geometry-and-trigonometry

[dense sky]

But I kept thinking

[gritty flare]

If you doubt yourself youre not sure

so

Youre not sure you know it well

[dense sky]

I got the answer

[gritty flare]

be sure in yourself!

[dense sky]

Thank you

[gritty flare]

okay anywaus

[dense sky]

Ok what now

[gritty flare]

For the next problem

I gotta go back to what i was explaining when you told me please 5

[dense sky]

We are in number 13

[gritty flare]

I know

this wilbe relavant for the next 6 problems

Okay so Length and Measure of arc differ

Length is how long it is and measure is how much degrees it is

The measure of the arc is determined by the central angle which is intercepting that arc

O is the center and AOC is the central angle

So measure of arc AC = measure of angle AOC = alpha

Since the full rotation in the full circle is 360 degrees

full rotation of the measure of a circles circumference is 360 degrees also

[dense sky]

Is the answer 23

[gritty flare]

now you should be able to solve problem 13

Wat.

lemme... solve it wait

qwq

Yes the answer is 23

5x+245=360

x=23

[dense sky]

The next one is 5

[gritty flare]

it is

Okay i wont explain next one then solve it yourself

[dense sky]

Ok now onto number 15

[gritty flare]

and ill check

[dense sky]

Is BAC 180

[gritty flare]

that is impossible

[dense sky]

Oh right

How do I find BAC then

[gritty flare]

Well central angle is equal to measure of the arc its intercepting

What is the central angle in our picture

Mark the angle by 3 letters

[dense sky]

Where is the central

[gritty flare]

The angle which is in center

if you look at the picture I sent its the angle AOC

the my drawing one

[dense sky]

This is for 15?

[gritty flare]

It is for 13-18

OwO

If youre about my drawing

[dense sky]

Is 17

144

[gritty flare]

No

[dense sky]

It’s 36

[gritty flare]

Incribed angle is Half the central angle intercepting the same arc

Yes 17 is 36

[dense sky]

18 is 44

[gritty flare]

yep

Okay now lets go back to 15 and 16

[dense sky]

19 is 112

I don’t think your drawing works for 15-16

[gritty flare]

Okay then

Now

ABC is the central angle intercepting the arc AC

So mABC =mAC=110

BA=BC. can you tell me why?

[dense sky]

They are similar sides :/

[gritty flare]

no

Think

B is the center

B is the center of the circle

[dense sky]

They are tangent

[gritty flare]

No

think

[dense sky]

Because they just are

[vale nimbus]

u have a point in the center of a circle connected to a point on the circle itself

whats the line connecting them called?

[gritty flare]

[vale nimbus]

(to put that into a drawing)

[gritty flare]

@dense sky would you like to take a rest? you seem tired

[mossy shoal]

a clock

[dense sky]

A bisector

[gritty flare]

@mossy shoal get out

[vale nimbus]

a clock

[gritty flare]

SK its a radius

[dense sky]

Oh right

[gritty flare]

pls take a rest

[dense sky]

No not yet

[gritty flare]

maybe grab a coffee

[vale nimbus]

respect to sorryful for doing all this tho god damn

[dense sky]

Thank you for the help

[gritty flare]

I have time but sks mind is just shutting down before me

[mossy shoal]

big takeaway is in any circle if u have a line from the edge to the center of circle

its the radius : )

[dense sky]

Their are more problems

[gritty flare]

how muxh time do tou have left sk

[dense sky]

Plenty

[gritty flare]

Then rest

sheesh

[vale nimbus]

go take a break then bro

[dense sky]

No

[mossy shoal]

this mans is determined

[dense sky]

[mossy shoal]

Go put on 10 hours of brain power on youtube

[gritty flare]

OKAY THEN

[mossy shoal]

Lets get it

[vale nimbus]

extreme math session inbound

[mossy shoal]

hit up a homie for some coke™

[gritty flare]

brb

[mossy shoal]

Hol up

Have u been solving the prev problems

[dense sky]

Yes

[mossy shoal]

what u confused about then

use the concepts from those probs

to solve these

[vale nimbus]

holdup can someone explain 29

[gritty flare]

Let us

actuallt

go back to number 15 pls

[vale nimbus]

@gritty flare can u tell me the answer for 29 in dm's please?

[gritty flare]

Not ywt

Okay so the drawing now

number 15

[vale nimbus]

[mossy shoal]

are these two homies in the same class

[vale nimbus]

no

im just lurking

[mossy shoal]

o gotit

[vale nimbus]

trynna make sure i remember all this

but i dont get 29/30

[gritty flare]

AB ans BC are equal cause they are radii

[dense sky]

Ok

[gritty flare]

so what kind of triangle is ABC SK?

