]

can someone explain this

[shy wedge]

Did you get anything?

[random widget]

Help please

<@&286206848099549185>

[signal swallow]

what is true about parallel lines?

[random widget]

they have teh same slope

[signal swallow]

and what's the formula for slope?

[random widget]

y2-y1/x2-x1

[signal swallow]

right

so the slope of one of the lines must be equal to the slope of the other line

and you're given 4 sets of coordinates

[random widget]

yep

so i get

-9/5 = n+4-3/5-2n

[signal swallow]

looks good

now solve for n

[random widget]

but i cant solve

idk how

[signal swallow]

have you heard of cross multiplying?

[random widget]

yea ig

but how would that work here

[signal swallow]

$\frac{a}{b}=\frac{c}{d} \implies a\cdot d = b\cdot c$

[somber coyote]

a disappointing son

[signal swallow]

and you currently have $$-\frac95=\frac{n+1}{5-2n}$$

[somber coyote]

a disappointing son

[signal swallow]

so you can cross multiply, distribute, and simplify

[random widget]

i got 23n=40

so n=1.73913043

is that correct?

[signal swallow]

i got something a bit different

what did you get after cross multiplying?

[random widget]

OHH

WAIT

1 SEC

I GOT 3.84615385

is that right?

[signal swallow]

is that 50/13 lol

[random widget]

yea

[signal swallow]

👍

[random widget]

tysm

[upper karma]

nope

[tidal copper]

https://media.discordapp.net/attachments/624319985027776514/909602313965010954/x.png

[smoky palm]

If I had

Cos(x)=y

How would I get x alone?

[wintry tundra]

Take the inverse cosine of both sides

[warm totem]

\

[warm totem]

pls help

[dark sparrow]

@warm totem do you still need help with this?

[warm totem]

yes

[dark sparrow]

have you made a diagram yet?

[dark sparrow]

@warm totem ??

[warm totem]

no

[dark sparrow]

is there any reason why you went silent for a whole 20 minutes?

but whatever

can you make a diagram now?

[wintry tundra]

hes busy being cool

[warm totem]

im sorry I got sucked up in some family matter

[wintry tundra]

come on man i set up a great excuse for you

social credit and all

[warm totem]

yep

[warm totem]

How would someone make a diagram?

[dark sparrow]

you have here a circle centered at T

[wintry tundra]

just draw a triangle with the stuff given to you

or whatever you need

[dark sparrow]

and two points on this circle named G and H

this is not your first time making a diagram for a geometry problem, right?

[warm totem]

it might be

[rough prawn]

helo

can someone help me with this

[warm thorn]

ope sorry

[prime vigil]

Gammaking im not understanding what this question is asking

[oblique hatch]

Try dividing by cos, which then makes sine/cos = tan and try going from there

[dapper marlin]

help pls

[upper merlin]

If you are not able to get any answer then its better to go through the material once again

[dapper marlin]

i just need help with 4 now

[upper merlin]

ohk

does x represent the arc QS?

if yes then you can mark the centre as O and join QO and SO

once you get angle QOS its pretty easy to get to the answer

@dapper marlin

[dapper marlin]

i have no clue

[supple bronze]

How could you calculate for the ?

[quiet blade]

Under the assumptions that the 40mm line is the diameter of the circle, first draw the circle's center. Then, connect the center with the endpoints of the segment with length 25.5mm.
If you find the shortest distance from the center to the segment, you might see how it will help you find the unknown length

[wintry tundra]

ive had to do this before for my robotics team... its trigonometry and pain

[mighty rune]

use your trig functions to find the other lengths

[oblique hatch]

If one side and the two adjacent angles are known then the perimeter of the triangle can be found using the formula:

p = a + (a ÷ sin(β + γ)) × (sin(β) + sin(γ))

And since angles of triangle must = 180, 90 + 17 = 107, so the other angle = 180 - 107

[wintry tundra]

if u need to ask why side lengths are the way they are feel free to ping me

[inner sandal]

for this i put

let C be a great circle so C = Π ∩ Sn, then since L is injective L(C) = L(Π ∩ Sn) = L(Π) ∩ (Sn) and as L is linear L(Π) is also a plane, and L(Sn) = Sn because L is surjective. So L(C) = L(Π) ∩ Sn is also a great circle

is that right?

