geometry-and-trigonometry

[paper vale]

then sine rule to work out QED, while also noting that the largest value will be 180-n, and that is answer

as sin(x)=sin(180-x) when x is acute

[sullen canopy]

Wow

[fathom root]

https://gyazo.com/d9f60fff51b3251e2ad59d86e24095f8

how is this wrong?

anyone know?

[acoustic jungle]

because that is not a subscript

[fathom root]

can you explain

im confused

[acoustic jungle]

x^2 is not x_2

x^2 means x squared

you are not squaring the x inside the brackets

[fathom root]

oh am i not?

[acoustic jungle]

you wrote an exponent instead of a subscript

you can search up the formula on wikiepedia

[fathom root]

https://gyazo.com/5a72cf0ecf3e8fcd4cc76358ac5a5481

so this is an exponent

not a subscript correct

[little osprey]

uh

so its meant to be a subscript

[fathom root]

oh

so its not supposed to be squared

its supposed to be sub?

[little osprey]

yes its not

yes

it is

[fathom root]

i got it

ty

[little osprey]

yes its not was for not supposed to be squared

yes it is was for sub

[fathom root]

quick question

so i ordered a ti84 +

if i want to graph y = (x + 3)2 + 4

can i just plug it in real easy and it will show me the parabola?

or is there steps to it

[little osprey]

I mean yes if you plug it into the y= thing

but then

You would have to fix the window of the graph

[fathom root]

ok ill experiment when i get it

how would you set this up?

Find x so the distance between
(x, 1)
and
(3, 2)
is

2

whoops

let me capture it

https://gyazo.com/f2528350e34df7c2d01e01c0eb00d128

[little osprey]

plug it into the distance formula

[fathom root]

so plug it in an then equal it to =sqrt2

?

[little osprey]

substitute the values yes and then solve backwards

[fathom root]

@little osprey could the answer be a square root?

i solved x = sqrt-7

nvm resolving

im going wrong somewhere, i got sqrt8/-x lol

[upper karma]

Can i see what you did? @fathom root

[fathom root]

Sure

Sry for bad handwriting

Oh I should’ve done divided by -1

So effectively I solved that x = sqrt8/-1

[fathom root]

anyone?

[paper vale]

Lol

Look at how u expanded (3-x)^2

[fathom root]

I'm aware

any tips

[paper vale]

(a+b)^2 doesn’t equal a^2 + b^2

[slow void]

need help with this?

[upper karma]

Consider $A_{\text{triangle}}=\frac12 a\cdot b \sin(C)$

[slow void]

how would i set up the numbers into the formula lol

[somber coyote]

Al𝟛dium:

[upper karma]

how would i set up the numbers into the formula lol
a and b are the 2 sides while C is the angle you want

[slow void]

63^2+62^2?

[upper karma]

Where do you see a plus sign there

[slow void]

i do not even know can you set it up for me and i can try figuring how to solve it?

[upper karma]

Uh sure $1600=\frac12(63)(62)\sin(C)$

[somber coyote]

Al𝟛dium:

[upper karma]

@slow void got it?

[slow void]

sorry fell asleep

C=0.96010613+2πn,2.18148652+2πn,

[upper karma]

What?

Did you just plugged it into wolfram...

[slow void]

mathway solve for c?

[upper karma]

Use your brain to solve it

[slow void]

lmao whats the point of using my noodle lmao

[upper karma]

We don't want that answer! You don't even know what it means, please try using algebra

[slow void]

@upper karma math is tough

[upper karma]

Definitely if you don't practice it, plus if you use a program to solve them.

I can help you solve it but don't use programs and start getting a pen and a paper to practice these.

[slow void]

my courses is anything goes

[upper karma]

What?

[slow void]

@upper karma Meaning we can use a wizard to zap the problem in by plugging it in the professor doesnt care

[upper karma]

That's very messed up

[slow void]

@upper karma my field is nursing we don't use much math same for English 101 were not going to write legal essays while performing CPR

I managed to solve 20 problems but stuck on those complex ones

[fathom root]

https://gyazo.com/8116d23569a63cd2e3d5990724890d3b

i think i fixed the first incorrect part

any idea how to get part 2, 3?

[smoky lichen]

Maybe it wants the numerator of the fraction after you square $\frac{\sqrt{3}}{2}$ since it gives you the denominator of 4

Bruh my latex is rusty

But anyways

@fathom root so I think they just want the numerator

After you square it

[somber coyote]

lovetolive:

[smoky lichen]

For part 2

[fathom root]

yea i got it im dumb

its 1/4 + 3/4

= 1

lmaoo

[smoky lichen]

Lolll

[fathom root]

still cant get this one

https://gyazo.com/e4182efe40fecc7b0bc7564ff08148c8

tough ass question

[smoky lichen]

Did you try using the distance formula

Or another method

[fathom root]

yea, i must have messed it up somewhere tho

cuz my answer was dumb and made no sense

[smoky lichen]

Oh ok

I can try it rlly quick

[fathom root]

ty, lmk what you get so i can try and work it out till i get it.

[smoky lichen]

Alrighty

[upper karma]

@fathom root didn't we told you that you did (a-b)²=a²-b² which was wrong

[fathom root]

i reworked it and got x^2-6x+9=2

[smoky lichen]

So this has 2 solutions

And you only have one of them as your answer I think

[fathom root]

2 and 4?

[upper karma]

i reworked it and got x^2-6x+9=2
Where did the +1 go?

[smoky lichen]

Yeah @fathom root

[fathom root]

damn it. I tried both of those didnt know i needed both

thanks.

should've read more closely.

[smoky lichen]

np

[fathom root]

can you find a point on the terminal side of 0 degrees?

1,0? im assuming?

[wheat swan]

Hello I am having trouble with this problem.

