geometry-and-trigonometry

[prisma hornet]

oh so if its negative

then the cast rule kinda flips

[obsidian harness]

Yes

[prisma hornet]

ohhhhhhhhh

[obsidian harness]

That's why I don't like CAST btw

[prisma hornet]

i get it now

our teacher didnt teach us unit cirlce yet

so if i ever see a negative

i just flip the quadrants in my mind right

[obsidian harness]

Well CAST is for people who don't know what the unit circle is ngl; if you know the unit circle you don't need CAST

This is literally it

[prisma hornet]

so instead of CAST being all possible postiive quadrants it becomes negatives

[prisma hornet]

i kinda dont get the x and y coardinate thing

[obsidian harness]

And also there's tan theta = sin theta / cos theta = y/x which is the slope of the line

[prisma hornet]

woah

i dont think weve learned that either

i was working

on this review

and 2) got me confused

[obsidian harness]

https://www.youtube.com/watch?v=57VrEiEPD1I
hope this helps

Yeah honestly you should be learning the unit circle the very first time you go past SOHCAHTOA

[prisma hornet]

ty ill check this out after my test tmrw

cuz we have trig part 2

[obsidian harness]

okok

[prisma hornet]

right aftyer it

and thats probably where it will come in handy

[obsidian harness]

ok

[prisma hornet]

trig part 2 is like all the identities i think

[obsidian harness]

Ah right yeah so sin^2 theta + cos^2 theta = 1

[prisma hornet]

btw

[obsidian harness]

Dividing both sides by cos^2 theta you get tan^2 theta + 1 = sec^2 theta

[prisma hornet]

is this the same thing

as what i was asking earlier

its just these ones i gotta have memorized

[obsidian harness]

Yes

Exact same type of question except you can't use a calculator

[prisma hornet]

ok ok

so like d) is tan 60

ok yeah i should be chilling

for that

i think im prolly ready for the test after a bit more review then

ty

oh wait

actually why is it negative

[obsidian harness]

mhm basically tan x is increasing and it can't be theta = 90 deg

So it has to be the next lowest one or 60 yes

There's ways to kind of guess at things
And if you don't know, just draw out the 30-60-90 or the 45-45-90 triangles

[prisma hornet]

yep thats what i got drawn out

[obsidian harness]

All of the angles you need to know are based on just these 2 triangles

[obsidian harness]

perfect

[prisma hornet]

this is liek this cuz of the

flipped CAST rule right?

[obsidian harness]

yeah

[prisma hornet]

ok

omg it finally makes sense

and there can only be 2 spots it can ever be im guessing

no suprise attacks from my teacher?

[obsidian harness]

yeah so sin, cos, and tan always occupy 2 quadrants each

no more no less, well except if you have special cases like sin theta = 1

Then it's debatable whether it spans 2 quadrants or if it's just a single point

[prisma hornet]

i see

i tried it on my calculator cuz i was curious

it just gave me the angle 90

lol

ok but tysm

ima keep reviewing

[obsidian harness]

No worries man

[hoary totem]

tbh idk what this convo is about but i find it sometimes useful to know the 37 - 53 of the 3 - 4 - 5 triangle

[dark sparrow]

you should be aware that those angles are approximate though

[hoary totem]

yes

[dark sparrow]

case in point:

,calc sin(37*pi/180)

[somber coyote]

Result:

0.60181502315205

[hoary totem]

or:

[dark sparrow]

not exactly 0.6 even if it is close

[hoary totem]

,calc arctan(0.75)*180/pi

[somber coyote]

The following error occured while calculating:
Error: Undefined function arctan

[dark sparrow]

atan

[hoary totem]

what

,calc atan(0.75)*180/pi

[somber coyote]

Result:

36.869897645844

[dark sparrow]

for calc it's called atan

[hoary totem]

ic

[dark sparrow]

or if you are a cool and sexy person like me:

,calc atan(3/4)

[somber coyote]

Result:

0.64350110879328

[hoary totem]

hah radians

[dark sparrow]

the smallest angle in a 3-4-5 is about 0.64 rad

[obsidian harness]

atan has a weird definition to make everything have the right value for the quadrant

[dark sparrow]

the other angle, naturally, is about 0.93 rad

[hoary totem]

i prefer atan2

for like

all uses

[obsidian harness]

I mean atan2

[dark sparrow]

you're thinking of atan2

[hoary totem]

ever

[obsidian harness]

ah yes

[hoary totem]

the clearly superior function

[obsidian hornet]

Also, one can perceive the tangent as the limiting position of the secant when two points converge and PBPD=PAPC turns into PB*PD=PA^2

[hoary totem]

well yes but the thing was about proving the secant secant theorem

the secant tangent theorem follows from the secant secant theorem by your reasoning tho

[obsidian hornet]

Yes, it's all just the power of the point P outside the circle and equal to the modulus d^2-R^2, where d is the distance from the center of the circle to the point, of course the sign of the expression changes when the point lies inside the circle(-), and on the circle, respectively, the degree is 0, since d^2-R^2=0

[hoary totem]

uh

yeah

[twin spindle]

Question, how do we find the values when it's jus given one value?

I used the trig identites, like sin=1/csc but it said that it's the wrong number apparently

[dark sparrow]

it said that it's the wrong number
who/what said that?

and show all your work, so we can tell where exactly you messed up

[twin spindle]

Oh wait nvm

It's right

My dumbass looked at the wrong page

I did screw up though by forgetting the negative, since its in Quad 2

Mb mb 🙏🙏

[shut gulch]

hello

problem with geometric mean theorem problems

where do i start with this problem

[dusk quail]

Did I get this right? (I only need to ifnd theta between 0 and pi)

[pliant current]

tf is even the question

[obsidian hornet]

What do you need to find out?

[north kindle]

I have no idea where to start with this problem

[grave pond]

Hmm, intuitively, the whole diagram would be stable under rotating 120° around the center of ABC.

[north kindle]

Yeah but that’s not really a proof

Maybe we could like make some congruent triangles with DH, FH and DF as corrosponding sides

[grave pond]

Then take it in several steps.
Triangles BAH = ACF = CBD by SAS.
Then triangles DBH = HAF = FCD by SAS too.

[north kindle]

How do you know those congruences are true

I don’t understand

[grave pond]

Do you know the SAS theorem?

[north kindle]

Yes

But you are just asserting it without showing that they satisfy it

[grave pond]

So BA = AC and AH = CF and angle BAH = angle ACF because each of them is 150°.

Thus triangles BAH and ACF are congruent.

