geometry-and-trigonometry

[mighty narwhal]

alterran. i just did a draft. chill :c

also the way I wrote it down is how you would get the diameter of a cuboid...

Javakid, cot(90) is not defined

$\tan{\theta}=\frac{1}{ctg{\theta}}$ is correct, but $\theta\not=0^\circ,90^\circ,180^\circ,270^\circ$

[somber coyote]

DarK:

[supple abyss]

cot(90 degrees) is 0

[mighty narwhal]

I mean tan 90 is undefined...

also if cot 90 is 0, you are dividing by 0

[olive sinew]

i forgot that at 90dg x=0

is there a rule where i can only ask a limited amount of question ? lol

[rustic dagger]

hi

[upper karma]

How would you copy a line segment with explaining it and justifying it

I’m new to geometry so I’m super confused

[hidden geode]

@upper karma do you mean 'how can you justify, from first principles, the construction where you copy a line segment from one place to another'? i believe this is constructed in Book I of Euclid's Elements

[upper karma]

Yes

Okay

[hidden geode]

(which is all i can talk about, since i once tried to study it & quit early on 😝 )

https://mathcs.clarku.edu/~djoyce/java/elements/bookI/propI2.html

try looking at that & seeing if it helps?

[upper karma]

What do I do for these on question 15?

[spark stag]

use trigonometric ratios.

"SOH CAH TOA"

[wheat hull]

Find the two angles the lines y=-5x -1 and y =2x+3 intersect at:

How would I solve this?

[supple abyss]

find the angles they make with the x axis

[wheat hull]

Is there a formula for this?

[supple abyss]

use trig

[wheat hull]

Can you tell me the whole formula

[spark stag]

"formula"?

i guess technically theres like

$\theta = \arctan \left(\frac{\text{rise}}{\text{run}}\right)$

[somber coyote]

Namington:

[spark stag]

but its not helpful to think of math in terms of "formulas"

unless you understand where they come from

[wheat hull]

equation

ok

can you go step by step for one of the lines plz

[spark stag]

but yes, the angle a line makes with the x-axis is the arctan of the slope

the concern in this case isn't the x-axis, it's between the two lines

but you can use these angles to find the angle at intersection

[wheat hull]

ok

[spark stag]

first, its helpful to make a rough sketch of what we're working with

just so we can see vaguely what we're looking for

[wheat hull]

ok

[spark stag]

we're looking for the values of the angles between the red and blue line

now

lets use this image as an example

[wheat hull]

ok

[spark stag]

in this case, theta is the angle the line makes with the x axis

recall that tan(theta) = opposite/adjacent

and when we make our triangle like this

opposite is our rise, adjacent is our run

so tan(theta) = rise/run

you might be familiar with 'rise over run'

[wheat hull]

yea

[spark stag]

or change in y / change in x

that's slope

[wheat hull]

yea

[spark stag]

so the tangent of the angle with the x axis

is the slope

in other words:

$\tan \theta = m \
\theta = \tan^{-1} m$

[somber coyote]

Namington:

[spark stag]

and again, theta is the angle with the x axis

so we can calculate that for each line

[wheat hull]

so -79 degrees for y=-5x -1

[spark stag]

y = -5x -1, the slope is -5, so the angle with the x axis is tan(-5)
y = 2x + 3, the slope is 2, so the angle with the x axis is tan(2)

yeah

anyway, these are just the angles with the x axis

[wheat hull]

63 degree for y=2x+3

ok

[spark stag]

but looking back to this diagram

we can see how we could use these

let me demonstrate

[wheat hull]

ok

[spark stag]

here

the angle of the lines is constant, since they're straight lines

so the angle with the x axis

will be the same

as the angle with the yellow line

[wheat hull]

ok

[spark stag]

so you'll note that

[wheat hull]

ok

[spark stag]

the whole green angle here

is equal to the red angle + the blue angle

but theres a catch

if you recall, one was negative - the blue line's angle

that indicates that, on the right side

it went below the x axis

that doesnt really matter here, though

so we use the positive values of both angles

that gives us 78.6900675 + 63.4349488

,calc 78.6900675 + 63.4349488

[somber coyote]

Result:

142.1250163

[spark stag]

so our bigger angle is ~142.1 degrees

[wheat hull]

ok

[spark stag]

finally, note that the pink angle here is the angle of a straight line

or half the angle of a circle

ie, 180 degrees

therefore, the remaining angle will be

180 - 142.1

which is 37.9

[wheat hull]

ok

[spark stag]

so those are our two angles

(round as appropriate)

[wheat hull]

ohh I understand so much right now

Thx for visually demonstrating it

[upper karma]

@spark stag gimme an example using the first problem.

[spark stag]

this is our triangle

we look at what we know:

we know the side 27

we know the angle 22 degrees

and we know that it's a right triangle

is the side of length 27 the opposite, adjacent, or hypotenuse?

[upper karma]

Hypotenuse.

[spark stag]

right; its the longest side, and on the other side of the right angle

so it's the hypotenuse

so, we know the angle = 22 degrees, and the hypotenuse = 27

next, we look at our unknown

k

is k the opposite or adjacent side of the 22degree angle?

@upper karma

[wheat hull]

is ross gone?

@spark stag Quick question

If the 5 wasn't a negative

I could still do the same right

[spark stag]

similar

[wheat hull]

what would be the difference?

[spark stag]

when we'd make our sketch, we'd have something like this

so now you'll note

this white angl

[upper karma]

Opposite side

[spark stag]

is the blue angle minus the red one

so we'd have to subtract for this one

[wheat hull]

ohh okay

[spark stag]

this happens when the slopes have the same sign, basically

[wheat hull]

ok thx for clarifying

[spark stag]

but yeah, we subtract the angles they make with the x axis

[upper karma]

Is it the opposite side?