I am on the roof of a 1000ft building, and I am looking at an even taller skyscraper. I have an inclinometer. When I look down at the base of the skyscraper, my inclinometer reads that I am looking down at an angle of depression of 65 degrees below the horizontal. When I look up at the top of the skyscraper, my inclinometer reads that I am looking up at an angle of elevation of 40 degrees above the horizontal. Find the height of the skyscraper in exact and approximate form.

[umbral snow]

It really does come down to:
Draw the diagram and label angles that you know, make right triangles out of those angles

Also the triangles should use lengths that you know

The "looking down" angle can be used to find how far the other skyscraper is away from you. That's the trick

[wheat swan]

Ok thanks I'll go with that and see where I get

[wheat swan]

Ok im stuck now

[fathom root]

https://gyazo.com/524aa195453a172e8960a75c637eed65

how would i convert these

[wheat swan]

So what I got for the height of the other skyscraper is 2985ft can anyone check my answer?

This is my problem - I am on the roof of a 1000ft building, and I am looking at an even taller skyscraper. I have an inclinometer. When I look down at the base of the skyscraper, my inclinometer reads that I am looking down at an angle of depression of 65 degrees below the horizontal. When I look up at the top of the skyscraper, my inclinometer reads that I am looking up at an angle of elevation of 40 degrees above the horizontal. Find the height of the skyscraper in exact and approximate form

[twin lodge]

Hey guys, so I knew that:
sin 60 degrees = sin 120 degrees
And this works for all supplementary angles
but why?

[upper karma]

I think symmetry around the unit circle is one way to think about it, if you know the unit circle

I do not know a triangle interpretation tbh

[twin lodge]

Is a unit circle that radius =1?

[runic beacon]

Yes

And centre 0,0

[twin lodge]

Hmm but how are these two related?

[runic beacon]

The angles 60 and 120?

[upper karma]

It is usually used to define the sine and cosine functions in high school

Points on the unit circle form right triangles with the origin

It’s basically ends up extending the definition of sin and cos to arbitrary angles

Rather than ones that can be made with a triangle

[twin lodge]

Something like this?

this is a unit circle

[upper karma]

No, I don’t think so

[twin lodge]

and a right angle triangle in it

[upper karma]

Although nothing is labeled here

So I can’t quite tell

You would need to place it on the Cartesian plane to get a useful interpretation

[twin lodge]

oh

[upper karma]

You can probably understand it btw

After a khan academy video or two

[upper karma]

Let me see if I can successfully draw something over this on my phone

[twin lodge]

Thanks

[upper karma]

Can’t say this is my best art

[twin lodge]

@upper karma Better than mine 😆

[upper karma]

The segment from the origin to (x,y) forms a right triangle in this way

So the angle between the x axis and the orange segment that was determined by (x,y)

We have

Well let’s call that angle t

Sin(t)=y and cos(t)=x

Because the x and y distances are really the triangle side lengths

That’s why this definition is “equivalent” (but not really, it’s an extension) to the triangle definition

Also keep in mind the hypotenuse is just 1 because our circle is a unit

If you are trying to reason through why this definition is consistent

[twin lodge]

am I correct?

[upper karma]

Uh, yes I think so, but again things aren’t labeled that clearly

[twin lodge]

does x means the whole x axis?

Or the adjecent for angle t

[upper karma]

No, (x,y) is a particular point

[twin lodge]

what point it is?

[upper karma]

Call it (a,b) if that makes it easier

[twin lodge]

any point? (as long as they form a right angle triangle)

[upper karma]

Yea, any point in the circle

Well, the right triangle is kind of induced by the point on the circle

If you pick a point on the unit circle, a triangle can be made

[twin lodge]

on the border of the circle? (circumference)

or inside it?

[upper karma]

Yea, not inside

[twin lodge]

ok

[upper karma]

Have you reasoned through the new sin and cos definition being consistent with your old triangle definition?

Notice that the 2 side lengths have length x and length y

[twin lodge]

I'm trying to get the sin (t) = a and cos (t) = b

as well as where is a and b i (x and y) in your diagram

[upper karma]

In mine, you can change the (x,y) to (a,b) if you want to switch to that

[twin lodge]

like this?

I simply choose two points

and let them to be a and b

[upper karma]

Uh sure but I’d switch their positions

For consistent naming

[twin lodge]

okiee

then why sin (t) = a and cos(t) = b?

isn't a and b two points?

how come trigo working on points?

[upper karma]

Yea, but they determine the triangle leg lengths

[twin lodge]

ok, so I have three sides (according to your location of three points), ob, oa and ab

so sin (t) = ab/ob

[upper karma]

The point on the circle is (a,b) right

[twin lodge]

the point of the circle should be O isn't it?

[upper karma]

That’s the origin

I mean the point the segment goes out to

It’s the coordinate (a,b) in the plane

[twin lodge]

this is what i'm thinking

[upper karma]

Ok no no this isn’t quite right

[twin lodge]

hmm how should I correct it?

[upper karma]

Send me a mostly blank, large image with a circle on a plane and I’ll draw it

[twin lodge]

here you go

don't care about the numbers in the graph

[upper karma]

Yes this works

[twin lodge]

okay, thanks very much!

[upper karma]

Does this make sense?

[twin lodge]

ok so a and b is basically the x and y coordinate for point (a,b)

makes sense

so sin t = b and cos t = a

am I right?

[upper karma]

Yes

So I’ll just say... this extends perfectly fine to the other 3 quadrants of the circle

You can learn more about that after

[twin lodge]

yep I can imagine

[upper karma]

But we are ready to answer your original question now

[twin lodge]

sure

about the supplementary angles

[upper karma]

If we reflect our segment over the y axis

We get a new angle and line and point and stuff

And these 2 angles sum up to 180 right?