<@&286206848099549185>

[upbeat helm]

you might wanna ask that in either #point-set-topology or #diff-geo-diff-top

[inner sandal]

which one

it's not differential geometry or topology tho

ill just ask in a help

[upper karma]

dis help yet?

also what is pi intersect Sn supposed to be

because my idea for a great circle is just k*Sn

where k is a real scalar

[inner sandal]

my definition of a great circle is basically taking a slice (plane pi) out of it and viewing that intersection

[upper karma]

what are great circles actually

oh

[inner sandal]

it's necessary the plane passes through zero so it's a great circle and not just a spherical circle

[upper karma]

ok i see what ur saying

L(Sn)=Sn?

what that supposed to mean lol

oh

because distance preserving

ig

but uh you should justify L(Sn) better

[inner sandal]

L(Sn) is the image of Sn under L

[upper karma]

because surjective doesn’t necessarily mean L(Sn) is Sn in R^n+1

[inner sandal]

why

[upper karma]

you can take a zero map

that is surjective too

L(x)=0

[inner sandal]

sorry i mean Sn → Sn: x |-> L(x) is surjective

specifically

[upper karma]

yes, but why is the image Sn.

thats what im saying you should justify

unless L(Sn) is defined to be Sn

oh bruh

i didnt see

[inner sandal]

im just showing Sn → Sn defined by L preserves great circles, not L: R^n+1 → R^n+1

[upper karma]

yeah i see now lol

yeah ur logic is good

ig if u really want to you can add that L(Pi) contains 0 because L is linear

but idk if necessary

u were just saying before that the planes were based at 0, but it can be implicitly understoood ig

[inner sandal]

yeah but L is linear so it's obv true

i see what you're saying to add it though, alright

[upper karma]

hopefully doesnt hurt but arg seems fine to me

was just confused at start because i lacked context

[clever fable]

why is it that horizontal asymptotes can cross while vertical asymptotes can't?

(also, I was not sure where to put this question, but I think since it can be considered geometry, this should be the right channel)

[livid moss]

If you sketch the function from left to right and you cross the vertical asymptote, you can't "go back" and get closer and closer to it again, because then it wouldn't be a function. So it has to do with the fact that every point on a function has a unique x value

Whereas, you can cross the horizontal asymptote, and then go back up/down and closer and closer to it

[hasty surge]

sirrrrr

need help

can anyone answer

pleaseeeeee

[upper karma]

no one can help if u dont send ur question

[hasty surge]

wait a sec

@upper karma here you go

[upper karma]

so much words

[hasty surge]

yeah!

[upper karma]

so we can say that AB is 12 and PQ is 25 and BQ is 70

[hasty surge]

yess

[upper karma]

I honestly have no idea

[hasty surge]

🥲

do you know anyone who can help? @upper karma

[upper karma]

if no one comes after 15 minutes you can @ helpers

[hasty surge]

okieee

@wild valley brooo
I need help

pleaseee

@river forge @ripe osprey answerr mane

it's urgent

<@&286206848099549185> guysssss

y'all aint responding

pls

respond

what are you people

who made you guys helpers

pathetic

[fair thistle]

Who knows

I need help

[hasty surge]

@topaz sorrel dude

I know youre watching it

I need you help

immediately

ma chudao saalon

[topaz sorrel]

I've said something in vc

And SOHCAHTOA

[upper karma]

that is the most useful tool in trig

only sad thing is, my man forgot all of trig

[cedar aspen]

i need a lot of help with proofs if anyone can

[bleak spade]

can someone please help me out 1 on 1 i just started geometry and need help with a couple questions please dm me anyone who can help thank you!

[wintry tundra]

it wouldnt make sense for the answer to be 236 because both angles are acute

so when you add them together they would be less than 180

[hexed tiger]

x and y are not acute

For sure

If the angles are x degrees and y degrees, x and y are the numbers

So just the sum of the angle measures

[wintry tundra]

my bad im bad at using my eyes

[mystic karma]

araam se bhai , teko doubt hote hai toh idhar aane ki jagah google lens use kiya kar

[upper merlin]

use the concept of similar triangles

you can express BS and SQ in terms of h

and please don't be so impatient and write stuff which is not meant to be there in this server

[stable maple]

ya the ans is 236 just use angle sum property to get x+y

[upper karma]

You can construct a straight line down from B and C to get two right triangles ABE and DCF. From that you can figure out <ABE = 28 and <DCF = 28. Since <EBC = 90 you have x = 28 + 90 = 118 and since <FCB = 90 you have y = 28 + 90 =118. So x + y = 236. You avoid assuming x and y are the same.

Not E and F should be on the line AD.

[stable maple]

Can't we just use x+y+62+62=360 ?

[upper karma]

Yeah that too. If you take to heart all angles add up to 360.

[upper karma]

This is just one way of doing it. The E is just the vertex on the line AD after you construct a line down from B to AD. I just need a label for it. Same for F. If you know all right triangle have an angle sum of 180 you understand how I got 28.

Yeah nvm forgot what I just said and do what Dragonclaw suggested. Making this unnecessary complicated.

