geometry-and-trigonometry

[sly dawn]

how do you prove that ABC is an isosceles?

[lusty girder]

well

AC=BC

[sly dawn]

oh wait

yea

my bad did read it properly

[lusty girder]

Aight

Idm

[sly dawn]

answers is 114?

[lusty girder]

Anyways once the angle of A in both triangles combined is solved, you solved the question

[lusty girder]

Yep

[sly dawn]

k

@tall ridge

[neon trellis]

Where am I going wrong? Why does the fraction change a lot? How do I know which one is denominator and which one is nominator? It’s messing up my scale factor value and that’s why I get the answer wrong. Either wrong or right, it’s just luck atp

Ahh I shouldn’t have tried to find the unknowns that way…

It changes based on which one is wanting to be found

UGH

It’s ok

[upper karma]

In these kinds of things, the order in which u write things is heavily heavily important
If u just want a tip to avoid such minor mistakes be extemely careful about the way u write stuff

[nocturne remnant]

you still have the wrong answer...

[dark sparrow]

well congratulations on spectacularly missing my point

[nocturne remnant]

anyone who isnt blind would realize that BAD is definitely not 90
degrees

so what did you mean to say?

[grave pond]

114 sounds right.

[smoky jetty]

do congruent sides must have congruent angles?

[smoky jetty]

ohhh bruh, this makes me disappointed how I typically miss the important details in the statement

[smoky jetty]

if so, what decent reasoning do y'all know about it? I'm quite confused with google

So far, this is how I understood it. If the measure of the adjacent sides of two angles of a polygon is congruent, then so is the angles' measure.

just combined some info from the web, and how I think of it

[smoky jetty]

does this works conversely?

[neon trellis]

thanks

[heavy adder]

I'm confused is this channel basses solely on university level geometry or any level?

[silent plank]

around high school level in this specific channel
though you're better off claiming your own channel
there may be more appropriate stuff under #get-advanced-access if you want uni-level help

[keen token]

3sin(x) + 4cos(x) = 5 find cos and sin

[nocturne remnant]

use the given equation and also sin^2 + cos^2 = 1 so obtain a system of simultaneous equations

and solve them

[upper karma]

U mean cos & sin of x ?

[keen token]

it just says, solve sinx and cosx

can someone solve it for me i wanna see the actual process

[somber coyote]

Pebble

[dreamy locust]

can yall give me the answer for this

[kind fjord]

@dreamy locust FDE = FED

Also, FDE = EDU

FDE = EDU which completes the proof(alt. interior angle)

[mossy stream]

Pls help

[sly dawn]

so basically, using pythagoras theorem-
720^2 + r^2 = (648 + r)^2
then you just solve the equation

[smoky jetty]

can u elaborate how u found it? (i.e how you knew that it would be the equation)

[sly dawn]

ok so, since MK is tangent to circle J, MK is perpendicular to JK (i.e. MKJ = 90 degrees). NJ = KJ as it is the radius of the circle, so we call it R. Then using the Pythagoras theorem which states that the hypotenuse squared is equal to the sum of the square or the remaining sides, we get-

720^2 + R^2 = (648 + R)^2

hope that answers your question.

I can also solve the equation if you want.

[meager skiff]

Am I Doing it right?

[smoky jetty]

ohh i see now. When u said pythagorean, i didnt know there would be variables included. Thx!

if I get this right, hypo as you have previously stated is = (648 + R)^2, as to complete the wholeness of the length of the hypo, the known value (i.e 648) and the other part (which is the missing value the variable R represents) would be added together, and so fort and so on

wow that's actually fun to realize, lol

[upper karma]

Can anyone help me

I have a doubt regarding trigonometry

[dark sparrow]

@upper karma post your doubt

http://dontasktoask.com

[upper karma]

@dark sparrow

@dark sparrow this is my doubt

[dark sparrow]

that's a problem. your doubt could be "i don't know how to begin" or "i tried something but got stuck" or "i got some nonsense"

also no need to double ping

[ebon sundial]

Anyone I need help with this one

[knotty pier]

i can't read it

3ctg [(pi / 6) + (x / 2)] > -sqrt3???

[ebon sundial]

Yeah ig

[sly dawn]

Glad you understood it. :)

[sly dawn]

Yeah its correct.
Btw you could also check it yourself by substituting the value of x and check if both sides are equal.

[sly dawn]

Why?

Like why here?

[silent plank]

<@&268886789983436800>

[next mantle]

<@&268886789983436800>

[still socket]

oh god

this message is literally in every channel

*almost

[high sequoia]

I solved it

[high sequoia]

Was this helpful???

[neon trellis]

how to do this

[solar frigate]

Sin ∅=4/10

[sly dawn]

0.4 is also equal to 4/10. so 4 becomes the perpendicular and 10 becomes the hypotenuse. then use pythagorean theorem to find the other side

[sly dawn]

Where'd you get that fancy looking theta?

[solar frigate]

Cos ∅ is equal to 6/10 also 0.6

Hold down on ur zero

[sly dawn]

um alright i'll try that when i am on my phone, thanks

[solar frigate]

Ok

Sorry it 8/10 not 6/10

0.8

Cos ∅ is equal to 0.8

[timber cargo]

Pythagorean identifies

Identities*

[high sequoia]

Is this correct??

[eternal phoenix]

ig yea

[timber cargo]

It is right

[eternal phoenix]

yes ; that is what i said....

[high sequoia]

Ok thanks

[silent plank]

depends on what you're told about theta

[high sequoia]

I don't get ?@silent plank

[silent plank]

in additional to the dodgy notation at the end
the sign of cos(theta) will depends on what quadrant theta is in and/or whether theta is an acute angle

[high sequoia]

Yep!

[smoky jetty]

meaning the ratio could be negative?

[brisk ferry]

yes

[high sequoia]

Ok

[upper karma]

what was it?

[still socket]

they deleted it

it was a server invite

for an adult tiktok server

[lusty needle]

[late heron]

Hell

[high sequoia]

Why hell?@late heron

[upper karma]

Hell o

[fiery cliff]

Who know how to do this

[grave pond]

The two angles at c (hmm, or is that perhaps an e?) are different in your proposal, but the problem looks like they're supposed to be equal.

And you have a triangle with angle sum 66+(58+56)+56 = 236.

And angles h and k which are clearly supplementary in the drawing, don't sum to 180° with your numbers.

[lilac wraith]

guys, If i want to learn geometry or trigonometry, from where do I need to start?

[severe orbit]

Hey guys

Need help with this problem

[smoky jetty]

What do you know about the properties of a quadrilateral rhombus?

[smoky jetty]

I'll give u 2 hints: Diagonals are perpendicular and bisect each other into two equal parts.

[glacial otter]

Khan academy, openstax, youtube,

[bleak blade]

Aops

[severe orbit]

Hey can someone help me with a question

The top proof

<@&286206848099549185>

[heavy shale]

plz send clear pic

[severe orbit]

Thats all i have man

[heavy shale]

also r u giving an exam?

[severe orbit]

i ripped the worksheet

[heavy shale]

oh lol

[severe orbit]

no its test corrections

[heavy shale]

its hard to read

[severe orbit]

i didnt know the answer

U can read if u zoom in

i mean i can

[heavy shale]

i am on pc so i cant sry

[severe orbit]

Oh ok

one sec

[heavy shale]

u gotta wait for someone else ig 😢

[severe orbit]

Better?