[dense sky]

Obtuse

[gritty flare]

Yes but what ELSE kind of triangle is it

whose two sides are equal

and the other one is a base

[dense sky]

Equilateral

[gritty flare]

Remember the terms

Its called ISOSCELES SK

ISOSCELES TRIANGLE

[dense sky]

[gritty flare]

so

now

[dense sky]

Yes yes

[gritty flare]

That mild panic needs rest but since you dont

What angular properties does an isosceles triangle have

If two sides are equal them the angles opposite to them?

[dense sky]

2 equal sides

[gritty flare]

Yes

so then the oppisite to the side angles are also?

also what

[dense sky]

EQUAL

[gritty flare]

correct

[dense sky]

So if the base is 110

[gritty flare]

So we have the peak angle or top angle or watever which is 110

nd then two equal angles

so

let those base angles be x

what is the sum of the angles in the triangle?

[dense sky]

AB + BC =110?

[gritty flare]

No

[dense sky]

Oh wait

[gritty flare]

angle mABC=110

bcs its thr central angle

also

mBAC=mACB=x

[dense sky]

Is it 70

[gritty flare]

bcs we want to find BAC

then

x+x+110=180

and what is x?

solve it

[dense sky]

35

[gritty flare]

yes

correct

same concept in 16

solve it and tell me the answer

[dense sky]

20

[gritty flare]

gj

Okay

[dense sky]

Ok on to 21

[gritty flare]

Do yoy alrdy know 19?

tell me the answer to it

[dense sky]

41?

[gritty flare]

nope

[dense sky]

139

131

[gritty flare]

Nope

Nope

[dense sky]

Bruh

Hold on

I know this

[gritty flare]

Its connected to inscribed angles

[dense sky]

112

[gritty flare]

yes

[dense sky]

Epic

[gritty flare]

If a quadrilateral is circumscribed thrn sum of the opposite angles is 180 degrees

cool to remember

Comes from inscribed angles its really easy to prove

same with 20

what is the answer on 20?

[dense sky]

131

[gritty flare]

yes

[dense sky]

Time for 21

[gritty flare]

yes

hey this actuallt i cant figure out

[dense sky]

We can skip it

And go to 23

[gritty flare]

Do yiu know anything connected to 21?

[dense sky]

Not really

[gritty flare]

oof 23 needs lotsa drawins

[dense sky]

Can it be explained without drawing

[gritty flare]

Nope not rly

[dense sky]

You gonna draw it?

[gritty flare]

Well ig you could remember this but i wanna show you how its proven

Ye im gonna draw it if its okay with you

[dense sky]

Sure

[gritty flare]

@dense sky okay

Look if one arc of circle is 220 what is the other arc

[dense sky]

140

[gritty flare]

Yep

Angle alpha is intersepting that arc so

as a central angle what is its measure

[dense sky]

360

[gritty flare]

No

central angle is intersepting arc which has measure of 140 degrees

[dense sky]

70

[gritty flare]

Central angle is equal to the measure of the arc its intersepting

[dense sky]

Boi

[dense sky]

So it’s 140?

[gritty flare]

Ye it is 140

Now

Look at the quadrilateral

since the angle x creates tangents which touch the circle

then the radii to those tangents will be perpendicular

so 90°

now we have a quadrilateral with angles
90°; 90°; 140° and x°
what is x

[dense sky]

130

Waited

Wait

40

[gritty flare]

yep

solve 24 yourself and tell me the answer

[dense sky]

130

[gritty flare]

yep correct

okay now back to 21

[dense sky]

Ok

[gritty flare]

Now this is a formula which you have to remember

This is
Angles of intersecting chords theorem

[vale nimbus]

what does this mean?

[gritty flare]

Arc measure

[dense sky]

We are doing it now

[vale nimbus]

oh

[gritty flare]

Okay so

[vale nimbus]

and this? :(

idk these signs

[gritty flare]

measure of angle QUS

m stands for measure

Anyways

Okay QUS and TUR are what kinds of angles

if you can remember that

what kinds of angles do we call those

[dense sky]

INTERSECTING

perpendicular?

[gritty flare]

tough luck

perpendicular means creating 90 degrees

so no

[dense sky]

I know they are related to each other

[gritty flare]

They are VERTICAL ANGLES

What is the property of vertical angles

do you remember?

[dense sky]

Yes I think

[gritty flare]

tell me then

[dense sky]

QR=ST

[gritty flare]

No

The property of vertical angles

Is that they are equal!

so Angle QUS and Angle TUR are equal

okay now the angle of intersecting chords theorem states that

[dense sky]

They have to equal 180?

[gritty flare]

mQUS=mTUR=(mSQ+mTR)/2

no no vertical angles are just equal

[dense sky]

So 85+65/2

[gritty flare]

Congruent=Equal

Ye mQUS=(65+85)/2

which is?

[dense sky]

117

[gritty flare]

nani

[dense sky]

Oh mb

Plugged in wrong number

It’s 75

[gritty flare]

Do you rly need calculators

[dense sky]

yes..

