geometry-and-trigonometry

[idle bloom]

doesn't matter

still is a number

so that square root 40 is the side length of the square right?

[bitter crater]

yea

[idle bloom]

so how do we get the area of that square

[bitter crater]

so just do square root 40 minus 3

[idle bloom]

just square it right?

[bitter crater]

to get the other sides right?

[idle bloom]

we don't need to

[bitter crater]

or you can just do square root 40 times 2

[idle bloom]

no -3

and it's not times 2

[bitter crater]

square

[idle bloom]

it's just the side length ^2

[bitter crater]

my bad

[idle bloom]

yea

do you get why we don't need that -3?

[bitter crater]

k thanks

yea I know

[idle bloom]

sweet

[bitter crater]

area = lw

thanks

is it possible to give me a basic review of geo?

[idle bloom]

that's a large order lol

[bitter crater]

yea acutally that's true

[idle bloom]

I say Khan Academy is great for all math up until calculus

so if you want to learn geometry go there

when you get to calculus there are wayyy better resources imo but until then Khan Academy is 👌

[bitter crater]

already passed calc but I gotta do this review quiz for another class lul

and I don't remember shit from high school

[idle bloom]

gotcha

[umbral snow]

Khan is good for calc 1 as well, but there's better sources these days

Professor Leonard being lit from what I've heard

[idle bloom]

Khan Calc is bad compared to PatrickJMT, Leonard, and Pauls Online Math Notes

[umbral snow]

Wow there's a picture for it and everything. Okay, I stand corrected

[dire rampart]

smh

[upper karma]

trigonometric identities is insane, that is all

[buoyant flicker]

https://cdn.discordapp.com/attachments/496872076569870340/551801488830103585/IMG_20190303_111955_4.jpg

does anyone have any tips on approaching 26 A and B? each of the points are supposed to be like complex points in the complex plane. I know there are several tools you can use to show this, I'm just not sure which ones.

[winged saddle]

,rotate 270

[winged saddle]

o boi

[north tendon]

,rotate 90

[winged saddle]

iM 2 l4zY

[buoyant flicker]

wtf

[idle bloom]

F

[buoyant flicker]

what

[buoyant flicker]

so I know for A, "meet at a point" implies that there exists some point say X so that AQRX, BRPX, CPQX are all circles. so ((A-R)/(A-X))/((Q-R)/(Q-X)) is a real number

((B-P)/(B-X))/((R-P)/(R-X)) is also real

and ((C-Q)/(C-X))/((P-Q)/(P-X)) is also real

but im not sure anywhere to go from this

<@&286206848099549185> anybody got any tips?

[hybrid tartan]

Are there any guides or examples on how to solve problems of the form "given these orientations, draw the shape on the isometric grid"?
I've tried searching on Google, but I've had barely any luck (and the guides I have found either use the wrong way of drawing orientations or are just examples that just show the finished result or the lines being drawn and don't actually explain anything)

(note: the way we draw orientations is by using solid lines to mark different depths and dashed lines to indicate holes)

[mental ether]

(sinx / cosx + siny / cosy) / (cosx/sinx + cosy / siny)

[hybrid tartan]

<@&286206848099549185> (for my question)

[winged saddle]

sorry im braindead

[upper karma]

Wish I could help but i dont even konw what isometric means

[pale wagon]

Hi, is there a way I can describe a rotation matrice in the traditional cartesian form?

[umbral snow]

Yeah, just multiply the matrix

x' = xcosθ - ysinθ
y' = xsinθ + ycosθ

@pale wagon

[pale wagon]

Okay, thanks

[winged matrix]

How much is there to learn in trigonometry in grade 10?

I've only done basic ones to find a certain side with one degree in a shape

<@&286206848099549185>

[upper karma]

There's a lot of trig that later math relies on

[buoyant flicker]

[upper karma]

That gets used in many different ways

[winged matrix]

what about in grade 10 itself

?

[cyan saffron]

Idk, Soh Cah Toa, sine rule, cosine rule,

Unit circle

Radians

idk

[buoyant flicker]

it depends on how your school teaches it

i didnt have that much

some sin, cos, tan stuff and manipulation

but it was just kind of tricky stuff

[cyan saffron]

I did, but I took the more "advanced class" in year 10

[winged matrix]

o

im doing quadratics now XD

i need to do more trig

that gonna be torture

[buoyant flicker]

its just a lot of practice

[winged matrix]

may you give me some problems to solve?

[buoyant flicker]

its not fun but if you get it well by now youll do better in like calc 2 probably

[winged matrix]

if you can

yeah

[cyan saffron]

Manipulations of Soh Cah Toa, Including using the inverse trig button on the calculator to find angles

[buoyant flicker]

idk havent had it in years

[cyan saffron]

Is what I did in year 9 extended math

[winged matrix]

I know soh cah toa

[pale wagon]

im in grade 12 u def need it for imaginary numbers

and vectors

[buoyant flicker]

just like proving manipulations of sin cos and tan rules

[winged matrix]

grade 11

I've skipped a lot of grade 10 trig

thats not good

[buoyant flicker]

think like prove that sin^2 + tan^2 = sec^4 or something.

that's probably not right.

but you get the idea

[winged matrix]

wow

thats part of grade 10 ?

great

[buoyant flicker]

all the trig stuff is related. youll get taught basic rules, and then youll manipulate them

[winged matrix]

alright

[buoyant flicker]

id recommend looking at some videos on khan academy or something. sometimes having a good visual of them can help a lot with trig

its honestly a bunch of bs lol

but its not bad if you practice

[winged matrix]

well its all branched out

I'll do another way I guess

[buoyant flicker]

do you want to major in math or just here for the visit?