Try to take an ss and see if u can zoom in

@heavy shale i can type it all

<@&286206848099549185>

[heavy shale]

ok so u need to prove CB=CD for it to be rhombus

so try thinking of ways to prove the triangles similar

[severe orbit]

Can y send like the statement and reasons, this thing is due in 7 mins

Please

I promise ill stay and listencto the explanation after

But i rlly need to submit this

[heavy shale]

ok wait

[severe orbit]

Ok thank you

[heavy shale]

Angle C = Angle C (Common)
Angle F = Angle E = 90 (Given they are perpendicular)
So, Triangle CEB ~ Triangle CFD (Angle-Angle)

BC/CD = CE/CF (Corresponding sides of a similar triangle are Proportional)
Since, CE=CF
BC/CD = 1
=>BC=CD
Since Adjacent and opposite sides are equal 
ABCD is Rhombus```

Sry cant send pic not on phone rn

[severe orbit]

Oh aight

Wym common?

[heavy shale]

they are the same

[severe orbit]

Oh ai

[smoky jetty]

nicee

does two = angles in both triangles automatically concludes to them being congruent?

[heavy shale]

Similar not Congruent

the property id AAA

[severe orbit]

Are you possibly on phone?

Or something

[smoky jetty]

nop

laptop

[severe orbit]

its hard to understand the proof

Like that

sorry

[smoky jetty]

ah ok

[heavy shale]

yea ik phone screen small

[smoky jetty]

wdym?

[heavy shale]

i cant take screenshot of it

[severe orbit]

Ok wait i have an idea

dont give me the givens

Because i alr have it

but

just give the info

one line at a time

like

a is congruent to d

reason: (put here)

[heavy shale]

Just use the image

[severe orbit]

i feel like i havwnt learned about the proportional part

Never heard it here

[smoky jetty]

me neither, lol, i mean when using it to proof things about quadrilaterals

[severe orbit]

Czn u like tell me row by row

like the statement and reason

like clearly

[smoky jetty]

wdym?

im confused, sorry

[severe orbit]

like tell me the statement

Then the reason

one by one

Not the givens

[heavy shale]

its like the reverse of SSS

similarity criterion

[smoky jetty]

what does sss mean here? eek me

[heavy shale]

Side - Side - Side

[smoky jetty]

yeah, but are u referring to another theorem?

[heavy shale]

its a similarity rule

no

[heavy shale]

its a similarity rule

[severe orbit]

What do i put so far

for the reason

[heavy shale]

wait have u learned similar triangles?

[severe orbit]

Dont think so

[smoky jetty]

oh i see, just googled it quick

[heavy shale]

i think then my proof is out of your syllabus

[smoky jetty]

hatin' not having textbook in geo rn

[severe orbit]

yep..

[heavy shale]

sry i dont know any other way

[severe orbit]

Because im in 10th grade

so yea..

[heavy shale]

i am also in 10th lol

[severe orbit]

Do you know how to do it

[severe orbit]

Oh wow lmaooo

[smoky jetty]

im in 9th, eek

[severe orbit]

nicee

[smoky jetty]

im gonna re-read this convo later, and see how much i could grasp, lol... more gaps were found

[heavy shale]

Ok so i got it

[tawny vector]

Is there a purpose to specifically naming the reciprocal functions

[upper karma]

reciprocal functions indicate perpendicular

[grave pond]

At least the cotangent does.

It varies a lot from country to county how much emphasis is given to secant, cosecant, and cotangent. It appears to be common in the US to teach and drill them, whereas in many European countries you never even see those names (and seem to do quite fine without them).

[tawny vector]

Hmm alr

Yeah had them briefly mentioned here in canada but there wasnt much emphasis

Thanks 👍

[high sequoia]

Really?

[grave pond]

To the best of my knowledge, yes.

[dark sparrow]

in russia you see sine, cosine, tangent and cotangent

secant and cosecant get at most a passing mention

[orchid agate]

In Ukraine as well

[dark sparrow]

wouldn't be surprised if that was also true for most other ex-soviet states tbh

[orchid agate]

Thankfully we are not post soviet now

But yeah

Geographically yes

[dark sparrow]

geography's all i was talking about, of course

[orchid agate]

Where are you from though

You also from Post Soviet country?

If you know that much stuff

About it

[dark sparrow]

i am from russia

[orchid agate]

I see

Gtg

[wise belfry]

Any one know about isometric math concepts?

[smoky jetty]

ight, so I re-read this again, and yeah, they all make sense to me. And I realized that we were just about to go to this topic next year in school, so I thought, what's better than learning it in advance, lmao

[versed vigil]

Hello, how can I solve for angle a?
k = sin(a) * cos(a)

0° < a < 45°
k > 0

[verbal frigate]

please help

[smoky jetty]

not sure, how about if k = 30deg?

[versed vigil]

k cannot be measured in degrees in this case since it is not an angle

I have since found out about double angle identity

[grave pond]

There's a constant factor missing there which disappeared on it way from the double-angle identity.

[versed vigil]

oh

oops

sin(a) * cos(a) = sin(2*a)/2

I really need to relearn trigonometry...

[smoky jetty]

same

[neat python]

How can I find trapezium area i only know MN=15 and height=10

[tawny vector]

Idk if this is right

But cant you solve for AB?

Oh wait im dumb sorry

[upper karma]

i made a fun problem a while ago

ABCD is similar to CXYZ. What is the perimeter of ABCXYZD?

[tawny vector]

Is it ||214.763|| or am i dumb

[upper karma]

I can check what I got

Ok so a year ago I got ||214.76|| so I think you are right. Today I tried it again and got 129.29 lol

I think I know what I did wrong this time

[tawny vector]

||22/11 = 2, so ABCD is increased by a factor of 2 (idk the right terminology lol). So that means that x = 18, 3x = 54. AD / AB solved by pythagorean, then divide by 2 for CX / XY||

[upper karma]

Yes

I think I was just lazy and used the length of 3x as the hypotenuse lol

I should have checked it but luckily it is not a big deal

[sick spindle]

In a trapezium, the line joining the mid points of non-parallel sides is Half of the sum of the parallel sides

So the area will be 150 sq units

[neat python]

I found area but Ty anyway

[sick spindle]

Alright cool

[smoky jetty]

shouldn't it be 100?

My equation was 10 = (b_1 + b_2)/2
(2)(10)= "
20 = "

Area = ((b_1 + b_2)/2)(h)
= ((20)/2)(10)
= (10)(10)
= 100?

[spiral marsh]

nope.MN=(b1+b2)/2=15,so the Area will@be MN X height=15 X 10=150 cm^2

15 multiplied by 10

[smoky jetty]

ohh ffs, i thought the median value was 10, eek, gonna wash these eyes thoroughly, lmao

[upper karma]

is locus is like a path of eqn? like eqn of circle is a kind of locus?

[grave pond]

"Locus" is a traditional word for "the set of points that satisfy some condition".

A circle is the locus of all points whose distance from the center is such-and-such.

[upper karma]

😭😭😭😭😭😭😭😭

[grave pond]

Is that so sad? There, there.

[upper karma]

yes hell sad

[upper karma]

so isnt that kind of eqn of line?

[grave pond]

No, the thing called a "locus" would be the line itself, not an equation.

[upper karma]

find eqn of locus of a centre of circle with some conditions?, does that imply to find eqn of of line that passes through centre with some " given condition" ?