[gritty flare]

Like

For this

[dense sky]

._.

[gritty flare]

okay anyways solve 22 by yourself

[dense sky]

65

[gritty flare]

ye correct

okay moving on to25

[summer spire]

@gritty flare what's what drawing app, btw?

[gritty flare]

Its uh Speednote it came on my phone

well tablet

Okay what can you tell me about this

[dense sky]

So EA=EC?

[gritty flare]

No

but can you tell me all the radii

of the circle

[dense sky]

5

Bc

CD

[gritty flare]

Ye and?

[dense sky]

BC

CD

[gritty flare]

BC; CD AND

[dense sky]

EC

[gritty flare]

yes

radii are equal so they are all 5

now what is AC

[dense sky]

21

[gritty flare]

hm

[dense sky]

Wait

[gritty flare]

is it?

[dense sky]

No

23

[gritty flare]

yes

okay so what is the property of tangent line

and radius

[dense sky]

They are equal

[gritty flare]

No!

When we have a radius on the tangent line

The radius is ???

CB is what to AB

[dense sky]

.-.

[gritty flare]

Blue is the radius

Red tangent

WHAT IS THE PROPERTY

look at it carefully

[dense sky]

Perpendicular

[gritty flare]

ye

it creates 90 degree angle

so CB is perpendicular to AB

then what kind of triangle is ABC

[dense sky]

90 degrees

[gritty flare]

Type of triangle

[dense sky]

Right triangle

[gritty flare]

yes

what formula can we use then?

theorem

[dense sky]

Pythagorean’s

[gritty flare]

ye

What lengths do we know?

[dense sky]

5 and 18

[gritty flare]

By names

pls

[dense sky]

BC EA

[gritty flare]

No

not EA

The triangle ABC

ignore the point E now. we found AC alrdy

[dense sky]

5

[gritty flare]

wat

[dense sky]

ABC?

[gritty flare]

Okay ill just tell you

we know BC = 5; AC=23 and we need to find BA

make the pythagorean theorem equation

[dense sky]

Ok

I’m getting decimal

[gritty flare]

ye same

lemme actually

hold on

6√(14)

is the answer apparently

A pretty ugly number

[dense sky]

Onto 27?

[gritty flare]

26

[dense sky]

Ok

Another decimal?

[gritty flare]

Ye well you will need exact form

tell me the decimal

[dense sky]

28.3?

[gritty flare]

ye sure the answer is 20√2

√(800)=√(400*2)=20√2

Okay 27 is easy

[dense sky]

28

Is it 28

[gritty flare]

What is the mXZW xentral angle

No

[dense sky]

How to do 27

[gritty flare]

Im telling you

step by step

What is the measure of angle XZW central angle

[dense sky]

85

[gritty flare]

dud

[dense sky]

95

[gritty flare]

keep this pic

inside your head

yes it is 95

angles YZX and XZW create together a straight line right?

[dense sky]

Yes

[gritty flare]

Then

[dense sky]

So it equals 180

[gritty flare]

yes the sum

[dense sky]

Meaning

The other part is 85

[gritty flare]

mXZY+mXZW=180

Yes

85

correct

Whats 28

[dense sky]

Same concept for 29?

45

[gritty flare]

okay gj

[dense sky]

Is 29 130

[gritty flare]

im guessing problem is talking about the smaller arc

Hm okay

Central angle = Arc measure and the reverse is true also

So if central angle is 100

the arc measure is?

[dense sky]

100

[gritty flare]

yep

Idk if this is this simple

are we doinf this correctly

[dense sky]

I think we are

[gritty flare]

30 should be 140

[dense sky]

People say this is an easy lesson

[gritty flare]

but dude it seems too easy compared to others

[dense sky]

It is really easy

I’m just really bad

[gritty flare]

okay then...

we done?

[dense sky]

NO

[gritty flare]

Well tough luck my phone's at 3 percent

sorry

[dense sky]

No ;-;

[gritty flare]

I could check tomorrow

if you have like 9 hours

[dense sky]

Wish me luck

[gritty flare]

you could dm me the answers

[dense sky]

I’ll try my best

[gritty flare]

!! good luck

maybe other ppl can also help if theyre online

[rich coyote]

how do i find the center of a rectangle that is not aligned with the axes

[vale nimbus]

@dense sky which ones do u ahve left?

i'd help but i still dont understand what this sign means

[olive plover]

m angle xwy

measure

[vale nimbus]

so just how big those angles are?

[silent plank]

minor angle

[brazen depot]

How would I graph something like this?

I understand sin and cos graphs, but not tan

amplitude - none
frequency - 1
period - pi

for this function

its super easy for me to find amplitude, frequency, and period but i have no idea how to graph tan graphs

ik they go up and down with asymptotes, but i seriously dont know how to graph

(pls ping if you respond :))

[feral wind]

I mighttry to see it as a rational graph.
Since its tan(sin theta/cos theta) or y/x

With how cos behaves with sin

[brazen depot]

a rational graph?

