geometry-and-trigonometry

[signal hornet]

2. I can find theta

which one should I do?

[gritty siren]

there's no r in this, there's no need to introduce one

[signal hornet]

Ok, I wil go from theta then

[frosty egret]

@signal hornet you dont need to find theta

just use trig identities

such as cos^2 + sin^2 = 1

[signal hornet]

ahh

thanks

[signal hornet]

Why did they pick cos^2x-sin^2x for the cos double angle?

why introduce sine?

[dark sparrow]

honestly idk i'd have gone with 2cos^2(2θ) - 1

and wouldn't be surprised if that was the next step

[signal hornet]

It is the next step

[dark sparrow]

yeah, you can go straight from $\cos(4\theta)$ to $2\cos^2(2\theta) - 1$

[somber coyote]

Ann:

[signal hornet]

I'm going to use the other double angle func

Yeah, that's what I got right now

[dark sparrow]

there's a wholly different way to solve this which i find more elegant

[signal hornet]

I did one method ealier

[dark sparrow]

using the sum-to-product formula $$\cos(u) + \cos(v) = 2 \cos\paren{\frac{u+v}{2}} \cos\paren{\frac{u-v}{2}}$$

[signal hornet]

It took me 15 minutes

and I only got 2/6 solutions

[somber coyote]

Ann:

[signal hornet]

Never heard of that identity, Never been taught it

[dark sparrow]

weird bc it's useful sometimes

[signal hornet]

Yeah, I imagine there are a ton

also we only lecture 2hrs a week

kinda screwed by design

[novel flax]

help

<@&286206848099549185>

[vivid wraith]

can you find the degree measure of angle(BAD) and angle(BCE)?

also the measure of angle(ABC)

[novel flax]

108 degrees

help

@mellow flicker

[vivid wraith]

how about angles(BAD) and (BCE)?

if you dont know, try searching up angles on regular pentagons

[novel flax]

is it 90

[vivid wraith]

its 72

for both

[novel flax]

ok thanks i got it

[vivid wraith]

welcom

[icy knoll]

How would I go about finding these? I'm a little stuck. The lesson did not help

[upper karma]

what do you know about the circle?

just from the pic

[icy knoll]

it's comprised of four 45-45-90 triangles

oh well, that's for the square oops

[upper karma]

id say its easier to work with it being comprised of 2 45 45 90 triangles but i guess that works too

how might you know one of the lengths of those?

[icy knoll]

1/2 bh? by counting the notches on the x and y axis?

[upper karma]

the side lengths are directly related to the radius/diameter of the circle in some way

what is the circle radius?

[icy knoll]

oh like (pi)r squared?

[upper karma]

no that would be to find the circle area

what is the radius?

[icy knoll]

radius is 3?

[upper karma]

yes

[icy knoll]

so radius=3 and diameter is 6 so far

uh

[upper karma]

you dont need the diameter but you could use that instead

and do it another way

but lets roll with radius

[icy knoll]

okay

[upper karma]

dont you know the lengths of these purple lines?

if you know the circle radius

[icy knoll]

well both are 3

[upper karma]

yes

so cant you find the hypotenuse? its a 45 45 90 triangle

and what is the hypotenuse of that triangle, in terms of what it is on the square

[icy knoll]

so wouldn't it be like leg times square root 2?

[upper karma]

yes

do you see how that answers question 1?

[icy knoll]

yeah I can see that, but what's confusing me is the square as a whole. Like I understand the triangle but I for some reason can't wrap my head around finding the whole square

[upper karma]

what do you mean? finding the perimeter and area?

[icy knoll]

yeah, I think it's the graph. I don't know why I'm having trouble. So I just use 3 square root 2 for the hyp and then plug in?

sorry

[upper karma]

the square is pretty independent of the circle at this point

for finding the area and perimeter

this is all you have to worry about

[icy knoll]

oh i get it now

cause perimeter is just 4 times a side and area is bh

[vague pagoda]

Yes

Now u can calc the ratio when u find out the answer

And other stuff that u need

[icy knoll]

okay i see it now. and is it a similar process for the octagon below?

[vague pagoda]

Yeah ig

[icy knoll]

Okay thank you guys. Sorry for making it difficult, I was having initial trouble understanding it.

[upper karma]

how are you going to do the octagon one btw? the best way i see isnt exactly a similar process

[icy knoll]

split a side into a triangle, if that works

[upper karma]

split how?

[icy knoll]

draw a radius from a corner to the center of O

treat the y axis as an apothem ?

[upper karma]

https://i.imgur.com/T0Ss0ts.png like this and solve for c?

[icy knoll]

yes

that's what i was thinking

[upper karma]

how do you plan on solving for c?

[icy knoll]

well, the length of both legs should be 3 right?

[upper karma]

yes i believe so

but its not a 45 45 90 triangle, is it?

[icy knoll]

probably bisect it

so you have two 45 45 90s?

[upper karma]

i think that gives you one 45 45 90 if you do that but not much info on c

well you can try to work it out

lemme know what you get

[icy knoll]

well, I tried adjacent/hypotenuse which would be cos90=x/3 or 3 times cos90=0 but that's obv not the answer, so i'm a little stuck

[upper karma]

im not sure where you got that

[icy knoll]

cause I know the length of the hypotenuse which is 3 and an angle of 90 degrees when i split it

unless I do 3cos60 which is 1.5

60 degrees being the angle adjacent to the hypotenuse and base

[upper karma]

how can you both split something and still have a hypotenuse of 3 somewhere in this problem

[icy knoll]

that original triangle I drew a bisector through, and i got a right triangle with the Y axis acting as the hypotenuse, which has a length of 3

[upper karma]

i cant tell what you mean

[icy knoll]

Like this

[upper karma]

what are you going to do with it now?