[grave pond]

I'm sorry, I don't understand that question well enough to suggest an answer.

[karmic forge]

can anyone help me? I have a math problem; there is a triangle with two perpendicular lines on the two sides ( AB and BC ), and i need to find the equation of the two perpendicular lines of AB and BC. you can look at the way that I did to find the line equation of the perpendicular line of AC, and the answer that i found is y=-2x+-1.5, but that real answer is y=-2x+2.5. what have I done wrong?

[storm portal]

Are e and f at certain points?

There are infinitely many lines that are perpendicular to a given line

[karmic forge]

whatever i solved it. i was trying to solve the perpendicular line of AB, which start at point E

it's the middle of AB

[narrow slate]

How do I solve this through algebra?

[somber coyote]

wintersavory

[wind compass]

Can someone explain direct inverse and joint variation please

[dark sparrow]

https://m.youtube.com/watch?v=v-k5L0BPOmc&vl=en

[smoky jetty]

khan all the way!!

[upper karma]

Why does trig functions need to be defined in a unit circle? Why not in a circle with a radius of 10 or 100? If its because its more simple to do calculations, does that trig functions wont work if it is defined in a circle with a radius other than 1?

Or is it arbitrary? Just like the 360 degrees measurement system

[nocturne remnant]

the unit circle is most natural... otherwise many trig identities would have unnecessary clumsy coefficients attached to them.
For example, if they are defined in a circle of radius 10, then sin(x+y) = 0.1(sin(x)cos(y) + cos(x)sin(y))

[hallow ferry]

Guys I have chapter this trignometry
Does anyone have any idea about Heights and Distances calculating with the help of Trignometry
Pls DM me and help me pls

[upper karma]

the actor or the teacher?

[smoky jetty]

I was tryna practice myself with geometry and found this problem which for me is ridiculously hard.. Anyone wanna give it a shot?

[grave pond]

There's a lot of irrelevant data there, so your problem is most likely that you're overthinking it. It is ridiculously much simpler than it looks.

(Note that the arrows near letters K, M, Q seem to say that IK and RQ are parallel).

[smoky jetty]

yeah, so far I was able to solve for D, C, Z, angle PDE, angle JPK, but idk how i could get to y and x

any other hints that you can give?

[grave pond]

It is much simpler than you think.

All of those are irrelevant for the relation between y and x.

[smoky jetty]

alright, i'll try to analyze it again

[grave pond]

By the time you say "analyze" you're already overcomplicating it.

[smoky jetty]

ah shoot

sorry, i really couldn't find it

could u show me how u got the answer?

[grave pond]

Hint: There's no computation going on.

[smoky jetty]

oh woww

hmm

[grave pond]

Hint 2: Ignore CJ and LM and all angles involving one of those lines.

[smoky jetty]

i still cant see how, coz for me i couldnt find their values

[grave pond]

You're not being asked to find their values in numbers.

[smoky jetty]

wdym?

[smoky jetty]

oh wait wha-

[grave pond]

"Express angle x in terms of y and z".

[smoky jetty]

if not, then basically x = y?

[grave pond]

Yes.

[smoky jetty]

omfg

bruhh moment for me

i was tryna find the numerical value the whole hour

[grave pond]

Sorry.

[smoky jetty]

oh gosh

cries in shame

anyway, thx for easing my sleep later, troposphere!!

[hearty iron]

Can someone help me with 3 or 4 please

[smoky jetty]

not very certain, but this is how I tried (for no.3):
70 and angle DOA are supplementary, as they are linear pairs of angles. Therefore,
70+ angle DOA = 180 so on...
Now, with the value found, the sum of angle A, DOA, and angle D must be = 180 (as for any triangle)
Since alternate interior angles are congruent (in a parallelogram), then ange D is = to angle DBC

if ever this is wrong, I'll try to revise/remove it to avoid confusion

[hearty iron]

Ah i didnt know the last rule

Thanks

I thought that but just wanted to be sure

[smoky jetty]

DC || AB, and diagonal DB acts as the transversal

thats how I often visualize it

[hearty iron]

You have an idea for number 4? What does it mean by the angle between the bisectors of the other 2

By guess is B

[smoky jetty]

im tryna see if i could

nvm i couldn't

[surreal obsidian]

Did I gaslight myself

I swear I remember reading a theorem where if you have four lines in a circle that meet at a single point, partitioning the regions by taking every other one divides the circle into two sets that are each half the area of the original circle

but I can't find it anywhere online

nvm found it

requires the additional restriction that the angle of each sector where they meet is equal

[wind compass]

thank you!

[smoky jetty]

if u were to guide a student, would u prefer them to learn triangle congruence first, before theorems on quadrilaterals? or the other way around?

[lost siren]

Ig Triangles and later Circles and start quadilaterals between circles from cyclic quadilaterals

[dark sparrow]

triangles first, definitely.

[smoky jetty]

oh now it made sense

our teacher first discussed quads before triangle congruence, thats why when she gave us some proving problems, some of the reasons for me were unfamiliar coz they were yet to be discussed. I wouldn't know if I didn't look them up on my own

when coming up with this problem (for reference 5x is the median of a trapezoid), I often get curious why our teacher (or even the cool guy on YT, michael van bizen) transposes the denominator to the LHS. Though I tried not to (i.e diving the numerator to it), I still got the same answer. Is there a particular reason why some do it?

[dark sparrow]

none in particular

this is a linear equation

there are multiple ways to solve those

there is rarely a big Reason why one solution path is chosen over another

[smoky jetty]

oh ok thx. hopefully when i got a pretty long free time, i could study basic algebra again

[smoky jetty]

I got this quiz problem before (unfortunately, at the time, I wasn't able to get the answer, and I got 0 points from it), though after reviewing kites and trapezoids (our current geometry lessons) for hours, I still don't have an idea on how to get the area, or even know what quadrilateral the said lot is. Any hints or guide would be appreciated!

[feral sedge]

Hello @upper karma I have solved this problem
Rectangle ABCD ~ Rectangle CXYZ =>
CZ/BC = YZ/CD
CZ = YZ =>
BC = CD
So, similarity ratio r is equal to
CZ = CD/2, BC = CD(similarity)
r = BC /CZ = CD/YZ = 2
........

[upper karma]

Yes the scale ratio is 2

[feral sedge]

yeah, that's right

have you solved it ?

[split galleon]

Yeah try solving this one

[sullen flame]

arctan(12/6) =

evaluate that

@split galleon

[split galleon]

Thats not the answer

Tho

??

[astral wadi]

I would advice drawing a picture of the scenario first and go from there

[deft fog]

Hi guys, help me solve this please, and what is the answer?

[stuck sleet]

Analytic and Coordinate geometry is the same thing right????

[rain oak]

Google: congruence theorem triangle SSA

Should be around 243.516

[soft nest]

arctan(2 )= angle (theta) = ---> theta = 63.43 degress

[split galleon]

Nope

[upper karma]

243?

[split galleon]

The answer is arctan(2/11)

[dark sparrow]

do not give out answers.

[soft nest]

ohh the angle subtended is with respect to the base of the statue not the ground

rigtht?

[split galleon]

Alright sorry I am actually new so 😅

[split galleon]

Shall I give you my diagram?

[soft nest]

na na i got it ty tho

[split galleon]

And you even got the answer?

