geometry-and-trigonometry

[dark sparrow]

"sec is cos" is misleading at best, or just false at worst

well if you meant reciprocal then you have to say reciprocal lmao.

sec(x) = 1/cos(x)

but you cannot just drop the "1/..."

[wispy jewel]

Still yet to understand Maths English, My teacher told me it's a whole different playing field 🤦‍♂️

[upper karma]

try using properties of triangles

[lime dune]

"maths english" no this is just "basic communication"

say exactly what you mean and mean exactly what you say

[faint pasture]

are A B and C angles of a triangle? it wasnt mentioned anywhere-

[wispy jewel]

So...

[lime dune]

there's nothing particularly deep to explain, it's just plugging into the definitions of your trig ratios

[upper karma]

huh lemme see

this question is from "properties of triangles" topic

oh a, b and c are sides of a triangle

[faint pasture]

💀 shoulda said so- yea so the problem is solved now if i consider it as an angles of a triangle

[upper karma]

and A is the angle opposite to side with length a

and B is the angle opposite to side with length b

[faint pasture]

iseee

[upper karma]

why tf do they make these questions ambiguous

[faint pasture]

tbh they gave us enough context imagine if they had to write everything there book would like a billion pages

[upper karma]

I realised this after looking at the solution

[upper karma]

they're just lazy 😔

[faint pasture]

lmao

ill try it again a bit later

[upper karma]

alr

[fast oriole]

forgot mos stuff

so i need recap

nvm

[strong hearth]

How?

[fast oriole]

changed question

[silent plank]

what's your issue here

[fast oriole]

idk anything about it

[silent plank]

look up geometry notation

also wacth their video guide

[dark sparrow]

apply difference of squares to num and denom separately, and you will get some tan(a+b) simplifications afterward.

[obsidian hornet]

Hi everyone, I would like to ask about one type of problem associated with power of the point, when you can rotate chord in circle and get another 4 points that lie on one circle. For example, I have problem like this: given a parallelogram with configurations as in the picture and you need to proof that 4 points (2 vertices of the parallelogram, the bases of the perpendicular and the continuation of the median) lie on one circle. So, does anyone know problems with a similar idea?

[dark sparrow]

what kind of file is that?

can you reupload properly

[proud obsidian]

yea I cannot open the file

[faint pasture]

Yall might have just downloaded malware-

[dark sparrow]

i did not download it

i saw that it is a file with no extension, and for this reason alone i raised the suspicion

[obsidian hornet]

Sorry, give me a second

[proud obsidian]

here is his problem

nvm he sended it

[faint pasture]

Nice

[grave pond]

(I removed one of the copies)

[faint pasture]

How do i find the maximum and minimum in problems like this one generally?

The official solution is just hit and trial

But there gotta be a better way

And like how would i do it if i had to find the minimum instead of just its sign?

[nocturne remnant]

you can convert it into a geometry problem

[proven summit]

Is it postive

[proven summit]

Not in his syllabus

[nocturne remnant]

?

[faint pasture]

sure but why would that help

[faint pasture]

its negative

[proven summit]

are u familiar with inverse trigonometry

[faint pasture]

also the angles can be negative

[nocturne remnant]

actually nvm now that i think about it doesnt really work

yeah

[faint pasture]

not jee level but enough yea

[proven summit]

oh fuck it said minimum

For max what I did is put 2 of them as pi/2 one as 0

[nocturne remnant]

well you can just prove that it is negative at some point, then you've proved that the minimum must be negative (since the minimum must exist as the expression is lower bounded by -3)

[faint pasture]

yea but that was more hit and trial

[proven summit]

for minimum it could be
π+π-π

Does that satisfy

[faint pasture]

(for min the book did -pi/2, -pi/2 and 2 pi but why)

[proven summit]

oh wait it doesn't

[faint pasture]

thats 0

[proven summit]

yea

I have done a similar question

But of invese trigonometry

[faint pasture]

hmmm

[proven summit]

Least possible value (-1) yeilds at -π/2

[faint pasture]

this was in amit m agarwal

[proven summit]

Really good book

[faint pasture]

like sure but is there some way to defintely prove that this is the most optimal combination

[proven summit]

I just have the calculus one

[faint pasture]

yes the calc one is jod, trig one on the other hand is just pain

[proven summit]

With Domain range I think

[faint pasture]

can we?

like sure the sign determination is ez

[proven summit]

Ye

[faint pasture]

but finding the exact value

howd we do that

[proven summit]

Lemme see it

[faint pasture]

:D thanks

[proven summit]

I think

As it's alpha beta gamma

U can replace it with sin^(-1)

[faint pasture]

nope

[proven summit]

so the limited domain becomes from (-π/2 , π/2)

for all

[faint pasture]

cuz alpha beta gamma dont have to fit the range of sin inverse

[proven summit]

But it says sinA= x

then A could be sin^-1 x

[faint pasture]

it would be one of the infinite possible values of A right?

[proven summit]

sin^-1 has a restricted domain

[-π/2 , π/2]

Sorry range

[faint pasture]

ye so thats why we cant use it in this problem ig

[proven summit]

Restricted range

[upper karma]

Domain expansion

[faint pasture]

since even the official solution included, -pi/2, -pi/2, 2pi

so ye

[faint pasture]

jjk

[proven summit]

Else if u don't consider that there could be infinite values of a b c

[proven summit]

Boxing ring

[faint pasture]

yea but we just care about the case of a b c when the resulting value is minimum

[proven summit]

Resulting value of sina + sinb+ sinc

[faint pasture]

yea

sure we could write c = pi - b- a and get an eq in 2 variables but i dont see what i can do further

[proven summit]

not gonna work

[faint pasture]

hmmm

[upper karma]

this shit is so confusing

sin(a+b) is sinAcosB + cosASinB

Sin C + Sin D is 2SIN(C+D/2)*cos(C-D/2)

[faint pasture]

remembering the formulas is like easiest part of trig, applying them is smthn else though

[upper karma]

i hate manipulations

fr

[faint pasture]

broo fr
try this if you havent tan(6)*tan(66)*tan(78)*tan(42)

like how am i supposed to think of that in an exam setting

[upper karma]

wait

its something related to

nvm

1

ig

[faint pasture]

yea 💀 what was you approach?

multiplying and dividing by tan54*tan18 ig?