[spark stag]

and then the other angle is

180 - that angle

as usual

[wheat hull]

okay

[spark stag]

@upper karma

not quite

note that the side k

is "touching" the 22 degrees

so we say its the adjacent

the "opposite" is the side across the triangle - the unlabeled side, in this case

anyway, so we have:

knowns: hypotenuse, angle
unknown: adjacent

if we look at SOH CAH TOA

which ratio includes the adjacent and the hypotenuse?

[upper karma]

TOA?

[spark stag]

nope

TOA is tan = opposite/adjacent

we're looking for the adjacent (A) and hypotenuse (H)

which will be CAH

this stands for

$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ \
$\theta$ just means the angle

[somber coyote]

Namington:

[spark stag]

so, plugging in our values

$\cos (22^\circ) = \frac{k}{27}$

[somber coyote]

Namington:

[spark stag]

does that make sense?

finally, we can solve for k by multiplying both sides by 27

this gives us

$k = 27 \cdot \cos(22^\circ)$

[somber coyote]

Namington:

[spark stag]

plug that into your calculator

and there's the length of k

[make sure your calculator's in degrees!]

[upper karma]

Hang on do you have like a template for these problems where it shows what I need to plugin?

[spark stag]

SOH CAH TOA is the entire "trick"

we look at what we know, and what we want to find

if we know or want the opposite, we look for the ratio with O
if we know/want the adjacent, we look for A
if we know/want the hypotenuse, we look for H

then we can write out the ratio, substitute in everything we know, and solve for the unknown

[upper karma]

Hmm.

So they're all the same but different like in some problems it's sine opposite over adjacent or opposite over hypotenuse?

[spark stag]

sure, SOH CAH TOA basically represents which trig ratio you use for which sides

$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \
\
\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \
\
\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$

[somber coyote]

Namington:

[spark stag]

hence, SOH
CAH
TOA

(and again, im using theta - that weird oval with a line in the middle - to mean "the angle")

[mighty narwhal]

Soh cah toa makes me cry

Why can't you just learn what is what ;-;
This way you are just memorizing it, nut understanding

[dark sparrow]

i mean... SOH CAH TOA is a definitional mnemonic.

it's just a matter of learning what name refers to what. for this particular thing there isn't much to "understand" as opposed to, say, FOIL

[mighty narwhal]

Yea, but I just find it weird when people write sohcahtoa on the top of their tests

[dark sparrow]

what if it helps them recall

or yknow

keep it straight what's called what

[cinder portal]

lmao i find myself writing some very specific ass formula on the top of the test i just studied for, sounds normal

[upper karma]

doesnt matter if im 12 or 80 ill still be writing SOHCAHTOA for life bitches

slight lie

I never write it anymore

but we can forget about that

but I still se it

use it

[mighty narwhal]

We don't have sohcahtoa in Hungary

We memorise it like

Szinusz= Szöggel Szemközti Szár / átfogó
(Sine= leg opposite angle / hyp)
Koszinusz= Szög meletti szár / átfogó
(Szemközti means opposite, meletti means next to)
Tangens= Szemközti/melletti

[upper karma]

sounds lamw

lame

[cinder portal]

when you work with polar coordinates a bunch, trig functions are justkinda nailed in ur head at that point

cos usually denotes x coordinate, sin y coordinate, from there u can figure out if its sohcahtoa or whatever it is

[mighty narwhal]

Yepp

Also after you help people from school doing trig homework, it also kinda burns into your brain

[lusty quest]

guys, why is this one wrong?

this is the angle

so the formula should be adjacent over opposite

the correct answer is apparently this?

[mystic shadow]

your answer is wrong because you did not simplify

$-\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\frac{\sqrt{3}}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$

[somber coyote]

MiracleMan:

[lusty quest]

so i didnt finish the equation? damn wtf

thank you for the insight

[mystic shadow]

simplifying means you can't have any square roots in the denominatior

[night karma]

I found this problem online and there's quit e a lot of variants on these. Yet, I do not know what steps to take with the equal sides. Is there a trick for it?

[devout shell]

There's a shared side and a common angle but I can't see how to use that to prove similarity

EBD would be SAS but DCB is SSA

[night karma]

Hmm

If you were to make up a length for the equal sides, would you use a variable or a number? Perhaps that could work

[teal epoch]

hello, can someone explain me this ? thank you !

[mystic shadow]

Try distributing the second expression

[teal epoch]

i mean

how i am supposed to guess this ? is there some idea behin this ?

[umbral snow]

@teal epoch
It's the same as
1 + x² = x(x⁻¹ + x)

[teal epoch]

well

i never saw that, thnk you !

[upper karma]

I have literally zero understanding of how to calculate this into any given answer, It just seems like complete nonsense to me to have division as part of the values I have to multiply

Like 4(2/3)*7(1/3) doesn't give me any answers that register on the same realm of my choices

why are u multiplying them`?

sorry for not including full context

what did u get when u multiply

6.2

the multiplication is wrong, 7x4 is 28 so it has to be more than that, no?

The division/fractions

2/3 & 1/3

they're what chop it down

Regardless even if they didn't it's not near the answer

i think they m ean 7 and a 1/3

not 7/3

So 7 + 0.3?

that seems random and weird

0,3 recurring

how is it weird?

Because none of the potential answers have decimals.

and it provides nothing to the problem to have them there if they're not part of the equation anyway

is any of the answers above 28

Yes and no, all the answers are in feet while the given variables are in yards

the answers range from 18 and 308

feet squared specifically

ofc its squared its an area

yeah

But yeah these are all the options provided

I've yet to get any real answers so I cant reasonably select none yet

ok ill give u a way to solve it

write 7 1/3 as (7+1/3) and the same with the second one

multiply the brackets,

sum, u get an answer

transfer to ft 2

and its definetly one of the answers

The only answer I've gotten is 2.0740740740740740667 ft^2 so far, heh

but I'll try that

youre multiplying wrong

It's a calculator???

