geometry-and-trigonometry

[remote radish]

And cb is 7

Is it supposed to have similar ratios for de as well?

Like de/15?

[honest bay]

well yeah

first you need to prove that the triangles are similar

we have angle A in common

and angle DEA is equal to CBA

because DE and CB are parallel and AB acts like a secant

is everything clear?

@remote radish ?

[remote radish]

@honest bay I guessss

[honest bay]

we have two angels in common so the third is also in common

so the two triangles are similar

does that makes sense?

[remote radish]

Yea

I'm trying to figure out what de is

[honest bay]

yeah now the ratios between the sides are the same

so AC/AD = CB/DE

u got it?

[remote radish]

Ok so that means it would be 7/de?

[honest bay]

20/15 = 7/DE

yeah

now u know how to solve that right?

[remote radish]

Yea

[honest bay]

is it all clear now?

[remote radish]

Yessir

[rustic dagger]

Cross multiply

[wooden crypt]

Is there a method for memorizing all the important angles for example Cot of 180 or sin of 90

[devout shell]

you would really only need to know the ones for sin(x) and cos(x) after that you use the identities to figure them out

tan(x)=sin(x)/cos(x)
sec(x)=1/cos(x)
csc(x)=1/sin(x)

cot(x)=1/tan(x)=cos(x)/sin(x)

[wooden crypt]

My summer assignment is to remember all 6 trig Identities and special angles

[devout shell]

like I said if you remember them for sin(x) and cos(x) then you have it all

just use the identities to get the answer

[wooden crypt]

Ok

[upper karma]

Hello someone can help with geometry?

A triangle ABC has a right angle at B. The length of AB is (3x - 2)cm and the length of BC is (x + 1)cm. Find the perimeter when the area is 14cm^2.

[spark stag]

Right, so, there's really gonna be two "parts" to this answer

We first need to figure out what x is

The only value we know with certainty is the area

So we'll be using that to find x

[upper karma]

Nvm fix it forgot tu multiply 3*2 😦

[spark stag]

Ah.

[upper karma]

But didnt understand it well

So if u want u can keep explaining it 😃

I used the b*h = area

And then used pitagoram

Me

A right-angled triangle has sides of lenght x, y^2 -1 and y^2 +1, where x and y are integers, and y > x.
(i) Find x in terms of y.
(ii) Find the lengths of the sides in the cases where y=2, 4, and 6.
(iii) Find the values of y, if they exist, for the right-angled triangles with sides. (a) 20,99,101 (b) 20,21,29

Okay I'm lost

How do I find x?

Ok, I dont know how to do it, guess im boosted af

And I have 1 week before the exams xdxd

[devout shell]

you want the helper role so you can tutor all the questions that come through lol? ask a mod for that

then you won't need to put tutor in the name

[upper karma]

can you guys help me with any of these?

it says algbra 2 but im pretty sure its trig

well unit circle is I think

[devout shell]

helper role then you can help all lol

fair enough then lol

[upper karma]

was I supposed to ask in the other one im confused

[bitter merlin]

3B

[mossy vine]

do you remember the rules for sin/cos/tan

[bitter merlin]

Uhh yes

[mossy vine]

ok so what are they?

[bitter merlin]

Sin (0)= O over H

Cos is A over H

And Tan is O over A

[mossy vine]

mmhmm good

[bitter merlin]

Can we do the command?

So we can have a better copy of it

[mossy vine]

idk what you mean by the command

[bitter merlin]

Brb

[mossy vine]

but ok which sides do you have?

[bitter merlin]

Like

$

[mossy vine]

the picture's clear for me

[bitter merlin]

@mossy vine

[mossy vine]

hi

for this question

work out which sides you have relative to the angle you have

i.e. which of opposite, adjacent andhypotenuse

[bitter merlin]

Ok

[mossy vine]

and then you can form an equation in terms of whichever sides it is, and the appropriate trig function

[rustic dagger]

Hi

[heavy verge]

Is my reasoning correct?

[umbral snow]

,calc tan(pi/12) - (2 - sqrt(3))

[somber coyote]

Result:

-1.1102230246252e-16

[umbral snow]

Looks right to me

[heavy verge]

nice, thanks

[upper karma]

Can I say that ΑΔ/ΒΓ=ΑΕ/ΝΜ?

[halcyon coyote]

yes

[upper karma]

Thank you

[fleet lake]

I want a analytic approach .

[fleet lake]

<@&286206848099549185>

[plucky marlin]

use the hint

35

its the number that comes after 34

the rest is pretty apparent

[dark sparrow]

@fleet lake what do you mean by "analytic"

[plucky marlin]

complex differentiable?

[dark sparrow]

no

definitely not

don't even bring this up

it's going to cause unnecessary confusion

[plucky marlin]

[fleet lake]

I mean I want to solve this using graph @dark sparrow

[dark sparrow]

okay

[fleet lake]

I can draw you picture if you want

[dark sparrow]

place the base of the short pole at the origin, and the base of the tall pole 12 feet to its left, at (-12, 0)

the tips of the poles will be at (0, 11) and (-12, 14)

[fleet lake]

Slope is -1/4

[dark sparrow]

the line connecting the tips will have equation y = -1/4 x + 11

[fleet lake]

Ok so what to next @dark sparrow

To do*

I got the answer it's 44 ft

Using the slope

Thanks for your efforts
Ann

[forest hazel]

I need all three qwq

[copper valve]

hmmm i am big geometry noob, but for the second one (q14) i think id wanna show that OC is parallel to AP

[forest hazel]

Oops don't need the second one anymore

Just the first and third one

[copper valve]

Oop

[forest hazel]

<@&286206848099549185>

[devout shell]

Why do it in pen lol?

