geometry-and-trigonometry

[snow karma]

prob some stuff related to modular equations as well

[warm shuttle]

actually that's probably in the Elements as well

i didn't even realise

[zenith warren]

any1 alive rn

?

??

[obsidian harness]

you are a figment of my imagination

[zenith warren]

i have a general doubt

let me eat then i tell

yeah

tan (360-θ) = θ
tan (360+θ) = θ
Now
I want to find the value of tan (1040)
So, first I take
tan (11π/2 + 50) = tan (50)
second
tan (1080 - 40) = tan (40)
[1080 = 6π]

@obsidian harness

[obsidian harness]

no that's not correct

tan(1080 - 40) = tan(-40)

now, tan(-x) = -tan x

[zenith warren]

ohk

[obsidian harness]

$\frac{\sin(-x)}{\cos(-x)} = \frac{-\sin x}{\cos x} = -\tan x$

[somber coyote]

southy

[obsidian harness]

sin is odd and cos is even

[zenith warren]

thx

[obsidian harness]

no worries!

[cinder plover]

Looking for sb to study with

Regularly when we both have time

[zenith warren]

sb

?

wdym by sb

salmaan bhai?

[tribal rose]

They mean a study buddy, I believe.

[zenith warren]

how can I forget this

[faint pasture]

antimony obv

[zenith warren]

No that Sb

[faint pasture]

Maybe their capslock key was broken

[zenith warren]

shift X 2

[faint pasture]

Maybe that was broken too

Maybe they were using speech to text

[zenith warren]

so he would have said antimony

[faint pasture]

Maybe he was just lazy

[zenith warren]

@cinder plover How can you be so lazy

@faint pasture I have a chem doubt

[faint pasture]

Bold of you to assume i know chem 💀 but ask away there like a 3.2% chance i can help

[zenith warren]

leave this

[faint pasture]

Electronegativity should do ig?

[zenith warren]

nah not att all no connection

[faint pasture]

Oh no wait electron affinity would

[zenith warren]

ik F most electronegative

but no

[faint pasture]

Well oxidizing agents get reduced so they gain electrons, and Cl- has higher electron affinity than F- iirc soo yea Cl- should be a better oxidizing agent

[zenith warren]

I don't understand, If there is a oxiding agent it reduce itself how is it gaining e-

[faint pasture]

Well reduction is just a reduce in oxidation state, which is just a gain of electron

[faint pasture]

Nvm i was tripping hard ignore this

[zenith warren]

soka

[faint pasture]

Tho electron affinity is what you want

[zenith warren]

ya maybe but I understand

[faint pasture]

(And by an extended means electron gain enthlapy)

[zenith warren]

yeah

do you know backward reaction, I wanted to know

[plucky abyss]

Endothermic reaction

[zenith warren]

wait backward reaction are endothermic

?

[faint pasture]

No i dont, sorry

[zenith warren]

but i know endothermic

[plucky abyss]

Idk

[faint pasture]

Endothermic is just those which take in energy backwards reaction are different at least definition wise

[zenith warren]

this is endothermic

?

[zinc epoch]

chemistry in maths discord

[plucky abyss]

Math to balance reactions

[warm shuttle]

i've heard chemistry is just physics + stats

[upper karma]

~~https://discord.com/channels/268882317391429632/1018703193473024061~~ i solved it, thanks

Is my reasoning correct?

geometry no bueno

[plucky abyss]

Math > Physics+ Stats

[shell lantern]

?

[plucky abyss]

Is calculator allowed

[exotic yarrow]

Hint: Cosine double angle formula: $$\cos(2\theta)=2\cos^2 (\theta)-1$$

[somber coyote]

Civil Service Pigeon

[shell lantern]

How?

[shell lantern]

One minute

Understood

[shell lantern]

Example 8?

[exotic yarrow]

Hint: Sine double angle formula: $$\sin(2\theta)=2 \sin(\theta) \cos(\theta)$$

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

Further hint: ||Consider what you can do with sin(pi/15)||

[shell lantern]

One minute

?

Not coming

[exotic yarrow]

try smt else

[shell lantern]

How?

?

[flint shale]

this gives me flash backs

[shell lantern]

Got it

Example 8 jee question?

Sorry example 9

[exotic yarrow]

Rewrite $\sin^2 \frac{\pi}{16}$ and $\cos^2 \frac{\pi}{16}$ using the cosine double angle formula

[somber coyote]

Civil Service Pigeon

[shell lantern]

One minute

Cospi/8 value?

[exotic yarrow]

uh

what did you do

[shell lantern]

Can u point out the mistake

[exotic yarrow]

I didn't read it that closesly

b/c it's all

basically irrelevant

[shell lantern]

Ok

One

Ok crct first step?

[trail tendon]

what crct mean

[exotic yarrow]

probably "correct"

[exotic yarrow]

this is relevant

the rest is useless

[shell lantern]

Then how to do can u teach me?

I am not expert as u

In trigonometry

Just learning

?

[exotic yarrow]

put $\theta=\frac{\pi}{8}$ and you get $\cos^2 \frac{\pi}{16}=\frac{1}{2} \left(\cos \frac{\pi}{8}+1 \right)$

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

wait hold on

is what you did relevant?

It's so messy I can't tell

💀

[shell lantern]

Ok i will try this way

[exotic yarrow]

ok just rewrite all the squared stuff using double angle and tell me what you get

[shell lantern]

Cos²tete-costeta/2+1

[exotic yarrow]

I can't read that.

[shell lantern]

1/2

[exotic yarrow]

did you mean to put 0 on the left hand side of the equal sign on the last two lines...

[shell lantern]

Why zero?

[exotic yarrow]

then why are there equal signs

[shell lantern]

Ok no zero

No equal sign

[exotic yarrow]

so now you've just written expressions

devoid of any context

[shell lantern]

?

?

[flint shale]

i'm having a bit of trouble with this one

i did this first but i can't find a value for x

[maiden brook]

that length wouldn’t be 5 (in red) unless it’s a trapezoid

[twin spindle]

I'm confused on what to do w this question

[flint shale]

but didn't they give 5

[maiden brook]

transformations of functions. These transformations apply for all functions

[maiden brook]

it’s unclear what the 5 is meant to mean precisely

[snow karma]

the lines arent said to be parallel

Thus u cant say the red line is 5 too

[flint shale]

oof

[twin spindle]

So would y=cos(4x)-3?

[flint shale]

is the question possible?

[maiden brook]

close by no

[twin spindle]

cos(4(x-1))-3?

[maiden brook]

it’s clear that this doesn’t exactly work cause it’s still cos(something), which will output a number between -1 and 1

so the range is the same, but u got the -3 part right

[maiden brook]

why?

