geometry-and-trigonometry

[silent plank]

cos is a function

cos isn't a variable multiplying those numbers

[glad shadow]

Okey thanks

[wind heart]

I got more of this done today, and I want to make sure that this is correct. Can somebody check to make sure? Thanks

There.

[coral sapphire]

Yep looks good

Try to be a bit neater though, like the H looks like an A, i assumed the last point was D etc.

Use arrows instead of segments to point towards things, making it more clear the line isn't part of the shape

[wind heart]

Thank you my savior

[upper karma]

quick question

what exactly is an inverse tangent

[thorn linden]

its what you use to find an angle if you are given the two sides of a triangle that is not the hypotenuse

[rich carbon]

$$\arctan(\frac{opp}{adj})= \theta$$

[somber coyote]

Matt_:

[rich carbon]

$$\tan(\theta)= \frac{opp}{adj}$$

[somber coyote]

Matt_:

[rich carbon]

@thorn linden

[sweet cosmos]

@rich carbon severely off topic but why the two $$? i thought the bot only took one. im super new to this server so haha

i get it texit

[rich carbon]

so, i think one $ is for inline, and $$ is for not inline

ie $\text{this is some inline latex }\frac{a}{b}$

[somber coyote]

Matt_:

[rich carbon]

and $$\text{this is not inline} \frac{d}{dx}$$

[somber coyote]

Matt_:

[sweet cosmos]

ohhhhhhhhhhhhhhhhhhh ok thanks. ive never used this bot or attempted to yet so ill have fun later

[rich carbon]

other way around my bad

[sweet cosmos]

yup

[rich carbon]

$$ is inline

[weary drift]

\dv{x} for future reference

[rich carbon]

i prefer \frac{d}{dx}

[night tiger]

how does one go about proving that the midpoint of a line segment is on a bisector of the segment

[sly marlin]

@night tiger How do you define "midpoint" and "bisector"?

[flint pelican]

For this problem the two values for t sal finds are 103 and 262 but

I have never seen that cos(2pi-theta)

I used cos(-theta) instead to get my second value

I got 103 like him but not 262

Why do we use cos(2pi-theta) here and not cos(-theta)

[flint pelican]

<@&286206848099549185>

[upper karma]

@flint pelican cosine has a period of 2pi

@flint pelican remember cos is even, hence cos(x) = cos(-x)

[flint pelican]

Hm wait for sine it is pi-theta though right

To find the other angle

Like if it was arcsin(-0.2)

That would have 2 values so you would get -0.201

And then to find the other value it's pi- -0.201 right?

But if it is cos it is arccos(-0.2)

Which gives 1.7

And then I do 2p - 1.7?

So for sine it is pi- theta and for cos it is 2pi-theta?

[flint pelican]

Ye actually that sounds right and it's giving me the correct answers

[blazing skiff]

Hey I need help with word problem

[upper karma]

Just ask 😀

[flint pelican]

Idk where I messed up I mean

I thought we were supposed to distribute the minus sign

Are we not because I’m certain I saw an example before that could only be solved doing that

And both sine and cosine should be negative since 195 is in Q3

Right?

So idek

[silent plank]

sign errors

in the angle sum identitity

you are doing cos(45°) which is positive

45° and 150° are in quadrants 1 and 2, so the sin of those are positive

[rigid siren]

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[upper karma]

Would it be safe to say that proving these figures congruent can be simply done by saying they’re a rotation or reflection

And if they’re not a reflection or rotation of eachother then they’re not congruent

[silent plank]

no

to justify that they're rotations or reflections
requires you to show that they're congruent first

(via you standard tests: AAS, SAS, RHS, SSS)

[upper karma]

they’re congruent because none of these questions say to prove wether they are congruent or not

they’re asking why they’re congruent

[vague berry]

it’s suppose to be done in about a minute anyone know a shortcut

please help

[silent plank]

wdym, that's essentially asking you to prove congruency.

@upper karma

it even says it on the page

prove triangle ### congruent to triangle ###

[upper karma]

If it is always congruent and it asks why they’re congruent then a proof would be they’re reflections or rotations of eachother

If they’re not, then the question is impossible

Not factoring in translations

[silent plank]

you would reach the conclusion that they're reflections/rotations from congruency.
but you can't go back and say that they're congruent because they're reflections/rotations because you somehow know they're congruent. because that would be circular

not sure what you're trying to do,
are you trying to bypass (AAS, SAS, RHS, SSS) proofs?

[upper karma]

Yes because imo it boils down common sense to a needless degree

Also

It’s a mix of philosophy and definitions

[silent plank]

well for these questions, you'd need to go through those proofs

[upper karma]

The point is to bypass the needless jargon

Also by the teacher’s logic, she uses vertical angles because she can look at it and see they’re vertical

By that logic I can look at something and reason why they’re congruent

[silent plank]

its true that 2 column proofs contain a lot of crap

[upper karma]

Thank you

[silent plank]

but you need to at least admit you would need to reach AAS, SAS, RHS, SSS to prove congruency in these questions

and fill the rest of the crap in to get the marks.
my other point still stands

[upper karma]

If it were to end between us having a battle

Boiling it down to those laws

Reflection and rotations does work

But besides the marks

I agree it is circular

Self supporting

But the philosophy of the question eliminates the uncertainty

[silent plank]

well,
you could apply the same philosophy to all prove questions...

its true because you're already telling me that its true

that's hardly a proof, (could also be a trick and/or error) and won't get you any marks

[upper karma]

But it’s the degree to which you’re boiling down why it is true

They ask to boil it down to such a point that there is no going beyond

I spare the pain and the reflections proof is already built upon it

And they don’t ask me to prove my proof

It’s just loopholes

[silent plank]

i mean you could try and write because "reflections" and get 0 marks for it if you really insist

[upper karma]

