geometry-and-trigonometry

[dull remnant]

yea its just plug in arctan(-1) into calc

subtract that number from 180

but one of the solutions in an answer was subtracting from the bottom number?

[dusky surge]

Always use the positive ratio for reference

[dull remnant]

im sorry if im really unclear

o

so we are starting with positive ratios first

ok

[dusky surge]

And then use the table or the picture I gave you to Solve it

[dull remnant]

ok

[dusky surge]

That's how we do it :D

[dull remnant]

sure

so

lets say the example is

arcsin (.18)

[dusky surge]

Sure

[dull remnant]

wait no

arcsin -.45

180<theta<270

[dusky surge]

first find θ when sin θ = 0.45

Then 180°+θ

Is the answer.

[dull remnant]

why is it the smaller number

instead of the bigger number like the other examples

[dusky surge]

Table table table~

The graphy table

[dull remnant]

ok

can u give me an example

ill try ur method

[dusky surge]

Lemme think

arccos(-0.2588), Quad3

[dull remnant]

ok

so quad 3 means 180-270

also

these angles can only be positive right?

[dusky surge]

Pick the positive ratio for reference :)

[dull remnant]

so i got 104.9988703 from arcos(-.2588

then according to the graph

im supposed to add it?

which means it is 284.9988 but thats not in the 3rd

quad

[dusky surge]

Positive one, which means you put 0.2588 instead of -0.2588 in your calculator

Always pick the positive ratio for reference

[dull remnant]

so

it doesnt matter what the ratio is

just make it positive

so answer is 255.0011?

[dusky surge]

Yep

Yea

Wait

Yea
Yes,

Got sleepy lol

[dull remnant]

ok

so

i think i get it

for quad 3

for cos and sin

u add 180 to it

but if its tan

u subtract from 270?

and for quad 2, u add 90 to the angle for cos and tan but subtract the angle from 180 if its sin

[dusky surge]

According to that graphy table, there should not be 270 plus or minus something

and for quad 2, u add 90 to the angle for cos and tan but subtract the angle from 180 if its sin
@dull remnant
Nor 90 plus or minus something

[dull remnant]

ok

lemme try again

for third quadrant, for cosine and tangent, you add 180 to the angle, but if it's tangent you ?
for the second quadrant, for cos and tangent you subtract the angle from 180, but for sin you ?
for the fourth quadrant for sin and tan you subtract the angle from 360, but for cos you ?

i dont know what goes in the ?

[dusky surge]

Stays the same

q1, no change
Q2, 180-theta
Q3, 180+theta
Q4, 360-theta or just -theta

[dull remnant]

what

for all of them?

but there is a sign

after the equal sign thats negative for sin and cos in quad 3

[dusky surge]

That means

The ratio

If the ratio is negative blah blah blah

Sine is positive for Q1,2
Cosine is positive for Q1,4
Tangent is positive for Q1,3

[dull remnant]

@dusky surge OO

so at the END once u get the angle from 180-theta

then u change the positive or negative sign

depending on the stats

wait thatwouldnt work because if the angle is from 180-270 you cant change the sign right?

[dusky surge]

wait thatwouldnt work because if the angle is from 180-270 you cant change the sign right?
@dull remnant
Hmmmm..... I don't really get what ya mean, do you have an example

[dull remnant]

ok

im just confused

on why the 180+theta

applies for everything in q2

because in the graph

it says sin stays positive

but tan and cosine dont?

[dusky surge]

Sin is positive on Q2

180+theta would be on Q3

[dull remnant]

yea

ok

so lets say

i get a problem

that restricts to q3

ik its 180+theta

for all of them

would that work for all of them?

because in q3 tan seems to be psotive

[dusky surge]

Yep, you're right, so when you get something like arctan(1) on Q3, we will have to first find the 45° then apply 180°+45° to find the answer

[dull remnant]

ok

but what do the signs

on the left side of the equation mean

[dusky surge]

It does something when you are NOT calculating the inverse trigonometric function

It's another big topic XD

[dull remnant]

ok

sure

alr

to finish the question

if the ratio is positive

u just enter in calc, and then do whatever the quadrant says

what if the ratio is negative?

[dusky surge]

Find the angle for positive ratio for reference.

[dull remnant]

i just turn it positive?

[dusky surge]

Do it in the draft.

[dull remnant]

wha tdoes that mean

[dusky surge]

Like for
sinθ=-√2/2 on Q3
Since for arcsin(√2/2)=45°
The answer will be 180°+45°

It means, we do not TURN it into positive, we just use the positive ratio FOR REFERENCE

[dull remnant]

ok

so for sin theta=-.45 on q3

arcsin(.45)=26.743

so 180+26.742=206.743

so that would be the answer?

[daring venture]

The length of a chord is equal to its distance to the center of the circle. A second chord in the same circle is twice as long as the first one. How far is the second chord from the center.

[dusky surge]

Correct! @dull remnant

[dull remnant]

1 last question

doesnt that mean

[daring venture]

[dull remnant]

that sintheta=.45 and sintheta=-.45 are the same?