[stable maple]

you know that in any quadilateral sum of all angles are 360 so use that and if you wanna do for a another figure like 6 sided figure then you can use that formula

[upper karma]

I'd just calculate the percentage of the circle

As the angle gives you a per-360 value

[upper karma]

Is it possible to find the locus of a point on a pentagonal prism's perimeter?

[smoky palm]

If I have X-Ycos(z)=0

how would I get z alone?

could I do

x=Ycos(z)

then

x/y=cos(z)

then

cos^-1(x/y)=z

would that work?

[humble pulsar]

yes

[oblique snow]

can anyone help me?

gotta prove this

[nocturne pebble]

Please help with 5

[manic shell]

well

first

u gotta convert

the units

to inches

to make the calculations easier

so

maple trees for tapping for syrup should be 18 inches

in diameter

and the circumference of this tree

would be

50 inches

now

[nocturne pebble]

wait

[manic shell]

you should know

how to

calculate

circumference

[nocturne pebble]

I did it myself can u tell me if I'm correct

[manic shell]

alr

[upper karma]

Expand LHS and divide numerator and denominator by cosx*cosy

[nocturne pebble]

Wrong one

[manic shell]

well the circumference

is

2πr

and yeah what u did is right

but

i would have first converted to inches

which would be

9*2π

which is 18π

which is very close to what u got

yeah ur answer is right

[nocturne pebble]

Did I do the steps correctly though

[manic shell]

but to be more accurate convert ot inches first

yeah

[nocturne pebble]

convert 4 and 2 inches into inches?

[manic shell]

no

1.5 ft

into inches

[nocturne pebble]

convert 1.5 feet into inches

ok

thank you

one more thing

[manic shell]

what u did for hte 4ft2in one was right

yep

[nocturne pebble]

I don't even know where to start with this one

Find the perimeter of the region

[manic shell]

right

well

thats gonna be the lengths we already know

which r

28x2 +6x2

ans

and

now you can see

that the curved parts are 1/4 of a circle

and theres 4 of them

which means

the perimeter will be

the circumference of that circle

[manic shell]

• this

ill tell u that the radius

is 3 in

[nocturne pebble]

is teh r 3

oh

lol

[manic shell]

u should be able to do it urself now

[nocturne pebble]

Ok thanks I think I got it

[manic shell]

no problem

good luc

k

[nocturne pebble]

Is the answer 143.4

[oblique snow]

help plz

[upper karma]

Heeellppp

[slate pewter]

random question

in the calculator on the phone

I can write
tan (90) and it gives me -1.9952

and tan of 90 is undefined

then I relized it have to be

tan (90°) to give me undefined

my question is

90° is okay a degree
but what is 90 here and why it have a answer like in what way does the calculator take just 90 as a number ?

[dark sparrow]

your calculator is probably in radian mode, which means it treats 90 as 90 radians

[slate pewter]

and what is that

I mean it's kinda solved i just have to write

the degree symbol for degrees

[dark sparrow]

radians are a unit of angle

1 radian is the central angle made by an arc whose length equals the radius

or to put it another way a radian is 1/(2pi) of a full circle

[slate pewter]

oh okay

but like

why not just use degrees

[dark sparrow]

you might change your mind by the time you learn calculus

[weary hearth]

[keen nova]

Radians are a much more natural way of talking about angles, it might not seem that way at first, but using degrees feels insane to me now

Radians are like describing to circles in their native language, degrees is like using Google translate

[slate pewter]

I am learning calculus

so yah

I guess I'll find out why

also i have a question about

the 6 ratios of the right triangle

it's kinda uhh but

In a right angled triangle if we know two out of the following three thing- (1). an angle other than the 90 degree. (2). length of side opposite to the angle (3). length of hypotenuse side. - then the third can be determined by using sine or cosecant of that angle. So they both are used for the exact same purpose.

Similarly cosine and secant are used for the exact same purpose. And same goes for tangent and cotangent.

I just wish to know why aren’t only three ratios used- either sine, cosine and tangent; or cosecant, secant and cotangent.

Are there situations where we couldnt use sine and could only use cosec and vice versa…and the same for the rest of two pairs. Or are they just used to make algebric equations look more complicated.

In mathematics we always try to reduce the work and effort; and find the easiest way. And i don’t understand why 6 are used instead of three.

I like to be able to explain stuff even for myself so I'll know that I deeply understanded it

I’d be really grateful to anyone helping.

uh this is long

[dark sparrow]

Are there situations where we couldnt use sine and could only use cosec and vice versa…
no

the reliance on csc, sec and cot is more of a historical happenstance than anything

and it also varies by region

i know that when i was at school only sin, cos, tan and sometimes cot were made use of, and everybody managed just fine

[slate pewter]

okay

[lapis fossil]

hello there im a little confused on how to do a question like this

like i know what to do with like my special triangles but this pi/12 is confusing me

[dark sparrow]

@lapis fossil this question refers to 'compound angle formulas'. do you know what that means?