(7+1/3)(4+2/3)

youre putting it in wrong sweaty

They're right next to each other that means they'd be multiplied not added, no?

[dark sparrow]

ok so

[upper karma]

if they were being added there'd be a +

[dark sparrow]

there's this horrible notation

called mixed fractions

[upper karma]

its just notation because they dont want to put 0,3 recurring

[dark sparrow]

where $5 \frac37$ stands, inexplicably, for "five and three sevenths", or $5 + \frac37$

[somber coyote]

Ann:

[dark sparrow]

also, decimals suck and should not be used honestly

this problem should not require the use of a calculator

also!

the problem asks for an answer in square feet, yet it gives the lengths in yards

[upper karma]

yeah

[dark sparrow]

consider converting the yard measurements to feet

[upper karma]

so divide any answer by 3

[dark sparrow]

no.

[somber coyote]

Hexedirl:

[dark sparrow]

y'all

seriously

convert to feet first

that'll get rid of the fractions

[upper karma]

oh

you still get the right answer

[dark sparrow]

it's going to be so much easier

[upper karma]

if you sum before then convert

[dark sparrow]

of course you could convert later but like

fractions are more error-prone to work with than integers, even once you're fluent in algebra

[upper karma]

So I divide the fractions by 3 then to eradicate them?

[dark sparrow]

what? no

[upper karma]

you said to make it into feet

yards are just 3 foot, no?

[dark sparrow]

yeah, so

1 yard is 3 feet

so (4 + 2/3) yards is (4 + 2/3)*3 ft

[upper karma]

so you multiply by 3

[dark sparrow]

or 14 ft

i mean honestly

[upper karma]

oh

[dark sparrow]

if you DID want to go about this without immediately converting

[upper karma]

that solved literally all my problems

[dark sparrow]

there's one thing i always tell people

and it's that mixed fractions should never, EVER be worked with w/o rewriting them as improper fractions

[quiet mason]

whats the temp in russia ann

[dark sparrow]

the what

[quiet mason]

its 25 degrees

alr

[dark sparrow]

Russia is a big country, the air temperature surely varies a lot throughout

[upper karma]

Btw what on gods name is a "dm", I've never heard of that measurement in my life

[dark sparrow]

i can tell you that it's 16°C where i am

dm = decimeter

0.1 m, or 10 cm

[quiet mason]

theres decameter

too

[dark sparrow]

decameter would be dam

nobody uses the deca- prefix though

like seriously it's so dated

deci- still finds use in decibel

[upper karma]

deca is used for decahedron etc

So .4 dm would be 4 cm?

[dark sparrow]

0.4 dm = 4 cm yes

don't write decimals without the leading zero, it's ugly to look at and that dot is incredibly easy to not notice

[upper karma]

mb

[dark sparrow]

@upper karma yeah but that isn't as a prefix on a unit of measurement

[upper karma]

yeah but u said nobody used the deca prefix

[dark sparrow]

i wouldn't consider it a prefix in decagon

or decahedron

[upper karma]

Okay so I multiplied 84 inches by 144 inches and got 12096 inches which converts to 1008 feet but that's not the answer

I'm very confused

[dark sparrow]

okay

[upper karma]

These conversions are actually breaking my brain rn

[dark sparrow]

so first off

when you multiplied 84 inches by 144 inches, you did NOT get 12096 inches

you got 12096 square inches.

and a square inch is NOT 1/12 of a square foot.

[upper karma]

oh

[dark sparrow]

1 ft^2 = (12 in)^2 = 144 in^2

likewise, for volume, 1 ft^3 = (12 in)^3 = 1728 in^3

[upper karma]

Is there a calculator setting I can use to do that for me? my brain is actually so slow rn

I don't even know how to reverse a squaring mentally

[dark sparrow]

wdym "reverse a squaring"

[upper karma]

like if you square 252 for example, I have no idea how to get 252 out of 63504

[dark sparrow]

that's called a square root

you aren't going to be able to take square roots of big numbers mentally unless you have some mad number-cruncher for a brain

[upper karma]

oh wait I'm braindead I learned this like 2 years ago and completely blanked

I remember how to do it now

At any rate thank you very much for the help

Plato'ing a sophomore class my senior year proves to be more of a start from scratch adventure than a relearning.

[quiet mason]

one question

are u ok?

[upper karma]

I haven't slept well recently but otherwise I'm probably fine.

[upper karma]

hi y'all can anyone offer me a hint to this prob?

[dark sparrow]

make a diagram

[upper karma]

think i must be braindead or something @dark sparrow

[dark sparrow]

but dis not rite

just reading that made me throw up mentally

[upper karma]

lol

[dark sparrow]

idk maybe the problem was asking for the other angle

[upper karma]

uh yup that works lel

ty

[upper karma]

I just had an epiphany of math that makes everything so much easier, half the stuff I do I don't even need to. If it wants the circumference of a circle I don't have to even look at the fractions or try to multiply by pi because ultimately the answer doesn't need either of those things to be solved to be the answer. In fact I feel completely stupid for ever bothering with that stuff in the first place lmao

TL:DR I realized I can just use completed formulas as answers instead of fully solved solutions.

[upper karma]

is E the centre of th circle

[mighty narwhal]

Wait

E isnt necessary the center of the circle

you can find angle BEA using law of cosines
BEA is equal to CED
using the law of cosines again you can find CD with CE, DE and angle CED

@torpid horizon @upper karma

E is definetly not the center of the circle
because then BD and CA would be both the diameter
but BD=/=AC

[torpid horizon]

ok

ill delete what i wrote

[novel olive]

Quick question of something I thought of in class

Would it make sense to compare a rotation to a homothetic function with a complex "ratio"

[olive sinew]

i need a hint for 72

a)

[spark stag]

what are the current angles of the square?