[feral wind]

I always seen the tan graph as a a rational since it has asymptotes

[brazen depot]

wym by a rational graph tho? (sorry im lost 😅 )

[feral wind]

Oh have you heard of them before?

[brazen depot]

no lol

i recently learned graphing these this week and i got a test on monday 😂

[feral wind]

They basicly like in the tan graph have "asymptotes".
Like imagine 1/x.
As the x values get larger and larger the y values gets smaller, like 1/1 =1, 1/4 = 0.25, 1/10 = 0.1 but as you can imagine it will never reach 0

[brazen depot]

ah yes

[feral wind]

Oh! But theres a asymptote at 0 as well since 1/0 is undefined

[brazen depot]

oh

oh yes

i get u

[feral wind]

And so thats why i believe tangent takes a rational form

[vale nimbus]

i draw faint lines on each asymptote value

[feral wind]

But im sure theres a actuall answer somewhere xD

[brazen depot]

i get why theres asymptotes, but I don't get where I can graph 😂

[vale nimbus]

and then just pray it looks somewhat decent when i give the line inbetween the asymptotes a tan function sort of look

[brazen depot]

so like

i dont understand the graphing part

[vale nimbus]

wait whats graphing u mean just drawing the graph?

not native so i could be on about something different

[brazen depot]

yes

drawing the graph part

😂

[vale nimbus]

what i'd do is literally just draw lines at every value where there's an asymptote

and then draw the typical lines of a tan function inbetween

but thats probs not the proper answer

[brazen depot]

but how do u know how long it is

horizontally

like

it could be skinny

or fat

[vale nimbus]

is it allowed to link vids in here?

[brazen depot]

u can dm me if not, but i think so

[vale nimbus]

https://www.youtube.com/watch?v=sKoR8KHx4uU

[brazen depot]

YES

[vale nimbus]

from my experience this guy's vids are pretty solid

[brazen depot]

U R ABSOLUTELY AMAZING

I WILL WATH THAT

thank you 🎬

woops

[vale nimbus]

[brazen depot]

👏

[vale nimbus]

goodluck bro

[brazen depot]

thank u bro

[upper karma]

What would i do to caculate for x?

would i make a theoretical line (y) and caculate pythagorean therom?

[onyx basin]

first convert mi to in

[upper karma]

aitee

[onyx basin]

love imperial measurement system

[upper karma]

lmao yea

its all pain

[onyx basin]

then 1 mi is your hypo

[cobalt turtle]

might be american but i sure dont

[onyx basin]

so you'd set up pythag thm

[cobalt turtle]

except for carpentry i like it there

[onyx basin]

a2 + b2 = c2, where a and b are legs and c is hypo

plug in values and solve

[upper karma]

c is curved tho

[onyx basin]

oh that's a good point

[cobalt turtle]

once again jumped right to tan kill me

[onyx basin]

i thought that was just a drawing error

[upper karma]

its the oppisite of a sector

so would i

[cobalt turtle]

darn it gonk

let me look smart

[upper karma]

G O N K

like

[cobalt turtle]

droid?

[upper karma]

ye

[cobalt turtle]

yusss

[upper karma]

im doing some big brain math

A+B = 180 degree right?

So since the angle of the sectore = (θ ÷ 360 degrees) × πr^2, then we could get anglle of the theta/a

ok

OK

so angle a = 0.05644314 degrees

im getting somewehre

[vale nimbus]

thats quite the small angle

[upper karma]

wait

i oofed on metric conversion

hol up

its 0.09093664746 degrees

ok

[digital gulch]

s=44\pi\cdot\cos^{-1}\left(\frac{3960}{3960+h}\right)

[upper karma]

im trying to caculate earths angle of delevation over a mile @vale nimbus and so i found the angle of the sector if the arc length is a mile so now in theroy i can begin finding other angles

[vale nimbus]

jesus goodluck

[upper karma]

my brAIN HURTS FROM trying to conceptualize this

In theroy a and be are supplementary

so b = 89.9090633525

I think i might have got it

wanna be sure tho

wait

the angle of delevdation would be constant as earth is round and thus you can start on an infinite amoutn of set start points go forward a mile and have the same length of delevation

WAIT I KNOW IT I JUSt need to caculate Angle of elevation per mile = 8inches/Circumfrence of earth

SO The angle of delevation over 1 mile of earths suface would be 5.07058457e-9 degrees @vale nimbus

[onyx basin]

i have literally no idea what you just said but go off

[vale nimbus]

id like to pretend i know whats going on but nah

[upper karma]

Idk

[rich wolf]

@vale nimbus lmao

[upper karma]

i dont know either

[vale nimbus]

cant u just google it

[upper karma]

I want to do it on my own and see how close i am

[vale nimbus]

well ur done no?

