geometry-and-trigonometry

[foggy parcel]

in other words the PQ' will be 10

since teh angle being 45 the two sides in the right tringle are eqaul

[gilded lion]

yep

[smoky jetty]

it should be 7.07 (rounded off) afaik

[foggy parcel]

u mean QQ' ?

QQ' will be 7

[smoky jetty]

no, PQ'

[foggy parcel]

Oh my bad

yea u r right

I meant the RQ' and PQ' will be equal

whatever it will be cos45 = PQ'/10 {PR = 10}

[smoky jetty]

it's also a 45-45-90 case, so 7.01 *square root = hypotenuse = ~10

then u can find the value of RQ' using Pythagorean theorem where RQ' = hypotenuse^2 - base^2

[foggy parcel]

having the value of PQ' u can get the value of QQ' & RQ' ..and with these u can get the value of RQ

[smoky jetty]

thus, leading us to using pythagorean theorem once more to finally find RQ, or the hypotenuse of te right triangle RQQ'

[foggy parcel]

Yea that's it @gilded lion

I hope u get it

[smoky jetty]

that's actually a pretty cool way, balls stealer, i only realized that now

but if ever u'd like the easier way @gilded lion, u can refer to this

[foggy parcel]

Well we can do it in anyway..

This is the fun part of math lol

[smoky jetty]

indeed!

another way i found to find PQ' is through the equation 10^2 = x^2 + x^2

as PQ'=RQ'

[gilded lion]

It did thank you!

[amber quartz]

Please anyone tell this

[opaque swan]

Does anyone get it?

[amber quartz]

Idk the answer...of my problem

[amber quartz]

Actually i did it guys

[dusk coral]

hi.. guys?

ok i have some formula

Inverse trigonometric functions and finding the angle of a right rectangle (as long as with sine): θ = arcsin(a\b)

[grand heron]

you only know the length of that diagonal line and nothing else?

[dusk coral]

I will try to solve it in the chat

[grand heron]

whats the problem?

[dusk coral]

for me - no

[grand heron]

I'm confused

[dusk coral]

ok

[grand heron]

no, I don't get what you were trying to say

[dusk coral]

something smart

[dusk coral]

a, b its 2*(a+b) =

[dusk coral]

h*2

[grand heron]

who are you talking to now=-=

[dusk coral]

with myself so that I understand to myself that I said something reasonable, and not necessarily correct

[dusk coral]

question?

[grand heron]

?

[dusk coral]

ok i'm just talking to you through a translator

[grand heron]

ah, its confusing to talk to you, I do not have a mathematical problem only here to help

[dusk coral]

ok

[dusk coral]

example of this:

θ = arcsin(a\b)

a^2+b^2=h^2

15^2+20^2=150^2=300

θ = arcsin(15\20)

θ = arcsin 0.75 \ n2

0.75\3.14*2

≈ 0.47

θ ≈ 0.47

I finished everything

[brisk trail]

so how did they find 2x

or get 2x

for question #4

[ebon echo]

can you please take a picture with zoomed in

[brisk trail]

@ebon echo

[ebon echo]

the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle

and arc EB (4x) is subtended by EDB here

[brisk trail]

what do u mean

by subtended

[ebon echo]

see the arc of length 4x between E and B

[brisk trail]

yes

[ebon echo]

we say that, the arc is made or subtended by EDB

[brisk trail]

so they divided 4x in half and got 2x?

it seems too simple

[ebon echo]

actually there is a theorem that if EAB subtends an arc of length say x, D being anywhere on the circle will subtend the same arc with angle EDB = x/2

incase I am just confusing you more, check this out,

https://www.youtube.com/watch?v=XUus6-9E9sQ

[rancid knoll]

its a ninth grade concept

[brisk trail]

ok thanks

[rancid knoll]

Ez pz

the question was quite easy essa

is anyone a calculus student here

[brisk trail]

thanks for letting me know...

[rancid knoll]

are you calc student

[brisk trail]

no im taking pre calc but my friend asked for geometry help

[rancid knoll]

oh

[brisk trail]

and i cant remember anything

and got stuck on that question

[rancid knoll]

you will just practice

[ebon echo]

i can feel the pain

[opaque swan]

It's not a diagonal...

[grand heron]

it isn't a diagonal but it is a line which isn't horizontal and also not verticle but diagonal

[civic elbow]

HELP

HELP ME

[timber cargo]

With?

[iron blade]

hey yall

anyone willing to answer a question?

[static magnet]

can someone please help me with this

how do they get this

[grand heron]

what exactly

[thin locust]

(ABC) is for the circumcircle of ABC right?

[limpid onyx]

Yes, the circle that passes through those 3 points

[thin locust]

ah ok ty

it was the thing which made the most sense but i wasnt 100% sure

[limpid onyx]

Is there a name for the group of shapes with n foci? (like created with n pins and a string)

[violet shadow]

For my working out here

Is the line supposed to be going through the circle at two points or is it a tangent?

Cause I don’t think this method works if it goes through two points? As the radius isn’t equal to the distance between the centre and the line then

[crude cedar]

that is correct

[crude cedar]

the lines when k = 0 or k = 15/8 are tangents to the circle but if k<0 or k>15/8 the line goes through two points on the circle

[violet shadow]

Ok

For my diagram the line is a tangent then?

[crude cedar]

If you are trying to draw the line where k=15/8 then yes it would be

As in y=15x/8

If you just drew a random line then no it would not be a tangent since it does cross the circle at two points (very close though)

[violet shadow]

Ye ok then

Ye I just saw the question mentioned two distinct points

Are u taking further maths right now

[crude cedar]

Yes

[violet shadow]

What topic are you guys on

[crude cedar]

I mean we are doing 4 different topics (4 teachers) rn: Linear transformations (with matrices), Trigonometry and modelling, Functions and graphs and distributions

Although we jump all over the place

Like we do the last chapter in the text book before the second

Year 12 currently

So a lot of the stuff I am doing rn is just normal maths year 13

[violet shadow]

Ye that’s fair

We just started matrices

[crude cedar]

I see

[violet shadow]

I’ve done most of the second year normal maths and further maths

But I skipped over matrices with transformations first year

[crude cedar]

Yeah they love skipping random things

[violet shadow]

Wasnt something I could self teach myself for some reason

[crude cedar]

Then going back later

Shame

[violet shadow]

Most of the maths I learn I can’t even apply to normal maths questions

[crude cedar]