[dusky surge]

Nope

[dull remnant]

hm

whats the difference

[dusky surge]

One is positive one is negative XD

[daring venture]

how do u do this The length of a chord is equal to its distance to the center of the circle. A second chord in the same circle is twice as long as the first one. How far is the second chord from the center.

[dull remnant]

@dusky surge the answer lol

the angle meaasure i mean

[daring venture]

[dusky surge]

The positive one is on Q1,2
The negative one is on Q3,4

@daring venture please use the questions channel :)

[daring venture]

ok

[dull remnant]

@dusky surge ok tysm for answering all of the questions

[dusky surge]

You're welcome!

[pastel zenith]

sin(x)=/=x

[vale nimbus]

sin(x) = x
when x is very small

:)

[pastel zenith]

and by very small, we mean, like, basically 0

[rich wolf]

Only in radians tho

If you say that sin(1°) ~ 1 ur gonna have a bad time

[vale nimbus]

sin(0°) = 0
==> sin(x°) = x

cmon now guys

[rich wolf]

,w sin(0.01 deg)

[somber coyote]

Display results online and refine query

Requested by AMD#4434

[rich wolf]

Off by a factor of 100

,w 180/pi

=tex 180/pi

[vale nimbus]

,w sin(0 deg)

[somber coyote]

Something went wrong!

An unexpected error occurred while processing your command!
The error has been reported to my dev team and should be fixed soon.
If the error persists, please contact our friendly support team at the support guild!

[vale nimbus]

,w sin(0 deg)

[somber coyote]

Something went wrong!

An unexpected error occurred while processing your command!
The error has been reported to my dev team and should be fixed soon.
If the error persists, please contact our friendly support team at the support guild!

[rich wolf]

,w sin(1°)

[somber coyote]

Something went wrong!

An unexpected error occurred while processing your command!
The error has been reported to my dev team and should be fixed soon.
If the error persists, please contact our friendly support team at the support guild!

[vale nimbus]

bot doesnt want me to prove how correct i am

[rich wolf]

,w sin(1°)

[somber coyote]

Something went wrong!

An unexpected error occurred while processing your command!
The error has been reported to my dev team and should be fixed soon.
If the error persists, please contact our friendly support team at the support guild!

[rich wolf]

,w sin(1)

[somber coyote]

Display results online and refine query

Requested by AMD#4434

[vale nimbus]

see it rounds up perfectly trust me

pi might aswell be 3 honestly

pi² = g

[raw tiger]

Sounds like we got an engineer

[eager kraken]

How do I tell what these signs should be if they are plus or minus? Is it just flipped? If cos(x+y), then cosxcosy - sinxsiny ?

[devout harbor]

Yes

[eager kraken]

I think the reason i struggle mostly in math is because there is so many small things like this that my trash teachers failed to teach me

[upper karma]

<+ on left corresponds to - on right and vice versa>

[eager kraken]

aight

[tame ore]

so if y’all know the app named sara

it’s hell

can someone help me with this

[silent plank]

what have you tried?

[eager kraken]

this is an identity

that is cos(x/2)

so why does it can't plug 106 into x?

otherwise i am just not sure where to start solving this one

[silent plank]

,w (1+cos(106 deg))/2 = cos(106/2 deg)

[somber coyote]

Display results online and refine query

Requested by ramonov#0921

[silent plank]

which specific identity were you trying to apply?

[silent plank]

whoops forgot the sqrt

,w sqrt((1+cos(106 deg))/2) = cos(106/2 deg)

you should be able to

[eager kraken]

oi vey

it was right it just didnt want it like that

it's asking for an exact value as a degree and yet the answer was cos53

thanks

[tame ore]

pls help i can’t get passed it

[visual mist]

anyone help

[dark sparrow]

what are you asked for

[versed river]

@tame ore f(x) is plotted on the y-axis, so where is the line in the negative part of the y-axis? (f(x)<0 implies that f(x) is negative, right)

[visual mist]

angle ABC

@dark sparrow

[dark sparrow]

ok so

1. why did it take you 45 minutes to answer that

2. what's giving you trouble as far as finding angle ABC goes

[hushed bison]

i have the coordinates of 1

i have the coordinates of 2

i have the distance x between the 2 coordinates

what's a formula i can use to calculate the coordinates of all 5 points of the star?

[rich wolf]

@hushed bison construct coordinate axes

Let point 1 be at the origin

Let the x axis go through the right most point

[hushed bison]

i dont study math, this is for a cs assignment

[tropic tide]

Anyone can spot the error here

[hushed bison]

im looking for a formula

[rich wolf]

@hushed bison are you fine with polar coordinates

[hushed bison]

nope

not at all

[tropic tide]

Trying to find solutions of sin and the like but I got an imaginary number

[rich wolf]

Why not lol

[hushed bison]

because i dont study math

i study computer science

i just need a formula so i can write it into my program

[rich wolf]

Learn math

Math good

[hushed bison]

yes

but can u help me with this problem?