[upper karma]

yeah

i think ivbe done this before

[faint pasture]

yea 💀 like how tf am i gonna think of that in an exam setting

[upper karma]

its some shitty

identity

[faint pasture]

our tchr gave that one to us in the class none of us solved it

[upper karma]

what grade

[faint pasture]

tan(a) tan(60-a) tan(60+a) = tan(3a)

[upper karma]

yea

[faint pasture]

11th

[upper karma]

im in 9th 💀

[faint pasture]

like trig can be fun but dont ask shit like that in an exam bruh-

[faint pasture]

th- @proven summit see this-

[upper karma]

i learnt trig when i was in class 8

like

identities

and all that

i forgot most of it

[faint pasture]

ahem lmaoooo nice, icse kid i assume?

[upper karma]

cbse

• kv

fucked up

franchise

[faint pasture]

bro tf lmao have fun

[upper karma]

i can solve trig equations

i struggle with application of trig

[faint pasture]

10th grade lvl?

[upper karma]

things that includes triangles

[upper karma]

i think 11th

10th is super basic

[faint pasture]

hmmmm solutions of triangles can be brutal sure

you will get thru it though you have plenty of time

[upper karma]

im only able to deal with questions including sin cos and stuff

when i get question with finding the length of the triangle

idk what to do

[faint pasture]

uh get some book read some illustrations then practice

[faint pasture]

true

[upper karma]

where can i use actual calculus?

[faint pasture]

@proven summit dont you ever dare pick on me again-

[faint pasture]

uh like have you done graphs?

and like proper coordinate geometry

[upper karma]

what?

[upper karma]

no

[faint pasture]

ignore that 💀

[upper karma]

who cares abt graphs

[faint pasture]

then you cant use calculus

main use of calculus is to extract information from graphs

areas tangents and stuff like that

[upper karma]

in physics

?

[faint pasture]

like in physics too you are basically finding the area when you are integrating a function

thats what integration fundamentally means

to find the area under the graph

[upper karma]

shit

i dont likee graphs

i dont like geometry

the whole thing is

[faint pasture]

💀 lmaoo thats why you dont just mug up formulae

[faint pasture]

i dont either but calc is fun

[upper karma]

ive mugged it all up

[faint pasture]

sure you probably might not know the significance of what you are doing

[upper karma]

in trigonometry

how tf am i supposed to know that i multiply the equation by fucking 3/2*2/3 to get the answer

[faint pasture]

but just the feeling of successfully solving an integral or a limit you spent hours on is satisfying

[faint pasture]

honestly get a tchr they would explain it much better than i can

[upper karma]

its embarassing 💀

[faint pasture]

no? thats where you are wrong

there are so many integrals you just cant solve in less than like 10 mins at least even if you know the idea behind them

[upper karma]

i cant solve integrals which includes more than 2 terms

[faint pasture]

wdym by 2 terms?

like you couldnt solve sin^3(x) cos^2(x) sec^2(x)cosec^3(x)?

[upper karma]

i dont remember trig functions

i do some

sin x is

[faint pasture]

dont rush.

[upper karma]

-cosx

wait

cosx is

-sinx

sec idk

[obsidian harness]

Nope

[upper karma]

maybe cosectan

[upper karma]

sinx?

[faint pasture]

nope

[obsidian harness]

(sin x)' = cos x

[obsidian harness]

yes

[upper karma]

yea

[faint pasture]

dont rush you got plenty of time-

you can slow down

[upper karma]

i remembered those two

cause i used them in differentiation

basic

idk others 💀

[faint pasture]

why is derivative of sin(x) equal to cos(x)

[upper karma]

never used

[faint pasture]

?

[obsidian harness]

Depends on what definition of sin x you're using

[upper karma]

idk it just does

[obsidian harness]

It follows pretty nicely from the series definition

[upper karma]

maybe graph

[faint pasture]

i am not asking as in for me, just asking so that he gets his concepts straightened out

[faint pasture]

nope

[upper karma]

sin/cos = tan

[faint pasture]

learn, dont mug up

[faint pasture]

💀 lmaoooooo

[faint pasture]

the complex number definition is prettier if you ask me

[upper karma]

inmaginary and real

is complex

simple

[faint pasture]

(but for him i was expecting him to use the basic limit defniition of derivative)

[upper karma]

i dont know limits

[obsidian harness]

ahhhh

[faint pasture]

and you know integrals, nice

[upper karma]

lim x => 0

[obsidian harness]

You definitely need to go over that again

[upper karma]

i forgot

[faint pasture]

you are moving too fast-

sure i get the feeling of wanting to be the best

[upper karma]

im in a valorant game

[faint pasture]

lmfao

[upper karma]

ok

now im free

lets talk

elderly talk

[faint pasture]

tf lmao

slow donw

get you basics clear

[upper karma]

wdym slow down

[faint pasture]

by basics i mean trig + limits

then try calc

if you have interest then continue

[upper karma]

i mug up shit

done

[faint pasture]

slow down as in dont go so fast that you complete entire calc in a week then next week you know nothing

[faint pasture]

you are fucked up beyond repair then-

[upper karma]

127.0.0.1

[faint pasture]

💀 what?-

why include localhost

[upper karma]

localhost

idk

[faint pasture]

send me your real ip

🗿

[upper karma]

1.618.69.420

[faint pasture]

what will i do with localhost-

[upper karma]

here

what do i need for ioqm

[faint pasture]

missile strike initiated

[faint pasture]

def not calc

[upper karma]

inter continental ballistic missile

wait i need water

[faint pasture]

idk i skipped all of calc and geo

number theory nd pnc are the simplest topics for olympiad math

but yea your choice

[upper karma]

i dont like geomtry

fucking

1000 rules

[faint pasture]

anyways you aint clearing ioqm by mugging up shit anyways

[upper karma]

on two fucking lines

[upper karma]

i can

i can solve some ioqm pyqs

[faint pasture]

us fr lmao like i can tolerate schl geometry but ioqm geometry fucking killed me

[upper karma]

whtg grde are u

[faint pasture]

give the latest paper as a mock

11th

[upper karma]

if i score 2 without any preps?

[faint pasture]

score 10 then we'll talk if you score 10 at home youd prolly get like a 5 there cuz of the stress

i was scoring in 20s in mocks

but in the real exam i got 15

so yea

[upper karma]

in 11th?

