geometry-and-trigonometry

[gray pivot]

Try solving

[upper karma]

dive it?

[gray pivot]

You can start by combining like terms

Sum of angles=180°

And the sum of angles is 4.8z+5.6z+7.6z

So 4.8z+5.6z+7.6z=180

[upper karma]

OHHH

I GOTTA FIND Z

I ET IY

wait

so

i gotta add

lets say

4.8+z?=180?

[gray pivot]

No

[gray pivot]

The 4.8 next to the z is multiplication not addistion

4.8z+5.6z+7.6z=180
z(4.8+5.6+7.6)=180

You can also factor out the z

@upper karma

[upper karma]

aohhh

THANK YOU SM

I BEEN SDOIN THIS FOR SO LONG

see this is how hard it was

and

ion even kno

thank yo uman

@gray pivot ur amazing

[bleak basin]

hey guys can i get some help for my sophmore geomotry please

[frosty kraken]

Sure

[stoic mesa]

hey y'all i joined just because of this HAHAHAHA 😭 pleaseeee i need help answering them

this is supposed to be our reviewer so like no need to worry for acad dishonesty i swear 🫡

[coarse wing]

Can y'all give me some tips as to where I can find corresponding angle practice problems to help me cram for the test that's due today?

[bright hill]

are you in the US?

[coarse wing]

Sure am

[bright hill]

you could try PMT but that uses UK exam boards

don't think the questions should be that different though

[coarse wing]

Alright ty! Is it free or?

[bright hill]

yeah

[upper karma]

could someone explain how i would go about solving this?

[faint pasture]

Midpoint/thales theorem

[frosty kraken]

QRS is similar to MNP with ratio 1:2

[merry knot]

The homothety centered at the centroid of MNP with factor -1/2 sends MNP to RSQ

[obsidian harness]

the 25 and 22 are not needed to find x

but yeah using the perimeter of QRS is a great idea

then multiply by 2

[gray star]

whats even the question

NEVERMIND !!!

[gray star]

who said that lines were parallel

[faint pasture]

You could prove it

[gray star]

ah

[faint pasture]

But yea my bad i didnt check it the first time

[unkempt gulch]

I know the answer is D but I really don’t understand why

[sudden sky]

Whats the equation for an arcs end x and y (carthesian) if I have the arcs length, radius, linear and angular velocities?

[gray star]

what grade

[gray star]

how " a " is in R and then they said its in ]0 ; 1[

bruh

[unkempt gulch]

It’s an alpha sign in R

[gray star]

bro why they complicate all the thing with doing similar stuff

why they just not do belta

so ur question is so hard bye

[gray star]

chatgpt says the correct answear is " c " idk

[coarse wing]

Am I wrong to tilt this to the side and treat it like a parallelogram with a transversal line through it?

Plus the triangle ofc

[solemn obsidian]

yo can anyone help me w a problem irl. i got a box that’s too big in l, w, and height and i need it smaller to fit a motherboard. lmk if u can help
so i can send dimensions 😭🙏

begging that there’s someone goated at geometry in here

[upper karma]

,rcw

[analog cargo]

,rcw

[analog cargo]

Pls help the theorem is "If two lines are intersected by a transversal and a pair of consecutive interior angles are supplementary, then the lines are parallel

😭

[pulsar hatch]

what do you want? a proof?

[analog cargo]

Yeah a two column proof pls explain

Like prove any two of the consecutive angles are supplementary 😭😭

[pure anchor]

Triangle RKT is isosceles
-> RKT = T
-> T = U ——(i)

If angle 1 = angle 2,
-> 180 - angle 1 = 180 - angle 2
-> R = C ——(ii)

Now, T + U + R + C = 360

From (i) and (ii),
2 (R + T) = 360
R + T = 180
So, RU and CT are parallel

Also, C + T = 180
So, RT and CU are parallel

Therefore, RUCT is a parallelogram

[analog cargo]

TYSM

[pure anchor]

Np

[faint pasture]

are coincident lines considered parallel

because two lines are parallel when their slopes are equal

are coincident lines obviously have equal slopes so they should be

but google and my textbook both disagree

[upper karma]

easier than it looks

[obsidian harness]

yes arcsin(x) + arcsin(y) + arcsin(z) = 3pi/2 is a very restrictive condition

[warped bronze]

Isn't it the axiom or sth?

[sage shoal]

Can someone recommend me any book about basic trigonometry?

[faint pasture]

||js 0||

[upper karma]

ya its 0

x = y = z = 1

[sudden sky]

Whats the equation for an arcs end x and y (carthesian) if I have the arcs length, radius, linear and angular velocities?

[obsidian harness]

you can use this formula to find the angle in radians

ah but you need the centre of the circle

you need that information for the coordinates of the arc endpoints

or just like any coordinate in fact

[sudden sky]

What I have is arc length, angular and linear vel, starting coordinates

Doing this in a robotics system

[obsidian harness]

yeah your starting coordinates

we need that

[sudden sky]

Trying to make location tracking for my robot using imu and encoders

The imu gives me angular velocity, linear velocity and all that

[obsidian harness]

there are actually 4 possibilities

ah so you actually need your linear velocity to tell which direction the initial velocity is, so there are 2 cases/direction * 2 directions = 4 cases in total

you somehow need to find if your robot is going counterclockwise or clockwise

hopefully your angular velocity has a sign or something (it should be, cause velocity is a vector)

[sudden sky]

I can determine that by seeing if my ang vel is positive or negative

Positive is turning towards right negative is turning towards left

So I got that part covered

[obsidian harness]

but one of the cases would be ending position = $(x_0, y_0) + r (\cos t, \sin t)$ where $t = \frac{l}{r}$

$(x_0, y_0)$ is the starting position

$r$ = radius, $t$ = angle, $l$ = arc length

[somber coyote]

south

[obsidian harness]

change t with -t to change clockwise/anticlockwise

oh and then change t with (pi - t) radians to get the other circle, so flip left/right

[sudden sky]

Does o represent starting point and t angular velocity?

[obsidian harness]

(x0, y0) are the starting coordinates

[obsidian harness]

no, t is the angle and t = l/r

l and r you know

[sudden sky]

L yes r I'm not sure

[obsidian harness]

ah right sorry

ah but linear velocity = radius * angular velocity

now you know the radius

$v = \omega r$

[somber coyote]

south

[sudden sky]

@obsidian harness Everything in radians right?

[obsidian harness]

yeah

[sudden sky]

So the base algorithm is that end position is xo, yo+l +r(cos(t), sin(t)),

[obsidian harness]

it's this

[sudden sky]

And if the arc is towards left we flip to -(pi-t)?

