geometry-and-trigonometry

[viscid ginkgo]

so it's -sin

ok ok there

[slate bay]

When I do sin(361/2 degrees), I get -0.008...

[viscid ginkgo]

lmao

it's -(sinA/2) here

ok

[upper karma]

you never calculate with anything like "361"

you always reduce it between 0 - 360

361 deg = 1 deg

[viscid ginkgo]

wtf

[upper karma]

if you subtract 360 from 361, you get 1

[viscid ginkgo]

but (361/2) is 180.5

[upper karma]

180.5 is negative sin, yes

[viscid ginkgo]

so it's both?

that doesn't make sense

and if not 361 then uhh

360?

[upper karma]

https://cdn.discordapp.com/attachments/326138757474680852/715342766112112640/unknown.png

see where "sin" is?

[viscid ginkgo]

no 360 is positive

[upper karma]

True or false: if sinA is positive, then sin A/2 is positive as well.
@viscid ginkgo

[slate bay]

you always reduce it between 0 - 360

I don't see why this is true.

[upper karma]

because you end up in the same spot

[slate bay]

No, I get that. But if A = 361, then A/2 = 180.5.

[upper karma]

@viscid ginkgo "if sin(A) is positive"

not "if angle A is positive"

sin(180.5) is negative

[slate bay]

I thought A had no restrictions?

[upper karma]

for it to be positive, it needs to be in this top half: https://cdn.discordapp.com/attachments/326138757474680852/715344919920771084/unknown.png

[slate bay]

Sorry, lemme check original statement.

[upper karma]

sin(A) "positive" means top half

[viscid ginkgo]

ok so if 361 = 1 degrees...

[slate bay]

That is where I disagree.

[viscid ginkgo]

hmm

[slate bay]

Is that a definition?

I mean, yes, sin(A) positive means top half.

However, I don't agree that you have to have angle A be from 0 to 180 for sin(A) to be positive.

That's my problem.

Like, if we just forget the unit circle and just treat sin(A) as a power series, for example.

[viscid ginkgo]

I don't get it

[upper karma]

180.5 is not in the top half

[slate bay]

180.5 would be A/2.

Not A.

[viscid ginkgo]

true

[slate bay]

Also, I said I don't agree that 0 < A < 180 for sin(A) > 0.

[viscid ginkgo]

sin(1) in Q1

sin A/2 in Q3

[upper karma]

why Q3?

[viscid ginkgo]

180.5 Q3

could maybe do Q4

but in this case, Q3

[slate bay]

Ok, let me go back to the beginning, just so we can start on the same page and remain on the same page.

[upper karma]

180.5, yes, Q3

but not A/2

[slate bay]

True or false: if sin(A) > 0, then sin(A / 2) > 0.

This is the problem statement, right?

[upper karma]

if sin(A) = positive value, yes

[slate bay]

Ok, so you qualified your answer.

Does that mean there's something wrong with the way I stated the problem?

[upper karma]

sin(A) needs to equal a positive number

sin(180.5) does not

[slate bay]

Let's not jump there yet.

sin(A) needs to equal a positive number.

This is equivalent to sin(A) > 0, correct?

[upper karma]

yes

[slate bay]

Ok.

So,

True or false: if sin(A) > 0, then sin(A / 2) > 0.

This is the problem, correct?

[upper karma]

more precisely, sin(A) = "something" > 0

ok

[slate bay]

Why is it necessary to insert "something"?

[upper karma]

sin(A) needs to equal a positive value (value greater than 0)

[slate bay]

What is the difference between sin(A) > 0 and sin(A) = "something" > 0?

sin(A) is a value, is it not?

[upper karma]

yes

[slate bay]

I don't see a need to say value = "something" > 0.

I feel it's unnecessary.

However, you insist on it, so there must be a reason behind it.

[upper karma]

ok, "sin(A) > 0"

[slate bay]

If there is a difference in the understanding of the problem, we need to iron it out as early as possible.

This is why I insist on making sure we're on the same page.

[upper karma]

True or false: if sinA is positive, then sin A/2 is positive as well.

[slate bay]

Yes, that's the original problem statement.

And I asked if it's equivalent to how I stated it or not.

If it is not, then I need to know why in order to correct any errors I make.

[upper karma]

True or false: if sin(A) > 0, then sin(A / 2) > 0.
@slate bay

it's equivalent, yes

[slate bay]

Ok.

I am just going to say I have a habit of misreading things.

So, although it may seem trivial, this is in fact a very important step.

True or false: if sin(A) > 0, then sin(A / 2) > 0.
So, I propose a counterexample.

Let A = 361 degrees.

The first quantity that we should agree on is sin(A).

,w sin(361 degrees)

[somber coyote]

Display results online and refine query

Requested by Red Herring#8090

[slate bay]

Ok, so #1, sin(A) is about 0.017.

Agree or disagree?

[upper karma]

agreed

I see where this is going

[slate bay]

Yes, but let's not jump to the future.

We need to make sure we go in sequence.

[upper karma]

they don't specify a range domain

[slate bay]

A range for what? As in there is no domain restriction?

If so, then yes, they do not, and that's what my counterexample hinges on.

Anyways, #2, the next quantity to agree on is A/2. A/2 = 361/2 degrees = 180.5 degrees.

If we agree on #2, then we would need to agree on a value of sin(A/2). I say, #3, that sin(A/2) = sin(180.5 degrees).