[twin spindle]

I'm jus confused abt what it means by "a factor of 4"

I kinda guessed that i had to factor out a 4 or sum.

[maiden brook]

as an example

this should be clear though

[snow karma]

Pretty sure theres a missing info

[flint shale]

oh ok

[snow karma]

Cause visually, its pretty impossible

Without defining that quadrilateral

[flint shale]

that is what i thought

[maiden brook]

(@lil Asian do u see what the numbers here do)

@twin spindle

[twin spindle]

Yeah

[maiden brook]

what do they do?

[twin spindle]

The higher the y value, the longer the graph stretches

And vise versa for a lower value

[maiden brook]

if a graph is stretched vertically by a factor of k, then for each point on the graph (x, y), it becomes (x, ky)

[maiden brook]

u see how to write the transformation though?

[twin spindle]

Yeah, I think so?

W the formula Acos(bx-h)+k

[maiden brook]

wdym “W”?

[twin spindle]

"With"

Mb mb 😭😭

[maiden brook]

I don’t understand why teachers teach about transformations of parabola, then separately like transformations about basic exponential functions and THEN SEPARATELY with the common trigonometric functions

but anyway

[maiden brook]

u should understand how that works u shouldn’t need that formula

[maiden brook]

do u see how to do the problem though?

[twin spindle]

I think so

[maiden brook]

what do u have?

[twin spindle]

The main thing I'm confused of is how do you find a sin equation, w a cos graph?

[maiden brook]

wdym? sine is cosine, just translated a little bit

[twin spindle]

Nah cause it's asking "find the equation of a sine wave that I'd obtained by vertically stretching the graph f(x)=cos(x)"

[maiden brook]

can u show me what it says exactly and like the choices?

or what it says

like a screenshot pls

[twin spindle]

This is the whole question

No multiple choice

That's the info it gave me and that's it 💀💀

[maiden brook]

oh that’s annoying it’s unclear if it refers to sine in particular or the trigonometric functions

if it doesn’t matter so much if u get this question right or wrong just use cosine I would say I’m not sure if sine wave refers to sine in particular (for example sinusoidal generally doesn’t not)

[twin spindle]

Alrr thank you 🙏🙏

Their's also one more problem that I'm stuck on

The horizontal shift is -5pi/4 and when I shifted the midline -5pi/4 from the x-axis it says I'm getting wrong 💀💀

[obsidian harness]

your horizontal compression is wrong

you've made half a period 3pi/2 - (-3pi/2) = 3pi

instead of half a period being pi

[half wave]

Look at these questions

How in did they get these answers? I cant seem to understand

[north kindle]

i do not understand this problem

what does the tire diameter have to do with the speedometer?

[tribal rose]

I had to Google this to make sure.

If the tires have a larger diameter, it underreports your speed because it calculates based on tire rotations.

[north kindle]

ah

so like, if i doubled the radius of the tires, it would half the reported speed?

[tribal rose]

That seems to be the case, but someone would have to correct me on that.

[north kindle]

alr makes sense

[tribal rose]

I think tire size is proportional to how accurate the speedometer is.

[half wave]

..

[obsidian harness]

If you add 360 degrees to an angle, the angle does not change position

Also 360 deg = 2 pi radians

[warm shuttle]

call vertices ABCD (top left, top right, bottom left, bottom right).
Choose coordinate system such that
C = (0, 0)
D = (2, 0)
Then we have
B = (2 + 5/tan(80°), 5)
we don't know angle ABD, but let's pretend that we do. Then we have
A = (2 + 5/tan(80°) + 6cos(ABD-100°), 5 + 6sin(ABD-100°))
We have law of cosines that says
c² = a² + b² - 2ab cos(angle)
we can rearange for angle to get
angle = arccos[(a² + b² - c²)/(2ab)]
Or in our case:
θ(ABD) = arccos[(6² + (2 + 5/tan(80°) + 6cos(ABD-100°))² + (5 + 6sin(ABD-100°))² - (2 + 5/tan(80°))² - 5²)/(2 * 6 * sqrt((2 + 5/tan(80°) + 6cos(ABD-100°))² + (5 + 6sin(ABD-100°))²))]

(260°-ABD)*

[snow karma]

Consider (C) a circle with center O and diameter [BC] and A a point of (C) / { B ; C }. M belongs to [BC] such that 3CM = BC
. (C') is a circle with center I' and diameter [CM] where (C') cuts (AC) in a point H. A' is a point of (C) \ { A ; B ; C } such that:
(A'B) // (AM) and I = A' * A where I is a ppint on [BC] \ { B ; O ; M ; C }
If S is the direct similitude with center H such that:
S: A •→ M, what is the angle of S and its relation?

[stuck sleet]

Why I cannot solve this?

The variable is always eliminated

yet

there is an intersection of line

[vernal pilot]

Do you know matrix row reduction

[stuck sleet]

No

[vernal pilot]

Fair enough

[stuck sleet]

Is there an alternative?

[vernal pilot]

What’s that meant to be?

[stuck sleet]

I forgot it what it is

[vernal pilot]

It looks like you’ve just subbed the equation into itself.

[stuck sleet]

But I use the two equations, where should I substitute 3x + 2y = 48?

[vernal pilot]

Here’s how I would do it: in both equation 1 and 2, move everything to one side. Set both equations equal then rearrange for y.

[grave pond]

The red-circled equation is 2y+3x=48 with the expression for y substituted in.

[vernal pilot]

Yes, which is where the error is.

[grave pond]

Ah yes. Sorry, I didn't read enough of the context.

[vernal pilot]

All good

[stuck sleet]

I still can't understand the problem...

[shell lantern]

?

80-83?

Jee pyq's

?

[cyan summit]

hwyyyyy

[rain glacier]

Does anyone know how to check ambiguous case

[pseudo nebula]

of what

[plucky abyss]

Person

[jovial wedge]

isnt the answer here wrong for c?

shouldnt the answer be only within the second and third quadrant where cos(x) is negative?

[frank patio]

yeah the answer for 2x should be

and it is

but since we divided by 2 for x, some stuff will go in the first quadrant

but the sec is evaluated at 2x so thats what matters anyway and that is indeed in the second or third quadrant so theres no problem

tldr: 2x needs to be in the second/third quadrant, x doesnt

[jovial wedge]

oh okay thanks

quite weird to understand at first though

trying to get through this haese green book before starting my IB math aa hl first school year xd

[narrow pasture]

How do we find the angles for these three questions? The reasoning also please

[pastel gate]

My mind is fried how do I solve this?