[obsidian harness]

yeah, if you want to flip left/right

up/down is -t

[sudden sky]

There won't be down so we can take that factor out

So to flip left and right t = pi -t?

[obsidian harness]

yes

[sudden sky]

K imma program that, do you know some python so u can verify?

[obsidian harness]

nah sorry about that

[safe yacht]

I'm not really sure how I would solve this problem, i do know simple things about the triangle like the base angles are equal and how to model it, but I'm unsure how I would get the measures of the angles.

[merry knot]

Set <A=x and express <B in terms of x

Then use the triangle angle sum property on triangle ADB

[exotic berry]

would anyone mind helping me out with some concepts i need to review for my geometry final (geometry of design). its basic stuff like arcs, circles, triangles, law of sin/cos, etc?

[frosty kraken]

Let ABC be a triangle with AB<AC. Let M be the midpoint of BC and let the bisector of <BAC intersects BC at D. Circumcircle of ABC and circumcircle of ADM intersects the second time at E. Prove the triangle EBC is isosceles

[maiden brook]

where’s this from

[chrome token]

when f(f(x)) = x?

[frosty kraken]

Idk they didn't write the source

[wary bluff]

if you're talking about the meme i posted then i found it on reddit

[faint pasture]

,w f(f(x))= x

Preeetty cool

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by enderdoesntmind

[kind rapids]

Which field of math is this?

[faint pasture]

Idk i just saw someones message in this channel saying this lmao

[faint pasture]

This one

[mellow merlin]

is my answer correct

[spiral lodge]

Yep, it's correct

[upper karma]

Lmao I finally understand this as an algebra 2 student

[merry knot]

Let the angle bisector meet (ABC) at L. Let the perpendicular bisector of BC meet (ABC) at E' not equal to L. You want to prove AE'MD is cyclic.

[frosty kraken]

Nice

[tulip willow]

Im in the same part of the book lol

[frosty kraken]

<E'MD=90 for obvious reason. Moreover, both E' and L are midpoints of arc BAC and BLC respectively. Then E'L is diameter of (ABC) which means <E'AL=<E'AD=90. This implies AE'MD is cylic so E'=E which implies E midpoint of arc BAC.

[merry knot]

True 👍

[mellow merlin]

how do i go on about this

im stuck

[junior trench]

XZP + angle Z = 180
angle Z = 180 - XZP
angle X + angle Y + angle Z = 180
angle X + angle Y + 180 - XZP = 180
angle X + angle Y = XZP

this is the proof for the
exterior angle = 2 opposite interior angles (in a triangle)

[mellow merlin]

help 💀

#dying!

wait wait wait i think i got it

.

kill me

[junior trench]

,w 180-(x+55) + 2x-20 + 30 = 180

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by milooo09.

[junior trench]

,w 180 - 55 -45

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by milooo09.

[prime tartan]

is there something i should add?

im working on a presentation about ellipses in mathematics, with the focus on a circle

[river abyss]

You can list how sine and cosine relate to it

Or polar coordinates

Or Euler's identity

[open vapor]

area of ellipse

[azure helm]

Circle $C$ has radius 10 cm. How many square centimeters are in the area of the largest possible inscribed triangle having one side as a diameter of circle $C$?

[somber coyote]

938c2cc0dcc05f2b68c4287040cfcf71

[azure helm]

can I get some help

[drifting flint]

Can anyone explain this <@&286206848099549185>

[last coral]

Are you aware of the trigonometric functions and their definitions according to Perpendicular Base and Hypotenuse of a triangle ?

[vapid cobalt]

Hello

I need help with how I divide by -4 and then do the trig identities

Question 3.

[rich elm]

Ok so for this find the second derivatives for both

Y1 is -4cos(2t) and Y2 is -4sint

Then set them equal to eachother

The negative 4’s cancel and now you have cos2t=sint

The only possible values for this are pi/3 and pi/4 and 5pi/3

[vapid cobalt]

Thank you!

Also I’m not sure how she did this

I’m not sure if it’s another trig identity?

[mellow merlin]

why what

good answer?

[junior trench]

hmm
QPR = 140 and PQR = 40
in supposedly triangle PQR, sum of angles should be 180
but QPR = 140 and PQR = 40 and those are 180 but thats not even including PRQ

bc theyre not parallel is js bc u can see theyre not parallel but i dont think thats rlly an ans

[limpid marsh]

How do I start this?

[upper karma]

Is it only me who doesn't like geo?

[obsidian harness]

https://www.youtube.com/watch?v=i3bjEOA5_zc

[tender jungle]

Help on question 6

[wheat roost]

The 2 semicircles on the legs have areas that sum to the are of the semicircle on the hypotenuse

Bc the area of a circle is proportional to the square of the radius

So it is just the Pythagorean theorem

[lament wagon]

hello

how would i solve question 6 when it asks for the perimeter of the quadrilateral

[wheat roost]

42?

[lament wagon]

thats what i put and my teacher marked it wrong on my quiz

[wheat roost]

Oh

[junior trench]

it is 42

im guessing its safe to assume the circle is perfectly inside the quadrilateral

the main rule here is tangents from an external point are equal in length

[wheat roost]

Yes

[flint kestrel]

someone please help with #4

maybe i’m just dumb but i can’t remember how to do this for the life of me

[wheat roost]

Umm just subtract

[junior trench]

start by marking the angles on the diagram and ull figure out what to do next

[plucky kindle]

I-

I think I’m missing something here

Because I did not get that

How is it not 62?

Bad handwriting, but how is it not this?

Also how is your writing so neat 😭😭

[upper karma]

tf

A complete revolution is 360 degrees

And they asked you for angle rqp

[plucky kindle]

Kqp

But ye and?

[upper karma]

Ah mb

I did it properly in the pic I sent tho

[plucky kindle]

Oh

My main question was how 164 is the bottom angle

Bc 164 is acute and in the picture that angle looks like it should be obtuse so idk

[upper karma]

It’s given rqp = 164 degrees

[plucky kindle]

Not acute

But

Less than 180

Not acute

[upper karma]

Ohh you are wondering if rqk + kqp = rqp

[plucky kindle]

Well yeah

It should right?

[upper karma]

Ohh shit yeah mb

This is the answer

[plucky kindle]

In the picture the line doesn’t look straight, therefore where you placed rqp should be >180

[upper karma]

62

Ur Right

[plucky kindle]

Oh okay, thanks for clarifying :D

(And how is your handwriting so neeat???)

[upper karma]

Apple Pencil 😎

[plucky kindle]

Ah I see

My finger on a phone screen at 1:30am is not very good at writing neatly lol

[hot wraith]

U call that neat..