[forest hazel]

@devout shell it's dead like my soul

qwq

<@&286206848099549185>

Sorry but it's real urgent

And I'm just left with the first question now

[fringe dirge]

Why is S in this problem?

[forest hazel]

No idea

Does it help with anything

I got 3 angles of the same size but I don't know how to continue from there

[fringe dirge]

Well how do you show that two lines are parallel?

[forest hazel]

Well this but none of the angles can be shown to be alternating or corresponding somehow

Oh wait

Do I use similar triangles

[fringe dirge]

That works yes

[forest hazel]

But I can't prove that they're similar?

Like I can only see one common angle

[tardy fossil]

fellas

I have some qualms with the quirks of arcsin

algebraing around, I get $\sin 2x = -\dfrac{\sqrt{3}}{2}$

[somber coyote]

Stract:

[tardy fossil]

but here's the thing

if I arcsin both sides, we know that $-\dfrac{\pi}{2} \leq \arcsin x \leq \dfrac{\pi}{2}$

[somber coyote]

Stract:

[tardy fossil]

so naturally I get -pi/3 as my arcsin and therefore x=-pi/6

so that x is my reference angle

so I get, for solutions, x=5pi/6 and 11pi/6

how am I supposed to get 4pi/3 and 5pi/3 as answers (which they are) by definition/restriction of arcsin

[dark sparrow]

don't arcsin both sides

that's how

draw a unit circle

[tardy fossil]

Yes I know that 4pi/3 also has a sin of -sqrt(3)/2

[dark sparrow]

and get ALL solutions to the equation. and then pick out the ones that fall between 0 and 2pi.

[tardy fossil]

but why does arcsin fail here?

I'm trying to see why this discrepancy exists

like, it feels less-right just pulling angles off the unit circle and then saying the argument of the sin is equal to that

it feels more proper to use arcsin

[dark sparrow]

because arcsin(sin(x)) = x doesn't work if x is not between -pi/2 and pi/2 that's why

"proper" my ass

[tardy fossil]

so I should just stick to the more intuitive method of checking against the unit circle?

I tried to involve arcsin because it seemed valid but I guess not

[brisk lake]

Can someone help me

[heavy verge]

Is there a way I can get rid of those ugly radicals inside radicals? Also I don't understand what to do in part c

[supple abyss]

no there isn't

also remember $\tan(x)\equiv \frac{\sin(x)}{\cos(x)}$

[somber coyote]

CaptainLightning:

[heavy verge]

yes, I was thinking about that however notice the question says to start by deriving an expression for tan^2(pi/12)

[umbral snow]

Check if
2 + √3 = (a + b√3)²
For some a and b

[heavy verge]

@umbral snow what is that for?

[umbral snow]

If that is the case, then √[2 + √3] simplifies

[heavy verge]

[umbral snow]

Here, let's do it together.
2 + √3 = (a + b√3)²
2 + √3 = a² + 3b² + 2ab√3

a² + 3b² = 2
2ab = 1

[supple abyss]

try ^2 maybe

[umbral snow]

,w solve
a^2 + 3b^2 = 2
2ab = 1

[somber coyote]

Display results online and refine query

Requested by Kaynex#7520

[supple abyss]

[umbral snow]

Sick, I'll take it.
2 + √3 = (1/√2 + √3/√2)²

[heavy verge]

that's cool

[umbral snow]

Therefore cos(π/12) = (1 + √3)/2√2

[heavy verge]

nice

and regarding part c? any ideas?

[umbral snow]

If that is simpler or not, up to you lel

[heavy verge]

definitely not simpler, just "prettier"

[supple abyss]

well for part (c) you can write

$\tan\left(\frac{\pi}{12}\right)=\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}$

[somber coyote]

CaptainLightning:

[heavy verge]

just dividing the sin by the cos?

[supple abyss]

and then consider tan²(pi/12) as it said

ye

[heavy verge]

ok that's the part I don't get, the tan²(pi/12) part

what am I supposed to do with that

[supple abyss]

it'll remove the outer radicals

after doing that I expect you'll want to rationalise the denominator

[heavy verge]

but isn't just $\tan\left(\frac{\pi}{12}\right)=\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}$

the answer?

[somber coyote]

bug:

[supple abyss]

yes

[heavy verge]

but you said "and then consider tan²(pi/12) as it said"

[supple abyss]

what you want to do will depend on what you got in (a)

is that what you got in (a)?

[heavy verge]

I got \tan\left(\frac{\pi}{12}\right)=2-\sqrt{3}

\tan\left(\frac{\pi}{12}\right)=2-\sqrt{3}

[supple abyss]

ok so show $\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=2-\sqrt{3}$

[somber coyote]

CaptainLightning:

[supple abyss]

squaring it seems like a good idea

to me

[mint sandal]

divide and multiply by sqrt(2+sqrt(3))

num becomes 1

then conjugate the same way

[heavy verge]

@mint sandal good approach, did it and showed they're equal

however I'm not totally sure, I think the question means something else by "deriving" and the tan^2 thing

[supple abyss]

what I meant is

$\left (\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\right)^2\= \frac{2-\sqrt{3}}{2+\sqrt{3}} \= \frac{7-4\sqrt{3}}{1}\= (2-\sqrt{3})^2$

[somber coyote]

CaptainLightning:

[supple abyss]

and then as they're both positive and have the same square they must be equal

that's likely what they were suggesting

although it's just conjugating with extra steps so a bit unnecessary

[floral magnet]

what would a 3d polyhedron made up of triangles be called?

deltahedron isnt quite what im looking for, thats equilateral triangles

and geodesic polyhedron isnt quite right either

those are convex

[upper karma]

https://i.imgur.com/G2kn8Dw.png

So, I have tried drawing an altitude from Z to line YX, but it does not appear to help in solving this triangle.