[twin lodge]

you mean the yellow and orange angles?

[upper karma]

Yes

Labeled as they are

[twin lodge]

why they sum to 180?

oh i see

okay

I imagined it in a geometrical way

[upper karma]

I... can’t word a good explanation for that

[twin lodge]

no problem

go on

[upper karma]

Ok if you got it then that’s fine

Now notice

What is the sine of each of the angles?

[twin lodge]

sin (yellow) should be b

but orange isn't in a triangle, is it?

[upper karma]

It is, but it’s formed in a different way

[twin lodge]

wait so the angle you're mentioing is (180 degrees - orange)?

[upper karma]

Yes, this may be a bit confusing

[twin lodge]

ok

then sin (180-orange) should also be b

[upper karma]

I’ve been trying to sweep the extension to the other 3 quadrants under the rug a bit lol

But yes, it works how you’d want it to

Yep, so the sines of each angle are the same

[twin lodge]

yup

but not for one angle

that the result is a and -a

[upper karma]

Sin is not dependent on a, remember?

Only b

[twin lodge]

wait but when we sin different angles it gives different results

[upper karma]

Well yea, that sure can happen

[twin lodge]

so the result is different for only the angle up there (that you didn't indicate) that the result will be a and -a

[upper karma]

But the special thing is

When the points are symmetric about the y axis

The y coordinate is preserved, so they have the same sin()

And they are supplementary

[twin lodge]

The y coordinate is preserved, so they have the same sin()
@upper karma ohhh I get it

but it won't work in this condition right?

[upper karma]

No

[twin lodge]

because their symmetric about the x axis

and we can know it geometrically

[upper karma]

The cosines of those 2 angles would be the same

[twin lodge]

oh yeah

btw, does sin (x) = sin (180-x) works in cosine too?

[upper karma]

Uh what

[twin lodge]

like if two angles are supplementary

a and b

does cos (a) = cos (b)?

[upper karma]

No

[twin lodge]

okay

[upper karma]

There should be something like the thing we just went through for sin that they satisfy though

Don’t know what it is formulaically off the top of my head, but something about symmetry over the x axis

[twin lodge]

Thanks anyways 😃

[slow void]

how would i solve these?

[earnest echo]

Surely, you have done part (a) and part (c), right?

[slow void]

@earnest echo how would i set it up first?

[earnest echo]

Can you right tan(theta) in sine and cosine?

[slow void]

like tan(theta)=7/12?

[earnest echo]

No

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$

[somber coyote]

The Godfather:

[earnest echo]

Like this

[slow void]

@earnest echo i see which numbers to plug in for that formula

there only 7/12

[earnest echo]

Forget about numbers for now

Can you write cot(theta) in sine and cosine

[slow void]

@earnest echo oh i have to fill in the theta values like substitue

[earnest echo]

First of all stop vanishing

Now,
You DON'T NEED THETA VALUES FOR NOW

[slow void]

@earnest echo i suck at this lol no clue how i got a A+ in precalc and college algebra

[earnest echo]

And stop tagging me, I'm here

[slow void]

Sorry

[earnest echo]

Can you write cot(theta) in sine and cosine

[slow void]

one moment

guessed those two

[livid moss]

Don't guess

[slow void]

@livid moss ive been up since 10 am lol and its 3 am now

[livid moss]

@slow void what is cot(θ) by definition?

[slow void]

-7sqrt95/95?

[livid moss]

What

I mean in general

How is cot(θ) defined?

[slow void]

its a right angle contagent something

[livid moss]

Are you seriously going to tell me that you did not learn that cotθ = 1/tanθ?

[slow void]

@livid moss this professor did not provide notes

very lost tbh

[livid moss]

So you are doing this without any idea what cotθ is?

[slow void]

@livid moss mostly mathway has been my savior lol

[livid moss]

Well now I told you

cotθ = 1/tanθ

[slow void]

oh okay

[livid moss]

Similarly, secθ = 1/cosθ

And cosecθ = 1/sinθ

[slow void]

what about the 7/12 where does that fall in place?

[livid moss]

You honestly only need it for the second question

[slow void]

@livid moss its 1 thanks 😄

[livid moss]

Yes

[slow void]

they could have said simplify the problem instead of leaving it like that for confusing

[livid moss]

Lol, no. Find is fine. It's obvious what they mean. Find the number it is equal to

And don't try to do questions again if you don't know the terms and functions being used in the question. You won't get anywhere

[slow void]

It's the use of English they put in to confuse people on purpose

[livid moss]

Lol no, you're being paranoid

[slow void]

Same for this one at least in the video tell the person to stack them up like this and plug this is first and this second.

this degree goes here and that goes there thats how you put it together to solve instead of making things confusing that YouTube channel mathispower4u is horrible and outdated

[livid moss]

I'm not familiar with the channel. But the question is not supposed to tell you how to solve it, lol

[slow void]

@livid moss once you know how to solve it you can apply the steps to certain problems

[livid moss]

You shouldn't memorize steps

You should know the theory and look to see what can be applied to each problem

[slow void]

@livid moss too much riddles and puzzles very tiring

[livid moss]

Go sleep

[slow void]

thx

[empty slate]

Can someone help me with this?

The answer is (3). I don't know how to obtain this at all - there's no similarity involved or anything. Here's how they've attempted it:

[empty slate]

Update: Got the answer by using the concept of Mass Point Geometry guys. A friend just taught me the concept

[slate saffron]

Can someone help me with this?
@empty slate Search Menelaus theorem

[empty slate]

Okay I will

[upper karma]

More like Ceva's

In this case

[little osprey]

agree

ceva

[delicate laurel]

how would I approach this?
'

[upper karma]

Recall triangular inequality

[delicate laurel]

so 13 < 6 + a

a>7 ?