,w sin(180.5 degrees)

[somber coyote]

Display results online and refine query

Requested by Red Herring#8090

[slate bay]

Then, we see that sin(A) > 0, but sin(A/2) < 0, concluding the counterexample.

[oak shale]

x^2+y^2+6x-24=8y

already asked about it before, but i didn't see the response until later and it was too late for me to ask a question about some stuff I didnt understand.

Anyone know how to convert the equation into a standard circle equation?

[versed river]

this channel is taken i think, maybe go #precalculus or #❓how-to-get-help

[slate bay]

@viscid ginkgo If the counterexample above is without flaw, then you can conclude that the statement is indeed false.

However, don't take my word for it.

[blazing panther]

Can someone help with 3

[eternal crag]

what have you tried?

[blazing panther]

I have no idea

What to do

I just have been thinking

[eternal crag]

are you familiar with the idea of soh-cah-toa?

[blazing panther]

yes

[eternal crag]

right, use exactly that

[blazing panther]

um

i dont see how that would work

for this

[eternal crag]

let's look at option (1)

tan35 = 8/x

[devout harbor]

publius is looking at the wrong question just like I did

😛

[eternal crag]

oh fuck.

[blazing panther]

um

what

o

[eternal crag]

nvm

[devout harbor]

$\sin2x = \cos(3x-10)$

[eternal crag]

i read the wrong question lmao

[somber coyote]

Sup?:

[eternal crag]

sorry

[blazing panther]

o

[regal shell]

Do u know sin2x formula?

[blazing panther]

no

[regal shell]

2sinxcosx

And then treat cos(3x-10) like cos(a-b) maybe

[blazing panther]

i am in regents geometry

well

im not

but my friend is

[regal shell]

What’s regents geometry

[eternal crag]

a test from new york

eh why are you bring that up?

[blazing panther]

because

i dont think that was used in geometry

at least

regents

[eternal crag]

,w sin(2x) = cos(3x-10) exact solution

[blazing panther]

what

[somber coyote]

Display results online and refine query

Requested by Publius#2314

[eternal crag]

i mean,

[blazing panther]

o

[eternal crag]

it's gonna be a nightmare

[blazing panther]

jesus

what the hell

oh my god

[eternal crag]

oh that's like 10% of the solution

i can scroll way further

[blazing panther]

BRUH

[eternal crag]

if this is a regent question i suppose they want to approximate the solution with a calculator

[blazing panther]

um

hold up

lemme just

what the hell

this is a painful question

[eternal crag]

well these two solutions seems somewhat reasonable

[blazing panther]

um

i dont know how this got onto a regents geometry

problem

[eternal crag]

it's probably a typo or some sort

unless regent up their game this year

[blazing panther]

no way

imma tell him to

ask wut happened wit dat

[pearl lava]

does determining if the orientation of a figure will change depend on the vertices or no?

[blazing panther]

Also

Can sum help with this

[silent plank]

let the height be h
and |AC| be x.
apply the appropriate trig function for the two triangles and solve the system.

[blazing panther]

wait so

should i do law of sines

[silent plank]

that could work, but consider using something simpler

[blazing panther]

ok

let me think

but here is my problem

if only i could find the middle line

in between my two triangles

but i cant because i only have that one side

without a right angle in that triangle

[silent plank]

find the length of that line isn't needed

consider that the height (h) is common in both triangles

and try to express in two different ways

[blazing panther]

hmm

how is h common in both truangles

wait

what do you mean by h

[silent plank]

both right triangles

the height of the tree

[blazing panther]

ok i got tan(43)=h/x

And

tan(32)=h/(x+100)

[silent plank]

parentheses

[blazing panther]

sorry

[silent plank]

isolate h in both equations

and it should be quite intuitive from there

[blazing panther]

Thank you so much

[silent plank]

note that there is a condensed formula, and this is a method to derive it

[supple gazelle]

Are you guys allowed to help on projects?

I only saw rules about helping on test being forhibited. Projects are okay, correct?

[rich wolf]

sure

if you have a specific question about how to do something

[viscid ginkgo]

my friends

the end is nigh

https://i.imgur.com/bC73EnV.png

[upper karma]

That L isn't capitalized.

What a heathen.

Also doesn't bother to say PM or AM.

But that's alright ig.

[rich wolf]

Maybe it's a FINAI exam

[viscid ginkgo]

XD

I keep getting confused about the fraction rule

symbolab is tripping me out

a/b/c = a/bc

a/b/c = ac/b

what's up with that?

a/b/c = a/b * 1/c

and a/b/c = a/1 * c/b

I'm confused which way to properly work with it

well

ac/b is one result

the one which is confusing me a lot

is

b/c/a = b/ca

[upper karma]

?

[somber coyote]

AASlavoj:

[upper karma]

@viscid ginkgo

I hope this makes sense

https://www.youtube.com/results?search_query=reciprocal+fractions

[azure reef]

nah

the issue is that parentheses are omitted

but a/b/c should be read as a/(bc) since we go from left to right

[onyx beacon]

someone please help

[upper karma]

You went for the RHS didnt ya? @viscid ginkgo

[viscid ginkgo]

I solved it

but thank you

[upper karma]

Nw lol

[viscid ginkgo]

I'm stuck

[obsidian ravine]

where?

[viscid ginkgo]

sorry

it needs to be = ([sqrt(6) + sqrt(2)]) / 4

[silent plank]

consider: $2 + \sqrt{3} = \frac12 \times (4+2\sqrt{3})$

[somber coyote]

ramonov:

[viscid ginkgo]

how?