I’m not going for the marks to be honest

[silent plank]

generally for these proofs, you need to start it without knowing whether its true or not

[upper karma]

The weights are in order the tests and quizzes outweigh classwork

Plus geometric proofs are the only completely unreasonable aids I’ve seen, and I’ve seen the alg2, precalc and physics course syllabus

Also the premise of these questions is logic, if I cannot use logic to conclude something then it is biased

The general acceptance of something I see invalid

It’s one of those small things I’m fighting but I stand by myself until proven otherwise

Either way, off topic

Well, good talk

[tawdry cape]

Hi, can anyone help me with this? The straight line y=mx+2 is tangent to the circle x^2+y^2+4x-6y+10=0. What are the possible values of m? so far i have plugged mx+2 into the circle and then used the quadratic formula to check when the discriminant = 0, but my answer is not correct. now i'm lost :^)

[silent plank]

can you show your working out?

[tawdry cape]

yes: x^2+m^2x^2-2mx+4x+2

which i factored as (m^2+1)x^2+2(2-m)x+2

$(m^{2}+1)x^{2}+2(2-m)x+2$

[somber coyote]

wexler:

[silent plank]

continue

*=0

[tawdry cape]

so the discriminant is: $2^{2}-8(m^{2}+1)=0$ ?

[somber coyote]

wexler:

[silent plank]

no

the coefficient of x is 2(2-m)

[tawdry cape]

ah

$(4-2m)^{2}-8(m^{2}+1)=0$

[somber coyote]

wexler:

[tawdry cape]

then i get $8-4m^{2}-16m=0$

[somber coyote]

wexler:

[tawdry cape]

then i solve this as a normal quadratic?

[silent plank]

yes

[tawdry cape]

ah

so after using the quadratic formula again i ended up with: $\frac{-4\pm2\sqrt{6}}{2}$

[somber coyote]

wexler:

[silent plank]

which simplifies to:

[tawdry cape]

$-2\pm\sqrt{6}$ !

[somber coyote]

wexler:

[tawdry cape]

that is not meant to be a factorial :^)

[silent plank]

all g?

[tawdry cape]

yes, thank you very much!

[tall dagger]

does anyone know the rule for this?

[umbral snow]

There's 2π radians in the entire circle

That's 2/3 π of them

[tall dagger]

so it's straight up 2/3?

[umbral snow]

Nop

[tall dagger]

hmm

[umbral snow]

It's definitely not more than half of the circle

[tall dagger]

i did this before, however I forgot the rule for it lmao

[umbral snow]

Let's say there's 20 jelly beans, you eat 4 of them. The fraction you ate is 4/20 = 1/5

In this case, there's 2π radians, you use 2π/3 of them. The fraction used is (2π/3)/2π = 1/3

[tall dagger]

OH I see

[coral sapphire]

Remember that the angle to 2π is proportional to arc length to circumference

[tall dagger]

so it's simply just, the arc length divided by the amount of radians?

[umbral snow]

Yussir

[coral sapphire]

Radians are not arcs?

[weary drift]

No, ratio of arc length to circumference

[tall dagger]

not the arc length, but central angle

central angle divided by radians

which would be in this case (2π/3)/2π

[umbral snow]

Fraction = amount used / total amount
For a question this general, I'm not sure you can say more

[tall dagger]

Haha yes, I see. Thank you very much

[upper karma]

hey, can anyone tell me if I’m missing anything on this proof?

[coral sapphire]

That is correct

[upper karma]

thanks muchly

[tired bison]

How would I go about solving sin(5pi/6) without memorizing the unit circle?

Or do I just have to memorize all the tables n shit

[mighty narwhal]

You have to memorize the first quadrant

Sin(x)=sin(pi-x)

So sin(5pi/6)=sin(pi/6)=1/2

[queen python]

erm

is this possible

[proven plover]

hmm idk how to find the area of the missing piece on the bottom left

looks like you can combine the pieces of the circle into one circle and subtract it from the area of the triangle

but it looks doable

[random glacier]

Doesn't look that hard imo

Find the area of the box, then find the combined areas of the circles. Subtract the area of the box from the areas of the circles, and divide that by 2.

That's how I'd go through with it

[proven plover]

you still have that little missing corner on the left

which isnt red

[random glacier]

Which is what I realized 10 seconds before you replied

[proven plover]

https://www.youtube.com/watch?v=xnE_sO7PbBs

[random glacier]

o shid

[queen python]

i see

i found that

its a fun problem

but its a bit advanced for what im trying to tutor

even i couldnt figure it out at first

😩

[proven plover]

Yeah that last little part is tedious

[umbral snow]

I'd be tempted to hit it with an integral

[brave zephyr]

So basically we have to describe the following transformation. The problem I had is that it is going from the original figure directly to a double prime. If anyone could help, that would be appreciated.

[rigid siren]

@brave zephyr now conduct the same transformation on the primed figure

rotatating it again

by however many degrees

[wind heart]

Hi, I have a question. I’m confused.

Why is that that? It makes no sense

If they are parallel, they should be the same

And those don’t add up to 180, yet I’m positive I got the right answer for x and y

[upper karma]

yes that seems incorrect

well what you could do is say that

4x+12=y-10

and y-10+3x=180

4x+22=y

4x+22-10+3x=180

7x+12=180

[wind heart]

Where’d you uhhhh

Get the 4x + 12 from?

[upper karma]

its at the top left corner

[wind heart]

Oh that’s 9x + 12

Sorry for my bad handwriting lol

[upper karma]

oh ok

[wind heart]

btw y = 13, it kinda doesn’t look like a 3 lol

[silent plank]

what were your calculations for y?

[dusky citrus]

hello, i have a question

im legit stuck with solving these two can anyone give me a hand

[silent plank]

what's the question?