[meager acorn]

Can anybody recommend an in depth video tutorial on trigonometry on yt? Most ones I find are quite basic and cover simple examples.

[sharp terrace]

can someone solve this

[obsidian harness]

You actually don't need the 63 at all

Search up parallel line properties

[plucky kindle]

Look at mine -_-

It’s neat

Terrible diagram, but I believe this is the answer, let me know if you have any questions!

Did it on paper to try to make it more readable, the highlighting of pink and yellow shows the angles used in the equation and the green is the Z, I hope this is better

[azure helm]

how do I find all the solutions to arctan(-1)

[trail tendon]

do you mean all the solutions to tan(x) = -1?

arctan(-1) only has one result

[violet blade]

as a trigonometric ratio or as a function ?

only -pi/4

[azure helm]

I also thought -pi/4 is only solution, maybe I'm tripping

Like was related to an exercise is was trying to solve like the other day

Dunno, ignore me i am 100% tripping

[mellow merlin]

x is 40 but i keep getting 20

i checked with a protractor

[jolly island]

The diagram is not to scale

You see BOC is not the 3.5 times of BOF

[junior trench]

,w (180-2x)2 = 7(2x)

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by milooo09.

[junior trench]

it IS 20

[timid swift]

can anyone help me w some trig pls 🙏

[left leaf]

I got

Plane geometry test in 48 mins

Tell me everything i need to know

[upper karma]

360°-71°?

[void hearth]

i hate trig

so much

[upper karma]

Lol

Where you from

[void hearth]

canada

[upper karma]

Idk the condition there

Here it's easy for school purpose

[void hearth]

i'm taking IB though cuz the education system there is kinda mid

[upper karma]

Lemme google really quick

[upper karma]

Explain what you meant

[void hearth]

i'm taking the IB program rather than the local education system there cuz it sucks

[upper karma]

Idk what IB is

[void hearth]

International Baccalaureate

[upper karma]

Ookay...

Lemme google really quick

This thing is cool bruh

No one ever told me about these things

[void hearth]

ye IB is really cool

its international as well

[upper karma]

But why only 1 sub

[void hearth]

?

[upper karma]

I mean... if students could get access for all 6 subjects, it'd have helped ig

[plucky kindle]

Angles and stuff

[haughty blaze]

Any tips

ping me

[obsidian hornet]

Count angles for example

[haughty blaze]

I did write quite a few angles , by using the fact that the sum is 180, but didnt get anywhere

[haughty blaze]

Any more specific tip

[obsidian hornet]

tangent angles

similar triangles

[mellow solstice]

which are similar?

[obsidian hornet]

||TXA and TYC||

[sharp night]

y=mx+b = the slope of a line

i’m genius

[trail tendon]

the slope is m

[gray pivot]

mx+b= the output of a line (or y value)

m-slope
x-input
b-vertical shift

[merry knot]

Wrong

[trail tendon]

wdym wrong :<

[merry knot]

You forgor infinite slope

[trail tendon]

thats not a defined slope :<

[maiden brook]

lines of the form x = a

[merry knot]

clownbait

[trail tendon]

whats the slope of that line

[merry knot]

infinite

ax+by+c=0 is the true line equation

[maiden brook]

undefined

[maiden brook]

but it doesnt matter. its a line still

[trail tendon]

indeed

[obsidian harness]

you got scammed bruv, provincial boards have decent education

kick back and get your 90%+ average

but no, your parents drank the IB kool-aid and now you have to suffer

[elfin sentinel]

hello

im here to get very active role

and also to get good at trig

[obsidian harness]

good luck!

[elfin sentinel]

can anyone provide any good resources for knowing stuff,

[obsidian harness]

Khan Academy and Paul's Math Notes (algebra section)

for textbooks Axler's Algebra and Trigonometry is pretty good

[elfin sentinel]

whats paul's math notes

[obsidian harness]

there are a lot of other resources

[obsidian harness]

a summary of lecture notes from a uni

like that's good if you want to understand one specific topic quickly or for review

doesn't have a ton of practice material

[undone quest]

If a chord makes an angle with centre of circle find the angle made by same chord in minor segment

Angle =70 deg

[obsidian harness]

https://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/TrigIntro.aspx

ah right they have everything except trig in more detail

sigh

[undone quest]

Can anyone solve the above

[obsidian harness]

angle at centre is twice angle on circumference

[undone quest]

Hmm

Then?

Answer

?

[obsidian harness]

bruh use your brain

[elfin sentinel]

140 or 35

i prooved it 🙂

before i could prove a^2 + a is always even

😦

[obsidian harness]

yeah question is ambiguous

[elfin sentinel]

no no

its easy

[obsidian harness]

oh wait nah they mean centre of circle = 70 deg

hence 35

[elfin sentinel]

my friend who helped just said place down whatever you want on there

and i did

and it was quite simple

and i'm proud of myself

[jolly island]

Here's the PROOF

Of the problem proposed by Shaolin Monk

[wraith dune]

I'm a high school student preparing for jee, I'm struggling WITH geometry
To me, geometry feels like doing random things and finding stuff and just hope you get the answer. Feels like there's no methodical way to solve qs

Any advice?

[obsidian hornet]

In geometry, there is no clear recipe for solving this particular problem (not coordinates, complex numbers, barycentric, etc., but even there you need to be able to count correctly), but there is a certain base with which you can solve most problems from various tests and exams. Usually, the basic knowledge of the textbook is always enough for this, but if it does not work out at all, then you can turn to the counting method of solving problems.

[wraith dune]

Countjng method?

[obsidian hornet]

Count in coordinates for example.

[jolly island]

Geometry is art of find the problem it's a problem solving skill with whom only those who are truly creative are gifted with we should appreciate it as an important part of the Mathematics.

[lilac wadi]

Yeah, sometimes you need to use creativity for solving problems that you don’t see a “direct” way to solve it, like drawing auxiliary lines and segments. This is an beautiful and important part of Math

[merry knot]

trivial by complex bash

[analog cargo]

GIVEN: QUADRILATERAL PEAR, PE is perpendicular to EA. PE is congruent to EA. PROVE QUADRILATERAL IS A SQUARE.

Basically if all you know is a quadrilateral has one pair of congruent perpendicular adjacent sides how tf do u prove it's a square

[idle saffron]

who knows similar books to zambak geometry or just geometry problem book with picture 7-9 grades?

[upper echo]

If there are no other data it may not be a square.

[analog cargo]

SO THEN

IS THE SYMBOL A PARALLELOGRAM

can I show u it

[upper echo]

no, not necessarily

The vertex R can be anywhere

[upper echo]

so, it is already given that PEAR is a parallelogram

[analog cargo]

SO THATS A PARALLELOGRAM SYMBOL

...