[north kindle]

How do you know BAH and ACF are 150

Oh wait I see

Fuck

[grave pond]

Each of them is 90° + 60° which makes 150°.

[north kindle]

I should’ve seen that

But what about the 2nd set, how do you know BH = AF

[grave pond]

Because BAH and ACF are congruent.

[north kindle]

I don’t understand

[grave pond]

That's what we just concluded!

BH and AF are corresponding sides in the two triangles that I just argued are congruent.

[north kindle]

Makes sense

Fuck I’m dumb

[hoary totem]

||ΔDBH and ΔHAF congruent, so ∠AHF = ∠HDB||
||by parallel lines, all the green angles are the same||
||the pink angles are 120 given by 360 - 90 - 90 - 60 so theyre the same||
||sides AH = CF||
||so by AAS we have ΔPAH ≡ ΔCFQ||
||we can see from ΔCFQ that yellow + green = 180 - pink = 60||
||but ∠DHF = yellow + green = 60||
||do this three times and u have all angles of the big triangle are 60 deg||

my very roundabout solution

probably much faster ways to do this

[tepid burrow]

does anyone know how i would use this formula provided in the website with a center latitude and longitude?
http://www.quadibloc.com/maps/maz0206.htm

[upper karma]

i need help with a task

square ABCD is divided into two parts - triangle AED with an area of ​​16 cm and trapezoid ABCE (as in the picture), point E is the midpoint of side CD

i need it more explained how do i do it then get it solved

[tepid burrow]

what are you trying to solve? the angles or area?

[upper karma]

ohhhh i forgot to mention

i need to have the trapezoidal circumference

ec + cb + ab + ae

[tepid burrow]

the perimeter?

[upper karma]

yh

i started with doing the traingle 16 = 1/2 a x b

but got stuck

[tepid burrow]

try to find the area of the whole square, and then you can get the perimeter of one side by using the area formula

[upper karma]

the square is 8x8 then

[tepid burrow]

yes!

[upper karma]

so 8+8+4 + 2√20 = 20 + 2√20

i got 2√20 by doing pythagoras

8 squared + 4 squared = C SQUARED

[tepid burrow]

yes nice bro

[upper karma]

NICE THX

[tepid burrow]

np

[upper karma]

🙏

[solar minnow]

Is this equation true based on the triangles area

Or does this solve for the triangles area

Correctly

[grave pond]

The expression does give the area of the triangle.

[solar minnow]

Thank you

[earnest trout]

When area of segment = area of teiangle and i need to show that theta= 2 sin theta

First i dont understand why 1/2r^2 is removed from the area of triangle, is it because i have to solve the equation in terms of theta only? So i just ignore it?

Secondly, i dont understand how the equation is solved with does sin (pi-theta) equals to sin theta? Where did the pi go? Is this an identity?

[amber hollow]

how are they getting -pi/4 from 7pi/4 and why are they using 7pi/4 and not 5pi/4

[hard moon]

Because -pi/4 and 7pi/4 are exactly one revolution apart

so the point they correspond to on the unit circle is exactly the same (which means the sine and cosine values of -pi/4 and 7pi/4 are the same)

[hard moon]

yes, sin(pi - theta) = sin(theta) is an identity

in addition, when you set the area of the segment equal to the area of the triangle, notice what all the terms have in common; how can you get rid of that thing that they all have in common?

[amber hollow]

i don't see -pi/4 on the unit circle, how do i get to it

[silent plank]

add/subtract integer multiples of 2pi to get coterminal angles

[hard moon]

So positive angles you go counterclockwise starting from the positive x axis, and for negative angles, you go clockwise starting from the positive x axis

[earnest trout]

Ahh ok but how is thelta determined in b?

[obsidian hornet]

Vector rotation of DAH.

[hasty karma]

Where can i see proof for trigonometric substitution?
like one that is t = tan(x/2)

[dark sparrow]

proof of what exactly

@hasty karma

[hasty karma]

t = tan(x/2), need to express cosx sinx tanx from t

need proof or how to get it

(sometimes it is same)

[dark sparrow]

you can derive these expressions yourself with some applications of the double angle identities for sin, cos and tan

[hasty karma]

obvious, but how exactly?

[dark sparrow]

cos(x) = cos(2*x/2) = cos^2(x/2) - sin^2(x/2) = cos^2(x/2) * (1 - tan^2(x/2))

play around with other stuff like this

[hasty karma]

How i know im doing it in a right order?

[dark sparrow]

you don't

[hasty karma]

So, each step i try all variants, is it enough to limit trigonometric ones involved with trig power, half angle and some basic stuff like sin2(x) + cos2(x) = 1?

It would be a good thing to throw some useless, right?

[dark sparrow]

sin^2(x) + cos^2(x) was equal to just 1 this morning, not 2

inflation has gone out of control!

but also i think Pythagoras and variants + double angle identities is really all you need

[hasty karma]

yep, i just switched to phone because of terrible internet

so typo

[dark sparrow]

i know

was trying to make a joke

[hasty karma]

yeah

[loud merlin]

can someone explain to me how to do analytical geometry??

[obsidian harness]

Your best bet is to find some YouTube videos on this

[loud merlin]

ohh ok thx

[obsidian harness]

no worries

Like try specific topics cause they are more useful then trying to learn from like a 5 hour video

And practice using Khan Academy or something in between

[empty yew]

Anyone help

(12th question)how to eliminate theta here?

[dark sparrow]

treat this as a linear system in sec(θ) and tan(θ), solve for those in terms of a, b, c, and d, and then plug all that into sec^2(θ) - tan^2(θ) = 1

seems to be the only method that doesn't involve great pain and suffering

[grave pond]

Hmm, my first instinct was to multiply through by cos(theta) so I could work with sin and cos instead of sec and tan. But I think that actually ends up being equivalent after the dust settles. I end up with a great big homogeneous quartic relation between a,b,c,d with 9 terms and no particular rhyme or reason ...

[shadow depot]

That’s smart af

[elder anchor]

is there a way to find the intersection points of f(x) = x and g(x) = 4sin( (π/12) (x(x-1)) ) without using a graphing calculator?

[dark sparrow]

probably not

why do you need those

[elder anchor]

I'm doing an exercise about finding the area between 2 intersection points (the exercise explicitly says to use a graphing calculator to find the boundaries) and I was just wondering if it was possible to do it algebraically

[grave pond]

Your graphing calculator will give you two intersection points that have nice enough coordinates to show manually that they work.

(And then three more whose coordinates are horrible).