[upper karma]

Yes

But also a < c

Because c is the longest side

[delicate laurel]

yeah so the possible solutions would be

between 7 < a < 13

So i guess 10 would be an answer?

@upper karma

[upper karma]

Sure

[delicate laurel]

alright, thanks a lot HoboSas

[upper karma]

Np

[slim wadi]

how would i do this?

[upper karma]

The lengths of MN and NP sum to the length of MP, which has its length given

[slim wadi]

yea but like i have no clue how i would solve for x im new to geometry and didnt do well in algebra

[upper karma]

Have you written the equation?

[slim wadi]

Nah.. thats why im trying to figure out

what*

i can solve it on my own, i just have no clue how to make the equation

[upper karma]

My first message was supposed to be suggestive of an equation

x - 5 + 3x +2 = ?

[slim wadi]

4x−3?

what would I do after i solve for x tho? @upper karma

[upper karma]

The ? is something to fill in

[slim wadi]

wait nvm lmao

[upper karma]

I’m guessing you got it, so after you solve for x simply plug it back into 3x + 2

[slim wadi]

for x i got 4x-3 tho

im p sure thats right bc i did it again and got the same answer

[upper karma]

Uh what

x - 5 + 3x +2 = ?

What did you write on the right hand side

You can’t just simplify the left hand side, that’s not enough information

You need to set this equal to something

The problem gives you another piece of information, what haven’t you used yet?

How about “length of MN plus length of NP equals length of MP”

[ionic girder]

Hi guys does anyone know how to do this question 🥺 if not no worries

#2

[upper karma]

@ionic girder okay you there?

[ionic girder]

Mmmhm

[upper karma]

How am i supposed to interpret that?

[ionic girder]

Yes??? 😭😭

[upper karma]

Okay where are you stuck at

[ionic girder]

My brain stopped working after quarantine so I don’t even know how to start with this question 😌

I got 2a

[fresh pike]

can someone explain to me where did c come from? i'm learning the properties of algebra for geometry

[upper karma]

@fresh pike occuppied

@ionic girder so have you tried b?

[fresh pike]

oh i got it

[ionic girder]

Um 😌 I have tried to the best of my ability but I don’t even know where to start

I haven’t done math in a v v long time, I’m so sorry

Like I’m pretty sure this is wrong

Or is it tanx=3

[steep coral]

can someone help me on the plotting, im not sure how to do it

[silent plank]

determine where each pair of lines intersect

[steep coral]

oh

is there an easy way to do that

[silent plank]

algebraically or graphically

[steep coral]

how do i do it algebraically

[silent plank]

solve some systems of equations

[steep coral]

is there a formula?

ooooo

how do i do that again

[silent plank]

using substitution or elimination

[steep coral]

i think i remember learning that now

thanks

[fervent wadi]

Can anyone help me with this problem? Here it is:

A regular pentagon D E F G H is drawn inside a triangle A B C such that vertices D and E are on side A B, F is on B C, G and H are on C A and angle E F B is 18 degrees.

I want to find the angles of triangle ABC and prove that C G = D H

[dark sparrow]

have you made a diagram yet

[fervent wadi]

yes

[dark sparrow]

can i see the diagram

[fervent wadi]

sure, hold on

just a rought diagram

D H is meant to look parallel to C B

Hold on will be back in 15 minutes

[silent plank]

apply properties of a regular pentagon
and angle sum properties of polygons and lines

[slow void]

How would I find the formula describing the zeroes? for y=sin(θ)?

[thorny bough]

@dark sparrow i hope you remember helping me. i was asking about sin(t)=sin(vt)

you suggested trying sin vt - sin t = 0.
can i also do sin t - sin vt = 0?
it feels like i get a different answer but maybe it just appeara that way

[dark sparrow]

it just appears that way.

[thorny bough]

true

i think i see how to.

thanks

[fervent wadi]

ok im back

[thorny bough]

if i have something like 2pi(1+2k) can i change it into anything else?

k is any integer

o think i see? one is 2pi(1+2k) and the other one has a negative 1 multiplied. I think since the parentheses is essentially (any odd integer) then if it were negative, then the negative sign would go away. So in both cases the same thing is beingn said. Any odd multiple of 2pi. Yes?

[thorny bough]

agh. k being an integer. derp. plus or minus.

[slate blade]

I have this question I am having difficulty solving. I did it a really weird way but it took way too long

[acoustic jungle]

You find a telescoping series

with cot I believe.

[green hazel]

using 1 + cot^2 = csc^2?

I wanna know the answer to this as well tbh

I've been going at this for the past half hour but I can't

[acoustic jungle]

cot x - cot(2x)=cossec (2x)

one sec I need to finish the question I don't want to lead anyone the wrong way

[green hazel]

huh, that's a thing?

I'm actually so bad at trig yikes

[acoustic jungle]

oh yeah it is right

I solved it

well actually not really

I vaguely remembered it from somewhere lol

@slate blade use cot x - cot(2x)=cosec(2x)

[green hazel]

dang

I'll use it then

[acoustic jungle]

It's probably very difficult to just derive that while doing the question. I wouldn't have thought if it either.

[green hazel]

I want my 40 minutes back

this is now not actually not that bad

😔

that formula was magical though

just poof from nowhere

[acoustic jungle]

I found where I remembered it from

[slate blade]

Bruhhhh ty

I did it a cheese way

[upper karma]

Wasn't the exercise guided?

[green hazel]

wait

[upper karma]

Like didn't that formula appear in previous tasks?

[green hazel]

I still don't get it

[slate blade]

Nope well my friend sent the question individually and she was like plz help solve

[green hazel]

I got csc(2pi/15) + cot(2pi/15) - cot(16pi/15)

[slate blade]

This is what I did

[upper karma]

Like didn't that formula appear in previous tasks?
I believe this is the case

[green hazel]

,rotate

[slate blade]

You goofed you the formula @green hazel

Yeah no my method is BS. I literally went hmm trig identities??!?

