geometry-and-trigonometry

[obsidian harness]

yeah, so you would need sin(90 - (-180))
= cos(-180)

[slim plinth]

Thanks

South do you focus on Geometry

[obsidian harness]

I can help with anything at a pre-uni level

[slim plinth]

Good

I am focused on Algebra

[slim plinth]

So yea on SAT I would have to know the identity

[young mist]

I know this is math server but can anyone recommend to one of Physics and Chemistry

[slim plinth]

sin(90-x)=cos
cos(90-x)=sin

[obsidian harness]

#old-network

[young mist]

thank you

[slim plinth]

I have an idea

I’ll also use #prealg-and-algebra to send out questions

[slim plinth]

You vs Ann

It’s up

South come solve this

[spiral lodge]

Why are you challenging people?

[young mist]

x2⋅t=x2+x

[spiral lodge]

Use ^2 for the squaring

Or even better ² if you have it on your keyboard

[young mist]

x^3=x^2+x

[obsidian harness]

you shouldn't use x again

use a different letter

[young mist]

i wii use p

p^3−p^2−p=0

[hollow lily]

show that 0 is a solution, then in the case x is not 0, devide both sides bys sqrt{x} to get quadratic

[young mist]

p(p^2−p−1)=0

p^3-p^2-p=0

then use the quadratic formula to solve

[slim plinth]

a≠0

[dark sparrow]

this is very rude.

[young mist]

whats very rude

[grave pond]

Pretending he can order helpers around to race for completing his tasks.

[shut thorn]

acosA-bsinA = c. Prove that asinA+bcosA = + or - sqrt(a^2 + b^2 - c^2)
how do i even. I dont even know where to begin. Doing from my older siblings book and its written that its similar to the question
"if sin^2(A) + sinA = 1. Prove that cos^4(A) + cos^2(A) = 1."
My mind is still blank

do i square? or somethign

[hollow lily]

take a look at what problem asks, it has sqrt and square in side the sqrt so I would say yes

[shut thorn]

alr lemme like

[hollow lily]

Square both sides then take sqrt seems like a good way

[shut thorn]

i see a bunch of sinA and cosA

i have a feeling pythagorean identity is goanna be used.

but like

how

the hell

do i

what if i just like

isolate a, b and c. And then take their squares and then like.

[hollow lily]

yeah, use it twice and you'll get the answer

[shut thorn]

how tho

oh also random but not random question

can i like

write -xsinA as -x*(-sinA)

no wait

thats a stupid question now that i think of it

😭

[hollow lily]

What did you get after squaring both sides

[shut thorn]

a^2(cos^2(A))+b^2(sin^2(A)) -2absinAcosA

= c^2

the variables are different

cant factor a and b

[hollow lily]

using pythagorean, cos^2A = 1 - sin^2A

[shut thorn]

a^2 -a^2(sin^2(A))+b^2(sin^2(A))-2absinAcosA

[hollow lily]

Do the same for b^2(sin^2(A))

1- cos^2A = sin^2A

[shut thorn]

b^2 - b^2(cos^2(A))

oh also

doesnt that like

put you back where you were

[hollow lily]

no, coefficient of cos A is now b

and for sinA is now a

[shut thorn]

ye

[hollow lily]

Exactly what we want here

[shut thorn]

a

sa

ah

[hollow lily]

asinA+bcosA = + or - sqrt(a^2 + b^2 - c^2)

[shut thorn]

what if i

a^2 - a^2(sin^2(A)) + b^2 - b^2(cos^2(A)) - 2absinAcosA
what if i like. a^2 + b^2 - a^2(sin^2(A)) - b^2(cos^2(A)) - 2absinAcosA

then i

-( a^2(sin^2(A)) + b^2(cos^2(A)) + 2absinAcosA)

which is uh

(asinA+bcosA)^2

oh wait wait

OH YES

a^2 + b^2 - (asinA+bcosA)^2 = c^2
-> - (asinA+bcosA)^2 = c^2 - a^2 - b^2

yikes

that took some brainpower. Now i must sleep.

[shut thorn]

you can also use middle term splitting and i must say that its far much easier.

wait no?

yeah it cannot be factored using middle term splitting.

mb

[sudden folio]

trig

[mellow moth]

(and geometry)

[verbal matrix]

just passed my trig final with a beautiful 97 🙏 now to move onto calc

[wanton yacht]

Woooo

[slim plinth]

I notice Law of Sin looks like pythagorean theorem

Really

a^2= something

b^2= something
c^2=something

[violet oasis]

?

[slim plinth]

Wait

I meant Law of Cos

[violet oasis]

Oh okay makes sense now.

[slim plinth]

Does it look like pythagorean theorem?

[verbal matrix]

theresactually a fun reason behind that. when the angle is a right angle, then cos(theta) = 0

[violet oasis]

It is an extended version of Pythagorean theorem.

[verbal matrix]

it is the pythagorean theorem (but big)

[slim plinth]

Also using (a+b)^2 I assume

[violet oasis]

Pythagorean theorem only works when $\theta = 90\degrees$.