[upper echo]

then there is no problem. It is obviously a square.

[analog cargo]

IM GONNA CRY

I WASTED ONE HOUR

SCREAMING OVER THIS

2 HOURS

JUST TO FIND OUT

but thx for telling me

:3

[upper echo]

Take it easy. That's life. It's always like that.

[autumn olive]

yo

[violet blade]

bro

[violet blade]

the note is saying that it is a quadrilateral

not necessarily that it is a parallelogram

[analog cargo]

Okay tt

Tyyy

[upper karma]

Should I try my hand at seeing if I can explain this myself

Actually I'll probably f it up

[upper karma]

Yeah I just realized I'm wrong not typing it

[white wedge]

hi amogh

[upper karma]

PEAR with 2 consecutive sides PE and AE are equal so PEAR is a rhombus and with PE is perpedencular to EA so angle PEA is 90 degrees so PEAR is a square

[blazing oriole]

can nsomeone look at my proof w me

this question

ans is 1

i got 1

and 0

ad i figured out what i did wrong in 0

pleasueeeeee

[elfin sentinel]

im not sure how many messages i need to send in here daily to get my role

but im trying 2 for now

[tranquil heart]

some one teach me ts please

[sharp night]

hey guys can someone explain sometjing to me

i need to find value of
m<1,2,3,6,7,8,9,10,11

[violet blade]

angle 5 is equal to 9 and 7

angle 4 is equal to 11 and 2

For the angle 10, you must apply the theorem of the sum of the internal angles of a triangle

[obsidian goblet]

i remember you getting somewhere last time, use product to sum on the denominator on wherr you left off

[plucky kindle]

m<1 = 180 - m<4
m<8 = 180 - m<5
m<6 = m<8
m<3 = m<1
m<7 = m<5
m<2 = m<4
So now you have 1, 2, 3, 4, 5, 6, 7 and 8
m<9 + m<10 + m<11 = 180
And the triangle in the middle has to have angles summing to 180
m<10 = 180 - m<5 - m<4
You can also find 10 using the parallel lines, so
m<9 = m<5
m<11 = m<4
m<10 = 180 - m<9 - m<11
This gives the same answer
I believe this is all accurate, feel free to correct me if I made a mistake

The answers (I suggest trying to work them out first but if you can't do it then the answers are listed here in ascending order):
||m<1 = 150||
||m<2 = 30||
||m<3 = 150||
||m<4 = 30||
||m<5 = 50||
||m<6 = 130||
||m<7 = 50||
||m<8 = 13||
||m<9 = 50||
||m<10 = 100||
||m<11 = 30||

[candid ibex]

hey could you please help me

i need someone to help me with these triangle proofs asap

i need to get this is to pass this semester and its due tomorrow at 8am

[plucky kindle]

Not sure about this but I can try to help

[candid ibex]

one sec i have help rn

[plucky kindle]

Can you send me it anyways? I’d like to try to figure out what it is for later on

[candid ibex]

sure

[plucky kindle]

Thanks

[blazing oriole]

Yey eye y eI EMR SLEPT SO MY HELP CHANNEL GOT CLOSED LMAO

[steep ocean]

Hi, I'm studying 'Trigonometric Identities,' but honestly, I'm not understanding anything. Could someone help me with that?

[sharp night]

I’m crying over geometry work

Someone help

[formal geyser]

What exactly interests you?

There are plenty of trig identities. If you don't understand how they work, you should watch proof of them or visualisation

Hi, is the radius of the small circle = 3.58?

Oh, it might be 2,965

[formal geyser]

I did a mistake

[junior trench]

js 3

perhaps u rounded smth up before final ans?

okay so we can tell the diameter of the larger circle is 24 so the radius is 12
consider the right angled triangle, the vertical line is 12-r
the horizontal line is 27-12-r
the hypotenuse is 12+r

,w (12-r)^2 + (27-12-r)^2 = (12+r)^2

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by milooo09.

[junior trench]

r cant be 75 bc that doesnt make sense

so r is 3

[jolly island]

Can we discuss P VS NP

[obsidian harness]

not here

[jolly island]

So where

[obsidian harness]

#discussion

[jolly island]

At the discussion channel they are always cracking jokes, I obseved you guys are a bit more serious

You in particular

[obsidian harness]

this is still the wrong channel

I mean #math-discussion then

if you're serious

[jolly island]

So I see it that way

[nova jungle]

how did he turn the equation on the left into the equation on the right i don't get it

i hate trignometry soo bad

[upper karma]

it is mostly just memorizing a lot of identities

cot^2 is a quotient identity that turns into cos^2/sin^2

which gives you the opportunity to divide out cos^2

1/sin^2 is a reciprocal identity that equals to csc^2

then there is a pythagorean identity of 1 + cot^2 = csc^2

which can be changed to be cot^2 = csc^2 -1

which can be substituted

[nova jungle]

I couldn't see that i could divide out cos^2 ,which confused me a ton, thank you soo much

[zealous pike]

$e^{i\pi} = cos(\pi) + i sin(\pi)$

[trail tendon]

= -1 :>

[somber coyote]

Pi, a future fluent jp speaker

[trail tendon]

fixdth :D

[zealous pike]

I saw your editing

xD

[trail tendon]

XD

cos(pi) = 1 💯

im so smort

[ionic walrus]

tangents are drawn to circle x^2+y^2=8 at the points where the line x+2y=4 intersects the circle. Find the point of intersection of tangents

[obsidian harness]

what have you tried

[ionic walrus]

i tried out assuming the points of tangents to he (alpha,beta)then substituted in the eqn for xalpha+ y beta =8 the solution says a1/a2 =b1/b2 =c1/c2 for the equations xalpha+ybeta=8 and x+2y=4

[obsidian harness]

you can just do x = 4 - 2y and sub into x^2 + y^2 = 8

then use the perpendicular lines to the radius through those points

[ionic walrus]

ye ik but can u explain what method they used?

[obsidian harness]

can you send the sol?

[ionic walrus]

i just wrote it above

[obsidian harness]

um......... send the picture pls

[ionic walrus]

they used a1/a2=b1/b2=c1/c2 for the lines xalpha+ybeta =8 and x+2y=4

[ionic walrus]

i don't have the image

wait

let me find

[ionic walrus]

Here

[obsidian harness]

oh yeah shortcut method, can't explain it tho

[ionic walrus]

why

[obsidian harness]

they're the same somehow, idk

[ionic walrus]

no they both represent the same eqn of the tangent

Smth like that ig

two equation represnts the same lines so making them identical

but how are they identical

x+2y=4 is the eqn of the line

and xalpha+ybeta=8 is the eqn of the tangent from a,b

[obsidian harness]

ah, consider the point (4 - 2t, t) on the line

then we must have x * x + y * y = 8

replace and get x (4 - 2t) + y t = 8

hmmm

[obsidian harness]

yeah (2, 4) is always a point on this line

since we could have two different values of t, for the two tangent points of the circle

these two tangent lines must always intersect at this point

two non-parallel lines only intersect once

very very clever

and that means the working above is completely nonsensical

[ionic walrus]

which one

[obsidian harness]

this

no reason for those to be the same line

[ionic walrus]

yea

but its made by a famous author

vikas gupta

do u think he'll make such errors

u must know vikas gupta right

for unique solution a1/a2≠b1/b2?