[tropic shard]

Honestly, grade 10 I took Geometry, and grade 11 I skipped to precalc. That class basically was a mix of Precalc and Trigonometry for me.

[buoyant flicker]

same

[tropic shard]

All I can say is, it's possible.

[winged matrix]

yes

but I can probs conclude that it's painfull

[tropic shard]

If you have a teacher who could teach, then it'll be really easy for you, relative to what you would do later on in math.

[winged matrix]

yeah

[supple haven]

Sorry to ask so late, but is this correct?

[dire rampart]

,rotate -90

[dire rampart]

uhh

whys there a sin2

[supple haven]

I dont know

I put it there because I thought that'd be correct?

[dire rampart]

not really

[supple haven]

can you elaborate please

[dire rampart]

oh thats a theta

why do they look like a 2 smh

yea looks good

now try write it all interms of cos

[supple haven]

its the letter a, but its used as a theta pretty much

I write my a's like keyboard font

okay lemme attempt that

I got stuck

gimme a sec

wait up hmm

[dark sparrow]

you can simplify this

like, a lot

currently you have $$\cos(a)(\cos^2(a) - \sin^2(a)) - 2\sin^2(a)\cos(a)$$

[somber coyote]

Ann:

[dark sparrow]

it is possible to express this entirely in terms of cos(a)

[supple haven]

almost done

okay sorry that took so long

had to to talk to my mom for a bit

but uh

2cos^2(a) - 3cos(a)

is that correct?

@dire rampart

[dark sparrow]

no

[supple haven]

oh no

[dark sparrow]

can you show your work?

[supple haven]

yes one sec

,rotate -90

[dark sparrow]

k first off

[supple haven]

[dark sparrow]

cos^2 - sin^2 is not cos^2 - 1 - cos^2.

mind your parentheses, ok?

[supple haven]

oh oaky

okay*

uh what else did I do wrong then

[dark sparrow]

i mean there isn't much of a point in looking through your work further

redo it all

[supple haven]

okay one moment please

uh

3cos^3(a) -cos^2(a) -1

?

@dark sparrow

[dark sparrow]

show work? this is still wrong

[supple haven]

agh okay one moment

frustrated in my own incompetence, not at you at all btw

[dark sparrow]

jeez what happened to the parentheses

you should have had cos * (cos^2 - (1 - cos^2)), not cos * (cos^2) - (1 - cos^2)

[supple haven]

which step did I mess up in

oh hold up

I noticed I also misdistributed a negative

dang man I really need to stop being dumb

oh well

@dark sparrow can you show me how it was supposed to be done? Im at a complete loss

[dark sparrow]

ok

[supple haven]

thanks

[dark sparrow]

...class, sorry

[supple haven]

its fine

can someone help me solve 2cosx + sin2x = 0?

im so ????

it says that I go from 2cosx + 2sinxcosx = 0

to 2cosx(1+sinx)

but how tf does that happen???

<@&286206848099549185>

[fair lily]

@supple haven do you mean to ask how $2\cos{x} + 2\sin{x}\cos{x}$ becomes $2\cos{x}(1 + \sin{x})$?

[dark sparrow]

()

[somber coyote]

inubakari:

[dark sparrow]

@supple haven 2cos(x) * 1 + 2cos(x) * sin(x) = 2cos(x) * (1 + sin(x)). distribution.

[supple haven]

uh

one sec

(2cosx)+(2sinxcosx)=0

becomes

2cosx(1+sinx)=0

[fair lily]

$2\cos{x} + 2\sin{x}\cos{x} = 2\cos{x}(1) + 2\cos{x}(\sin{x}) = 2\cos{x}(1 + \sin{x})$

[somber coyote]

inubakari:

[fair lily]

$\because ab + ac = a(b+c)$

[somber coyote]

inubakari:

[fair lily]

@supple haven do you understand?

[supple haven]

uh

let me process for a sec

[fair lily]

hmm

[supple haven]

AHA

I understand now

2xy = 2yx

im big dumb

ty so much

[fair lily]

it's ok

😄

[supple haven]

you're a life saver

I need to learn an entire chapters worth of work

in uh about

3hrs 20 mins

[fair lily]

oof

[supple haven]

uh so

I have sin2xsinx=cosx

and um

like

after converting it to 2sin^2(x)cosx=cosx

the guy subtracts 2sin^2(x) from both sides

rather than just dividing and getting 2sin^2(x)=1

why is that?

[hard gale]

you're losing potential solutions by dividing by cos(x) (namely those where cos(x) = 0)

[supple haven]

oh jeez man

is there any definite way to know you're losing solutions?

