geometry-and-trigonometry

[upper karma]

I think its B without a really solid valid reason other than 90 + (1/2)AB should be 45?

[weary drift]

90 + (1/2)AB should be 45
what

[upper karma]

You gonna help or just meme? cuz really not in the mood for that right now

[weary drift]

what makes you think i'm here to mess around?

90 + (1/2)AB should be 45
i don't know where this came from

[upper karma]

That is so wtf. This program can't even add a picture.

@upper karma

[glossy jewel]

@dark sparrow what does it mean when we say without any lose of genrality assume a>b

[dark sparrow]

you mean without LOSS of generality?

"assume without loss of generality that a > b" = "if b > a then we'll switch them around and what follows will work exactly the same way" pretty much

[glossy jewel]

yes

ok

So its like a assumption that doesnt change the result

[dark sparrow]

well... yeah pretty much

[glossy jewel]

$\cos{x}+\cos^2{x}=1 \implies \cos{x}=sin^2{x}$

[dark sparrow]

what's that x doing there at the very end

[somber coyote]

Krishna:

[glossy jewel]

how do i find x

[dark sparrow]

you don't need x itself

[glossy jewel]

wait

i am such a dumbass

anyways thanks for the hint

[glossy jewel]

,w convert 65degrees30minutes into radian

[somber coyote]

Display results online and refine query

Requested by Krishna#4460

[dark sparrow]

65°30' = 65.5°

also, #bots

[quiet mason]

^

[glossy jewel]

yes sorry

[glossy jewel]

Eliminate $\theta$ from $2x=3-4\tan{x},$3y=5+3\sec{x}$

?

[dark sparrow]

\theta

[glossy jewel]

$\frac{2x-3}{4}=-\tan{x}$ and $\frac{3y-5}{3}=\sec{x}$

[somber coyote]

Krishna:

[glossy jewel]

square and subtract?

$(\frac{3y-5}{3})^2-(\frac{2x-3}{4})^2=1$

[somber coyote]

Krishna:

[glossy jewel]

is this what they meant ? @dark sparrow

I havent read the textbook thats why asking

[dark sparrow]

tan(θ) and sec(θ)

[glossy jewel]

?

[dark sparrow]

not tan(x) and sec(x)

[glossy jewel]

yes sorrry about that

[dark sparrow]

but yes this is what was meant

up to algebraic manipulations this is the answer

[glossy jewel]

ok thanks . I have a exam tommorow

do i need to expan dfurther

simplify further?

[dark sparrow]

if you want

[glossy jewel]

help

[silk crown]

I think writing all the terms wrt sin and cos will work there

[dark sparrow]

is this an r

[glossy jewel]

yes r

@dark sparrow

[silk crown]

I highly doubt it's pγove

[glossy jewel]

lmao

[dark sparrow]

this way of writing r still weirds me out

[glossy jewel]

i am not ables to prove that

[dark sparrow]

which one

#0 or #1?

[glossy jewel]

#0

[silk crown]

Oh that one

[glossy jewel]

yes u told to write evrything in terms of cos and sin that didnt worked

[silk crown]

Yeah I was tellin u for #1

I think for this one you just have to use the double angle formula

[glossy jewel]

[silk crown]

Yep works out

[glossy jewel]

i think the result we are required to prove is wrong

cos^2x-sin^x=2cos^x

[quiet mason]

its actually2cos²(theta)-1

cos(2*theta)

[silk crown]

it is

innit

well rip I'm now not sure why it worked out for me

[quiet mason]

show work

and also

cos^2x-sin^x=2cos^x

[silk crown]

sin^x?

cos(2x) + 1 = 2cos^2(x) is the identity not sure where sin^x comes in

[silent plank]

start from the rhs instead of the left

[silk crown]

I apparently have a superhuman ability to prove incorrect statements

[patent forge]

Hello ...I need to find the value of arctan(tan(3)) but not sure is this right or wrong ...

[dark sparrow]

tan^-1(tan(3)) = 3
already wrong

[patent forge]

but I saw a site said f^-1(f(x)) should be x ? so it doesn't apply on arctan?

[dark sparrow]

arctan is not the true inverse of tan

[glossy jewel]

Help with #1 plis

No thanks done

[silk crown]

@patent forge tan^-1(tan(180°))=0°

And 180°≠0°

so tan^-1(tan(x)) will have infinite solutions, and you can write an expression in terms of n as a general solution for it

[patent forge]

hmmm ok I will try to understand it ..thanks

[silk crown]

Just take a look at the graph of tan(x) if you want to know why I say there'll be infinite solns

[silent plank]

tan(x) = tan(3) will have infinite solutions for x
but
arctan (tan (3)) will only have one possible value

[silk crown]

that isn't what they're trying to find tho is it

[patent forge]

hmmm it is what I am trying to find lol

[silk crown]

oh shid sorry I was blind

[patent forge]

It is fine 😄

[silk crown]

well basically it's asking, what angle in the range -π/2,π/2 will yield the same value of tan as tan(3)

And the fundamental period of tan is π

Figure out which quadrant tan(3) lies in and hence its sign, and then try to find an angle in the range of tan^-1 that will have the same sign and value

[patent forge]

so tan(3) is -0.1425 and tan( -pi +3) is the same ...so I am kind of right? even tho my method might be wrong...

[silent plank]

that's fine. for tan, add k*pi to you angle so that it is in (-pi/2, pi/2). (where k is an integer and in this case k = -1)

arctan ( tan (x)) = x when -pi/2 < x < pi/2
arctan ( tan (3)) = arctan ( tan (3- pi))

[patent forge]

oh thanks all 😄

[upper karma]

I know it's all a bit over the place, but my problem - and my solution is in the box off to the right. I know it is wrong, because the answer is 6√2

I said it was 14 but I don't know how I can go about doing it

[silent plank]

don't round 5sqrt(2) to 7

leave it as an exact value

[upper karma]

Where is that Ram?