[somber coyote]

Restarter
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[slim plinth]

And also how come Law on Sin is a proportion

[violet oasis]

Law of Sines is basically comparing the sides of the triangle with its altitude.

[upper karma]

Is this 88?

Or 53

Its not exterior angle thereom since that needs the shape to be defined as a triangle

[exotic yarrow]

In the future, please use the ,rotate command.

,rotate

[exotic yarrow]

,w find 180-(5x+10)-(12x+32) if 5x+10=3x+20

[somber coyote]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[exotic yarrow]

this

[upper karma]

Why is exterior angle theorem used if its not defined as a triangle

[exotic yarrow]

you could use the exterior angle theorem, but you don't have to

I just did alternate interior angle spam

[upper karma]

Oh

[shut thorn]

god damn i was breaking my neck in order to see the picture and just forgot to scroll down a bit 😭

[exotic yarrow]

[shut thorn]

i dont have nitro so i just made a photo

[exotic yarrow]

mood

me going to disc emoji downloader whenever I need to rip an emoji

[upper karma]

If im given 2 angles in a quadilateral, can i find x using only the 2 angles

But i know its a parallelogram

[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]

NA II AL

Would i just double both

[shut thorn]

how are they both parallel?

shouldnt it be NA||LO?

or is ANOL a ||gm?

[upper karma]

Thats whats given

[shut thorn]

ok so that means that 4x^2 - 13x + 122 + x + 58 = 180. (co-interior angles)

bring that 180 to LHS

[upper karma]

From theres its just quadratic formula right?

[shut thorn]

no you can do it by factoring too

what do you get after simplifiying 4x^2 - 13x + 122 + x + 58 - 180?

[upper karma]

4x^2-12x+180-180

[shut thorn]

which is

[upper karma]

4x^2-12x=0

[shut thorn]

yes

now factor out common from both of them

[upper karma]

x^2-3x=0

[shut thorn]

no no

you can also factor out an 'x'

[upper karma]

3x=0

[shut thorn]

no.

from 4x^2 - 12x

you can factor out a 4x

4x*

[upper karma]

Ohh 4x(x-3)

[shut thorn]

ye

now 4x(x-3) = 0

right?

[upper karma]

So that would be the answer for <NAL or z

X

[shut thorn]

oh you have to solve for it?

[upper karma]

I don’t know, it leaves it vague

[shut thorn]

its an MCQ?

or what is the objective of the question?

[upper karma]

“Given the image below of a quadrilateral nola, with NA II AL, m<NAL

[shut thorn]

so find angle NAL?

[upper karma]

I think

[shut thorn]

oh also

NA literally cant be parallel to AL

they literally MEET IN THE FIGURE 😭

[shut thorn]

ok so for that lets continue from 4x(x-3)
here. We know that 4x(x-3) = 0. Which must mean that either 4x = 0 or x-3 = 0

because the answer of a product is zero if and only if one of their term* is zero.

variable?

idk i forgot the name.
i think its like a uhh. TERM

fuck

[upper karma]

X-3
0+3
X=3

[shut thorn]

or.

do the same with 4x=0

[upper karma]

It would just be undefined

[shut thorn]

no

it isnt.

divide both sides by four

[upper karma]

X=0

[shut thorn]

yeah so x = 0 or 3

now plug in x = 0 into 4x^2 - 13x + 122

and then do the same but plug in x = 3

[upper karma]

its prob 3

This questions was so stupid since we weren’t taught anything about quadilaterals yet

[shut thorn]

no not even that

it has two answers.

yeah it has two answers

br

[upper karma]

Thanks

[shut thorn]

np

[violet oasis]

Daily calculation of area of famous logo.

Day 1.

This is a blueprint of a logo of a famous company.

In the blueprint, there was an error and it needs to be fixed.

Find the error in the blueprint.

Hence, calculate the area of the shaded region with the corrected version of the blueprint.

[kindred brook]

Turkish exams will be funny here

[drifting kindle]

is trigonometry easy

[quiet vigil]

VERY EASY FR MAN 🥲

[shut thorn]

yesnt.

[shut thorn]

it's easy until IIt isn't.

[ornate kite]

bro i need help with this one questiion

wait what is a eculidean geometry

[quiet vigil]

Fr man

[quiet vigil]

Geometry used by euclid

Ig it involves mostly the basic.2d shapes

And angles

[sly urchin]

Can someone explain exsimilicenters to me in simple language?

[spring lion]

Help 🫠

[obsidian harness]

where are you stuck?

[spring lion]

Idk how to do irl problems at all

[obsidian harness]

didn't the videos teach you how to do right-angled trig though?

[spring lion]

I can do normal questions but when it's an irl one I just can't

So um

[spring lion]

Is this sine?

[obsidian harness]

no

which length of the triangle do they want you to find?

[spring lion]

Um

[spring lion]

Not h?

Oh wait

[spring lion]

Here?

But

Idk

😭

Cos then?

Cos 37??

Well inverse cos 37.