[hard gale]

that's why you usually don't divide your way out of things with equations

try factoring first

[supple haven]

okay, so avoid dividing out of equations before factoring and stuff

got it

[hard gale]

well if you're dividing by something, it's pretty obvious that you're excluding the case when the thing you divided with = 0

you can divide, it's just that you have to check whether the case you excluded by dividing is solution of the original equation or not in the end

[supple haven]

ah okay

tyyyy

uh

lets see if I do rad sin^2(x)

it will be +-sinx right?

[dark sparrow]

|sin(x)|.

[supple haven]

isnt that the same thing?

woo I got the thingy right

onto the next

[dark sparrow]

no, it's not.

[supple haven]

plus or minus sin is different from the absolute value of sin?

[dark sparrow]

i wonder why people even teach this idea that ±a and |a| are the same thing

[supple haven]

I dunno

I just know that thats what my math teacher said

[tropic shard]

I just realized what you guys were talking about lmao

[supple haven]

maths hard man

[tropic shard]

Math's a different way of thinking.

[supple haven]

yeah it really is

I wish teachers would understand that and go over something in a few different ways to ensure everyone understands how you ought to thing about solving an equation :/

@dark sparrow erm are you going to explain the cos thing?

[tropic shard]

Where are you guys right now?

[dark sparrow]

oh yeah sorry let me write it out on paper now

[supple haven]

thanks

im in cali

its 5 am here

i've been up all night studyingh

[supple haven]

how would I go about solving

cos150 + cos45

no calc

[tropic shard]

Know: cos(45°)=√[2]/2 (gotta memorize from unit circle

[supple haven]

using the sum and difference formula

[tropic shard]

oh

[supple haven]

im sorry im like really out of it so im forgetting to mention a lot of stuff

im hoping the caffine will help during the test

[tropic shard]

lol it's okay

,, \cos(x)+\cos(y)=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2}) ?

[somber coyote]

𝓗𝑒𝓃𝓇𝓎 𝓒𝒶𝓈𝓉𝓁𝑒:

[tropic shard]

Using that formula?

I'm not a master at the names, lol.

[supple haven]

uh

cos(a+b) = cos(a)cos(b) - sin(a)sin(b)

[hard gale]

decompose 150 into 90+60 tbh

[supple haven]

ah thats what I was thinking but was worried it'd be wrong

[tropic shard]

Oh so yeah, so could we use what I said earlier cos(45°)=√[2]/2

[supple haven]

radical thanks man

i'll try to work on it a bit

I need to finish this practice test right quick to make sure that im at least decently versed in what the test will be on

[tropic shard]

Forget what I said unless Sum to Product is on the test as well

[hard gale]

i don't know cos(195/2) but ye

[supple haven]

sum to product?

[tropic shard]

idk I'm just silly. I thought they didn't want us to use our knowledge of unit circles lmao.

Well, the formula I used on top changes the sum of 2 cosines into the product of 2 cosines, hense the name.

[supple haven]

ooo

[tropic shard]

But that could be very extra if you haven't gone over it in class, and highly unnecessary

[supple haven]

I might ahve learned it and am just blanking atm

but thanks for the info amigo

i needa absorb all this math knowledge

[tropic shard]

It's okay, just don't stress out xD

Eat some tasty food now to stimulate your sense and better your memory-ish

[supple haven]

good idea

I think my moms making breakfast in a bit

im just stressing out bc I failed the test once already and cant afford to fail again

[tropic shard]

The more relaxed you are, the less likely you are to doubt your memory and answers on the test. Last thing you want is to say to yourself that you're wrong when you're actually right.

[supple haven]

o o f

thats me every test man

[tropic shard]

I feel you

[supple haven]

thankfully I studied the basic stuff really hard and memorized the derivations

so as long as I can remember the original 3 all fine and dandy

then I can find the other ones ill need

[tropic shard]

What are the 3?

[supple haven]

technically 6

uh

lemme take a pic of the sheet

[tropic shard]

[supple haven]

gk;dfgkdslf

im going to yeet my phone across the room

its a taking a bit of time to send

[tropic shard]

lol

[supple haven]

oh fudge

I needa pay my phone bill

dang no wonder

one sec

[tropic shard]

oof

[supple haven]

can someone please explain to me how the bottom disappeared when they multiplied by the reciprocal ?

[upper karma]

Does anyone know how to find coordinates of D plz?

[dark sparrow]

what have you tried so far?

[supple haven]

o h

t h a n k s y ' a l l

okay

[dark sparrow]

i'll just take that down

[supple haven]

alrighty then so im just going to assume a reciprocal will cancel out the opposite and not for the other things

I hope to god I manage to remember that

[robust socket]

https://www.wolframalpha.com/input/?i=5(3%2F(3%2B2\sqrt(3)))+%3D+5(2%5Csqrt(3)-3)

can someone please explain me how to get to the second part of the equality starting from the first one?

blonde moment

[valid pike]

How do I do this

[limpid basin]

If you label the vertex at the top D, do you know what is the measure of angle ABD?

in terms of x ^

[solar shale]

I need help with using tan cos and sin on the calculator

[upper karma]

what you need fren

@solar shale

[solar shale]

So on my paper it’s say to sin 0.42 and the answer was 24.8 but I can’t get it and also it says round to one decimal place

[upper karma]

are you in radian or degrees?