[marble trout]

^

[spice pawn]

Why does trigonometry only work on right triangles? By trigonometry I refer to the trigonometric functions.

[upper karma]

can I call this sine rule for quadrilateral?

[dusty coral]

@spice pawn It works on any triangle, right triangles are a special case

[upper karma]

does (x)(cos^2(x))=0 have how many solutoins in the domain of -pi to pi

[umbral snow]

@upper karma
If ab = 0, then either a = 0 or b = 0

[upper karma]

@umbral snow yeah but it changes when you have restricted domain

to -pi to pi

[umbral snow]

What I said still holds

Then
x = 0 or cos²(x) = 0

Are there any x that cause cos²(x) = 0 in that interval?

[rigid siren]

how do you solve log base 2 (2x+5) < 0

[jovial axle]

you need to change base my man

change base to e

[rigid siren]

how do I do that

[jovial axle]

then ln that argument out of the inside

[long sapphire]

what's the need for that

[jovial axle]

,w change of base

🤷

[somber coyote]

Display results online and refine query

Requested by jan Niku#2119

[long sapphire]

$b^{log_b{n}} =n$

[jovial axle]

lol

[somber coyote]

Jiramide:

[long sapphire]

beauty

[jovial axle]

[rigid siren]

ohok

thx

[willow oak]

How did they get these answers?
I set up the equation as theta + pi/3 = pi/6 + 2npi
I then got theta = -pi/6 + 2npi
However, every time I try to plug values into n I get numbers not included in the interval. Did I set up my equation wrong?

[silent plank]

sin (pi/6) isn't -0.5

also sin is negative in 2 quadrants

[steep sedge]

Hi. can someone help with this? the answer is A but I'm not certain how to get to that answer

[silent plank]

what have you thought about? what trig rules/laws can be applied here?

hint: ||ignore what they tell you about cos(60°)||

[steep sedge]

Sin 45 / root 6 ?

[sly marlin]

that's a number

not a trig rule/law

[silent plank]

also degrees °

[steep sedge]

I'm asking for a friend. someone told him this and he's confused as to how they get to it https://images-ext-1.discordapp.net/external/ArXmS9EPIxe0iF6VNzS3IDGMlqUQo_i248aSPKGplVc/https/media.discordapp.net/attachments/628659981318881306/637666656344342528/unknown.png?width=469&height=473

[sly marlin]

maybe more effective for your friend to be here

and does your friend know what the first line is?

the first to second line is just calculations

[static lodge]

One message removed from a suspended account.

[silent plank]

sin(60°) has an exact value of sqrt(3)/2

@static lodge where are you even getting Sin(60) x 0.5 = 2.1
that's not even close.

[static lodge]

One message removed from a suspended account.

One message removed from a suspended account.

[silent plank]

why are you multiplying those two values?

[coral solar]

hi can someone please explain 3c to me

[upper karma]

iT'S ASKING THE Y-INTERCEPT BRO

pUT IN 0 FOR X

AND SOLVE FOR Y

FROM THE EQUATION U DERVIED IN PART b

[coral solar]

why did they square x+1

[upper karma]

@coral solar

they want f(x+1)

so u must replace eveyr x with x+1

x^2= (x+1)^2

[coral solar]

ohh they inserted that into the first equation

thank you so much

[gentle fog]

Not sure which channel this belongs in but does anyone have a good source for graphing complex numbers beside the standing (a+bi)? For example, where you'd place e^i on the complex plane

A good source to learn more about it*

[fringe crater]

@gentle fog , this is a great resource for learning about complex numbers, this guy has a whole series of articles on them that are great and visual: https://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/

Also, check out 3blue1brown's videos on euler's formula:
https://www.youtube.com/watch?v=v0YEaeIClKY

[thorny path]

cos^2(x) = (cos(x))^2

this is the same right

[marble topaz]

Yep

Stuffing the exponent "between" the cos and whatever theta is is how you'll normally see it btw

[gentle fog]

Oh 3blue1brown! I love it. I haven't seen that video, so I'll check it out. Thanks!

[merry totem]

can someone help me

im a bit unsure how to derive

the magnitude of the cross product of two vectors

yes

but does it also equal

|v||w|sin(theta)

i dont really get how u can get dat from sine rul

i know at the start of the proof

u will draw a traingle between v, w, and v-w

and then take the cross product of v with w, w with v-w, and v with v-w

then u can produce the same triangle but in another plane

but then im lost

i mean the magnitude of the vector

magnitude of the vector = |v||w|sin(theta)

its ok

i just dont know what this means

wait dw

i get it

wait i dont

i dont get how the second line related to the 5th line

[sudden locust]

https://youtu.be/c8u-ur791Os In this video, I will take a look at the definition of acute angle, as well as looking at a few examples! Hope you have a great Sunday so far peeps! 🥰👍

[quiet mason]

4 min video for a name given to an angle

[signal hornet]

how come that this is not undefined?

$\arcsin(\sin(\frac{9\pi}{8}))$

[somber coyote]

Jason_Bjorn:

[signal hornet]

$\frac{9\pi}{8} = 3.53$

[somber coyote]

Jason_Bjorn:

[signal hornet]

3.53 is not in the domain of $Sin(\frac{9\pi}{8})$

[somber coyote]

Jason_Bjorn:

[serene field]

@signal hornet What is the domain of arcsin?

[signal hornet]

[-1, 1]

same as the range of sin

[serene field]

3.53 doesn't fall in the domain, thus is undefined.

[signal hornet]

hmm

why is my calculator still able to produce an asnwer then?

[marble topaz]

Photomath gives -.39

[signal hornet]

same as my calc

[weary drift]

it is not

[marble topaz]

Okay no I misread

[serene field]

Your calculator evaluates inside out, it would evaluate the sine of 9pi/8 to about -0.38, then find the corresponding theta for y = -0.38, yielding -0.39, but that would be the Q1 solution.