Is this also false? 💀

[spring lion]

Did I at least mark the side right?

[orchid rampart]

sin

[spring lion]

Bruh he said it wasn't sin tho

[orchid rampart]

opp/hypotenuse

oh right mb

cos

[spring lion]

Okay so

Cos37

Do I take the inverse of it?

[orchid rampart]

no

let the distance to be found is x
cos37 =x/2.1

[spring lion]

Why is that?

[orchid rampart]

cos = adjacent/hypotenuse

[spring lion]

Okay

Soh cah toa

[orchid rampart]

yeah

[spring lion]

And

x is

Right?

[orchid rampart]

substitue value of cos 37

[orchid rampart]

yes

[spring lion]

So um

I multiply cos37 by 2.1?

[orchid rampart]

yes

[spring lion]

Okayyy

Omg it's right

Ok

Nahhhhh 😭😭😭 how in the hell am I supposed to solve these???

Is this tan?

Tan(6.6/10)?

Nvm

[spring lion]

This is opposite/adjacent though no???

Wow um...I did it?

It was inverse tan!!

[orchid rampart]

yes

[spring lion]

So there's this

Let me think.

I'm gonna add a 3.5

At the end

When I finish the calculations

Right?

[spring lion]

Opposite/hypotenuse

Sin 23

Sin23=x/4

Sin23*4=x

1.56+3.5

5.06

Is this correct??

Did I make a mistake somewhere?

Oh it was false.

Welp...damn it.

It was cos???

😭 how tf did I mix that up???

Omg I'm dumb as hell...

Hell yeahhhh I did this!!!

Ok guys I need your help 🙏 I want to do this last question right.

[slim plinth]

Use SOHCAHTOA

[spring lion]

Right

Sooo

59°

But is it sin cos or tan...

Let me think.

[slim plinth]

Is it the shaded

[spring lion]

Tan?

[slim plinth]

Or the grey part

[spring lion]

Wdym?

[slim plinth]

Is the whole right triangle

[spring lion]

Opposite/adjacent no?

[slim plinth]

Yes

[spring lion]

Oh

Okay

One second

[slim plinth]

tan(59)=5/x

[spring lion]

Yeah yeah

Wait

Not 59?

[slim plinth]

Just swap 5/tan(59)

[spring lion]

Ohhh

Right right

Sooo

I'll go with 31

[slim plinth]

Oops

[spring lion]

Tan(31)*5=x

[spring lion]

Is this correct?

[orchid rampart]

nah

[slim plinth]

I thought it’s 5/tan(31)

Yea

[spring lion]

Okay can you explain why? 🧐

[slim plinth]

Then you multiply the x to tan(31) and divide tan(31) into 5

[spring lion]

Isn't this green angle 59?

Why not go with tan 59?

[orchid rampart]

you can

tan 59 = x/5

[slim plinth]

Where did you get the tan 59 from

[spring lion]

90+31+59

[slim plinth]

Oh

How about the 4

Is it the total triangle if I’m correct

[spring lion]

4?

[wide jetty]

Why is it tan 59? Isn’t it tan 31?

[slim plinth]

Oh it’s a typo

[wide jetty]

Oh

[slim plinth]

That’s what I wonder too

[wide jetty]

lol

[spring lion]

Damn

I made a mistake then

But can y'all explain why?

[wide jetty]

Uhhhh

[orchid rampart]

plug 59 instead of 31

[spring lion]

I mean 31 isn't adjacent, it's hypotenuse

[spring lion]

.

OHHH or wait

We do the

Wait let me draw

[wide jetty]

Isn’t that the answer

[orchid rampart]

yes

[slim plinth]

Use tan 31 =5/x and you get tan31x=5 and divide tan 31 to the 5 to get x=5/tan(31)

[spring lion]

I'm trying to understand how 😭

[wide jetty]

Ohhhhh

[slim plinth]

Use your calculator

Make sure it’s degree mode

[spring lion]

One second

[spring lion]

So from this

[slim plinth]

Okay

[spring lion]

Are we not trying to find the yellow dotted line?

[slim plinth]

Just ignore

Once we get the angle

Use Pythagorean theorem

[spring lion]

5/tan31 let me see

8.32

[slim plinth]

To solve its hypotenuse

[wide jetty]

I don’t think so… we should the finding the line like that’s in the ground (fully)

[slim plinth]

Yes

≈8.32 m

[spring lion]

The line in the ground is 8.32?

[wide jetty]

Yeah… I think

[slim plinth]

Okay we can now use Pythagorean theorem

[spring lion]

Right

[slim plinth]

a=5, b=8.32, c=?

[spring lion]

5²+(8.32)²=hypotenuse²?

[slim plinth]

Yes

[spring lion]

Yeah

,calc sqr(8.32)

[somber coyote]

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

[spring lion]

Damn

[slim plinth]

This is my first time tutoring you geometry

[spring lion]

How do I use the bot here?