[green hazel]

oh yea

i could conver tthe last csc into that

bruh

I was working with csc(u) where u = 2pi/15

so I didn't think about that

[slate blade]

Well thanks a bunch people

[green hazel]

god damn

[slate blade]

This chat is super helpful lol

[green hazel]

Doodaide you're big brained

conjuring up 2cos^2(x) out of nowhere

what a gamer move

[slate blade]

Yeah, cuz if I add then subtract it’s 0 right??

[green hazel]

yea

how long have you been working on this

[slate blade]

Time is relative

Glory is eternal

[green hazel]

ah yes

[slate blade]

Let’s not talk about that.

Anyhow I’m gonna sleep now toodles

[green hazel]

nighty o/

[narrow storm]

How would you start this?

[dark sparrow]

there is an unmarked downward force of 50 newtons acting on the suspension point

it, F1 and F2 sum to zero

[narrow storm]

As in zero newtons

[paper vale]

you need to equate vertical and horizontal forces

[dark sparrow]

as in the zero vector

@paper vale that spelling

hotricontala

[paper vale]

equate in magintude i mean, ie summing to zero

[narrow storm]

So zero newtons zero angle

[paper vale]

what

[narrow storm]

She said zero vector

No magnitude no angle

A point

?

@dark sparrow

[paper vale]

the vertical component of f1 is -sin(alpha)35

[dark sparrow]

yes

[paper vale]

and vertical of f2 is -sin60f2

[dark sparrow]

and then, following kane's initially grossly misspelled suggestion, draw out a pair of axes

and equate the components along each

[narrow storm]

Does this make any sense?

Ann

[paper vale]

no

[narrow storm]

She said the sum of all of the forces should equal 0

Why doesn't this work?

[paper vale]

it is because it is in equilibrium, but as i said the forces need to be component form

[dark sparrow]

at no point did i say the magnitudes of the forces had to sum to zero

in fact there could be no way they could do that unless the forces were all individually zero

which they are most definitely not

[narrow storm]

Ok

[paper vale]

well if there were parallel and have no torque then you could

but in this case it is not

[uncut rock]

Can someone help with these these 2 problems?

[paper vale]

write down the pythagorus's theorem of this triangle

then solve quadratic

both have the same idea

[acoustic jungle]

or use power of a point

[paper vale]

power of a point is proven by pythag

but i guess u can still use that

[rain steeple]

Hey

Do you need help?

[upper karma]

Hi I need help with a step

this is the question

I don't understand how that =4*73^h

[slate oar]

0.1^(1/7.3) happens to be close to 0.73

,w 0.1^(1/7.3)

[somber coyote]

Results provided by WolframAlpha

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

Requested by Look At Me#8623

[upper karma]

omg

im an idiot

I was doing 0.1^1/7.3

i didnt put parenthesis

[slate oar]

oof

[upper karma]

wait i have another question

how come it changes to 7.3

when it was 0.73 originally

when you're multiplying by the 10% you need to multiply both the 4 and 0.73?

[slate oar]

wdym?

[upper karma]

wel lthe equation was L= 4*0.73^h

but it became 4*0.1^h/7.3

[slate oar]

they made the base 0.1 so that when the exponent is 1, the initial value is multiplied by 0.1, becoming 10% of what it was originally

[upper karma]

so the h changed to a 1?

[slate oar]

nono

the exponent is h/7.3

it's just that, the initial value will be multiplied by 0.1 when that expression is exactly 1

so when h = 7.3, h/7.3 is 1, and then you'll get 4*0.1 which is 10% the initial value

[upper karma]

and it would change to a 2 if you lose 80%

[slate oar]

yeah they would use 0.2, but you'd need to find a different exponent

[upper karma]

why

[slate oar]

because 0.2^(1/7.3) is not 0.73. you'd need to find x such that 0.2^x = 0.73 to have the two expressions be equivalent

,w log(0.73)/log(0.2)

[somber coyote]

Results provided by WolframAlpha

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

Requested by Look At Me#8623

[slate oar]

you could use that, for example

since 4 * 0.2^(h0.19) = 4 * (0.2^0.19)^h ~ 40.73^h = L

[upper karma]

but how come here it shows 1/7.3

when you said h/7.3 = 1

so wouldnt that mean h = 7.3

[slate oar]

i didn't say it equals that. i said it will equal that when h = 7.3. that's why it "highlights" the amount of hours that need to pass before 90% is lost

[upper karma]

not 1

[slate oar]

L = 4 * 0.73^h
this highlights the amount of hours that need to pass before the luminosity is reduced to 73% of its original value (one hour). but we need to find something that highlights when it is reduced to 10% to its original value. so we need to change the base there to 0.1. we've shown that the expression is approximately equal to
L = 4 * 0.1^(h/7.3)
what this means is, once h/7.3 is 1, the luminosity will have been reduced by 10%. in other words, after 7.3 hours. that's why it "highlights" the amount of time needed for that reduction

[wheat swan]

I am on the roof of a 1000ft building, and I am looking at an even taller skyscraper. I have an inclinometer. When I look down at the base of the skyscraper, my inclinometer reads that I am looking down at an angle of depression of 65 degrees below the horizontal. When I look up at the top of the skyscraper, my inclinometer reads that I am looking up at an angle of elevation of 40 degrees above the horizontal. Find the height of the skyscraper in exact and approximate form.