It would be the angle opposite that.

[signal hornet]

3.53 is not in the domain of Sin, note the big S

[weary drift]

$\sin(x)=\sin(\pi -x)$

[somber coyote]

RokettoJanpu:

[signal hornet]

isn't it 2pi?

[weary drift]

no

[signal hornet]

for one who rotation

[marble topaz]

Oh so just add pi for the "correct" solution

Which shows it doesn't work

I'm really fuzzy on the inverse trig functions so I'll bow out now

[weary drift]

use $\sin(x)=\sin(\pi -x) \\ \text{to rewrite }\sin(\frac{9\pi}{8})$

[somber coyote]

RokettoJanpu:

[signal hornet]

-pi/8

[weary drift]

no

[signal hornet]

-pi/8

which would be in the domain of Sin

[weary drift]

the domain of f(x)=sin(x) is all reals

what's relevant here is the dom/range of arcsin

[signal hornet]

I am talking about Sin though

sin != Sin

[hard gale]

what's big sin then

[signal hornet]

sin with a restricted domain

that's how the prof taught it

because just sin(x) it not 1 to 1

[hard gale]

that notation is like not standard at all

[signal hornet]

Only 1 to 1 over [-pi/2, pi/2]

[hard gale]

hence the confusion

[signal hornet]

sorry

[hard gale]

ok yea

[signal hornet]

is there a standard notation for such a thing?

[marble topaz]

Oh, that's what that means

[weary drift]

your prof probably invented that notation so students wouldn't have to manually restrict the dom of sin everytime they want to solve this kind of q

[marble topaz]

I've seen big-S-sin like one other place outside of here

[signal hornet]

so what's the "proper" way of doing it?

[hard gale]

i mean it's ok

we were just like wtf what's Sin

[signal hornet]

how are you supposed to write it?

[weary drift]

$\text{Sin}(x)=\sin(x), -\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$

[somber coyote]

RokettoJanpu:

[weary drift]

[serene field]

As far as I know, we just use sin(x) and restrict the domain to [-pi/2, pi/2] rather than writing Sin.

[signal hornet]

@weary drift thanks

[weary drift]

@signal hornet solving arcsin(Sin(9pi/8)). x=9pi/8 isn't in Sin's dom, but recall sin(x)=sin(pi-x), so sin(9pi/8)=sin(-pi/8)

[signal hornet]

right

but this homework problem says we might need undfined at some point

so I don't think it wants us to do that

[weary drift]

then don't use this newfangled notation for sin with a restricted dom, restrict it yourself

[signal hornet]

I don't see how that would yeild a different answer

[weary drift]

you just need to remember sin doesn't have a true inverse. arcsin is the inverse of sin restricted to the dom -pi/2 \leq x \leq pi/2

[signal hornet]

what is \leq?

[weary drift]

$\leq$

[somber coyote]

RokettoJanpu:

[signal hornet]

ahh

[weary drift]

then apply the identity sin(x)=sin(pi-x)

[signal hornet]

so I can just shift that window around to include 9pi/8?

[weary drift]

no

then apply the identity sin(x)=sin(pi-x)

[signal hornet]

hmm

no idea what the prof wants at this point

either he wants me to recognize that 9pi/8 is outside of the [-pi/2 , pi/2] domain and/or he wants use to use sin(x)=sin(pi-x)
to make it work

[weary drift]

Both

[signal hornet]

but then will any of the problems ever by undefined?

I don't see why they would have included that if non of them will be undefined @weary drift

[weary drift]

I’m not gonna look at each q to see which one’s undefined

[signal hornet]

but none of them will be as long as you just move the domain

[weary drift]

Don’t know what move the domain means

[signal hornet]

well for sine and cosine to be 1 to 1 you have to make the domain have a length of pi

but it doesn't matter where you slice it

as long as it is only pi long

it could be from [-pi/2, pi/2] or it could be [pi/2, 3pi/2], etc.

[weary drift]

arcsin is the inverse of sin restricted to the dom -pi/2 to pi/2, that’s it

[signal hornet]

oh

you're right

oh so that makes it a little easier

[jaunty sail]

How do I find the area of a regular hexagon that is circumscribed about a circle of a radius of 12 inches

[torpid torrent]

whats 2

so confusing

[silent plank]

what's the sum of the 2 angles of the quadrilateral at the straight line?

[torpid torrent]

what

which straight line

@silent plank

is it 170

because x+w is 80

and y+z is 90

[silent plank]

you aren't told that the lines are parallel

[torpid torrent]

fricc so what do i do

consecutive angles on a line do sum to 180

360 for top 3 and bottom 3

[silent plank]

label your quadrilateral ABCD anticlockwise from the top.
what is angleBCD + angleBAD?

[torpid torrent]

360

thats just splitting the quad in half

[silent plank]

no.
you are given 2 angles of the quadrilateral.
what's the sum of the 2 remaining angles

[torpid torrent]

190

so would the answer be 170

[silent plank]

yes

you can apply external angle theorems but you should know where they come from

(in this case angle sum of quad and straight lines)

[silent plank]

what have you tried?

[lime flame]

Find all solutions of the equation in the interval [0,2pi).
cos/theta = 0.7432

[upper karma]

hello

i need help with proofs

im only 12

[median crown]

Sorry, this is a 13+ discord

[woven hound]

I solved this for 60/17 using the long method of assigning variables and solving for them

however I noticed that:

$\frac{60}{17} = \frac{1}{\frac{1}{5} + \frac{1}{12}}$

Is this a coincidence? Or is there a theorem behind this?