[slim plinth]

,calc (5)^2+ (8.32)^2

[somber coyote]

Result:

94.2224

[spring lion]

Ohh

[slim plinth]

Take their sqrt root

[wide jetty]

Gng why do we need the hypotoneuse

[slim plinth]

,calc sqrt(94.22)

[somber coyote]

Result:

9.7066987178958

[spring lion]

9.70

[slim plinth]

Length on the yellow line

[spring lion]

Okay.

[spring lion]

Yellow dot

Line

So the answer is 9.70?

[wide jetty]

I think the question means the horizontal along the line

[slim plinth]

So ans is he slept in 8.32 m

[spring lion]

Hmmm

[slim plinth]

8.32 m is the final ans

[wide jetty]

Yeah

[spring lion]

.

[slim plinth]

Yes

[spring lion]

Ahh

Okay okay

[slim plinth]

Yep

[spring lion]

The yellow dot wouldn't make sense lol

[slim plinth]

Yes

Did it mark correct

[spring lion]

It did!

Thank you all man

[wide jetty]

Yay

[spring lion]

I did 2 on my own 2 with help that's not bad

[slim plinth]

What’s the work

[wide jetty]

Ye

[slim plinth]

That khan wants

[slim plinth]

Friend me and I’ll try to assist you

In dm

[spring lion]

Okay sure

[spring lion]

I'm just doing all of math

Started from 6th grade

[slim plinth]

Ok

[slim plinth]

Btw I’ll majority will be in algebra section

Help people that is in trouble of algebra

[ornate linden]

can anyone help prove LHS=RHS

[spring lion]

Okay

[slim plinth]

Use sin^2(x)+cos^2(x)=1

Try that

[orchid rampart]

try using 1+ tan^2 θ= sec^2 θand 1 + cot^2 θ = cosec^2 θ

[ornate linden]

ik the identities, i did use em but it got very confusing

[orchid rampart]

show your work

[ornate linden]

wait

i cant rn

ill show in a while

[slim plinth]

How about common base

Use (1+tan^2(theta))(1+cos^2(theta))

[orchid rampart]

mm.. that will be messy

[sudden folio]

trig trig trig trig

[shut thorn]

our trig has hints on exercise questions

and one of the hints is just

"see example 3"

like what

😭

just show it here br

[slim plinth]

Can you foil that

[devout gorge]

start by putting 1+tan^2(theta)=sec^2(theta) and 1+cot^2(theta)=cosec^2(theta)

[acoustic walrus]

Can someone help me

[foggy swallow]

its quite simple basically all you need to do is

first convert into sin cos [as the rhs is in sin cos form]

then solve the fraction

you will end with something like sin4 + cos4/sincos

convert the sin4 into [sin2]2 and same with cos

and use the identity

and the answer should come

[hint separate the fraction at the end]

[ornate linden]

ohh alright thankss!

[ornate linden]

thank you

[livid ruin]

Anybody up for helping me please?

[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 echo]

You can use Menelaus' theorem for ABC intersected by DE and the fact BD=BE.
https://en.wikipedia.org/wiki/Menelaus's_theorem

[livid ruin]

we have not been tought that yet in grade 10

[upper echo]

just look up its proof and apply it to your problem. It fits grade 10 pretty well

[livid ruin]

I tried the solution. Is this correct tho?

[upper echo]

I would draw CG parallel to AB.

[livid ruin]

But the same concept right?

[upper echo]

because you don't need DF in your formula

[livid ruin]

Where is DF?

[upper echo]

yep, drawing parallel line and using similarity is a standard way to prove Menelaus

[upper echo]

in your intermediate formulas

[livid ruin]

basic proportionality theorem

[ocean zodiac]

What is the least amount of equations needed to contruct a perfect donut in a 3D space?

[livid ruin]

This is the improved version @upper echo

[livid ruin]

Bruh

[orchid rampart]

1

[ocean zodiac]

Can you write it?

[hot gazelle]

I have been struggling to do the last geometry problem, i need some assistance and ideas to keep going but i have found nothin so far.

Given a circle (O; R) with diameter AB. Draw the tangent line Ax to the circle (O). Let C be an arbitrary point on the circle (O) such that AC < BC. The tangent to the circle (O) at C intersects Ax at D.
​a) Prove that the four points O, A, D, C lie on the same circle (are concyclic).
​b) The line passing through O and perpendicular to BC at M intersects the line DC at E. Prove that OM \cdot OE = R^2 and that BE is a tangent to the circle (O).
​c) AM intersects the circle (O) again at a point N (N \neq A). Let K be the midpoint of EM. Prove that B, N, K are collinear

If anyone is free, i will be very appreciate to hear some thoughts about the problem, thanks.

[shut thorn]

and for the concyclic paret

part

[upper echo]

that's obvious

[shut thorn]

fair enough

[hot gazelle]

Thank u so much 😄 i think i did overcomplicate the problem a bit 😅

[shut thorn]

if cosecA - sin

fu

wait

if cosecA - sinA = a^3 and secA - cosA = b^3
prove that a^(2)b^(2)[a^(2) + b^(2)] = 1

my mind is blank.

i tried to like. simplify cosecA - sinA and secA -cosA into cos^2(A)/sinA and sin^2(A)/cosA repectively

then took the cbrt

then like.

whta do i do

[orchid rampart]

you mean this?
If $\csc A - \sin A = a^3$ and $\sec A - \cos A = b^3$, prove that $a^{2}b^{2}(a^{2} + b^{2}) = 1$.