[devout shell]

are we allowed law of cosines?

[upper karma]

Yes.

[devout shell]

we can knock this one out then. so if we just find angle ZYX then the angle in question is what?

[upper karma]

Wow.

I was so fixated on SAS.

Nevermind, thank you!

[devout shell]

no problem lol

[bitter merlin]

@devout shell

[devout shell]

yea?

[bitter merlin]

I need help

[devout shell]

you have to post something

[bitter merlin]

Question 6 every second question

Question 7 i need to do all

Can you help me do a few exp?

And maybe for the rest i just send my working out and yeah

[devout shell]

you know what sine, cosine, and tangent are?

[bitter merlin]

Yep

[devout shell]

ok very good

[vagrant elk]

seems mostly like just finding arcsin, arccos etc

[devout shell]

yes

we need to use the inverses to get the angles

you have calculator with you?

[vagrant elk]

basically just pressing sin-1 and cos-1 in your calculator

[bitter merlin]

Yes

[fresh wind]

Just need to know SOH CAH TOA

[bitter merlin]

I know

That

[devout shell]

inverse functions are the sin^-1 cos^-1 and tan^-1 buttons

[bitter merlin]

How do I do question 6B?

[devout shell]

let me see then

[bitter merlin]

I sent a pic

Look up

[devout shell]

I know

sin(x)=1/3

so then take the inverse of both sides

[vagrant elk]

just do sin-1 of 1/3 basically

[devout shell]

then you have x=sin^-1(1/3)

[bitter merlin]

As I said before amph

Idk how lmo

Can someone if possible write it on a sheet of paper and show it to me?

[vagrant elk]

try using the sin-1 key on your calculator

[devout shell]

I can't teach you how to type on a calculator

[bitter merlin]

oh on the calculator

My bad lmao

So

[devout shell]

so now it's clear what to do?

input sin^-1(1/3)

[bitter merlin]

What do i type in again

Ok

[devout shell]

your book even has a calculator symbol meaning that it should be done with one lol

[bitter merlin]

@devout shell

I finished q6

What do I do for q7

[devout shell]

you use the correct inverse trig funtion

the choice is based on whether you have opposite, adjacent, or hypotenuse given

then you take inverse to get angle

like 7a

you have adjacent side and hypotenuse

so use?

[bitter merlin]

Yes

Uhh

[devout shell]

use which trig function?

[fleet lake]

Can anyone explain me why the solution is correct ?

[spark stag]

when we have straight lines, there are three options for how many times they intersect:

1. if they're both parallel, and not the same line, they never intersect

2. if they're both parallel, and are the same line, then they intersect at every point (ie, an infinite amount of times)

3. if they're not parallel, they intersect once

so, we compare the three lines to each other

the first and second line are parallel and not the same, so 0

the first and third line are not parallel, so 1

the second and third line are not parallel, so 1

add it together: 0 + 1 + 1 = 2

[fleet lake]

Alright thanks namigthon

[obsidian sierra]

I came up with a formula for the area of any regular polygon given the number of sides n and side-length x

i know it's nothing but thought i'd share here

[dire rampart]

epic

[obsidian sierra]

ok

[dire rampart]

what if it's a circle

[obsidian sierra]

maybe if n goes to infinity
and x goes towards 0 or somethin'

i'm just a 2nd semester fetus i don't know

[fiery cargo]

I'm trying to help my daughter understand why she got a problem wrong on geometry homework.

The question is:
The right rectangular prism below is made up of 8 cubes. Each cube has an edge length of 1/2 inch. What is the volume of this prism?

Her answer was:
one cube = 1/2 * 1/2 * 1/2 = 1/8 cu in
8 cubes = 8 * 1/8 cu in = 1 cu in

Could someone point out where the mistake is? I tried in HELP and they thought it looked ok

[mystic shadow]

Fun fact: the people answering at #❓how-to-get-help are the same people who are gonna look at this channel

Don't post the same question in multiple channels

[fiery cargo]

in theory yes, in reality not so much

[hidden moon]

I need help

I think I got iy

[cinder portal]

wait wats wrong

looks right...

[sharp fern]

i'm struggling to understand this

[devout shell]

this cat picture, yes lol

[sharp fern]

i understand tan beta = slope, and that when x approaches pi/2, tan approaches infinity

but im not following how they graph the phase shift

[spark stag]

the phase shift is the horizontal translation

(well, its additive inverse)

yes?

do you know how to find a function's horizontal translation?

[sharp fern]

when a number is added or subtracted to the x value?

[spark stag]

well, almost

but we have to factor out our b value

in this case, we have 2tan(3x + pi)

so we factor out the 3

2tan(3(x + pi/3))

thus, our translation is pi/3 units left

so our phase shift is -pi/3

your guide phrased this in a slightly different way:

it said "solve bx - c = 0"

so, if b = 3 and c = pi

thats bx - pi = 0

or x = pi/3

[sharp fern]

ooh that makes a lot more sense, thank you

[keen glen]

can anyone help me with a question

nvm

[upper karma]

In the cuboid, the diagonals of the side walls coming from the same vertex form the angle alpha = 60 degrees and each of them has the length d = sqrt(5). Calculate the area and volume of this cuboid

[upper karma]

Hi, got a question

The graph of y = 5^x is transformed by a stretch in the y-direction, scale factor 5
State the equation of the transformed graph.

How i resolve this?

I tried but dunno how to do it

[spark stag]

its stretched by a factor of 5 in the y

that means, for each point

you multiplied the y value by 5

how can you reflect this in the function?