[hollow finch]

Oh

I found the problem

$\cos^{2}y=1-\sin^{2}y$

[somber coyote]

Lσνιηg✧Sσνєяєιgη:

[hushed bison]

the angle between each vertex is 144 degrees

[hollow finch]

$\cos^{2}y \neq 1-2\sin^{2}y$

[somber coyote]

Lσνιηg✧Sσνєяєιgη:

[hushed bison]

need a formula that can calculate all 5 vertices with the given information

[rich wolf]

Do u need exact answers or approximations

[hushed bison]

exact coordinates

[hollow finch]

$\implies (\sin{y}-1)(3\sin{y}-2)=0$

[somber coyote]

Lσνιηg✧Sσνєяєιgη:

[hollow finch]

@tropic tide

[tropic tide]

Oh, I got confused with double angle lol

[rich wolf]

Construct coordinate axes

Trust me

Nvm lol

[viscid ginkgo]

https://i.imgur.com/HBJBKx6.png

[supple wedge]

what the fuck is that

is that sec x but gay

,w graph sec x

[somber coyote]

Display results online and refine query

Requested by Hmm#2554

[jaunty ginkgo]

how did this substitution work exactly?

am i dumb

[dark sparrow]

no substitution is being made here

[jaunty ginkgo]

well whatever this process is called

[jaunty ginkgo]

@dark sparrow THANK YOU

im dumb

[viscid ginkgo]

I'm having trouble finding the horizontal shift of a cosine graph that's shifted -3pi/8 to the left from 0, ends at 5pi/8 so it's a cycle (or period?) of pi. The horizontal shift is apparently 3pi/4 and idk how

I tried doing 0 <= 2x - 3pi/8 <= pi (and 2pi)

but that's wrong

gets me -3pi/16 <= x <= 5pi/8...

<@&286206848099549185>

https://i.imgur.com/wzy5mU0.png

how do you get C when this graph is

-3cos(2x+3pi/4)

[devout harbor]

C?

[viscid ginkgo]

3pi/4

it's C right?

-Acos(Bx+C) + d

I believe it is

I'm supposed to get the equation from the graph

[devout harbor]

Looks something like Acos(bx+c)+d

if a was -ve then it would be inverted

[viscid ginkgo]

I just don't know how they got 3pi/4

right it is positive my bad

[devout harbor]

hm

one way is, you have 4 known points by looking at the graph, you can make 4 equations and solve for a,b,c,d but this is a long method

[upper karma]

What do you get from: (cos(20°)/cos(10°)) - cos(10°).
I get -sin(10°)^2

[devout harbor]

should be -sin(10)^2 / cos(10)

[upper karma]

Hmm. I did = (cos(10°)^2 - sin(10°)^2)/cos(10°) - cos(10°) = cos(10°) - sin(10°)^2 - cos(10°) = -sin(10°)^2

Is there a link to the bot script so i can illustrate?

[devout harbor]

you're forgetting about the denominator of sin(10)^2

[viscid ginkgo]

damn no one can help me with my problem I've been stuck on for 1 hour

[upper karma]

Ok. I got it. It's -sin(10°)tan(10°)

[hollow finch]

Ya

Same same

[viscid ginkgo]

Pretty confusing graph for 31 not sure where you find amplitude at

Or vertical shift

I know it's -3-3sin2x but it's a weird graph ends in a weird way

I see it intercepts at -3

can't tell if that's vertical shift or amplitude

and apparently 33 has a vertical shift of 3 now that's not intuitive

[vale nimbus]

amplitude is 3 since the graph is shifted down 3 units (so the line "about" which the values of y fluctuate is -3)
graph touches the x-axis at its peak aka goes up 3 at most

shifted down because theres a -3 outside of the sin() stuff

isnt vertical shift the moved down part or am i dumb

[viscid ginkgo]

up or down

I guess I see it now

[vale nimbus]

@viscid ginkgo can you wait 10min? ill try explaining it better when im on pc

[viscid ginkgo]

ok

[upper karma]

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[viscid ginkgo]

yeah but I'm getting the equation from the graph idk if I'm misinterpreting

[upper karma]

One message removed from a suspended account.

[viscid ginkgo]

I just didn't know where the amplitude is found

and where you find the vertical shift

since they're both the same number

[upper karma]

One message removed from a suspended account.

[viscid ginkgo]

like I know there's a y intercept of -3

that could be amplitude or v

[upper karma]

One message removed from a suspended account.

[viscid ginkgo]

yeah

but from the graph, I need to find the amplitude

[upper karma]

One message removed from a suspended account.

[viscid ginkgo]

and vertical shift lol

[upper karma]

One message removed from a suspended account.

[viscid ginkgo]

chapter is "Get equation from the graph"

[upper karma]

One message removed from a suspended account.

[viscid ginkgo]

so I'm looking at the graph and finding the properties of the graph

well I have the answer from the back of the book

but I want to find it out myself

[upper karma]

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[viscid ginkgo]

that's B you get it from the cycle = 2pi/B

so pi = 2pi/B but

it's actually 3-3sin2x - pi/2

[upper karma]

One message removed from a suspended account.

[viscid ginkgo]

oh lmfao

I'm on another problem

it's -3-3sin2x though

[upper karma]

One message removed from a suspended account.