[faint pasture]

9th

[upper karma]

u got the certificate

where did u rep

prep

[faint pasture]

mmm

[faint pasture]

the math enthusiast in me wont let me tell that to someone who think maths can be done by mugging up-

[upper karma]

kids write shit in their bio to look cool

ffs

[upper karma]

i was kidding

i dont mug

[faint pasture]

💀 like i believe you?

[upper karma]

i know from where sin(a+b) came

[faint pasture]

doesnt matter :>

[upper karma]

tell

bro

[faint pasture]

sin(a+b) proof is weird enough anyways lmao

[upper karma]

u dont want me to progress

thats why u arent telling me

[upper karma]

yeah confusing

[faint pasture]

yea i dont- slow down

[upper karma]

pls tell

[faint pasture]

give the 2022 paper

tell me how much you score

then ill see

[upper karma]

ok

[faint pasture]

(2022 cuz i was in 9th when i gave it)

[upper karma]

i onkly find paper with answers

[faint pasture]

uh dont cheat ig?

if you cheat then whats the point

[modest garnet]

Hello

Anyone has some trigonometry questions

[worldly swan]

you need questions?

[modest garnet]

Yes please

Normal questions

[worldly swan]

https://www.scribd.com/document/259440520/HSC-Exam-Questions-Trigonometric-Ratios

you will find some here

[upper karma]

yeah

[proven summit]

What

[faint pasture]

that kid is doing calc in 9th grade- im still in 11th

[trail tendon]

i mean its not that hard

but also i wouldnt know if i would consider knowing how to solve basic integrals without knowing what limits are "doing calc"

i mean i guess technically

xD

[faint pasture]

its not ik 💀 but he was picking on me for liking it soo yea

[faint pasture]

💀 lmaooooo fr

[trail tendon]

when did he do that 🗿

[faint pasture]

a lot of times in #calculus

[obsidian harness]

If you spent as much time helping others as hanging around in discussy

You could be green by now

[faint pasture]

i dont even hang around in discussion that much 💀- i have like less than a 100 msgs there im sure

(also im dumb enough cant help others-)

and dont have enough patience to solve the easier problems

i help ppl in calc tho sometimes

[empty yew]

Is the domain of the equation $\frac{\sin\theta}{1+\cos\theta}$ $\mathbb{R}-{2n\pi\pm\pi}$?

[somber coyote]

KingDanger

[faint pasture]

that +- is redundant you could just use 2npi + pi
also yea

(hence why i said rare wolfram l)

[empty yew]

Ok

[obsidian hornet]

If you don't mind problems with geometric overtones, then I can suggest proving that the perimeter of an orthotriangle is 2Rsinasinbsinc, where R is the radius of the circumscribed circle of the original triangle and a,b,c-angles.

[faint pasture]

pa = ae, and ae = ba

[trail tendon]

is it just a bad picture?

[faint pasture]

yea

[trail tendon]

it says "regular" pentagon, doesnt that mean the angles/sides supposed to be

yeh

☠️

[faint pasture]

i mean it usually says the figures arent gonna be to scale

[trail tendon]

it looks like it was drawn by a 5 year old

or me

💀

[faint pasture]

lmaoooo my drawing would be worse

there is no way i could draw a straight line

[inner vapor]

how did you got that

[trail tendon]

oh yeah mb

[inner vapor]

[faint pasture]

equilateral triangle given

and regular pentagon

also given

[trail tendon]

ae = ba because they the side of the pentagon

and all sides are same

even though they cant draw

[faint pasture]

(you really need to work on your definitions before attempting amc)

[inner vapor]

ae=ba???????

[trail tendon]

regular pentagon means all sides of the pentagon are equal

ae and ba are sides of the pentagon

[inner vapor]

okay

,w 108-60

[somber coyote]

Results provided by WolframAlpha

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[upper karma]

screaming crying

[somber coyote]

Results provided by WolframAlpha

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

Requested by xbz.24

[twin spindle]

Is someone able to help w using the period, for evaluation?

Cause all of these examples are confusing me tbh 💀

[hard moon]

for any number x, sin(x + 2π) = sin(x)

Similarly for any number x, cos(x + 2π) = cos(x)

So for example, let’s say you want to figure out sin(9π/4). Then this is an example where this rule is useful. To see why, we know 9π/4 = π/4 + 2π. Thus, sin(9π/4) = sin(π/4 + 2π) = sin(π/4) = sqrt(2)/2

[hard moon]

does what I said above make sense?

[twin spindle]

Ohhhh i see mow

Now*

Because 9pi/4 is jus equivalent to pi/4

Since you do a whole revolution and then a pi/4 of a revolution

Right?

[hard moon]

yeah, so the problems you probably encountered before these ones were evaluating sine and cosine at angles between 0 and 2π

But these rules allow you to deal with angles outside of that range, so you can evaluate sine and cosine of angles less than 0 (say sin(-π/3)) and sine and cosine of angles greater than 2π (say sin(17π/4))

So say you want to find sin(17π/4)

As you said, you basically just do a whole revolution of 2π, and the sine value will remain the same

So putting this rule in action, we have sin(17π/4), but the angle is too big, so let’s move 2π backwards - - so we see this is the same as sin(9π/4) - - still too big, so we move 2π backwards again - - this is the same as sin(π/4), and since π/4 falls within the range of 0 and 2π (and it’s one of the nice angles on the unit circle), we surely can evaluate it

[slow sundial]

how do i calculate the angle of this

or simply whats the angle

[hoary totem]

doesnt look like a regular pentagon lol

[inner vapor]

?

[rustic grove]

I doubt answer is wrong

Anybody?

[deft sail]

If you add all the angles of a pentagon, it’s 540° .two of them are ninety. If you subtract them it’s 360°. There is a triangle at the bottom. Use the sine or cosine theorem. It’ll find the angle degree

That’s what I think

[north kindle]

This problem is giving me a lot of trouble, I’ve tried messing around with it but can’t find anything

[lime dune]

see that perpendicular that stops suddenly?

what if you extended it to hit WV?

[hoary totem]

^

then what do you know about the angles of a parallelogram?

[north kindle]

?

[hoary totem]

did u manage to solve it?

once u extend the perpendicular line like so

you can solve it pretty easily

[north kindle]

I didn’t

[hoary totem]

what can you say about angle PYH?