[upper karma]

so the answer is y= 5 * 5^x

[dark sparrow]

yes

[upper karma]

aswell

the gradient formula is m=y2-y1/x2-x1

How do i differentiate the points of x1,y1 and x2,y2?

I saw many videos and they don't explain it

[cinder portal]

you mean slope?

[upper karma]

the points

[cinder portal]

what are you trying to do?

[upper karma]

trying to know how to correctly put in the equation y2-y1 and x2-x1

like if u see in the img u can see he named x1,y,1 and x2,y2 but i'm trying to understand why he did it

he didn't explain it

[cinder portal]

sounds like you're trying to find slope

[upper karma]

dunno what's slope

[cinder portal]

or in other words, how steep the graph is

slope is defined by how much your y changes based on how much your x changes

[dark sparrow]

slope == gradient

[upper karma]

yep

[dark sparrow]

@upper karma if you're asking about which point should be (x_1, y_1) and which one should be (x_2, y_2)

[cinder portal]

o rly? i always thought gradient was used more in higher level maths as in multi vari

[dark sparrow]

it doesn't matter

[upper karma]

it dosnt? oh

ok

❤

Was hardstuck in my mind a lot of time

[cinder portal]

back on track, yup, two points, how your y changes divided by how much x changes

change is usually denoted as delta

that small triangle

[upper karma]

yes

I see

this would give me the gradient

[cinder portal]

yup

or if you want to think about it more application like

how steep is it?

[upper karma]

hmmm

I don't understand

oh

so is the middle point no?

[cinder portal]

think of a mountain

if you're on one point of the mountain

and another point on the mountain

gradient, or slope, measures how steep that mountain is

the larger the value, the steeper it is

the smaller the value, the less steeper it is

[upper karma]

by value u mean point between a and b?

[cinder portal]

value as in the slope value

when you take delta y divided by delta x

delta y divided by delta x, or (y2 - y1)/(x2-x1) gives you a numerical value

[upper karma]

so the gradient

is the total

between dose 2 points

no?

[cinder portal]

lets do an example shall we?

[upper karma]

like what u need to go from a to b

[cinder portal]

the gradient just measures how steep things are

thats about it

don't think too much about it

[upper karma]

oh

is the pending

[cinder portal]

for now, just know, that the gradient just measures how steep it is between two points, (x1,y1) and (x2,y2)

[upper karma]

the pending

[cinder portal]

what is pending?

[upper karma]

oh i see

hahaha

[cinder portal]

sorry i use like different math vocabulary, so might have to be a bit more descriptive LOL

[upper karma]

used google translate

srry

[cinder portal]

no problem

[upper karma]

pendiente in spanish is like steep

okay gotcha

[cinder portal]

ahhh i c i c

[upper karma]

I understand it, thx

[cinder portal]

does it make sense though?

alright awesome

[upper karma]

yep

thx 4 ur time ❤

[cinder portal]

np 👍

[upper karma]

2x+5y=7

what would be the gradient

i did y = (7-2x)/5

should i deduce that the gradient is -2/5

[devout shell]

correct

[upper karma]

so how do i found it? or i just need to deduce it?

find*

[cinder portal]

given an equation of the line, if you solve for slope-intercept form

[devout shell]

you did find it though

[cinder portal]

you would always get an equation in the form of $y = mx + b$
in this case, m = -2/5, so you should be fine

[somber coyote]

MemesPlease:

[upper karma]

oh

so b is 7

[cinder portal]

b is 7/5

[upper karma]

and mx is -2/5

[cinder portal]

be careful with division

[upper karma]

ups

I see know

[steady lily]

when you accidentally say cos instead of cosine: forgive me god for I have sinned

ha

ha

ha

crickets

[patent ibex]

How do I do this

[devout shell]

,rotate

[upper karma]

How do I solve this?, i know the answe is 4

But don't know the method

[devout shell]

period is: $\frac{2\pi}{b}$

[somber coyote]

⚡Amphy⚡:

[devout shell]

general formula for sine function is A * sin(bx+c)

so in this problem you see that b is?

[upper karma]

is 2?

or n?

[vagrant elk]

b is $\pi/2$

[somber coyote]

Darkrifts:

[vagrant elk]

since n is the same type of thing as x since they're the variable

[upper karma]

oh i see

[vagrant elk]

yeah

[upper karma]

thx ❤

[vagrant elk]

it's the constant part

[devout shell]

I thought you were going to get tripped up with n vs x lol

it's just a variable lol

[vagrant elk]

Yeah

[upper karma]

though was other thing mb

[upper karma]

okay so i tried and tried to solve it but i didnt got the 4

[devout shell]

$\frac{2\pi}{\frac{\pi}{2}} = 2\pi \cdot \frac{2}{\pi}=4$

[somber coyote]

⚡Amphy⚡:

[upper karma]

Oh i see thx

[patent ibex]

How do you find an angle in which a projectile should be released

I have a radius of 3.3

And my target is (8.7, 2.6)

[upper karma]

How do i solve this?

[devout shell]

$x=e^{y-4}$ Solve for y

[somber coyote]

⚡Amphy⚡:

[devout shell]

that will give the inverse of f(x)

[upper karma]

so i can inverse y and x

and say

y=e^(x-4)

and then e = ln so i can say ln y= x-4

[devout shell]

what are you doing now?

you started with y=e^(x-4)

to get the inverse, change the places of x and y and then solve for y
x=e^(y-4)

now solve that for y

[upper karma]

i change it no?

[devout shell]

we already swapped the places of the variables

don't swap again, because that would be useless

[upper karma]

hmm okay

[devout shell]

solve x=e^(y-4) for y, that's all you need to do to get the inverse

[upper karma]

x= ln ^(y-4)

ln x = y-4

y = 4 + ln x

[devout shell]

that is the correct inverse funtion

[upper karma]

so then i can say

[devout shell]

domain for this is the domain of ln(x)

[upper karma]

f^-1 (x) = 4 + ln x

[devout shell]

yes

[upper karma]

so x > 0

How do I solve this?