[viscid ginkgo]

yeah I still don't know hwo to find amplitude and vertical shift from that graph

spent like an hour on it

[upper karma]

One message removed from a suspended account.

[viscid ginkgo]

yeah

[upper karma]

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[viscid ginkgo]

I see v because the graph should be halved

now

[upper karma]

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[viscid ginkgo]

like the wave goes half way

[upper karma]

One message removed from a suspended account.

[viscid ginkgo]

idk about amplitude though

[upper karma]

One message removed from a suspended account.

[viscid ginkgo]

well like the wave goes from -3 to 3 for example

[upper karma]

One message removed from a suspended account.

[viscid ginkgo]

yeah

[upper karma]

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[viscid ginkgo]

ah I see it so the bumps go up 3

makes sense

[upper karma]

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[viscid ginkgo]

and I think the graphi is reflected about the x-axis whatever that means exactly

[upper karma]

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[viscid ginkgo]

I think I read that sin equals cos when it's negative or someshit like that

lool

https://i.imgur.com/yA5LnXG.png

[upper karma]

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[viscid ginkgo]

yeah idkl

now I'm wondering why tf it's negative

[upper karma]

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[viscid ginkgo]

here is the difference

https://i.imgur.com/m8Nutai.png

basically the sin starts a certain way

[upper karma]

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[viscid ginkgo]

well it's upside down

[upper karma]

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[viscid ginkgo]

you see it points upwards when negative amplitude

and starts from the bottom when positive amplitude

"baseline" sine I guess that's how I'm gonna think about it

[upper karma]

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[viscid ginkgo]

yeah, thank you

[rich wolf]

@scenic scroll if EAF is 45° what does that tell you about the two angles around it

If the whole angle of that rectangle is 90°

[upper karma]

Maybe you can use letters instead of numerical values?

[upper karma]

@scenic scroll think about the angle that you are given, and look that there are two equal triangles

And that they are inside a 90° corner

[bitter jetty]

Can someone help me with this?

[upper karma]

@bitter jetty sure

Okay

Have you tried drawing it?

[bitter jetty]

no

[upper karma]

Drawing is a must in geometry

Imo

Try

[vale nimbus]

isoceles is what again? (not native )

[bitter jetty]

ok

[vale nimbus]

and yeah drawing is a must

[upper karma]

@vale nimbus tf

Yeah

[vale nimbus]

wtf do you mean "tf"

[bitter jetty]

@vale nimbus its a triangle with 2 sides of the same length

[vale nimbus]

i not native therefore i dont know english triangle names lmao

thanks @bitter jetty

[upper karma]

@bitter jetty have you drawn it

[bitter jetty]

im tryin

[upper karma]

I was helping lol

[vale nimbus]

wdym?

[upper karma]

Its okay

Let him learn by himself

[vale nimbus]

bro i asked what an isoceles triangle is 😭 ur tripping

[upper karma]

Oh lol

Srry @vale nimbus xD

[vale nimbus]

no worries :p

[upper karma]

If O is the center of the circle and M is a point then the circle equation will be MO = ... right?

[bitter jetty]

i cant seem to draw it

[upper karma]

MO = ...?

[vale nimbus]

i'd just draw the triangle and then draw a circle around it

[bitter jetty]

how do i

[upper karma]

Like

[bitter jetty]

it only gave me endpoints for the inscribed traingle

[upper karma]

Yes

[vale nimbus]

they pretty much gave you the lenght for the hypotheneus

[upper karma]

MO = the rayon of the circle

[vale nimbus]

radius

man went full french mode for a second

[upper karma]

Have you drawn the axis

[bitter jetty]

no

[upper karma]

@vale nimbus dudeee

[bitter jetty]

how do I

[upper karma]

I said let him learn

Okay

[vale nimbus]

i just gave him a hint?

[bitter jetty]

o

[upper karma]

Uggh

[vale nimbus]

ima just go back to anime man ur

[upper karma]

Axis

[bitter jetty]

o ya i have that

[upper karma]

Okay sorry its rotated lol

[bitter jetty]

im drawing it on there rn

[upper karma]

Have you drawn the line of the hypothenuse

[bitter jetty]

yes

[upper karma]

So

That ''(4, 4) " what does it tell us?

[bitter jetty]

?

wdym

[vale nimbus]

P(4,4)

[upper karma]

The point (4, 4)

Of the hypothenuse

[bitter jetty]

the radius is root(32)

[upper karma]

Okay I just saw what a circle equation looks like

It tells us that the hypothenuse ...

[vale nimbus]

no it isnt

[bitter jetty]

im confused

[upper karma]

Its okay

MO = r is just an intermediate step

[bitter jetty]

wth is MO=r

[upper karma]

So (x, y) , what does x mean in a parenthesis like these

We want that all the points of the circle has the same distance with the center

[bitter jetty]

how much u move on the x axis

[upper karma]

Yes

And same with y

[bitter jetty]

@upper karma ik that

[vale nimbus]

@upper karma just let al3dium explain he's doing a good job

[upper karma]

Ty

[vale nimbus]

[upper karma]

I would gave a better explanation in French...