[elder anchor]

yeah, I was trying to find the horrible ones without graphing.. but it seems like it can't be done

anyawy thanks for your replies Ann & Troposphere

[grave pond]

Equations with the unknown both inside and outside a trig functions generally don't have nice solutions other than by accident.

[deft sail]

Can someone help with this

[wild ravine]

k

[deft sail]

So

How do I split it up

[twilit ridge]

https://youtu.be/VHeWndnHuQs?si=JOZz6E-CcCUFTdrk

[spice flare]

Can anyone tell me how do i solve for x?

[trail tendon]

algebra (most likely)

[spice flare]

No

e1//e2

[trail tendon]

._.

[trail tendon]

what?

[spice flare]

Parallel

[trail tendon]

oh

do you have any information about the angles

[spice flare]

No....

It just wantse to solve x

[trail tendon]

do you have any information of the length of E to delta

it doesn't seem solveable to me rn

it seems that x could be anything depending on the angle

[spice flare]

Oh ok

[trail tendon]

i mean you see what i mean?

[spice flare]

Yeah but we have a word in my country to describe these angles

[trail tendon]

like theoretically you flatten the line and none of your information would change

[spice flare]

Idk if u have the same

Its like an X

The lines

[trail tendon]

ohh wait am i stpuid

hmm

[spice flare]

Other guys said to me that 32/24=x+5/x

Which might be right but i cant solve it neither with that information

[trail tendon]

oh

these are similar triangles

i'm dumb

[trail tendon]

ok, multiply x by both sides

[spice flare]

Similar yeah

[trail tendon]

yeah

all the sides are proportionate i think

[spice flare]

So i multiply x

[trail tendon]

yeh

[spice flare]

With 24 and 32

[trail tendon]

uh

actually try multiplying by 24x

[trail tendon]

i mean yes

[spice flare]

And then

[trail tendon]

but you'd also want to multiply by 24

to get rid of all the denominators

and then get all the x on one side

and all the constant terms on the other side

[spice flare]

Can u dm me

[trail tendon]

sure

[spice flare]

Thanks

[empty yew]

Nvm

@dark sparrow Is this correct?

If correct what to do next?

@trail tendon do you know about this?

[trail tendon]

which problem r u doing

[empty yew]

This

[trail tendon]

9,10,11,12...?

[empty yew]

12

Sorry

[trail tendon]

a

[empty yew]

Is my approach correct?

[trail tendon]

im tryna figure out gimme a sec lol

i... don't think that gets you anywhere....

[empty yew]

What to do now?

[trail tendon]

good question 😂

i've never done something like that before

lemme see

im thinking of multiplying the first equation by b on both sides, and the second equation by a on both sides, then subtracting them to get rid of the secant term, then divide and take the inverse tangent

not arctan

inverse tan

[wheat canyon]

for some reason (maybe bc my brain is out of fuel) it take me hours to realize i could just add "2"to the equation and find the intersection between the plane and the paraboloid

not asking any questions, i just find it funny the amount of time this took me :/

[trail tendon]

nah

i was solving for theta instead of cancelling it out 😂

oh but then i guess

could sub it back in? :l

irdk lol

does it want you to write the equations but without using theta?

[empty yew]

,, \tan\theta=\frac{ac-b^{2}}{bc+ad}

[somber coyote]

KingDanger

[empty yew]

Yes

I want to make the equation exists without the theta

One min i'll send you an example

[trail tendon]

ok

then uh

[trail tendon]

take inverse tangent, then sub theta back into the og equations

boom

lol

it prolly wants you to simplify that too

[trail tendon]

but thats not hard

oo

maybe they wanted you to square it or smthn but oops 😂

[trail tendon]

do you know what i mean tho

[empty yew]

Nope 😁

[empty yew]

I tried that but idk what to do there!! Lol

[trail tendon]

like ctan(theta) = ctan(inversetan( (ac-b^2)/(ad-bc))
tan(inversetan(x)) = x
so, ctan(inversetan( (ac-b^2)/(ad-bc) ) = c*(ac-b^2)/(ad-bc)

and since sec(theta) is 1/tan(theta) you can do the same thing, but it will be flipped

[empty yew]

What's a ctan and inverse tan?

[trail tendon]

oh you don't know what inverse tan is?

[trail tendon]

ctan is c * tan

c the variable

dang uhh

oh i mean

you don't even need to do that

they have tan(theta) in the equation

[empty yew]

,, \tan^{-1}\theta

[somber coyote]

KingDanger

[trail tendon]

and sec(theta) = 1/tan(theta)

[empty yew]

Is it the inverse?

[trail tendon]

yeah

[trail tendon]

thats it

but now tha ti think about it

you don't need to do that

a * 1/tan(theta) - c * tan(theta) = b

you can sub in tan(theta) = (ac-b^2)/(ad-bc)

right?

and same with other equation

[empty yew]

Sec(x)=1/cos(x)?

[trail tendon]

yes

oh

LOL

i was thinking of derivatives 🤣

ok anyway uhh

yeah i guess you just gotta plug in theta = inverse tan(allat)

you can use a triangle to find secant i guess

[faint comet]

Hello guys, how do I find the y-inter of a trig graph ?

[empty yew]

🤔

[trail tendon]

set it equal to 0?

[trail tendon]

thats mistake XD

[faint comet]

Set what equal to 0 to trig ?

[trail tendon]

but you can plug in theta = inverse tan(blah blah blah)

[trail tendon]

can you give an example?

[faint comet]

Sure

[faint comet]

4sin(X+3pi/2)

[trail tendon]

well

you want to set it equal to 0 to find y intercepts right?

so

4sin(x+3pi/2) = 0

you can divide by 4

sin(x+3pi/2) = 0

for what values of x + 3pi/2 is this true?

[faint comet]

• of that ?

Negative

[trail tendon]

uh

sin(u) = 0, what are the possible values of u?

[faint comet]

0

[trail tendon]

but also pi, right?

[faint comet]

Yeah

[trail tendon]

and 2pi

and actually all k*pi for all integer values of k

that just means like 0pi, 1pi, 2pi, 3pi, 4pi, ect.

yeah

so

x+3pi/2 = allat

so you can solve for the different values of x

[faint comet]

But it says the y inter = -4

[trail tendon]

by subtracting each value by 3pi/2

[faint comet]

Yeah

[trail tendon]

hold up

💀

LOL

[faint comet]

Yeah

[trail tendon]

i was finding x intercepts 😂

for y intercepts you just plug in x = 0 right

lol

[faint comet]

Ok

[trail tendon]

mb 😂

[faint comet]

All good

[dark sparrow]

this doesn't follow my instructions at all

[faint comet]

But how does y=4sin(3pi/2) =-4

[trail tendon]

well do you know how to calculate sin(3pi/2) ?