[trail tendon]

DO/CO = BO/AO, pretty sure its just a geometry rule

[silver mason]

Anyone want to solve the hardest problem from a Russian common 7th grade geometry textbook?

[snow karma]

Thales theorem

Easy

[silver mason]

Inside the isosceles triangle ABC with base BC, a point M is taken such that ∠MBC = 30°, ∠MCВ = 10°. Find the angle AMC if ∠BAC = 80°.

[snow karma]

Is M over or below segment [BC]?

[trail tendon]

above

it says "inside"

well

[trail tendon]

i guess it depends on how you draw it

but its inside the triangle yeh

[silver mason]

M is inside the triangle ABC. M doesn't lay on the line BC of the segmant [BC]

[snow karma]

Oh yeah it says inside

im a dumbass

[opal ibex]

Would anyone be familiar with the formula "Acos(cx)+B"?

[trail tendon]

it's a cosine function without a phase shift

[opal ibex]

Thank you, would anyone also be familiar with determining amplitude using a point on a graph?

[rough crater]

“Express in terms of sine only” is this correct?

[rough crater]

What points do you have

[warm shuttle]

https://en.wikipedia.org/wiki/Cyclic_quadrilateral
https://en.wikipedia.org/wiki/Circumcircle#Angles

Also see inscribed angle theorem, Sum of angles in a triangle, and sum of angles that make a straight line

[mild dagger]

can anyonde help with this problem I'm having trouble with what steps I should take next.

this is my steps so far

[opal ibex]

P(53.2,3802)

[opal ibex]

Square both terms, add them together, and the square root of that will be the first half of the answer. The OTHER half will be the cis(arctan(b/a)).

[rough crater]

Sorry thought you had the max and min, dont know any other ways

[opal ibex]

That's alright, thanks for offering to help

Gosh, I hate signature assignments with a passion

[warm shuttle]

this seems pretty impossible

[opal ibex]

Agreed. This is clearly one of those problems that expects you to be more intuitive. I don't know why project assignments are made this way.

[warm shuttle]

kinda crazy

If G is centroid, then you can drop height AG

so you have triangle AHC and you know 2 of its angles

[opal ibex]

This is definitely advanced

[warm shuttle]

If G is centroid then it divides the medians 2:1 is the property i think

if you call point at angle x M, and point below that N; then GM=AM -> BM = 3AM -> BM:AM = 3:1 -> MN:AN = 3:1 -> tan(x) = 1/3

I don't think there is correct answer

If my reasoning is solid it should be closer to 36.86989.../2

[hoary prism]

find the height

[pastel gate]

why is the answer a circle

[grave pond]

The whole configuration is symmetric under rotations about the line that connects the two sphere centers. Surely that includes the intersection.

Why do you think it should be an ellipse?

[pastel gate]

Oooh I get it now

I wasnt sure of the answer I was thinking in 2d even thought I was told it was three dimensional my bad

thank you

[charred furnace]

The square ABCD has a side length of 3. A point E is selected on ray AD (through D) such that DE = 4.
Draw circle O which is tangent to both the midpoint of EC and a point F on line AB. The radius of O may
be expressed in the form r/s where r and s are relatively prime positive integers. Find r + s.

[warm shuttle]

@charred furnace Did you solve the folding in equilateral triangle problem? What was the solution?

i wanna know

[charred furnace]

I didn't

I was clueless and no one waa responding so I just asked for help for other simpler geom problems I couldn't solve

[pastel gate]

the answer is said to be a) but why is that?

[pastel gate]

nvm found itt

[grave pond]

Would it have killed them to include a sketch instead of just a description in words?

[pastel gate]

ikr hahaha

whats the logic behind this one

[steep thistle]

we are evaluating which graphs look similar only under the given interval

observe the the variable x is only within the arguments of the trigonometric functions given. so it can be thought of as an angle value input into these functions

[obsidian harness]

note that cos |x| is the graph of cos x where x > 0, and then reflected across the y-axis to replace the part when x < 0

because cos |-x| = cos x for x > 0

[steep thistle]

focus on C1:
it might be easier to break apart the questions into smaller pieces...
start by evaluating for when x>0
well when x>0, x = |x|, so cos(x) and cos(|x|) are the same here
now look at x<0
when x<0 |x| = -x, so cos(|x|) = cos(-x), however we can recall that cosine is an even function, hence cos(x) = cos(-x), thus cos(|x|) = cos(x) over this interval too.
Lastly, we can affirm that cos(0) = cos(|0|).
We have shown that these graphs are the same over the entire interval.
We can apply this process for C2 and C3 to evaluate if they are the same or not

[obsidian harness]

the other two you should be more comfortable with: for C3 for example, try sketching out $\sqrt{x^2}$

[somber coyote]

southy

[obsidian harness]

pay attention to what happens when x is negative

so that applies to when tan x is negative

[steep thistle]

I believe the correct answer is only C1, answer choice A

[obsidian harness]

ah wait I mixed up sin and cos

[obsidian harness]

yes A is correct

[steep thistle]

C2 equivalence counterexample:
let x = -pi/6, sin(x) = -1/2, |sin(x)| = 1/2. Thus when x = -pi/6, the graphs are not equivalent.
C3 equivalence counterexample:
let x = -pi/4, sqrt[(tanx)^2] = sqrt[(-1)^2] = sqrt(1) = 1,
tanx = -1, hence the two graphs are not equivalent at x = -pi/4

[steep thistle]

hope this helps @pastel gate

[pastel gate]

Yess thank you!

[sick walrus]

I need help with these two

[trail tendon]

for the second one, can you make one equation?

xD

[sick walrus]

I was js not entirely sure on how to solve it

as in how to write it out

[trail tendon]

well

the angles in a triangle add up to 180 right?

[sick walrus]

yes

[trail tendon]

what are the three angles

[sick walrus]

86, (4x-7), (7y-1)..? but I didnt solve them yet

[trail tendon]

oh i was talkin about the second one

u wanna do the first one first?

[sick walrus]

oh sure

[trail tendon]

ok

[trail tendon]

so yeah this is true

so 86 + 4x-7 + 7y-1 = 180, right?

[sick walrus]

yes

[trail tendon]

do you also know how uh

this is true

[sick walrus]

so do I js use that equation?

[trail tendon]

forgot what this is called but 💀

[sick walrus]

uhh supplementary angles?

[trail tendon]

yeah

do you see the supplementary angles in ur picture

[sick walrus]

yea

I see it

[trail tendon]

what are the angles that add up to 180 deg

[sick walrus]

wait r they js the same ones u said..?

86, 4x-7, and 7y-1

[trail tendon]

nah, theres only 2 angles this time

[sick walrus]

so 9x+4 and 7x-1?