[devout shell]

it just gives eqautions

is that all you are given in the problem?

[upper karma]

show that dy/dx = -3/4 * 2^(2t)

anyway gonna go to sleep better, i studied for almost 8 hours and I'm really tired...

When i wake up would keep asking ❤ thx 4 your time cat cat ❤ I appreciate it

[devout shell]

I ain't going to wake up to answer questions all day, there are things I need to do as well

though you should be able to answer that one above from formula in your book

[upper karma]

okay, thx will give a look later

❤

[stoic wraith]

Anyone have any nice resources to quickly learn how to find the length of a circle's arc based on the radius and a subtended angle? I'm taking trig this summer and it's compressed into 10 weeks so I'm having a bit of trouble keeping up. Gotta learn this stuff fast.

[cinder portal]

mmmm a picture would help, based on what's given

there are a bunch of circle theorems, and i need an example to show you a specific circle/arc theorem

[dark sparrow]

if you measure your angles in radians, then arc length = radius * central angle

[stoic wraith]

@cinder portal

That's the example problem I'm presented with at the moment.

I tried converting the angle to radians but came out with a wonky answer which was apparently far from correct.

I did what Ann suggested with the central angle's value in radians but when multiplied by the radius my answer was not marked correct by the system.

[cinder portal]

Oh, for this problem, I would probably just multiply the circumference of the circle with the fraction of the angle out of 360 degrees

does that make sense?

lets say I have a circle. The circumference would just be 2pir right? if you wanted, for example, an arc that was half the circumference, then it would be (1/2)2pi*r

[stoic wraith]

As in the fraction of the angle that 75 degrees is? 75/360 reduced?

[cinder portal]

correct1

take that fraction

[stoic wraith]

Sure.

[cinder portal]

and multiply it with the circumference

and then you get the arc, which is really just a fraction of the circumference

[stoic wraith]

How would I go about determining the circumference from the radius? I've totally forgotten the formula for that.

[cinder portal]

$Circumference = 2 * \pi * r$

[somber coyote]

MemesPlease:

[stoic wraith]

Thanks a ton @cinder portal I'll let you know how it goes.

[cinder portal]

np

[dark sparrow]

75° = 5pi/12

[stoic wraith]

13.08996939 Thanks so much @dark sparrow and @cinder portal You're lifesavers.

[cinder portal]

👍

[rancid mantle]

hmmmmm

how do i solve

{-x + 4y =8 , 2x-7y=-5

it says " find the inverse matrix using the system on liner equations

[dark sparrow]

it says what?

can you post a screenshot of exactly what the problem says

[rustic cairn]

How do Solve this

[rustic dagger]

Finding X y and z?

@rustic cairn

[rustic cairn]

Yes

[rustic dagger]

Which one do u want to start with

Try to find y

[rustic cairn]

Y

Is it 52

[rustic dagger]

It’s like the same case as x

Yea 52

[rustic cairn]

Than what’s z

[rustic dagger]

It’s an angle inside the circle right

So the formula for it is the 2 arcs vertical to the angle added divided by 2 is Z

104+100 divided by 2 is z @rustic cairn

[rustic cairn]

Thank you

[rustic dagger]

Np

[lilac wraith]

It's just pythag

[shut crest]

Is there any use for the orthocenter of a triangle?

[bright parrot]

drawing things

this is just a bet of mine based off of one talk i sat in at the the jmm but there's probably a use for folding strange things considering how many ways there are to divide planes into triangles

and then you can have several colinear points on a single plane based on how you folded from knowing that kind of geometry info about a triangle

but that are not on a line on when the plane is unfolded

other random bet, there's probably a fun way to get optimal walks on graphs based on the triangles on planes

probably even more fun on a toric

but "use" is as in useful? ask an engineer?

[shut crest]

Huh

[bright parrot]

yes?

also graphics

graphics is a use

[shut crest]

Graphics?

[bright parrot]

you can probably average shapes and pixels that way

and in a "nice" way

that's just art intuition though. pixels are squares, they are less "natural" than triangles. being able to average across triangles would be "nicer" than average across squares

https://www.tandfonline.com/doi/full/10.1080/00207390701734242

looking for use in math might be too much sometimes. look for "oh hey that looks neat"

[shut crest]

I guess that true, I just find it weird that they teach so much usless stuff in scyiil

[bright parrot]

it's not useless, YOU just haven't learned the use yet

finding the orthocenter of a triangle means you can find the centroid

which means on some object with the same thickness and even density across it you can find the center of gravity

so say the human shoulder, it's a triangle, you can average human shoulders and perhaps build an artificial shoulder

or build padding for a shoulder

patella is also triangular but more like a little tetrahedron

but that's four triangular faces, all with an orthocenter, maybe a little skew, you can find the centers and weight bearing properties

lots of stability is about having three points of contact, triangles are important in the the physical world

being able to find critical points of triangles means you can find when things might tip over or crack

[shut crest]

I'm abit confused

[bright parrot]

which part

[shut crest]

Well firstly you can find the centriod alone, and it'd easyire then the orthocenter, so why would you use the orthocenter?

[bright parrot]

http://mathforum.org/library/drmath/view/55274.html

[shut crest]

Oh neat

[bright parrot]

it's really hard to say what part o fmath will be useful

there's something my advisor was working on that couldn't even be done in the 1850s because we didnt' have computers capable of doing polynomial shit in a reasonable amount of time

[shut crest]

So the ortho center is the part with the smallest distence from eachside

[bright parrot]

but now he can code it in C

[lilac wraith]

Knowing properties of special points in a triangle can massively simplify and give different perspectives and insight on certain problems.