[silent plank]

i factored out 1/2

from 2 + sqrt(3)

[viscid ginkgo]

ok

so now we have an 8 in our denominator?

[silent plank]

the idea behind it is to generate the 2\sqrt{3} term which may make it easier to apply binomial theorem to calculate square roots

[viscid ginkgo]

alright soo

we have

sqrt(4+2(sqrt(3)) / 2 )

[silent plank]

$=\sqrt { \frac{4 + 2\sqrt{3}}{8}}$

[somber coyote]

ramonov:

[viscid ginkgo]

yes

[silent plank]

now also consider that 4 = 1 + sqrt(3)^2

[viscid ginkgo]

yes

umm

^2

^2

I mean

yeah that's rtight

[silent plank]

$=\sqrt { \frac{1^2 + 2\sqrt{3} + (\sqrt{3})^2}{8}}$

[somber coyote]

ramonov:

[silent plank]

can you continue from here?

[viscid ginkgo]

ill try

I factored out 1/4

=sqrt(2+ sqrt(3))/4

-_-

alright what's confusing me is how fractions work

like I don't know what to factor out first, you know?

[silent plank]

ok. do you understand how we got to this point?

[viscid ginkgo]

yes

[silent plank]

applying the property of square roots, we can split it since both the numerator and denominator are positive

$=\frac{\sqrt{1^2 + 2\sqrt{3} + (\sqrt{3})^2} }{\sqrt{8}}$

1 sec

[somber coyote]

ramonov:

[silent plank]

you should be able to simplify the denominator relatively easily

[viscid ginkgo]

yes

[silent plank]

and the radicand in the numerator is in the form: a^2 + 2ab + b^2

(where a=1 and b=sqrt(3))

[viscid ginkgo]

(1 + 2sqrt3 + sqrt(3)^2 /sqrt 2) / 2

[silent plank]

missing the sqaure root on the numerator

$=\frac{\sqrt{1^2 + 2\sqrt{3} + (\sqrt{3})^2} }{2\sqrt{2}}$

[somber coyote]

ramonov:

[silent plank]

and consider what i said and try and simplify it

[viscid ginkgo]

sqrt((1 - sqrt(3)) / 2sqrt(2)

oops

(1-sqrt(3))^2

(1-sqrt(3)) / 2sqrt(2)

[silent plank]

should be + not -

[viscid ginkgo]

rea;;y

o.o

ok

[silent plank]

yeh, the middle term is positive

[viscid ginkgo]

(1+sqrt(3))/2sqrt2

[silent plank]

parentheses around the denominator

when writing in plain text

(1+sqrt(3))/(2sqrt(2))

[viscid ginkgo]

ok

[silent plank]

do you know how to rationalise denominators?

$=\frac{1+\sqrt{3}}{2\sqrt{2}}$

[somber coyote]

ramonov:

[viscid ginkgo]

so I see

that was really complex

idk why the book doesn't explain stuff like u did

ty so much

[silent plank]

were you able to reach the simplfied answer from here?

[viscid ginkgo]

yes

I appreciate your help

[merry smelt]

Is cos^-1 the same as the cosecant? If not, how do I calculate the cosecant with the hypotenuse and opposite sides? I know the cosecant is found with hypotenuse over opposite side, and I know if I had cosine it would be cos(adj/hyp), but my scientific calculator only has sin, cos, tan, and sin^-1, cos^-1, and tan^-1

[dark sparrow]

no, cos^-1 is not the same as csc.

[merry smelt]

Okay then how do I calculate that

[dark sparrow]

csc(x) = 1/sin(x)

[merry smelt]

Ooooo okay ty

[dark sparrow]

also sec(x) = 1/cos(x), in case you need that too

[merry smelt]

So if I want to find csc(17/8), would that be 1/sin(8/17) (since 8 is opposite and 17 is hypotenuse) or 1/sin(17/8)?

(I don’t know theta, just the sides)

And ty

This is the problem with my work so far

@dark sparrow

[dark sparrow]

uh

"csc(17/8)"

what

[merry smelt]

Did I overthink it

[dark sparrow]

you

well

yeah you kinda did because you are not taking the cosecant of an angle measuring 2.125 degrees

or radians for that matter

csc(θ) = 17/8

that's it really

[merry smelt]

Oh I’m stupid

Yeah I put those in as if they were theta 🙈

[merry smelt]

@dark sparrow You said csc(x)=1/sin(x) and sec(x)=1/cos(x). What is cot(x)?

[silent plank]

1/tan(x)

[upper karma]

wait csc is 1/sin(x)? always thought that that was 1/cos(x) and sec(x) = 1/cos(x)

[wooden current]

@upper karma keep it family friendly

[upper karma]

lol oops

didnt even notice that

[wooden current]

😆

[red isle]

@upper karma cscx=1/sinx and secx=1/cosx.

[upper karma]

wait csc is 1/sin(x)? always thought that that was 1/cos(x) and sec(x) = 1/cos(x)
@upper karma why would you have 2 terms for the same thing

[upper karma]

dude wtf is up with my grammar today

[rich wolf]

Yeah for real

[upper karma]

i meant to say i thought sec(x) = 1/sin(x)

[shrewd edge]

is this legit?

[rich wolf]

@shrewd edge wdym

[upper karma]

is this true?