[olive sinew]

60

[spark stag]

uh

[olive sinew]

degrees

[spark stag]

they're right angles

which are 90 degrees

anyway, when we draw a diagonal across the square

it should be clear that we cut these angles cleanly in half

ie bisect

[olive sinew]

ok makes sense

now for b

the diagonal should be V2k

[spark stag]

consider viewing the square as two right triangles

uh

do you mean

$\sqrt{2k^2}$?

[somber coyote]

Namington:

[olive sinew]

the sqr root of 2 times k

[spark stag]

not quite

consider, we have

a^2 + b^2 = c^2

k^2 + k^2 = c^2

2k^2 = c^2

c = sqrt(2k^2)

you need the ^2 on the k

and from there

we can simplify

can you see how?

[olive sinew]

yes

[spark stag]

right, so we get

length of the diagonal = $k \cdot \sqrt{2}$

[somber coyote]

Namington:

[olive sinew]

thats what i said

[spark stag]

which should be enough to answer c as well

ohh wait

i see

when you said

"the sqr root of 2 times k"

i thought this meant

"the sqr root of (2 times k)"

not "(the sqr root of 2) times k"

[olive sinew]

my bad

[spark stag]

dont worry, its ambiguous

but yeah, thats right then

[upper karma]

Which is the formula to know the are of the circle?

[mighty narwhal]

$A=r^2\pi$

[somber coyote]

DarK:

[upper karma]

A is 9? and r is the x-4^2 and y+4^2

[novel pecan]

How to find the period of y = 1/3tan(x/3)-1/5cos(2x/5). I know the period of individual terms but what about the whole function?

[mighty narwhal]

The period should be the least common multiple of the individual periods, I believe

because the LCM is when both has their "resets"

So the whole function also has it at the LCM

I hope you could understand what I just said

,w y = 1/3tan(x/3)-1/5cos(2x/5)

[somber coyote]

Display results online and refine query

Requested by DarK#2404

[mighty narwhal]

So appearently Im wrong

sorry!

[novel pecan]

But LCM(3, 5) = 15. So I think you are correct

Can someone confirm?

[mighty narwhal]

Oh

[tawdry pivot]

,rotate

[lusty quest]

is this asking if x = 0 and y = -1?

[pure pivot]

So epic how do you do it?

[fringe dirge]

thank god people can't use that ping

[tawdry pivot]

@Helpers @Moderators

[timber hinge]

[left ermine]

lmao

[timber hinge]

no help 4 u

[worthy igloo]

@lethal swift don't do that

[left ermine]

^

[fringe dirge]

The more of a dick you are, the less inclined people are to help you

This might be a nice lesson for life in general

[timber hinge]

tfw trying to ping 14k members for 1 trig problem

[worthy igloo]

waited 15 minutes at least

[fast tinsel]

@pure pivot I'm no expert but I think I can take a crack at that. <DBC looks like a right angle, but since it doesn't have the symbol I'm going to assume that it is not and you will have to use the law of cosines. It states that a^2 + b^2 - 2ab x cos(C) = c^2, where a and b are the lengths of the sides that include the angle you're trying to find, C is that angle, and c is the opposite side.

For your example, you would solve 20^2 = 32^2 + 28^2 -2(32)(28) x cos(C) for C. I had to do this by hand on a whiteboard with some calculator help, but your answer (unless I'm terribly mistaken) should be about ||38.21 degrees||.

I also changed the numbers to solve for angle <DBC and found that this is not a right triangle. Never assume 😆

[twin heron]

its between 35 and 12837, ninja

[left ermine]

he only waited 15 minutes because the first ping didn't work

[twin heron]

yeah

[left ermine]

lmao "explanation" "yeah"

[worthy igloo]

@lethal swift moderator pings are for things that need moderator attention only. Helper pings are only if no one reacted to your question in some time. Still mind that no one here is obligated to help you, so demanding answers isn't gonna get you very far.

[fast tinsel]

How can I prove that <A is congruent to <C?

[fringe dirge]

Can you write B in terms of A?

[tawdry pivot]

@fast tinsel @pure pivot

the question is for $\Delta BCD$ not $\angle BCD$. We see $\Delta BCD\sim\Delta BVU$ since $\frac{BC}{BU}=\frac{BD}{BV}=\frac{DC}{VU}=4.$

[somber coyote]

EpicGuy4227:

[pure pivot]

Wait so it’s bvu right?

I got bvu

[tawdry pivot]

do you understand why

yes if you got it then good

[pure pivot]

Yea

Thank you guys

[lusty quest]

No one helped my question 🥺

[pure pivot]

Oh yea

[lusty quest]

@modz

[pure pivot]

😂

[lusty quest]

@MODSSSSSSS!!!!!

[fringe dirge]

No 👏 one 👏 is 👏 obligated 👏 to 👏 give 👏 you 👏 help

[left ermine]

@everyone is obligated to give you help

[lusty quest]

Perfect

I will pay you 2 of my finest rupees for your service

[fringe dirge]

The more of a dick you are, the less inclined people are to give you help

You should take that as a life lesson

Be polite and nice and people will help you

[pure pivot]

He didn’t do anything?

[lusty quest]

Sir, i need to talk to your manager

[left ermine]

2 rupees = $0.029

pass

[lusty quest]

This behavior is unacceptable, ill get you all fired

[pure pivot]

Lol

[fringe dirge]

lmao, we all do this because we like helping people

helping people is nicer when they're decent humans

[lusty quest]

Ok, do you KNOW who my father is?

Oh boy when he gets the news of this...

[left ermine]

@fast tinsel

[fringe dirge]

Giving people answers is not teaching either

[left ermine]

yes but sometimes it's easier

[fringe dirge]

Doing the easiest thing isn't always a good idea?