As do most abstract concepts in math.

[bright parrot]

no one thought "oh hey fields" and went "and thus quanta"

but the math of it existed before the discovery of atoms

[shut crest]

So what is a triangle plane?

[dark sparrow]

you're getting... cos(Y) = 121?!

how are you getting that the cosine of an angle is greater than 1

[cinder portal]

$c^2 = a^2 + b^2 - 2(a)(b)cos(C)$

[somber coyote]

MemesPlease:

[cinder portal]

in this case

$11^2 = 9^2 + 10^2 - 2(9)(10)\cos{Y}$

[dark sparrow]

\cos

[cinder portal]

ah thank you

[somber coyote]

MemesPlease:

[cinder portal]

you should NOT be getting cosY = 121

oh boi

solve for cosY first

let me know what you get

do algebra

would you like me to walk you through it?

okey

[dark sparrow]

solve for cos(Y)

that's EXACTLY what @cinder portal told you to do

$121 = 181 - 180 \cos(Y)$

[somber coyote]

Ann:

[dark sparrow]

how about now

and most importantly do you understand what i just did

[cinder portal]

Let me know what you got when you solve for cos(Y)
i.e. $\cos{Y} = value$

[somber coyote]

MemesPlease:

[cinder portal]

what the

[vagrant elk]

bruh what

[cinder portal]

still shouldnt get that value

[vagrant elk]

i'm still dumbfounded as to what you did

[cinder portal]

ok shall we walk through this together?

nooOOOOO

Ok im going to run through the math with you

$121 = 181 - 180 \cos(Y)$

[somber coyote]

MemesPlease:

[cinder portal]

this is our starting point

right?

oh

uhhh

whats cos(Y) ?

not Y, cos(Y)

wanna make sure you did your algebra correctly

ok 60/180 is right

cos(Y) = 0.3333333

where r u getting 0.9999?

is ur calculator in degrees or radians

what did it spit out?

o nothing

if its in degrees, u good

[vagrant elk]

wait why would you do cos of 1/3?

[cinder portal]

wait shit no

cos(Y) = 60/180

have you worked with inverse trigonometry functions?

[vagrant elk]

[cinder portal]

what you're really trying to solve for is Y

we know cos(Y) = 1/3

how would you solve for Y?

Y = arccos(1/3) correct

[vagrant elk]

cos^-1 of 1/3 yes

[spark stag]

you can think of this as

taking the inverse cos of both sides

inverse cos "undoes" cos

so they "cancel" so to speak on the left side

leaving you with y = arccos(1/3)

(or cos^-1(1/3) if you prefer that notation)

[cinder portal]

same thing as inverse cos

[spark stag]

$\arccos{x} = \cos^{-1} x$

[somber coyote]

Namington:

[spark stag]

just alternate notation

[woeful acorn]

What are the hyperbolic versions of the trig functions for?

[spark stag]

well, obviously in hyperbolas

but they'll also come up as solutions to certain classes of problems, such as certain integrals

as $\dv{x} \text{arsinh}(x) = \frac{1}{\sqrt{x^2 + 1}}$, for example

[woeful acorn]

I mean what in hyperbolas

[somber coyote]

Namington:

[woeful acorn]

Like what does it do

In the hyperbola

[spark stag]

well, they're analogous to the unit circle in that

if you draw an arm with a certain angle

then the y coordinate of that angle's intercept with the unit hyperbola

will be sinh(that angle)

[woeful acorn]

Oh

[spark stag]

and the x coordinate will be cosh(that angle)

[woeful acorn]

So it's like replacing the derivation of the trig functions with a hyperbola

[dark sparrow]

it's not an angle @spark stag

[spark stag]

where "regular" trig deals with circles, hyperbolic trig deals with hyperbola

[woeful acorn]

Yea, that's what I meant

[spark stag]

er yeah, fk

not angle, 2*area

[woeful acorn]

Oh okay

[spark stag]

between the arm and the hyperbola

[woeful acorn]

How do you graph a hyperbola

[spark stag]

thats my brain fart

,w plot x^2 - y^2 = 1

[somber coyote]

Display results online and refine query

Requested by Namington#1983

[spark stag]

...thats a shitty plot

one sec

like such

note that a hyperbola is comparable to a circle in that

in a circle, x^2 + y^2 = r^2

and in a hyperbola, x^2 - y^2 = c^2

[woeful acorn]

What is c

[spark stag]

length from the origin to the closest point

[woeful acorn]

Thats pretty cool

[spark stag]

c is just some arbitrary constant

(i should've used some other one, though, as c often refers to the focal point)

[woeful acorn]

So I just accidentally graphed an ellipse by multiplying one of the coordinates by -z

In an ellipse, what number determines the distance between the foci

[spark stag]

that'll be twice the focal distance

again, we usually notate that c, so this'd be 2c

[woeful acorn]

Thanks

[upper karma]

anyone get this

[fringe dirge]

Try to factor both the top and the bottom of the fraction

[upper karma]

I tried

Wait

@fringe dirge this is what I did so far

[fringe dirge]

How were you able to just remove the bottom of the fraction?

[dark sparrow]

[upper karma]

i was just simplfuing the top

but the step i got to would be that over the sin(sin-cos)

[shut crest]

Is there any tricks to simplify complex trig things

[cinder portal]

got an example?

[shut crest]

This is one

[cinder portal]

uhhh

what the

cursed image, this is just as bad as @devout shell stupid riemann sum

[shut crest]

What's wrong with it?