[shrewd edge]

@rich wolf did I do my work right 😅

[shrewd edge]

@upper karma https://www.geogebra.org/classic/rgegcugv

I'm shit at geogebra/geometry in general, so my construction lines are probably horrid, but yep

@upper karma did you prove this? how?

@upper karma here's a gif

[upper karma]

phew

hold on, i'll show you how i proved it

so, a full rotation has 360 degrees

and if you draw 3 lines from the center of the circle to where the circle is touching, you'll have 3 equal angles

so you have 120 degrees on each

and then you connect the lines

since it's an isosceles triangle, the other two angles will be equal, so it's 120 + 2x = 180, and x is 30°

solving for the missing side, you get $r\sqrt{3}$

[somber coyote]

Mofumofu:

[upper karma]

so $\frac{x}{2}$ is equal to $r\sqrt{3}$ and then $x = 2r\sqrt{3}$

[somber coyote]

Mofumofu:

[silent plank]

don't use x to denote both the side and the angle

[upper karma]

oops

forgot that i had used x for the angle

imagine it's y then lmao

i use x for small problems within a larger problem

[silent plank]

that seems like a complicated approach but it works.
consider the right triangle formed from

[upper karma]

that's how i did it 🤷‍♀️

[silent plank]

the centre, a vertex and an appropriate point of tangency

[upper karma]

uhh

i don't know much about geometry

i guess i just know the basic things

[silent plank]

this is less complicated than what you did

[upper karma]

i don't even know what a vertex or point of tangency is, so, i guess that's why i didn't do it like that

[silent plank]

vertex here refers to a corner of you triangle

[upper karma]

oh wait

[silent plank]

a point of tangency is a point where a line is tangent to a curve (in this case, the circle)

[upper karma]

this?

[silent plank]

yes

[rotund yew]

I feel like an idiot, but I don't understand

[upper karma]

well but how is it going to solve the problem? you don't know the hypothenuse @silent plank

[silent plank]

properties of an equilateral triangle, bisection theorems

that will be a 30,60,90 triangle

[upper karma]

well idk that

[silent plank]

it's pretty much the same concept as how you got 120° in your work

[upper karma]

🤔

but my 120° was simpler

i also don't know how you can create a triangle and find the missing side just by telling the measurements of 2 sides. like, i don't know what angles they have

without drawing it to scale

[silent plank]

if all the sides have the same length x, you have an equilateral triangle

[upper karma]

what about m1n1's problem

i don't know if it's isosceles

[silent plank]

axiom of a triangle

sum of lengths of 2 smaller sides > third

[upper karma]

you don't know if these are the smaller sides

[silent plank]

determine the length of the 3rd side from the options and check whether a triangle can actually be formed

[upper karma]

i don't know if it can be formed

[silent plank]

checking a) if the total length was 34m, what would be the length of your 3rd side?

are you referring to your problem or m1n1s atm?

[upper karma]

m1n1

[silent plank]

there's an even better way with inequalities

@rotund yew

sum of lengths of 2 smaller sides > third
try applying that to determine the minimum and maximum length of the third side

[upper karma]

well, applying that you'll only get a 50% chance of getting it right. A or B

[silent plank]

consider all cases

it must be greater than (18-13)
but also less than (18+13)

[upper karma]

Ah so its B

ahah

hahahahahahahaahaahahhaah

[upper karma]

🤔

[sullen pollen]

would like some help on this problem totally confused

instruction is to solve for variables. Write the equations, and any segments that appear are tangent are tangent to the circle

[silent plank]

apply the tangent secant theorem

[sullen pollen]

ok

what theorem would i use for this? @silent plank

[silent plank]

,rotate 270

rotate 180

,rotate 180

[silent plank]

intersecting chords theorem is what is being described in that chunk of text

[sullen pollen]

thx

that should be correct ? @silent plank

[silent plank]

incorrect product of segments

AD * BD

[sullen pollen]

okay

[analog mantle]

Hi, I am having a test now for one hour, who would be able to help? message please

[dark sparrow]

what

<@&268886789983436800> we got another test cheating attempt here

from @analog mantle

[worthy igloo]

ok

[dark sparrow]

thank you

[wooden current]

people are wilding these days

[upper karma]

LOL

[upper karma]

What are some common ways of remembering the pi values of the unit circle?

(I can go around the circumference npi times and then chop it into the fraction, but this takes a while...)

[graceful narwhal]

do you mean cos(pi/4) = 1/sqrt(2)?

[upper karma]

No, none of the sqrt values, just basic pi values

[silent plank]

pi = 180°

[graceful narwhal]

wut?

pi = 360 and then just divide

there is nothing remember

[silent plank]

180

[graceful narwhal]

ya sorry

[upper karma]

No I meant, e.g. 7pi/4, however, I just realized the pattern too tired. Thanks.

[graceful narwhal]

$7 \cdot \frac{\pi}{4}$

[somber coyote]

deekaan:

[bitter jetty]

thats right right

?

[upper karma]

@bitter jetty

[bitter jetty]

ok

thx

thats what i was thinkin to

[upper karma]

anybody around?

[wooden current]

yes

[upper karma]

loll

any help would be appreciated on this question

[wooden current]

You can see that the object is composed of a cylinder and two equal parts of a sphere which makes up a sphere.

So the volume is Volume_cylinder + Volume_Sphere

If you know the formulas for the volumes of these you are good to go

[upper karma]

@stone spoke is it an exam

well, is it?