Oh yeah

|x| = 7

It was an ugly decimal but my calculator didn't show it as a surd

I didn't know it was 5 sqrt 2

[silent plank]

you should know your special angles

[upper karma]

What are special angles?

[silent plank]

30,60,90. 45,45,90

[upper karma]

Also, so if I just swapped x = 7 out for 5√2 it would be right?

[silent plank]

find the correct area of l and continue from there

don't know what you were trying to do with m at the top right

[upper karma]

Oh that?

See the triangle in the bottom right of the square

where i labelled m

Actually I'm kinda lost too lol

[silent plank]

yeh. Cross that part out. what's you area L now?

[upper karma]

I worked it out as 10.5cm^2

[silent plank]

that's using the wrong value you had before

[upper karma]

So,

L = 0.5*(5√2)^2?

Which gives us 5cm^2

[silent plank]

no,
L = 35 - 0.5*(5√2)^2

[upper karma]

Whoops

Yes

30

[silent plank]

no.

what is (5sqrt(2))^2?

[upper karma]

50

[silent plank]

what is half of that?

[upper karma]

25

[silent plank]

so what would
L = 35 - 0.5*(5√2)^2
be?

[upper karma]

10?

[silent plank]

yes

Are you able to determine the length of GF?

[upper karma]

I don't think I have enough information to determine it

[silent plank]

when you split the trapezium into a rectangle and triangle, what type of triangle was it?

[upper karma]

A right angled triangle

AH WAIT WAIT!

Okay.

So where x is, it can be replaced with 5√2.

Now,

This means that beta = √100-(5√2)^2

Which gives you

9.48683298051

(I can't find it in surd form hold on)

[silent plank]

no

[upper karma]

3√10

[silent plank]

no

[upper karma]

Hmm, why not

[silent plank]

nfi what ur even doing

[upper karma]

nfi?

[silent plank]

no fkn idea

[upper karma]

Oh

You see the right triangle

with the hypotenuse of 10

Well, where x is, we worked that out to be 5√2 yes?

Meaning that using Pythagoras we get an answer of 3√10 when we work out beta

[silent plank]

beta or GF isn't even part of that triangle

[upper karma]

I know GF isn't but beta is

I wrote beta as that sector

NO OOPS

I mean alpha...

🤦

Alpha, yes

Alpha equals 3√10 yes?

[silent plank]

no

What type of triangle has angles 45,45,90?

[upper karma]

right angled

[silent plank]

what else?

the angles the legs form with the base are equal which means it is also an:

[upper karma]

One of the legs is orthogonal

The other one forms a 45 degrees angle

[silent plank]

what types of triangles do you know?

[upper karma]

scalene triangle

equilateral

and right angled

a long time ago a friend i had wrote me this

but i wont simply just accept his answer

i also don't understand it xD

[silent plank]

have you ever heard of isosceles?

[upper karma]

Oh yeah isoceles

isosceles

i forgot about that one

[silent plank]

would this be an isosceles triangle?

[upper karma]

no because its right angled

[silent plank]

being right angled doesnt stop it from being isosceles

it could be both

[upper karma]

Huh?!?!

[silent plank]

what's the definition of an isosceles triangle?
does it have anything to do with right angles?

[upper karma]

Oh yes wait

It does

You can split an isosceles in to two right angled ones right?

[silent plank]

I meant direct a direct relation.
but the first question?

[upper karma]

Oh it's a triangle whose two um

Wait

yeah

whose two legs are equal right

the line through their leg whatever that shows i can't remember

[silent plank]

(the angles the legs form with the base are also equal in isosceles triangle)

in this case, the angles are 45 and 45, therefore it is an isosceles triangle and the legs are also equal

x=a = 5sqrt(2)

Pythagoras could've worked if done properly
or you could've also used trig again.

[upper karma]

Hmmm 🤔

I will solve this again soon, but can you help me to understand what my friend said?

It's gibberish

https://cdn.discordapp.com/attachments/326138757474680852/632994668904316943/Screenshot_20191013-015316_Discord.jpg

Such weird wording lol

[silent plank]

mostly similar to what were going through

you're nearly done. it's just 2 small steps away

[upper karma]

Okay! Just give me a second I need to do something then I will be back.

[dense pollen]

In number 3 Sin has a - sign in front of it.. what does this change for me? Trying to find the quadrant it will be in and it has me stuck.. https://gyazo.com/e04ea5aa430267caa5be6ee2f021c387

[dark sparrow]

find sin(-117°) then multiply by -1

@dense pollen

[dense pollen]

So this will be in Quadrant 2 and positive?

@dark sparrow

[dark sparrow]

no, -117° is not in quadrant 2

it's in quadrant 3

[dense pollen]

And because -sin the wave is flipped so usually q3 is negative, now its positive?

@dark sparrow

[dark sparrow]

wording but yes

[dense pollen]

Okay thanks

[dense pollen]

Two ropes support a 78.3 lb crate from above. The tensions in the ropes are 50.6 and 37.5 lbs. What is the angle between the support ropes? Hint: the support ropes are like sides to a triangle where the weight is a side that is straight down... Can someone help me out with this?

[umbral snow]

First step would be to draw a picture

[dense pollen]

Are either of these right?

[umbral snow]

You have three vectors involved here, and since they're not in motion, they must balance out. That is, they sum to (0,0)

No, the two ropes hold the box to a ceiling

[umbral snow]

That works!

As mentioned, those vectors have to sum to the zero vector.