Like if you don't want to put in effort to teach

Then just let someone else do it?

[left ermine]

i find sometimes that if the person sees the answer they can see logically how to arrive at the answer

[tawdry pivot]

wut

[fringe dirge]

I mean sure, but it usually isn't better than having them arriving at the answer themself

[lusty quest]

@pure pivot they dont seem to get sarcasm

[tawdry pivot]

noo don't delete the mother whipping you msg

do you prefer horse hair or belt

[lusty quest]

I dont wanna troll too much 😣

It really depends on the occasion

[zenith ember]

You shouldn't be trolling at all in here.

[lusty quest]

Cant help it, im a bad boy 😏

[fast tinsel]

Not sure how I didn't see that before lol

[pure pivot]

😂

[fast tinsel]

I've been doing algebra for 12 hours I forgot how triangles work sorry xD

[upper karma]

For the future people that come here: Take all non-mathematical banter to #chill pleaae.

[olive sinew]

Ok found a and b

I just want to confirm that this is right

If*

[devout shell]

yes

[olive sinew]

Thanks

so for d must be same as b

and finally for c i got 12(V6)

am i correct ?

[devout shell]

yes

[ornate lodge]

If I have two circles with r = 4 that are touching, there is a third circle that touches both of them and their shared tangent line, I have to find it's radius. They will form a triangle where one side is 4, the other is 4+x, but why is the last side 4-x?

[mighty narwhal]

picture please

[ornate lodge]

I think it's like this

where the small circles radii are 4

[plucky marlin]

its not touching their shared tangent line is it ?

its intersecting it

https://prnt.sc/o1tc5y

maybe something like this ?

[ornate lodge]

It could be like that as well I'm not really sure but the problem is worded exactly how I put it above

[plucky marlin]

https://prnt.sc/o1telu

makes sense ?

[ornate lodge]

Yeah I see it now, thanks. I thought it was a bigger circle, not smaller

[plucky marlin]

np

[upper karma]

Help

[wintry mural]

,rotate

[wintry mural]

Oh

[upper karma]

Well i suppose the area of the medium circles is 6

Dunno how you would demonstrate that without more info

Small ones look like 4 but again just by looking at them

[night karma]

😜

[upper karma]

I really need help

How to justify copying a line segment and justify copying an angle

Please

[upper karma]

You are allowed to measure with a ruler?

[pseudo robin]

Hi so just did a big exam on Monday and this question stood out as "I don't have a clue how to do this," could someone help me with this. It was designed for 15 year olds in Ireland.

[supple abyss]

(b) I'm guessing?

[subtle gate]

first one is just the tangent or cotangent

so i guess it is b

[supple abyss]

a=c cos(y)

b=c sin(Y)

so c(cos(y)+sin(y))>c

then as c>0 you can divide by c to get the required result

[subtle gate]

i mean this was a 10 minute question what did i expect

[pseudo robin]

Yes it was (b), (a) was quite easy.

[zenith ember]

Use SOH CAH TOA

She says, late to the party

[spark stag]

and unhelpfully.

[zenith ember]

Um

Ok...

[night karma]

a + b > c, Isn't that always the case? Or is it meant as a hint?

[zenith ember]

Both

[night karma]

That explains 😛

[astral patio]

<@&286206848099549185> Hello. Is anyone privvy to trigonometry available to help me go over some problems please?

[olive solar]

also, ask your question instead of asking if you can ask

read rules and #❓how-to-get-help

[astral patio]

These are the types of problems I'm working on now. I'm having issues with 37 and 38.

[cinder portal]

Let's do 37 first, you cool with that?

[astral patio]

Yea, of course. Thanks...do I need to hop in a voice chat?

[cinder portal]

naw no need

What did you try first

I wanna see what you attempted so far

[astral patio]

I'm getting hung up somewhere.

[devout shell]

Took square root

[cinder portal]

mmmmm here's what I would recommend, try converting

tan^2(x) to 1 - sec^2(x)

[devout shell]

But did you make sure everything got the square root applied to it

[cinder portal]

or mb

the other way

tan^2(x) = sec^2(x) - 1

[devout shell]

And seeing tan and sec, I think that identity first too

That way would could factor the expression

[astral patio]

Hmm, I don't see that identity on my cheat sheet. 😮

[devout shell]

Time to add it

[astral patio]

Oh, I think I did find it, but...

Bottom right

is the sec^2(x)-1 an identity or just a mathematical equivelant?

[zenith ember]

What's the difference?

[astral patio]

I thought that after i asked the question, but I figured the difference would be identities are given in the book and the equivelants are learned from experience/knowledge?

[zenith ember]

It's a little of both then

I've seen it listed but it's not an S tier identity

[astral patio]

Ok, I got secx=1/2 and 2

@cinder portal

[whole plover]

Hey guys anyone know what shape this is?

[spark stag]

looks like a mobius strip.

actually, wait

its weird on the back

hmm

yeah, nvm

[whole plover]

it looks like some hyperbolic paraboloids and a combo of 4 of them to make an impossible shape

been searching up the internet for this image but all that comes up is "cool 3d printed art"

ohh ok i got it

its called a honey comb borromean

[spark stag]

that... doesnt seem like a borromean to me

but maybe its just the angle

[whole plover]

https://www.youtube.com/watch?time_continue=28&v=nIO4HNNURFI

[wary vector]

oh lol, when you said shape I thought you meant like plane shapes

[cinder portal]

its a triangle i think

[olive sinew]

Is anyone up

?

[sudden swallow]

@everyone dp u know how to do this find the theta value with all circle geometric reasoning for 4 marks

[dark sparrow]

1. don't ping everyone
2. don't multipost

[upper karma]

39.33°?