[somber coyote]

oppenheimer

[shut thorn]

ye

wait csc means cosec?

[orchid rampart]

yeah

[shut thorn]

huh

[orchid rampart]

go on

substitute

[shut thorn]

after substituting and simplifying. It becomes (sinAcosA)^(2/3) [(cos^2(A)/sinA)^(2/3) + (sin^2(A)/cosA)^(2/3)]

ok no i got it

why was i worrying about this again 😭

[orchid rampart]

👍

[shut thorn]

a^(2)*b^(2) = (ab)^2 right?

i forgot basic identities and have to double check

😭

ye

[orchid rampart]

yes

[upper echo]

Express a^3 and b^3 in terms of cosA and sinA and divide them. So, you get b=a*tan(A). Then just substitute it into that expression with a,b

[violet oasis]

Yeah they're the same thing.

[shut thorn]

I mean I did it from aan easier way bubut this works? Thanks ig

[shut thorn]

HHuh

[sly urchin]

Given quadrilateral ABCD. E,F are midpoints of AC,BD accordingly. Prove AB+BC+CD+DA≥AC+BD+2EF

[orchid rampart]

Yeah

[shut thorn]

alr br how do you even. 😭

YYe but

how does one decide that from one. ONE question

😭🙏

[shut thorn]

dk tell me that the next time I see u

nah I'm like.

saving the link to this message so I ping you with this message whenever I see u online.

[spring lion]

I'm trying it

-2, 9

Radius is 4?

Or 2?

[next mantle]

All wrong

[spring lion]

💀

[next mantle]

jk

[spring lion]

What do you think?

I think radius is 2?

Ok it was right

How's this?

(x+2.3)²+(y)²=36/49?

[next mantle]

looks right

[spring lion]

Ok

Ok ty

[true sluice]

Find an example of two triangles $ABC$ and $XYZ$ such that $AB : XY = BC : YZ$, $\measuredangle{BCA} = \measuredangle{YZX}$, but $\triangle{ABC}$ and $\triangle{XYZ}$ are not similar. (EGMO by Evan Chen, problem 2.2)

[somber coyote]

OSHOHAY_BALOK_:(

[true sluice]

I can't find any

[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 echo]

I don't think it's too hard

[long geyser]

I think I overcomplicated this too much :((( I was trying to visualize a 120 degree rotation but I couldn't, ultimately only a 360 degree rotation would be a rotational symmetry because one of the side lengths isn't the same as the others. I spent like 30 minutes trying to solve it.

I suppose that one way to imagine this for myself intuitively in hind-sight is that I know that if a square isn't turned in divisions of 4 of 360 degrees or just respectively 90 degrees 180 degrees 270 degrees and 360 degrees that it won't be symmetrical, and I think that an isosceles triangle shares at least 3 vertices with a square right? So if it's not turned like that then it won't be symmetrical and because 180 doesn't work because the vertex of the right angle or whatever it's called wouldn't have another vertex where it would switch places with to become symmetrical.

[lone panther]

alternatively, identify the corner that has two sides of equal length attached to it
theres only one kind of point like that in an isosceles non-equilateral triangle
if it had any rotational symmetry, there would be more of those points, which isnt true
or, that point is the center of rotation, which when rotated shows the shape has no rotational symmetry whatsoever

[long geyser]

Thank you, yeah!

[lone panther]

np

[long geyser]

For an isosceles equilateral triangle, would rotating it 60 degrees by any multiple such as 60x2 make it rotationally symmetric?

[lone panther]

120 degrees

not 60

[long geyser]

I see, thank you! 🩷

[lone panther]

np

[long geyser]

Yeah that makes sense!

I assume a shape with 6 vertices or just a hexagon would be able to do that?

[lone panther]

yea

[long geyser]

Wow, thank you. That's really cool!

[sterile beacon]

If you shear a parabola, does it always remain a parabola?

[exotic yarrow]

yes

[sterile beacon]

That is so weird

Seems like no linear transformation can transform a parabola into a non-parabola, except the obvious edge cases of scaling by a factor of 0, or some kind of "infinite shear"

[exotic yarrow]

You can verify it algebraically tbf

[true sluice]

Wait

Here $\frac{BC}{YZ} = \frac{3\sqrt{2}}{2}$

[somber coyote]

OSHOHAY_BALOK_:(

[upper echo]

Nope, just count the cells

[true sluice]

eh what

Shouldn't I use the distance formula from coordinate geo

[upper echo]

BC is a diagonal of a square 6x6 cells, YZ is the diagonal of 4x4. So, BC/YZ=6/4=3/2

[true sluice]

I saw

But what about the distance formula

that shows this

[true sluice]

BC= 6\sqrt{2} and YZ=4

[upper echo]

Ok, use it, and get the same ratio

[upper echo]

YZ=4sqrt(2)

[true sluice]

It shouldn't be this simple😭

[true sluice]

wait, wth is wrong with me, I just added 2 and -2 instead of subtracting

tnx and sry

but yet

[true sluice]

They said Directed angle will fail here, that means normal angles must work

but you used 45 in both cases

Whatever tnx

[upper echo]

I don't know, there was nothing like that in the statement.