,w 0.42 radians to degrees

[somber coyote]

Display results online and refine query

Requested by Babowwz#1394

[upper karma]

lemme see the whole problem

[solar shale]

Ok

[umbral snow]

So you are not plugging in sin(0.42) into your calculator

[solar shale]

Wdym

[upper karma]

Its sin inverse

[umbral snow]

You are looking for a number A, that when you take the sin of it, you get 0.42

[upper karma]

So look for sin^-1

[solar shale]

But like how would ik that if it were to come up in a quiz

[upper karma]

On your calculator

[solar shale]

Like is there a rule for acute angles?

[upper karma]

Its all calculator

[umbral snow]

sin(0.42) = A
As opposed to
sin(A) = 0.42

[solar shale]

Oh ok

[upper karma]

Its sin inverse of (0.42)=A

,w asin(0.42)

[somber coyote]

Display results online and refine query

Requested by Babowwz#1394

[upper karma]

,w 0.433445 radiant to degrees

[somber coyote]

Display results online and refine query

Requested by Babowwz#1394

[umbral snow]

You can move the sin from one side to the other, but you have to use sin inverse if you do:
sin(A) = 0.42
A = sin¯¹(0.42)

[solar shale]

Ah I get it tnx for the help

How do I do this

[thin oriole]

I need some serious help

I need 2 equations that determine the perimeter and area of of the stages without having use the number of previous triangles

I cant figure it out

The paper is due tomorrow so uh 🤣

[solar shale]

<@&286206848099549185>

How do I do this

[dark sparrow]

do what

[solar shale]

I don’t understand how I answer the problem

Like matching the ratios

[dark sparrow]

there's a mnemonic some people use to remember which ratios are called what

SOH CAH TOA

[solar shale]

Yeah I’ve heard of it a bit don’t know too much tho

[thin oriole]

Does anyone have any idea for my request

🤣

I’m kind of desperate right now

[dark sparrow]

sin = opp/hyp
cos = adj/hyp
tan = opp/adj

opp, hyp and adj stand for opposite, adjacent and hypotenuse respectively

[solar shale]

👍🏼 I get how it works

Tnx

[upper karma]

[wild flame]

does anyone know how to use double angles and half angles?

[wild flame]

uhh

[upper karma]

[wild flame]

i guess more specifically how do I use the triangles

[upper karma]

wut is that

@wild flame wdym by double/half angle?

[wild flame]

double angle identities and half angle identities

[upper karma]

Oh

Do you know cos/sin addition

If then it trivial

[wild flame]

no idea what that means

[plucky marlin]

sin(a + b) = ?

[wild flame]

no thats sum and difference

[plucky marlin]

yeah

[wild flame]

oh then yes

[plucky marlin]

you know that ?

sin(2a) = sin(a + a) now use sine addition formula

[wild flame]

i mean how do I use triangles to solve a problem dealing with double/half angle identities

[plucky marlin]

[wild flame]

let me take a pick of my problem

[low veldt]

sin(a+b)=sina×cosb+cosa×sinb

[wild flame]

that

i dont know where to start or how to start the problem

[plucky marlin]

$\sin(2 \theta) = 2 \sin(\theta) \cos(\theta)$

[somber coyote]

soap:

[plucky marlin]

https://prnt.sc/mtcb2f

so now u can see sin(theta) = 4/5

u have cos(theta) and also sin(theta)

have fun

[wild flame]

where did u get the 25 and 9

[plucky marlin]

5^2 - 3^3

pythagorean theorem

[wild flame]

ohh

[plucky marlin]

or u can just notice its a 3-4-5 right triangle

[wild flame]

ty for those

[low veldt]

PQ²=1²+1²-2×1×1×cos(A-B)
PQ²=(cosA-cosB)²+(sinA-sinB)²

🤔

[royal lichen]

https://gyazo.com/ddee155ac72442c5d0fedbbc2fdfcc07 can anyone help me

[dark sparrow]

what have you tried so far and where are you stuck?

[low veldt]

@royal lichen
30-gon
circumscribed?
inscribed?

[low veldt]

30×(1/2×r²×sin 12°)

30×(1/2×(r/(cos6°))²×sin12°)

252.61
255.40

[upper karma]

I know sin sum and sub
can't use sin(2a) = sin(a+a)

[atomic fable]

Lol what are you doing

[upper karma]

I mean that was what literally happened here

[buoyant flicker]

do you guys have any advice on showing this?

I'm trying to prove that the if you have a triangle, with a circle inscribed and a circle formed by going through the midpoints of the triangles. I'm wanting to prove that a circle that is outside tangent of the inscribed circle can not go through all 3 points.

some sort of computational proof is what I'm trying to see

[upper karma]

Do you mean applying cartesian coordinates, or beyond?

[buoyant flicker]

anything really, but I've been using complex plane coordinates in the class @upper karma

[rotund orbit]

@steady sleet hello friend quick question. if we have triangle ACD and B is a midpoint of CD then how can we show that AB < (AC + AD)/2?.

am getting AB < (AC + AD + CD)/2.

[buoyant flicker]

are you talking about the angles @rotund orbit

[rotund orbit]

nope sides.

[buoyant flicker]

nvm

hmm

[rotund orbit]

it says

But then AB is the median of the triangle ACD and hence (by a well-known result in plane
geometry)
AB < (AC + AD)/2

am lost here.

hmm this well known result doesnt seem to be that well known.