[obsidian harness]

Right answer wrong reasoning ig

[ionic walrus]

update: i found out that the soln was correct

it was the formula for chord of contact of a point with respect to a circle eqns

So the method used was correct

bruh

[faint pasture]

Do yk about the chord of contact?

[faint pasture]

Nvm

[faint pasture]

Nah

[elfin sentinel]

so

how do you prove that sin(a+b)=sin(a)cos(b)+sin(b)cos(a)

[abstract totem]

can u express sin^2 using the double angle identity or does it not work when sin is squared?

[maiden brook]

here’s one proof

here. it isn’t too hard to extend this so that it works for all angles

I’ll let u figure it out!

[open vapor]

you can try with the definition of dot product of vectors and derive it

[obsidian harness]

I will also avoid spoiling the answer

but try rearranging the identity

$\cos^2 x = \frac{1 + \cos 2x}{2}$, see what you get

[somber coyote]

south

[spark flare]

Why was i never taught that the slope of a line is the tangent of the angle the line makes with a horizontal?

[obtuse field]

Not completely related but have you ever seen this cool graphic with all the trig functions, plus some ones that are no longer used (versin, exsec, etc.)? it's pretty neat

[spark flare]

Yes I love that

I wasn't shown that in school either

I had to learn it on my own

It's very pretty

I've never ever seen versin or exsec or excsc before!

I drew this the other day

[ionic walrus]

chord of a contact of tangents drawn from a point on the circle x^2+y^2=a^2 to the circle x^2+y^2=b^2 touches the circle x^2+y^2=c^2(a,b,c>0) prove that a,b,c are in a gp

I solved this by similarity but need to solve via concepts of circles

[faint pasture]

Just draw the chord of contact from acostheta, a sintheta to x^2+y^2=b^2

Then set the distance of the line from (0,0) =c

[warped edge]

Any have a good playlist or a nice textbook in affine and Euclidean geometry
I need it so much

[ionic walrus]

why acostheta and asintheta

[faint pasture]

Parameteric point of x^2+y^2=a^2

[ionic walrus]

wait i forgot

sry

[elfin sentinel]

aw hell naw, i cant prove nothing

[maiden brook]

ok I can show u. but it’s good for u to figure out (as both a good exercise, and so it’s memorable)

[elfin sentinel]

ill give it a go

when im not playing mc 😛

[slender rain]

Why the discriminant of of the substituted equation is = 0?

[cunning lion]

there's only one point of intersection which would correspond to a double root of the quadratic equation

[slender rain]

What supposed double root(x¹ = x²) have to do to 1 intersection?
I still can't understand

[cunning lion]

can you clarify what is the substituted equation?

[slender rain]

y (of the tangent line) = y (of the parabola)
2x - 1 = x²
0 = x² -2x + 1

So i can create new parabole with the new equation that the roots will be same?

[cunning lion]

the new parabola will have its roots at the x-coordinate of the intersections of the original parabola and line

a quadratic with discriminant 0 has only one root, which corresponds to only one intersection between the line and parabola

[slender rain]

I SEEEEEEE

THANK YOUU

[upper karma]

Does "Sin n(theta) = 2sin (n x theta/2) x cos (n x theta/2)" exist as a formula ? . Refference -> Sin 2(theta) = 2sin(theta)xcos(theta)

[worn scroll]

you can substite u = ntheta and yes it'd work

can be anything basically as long as u split it in half

[upper karma]

Okay! So the thing that I am interested in is, whether it would be the same as sin 3theta which is a triple angle identity. ref -> sin3theta = 3sintheta - 4sin^3theta.

[upper karma]

since sin 3 theta according to this would be Sin 3theta = 2 sin(3theta/2)cos(3theta/2).

Are they equal mathematically?

[cedar stratus]

no it doesnt work like that

the multiple angle formula for sine is a little more complicated than that

$\sin^n \theta = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} (-1)^k \cos((n-2k)\theta)$

[somber coyote]

ashy!

[trail tendon]

i wonder how they derived that

[cedar stratus]

ask daddy euler

[maiden brook]

it does

sin(3 * 17) = 2 sin(3 * 17/2) cos(3 * 17/2)

[cedar stratus]

wait BRUH i thought they meant sin^2 theta

cos the 2 was outside the bracket

[trail tendon]

this is what u get for not reading closely

(me not reading it at all)

[cedar stratus]

soent all that time typing ts for nothing

[maiden brook]

de Moivre’s, probably

[chrome token]

what are all th trandformations which take circles/lines to circles/lines?

inversions + homothety + rotation + reflection is all ig?

(translation is a special case of rotation)

[golden marten]

Hello. I’m taking calc 2 next semester and I’m worried about trig and algebra. Does anyone have any resources where I can practice/learn in order to succeed? I’m super overwhelmed with all the resources available and I’m not sure where to start

[worn scroll]

trig is not a big deal you just need to remember the classic values and the identities

[slender rain]

Is the smaller cone congruence to the larger cone?

[junior trench]

nope

[slender rain]

So for example if the height of larger cone is H and the height of the smaller cone inside of the larger cone is h, I can't use comparison like H/h = 1/2 right?

And for this why is it x¹ + x² > 0 or the x¹ . x² > 0? Is it because of the discriminant is > 0?

[junior trench]

thats similarity not congruency

[junior trench]

can u translate the ques?
discriminant > 0 doesn’t necessarily mean x1 + x2 > 0

im guessing its smth to do with the graph on the right

[slender rain]

Ahhhh yes, that is what I mean, is the smaller and the larger cone is similiar?

[slender rain]

Yes, so I need to evaluate whether the condition in (1) to (4) is determining the parabola on the right side

I can't evaluate wether the (3) and (4) statement is right

[obsidian harness]

Sum of roots and product of roots

[hollow marsh]

how is this pre university

[faint pasture]

Just get to #calculus lmao

[open vapor]

isnt this the pi equation where an indian godess reveal to ramanujan in his dream?

its like half of his equation are from his dreams

[slender rain]

How did it get the > 0 in -b/a & c/a?