[true sluice]

not there, while stating similar triangles properties in the previous section of the book

Well it did use directed angles

[true sluice]

Oh it worked maybe bczz its not SAS but SSA

[quiet vigil]

Ssa is a thing!?

[obsidian harness]

SSA is not true in general

but when you have a >= b, then SSA really does imply congruent

[quiet vigil]

Can I get an example of such case?

[obsidian harness]

the picture explains it

sorry it's a bit small though

[quiet vigil]

Oh wait yeahh now I get it , ty :)

[quiet vigil]

Nah np :)

[edgy phoenix]

$$
\text{The vertices of a triangle are }
A(x_1,; x_1\tan\alpha),;
B(x_2,; x_2\tan\beta),;
C(x_3,; x_3\tan\gamma).
$$

$$
\text{If the circumcentre of } \triangle ABC \text{ is the origin and }
H(a,b) \text{ is the orthocentre, then } \frac{a}{b} \text{ is equal to:}
$$

$$
\text{(A)}\quad
\frac{\cos\alpha+\cos\beta+\cos\gamma}
{\cos\alpha\cos\beta\cos\gamma}
$$

$$
\text{(B)}\quad
\frac{\sin\alpha+\sin\beta+\sin\gamma}
{\sin\alpha\sin\beta\sin\gamma}
$$

$$
\text{(C)}\quad
\frac{\tan\alpha+\tan\beta+\tan\gamma}
{\tan\alpha\tan\beta\tan\gamma}
$$

$$
\text{(D)}\quad
\frac{\cos\alpha+\cos\beta+\cos\gamma}
{\sin\alpha+\sin\beta+\sin\gamma}
$$

[somber coyote]

BlackidoZΣ

[edgy phoenix]

a/b is equal to?

[exotic yarrow]

Recall that for a triangle $ABC$ with circumcenter at the origin, we have that $$\overrightarrow{OH}=\overrightarrow{OA}+ \overrightarrow{OB}+ \overrightarrow{OC}.$$

[somber coyote]

Civil Service Pigeon

[spring lion]

Help

Prism is base area times height

Let me think

1/2*49π

*20

24.5π

*20

490π

LOL I JUST SUDDENLY GAINED INSPIRATION AND SOLVED IT ON MY OWN NVM

[spring lion]

Radius is 4

Height is 10

[next mantle]

Peak 🗣️ 🔥

[spring lion]

1/3*3.14*4*10

Right?

Times 3/4

Dangit I got it wrong...

Rip.

I was afraid they'd ask real life problems

And they DID.

I have to overcome this...

[spring lion]

How tf...

Do I just add the values together here?

1/3bh

Base is

,calc 64*64

[somber coyote]

Result:

4096

[spring lion]

4096+(?)

Or wait.

MULTIPLY?!

But there are 2 pyramids

Uhhh

[twilit scaffold]

uhh yuo should js use formula find overall pyrmaid and just subtract the other pyramid

[spring lion]

Oh

Let me see

4096*48?

(Times 1/3)

[twilit scaffold]

so dyk forumla?

so its

[spring lion]

Yep

[spring lion]

.

[twilit scaffold]

1/3 times base tims height

[spring lion]

Yeah

[twilit scaffold]

yeah and then

[spring lion]

Ik

Base area is 4096 no?

[spring lion]

.

[twilit scaffold]

shld be?

[spring lion]

Ok

[twilit scaffold]

4096 cos 64^2

[spring lion]

Multiply by 48?

[twilit scaffold]

yes

[spring lion]

Ok

[twilit scaffold]

and then subtract the other pyramd

[spring lion]

Wait wait

I first gotta divide by 3 no?

,calc 4096*48

[somber coyote]

Result:

1.96608e+5

[spring lion]

Bruh

196608

[twilit scaffold]

uhh

lemme checl

[spring lion]

Check this

[twilit scaffold]

no..?

so volume u js do

4096

minus the other pyramid

[spring lion]

But 1/3?

[twilit scaffold]

wait leme ge my calv

calc

[spring lion]

Plus you gotta multiply by height.

[spring lion]

Ok

,calc 16^2

[somber coyote]

Result:

256

[twilit scaffold]

so

[spring lion]

,calc 256*12

[somber coyote]

Result:

3072

[twilit scaffold]

you get 65536

then

[spring lion]

Yupp

[twilit scaffold]

u subtract

[spring lion]

3072/3

[twilit scaffold]

12^2 times 1/3 times 12

yeah

[spring lion]

,calc 3072/3

[somber coyote]

Result:

1024

[spring lion]

,calc 65536-1024

[somber coyote]

Result:

64512

[twilit scaffold]

64960

isnt it

[spring lion]

Huh

[twilit scaffold]

1/3 times 12^2 times 12

[spring lion]

16

[twilit scaffold]

oh crao

im blind

[spring lion]

Lol dw

[twilit scaffold]

you get 63936

[spring lion]

Really?