[steady sleet]

why did u ping me

[rotund orbit]

because cant understand how they got the inequality.

because you iz mah friend.

T-T sry.

[dire rampart]

lol

[rotund orbit]

<@&286206848099549185> .

[rugged plume]

tan theta -1 =0 find all solutions when doing this i only seem to get 1 which is pi/4 however my professor tells me there are 2

[gritty siren]

There are infinitely many solutions actually

tan is π-periodic

[rugged plume]

sorry let me reprahse, given 0<theta<2pi hes asking for solutions for example sqr(2)sin(theta)-1=0 comes to the solutions being pi/4 and 3pi/4

[gritty siren]

Well then, in ]0,2π[ you also have 5π/4 as a solution to tan(θ)-1=0

[rugged plume]

can you explain how you came to this answer

[gritty siren]

tan(θ)-1=0 means tan(θ)=1 means sin(θ)/cos(θ)=1 means sin(θ)=cos(θ)

this happens twice

[rugged plume]

i see what you're saaying

thank you

i didnt look at it in terms of sin/cos

[gritty siren]

You could also memorize special values of tan

not very pleasant though

[rugged plume]

I looked into doing that and its quite daunting however I might give it a go to make life easier in the future

[zinc grail]

could someone

help

a desperate lost soul

who has a big boi test tomorrow

[buoyant flicker]

just say your question

[zinc grail]

so just arc length/ raduis?

then convert to degs?

[edgy vortex]

,, \frac{x}{360} = \frac{AB}{2OA \pi}

[somber coyote]

CraftMechanics:

[edgy vortex]

The angle of an arc is proportional to its its length

[glad ocean]

that profile picture

sorry

[umbral snow]

@onyx basin
Let D be the origin, we can place everything on the Cartesian plane. What's the equation for AC? For DB? Where do they intersect?

[onyx basin]

to be frank with you, i haven't the slightest clue

[umbral snow]

With which?

[onyx basin]

in class, we weren't taught to place diagrams onto coordinate planes and solve that way lol

the teacher provided this in the answer key, and i'm having issues figuring out how she reached that

[umbral snow]

I'm not sure how she did it. I tried my own method, by assuming DC was a specific value

[onyx basin]

Alright then. Could you try explaining your Cartesian plane solution then?

[umbral snow]

Assume the length is L

AC:
y = 40 + (-40/L)x

DB:
y = (25/L)x

Solving them simultaneously gives the intersection, y = 200/13
Suprisingly, L divides out!

[onyx basin]

so we'd set 40 + (-40/L)x = (25/L)x?

[umbral snow]

You could do that to solve for x. Then, solve either for y

[onyx basin]

right, thanks

[upper karma]

0. Two distinct points uniquely determines A Euclidean line
1. Two Euclidean lines meet at 0 or 1 point
2. Given a point, there is exactly one line passing by the point which doesn't meet with a given line

I wonder if this condition is special

Maybe there are cases where there is no line passing by the two points?

Or..
Given a line, there is exactly one point on the line which doesn't form a line with a given point

I think Linear algebra also agrees with the euclidean geometry

Hmm there should be some related property

[dark sparrow]

what?

[upper karma]

Why?

Is there any problem

[dark sparrow]

i don't get what you're asking

[upper karma]

I don't get where you got the impression that I'm asking

I was just thinking.

(And maybe to get some comments on it if possible)

Hm I wonder, how did Euclid know there is exactly one parallel line passing the given point?

(Now this is the question)

[upper karma]

how are the two lines that intersect with each other in such a way that a 90 degree angle is created, called?

[dark sparrow]

such lines are called perpendicular

[upper karma]

ty

so there are perpendicular, parallel and what lines

those that intersect with each other but don't create a right angle

[dark sparrow]

those don't have a special name

there is no special word for a pair of lines that intersect at a non-right angle

[vale inlet]

@upper karma After getting out the good ol fashion paper & pencil, it's not too bad to prove the uniqueness of parallel lines passing through a said point, using the a + __v__t thing

[spring marsh]

Am i dumb or what? How can i get the length of EF ? Please help.

<@&286206848099549185>

[spring marsh]

anyone?

[dire rampart]

similar triangles

[spring marsh]

Solved with proportion.

[dire rampart]

epic

[shell garden]

h

hi

so,

i have a big issue with visualizing the word problem

because im dumb

its getting angles

which ik how to do ez

[upper karma]

what problem

[shell garden]

but making the triangle i cant visualize

A tow truck raises the front end of a car 0.75m above the ground. If the car is 2.8m long what angle does the car make with the ground?

ialready know the answer

i just need help

with making the actual triangle

since you dont get it u need to make it

[sweet root]

In a circle inscribed in a triangle, is AB perpendicular to CD?

[tender adder]

uuum

I think so, if the line makes 2 right angles

[sweet root]

@tender adder if you connect the points of tangency and the apex of the triangle

[supple haven]

vectors are kinda hard man

[tender adder]

How do I solve these with the law of cosines?

[vernal tusk]

well whats the law of cos?