Where the > 0 comes from

I only know x¹ + x² = -b/a & x¹ . x² = c/a

[obsidian harness]

Both roots are >0

From the graph

[slender rain]

Ahhhh i seeee

Didnt see that

[wise zodiac]

is it clear what definition of trapezoid from this alone?

inclusive or exclusive

[maiden brook]

if u mean about a shape having more than one pair of opposite sides being parallel, there is not uniform agreement if it should be called a trapezoid. it seems this source would call such a shape a trapezoid

(a square is one obvious example - but yeah sources disagree if a square is a trapezoid for the above reason)

[wise zodiac]

yeah i know there’s disagreement

i want to know what this source thinks

[maiden brook]

at least one pair of opposite sides are parallel, i would think

[wise zodiac]

is it phrased ambiguously?

[maiden brook]

slightly. but a shape with two pairs of opposite sides being parallel does have "two opposite sides [which] are parallel"

[wise pawn]

personally once I came to terms with accepting that squares were rectangles as a kid, seeing this I feel totally fine accepting that rectangles are trapezoids.

[wise zodiac]

hmm fair enough, i thought so too

[wise pawn]

maybe as just a reassuring thing, if the length of the bases are b and B respectively and it has height h, the area of a trapezoid is h*(b+B)/2 which when b=B makes bh. So the area of a trapezoid formula really does give you the area of a rectangle as a special case

[wise zodiac]

i need 100% confirmation cuz i need to use whatever definition is here on my exam

lol

[coral cobalt]

depends, what exam

[wise zodiac]

yeah that’s fair enough

[coral cobalt]

i don't see any exam that wouldn't bother clearing the ambiguity caring about it

[wise zodiac]

no the handbook already provides this definition

which is unclear

but yeah so they won’t repeat the definition

😭😭

[coral cobalt]

which handbook

[wise zodiac]

i think the mainstream definition is the inclusive one

[wise zodiac]

some random exam

[coral cobalt]

which is

[wise zodiac]

yep fair enough, thanks

[wise zodiac]

school exam

its not jee

[coral cobalt]

oh

man

you could call a triangle a dorito for all they care

[wise zodiac]

i’m fine with all that but i need to know whatever this definition deciphers to cuz this is all that matters

[wise pawn]

you might want to ask your teacher real quick before or after class starts, "Hey prof! Is the definition of trapezoid is exclusive of rectangles?" if they say yes or no is what really matters since they're grading it! 🤣

[wise zodiac]

too late for that 😭also it’s a bit of a standardized exam

[coral cobalt]

i mean i dont see any question on a school exam asking this, and moreover its disputed

[wise zodiac]

whatever u guys interpret that thing means

[coral cobalt]

boards eh

[wise zodiac]

is probably how it’s supposed to be interpreted

[wise zodiac]

it’s not disputed bc they just provided that

this sounds like the inclusive definition of a trapezoid

[coral cobalt]

then take it at face value

[maiden brook]

also squares and rectangles fit under the definition of isosceles trapezoid, so it definitely makes sense

[median trail]

Guys can you give me a geometry question

[tough lotus]

Prove that the shortest line segment that can be drawn from a point to given line is Perpendicular to the given line.

[wise zodiac]

intercept form of a plane

we define the intercept form for a plane as x/j + y/k + z/l = 1, where the coefficients of j,k, and l are to be determined.

you know ax + by + cz + d = 0 is the general form of a plane

and then proceed from there

that's for a line

https://byjus.com/maths/intercept-form/

u can see a bunch of derivations there

[maiden brook]

||just Pythagorean theorem||

[maiden brook]

Prove Feuerbach’s theorem: that the nine-point circle is tangent to the three excircles of a triangle as well as its incircle

[obsidian hornet]

There are two triangles ABE and ECD with a common escribed circle- W, which touches the lines CE and BE at points B1 and C1, respectively. Let I and J be the centers of the inscribed circles ABE and ECD, respectively. It is known that the lines IC1 and JB1 intersect at a point S lying on W . Prove that the circumscribed circle of triangle AED touches W

[median trail]

Can i use coordenetes

[obsidian hornet]

There is a different ways to solve this, you can try

[quiet sonnet]

Would someone like to help me with this? It’s proving with trig identities. I get to the part with cos/1-sin but I’m having a brain fart on how you can multiply it by 1+sin/1+sin. Wouldn’t you have to multiply it by 1-sin/1-sin? This may be a stupid question but bear with me

I get to the part with cos/1-sin • 1/1 *

[prime musk]

no, u have to multiply and divide by (1-sinx) to convert the denominator to (a+b)(a-b) form- which gives u a^2-b^2..... So u get 1-sin^2x which is equal to cos^2x

[quiet sonnet]

Sorry, I’m still not getting it. Would doing that not end up with this?

Wait no I see what’s wrong with that

[prime musk]

u can't multiply different factors in the numerator and denominator

it has to be the same such that it reduces to 1

[prime musk]

and multiplying with (1-sinx) will give (1-sinx)^2

[quiet sonnet]

Yes I realized that after I sent it

[quiet sonnet]

Correct me if I’m wrong but you use 1+sinx just so you can get to the (a+b) (a-b) form right? I see that’s what you were saying now

[prime musk]

yes

[quiet sonnet]

Ok thanks for the help

[prime musk]

so that it becomes cos^2x eventually

[quiet sonnet]

Honest opinion. Do u think im cooked in calc 2 next semester if I had that question

This is apart of a refresher I’m making before I start classes again

[prime musk]

I mean depends on what's the syllabus-

but basic trigonometry just requires a bit of patience and practice- not that big a deal

[quiet sonnet]

Alright thanks

[tiny burrow]

js use yt

[coral cobalt]

hi guys! does anyone know any sources to learn olympiad geometry from scratch?

i mean i do know like id say 60% of the theorems and stuff but my mind goes blank the second i see a geometry question

[upbeat compass]

My teacher gave me like 100 math problems over break lol

i fucking love algebra and calc

but i dont really like geometry

cant draw for shit

its just not me

i like trig tho

its like the best parts of geometry

combined with all of algebra

to create a beautiful combinatio

[trail tendon]

fr

[upbeat compass]

graphing is shit

you have to be precise

but thats all my teacher gives me for extra credit

also

my teacher quit mid school year

so now we have an asian teacher as a replacement

like 5 kids got detention cuz mfs were making the stereotypes

i feel genuinely bad for her

she already getting hate

she aint even teaching

idk how she teaches

she talked to us

she is going to be faster than her

but i keep up

[gritty topaz]

yt

filled with great problems with great solutions by great teachers

[obsidian harness]

Highly recommend Evan Chen

Specifically, EGMO book

[coral cobalt]

any channel recs?