[spring lion]

.

[twilit scaffold]

65536-(16^2x12x1/3)

[spring lion]

16*16*12/3

Lol one of us made a mistake

One sec

[twilit scaffold]

wait mb cuh i put in calc wrong

i js studied for too long i cant read rn

[twilit scaffold]

you should be write

[spring lion]

Dw gng

[twilit scaffold]

right*

cant even spell yo

[spring lion]

Ok o

K

[twilit scaffold]

and that shld be answr

[spring lion]

Lmaoooo

[spring lion]

Yeah man thank you.

[twilit scaffold]

algs man take care

[spring lion]

It was right

[twilit scaffold]

eyy lesgo

[spring lion]

👍

36^2

Times 36

/3

15552

Umm

15552*0.75?

DAMN it was wrong

RIP

Ahhh I want to be able to do it on my own man

Ah damn. I found 11664 right but

I forgot one calculation...

😭

1/3*10^3

333,33...

2 cm from the top...

So is it 8 cm??

[spring lion]

I found 120

But it's false???

333-213 bruh...

Oh damn. The length was also 8...but why?

[next mantle]

You mean the side length of the base square?

[spring lion]

Yes

The square side length was 8

[next mantle]

Well try visualizing it

how could the side length of the base square be the same?

Try imagining you filling up the inverted pyramid with water

and stopping at 8 cm

[spring lion]

Oh

Ok

I just wanna get this one right on my own what do I do?

Don't help yet

[spring lion]

So I found a 54.

There's a semi-cylinder...

What the hell is the radius here??

3?

Damn

[silent plank]

what's your definition of radius of a circle

[spring lion]

Distance between center and a point

Of the circle

[silent plank]

yeh

[spring lion]

Ok

[next mantle]

It's also half the diameter right? So if you can find the diameter, radius is just half of that

[spring lion]

Yeah

4.5?

My last guess

[silent plank]

don't guess

[spring lion]

😭

[next mantle]

stop guessing

[spring lion]

Bro 9/2

[next mantle]

just look at the picture and think 🙏

[silent plank]

why 9?

[next mantle]

How is 9 the diamater of anything 🗿

[silent plank]

9 doesn't have anything to do with the (semi)-circle

[next mantle]

Just take it slow man

[spring lion]

Idk

[spring lion]

Right

[silent plank]

focus on this part of the pic

[spring lion]

Is that 3

[next mantle]

Is what 3?

[spring lion]

The side length

[next mantle]

You're asking about this red line on top?

[silent plank]

if you mean the top side of the rectangle,
the length of that line (that i've highlighted in red) will be 3m
(note that this is not the radius of the (semi)-circle)

[spring lion]

Ok

So at least I got the 3 right but how do I find the radius?

[silent plank]

it may help if you complete the full circle
and identify what type of line you've just determined the length of

[spring lion]

Is 3 the diameter then...?

[silent plank]

yes

[spring lion]

Ahhhh

Radius 1.5

Ok

πr²h

π*2.25*2?

[silent plank]

no

[spring lion]

/2

[silent plank]

no

[spring lion]

Semi cylinder

[spring lion]

😭

[spring lion]

Why not

[silent plank]

well you'd want to divide by 2 because you have a semi-cylinder
but you made a mistake before that

π * 2.25 * 2
why multiply be 2 there?

i.e. why did you use 2 for the height?

[spring lion]

OHHH

that's

[silent plank]

has got nothing to do with the semi-cylinder

[spring lion]

Not

😭

Ok

How do I find the height?

We only have a 1.5

That is the radius

Idk what numbers we can go off of rn.

[next mantle]

The height is given in the picture

[spring lion]

🤨

[silent plank]

don't think of "h" in formulae as vertical

[spring lion]

I'm blind

Idk boss

[silent plank]

consider the direction you'd extend that semi-circle at the front
to get to the other semi-circle

[spring lion]

Um

[silent plank]

and draw something like an arrow

yes

[spring lion]

9??

[silent plank]

yes

[spring lion]

I could NEVER GUESS THAT IN A MILLION YEARS 😭

How did you think that?

[silent plank]

you shouldn't be guessing

[next mantle]

It's not about guessing

[spring lion]

It would never cross my mind man

[silent plank]

its about recognising what each variable represents in formula

[spring lion]

Ok

I guess I just need practice

Because I couldn't know that

Soo

π(1.5)²*9?