[trail tendon]

teah, 9x+4 and 7y-1

so 9x+4 + 7y-1 = 180

[sick walrus]

yea

how abt 2nd one?

[trail tendon]

86 + 4x-7 + 7y-1 = 180,
9x+4+7y-1 = 180

two equations, two unknowns

[sick walrus]

ohh

[trail tendon]

yea

u know how to do from there?

[sick walrus]

yea

[trail tendon]

ok kewl

[sick walrus]

so recap: when there is a figure like that, then u would spilt to two eq??

or js whenever there is a value given..?

[trail tendon]

if you have two unknowns you should try to find two equations

[sick walrus]

kk

got it

[noble knot]

is it possible to calculate the circumference from a circle segment, but only the circumference from the segment itself.

nvm got it, its 2* r* pi * theta/360

[vernal pilot]

are you still having trouble with question 1?

[plucky abyss]

[harsh pond]

How can i find the initial speed in x

[warm shuttle]

oh there is a cool solution

[warm shuttle]

but it's not geometry

[harsh pond]

Did you know any server that i can ask about this physic nightmare?

[grave pond]

At 30°, the vertical part of the initial velocity is exactly half of the launch speed.

[upper karma]

How did "50x" get there?

[warm shuttle]

magic

but also distributive property

[void mortar]

(a-b)^2 = a^2 + b^2 - 2ab apply this identity by replacing a=25 and b=x

I hope that helps @upper karma

[hasty dew]

@harsh pond

[upper karma]

It does, thank you!

[worn stag]

Wtf is that notation

[snow karma]

Solve for his instagram lmao

[hasty dew]

U : initial velocity
Ux: initial velocity along x axis
Uy: initial velocity along y axis
@: is just the angle of initial velocity

[warm shuttle]

i was gonna say vertically:
(1/2)mv² = mgh -> v = sqrt(2gh) ≈ 16.573 m/s
total v ≈ 33.15 m/s

[hasty dew]

Great solution 👍
(kinetic energy at ground = potential energy at maximum hight)

[zealous pike]

Hey

How do I prove that the number of diagonals of a point in a polygon is n-3?

To get the total number of diagonals of a polygon of n sides I would use the formula

$\frac{n(n-3)}{2}$

[somber coyote]

Pi, a future fluent jp speaker

[zealous pike]

n-3 is the number of diagonals per point

[obsidian harness]

oh you should draw some polygons out to see

one of the points is the point you chose

two more points are given by the sides of the polygon

so that's n - 3 possible points left to make diagonals

[warm shuttle]

oh

[peak nymph]

Can anyone help me with this?
343. Scrooge McDuck's money bin is surrounded by a 4.0-meter-high fence at a distance of 12.0 meters. The Beagle Boys are trying to climb the fence from the outside using a ladder to reach the wall of the money bin. How long must the ladder be at least to reach the wall of the money bin?
Hint: Let x be the distance from the bottom of the ladder to the fence and y the distance from the top of the ladder to the ground. You can find the relationship between the variables x and y by using similar triangles.

[brazen epoch]

They cant have a diagonal with him and with the 2 neighbors points so there remains n-3

[trail tendon]

what does "at a distance of 12.0 meters" mean

does it mean the ladder is 12.0 meters away from the fence?

or wut

or does it mean 12.0 meters away from the money bin

i feel like this is a stupid problem either way 💀

[plucky abyss]

You have to go deeper to understand this question

Atleast 12 m deep

[peak nymph]

it means the fence is 12 meters away from the wall of the money bin

[trail tendon]

ok then this is a stupid problem 💀

[peak nymph]

ik it's annoying

[trail tendon]

if you have a triangle

with a height of 4

a length of whatever you want

whats the minimum hypotenuse length

[peak nymph]

ye

[trail tendon]

u know?

[peak nymph]

the answer should be 21.6 meters but I have no idea how to get to it

[trail tendon]

uhm

now i dont understand again 💀

this is what i interpret the problem as

the 12m is completely irrelevant, we're trying to minimize y

[peak nymph]

n

no

[trail tendon]

whats the square box?

[peak nymph]

like y is the height on which the ladder touches the bin

[peak nymph]

the money bin

[trail tendon]

what happened to the wall?

[peak nymph]

what wall?

wdym

[trail tendon]

the fence?

[peak nymph]

it's the small one

[trail tendon]

the small what?

[peak nymph]

that the arrow and 4m point at

line

[trail tendon]

oh

[peak nymph]

here's the one from the book if it helps

[trail tendon]

ohhhhhh i thought "the wall of the money bin" meant the fence 😭

[peak nymph]

ohh

no not like that

[trail tendon]

hmm i see now

[kind rapids]

You should just post this, tho

[peak nymph]

ye I forgot sry guys

[plucky abyss]

[peak nymph]

and because there are like two monomorphic triangles you should probably find x and y with the help of them or smth

the hard part is how tf do I get an equation of it to see the shortest possible ladder

[trail tendon]

ah

hmm

[peak nymph]

like because it is in a logarithm and exponent book with derivatives I would think you need to use them. I think with an equation you could from its derivative form find out the minumum and that is the smallest possible ladder?

[kind rapids]

So we have: $y = \frac{4(x+12)}{x}$ right?

[somber coyote]

Closer

[trail tendon]

oh LOL i thought we were supposed to do it without that 😂

[trail tendon]

how did you get that?

[peak nymph]

I'm not sure maybe?

[kind rapids]

Similar triangle

[trail tendon]

x/4 = (x+12)/y
y = 4(x+12)/x gotcha

[kind rapids]

Using Pythagorean theorem we have: $L = \sqrt{x^2 + (\frac{4(x+12)}{x}})^2$

[somber coyote]

Closer

[trail tendon]

😔

[kind rapids]

Wait

[peak nymph]

oh now I see

[trail tendon]

(x+12)^2 but yea

[kind rapids]

$L = \sqrt{(x+12)^2 + (\frac{4(x+12)}{x}})^2$

[somber coyote]

Closer

[trail tendon]

yeah

and then derivative :l

fun . . .

[kind rapids]

Imma use Wolfram Alpha

[trail tendon]

ok 💀

[kind rapids]

Ok. It is 21.6 so looks right

[peak nymph]

how did u get that

[trail tendon]

me over here doing it manually:

[peak nymph]

when I took the root of the derive I got something very different

[kind rapids]

That is just the value of x

You need to plug x to L to get the length of the ladder

[peak nymph]

oh ye that also 🫠

It doesnt work to do it on this calculator because it is only for derivatives

so gotta do it again on another one

[trail tendon]

yea mk im not doin this without a calc 💀

[peak nymph]

ye lol

finally thanks

[median tartan]

Are there only 4 machin like formula which in a form of 2 terms, and both arguments are 1 over some integer? or there are only 4 which were found?