[soft nest]

yes just use differnce of angles formula from there

[split galleon]

Yup

[soft nest]

and u get it

[split galleon]

Right

The question was actually asked in JEE mains

Like one of the toughest exam in india

[soft nest]

fr ahh , i am glad i got it 💩

[soft nest]

i am indian only

[split galleon]

Yo boi 😂

[soft nest]

i am in 10th so its pretty nice that i got it

[split galleon]

Great for you!

[runic beacon]

That ‘only’ being another proof of that

[split galleon]

Try this

[runic beacon]

Who?

Do you need help with that or?

[split galleon]

No I don't need help

U can just solve it for fun.

[soft nest]

sin theta = 1/2.6

[split galleon]

No

[soft nest]

whaa it shud be

i will show working

csc x + cotx = 5 ---- 1
1+ cot x^2 = csc^2 x
1 = (csc x + cot x ) (cscx- cotx)
csc x - cot x = 1/5 ---- 2
adding 1 and 2
2 csc x = 5 +1/5
csc x = 2.6
sinx = 1/2.6

[soft nest]

whats wrong

anyone?

[upper karma]

How did you get the second line?

[soft nest]

diff of squares

csc x^2 - cot^2 x

[upper karma]

No, the line above that

[soft nest]

thats a trig identity very common

[upper karma]

Oh

I should study trigonometry lol

[soft nest]

\

[soft nest]

its the same as 5/13 , 5/13 =1/2.6

[split galleon]

Yup i got u

I got confused with that 2.6 of yours

Good going.

[lunar smelt]

Hello there. I wish to get a head start in my high school studies and have been considering learning geometry from either Khan Academy or the AoPS Geometry textbook. If it is of any help, I also wish to pursue mathematical and computer science competitions. If there is also any other resources that may help me, I would be greatly appreciative to know about them.

[keen lintel]

Urm for geometry, if u wish to have a good basics, then EGMO by Evan Chen would be great, and also A beautiful journey through Olympiad geometry by Stefan is a good book too. But both of these are more oriented towards math olympiad styles. If u wish to pursue geometry that is around university standards, then sadly I cant help with that.

[lost vector]

Help, please!

[smoky jetty]

interesting problems, though I cant help

[finite nebula]

Which number specifically?

[smoky jetty]

I got this quiz problem a few weeks ago, and I got 0 points because I didn't know how I was going to solve it. Trying to solve it yet again (just to practice), but I still don't know. By making the illustration of the kite vertically, I couldn't find the answer. Though, I could, if I made the illustration horizontally, but, the value doesn't seem to match how it's represented by the illustration I made (i.e. the value of supposedly half of the shorter diagonal is greater than the hypotenuse of the larger isosceles triangle

assum the diagonal IT is longer, coz I suck at drawing online

is there something wrong with my diagram? Or there's a concept I must know in order to find it

[finite nebula]

4: SY = 3 cm

[grave pond]

Hmm, yes, the problem is only solvable if you assume IX=TX. It looks like the author of the problem assumed that there's a standard order to refer to vertices of a kite, such that "LIYT is a kite" would give that information, but I don't think any such convention exists.

[obtuse quiver]

yeah if ix=tx use the pythagoras theorem and double it

[smoky jetty]

yeah thats what I thought as well

is there a way to solve an isosceles trapezoid's dimension with only the diagonals & the area given? I've searched on google, though I couldn't find any

[grave pond]

No, since there are more degrees of freedom than givens. For diagonal length of 1, as long as the given area is 1/2 or smaller, you can always find a rectangle solution -- but obviously not every isosceles trapezoid with a small area is a rectangle.

[smoky jetty]

hmm, i wonder what's up with the quiz questions given to us...

can u take a look at this, maybe my comprehension was wrong

it didn't specify what quadrilateral or even shape it was, but I just assumed that it was an isosceles trapezoid as it was our previous lesson. I did get the value of the diagonal from it

[grave pond]

Yeah, that looks underspecified. You can get the perpendicular distance from the length-8 diagonal to each of the two other corners, but not x itself.

[rain wharf]

How do I solve for x? Knowing all 3 lengths of the triangle

[upper karma]

um

tbh i am also confused on how to even explain it

[upper karma]

Could you add more information about those segments?

[timber cargo]

If that's an equilateral triangle, then x would be of the same length as halve of a side length of the triangle

[real wyvern]

Just a question but this is the right identity right

[somber coyote]

Jagteshwar(JPS838898)

[rain wharf]

All sides are different

[heady cloud]

hey so im not the best with geometry can someone tell me if i have done this assignentment right?

(i measured my desk and those are the actual dimensions of it... is there a part im missing or didn't add.

[potent flower]

just multiply every dimension with given factor..

[dusky stratus]

how do i solve this

[pliant roost]

@dusky stratus to what topic is this problem even related?

in what chapter is it

[dusky stratus]

circles

[pliant roost]

maybe it's in some chapter

or wherever

maybe you should make up some right triangles, m?

ohh, wait, I think I've gotten an idea

I'll answer in #competition-math

[glad geyser]

if the normal to the ellipse x^{2} + 4y^{2} = 4 at the extremities of chords ax + by = 1 and px + qy = 1 are concurrent then ap + bq = ?

[grave pond]

By the way, don't post the same question in several different channels like that. It is inconsiderate to people who reply in one channel without having seen the discussion already in the other one.

[upbeat latch]

Could anyone tell of their approach to tackle this Question

[grave pond]

That's not a question, just an expression.

[dark sparrow]

well, the question might implicitly be "evaluate this"

[upbeat latch]

yes, it is EVALUATE the following

[grave pond]

Well, the inner series at least should be simple to evaluate -- that's just a finite arithmetic sequence.

[pliant roost]

since it's finite, you could probably die writing out this immensely long sum of cotangents

I wonder if there's something more elegant

[grave pond]

The cotangent of k is the argument of k+i, so the sum of cotangents here is the argument of the product (3+i)(7+i)(13+i)···(381+i). But for that product I can't think of anything better than brute force.

Oh, wait -- the factors are 1+n(n+1)+i which happen to be (1-ni)(1+(n+1)i).

So if we're only interested in arguments, the series telescopes!

And we're left with cot(argument of (1-i)(1+20i)).

[pliant roost]

forgot that we can use complex numbers! then of course the problem reduces to something simpler...

is the answer 21/19, then?

since (1-i)(1+20i)=21+19i

[grave pond]

That's what I conclude. But I might have missed something along the way.

[upbeat latch]

yes

had not thought of Complex Numbers while solving, so was stuck

Thanks👍

[keen yoke]

Is there anybody who had decent knowledge of Python and knows 3D trigonometry that can help me out with a small problem that I have in my code? I am trying to determine the vertices of a rotated cube across all axes, but something is breaking and it is telling me the y value when rotated by 90 degrees is 2, which is impossible

[noble crest]

Hey can anyone give me the AOPS geometry pdf please

[bleak blade]

It’s on libgen

[foggy ledge]

Just need the blanks

[timber cargo]

Sin of B is b/a

and Cot (A) is 1/tan(A)

[timber cargo]

You can set a to x and b to 2x

Then pythagorean theorem to the answer

[silk jacinth]

if a circle is divided into subsections that are bounded by perimeter(s) of smaller concentric circle(s) and evenly spaced radii, do these subsections have names? examples of two such subsections are colored red and blue in the image

[wise pawn]

I've heard the blue region referred to as a polar rectangle and the red region is a sector, but this is without the evenly spaced radii constraint

[silk jacinth]

the evenly spaced constraint might not be relevant here, just thought that there would be some radii bounding the sections

yeah the red would be the sector of the smaller circle, but it has a distinct relation to the larger circle

polar triangle?