Further maths can be quite annoying

5+5

[violet shadow]

Our teacher is making us to discrete and no one wants to do it

[violet shadow]

If u use complex analysis and calculate the mass of the sun I think it’s 55

[crude cedar]

Na it’s actually 1 solar mass

[violet shadow]

💀

Damn never seen that thing

Like some new sticker on discord or something

[crude cedar]

Mortifying

Sleep schedule is in the gutter

Although that goes for all further maths students

[violet shadow]

Not really

Chemistry takes up more time

And it’s like not difficult to pick up the further maths concepts

I wanted to do a past AS paper for normal maths and use further maths for all the questions and this and a binominal expansion was the only question I could actually try to do so

Cause u can use maclaurin series for binominal expansion

[crude cedar]

I have heard things about chemistry…

I do Maths, Further maths, Physics and Computer science

So I do not know the horrors of chemistry that I have been told about

[crude cedar]

I would answer but I do not know what the maclaurin series is

[sinful remnant]

So I got a big question about solving a triangle

Say I need to solve this triangle. So the first obvious step is using the law of cosines to find the missing side. Afterwards, I would use law of sines to find the angles.

So rounding up my numbers, I would get that a=11.1

I have like three questions about finding the measure of the missing angles.

1. Should I use law of sines to find the missing angles or the law of cosines?
2. If I use the law of sines, say I want to find first angle B so that would be B=62.4°, but if I also consider the supplement I would have B=117.6°. And so I gather that there are two solutions. However, if I first find angle C, then I would get that C=79.6° and if I get the supplement I would get C=100.4°. So is it correct to say that there are four possible triangles?
3. Should I use the law of cosines instead? Or should I valid the law of cosines for the four solutions I got and then discard the ones where the law of cosines does not hold?

[nocturne remnant]

in this case it is preferable to use the law of cosines because there are no risk for ambiguity of which angle is correct

there is only 1 possible triangle; the reason that sine rule produces extra possibilities it because you are putting in two sides and a non-included angle (SSA), while for cosine rule you are putting in three sides (SSS)

here SSS implies congruence; SSA does not

[violet shadow]

Like u would make f(x) the thing your expanding then evaluate it at f(0) for the first term then find f prime(x) and evaluate it to 0 and get the second term and for the rest of the terms u do the same but divide the f(0) by 2 factorial the 3 then 4..

[somber coyote]

ed1r0n

[smoky jetty]

could u elaborate on this?

[vivid seal]

i don't know where to start and what to do, please help me

[smoky jetty]

Segment ED intersects the two sides of triangle ABC and is parallel to the side AB of triangle ABC, hence ABC and triangle EDC are similar, and the ratio of their corresponding sides and angles are in proportion

basic proportionality theorem, or think of the triangle ABC as a version of triangle EDC scaled-up proportionally

[crude cedar]

EDC = 180-4x (angles on straight line sum to 180)
2x + (180-4x) + 78 = 180 (angles in a triangle sun to 180)

Then solve for x

And then because the two triangles are similar the angle y = the angle 2x

y=2x

Solve for y

[smoky jetty]

yeah this was my equation as well.. just got skeptic of x from this coz the measure of the 3rd interior angle of triangle EDC is quite not the way presented in the diagram

[crude cedar]

True since you would get 24 degrees for it but the diagrams are sometimes very not to scale

And I don’t see anything wrong with what I have done

[smoky jetty]

i guess so.. but from the equation, then the triangles are isosceles

[crude cedar]

True but that looks about right

Very roughly

Very very

[smoky jetty]

seems like it

[crude cedar]

I do see where you are coming from but I am going to say the diagram is very not to scale

Since the maths we have done seems to be correct

[smoky jetty]

wdym by where i am coming from? 😅

[crude cedar]

Means that I understand why you would be skeptical about the value calculated for x

[smoky jetty]

ahh ok

[sinful remnant]

Why would it be preferable to use the law of cosines? I guess that if I used the law of sines, I would need to use the law of cosines to confirm my result?

[floral shale]

You're given an angle between two sides

All 3 of which is given

[nocturne remnant]

Have you read my entire message🤨

[sinful remnant]

Yes, I did. I still understand that law of sines is viable. If I know the concept correctly, congruence has to do when two triangles are exactly the same, so I don’t get that part

[nocturne remnant]

If you use the law of sines it is not immediately clear whether the desired angle is acute or obtuse

You are given two sides and an included angle, so there is only one possible triangle (because SAS implies congruence))

[sinful remnant]

Just to ask again about congruence. Which two triangles would be congruent? Or is there a second meaning for congruence?

[thin locust]

need a hint (like where to look at)

[keen lintel]

Urm for me I would just look at the nine point circle but that maybe a circular proof as you probably get the 9 point circle from this

My hint would be to try to prove BHC and BXC are congruent

[thin locust]

oh ok ty

ill try it

[thin locust]

oh i had done it and tried to use it to proof that ABXC is a cylic quadrilateral

but i wasnt sure if that is enough

[keen lintel]

Oh you did it the wrong way, do you know reverse construction?

[thin locust]

oh no

wait ill look it up

[keen lintel]

Wait actually you dont need it

Dont worry

Urm so BHC and BXC are congruent immediately by the fact that X is the reflection of H over BC

[thin locust]

yea

thats what i also thought

[keen lintel]

Then you just need to prove angle BCX= angle BAX,

[thin locust]

oh

[keen lintel]

Which is obvious by angle chasing

[thin locust]

ah oh

oh

yea

damn

i forgot about that rule

ty

[keen lintel]

ah dont forget that lol, cyclic quads are important especially if ur looking at egmo

[thin locust]

trying to not forget it but im having a hard time remembering it

i often do it wrong

but ty

[keen lintel]

Ah no problem

[upper karma]

kinda need help

sorry 😅

[crude cedar]

The linear scale factor (for each line) is 9/8

So the area scale factor is (9/8)^2

=81/64

So 81/64 x 320 should get you the answer for a)

Then for b) 24 x 9/8

Then for c)

Get the volume scale factor: (9/8)^3

729/512

So 1536 x 729/512 should get you the answer

[crude cedar]

This good?

[upper karma]

yeah thank you very much

[crude cedar]

np

[flint wagon]

Hi everybody, I think I found a proof for the area of a triangle without having to know the height of the triangle

For a isosceles triangle:

and for a equilateral triangle:

[pliant roost]

heron's formula? :)

[flint wagon]

no, bc u don't need s

i think right?