[rough crater]

“Find the acute angle formed by the two lines”
y=1/3x + 1
and
y=-1/2x - 3
How to find this very fast without graphing it?
Since the slopes are known do I simply use the fact that tan(theta) = slope?
Then solve for theta on both expressions and add the angles together?

[exotic yarrow]

https://www.youtube.com/watch?app=desktop&v=NWgC6UhHbSM

[rough crater]

So thats using the addition formula for tan

Thank you

[sage skiff]

So I am starting Geometry Honors this upcoming school year for my freshman year, would there be anything good for me to know before the school year starts?

[twin spindle]

For this question, why do you know that 5pi/3 is the "second time" Luiza reached 1.2 m?

[warm shuttle]

(5π/3) is not the correct answer

5π/3 ≈ 5.23

[twin spindle]

That's what the vid says 😭

I see thoo

Well actually, I didn't send all the work

This is everything

I was jus questioning, about why 5pi/3

And not any other value that equals to 1/2

Is it cause 5pi/3 is between the values [-pi/2,pi/2]?

[trail tendon]

they said "the second time"

[twin spindle]

Oh wait im stupid

1/2 only occurs twice on the unit circle and 💀💀

Mb mb 😭😭🙏🙏

[plucky abyss]

[warm shuttle]

can arcsin[(1-cos(rational))sqrt(2 - 2/sqrt(5))] ever be rational?

that root = tan(36°) + tan(18°) if that helps

except for the trivial solution where cos = 1

rational in degrees*

i tried plugging specific values into cos and wolfram alpha seems to know that the result is irrational for them somehow
edit: it was in radians mode

[warm shuttle]

Alternatively:
Draw your regular trig diagram with angle A, move (cosA, 0) towards origin such that you are effectively scaling the segment (cosA, 0) <-> (1, 0) by a factor of X keeping (1, 0) in place), Draw the perpendicular up from until it intersects circle. A new angle B is formed. Is it possible A/π, B/π are rational and X = sqrt(2-2/sqrt(5))

[plucky abyss]

If you know trigonometry no one can make you a fool
If sin rule doesn't work use cosine rule

[warm shuttle]

idk if that helps dude

[plucky abyss]

It doesn't

I sorry

[nocturne remnant]

This looks very nontrivial

[mild dagger]

Can someone pls help me with number 9 I've been working on it for an hour and its due today at midnight.

[faint pasture]

Does sinA cos B + sinB cosA ring any bells

[mild dagger]

Yes

[faint pasture]

Can you identify A and B for your problem?

[mild dagger]

This is the example problem given.

[faint pasture]

Well answer what i asked you

[mild dagger]

sin A is sin x

[mild dagger]

I got it thanks for helping.

[slate pasture]

Solve: cosA+cos2A+cos3A=0

[faint pasture]

Do you know cosA + cosB identity?

[plucky abyss]

Yes

2cos((A+B)/2)cos((A-B)/2)

[faint pasture]

💀 i asked the one who asked the que but sure

[plucky abyss]

[twin spindle]

Can someone help solve this equation

[plucky abyss]

Write cot in sin cos

[twin spindle]

I did that, and got to here, and this is the part where I got stuck

[silent plank]

factorise

[twin spindle]

I got 0 and 3/4 but it says it's wrong 😭

Oh shi wait, I forgot to take the arc cos

[burnt lagoon]

you cannot cancel out functions, it is prohibited

[grave pond]

It looks like he's not cancelling a function, but a number that happens to be produced by a function.

[burnt lagoon]

he cancelled sin(t), now he have to put sin(t)=0

[maiden brook]

but that’s not a solution

[maiden brook]

f(x)/f(x) = 1, f(x) - f(x) = 0

[burnt lagoon]

yep right way to do it

[grave pond]

If sin(t) were 0, cot(t) wouldn't exist in the first place.
After getting numeric solutions at the end, he'll need to verify that they don't have sin(t)=0, of course.

[maiden brook]

if u know sin(t) != 0 u can cancel

I typed cance and that corrected to Vance lol

[burnt lagoon]

that will give t = pi/2

[burnt lagoon]

yes

but it is prohibited to cancel functions

[undone quest]

Wht is the geometrical difference between inverse and non inverse trigo

[obsidian harness]

wdym by the geometrical difference

[grave pond]

There's no such prohibition that I know of.

[undone quest]

@obsidian harness Like constriction wise

How they r different like one who invented it must have though of some sort of inverse triangle or something

[jovial wedge]

why is \frac{\pi}{3} (60) not in the answers?

shouldnt it be in the needed 1st and 3rd quadrant and suffice?

[silent plank]

where's pi/3 even coming from

[jovial wedge]

i mean the principal value is 30 since the inverse tangent of \frac{1}{\sqrt{3}} is 30

30+30=60

[silent plank]

why are you doing 30+30

[jovial wedge]

you can begin from 270 too

the end of thge third quadrant

you get to 60

is this somethign to do with the domains?

[silent plank]

for what you're asking, no

I just don't know why you're doing stuff like 30+30

tan(30°) is 1/sqrt(3)
tan(60°) is not that, why do you think 60° is relevant here

[silent plank]

also wdym by begin from 270

[jovial wedge]

the end of the domain

im subtracting the end of the domain using the principal value

[silent plank]

that's not how reference angles work

[jovial wedge]

how woudl you solve this exercise

im selfstudying so i make some mistakes

ive gotten the gist of it

[silent plank]

you add/subtract from the horizontal axis depending on your quadrant

and I still have no idea why you did 30+30

the reference angle to
tan(t) = 1/sqrt(3)
is pi/6 or 30°
since tan is POS in Q3,
a Q3 solution would be pi + pi/6, which they've simplified to 7pi/6

and then add 2pi to each one for more solutions

[jovial wedge]

yeah i got 7pi/6

reference angle = principal value

[silent plank]

or just add pi (since it's tan)

and I still have no idea why you did 30+30

[jovial wedge]

i thought by addign the reference angle together and checking if the angles are within the domain of the quadrants, you could find the angles needed

[silent plank]

that's completely wrong

[jovial wedge]

why 2pi

[jovial wedge]

so if you subtract from the 1st quadrant, 90-30=60

[silent plank]

no

keywords: horizontal axis

if you look up reference angles trig,
you should see images where they're measured from the horizontal, NEVER vertical

[jovial wedge]

oh i got it

60 cant be because of 2x in the equation

[jovial wedge]

you mean this?:

[silent plank]

yeh

[twin spindle]

Are these the only two solutions? Because if I got it right, it would show a green checkmark, so im not 100% if these are the only solutions

[silent plank]

show work

[exotic yarrow]

$\cos t=0$ has the additional solution of $t=\frac{3\pi}{2}$ and $\cos t=\frac{3}{4}$ has an additional solution of $t=2\pi-\cos^{-1} \left(\frac{3}{4} \right)$

[somber coyote]

Civil Service Pigeon

[twin spindle]

Interesting

How'd do you get the 3pi/2?