[coral cobalt]

thanks!

[gritty topaz]

my special interest was angle problems

but mainly in turkish

https://www.youtube.com/@GeometryKR/videos

and azeri

[reef plover]

Hi guys wsp

Anyone could help me w that?

[reef plover]

Found it

[upper echo]

Since two white triangles on the left are similar then (s-sqrt(b^2-s^2))/a=s/b.

[reef plover]

Did it already. Still thanks tho

[reef plover]

.

[obsidian harness]

https://math.stackexchange.com/questions/4839673/four-rational-pythagorean-triples-in-a-square#4839710

You might be interested in this

[obsidian harness]

Take cos of both sides using the identity for cos(a + b)

Also cos sin x, draw a right triangle for tjis

,w 2 arccos(1/4) + arcsin(7/8)

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by sooouth.

[obsidian harness]

Wait what did you even write

If that's not 7/8

[upper karma]

it is 7/8

sorry handwriting shit

[upper karma]

I don't understand sir :(

[obsidian harness]

The question is wrong

[upper karma]

oh no

wait how

[slender rain]

In ax² - bx + c = 0, what does a and -a have to do with the open up and open down of a parabola. Ik if a > 0 then it is open up and vice versa, but why??

[trail tendon]

for bigger x values, the ax^2 term is much bigger than (-bx+c), so the ax^2 makes the big changes. Also, we know that x^2 is positive, so the sign of a determines the sign of ax^2
So if a>0, then the biggest term is a big positive number so the parabola opens up. If a<0, the biggest term is a big negative number so the parabola opens down.

not the best explanation but idk how to explain it better 💀

[slender rain]

Dw I understand it, it's like ax^2 is the president and (-bx+c) is the citizen right

[trail tendon]

frr 😂

if you're close to the vertex then (-bx+c) and ax^2 are closer, but once u stray far enough away from the vertex ax^2 just completely overweighs (-bx+c)

just made up a new word overweighs

unless thats a word

idk

[maiden brook]

dominates is what u r looking for

but this looks like the wrong channel for that

[trail tendon]

[trail tendon]

math be like

[maiden brook]

its an actual term yeah thats used

[errant veldt]

https://youtu.be/kjol1py687M?si=SFhv5uqJhC3lgWEO Angles

https://youtu.be/a-rD7Pbnbnk?si=YF7dIWLijL9oIYmf

[slender rain]

Alright ty

[flat coyote]

Can anyone solve

[azure helm]

can someone help me

[craggy nest]

Are they rectangular ?

[flat coyote]

Yes

[junior finch]

Does anyone know how to solve this one

[pulsar hatch]

Turn the two triangles into right triangles with side length 2 as hypotenuse, then use the fact that regular hexagon has each angle 120º to determine that the triangles are all 30-60-90, this means that the leg of the triangle that is tangent to the square is equal to √3. Then, since the total height of the hexagon is 2√3, take 4-2√3 as the height of the shaded region and multiply by the width, 2.

[boreal shard]

I had that in a past paper a while ago lol

iirc you use the interior angles of a hexagon to construct a triangle so that you can find the height of a line to the middle of the hexagon, then because you know that the squares have side length two you can work out the area of the overlap

[azure helm]

Triangle $ABC$ is a right triangle. If the measure of angle $PAB$ is $x^\circ$ and the measure of angle $ACB$ is expressed in the form $(Mx+N)^\circ$ with $M=1$, what is the value of $M+N$?

[somber coyote]

938c2cc0dcc05f2b68c4287040cfcf71

[azure helm]

can I get some help

[median trail]

=1

[noble zinc]

try using the fact that the sum of the internal angles of a triangle is 180deg

then try finding N by seeing if any unknowns can get cancelled

since you know M = 1, you would reach the answer M + N, whatever it may be

[median trail]

Do you have the answer

?

[noble zinc]

I can work it out but it might be best to let the one who questioned figure it out themselves

I'm just posing one approach he/she can take.

[azure helm]

180 = x + bac
bac = 180 - x
180 = bac + 90 + bca
180 = (180 - x) + 90 + bca
acb = bca = Mx+N
180 = (180-x) + 90 + Mx+N
90 = 180 - x + Mx+N
-90 = -x + (Mx + N)
"acb = Mx + N with M = 1"
-90 = -x + x + N
N = -90
M + N = ?
1 - 90 = -89

[median trail]

Yes i got it right W

[azure helm]

good job 💪

[median trail]

Have you got any good other question

[azure helm]

I got a ton

[median trail]

Give other one

[azure helm]

but hopefully they are not too easy

[median trail]

It ok

I will try

[upper echo]

its 15 degrees

[azure helm]

elaborate

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[quaint obsidian]

helloo, how should I start learning trigonometry? Like, from where? Learning trigonometric functions maybe?

[azure helm]

what in the world

[upper echo]

yep, I overcomplicated. This equalateral is not necessary

[topaz marsh]

can someone please teach me

[azure helm]

I managed

[pulsar prawn]

$Happy:\left(\sum_{i=1}^9i\right)^2$

[somber coyote]

Aestusy

[lusty heath]

(20+25)^2=2025😋

[somber coyote]

occasus

[open vapor]

what kind of identities are you looking for?

[open vapor]

i mean it can be rewritten as $(\tan x + \cot x)^2$

[somber coyote]

世界最強の暗殺者、豹のように迅速の勾陳一

[azure helm]

can i get some help

[open vapor]

well first connect CG

and notice that BC = BG

see what that leads to

[upper karma]

ohh

missed that :p

[azure helm]

why do you say BC=BG?

[open vapor]

well BEFG is a square and BCE is an equilateral triangle

[azure helm]

ye ye my bad

you are right

if one of the sides of the equilateral is equal to one of the sides of the square then BC=BG

[open vapor]

yes

[azure helm]

I missed that

150 + 2x = 180

2x = 30

x = 15

BCG = 15

BGC = 15

[open vapor]

and how would you find GCE?

[azure helm]

GCE = 60-15 = 45

[open vapor]

exactly

[azure helm]

I appreciate the help, really

[torpid mulch]

Yo hey guys

Cab u tell me the relation bw sec and tan and secant and tangent in a circle?

[ancient fern]

Y'all got me feeling like I need to see some Khan academy 😭

[worldly juniper]

https://tenor.com/view/matematica-gif-7322246

sec is just the inverse of tan im pretty sure

[torpid mulch]

Isnt that cot?

[worldly juniper]

well, yeah. my bad

[torpid mulch]

Its k man

Can u help me in a sheet tho?