[somber coyote]

Gale:

[weary drift]

take the reciprocal of both sides and you get the original equation

[woven hound]

Yeah, I know

I meant the solution to the problem is the reciprocal of the sum of the reciprocals of the side length

Wrote it in the second form because thought that was also nice

[upper karma]

hey guys

i need help badly

for proofs

[sly marlin]

@upper karma geometry proof question?

here or #❓how-to-get-help

[jaunty sail]

Hey can someone help me find the area of a reg. hexagon that’s circumscribed about a circle with a radius of 12

[silent plank]

have you drawn a pic ?

[jaunty sail]

Yes

I have one

[silent plank]

what else have you done?

[silent plank]

have you done trig yet?

[jaunty sail]

I’ve done geometry and I’m in pre calc rn

I just did a hexagon inscribed in a circle

[silent plank]

trig as in stuff involving sin, cos, tan

[jaunty sail]

This is the section of law of cosines

This chapter is called analytic trig

[silent plank]

so that would be a yes?

[jaunty sail]

Yes

[silent plank]

draw lines from the centre to the vertices of the hexagon

[jaunty sail]

Alright

[silent plank]

can you do the rest?

[jaunty sail]

I’ll try

I’m not getting the right answer

[silent plank]

what's your working out?

and/or what did you get?

[silent plank]

wrong trig ratio

which one should you be using for opposite/adjacent?

[jaunty sail]

Oh shit I’m supposed to be using tangent

Wow I’m a retard

Then I get the area of the right triangle right?

[silent plank]

you could get the area of the bigger triangle directly.

note that "x" is already half the base

[jaunty sail]

I’m here rn

X= 6.928 btw

I mis wrote that

I got the answer now though

Calculator error

[silent plank]

use exact values

[jaunty sail]

4sqrt3

((4sqrt3)(12)/2)(12)

[silent plank]

yep which simplied to

[jaunty sail]

498.83

[silent plank]

(the exactvalue)

(generally, unless told to round to dp, use exact values)

[jaunty sail]

576sqrt3/2

The book gives the answer as 498.3

288sqrt3

[silent plank]

did the book ask you to round to decimals?

[jaunty sail]

Yes

[silent plank]

ah ok then

[jaunty sail]

Thanks for always being such a great help ramonov

[silent plank]

np

[jaunty sail]

I’m working with the ambiguous case with law of sines

And I’m confused on how to find the values of everything if there’s two triangles

[silent plank]

@spring oyster are you given any more information?

[spring oyster]

na

[jaunty sail]

Is it possible to find more values than I already have listed

Little b2? Right?

[silent plank]

what's B_2 supposed to be?

[jaunty sail]

I’m having trouble visualizing what’s it’s supposed to be too but isn’t it an angle in my second triangle

[silent plank]

180° - angleB would give the external angle

[jaunty sail]

Ya that’s my b_2

[silent plank]

what did you mean by if there's two triangles?

[jaunty sail]

The instructions are “state whether the given measurements determine zero, one, or two triangles”

I was given the values C=30 degrees, c=10, b=16

Then I found B, which was 53.13 degrees

I did 180-53.13 and got 126.87 degrees which when you add 30 degrees to that it’s less than 180

So I would have two triangles then, right?

[silent plank]

ah, yeh i misread the question.

[jaunty sail]

So I’m confused on how to find all my values when I have two triangles

Are there A values to find

And is b_2 found by doing 10/sin30=b_2/sin126.87

[silent plank]

1st triangle: A = 180 - C - B
2nd triangle: A_2 = 180 - C - B_2

[jaunty sail]

And then I can find my a values

So C is a sort of fixed angle right

[silent plank]

b_2 would be the same as b which was 16

[jaunty sail]

I only have one C

[silent plank]

yes C is fixed as 30°
c was also fixed as 10
and b as 16

[jaunty sail]

Oh alright

[short gust]

I am here

@upper karma

[upper karma]

great?

[short gust]

Uh you wanted to talk

?

Err oh nvrmind.

[upper karma]

no i wanted u to stop posting circles in chill

[short gust]

Oh I'm sorry.

[upper karma]

its fine lol

[quiet mason]

lmao

[upper karma]

what is the formula for adding two different Arccos values to each other?
more specifically Arccos(27/48) + Arccos(60.75/108)

[silent plank]

simplify 27/48 and 60.75/108

[silent plank]

@upper karma this is what happens when you post in multiple channels. you miss out on help that was clearly given over 15 minutes ago.

[sudden locust]

@quiet mason Thanks for your feedback yesterday, I missed it, ❤

[mossy narwhal]

,rotate

[quiet mason]

what do you think it is

the line with the 100° is transversal to the two parallel lines

and the line with 70 too

[fast spindle]

Just say the rule F and Z form and angle 180 degree

[hallow rose]

@somber coyote it’s 30

,rotate

[upper karma]

70 plus 80

150

180-150

Is 30

@mossy narwhal just remember a straight line is 180degree

[maiden rain]

True ? or False ?
I said think it's True...

[silent plank]

why do you think it's true?

expand
(cos x + sin x)^2

[maiden rain]

uh shit now that I think about it, it's Actually false..

cos x ^2 +2cos x sin x + sin x^2

right?

[silent plank]

yeh

[wind heart]

Are 3,4,6 & 7 correct?

Oh, 5 too

<@&286206848099549185>

[upper karma]

Hmm for #3 it seems like you assumed those are all right angles

It looks like the shape in #3 is an irregular trapezoid, and the only condition you're given is that the right and left sides are parallel

[wind heart]

Yeah I kind of did

[upper karma]

We shouldn't assume that all the angles are equal to 90 degrees. Instead, we know that any quadrilateral's angles will add up to 360 degrees

So 9y + 6y + 6x + 90 = 360

[wind heart]

oh we never learned that

We only learned about supplementary and complimentary

Let’s see

I got uhhh

15y + 6x = 270

And I don’t really know how to find it when like

theres two different variables in one thing

[upper karma]

Yes, but that's where supplementary angles will help us

The angles 6y and 9y are supplementary, meaning they add up to 180

So 6y + 9y = 180

15y = 180
y = 12

[wind heart]

Let’s seee

Would x be 30?