[wise pawn]

no

I think if you want to get more descriptive and you're already using polar coordinates, people will just refer to the coordinates themselves

[silk jacinth]

I see

I think I'll get started with the polar rectangle stuff though 🙂 thanks!

[wise pawn]

like the region between theta_1 and theta_2 and r_1 and r_2

yup you're welcome

[errant lake]

Really not sure how to solve them I tried drawing triangles and the exterior angles but I couldn’t solve anything

Any help?

[south vessel]

For Problem two, I think you need to find all the exterior angles using this rule x+y+z = 360. Sum of exterior angles of triangle equal 360 degree

[half cosmos]

is it possible to solve a triangle given two sines?

[errant lake]

Yeah I found problem the answer to problem 2 already but I still wanna know the relationship in between exterior angles and question 1

[smoky hearth]

i need help

asap plz

[pine sequoia]

hmm?

[upper karma]

geometry is so annoying dude proofs are so stupid

like why do i have to show 48 steps when i can give the` answer in 1

[pine sequoia]

hmm

[oak matrix]

are terminal and initial sides only found in standard form angles?

also

can yall answer and break down all the possible answers of this question

if costheta > o and sintheta < 0 then theta lies in what quadrant?

my answer is quadrant 1

[dark sparrow]

costheta > o

do you really mean the letter o there?

and NOT the number zero

[oak matrix]

sorry AWUDHUAWHd

its zero

misclick

[dark sparrow]

right

so cos(θ) is positive while sin(θ) is negative?

that cannot be quadrant 1.

[oak matrix]

oh shit i got it the other way around

[oak matrix]

okay i get it now

quadrant 1 only happens when all funtions are positive

(dont mind, im just gonna discuss so i can remember it)

quadrant 2 is when the adjacent is negative

(all standard angles)

[oak matrix]

wrong

when x is negative

quadrant 3 when tangent and cotangent is negative

4 when cos and secant are positive

@dark sparrow idk if i can ping u but how did he know that that was y without memorizing the unit circle?

or the question is purely based on the unit circle?

[dark sparrow]

the 45-45-90 triangle

or memorized trig values for special angles

er. wait.

are you asking why the letter y was chosen to denote an angle here?

[oak matrix]

https://www.youtube.com/watch?v=Tfc6VdfHMBM&ab_channel=TheMathSorcerer

he said in the video

[dark sparrow]

timestamp?

[oak matrix]

that the range of that equation is the line i pointed

[oak matrix]

:30

[dark sparrow]

"the range of arcsine"

he is talking about the range of the arcsine function, not "the range of the equation"

[oak matrix]

what is arcsine

[dark sparrow]

sin^-1

also denoted arcsin

it's the function that, on the input of a number between -1 and 1, returns the angle between -pi/2 and +pi/2 whose sine equals that number.

[oak matrix]

ooooooooo

[dark sparrow]

the range of sin^-1 is [-pi/2, +pi/2], and this is part of the definition of sin^-1

[oak matrix]

okay now i understand the video better

[oak matrix]

where is -pi/2 here

[dark sparrow]

3pi/2.

3pi/2 and -pi/2 are one full turn apart.

[oak matrix]

OOOOOOO

okay

man im stupid

wait im gonna draw

[oak matrix]

right?

[dark sparrow]

3pi/2 not 3pi/3 but yes sure

[oak matrix]

ah yes miswritten

@dark sparrow question again t-t

nvm

wait

sorry

why is it like this

[dark sparrow]

sin^-1
also denoted arcsin
it's the function that, on the input of a number between -1 and 1, returns the angle between -pi/2 and +pi/2 whose sine equals that number.

also technically that is incorrect

it should be $y = \arcsin(x) \iff (\sin(y) = x \text{ AND } y \in [-\pi/2, +\pi/2] )$

[somber coyote]

Ann

[rustic hazel]

konse states h tu ?

[upper karma]

delhi

[rustic hazel]

is delhi a state?

[upper karma]

union territory

[upper karma]

@main zenith

[upper karma]

@upper karma

[thorny ruin]

How can someone master geometry?

I mean the high-level questions

[main zenith]

D

[rugged dew]

hey people, why does the paraxial approximation map theta, instead of 0, to sin?

I cannot figure out the reasoning that leads to that, which implies I may also be incorrectly understanding why paraxially-approximated cosine = 1.

[knotty bough]

Hello and thanks for the help in advance, very cool to have a place like this for people like me that have 0 idea how geometry works 😄 I have a picture of a shape that I'd like to define and know the measurements of, since the shape grows smaller in size due to the distance of it in the picture it's hard for me to know what the actual shape is (a square, a rectangle, or whatever). Anyone free to receive a PM of a picture of it willing to help out, hit me up! Thank you.

[rugged dew]

I think that if you don't know the distance between the lens (perhaps virtual) and the shape, you can't calculate this.

[knotty bough]

Yea I was afraid of that.

Still willing to try or estimate?

[rugged dew]

do you know the distance along the focal axis between the lens and some object a, and the distance along the focal axis between a + the shape you want to measure? Other than constructing, via geo/trig, a path along which you can figure out axial distance, you could use something super advanced. If you have a very high resolution image and you could train it (an ai) to recognize any similar textures or patterns between the object you want to measure the perceived distance of, for instance, you could train an ai to detect the level of relative downsampling, and thus get a proportionate distance to something you can estimate.

[knotty bough]

I hve no clue what a focal axis is 😄

I don't have any tools sadly, I'm extremely nooby.

[rugged dew]

probably the center of the image, If it's taken with most any camera

[knotty bough]

I don't know the distance sadly. 😦

It's just a picture from the middle of the room basically.

The shape does appear in multiple scenes in the movies I'm using to get the image from but I guess that doesn't help?

[rugged dew]

I dunno then man. I worked with ai on image analysis for a year and it's not practical to get into as anything short of a major project for a university student or a professional.

well, you don't have distances between the lenses and objects, so your frame of reference is unknown

[knotty bough]

Yea exactly, sigh. 😦

[rugged dew]

it's like seeing dots on a grid and no x or y axis. what you gonna do?

say they're a thumb apart? I load that picture on my bigass monitor, and they're two thumbs apart.

[knotty bough]

Is there a way to identify the shape though, instead of only the measurement.

[rugged dew]

yeah definitely

[knotty bough]

Yea that would be a start.

Any idea how?

I sent u a PM better to do it there maybe as to not clutter up the entire section here. 😄

[rugged dew]

I come from bioinformatics so, ImageJ or CellProfiler

is there either 1. a texture or 2. a color difference between the object you want to outline and the background?

p.s. ImageJ is a generic tool. I'd use Fiji, btw, which is just a convenient UI wrapper around ImageJ

the generic thing you're looking for is called partitioning, in imaging, btw.

if the color is distinctly different, you may be able to use thresholding alone to get your shape

[upper karma]

lol

[upper karma]

yeh hai trigonometry bhai. barabaar trigonometry hogaya. kiya aap jee ka prepare kar raha ho?

[upper karma]

nope

why?

[upper karma]

just asking

[limpid onyx]

Is it always possible find the exact value of $\sin(\frac{2\pi}{n})$ where $n \in \mathbb{Z}$?