[wooden current]

breh

[upper karma]

Seems like so if he doesn't answer lol

[wooden current]

oof

[tight tide]

ok is it just me is the trig material on Khan Academy kinda trash ?

[upper karma]

why do you think that?

[tight tide]

there are sections on here that are just straight up not explained

[upper karma]

you have to go through the lessons first, duh

[tight tide]

... my literal point is that the lesson isn't there

[upper karma]

https://www.khanacademy.org/math/trigonometry

[tight tide]

as in I'm going through Graph sinusoidal functions: phase shift

and there is no details on there on how to deal with it

[upper karma]

🤔

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:trig/x2ec2f6f830c9fb89:sinusoidal-models/v/modeling-periodic-function-with-shift

personally, I didn't have a problem understanding the "phase shift" (I call it horizontal offset)

what I didn't like about their trig lessons was how they mangled the periodicity coefficient

[tight tide]

I mean I probably wouldn't either if any of the material on there actually explained it

(that video didn't either )

[upper karma]

I had to learn it the "proper" way from this guy: https://www.youtube.com/watch?v=NtNIGO9FHIk&list=PLX2gX-ftPVXWiPVqdVFdCvLDGEtCj8iyy&index=32

[tight tide]

ok for real thanks for this video

looks like exactly what I need

[upper karma]

sure thing

I highly recommend that entire playlist

and always read the comments

because sometimes other people point out some mistakes

[tight tide]

will do

[wooden current]

what do you mean?

b/c = p/q => b/(b+c) = p/(p+q) <=> b/(b+c) = p/a <=> p = CD = (ab)/(b+c)

The same for DB

[upper karma]

which one is lowercase "c"?

"Thus, we have-"

thus we have what?

[silent plank]

by convention the lowercase letter is the side opposite the respective capital letter denoting the point/angle (where it isn't ambiguous)

i.e. lowercase c is AB

[upper karma]

nah, I don't think so

maybe it's like this

b/c = p/q

[silent plank]

no

[upper karma]

ok, then which one is lowercase "a"?

[silent plank]

BC

[upper karma]

p/q = a ratio

which should be identical to b/c from the image I posted above

since they split those edges by exactly half

(bisector)

[silent plank]

why don't we just throw labelling convention out the window

[wooden current]

what

[silent plank]

since they split those edges by exactly half
the angle bisector doesn't necessarily bisect the edge

it would if the base angles were equal (which isn't the case here)

[upper karma]

yeah, you're right

[paper vale]

yeah thats a median

[wooden current]

what to heck is going on 😭

[silent plank]

Dspider spouting nonsense

[wooden current]

D:

[paper vale]

when it says bisector it relates to angles

in this case

[upper karma]

https://tenor.com/view/nonsense-the-office-jim-halpert-dwight-schrute-rainn-wilson-gif-4636300

[wooden current]

perpendicular bisector on the other hand

😏

[paper vale]

O

(circumcenter)

[wooden current]

epic

ok but why the intersection of perpendicular bisectors of a triangle is the center of the circle circumscribed about the triangle

wait nvm

[upper karma]

😂 👍

[molten holly]

ZY is tangent to both circles (hence right angles), need to find x. Any help?

[silent plank]

consider splitting the trapezium into a rectangle and triangle

[molten holly]

Ohh I see it now

Rectangle with width 9 and length 25, then I can find hypotenuse with Pythagorean

Yeah it worked much thanks

[lavish umbra]

am I retarded because I keep getting ~1065units^2 for this

[upper karma]

I have no idea

[lavish umbra]

Can someone help?

I think its g

[upper karma]

@molten holly so what was x?

(or side OP)

well?

@molten holly

[upper karma]

@lavish umbra you forgot to subtract the inside of the shape

https://discordapp.com/channels/268882317391429632/326138757474680852/716047659923800074
F and H seem plausible to me

wait, spherical geometry

still F and H

[molten holly]

@lavish umbra You are correct, it is ~25.17, though I didn’t use the same method as you

Sorry wrong @ @upper karma

[upper karma]

25.179 rounds off to 25.18, tho 🙂

[molten holly]

Right right I just used ~25.2

[upper karma]

if they don't ask you to round the result, keep it using the square root

sqrt(634)

[molten holly]

Yeah it’s just a practice sheet and the key didn’t have exact form so

But you’re right got to look out for that

[upper karma]

yep

anyway, I'm off to bed

[molten holly]

Alright thanks for taking the time to do out the problem have a good night

[upper karma]

you're welcome

g'night

[upper karma]

I'm having a really hard time grasping the concept of degrees. My teacher keeps telling me to just think about it as a slice of a circle - which does not make sense at all. Does anyone have a good explanation of what degrees are?

Yes

Would you please explain?

https://youtu.be/JJZDgjFRPXo

https://youtu.be/_n3KZR1DSEo

https://youtu.be/92aLiyeQj0w

https://youtu.be/yzUqxWgtLyY

Thank you, I will watch these vidoes now.

can someone help me understand this task im reading it and having a headache

Yea

The ship is sailing 150km from point A to point B. Than it turnes towards north-west for 34 degres and sails 275 degres to point C. How far is point A from point C?

Take picture of whole problem?

@upper karma

it is in croatian so

and there isnt a photo just text

I can understand Croatian

Just take a picture of it

sec

and is 1. b) even possible

Gimme a sec

[keen aspen]

Yeah with law of sines

[upper karma]

@upper karma yea with law of sines

See how it forms a triangle

Just use the law of sides to find the other side

Which is point á to point c

but i get errors for 1.b

and i dont get how should sketch look for 4.