Break them into their components, sum them, solve for unknowns

[dense pollen]

Can I mention this should be solvable using Cosine or Sine law

[umbral snow]

Okay, then put all three tip to tail

They should make a closed loop

[dense pollen]

Then use cosine law

Ffs so simple, need to get better at drawing pictures

[unborn jacinth]

All three of them have a hole/discontinuity at x = 1 right

[umbral snow]

No, not the third

@unborn jacinth

,w graph (x - 1) / (x - 1)^2

[somber coyote]

Display results online and refine query

Requested by Kaynex#7520

[unborn jacinth]

@umbral snow

im confused

ohh wait

cuz there is an asymptote there

so no hole

[umbral snow]

Yup

[unborn jacinth]

so then @umbral snow this wouldn't have any holes either right

[umbral snow]

No it doesn't

[unborn jacinth]

im trolling

yeah

[umbral snow]

Yeah

[unborn jacinth]

ok 😄

[umbral snow]

Trolling

Yup you got it

[unborn jacinth]

are u a math genius

[umbral snow]

No

I just did this for a while

[unborn jacinth]

oh

are u in college(don't have to answer)?

[umbral snow]

I was! Graduated engineering

[unborn jacinth]

oh

nice

[winged halo]

yo quick question if U is an open subset of R^n and the 0th de rham cohomology is H^0(U) = R then does it follow that U must be connected

[fringe dirge]

Uh I'm not sure that this is the right channel for this

[upper karma]

preuni cohomology

[woeful obsidian]

Can someone guide me on how to find the variables in the following? I got a feeling it's super simple and I'm just forgetting something we did:

[sterile mason]

How do I find what -60° is from a scale of 0° to 360° again? sorry I forgot a lot about maths

[silent plank]

since you don't know any of the lengths of the big triangle,
lets start with the smaller 60° triangle first
what would you use to find m?

[woeful obsidian]

No clue

[silent plank]

are you familiar with your trig ratios
soh cah toa?

[sterile mason]

you have a right angle

[woeful obsidian]

I was thinking sin 60 = x /7

For a originally

Wait, I can just use algerbra, 7sin60 =a

Correct?

[silent plank]

not quite,
firstly you used x instead of a
and the ratio of sin(60°) = opp/hyp isn't a/7

which side is opposite to 60°?

[sterile mason]

How do I find what -60° is from a scale of 0° to 360° again? sorry I forgot a lot about maths

can anyone help with this?

[woeful obsidian]

I need to wake up lmao it's 7/a

[silent plank]

yes

[woeful obsidian]

a=sin(60)/7

[silent plank]

nope

[woeful obsidian]

a=7/sin60

[silent plank]

a =

[woeful obsidian]

I'm making the silliest mistakes right now

[silent plank]

are you able to get an exact value for a?

[woeful obsidian]

14sqrt(3)/3

[silent plank]

very good

do you think you can handle the rest?

[woeful obsidian]

Think so, M would be m = 7/tan(60)

[silent plank]

yeh

[woeful obsidian]

Gotta remember the algebra before advanced 😂 I learn these new things, but forget the basics

Thanks man

[upper karma]

I am stuck on question (b)

AED is 75.5 degrees

[scenic trench]

Alright, if AED is 75.5 degrees, what other angle is also 75.5 degreeS?

AD = ED which means that triangle is an isosceles. So 2 angles will be equivalent.

[upper karma]

EAB is 75,6

.5*

@scenic trench

[scenic trench]

Right, so 45 + 75.5 = some number. If you subtract that number from 180, you’d get another angle of the triangle. Plus the other angle near B.

And everything in the triangle adds up to 180.

[novel lintel]

Hello fellas. Any idea how to solve cos(x)=1/5?

I think that x might be expressed in terms of Pi but not quite sure how.

[umbral snow]

,w arccos(1/5)

[somber coyote]

Display results online and refine query

Requested by Kaynex#7520

[umbral snow]

@novel lintel
I'm pretty sure very few values are in terms of π and 1/5 isn't one of them

[rigid sinew]

In a right angle triangle if your using cos for it which side do use for the adjacent side?

[silent plank]

the side that is adjacent to your angle

[rigid sinew]

So does that mean either one?

[wary turtle]

Yes, depending on the angle you're referring to.

[silent plank]

the angle is between 2 sides. the adjacent side (to that angle) refers to the side that isn't the hypotenuse

[brave zephyr]

I’m stuck on this question. Can anyone help me?

[chrome fiber]

read #❓how-to-get-help

[brave zephyr]

Oops, sorry

[brave zephyr]

15 minutes is up

<@&286206848099549185>

[upper karma]

So what happens from f -> f'

Let's just say the triangle is f.

@brave zephyr

[brave zephyr]

It translates.

[upper karma]

You sure?

Are the points ABC moving the same amount?

We're talking about the 1st transformation btw.

[brave zephyr]

Reflects?

[upper karma]

Exactly.

But be more specific.

Reflecting over what?

[brave zephyr]

The x-axis

[upper karma]

Okay so from f' -> f'' what happens?

[brave zephyr]

It reflects over the x-axis.

ohhhh

[upper karma]

No, I mean from the 2nd transformation to the 3rd.

[brave zephyr]

Oh

[upper karma]

From triangle A'B'C' -> A''B''C''

[brave zephyr]

It translates?

[upper karma]

Exactly.

So in the first transformation, it reflects over the x-axis and in the 2nd transformation it translates.

Makes sense?

[brave zephyr]

Yes.

Thank you so much for helping me.

[upper karma]

Np.

[marble parcel]

How would i draw this figure

[normal epoch]

i need help with this

[silent plank]

what have you tried?

[normal epoch]

Im not exactly sure what the question is asking

[silent plank]

find the angle s in the first quadrant (in radians) where
cot(s) = 0.2994

[normal epoch]

and cot would be x/y ?

[silent plank]

ig
what's the relation between tan(s) and cot(s)?

[normal epoch]

y/x and x/y so cot s would be its reciprocal

[silent plank]

so if cot(s) = 0.2994
what is tan(s)?

[normal epoch]

wouldnt it be 3.34?

or

am i wrong

[silent plank]

you shouldn't round until the end
leave it as tan(s) = 1/0.2994

are you able to find s now?

[normal epoch]

oh alright

would i convert that to radians?

[silent plank]

calculators have a radians setting
also do you know how to convert degrees to radians?

[normal epoch]

multiply by pie/180 right?