[somber coyote]

kappa

[brittle lotus]

no

[rapid sigil]

how do I make line segments with specific length like maybe 7 cm or similar in geogebra?

[floral shale]

Taylor series goes crazy

[upper karma]

<@&286206848099549185>

[quiet hinge]

please help me with this question

also correction in the equation of the ellipse
its a^2

[brittle lotus]

so let's first represent secA-tanA also in terms of x

the pythagorean identity sin^2x+cos^2x=1, divide everything by cos^2x gives you tan^2x+1=sec^2x, moving the terms make it sec^2-tan^2x=1. This is equivalent to (secx+tanx)(secx-tanx). We know from the question that the first part is equivalent to x. So it's x(secx-tanx)=1. This is 1/x=(secx-tanx).

Now we have two equations. 1)secA-tanA=x; 2)1/2=secA-tanA

we add up the two together, giving us 2secA=x+1/x

and then divide by 2 gets you the answer

[brittle lotus]

this is (x^2+1)/x

so it's answer D

[brittle lotus]

dyk whats eccentric angles or parametrization of the coordinates?

[quiet hinge]

yes

[wispy lance]

i need help idk how to do this

[brittle lotus]

circumcentre

[brittle lotus]

orthocentre

learn the definitions of all first

theres also incentre, excentre, and centroid

[limpid onyx]

Can you elaborate?

[upper karma]

I think i can answer your question

[floral shale]

nah dw about taylor series

unless you know calculus

[upper karma]

Yh I know about Taylor series

Do you want the answer in closed form? Or by cases

The only form I can think of is an infinite product

@floral shale

[floral shale]

Product moment

[fickle leaf]

Is there any reasonable way to evaluate inverse trig functions without knowing the unit circle? I didn't memorize it at first because of the 30/60/90 and 45/45/90 tricks so now I'm wondering if there's another trick or if I just need to bite the bullet

[snow kite]

just learn the unit circle

[fickle leaf]

ty tho appreciate it at least I know now

[mellow quiver]

you can find (using properties of 45-45-90 and equilateral triangles) that the longest diagonal is length 1/4*(sqrt6+sqrt2)

i was wondering if there was any reason why it happened to be sin(75 degrees) or if that was coinkydink

[snow kite]

with trig the real value is learning demoivre's theorem. infinitely useful

[fickle leaf]

I'm still learning trig tbh. High school drop out returning to college trying to get a CS degree and learning all my HS math

[snow kite]

nice man

what are the prerequisites for the CS course?

[upper karma]

lmao

[fickle leaf]

Everything up to calculus with an actual letter grade, so I need to take remedial courses as I have no transcripts. The course I'm trying to land in is basically a trig course, but I need ot know some trig to get into it, it's lowkey kinda dumb but such is life

[fickle leaf]

?

[snow kite]

yeah. i highly suggest looking at the precalculus course on khan. the trig sections are good

[fickle leaf]

oh yeah, mr khan himself has been a huge help. That and organic chem tutor. The school uses Aleks to let people test out of courses, so I've been using that to study as well. Planning on doing a retake tonight

[snow kite]

what type of questions do you get on those tests?

[fickle leaf]

With aleks it's basically 1 or 2 questions per topic. It's not looking so much for an overall score, it's looking for mastery of subjects. If you answer say a simple factoring question correctly, it'll then try and give you a harder one, if you answer that right it'll jump up a few topics and ask one that if you answer correctly proves you have mastery over several topics in factoring, so it gives you credit for all of them. It basically covers everything in US high school and middle school mathematics, everything you need to get to Calc 1

[snow kite]

oh ok. hard to prepare for that

[limpid onyx]

So when is this not transcendental?

[snow kite]

it's all of highschool math combined?

[fickle leaf]

yeap. And it can ask you any question in the set that you haven't already proved mastery over

[limpid onyx]

Ik basic calc, like Taylor series and stuff

[fickle leaf]

Around 290 topics. I have about 50% of them done in live testing

[snow kite]

then i'd learn as generally as possible

maybe a bit of linear algebra?

do you get a formula booklet

[fickle leaf]

Nah it's all built into the website itself, it walks you through stuff and teaches each thing individually. I've run through almost all of the topics, I have less than 20 left and they're mostly trig and more difficult functions and logarithms stuff

It's been a JOURNEY to say the least, I'm on a very short timeline and had very little time to prepare

[snow kite]

gotcha

[fickle leaf]

hence why I was asking aobut the unit circle, it's more time I don't have to memorize something, but I'm sure there's some easy way to like everyting else

[snow kite]

yeah cramming all of HS even into 2 years can be tough

[fickle leaf]

bro I wish i had 2 y ears

I have 2 weeks

[snow kite]

oh shit

how good are you?

[fickle leaf]

I guess we're gonna find out LMAO

[snow kite]

alr man

do you need to memorize the unit circle?

[fickle leaf]

Basically there's different modules thast you can be put in for Aleks, when I took the test initially in summer I scored so bbad they put me into a different module. About 2 weeks ago I asked why my score wasn't improving much even after months of study and they went "oh oopsies you were in the wrong one" and almost all of my progress was wiped, and I only had 30 topics left. So I've learned about 150 topics in the last 2 weeks, but I feel like it's sticking, after the retake tonightI 'll know for sure.

[fickle leaf]

Hopefully not. I don't need to get all of trig right, only about 40% of it in addition to 70% overall with all the topics combined. After the retake tonightI 'll know for sure what I need to know trig wise for my last retake which I'll probably take on saturday or friday night

I'll have the entire sem to really cement trig, I just need to know enough to skirt by on the test

And this test doesn't go on my record, it's just placement

Technical my knowledge checks they have you take every 20 topics has me placing all the way in calculus, not just trig, so I'm hopeful

[misty forge]

I can't continue from here, could someone please have a look at this and share some things I may be missing please?

[grave pond]

Hmm, a first observation is that the $3\sqrt2$ is a bit of a red herring, since that is the only explicitly given length. The real question is, with a configuration like in the diagram, what is $\dfrac{AE+CE}{BE}$? This ratio turns out to be independent of where on the arc ADC we put E!

[somber coyote]

Troposphere

[grave pond]

I got through with a proof of that by choosing a coordinate where the coordinates of the points are A(1,0), B(0,-1), C(1,0).

Then what we want is the value of $$\frac{\sqrt{(x-1)^2+y^2}+\sqrt{(x+1)^2+y^2}}{\sqrt{x^2+(y+1)^2}}$$ under the assumption that $x^2+y^2=1$ and $y>0$. Took a few false starts, and it's still a bit ugly, but \emph{possible}.

[somber coyote]

Troposphere

[sonic willow]

Is it possible to solve for area
Using some basic geometry like pytagorean theorem

[grave pond]

If you draw an inscribed square (turned 45° with respect to the outer one) it is easier to see that the sum of opposite pairs of quadrilaterals must be the same.

[sonic willow]

I don't know what to do next

Ik that there are 4 45 45 90 triangles, and they have half area of the big square, but I have no Idea what to do with that fact

[grave pond]

The 45-45-90 triangles are all equal, so now you only need to see that the irregular triangles on the inside have the same pairwise area sum.

(Hint: declare the red edges to be their bases).

[sonic willow]

kinda stupid, but I got ?=80 and I think it is correct

but it doesn't look like that on a picture, so I probably got a lot of things wrong

[austere socket]

Is there a way to describe the process of finding the smallest polygon that encapsulates a list of coordinates ?