I am still having trouble with this problem

My professor told me that It could be done with one trigonometric function

[slate blade]

so you draw your diagram right?

therefore

$$ tan65 = 1000/x $$

[somber coyote]

Doodaide:

[slate blade]

eh I'll learn soon

and $$tan 40 = y/x $$

[somber coyote]

Doodaide:

[slate blade]

therefore
$$ x = 1000/tan65
tan 40 = y/(1000/tan65)
y = tan40 \times (1000/tan65)

[somber coyote]

Doodaide:
Compile Error! Click the reaction for details. (You may edit your message)

[slate blade]

which equals approx 391 ft. add that to 1000 and you get 1391 ft as your answer

@wheat swan hope this helps

[carmine sundial]

eyy, i have a confused friend who needs help and i want to clarify something.

does this still count as a parallelogram?

[silent plank]

no

[carmine sundial]

aight just as ive thought

[silent plank]

both pairs of opposite sides need to be parallel for it to be a parallelogram
and it can be proven that the unmarked pair are not parallel

[carmine sundial]

danke, Ramonov!

[junior light]

Also, opposite angles of a parallelogram are equal. If this were to happen in case of the given quadrilateral, the sum would be 380 degrees instead of 360.

[silent plank]

i'd just go with co-interior angles

[junior light]

From Paul Zeitz's Art and Craft of Problem Solving
A great circle is a circle drawn on a sphere that is an "equator;" i.e., its center is also the center of the sphere. There are n great circles on a sphere, no three of which meet at any point. They divide the sphere into how many regions?

For n=1, 2 and 3, the number of regions formed is 2, 4 and 8, respectively.

This seems to give the idea that perhaps the number of regions for a certain n is given by R(n)=2^n

Interestingly, when I tried drawing these great circles on a sphere(using Microsoft's Paint 3D), I could not draw more than 3 great circles without violating the no-three-of-which-meet-at-any-point condition

So is n=3 the limit here, or am I missing something?

[junior light]

<@&286206848099549185>

[twin prawn]

You are missing something:

When you draw a spherical n-gon, its sides are great circles, no three of which meet at a single point

in this case n = 3, i.e. the depiction of a spherical triangle

[junior light]

So this works for any arbitrary n?

I'm actually having a hard time visualising more than 3 great circles, and I'm not aware of spherical n-gon's tbh

[acoustic jungle]

have you tried multiplying by the conjugate

Why?

What is happening

[junior light]

Ummm can anyone help with my question above?

[wheat swan]

@slate blade Thank you very much

🙂

So why is it tan40 = y/x?

[rich wolf]

sohcahtoa

tan = opp/adj

[eager ocean]

Name the intersection of plane P and line m.

How do I do that?

[upper karma]

Can you see where both are intersecting?

[eager ocean]

Plane p though isnt that the whole box thing

[rich wolf]

name it gary

[eager ocean]

so wouldnt the intersection be the boundrary

[upper karma]

Lmao

Plane p though isnt that the whole box thing
Box?

[eager ocean]

yes this is very wobbly box

[upper karma]

Oh yeah

[eager ocean]

[upper karma]

I didn't recognise it as a box lmao

Do you know where the line m is at?

[upper karma]

Correct.

Can you notice which point does the line m and plane p have in common?

You can imagine it irl like having a paper horizontally and then putting a pen vertically breaking the paper

[eager ocean]

a point?

[upper karma]

Yes

[eager ocean]

point A?

[upper karma]

Yes.

[eager ocean]

but wait

isnt Point E also

on the plane

[upper karma]

No because remember this is three dimensional

[eager ocean]

three dimensional

isnt it flat

[upper karma]

Can you maybe give more context

[eager ocean]

Im in geometry, right after Algebra 1

let me give u the answer choices one sec

point A, line n, AC, segment AE

so like we are learning linear pairs, concave and convex, stuff like that

so pretty sure the plane isnt 3d

[upper karma]

I didn't say the plane is 3d

The plane is flat as you said

But we are in a space 3 dimentional

Not sure if i'm using the english native terms correctly here wait

[heady parrot]

I’m retarded and lazy

Solve

[upper karma]

@heady parrot

1. Not the place if you are asking for "solving"
2. This channel is occupied.

@eager ocean we are talking in the 3 dimentional space

Imagine it by how i described above

You can imagine it irl like having a paper horizontally and then putting a pen vertically breaking the paper

The point E is on the plane P but not the line m

[eager ocean]

oh that what the dotted line means?

[upper karma]

Because of the discontinuous lines

Yes

[eager ocean]

so wait where is point E then

[upper karma]

It means it's going like behind

[eager ocean]

is E just a ray no where?

[upper karma]

E is a point

On the plane P

Just like point D or B

[eager ocean]

Sorry that I'm a bit slow lol. Whats the arrow then below Point e going to?

if E isn't a ray

[upper karma]

It's the line m

Don't worry about being slow

Remember it is going behind the plane

Really

You can imagine it irl like having a paper horizontally and then putting a pen vertically breaking the paper
I'm sure this'll help

A lot to visualising

[eager ocean]

OH okay so line m only has one point, A- the E has nothing to do with line m and isnt even part of a ray or line segment or line

[upper karma]

Exactly, E is just on the plane P

The discontinuous lines of line m means it's going behind the plane

Remember

[eager ocean]

ohhh, okay so then the intersection of Plane P and Line M would be A, since A is the only point that Line M and plane P have in common

[upper karma]

Yes.

But line m not M

[eager ocean]

oh

yeah my bad

[upper karma]

Dw

[eager ocean]

Thank you so much

[upper karma]

Yw!

[eager ocean]

Sorry, one more question lol. Is a regular polygon always convex?

[upper karma]

Yes. By definition

[eager ocean]

All right, thank you

[upper karma]

why does this turn positive

[umbral sparrow]

you can divide both sides by negative one and keep the eq the same

and if you divide them by negative one their signs switch

[upper karma]

ok thanks

[kindred fox]

Been trying to figure this out for 3 hours.