[empty yew]

Ohhh!

[faint comet]

Special triangles ?

[trail tendon]

is there a better way than solving for theta 😂

[trail tendon]

actually i was thinking of the unit circle

[dark sparrow]

i didn't say to solve for theta

[faint comet]

OHHHHHHH

[dark sparrow]

i said to solve for sec(θ) and tan(θ) treating them as two variables in a linear system of equations

[empty yew]

What to do now?

[faint comet]

So for sin 3pi/2 = 0 so we’re left with 4 and then take 4 to other side and -4?

[trail tendon]

oh yea thats what we did tho

[empty yew]

Ohhh

[faint comet]

I mean 3pi/2 = -1

[dark sparrow]

you would get sec(θ) = ...
and tan(θ) = ...

[trail tendon]

yeah

[faint comet]

Thansk bir

Have a good day

[empty yew]

👍

[trail tendon]

oh right, coulda just solved for sec(theta) 💀

[fast oriole]

i dont have a protactor + ruler

[faint pasture]

def not something we can help you with, i mean you could make a makeshift ruler out of anything really but protractor would be tough

[fast oriole]

i mean my teacher said if i dont have the equipment i wont get detention and its due tmr

[faint pasture]

eh eh the tchr wont measure it anyways lmaoo just make it good enough and get your lines straight using anything with a straight edge

[dark sparrow]

then... just tell them you don't have the equipment?

your teacher told you that you evade detention because of that

[fast oriole]

is there any way i could solve it

or do i just leave it

[dark sparrow]

leave it

i mean ok like

if you know trigonometry, there is a way to get the exact value of x

but i don't think this is accessible to you atm

[fast oriole]

oh

aight

[clear anvil]

you can use drafting tools

[fast oriole]

whats tha

[clear anvil]

something like autocad?

i think geogebra/desmos does it as well

it gives you the dimension automatically because the triangle is inherently fully defined with the features you've fed it

[fast oriole]

wheres the website

or how do i access it

[clear anvil]

well like i said you can try desmos/geogebra

desmos geometry

google it

[remote fox]

Can someone help with this?

[dark sparrow]

!status

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[dark sparrow]

draw the center of the circle

maybe call it O, since that's a common letter to call the center of a circle

also draw the lines OA and OB.

[remote fox]

Now what?

Is C 120°

And the other 2 are 30°?

🤞

[dark sparrow]

no to both.

show me what you have so far.

[remote fox]

Nothing

[wooden silo]

hi

[remote fox]

I just drew the lines in my head

[dark sparrow]

well don't draw them in your head

get yourself a piece of paper, recreate your diagram on it, and do as i said.

[remote fox]

Alr

Done

Now what?

[dark sparrow]

show me

[dark sparrow]

ok

look at triangle OAB and also your data

[remote fox]

Is AOB equilateral?

[dark sparrow]

is it?

you should be able to answer that yourself.

if yes, why? if not, why not?

[remote fox]

Well, all sides are length 3/4, right?

[dark sparrow]

indeed they are

[remote fox]

So, does a triangle with same side lengths have same angles?

[dark sparrow]

you tell me. does it?

[remote fox]

I'm asking you, I think they do but I want clarity

[dark sparrow]

yes, of course.

[remote fox]

Ok, now I'm stuck

[dark sparrow]

what is angle AOB?

[remote fox]

180°

[dark sparrow]

are you sure?

[remote fox]

Yeah

[dark sparrow]

does AOB look like a half-turn angle to you?

[remote fox]

Wait, what is AOB, i thought you meant the whole triangle

[dark sparrow]

no

if i wanted to ask you about the sum of all angles in triangle AOB, then i would have said all those words

[dark sparrow]

angle AOB means the angle with vertex at O and sides OA and OB.

[remote fox]

So, 60°

[dark sparrow]

yes

thus the minor arc AB is also 60°.

so how many degrees is the major arc AB?

[remote fox]

I don't know

[dark sparrow]

how many degrees make a full circle?

[remote fox]

360°

[dark sparrow]

right

so when you take away the minor arc of 60°, how much is left?

[remote fox]

Ohhhhhh, 300°

[dark sparrow]

right

[remote fox]

How do we know the minor arc is 60° though?

[dark sparrow]

it's cut out by a central angle of the same size.

[grave pond]

Just to be on the safe side @remote fox, do you know the central angle theorem?

[remote fox]

Maybe, but not the term you're using, what is it?

[grave pond]

The theorem that says (in this case) that angle ACB is two times angle AOB.

[dark sparrow]

it isn't tho

they don't cut out the same arcs but complementary ones

[remote fox]

I know the theorem

[dark sparrow]

also you got that mixed up

@remote fox what i was gonna say now is: angle ACB is an inscribed angle which cuts out this 300° arc. what is the size of angle ACB?

[remote fox]

It's the other way round, isn't it?

[grave pond]

Ah yes, half of. Sorry.

[dark sparrow]

tropo got two things mixed up

[grave pond]

I'm also assuming that we're measuring AOB the long way around.

[remote fox]

Would it be 150° because of the central angle theorem?

[dark sparrow]

yes

[dark sparrow]

that merited clarification imo

[grave pond]

Yeah, sorry.

[remote fox]

So we're comparing those?

[dark sparrow]

yes

[flint ingot]

for question 1b does it matter where the centre of enlargement is when i enlarge it?

[dark sparrow]

it was left unspecified

so it looks like it does not matter so long as the enlarged shape is still within the grid

[flint ingot]

ok thanks

[obsidian hornet]

I came up with a problem, but I did not come up with a simple solution, maybe someone will find: an acute-angled triangle ABC is given, points B and C are selected on the sides of AP and AQ such that ABC is an isosceles triangle, a straight line AE is drawn intersecting BC and PQ at points E and R, respectively, it turned out that the quadrilaterals PBER and RECQ-inscribed, prove that the centers of the circles inscribed in the triangles ABE and AEC lie on the same circle with the centers of the inscribed circles of the quadrangles.

[upper karma]

Status 1

I didn't understand the solution either

[pliant current]

is it 4unit distance from point p?

[upper karma]

That is what the question says

But the next few words doesn't make any sense

[nocturne remnant]

this can be done with angle chasing only

[obsidian hornet]

Show me please, p.s. I solved it with angle chasing, didn’t notice.