[pliant roost]

no ik

I just think heron's simpler

[flint wagon]

oh ok

but, what is s?

[pliant roost]

semi perimeter

(a + b + c) / 2

[flint wagon]

oh

sad life

i thought i was genius

Idk if it's simpler tho

my formula is shorter right?

[flint wagon]

is there a proof for herons equation?

[pliant roost]

never hurts to feel like. :)

[pliant roost]

ofc, it's ancient

[flint wagon]

haha

but, why isn't this being taught in schools?

[pliant roost]

idk

not in the curriculum

how had you derived your formula? @flint wagon

[flint wagon]

I have a proof, but i just have to translate it

Do u want to see it?

[pliant roost]

I don't really need the proof, just the outline of it

I can fill in the details

[flint wagon]

oh well, it's based on two circles with the radius of the sides look:

then u just use this formula for the two circles

and then you get a solution for where the two circles come together

then u get your coordinates

thus the height

then use h*b/2

and then you have your area

also, the height is this

you get it?

[pliant roost]

wait lemme read

[flint wagon]

oh alright

[pliant roost]

oh, that

so you basically constructed a triangle, yeah

kinda clever, but I'd just solve an equivalent system of eq. lol

[flint wagon]

yeah, i get it, so you don't think it wouldn't be useful for mathematicians?

[pliant roost]

heron's formula. :)

also height = sqrt((c - (a^2 + b^2 + c^2)/2b)(c + (a^2 + b^2 + c^2)/2b)), may make it look simpler

[flint wagon]

yeah, i agree

[pliant roost]

sqrt((cb - (a^2 + b^2 + c^2)/2)(cb + (a^2 + b^2 + c^2)/2))/2 = A

[flint wagon]

oh also, from this formula, i found a formula for a quadrilateral

[pliant roost]

any quadrilateral?

[flint wagon]

yeah

but d is a diagonal

so you do need that

and for a parallelogram

[pliant roost]

well then that's kinda trivial for sure

parallelogram = A_triangle * 2

lol

[flint wagon]

bro take a guess how old i am

@pliant roost

[pliant roost]

15

[flint wagon]

correct how tf

[pliant roost]

you usually study those in 9th grade in my country, and I've just extrapolated it, so 15

[flint wagon]

i know, but im in 12th grade tho 😂

[pliant roost]

also you're usually interested in that at 15 years old

also my grade system is different

not that different for sure, but still

you jumped a few classes, didn't you?

[flint wagon]

haha ye, 3 to be exact

[pliant roost]

lmao

interesting how you considered it something valuable. I also would for a second ngl, but it would just not make sense if there was no such formula alr

@flint wagon you can look up a geometric proof of Heron's formula and Brahmagupta's formula respectively, as they exist. they are beautiful in their sort

[flint wagon]

yes it's true, mathematics has been in existence for a while, so someone must've considered finding a formula for this, which indeed happened to heron

[flint wagon]

I'll do that

ayo @pliant roost check this out

this does't happen with my formula

so maybe mine is kinda better?

[pliant roost]

no, it's exact

just computation may bring issues

which is fair

but your formula will happen to do the same :(

[flint wagon]

oh, why?

[pliant roost]

because that's how computation works

I am not a numerical stability expert

but the deviance grows, as one side tends to s, and other to 0

so the value jumps around

[flint wagon]

oh ok

but, i'm wondering, what would happen if you'd equate the two formulas to each other (mine and heron's)

[pliant roost]

try it

[floral shale]

Your equation and heron's are similar

Heron beat you

[elder sorrel]

can someone show me how to start

[quick viper]

RT/PR = QS/PQ = QR/ST?

I think

[smoky jetty]

could u elaborate?

[undone moat]

You can say that assuming that sides ST and QR are parallel.

Which I can't find as an explicitly mentioned property in the question.

[flint wagon]

lmao

[flint wagon]

wow

but look how long his formula is compared to mine

[pliant roost]

that's why everyone subs s for semi-perimeter lol

it's just easier to remember

[flint wagon]

true

[bright olive]

PT = PR + 2.4
PR = x
PT = x + 2.4

and if you are confused about the 3ST = 4QR
then lets say at 3 ST is 36 then 1 ST = 12 (it doesnt really matter what number it is as long as the ratio is the same, but i would suggest to use numbers that can be devided easily(you know or memorize))
if 3ST is 36 and 3ST=4QR then
QR = 36/4
QR = 9

Then you can

12/9 = x+2.4/ x to find the answer

[pliant pumice]

Hello! Isosceles triangles have 2 congruent angles. I found a proof for it. Is it correct?

[smoky jetty]

why not 36 instead in the ratio of 12/9? coz I thought 36 would be substituted to 3ST

[grave pond]

Looks convincing enough.

[smoky jetty]

oh wait im kinda getting it now

coz if it were 36=9 then ig this isn't equal

[pliant pumice]

Ok, thank you!

[grave pond]

Euclid's proof of the same claim is rather more complicated than this; the commentary at http://aleph0.clarku.edu/~djoyce/elements/bookI/propI5.html suggests some reasons that might be.

See also https://en.wikipedia.org/wiki/Pons_asinorum for more discussion.

[pliant pumice]

Yeah, I saw it already. But Euclid's proof looks a bit too complicated. If the proof I sent is right, it's much easier to understand and memorize

[grave pond]

Yeah, it's what both of the sources credit to Pappus.

[smoky jetty]

There has been much speculation and debate as to why Euclid added the second conclusion to the theorem, given that it makes the proof more complicated.
the second conclusion sounded reasonable tho

[grave pond]

Nobody disputes the second conclusion is true, as far as I understand.

[smoky jetty]

were the auxiliary lines of the same distance from the base?

[grave pond]

Yes.

[crystal oasis]

Hi

So there is this funny thing that is called "volume of partial cone/triangle" and well i want to learn it, but i do not know where to learn it.

Do any of you have some sort of website for learning and doing exercises?

[thin locust]

i only need to understand what (as L is on the angle bisector) means?

i dont understand the sentence/wording

i understand the rest

but what is meant by "as L is on the angle bisector"

is it just like: L is on the ray AI which is the angle bisector of angle BAC

[grave pond]

Yes, plus the central angle theorem for the arcs BL and LC, which says that AL also bisects angle BAC.

[thin locust]

is central angle theorem just inscribed angle theorem but the other way around?

oh wait

[grave pond]

Yes, that's another name for the same theorem.