[exotic yarrow]

Think of it geometrically: What does cos(t)=0 tell you?

[twin spindle]

Ohhhhhh I get it

Wait nvm

I get the part where cos(t)= 0, which makes its pi/2

But why the 3?

[exotic yarrow]

Sooo ... can you answer the question?

[twin spindle]

No, their's still sum im not getting 😭

[exotic yarrow]

I'm trying to scaffold the explanation

But if you're not going to cooperate and leave me talking to myself

then there's rlly no point in me continuing to engage

[twin spindle]

What 💀

I asked "but why the 3?"

Like why do we add the 3 in

I get that cos(0) is pi/2 but im still not sure why the 3 is being added

[maiden stirrup]

I am having an issue with the use of the inverse trig functions in linear and quadratic trig questions because it says in my teachers note to "use inverse cos" but the number im getting in not even close to his despite me putting in the exact question

[plucky abyss]

Show the question

[warm shuttle]

switch degree/radians mode

[fallow palm]

are you sure

[warm shuttle]

arccos(0)?

[pliant grotto]

is the same but it will depend

on the context question

[sinful flame]

i got question.

How do i make Parallelogram with these values (draw)

a = 6,5 cm ; e = |AC| = 8 cm ; Angle Beta 75 degress

small letter is wall and point is caps letter

[silent plank]

what tools do you have

[grave pond]

And do you have a diagram? Doesn't need to be to scale, but it's hard to respond without knowing how those letters relate to each other.

[sinful flame]

yeah wait

[sinful flame]

i got scale , set square , protractor divider

[sinful flame]

i got construction procedure ( i used google translator )

and sketch

does this help 😭

lmk if u read allat

[grave pond]

Cleaned up a bit.

I would start with solving triangle ABC, where you already know two sides and an angle. Do you happen to already know an approach for solving SSA triangles?

Wait, no, you wanted to draw it rather than compute it.
Then just start by drawing AB of the right length, then angle B with a protractor, and finally point C where an 8 cm circle centered on A intersects the right leg of angle B.
Finally draw paralell lins from A and C to met in D.

[sinful flame]

Yeah thats pretty much what I did thank you i needed to be sure if I got it right

Maybe we do or dont calc it but basically goal was "draw" it and write down step by step what I done

[sick walrus]

can someone check my work on whether or not the steps are fine?

[maiden brook]

if the circles are internally tangent and the points J, K, and G are collinear, and so are points J, I and F, why must IK and JG be parallel?

i need this as part of something

[maiden brook]

where are E and F? and what is that odd notation

[maiden brook]

thanks!

[grave pond]

Because the two circles are related by a simple scaling of the plane centered on J, and this scaling also preserves the two lines out of J. So the triangles IJK and FJG are related by the same scaling and are therefore similar.

[grave pond]

Oh, that's actually neat. (Sorry didn't see that before I typed).

[mellow quest]

I hate homothety

its soo boring

[grave pond]

Well, move to the hyperbolic plane, then.

[mellow quest]

😨

[sick walrus]

it didnt ask for more points but wasnt sure and which notation do u mean?

[broken garnet]

everything is fine

[sick walrus]

kk I submitted it

[broken garnet]

wait

are you supposed to prove the triangles congruent of the line segments

[sick walrus]

uh yea

[broken garnet]

if it’s the lines you need to put them congruent by cpctc

[sick walrus]

ohh

[broken garnet]

since the triangles are equal the lines should also be equal

[sick walrus]

uh I cant unsubmit..

[broken garnet]

oh

you need help with anything else?

[sick walrus]

but I can till the teach grade it

[broken garnet]

ah

ok

[broken garnet]

it’s kinda blurry (the triangles)

[sick walrus]

its js sm going on

[sick walrus]

sorry here

[broken garnet]

easy, proportions

ima do this on my ipad rq

[sick walrus]

alrr

u mind if I friend u too? since rn the only help I need is geo

[broken garnet]

yes, geometry is my specialty

[sick walrus]

yea I read ur status lol

[broken garnet]

yes, i think I was the teachers favorite cuz my average

[sick walrus]

oh wow thats nice

[broken garnet]

yuh

ima make a gc and add my ipads discord

[sick walrus]

kk

[broken garnet]

dude ima be in this channel for the rest of my life

[modest forge]

Idk where else to ask this so what's the definition of lines/line segments in circular geometry?

[snow karma]

could be wrong cause idk what is 'circular geometry'

isnt a line js a set of points that are all aligned

To define one line in the plane, js put two points

[modest forge]

Basically just geometry on a circle is the simplest way I'd try to define but it can go to higher dimensions

[snow karma]

Oh

Forget my definition then

though whats the point of geometry on js a circle

[modest forge]

Yeah cuz normally in euclidean geometry afaik we interpret the first axiom of euclidean geo to mean if you give me two points I can give you a unique line segment that may not be true on a circular geometry

[modest forge]

Geometry on earth

Would be similar

[snow karma]

so geometry on a sphere or on a circle

?

well actually thats similar shit

[vivid karma]

is there a word for a general sphere? Ie “ A set of points that are all the same distance from r In any dimension given any metric space” so segments and points are included

[modest forge]

N-balls, n dimensional circles and so on all work I think

N dimensional circles/spheres are probably the best

[modest forge]

Maybe but lines on a sphere don't behave as nicely as they do on euclidean geometry

[plucky abyss]

I like lines inside a sphere

Hollow sphere

[formal belfry]

Is there a way to understand trig identities

Or do I just have to remember them all

[plucky abyss]

Both

[pliant kettle]

You could always derive them on your own whenever necessary

So yes you do have to remember the basic ones

[formal belfry]

There’s a way to derive like sin(A+B) and those????

[faint pasture]

imo the basic ones are easier to derive than not so basic ones

[faint pasture]

there is a way but you cant do that whenever you need it cuz the way is fairly long

[formal belfry]

Damn

[faint pasture]

there is a complex numbers way though

if you know those then its simpler

[formal belfry]

Elaborate

What way

Cis?