[dark sparrow]

@upper karma here is what this question says in slightly saner wording:

• on a certain circle, call the points closest to and furthest from P (2,1) by the names A and B respectively.
• PA = 4.
• B is the point (6,5).

does that make more sense?

[upper karma]

A lot

Got it

Thanks

[ruby wigeon]

did anyknow know about solid geometry ?

I'm so fucked rn

[upper karma]

Say for a circle with centre as origin(for simplified version) and radius r
Let there be a point (x1,y1) and let S11 be x1^2+y1^2-r^2
Now my doubt is what exactly does s11 represent

I know that we can find the position of a point wrt circle by finding the sign of s11

[dark sparrow]

s_11 is kind of a strange name for that quantity

and i dont think it means anything geometrically

$s_{11} > 0$ (resp. $<0$) if and only if $x^2+y^2 > r^2$ (resp. $<r^2$) if and only if your point lies further (resp. closer) than $r$ units from the origin

[somber coyote]

|Ann⟩

[dark sparrow]

idt it's any deeper than that

you might as well just compare $x_1^2 + y_1^2$ with $r^2$

[somber coyote]

|Ann⟩

[upper karma]

Okay,thank you Ann

[upper karma]

I am aware of these

[ruby wigeon]

I've got a question from class

now you’ve got a union of a quadrangular cone with a rectangular base and one lateral prong perpendicular to the base and a triangular cone with four faces at right angles as the figure shows,$PA \perp ABCE$,$ABCD$ is a square,$AD=2,ED=1$,if the external ball volume of $P-ADE$ is $\frac{7\sqrt{14}\pi}{3}$,then the surface area of $P-ABCD$’s external ball is ()

[somber coyote]

e_waste

[dark sparrow]

quadrangular cone?

[ruby wigeon]

this is the figure

mention me when you guys firgure it out thx

I have to sleep cuz I only slept for 6 hrs in the last two days

[dark sparrow]

oh, you're looking for the word pyramid.

"cone" is usually reserved for when the base is a circle.

[wild ravine]

hi

[ruby wigeon]

oh yes

sorry about that,dunno much about English

[wicked pike]

it looks like you get the volume of the triangular pyramid and the square pyramid

then add them up together

[spring pagoda]

How do I draw a perpendicular bisector

[grave pond]

Sounded like they need the surface area rather than the volume.

[grave pond]

Draw circles of the same radius around each endpoint; connect the two points where the circles intersect.

[spring pagoda]

Thanks

[wicked pike]

oh mbmb

[grave pond]

Or perhaps the area of a circumscribed sphere? There's quite a bit of translation noise.

[amber hollow]

any clue what this question is asking for?

[faint pasture]

Well yk how to interconvert between inverse trig funcs?

[amber hollow]

wdym by convert

[dark sparrow]

do you want to know "what is this question asking for?" or "what is this question asking for, and how do i get there?"

[amber hollow]

That depends on what the question is asking for tbh, but probably the second one? I kinda understand how to solve questions like it but I don't really get what this question in particular is asking for and I might not understand it

[dark sparrow]

it's asking you to rewrite cos(arctan(x)) in a way that doesn't contain any trig functions (regular or inverse)

[amber hollow]

okay i got it, ty but i am confused why adjacent is just 1 by default?

i do know that cos^2 + sin^2 = 1 but that would make me think that just the hypotenuse is 1 by default?

[dark sparrow]

it can be anything you want

you can have it be 42069 if you so desire

it is just convenient in this case to have it be 1

any one side of the triangle can be set to 1

[amber hollow]

oh i see, thx

[amber hollow]

cant think of a reason why im wrong here

[dark sparrow]

redundancy

4pi/3 is already covered by pi/3 + kpi

[amber hollow]

wut

[grave pond]

pi/3 + 1·pi = 4pi/3

[amber hollow]

i guess that's fair, thanks

[tepid burrow]

for page 5 in the pdf does anyone know the correct formulas for Rt and Ct? the first formulas work very well but the Ct is not the correct number.
https://apps.dtic.mil/sti/pdfs/AD0620520.pdf

[amber hollow]

but this one isnt redundant?

[dark sparrow]

indeed it is not

[trail tendon]

no?

pi/12 + (1)pi/2 = pi/12 + 6pi/12 = 7pi/12

wouldn't that be redundant?

pi/12 and 7pi/12 are pi/2 away from each other anyway...

where am i thinking wrongly XD

[spare lagoon]

can i ask you a question since you are a teacher?

[trail tendon]

what kind of question? 👀

[spare lagoon]

i asked my teacher a question today "why lim x->0 sin(x)/x = 1" and he said "when you will go to univeristy and get deep into limites you will understand" so i want to know why?

[trail tendon]

oh there's a geometric proof using the squeeze theorem, or you can use l'hop but that may not be too proofy

lemme see if i can find a link to the proof

if not then someone can prob help XD

[spare lagoon]

tnx are you also a teacher?

[trail tendon]

no, i don't think she is either

[spare lagoon]

she is about to finish her master so sort of she is a teacher

[clear anvil]

you're not wrong

[trail tendon]

i mean being knowledgeable about something doesnt mean ur a teacher, maybe it means u can teach

[spare lagoon]

ok ...

[trail tendon]

https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-new/ab-1-8/v/sinx-over-x-as-x-approaches-0 khan academy vid on it
https://www.youtube.com/watch?v=0flNdU7Mrws&ab_channel=TamásGörbe random channel proof of it

it uses squeeze theorem i think

[spare lagoon]

tnx

[clear anvil]

,,\frac{7\pi + 6\pi k}{12} = \frac{6 \pi + \pi + 6\pi k}{12} = \frac{\pi + 6\pi\overbrace{(k+1)}^q}{12} = \frac{\pi + 6 \pi q}{12}

[somber coyote]

nyxie9151

[static hearth]

I asked this in a help channel. nobody can figure out why the process of #2 is incorrect.

calculator issue?? I keep getting the same answer.

(focus on far right side)

sorry for the ugly handwriting btw. did this test with shaky hands

[trail tendon]

44.524

ur decimal off on B because of rounding

still that would be a stupid reason to mark down

who in their right mind would write arcsin(8sin(24 deg)/(sqrt(80-64cos(24 deg)))) into a calculator unless accuracy was THAT important

and then A would round to 111.48 ( from 111.475577...)