[icy knoll]

I tried using the adjacent over hypotenuse and got 1.5 for the supposed base at C, which means the whole base is 3?

[rich carbon]

Can’t figure out how to evaluate this.

I tried finding a reference angle with no success

[upper karma]

tan and tan^-1 are inverse processes

[rich carbon]

Yeah

[upper karma]

so what does that suggest?

i dont think the whole base is going to be 3

[rich carbon]

Well $f(f^{-1}(x))$ is not equal to $x$ in this case.

[somber coyote]

Matt_:

[upper karma]

hm these work funny when the values are not in the usually accepted domain of arctan

[rich carbon]

Yea exactly.

So how can I find a coterminal angle to 27pi/8?

[upper karma]

actually

[rich carbon]

That is in the range of arc tan

[upper karma]

the domain of arctan is all real numbers anyway

^Ye.

[rich carbon]

So

[upper karma]

they just work funny outside of pi/2

[rich carbon]

Exactly

So I can’t figure it out

[upper karma]

Huh.

Uh-

ok i think you can

subtract 3pi

[rich carbon]

Ahhhh

[upper karma]

or any other multiple of pi (the period of arctan)

[rich carbon]

Ohhh

[upper karma]

Yeah I think that'll work great.

[rich carbon]

So now I have 3pi/8 which is in the range

[upper karma]

Wait does arctan have a period?

Aren't inverse trigonometric functions like, non-periodic or some shit?

[rich carbon]

Yeah

They’re not

[upper karma]

arcsin and arccos dont

[rich carbon]

I’ll figure it out thanks for the help though

[upper karma]

but arctan kinda does

[rich carbon]

Meh

It’s arbitrary

[upper karma]

reducing it to 3pi/8 works for some reason

not exactly sure why tho

Oh wait.

It makes sense because tan is periodic.

arctan is the non-periodic one.

We just need to find a coterminal angle of 27pi/8 in the domain of (-pi/2;pi/2) and get it (I think??)

Because that's the restricted domain of tan used for arctan if I recall correctly.

in a graphing calc
arctan(tan(27pi/8)) = arctan(tan(3pi/8) = 3pi/8

Yeah.

Makes sense.

[rich carbon]

Ya. Got it now

[upper karma]

Good luck with the rest mate!

[signal hornet]

I need help figuring out which sign to give to a and b

I found the domain for theta/2, and I think it is relevant when trying to find the sign of a and b. I just don't see how

[dire rampart]

a and B?

[signal hornet]

yes

labeled on the left

problem a and problem b

I got +- sqrt(5)/5 and +- 2

But each answer only has one sign

[dire rampart]

Oh right

So u know the range of theta

What's the range of theta/2

[signal hornet]

you mean domain?

I mean range is [-1, 1]

[dire rampart]

Eh range is a more suitable word here but sure

No that's the range of sin/cos

Anyway

What values can theta/2 take

[signal hornet]

pi/4 to pi/2

[dire rampart]

Right

So what quadrant is it in

[signal hornet]

I

[dire rampart]

So what does that tell u about the cos and tan of theta/2

[signal hornet]

also positive

[dire rampart]

There u go then

[signal hornet]

But what if theta/2 was from pi/4 to 3pi/4 then how would I figure it out

In my case I got lucky because the domain of theta/2 falls entirely inside of quadrant 1

[dire rampart]

Then you would need slightly more info

[signal hornet]

Ok

As long as I am not missing some

thanks

[signal hornet]

How come WA is saying only 7pi/4?

https://www.wolframalpha.com/input/?i=sin(theta)-cos(theta)%3D-sqrt(2)%2C+0<%3Dtheta<2pi

[signal hornet]

how does this make sense?

sin of 3pi/2 is -1

[vivid wraith]

its sin(3pi/4)

[fringe dirge]

@upper karma Please stop this nonsense

[signal hornet]

@vivid wraith that's what I typed

[vivid wraith]

your message says "sin of 3pi/2 is -1"

[signal hornet]

well 2theta is 3pi/2

so theta is 3pi/4

[signal hornet]

<@&286206848099549185>

[upper karma]

what is the questions

[signal hornet]

why is 3pi/4 not a solution

https://cdn.discordapp.com/attachments/326138757474680852/655921519574253588/unknown.png

@upper karma any ideas?

[upper karma]

ok lets see here

thats because when you squared the sqrt(2) you got rid of the negative

so you found an extraneous solution

imagine that the statement was sin(theta)-cos(theta)=sqrt(2)

using your method you wouldve got the same solutions

[signal hornet]

I follow

[upper karma]

so 3pi/4 will make the equation equal sqrt(2)

if you plug it in

not negative sqrt(2)

[signal hornet]

so how do I know which of the solutions is right, without using the calc?

[upper karma]

well it is possible to find the sin of 3pi/4

its on the unit circle

or at least it makes a special right triangle

sin(3pi/4) is sqrt(2)/2

this might help

pi/4 is 45 degrees

oh thats small

[signal hornet]

am I supposed to look only at the right triangle?

[upper karma]

no

[signal hornet]

I got a unit circle, don't worry

[upper karma]

ok so that means that you should know what the sin of 3pi/4 is

and cos of 3pi/4

if you subtract the two, you get positive square root of 2

[signal hornet]

oh

I see now

any way to solve this problem realistically whithout squaring?

ALthough, I think we are supposed ot square

[dark sparrow]

what was the original

sin(θ) - cos(θ) = -sqrt(2)?