[silent plank]

pi not pie

[normal epoch]

yeah

[silent plank]

also better to think of tan(x) as sin(x)/cos(x) and cot(x) as cos(x)/sin(x)

what did you get

[normal epoch]

i think i did something wrong

i understand how to convert to radians and degrees

but im confused on tan(s)= 1/.2994

how would i convert it to radians

[silent plank]

well first you need to find s first

(depending on your calculator settings, the value you get may already be in radians)

[normal epoch]

how would i put tan(s)= 1/.2994 into my calculator?

[silent plank]

how would you solve for s?

[normal epoch]

yes

[silent plank]

arctan?

[normal epoch]

tan of angle = 1/.2994 is what its asking right?

so

what angle when you take the tangent of it is equal to 1/.2994

?

[silent plank]

yes

[normal epoch]

and itll be in the range of 0 and pi/2

so the first quadrant

okay i think i understand

im going to try and work it out and ill tell you what i get

thank you sm

[dark sparrow]

π is not "pie"

[silk crown]

it can be if you believe it is

[silent plank]

pie = pi * e

[dark sparrow]

no

god

[silk crown]

e≈π≈2

[weary drift]

found the engineer

[fervent raft]

$\pi \approx e \approx \sqrt{g}$

[somber coyote]

agentnola:

[mighty narwhal]

g?

[weary drift]

,calc sqrt(9.81)

[somber coyote]

Result:

3.1320919526732

[maiden rain]

Can someone explain to me this question.. I know what what symmetry is..

[dark sparrow]

I know what what symmetry is..

was there a typo in there

[graceful hemlock]

@maiden rain in order to determine whether something is symmetrical or not, we would have to find some feature of the graph and its axes to look at. Sometimes it might even involve a simple function where it returns one from an input of one. All you have to do is judge from something that cuts it in half or does something similar (one of the four options) and figure out whether the other side is a mirror reflection of the other.

[maiden rain]

@dark sparrow yes ops

[sinful isle]

How do I do these 2

I need help

[silent plank]

what have you tried?

[sinful isle]

720-all the angles inside the first shape of b

but I don't know where to go from there

[silent plank]

try finding x first

[sinful isle]

ok so for the other shape

uh

180= 6x?

[silent plank]

no.

are you familiar with external angles theorems
or (angle around a point, angle sum of quadrilateral)?

[sinful isle]

oh

360

[upper karma]

I need some assistance

[dark sparrow]

what is holding you up here?

[upper karma]

I don’t know how to solve turning point

[buoyant oasis]

at the turning point dy/dx = 0

you just gotta complete the eq

find b n c

[marble parcel]

Need help proving this

[dark sparrow]

make a picture

[marble parcel]

sorry its hella messy

Wait

E and G are supposed to be switched woops

[upper karma]

there are the answer choices
m∠SQV + m∠VQT = 180°
m∠SQV + m∠SQT = m∠VQT
m∠SQV + m∠SQT = 180°
m∠VQT + m∠ZRS = 180°

I picked the fourth option am i right?

^ thats the figure for the question

[coral sapphire]

Ping helpers after 15 minutes

[upper karma]

oh ok i thought it was 10

[coral sapphire]

Also yeah your answer is correct

[upper karma]

oh ok thx

Which of the following accurately completes the missing statement and justification of the two-column proof?
m∠ABC = m∠CED; Corresponding Angles Theorem
m∠ABC = m∠CED; Alternate Interior Angles Theorem
m∠ABC = m∠BED; Corresponding Angles Theorem
m∠ABC = m∠BED; Alternate Interior Angles Theorem

I am confused on the difference of corresponding angles and alternate interior

[coral sapphire]

What is the difference between those theorems? Or do you not know?

[upper karma]

I do not know

If i know that i am sure i can answer it correctly

[coral sapphire]

Just Google it then, it will do a better job of explaining than me

tldr, corresponding would be DEA and CBA are congruent, while alternate interior would say CED and BCE are comgruent as examplea

[upper karma]

oh ok i will see those examples and will give my best guess

[coral sapphire]

Don't guess and try to learn it

These are important theorems

[upper karma]

k i mean't I will see the two examples u gave and learn it and then check if it was right

I would say third option has the common endpoint so that is right?

common endpoint meaning same vertex

[upper karma]

Which theorem accurately completes Reason A?

Alternate Interior Angles Theorem
Corresponding Angles Theorem
Alternate Exterior Angles Theorem
Same-Side Interior Angles Theorem

[coral sapphire]

this is the same problem as before?

[upper karma]

No now it asks for Reason A not Reason C

I think it is Same-Side Interior angles theorem am i right?

[regal void]

same side interior angles postulate

[upper karma]

but all of them are theorems not postulates @regal void

[regal void]

Oh rip

theorem then

[upper karma]

What is the missing statement for step 5?
m∠SQV = m∠SQT
m∠SQV + m∠VQT = 180°
m∠SQV + m∠SQT = m∠VQT
m∠SQV + m∠SQT = 180°

This one I am totally confused

[regal void]

First one can be ruled out b/c sqv is less than sqt

Third one can be ruled out b/c vqt is less than sqt

Remember that sqt equals 180 degrees

I think using that info, you may be able to deduce the right answer

[upper karma]

k thx so it is the second option

[upper karma]

When constructing an inscribed square, what step comes after both diameters of the circle have been created?

Connect every other intersection of an arc and the circle with a segment.
Connect the intersections of the diameters and the circle with a segment.
Connect every intersection of an arc and the circle with a segment.
Connect the midpoint of the radius to the endpoint of the diameter.

I think it is the second option am i right?

<@&286206848099549185>

[coral sapphire]

15 minutes to ping helpers & technically go to empty #❓how-to-get-help channel for this

[upper karma]

It has been 15 minutes I posted my question at 7:11PM and It is 7:29PM

[coral sapphire]

Says 7:11 & 7:23 on my screen?

[upper karma]

yea 7:23 was what i thought the answer was in my question

only the question counts tho

[chrome bone]

hihi! i hope this is the right channel -- my friend needs help with some homework, and im not the best at math, could someone help with this?