[faint pasture]

https://math.stackexchange.com/questions/713747/neat-way-to-prove-sin-alpha-beta-using-complex-exponential

[formal belfry]

I think i should just remember them

Until I know more

[faint pasture]

lmao thats better

[plucky abyss]

[modest forge]

the other option is rotation matrix but those are similar

if you dont know understand matrix multiplication its still annoying

[modest forge]

i just read this and it isnt really that helpful but ok

the answer here is right imo

[faint pasture]

well the hint's pretty much enough

[mellow quest]

what should I do next?

I proved that EH1 = EH2

Sade / Sbec = AD/ BC

[mellow quest]

In a quadrilateral ABCD inscribed in a circle, the bisectors of angles A and B intersect at the point E lying on the side CD. It is known that CD : BC = 3 : 2

[obsidian hornet]

Prove that CD=BC+AD

[mellow quest]

how

[obsidian hornet]

For example, draw line parallel AB, going through E

[mellow quest]

motivation?

[obsidian hornet]

Get two equal triangles

[obsidian hornet]

Here is full solution, final part just counting segments

[mellow quest]

thx

[obsidian hornet]

No problem

[grave pond]

How do you conclude that DE and EC have the lengths you claim?

[broken garnet]

it’s a chord and it’s being cut in the center I think idk how to explain it

i’m rusty in this subject lol

[grave pond]

Oh, I think I see. Since |H1E|=|H2E| together with the two black angles being equal, there are two congruent triangles making |EC| equal to its counterpart on the left.

[shell lantern]

How to deal with trigonometry?

Can u give me some advice?

[hybrid laurel]

hello

[obsidian harness]

hi just ask your question

[hybrid laurel]

I need help with geometry. I can't find any geometry book in Spanish that explains it well. I can read English more or less well. If someone knew of a book/course that explains it well in English, I wouldn't have a problem.
PS: I'm using Google Translate. If that's why I have bad diction.

[obsidian harness]

I have a problem if you want to try btw

[obsidian harness]

https://i.warosu.org/data/sci/img/0147/14/1659324540793628.pdf
don't know your level but this Olympiad geometry book is really well-written

or if you're looking for easier stuff, Khan Academy (a website with videos and interactive practice problems) is good

[shell lantern]

Needed help with trigonometry.

[obsidian harness]

Khan Academy if no one else has anything better

besides aren't there tons of YouTubers who explain JEE stuff

just type in JEE trigonometry into YouTube

[shell lantern]

Khan academy is there ah

[broken garnet]

@long temple

wanna be friends

[long temple]

lol yeah lets go

[broken garnet]

friend me

[long temple]

alralr

[formal belfry]

Anyone got like an ultimate trig identity table

With everything on it

[kind rapids]

Bruh there are hundreds of trig identity

[broken garnet]

do you want what like

[formal belfry]

Main ones ig

[broken garnet]

everything does and stuff

[broken garnet]

use sohcahtoa

[formal belfry]

I rlly don’t know trig

[shell lantern]

Need help in trigonometry

[formal belfry]

No no

[kind rapids]

But for the most used one you can search google or download an app

[formal belfry]

That I know

Like

[broken garnet]

sohcahtoa?

funny word

[formal belfry]

Sin(A+B)

[broken garnet]

oh

no idk

[formal belfry]

Sin 2theta

And stuff

[broken garnet]

nope don’t know that

[shell lantern]

Let's solve some jee questions?

[broken garnet]

only know

geo stuff

geo trig

[shell lantern]

Sinacosb+sinbcosa

[formal belfry]

Yeh yeh

But like the whole table

[shell lantern]

What whole u want?

[kind rapids]

Download a formula app might be a choice

[shell lantern]

11?

Give a just basic idea how to solve

I will try

[obsidian harness]

you should learn how to derive those equations from what you know already

for instance, sin(a + b) = sin a cos b + cos b sin a

what happens when a = b = x?
you get the formula for sin(x + x) = sin 2x ofc but what's on the RHS?

how about cos(a + b) = cos a cos b - sin a sin b

what happens when you replace b with -b?

[faint pasture]

There are, you just dk them

[fallow palm]

I just use this one.

(Taken from HKDSE M2 paper.)

Of course, it's better to use these identities without having to look up a table. Simply memorising the first two in the above table, and you can derive the other 8.

[exotic magnet]

does minus-plus just mean that the first answer is from subtracting

[fallow palm]

yes

[faint pasture]

rest sure but the tan(a+b) will be calculative to derive so id say remember that too

[north kindle]

im having trouble proving every cyclic trapezoid is isocoles

[maiden brook]

angles

[north kindle]

ive tried using congruent triangles, stuff with the fact that means all the circumcenters of any set of 3 vertices of the trapezoid would be the same point, but i cant get it

[north kindle]

ive tried that

[maiden brook]

well u should get it directly from there

[north kindle]

well maybe im just fucking stupid then but im not

so i would like some help

[maiden brook]

what’s true about the angles

[warm shuttle]

trapezoid would have the sides perpendicular to a radius, no?

[maiden brook]

ok?

[north kindle]

i know theres a bunch of isocoles triangles i can make

[maiden brook]

that’s way more than u need

[north kindle]

wait i might have something

i think i got it

[maiden brook]

yeah?

@north kindle

[north kindle]

yes

im pretty sure i did

[maiden brook]

what did u do?

[north kindle]

using the fact BAD + ADC = 180 and the same for the other two angles

[maiden brook]

u can do it in like line: Since opposite angles are supplementary and so are the “same-side interior angles,” we get that the base angles are congruent

therefore it is isosceles

[north kindle]

how do you know opposite angles are supplementary

[maiden brook]

it’s cyclic

angle D is half of arc ABC and angle B is half of arc ADC

what do u get when u sum them

[north kindle]

oh

makes sense

[maiden brook]

yeah

that’s something u should really know about cyclic quads (in fact if the opposite angles are supplementary, u can show by contradiction that the quadrilateral must be cyclic)

[north kindle]

i did know that i just didnt think to apply it here

[plucky abyss]

[fallow palm]

<BAC = <ACD because parallel, so chords BC and AD have equal length, done.

[main chasm]

\int x^2 dx

[maiden brook]

BAC and ACD aren’t similar so wdym?

[fallow palm]

so what

parallel lines

who said anything about similarity

[maiden brook]

so why does that imply BC = AD

@fallow palm

[fallow palm]

property of circles, equal angles => equal chords

[hoary gate]

If four sides of parallelogram sides touches a circle like this then it will be

[gloomy eagle]

?

[hoary gate]

??

[gloomy eagle]

ye i think If four sides of parallelogram sides touches a circle like this then it will be 👍

do u mean what shape will it be ?

[hoary gate]

Yeah

Or what else

[gloomy eagle]

oh then it will be a rhombus

[hoary gate]

Why?