[devout shell]

at least it keeps exact values

and do not insult Reimann

you only hate it because you ||can't do it||

[cinder portal]

😦

[devout shell]

gottem

[upper karma]

i'm crying what is that

Did ur teacher assign some long assignment

Guys when is a trig function neither odd or even

[shut crest]

I'm doing it for fun.

[upper karma]

oh

[cinder portal]

i would probably

just simplify each one, i dont see a pattern

[shut crest]

Working on finding all the circles centers, it's less complex then it looks, but I need to simpify

[upper karma]

@cinder portal @shut crest when is a trig function neither odd nor even

[shut crest]

What does that mean?

tan(90)?

[upper karma]

like odd and even functions

[cinder portal]

the parent functions, are all either odd or even, but if you start messing around with them like sin(x+1), then it becomes neither even or odd

[upper karma]

oh

is this odd or even: sin(x)+c0s(x)

[cinder portal]

shud be neither

[upper karma]

k

thx

[cinder portal]

if you're unsure

do the standard odd even function check method

plug in -x, and see what you get kinda deal

f(-x) = f(x) even function
f(-x) = -f(x) odd function

[zealous trail]

looking for help on my hw anyone around?

[cinder portal]

[devout shell]

posted it in #help-2 already

it's about vector addition

[river vale]

could anyone explain how this works?

never seen anything like it before

[dark sparrow]

$\sin(u) \sin(v) = \frac12(\cos(u-v) - \cos(u+v))$

[somber coyote]

Ann:

[river vale]

phew, where does that come from

well, thanks!

[mystic shadow]

Euler's identity on one of them and fourier transform

Just kidding it's product to sum identity

Which you can derive a couple ways, been doing fourier series/transforms so that's how I recognized it at first

[upper karma]

what does the range of n mean??

(circled part)

nvm got it

[woeful acorn]

Are two lines with different slopes considered tangent to each other?

[hard blaze]

I'd say no, but as I'm not an english native speaker, I'd wait so else point of view ^^

[woeful acorn]

@hard blaze https://i.imgur.com/w4RtZ6A.png

[hard blaze]

No

Let me check sth, wait a sec

[woeful acorn]

Okay

[hard blaze]

They are secant

[woeful acorn]

They only intersect at one point though

Secant = 2 points

[hard blaze]

Or french definition isn't the same than English one, or it's a bit more different than this

And it would by the way mean that 2 lines can't get secants

[woeful acorn]

Tangente signifie qu'ils ne se croisent qu'en un point

~Google Translate

[hard blaze]

Yeah, don't worry, I understood this

But in France it's a bit different

In France, what u showed me are two secent lines

[woeful acorn]

Thats really weird

Why would the definition change based on language

[hard blaze]

And those, are two tangent curves (french definition)

[woeful acorn]

Those are also tangent in english

[hard blaze]

But they're not crossing each other

While ur two lines are

[woeful acorn]

https://i.imgur.com/cPIDQBY.png

Thats secant

[hard blaze]

And, to me that's the difference

[woeful acorn]

This is tangent https://i.imgur.com/9gtPwLV.png

[hard blaze]

Yeah, they are secants in two points

And on second pic, they are tangent

[woeful acorn]

I agree with that

That's true in english

These two lines touch at one point https://i.imgur.com/NHkNcTv.png

[hard blaze]

The thing is, tangent means they just "touch each other", while secant means they are "crossing" each other.
Like a hug and a pal-supplice stuff, u see ?
(Relating to the French domain)

[woeful acorn]

Okay so this is rooted in my misunderstanding of the definition

And I like that analogy lol

[hard blaze]

Thanks xD

Any other questions ? ^^

[woeful acorn]

Nope, thanks.

[hard blaze]

U're welcome !

[woeful acorn]

@hard blaze Actually, do you know about derivatives

[hard blaze]

I know a bit yeah, depending of what u need, but if you need help, just switch to the good channel, I'm following you in it ^^

[minor arch]

Is it possible to find out if there are 0,1 or 2 triangles using law of cosines?

[dark sparrow]

context?

[shut crest]

You need the angles and sides of a triangle for the law of Cosines, so using it to determen the number of triangles dosn't really make scene

Unless it's like a math problem dealing with how many possible triagles thee are

Tho I'd use this http://www.coshoctonredskins.com/Downloads/8-5 Proving Triangles Similar.pdf

[minor arch]

TY. I asked because I had a problem where the only possible solution was using law of cosines.

and it asked me how many triangles could be formed.

I just assumed 1 and was right.

[shut crest]

Interesting

[minor arch]

https://i.imgur.com/Mn1b5EH.png

I keep getting this fucking question wrong.

Can someone give me some guidance?

this chapter so far has covered law of sin and cosign. So I imagine of one those two are instrumental in solving the problem.

But when I search online, solutions seem to involve the use of tan

[dark sparrow]

h cot(40.6°) - h cot(47.12°) = 150

[minor arch]

Wow. That's so odd to me that a chapter revolving around sin and cosign would need cot/tan to be solved.

I'll examine that and try again.

[dark sparrow]

sine*

cosine*

[minor arch]

that too.

[idle bloom]

I mean

Sine and cosine are the 2 "basics" and everything else can be written in terms of sine and cosine

[upper karma]

Anybody wanna help?

I need someone to explain how to do this, I'm not looking for answers I just want to know how to do it

[devout shell]

If AB is a diameter, then the inscribed triangle is a right triangle

Knowing that, parts a,b, and c should be easy enough

Let me get this THM for part d

[upper karma]

I'm doomed

It's hard when you don't have degrees for B I know a circle adds to 360 degrees but

Wait DB makes up a right triangle right? @devout shell

[olive solar]

it's a right triangle, yeah

because AB is a diameter, it has arc angle of 180, meaning the inscribed angle of C is half of that

(1/2)180 -> 90 degrees

[upper karma]

Alright well we're getting somewhere

[olive solar]

you can find AB, with pythagorean, and radius is just half of AB because AB is the diameter, you can find ABC using sin properties, and you can find AC using ABC

[upper karma]

Alrighty, what about this one, I just need to know what to use, Tangent? Cosine or Law of Sines?