[lapis fossil]

yeah, i have a sheet with a bunch of formulas on it. Is the compound angle formula sin(x+y)=sinxcosy+cosxsiny ?

[dark sparrow]

yes, that's the one for this question.

[lapis fossil]

ah ok so that so i get that solves the first part of the question right

but what about the exact value part

hold on i'll try to send a picture

[dark sparrow]

sin(pi/4) cos(pi/12) + cos(pi/4) sin(pi/12) = sin(pi/4 + pi/12)

you can work out what pi/4 + pi/12 is, right?

[lapis fossil]

oh wait does it equal pi/4?

[dark sparrow]

no, pi/4 + pi/12 is not pi/4.

[lapis fossil]

oh my mistake, i'll try again

oh is the answer pi/3?

[dark sparrow]

the answer to the problem is not pi/3, but it is what i wanted you to work out.

[lapis fossil]

ohh yeah mb

[dark sparrow]

your expression is sin(pi/3).

[lapis fossil]

ahh i see why thats important

can i take a gander on what to do next?

to fully solve this problem do i use the speical triangles?

the one specifically with pi/3?

[dark sparrow]

i mean, now we've reduced this problem to calculating the value of sin(pi/3).

you either know that off the top of your head or you draw a special triangle to help yourself remember.

[lapis fossil]

ahh ok

is the answer root3/1?

for the exact values part

[dark sparrow]

no

sin(pi/3) ≠ sqrt(3)

[lapis fossil]

Dangit, here’s my work maybe I did something wrong here

[dark sparrow]

what's the hypotenuse?

[lapis fossil]

ohgod, im cringing so hard rn

that's embarassing

yeah it's root3/2 tysm, i'll try not to make a stupid mistake like that ever again 😅

[onyx iron]

hii how do i illustrate this?

1. There exist exactly three distinct lines in the geometry.
2. Each two distinct lines are on exactly one point.
3. Not all lines of the geometry are on the same point.
4. Each two distinct points are on at least one line.

[dapper marlin]

Find the length of the bolded arc & the area of the bolded sector.

[oblique snow]

can anyone help me?

[wise pawn]

what have you tried

[potent orbit]

could someone check this with me?

[oblique snow]

i only found 1 radian answer

teacher said there should be 4

[wise pawn]

show your work

[oblique snow]

@wise pawn like i found one radian only but the teacher said there should be 4 radians in total

total

[wise pawn]

looks like you should have at least 2 from the + and - cases right?

[oblique snow]

yeah

+2 and -2

[wise pawn]

ok good, now why do each of these give us 2 more answers

[oblique snow]

is it bc its sqaured?

or is it bc theres a 4 infront?

[wise pawn]

nope, that's where the first split came from

the 4 in front doesn't change anything

[oblique snow]

oh

[wise pawn]

it might help to think about the graph of tan(x) and the period of tan

[oblique snow]

can we do a voice call?

[wise pawn]

nope, sorry

[celest cloud]

any geometry chads here?

[storm portal]

#❓how-to-get-help message

[celest cloud]

so are there ?

[storm portal]

Just ask your question.

[celest cloud]

asked it above

right let me lay it down

1) find sin(A+B) (no calculator!!!!!!)```

[storm portal]

Start by drawing a diagram

[wise pawn]

have you heard of angle sum identities

[celest cloud]

apparently we have to use trig identities fellas

[storm portal]

Yup - angle sum identities for trig functions

[celest cloud]

my working out so far:
we know sin(A+B) = (SinA)(CosB) + (CosA)(SinB)
so i did:
5/13*4/5 = 20/65
Sin(A+B) = 20/65 + (CosA)(SinB)
this is where im at right now ```

any advice fellas?

if any geometry chad is capable of doing this then id appreciate the advice 👍

[signal swallow]

draw triangles that satisfy sin(A)=5/13 and cos(B)=4/5

you can get cos(A) and sin(B) this way

[celest cloud]

im trying to do it through trig identities tho

its a skill i need to learn

[signal swallow]

the trig identity was the sum one you used

[wise pawn]

you can use the pythagorean identity

[celest cloud]

ye and im not using it properly

[signal swallow]

yes you are lol

[celest cloud]

i didnt get the answer

??

[wise pawn]

to solve for sine or cosine of the other

[signal swallow]

that's where this comes in

[celest cloud]

so is A the angle?

[signal swallow]

A is an angle