[thin locust]

ah ok

thank you ❤️

[silent snow]

Taco

[weak flare]

I'm pretty sure 2 and 3 are wrong

And then like all of the 2nd page is probably wrong

[weak crypt]

am i on the right track for these two

[keen dirge]

is it possible to rewrite the function y=xsin(x^2) + 1 in terms of y? (solving for x)

[grave pond]

Not in general -- the function isn't even injective, and y=1 could lead to x=±sqrt(n·pi) for any natural n.

Even restricted to small intervals where the function is injective, it looks too cursed to be likely to have an inverse with a nice closed formula.

[snow crystal]

How to approach this?

I have to evaluate the alpha angle

[dark sparrow]

angles BCO, ACD, EAO can be calculated immediately

[snow crystal]

How? I dont have that knowledge

[dark sparrow]

triangle BCO is isosceles, triangle ACD is right.

[snow crystal]

Ohhj right, because of the radius

Also, how do you know that triangle ACD is right?

[dark sparrow]

multiple ways to show that

for example from the fact that triangles AOC and COD are both isosceles

and so angle OAC + angle DOC make up exactly half the total of internal angles in triangle ACD

[snow crystal]

But if the angle OBC is 32° and triangle BOC is isosceles, then the angle OCB is equal to 32°, triangle OCD is also isosceles so the angle OCD is 67°, 67 + 32 is 99°

That's the ACD angle

Which would mean that triangle ACD is not right

[dark sparrow]

angle OCB + angle OCD is not angle ACD

it's angle BCD

which need not be right

in fact it differs from the right angle ACD by none other than the very same alpha you seek

[snow crystal]

Ok I see it now, thank you

[silent plank]

ty

[upper karma]

Is geometry hard

I never actually studied geometry

I skipped to trig

[foggy parcel]

well to understnad trig well u need to have a basic knowledge of geometry

Geometry isn't hard

Tip: If u wanna master trig right now, the geometry of triangles is crucial

[upper karma]

I think geometry can be pretty hard. I skipped to trig too, now in calculus and there’s a lot of problem solving that has geometry related stuff, it’s hard if you didn’t do geometry haha. Trig is easier imo but ofcourse you need very basic geometry understanding for trig.

[sick hemlock]

Honestly geometry requires alot of critical thinking I guess?

Like even just using all the concepts you've learned could be pretty hard

[timber cargo]

The geometry that’s taught in schools require memorization. But on the SAT math section, that requires critical thinking

[dark sparrow]

that is a lot of problems.

[timber cargo]

Those problems really test your memorization

[pliant pumice]

Hello! As I know, every triangle can be divided into 2 right triangles. Should I prove this statement, or it is too obvious?

[hollow fulcrum]

You can say something like "We draw an altitude to the side AB which divides the triangle into two right triangles"

Nothing really to prove...

[pliant pumice]

I thought it needs a proof because if you have an obtuse triangle you can draw an internal perpendicular not to every edge

[pallid vine]

Every triangle must have at least one angle which is either right angled, or obtuse. If it's obtuse, you should make the base of the triangle be the side opposite the obtuse angle, and then you can draw an altitude down, as you've shown above. For right-angled triangles, it's a special case where you can make two smaller similar triangles

Hopefully that kind of makes sense in words

Oh wait shoot take that back

[pliant pumice]

Yeah, I understand. But how do I explain it? Is it an axiom or what?

[pallid vine]

I wouldn't say it's an axiom, but definitely quite a common fact. If you wanted to prove it, you can probably continue to think about the properties of triangles, and their internal angles

[pallid vine]

Only flaw in this argument is if it's an acute triangle, but that should be obvious from a sketch

[pliant pumice]

If I am writing a proof, can I use this "fact" without proving it?

[pallid vine]

Depends how rigorous you want to be in your proof. If it's a relatively rigorous proof, then probably start from scratch and prove the fact before using it.

Although, I think most people just use the fact as a given, so you should be alright if you wanted to just use it

[hollow fulcrum]

You can always draw an altitude to the longest side of any triangle

[pliant pumice]

Ok, thank you!

[hollow fulcrum]

You can construct the point the altitude will go to using the pythaogrean theorem which is a proper proof

[pliant pumice]

Using the formula and lengths of sides?

[hollow fulcrum]

Yeah

[pliant pumice]

I will try, thanks

[hollow fulcrum]

np

[pliant pumice]

I tried. It helps to find the point where perpendicular touches the edge, but how does it prove that any triangle has such point?

[hollow fulcrum]

you have a point so you proved it exists lol

now you need to show that it is positive

[pliant pumice]

What exactly?

[hollow fulcrum]

c1

needs to be between 0 and c

[pliant pumice]

And if it's not? Should I just try another edge?

[hollow fulcrum]

yes

it is guaranteed to work if the edge is longest side

so use that hypothesis

[pliant pumice]

Ok, it makes sense. Thank you

[sick hemlock]

Yeah I was more talking about the SAT sections or math olympiads

[timber cargo]

Have you taken the SAT

[sick hemlock]

Nope, but I have seen some papers. It's not really as prevalent here as it is in US

Where I live it's mostly just olympiads or competitions

[pliant pumice]

Using my previous formula for c2 and c1, I tried to prove that if c is the longest side in triangle, c1 and c2 will be positive. Is it right?

[trail lily]

If I solve a trig quiestion like 7tan30 how do I make it radical form, pls help

[floral shale]

tan(30) = sin(30)/cos(30)

[trail lily]

i know what 7tan(30) is as a decimal, how do you find radical

[lyric sonnet]

I love trig and geo.

[trail lily]

can you help?

so to find 7tan30 you just put in a calc but the calc at school only gives decimal form, 4.04145188433

how do i keep in simplified radical

[lyric sonnet]

firstly tan 30 = $((1/2)/(√3)/2)$

[somber coyote]

Ishigami Senku

[lyric sonnet]

therefore: tan 30 = $1/√3$

[somber coyote]

Ishigami Senku

[lyric sonnet]

then multiply by 7 and you get $7/√3$

[somber coyote]

Ishigami Senku

[trail lily]

the calc only gives decimal number how did you get that radical

[lyric sonnet]

I learnt surds ⚔
and .....

[trail lily]

what is surds

what-

[lyric sonnet]

Just send a DM and I'll explain
further.