[north kindle]

It’s a right angle?

[hoary totem]

PYH? check again

[north kindle]

I don’t know

I don’t even know what angle you are talking about

[hoary totem]

this angle here

angle PYH

thats how angles are named, with 3 points, the vertex of the angle is the middle letter

[north kindle]

Ah

I don’t know

[hoary totem]

so for example in the diagram they give u that angle ZWV = ZWP = PWZ = VPW = 140 degrees

[hoary totem]

a thing about angles is that if u have an X of angles

opposite angles will be the same

[north kindle]

I know

Oh it’s x

[hoary totem]

yes

[north kindle]

I didn’t see that there

[hoary totem]

ok so now
we know PYV is 3x
and PYH is x
so HYV is?

[north kindle]

140-4x?

[hoary totem]

not quite

[north kindle]

Or wait

140-5x?

[hoary totem]

check what angle im talking about again

[north kindle]

Oh

2x

[hoary totem]

yes

we can express HYV in another way

and that allows us to solve for x

do u think u can try to find it?

[north kindle]

Oooh I have an idea

Wait no, that would require a theorem I haven’t proven yet

[hoary totem]

what theorem would that be?

what was your idea?

[north kindle]

The sum of the angles of a triangle being 180 degrees

I was gonna do something with the triangle HVY

[hoary totem]

cool idea but u dont need that here if u dont wanna use that

u can do it solely within the parallelogram

we're given that angle ZWV is 140 degrees
and also that angle ZYH is 90 degrees
try to use that in order to express HYV in another way

[north kindle]

Oh wait

It’s 140-90

[hoary totem]

exactly good job

so now x is?

[north kindle]

25?

[hoary totem]

yes 👍

[north kindle]

Nice

These are hard lol

[hoary totem]

they become easy as you gain intuition and practice

and im sure u will gain that intuition

fast

lmao

[north kindle]

Nahhhh

[hoary totem]

nah trust

ur actually a genius

i think ull find geometry pretty fun

[north kindle]

I’m not escpecially smart I don’t think

[hoary totem]

ur pretty smart

[north kindle]

Nahhhh

[crimson arrow]

how do I find the reference angle of cot(180)?

[hoary totem]

well cotangent of 180 degrees is not an angle

its.. undefined

[crimson arrow]

okay and what about the exact function value?

[hoary totem]

its undefined

notice that cotangent is given by cosine/sine

and cosine 180 degrees is -1
sine 180 degrees is 0

so cotangent 180 degrees is -1/0

ur dividing by zero here which isnt allowed

[crimson arrow]

okay, draw the reference angle and find the exact function value if it exists of Sec(-270)?

[crimson arrow]

If sina= 5/13 and sinB= 12/13, where a and B are between 0 and π/2, evaluate sin(a-B) and cos(a+B).

[hoary totem]

i tried proving it, this is my proof:

• let:
  A, B, C be points on a sphere with center O and radius OA = OB = OC = 1

• draw:
  CA' and CB' to be tangents of the sphere at C in direction of A and B respectively
  then A'CO and B'CO = 90 deg.

• draw:
  extend OA and OB until they intersect CA' and CB' respectively.
  call the intersection points A' and B' respectively.

• denote:
  A = CAB B = ABC C = BCA
  a = BOC b = AOC c = AOB
  a' = CB' b' = CA' c' = A'B'

• planar law of cosines on triangle A'CB' gives:
  c'² = a'² + b'² - 2a'b'cos(C) [1]

• from definition of tan(x), and since A'CO = B'CO = 90, and OC = 1:
  a' = tan(a) [2]
  b' = tan(b) [2]

• [1] and [2] together make:
  c'² = tan²(a) + tan²(b) - 2tan(a)tan(b)cos(C) [3]

• pythagorean theorem on triangles A'OC and B'OC along with [2] give:
  A'O² = 1 + tan²(b)
  B'O² = 1 + tan²(a)

• apply the trig identity 1 + tan²(x) = sec²(x):
  A'O² = sec²(b) [4]
  B'O² = sec²(a) [4]

• square root both sides:
  B'O = sec(a) [5]
  A'O = sec(b) [5]

• planar law of cosines on triangle A'OB' along with [4] and [5] gives:
  c'² = sec²(a) + sec²(b) - 2sec(a)sec(b)cos(c) [6]

• equate [3] and [6]:
  tan²(a) + tan²(b) - 2tan(a)tan(b)cos(C) = sec²(a) + sec²(b) - 2sec(a)sec(b)cos(c)
  cos(c) = tan²(a) + tan²(b) - 2tan(a)tan(b)cos(C) - sec²(a) - sec²(b) / -2sec(a)sec(b) [7]

• apply the trig identity tan²(x) - sec²(x) = -1 and obtain:
  cos(c) = -1 - 1 - 2tan(a)tan(b)cos(C) / -2sec(a)sec(b)
  = 2 + 2tan(a)tan(b)cos(C) / 2sec(a)sec(b)
  = 1 + tan(a)tan(b)cos(C) / sec(a)sec(b)
  = 1/sec(a)sec(b) + tan(a)tan(b)cos(C) / sec(a)sec(b)

• apply trig identities 1/sec(x) = cos(x) and tan(x) = sin(x)/cos(x) and obtain:
  cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(C) [8]

• [8] is the spherical law of cosines. QED.

wasnt actually too difficult, only took like 10 minutes

it was quite fun seeing it all come together

esp cuz it was much much shorter than what i had imagined it to be

[faint pasture]

The proof can be easyy if you know that alternate angles are equal

[north kindle]

I know

[faint pasture]

This definitely much simpler than the proof on wikipedia
Or maybe its just less formal idk but yea

[hoary totem]

well /shrug

it works

[faint pasture]

XD yeaa

[twin spindle]

Tips on proving trig identites? I jus started learning trig, and im stuck on ways to figure out how to prove the identites

[cunning lion]

pick some side, use known identities and algebra until it matches the other side

[twin spindle]

In most scenarios is it typically jus proving, by changing on identity?