[signal hornet]

yes

[dark sparrow]

$\sin(\theta) - \cos(\theta) = \sqrt{2} \sin\paren{\theta - \frac\pi4}$

[somber coyote]

Ann:

[dark sparrow]

¯_(ツ)_/¯

special case of what i believe is sometimes called the "R formula"

if you want a search term

[signal hornet]

this one @dark sparrow https://brilliant.org/wiki/trigonometric-r-method/ forgive the brilliant link

[dark sparrow]

yes precisely this

[signal hornet]

I will bookmark it and look at it again after the final

I don't want to confuse myself

@dark sparrow thanks I appreciate insight into these weird formulas

one day I might use them

[onyx cypress]

Could someone walk me through this? Having a lot of trouble

Am I gonna use the values on the trig circle? or leave it as sin(pi) and tan(pi)?

[dark sparrow]

no and no

well ok

i guess you CAN make use of the values of sin(π) and tan(π)

and in fact you SHOULD because those aren't the kind of values that cannot be simplified

[onyx cypress]

Ok I'll use the exact values then

Like this?

[dark sparrow]

no

[onyx cypress]

ok

I tried again but got up to this

[signal hornet]

can't you immediately re-write tan(pi + x) as tan(x)?

[onyx cypress]

Yeah. Then I rewrote it as (sin(x))/(cos(x))

[silent plank]

check your identity for sin(pi-x)

@onyx cypress

[onyx cypress]

Like sin(pi-x) = sin(pi)cos(x) - cos(pi)sin(x)?

[silent plank]

you can go that route if you want

[onyx cypress]

Is it also sin(pi-x) = -sin(x)?

[silent plank]

no

[onyx cypress]

Ok i better go the other route

[silent plank]

do both

the result should be the same

[onyx cypress]

What's the other way if sin(pi-x) = -sin(x) is not good?

[silent plank]

well first show me your work for
sin(pi-x) = sin(pi)cos(x) - cos(pi)sin(x)

[onyx cypress]

ok

[silent plank]

sin(pi) = ?
cos(pi) = ?

[silent plank]

that's better

you can apply
sin(pi-x) = sin(x) and
tan(pi + x) = tan(x)
directly

[onyx cypress]

Hahaha yeah only now do I see it after working it out

[silent plank]

from the properties of the functions, periods etc...

they should also be in you list of identities

some weird stuff also happened around your tan(x)

[silent plank]

check you angle Identity for tan

(you made 2 errors that cancelled each other out)

[onyx cypress]

was it the negatives?

[silent plank]

tan(a**+**b) = ?

[onyx cypress]

this here?

[silent plank]

yeh. there are negatives that should be there

[onyx cypress]

Oh yeah

on the tan(x)

How is this?

[silent plank]

, rotate

[silent plank]

it's a bit unclear,
are you doing
-tan(pi + x) OR tan(pi + x) in the top right?

[onyx cypress]

oh that's -tan(pi+x)

[silent plank]

I can see the - sign there
also why not just do tan(pi+x) and then distribute the - after

[onyx cypress]

Oh I didn't think about that

Do both versions work or do I have to distribute the - after

[silent plank]

if everything is mathematically correct, you will get the same result

tan(pi+x)=tan(x)
-tan(pi+x)=-tan(x)

[onyx cypress]

Okay

Other than that does everything make mathematical sense?

Am I allowed to multiply both sides by cosx to get rid of it in the denominator

[silent plank]

not a fan of manipulating both sides like that,
get a common denominator,
subtract the fractions to get the rhs

[onyx cypress]

I'm not sure how to get a common denominator on the LHS

[silent plank]

what's the denominator of sin(x)/cos(x)

[onyx cypress]

Cos(x)

[silent plank]

what do you need to do to the sin(x) term to get a denominator of cos(x)?

(while keeping its value the same)

[onyx cypress]

Multiply by 1/cosx?

[silent plank]

no.

multiplying by that will change the value

[onyx cypress]

Yeah that don't really make sense

[silent plank]

eg what would you do if it was
1/2 + 1/3

[onyx cypress]

Like multiply the 2 by 3 and the 3 by 2?

[silent plank]

be more specific

and show the work to get the result

[onyx cypress]

So the common denom of 1 and cosx

[silent plank]

👌

[onyx cypress]

Thank you very much for your help

[signal hornet]

why am I getting a different answer

I even added pi to get the angle into quadrant 3

[silent plank]

calc is in radians

-1.93478 + 2pi ~ 4.34

[signal hornet]

so I am right, calc is wrong?

[silent plank]

both are alright

[signal hornet]

Ok

[silent plank]

depends a bit on how they want the angle

[signal hornet]

didn't say

I'm just using this to convert

I wish they had drawn parantheses though

thanks you ramonov

[upper karma]

How would you do this question without sine rule or cosine rule?

An aeroplane is flying horizontally directly towards a city at an altitude of 400 metres. At a given time the pilot views the lights of the city at an angle of depression of 1.5 degrees. Two minutes later the angle of depression of the city lights is 5 degrees. Find the speed of the aeroplane in km/hr correct to one decimal.

My textbook included this in a section with angles of elevation and depression, although when I searched this question online, all of the solutions included the sine rule

[quiet mason]

why not use elevation and depression

[upper karma]

Oh no I've tried using the angle of depression and couldn't figure it out, that's why I went online to search it up

[quiet mason]

what didnt you understand

either post what the book did

or what you couldnt understand

[upper karma]

The textbook doesn't show any solution besides the actual answer. I understand what the question is asking, however the diagram which I'm setting out seems to lead me to the wrong answer

[quiet mason]

show

post

[upper karma]

Of how I drew the diagram?

[quiet mason]

yea

[upper karma]

6.5 combining the 1.5 and 5 degrees together

I know I've probably done it wrong, however I'm trying to figure out how this problem can be done without the sine rule

[quiet mason]

see whats an angle of depression

[upper karma]

The angle between the horizontal and another line that's formed when you look down

[silent plank]

are you able to use that and redraw the diagram?

[upper karma]

Use what?