[upper karma]

im trying to find if some1 has an angle measure

and i cant find 1 yet

i think im right

all myt caculations came out the same on caculators distance wise

so unless i oofed on pressing buttons then it should be right

[vale nimbus]

be sure to let us know how close u were

[upper karma]

i dont think ima find anything

😦

[vale nimbus]

cant u find the delevation angle online?

[upper karma]

no

not rly

you can find the distance earth curves over a mile

which is the basis i built of of this

[olive plover]

for conterminal angles do I just take the given angle and +-360 to find both positive and negative?

[visual mist]

find the coordinates of the point that divides the segment joining (-2,0) and (3,4) into the ratio of 1:2. Ratio is to be considered from left to right. can someone help me?

[rich wolf]

@visual mist ok

You asked this yesterday

[visual mist]

yeah but u didnt say how to do it

[rich wolf]

U never responded lol

[visual mist]

i did

i said horizontal is 5 and vertical is 4

[rich wolf]

Thats wrong

You got it switched

Horizontal is 5

[visual mist]

i edited it lol

so whats next

??????

[rich wolf]

@visual mist sorry for late reply i got caught up in something else

Still need help?

Basically its asking for a 1:2 ratio

So that means finding the length of the segment that is 1/3 the original length

Since you found the vertical and horizontal components you just add them to the original point

So the point is at (-2+5/3, 0+4/3)

Since a third of the segment is a third of the horizontal and vertical components

[visual mist]

why do u need to -2+5/3?

[rich wolf]

@visual mist because you start at -2

And you are adding a third of the horizontal distance to the original point

[visual mist]

oh okay

thanks

[west robin]

Dm me if you wanna make some money and understand college algebra

[long venture]

can you tell give me a hint of how to prove

2SinxCosx=2Tanx/1+Tan²x

from the LHS

[silent plank]

parentheses and lowercase

2tan(x)/(1 + tan^2(x))

its easier to start from the right side.
figure that out first then work backwards

[long venture]

yeah from RHS I proved

So I was trying to prove

Sinx+Siny=2Sin[(x+y) /2]Cos[(x-y)/2]

I used a trick like

but i actually proved that

Sinx+Siny=2Sin[(x-y) /2]Cos[(x+y)/2]

now I don't understand what did I do wrong

🦧

plox help

I proved
Cosx-Cosy=-2Sin[(x+y)/2]Cos[(x-y)/2]

from the same method without trouble

hmm it doesn't seem to work for proof of
Sinx+Siny
and
Sinx-Siny

nvm it works

[vale nimbus]

@long venture

2SinxCosx=2Tanx/1+Tan²x

i proved this today aswell lel

write the tan²(x) in the bottom part of the fraction in sin and cos

multiply 1 by cos²(x)/cos²(x)

and from then on its pretty straight forward if you keep in mind that sin²(x) + cos²(x) = 1

[mint drift]

Yo anyone know if upper triangulation deteriminant method works for a matrix that was put in row echelon form?

If the n * n matrix isn't original in row echelon form

[vale nimbus]

are you talking about the basic sin(x) and cos(x) functions or also transformed ones?

[dark sparrow]

what's "it"

[eager kraken]

oh just basic ones

I am kinda lost on how to graph sin, cos, tan functions...

like where do i start the parabola on this graph?

the peak of the bottom of the parabola be at positive pi/2?

[dark sparrow]

i mean i'd rewrite this as 1/cos(x - π/2) and then as 1/sin(x) ¯_(ツ)_/¯

if you can graph csc(x) then you can graph that thing bc they're lit the same

[quiet mason]

is it always possible to draw a circle which passes through an external point from which through two tangents to another circle are drawn,the centre of the other other circle,the tangent points

[upper karma]

I have no idea 🤷‍♂️🤷‍♀️

[dark sparrow]

uhhh

can you say that again but in a less confusing manner @quiet mason

[quiet mason]

ok

suppose i have a circle with centre O

i have an external point P from which i draw two tangents PA and PB to circle with centre O

now can i always draw a circle such that the centre O,point P ,point A,point B lie on the cirumference of this circle

Anna

[dark sparrow]

yes, OAPB will always be a cyclic quadrilateral.

[limber anchor]

Hey fellas I was just wondering a quick thing before I do a math test, would I do the pythagorean theorem to find that missing side or could I do something with similar triangles?

[languid ingot]

definitely not pythagorean theorem because it needs to be a right triangle

[limber anchor]

that’s right i forgot about that

[languid ingot]

hm. i think i remember this from geometry. i think the two triangles are similar but im not sure

oh yeah. it's the the Angle Bisector Theorem

[quiet mason]

are those top two angles equal

[limber anchor]

ohhhh i think i remember what to do

so it’s like X over 4 equals 1 over 2

and then i cross multiply

Right?