[upper karma]

Well if 15y + 6x = 270 and we know that y = 12, then we can solve for x

15(12) + 6x = 270
180 + 6x = 270
6x = 90
x = 15

[wind heart]

Oh

I see

Did I get the others right?

[jolly bear]

How do you set up number 1

[upper karma]

@wind heart everything looks right, I'm a little confused with #7 though

Does that say 10x degrees?

[wind heart]

Let’s see

Oh wait, the 10x is a 90x and the 90x is just an x

The 90 was just my handwriting

The x is what was actually there

I have one more question

I have like, no idea

How to find this stuff

It’s so mind boggling

I think I MIGHT have figured it out. I got. X = 17 y = 1 z = 25

[upper karma]

Hmm well (3y + 2) + 85 = 180, which we can simplify.

3y + 2 = 95
3y = 93
y = 31

We also know that 7z + 138 = 180

7z = 42
z = 6

Also let's label this hidden angle as a

We know that a and 85 are supplementary, so a + 85 = 180

Angle a = 95 degrees

And because angle a and 5x are also supplementary, we know that 95 + 5x = 180

5x = 85, so x = 17

@wind heart

Sorry, I was multi tasking

[wind heart]

huh

Let’s see

Wait so like

For angle a, would that be apart of the second parelel angle? (The one in the middle)

Oh

I get it now

Jeez thanks a bunch

That was stresssful for my brain to bare

[jolly bear]

Hello guys

Can anyone help me to set up number 1 pls

[median crown]

Wdym

Just du what it says

8 km apart

A is on the left B is on the right

Make a triangle using those degrees

Use trig, law of sines or cosines, solve done

S 32 E means rotate 32 degrees south of the line pointing in the direction of east

In regular geometry, nothing is really abstract enough so that you can’t draw everything written down

[marble parcel]

Can someone walk me through this

Bottom letter is C

[silent plank]

let x = BD
let y = AD
calculate AB and AC in terms of x and y.

[marble parcel]

How would i draw this figure

[upper karma]

<@&286206848099549185>

[forest dove]

@eternal orchid

[upper karma]

Is that who I message to help me?

[marble parcel]

How would i draw this figure

[summer spire]

have you drawn any part of it?

[marble parcel]

Would QT be the 2 bottom points of the plane @summer spire

[upper karma]

Perpendicular implies 90 degrees.

So you draw a line

Then you draw a line on the base of said line going straight upwards

And you have a perpendicular bisector QT

[marble parcel]

like the top or bottom @upper karma

[upper karma]

Wot the fuck

Oh okay, your square is the representation of a plane

@marble parcel let your notebook be the representation of the plane. Just draw the line and the perpendicular bisector. It looks like the bottom one, yes

[marble parcel]

okay

[marble parcel]

@upper karma how can i prove that PR+RT = PT

[marble topaz]

Oh whoops

Sorry

[upper karma]

I want you to think of what this is saying @marble parcel

And I want you think about what a bisector is

Whats a bisector? Its very specific.

[marble parcel]

splits into 2 equal parts

but it doesnt say L is the bisector of PT so wouldnt PR+RT=PT because of definition of between @upper karma

[upper karma]

@marble parcel this is how I see it

Now its saying point P is to the left of R

[marble parcel]

oh

[burnt pendant]

help

pls

[dark sparrow]

@burnt pendant do you still need help with this?

[burnt pendant]

yes very much

hope im not too late

@dark sparrow

[dark sparrow]

ok well

consider that $\frac25\pi + \frac{1}{10}\pi = \frac12\pi$

[somber coyote]

Ann:

[burnt pendant]

wait

why

[dark sparrow]

alright let me restate this

consider that $\frac25\pi = \frac\pi2 - \frac{1}{10}\pi$

[somber coyote]

Ann:

[burnt pendant]

sorry i dont know why we're considering this

[dark sparrow]

2π/5 and π/10 are complementary angles

[burnt pendant]

oh right

[dark sparrow]

$\cos\left(\frac\pi2 - x\right) = \sin(x)$

[somber coyote]

Ann:

[burnt pendant]

is that something that i needed to know before this question

[dark sparrow]

i mean

the equation given in the question is basically just a special case of that

[burnt pendant]

1sec

ah ok

do you mind explaining how them being complementary angles will lead to this equation

actually no

ill try and figure it out first

[dark sparrow]

complementary angles are angles whose sum is π/2

[burnt pendant]

since im taking up a lot of time right now, do you mind explaining it because i need to move on and figure this out properly later

[dark sparrow]

explaining what

[burnt pendant]

@dark sparrow

how they are equal

[dark sparrow]

"they"?

[burnt pendant]

cos(2pi/5) = sin(pi/10)

i dont understand why cos(pi/2 - x) = sin(x)

[dark sparrow]

do you know the right-triangle definitions of sin and cos

[burnt pendant]

like sohcahtoa?