@sweet root yes, ab is perp to cd

the radius of a circle is perpendicular to the tangent line that intersects the circle at that point

you can prove it by contradiction (assuming the opposite)

turn your paper sideways

so that the triangle is sideeays and the circle is "lying on" its right leg there

draw a random point off that leg that isnt the center of the circle but is near to it

so like if you just drew it east of the center, or west, or wherever

[sweet root]

Thank you

[vernal tusk]

oh wait no, the easier proof is

pick a random point on the tangent line besides the pt on the circle

connect it to the center

then draw a perpendicular from the center

[sweet root]

Ok

[vernal tusk]

perpendicular really just means that its the closest point to a given line, really, if that makes sense

because if you draw any other point

you can drop a perpendicular and now that other point forms the hypotenuse of a right triangle, which is always longer than its legs

hold on

horrible, horrible edplanation

explanation

you're welcome and you are right

i just like to teach the proof for stuff instead of just blindly saging it

watch this video if you want to understand the proof

it takes a sec but then you go "ohhhh" after thunking about it a little

https://youtu.be/St3MU5i3chc

[tender adder]

@vernal tusk this

[vernal tusk]

ik what it is

i just meant to remind you of it

[devout iron]

need a lil help

[glad ocean]

6

x·x = 9·4

[glad ocean]

x^2 = 36; x = 6

[devout iron]

i finished that

ty

@glad ocean

[glad ocean]

oo

76°

x = 76

PHEW it finished

well that's all my DMs with you and Matt down the drain ive played this out well

[devout iron]

@glad ocean this one pls

last ohe

[glad ocean]

k

[devout iron]

?

@glad ocean !! due in 7 mins 😭

[glad ocean]

radius =
CU = 36

radius = (hold up im solving it)

[devout iron]

oh ok

[glad ocean]

radius = 20

i think

ur welcome

[devout iron]

ty

it's wrong

[glad ocean]

try 8

or 16

im doing this all in my head asdf

[rancid mantle]

plz help

[twin prawn]

[upper karma]

,rotate

[upper karma]

Q38

[keen aspen]

To find the period of cosine, you do 2pi/b

where b is the coefficient in front of the x

so 2pi/(8/6) -> 2pi*6/8 -> 3pi/2

[upper karma]

Thank you!

[halcyon pumice]

Hey, how would I go about solving this:
$\int_0^\pi \sqrt(2-2cos \theta)d\theta$

[somber coyote]

Thouv:

[halcyon pumice]

Trying to find the average distance between two points on a circle, not sure if this should go in here or calc

Ik the numerical answer is 4 but I'd like an "algebraic" solution

[vale inlet]

@halcyon pumice probably worth posting an integral question in calculus, but to solve that use the substitution u = cos(theta) & that fact 1 - u^2 = 1 - cos^2 = sin^2

[upper karma]

Hellppp

[shell garden]

can someone explain to me how to solve this ? a tow truck raises the front end of a car 0.75m above the ground. If the car is 2.8m long what angle does the car make with the ground?

[upper karma]

Hey

[shell garden]

hey

[upper karma]

I need to learn how to do basic trigonometry for a test tomorrow

never learnt it before

i need to find out how to do a question similar to this https://cdn.discordapp.com/attachments/549252850744819712/553321614511702035/Obtuse.png

[upper karma]

@upper karma

talk here

Okay

I found a website that explains the law

http://www.softschools.com/math/calculus/using_the_law_of_sines_to_find_an_unknown_angle/

And you can use it to solve the question above

Ask me if you don’t get something?

👍*

dont get it

Which part

having clicked it yet

havent*

XD

Ok i am not going to bother learning this

Atleast for now

"cross multiply"

XD

How much of trig do you need to know?

Also cross multiplying is real simple

all i need to know is how to do the qustion i sent you

Like thats literally it

I dont need trig for anything else anytime soon

Just for that question and thats it

Cross multiplying is just multiplying the highlighted parts

And putting them on different sides of the equal sign

ah right

Want to try solving the question above?

sure

https://cdn.discordapp.com/attachments/549252850744819712/553321614511702035/Obtuse.png

@upper karma

had to go for a sec back now

Okay let’s see

First we gotta fill in the law of sines for this triangle

Thats what i dont get

How do i know which is sin a

ect

So to help you can label each side of the triangle a b and c

does it matter which side i label a

or b or c

Nope

Aight so ill do top left corner a

bottom left b

bottom right c

Right

https://cdn.discordapp.com/attachments/549252850744819712/553321614511702035/Obtuse.png

So we can now say 16/sin(a) = 10/sin(b) = c/sin(c)

Just plugging in the lengths where the letters are

Yep i get that

Progress

Now we look at the angles across from the numbers

Example 16 and angle y

and 10 and 34

Yep

We're getting somewhere

Now we know that 16/sin(y) = 10/sin(34)

Plugging in the angles across from the lengths

ok theres where im losing it a little

Right

So 16 is just a. then the a after sin below the fraction is the angle opposite

https://cdn.discordapp.com/attachments/326138757474680852/553334452365950987/image0.png

Yep

You just put the sin before it

So A = 16 then sin(y)

Yep

Then B = 10 and Sin(34)

Yep

You got it

Aight got it

And then we can do that cross multiplying thing

16/sin(y) = 10/sin(34)

16 • sin(34) = 10 • sin(y)

Make sense?

so you take the length from A and times it by the angle from B?

and vise versa

Yes

what next

16 • sin(34) = 10 • sin(y)

We can divide both sides by 10

16 • sin(34)/10 = sin(y)

So now we have sin(y) on one side

[tropic shard]

isolate it more, until it's as lonely as I am.