[fast spindle]

Like CA = FD?

[chrome bone]

yep, i believe so ^^;

[fast spindle]

Just turn the grad

I meant you know s, a, h?

Wait

Like sin = soh

Example you can see that BC smaller than AB and EF smaller than ED

[chrome bone]

ah alrighty

[fast spindle]

But actually the answers are above

Lol

[upper karma]

am i right?

I think it is second option because it has the common endpoint/vertex

[thorn linden]

yes

[somber coyote]

A regular kugelblitz:

[balmy pelican]

I've been getting different definitions and now I'm confused

[upper karma]

Hey

Can you solve a problem that I have no idea how to solve?

Please

[olive solar]

[upper karma]

Okay

It'd take me some time to translate the problem into English

I worked hard on the translation so...

[olive solar]

gimme a sec to read it all

[upper karma]

It's an 11th grade problem in Israel

It should be easy, I just have been going in and out of the hospital for the past month, so I haven't got the chance to listen to all the explanations in class

It may combine Geometry and Trigonometry, and may have something to do with circles

[olive solar]

gimme a lil bit of time

definitely been a while since ive done geometry ahaha

[upper karma]

Btw, I just got the idea, maybe each one of the triangles is an equilateral triangle, it depends. Is there a rule that if a geometrical shape is bounded within another, then the bounding circle of the little shape must also be bounded within that big shape? (E.g. the circle bounding the little triangle must also be bounded within the big triangle (because within the big triangle there bounded a little triangle DEF))? Is there any rule like that? Because if there is, I think we can say the the angle between the tangent line of the circle (which is the side of the big triangle) and the chord, is equal to the angle that is opposite to the chord. And that gives us that all triangles are equilateral

I can explain with a pic

(Ofc if there's any theorem like that)

[olive solar]

each one of those triangles is, in fact, an equilateral triangle, i just don't remember how to prove it lol

[upper karma]

I think what I said is a prove (ofc if there's any theorems that says what I wrote in the begging of the message)

Btw

Maybe it would help if you see the original painting(?) Of the problem

[upper karma]

I'm just re-sending the problem if anybody else wanna help

[sharp fern]

find I?

also how are you suppose to find the length unless size of the triangle has been stated?

because idk th scale

unless its like finding A = A+B??

[upper karma]

Find I as in question I

The size of each side of the big triangle is stated... It's a @sharp fern

Yeah it is like that

Look maybe what we learn in Israel isn't the same as the USA

Prolly you learn smth else

Looking at Khan Academy's explanations, it's definitely a way lower level but Idk, is Khan academy covering everything learned in the USA?

[sharp fern]

each side would of DEF would be a/2

[upper karma]

I know

But I need to use α

Idk how to do that

[sharp fern]

law of sines maybe?

wait just normal socahtoa, i dont think u need law of sine/cosine

[upper karma]

I managed to solve it

Thanks for the help

[balmy pelican]

hi

would the superposition of 4sin(x) and 2sin(x) be equal to 2sin(x) or 6sin(x)?

[weary drift]

what do you think?

[balmy pelican]

I think it should be 6sin(x) because there's no displacement, so it would be constructive interference, however I was given this formula: sinα+sinβ=2sin[(α+β)/2]cos[(α−β)/2]

and these are the options I was given

[weary drift]

the two sinusoids are perfectly in phase. your gut feeling is right

[balmy pelican]

ok thanks

what would that formula be used for then?

[weary drift]

$ac+bc=(a+b)c$

[somber coyote]

RokettoJanpu:

[balmy pelican]

yeah, that makes sense

I mean this one though: sinα+sinβ=2sin[(α+β)/2]cos[(α−β)/2]

what's it for?

[weary drift]

left side tells you this is for adding sinusoids of any two phases/frequencies

[balmy pelican]

oh

lol ok

thank you :)

[weary drift]

no problem

[upper karma]

am i right?

<@&286206848099549185>

[mighty gull]

is there something that makes you doubt your answer?

[upper karma]

No but i want to check it cause i am 75 percent sure

@mighty gull so am i right?

[mighty gull]

yes

[upper karma]

oh ok thx

[mighty gull]

prolly be more confident in your answers tho

a big part of math in your level is having the confidence to do stuff

[upper karma]

oh ok

[unborn jacinth]

is -sqrt(x) a one-to-one function

if so, what is it's inverse

because I think it is

but I get x^2 as it's inverse

and the domain and ranges don't match

like swap

so im confused

[upper karma]

you are right

the inverse is x^2

but only the left side of the parabola

because the domains and ranges swap as well

this means that if -sqrt(x)<0 then the domain of the parabola will be less then 0

[unborn jacinth]

oh got it

ok

thanks!

[maiden rain]

What's the difference between: Y= - sin (x) And Y= sin (x)

Graphically

[dark sparrow]

y = f(x) and y = -f(x) are reflections of each other about the y axis

[upper karma]

Isn’t the answer to that reason wrong

Shouldn’t that be transitive

Oh sry for interrupting

[dark sparrow]

technically it should be two applications of the transitive law

[upper karma]

Wait how

[dark sparrow]

just to take the rigidity of this bureaucratic "proof format" to the absurd

oh and you also have to use the symmetric law once

statements 2 and 3

[upper karma]

Isn’t transitive where you take the whole side of two equations and put them together

Our teacher said we don’t need to wry about symmetric

[dark sparrow]

AD = CB (i)
AC+CD = AD (ii)
CD+DB = CB (iii)
from (ii), (i) and the transitive law, derive AC+CD = CB (iv)
from (iii) and the symmetric law, derive CB = CD+DB (v)
from (iv), (v) and the transitive law, derive AC+CD = CD+DB

this is highly formal

bureaucratic even

this obscures the underlying logic, which otherwise is rather obvious

but not much more so than the two col proof format does anyway

[upper karma]

Ok

But how is the step I highlighted not transitive

[dark sparrow]

it's TWO applications of the transitive law.