[thin locust]

when you guys have to solve a geometry problem which is like this, do you draw the figure on a paper or are you able to "see" the figure

idk how to describe it

but, do you just see how the stuff is going to look like?

[silent plank]

draw, always draw

regardless of whether you think you can imagine it

[keen lintel]

Always draw lol, Geogebra is a good app if you need it,

https://web.evanchen.cc/handouts/Constructions/Constructions.pdf

[smoky jetty]

could be coz of the 30-60-90 triangle case, where the length of the longer side is (n) sqrt3 times the length of shortest side (n), and the longest side (hypotenuse, n) is twice the length of the shortest side (n)

[sick hemlock]

damn me too 🤝

[upper karma]

If I knew theta (written in black), does that mean I know theta (written in red)?

[topaz grove]

i believe so

[sick hemlock]

Only if the lines are parallel

[upper karma]

what lines

?

[celest osprey]

pls name the points of the triangle so it would be easy to describe

[upper karma]

call top point a and bottom right b

[celest osprey]

here

[upper karma]

yeah

[celest osprey]

if de and mc

are parallel

then theta will be equal

[upper karma]

they are parrellel, but obviously mc is longer. does that matter?

[celest osprey]

no it doesnt

[upper karma]

thank you so much

[celest osprey]

my pleasure

[sick hemlock]

In general, if 2 parallel lines are joined by another line, then the angle made by the line joining the parallel lines will be equal

I'll send a figure later

[smoky jetty]

yep, hence, <E & <C are corresponding angles

[upper karma]

Here is the actual full question and my working out. I can't see anything i did wrong but the numbers seem a bit off. Can you check?

[fallow dock]

final answer looks correct to me.

[edgy panther]

can anyone help me solve these trigonometric ecuations?

[sick hemlock]

For most of those you should try taking a term to the right side

That should provide some clarity

for example write sin2x - sinx = 0 as sin2x = sinx

[sick hemlock]

You know that sinx = cos(90-x) right?

The 2nd one is a bit of a tricky one.

When do sine and cosine have opposite signs?

[edgy panther]

when they are in the second and fourth quadrant

[sick hemlock]

Yep, so your answers lie there

I'm not really sure how you'd go about showing your work

[edgy panther]

ahhh i think i kinda get it now

[sick hemlock]

You could also use the formulae for sin2x and sin3x if you want

Yeah you can do like 2 of those with complimentary angles, and the rest with the addition formulae

[violet shadow]

I’m not 100% sure on part b

But I got 7.67

I find the equation of the line passing through The centre that is partake to other line

Then I found the equation of the normal to that that goes through circle

And find where it intersected the circle

Then found the distance between that point and point where the line given crosses x axis but idk if this is right cause it seems very long winded

[sick hemlock]

@edgy panther Hey I found out how you could go about showing your work for the 2nd one

I didn't notice the sin(-x) and was confused the entire time

But you can write -cosx as sin(90+x)

@violet shadow Did you find the coordinates of the centre?

[violet shadow]

Ye that’s 5,4

[sick hemlock]

I guess you could go about finding the perpendicular distance between l and centre then

I don't really understand what you meant so you might have done this only

Lol

[violet shadow]

Ye but it wants shortest distance isnt that from the edge of the circle?

[sick hemlock]

First try the shortest distance from the centre, then subtract radius ig

It seems right for some reason

Shortest distance between the centre of the circle and the line is the perpendicular distance from the line

[violet shadow]

Idk if that’s what the question wants tho

But ye I think your right

Find perpendicular distance and then subtract the radius should work

[sick hemlock]

Yeah, I just can't find a good explanation for this

Maybe something to do with right triangles

I'll brb

I've got it I think

I'll send a clean ss

[violet shadow]

I got 5.497

Perpendicular distance is 19root 5 all over 5

Subtract 3 which is radius

[sick hemlock]

@violet shadow

I guess this counts as a proof?

Line n is a tangent to the circle at the point where perpendicular distance cuts the circle

And it's parallel to the line l due to angles

So the distance between the parallel lines is the same throughout

And n only touches the circle at one point so that means that's the minimum distance

[violet shadow]

Well u have to find that line then

U have to find the tangent to the circle your way then

Which is long

[sick hemlock]

Yeah I wasn't suggesting a solution

I was suggesting a proof to that thing I said

[sick hemlock]

^

[sick hemlock]

Let's go, smaller than your previous answer

[violet shadow]

U got it aswell?

Ye I think my 7 was wrong cause I did it a long wounded way

I think this is correct tho

[sick hemlock]

Yeah

Well we learned something new today

[violet shadow]

The one time I didn’t use further maths I got it wrong

Ye

[sick hemlock]

Something about shortest distance from a circle

[violet shadow]

I usually use further maths for geometry

Cause there is a formula for perpendicular distance between a point and a line

[sick hemlock]

Honestly geometry is scary sometimes

You don't know if the question wants you to do it a long winded way

Like say, finding the orthocentre of a normal triangle

[sick hemlock]

Oh there is?

Yeah I vaguely recall something similar

[violet shadow]

It’s in the vectors section

Trying to send a pic but it’s slow

[sick hemlock]

I saw a formula for perp distance from the origin nvm

[violet shadow]

K

[edgy panther]

yeee you re right now it makes sense

[sick hemlock]

eyy found the formula for perp distance from a line

I hate it

[solemn ivy]

Can someone help

[sick hemlock]

help with...?

[upper karma]

Do you think I could of used law of sines after I found the angles?

[sick hemlock]

Honestly, you didn't need to calculate tan inverse of 10 either

[upper karma]

Then how to get the angle?

[ruby oak]

I have no clue how to do this and need a walk through

[crimson pawn]

ask a help room

[exotic yarrow]

(answered in another server)

[upper karma]

How would you do it?

[meager mango]

I’m so dumb

[sick hemlock]

Sorry for not responding, I was asleep at the time lol

But since you found the 2 angles to be equal, you could have simply set tantheta = tanalpha. And since the question is just asking for the diameter you don't need to calculate the angles @upper karma

[upper karma]

help

[sick hemlock]

help with...?

[upper karma]

this center

so confusing

[sick hemlock]

Okay, what would you need to write an equation for the circle

[upper karma]

coordinates

ig

[sick hemlock]

Just to clarify, equation of the circle means equation for points lying on the circle

[upper karma]

these were easy to find

idk bout that first pic tho

[sick hemlock]

So you found coordinates of 3 points lying on the circle I assume?