[gloomy eagle]

never a parallelogram

[hoary gate]

A special rhombus? Like square

Or all rhombus

[gloomy eagle]

given the 4 sides r tangents to the circle, u can use the tangent property to prove it

[gloomy eagle]

either a rhombus or a square

[hoary gate]

What if both options in the answer?

[gloomy eagle]

the ans is wrong maybe..

lemme find u a picture

like this one

its a rhombus

square

[hoary gate]

In my question both option are persent in the answer?

[gloomy eagle]

idk

wrong ques maybe

💀

[hoary gate]

No further condtion given

[silent plank]

you'd answer rhombus

[hoary gate]

But why?

Any specific reason

[silent plank]

as all squares are rhombuses
but not all rhombuses are squares

[gloomy eagle]

sounds but maybe thats the best fit

[hoary gate]

Given answer is rhombus

[gloomy eagle]

[silent plank]

as demonstrated above, such a parallelogram can be drawn that isn't s square so that can't be correct

[hoary gate]

I want to read more about this topic

But not finding particular

[obsidian hornet]

Just consider that square is special type of rhombus, when u move rhombus diagonals in equal position(i mean length of diagonals)

[hoary gate]

Can I draw circle inside rectangle?

[obsidian hornet]

No, only ellipse

If sum of the opposite sides of quadrilateral is equal than u can inscribe circle in it

Ol less trivial sign, If there are four triangles into which the diagonals of a convex quadrilateral divide it, the following relation holds for the radii of their inscribed circles: 1/r1+1/r3=1/r2+1/r4.
Another sign: if u drop a perpendiculars on sides from diagonal intersection point(a,b,c,d respectively ) and for opposite heights will be following that relation: 1/a+1/c=1/b+1/d

And, by the way, 4 incenters will lie on one circle

[paper spoke]

I had this question in my math test today. WTH ARE THEY TRYNA DO MAKING ME FIND THE Y INTERCEPT OF THE PERPENDICULAR BISECTOR OF 6 PAIRS OF POINTS??? It's not hard it's js teadious.

There were 5 questions the other 4 were not as bad, this one is just crazy...

[trail tendon]

😔

[pseudo nebula]

you can do it with some calculator tricks, annoying but not that bad

[warm shuttle]

i assume calculator not allowed?

[paper spoke]

calculator was allowed, how do you do it fast with calculator?
The only way I know how to do it was to solve for the slope, then take the negative reciprocal, then solve a linear equation for a y intercept

[paper spoke]

how? I only know what I said above, what can you do with the calculator to speed things up (other than doing some basic arithmetic instead of doing that yourself)

[pseudo nebula]

depends on the calculator, what do you have access to?

[paper spoke]

I use desmos scientific calculator on my phone

[pseudo nebula]

i don't think you can do anything too fancy with a scientific calculator unfortunately

[silver mason]

Any ex-USSRer here? If I read: △ABC = △DEF, can I be sure that, in particular, ∠A = ∠D or AB = DE?
(I assume that Russian tradition about this question is different from Western)

[warm shuttle]

I hate this

[warm shuttle]

if we are thinking of the same app, you can say
x1 = -4
y1 = 5 ...
then write all your equations in x1, y1 ... and it will give you the answer, then just change the values on the top and the answer will change

even if you don't have a calculator, you can actually solve the general case and the final expression simplifies a lot (not as much as i would've liked tho)

[obsidian hornet]

For us, this usually means that each vertex corresponds to each(A==D, B==E, C==F)

[warm shuttle]

yeah i have no clue what a triangle is

or what triangles satisfy relation =

[paper spoke]

oh? You can do this? Let me try... OMG YES YOU CAN SET VALUES FOR VARIABLES TYSM I CAN PUT FORMULAS AND STUFF YESSSSSSSSSSSSSSS

[upper karma]

dumb question but this is 10 - 2 and 7 - 4 right?

[hard glen]

slope = rise/run = change in y/change in x = (10-2)/(7-4) = 8/3

[upper karma]

aight, thanks easy

[rough shoal]

how to make polygons with x vertices and fill them with triangles

[fleet summit]

What is the theta angle

[rough shoal]

is that 3 degrees?

[fleet summit]

30

[rough shoal]

this

[fleet summit]

Oh

Those two are 60 degree

[rough shoal]

theta + x = 180

idk above that

SKUL

[vagrant snow]

Wait no way is that oreminer

From hd

[vagrant snow]

Believe it's 30 degrees

Because in that little right triangle in the box

If the line is the normal then it's perpendicular to the slope so it's just 180-90-60

[grand ginkgo]

Would this be an understandable geometric proof that (a-b)(a+b) = a^2 - b^2 ? (the vectors are there to be arrows)

[grave pond]

I would recommend losing all of the irrelevant names for points. (Perhaps find a more suitable drawing software than a geoetry sketching app?)
Also, it will be easier to follow what's going on if you make b more clearly different from a-b.

[grand ginkgo]

thank you.

[tired valve]

Currently doing a taster for vectors, coterminal angles and mathematical logic in Year 10 math. Any general tips or diagrams I should be referring to? I'm gonna need all the help I can get if I'm gonna do really well in the subject next year. (plus, im trying to get a really good teacher employed at my school)

[burnt lagoon]

30

[undone quest]

Must be 30

[undone quest]

2theta +120=180 theta=30

Sum of angle on a 2d plane

Is 180

Sorry lineqr

Wrong method but correct answer

Hh

[reef dragon]

guys I'm doing geometry, what's the hardest concept

and any tips about where to study for them

[fallow palm]

there's a lot of hard stuff in geometry

[reef dragon]

I heard from a lot of people that proofs is the hardest?

[fallow palm]

also difficulty is subjective, it doesn't make sense to say what the hardest is

[fallow palm]

No, that's the basics

[reef dragon]

true

do u know any place to study for that

[fallow palm]

are you studying this for school or for competitions?

[reef dragon]

both

but mostly school

[fallow palm]

I'm not very familiar with this, but probably try Khan Academy

[reef dragon]

k got it

ty

[sly crown]

i generally struggled with more advanced circle problems because at first i didnt fully understand the constructions i feel like that is a pretty common issue

[undone quest]

Lets see who can solve this

Radius != 21

[upper karma]

Ima solve it

[upper karma]

15

The minimum number of vertices in the triangulation that satisfies the given estimate is 15 I think

[trail tendon]

12

the answer is clearly 12

[upper karma]

thats what I was thinking

[trail tendon]

definitely

the answer is definitely 12

100%

no way around it

[upper karma]

its atually?

[trail tendon]

i have no idea, what am i solving for 😂