[silent plank]

that definition

if you stil aren't completely sure, post what you have after drawing
the plane is at an elevation of 400m
the city is at an angle of depression of 1.5°

[silent plank]

can you label where the plane and the city are?

[silent plank]

The angle between the horizontal and another line that's formed when you look down

the line drawn from P to the left is your horizontal line right?

[upper karma]

Yes

Ah now I see

[silent plank]

yes that's better

and then. you travel a certain distance towards the city and will have a new angle

[silent plank]

no

you are travelling. AWAY from the city in that diagram

you are further from C than when you started

[upper karma]

How would I draw it then

oh wait

[silent plank]

in your diagram, the plane flies to the left

[silent plank]

yeh

[upper karma]

What do I do after this?

[silent plank]

find the distance the plane travelled. tan will be the most useful here

[upper karma]

I've tried but got nothing near the answer

[silent plank]

show your work

[upper karma]

400/tan1.5+400/tan5

[silent plank]

why are you adding?

[upper karma]

Originally I thought it was 400/tan1.5 - 400/tan5 but I think that's incorrect too

[silent plank]

why did you think that would be incorrect?

[upper karma]

Because the answer to that doesnt come close

[silent plank]

also remember this is the distance travelled in 2 minutes

and they want speed in km/hr

[upper karma]

So change the answer to km and times by 30?

[silent plank]

yes

[upper karma]

Ah got it, thanks a lot. I don't really want to come back here that much for answers for questions so would there be any advice you'd be able to give on trig in terms of the more visual stuff like this

Because trig really isn't my thing and I'm just trying to improve

[silent plank]

learn definitions / how to apply them,
read what the question is asking for

you'll get the hang of it. a lot of people find word problems a pain

[upper karma]

Aight, thanks for the help. I'll just have to attempt more of these harder kinds of examples.

[silk crown]

how can I represent (cos(t), cos(t)) as an equation in x and y

wait no

[dire rampart]

[silk crown]

(cot(t), cos(t))

[rocky niche]

What's cot(t)

[silk crown]

cotangent

[dire rampart]

1/tangent

[quiet mason]

black hole escape velocity

no just his name

2GM/c^2

[dire rampart]

its not vel

units dont match

[weak shoal]

Lmao that’s the schwarzchild radius

Anyways, the newtonian derivation is wrong

[dire rampart]

@silk crown x=cot(t) y=cos(t)

[weak shoal]

You gotta go to GR for the correct proof

[silk crown]

yes

how do I write that in a single equation

as in, x=sec(t) y=tan(t)
-> x²-y²=1

Can I do something similar here

[dark sparrow]

cot^2(t) + 1 = 1/sin^2(t)

so 1 - 1/(cot^2(t) + 1) = cos^2(t)

[silk crown]

thx

[rocky niche]

My name is the radius of a black hole

[weary drift]

i'd rather you leave the asterisk out, schwarzschild

[rocky niche]

Also know as schwarzchild radius

Like that?

[weary drift]

better

[rocky niche]

Btw i was playing around with isoceles triangles and I came to the conclusion that with only one side or angle ou could find any other side or angle which is obsiously false.
What i did was that i have 2Θ + Θ' = π => Θ'=π-2Θ
And then just apply the generalised pythagorean formula:
c²=r²+r²-2r²cos(π-2Θ)
r²=r²+c²-2rc*cos(Θ)

Which means that if you do an system of equations and know either Θ,r or c you can calculate everything which obsiously fasle :/

[weak shoal]

[rocky niche]

I know im a bit lost too xD

[visual mist]

So what do u need help with

[silent plank]

did you actually try solving that system of equations?

[rocky niche]

I tried putting it in the calculator but it just gave me lots of different values

Sushi i am trying to find my error

[silent plank]

solving those two equations will get you true statements

they pretty much represent the same thing and won't provide any new information

eg.
c - 2r(cos(Θ)) = 0 → r = c/(2cos(Θ))
c^2 = 2r^2 (1 + cos(2Θ))
c^2 = 2c^2/(4cos^2(Θ)) * 2cos^2(Θ)
c^2 = c^2

[rocky niche]

Yep that was what i was thinking because theses equations come from the same formula

[silent plank]

also never heard this referred to as the "generalised pythagorean formula"
this is an application of the cosine rule

[rocky niche]

I just didnt know the name of the rule xD

I am really bad at remembering stuff ):

[silent plank]

its a rule that only involves the cosine trig function
chances are: its the cosine rule 😄

[silk crown]

cosecant=cossecant=1

[quiet mason]

Wow

[idle bloom]

It's an isoceles triangle so if you know one angle you know the other 2 immediately

[brisk spruce]

How am I supposed to do this scale factor crap

Does it matter if I do AC/DE or DE/AC

[silk crown]

what

what scale factor crap are we talking about

[deft ingot]

Similar triangles?

Then yes, they matter

[brisk spruce]

So how do I know which one goes in the num and which one in the demon

Demoniator

[silk crown]

@deft ingot how did you figure that out

[deft ingot]

It is always A to B

You cant switch things up for no reason

Triangle ABC ~ Triangle XYZ <=> AB/XY = BC/YZ = AC/XZ

[brisk spruce]

So it matters if I put AB/DE and AC/DF or DE/AB and DF/AC?

Like i have to commit to a solid pattern right

[deft ingot]

Yes

[brisk spruce]

Like I can say DF/AC but I have to commit that format throughout?