[upper karma]

Because u already know that ad = can

Why is it two applications

[dark sparrow]

read what i sent above

if you have x=y, y=z and z=t, then the get x=t, FORMALLY you have to first apply the transitive law to x=y and y=z to get x=z, then once again to x=z and z=t to get x=t

[maiden rain]

@dark sparrow U said it's a reflection about the Y-Axis... But in The Book It says "It's a reflection About the X-Axis".

[upper karma]

Ohhhh

[dark sparrow]

did i

[upper karma]

I get it

[dark sparrow]

oops

@maiden rain that was a typo, my bad

[upper karma]

Tysm @dark sparrow

So just to clarify

When you use transitive twice, you get substitution?

[dark sparrow]

what

no?

two applications of the transitive law is two applications of the transitive law

[maiden rain]

@dark sparrow I don't know man. I think the book is wrong.

[dark sparrow]

no, the book is correct

i was the one who typo'd

[maiden rain]

@dark sparrow Shouldn't these two functions be the same when it comes to reflection?

[median crown]

yes

both are over the x axis

[maiden rain]

@median crown But When I see the graph, It actually looks reflected about the Y-axis..

[median crown]

thats cus its odd

[maiden rain]

What does that have to do with anything ? @median crown

[median crown]

thats why it looks like its over y axis

[maiden rain]

So if it's an even function it will be reflected downward ?

[median crown]

over the x axis

its inherent that odd functions will have the same graph if reflected over the x or y axis

y=x is another one that does that

y=x^3 is another example too

if you reflect those functions over the x axis or the y axis, it doesn't really matter

you'll still wind up with the same thing

[maiden rain]

Oh my godness..

[grand shard]

hi

does anyone know how i can get the position of the dot located on the outline of the circle using some sort of an equation? sorry if u cant see i suck at drawing and writing

[median crown]

You might be able to use slope

If you have the radius

[upper karma]

At the bottom it would be -cot^2 as the first step correct?

[umbral snow]

,rotate

[umbral snow]

You want the velocity of the satellite. That is, the distance/time

[median crown]

how is that going to help?

[umbral snow]

Oh fuq I assumed this was a physics question

Literally looked at the picture and assumed it was a "geostationary satellite" question

Mb ignore me plz

[median crown]

idk do some trig to solve it

find the length of the sides of those triangles

like

the half triangles

here's a clue

a^2+b^2=c^2

[dusty coral]

sin(0) = sqrt(0)/2
sin(pi/6) = sqrt(1)/2
sin(pi/4) = sqrt(2)/2
sin(pi/3) = sqrt(3)/2
sin(pi/2) = sqrt(4)/2

[mild grotto]

Nah I figured it out nvm lol sorry

[unborn bison]

@grand shard I think x squared + y squared = r squared

and it can be (x - a number) (y - a number) = r squared if you need the coordinates to get the circle center

google circle equation

[maiden rain]

What's the domain of this function? I know It should be {R} except the zeros of the denominator..

[dark sparrow]

{R}
no

R is not the same thing as {R}

and i mean

yes that's exactly what it is

[maiden rain]

And I know Sin equals zero at 2pi

or pi

but that 1 confuses me

On the denominator

[silent plank]

solve: 1 - sin(x) = 0

[maiden rain]

I did,, sin(x) = 1

[silent plank]

you haven't solved for x yet

[maiden rain]

Sin(+-pi/2, +-3pi/2 etc...) = 1

[silent plank]

nope, sin(-pi/2) and sin(3pi/2) aren't equal to 1

also not a fan of that notation

[maiden rain]

sin(90)=1

[dark sparrow]

,w sin(90 radians)

[maiden rain]

so sin pi/2 =1

[dark sparrow]

,w sin(90 rad)

[somber coyote]

Display results online and refine query

Requested by Ann#0413

[dark sparrow]

there

this is not 1

[silent plank]

yes. sin(pi/2) = 1
x = pi/2 is a solution
how would you write the general solution in terms of k where k is an integer?

[maiden rain]

Wait so when sin is equal to 1 it means that the function is undefined right?

[silent plank]

when sin(x) = 1, yes

[maiden rain]

Ok I know sin(90) is 1 and cos(0) is 1

But Ann says otherwise.

[dark sparrow]

sin(90 radians) is not 1

as i just

demonstrated

to make a point

[silent plank]

90 != 90°

[dark sparrow]

but the point went right over your head

so what do i care

[maiden rain]

Yes 90 degrees

= Pi/2

[silent plank]

without degrees, radians is implied

[maiden rain]

radians

[silent plank]

x = pi/2 is a solution
how would you write the general solution in terms of k where k is an integer?

[maiden rain]

I don't know

[silent plank]

what's the period of the sine function?

[maiden rain]

2pi

[silent plank]

does that give you any ideas?

[maiden rain]

nvm

[silent plank]

do you know what it means to have a period of 2pi?

[maiden rain]

Yeah it means for every 2pi sin = 1

if it was at pi/2

[silent plank]

what would be the next solution?

[maiden rain]

3pi/2

[silent plank]

how are you getting from pi/2 to 3pi/2?

[maiden rain]

multiplied by 3

[silent plank]

why?

[maiden rain]

To get to 3pi/2

[silent plank]

but why did you multiply by 3? what was it supposed to do?

what does 3pi/2 represent

how does it relate to every 2pi?

[maiden rain]

3pi/2 is the other solution that makes sin equal 1

[silent plank]

who do you think multiplying by 3 will get you another solution?

is what i'm trying to ask

you said yourself not so long ago that every 2pi will get you another solution

[maiden rain]

multiplying by 3 = 2pi ?

[silent plank]

no

the sine function repeats every 2pi, so if i added 2pi to pi/2, would that be another solution?

[maiden rain]

I know that every 2pi sin will equal to 1

Yes @silent plank

[silent plank]

ok.
if i added 2pi again, would that be another solution?

[maiden rain]

Of course.