[languid ingot]

ye.

[limber anchor]

alright sweet thanks fellas

[quiet mason]

no

you have to prove theyre similar

@limber anchor @languid ingot ??

[limber anchor]

What do you mean

[quiet mason]

how did you even prove the two triangles are similar

[limber anchor]

i didn’t need to?

i’m not entirely sure what you’re talking about

[languid ingot]

Angle Bisector Theorem @quiet mason

basically states that: (image taken from internet)

[quiet mason]

was it given that the two angles are equal

[limber anchor]

i mean that’s what the little arcs imply doesn’t it

[quiet mason]

mb idk

[sly marlin]

Just similar triangles

Consider extending b up and left by a.

[quiet mason]

yes this will all work if the angle is actually bisected

but im not familiar with that notation

[grave python]

Is volume d^2cosacosb(sqrt(cosa)) correct?

[sly marlin]

hmm...

dimension is already wrong

needs d^3

@grave python

height is d cos beta
width is d sin alpha
depth is just d sqrt(1-cos^2 beta - sin^2 alpha) = d sqrt( sin^2 beta - sin^2 alpha)

[weary drift]

[devout harbor]

can u show ur working

[dark sparrow]

where is the angle

is it inbetween the sides

[somber coyote]

nebula:

[devout harbor]

It should be $8^2 = x^2 + 10^2 - 2(x)(10)\cos32$

I remember it like this: The one opposite to the angle is on the LHS

The ones adjacent to the angle are on the RHS

no worries, you can perform the next test better

np

[somber coyote]

Sup?:

[upper karma]

Can someone help me with number 1 please?

[upper karma]

Is the answer the third one?

[brazen depot]

How do I do a question like this

[onyx basin]

can't you use arccos to find theta

[umbral snow]

@upper karma
There's no problem. That's a proof of the Pythagorean Theorem

[onyx basin]

oh u got the cos -1 at the bottom

[upper karma]

@umbral snow so what’s the answer?

Those are my only options as shown

[brazen depot]

@onyx basin wym arccos?

[onyx basin]

arccos is inverse cos

i didn't see the cos-1 you had at the bottom

[brazen depot]

u mean cos-1?

oh i see

yeah i have no idea how my teacher got 5pi/4

all i saw her wrote down was the 225 degreees times pi/180=5pi/4

but idk how she got 225 degrees

[onyx basin]

dunno how you got the arccos to be 180

[brazen depot]

i think i just put into my calc wrong, as i tried it again and now its 135 degrees

but how would u get 225 degrees?

(non related but ur music taste is awesome)

[onyx basin]

lmfao

i mean, theta is defined as greater than pi, pi = 180deg

3pi/4 < pi

so they'd need to find an equivalent of 3pi/4 that equals 180?

honestly my brain kinda crashing rn

[brazen depot]

360-225=135 right

but why do u subtract it

[onyx basin]

ohhhh

i got it

think about it

cosine finds the x value, right?

so that means the x value must be -sqrt2/2

which limits us to the second and third quadrant

by doing arccos, we find that theta = 3pi/4

but we're also told that pi < theta < 3pi/2, which restricts us to the third quadrant

so we need to find the next possible theta value that satisfies cos(theta)=-sqrt2/2

which would be 5pi/4

[brazen depot]

oh so cuz

3pi/4 is too small

we need to find something bigger

[onyx basin]

mhm

[brazen depot]

so u would have to memorize the unit circle for this?

is there a way u can solve mathematically?

[onyx basin]

well it's good to memorize the unit circle when dealing with trig

[brazen depot]

😓 u r totally right... but i wont be able to memorize in time for my test tmr 😂

[onyx basin]

it will be extremely useful to you if you memorize the sin and cos of common values in the unit circle

unit circle is pretty easy to understand

cosine is x, sine is y

do you know how to convert deg to rad and vice versa

[brazen depot]

yes

[onyx basin]

mkay

so a circle is 360deg

aka 2pi rad

[brazen depot]

yes

[onyx basin]

im trying to think of how to explain this to you in a fashion that would make it easy to memorize

honestly i only know these because of repetition

[brazen depot]

haha thank you

yes :/

but basically

[onyx basin]

https://www.youtube.com/watch?v=1m9p9iubMLU

[brazen depot]

the only way is to memorize the unit circle?

[onyx basin]

https://www.youtube.com/watch?v=ZffZvSH285c

well i dont think it's the only way

there's probably another way

i was just taught through repetition

[brazen depot]

okay

i will watch both vids!

thank u so much!!

[onyx basin]

np

gl

[gritty sail]

i think this is geo??

but idk

im terrible at clock problems

[robust crane]

when told to find an area but not given units is it better to say, "thus the area is 26 units squared" or just 26?