[faint pasture]

Just play around with it until it eventually works out

[faint pasture]

The solution said the area around the corners will bein the form of sector of a circle with radius 3 and central angle 120 degrees
But why do they assume that central angle is gonna be 120? Why not smthn else like 90 or smthn

[grave pond]

Have you drawn a diagram?
The immediate surrounding of each vertex of the original triangle spits into:
60° - the original triangle
90° - one of the rectangular parts of the desired area
90° - another rectangular part of the desired area
the rest - circular sector of the desired area

[faint pasture]

Sure but like how am i gonna be sure its a rectangle i mean yea it looks like one but it might end up being a trapezium for all i know

[grave pond]

We agree that all points in this particular rectangle and sector are in the shape you're looking for, right?

[faint pasture]

Yea

[grave pond]

And obviously the radius as well as the width of the rectangle both have to be 3.

[faint pasture]

Yep thats true

[grave pond]

So the point where the the straight part of the outer perimeter turns into a circular arc is both on the distance-3 parallel, and exactly 3 units from the vertex of the triangle.

(It would really help to have a diagram to refer to here).

[faint pasture]

This is the diagram from the official solution

[faint pasture]

Oh i get it now, thank youu!

[flint ingot]

Are there any special properties for golden shapes apart from the long side divide short side is golden ratio? For example in 30,60,90 degree triangles the hypotenuse is double the length of the shorter side

[grave pond]

"Golden" is not really a technical term in and of itself.

[flint ingot]

ok

then iscosles triangle withh 36,72,72 degree angles

[flint ingot]

do u need help?

@deft sail draw a line from the bottom right angle to the side with length 5 which is perpendicular to that side

[faint pasture]

what are you stuck at?

[deft sail]

I just need to know how to solve it

[faint pasture]

are you aware of the law of cosine?

[somber coyote]

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

[deft sail]

I keep getting the wrong answer some how

[faint pasture]

yea, so can you identify what sides should we call, a, b and c for this case

[deft sail]

Ow

[faint pasture]

nvm

im dumb

[faint pasture]

no

you were right

[deft sail]

Everyone makes mistakes

[flint ingot]

@deft sail u dont need law of cosines

[faint pasture]

what are you getting?

[deft sail]

Huh?

[flint ingot]

idk what law of cosines is but i figured it out

[faint pasture]

but the question specifically asked him to use it

[flint ingot]

oh

ok

[deft sail]

Wait

[flint ingot]

sry

[deft sail]

How do I find out without it?

[flint ingot]

so

if u draw a line from the bottom right angle to the side with lenght 5

which is perpendicular to that lenghth 5

[deft sail]

Can you show me. I’m confused?

[flint ingot]

you see you create a 90,60,30 degree triangle

ok ill draw it wait

[deft sail]

Ur saying to split the triangle

[faint pasture]

(this is getting so close to the proof of the law of cosine btw if u use variable instead of values you just end up proving the law)

[deft sail]

I see it

[flint ingot]

idk even know what cosine law is

[faint pasture]

it is exactly what you did rn

[flint ingot]

oh ok

[faint pasture]

except it has been generalised

so we can just use it as a formula

congrats for figuring out the proof on your own!

[flint ingot]

thanks

[deft sail]

Am I gonna find the digonal lines lenght?

(I don’t know how to spell)

[faint pasture]

yea

[deft sail]

The 5 will be split up evenly.

[faint pasture]

noo, use the value of cos(60) to figure out how 5 will be split

[deft sail]

Hmmm

The line I’m making I got 3sqrt(5)

C is sqrt(78)?

Nvm

Here’s another…

[faint pasture]

just use the formula

[lilac zinc]

hello their

[naive shale]

can someone help with this questiion

dont tell solution

[primal jasper]

((1/2)•7•9)sin(120) : (1/2)absinC

[dark sparrow]

!nosols

[lime crown]

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

[dark sparrow]

but also that's overcomplicated lol

[clear anvil]

why?

oh wait lol i thought A = area

[slender flint]

https://math.stackexchange.com/questions/2987659/calculate-sum-n-1-infty-arctan-bigl-frac2-sqrt2n21-bigr?noredirect=1

use the strategy here

[obsidian harness]

I knew it

Math SE

[clear anvil]

Lol

[naive shale]

thanks

[empty yew]

How to prove the 4th question.
I have got ```latex
$\cos^{6}\theta+\sin^{6}\theta=1-\sin^{2}\theta\cos^{2}\theta$ and
$\frac{4-3(m^{2}-1)^{2}}{4}=1-3\sin^{2}\theta\cos^{2}\theta$(when substituting $m=\sin\theta+\cos\theta$)

[somber coyote]

KingDanger

[empty yew]

I am confused what to do next

Upto this step am I correct?
Anyone help me

<@&286206848099549185>

And also i got $-\sqrt{2} \le m \le \sqrt{2}$

[somber coyote]

KingDanger

[faint pasture]

square the original equation

[faint pasture]

the first eqn is wrong i think shouldnt it be 1-3 sin^2 cos^2

[lime crown]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[hoary totem]

that was fast

[faint pasture]

did yall check who was it?

i definitely need to start looking at names before clicking at a link-

[empty yew]

Which equation?

[faint pasture]

sin^6 + cos^6 thing

[empty yew]

It's 1-sin^2cos^2

[faint pasture]

uh no it should be 1-3 sin^2 cos^2

[empty yew]

@faint pasture How could you please explain?

[faint pasture]

well sin^6 + cos^6 = (sin^2 + cos^2)(sin^4+cos^4-sin^2cos^2) we agree on this ?

[empty yew]

Yes

Should I send my work?(for checking of mistakes)

[faint pasture]

yea it will be much faster that way-

[empty yew]

One min

[faint pasture]

you used the identity wrong in the second step

this should be a minus

[empty yew]

Oh yes

Mb

Then it will equal

It's proven without the condition m^2 <= 2

@faint pasture Is the condition no need?

[twin spindle]

How come the Sin²x turned to a Cos²x?

[hoary totem]

do you know the identity
sin²x + cos²x = 1?

[empty yew]

It's an identity

[faint pasture]

yep

[twin spindle]

Yeah?

[empty yew]

Thank you so much

[twin spindle]

But ion see the identity on here

[hoary totem]

rearrange it

subtract sin²x from both sides

what do you get?

[empty yew]

Gone!!! 😁

[formal geyser]

Can anyone help to find x?