[silent plank]

and so on,
how would you represent all those solutions in terms of an integer k

(adding a multiple of 2pi to pi/2)

[maiden rain]

I get it, but I don't know how I would represent that in a formula.. @silent plank

[silent plank]

if you get it, you should be able to represent it

[maiden rain]

It's like 3+3 = 2*3

[silent plank]

what are the multiples of 2pi in terms of an integer k

don't overthink it

[maiden rain]

4pi?

[silent plank]

where's the k?

[maiden rain]

I can't represent those repeatable solutions n terms of an integer k. I just know that sin(pi/2 *2pi) =1

And it's repeatable

[silent plank]

*****2pi?

the multiples of 2pi can be represented within 4 characters

[maiden rain]

Are you saying that pi/2 * 2pi = 2kpi ?

[silent plank]

no

i'm saying that
2k pi (where k is an integer) represents all the multiples of 2pi

and when you add that to pi/2
i.e pi/2 + 2k pi
will get you all your solutions of x to sin(x) = 1. (the general solution)

[maiden rain]

Wait so you're saying that 2k pi = infinite multiples of 2pi?

[silent plank]

2k pi where k is an integer, represents all the multiples of 2pi

[grand shard]

hello

i need to talk to someone who is good at geometry

[dark sparrow]

[grand shard]

ah

ok

let me just draw real quick

[dark sparrow]

what am i looking at

[grand shard]

i dont know how to explain but

i need to know how to get the position of the outline of a circle

that was very poorly explained but, for an example to get the middle point of this circle you set x to width / 2 and y to height / 2

[dark sparrow]

...that needs way more context

what are x and y

what is width

[grand shard]

width of the circle

height of the circle

[dark sparrow]

...what

[grand shard]

ok fuck

[dark sparrow]

i guess i do

what do you mean "work with x, y"

can you tell me what you're trying to accomplish w/o trying to abstract it, bc it seems you're abstracting too much or doing it improperly, and as a result nobody can understand what you're after

[grand shard]

im bad at english

[dark sparrow]

what's your first language

alright nevermind then

[grand shard]

but thats even hrader

to explain in

[dark sparrow]

ok but... what are you doing

[grand shard]

i will try to make a drawing of what i am trying to achieve

[dark sparrow]

maybe the issue is that you don't have a problem statement in front of you

[grand shard]

ok idk if this is gonna make sense but

this is what i want

[median crown]

you want what you gave yourself?

[grand shard]

wat

no like

i want an explanation of how to get that point on the circumference because i dont know what any of that means

does that even make any sense

[silk crown]

x=a + rcos(t)
y=b + rsin(t)
you have the point then don't you

(x, y) is the point

[grand shard]

im very dumb ok

[silk crown]

I mean

What do you want to know

You have the point

If you want it in a "single formula"
$P = (a + r\cos \theta, b + r\sin \theta)$, where P is the point on the Cartesian plane, a is the x coordinate of the centre of the circle, b is the y coordinate of the centre of the circle, r is the radius of the circle, and $\theta$ is the angle made by the line PO with the +x-axis(O is the centre of the circle)

[somber coyote]

CoolShot:

[grand shard]

waaaaa

[silk crown]

I mean

it's kinda unclear what you want lol

[torpid torrent]

im confused

like

i know auxilliary lines would help but through x i guess?

im pr confused

[weary drift]

sure, extend some lines, apply some theorems, and see if that gets you somewhere

[torpid torrent]

wait so

hear me out

if i make an aux line through x

thats parallel to both ce and bg

and i take the supp of 160 so 20

and thats congruent to one part of x because alternate interior

and then 120 vertical to that angle across from it

and then

so 80?

[weary drift]

very naisu

[pure mason]

Applying trig identities problem?

I'm not good at math and they use different terms here in Canada

[umbral snow]

You should have a formula for sin(A + B)

Just let A be θ and B be π/6 @upper karma

[grave notch]

Hi ! I'm trying to solve an interesting problem I have in a space game. The game doesn't give access to the coordinate system. All I know is the distance between "here" and any well known waypoint. So once I get somewhere which is not a waypoint, I triangulate with the distance from this place to other waypoints. But when I need to get back to it, it's another problem...

I began to solve it in 2D

D is the objective, and A / B / C are known waypoints I can get to easily

the path I tried to solve is A,E,D : get to A easily, aim C and get closer to C until distance to C = EC (to solve), then aim B and get closer to B until the remaining distance is DB (DA, DB, DC are known)

I used the Al-Kashi theorem

so I'm (slowly) getting somewhere, but this way to pose the problem is not compatible with what I'm looking for

because in the game, it's 3D... so to extrapolate the problem, I guess I have to use 4 distances to waypoints

like if the objective was somewhere inside a pyramid

and knowing all 4 distances to A,B,C,D I guess it could be possible to easily go to A, aim B and advance until x, aim C and advance until y, aim D and advance until z ; or somthing like that...

but it's far above my knowledge to solve it

[grave notch]

ok I think I'm on to something... because the 1st part of the problem in 3D seems to be the same problem in 2D, with an objective that matches the intersection between D and the ABC plane

formulas to compute intersection surely exist for 3D

[rigid wharf]

Figure ABCD is a rhombus, and m∠BAE = 9x + 2 and m∠BAD = 130°. Solve for x.

Rhombus ABCD with diagonals AC and BD and point E as the point of intersection of the diagonals.

3.4
7
14.2
65

Ngl I have not a clue on this one

[tight mural]

@rigid wharf Remind me, what is that m notation?

It’s been years since I’ve done geometry haha

[rigid wharf]

m notation?

The m in front of the angle notation refers to the measure of the angle labeled A, B and C (with vertex at B). By definition, the term congruent means "having equal length or measure".
This is what it says online

[tight mural]

So angle at A is 130?

If you’re taking that whole triangle

[rigid wharf]

BAD=130

[tight mural]

Ok gotcha

So it’s a rhombus

So what’s BAE

Knowing BAD is 130?