I have been doing everything by hand til I found this problem, any idea what it could mean?

[austere mica]

Well the theta when T is least is the value on the x axis that gives the lowest point

idk exactly what else they mean

[kindred fox]

It sucks, I'm 21.60/22 in this homework and I can't figure out a technical detail lol.

[austere mica]

i mean from the picture, it looks like T decreases

so the further you go along, the smaller T is

so just look at the vaue at the end

idk

[knotty gulch]

In the figure below, m<PRS = ( 5x - 2 ). Find m<QRS.

I don't understand how to solve it.

[silent plank]

angle addition postulate with some algebra

[knotty gulch]

Thank you 🙏

[upper karma]

fact check x= 25

right?

[paper vale]

yea

[upper karma]

x=16
F=74

[astral maple]

@upper karma Why asking if you've already solved it

[upper karma]

can i dm you

i dont wanna talk about it in public

[astral maple]

Hm

Okay...

[upper karma]

heeeeeeeeeelp

[earnest echo]

First step: name all the goddamn angles

[upper karma]

@earnest echo then @!@

[earnest echo]

Send the pic with angles marked then we'll proceed

[upper karma]

its doesnt matter

just find the value of x @!@

@earnest echo

[earnest echo]

What do you mean, it doesn't matter

I'll just guide you

[dark sparrow]

i feel you might need some trig to solve this

[earnest echo]

It's impossible to do that without angles marked

[dark sparrow]

the answer is clean but i arrived at it in a way that isn't pure angle-chasing

[upper karma]

i feel you might need some trig to solve this
@dark sparrow need just elementary method @!@

[dark sparrow]

is @!@ meant to be a face?

[upper karma]

yah

lmao

[dark sparrow]

i've literally never seen it before

[upper karma]

lol

its just like (wow )

[dark sparrow]

this is what can be arrived at through simple angle sum considerations, i think

[upper karma]

yah

actually its an india qs from jee exam

lol

[dark sparrow]

oh, so it's one of those "solve this in under 5 seconds" questions

[earnest echo]

Then you can use trig

[dark sparrow]

if trig is allowed: ||tan(x)tan(80°)tan(40°)tan(20°) = 1||

[upper karma]

nah i need elementary mehod without trig @!@

Unfortunately, it is not allowed @!@

[earnest echo]

How did you come up with that relation, Ann?

[dark sparrow]

the diagonals meet at right angles

[earnest echo]

Yeah

[dark sparrow]

therefore you have four right triangles

[upper karma]

yah

[dark sparrow]

label the diagonal intersection point O and the outer vertices A, B, C and D clockwise such that A is the vertex of angle x

[upper karma]

then /

[dark sparrow]

tan(x) = OB/OA, tan(80°) = OC/OB, tan(40°) = OD/OC, tan(20°) = OA/OD

multiplying those together gives the relation i posted behind the spoiler tag

[earnest echo]

Okay, okay

[dark sparrow]

anyway, i can't think of a non-trig method right now

hm hold up

what if you extend AD and BC until they meet

will that produce anything nice

[upper karma]

ion know @!@

[dark sparrow]

i'm grasping at straws here

[upper karma]

x=30 yay

[dark sparrow]

oh wow

[little osprey]

Reminder that Russian math is as scary as Indian math

[abstract narwhal]

thanks for the reminder

[oak citrus]

Indian math is terrible... pls don’t compare

[carmine sundial]

how jc? any math is damn right scary to me

[junior light]

I'm fairly certain that nothing beats the horror of Soviet math

[dark sparrow]

@little osprey what do you consider as "Russian math"

[upper karma]

One of my professors said that Russian math books are written at the most difficult level possible because Russian professors only want to teach you if you are good enough to read high level textbooks, so this process thins out the weaker students

But I am guessing he was pulling my leg

[dark sparrow]

that's an over generalization

[junior light]

I guess Soviet math books might be the better generalisation

Lack of paper and a habit of sticking to the point made some texts extremely terse to follow

[earnest echo]

Don't know about Russian maths books
But once I was told to solve IE Irodov since my physics was weak
I really struggled and could only solve it partially

[gray turret]

some teachers here in india recommend irodov too for jee exams
a few toppers in our year used to solve irodov in grades 9 and 10 and i still struggle to do questions from it in grade 12 :/

[supple wedge]

hello its been a while

is $sin(4x)(tan²2x+1)=2tan2x$

[somber coyote]

Hmm:

[dark sparrow]

bruh bad tex

$\sin(4x)(\tan^2(2x) + 1) \overset{?}{=} 2 \tan(2x)$

[somber coyote]

Ann:

[dark sparrow]

yes

[supple wedge]

what is overset

[dark sparrow]

the command for putting one thing on top of another

[supple wedge]

why tho

[dark sparrow]

in this case the question mark over the equals sign

[supple wedge]

so is the eqn true

im solving some trig eqn and this is 1 side of it

fuck it ill send the whole thing

[dark sparrow]

yes the thing is true

[supple wedge]

(i) done

2tan2x=2cotx

right

@dark sparrow

[dark sparrow]

bruh

but...yes

[supple wedge]

cant solve that

[earnest echo]

Convert tan(2x) into tan(x) and do the same to cot(x) and try to simplify it

[supple wedge]

i got

$\frac{2tanx}{1-tan²x}=\frac{1}{tanx}$

[somber coyote]

Hmm:

[supple wedge]

then shit went south

from there

[earnest echo]

Simplify it

[supple wedge]

i got 2tan³x+tan²x-1=0

[earnest echo]

How do you have a cubed term?