[deft ingot]

Yes

[brisk spruce]

Ok but like it doesn’t matter which way I start off with right

[deft ingot]

Yeah

[brisk spruce]

Okie tyvm

[deft ingot]

You can just take the inverse in all the equalities

It is the same thing

[brisk spruce]

Ah ty

[winged narwhal]

So I have to form a Sinusoidal Equation given a max of 5pi/2, 14 and minimum of 15pi/2,6

I have Y=4sin(Bx) + 10

[upper karma]

Ngl trig makes me bawl

But I still need to learn it

@brisk spruce

I think we have the same textbook

[winged narwhal]

<@&286206848099549185>

[upper karma]

in this case the only thing we need to do is find the period

[winged narwhal]

Yea
I got the 2pi over B
but idk what to do after that

[upper karma]

we know that the max and min of a sinusoid are separated by 1/2 the period

that can be seen on a unit circle

since the distance from 1 to -1 is pi radians

the distance between the max and min here is 5pi

so the period is 10pi

2pi/10pi is 1/5

and if you plug these numbers in, you will get your answers

of course this is only 1 of the infinite solutions though

because the distance from 1 to -1 could also be 3pi radians

since 3pi = pi

so it could 3/2 of the period

and then the period would be 2/3 of 5 multiplied by pi

you could keep going

as a general rule, the period is 10pi/(1+2n)

where n is a natural number

but this is more then the question is asking for

[winged narwhal]

so wait
I literally just count 5pi to 15 pi
or is it more complicated than that and that's just coincidence

[upper karma]

well not really

but the outcome is the same

you just need to understand that the max and min are seperated by pi radians on the unit circle

and that it would be 1/2 of the period to get from the maximum to the minimum

Can I ask something?

yeah

is it related to this question?

Kinda
So if a pendulum were swinging from one end to the other, and the distance were measured to the vertical

Would the total distance be halved or quartered?

[winged narwhal]

I could've just doubled 5pi, since max to max is a period
im dummy head

So final answer would be Y= 4sin (10pi) +10
How would I put θ in there though

[upper karma]

no that wouldn't be final answer

you said it yourself

B=2pi/T

and T = 10pi

[winged narwhal]

so 2pi over old period = new period

[upper karma]

not the new period

just the coefficient

[winged narwhal]

ok
so Y= 4sin (1/5θ) +10

[upper karma]

yes

the period is 10pi

If this were the case and I had to graph it

Would the amplitude be 50, or 25 if we had to measure the distance to the vertical

Because the distance between the two is the same

But the function wouldn’t tell you on which side the sandbag is hanging?

[winged narwhal]

Amp is just |Max-min|/2 no?

[upper karma]

Yes

But

I’m confused if maybe negative distance were distance towards b

[winged narwhal]

Right side is positive

I remember this problem

[upper karma]

But what does that mean?

[winged narwhal]

So the dash is your base y- axis right?

[upper karma]

No

The dash is a vertical line

And the point of the function is to measure the distance from the sandbag to the vertical line

[winged narwhal]

Then it is your y-axis
left of that is negative, and right, positive

[upper karma]

But it’s not negative

Because you can’t have negative distance

But the person who is sit next to, disagrees

And says it’s negative

But I don’t understand

Why

[winged narwhal]

Ok, so on paper its negative
But just count the absolute value of it

so it counts as distance

the negative just means moving left

[upper karma]

Ok

?

[winged narwhal]

-3 = Move left 3

[upper karma]

Ohhhhh

Alright, I get it

Thank you

[dusty coral]

shit ass drawing

but whats the equation of this circle touching

wait nvm

[signal wyvern]

5x - 3y - 5 - 3 = 0
3x + 5y - 8 = 0

how ?? is there a mistake there? dosent make sense to me

[weak shoal]

Uhh yea that last line doesn't follow from the 2nd last line

I think they just mixed up the coefficients

[dark sparrow]

comic sans for math formulas?

[twin prawn]

finna write my phd thesis in comic sans

[signal hornet]

I need help, I don't think I did this one right

[signal hornet]

<@&286206848099549185>

[dire rampart]

y is r sintheta not just sin theta

[hybrid jasper]

why do we use -cos

[signal hornet]

@hybrid jasper because otherwise the hieght would be -52, and Height can't be negative

[hybrid jasper]

and for b where it says t = 4 sec wouldn't it be on top of the circle

at 90 degrees

[glad timber]

Can I get some help with this

[solar hearth]

Draw line hg such that hg is perpendicular to eg

[wind heart]

Once the other persons question is andwered, can someone make sue the circled problems are correct and if not explain?

[glad timber]

I’m not that far into geometry

I’m in a unit with similar polygons and ratios

[solar hearth]

Just a sec gotta eat ill explain later

[silent plank]

why isn't SSS listed as an option for congruency?

[solar hearth]

Wut?

😂

@glad timber triangle fhg is similar to triangle egh

[solar hearth]

Edumecation

[glad timber]

Thanks @solar hearth

[solar hearth]

Do you know how to solve it now?

Wait I'm retarded

[glad timber]

Wait they aren’t similar

They only have 1 congruent angle

[solar hearth]

They are you just wrote it in the paper

And I didn't see

[glad timber]

I know but I had that wrong

[solar hearth]

They share angle fgh

[glad timber]

That’s it

[silent plank]

pic1 no:8 is a bit ambiguous without seeing the whole question (since there are 3 triangles, two of which are congruent)
pic2: no:8 was left blank
pic3: no:22 what do you know about the two angles at the centre?

[glad timber]

Oh nevermind

Your right lol

[solar hearth]

Yeah you still stuck?

[glad timber]

Haven’t started it yet

I am now

[wind heart]

@silent plank is 8) none?

It’s like

SSS AAS SAS ASA HL or none

For each problem

[solar hearth]

Ok

So which side is shared by both triangles

[silent plank]

which 8 are you referring to?

[wind heart]

Number 8 for the first picture

[silent plank]

as mentioned, its ambiguous what they want since there are more than 2 triangles there

but two of them would be congruent by SSS

[solar hearth]

Since you have the two triangles find the ratio between to two then apply that to a side you know and eh to find the answer

[wind heart]

Oh ok

Would 13 be HL?