[gritty sail]

depends on teacher

[robust crane]

what about in general pure math world of communication

is it technically wrong to declare an area which is unitless

[gritty sail]

idk if anyone cares

except you apparently

[robust crane]

ok then

[dark sparrow]

this problem is ambiguous

it could be 01:30, 10:30, 13:30 or 22:30

[devout harbor]

@robust crane Yeah I always wrote unit squared and I still do!

[ember maple]

Hello, why does cos(27,5°-(x/2))=sin(62,5°+(x/2)) ?

As in why the -(x/2) changes to +(x/2) ?

[upper karma]

Guys

So I memorized the unit circle

But tbh, Idek what it really does

Can someone please explain its purpose

[devout harbor]

@ember maple $\cos\theta = \sin(90 - \theta)$

[somber coyote]

Sup?:

[devout harbor]

Think of 27.5 - (x/2) as theta

Then plug it in the identity above

[ember maple]

Oh I see now

[tropic sedge]

i got 2tanx for this, anyone concur?

[long venture]

How to prove Cos20Cos40Cos60=1/8

I did like

now that Cos20 term is extra

[silent plank]

,w cos(60deg)cos(40deg)cos(20deg)

[somber coyote]

Display results online and refine query

Requested by ramonov#0921

[silent plank]

that's not 1/8

also you made an algebraic mistake in the last line

[long venture]

Damn, Yeah the algebraic mistake I got lol

My school gave me that question in exam such idiots

[uneven kestrel]

Do functions sin(x) and cos(x) give absolute extremum? Basically if there is a max/min point reached multiple times is that still considered an absolute extremum

Is uniqueness a quality of absolute extremism

[uneven kestrel]

<@&286206848099549185>

[knotty merlin]

well in the complex world

sin(z) can be arbitrarely large i think

same with cos(z)

[surreal ingot]

Is this looking correct?

[surreal ingot]

<@&286206848099549185>

[silent plank]

Q4?

[surreal ingot]

Yes sir

[silent plank]

looks very messy

your work should include:

[surreal ingot]

Sorry, I should done on a piece of paper.

[silent plank]

$\tan(\alpha) = \
\tan(\beta) = \
\text{compound angle formula for} \tan$

[somber coyote]

ramonov:

[surreal ingot]

Like this?

[silent plank]

no, there's something very dodgy with that

[surreal ingot]

Where did I make mistake?

[silent plank]

tan(a) isnt tan(3/4) etc

[surreal ingot]

But I got from doing Pythagorean

[silent plank]

its just 3/4

[surreal ingot]

Thanks

[silent plank]

and how you even got those radicals is very questionable

[surreal ingot]

Tan 5/12 = tan 75 = tan 40 + tan 30

Then I find each value and sum the up

Tan 5pi/12

[silent plank]

Tan 5/12 = tan 75
???

[surreal ingot]

I use as tan 5pi/12
Tan 5(180)/12
Tan 75

[silent plank]

yeh, multiple levels of bad there

1. not tan(5/12)
2. mistaking 5/12 for 5pi/12

[surreal ingot]

I was looking at the angle.

[silent plank]

$\frac{ \frac34 + \frac5{12}}{1 - \frac34\cdot\frac5{12}}$

[somber coyote]

ramonov:

[silent plank]

after proper notation and subs, you should get something like that

[surreal ingot]

Yea, I got it. Before I was looking at angle instead of value that was my fault.

[vale nimbus]

since u have sin and cos of the angles cant u just use sin²(x) + cos²(x) = 1 to find everything u need?

or is there a problem with that

[loud geyser]

so, i can't figure this out in my textbook itself

given two planes in the form of ax + by + cz = d

is the norm <a,b,c> regardless of d?

[finite osprey]

I'm currently doing trig identity proofs and in need help with this one

[silent plank]

consider double angle identites

[finite osprey]

well i can get it down to about this and then i start to get confused

[languid ingot]

i thought when you're proving trig identity stuff you're not suppose to manipulate the other side

[somber coyote]

Dog:

[upper karma]

im taking an algebra 2 trig placement test soon, what is osmething that is prob gonna show up a lot?

i rly want to get into pre calc

[surreal ingot]

Does anyone know how to solve that?

[dark sparrow]

$-1 = \sin\paren{x + \frac\pi2} - \sin\paren{x - \frac\pi2}$

[somber coyote]

Ann:

[dark sparrow]

this?

[surreal ingot]

Yes

[dark sparrow]

what have you tried so far, where are you stuck, and what is giving you trouble?

[surreal ingot]

I try to use cofunction identifies

But it won't works

[dark sparrow]

... what

[surreal ingot]

Wait

Ohhh

I see why

Can't I use sum and difference formulas?

Wait

[dark sparrow]

well you can also just write sin(x + π/2) as cos(x) and sin(x - π/2) as -cos(x) and end up with 2cos(x) = -1...