[dark sparrow]

yes that

[burnt pendant]

yeah

[dark sparrow]

well

the leg that is opposite to one of the acute angles in the triangle (x) is the very same leg that is adjacent to the other acute angle (π/2 - x)

[burnt pendant]

is that common knowledge or can i prove it

since i understand that part

[dark sparrow]

i mean

[burnt pendant]

but im not sure if it needs proof

[dark sparrow]

it should be obvious

just from looking at a right triangle

[burnt pendant]

well maybe im just looking at it wrong

how is it obvious

[long sapphire]

if you have a right triangle with an angle theta, the other opposing angle (which isn't the right angle) is 90 - theta

[burnt pendant]

yeah i get that

maybe im just overthinking

one more question

[dark sparrow]

do you agree that cos(π/2 - x) = a/c and that sin(x) = a/c

[burnt pendant]

ohhhh

yes

ok

thank you

i am in fact very dense

but

how do i find the exact value of sin(7pi/8)

without a calculator

[dark sparrow]

using a half-angle identity

[burnt pendant]

i do not know what that is

why do i feel like i'm not absorbing anything i learn in class

go on

[dark sparrow]

there can be many reasons

...ok what about double-angle identities then

[burnt pendant]

nope

[dark sparrow]

...

okay what identities DO you know

[burnt pendant]

i probably dont know them by name

if you explain one of them i might know

[dark sparrow]

$\cos(2x) = 2\cos^2(x) - 1$

[somber coyote]

Ann:

[burnt pendant]

ok not that

[dark sparrow]

well yikes

[mild grotto]

Big oof

[dark sparrow]

$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$

[somber coyote]

Ann:

[dark sparrow]

$\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$

[somber coyote]

Ann:

[dark sparrow]

what about these

[burnt pendant]

not taught yet but ive seen them

[dark sparrow]

uhhhhhhh

[burnt pendant]

yeahhh

[dark sparrow]

how tf are you expected to calculate sin(7π/8) if you don't even know these???

let alone any of the others???

[burnt pendant]

idk

maybe i missed them but i doubt it

we didn't go over it in class because we were doing the same worksheet for the entire class

ok for now

can you teach me the half angle identity

[dark sparrow]

not if you don't know the double angle identities or even the angle sum identities

[burnt pendant]

uh

ok ill skip a few questions then

thanks for the help tho

[mossy narwhal]

@quiet mason @hallow rose @upper karma @fast spindle honestly I didnt much understand ur explanation but as I was reading ur answers I found a very simpler way (or simpler to me). 70 transverse and for 100 itll be 80 to make it to the other side of the line (cause they are line it should form 180). Tada! we got angles inside a triangle. Triangle has 3 angles combined equal 180 80+70+X= 180 —-> 150+X= 180 —-> X=180-150=30😄

Thanks anyways ❤️

[upper karma]

Erm ok

I think I did it the same exact way but simplified

[coral dagger]

Can someone please solve for 38 I kinda need help

[silent plank]

where's #38

[coral dagger]

*oops

[silent plank]

where are you getting 9 ft^2?

the question says the area is 28 (yds^2)

[coral dagger]

Damn

Now I realised

I messed it up with numbers 37

Lmao

Thanks

Can you check again if you don’t mind? This should be the correct answer

[silent plank]

uh...

** = 28**

and you can't make the 2 in the denominator disappear like that

you should be getting a nice round number for x.
if you don't, you did something wrong.

[coral dagger]

I got 4 this time

[silent plank]

yeh that's better

4, is the base.
and you should be able to get the height from that

[coral dagger]

Do you mind helping me with 44 as well

[silent plank]

what have you tried so far?

[coral dagger]

I’m actually stuck to begin with. I know I need to find the dimensions for area, but idk how to begin with my length and width

[silent plank]

what would be the expression for the amount of fencing needed in terms of length(l) and width(w)

[coral dagger]

Wait “65” has no relevance in the context right?

[silent plank]

yeh, that's irrelevant

just the name of the highway

[coral dagger]

Idk shouldn’t the expression just be I*W in order to produce an area of 40k ft?

[silent plank]

that would be for the area. and
l * w = 40000 would be one of your equations

you get your other equation from the amount of fencing

normally for a rectangle it would be 2w + 2l, but here you have 1 side removed.

(and you are told you have 500 ft of fencing)

[coral dagger]

So is it just 2w*I if one side was removed

[silent plank]

2w + l = 500

[coral dagger]

*sorry messed up my perimeter and area

[soft gulch]

Which question are you talking about

[coral dagger]

44

[silent plank]

so now you have
l * w = 40000
2w + l = 500
which is a system of equations

[soft gulch]

Yes

[silent plank]

any issues with solving those?

[coral dagger]

let me try

[soft gulch]

l = 40000/w

2w + 40000/w= 500

[silent plank]

don't post solutions especially if they said they're going to try it themselves

[soft gulch]

Oopsie

[coral dagger]

I might be hallucinating but where does it say 500 feet of fencing

[soft gulch]

This question is a grade higher than mine

Therefore i am unable to solve the equation

The first line blackmarket

[coral dagger]

Fuck

why do they have to put that in alphabetical words lmao

[soft gulch]

AND WHY USE FEET

[silent plank]

i nearly missed it too heh

#wordproblems

[coral dagger]

can I make W and L one single variable like X?

[soft gulch]

Yes

[coral dagger]

Uhm I got 200 for dimensions

Max area I got 166.66

[silent plank]

dimensions consist of l and w
also it doesn't make sense for area to be 166.66 << 40000
optimisation of area involves separate calculations

[coral dagger]

So do I solve this like

Dual equations?

Like you substitute

One into the other

solve for the other

[silent plank]

yes

[coral dagger]

I haven’t done dual equations since Algebra I, do you mind reviewing me on the steps

[silent plank]

rearrange to isolate l
l = 40000/w
2w + l= 500
substitute into other equation
2w + 40000/w= 500
multiply by w to get rid of fractions
2w^2 + 40000 = 500w

simplify and "attempt" to solve

"attempt" is in quotes because...you'll find out when you try

[coral dagger]

Uhm

why do I have

imaginery

numbers

not imaginery

[silent plank]

yeh. because there aren't any real solutions

[coral dagger]

wait

what

[silent plank]

its impossible to get 40000 ft^2 area with only 500 ft of fencing

[coral dagger]

So the question is wrong?