Just follow steps on what it says

It forms a triangle

is 275 degre angle CAB? than

or BCA?

[upper karma]

Hi can someone please help me?

[versed river]

well do you know anything about the angle a tangent makes with the radius?

[upper karma]

No... :( this is an online course and I'm terrible at learning math just visually.

[versed river]

well, i suggest you read up on circle theorems. I would recommend khan academy personally

the important theorem here is this one:https://youtu.be/St3MU5i3chc

[brisk ginkgo]

I NEED HELP WITH MATH

<@&286206848099549185>

[versed river]

this channel is occupied

[brisk ginkgo]

OK SORRY

[versed river]

also read #❓how-to-get-help

dont ping helpers for 15 minutes

anyway @upper karma once you have your head around this theorem, you can use it to deduce what the measure of angles ONM and OPM are, then you might see how you could figure it out

all it means is that any tangent to a point is perpendicular to the radius to the same point (in a circle)

[upper karma]

Hmm okay. Thank you

[versed river]

if any of this doesnt make sense to you or you still arent sure feel free to come back

[rotund yew]

I don't understand how to do this

[viscid ginkgo]

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

am I right that 30 degrees is a corresponding angle to C?

[regal shell]

We don’t know that bc is parallel to e, so I think you can’t technically

But it looks like it

[viscid ginkgo]

https://i.imgur.com/GnYIo24.png

ok so I redrew it, and now I see that there's a right triangle right?

so we have a 30-60-90 triangle

alright I think I solved it

I believe x = 55, and y = 125

[regal shell]

If bc is parallel to e, then yeah ur right

[viscid ginkgo]

https://i.imgur.com/P8bGYc8.png

triangle ABC is obviously 60-60-60

so then angle 1 is supplementary, so it's 120 degreesd

since side AC and CD are congruent, ITT tells us angle 2 and 3 are congruent

[regal shell]

So then other angles are 30 each

[viscid ginkgo]

so can I deduce that 30

ok

[upper karma]

in an equilateral triangle, draw 3 lines from each vertex to a side such that the line is perpendicular to that side. will the crossing point of the 3 lines be the the same as the center of the circle inscribed in that triangle?

[dark sparrow]

yes all centers coincide in an equilateral triangle

in your case, the orthocenter and incenter

[upper karma]

how does one prove that though?

[dark sparrow]

in an isosceles triangle the altitude and bisector (and median) originating at the apex coincide

an equilateral triangle is isosceles but any point can be its apex

[upper karma]

ahh i see

[viscid ginkgo]

what am I doing wrong?

I don't know what to put as a reason for why CH is parallel to AN, since I already did so.

[dark sparrow]

yikes, two col proofs

[viscid ginkgo]

I've spent over an hour on it, it should be easy

but no

[silent plank]

is this the original problem with all the information?

why did you label those angles as 60° when that didn't seem to be mentioned?

[azure reef]

yes i now trying to understand how he derived it

if it was given that CH = AN it would be trivial

but without looks like meh

like if we assume that NH is parallet to CT then it follows

but nah

[viscid ginkgo]

sorry

I was working on it

[silent plank]

wait

[viscid ginkgo]

@silent plank the original doesn't have the 60 degrees written.

[silent plank]

read the wrong points

[viscid ginkgo]

I assumed that it's 60 degrees

cuz of ITT

[dark sparrow]

what deos ITT even stand for

[viscid ginkgo]

isoscoles triangle theorem

[silent plank]

not all isosceles triangles are equilateral

[viscid ginkgo]

equiangular?

[azure reef]

ah

i got the proof

[viscid ginkgo]

cool!

[azure reef]

external angles are the key

[silent plank]

you can conclude that <1 and <2 are equal but they're not necessarily 60°

[azure reef]

what is external angle of triangle, aaron?

[silent plank]

don't need that

[viscid ginkgo]

4. m(3) + m(N) = m(CAN)

actually

first you have to say def. of exterior angle

[silent plank]

ditch 4 and 5

[azure reef]

i am not understanding that notation meh

[viscid ginkgo]

sorry it's hard to right angles

I was writing measure of angle

[azure reef]

just what equals exterior angle of triangle?

[silent plank]

conclude that <1 = <3

[azure reef]

what

[viscid ginkgo]

measure of angle 3 + measure of angle N = measure of angle CAN

[azure reef]

conclude that <1 = <3
well that angles near base are equal is trivial

[silent plank]

and apply converse of corresponding angles on parallel lines

[azure reef]

measure of angle 3 and measure of angle N = measure of angle CAN
i ask general case

[viscid ginkgo]

the remote interior angles of the exterior angle

the 2 remote interiors

[azure reef]

yes

[viscid ginkgo]

added together =

EAT

[azure reef]

so ok

in triange CHA you have angles

1=2

and angle CHA

which equals?

[viscid ginkgo]

60

[azure reef]

nah

[viscid ginkgo]

reallyl

well

1 and 2 are not 60?

[silent plank]

not all isosceles triangles are equilateral

[azure reef]

given two angles $\alpha$ what equals third angle?

[viscid ginkgo]

o.o

[somber coyote]

Commander Vimes:

[viscid ginkgo]

2a

[azure reef]

how

what is the sum of angles of triangle?

[viscid ginkgo]

180 - 2a

[azure reef]

yes

and exterior angle of angle 1 is?