[upper karma]

👌

And after this long shit hole we get the final step

XD

[tropic shard]

Yerp

[upper karma]

You better hope you know your calculator buttons

[tropic shard]

Wow, I see you've done a lot of work on this xD

[upper karma]

XD

oh my

Ok so 16 times sin(34) i get that

cross multiply

thats cool

ah so you've arranged it into an equation i see

Yeppppp

then basic algebra from there

Yeh until this part

16 • sin(34)/10 = sin(y)

i dont use the dot as a multiply from where im from

Yehhh I just don’t want to confuse with x variable

[tropic shard]

× can be used, but reserved for the variable x.

[upper karma]

Okay so lastly

Sin^-1(16 • sin(34)/10) = y

And boom

(Sin^-1 is arcsine or some shozzle)

[tropic shard]

sin⁻¹(16·sin(34°)÷10)=y

[upper karma]

Fuck I want that shit

[tropic shard]

lol

[upper karma]

SO BOOM YOU JUST DID LAW OF SINES THING

[tropic shard]

You're ready to be a lawyer ;)

[upper karma]

you gotta slow it down for me honestly

Im sorry but im not used to thhis complexity xD

[tropic shard]

But you do know the whole arcsin(sin(θ))=θ deal, right?

[upper karma]

So go back to the start of the cross multiply

no i dont

never done trig in my life

[tropic shard]

Ah

[upper karma]

I wont be learning it for another year. just popped up on my test and need to know how to do it

[tropic shard]

So you mean at 16×sin(34°)÷10=sin(y)?

Oh, luckily it's easy to do once you learn it

[upper karma]

XD

So after cross multiplying

16 • sin(34) = 10 • sin(y)

[tropic shard]

(oh)

[upper karma]

We can divide both sides by ten to make sin(y) Lonely af

Hence

16 • sin(34)/10 = sin(y)

Henry make it fucking beautiful

so put it in an equation first right

Yep

and the equation would be 16 x 34 = 10 x sin(y)

Sin(34)

or is that wrong

[tropic shard]

(oh hi back) 16·sin(34°)÷10=sin(y)

[upper karma]

what does the sin mean

Why does it need to be infront of the 34

It’s a function that your calculator will do something with later

[tropic shard]

It's a function like f(x)=(g(x)²+2)

[upper karma]

16/sin(y) = 10/sin(34)

So we had this from law of sines

Henry make it beautiful

and simple

[tropic shard]

In trig they'll go in detail about it being a ratio between sides (SOHCAHTOA) but it'll seem random now so trust the Leonard, for now.

[upper karma]

pls

XD

[somber coyote]

𝓗𝑒𝓃𝓇𝓎 𝓒𝒶𝓈𝓉𝓁𝑒:

[upper karma]

Yay

yayayy

then do what it would look like in cross multiplication

I get the whole SOH CAH TOA concept. Like its a different method depending on what angles or sides you've been given

TOA Is like opposite and adjacent and something else

So you multiple the highlighted

(Cross multiplying)

Aight i get that part

16 • sin(34) = 10 • sin(y)

Make it beautiful

[tropic shard]

Un.

Umz

[upper karma]

xd

[tropic shard]

16·sin(34°)=10·sin(y) lmao

[upper karma]

XD

So now we divide everything by ten

[tropic shard]

oof brb

[upper karma]

ayyy

its gone 😮

[tropic shard]

,, \frac{16sin(34)}{10}=\frac{10sin(y)}{10}

[somber coyote]

𝓗𝑒𝓃𝓇𝓎 𝓒𝒶𝓈𝓉𝓁𝑒:

[upper karma]

Yeeeee

Would i enter that into my calculator

[tropic shard]

Not yet

[upper karma]

Not yet

ah right as we dont have y

Now we have

Do the honors

[tropic shard]

,, \frac{16*sin(34)}{10}=sin(y)

[somber coyote]

𝓗𝑒𝓃𝓇𝓎 𝓒𝒶𝓈𝓉𝓁𝑒:

[upper karma]

how do you remember all these commands lmao

XD

human calculator

[tropic shard]

Y'all can do it too one day

[upper karma]

ok so what next

Okay so we don’t want sin(y)

We just want y

divide by sin?

[tropic shard]

Nope,

[upper karma]

oh wait wtf

xD

how would that ever work

We gotta make shit complex dog

ok cat

https://cdn.discordapp.com/attachments/326138757474680852/553342283488034826/171055699479101450.png

We gotta us sin^-1

Or arcsine

To the power of negative 1 right?

Yep

[tropic shard]

Nope

[upper karma]

um

[tropic shard]

It's notation.

[upper karma]

Same thing ;-;

and what is notation sir?

[tropic shard]

Doesn't mean the same thing.

[upper karma]

Shhhh

He don’t need to know da shizzle

Just what da shizzle do

[tropic shard]

Was that why you were saying "divide the sine"?

[upper karma]

OKAY

So

Arcsin(16 • sin(34)/10) = arcsin(sin(y))

Good flipin luck Henry

Not a clue what acrsin is

arc

It basically gets rid of any sins

ok lets pretend i have any idea what this means

continue

But you can type this into your calc

Arcsin(16 • sin(34)/10) = arcsin(sin(y))

Is y = arcsin(16•sin(34)/10)

And then you can type the arcsin(16•sin(34)/10)

To get an answer

Right...