[mighty narwhal]

$\angle BAC$ is the half of $\angle BOC$

You know $\angle BAC, \angle ABC$ and $\angle ACB$

These tree add up to $180^\circ$

[somber coyote]

DarK:

[mighty narwhal]

,w 3x+(124/2)=180

,calc 118/3

[somber coyote]

Result:

39.333333333333

[mighty narwhal]

Yepp

[olive sinew]

Any idea on how to work number 73

[devout shell]

did you take their hint?

[olive sinew]

Yes

Hold on let send you what I got so far

[olive sinew]

Since 180-150= is a right triangle of 30 dg

I would say that y =1 and x = -V3

But I don’t know what the 6 means

[devout shell]

6?

[olive sinew]

I think 6 is the side of one of the triangles

[devout shell]

I'm not seeing what you are saying

[devout shell]

radius 6

[olive sinew]

Ohhhhhh

[devout shell]

so just take the radius as 6 then

[olive sinew]

Ok

Thank you bro

[olive sinew]

How do you work out 78

@devout shell

[devout shell]

in the interval of 90 < x < 180

so the sign of an angle in quadrant 2 is?

[olive sinew]

Positive

[devout shell]

do you have the answer key? I'm not so sure that it's simple like that lol
It just feels too simple

[olive sinew]

I know but I just did 79 and I saw the answer in the back and I got it right

[dark sparrow]

@devout shell sine

not sign

[novel pecan]

Can you turn a circle into a square by dividing it in 4 parts and stretching it out? Would the square have the same area?

[spark stag]

question 1: sure, if you can define what you mean by "stretching it out"
question 2: not unless your stretches are area-preserving, which is unlikely

[novel pecan]

When you unroll a cone it's two lines and an arc, but you can look at that as a triangle, right? So if you have a circle with circumference 4, why isn't that the same as a 1x1 square? Isn't the circle arc of 1 the same as a straight line of 1 length

[spark stag]

"if you have a circle with circumference 4, why isn't that the same as a 1x1 square?"

they're... different shapes

and if you preserve this "circumference"/"perimeter"

you'll have to change the area

consider these two shapes

the right shape has a larger perimeter, but a smaller area

[novel pecan]

Why can you do that with an unrolled cone though

[spark stag]

im... not sure what you mean

[novel pecan]

The circle and square would have the same perimeter/circumference

[spark stag]

sure, but they wouldn't have the same area then

the area of a circle with circumference 4 is ~1.27

the area of a square with perimeter 4 is 1

moreover, you cant look at the lateral surface of an unrolled cone as a triangle

because its not

[novel pecan]

How do you preserve the area then if you strech the circle to make a square

[spark stag]

calculations.

theres no way with compass-and-straightedge to do this

it's known as 'squaring the circle', and has been proven impossible

the simplest way, though somewhat unsatisfying:

measure the circle's radius, calculate the area

take the square root of that area

make a square with those side lengths

this will require changing the size of the boundary (the circumference will be different than the perimeter), but it works.

if you dont care about preserving area

then sure, your quadrant method works

cut the circle into 4, fix (hold down) the ends of each line segment

and then just stretch them into a corner

and glue them back together

[novel pecan]

I was just thinking about whether that would be useful in some problem the future, but seems unlikely

[olive sinew]

So I checked my answer at the back of the book I got 68 dg but there’s another answer next to it

[devout shell]

were you given an interval initially?

[olive sinew]

From where 112dg is coming from

Yes but my professor told me to not worry about, let me send a picture of the original problem

63

[devout shell]

the interval is important though lol

because that opens the possibility for two solutions

an sine of an angle in quadrant 2 has the same sign as an angle in quadrant 1

[olive sinew]

I know it is but she said in class that we’ll discuss the intervals later in the future

[devout shell]

so that means there is going to be another angle that is the same as 68 dg in quadrant 2

90-68=22
90+22=112

[olive sinew]

Oh I see

[subtle gate]

Is there a point in every non-complex polygon where if you connect every vertex to that point, the links are all inside the polygon? (I couldn't find it on google).

[dark sparrow]

what's a "complex" polygon

also please get a nickname that isn't fucked up

[subtle gate]

if any sides of a polygon intersect another side of the polygon it is considered complex also

i cannot type your name symbols either

[dark sparrow]

ok so self-intersecting?

my discord username is Ann, so i can be pinged no problem

[subtle gate]

if any sides intersect, it is complex and i don't care about it. i am looking at non-complex ones

[dark sparrow]

you cannot

ok, so you want non-self-intersecting polygons

[subtle gate]

yes

why can i not nick myself re

(i am sorry bot for now i can't change it)

[dark sparrow]

<@&268886789983436800>

[rugged wind]

I can change it if you'd like.

Have it like that as the default to prevent username abuse like zalgo text.

[dark sparrow]

anyway

p sure this polygon doesn't have any point of the kind you want

[subtle gate]

i guess that is good enough to show it's not true

also just please change my username nick to "Glitched Pixels"

this is the first server asking to change the base name, where in others people just asked to have a diffrent nick

[dark sparrow]

different nick is fine

[rugged wind]

No it's fine, I changed your nickname

It's more so that if we force change someone's username for being deliberately annoying they can't just change it or their default name visibility

[dark sparrow]

why tf do people even do this with their names

'tis a mystery

[subtle gate]

why do people make nicknames weird so that you have to click on their profile to get their name

[rugged wind]

Nah it's because you can make your nickname using text like zalgo where it either does something funky and lags peoples' clients, or is just really long and goes out of the normal name space

[subtle gate]

please don't take offense and let's not get too off-topic

(but thanks dGhost c:)

[rugged wind]

all good mate

[subtle gate]

i can just now see that i don't actually need that property because cutting a polygon into lots of triangles isn't that miraculously difficult

[upper karma]

Hey I'm back to brag about how bad I am at using formulas, How would one go about obtaining the answer to this problem? because I can't figure out how to reverse engineer this at all

[mighty narwhal]

Um

ACX is 30deg

[upper karma]

The formula it recommended me use was A^2 + B^2 = C^2 but I have no understanding of how to do so

[mighty narwhal]

Oh, if you can use that

AB and BC are also 8 miles

AX is the half of AB (because CX bisects ACB and this is an eq. triangle)

So we have right triangle ACX

Let the 3 sides be A B and C

A=4
B=x
C=8

$4^2+x^2=8^2$

[somber coyote]

DarK:

[upper karma]

wait why is A different from C aren't both just one point?