[silent plank]

i'm assuming HL is the same as RHS, then yes

[wind heart]

And would 23 be HL?

[silent plank]

yeh

[wind heart]

Hmmmm

2 more questions

22 & 30

Did I do those correctly

For 22 I got none because of 3 congruent sides and congruent set of angles

[silent plank]

22: no,
30: yes

[wind heart]

Because like

Idk if it can have 4 things

[silent plank]

well from that, you can already prove congruency two ways

since you are clearly already given 3 sides, then just state SSS
if SSS wasn't an option then, apply vert-angles and use SAS

either should be fine

[wind heart]

Oh

I seee

Thanks for the help

[solar hearth]

Hg/10 = 6/hg may be helpful

[wind heart]

Oh wait one more question oof

Is that correct

[silent plank]

back to what i stated a few posts ago,
is SSS being omitted as one of the options?

[wind heart]

I mean

It’s not if there’s 3 congruent sides

[silent plank]

what's not?

[upper karma]

how are you getting 8 for square

[silent plank]

i'm assuming these are triangles formed from connecting the diagonals?

8 for square seems alright.

how are you getting 8 for rhombus?

the diagonals of the rhombus aren't necessarily equal

well the diagonals do bisect each other

yeh

(so i believe you miscounted those 4 small triangles)

[upper karma]

this question seems not well defined, are there certain rules to follow?

i think you could make infinitely many isosceles triangles

in a square

w, 4.74 * sqrt 55

,w 4.74*sqrt 55

[somber coyote]

Display results online and refine query

Requested by Acash#3166

[upper karma]

why not 32

what are the restrictions on forming the triangles?

[silent plank]

it was already established that

triangles formed from connecting the diagonals

[upper karma]

o i see now

didnt notice the yes reply to that

[silent plank]

are you able to find the probability of 0 right triangles?

[dire obsidian]

Can someone help w 47

That’s the correct answer for reference

This is what I’ve tried

[upper karma]

@dire obsidian you can find the dimensions of a rectangle that surrounds the path

and its mostly just right triangle geometry to solve for the rectangle and its diagonal

[dire obsidian]

@upper karma uh what

It’s a plane

Triangle

[upper karma]

yes

ok one sec lemme draw it

[dire obsidian]

lol okay

[upper karma]

https://i.imgur.com/6JDeND1.png

cant you solve for the width and height of this rectangle with this much information

and a diagonal from the bottom left to the top right would be the distance, if you could solve for it

[dire obsidian]

Holy shat

I think this is what they’re looking for

[upper karma]

well yes thats exactly what i drew but with a rectangle around it

[dire obsidian]

Could you try doing it

Im@gonna try again

But if you get it, I’m gonna need to compare

[upper karma]

yes i just worked it out and the answer was correct

[median crown]

u sure that's 12 degrees?

[dire obsidian]

not drawn to scale 😖

[median crown]

there isn't even a label on it

its just 12

that could be 12 radians

12 feet

idk

[upper karma]

excuse my sharex drawing

[dire obsidian]

@upper karma uh yeah im@not getting it

This would be the correct equation set up right

[median crown]

,rotate

[upper karma]

i dont have law of cosines memorized but yea you can probably do that

https://i.imgur.com/gLjMcJx.png but this is what i did

[dire obsidian]

I feel like I messed up somewhere in my calculator

[upper karma]

and solve for x and y with trig

then use the pythagorean theorem to solve for the green diagonal

[upper karma]

again i unfortunately do not have law of cosines memorized so i dont know if it was set up correctly

[dire obsidian]

alright thank you for helping tho!

[upper karma]

my way comes out to the right answer though

just use trig for x and y and throw them onto the second drawing

[dire obsidian]

O ik what’s wrong

I keep plugging in for cos12

But it’s cos168

[upper karma]

👍

what i did was probably how law of cosines works

[upper karma]

Hey guys

In 3rd picture

If i were to tilt the intersection sp that it is no longer perpendicular to base

Will the intersection points still make up hyperbole?

[dark sparrow]

yes

hyperbola*

[upper karma]

Thanks

[silk crown]

it wouldn't be of the form x²/a² - y²/b² = 1 after that would it though

[polar sable]

About euclidian geometry
Let's consider ABC a right triangle as shown here (using geogebra)

Let's now move C up in the line it belongs to.

Also, this line is perpendicular to (AB) so it can't move

If we move C very far up, its angle starts decreasing till it reaches 0°

Which means that the angle A reaches 90°

So we have got a triangle with two right triangles.

Not even sure if we can call this a triangle and I don't know what is happening here. It's impossible to visualize since the angle C is equal to 0

Can anybody clarify this?

[dark sparrow]

the further up AB you move your point C, the closer the angle at A will be to 90°

but it will never actually be exactly 90°

[polar sable]

oh

is there an actual proof for this? @dark sparrow

[dark sparrow]

angle ACB will always be greater than 0 because at no point do A, B and C become collinear

[devout vine]

<@&286206848099549185> help pls i have no idea and i need this

Median is basically bisector right

So ae is congruent to ad

Nvm

[dire rampart]

[brave zephyr]

Can someone explain to me what Postulate is and how you would use it?

[weak shoal]

[rigid siren]

how do you find the zeroes of a piecewise function

[silent plank]

find the zeroes of each of the pieces

[silk crown]

@brave zephyr you should probably search the word up in a dictionary

[solar hearth]

Parts b) and c)

[silk crown]

Have you made a diagram

[solar hearth]

I have its kinda messy

[supple kraken]

how can i get the height of a triangle with 3 angles and 1 side

[dire rampart]

What did u try

[supple kraken]

well i know this may be wrong but I tried to do (300cos(31)) - (300cos(23))

and subbed in the difference like this

19sin31 = 9.8 metres

@dire rampart sorry for the ping, im in a hurry cuz class ends in 6 mins

[dire rampart]

Uhh

Not sure why u did that

[supple kraken]

was just an attempt

i knew it was wromg

[dire rampart]

But essentially u wanna find equations for the base of both triangles in terms of h

Add them up

And set them equal to 300

[supple kraken]

how would I find the base of the triangles

[dire rampart]

Well they're both right triangles

That fact will help you

[upper karma]

What does it mean for a figure to map to itself?