[rigid fog]

How do i know when should i divide the equation with the trigonometric function and when not? IN the first image i have to divide the equation with the sine in order to get the right answer while in the second image i musn't do that if i want the right answer, how do i know when do i divide it and then solve it and when do i just solve it

[trail tendon]

if i understand your question correctly, i think it depends on whether you're solving for the angle variable inside the trig function, or if you are solving for a variable outside the trig function

[rigid fog]

its for the same thing except that in the first image its for the c side and in the second for the a side

[trail tendon]

what does this say

[rigid fog]

sin alpha

[trail tendon]

oh

[rigid fog]

the order in the first pic is not the best

[trail tendon]

whats this?

[rigid fog]

centimeters

[trail tendon]

i'm just trying to understand XD

oh

a/sin(a) = 6?

[rigid fog]

c=a/sin(alpha)

[trail tendon]

= 6?

[rigid fog]

yes

[trail tendon]

ok

i'm still not sure i understand your question

[rigid fog]

basically in which cases do i have do divide the equation with the fuction (for example sine) and in which cases do i have to not do that

[trail tendon]

you have to divide if your sin is on the same side as the variable you're trying to solve for

a/c = sin(x), you want a alone? multiply by c because its on the same side as a and ur trying to get rid of it
a = csin(x)

a = c/(sin(x)), you want c alone? multiply both sides by sin(x) because its on the same side as c and you're trying to get rid of it
asin(x) = c

ik the problems are a bit off from ur problems but

i also dont know fs if thats what ur asking

[rigid fog]

yeah i kind of get it, thank you very much

[static hearth]

OH MY GOD. THIS IS SO FREAKING LAME

I really appreciate it though.

[trail tendon]

ikr 😭

like you obviously knew how to do it

[static hearth]

this will save me a couple points on my final for sure😭

[trail tendon]

the prof didn't have to do that 😭 🗿

[static hearth]

thing is he’s such a lax and funny prof😂😭

he literally lets us have airpods during exams!!

[trail tendon]

harsh grader tho 😂

[static hearth]

absolutely 😂 i’m gonna tp his classroom for this

[empty yew]

@graceful wigeon do you know trigonometric identities(especially Pythagorarean identities)?

[upper karma]

Ignore the marking with the blue pen

Status 1

[static hearth]

google

[uncut marsh]

Easy

[dark sparrow]

this is kind of bashable

[upper karma]

Washable?

[dark sparrow]

bashable

[upper karma]

B

[dark sparrow]

you can bash it

[upper karma]

Means?

Meaning*

[dark sparrow]

bashing means solving a problem through symbolic manipulations, usually with coordinates and usually unpleasant

[upper karma]

It is from jee advanced (mock test)

[dark sparrow]

oh bother

the centers of your circles lie on another line through the origin, say y=kx, which bisects the angle between y=mx and the x-axis

• call their radii r_1 and r_2
• write down the circles' equations, noting that the y coords of the centers are equal to r_i
• put x=6 and y=4
• eventually arrive at a quadratic of which r_1 and r_2 are both roots
• use vieta on that (knowing r_1*r_2 = 52/3) and get k

[upper karma]

I will be back

[static hearth]

does trig get fun eventually? or is it going to be uphill battle for the next 2 years?

[near bridge]

who can prove it?

[rigid fog]

how to calculate cotangent on a calculator? my math teacher was talking about how there is no cotangent on calculator cause its too simple and i have to reverse the tangent but how do i do that and i think tan-1 isnt the cotangent

[dark sparrow]

it's 1/tan(x).

tan^-1 is arctan, which is indeed different.

@rigid fog

[rigid fog]

ooh alr thanks

[twin spindle]

Hey, I'm pretty confused on this section on the exercise, I'm not 100% sure on how to find the functions w points like this

[dark sparrow]

if you were just given the coords of one point on the terminal side, could you do the problem

[twin spindle]

Yeah

[dark sparrow]

ok then find the coords of one point on each line in the right quad

and do it that way

well, that or there's another way

read off tan(θ) directly (-ish for #30) and work out everything else from there

[rigid fog]

does it apply for other funcions aswell? (tan, cos, ctg...)

[obsidian harness]

Yes

Usually the purpose of this is to conver cosecant to sin, secant to cos, and cotangent to tan

So if we have 5 cot x = 1

That's just the same as 5 * 1 / (tan x) = 1

So multiplying both sides by tan x gives just 5 = tan x

And also 5 cot x = 1 is the same as cot x = 1/5

[upper karma]

This made all the difference

Got it ,thanks

[obsidian hornet]

If I understood correctly, then we need to find the length DM, it turns out that due to the equality of the triangles ADM and DMC (counting the corners, given that M is equidistant from AECF) DM lies on the diagonal BD. Then MD=BD-BM(BM can be calculated by Ptolemy's theorem (Calculations in the figure).

[upper karma]

How to visualize a tetrahedron ?

[dark sparrow]

a tetrahedron is a triangular pyramid

@upper karma

[upper karma]

Aye

I am having trouble with solving questions related to tetrahedron

[dark sparrow]

!xy

[lime crown]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[upper karma]

One min

,rotate

[upper karma]

Don't mind the blue pen marking

!status

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[upper karma]

3

Jee advanced (mock test) again

[tepid burrow]

for page 5 in the pdf does anyone know the correct formulas for Rt and Ct? the first formulas work very well but the Ct is not the correct number.
https://apps.dtic.mil/sti/pdfs/AD0620520.pdf

[dark sparrow]

@upper karma well, one thing i can say immediately is that faces ABC and ABD are right trangles

[upper karma]

Yep

[dark sparrow]

also the other two faces are isosceles

[upper karma]

What I did is , volume of tetrahedron is area of ABC times CD

[dark sparrow]

that is incorrect

not only bc you're missing the 1/3 factor, but bc CD is not known to be perpendicular to the plane ABC.

[upper karma]

I am imagining it as rotating triangle ABC along CD

Along AB*

[dark sparrow]

what does rotating along a line segment mean

also again

[upper karma]

A plane

Rotating a plane right ?

[dark sparrow]

you CANNOT use the V = Bh/3 formula with a line segment that isnt the height

like

CD isnt perpendicular to plane ABC

idk what the hell you are imagining, but it is definitely leading you directly into bullshit

[upper karma]

Yeah

[dark sparrow]

let H be the midpoint of CD (so that BH and AH are altitudes of the faces they belong to),
then our tetrahedron has ABH as a plane of symmetry
and in fact its volume is equal to 1/3 * S(ABH) * CD.

[upper karma]

Something new

Okay

What's the right method

I have one but it's related to planes

[dark sparrow]

what made you dismiss my method as not right?