[mighty narwhal]

Can you solve this?
The x will be the heigh
$x=4\sqrt{3}$

[somber coyote]

DarK:

[mighty narwhal]

?

??

[upper karma]

oh wait I'm dumb

[mighty narwhal]

A B and C means sides

Not the points

AB = c
AX=a
BX= b

[upper karma]

Okay so 4^2 + x = 8^2, so X would just be 4^2 wouldn't it? why is it 4 sqrt(3)? where does the three cme from

[mighty narwhal]

Um no

16+x^2=64

[upper karma]

so x = 48?

[mighty narwhal]

Yes

No

x^2=48

[upper karma]

wait no sqrt(48) = 6.9 and that equals 4sqrt(3)

right

[mighty narwhal]

Yes

Also

The height of an eq. triangle is $\frac{a\sqrt{3}}{2}$ with a being the side

[somber coyote]

DarK:

[mighty narwhal]

$\frac{8\sqrt{3}}{2}$

[somber coyote]

DarK:

[mighty narwhal]

$4\sqrt{4}$

[somber coyote]

DarK:

[quiet mason]

can any of you help me

solve this crazy boi

take your time and if you solve @ me

prob 22

[mighty narwhal]

Try #competition-math

Im not able to solve this rip

[subtle gate]

no means?

no cauchy?

[quiet mason]

nothing lmao its a competition problem

from india

i think i need to stop trying to solve it

[subtle gate]

i can't believe there is neither means nor cauchy schwarz

[upper karma]

I find it funny that I don't know how to solve these but the question answers make it easy for me to get the right one through simple elimination of impossible answers, almost as funny as the fact that I don't know this stuff after taking 4 years of math in high school

[mighty narwhal]

Did you learn trigonometry yet?

[upper karma]

I learned it last year, or at least I was taught it, didn't appear to retain anything though

except the sohcahtoh thing or whatever, I can probably still do that stuff

[mighty narwhal]

Well nevermind

We have a formula that

$h^2=xy$

Where x and y are 3 and ? in our case

[somber coyote]

DarK:

[mighty narwhal]

x and y must be the two part of the hyp which is divided by the height

$7^2=3\cdot?$

[somber coyote]

DarK:

[mighty narwhal]

Is this clear enough? @upper karma

[upper karma]

I'm trying to understand it, idk

I feel like a computer with 4g of ram trying to run multiple tabs of google

I can't piece it all together effectively

So x and y are 3*?, and xy = 12 yeah? which means ? must be 4?

So how was the 3*? gotten from 12 to begin with? or is it always just 3 * ? to solve for this kind of thing

sorry for having the mental equivalence of a 2nd grader when it comes to math

[tender token]

Hello everyone

tomorrow i have exam can anyone please help me to solve this question?

Thank you so much in advance

[mighty narwhal]

Okay, first you might want to complete the square

$(x+3)^2-9+(y-1)^2-1=7$

[somber coyote]

DarK:

[mighty narwhal]

$(x+3)^2+(y-1)^2=17$

[somber coyote]

DarK:

[mighty narwhal]

Is this clear?

[tender token]

yes

[mighty narwhal]

brb

So lets focus on l1

[tender token]

Okay

[mighty narwhal]

we know that it passes tru point (0,6)
6=m0+b

From this we know that b=6

We only need to figure out m

[tender token]

its clear till here

[mighty narwhal]

Point P must be on the circle's circm

So point P is \sqrt{17} away from the centre

$P(x_0,y_0)$

[somber coyote]

DarK:

[mighty narwhal]

$(x_0-3)^2+(y_0+1)^2=17$

Wait

[somber coyote]

DarK:

[mighty narwhal]

Errrmmm

[tender token]

what?

[mighty narwhal]

Ummmm

I may have messed up

[tender token]

do you have time to solve it in paper and send it to me, pleaseee?

i have tried to solve it but i could not

[mighty narwhal]

I was trying to get point P in terms of x_0 only

But I realized that this doesnt guarantee that the equation I get will be a tangent

[tender token]

like i spend more than 1 hour to find the solution of this but i could not

[mighty narwhal]

Did you learn linear algebra yet?

Especially vectors

I think they might be useful here
but sadly I didnt learn any of that stuff

Either is this a really hard question, or we dont know something that would make this easy?

[tender token]

well i used to but the thing is that tomorrow i have a pre-exam for entering to a course so i had to know the answer of these questions

maybe the question is so easy but i do not know how to solve it

[mighty narwhal]

Well, im not sure

I saw some problems similar to this, but I dont think there is a trick to this one

You just have to write down everything you know, get a lot of equations with a lot of variables and then solve them all...

I dont have the brain for this, but try this:

You have 4 lines

PA, QA, PR and QR

write the equation down for every line

PR and QR will have b=6

and the m for PA and QR will be the same
and the m for QA and PR will be the same (they are parallell)

PA and QA passes tru point (-3,1)

And you may be able to get things like this
Then you will have a lot of equations and you can supposedly solve them

[tender token]

Thank you so much @mighty narwhal for you help i will try to solve it

Thank you so so much

[mighty narwhal]

Your welcome, but idk if this will work tho

[tender token]

Thank you for you time i really do appreciate

[mighty narwhal]

[olive sinew]

Is this right?