[dark sparrow]

it's not sin^2(x) that became cos^2(x)

it's 1 - sin^2(x) that became cos^2(x)

[hoary totem]

try drawing a horizontal line that cuts the diagram in half

[twin spindle]

Wait im actually so dumb, I thought you meant the equation in the problem, not the identity itself 💀💀

So basically you're able to change the identity to make it equal another identity?

I said that so badly but idk how to explain it 💀💀

[empty yew]

@hoary totem Are lines from the centre joining the tangent equal?

[hoary totem]

yes

[grave pond]

... by the SAS theorem on the horizontal line, the two-ticks lines, and the 20° angles the 40° angle gets bisected into.

Angles CAO and BAO both have a sine of (radius)/|AO|, so they are equal.
Therefore AHO is congruent to AFO by the SAS theorem.
In particular HO = FO.

[hoary totem]

yeah

now u can express AHO in terms of x and constants

and solve for x

[grave pond]

(Even if we didn't know that AH=AF, we'd still get the same x, but it would need more work).

(Hmm, not that much more work, in fact).

[hoary totem]

draw a diagram and work out the algebra

or i guess that would be the procedure for proving formulae not identites

oops

[grave pond]

HOM = HOC and FOM = FOB

[formal geyser]

So what's the actual approach to my problem?

[grave pond]

I've shown you two different ones now.

[obtuse quiver]

Gotta simplify this

Can anybody help me w this

[hoary totem]

try splitting the sin⁷θ into more manageable things

[obtuse quiver]

That the wrong q

Correct one in help

[formal geyser]

Is it sin 70 + sin 70 in the numerator?

Or sine to the power of 7?

[hoary totem]

its sin(7θ) + sin⁷(θ)

[upper karma]

I need someone to help me and explain to me how to solve my trigonometry homework

[hard moon]

okay

[autumn pelican]

What is SinB? I figured it was just B1

[hard moon]

Hint: law of sines

oh wait

i see you gave the “measure of angle B”

they just want sine of the angle

[autumn pelican]

Oh I see... I was meant to give the radian not the degree :/

[hard moon]

actually they want sin(56.3°)

because sinB

[autumn pelican]

I think my brain is fried, I for a minute thought that was the radian conversion

What am I looking at here?

[obtuse quiver]

The q was wrong it's sin7∅+sin3∅/cos7∅+cos3∅

[crimson thunder]

how do i solve this

[atomic axle]

A shark fin I believe ❤️

[autumn pelican]

Yes... although I think seeing this IRL while swimming would be less insanity inducing than math

[maiden brook]

one way is to construct AB and CD

[hard moon]

One way to do it is to draw triangles AB and CD (like hockey said), then use the inscribed angle theorem, for example, mBDC = 1/2(161°). Then you can find m(DEC) (why?). I’ll leave out the rest

[maiden brook]

yes u can also use exterior angles to triangles

and that

[crimson thunder]

._.

[maiden brook]

what?

[crimson thunder]

im stupid and forgot my notes whats the inscribed angle theorem

[crimson thunder]

ooooooooh

[maiden brook]

not stupid for not memorizing a theorem

[crimson thunder]

wait now my math ain't mathing isn't 1/2 of 161 = 80.5

how is the angle 93

[maiden brook]

so why is E the center? does it look like it is

[crimson thunder]

e isn't the center its the point where line BD and line AD intersect

[maiden brook]

yeah so what’s wrong then? The inscribed angle theorem doesn’t talk about the 93 degree angle

[crimson thunder]

;-;

how do i find measure of arc AD

[maiden brook]

what is the measure of angle BAC

[crimson thunder]

is it 93?

[maiden brook]

why?

[crimson thunder]

because of coresponding angles?

[maiden brook]

look at my diagram

[crimson thunder]

oh

[maiden brook]

corresponding angles? which sides are parallel

[crimson thunder]

87

[maiden brook]

why?

[crimson thunder]

because of linear pairs

WAIT

OOPS

[maiden brook]

if things dont even look parallel and u have a good diagram and u have no reason for justifying it then those things very likely aren’t parallel

anyway, what is angle BAC?

[crimson thunder]

i can't figure it out

[maiden brook]

what’s the theorem we just talked about?

[crimson thunder]

inscribed angle theorem

[maiden brook]

so …

[crimson thunder]

80.5.....

[maiden brook]

yes

finally

but good

[maiden brook]

now what is angle ABD

[crimson thunder]

12.5?

[maiden brook]

Good yes

Now what is arc AD?

[crimson thunder]

25

25

OH

THANK YOU SO MUCHHHH

[maiden brook]

yes no problem

one more thing though @crimson thunder

what is the average of arc AD and arc BC (minor arcs) in terms of their measures

[crimson thunder]

180....

wait don't they add to be 180

[maiden brook]

what’s their average?

[crimson thunder]

less than 180?

[maiden brook]

bruh

what specifically

[crimson thunder]

im stupid idk

[maiden brook]

hopefully u can do that by this point

bruh

[maiden brook]

do u know what the average means in this case?

[crimson thunder]

no

[maiden brook]

the sum of the numbers divided by the number of numbers

[crimson thunder]

so wouldn't be 90?

[maiden brook]

e.g. the average of 1, 4, 10 is (1 + 4 + 10)/3 = 5

[maiden brook]

what’s 25 + 161

[crimson thunder]

186

[maiden brook]

is 186 = 180

?

[crimson thunder]

no

[maiden brook]

180?

[maiden brook]

so not this

then what’s their average

[crimson thunder]

93

[hard moon]

oh yeah I forgot about that one

[maiden brook]

93? have u seen that before?

[crimson thunder]

yes

it was the original number

[hard moon]

so our problem (with 93° angle and 161° arc) is just like this one, the difference is instead of given two arcs and we need to find the angle, we are given one angle and one arc and we need to find the other arc

[crimson thunder]

that sounds complicated and my brain isn't processing it

[maiden brook]

it turns out whenever u have intersecting chords like this the angle between the chords is the average of the two arcs it subtends (the other arc is separated by a (congruent) vertical angle)

[crimson thunder]

i understand how to solve it more simply

[maiden brook]

wdym?

[crimson thunder]

isn't 93 = 1/2(161 + x)

[hard moon]

that’s exactly it

[maiden brook]

yes but why?

[crimson thunder]

so multiply 2 on both sides and then you subtract 161 from both sides to get 25...