[worldly juniper]

huh

[torpid mulch]

A math sheet

Class 9th and 10th

[worldly juniper]

16. is 1?

not sure

[torpid mulch]

Howd u do it?

[torpid mulch]

Lemme see

[worldly juniper]

sure

you just do polynomial division

long division*

[torpid mulch]

Nope

[worldly juniper]

dang

idk then

[torpid mulch]

f

5*

[craggy warren]

guys can you recommend me a good book about Trigonometry geometry?

[worldly juniper]

is 17. 2 by any chance

@torpid mulch

[torpid mulch]

Lemme see

[worldly juniper]

https://artofproblemsolving.com/store/book/intro-geometry maybe this but idr if ive used it or not so take my advice with a graain of salt

[torpid mulch]

BINGO!!

[craggy warren]

thank you

[worldly juniper]

yayy

[worldly juniper]

ofc

[torpid mulch]

Howd ya do ot

[worldly juniper]

you set the equation equal to 0

and factor out x

then you can assume 1 is a root

and go from there

[torpid mulch]

OMG IM AN IDIOT

TYSM MAN!!

[worldly juniper]

huh

you sent the sheet

[craggy warren]

it costs 65$ (
Do you have free version of this book?

[torpid mulch]

Yea

[worldly juniper]

ok yw

[torpid mulch]

Bcz i had doubts in it and u helped me

Can u solve q20?

Its pretty good

[worldly juniper]

https://openstax.org/details/books/algebra-and-trigonometry-2e

dont think they have a geometry one unfortunately

personally geometry was like, my least favorite

so i never cared

[worldly juniper]

uh ill try

[torpid mulch]

Ok

[worldly juniper]

is it 38

completely random guess

@torpid mulch

[flint ingot]

Show that you can split any convex quadrilaterals into any finite amount of pieces and rearrange them to form a rectangle.

[slender rain]

I see that parabola is opening up infinitly, does that mean the green line would intercept the parabola at one point?

[trail tendon]

no

the parabola grows much faster and the green line grows much slower

[slender rain]

I didn't quiet understand what is this means

[trail tendon]

as you go farther along the positive x-axis, the distance between the parabola and the line gets greater

and it continues to do that forever

since the distance between them grows, they will never intersect

it would have to shrink for them to intersect

if the distance between them was decreasing like this, you know they would intersect

as long as they continue on their implied path lol

[wise pawn]

I suppose you could also try solving y=x^2 and y=x-1 and check the discriminant

[trail tendon]

as a way to prove that they dont intersect?

[open vapor]

yeah that could be a good way

[slender rain]

I see, make sense , but i was thinking that a parabola is going to open up widening, so it still doesn't have the chance to intercept the line in x positive? Even when the line goes to a place further away from the parabola, while parabola is widening?

[open vapor]

well if the two graph intercept

there must be a point where one x satisify both equation

the x does not exist

[trail tendon]

it could, but it doesnt seem like it will based on the trajectory that you drew

[open vapor]

hence they dont intercept

[slender rain]

What discriminant have to do with the interception? I think I also still didn't quite understand this one

[slender rain]

So x1 = x2, and y1 = y2? I seee

[open vapor]

yes, there is no such x that can result in same y

[slender rain]

Why??

[trail tendon]

because its path is no longer pointing towards the line (right) but rather up more

[open vapor]

and for the discriminant, it is essentially saying that y = x^2 = x-1 can be rearrage into x^2 -x + 1, this equation does not have real roots by checking discriminant

[trail tendon]

the arrow that implies the path is only going to turn in the left direction basically

[slender rain]

What is the topic of this lesson?

[open vapor]

quadratic equation

[slender rain]

Could you also show me why that line will be going to the left direction?

[trail tendon]

i cant draw very well but the more you 'zoom out' it will keep going up 💀

[slender rain]

Is it because the increase in y value will be greater than x value in higher number that causes the parabola to have steeper graph? For example in x² - 2x - 5, when x = 1.000 and when x = 1.001 have bigger difference in the y value

Im trying to process this one

[trail tendon]

essentially yea
the difference between y=x^2 and y=x-1 increases as x increases

[slender rain]

I see I see

[torpid mulch]

Nah its ||8||

[maiden brook]

or alternatively x^2 - (x - 1) is (x -1/2)^2 + 3/4 so always positive

[willow sinew]

Could someone please help me understand the differences and utilization of degrees and radians? I’m really struggling to understand. Thanks in advance for any help!

[cunning lion]

1 full turn = 360 degrees = 2π radians

degrees have easier arithmetic but radians have a nice arc length formula and become extremely important in calculus

[worldly juniper]

hmm ok

[slender rain]

what is the sum of the total angles of this triangle?

[void raptor]

the sum of all angles in ANY triangle should always be 180 degrees

[azure helm]

interior angles but yeah

The medians $AD$, $BE$, and $CF$ of triangle $ABC$ intersect at the centroid $G$. The line through $G$ that is parallel to $BC$ intersects $AB$ and $AC$ at $M$ and $N$, respectively. If the area of triangle $ABC$ is $810$, then find the area of triangle $AMN$.

[somber coyote]

938c2cc0dcc05f2b68c4287040cfcf71

[azure helm]

can i get some help

I tried doing a drawing, let me fix it and send it

[azure drift]

What is the value of y?

[bright parrot]

I think 65° because we know that the angle OQR is 25°. Moreover, we also know that POQ is a isosceles triangle because OP = OQ. So, the angle OQR is the same as QPO (25°). We can see that the angle POQ = 180 - 50 = 130°. It's also a central angle with the same arc than the angle y. We know that a central angle is 2 times an inscribed angle of a same arc. We have : y = 130/2 = 65°.

[azure drift]

Yeah it's probably that thanks

[worn scroll]

yeah definitely

[azure helm]

can I get some help?

[worn scroll]

do you notice anything about the triangles ABC and the folded triangle?

[azure helm]

similarity

[worn scroll]

exactly! what do you know about the "ratios" of the triangles ABC and CDE? (CDE in the end is just the folded part)

well you got some help on the other channel

but basically in similar triangles a side of a triangle is, let's say "scaled up a factor k" in the other similar triangle

same thing with the area, but it is scaled a factor k^2

makes sense, since we are talking about an area which would be base * height

and in the similar triangle it would be (base * k) * (height*k) which would be the same thing as (base * height) * k^2

so it's just the area of the first triangle * k^2

note the k is the same as the k for the sides

@azure helm

so if you know that CDE = 16/100 ABC

you can try to find the scale factor

and then since you have k^2

you just want k because we want to find a side

so square root of that

and then you just multiply it by the side of the first triangle

[azure helm]

I don't get it

[worn scroll]

whenever u have similar triangles