...i have to go now

[ebon minnow]

How to get the height when the only things I know that DC is ✓2 and <SCO is 30°

[dark sparrow]

you can't

unless you have omitted something

[ebon minnow]

SC = 2x and SO = x, because 30° is in front of SO

You get that AC is 2 from the pythagorean theorem

[rich wolf]

@ebon minnow i am assuming that this is a square pyramid

Yes you are right on that first part

AC is 2

OC must be half of that because the base is a square

It's very easy now

[ebon minnow]

Yeah, forgot to mention it's squared

[dark sparrow]

yeah so you did

[tepid rapids]

hi guys

so i just wanted to ask, why arent trig functions defined for scalene triangles or isoceles triangles?

why right triangles only?

[dark sparrow]

what

[tepid rapids]

even the unit circle definition of trig functions that extends it to the real line is still defined in terms of right triangles (just in different quadrants)

[dark sparrow]

trig functions are not defined for triangles

the input of a trig function is an angle

[tepid rapids]

right but when defining a trig function, you use a right triangle

why is that necessarily the case?

[dark sparrow]

i mean i could just say "that's just how the definition is"

[tepid rapids]

i wouldnt be satisfied xD

[dark sparrow]

the reason you can't use an arbitrary triangle is because if you only fix one angle (the one whose sin or cos you are trying to determine) this isn't enough to pin down the shape of the triangle and thus the ratios of its sides

[tepid rapids]

ah right

so if a trig function took two args maybe we could define it for arbitrary triangles?

[dark sparrow]

...

......

..........

that's meaningless

[tepid rapids]

hypothetically

[dark sparrow]

i don't even know wtf you're on about or what you're smoking

[tepid rapids]

lol

there are no stupid questions 😉

[dark sparrow]

did i say your question was stupid

no, i did not

[vale nimbus]

hey could i get some help with the usage of simpson's formulas in triangles to prove certain identities that are valid for those triangles? (will send an example in just a sec)

(they're asking to prove that the triangle has 3 equal sides)

is there anything more to doing exercises like these on your own other than just playing around with the formulas hoping it somehow fits?

[vale nimbus]

<@&286206848099549185>

[twin prawn]

da's wel een pittig bewijsje op zich

[vale nimbus]

ik kan veel van de goniometrie dingen maar die driehoekstoepassingen lukken echt niet

[twin prawn]

Da's altijd een minpuntje van dit soort bewijzen doen, soms zijn ze niet echt makkelijk om te "zien"

Enige manier die ik me kan bedenken om beter te worden is gewoon meer en meer opdrachten en sommetjes doen

[vale nimbus]

had een week vakantie en heb daarin de andere soorten bewijzen ingeoefend en deze proberen uit te stellen omdat ze er moeilijk uit zagen, niet mijn slimste plan

de toets is namelijk morgen

[muted jacinth]

Can someone check me with this solution ? I have used extra things to solve the problem.

<@&286206848099549185>

[devout harbor]

@grave hatch We know AC = 4, so you can break it into two equal parts and get that OC = 2

Now that you have OC and BC, figure out the angle OCB

[grave hatch]

cos^-1(2/4) yea

60

degrees

why do you need that angle

i got the rhombus area as 8root3

[devout harbor]

Have you done sectors before?

Area of sectors, etc

lol

[grave hatch]

yes

but

the diagram looks wonkey ponkey to me

[quiet mason]

ok if its an ellipse

you havr both the axes or whatever

and you can find the ar of the rhombus

[devout harbor]

You need that angle because we can find the area of the sector with it. OCB = 60 degrees or pi/3 radians, then angle DCB = 120 or 2pi/3 radians

[quiet mason]

cant you use ar of ellipse = pi*product of axes lengths

and then ar ellipse - ar rhombus

[grave hatch]

ok what do i do with 120

im not seeing it yet

[devout harbor]

i'll show u

that's the area you get

[grave hatch]

ohhhh

hmakes sense now

[devout harbor]

subtract the area of the rhombus

and you'll get the two shaded parts left

[grave hatch]

yeah

[devout harbor]

but there's actually four

[grave hatch]

thank you so much omg

[devout harbor]

so multilply by 2

np

[primal dune]

Could anybody help me understand a geometry problem?

[dire rampart]

[primal dune]

https://www.whiteplainspublicschools.org/cms/lib/NY01000029/Centricity/Domain/1689/GeoH - Volume Day 1 HW ANS 2015 ML.pdf

Question 8

I can't understand where the 19 or the 12 come from

[crisp grove]

@primal dune 12 is base diameter (inner diam. + outer diam. * 2)... 19 is inner height (total height - base height thickness)

hmm I should rephrase that but I guess you get the idea, right?

[primal dune]

How did you find the outer diameter?

The 10 is given, I know