[ornate lodge]

So the solutions to cos2x=0 are x = pi/4 + pi(n), x = 3pi/4 +pi(n). How can I find the number of solutions in the interval [20,50], without just listing them individually?

[dark sparrow]

well a good start would be to realize that these two families can be combined into one

namely, $x = \frac{\pi}{4} + \frac{\pi}{2}n$

[somber coyote]

Ann:

[dark sparrow]

and then you'll need to find the number of integer solutions to the inequality $20 < \frac{\pi}{4} + \frac{\pi}{2}n < 50$

[somber coyote]

Ann:

[ornate lodge]

I see. Thanks

[night karma]

I made my own math problem! 😬 ( edit: not 100% on scale )

[lusty quest]

perfecto mama

[olive sinew]

I need help

With my homework

83

What I got so far

I tried to find the inverse of sin(theta_2) and I did not get the answer I was expecting

Can anyone please take the time to look at what I got, thank you

[olive sinew]

@devout shell

[fringe dirge]

Read the rules in #❓how-to-get-help please

[olive sinew]

I can’t

Right now I’ll read it some other time

[night karma]

Anyone tried out my math problem already? 😋

[austere haven]

It's...hard

(tbh i haven't tried it, I'm in algebra 1 rn lol)

[night karma]

Not sure how the gradesystems work there, but I'd like to believe!

[floral igloo]

is that 60° 80° and 20°

[fast tinsel]

@night karma I'm gonna take a crack at it, looks awesome

[night karma]

It is, yes

Good luck!

And thanks 😉

[fast tinsel]

All my friends and I are stumped somehow so good job lol

Maybe Imll have an answer in a bit

[night karma]

Good to hear, haha. There's some symmetric parts though

[upper karma]

sup fam

i have a questino about trig operations. not a problem to solve. im rather looking for explanation

is this the right channel?

or do i go to a problem solving channel?

[nimble mason]

@upper karma Ask

[upper karma]

cooollios

ok so

[nimble mason]

Not that I can necessarily answer it

[upper karma]

lol

ill go back to the question channel 😛

[nimble mason]

Ok

This is still a pretty good channel for that though

[upper karma]

Can someone pls help me with 819? You need to proove the equality between the two sides

I tried this but it didnt go well

[chrome fiber]

third line

it's +tan

[upper karma]

k

so what do i do after i correct that?

@chrome fiber

do i just use those formulas like tg(a+b)=...?

[chrome fiber]

yeah

should be simple

use it on the first two terms instead of the third one

$\tan A + \tan B = \tan(A + B)\cdot(1 - \tan A\tan B)$

[somber coyote]

.jun:

[upper karma]

tf i dont have this in my math textbook

oh god

im dumb

[vagrant elk]

you know tan(a+b) yes?

just

[upper karma]

yeah yeah

[vagrant elk]

slide the fraction up

[upper karma]

i just realized that too

tnx

[chrome fiber]

haha np

[gloomy pond]

can anyone explain this?

[dusty pagoda]

^ pulling this up from the internet as an example

Can I say ∠6 and ∠3 are alternate interior angles?

[zenith ember]

Yes

[dusty pagoda]

Thank you!

[spark stag]

,rotate

[zenith ember]

My instinct says start with the smallest value cos takes

But that doesn't work

Next instinct is to convert to ratios of related triangle sides.

Ok..

[upper karma]

Ok..

[fringe dirge]

Giving up on things that are hard is the best way to stay bad at something

[zenith ember]

To be fair, that's not a beginner's trig problem.

And I think it's totally ok to recognize that you're not up to that level yet.

[upper karma]

what about some x where cos(x) = sin(x)?

so like 0

sounds like a good value to me

woops I mean

pi/4

shit idk

yuh pi/4

[worldly junco]

can someone check this for me

[mental wharf]

looks right

[worldly junco]

thanks

[spark stag]

And you sure are nice.

(also, this clearly seems like some sort of competition-esque problem, or at least not just jamming calculations; what good does doing problems you already know do?)

[fringe dirge]

@upper karma if you really are doing competition math, then take it from someone who's made usamo that you really should be doing one problem for the next 3 hours

do you know what the AIME is

top 250 students on aime make usamo

[spark stag]

It's the final qualifier round for the US' IMO team

[fringe dirge]

of course

[spark stag]

I'd be surprised if someone in competition math didn't

[fringe dirge]

@spark stag kinda not true, there's a TST at MOSP

@upper karma In terms of books?

Idk it's been like two years since I was in high school

If they haven't released anymore books in teh last two years

I read every single one, did every single exercise

idk what that is, I think that's new

I did the WOOT class one year too

yeah that's new

yeah his abstract algebra book lmao

It's not about the quantity of problems at all

Like my advice earlier that you were rude to, it's about doing problems that are outside your comfort zone

No because new ideas will come to you if you keep thinking

there's a fine line between doing problems that are too hard and problems that are too easy

So it's safer to just play on the side of harder questions

I didn't do amazingly well on usamo

I was studying with a friend who made mosp and knew that I couldn't get to that level in the the time that I had

a little less than 200 iirc but yeah

yeah some

most of which I can't remember

some of andrescuu's books

think the algebra problems one

"think the algebra problems one"

one of them can't remember

damn mb

too busy doing research sorry

yep

sure how much you want to pay me

There's a competition math discord you might be interested in

It's in the channel description for #competition-math

depends what you want me to do

Eh I mean, having someone teach you the nice approaches can help sometimes

but figuring it out yourself can help you get better at solving problems in the long run

anyways i don't really do competition math anymore

there is a college competition math, but I'm not super into it

if you really want me to, there are probably better people for what you want but I"ll do it I guess

[upper karma]

this is the last question on my exam

does anyone know how to solve it?

[rustic dagger]

I do owo

But I can’t help u cheat on an exam

Even though it’s prob over by now lol

[upper karma]

Can I send my problem here now?