[silent plank]

and don't forget to divide by 2, because you have a semi-cylinder

[spring lion]

Then divided by two

[spring lion]

Yes

Can I do the calculation with this bot? Let's see

,calc π*(1.5)²*9

[somber coyote]

The following error occured while calculating:
Error: Syntax error in part "²\*9" (char 8)

[spring lion]

Rip

[next mantle]

I think pi is just pi

,calc 2 * pi

[somber coyote]

Result:

6.2831853071796

[spring lion]

Oh

[next mantle]

not the unicode pi

[spring lion]

Do I need a ?

\

[next mantle]

No

[spring lion]

But

It'd be like

this

[next mantle]

Just try in #bots see what works

[spring lion]

Eh I'll just google the calc

One sec

Ok so

31.80

[spring lion]

Don't 🗿 me bruh

Fine I'll do it there as well

Result:

31.808625617596

There we go

[spring lion]

Is the rectangular prism

54 at least?

I can't believe I'm failing miserably at this 😭

[next mantle]

You're not failing, you're learning

[spring lion]

Ok

[spring lion]

So do I just add these two together?

[next mantle]

Ya

[spring lion]

Ok

54+31.80

85.80

Round

So

Ok sheesh.

I'll take a 30 min break.

[spring lion]

Bro this is so hard 😭

I'm absolutely cooked I can't solve these problems...

So...base area times height

1.2*x/6

?

Wait wait wait.

1.2x/6>1.2y/4?

[spring lion]

If this isn't the right way then idk.

It's definitely incorrect...yeah.

[opal raptor]

yo

need a hand?

[spring lion]

Hi

[spring lion]

Yeah

[opal raptor]

yeah alright

let’s see

[spring lion]

I always suck at problems

[opal raptor]

alright, let’s start simple and define variables.

[spring lion]

Ok

[opal raptor]

1.2m^2 base, 6 fish, 0.25m^3 per fish

would you agree?

[spring lion]

Yep

[opal raptor]

alright.

[spring lion]

Wait

How did we find 1m^3 per fish

[opal raptor]

Oh wait hang on

i misread that

whoops haha sorry

[spring lion]

Dw

[opal raptor]

We have $6$ fish, and we need $0.25m^3$ per $1$ fish.\
In this case, how many $m^3$ are needed for $6$ fish?

[somber coyote]

WanderingFurr

[opal raptor]

(i’m still new to TeXit so sorry if i screw smth up lol)

[spring lion]

1.5?

[opal raptor]

alright, how’d you get that?

[spring lion]

I found 1.5 already but not sure how to use it.

[spring lion]

.

[opal raptor]

One moment.

[spring lion]

This gave x>1.5y

[opal raptor]

Ok, so we now know that we need $1.5m^3$ of water.

[somber coyote]

WanderingFurr

WanderingFurr

[opal raptor]

So, how do you find volume?

[spring lion]

Base area times height

[opal raptor]

Now we’re cooking.

So we have..

$1.2 \times h = 1.5$

[somber coyote]

WanderingFurr

[opal raptor]

Think you can solve for h?

[spring lion]

Give me one sec

Ummm

My brain got fried 😭 sorry I'll solve it

[opal raptor]

It’s fine haha

Take your time

[spring lion]

1.25?

[opal raptor]

I would believe so

[spring lion]

Ok it's 1.25.

Is that the height then?

[opal raptor]

And the unit of measurement is..?

[spring lion]

m³

[opal raptor]

Yep

So let’s work it backwards to check it’s right

[spring lion]

Ok

[spring lion]

Is it?

[opal raptor]

$1.2 \times 1.25 = 1.5$

[somber coyote]

WanderingFurr

[opal raptor]

I would say so.

[spring lion]

Okayyy.

[opal raptor]

A fish tank of base $1.2m^2$ can fit $6$ fish in with $0.25m^3$ per $1$ fish given that the water is filled $1.25m$

[somber coyote]

WanderingFurr

[spring lion]

Right.

[opal raptor]

That’s how you would answer, as this is a worded question

worded questions always require worded responses

so keep that in mind on exams

[spring lion]

So the answer is indeed 1.25 meters³?

[spring lion]

Okay

[opal raptor]

Not cubed, as we’re just talking about height here

[spring lion]

Right right

[opal raptor]

$1.25m$

[somber coyote]

WanderingFurr

[opal raptor]

should ultimately be your final answer

Got any other problems to do?

[spring lion]

Bro fuck me the site gave an error 😭

So it resetted into a different question

[opal raptor]

the fuck-

[opal raptor]

bruh alright let’s hear it

[spring lion]

It happens a lot with me

[spring lion]

Ok

[opal raptor]

uh alright

What formulae will we need?

[spring lion]

Give me one sec

[opal raptor]

Take your time.

[spring lion]

1/3bh?

[opal raptor]

-# still trying to solve it myself..

[spring lion]

Pyramid volume

[opal raptor]

Yeah alright

Anything else or you think we good?

[spring lion]

Ok 👍

Let me see.

Yeah I think we good it doesn't mention any other shapes!

[opal raptor]

So

What info do we have?

[spring lion]

Let me think.

Base area 120 cm².

[opal raptor]

I already compiled a small list but I want to see what you find

[spring lion]

Height=x