[silent plank]

probably intentional

actually wait

can't take shortcuts for max area
.

ignore that stuff i deleted

instead of lw = 40000
your equations would be A = l * w
and 2w + l = 500 → l = 500 - 2w

still here? @coral dagger

[coral dagger]

ye

so you find Length

then you plug to find width

then you find the area

[silent plank]

you'll get Area as a function of w

and then use you knowledge of roots/symmetry/vertex to find the value of w to maximise A

[coral dagger]

Thanks

[silent plank]

your max area should be over 30000
check with me when you get your values

[coral dagger]

Aight

I got 93750

[silent plank]

that's way too big.
what was your w?

[coral dagger]

I plugged in L for -2w+500 then multiplied by W then found the max value

-(-2)/2(500) then plugged that answer to find the max

[silent plank]

you should get something like:
A = w * (500 - 2w) = 2w (250 - w)

-(-2)/2(500)
don't know where you're getting that

[coral dagger]

that’s to find the axis of symmetry. Axis of symmetry is the same as your X in a parabola. So I subsitude that to find Y, Y would be maximum/minimum value

[silent plank]

ah i don't think you applied that properly

from
A = 2w (250 - w)
what would be the zeros / (w intercepts)

(i factored out 2 from (500 -2w) to make it simpler)

[coral dagger]

Ah

Wait so from here, what do I do?

[silent plank]

well what are the w intercepts?

and can you determine the axis of symmetry from those two values?

[coral dagger]

my axis of symmetry is still 125

[silent plank]

ah. yeh thats correct.

-(-2)/2(500) represented something completely different

since it is a concave down parabola, you'll know that
Area will be a maximum at the axis of symmetry i.e. when w = 125

what will be the value of your length?

(from 2w + l = 500)

[coral dagger]

250?

[silent plank]

yep

and then what will your max Area be?

[soft gulch]

I came looking for copper

I found gold

[coral dagger]

lemme just plug it in

I got 28750

[silent plank]

w = 125
l = 250
l * w = ?

[coral dagger]

31250

[silent plank]

yeh. that's better. (in ft^2)

[coral dagger]

Holy crap

[soft gulch]

Any geometry problems for 9th graders?

[coral dagger]

Thank you

[silent plank]

do you understand all the steps we went through?

[coral dagger]

I think I got it down

[silent plank]

k

[sly ermine]

Ye just find A in terms of w (width)
Then differentiate it
Find the value of w when dA/dw is 0
Then use value of of the lengths in w
Sub in the value for w and divide by 2
Answer should be 2w plus l

[soft gulch]

I have a question

Consider a Right Angled isosceles triangle

And then name it

ABC clockwise

AC being a straight leg

Instead of considering BC as base

And AC as height

What if we considered BA as base

Will the height 'line' be a median?

It maybe a stupid question

But I'm dumh

[fast spindle]

Hmm

Is angle ACB =angle CAB?

[soft gulch]

No

The triangle is right angled at ACB

[fast spindle]

Can you draw your triangle?

[soft gulch]

A is at the top

ABC clockwise

[fast spindle]

Oh AC=BC?

Lol

Something wrong

[sly ermine]

The height of that triangle would be AC wouldn't it

[fast spindle]

Is that your question?

[languid dock]

The height is one of the legs

[coral solar]

can someone please explain how the person got the answer for 5a (bearing)

[fringe crater]

Okay, so you figured out that the resultant vector was 2i - j

Draw it out in the plane

2 units East and 1 unit South. Now you have a triangle, with the origin, (0, 2) and (2, -1) as your three points

Look at the angle at the origin point. It's tangent is -1/2

Now, plug in -1/2 to the arctan function and you get the angle: -26.6 degrees

By convention, angles are written so that 0 degrees is the positive x-axis, or East, and counterclockwise is positive while clockwise (swinging to the South) is a negative angle

But if you want to see your bearing from North, or positive y-axis

With clockwise defined as the positive direction, then you have to subtract your East bearing from 90

So 90 - (-26.6) = 116.6

Let me know if you have any questions

[coral solar]

thank you so much for explaining it

[fringe crater]

You're welcome

[coral solar]

so tan(30) = root3/3 , how do i simplify that to make 1/root3?

[marble topaz]

Factor out root3/root3

[mossy narwhal]

,rotate

[unborn jacinth]

so we learned this theorem

that states that these two are the same thing

but when I graph them

they are different graphs

[steel pawn]

domains are different but outputs for positive numbers should be the same

[silent plank]

the outputs where X is defined is the same.

[steel pawn]

when you square x you can have a negative value because you always turn it into a positive before you put it through the log function

[unborn jacinth]

ohhh

[silent plank]

also wrong channel

[unborn jacinth]

why

im in trig

well alg2 part

of trigh

[steel pawn]

log of a negative number is undefined tho

[unborn jacinth]

ok

but when you actually know what x is

it always works

right

[silent plank]

@mossy narwhal what have you tried?

[unborn jacinth]

if x is defined

@steel pawn

o wait shoot no ping

sorry

[steel pawn]

ack

[unborn jacinth]

sorry

sorry

[steel pawn]

no its cool

[unborn jacinth]

oops

[silent plank]

it should actually be
2log_4 ( |X|)

[unborn jacinth]

that makes sense but why

[steel pawn]

I know the abs value comes out when you do integrals, but where does it come from really

[unborn jacinth]

isn't the theorem that you just take out 2

[mossy narwhal]

@silent plank nothing cause i have no clue. I can find out the intersecting point from 2 linear equations. But not 3 points (to form a triangle) T - T

[unborn jacinth]

y axis

it says

[silent plank]

only when the base is positive

[unborn jacinth]

the base can be negative?

ok but in this case the base is +

so why is it absolute value

[steel pawn]

like i know the sqrt of a number is technically plus or minus. so doing that transform forces the domain to still be relevant. Like does it happen because the domain of the function was already valid for all positive and negative numbers except zero.

[silent plank]

in this case, the base was X
and you don't know whether it's pos or negative