[vale nimbus]

je m'excuse

[upper karma]

So if you know that its moving 4 to the right and 4 up

[bitter jetty]

so from the origin to (4,4) that's the radius right

[upper karma]

Its actually giving you the

Yes

[vale nimbus]

is it?

[bitter jetty]

so the length of the hypotenuse is sqroot(32)

[upper karma]

The radius its 4

I wanted him to say length and then get some coordinate geometry things

[vale nimbus]

no it isnt is it?

[upper karma]

But ig its same

[vale nimbus]

cuz they didnt say the center of the circle is the origin

[bitter jetty]

so how do u get the length of the radius

[vale nimbus]

@upper karma can i say how i'd do this?

[upper karma]

Yes

@bitter jetty it is the length

Of the radius

Do you know the distance formula

Wth is the hypotenuse of an isocele triangle?

[bitter jetty]

so sqr root 32

[upper karma]

That's why i wanted this

[vale nimbus]

draw a line from the origin to P(4,4)
draw a line from the origin to Q(0,4)
draw a line from Q(0,4) to S(4,4)
now u have ur triangle
construct a circle around the triangle
measure how long the radius is

or calculate the radius if it isnt easily measurable

[upper karma]

@bitter jetty yeah

[vale nimbus]

u could also make Q have cords (4,0) and do the same thing

[upper karma]

I dont know what piece is doing lol

[vale nimbus]

??

[upper karma]

Oh the triangle is also right

[vale nimbus]

isoceles in a circle

how exactly are any of these sides the radius?

its not given that the center of the circle is on the hypotheneus either

[upper karma]

The triangle is right

@vale nimbus oh you were talking about his problem?

[vale nimbus]

yes

[upper karma]

Lol

[vale nimbus]

XDD

the triangle aint right is it??

im being trolled heavy as fuck

@upper karma who knows maybe it looks like this even?

the sqrt(32) thing is the lenght of the hypotheneus

which can never be the radius of the circle since both ends of it have to touch the circle

at most it'd be the diameter

[upper karma]

Actually the triangle is right

[vale nimbus]

??

isoceles triangle tho

[upper karma]

Yes

[vale nimbus]

wait yeah wtf

its a right triangle

[upper karma]

Yes

[vale nimbus]

yes

[upper karma]

Also

Wtf

I can't solve this problem

[vale nimbus]

pretty sure u can solve with area of a triangle in function of sin or whatever

[upper karma]

@upper karma okay

Did you saw my post?

I tried to set litteral values then solve an equation

Yes I saw it

I feel so dumb

[acoustic jungle]

@upper karma wait so how did you solve it?

I can't solve it

[upper karma]

I didn't

[acoustic jungle]

oh.

[upper karma]

I tried to solve an equation but that don't work

[vale nimbus]

it seems so tedious to solve wtf

[acoustic jungle]

if there is no big brain way to solve it and requires 10 different equations then it is not a very good question

[vale nimbus]

also @upper karma my bad i was wrong the radius is 4

[acoustic jungle]

maybe there is a less tedious way to solve it

[vale nimbus]

well not really tedious just

i cant actually draw the problem down on my paper properly

so thats tedious xd

[upper karma]

Geometry sucks

[acoustic jungle]

geometry is intuitive though

[bitter jetty]

ive been trying this for so long

how do i do it

[upper karma]

Okay so

First

Calculate the third side of the violet triangle

@bitter jetty yeah, get the AC

[bitter jetty]

ok

[upper karma]

What you got

[bitter jetty]

1.89

[upper karma]

Okay so

[bitter jetty]

?

[upper karma]

Dang i hate not being able to draw

Rn

@bitter jetty have you drawn

It

[bitter jetty]

yes

its 1.89

[upper karma]

Hmm

Okay, so draw a perpendicular line from C

(I haven't tried this so, i might be wrong)

[bitter jetty]

ok?

[upper karma]

Have you drawn it

[bitter jetty]

yes

[upper karma]

Okay so lets name that point D

So we can talk

[bitter jetty]

ok

so u get rectangle with a diagonal of 1.89

[upper karma]

you see that BC = AD , right?

@bitter jetty yeah

[bitter jetty]

yes

but we dont have ba or cd

[upper karma]

Okay so draw this, continue from C a line until the end of the violet triangle, so you get rectangular triangle

[bitter jetty]

u mean a triangle inside a recatngle

[upper karma]

Sorry, from C, draw a line to the right

[bitter jetty]

ok

if iu call that point e

[upper karma]

And stop when it gets to the perpendicular of the vortex of the violet triangle

[bitter jetty]

i have be and af

[upper karma]

Idk if im explaining lol

Can you show what you've drawn

[bitter jetty]

no sry

[upper karma]

Idc if its horribly drawn

[bitter jetty]

im on computer

[upper karma]

Oh

[bitter jetty]

i dont have a phone

wait let me see

di u get the idea

sry lines arent straight

[upper karma]

Yes

Perfect

So EC is...

[bitter jetty]

idk

[upper karma]

Look at AF

[bitter jetty]

its 3.04 - cb

[upper karma]

Yes

Yes

Better than AF yeah

And EF is..