Cuz I can't seem to get good google searches for a solution

I think it's called minimum bounding geometry but I can't find much online

[nocturne remnant]

By “smallest polygon that encapsulates a list of coordinates” I think you mean https://en.wikipedia.org/wiki/Convex_hull?

[austere socket]

I should have also specified that I want only edges that go to and from points of that list

idk if that changes your answer

[nocturne remnant]

Well, that’s pretty obvious. The convex hull of a finite set of points would “tightly fit” the points if that makes sense

There are some algorithms on finding the convex hull but I haven’t looked much into it

[austere socket]

I also forgot that they have to be straight yeah

so no weirdly fitting shapy

[nocturne remnant]

. yeah that’s what I was trying to say

Anyways

https://youtu.be/B2AJoQSZf4M

Maybe this will be useful?

[austere socket]

thanks

[grave pond]

But 16+80 does not equal 20+28.

[lapis vigil]

can someone help

U mean this?

yea cool so i thikn i got sin and cos down but ned help on the rest

ts what i got so far

[sinful remnant]

What’s a good textbook for trigonometry?

[undone moat]

$x=1/3(((y^2x)/4pi^2)-5)

$x=1/3(((y^2x)/4pi^2)-5)$

[somber coyote]

Professor Wumpus.

[dark sparrow]

[graceful matrix]

savvas envision

[timber cargo]

Tan is sin/cos

csc = 1/sin

sec=1/cos

cot=cos/sin

[grand cargo]

can someone help

[floral shale]

Similar triangles

Find BD

[feral galleon]

Can someone teach me trigonometry

Pls

[timber cargo]

Go to Khan Academy.

[feral galleon]

No

[tribal rover]

Can anyone solve and explain this grade 10 question to me?

[timber cargo]

Sorry pal, but I don't think anyone here will waste their time teaching trigonometry to you when you have access to so many websites and opportunities to learn it yourself.

[tribal rover]

That’s sad

[golden steppe]

can't you do the first question using a pythagorean triplet?

the formula m^2 + 1, m^2 -1, 2m always forms a pythagorean triplet

with m2 +1 being the hypotenuse

I'm just a silly 9th grader but I'm sure you could input into that to find the dimensions

[glossy basalt]

for the first u could do
x^2 + (x+8)^2 = (x+11)^2

solve it

x = 3+r66

[dark sparrow]

don't give out answers (and especially not in this bad format)

[glossy basalt]

whoopsie so sorry

[tribal rover]

UGHHHHH

[worthy nimbus]

nice geometry

[dark sparrow]

wrong server?

[tribal rover]

No❤️

I wanted to share a price of my life event with u guys

Piece*

[dark sparrow]

@tribal rover this sort of stuff belongs in #discussion or #chill

[upper karma]

https://www.examstack.in/2022/12/cengage-mathematics-g-tewani-1.html

go to trigonometry book and download it. it is free and very challenging

[floral ginkgo]

How do I calculate a triangle that’s inside a circle?

The distance between AB

In both circles

[solemn goblet]

Or just

Ok

So

The area of the first should be 3sin 50

And the second should be 2sin angle m

[solemn goblet]

For the sides use the laws of cosines

[pliant roost]

the second circle is superfluous

[floral ginkgo]

Yeah but i need to know Wich circle creates the least distance between A and B

And my answer book says I need to divide 50 trough 360

Then times 2pi

Times 3

[pliant roost]

it's equal?

AB = AB

[solemn goblet]

Those are the circular measure formulas

They are another way to do then

M

[floral ginkgo]

Yeah but Both circles go trough em

And they wanna know Wich circle has the least distance

[solemn goblet]

U can just look up circular measure formulas or I can tell you

[pliant roost]

the one with the lesser angle?

or you can just measure heights

[floral ginkgo]

The one with the lesser circle distance

Like the circle goes trough A and B

And they wanna know how long it takes for it to do so

Like the distance in the circle

[floral ginkgo]

@pliant roost @solemn goblet I solved it!

It was a lil confusing but I made it

Idk how it worked though

[grave pond]

Uhm, if you don't know how it works, the it doesn't sound like you have solved anything in a mathematical sense.

[floral ginkgo]

Well I got the answer

And I sort of understand how it works

But like 2 steps I don’t understand

[floral ginkgo]

I understand the math

But I have a lot of trouble getting to the point where I know what to use

[upper karma]

can someone help me please

SOMEONE HELP PLZ

@everyone ik this ping does nothing but help

Calculate the slope of each line

If the slopes are equal, the two lines are perpendicular

If $m_1=-\frac{1}{m_2}$ the two lines with slope $m_1$ and $m_2$ are perpendicular

[somber coyote]

HPH_Bulko

[upper karma]

thank you so much

i rly appreciate

[potent forge]

hi how can i tell if a radian is between 0 and 2 pi

[junior grove]

What do you want to say?

like 45 degree

[potent forge]

umm

am i allowed to ask for helpp in here

[junior grove]

sure, probably 🙂

[potent forge]

so my question is

-7pi/3

Find an angle between
0
and
2
π
that is coterminal with the given angle.

[junior grove]

pi is 180 degree

[potent forge]

2pi is 360 yes

[junior grove]

hold on

[potent forge]

and I would add 2pi over 1

times the denominator

of 3

the first answer is -pi/3

but i noticed that is a negative so its not between 0 and 2 pi

so I did it again

[junior grove]

-7/3 +2pi+2pi =5pi/3

300 degree

[potent forge]

yes I got that

okay so basically I could tell if its between 0 and 2 pi of I convert the radians to degree

gotchu

makes sense

okay the next one is this

[junior grove]

Trust me question is wrong

[potent forge]

77pi/2 how can I do this fast

yah I tried that and it was wrong so I found out it wasnt between that

[junior grove]

77/2 is = 38 right?

[potent forge]

38pi/2

and that seems to be out of range

so then 38/2

19pi/2

still out of range

[sonic willow]

I have a question
Does pythagorean theorem work in non-euclidian geometry too?

[timber cargo]

I believe it's Riemann Curvature Tensor that describes the bending of space

The pythagorean theorem fails at that

[bright elm]

An observer (O) is located 500 feet from a school (S). The observer notices a bird (B) flying at a 39° angle of elevation from his line of sight. How high is the bird flying over the school? Please show all work.

[timber cargo]

tan(39 degrees)=x/500

But that question is impossible to answer without knowing the height of the school.

Only when you know the height of the school will you know how high the bird is flying over the school

[upper karma]

and the height of the line of sight of the person

[timber cargo]

And that too

[worthy nimbus]

lol

[tribal rover]

GUYS HELP

[frank moth]

Just use the distance formula for all four segments and verify they work. Then find the slope of the two diagonals and check if they are negative reciprocals.

[dark sparrow]

@tribal rover do you still need help with this?

[prime musk]

Guys help me with this sum…

If two internal angle bisector of a triangle to the opposite side are equal in length, then prove that the triangle is isosceles.

[snow kite]

do you use vectors for that?

[prime musk]

Umm I don’t know how to solve…. Would be very grateful if anyone can show a detailed proof

[snow kite]

I don't know how.

[civic folio]

does anyone know what formula needed for ray casting?

[tribal rover]

YES

[dark sparrow]

okay, have you made any progress thus far?