[viscid ginkgo]

120

[azure reef]

[viscid ginkgo]

120+60 = 180

[silent plank]

not all isosceles triangles are equilateral

[viscid ginkgo]

o.o

cuz it's a straight line

[silent plank]

don't use 60° anywhere in your problem

[viscid ginkgo]

I don't understand, it's a linear pair

[azure reef]

there is no paint on my ubuntu, can you please redraw the task?

[silent plank]

because you are trying to prove it works for ALL isosceles triangles and not just for the special case where it is equilateral

[viscid ginkgo]

to the ... what do I call it? x-axis, of line ... CT?

I don't know the term

and it's kinda outside of CT

but I think my prof said you can call a line whatever you want

as long as it's a "member" of the line

hmm

[silent plank]

use conventional labelling systems,

and also this is starting to overcomplicate it

[viscid ginkgo]

my book says that if the 2 sides are congruent, then the base angles are also congruent./

[silent plank]

yes. but not 60°

[viscid ginkgo]

ok

right

I assumed that angle 1 is 60

[silent plank]

if you introduced those labels,
<1 = < 2 is fine

[viscid ginkgo]

it can be any angle.

I think

lol

[silent plank]

and <2 = <3 is given is also fine

[viscid ginkgo]

yes

ok

[silent plank]

from which you can get <1 = <3

[viscid ginkgo]

so angle 1 = 180 - angle 1

is this what u wanted me to say earlier

[silent plank]

wasn't what "I" wanted...

[viscid ginkgo]

@azure reef yeah

umm u want me to send u another link

ill redraw it

fuck it

[silent plank]

from which you can get <1 = <3

[viscid ginkgo]

yes im trying to understand sorry

im redrawing

[silent plank]

in your new diagram, don't make it seem that HN and CT are parallel

[azure reef]

we are particularly interested in angles c and b

[silent plank]

make triangle ANT distinctly scalene

[azure reef]

so aaron, between angles a, b and c there is a marvellous relation

which yields AN || CH

do you understand why there is 2 b's?

[silent plank]

this is overcomplicating it

[azure reef]

ramonov why overcomplicating?

[silent plank]

in your new diagram, don't make it seem that HN and CT are parallel

[viscid ginkgo]

I actually wasn't

I'm just a shitty drawer

lmfao

[azure reef]

literally 3-4 operations which yield parallel lines

[silent plank]

like draw N higher

and apply converse of corresponding angles on parallel lines

[viscid ginkgo]

so 4. angle 1 = angle 3

[silent plank]

yes

[viscid ginkgo]

we are particularly interested in angles c and b

there's no angle b

oh

in your diagram

[azure reef]

ye here

[viscid ginkgo]

c = angle a and b

added

umm

[azure reef]

c = a + b ok

[viscid ginkgo]

180 - angle a

[azure reef]

and CAN = ?

[viscid ginkgo]

= angle a + angle b

[azure reef]

if angle c = angle CAN it yields?

but do you understand why there are two b's?

[silent plank]

fine do extra stuff and completely ignore me and make the 2 column proof longer than it needs to be

[viscid ginkgo]

im not sure

[azure reef]

oh lol i now got what ramonov meant

[viscid ginkgo]

im confused as fuck right now

like

who do I listen to

and I don't know what angle c =

if it = angle CAN

ok angle 1 = angle 3

for #4

[azure reef]

they are bot equal to 180 - a

that what was ramonov saying

and what i did by long route

and if two lines cross third line and make equal angles it yields

[viscid ginkgo]

fuck I gotta brb

hold a sec sorry

[silent plank]

so yes,
#4. angle 1 = angle 3 (transitive)

and apply converse of corresponding angles on parallel lines

[viscid ginkgo]

sorry

I had to take an exam lmao

[silent plank]

corresponding angles on parallel lines are equal
converse: if corresponding angles are equal then the lines they're on are parallel

[viscid ginkgo]

oh yeah I did that one earlier

yeah

makes sense

so

5. angle 1, angle 3 are corresponding angles -- reason: Def. of Corresponding angles

6. parallel

Reason: if corresponding angles are congruent, then the line is parallel

the line is parallel part doesn't sound right

what line, you know?

[silent plank]

the lines they're on

[viscid ginkgo]

then their lines are parallel?

or can I say

Thm: if lines are parallel, then corresponding angles are congruent.

turned around

Thm: if corresponding angles are congruent, then the lines are parallel.

idk if it matters?

[upper karma]

Sure

[viscid ginkgo]

thanks so much ramonov

and other guyh

[azure reef]

[regal shell]

Other guy ftw 💪

[stark wraith]

how can i prove that for any a>0 tan(a) x tan(a+90)=-1

[dark sparrow]

did you use x for multiplication there?

please don't do that

anyway, "proving tan(a)tan(a+90°) = -1 for any a > 0" is a bit of a tall order given that there are many values of a for which tan(a)tan(a+90°) is undefined

for example, if a = 90°

how are you going to take tan(90°)

[stark wraith]

i have two Z numbers that i know have 90 degrees difference between their angels , and i need to prove that the lines that go from (0,0) to each of them are vertical

[dark sparrow]

"Z numbers"?