[rigid wharf]

9x+2

[tight mural]

There’s a number

Without needing the equation

BAE is half of BAD

Since it’s symmetrical

[rigid wharf]

65?

[tight mural]

Yes

So now 9x+2 = 65

Solve for x

It comes out to a clean number

[rigid wharf]

Wait I thought X was 65

[tight mural]

Brb

[rigid wharf]

Oh wait its 7 nvm thanks

Do you know this one by any chance?

A land surveyor places two stakes 500 ft apart. He locates the midpoint between the two stakes and creates a perpendicular to the line that connects these two stakes. He needs to place a third stake 100 ft away along this perpendicular line. To apply the Perpendicular Bisector Theorem, the land surveyor would need to identify

the location of the third stake as equidistant from the original two stakes
the location of the third stake as closer to one of the original two stakes
a line parallel to the line connecting the two stakes
a line congruent to the line connecting the two stakes

I think its either A or B but im not 100% sure

[median crown]

Lol

[rigid wharf]

Bruh

[median crown]

Draw it

Have u tried that?

[rigid wharf]

Ya

Imma just stick with A

[median crown]

Then look at the answers and answer

[rigid wharf]

The wording man

If they made it just not so damn complicated it would be a breeze

[median crown]

It’s literally in regular English

Not even in geometric terms

Here let’s go through them

[rigid wharf]

okay

[median crown]

Will the guy need the stakes to be equidistant?

Look at your drawing

[rigid wharf]

Yes?

[median crown]

Will the stakes not be equidistant?

[rigid wharf]

The first 2 will be but the 3rd wont

[median crown]

Wdym

The first two have to be equidistant

And the third is equidistant

[rigid wharf]

Oh

[median crown]

you have all 3 points no?

[rigid wharf]

Yea

[median crown]

Draw a triangle

then cut that triangle in half, where the height of the triangle is that perpendicular line

[rigid wharf]

Wait ya it is A

[median crown]

the resulting two triangles are congruent

By SAS

therefore the hypotenuse of those triangles is the same

Hence they are equidistant

[rigid wharf]

Alright, lit

Thank you

[flint pelican]

Guys for this one sal did tan(195)=tan(225-30)

But I did tan(195)=tan(135+60)

And got 1+sqrt3 / 1-sqrt 3

But that isn't an option and he got 3- sqrt3 / 3+sqrt3

[gritty sail]

they are the same things

really

this problem did not write it very well

but if u simplify by multiplying by conjugate you get the same thing

[vestal oak]

You could do 150 and 45

Tan((180-30) + 45)

[median crown]

You might be able to do tan(195+0)

[marble parcel]

Can someone help me with this proof

[waxen gorge]

Angles

Sides

[buoyant spruce]

Hello so I need some help with this question

[upper karma]

@marble parcel you can prove line segments da and be are congruent making triangle acb an iscoseles triangle and prove angle daf and ebf are congruent proving angles Fab and fba are congruent making the triangle iscoseles

It's up to you to fill in the blanks

Maybe you solved it by now

Also @buoyant spruce it's the answers c

[maiden rain]

What is this question ? It doesn't make any sense,, I know cos 2(x)= cos^2 - sin^2

And I don't see that answer there.

[dark sparrow]

cos(2x) = cos^2(x) - sin^2(x) sure but you there are other ways you can write that

for example,

what if you replace cos^2(x) with 1 - sin^2(x)

also, it's cos(2x), not "cos 2(x)"

[maiden rain]

Then it will be 1-2sin^2x

[dark sparrow]

exactly

$\cos(2x) = \cos^2(x) - \sin^2(x) = 2\cos^2(x) - 1 = 1 - 2\sin^2(x)$

[somber coyote]

Ann:

[upper karma]

What for function would you graph this with?

[gritty sail]

2 is right

3 is j any other line that touches the line MP

[dark sparrow]

@upper karma what?

[balmy pelican]

hey quick q: is A in Asin(Bx+C)+D always the amplitude? or does that only apply to arguments of sin() that are in radians?

[gritty sail]

Always amplitude

[weary drift]

@balmy pelican doesn't matter if the arg is in deg or rad

[balmy pelican]

hmmm that's kinda weird then

I converted the argument of a sin to degrees and its amplitude changed

[weary drift]

show us

[balmy pelican]

oh it's inverted

well you get the idea

[weary drift]

put only the arg into degrees(), not the entire sin() function

[balmy pelican]

so like this?

[weary drift]

👍🏽

[balmy pelican]

weird thing

thanks

[weary drift]

np

[wind heart]

Is 1-5 correct?

[wind heart]

<@&286206848099549185>

[proven plover]

cant read it

[wind heart]

Oh

I’ll zoom it in

.

[proven plover]

still cant read the first picture

[wind heart]

Oh it’s in pencil that’s why

I’ll trace it

There we go

[proven plover]

check #6 and do 7

you didnt do it

[wind heart]

Oh I know, I just want to know if the first five are correct

[proven plover]

looks fine to me

[wind heart]

Ok cool thanks

[flint pelican]

There

I’m confused on which ones should be negative I used 45 instead of 135

And both the sin and cos of 45 are positive sqrt 2/2

And the Sin of 300 is negative while cos is positive

So I got sqrt2- sqrt6 /4

Could someone help thank you

[proven plover]

maybe you forgot about the negative?

-sqrt 2 *-sqrt 3 = + sqrt 6

the numbers seem right

[flint pelican]

Oh was I supposed to distribute the negative

Ok

[unborn jacinth]

@dark sparrow so 2^x vs 2^-x

its not necessarily a reflection over the y axis right

its just a reflection over the x value that makes it 1

[weary drift]

sure, reflect about x=3

[unborn jacinth]

ok

[marble parcel]

Can someone please help

[twin prawn]

Tf are congruent bisectors

[glad shadow]

If you divide cos of a number with the cos of another, different, number, can you get rid of both cos and be lett standing with the first number divided by the other?

[silent plank]

that would be a big NO