[upper karma]

Another topic I am not good with

I was talking about mine

[upper karma]

5 min

[tepid burrow]

does anyone know the correct formulas for the airy projection?
i found this website http://www.quadibloc.com/maps/maz0206.htm but the formulas are very wrong and do not work

[upper karma]

AH and BH are perpendicular right?

[dark sparrow]

to each other?

[upper karma]

Yes

[dark sparrow]

i'm not sure

[upper karma]

I can't visualise this

[dark sparrow]

but it doesn't seem to matter that much

ok let me make a sketch

[upper karma]

Thank you

OHHH

I get it now, why my method is wrong

[dark sparrow]

my CD is a bit wonky relative to the square grid even though i intended for it to be vertical

you know the convention of "draw unseen lines as dotted" for 3D drawings right

[upper karma]

Yes I do

[dark sparrow]

yeah, ok

that alone can make your 3D diagrams a lot better.

picking the best viewing angle is a bit of an art, but still

[upper karma]

I am still not able to understand it,
I will try it tomorrow again

Thank you for the help though

Can you send me any video or pdf explaining everything about tetrahedron

[dark sparrow]

no

[upper karma]

Oof,okay

[upper karma]

I have the values of alpha beta and gamma

Status 2

[dark sparrow]

alpha, beta and gamma sound like they might be the roots of some nicely expressible polynomial

esp. given that z is a seventh root of unity

[ebon dune]

Pls help me

[dark sparrow]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[upper karma]

Yes but how to get that polynomial ?

[dark sparrow]

good question

i don't see it yet

[upper karma]

I have the polynomial, if you want it

[ebon dune]

Ok sry

[dark sparrow]

what?

this was an invitation for you to post the question(s) you need help with.

[ebon dune]

K

[dark sparrow]

cause thats how we do things around here

[ebon dune]

Ok ok

I understand

Can i ask what i want now?

[dark sparrow]

always could have

[ebon dune]

How to solve that thing

[dark sparrow]

...what do "gonia" and "aitiologisi" translate to?

well ok "gonia" means "angles" i think

but i don't know enough greek to translate the other word

[ebon dune]

It means

Justification

[dark sparrow]

ah hm

ok

ok, so you need to find all unknown angles and give reasons for each.

let's go over a few basics

do you know what the three angles in a triangle add up to? (i.e. what their sum is?)

[ebon dune]

Yeah

It is the moirés eg 70⁰ right?

[dark sparrow]

moirés?

also i think you didn't answer my question.

[ebon dune]

Fates

The ⁰

[dark sparrow]

... degrees?

[ebon dune]

Idk maybe that

On translate it says fate

[dark sparrow]

oh, it is μοίρα in greek

[ebon dune]

Ye

[dark sparrow]

learn something new every day...

anyway

in english we call the angle unit "degree"

as in "360 degrees make a full circle"

[ebon dune]

Yeah ik that

[dark sparrow]

ok so anyway

if you add up all 3 angles in a triangle, what do you always get?

[ebon dune]

3 degrees

[dark sparrow]

no.

not three degrees at all.

[ebon dune]

In each angle

[dark sparrow]

3° is a tiny angle.

[ebon dune]

Ik

[dark sparrow]

the three angles of a triangle add to 180°.

[ebon dune]

Yeah

[fallen veldt]

So it's correct to say that "the greater the absolute value of the difference between onside and the sum of the other ones in a triangle, the less the measure of the angle opposite to the first side" and vice-versa?

[dark sparrow]

i think there is a language barrier between us @ebon dune

[ebon dune]

Uhm

Yeah

[dark sparrow]

i am having trouble explaining things to you

but there are two more things i had plans for:

• angles in an isosceles triangle
• angles made by two parallel lines cut by a third

[ebon dune]

I'll ty doing the half of them

But

Thanks for the help anyways

[upper karma]

^

[upper karma]

The equation is 8cos^3(theta)+4cos^2(theta)-4cos(theta)-1

But I didn't understand how to get there

Roots of the equation are Cos (2r(pie)/7),where r equals 1,2,3

[tall coral]

I may just be stupid, but I don't fully understand a b and c

is this the diagram??

[fallen veldt]

Yep

[tall coral]

are you using square brakets for absolute values?

or is there just some notation I'm missing

[fallen veldt]

I tried to formalize an intuitive notion I had that the angle opposite the BC side of ΔABC is smaller than the one opposite the BC side of ΔPBC, where P is an inner point of ΔABC.

Since PB+PC<AB+BC, I had the intuition that the angle opposite to BC of ΔPBC is grater than the one of Δ ABC.

[tall coral]

ok so are we comparing angles or sides?

[fallen veldt]

No, I do this so as not to pollute the reading with too many parentheses. But, yes, I'm looking at their absolute values.

[fallen veldt]

Angles

[tall coral]

ah\

[tall coral]

please just use parentheses or abs bars to describe their respective function

[fallen veldt]

Got it

[tall coral]

there was no mention of a P in the original question

can you please restate the question with all relavent ideas

oh I just realized English isn't your first language, maybe a bit of a language barrier then

[fallen veldt]

I think so

I'll try to be clearer

[tall coral]

what is P?

[fallen veldt]

I just want to prove that the angle CPB is grater than CAB using that idea: "the greater the absolute value of the difference between one side and the sum of the other ones in a triangle, the less the measure of the angle opposite to the first side". I tried to prove that idea on the first message and I'd like to know if it's correct.

So, if it's the case, since PB+PC< AB+AC, and BC is a common side of the triangles ΔABC and ΔPBC, then I'd like to know if I can conclue that angle CPB is grater than angle CAB.

[fallen veldt]

That one in the diagram

[fallen veldt]

First message

[tall coral]

P is not on that

[tall coral]

how did you construct P

how does P change with the other points

[fallen veldt]

What I tried to prove in the first message is a generalization, for any triangle. I think "lemma" is the exact word for what the first message really is.

So I tried to apply that lemma on that exercise where I need to prove that angle CPB is grater than CAB.

[twin spindle]

Question, for graphing the sin graph, do I have to memorize the key points?

[trail tendon]

it would be useful to know that sin(pi/2 + kpi) = 1, and sin(0 + kpi) = 0

thats bout it tho

and knowing when its concave down or concave up

like the basic shape

[twin spindle]

I see, how do you figure out were each point on the coordinate would be at tho, so like pi/2 as an example, how would ik where to graph that?

[trail tendon]

sin(pi/2) = the y coordinate

you'd prob use the unit circle

for the values

[steep mica]

Guys can you help me find the height of this triangular pyramid?