Nvm

[upper karma]

for this no

nvm

I meant

for this no

[sick hemlock]

okay, what's a speciality of circles that you can use to your advantage?

[upper karma]

they form diameter?

[sick hemlock]

or, each point lying on the circle is equidistant from the centre

[upper karma]

radius

[sick hemlock]

Though I guess you don't even need that

The question isn't asking for the equation of circle

[upper karma]

it is

[sick hemlock]

It's just asking you to find the diameter

[upper karma]

no?

it says equation

[sick hemlock]

But that's not what an equation of circle is, atleast where I've read

That's mostly just asking you to apply distance formula

[upper karma]

wait for this??

[sick hemlock]

by finding 2 points of a diameter

[upper karma]

I thought distance formula was for this

I'm soo confused

[sick hemlock]

Distance formula is for finding the distance between 2 points

Could be used wherever convenient

[sick hemlock]

Didn't you basically apply distance formula here?

[upper karma]

for my question, we just finding the center

[upper karma]

nope

we got the points from the graph itself

[sick hemlock]

You found the radius of the circle

[upper karma]

cuz I just counted the boxes

[sick hemlock]

Alright, sorry I was confused by the wording of the question

[upper karma]

it's alr

[sick hemlock]

Been a while since I touched coordinate geometry

So yeah, these questions require you to locate the centre

Find the distance between the centre and a single point lying on the circle

and then apply distance formula for all x,y that are the same distance from the centre

[upper karma]

-3,4

BUT SHE SAID WE AINT GOTTA USE ANY FORMULA FOR THE QUESTIONS THAT HAVE GRAPHS ON

[sick hemlock]

the question is literally asking you to use a formula

You can't really not use a formula there

[upper karma]

wait this is confusing

imma brb.. I'll call my teacher hol up

[sick hemlock]

Let me confirm: You did this by calculating the radius, and then applying distance formula for all x and y that are the same distance away from the centre right?

Just do what you've already done

[viral cave]

hello

i want to have a discussion about 4th spacial dimension

this word "space" is important here

because time is also known as 4th dimension but I am talking about space specifically

does anyone have an example of a 4th dimensional object or thing which exists in real life?

which can be observed?

or measured.

Don't tell time.

[upper karma]

wait actually I get it now

tysm anyways!

[obtuse patio]

no lol what

we live in a 3d world

[viral cave]

but does that imply we can't observe or measure it?

[obtuse patio]

yes

we cant even imagine it

[viral cave]

well

a circle can cut a line

so a point in line can see those atmost two points of the circle

a sphere can cut a plane

so a 2D being can see those curves

similarly

[obtuse patio]

yes

but we cant see the 4d

we can see the cut/shadow

[viral cave]

why can't something 4D come in contact in our space?

[obtuse patio]

it could

we cant prove it though

[viral cave]

where is it

[obtuse patio]

it would be a 3d shadow

[viral cave]

u have any example please?

[obtuse patio]

yeah one sec

[viral cave]

not necessary

[viral cave]

no

i said real life example

but thanks for this teseract image

[obtuse patio]

theres no real life example wym

4d is theoretical

[viral cave]

and thats what all I asked for

an s or no question

[obtuse patio]

i said that though alreayd

w/e

[viral cave]

and then, what if 4D doesn't exist at all?

[obtuse patio]

it may not

[viral cave]

i feels crazy

i know

i seem crazy

[obtuse patio]

if 4d exists maybe 5d exists too

who knows

[viral cave]

but what if the nature of space itself is like that

such that space can only have 3 dimensions

[obtuse patio]

maybe

we cant say

[viral cave]

i hope a mathematical proof or disprove in the future

[obtuse patio]

how

we cant

[viral cave]

why

[obtuse patio]

anything we do is 3d

how can you prove for example parallel unvierses exist

[viral cave]

well thats another thing

[obtuse patio]

its the same idea to me

its so far out

[viral cave]

this doesn't stop us from observing it if it ever came in contact with our dimension

[obtuse patio]

thats not a proof

[viral cave]

so

[obtuse patio]

so what

[viral cave]

it was never meant to be

[obtuse patio]

if we can somehow find it existing yeah

but Im saying you cant prove it with like math

[viral cave]

hm....

[obtuse patio]

itd have to be shown through experiment or smth

maybe using dark matter

who knows

many people are actively working on these kind of questions

maybe you will too one day

[viral cave]

saying "we can't" is just as arrogant as saying "we can" I guess

it is just left there

thats why I said hope

[obtuse patio]

it may not be a good thing

if we find out it exists

[viral cave]

oh

why so

[obtuse patio]

we could mean nothing

[viral cave]

well we already are

[obtuse patio]

eh

i think it would make it worse

[upper karma]

That is definitely wrong.. unless I’m not understanding what you mean. Have you looked at the question properly?

[sick hemlock]

Allow me to explain properly

So, when you look at the 2 right triangles, they're similar

That also leads to the same conclusion that the sides are proportional

Similarly, since the angles are equal, the tan of those angles should also be equal. Therefore tantheta = tanalpha, which leads to ratio of the height from the ground to the radius of the cone formed is 10.

Even then, this question does not even require trigonometry, it's basic similarity

All I said is that if the angles are equal, then the tangent of those angles should also be equal

And that leads to the same conclusion you'd get by going the similarity route

@upper karma

[upper karma]

Ok yeah I understand now regarding setting tan theta and tan alpha equal to each other. Thanks for explaining. How would you approach it by similarity route if you don’t mind explaining?

[sick hemlock]

Alright let me name the vertices one second

So, we're considering the triangles ADE and ABC here

Angle A is common to both the triangles, theta and alpha are equal, and Angle D and B are right angles

So all 3 angles are equal

A=A
D=B
E=C

So triangle ADE~triangle ABC

Now according to CPST, AD/AB = DE/BC = AE/AC

Using this, we can say that AD/AB = DE/BC

which gives you the same equation as above

[upper karma]

What is CPST? Yeah I never would of known how to use symmetry like that. I know basic symmetry but thats all.