[bitter jetty]

= ba

[upper karma]

Yes

And

[bitter jetty]

that doesnt rly help tho...

[upper karma]

Yes

See you have a system of equations

You can solve it

Do you see the system?

[bitter jetty]

yes kinda

[upper karma]

See how it helped

[vale nimbus]

@upper karma did u check my solution

[upper karma]

No xD

[vale nimbus]

then why'd you wanna see it dafuq

[upper karma]

Im dealing with this first

[bitter jetty]

its not a system

[upper karma]

@vale nimbus bc i thought i wouldnt be able to do it first lol

@bitter jetty why not

[bitter jetty]

its ef = ba and eb = 3.04-cb

how do i solve that

[upper karma]

What

Nonl

Sorry

Try doing pythagorian theorem

Idk if its written like that lol

With both triangles

With ABC and CEF

[bitter jetty]

hold on let me try using law of cosines

[upper karma]

Oh

[bitter jetty]

sins*

[upper karma]

I dont know that one lol

Lmk when you are done

[bitter jetty]

ok so for ba i got 1.48 and for bc i got 1.18

does that seem righ

ok i think it is

this one is confusing

ive been working on it for like over 30 min

[vale nimbus]

@bitter jetty how are you going to use law of sin when only 1 angle is given?

[upper karma]

@vale nimbus idk what that is so

[vale nimbus]

a/sin(A) = b/sin(B) = c/sin(C)

[bitter jetty]

i used it to find the angles for acf and then i did it for abc

[vale nimbus]

thats a roundabout way of doing it but i respect it alot

goodjob man

[bitter jetty]

can u help me with the other one

its super confusing

[upper karma]

@bitter jetty dang this one is weird af

[bitter jetty]

yes ikr

like wth

[upper karma]

I was about to do Thales theorem lol

Wait maybe we can

[acoustic jungle]

it's similar triangles

for this question

[upper karma]

Oh got it

[bitter jetty]

so like de = ab/2

[upper karma]

The key is at midpoint

Yeahh

And

[bitter jetty]

k

i got it

this is my last q

the volume should be directly proportional to the side length right

but when i calculate it it oesnt add up

[acoustic jungle]

no

that is not how it works

[bitter jetty]

?

[acoustic jungle]

it would be ^3 porportional

[bitter jetty]

rly

[acoustic jungle]

for example try similar triangles with area

[bitter jetty]

like what

so it would by like 4.5^3 = x^3

[acoustic jungle]

lets say area of 1 triangle is 3 and the ratio between the lengths are 5

the other triangle would not have an area of 5

but rather 25

try the question and let x be the ratio

so $2x4.5x4x$

[somber coyote]

Fishraider:

[acoustic jungle]

then it should be pretty nice from there

[bitter jetty]

is equal to 36 right

then find x

imean 121.5

[acoustic jungle]

are you supposed to round

the ratio isn't a very nice numbe

[bitter jetty]

i dont think so

i think its supposed to be a clean number

i got it tho

@acoustic jungle thx

[acoustic jungle]

@bitter jetty did you remember to cube the x

[viscid ginkgo]

In order for sin^-1 (sin x) = x, x must be a value between () and ()

I don't really even get the question, I mean sin^-1 and sin x?

I really don't know maybe -pi/2 to pi/2

[devout harbor]

Yes that is correct

[viscid ginkgo]

ok ty

why do they say sin^1 and sin x?

[devout harbor]

https://www.youtube.com/watch?v=cPb5akXwXL0

[viscid ginkgo]

oh

[devout harbor]

It's arcsin(sinx)

[viscid ginkgo]

interesting

I thought sin had a range of 2pi at least

[idle bloom]

Sin has a range from -1 to 1

I think you meant to say you thought arcsin(x) had a range of 2pi

which it doesn't because it then wouldn't be a function

[slow tinsel]

hey can someone help me wiht this problem?

[versed river]

try drawing a right triangle with PQ as its hypotenuse

then label its sides in terms of the length of ab

[ancient jacinth]

ping me!

[rich wolf]

@ancient jacinth i want you to draw a picture of the complete circle

[ancient jacinth]

i just did.

[rich wolf]

send a pic

[ancient jacinth]

uh

it just looks like that but with a circle

like

here one sec

[rich wolf]

@ancient jacinth nvm

i just whipped up something in ms paint

[ancient jacinth]

ok

tysm i thnk i have a photo sending lmfapo

[ancient jacinth]

bok

ok

oh wait

nvm

i FSYIHUJWKA

THANKS

I GOT IT

[rich wolf]

you know what to do?

[ancient jacinth]

yes

now i feel dumb

[rich wolf]

intersecting chords theorem?