[stark wraith]

composite number

s

[dark sparrow]

anyway, it'd be a very different story if you asked how to prove tan(a)tan(a+90°)=-1 for all a where the left-hand side makes sense rather than "for all a > 0"

because then i could tell you that for all real a you have sin(a + 90°) = cos(a) and cos(a+90°) = -sin(a)

and so as long as sin(a) and cos(a) aren't zero you can write tan(a)tan(a+90°) = (sin(a) sin(a+90°))(cos(a) cos(a+90°)) = (sin(a)cos(a))/(-cos(a)sin(a))

anyway, it looks like you're heavily misusing mathematical terminology

and i'm not sure if it's due to a language barrier or a lack of understanding what means what or both

[stark wraith]

i didnt get your way of getting to sinacosa/-sinacosa

[dark sparrow]

for all real a you have sin(a + 90°) = cos(a) and cos(a+90°) = -sin(a)

[stark wraith]

oh ok thank you

[dark sparrow]

did i really have to repeat myself

or did you just choose not to read what i said the first time round

[upper karma]

Sorry

[stark wraith]

you didnt have to repeat yourself either way

its your choice if you want to help or not

[upper karma]

@stark wraith Yep nice comeback

[viscid ginkgo]

in my proof if I assume in an indirect proof that rays BA and BC are perpendicular, given that they're not, and then reach a contradiction

what do I put for reject and therefore?

Reject: rays BA and BC perp

Therefore: Assumption is false. rays BA and BC not perp

[night wind]

https://www.youtube.com/watch?v=r-qNZN_JzRQ&t=6s

[wooden current]

um

[upper karma]

Some Overwatch short film lmao.

[wooden current]

why here

[upper karma]

Idk.

Idk

[upper karma]

yo

can anyone help

sin^2(-x)/tan^2(x)=cos^2x

[keen aspen]

reduce tan down to sin and cos

[upper karma]

so this is what i got

idk how it becomes cos^2x

it should be -cos^2x

[keen aspen]

no

ok so you agree that sin(-x) is -sin(x) right

[upper karma]

yea

[keen aspen]

ok so when you square it what happens

sin(-x)*sin(-x)

[upper karma]

mhm

ah

wait what the fuck

i swear to god

I was taught something

[keen aspen]

the negatives cancel

[upper karma]

that

sin^2(-x)=-sin^2x

or is that

just

with

a

regular

sin

[keen aspen]

not negative

[upper karma]

is

[keen aspen]

sin^2(-x) is sin(-x)*sin(-x)

[upper karma]

sin(-x) just -sinx

[keen aspen]

which is -sin(x)*-sin(x)

the negatives cancel

[upper karma]

gotcha

[keen aspen]

so its just sin^2(x)

[upper karma]

but sin(-x)=-sinx

[keen aspen]

yes

[upper karma]

kk

ty'

[graceful anchor]

how do you rearrange to find x

[keen aspen]

You cant really do so

[graceful anchor]

sad

[keen aspen]

Its ok

[umbral snow]

Newton's method af

[brisk holly]

why is the period near 500?

shouldnt the period be 1/365?

[devout harbor]

if it's sin(kx), that means an amount of k sin curves in an interval of 2pi, or each sin curve has an interval of 2pi/k. so you have sin(2pi / 365) which means an interval of [(2pi) / (2pi / 365)]

which is just 365

@brisk holly

for simplicity, take sin(2x). in an interval of 2pi, you have 2 sin curves

or each sin curve has an interval of 2pi / 2 which is just pi

[brisk holly]

oh i just forgot multiplying two fractions

I thought 2pi/2pi/365 simplified to 1/365

[neon jolt]

Hello, I have a question regarding this expression

vA dot vA = Theta = 0 therefore va dot va = a^2cos(0) = a^2?

where is this a^2cos(0) come from?

is it |a||a|cos(0) = a^2?

hence aa1

[weary drift]

where is this a^2cos(0) come from?
the geometric definition of the dot product

[neon jolt]

Perfecto

[upper karma]

What are the basics of Geogebra

[upper karma]

cheese

[azure reef]

basics of Geogebra is alge-bruh

[wooden current]

guys imagine if algebra was spelt algebruh 😂 😂 😂

[upper karma]

Aight Cool

[regal shell]

U can treat that like a regular cylinder

[upper karma]

^^^^

^^

@lavish glacier You got it

?

[lavish glacier]

I got 14071.26?

[upper karma]

Hold on son

[lavish glacier]

Ok papi

[upper karma]

@lavish glacier Well did you do everything correct?

[lavish glacier]

Let me try again

[upper karma]

Ok

I’m not saying you got it wrong

I haven’t even tried it myself

What makes you doubt your wrong?

If you did everything correct then sure

[lavish glacier]

Its not 1 of the multiple choice answers

[upper karma]

Oh

Try again or something

What exactly are you doing

To find the volume?

[regal shell]

I got one of the answer choices

[upper karma]

Do u see how 26.1 represents the height?

[regal shell]

What formula did u use?

[lavish glacier]

V = pi r²h

[upper karma]

Ya good

[lavish glacier]

Oh i used pi instead of 3.14

[upper karma]

Ya it specifically says just to use 3.14

[regal shell]

O ok u got it right then if u use 3.14

[lavish glacier]

D!

[upper karma]

Congrats!

👍

[lavish glacier]

Thanks papi!

[upper karma]

Yes son! 👏

👌

[regal shell]

Xi can you be my papi too?

[upper karma]

UwU

Alright son

👍

[regal shell]

Yay

[lavish glacier]

Brutha?@regal shell

[upper karma]

👨‍👦

[regal shell]

O

[upper karma]

👨‍👦‍👦

👍

[regal shell]

Who’s the one on the left