[upper karma]

so, the lowest amount of legs a polygon can have is 3, right? because you need 3 legs to close a triangle. now in 2 dimensions you need 6, because a pyramid is shaped like a polygon and has 6 legs. now does that mean that the lowest 4 dimensional polygon-like object has at least 12 legs?

[upper karma]

Find Missing Measures Using the Pythagorean Theorem but how

[vagrant crystal]

@thorn dragon check dm

[prime musk]

Umm anyone who can help me…..

[neat steeple]

Is this right?

[wanton fog]

HELP

please

[upper karma]

can someone help with question 1

[royal cedar]

Compute the angles CDF and FDE using the isosceles triangles

[upper karma]

and also refer to properties of kites

[smoky jetty]

diagonals of a kite are perpendicular (i.e they intersect at 90 degrees), therefore you can refer to Pythagorean theorem with that

another thing is that the longer diagonal bisects the triangle DEF which is an isosceles triangle, as part of the kite divided by the segment DF

[dark reef]

kites have a pair of opposite congruent angles

also use the fact that all the interior angles add up to 360

[keen lintel]

Its trivial by angle bisector theorem, unless you require the proof of angle bisector theorem

[keen lintel]

Consider two right angle triangles, ABC and ABD, Let angle BAD=y
,notice cos(y)=3/x=x/11 , Thus x^2=33 and just take the square root

[flat moth]

How did we find the second equation

From i+1 to the ones with square root of 2 and all

I do understand that they're equal to one another but idk how they did it

[dark sparrow]

|1+i| = sqrt(2)

their idea was to separate the modulus and argument but without writing out exp(i*arg)

so that it's just modulus * unit

[flat moth]

Whats unit

Idk the english terms that well

As in the whole thing

[dark sparrow]

i'm using it a bit informally

[flat moth]

1+i is the unit right?

[dark sparrow]

when i said "unit" i meant "complex number whose magnitude is 1"

[flat moth]

Okay I get it

Thanks

[old nacelle]

can someone help please. For the 2nd pic both 86 and 92 are wrong.

[near harness]

Yea they are wrong. Angle T and angle V should add up to 180° but only add up to 178

[old nacelle]

yeah that’s what i thought so is 88 the right answer?

[near harness]

Well the question asks measure of U

So U + V = 180
U = 180 - V

Which is 92 if we take V = 88

Otherwise 94 if we take V = 86

[old nacelle]

oh ok thanks

[smoky jetty]

could the value be uncertain inthe problem?

coz the only thing that I think would have a fixed value would be Angle S

[upbeat latch]

the simplest approach that I saw in the question is that, I did this sum by using Pythagoras in the three TRIANGLES-which leads to x= sqrt(33)

[upbeat latch]

Like, I first took Triangle BDA, then Triangle BDC, and then, Triangle ABC, and after applying Pythagoras to all of them, you would come up with some Commons that you could use in one of the Equations where the Equation becomes a Pure in "x" and a constant Numeric which leads to the final Answer of sqrt(33) which approximates 5.74

[shrewd idol]

BD^2 = AD*AC if i remember correctly

[pure cobalt]

someone speaks Spanish? I need help, urgently

[wicked narwhal]

hi, I would say (36+x2)/2=7x-6 but I don't know explain

It is a property of median

[novel axle]

Anyone know how to solve this?

[snow kite]

the horizontal line is the "image"

so part A would be 1 unit

[wanton fog]

thank u sm

[junior heath]

First question is a simultaneous equations problem in disguise. The two trapeziums created by the median are similar, so a*(x^2)=7x-6 and a(7x-6)=36 for some scaling factor a. You can then solve for x

[cerulean fiber]

Evan and Ryan each chipped in $1000 to buy an old boat to fix it up. Then Evan spent $825 on materials and Ryan spent $1650 on parts. They worked an equal number of hours and sold it for $6800. How should they fairly divide the $6800? Round to the nearest cent.

[narrow condor]

can someone help me with this?

[dark sparrow]

,rccw

[dark sparrow]

@narrow condor do you still need help with this?

[narrow condor]

yea please

[dark sparrow]

have you made a diagram?

[narrow condor]

i got up to here

[dark sparrow]

actually, come to think of it, the problem is somewhat ambiguous as stated if we view it in 3 dimensions...

,rccw

[narrow condor]

but i’m confused about the ambiguous case

[dark sparrow]

ok, i take it that B and C are the fishing boats while A is the school of fish

[narrow condor]

yes

[dark sparrow]

what ambiguous case? are you talking about the ambiguity i alluded to just now, or what?

[narrow condor]

yea that

[dark sparrow]

when you say "the ambiguous case", do you mean the possibility that can arise when solving a triangle that you can't tell if a particular angle is acute or obtuse despite knowing its sine?

because that's not at all what i was talking about.

[narrow condor]

yea i was talking about how it doesn’t state that the school is right in between them, but it could possibly be located on another side. here i’ll draw it out

[dark sparrow]

right

i viewed this problem from a 3-dimensional standpoint and reasoned thusly:
the set of all points at an angle of depression of 35° from the first fisherman is a cone with its apex at the fisherman's boat, extending downward. similarly for the other fisherman.

so the set of all possible locations for the school of fish is in fact the intersection between these two cones

wait hold on somethings wrong here still

ok fuck it im geogebraing this shit

ok, so the fish apparently have an entire ellipse of points in the water that they could be in lmao

[narrow condor]

great😃

[dark sparrow]

i placed the first fisherman at (0,0,0) and the second at (18,0,0)

and then plotted the cones of angle-of-depression 35° and 51° for them

yeah we in fact don't know that the angle between the first fisherman's line-of-sight to the fish and his line-of-sight to the second fisherman is 35°.

[narrow condor]

this question is ridiculous

[dark sparrow]

it sure is!

oh also i think i misspoke - idt the curve of intersection is an ellipse at all

[smoky jetty]

update on this, which I found unsolvable at first. Apparently, I don't know if it was my mistake for the labeling, or this just depends on how one labels a quadrilateral based on the 4 vertices which makes it solvable. Basically, if you label it as LIYT, but the letters are shifted to the right by 1 position, then you'd get the value. What do you guys think of this?

i waited for almost a month to know whats up with these problems coz i thought theres a property i didnt know

[smoky jetty]

ok now with this problem, the shape was said by our teacher, and upon knowing, the answer was easily found. But what I found quite confusing is how the shape was assumed to be kite, because of what our teacher said, "the given values are the diagonals and area, to find x, thus is a hint (translated). What i found ambigious with this is I think there's no logic behind it. yes, the x can be found if it was a kite, but quadrilaterals has area and diagonals as well, so it can be other quads as well, though may not be solved. But still, do u guys agree with my teacher's reasoning here? I just wanna know if it would make sense to rely on such "hint" if it were based on the previous lesson

coz idk if I did'nt do enough analyzing the paragraph per problem, or there's something wrong with it.

[near harness]

angle T and V have to be supplementary but they are not in the question since they add up to 178 deg not 180.

[smoky jetty]

i think it didn't state that the given is an isosceles trapezoid, as opposite angles would be =180 if it is an isosceles trapezoid

[royal dawn]

am i right with D?

[coarse lance]

am i meant to google this or is it easy to enough to create it with the information given?

(the second problem)

$2d^2$ i made this

[somber coyote]

Price

[coarse lance]

cool

that was right

[wicked narwhal]

No, it is C the correct answer

[royal dawn]

thank you so much for this. i jus figured it out like 10 mins ago 😭

i have like 5 more questions if your free can u help?