It’s all I need for this thing to end

[next peak]

how do i find the missing adjacent side of a triangle

Im on the verge of having a breakdown🥰

😝😝

[steep mica]

Fake math

[next peak]

what

[steep mica]

Exactly

[next peak]

help plsi…

[cunning lion]

difficult to help until you post a picture of the question

[cunning lion]

what formula relates the 3 sides of a right triangle together?

[next peak]

I figured it out😝🥰

thank u

(It took like 20 minutes of me staring at my paper)

LETS GOOO🐺🐺🐺🐺

sory i got excited

[steep mica]

Please guys 🙏🙏🙏🙏

[next peak]

I would but I havent learned that

I havent learned what im practicing

[tall coral]

sorry for the long response time. You proved the lemma, but you never applied it. Is this an excersize in lemma in specific?

[fallen veldt]

No problem, I appreciate any help.

I tried to formalize that lemma by myself to solve that specific problem using basic geometry theorems. My first idea was to argue that if PB+PC<AB+AC, then - (BC - (PB+PC)) < - (BC - (AB+AC)), hence, applying the lemma for the commom side, BC, of the triangles ΔABC and ΔPBC, angle BAC > angle BPC.

I've been wondering a bit and I feel like there's something wrong with that argument, but I'd like to be sure and figure out any mistakes.

[tall coral]

why not say BC - (PB+PC) > BC - (AB+AC)

[wicked pike]

hey kirby

[tall coral]

I AM THE REAL KIRBY

[tall coral]

I'm not super sure what the innequality is even trying to state

and there is no explemation as to how the sides relate the angles

unless I'm just bad at geometry, which is also an option

[fallen veldt]

I'm thinking about that difference in terms of absolute velues, so I just say in that way to emphasize that aspect, since 0>BC - (PB+PC) > BC - (AB+AC).

There's a theorem that states that in any triangle, the greater side subtends the greater angle (and vice-versa); in other words, the grater the side, the grater the angle. But, with that lemma I want to interpret it as "in a triangle, the smallest absolute value of the difference between one side and the sum of the others subtends the grater angle opposite to the first side (and vice-versa)"; i.e., the less the absolute value of that difference, the grater the angle opposite to the first side.

The advantage I seek by putting words in that way is to use that lemma on the common side of the triangles ΔABC and ΔPBC, since |BC - (PB+PC)| < |BC - (AB+AC)|

[fallen veldt]

Now I see: as the real numbers are dense, it's possible to always construct a triangle with a fixed side length, call it BC in a ΔABC, such that other side's length, AB, for example, is greater or less than BC, so it's opposite angle is greater than the one opposite to BC. Hence the lemma can be applied.

Cool

[hallow spade]

Is anyone here good at trigonometric functions 🥲

[dark sparrow]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[upper karma]

The answer is not matching

[dark sparrow]

show your work

[upper karma]

For the area of the triangle I used bing AI

[dark sparrow]

!nogpt

[lime crown]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[upper karma]

Herons formula with sides having root values

[dark sparrow]

do it without AI

[upper karma]

So now I have the sides of triangle ABH

[dark sparrow]

one moment

[upper karma]

How about a calculator ?

[dark sparrow]

,calc (6*sqrt(5)/2)^2 + (sqrt(153)/5)^2

[somber coyote]

Result:

51.12

[dark sparrow]

your AH is wrong

,calc (6*sqrt(2)/5)^2 + (sqrt(328)/5)^2

[somber coyote]

Result:

16

[dark sparrow]

do you think that me telling you not to use AI also means you cannot use calculators?

[upper karma]

No

[upper karma]

You entered it wrong

Sqrt 2 /5

[dark sparrow]

augh

ok yeah typo on my part

,calc (6*sqrt(2)/5)^2 + (sqrt(153)/5)^2

[somber coyote]

Result:

9

[dark sparrow]

ok now it checks or

out*

[upper karma]

Now I have the find the area of ABH

Herons formula ?

[dark sparrow]

yeah

hm

i can't think of anything less unpleasant

[upper karma]

I will be damned

[dark sparrow]

this isn't a 10-millisecond formula

[upper karma]

I have another solution do you mind explaining it to me?

[dark sparrow]

you have somebody else's worked solution for this problem?

[upper karma]

No

The test paper comes with a soltuion

[dark sparrow]

ok but it was still written by somebody else, in this case the paper authors

show it here

[upper karma]

Q min

1*

Is that okay ?

[dark sparrow]

??

[dark sparrow]

is this still that tetrahedron volume problem??

also is there anything above this?

i think you gave me like the last 30% of the solution

[upper karma]

No it's the full solution

[dark sparrow]

so there is nothing above?

[upper karma]

No

Just the video explanation of the solution

Where the teacher just reads out the solution xD

[dark sparrow]

ok so it looks like they took A as (0,0,0), B as (4,3,0), C as (0,3,0) and then just solved for the coordinates of D...

and they deliberately chose not to tell you anything about A, B and C

[upper karma]

I got that

But the last part

[upper karma]

They probably assumed that we can figure out ourself

[dark sparrow]

expand the first and last equations and substitute the second into both

you will get two linear equations in x and y

[upper karma]

Subtract 2 and 3 to get value of y

1 and 3 for x and sub x and y values in 3

We get z

What I didn't understand is how did they come to the conclusion that D lies on the plane pe4pendicular to PLANe ABC

[dark sparrow]

??

what is that supposed to mean

they didn't assume anything about D, they just basically found the altitude dropped from D onto plane ABC.

[upper karma]

In the volume of the tetrahedron, the height is the perpendicular distance between opposite vertex and plane right ?

[dark sparrow]

yes

[upper karma]

That clears it up

Got it

Thanks

[tame flame]

Hello, I'm taking the Applied Geometry course, where we're studying affine space and projective space. However, I'm having trouble understanding the guidebook we're using (Linear Geometry, Rafael Artzy). I would greatly appreciate it if someone knows of any text that covers these topics or any YouTube video. I really would appreciate it; I feel very lost in this subject, and the fact that there are 2 exams, each worth 50%, doesn't help.

[lime crown]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[fading hatch]

Try to ask there too

[hollow orchid]

hey am kinda having a geometry exam deciding my future for the next 5 years in a week and a half and the teacher who places it uses annoying circle shapes any tips to solve cyclic quadrilatirals/circle arc problems?

[obtuse quiver]

try to understand and know the circle theorems like the back of ur hand so you can quickly solve them