[maiden brook]

and this is what this is saying

[crimson thunder]

oh

[maiden brook]

but where does that come from

[crimson thunder]

where does what come from

[maiden brook]

do u understand how to derive it?

[maiden brook]

so then u shouldn’t be using it

[crimson thunder]

i understand how to solve it

and it came from my notes that i forgot to bring home with me.....

[maiden brook]

if u don’t understand the formula but know how to solve it the way I showed or a similar way, that’s much better

it’s not great if u don’t know how to derive it

[crimson thunder]

now i know that the angle comes from the average of the 2 arcs (big one - little one / 2)

so thank you

[maiden brook]

that’s not the point

[crimson thunder]

._.

[maiden brook]

it solves it faster but if u had a harder problem where u WERENT given the formula before and it was on an exam, what would u do?

and no, it would not be trivialized by a formula like this

sorry if I sound like this but it is actually very very helpful to understand how to prove things urself

I used to not understand this but for harder problems/some actual problem solving u rlly need this

[crimson thunder]

i memorize the formula and use the info given to me and taught to me to solve the problem which may include using an answer from another formula and putting into another formula

[maiden brook]

that’s not good

schools fault most likely though

[crimson thunder]

uh oh

[maiden brook]

what?

[crimson thunder]

i did something wrong if you said that's not good ._.

[maiden brook]

I don’t think u do but the problem is if u come across a harder problem that can’t be solved with one of ur formulas, what can u do?

the solution I showed (which problem isn’t the best one) only relies on basic techniques, not rlly any specific formulas

(yes the inscribed angle, but that’s like the most basic theorem relating to angles in circles)

[crimson thunder]

so i do basic stuff

[lavish dew]

quick question; my geometry project is asking to list a sample space for probailities of a name spinner. Do i have to list off of the possible 276 outcomes or just 20 random outcomes

[twin spindle]

Do yk how this person got cos^2/cos^2

[hard moon]

It appears because 1 - sin^2 (θ) = cos^2 (θ), because of the trig identity sin^2 (θ) + cos^2 (θ) = 1

[trail tendon]

i don't understand that last part

1=1

what does it mean??

🗿

[hard moon]

it seems it might just be scratch work idk

[twin spindle]

Ohh yeaah I get it now

Like this right?

[hard moon]

yeah

[twin spindle]

Bruuvv

How did I not notice that 😭😭

Thank you 🙏🙏

For the Pythagorean identites, does it have to be in the order its said to be?

So can this equal to 1?

[hard moon]

the Pythagorean identity is “sin^2 (θ) + cos^2 (θ) = 1”, and we know this is true; any simple algebraic manipulation will also result in an identity that is true, so what I mean is “sin^2 (θ) = 1 - cos^2 (θ)” is also an identity and so is “cos^2 (θ) = 1 - sin^2 (θ)”;

As for “cos^2 (θ) - sin^2 (θ) = 1”, this isn’t an identity (because both sides are not identical!)

We only have the first three “Pythagorean identities” I mentioned which have sin^2 and cos^2 in them

so yeah it’s just first three

[twin spindle]

I see I see

And we can manipulate the identites when needed, right?

[hard moon]

yes

[twin spindle]

Mmmm I see

Damn that can be confusing at times 💀💀

[hard moon]

In fact, “cos^2 (θ) - sin^2 (θ) = 1” is not an identity, but “cos^2 (θ) - sin^2 (θ) = 1 - 2sin^2 (θ) is an identity; you get this from subtracting “2sin^2 (θ)” on both sides of “sin^2 (θ) + cos^2 (θ) = 1”

[twin spindle]

Oh wrd, yk what problem I'm doing?

[hard moon]

no I don’t

[twin spindle]

Oh

Cause that's littrially the next problem that I'm doing rn 💀💀

Which im actually confused

[hard moon]

💀💀💀💀💀💀💀💀💀

[twin spindle]

I'm confused tho, cause im not sure how to use identites w powers

And when a sin is being multiplied by 2

Or jus any number being multiplies by a trig function

[hard moon]

alright, well I already gave you a big hint, although it’s not the solution: (the identity comes from subtracting “2sin^2 (θ)” on both sides of “sin^2 (θ) + cos^2 (θ) = 1”)

To establish that they’re identical, you would ideally want to establish a chain of equalities showing cos^2 (θ) - sin^2 (θ) = … = … = … = 1 - 2sin^2 (θ)

[twin spindle]

Subtracting?

But its alrdy neg?

Or did you mean add?

[hard moon]

so since we know sin^2 (θ) + cos^2 (θ) = 1 is an identity, the equation would remain an identity if we subtracted 2sin^2 (θ) on both sides

because it’s still identically true

[twin spindle]

Oh wait you meant the pytagorean identity

I thought you meant the problem itself 😭

[hard moon]

oh

[twin spindle]

Ohhh wait

I see what you mean now

Bruv when it involves the pytagorean identites, it always confuses me 😭

[hard moon]

that’s on the right track; however ideally, I think what they are looking for is this:

cos^2 (θ) - sin^2 (θ)
= (1 - sin^2 (θ)) - sin^2 (θ)
= 1 - 2sin^2 (θ)

Since it’s a chain of equalities showing that they are equal; this is because in the beginning of your work, it looks like you’re assuming they are equal to start, which is what we’re trying to prove

it takes time; eventually you’ll have no problem recognizing the pyth identity (with practice)

[twin spindle]

And on the left too, cause we're proving it, no?

[twin spindle]

I hope bro 🙏🙏

I jus started learning Trig, so it's kinda confusing 💀💀

[hard moon]

basically if you removed everything on the right hand side of what you have written, you would have exactly what I wrote

you can think of it like we are taking what you wrote, and only modifying one side of the equation so that the left hand side turns into exactly what is on the right hand side

[surreal escarp]

@cunning lion hey i hope im not bothering u but i still dont know what u meant by this? im pretty sure i got the answer anyway but how is the highest/lowest the same as to get exactly the number #help-24 message

[trail tendon]

the biggest x for x <= 92 is 92

<= is greater than or equal to btw

idk how to write

[surreal escarp]

so for the 2nd one the lowest x would be x<=84 or something???