[shut crest]

13 or 14?

[upper karma]

13

[shut crest]

Is there a diagram to?

[upper karma]

@shut crest

[shut crest]

The last on is 55.8505360638

[upper karma]

I was talking about A on 13

[olive solar]

can you post the image again but rotated

[olive solar]

,rotate 270

[olive solar]

@upper karma is there anything you notice about AB

[upper karma]

Well they talk about AB = 5 when it's not m

[olive solar]

right but

it passes through the center of the circle

what does that tell you about it

[upper karma]

That it bisected the circle and that one side = 180

[olive solar]

both sides would be 180 yes

now with this information, find angle C

[upper karma]

60?

[olive solar]

sorry, angle D

it was a little hard to see

find angle ADB

given what you know of arc measures and the inscribed angles

[upper karma]

Hmm, I feel like I'm getting an idea

[olive solar]

https://cdn.discordapp.com/attachments/326138757474680852/585135691005034515/image0.png

[upper karma]

So I would want to be 28 degrees in there?

[olive solar]

?

[upper karma]

I'm looking at problem A

[olive solar]

yeah

to find angle y, you need to find angle x and angle ADB

x is given

[upper karma]

76?

[olive solar]

?

where did that number come from

[upper karma]

180-28x1÷2

[olive solar]

???

oh

wait one moment

okay so

you seem to be doing a different approach than what i expected

but that's fine

if you know the other arc measures you can indeed just find the arc measure of AD

so, given that the total for a circle would be 360

you can remove 180 immediately, representing the left half of the circle

after which, you would have to convert angle x to an arc measure

and subtract that from the remainder

$(360 -180) - 2*(28) = mAD$

[somber coyote]

G6PD-efficient:
Compile Error! Click the reaction for details. (You may edit your message)

[upper karma]

That's the answer?

[olive solar]

yeah

[upper karma]

So it was minus 2?

[olive solar]

???\

minus (2*28)

[upper karma]

So 124

[olive solar]

oh

yeah

sorry, i was thinking of the angle

this should be correct

[upper karma]

Next one has the sides and wants me to find the area of c

[olive solar]

given that it's a right triangle

find the hypotenuse

[upper karma]

Don't you just need to find the area I mean, the legs are already given

Yeah I found 20

8x5
---- = 20
2

I am just gonna stop here

[prime jasper]

dunno if this is the right chat but

whats the method of this

[olive solar]

do you remember what vertical angles are?

[prime jasper]

nope

is that whats its called?

i thought they called reflex

angle

[olive solar]

well, it's how i learned it

maybe a different name for you

[prime jasper]

ye maybe

[olive solar]

it's using vertical/reflex angles paired with the idea of supplementary angles

[prime jasper]

whats that?

[elfin heath]

Which question? 8?

[prime jasper]

ye

idk too much terminology

btw this isnt my math homework this is just some recapping just saying it so i aint breaking rules

[olive solar]

er, you're actually allowed to ask for help on math homework lol

just not on exams/quizzes

[prime jasper]

i thought that was one of the rules

[elfin heath]

h and i are 180° - 166° = 14°, since angles on a straight line add up to 180°. g would be 166° because of vertically opposite angles.

[olive solar]

6. Do not ask for help on tests. Any violation of this will lead to appropriate action being taken at mod discretion.

7. When asking for help, do not insist on getting just the answer; we are here to help you learn, not cheat. Likewise, if you are providing help to others, try your best to explain and elaborate instead of simply giving away the answer.

[prime jasper]

oh i see

@elfin heath btw thanks i just looked at the answer

i will need to remember that

wow i just worked out the second one and got it correct

yeet better write it down

[shut crest]

Can you use sohcahtoa to simpify trigfuncions? Like sin(x)/cos(x) = (o/h)/(a/h) = o/h ?

[olive solar]

well

[fossil lotus]

Yes

[olive solar]

first, it would be simplified to o/a

and yes overall

sin/cos = tan

[fossil lotus]

Oh I didn’t even look at what they wrote

But yeah

sin/cos = tan and cos/sin = cot

[olive solar]

(just in case you haven't heard of the term before, cot -> cotangent = 1/tan)

[fossil lotus]

Very useful for solving trig problems

^

[shut crest]

there are so may trig funchions

[fossil lotus]

Just six main ones

There’s also sec (secant) = 1/cos and csc (cosecant) = 1/sin

[vernal tusk]

there's really only two

if you want to be as minimalistic and loose as possible (which is a nice thing when dealing with high schoolers usually)

cos is just shifted sin

and tan is just sin over shifted sin

then sec is just 1/shifted sin

cosec is just 1/sin

cotan is just (shifted sin)/sin

and don't you dare say haversin or i will hurt you

[shut crest]

But it can be usfull to use the other ones to males a shorter equation

[vernal tusk]

oh of course

i just like taking the mindset of "you only need to know like one thing"

it's june, so people are cramming for exams

i like to teach "know this and just derive or remember everything else related to it", and "derive everything"

i dont remember how asin(bx+c) + d works, for example

ik d is the midline

i think c is shift, yeah

b is probably period

a is probably amplitude

i guess i do remember then lol

but i had no clue as i started the final test on trig on khan academy

i just played with my calculator

(ik thats not a derivation at all, but for test yeeting it is)

[forest niche]

Guys I’m confused

I keep getting the wrong answer