[civic basin]

SOH CAH TOA only works for right triangles right?

[upper karma]

<@&286206848099549185> ?

[weak shoal]

A figure to map to itself? What do you mean? 'Map' has a very specific meaning

@civic basin I mean, that's just a way to remember the geometric definitions of sines and cosines. But yes, you'd have to draw a right triangle to visualize what soh cah toa is referring to.

[civic basin]

so non right triangle has no hypotenuse ? that info only applies to right triangle?

[weak shoal]

I mean, you could draw any random triangle and draw extra lines such that it consists of right-triangles. Then, you could define a 'hypotenuse' for each of those right triangles. The hypotenuse, really, is just a label given to the longest side of the right triangle.

[civic basin]

ok understood

thank u

[upper karma]

@weak shoal I don't know?

Say with a reflection, a figure maps to itself?

[weak shoal]

Surely there must have been some motivation for your question?

[upper karma]

@weak shoal If a figure maps to itself under a reflection over a certain line, that line is called a line of symmetry of the figure.

[white cradle]

domain = range ig

[dark sparrow]

[upper karma]

@dark sparrow ?

[dark sparrow]

nvm

uh

no, i'm gonna keep my thonk. i don't know what you mean. but you should probably give some context

where did you see talk of "figures mapping to themselves" exactly

[upper karma]

That's the formal definition of a line of symmetry of some figure.

My book.

That's the full sentence.

[dark sparrow]

can you show the sentence

[upper karma]

@dark sparrow

[dark sparrow]

ok so

[white cradle]

so if you define a mapping to reflect points about a line, and it maps all the points in the domain to the domain set itself

[dark sparrow]

what that means is that if you reflect your shape across that line, you get the same thing as the original

[white cradle]

that's what I got from it^

[upper karma]

@dark sparrow I know what that means. I'm asking about the mapping?

[dark sparrow]

the mapping would be reflection

[white cradle]

you can get an expression for reflection by some simple 2d coordinate geo

[dark sparrow]

i.e. it's a function defined on the plane which sends every point to its reflection across your axis.

[upper karma]

But what does "figure maps to itself" mean?

[dark sparrow]

it means that if you apply the reflection map to your shape (i.e. to its every point), you get the same thing as you started with.

[upper karma]

Ignore reflections for a moment

What does "figure maps to itself" mean?

[white cradle]

domain = range imo

[dark sparrow]

no

[white cradle]

why not

[dark sparrow]

"the figure maps to itself" means that if you apply your map (be it a reflection or something else) to your shape (i.e. to its every point), you get the same thing as you started with.

[white cradle]

isn't that the same

as the range set being the same as the domain

[dark sparrow]

the domain is the entire plane, flynn

[white cradle]

not if you're considering a specific figure

like a triangle or smtn

[median crown]

[white cradle]

in which case youre only reflecting the points in that figure

[dark sparrow]

you're considering the image of a figure under a map defined on the whole plane.

[white cradle]

oh kk

[upper karma]

@dark sparrow But do you get the same thing? Like the shape is congruent to the original one

But the location is completely different

[white cradle]

no

it needs to be the exact figure

like reflecting about the diagonal of the square

same figure, same location

[dark sparrow]

it needs to be that exact figure in the exact same location in order for the line to be a line of symmetry.

[upper karma]

How can something be reflected and be in the exact same location?

Then it wouldn't be reflected?

[weak shoal]

It can

If it's on the line of symmetry, it can undergo a reflection and be in the exact same location

[upper karma]

I'm talking about the image of the figure that's now reflected

So if figure A is reflected over l, then let it's reflection be A'

A' is not in the same location as A

[white cradle]

then that line isn't a line of symmetry

that's all

[upper karma]

???

https://i.imgur.com/3efKv2s.png

Do you think these two figures are in the exact same place?

[white cradle]

nope

hence the line of reflection is not a line of symmetry

[solar hearth]

nope

:smart:

[upper karma]

What do you mean that's not a line of symmetry???

@dark sparrow Can you clarify what he is saying?

[white cradle]

it's only a line of symmetry if after reflection you get the same figure at the same location with the same orientation

as if you've done nothing at all to the figure

[dark sparrow]

if your figure is the union of these two triangles

then that line is a line of symmetry for it

[white cradle]

for example draw a line perpendicular to the hypotenuse passing through the third vertex

now reflect it

[dark sparrow]

if it's only one of those triangles

then it is not a line of symmetry

[upper karma]

Yes, it's one of those triangles

Let the triangle on the left be A

Let the triangle on the right be A'

How is the line between A and A' not the line of symmetry?

Because FlynnXD is saying it's not.

[dark sparrow]

the line is not a line of symmetry for A nor a line of symmetry for A'

reflecting A across it does not give you A back

[white cradle]

listen to this

for example draw a line perpendicular to the hypotenuse passing through the third vertex

[upper karma]

@dark sparrow Can you reflect A and show how it would be?

[white cradle]

now reflect it

[upper karma]

In my example?

[dark sparrow]

reflecting A gives A'

[white cradle]

^

[upper karma]

...

Isn't that what is illustrated?

[dark sparrow]

...

yes??

[white cradle]

what even

[dark sparrow]

what's the problem

[upper karma]

???