[surreal escarp]

did u mean less or greater for this

[trail tendon]

oops XD

[trail tendon]

i meant less yea

[trail tendon]

uh

its hard to swap chats XD

[trail tendon]

i dont think the 2nd one is set up correctly

there should be 5 test scores in total

oh wait

[surreal escarp]

oh yeaa sorry thats a old pic, rn i have

sorry let me get a pic rq

[trail tendon]

unless x was both the tests lol

[surreal escarp]

https://cdn.discordapp.com/attachments/852951684157931580/1237234824633253968/IMG_7386.jpg?ex=663ae838&is=663996b8&hm=cb72bbbf0c543b30343465b127123ddea168936666523c5158e8250e97cb7b8a&

[trail tendon]

oh alr

[surreal escarp]

but apparently the x might not be the same so i have to do something else not 2x

[trail tendon]

hmm ok

x+y

[surreal escarp]

n find them seperately?

[trail tendon]

i mean i think you have to minimize their sum

[surreal escarp]

wdym

[trail tendon]

x+y is the lowest total score right

[surreal escarp]

yea

[trail tendon]

so you want x+y to be the minimum it can be

but also x+y >= 164 right

[surreal escarp]

how is x+y 164?

oh wait

nvm read that wrong

hows it great than or equal to 164

[trail tendon]

you solve 2x = 164 right

but its not actually equals

its an inequality

and its not actually technically 2x

(even tho it can be)

but its x + y

and your variables have to be greater than or equal to 164, but they also have to be the minimum that they can be

[surreal escarp]

so i did 256+x+y>=420 then subtracted 256 on both sides so now i just have x+y>=164

so then id just divide that by 2 to get the final two scores as 82 (but the total score is just 164 anyway)

[trail tendon]

yeah

[trail tendon]

i mean the two final scores could be anything that sums up to 164, technically

but yeh

[pearl pecan]

can anyone help

[pearl pecan]

im tryna find measurement z, arc wz, measurment w, and arc wx.

ik how to find the angles but not the arcs

[surreal escarp]

oh so we arent necessarily finding x or y

[maiden brook]

measure of angle z? wdym measurement z

[trail tendon]

nah, you cant

[pearl pecan]

yeah

[maiden brook]

u know cyclic quadrilaterals right?

[trail tendon]

you can only find the sum

[pearl pecan]

like

integrated quads?

yeah

[maiden brook]

integrated quads?

[pearl pecan]

inscriibed

inscribed

not integrated mb

[surreal escarp]

so im interpreting this as like if its asking for the highest/lowest number of an average, whatever x or y is has to be result to the said average cuz u cant rlly get any higher or lower than what is asked (respectively)?

sorry im not that great with explaining math either

[pearl pecan]

maybe ill find the answer one day

[trail tendon]

...sure...?

i'm not sure what you said XD

[surreal escarp]

so like for example question 3 asks for the highest score she can get for the fourth round to have an average of 92,, x is 86 because it results to exactly 92

it cant be 85 cuz thats not the highest but not 87 either cuz its a little more than 92?? idk what im even saying at this point lol

or in question 4, x+y has to be 164 cuz anyything else wouldnt be the lowest or itd be too low (like beyond the B grade)

idk but u helped me get the answers anyway so thank you

🫡

[oblique hull]

@vast skiff

[lime dune]

!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

[quick skiff]

can angle φ inbetween two diagonals of a paralellogram be 90°

without it being a rhombus

for instance if angles in a paralellogram are 60° and 120°

wouldnt the diagonal split them into 30° and 60° and you would get 4 congurent triangles

each with angles 30 60 and 180 - 90 which is 90

but is that possible for a parallelogram, for example sides 4 and 9

[dark sparrow]

no

[quick skiff]

thats what i thought

but don't diagonals split the angles into two congurent angles

θ/2

unless i had a brainfart

[formal geyser]

Can anyone help to solve it

I have no clue what to begin with

[dark sparrow]

give names to all relevant points, for a start

[quick skiff]

do you know if it passes through half of the first square

[formal geyser]

Okay, i understand it

[quick skiff]

if not you have similar triangles

[formal geyser]

Do you mean this line?

[quick skiff]

yes

but i don't think we're allowed to assume it passes through that half point

[formal geyser]

I got you now

[formal geyser]

Seemingly, it's not safe to assume so

[quick skiff]

use similarity of triangles

10 + x / x = 10 / √(10^2 - x^2)

$10 + x / x = 10 / sqrt(10^2 - x^2)$

ok idk how to use

$\frac{10 + x}{x} = \frac{10}{\sqrt{10^2 - x^2}}$

[somber coyote]

leon

[quick skiff]

@formal geyser

[pseudo yacht]

Hello everyone just now joined in this server. need trigonometry formulas from basics

[dark sparrow]

ochem tutor & khanacademy

look up their videos on trigonometry

[past geyser]

this line goes hard

i rapped it in my head

[dark sparrow]

??

[past geyser]

🎵 khan academy look up their vids of trigonometry 🎹

it rhymes

[dark sparrow]

it's a very big stretch of a rhyme

[past geyser]

in rap you can always stretch things

you can even rhyme "ryan" with "entitled" if you say it well enough

[trail tendon]

if you mispronounce them badly enough 👍

[lilac zinc]

what is the best visualization for math?

[past geyser]

hwhat 🤠

visualization of what

[formal geyser]

Sorry, i dont understand it

[formal geyser]

Did you mean AC/BC = DC/DB?

[hard moon]

i found a plausible answer, but the work I had was awkward, since I ended up having to “solve” a quartic equation (which I just used a calculator to obtain the solution)

[quick skiff]

yes

[formal geyser]

But why?

[hard moon]

if you call another side length “y” you can obtain a system of equations then solve for x and y (but it became strange as it became unreasonable to do so without an electronic device)

[formal geyser]

Sorry for such a big f

But

I mean it is more reasonable to say that AC/BC = FC/DC

[hard moon]

yes that’s true

the reason is because BD is parallel to AF

wait

actually I thought you said “AC/BC = FC/DC”

sorry

[quick skiff]

can you?

maybe

but its just simpler to go with similarity

[quick skiff]

similar triangles

[hard moon]

it is simpler if you do it with “AC/BC = FC/DC”

you still obtain a quartic equation, but yeah I got the same answer either way

[quick skiff]

OH WAIT NO

i meant FC

@formal geyser

$\frac{10 + x}{x} = \frac{\sqrt{(10+x)^2 + 10^2}}{10}$

[somber coyote]

leon