[sick hemlock]

Corresponding Parts of Similar Triangles

Similar figures are figures which are basically the exact same shape as each other, but differing in size

It's basically congruency but without the figures being the same size as one another

The sides of similar triangles are proportional to each other and the angles of vertices are the same (hence the same shape)

In determining similar polygons you need to see if the angles are the same

Hopefully that provides some context

[smoky jetty]

A yardstick casts a shadow of 24in. at the same time an electric post cast a shadow of 20ft. 8 in. In inches, what is the height of the electric post? Let h be the height of the electric post
I had this problem recently and Im somewhat skeptical of the triangles being similar. Because the statement did not say anything regarding the angles in the triangle being congruent to one another or state that the triangles are similar. But, if you put the given measures in proportion, you'd find h. But idk how I could make the statement stating that both are similar unless I assume that the interior angles are equal or, they're both right triangles.

[sick hemlock]

What's a yardstick

Okay

[obtuse patio]

Imperial

[sick hemlock]

@smoky jetty have you drawn a figure?

[upper karma]

whats a inch? the rest of the world uses SI units..

lol

[sick hemlock]

Just calculate in inches lol

Another similarity question

[upper karma]

I know duh.. just being smartass

[sick hemlock]

1ft=12in isn't it?

[smoky jetty]

yeah i have, but the problem did notstate that they are similar, so I could solve it if, i assumed that both have the same angles

[sick hemlock]

You have to

[smoky jetty]

i have the answer, only if they are similar, but idk how I could prove that they are similar in a statement

[sick hemlock]

Otherwise it's not really solvable I don't think

[smoky jetty]

indeed

[sick hemlock]

Okay, so it's a yardstick and an electric pole

[smoky jetty]

otherwise, im having a geo concept gap, lol

[upper karma]

Do you know much about secondary bonds im chemistry, such as van der waals?

[smoky jetty]

yes, you'd get the answer if u assme they're similar, but i kinda find the given being insufficient with info

[sick hemlock]

Whenever you are given pole like objects

[smoky jetty]

what are your thought?

[sick hemlock]

You kinda have to assume they're perpendicular to the ground

[smoky jetty]

yeah i thought so as well

[sick hemlock]

Unless stated otherwise

[upper karma]

right angles baby

[sick hemlock]

Yep

[upper karma]

we love that shit

[smoky jetty]

but i considered other factors being that one may be inclined as the other is perpendicular to the base

coz the problem didn't state such

[sick hemlock]

I mean, we are at relatively lower levels in geometry

So I doubt the question wants us to consider those possibilities

[smoky jetty]

i see

[sick hemlock]

So for now, rest easy and whenever you're given poles or similar objects, just assume they're at right angles

[smoky jetty]

i kinda feel bad that i have to assume their angles just to prove they're similar, lol

[sick hemlock]

Yeah I know what you mean

You could also think practically, why would someone build an inclined electric pole?

[smoky jetty]

do u think assuming it would prove their similarity otherwise?

[smoky jetty]

lmao, i've had previousproblems before with inclined poles

[sick hemlock]

Yeah but only sadists build those

[smoky jetty]

lol i see

[sick hemlock]

Apparently engineers are infamous in the math community

Why would they make stuff hard on themselves by making poles on slight inclinations

[smoky jetty]

well ig in short, to solve the given problem, one must assume they're perpendicular or have the same interior angles?

[sick hemlock]

Assume they're perpendicular for now

In later classes that might change, I don't really know

[smoky jetty]

alright, thank u very much.

[sick hemlock]

Eh I didn't really help

[smoky jetty]

how come? lol

[sick hemlock]

I just acted like those teachers who say "That's just the way it is"

[smoky jetty]

ig my teacher gave that problem with that attitude, lol

[sick hemlock]

Yeah, though it makes life easier for us as well

Similarity won't work always

Sometimes you'll have to use trigonometry

And in those cases the poles being perpendicular helps a lot

[smoky jetty]

yeah

what grade are u in btw?

hs before senior higschool?

ig im 9th grade relative to yours

[sick hemlock]

Damn, similarity in 9th grade

[smoky jetty]

idk if we'll ever reach congruency&heron's, prob due to the curriculum

btw thx for the responses earlier, I thought I was the only one who had to assume the triangles were similar to solve it, or who knows

[sick hemlock]

No you do need to assume them to be perpendicular

You know why the triangles are similar right

The sun will cast a shadow at the same angle for any perpendicular distance at a given time

Since angle of elevation of the Sun does not change from the ground

[celest osprey]

they must be perpendicular to the ground

[smoky jetty]

yeah I thought so as well

does that need to be explicitly said in the problem?

otherwise it's just assuming both are perpendicular to the ground?

[celest osprey]

wait i messed up

[near harness]

You can prove the triangles to be similar by the Angle-Angle similarity criterion.

One pair of angles will be 90°.
The other pair of angles will be equal as the sun will caste light at equal angles.

You can also use trigonometry to solve this problem.
Either way proving the non-perpendicular pair of angles is necessary.

[sick hemlock]

That wasn't his question though

He wanted to know why we had to assume the electric pole and yardstick to be perpendicular to the ground

[smoky jetty]

yes, but the problem didn't state any given angle, so i just need to assume

[near harness]

It's not an assumption but more like ideal.
For eg: a tree. It can be bent but ideally we take it to be straight.

[near harness]

Take the angle to be ∅.
Apply tan∅ in both triangles.

Since ∅ is equal in both triangles, tan∅ is also equal.

Equate both of their values and solve for x.

Where x is height.

Or in your case it is "h"

[smoky jetty]

u mean the ratio of tan ∅ in both triangles ?
i.e let's say, 36/24 = x/248 ?

[jagged galleon]

Does this shape have a name, and if so what is it called?
Just a cube with a cylinder cut out from all axes.

[near harness]

Yes

[smoky jetty]

ah yea, then you'd get x then

would u say that the problem should've explicitly said that the post and stick were perpendicular?

[near harness]

Um.. that's hard to answer. Mentioning it does make it more clear. So yes.

[sick hemlock]

My guy, that's what an assumption is

[uncut panther]

Given the vector of coordinates v=(-3/4,k) it is requested that for k=1, express v as a linear combination of a=(2,3) and b=(5,-1)

I managed to resolve it but I don't know if it's right or if I misunderstanded the exersice

I thought v=(-3/4,k) was w and did w=x(2,3)+y(5,-1)

[obtuse patio]

Break it into components

v = Ca + Db

-3/4 = C*2 + D*5

And then the y component

Two unknowns two equations

Solve

[uncut panther]

The answer is x=1/4 and y=-1/4 right?

[obtuse patio]

Idk lol I didn’t solve it

Plug the values back in and check

[prime umbra]

can someone please help with this 😭?