[white cradle]

is your question

[upper karma]

The problem is

FlynnXD : hence the line of reflection is not a line of symmetry

[white cradle]

yes it isn't

[upper karma]

(Talking about the two triangles)

Lol

[dark sparrow]

yes??? it's not a line of symmetry for either one of those triangles??? what's the problem???

[white cradle]

you're taking both triangles together as a single figure???

or just one triangle seperately as a figure

[upper karma]

So then draw a line of symmetry?

[white cradle]

I fucking told you, you're not reading my messages

for example draw a line perpendicular to the hypotenuse passing through the third vertex

[upper karma]

But what if I want to reflect them back to back

Like in my illustration?

[white cradle]

ok by all means go ahead

[upper karma]

I did

[white cradle]

but its not a line of symmetry then

[upper karma]

You are saying it's wrong

[upper karma]

Why is it not symmetric???

[dark sparrow]

it's not perfect but this is a line of symmetry for A

[white cradle]

LINE OF SYMMETRY DOESN'T MEAN IT'S SYMMETRIC

[dark sparrow]

best i can do with my finger on a phone screen

[white cradle]

It means you get the same figure

the EXACT SAME FIGURE

you dont get two different figures like this

[upper karma]

How are they different?

[white cradle]

after reflection it seems as if you haven't changed the figure AT ALL

only then that line is a line of symmetry

here it's obvious you've made a reflection since the figure literally translates to the other side of the line

[upper karma]

https://i.imgur.com/3efKv2s.png

So the line in the middle is not a line of symmetry?

[white cradle]

nope

by definition it simply is not

[upper karma]

How can I make a line of symmetry?

[white cradle]

that line cannot be a line of symmetry

draw the line Ann drew

[dark sparrow]

i just drew you one

[white cradle]

it's the only line of symmetry that triangle has

[upper karma]

Yes, but that's through the hypotenuse

[white cradle]

yes

[dark sparrow]

SO WHAT

[upper karma]

I want it to be one of the sides?

[dark sparrow]

NO

THAT'S NOT HOW IT WORKS

[white cradle]

you can't just draw a random line and MAKE IT a line of symmetry

you're not understanding what this means lmao

[upper karma]

Then explain

[dark sparrow]

LINES OF SYMMETRY ARE THINGS INHERENT TO THE SHAPE

[white cradle]

@upper karma what do you think a line of symmetry is?

[upper karma]

That both figures are the same on both sides of the line

Simply put

[white cradle]

no

that's not what it is

[upper karma]

Ok

[white cradle]

after reflection the figure must remain the same place, same location, same orientation

Read what I'm saying now closely

AFTER REFLECTION, it must seem as if you've actually done NOTHING to the figure

that is when it's a line of reflection

[upper karma]

So then this is not correct, right?

[white cradle]

also not

yeah

[upper karma]

It's from the book

[white cradle]

not a line of symmetry

[dark sparrow]

define correct

k is not a line of symmetry for ABC

[upper karma]

A'B'C' is the image of ABC upon reflecting over k

[white cradle]

yes it is the image

[weak shoal]

Okay, there's no reason for you to call someone a donkey

[white cradle]

THE LINE IS NOT A LINE OF SYMMETRY

[weak shoal]

Especially when they're attempting to help you

[dark sparrow]

because the reflection of ABC across it, which is A'B'C', DOES NOT COINCIDE WITH ABC.

[white cradle]

and don't you fucking blame your inability to explain your doubt on me, alright?

[dark sparrow]

the reflection of ABC across k, which is A'B'C', DOES NOT COINCIDE WITH ABC.

[upper karma]

I'm so confused now

What's k then?

In that image

[dark sparrow]

it's a line!

[white cradle]

it's just some line of reflection

[dark sparrow]

that happens NOT TO BE A LINE OF SYMMETRY FOR ABC!!!

[upper karma]

Ok so in

https://i.imgur.com/3efKv2s.png

The middle line is a reflection of A

?

[dark sparrow]

NO

[upper karma]

..

[white cradle]

the right triangle is a reflection of the left triangle

the line in the middle is the LINE of reflection

[dark sparrow]

A' is the reflection of A across that line

[white cradle]

that is all

[upper karma]

Ok

So when they say

If a figure maps to itself under a reflection over a certain line

They mean that for example

[white cradle]

which

[upper karma]

https://i.imgur.com/HkoyF6c.png

[white cradle]

yes

💯

[upper karma]

So

https://i.imgur.com/eHjzk5l.png

Highlighted points are the mapped to themselves?

[dark sparrow]

no they're mapped to each other

[white cradle]

the top red point is mapped to the bottom one and the bottom one maps to the top

[dark sparrow]

not to themselves

[white cradle]

they just map to some other point lying on the same figure

they don't map to themselves

[dark sparrow]

if every point is mapped to itself, you're not doing a reflection at all

[upper karma]

Ok

[dark sparrow]

every point is mapped to itself only when your map is the identity map

aka the do-nothing map

[white cradle]

yeah you'd get that if you reflect twice

[upper karma]

What is the identity map?

Like f(x) = x?

[white cradle]

the original state of your figure

f(x, y) =(x, y)

yeah

[upper karma]

Ok, thanks @white cradle @dark sparrow

[white cradle]

maybe next time don't resort to name calling people because you're unable to understand..

[upper karma]

If 4 congruent semicircles are connected only at then end of them and from A to B the diameter is 100 what is the distance from A to B by only traveling on the semicircles?

[shy cape]

Am I tthe only person that hates proofs?

[upper karma]

Probably not but they are kinda cool.

[shy cape]

I really hate them 😦

Like I guess I can do them but I dislike them a lot. I got a 93 on my geometry midterm weird flex but okay

[upper karma]

I just took a final and that was one of the problems above. I got 50 pi