[ancient jacinth]

mhm

ab=cd

[rich wolf]

yeah lol

[ancient jacinth]

tysm

i have no idea how that helped

[rich wolf]

i had no clue either till i drew it lol

visualization is key

in geometry

[ancient jacinth]

tysm

@rich wolf cani ping you, i feel like i will have more problems in about 15 mins lol

[rich wolf]

sure

[ancient jacinth]

tysm

@rich wolf

last probem

i thought you just had to divide 20/3.2

[rich wolf]

what is the formula for the area of a parallelogram

[ancient jacinth]

uh like bh

wait

yes

[solid topaz]

hey guys pretty new here and i have an assignment due tonight and was wondering if anyone would spare some time to help? if not its cool, im just having a tough time with this

[rich wolf]

@solid topaz just ask

[solid topaz]

its more so algebra 2 w/trig

[crisp patio]

#3

the angle is 60 degrees and i need to find the angle measure of z

wait is alright if i ask question shere instead of the question channel

[rich wolf]

@crisp patio find the other two anglws of that triangle

[crisp patio]

oh i got it figured out, thanks for helping though

amd as in advanced micro devices?

[fading stump]

Hello peeps! 😉

I’m new to the server

May someone explain how I evaluate a logarithm of -> log (base of 10) 5

Thank you 🙂

[versed river]

generally you don't, you just leave it as log 5 because its irrational. sometimes it may be appropriate to use a calculator, or use some technique to estimate the value of the logarithm, but otherwise, theres no real way to evaluate it that im aware of.

[upper karma]

@worthy oxide the hint is really the answer here

[worthy oxide]

Ok

[upper karma]

Is this the answer?

The inner term has 2 raised to power of 1/2 raised to the power i

@Cliff#0001

@Cliff#0001

Cliff you son of a beach

[versed river]

@worthy oxide

[dark sparrow]

they left

[visual mist]

anyone help?

[silent plank]

what are you allowed to use?

assuming you can't simply state circle theorem(s), apply properties of isosceles triangles

[visual mist]

yeah i cant mention theorems

are they isosceles tho

[silent plank]

||radii||

[visual mist]

but still they intersect at different places

[silent plank]

what intersects at different places?

[visual mist]

the radiis

[silent plank]

the construction of the red dotted line OC creates 2 isosceles triangles:

triangle AOC and triangle BOC right?

[visual mist]

oh yeah i thought u were talking about the other ones

[silent plank]

can you continue from there?

[visual mist]

let me try

[dark sparrow]

it's radii, not "radiis"

one radius, two radii

pronounced "ray-dee-eye"

[visual mist]

okay

[leaden belfry]

Not sure if this is the right place to ask, but does anyone know if there's software that can determine the length of lines in an image, if provided with known reference points?

https://puu.sh/FrGkE/e36311a1a2.png

This is an example of something I did manually and it's something that is quite time consuming in my job, so I'm looking for a way to automate it (in the image given I knew the total height/width/length of the table and used that to roughly calculate pixel:cm ratio)

[visual mist]

i couldn't work it outtt

[surreal bolt]

30 is the right answer

assuming it's a right triangle

oh duh it is.

cosine is a nice thought, but that's adjacent over hypotenuse. Sometimes at the beginning students have trouble deciding if a leg (cause adjacent is NEVER the hypotenuse) is opposite or adjacent.

Its adjacent if the leg and the hypotenuse form the angle.

if the leg in question does not form the angle (as it is in this problem) then it's opposite. (note that for ONE angle the opposite is the adjacent for the OTHER angle. Some identities later rely on this fact.)

yup np

yes.

your friend would be right. about X and the angle in the cosine would be 60 degrees.

[silent plank]

@visual mist
tri BOC is isosceles. determine angle OBC
then determine angle BOC
tri AOC is also isosceles. determine angle OAC
then angle AOC
what's the relation between angleBOC, angleAOC and the angleAOB?

[visual mist]

well OBC would be x+y? and BOC would be 180-2(x+y)
OAC would be x and OCA would also be x and AOC is 180-2x

[silent plank]

yep. just need that last part now

[visual mist]

the relation.. let me see

they r all the same

wait no

r they? im not sure abt their relation

[silent plank]

try not to overthink it. mark them on your diagram and it should be quite clear

[visual mist]

okay wait

i drew them out but i still dont know

[silent plank]

c = ?

[upper karma]

Why do my answers not add up?

To 180

[silent plank]

you've got quite a bit of notation issues

but how did you get 59.0° for your theta'?

[summer spire]

@leaden belfry some computer vision library maybe? I don't know of any super reliable way myself, but I'm sure something exists

[upper karma]

Umm idk

What is it supposed to be

[silent plank]

mixing $\theta'$ and $\theta_2$ notation is not recommended. \
the equation: $\theta' = \text{TOA}$ \ doesn't make sense
and for some reason that $\theta_2$ became $\theta^2$ in your work.