XD

I'll just write some random things down and hope I get a mark from. It

👌

This stuff is complicated

I get up to the arcsin

Once you get to arcsin

Just remove the whole left side of equation

So arcsin(sin(y)) = arcsin(16•sin(34)/10)

Say fuck you left side of equation

And keep arcsin(16•sin(34)/10)

And just put it in calculator

That is domestic abuse

Math likes it

And boom you got y

My brain hates me

Math will do that

[tropic shard]

,w calc(arcsin(16*sin(34)/10))

[somber coyote]

Display results online and refine query

Requested by 𝓗𝑒𝓃𝓇𝓎 𝓒𝒶𝓈𝓉𝓁𝑒#3049

[tropic shard]

Incase your teacher wanted up to the 99th significant figure.

[upper karma]

Yep

Cheers

XD

Needed that

Also make sure your calc

In some mode

Are you trying to kill me or something

What mode

Radian mode

Holy shit that thing really not drawn to scale XD

[tropic shard]

wait

[upper karma]

Wait

[tropic shard]

is it 34°?

I bet it is,

[upper karma]

Yep

[tropic shard]

So not radians

Put your calc in degrees

[upper karma]

WAIT YEH

XD

Fucc

[tropic shard]

Wolfram Alpha, you lied to us

[upper karma]

Damn it

,not radianscalc(arcsin(16*sin(34)/10))

👌

[tropic shard]

,w calc(arcsin(16*sin(34)/10)) degrees

[somber coyote]

Display results online and refine query

Requested by 𝓗𝑒𝓃𝓇𝓎 𝓒𝒶𝓈𝓉𝓁𝑒#3049

[tropic shard]

df

[upper karma]

??

[tropic shard]

Should be 63.5°

[upper karma]

not drawn to scale

Don’t question my DMs

O

O

Well I have to night night

Wait so what's the final answer

To that questiob

Question

@tropic shard

[tropic shard]

@upper karma

Here

oh no am i too late

[upper karma]

Huh

@tropic shard

[tropic shard]

Hi

Oh

Sorry I misread that,

The angle y should be ≈ 63.5°

[upper karma]

Aight ty

It says the angle is obtuse

[tropic shard]

oh

One moment,

116.5°, but I gotta think of how to organize the reasoning

@upper karma Idk how to without referencing the unit circle, sorry man.

(a) can be found by doing 90° minus the angle (63.5°). Since we see there're 2 a's, we multiply a by 2, and add that to our original angle to get 116.5°.

[tacit swan]

https://cdn.discordapp.com/attachments/269573202018041856/553357984714194954/image0.jpg I'd like some leads onw how to start off this

[tropic shard]

A=½a·c·sin(B) may be more helpful.

brb

[tacit swan]

danke

[solar shale]

I need help with finding side lengths and angle measure

How would I know when to use cos, tan or sin

And I get confused with identifying the adjacent and opposite placement in the triangle <@&286206848099549185>

[mystic shadow]

Given diameter d and height h, with 0<h<d, what is the value of x?

geometry was never my strong suit

thanks guys 👍

[forest niche]

Anyone help?

Elevation and depression?

[umbral snow]

Sure, I can definitely help

[forest niche]

.>

Alright thanks

[umbral snow]

Can you get that far?

[forest niche]

Thank you!

I know the rest

[umbral snow]

Kk, cool!

[forest niche]

Really appreciate it

[solar shale]

I need help

B and c are the answers I don’t understand how to solve it

[forest niche]

It’s b

[solar shale]

Read above I got the answers I don’t understand how he got it tho

[forest niche]

Depression is the same as elevation

That 23 depression would be used as tan 23

[solar shale]

I don’t understand how I’m supposed to know if x is above or below of fraction

[forest niche]

Well since it’s Tan

You do Opposite hypotenuse

[solar shale]

Isn’t opp/adj

[forest niche]

Since the opposite of 23 is 210

OOps lol

Yea

Mb

You put 210 on top

[solar shale]

Wait so 23 is the hypotenuse right?

Or which side is

[tacit otter]

oi

Can someone help me?

[mortal stirrup]

so i figured out that CA=15

thats about it

[dark sparrow]

that is correct

as in

AC is 15, yes

[mortal stirrup]

im not sure how to find the things on the right, though

[dark sparrow]

well

have you ever encountered those things before

[mortal stirrup]

yeah

[dark sparrow]

yeah ok

[mortal stirrup]

but this one is confusing for some reason

[dark sparrow]

there's a mnemonic some people use to remember what ratio is called what

it's a definitional mnemonic so it's one of the few i personally am okay with

[mortal stirrup]

is it the sohcahtoa thing?

[dark sparrow]

SOH CAH TOA, yes

[mortal stirrup]

ok

so would sin A=8/17

[swift lava]

Yes

[mortal stirrup]

i got around 28 degrees

then cos A=15/17

i got 28 too....

[dark sparrow]

you don't need to find angle A itself

only its sin, cos and tan

[mortal stirrup]

on the worksheet it says "round to the nearest degree"