[timber cargo]

There’s a beautiful formula for this: $(n-2)*180°$

[somber coyote]

Daniel S.

[timber cargo]

n being your total number of sides

and the result being the summation of all angles

[prime umbra]

• 2 and then * 180 to get the degree / sum

?

Thanks

[timber cargo]

You’re given that the polygon has a total of (x-8) sides, that’s your n value right there

So substitute that for n

and you’ll get what

@prime umbra Come back, don’t ghost me

[obtuse patio]

I think they already got it

[timber cargo]

I hope so

[prime umbra]

sorry 😭

what’d you say?

[timber cargo]

What is your answer

[prime umbra]

if it’s x-8

is my x what i get after 4320/180

[stray spruce]

pls help

[still pelican]

yo

nvm

[obtuse patio]

Where’s the line

[near harness]

My bad 😓

[wise pawn]

https://en.wikipedia.org/wiki/Steinmetz_solid#Tricylinder

[obtuse patio]

Origami ball

[lost sky]

Is that the Cartesian Plane? Why is there no line?

[short wyvern]

use the distance formula

$/sqrt(x2 - x1)^2 + (y2 - y1)^2$

but what do they mean by "the line"

?

$/sqrt{(x2 - x1)^2 + (y2 - y1)^2}$

[somber coyote]

Sanaltsalt

[floral zealot]

@short wyvern

[short wyvern]

yessir

[floral zealot]

backslash not forward

[short wyvern]

oof

[floral zealot]

\sqrt not /sqrt

[short wyvern]

alr

thanks again

[floral zealot]

good luck

[short wyvern]

alr

$\sqrt{(x2 - x1)^2 + (y2 - y1)^2}$

[somber coyote]

Sanaltsalt

[short wyvern]

this is basically displacement

but this is maths

just fill in y with "the line" and you should be fine

@stray spruce

[zenith forge]

first multiply out the areas of both rectangles

[timber cargo]

Refer back to your understanding of inverse trig functions

[obtuse patio]

Do you have a calculator for this

[snow crystal]

"What's the sum of all the six angles marked on the picture", how to approach this?

[violet shadow]

Is there a way to do this question part C without the discriminant because it’s very long that way

[trail lily]

Somone draw this out

PLS

[desert anvil]

hello, can someone help me with this? just the solution would be fine.

[celest osprey]

4x-8=23

opposite sides are equal

in a parallelogram

and 47+2y-3=180

adjacent sides are supplementary in a parallelogram

[waxen scaffold]

A water park has hired Sharnee to build part of a ramp for a new water slide. She builds a 12-metre-long
ramp that rises to a height of 250 cm. To meet safety regulations, the ramp can only have a gradient between
0.1 and 0.25. Show that the ramp Sharnee has built is within the regulations.

can anyone show me how to do this i cannot understand

[dark sparrow]

do you know what gradient means?

in this context it is also commonly known as slope, though it depends on where you are from.

[waxen scaffold]

yeah i know but idk how to find gradient in that question

[dark sparrow]

how would you find the gradient of a line on a coordinate grid?

[waxen scaffold]

y2-y1/x2-x1

[dark sparrow]

missing parentheses...

[waxen scaffold]

whats that

[dark sparrow]

brackets

anyway

the phrase "rise over run" should have been said at least once in your class

[waxen scaffold]

yeah

[dark sparrow]

She builds a 12 m long ramp that rises to a height of 250 cm.

what are the rise and run here?

[waxen scaffold]

ohh i get it now

i didnt consider that

thanks

[short wyvern]

find the gradient...

oh sorry

it's already answered

[smoky jetty]

give it a try

[lost sky]

Rise over Run. Rise/Run correct? And the RS/RN is the mx in y=mx+b. Correct?

[smoky jetty]

I made a very rough illustration regarding the problem if u need to solve it asap, hopefully u know the nexts to do

[dark sparrow]

rise over run is just the m, not mx

[smoky jetty]

the problem didn't specify if it's right triangle, i hope it represented the measures correctly

[lost sky]

Oh yeah I forgot.

[lost sky]

I would say it would be a right triangle if the little box indicator thingy is on the straight leg.

[smoky jetty]

yeah, but the problem didn't state it so

[keen fiber]

someone solve this: sinx+siny=a and cosx+cosy=a, what is sinx+cosx interms of a

[violet raft]

if there is the square in a corner that tells you it’s a right triangle. they just didn’t list it in the problem.

[smoky jetty]

i mean it's pretty hard to assume that it's a right triangle here

[lavish portal]

Help this question, find alpha in terms of a, b, c, d
Given this is a rectangle where the intersection between four lines is randomly picked

[fossil lava]

Guys can anyone explain why the unit circle is centered at (0,0)

[sick hemlock]

to make life easier

[sick hemlock]

Fr

[tardy thicket]

law of cosines

[dark sparrow]

so as not to have the coordinates of its center get in the way everywhere.

[cedar grotto]

You helped lots thank you for the explanation

[austere blade]

..?

I already finished that whole assignment

Yesterday

10th

I’m not the best at math

[violet raft]

u still need help?

[upper karma]

yut

yur

can someone help me

<@&286206848099549185>

[granite void]

isnt AOB 80 too?

since O is the center

and since its a circle you know 2 sides of the triangle are = so know more about the triangle

then ACB is just the whole circle missing a part, which you know

[rapid wraith]

<@&286206848099549185> can anybody help with this question?

i need to find the probability of p(A|B) in order to see if probabilty of a and probability of B are independent or not. im looking to see of P(A)(B) is equal to P(A|B)

p(A)= 11/50
p(B) = 5/11

[granite void]

are you given more info?

[rapid wraith]

yes

[granite void]

might be handy

[rapid wraith]

the meaning of p(A) is the probability of picking cheerios

11/50 chance

p(B) is drinking the milk after eating the Cheerios

5/11 chance

just trying to see if they are independent

[mint kite]

multiply it

[granite void]

is p(B) the chance of drinking milk after eating cheerios or the chance of drinking milk after eating cereal?

[rapid wraith]

Its the probability of drinking the milk thats left in the bowl after eating cheerios

[granite void]

So thats saying the prob of drinking the milk given the fact you ate cheerios

[rapid wraith]

Yes

[granite void]

oooh

[upper karma]

sssno more info is given

[rapid wraith]

any way to figure that out?

[granite void]

i meant with the probability

[upper karma]

oh

[rapid wraith]

i know p(A)(B) is 1/10