[somber coyote]

ramonov:

[upper karma]

Theta2 i just wrote it differently but it means the same

Hi

[silent plank]

no it doesn't

[upper karma]

Its just to differentiate it from the other abgle

[silent plank]

depending on the context its ambiguous

[upper karma]

Like i didnt actually square it it just means second angle

[silent plank]

and may be interpreted as $\theta \cdot \theta$

[somber coyote]

ramonov:

[silent plank]

which is an issue

[upper karma]

Oh

Ok

[silent plank]

if you want to use the same variable

you should be consistent

and use subscripts for both

[upper karma]

But what did i do wrong in my calculations

[silent plank]

i.e. $\theta_1$ and $\theta_2$

[somber coyote]

ramonov:

[silent plank]

anyway input:

$\tan^{-1} \left(\frac{5}{8.3}\right)$ into your calculator again

[somber coyote]

ramonov:

[upper karma]

Cool

Alsso hi xi

[silent plank]

that looks alright and is completely different from what you wrote down

[upper karma]

Yep must have done something wierd with my calculator before

What about my second calc

[silent plank]

same process, calculate it again.

[upper karma]

Tan-1 x 8.3 divid by 5

It’s good 👍

[silent plank]

um what's that "x" doing there?

[upper karma]

I have no idea

[rich wolf]

dude it's not multiplication when you take a trig function of something

[upper karma]

Sorry me being stupid

[silent plank]

how are you getting that?

[upper karma]

Very very confused

[silent plank]

what exactly are you entering into your calculator?

[rich wolf]

dude don't use android calculator it kind of sucks. you're probably not doing order of operations right either

[silent plank]

because the value you initially got for theta_2 was actually ok

[upper karma]

That’s some bad luck 👎

[visual mist]

sorry i went to shower c=a+b

[upper karma]

??

Good do it

Yep

Correct

Same thing as awhile ago

[silent plank]

nfi how you got ~70 in the previous image

[upper karma]

He just had some bad luck that’s all

[silent plank]

sorry i went to shower c=a+b
@visual mist
yes. now apply that to what you have in your original question

[upper karma]

Idk im using my phone calculator its confusing compared to a normal one

[visual mist]

okya

[upper karma]

So 31 and 58 are the correct angles

Ya but what did you insert in ur calculatorrrr

Bro idek

Ok

[silent plank]

not 58

[upper karma]

Yes it’s 58

👍

[silent plank]

no! its not

[upper karma]

???

It add up to 90

[silent plank]

,calc 31 + 58

[somber coyote]

Result:

89

[upper karma]

He just didn’t write the whole thing

It’s 31.6

And probably 58.4

Idk

[silent plank]

its not either of those

[visual mist]

?

[silent plank]

read carefully

[upper karma]

31.06

58.9

[silent plank]

,rotate

[upper karma]

Umm 31 plus 58 plus 90 is 180 right? Yeah it had a few decimals but teach said round it

[silent plank]

round consistently

[upper karma]

Ah yes

Soz

Running on no sleep equal big stupid

Thank u for all the help aha

Yes

[silent plank]

@visual mist not applying it properly

[upper karma]

Thank you too

[silent plank]

identify your smaller angles

[visual mist]

ok

[upper karma]

Geometry and Trigonometry Is Awesome!

👏

[visual mist]

okay i did it again

[silent plank]

,rotate

[silent plank]

looks alright now

[visual mist]

okay

thank you so much

[upper karma]

can anyone help me with this?

[quaint sinew]

you just multiply them

[upper karma]

im not sure what to multiply

[quaint sinew]

the only two numbers in the question

[upper karma]

ah gotchu thanks

[strong pilot]

idek where to start

[silent plank]

AHKW is a cyclic quadrilateral

[strong pilot]

so everything has to add up to 360?

[silent plank]

if by everything you mean internal angles, that would apply to all quadrilaterals

do you know any properties that are specific to quadrilaterals that are also cyclic

[strong pilot]

each pair of opposite angles = 180?

[silent plank]

um can you write a better description

[strong pilot]

for example: m<H + m<W = 180

[silent plank]

ok that's a bit better.

[strong pilot]

because they are opposite angles

[silent plank]

opposite angles add up to 180°

NOT each pair IS 180

anyway, consider applying that here.

[strong pilot]

alrighty ill try--how would i get the measure of the external angles with that though?
do i divided 180 by 5 to get H?

[silent plank]

you shouldn't need to apply external angles here

a quadrilateral has 2 pairs of opposite angles

m<H + m<W = 180
(is one of the equations you can get)

what is the other?

[strong pilot]

m<a + m<k = 180?

[silent plank]

capitals A and K but yes

and those 2 angles are also given (in terms of m)

[strong pilot]

so should i add 8 (m<A) and 10 (m<K) and then subtract that value from 180?

[silent plank]

no.

i can't be bothered writing the degrees.
m<A = 8m right?

(i'm expecting quick yes/no responses for these)

[strong pilot]

mhm

[silent plank]

what's mhm?

[strong pilot]

yea, m<A does = 8m

[silent plank]

and m<K = 10m right?

[strong pilot]

yes

[silent plank]

can you make those substitutions into the equation:

m<a + m<k = 180
you wrote earlier?

[strong pilot]

8 + 10 = 180

[silent plank]

where the the ms go

m<A is 8m NOT 8
m<K is 10m NOT 10

8 + 10 = 180 is clearly a false statement

[strong pilot]

my bad hecc

[dark sparrow]

but ramonov don't you see that the lefthand side has a 1, and 8 and a 0 and so does the right and so they must be equal /j

[strong pilot]

1?