[supple wedge]

oh my fuck i am blind

oops

is it $tan(x)=\pm{\sqrt{\frac{1}{3}}}$

[somber coyote]

Hmm:

[supple wedge]

@earnest echo

[earnest echo]

Yep

[supple wedge]

answer sheet is wrong

then

it says x=35.3

for the first solution

[earnest echo]

And the second solution?

[supple wedge]

144.7

mine is x=30

[earnest echo]

,w sin(4x)(tan^2(2x)+1)=2cot(x)

[somber coyote]

Results provided by WolframAlpha

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

Requested by The Godfather#1908

[supple wedge]

lolno it wont work

idk what z is

[earnest echo]

n belongs to Integers

[supple wedge]

ik

[earnest echo]

Z is representing integers

[supple wedge]

oh

[earnest echo]

Insert 35.3 in the given solution

See if the equality holds

[supple wedge]

yeah it doesnt

stupid answer sheet

[earnest echo]

Good, your answer is correct

[supple wedge]

out of alg, calc and geometry i hate geometry the most

everything else good

i like

[regal meadow]

How would you draw this diagram? Sorry if this isn't the right channel or something, first time here haha.

I don't think it's possible

[silent plank]

the problem lost all credibility when it stated "AO as diameter"

[dark sparrow]

there are two circles here i assume

one big circle centered at O, and one small circle inside it which has AO as its diameter

[regal meadow]

I see

[silent plank]

that makes more sense

[regal meadow]

But how can AO cut AB?

Am I just going insane?

[silent plank]

1 sec

[upper karma]

For some reason I am stuck on this

[upper karma]

A hint please

[regal meadow]

Thanks

[upper karma]

Sorry for interrupting btw didn't realize this channel was occupied

[silent plank]

whoops forgot to draw in line segment AO, indicating its the diameter of the red circle

[upper karma]

@upper karma go to iota

iota as in?

#help-9

Okay

[dreamy sinew]

Hi, does anyone understand how to do no. 9b and c?

I'm assuming theres a typo for the second angle, Beta, should be cotan Beta = 5/12 ?

[steel nimbus]

hey guyz

does anyone understand how to solve differential equation of specific general solution

?

[upper karma]

What else do I need for proofing angles?

Properties of Angles?

Anything else?

[upper karma]

@steel nimbus wrong channel

And properties of angles is probably all you need

• a lil intuition

[fluid crater]

dumb question, an n dimensional cube has 2^n corners, yea?

so if its a hypercube, the volume is also 2^n, so volume = number of corners in all dimensions?

and i am referring to corners, not edges

where all corners are infinitesimal, and the same number of dimensions as the cube itself

if that makes any sense

[noble heath]

That is true only for hypercubes of side length 2

[fluid crater]

very interesting, thanks for confirming!

[queen pond]

Ray NP is an angle bisector of <MNQ and m<PNQ = 2x + 1

Find x if m<MNQ = 42

[paper vale]

well just do 42=4x+2

[dry magnet]

ok so i have a 3d vector that is the distance between 2 things, i am using the Pythagorean theorem to find the distance between them, and i want to compare that with a number that is a distance, but can i instead remove the sqrt in the Pythagorean theorem and compare it with the distance i compare it to squared?

[steady brook]

I have no clue what you're trying to sya

but if you're asking if you can verify sqrt(x) = y

[dry magnet]

:(

[steady brook]

by checking if x = y^2

then yes

as long as y isn't negative

[dry magnet]

well distance well always be positive

[steady brook]

Then you should be fine

[dry magnet]

but just to make sure what ur saying applies to what im saying

i am finding x with the pythagroean theorom

[steady brook]

and by "I have no clue what you're trying to sya" all I meant is that your statement is impercise

[dry magnet]

and by "I have no clue what you're trying to sya" all I meant is that your statement is impercise
@steady brook how so?

[steady brook]

And there's obviously context there

that in your head this makes sense

but not to some random person reading it

[dry magnet]

ok

[steady brook]

you say a 3d vector that is the distance, distance is a scalar

[dry magnet]

whats a scalar

[steady brook]

I assume you mean you have two points and took the difference vector

a number

= scalar

[dry magnet]

ok

[steady brook]

you're saying you use the pythagorean theorem to find the distance and want to compare it to a number that is a distance

what does it mean for a number to be a distance?

[dry magnet]

ok

[steady brook]

do you mean a number that's supposed to be the distance to the two points?

[dry magnet]

i want to see if the distance between 2 objects

is the same as a number

[steady brook]

I mean

you can just compute the distance then

and then compare it no??

[dry magnet]

yes but i dont want to use sqrt

unless i have to

[steady brook]

I mean you could do it the way I said but

if you have the distance as sqrt(x)

and want to know if this is equal to y you can square both sides

[dry magnet]

so u know what the magnitude is right

[steady brook]

but then you have to know that y was positive to begin with

[dry magnet]

of a vector

[steady brook]

yes

[dry magnet]

ok

[steady brook]

I know the formula and what you're referring to

[dry magnet]

ok

[steady brook]

but this seems like you could just use a calculator

tbh

Like I don't see how

verifying x = y^2

is gonna be any easier than sqrt(x) = y

IMO

[dry magnet]

its a programming thing

and sqrt is kinda slow

[steady brook]

I guess

then yeah just square both

but you have to check if y is positive first

or well

non-negative

[dry magnet]

ok

[steady brook]

since if y = 0 then you're still fine

[dry magnet]

ill just square both

and y will always be positive

same with x

[little osprey]

I see...

[upper karma]

Well

[little osprey]

@upper karma btw its a discontunity

the graph

[upper karma]

What?

[little osprey]

the graph

uh wait

[upper karma]

Yeah ik lol

The weird thing it was that on plotting programs, the discontinuity didn't show up

[little osprey]

The weird thing it was that on plotting programs, the discontinuity didn't show up
@upper karma it does on desmos

if you keep zooming in on the 3