geometry-and-trigonometry

[upper karma]

do I need to use the arccos function?

[serene field]

Yes, use inverse cosine to find the angle in radians/degrees for each of your solutions

Then add a constant multiple to reach all solutions.

[upper karma]

inverse cos is arccos?

also the constant multiple is 2 pi right?

[serene field]

Inverse cosine, cos^-1, arccos

Same thing

[upper karma]

ok

or -2pi

[serene field]

It'd be a variable times pi

[upper karma]

thx

[serene field]

Where the variable can be any whole real number

[upper karma]

2kpi with k belonging to the real set

[serene field]

👍

[upper karma]

@serene field I can't get an arccos calculator online 😦

they only accept whole numbers

[serene field]

Just use desmos

[upper karma]

k

[serene field]

You'll need to use cos^-1 though

Same thing.

Know that your result will be in radians

Although that's probably what you're looking for anyway

[upper karma]

0.52 and 2.62

[serene field]

Check for extra solutions.

A hint, make a unit circle and two lines through them, one at sqrt(0.75), the other at -sqrt(0.75)

[upper karma]

Why is cos^2(pi/8) = (1+cos(pi/4))/2 Please ?

[keen aspen]

Half angle for cosine

[upper karma]

?

I didn't know

Those formulas

Thank you 😃

[keen aspen]

$ cos\left(\frac{\theta}{2}\right) = \left\sqrt{\frac{1+cos(\theta)}{2}\right} $

[somber coyote]

PJS:

$ cos\left(\frac{\theta}{2}\right) = \left\sqrt{\frac{1+cos(\theta)}{2}\right} $
```Compile error! Output:

! Missing delimiter (. inserted).
<to be read again>
\let
l.10 ...\left(\frac{\theta}{2}\right) = \left\sqrt
{\frac{1+cos(\theta)}{2}\r...
I was expecting to see something like (' or {' or
\}' here. If you typed, e.g., {' instead of \{', you should probably delete the {' by typing 1' now, so that braces don't get unbalanced. Otherwise just proceed. Acceptable delimiters are characters whose \delcode is nonnegative, or you can use \delimiter <delimiter code>'.

[keen aspen]

Np

[upper karma]

@serene field it didn't work 😦

[upper karma]

I do cos(arccos(sqrt(0.75))) and it's not equal to sqrt(0.75)

[serene field]

As you can see, your cosine inverse gave you two values.

[upper karma]

yep

[serene field]

However, there are 4 such values that will satisfy this.

*and their coterminal angles

[upper karma]

idk how to solve the equation I got

do you speak french?

[serene field]

I don't, but a few people around here do. 🤔

[upper karma]

ok

maybe we can talk on the voice channel?

it'll be easier to explain

[serene field]

A lack of a microphone makes this somewhat difficult

[upper karma]

ok

[serene field]

We're looking for those 4 angles and all coterminal angles.

The inverse function gave us the angle (in radians) of those in the 1st and 4th quadrants.

[upper karma]

the initial equation is 4cosx^4 - 11cosx^2 + 6 = 0

not sure about that though

I do cos(arccos(sqrt(0.75))) and it's not equal to sqrt(0.75)

is that normal?

[serene field]

Wait what.

[upper karma]

oh actually that's just my calculator that's bugging

[serene field]

Okay um.

Let cos(x) = any variable.

[upper karma]

yep

[serene field]

I'll use u for ease.

[somber coyote]

Yue:

[upper karma]

ok

[serene field]

We can rewrite it with v = u^2

[upper karma]

yep

[somber coyote]

Yue:

[serene field]

Which yields solutions 2, 3/4

Plugging back into u, we get solutions +-sqrt(2), +-sqrt(3)/2

sqrt(2) is > 1

Cosine inverse will be undefined for +-sqrt(2)

So we check for all values of theta that yield a cosine of +-sqrt(3)/2

Which I think you can do yourself. 👍

[upper karma]

yes I did all that!

so we find the values of arccos(sqrt(3/2)) and arccos(-sqrt(3/2)) ?

[serene field]

It's basically asking what angles will yield an x-value of +-sqrt(3)/2

We can do that without a calculator, as the unit circle is awesome.

[upper karma]

yep

[serene field]

Able to find the solutions?

[upper karma]

got them but it doesn't work when I replace x in the initial equation

-2.62, -0.52, 0.52, 2.62

the initial equation is 4cosx^4 - 11cosx^2 + 6 = 0

[serene field]

Lazy to do negatives. It'll work for those too, the powers are even.

[upper karma]

right

how did you find these though?

I replaced cos^2(x) by X

and then solved a polynomial equation

got 3/4 and 2, so 2 doesn't work...

[serene field]

Yes, we solved the polynomial via substituting twice.

So we reverse substituted to get back to the original.

[upper karma]

wdym substituting?

I used the delta method b^2 - 4ac

[serene field]

It's another method of solving

[upper karma]

ok

[serene field]

Our original:

[somber coyote]

Yue:

[upper karma]

yep

[somber coyote]

Yue:

Yue:

[serene field]

Our solutions for v, we plugged into u

Our solutions for u, we plugged back into cos(x)

[upper karma]

ok

[serene field]

The final solutions go back into the original equation.

[upper karma]

that's what I did 😛

not all of them

[serene field]

Try not rounding your values.

[upper karma]

I didn't too much

[upper karma]

$$-3.68*10^-11$$

[somber coyote]

luka:

[serene field]

We're asked for the angles at which we get an x-value of +-sqrt(3)/2

[upper karma]

gives me this

[serene field]

You shouldn't need to do any calculation at all.

Look at the unit circle

[upper karma]

(for one of the solutions)

sqrt(0.75) or -sqrt(0.75)?

how do you write this with pi?

[somber coyote]

Yue:

[upper karma]

oh ok right thx

[upper karma]

ok

I get it now

thanks!!

[serene field]

No problem.

[vital ermine]

anyone able to help with this?

I thought I had the answer with (6^2+8^2-6^2)/(2x6x8)

but answer was wrong, i tried using cosine rule to solve for Cos(O') EDIT: Forgot to rearrange to solve for O' it had to be (6^2+6^2-8^2)/(2x6x6) not the above. Solved now!

[nova jewel]

well i joined the server while yue (the boss ?) was helping Luka and i must thank him/her to the big help

[upper karma]

If we have a piece-wise defined function of two arbitrary polynomials, would it be possible to add together the center of each individual function in order to obtain the center of the piece-wise defined function?

and with center I mean midpoint

[upper karma]

Q1 pls

[keen tangle]

when it meets the x-axis, y=0

[upper karma]

Hint: what's true on the x axis?

Yeah, what he said :)

[keen tangle]

sorry for spoiling😁

[upper karma]

No you're good, I always try to goad other people into helping :)

[upper karma]

answer is D, anyone know how to do this?

[dire rampart]

split the fraction in two and try simplifying

[hybrid steppe]

Ok

[keen tangle]

yeah could you post the picture for reference?

[keen tangle]

Thales' Theorem says that if if AC is parallel to RP, then AR/BR=CP/PB

[hybrid steppe]

So that would make PB = 5?

[keen tangle]

yes

[hybrid steppe]

Ok, so then 15+18+21 would give me my answer?

[keen tangle]

right

[hybrid steppe]

Ok thank you!!

Actually, could you help me with one more?

[keen tangle]

sure

[keen tangle]

do you know what similar triangles are?

[hybrid steppe]

Yes, that’s where triangles are proportional but not the same size

[keen tangle]

you have equal angles at B&D and at Q from left and from right

[hybrid steppe]

Ok

[keen tangle]

Do you know how to continue?

[hybrid steppe]

Not really

[keen tangle]

BQA is similare to DQC

[hybrid steppe]

How would that be set up as a fraction equation?

[supple haven]

5/13/12/13 is = 5/12 right

[keen tangle]

@hybrid steppe AB/CD is BQ/QD

@supple haven depending on what order the fractions are

$$\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}$$

[somber coyote]

Gonzo17:

[hybrid steppe]

Ok thank u

[supple haven]

ohh okay so I was right then

thanks 😃

sometimes I have brain farts for things like that

what about 1/5/13?

[keen tangle]

use parenthes

[supple haven]

$$\frac{\frac{1}{5}}{\frac{13}{1}}$$

fuk

[hybrid steppe]

@keen tangle Thank you for ur help!!!

[keen tangle]

one more brace at he end there @supple haven

[somber coyote]

Kira:

[supple haven]

aye thanks

uh yeah so that thing

[keen tangle]

that's 1/65

[supple haven]

how did you get that?

ohhh 13/13 times the equation right?

[keen tangle]

$$\frac{\frac{1}{5}}{\frac{13}{1}}=\frac{1}{5}\cdot\frac{1}{13}=\frac{1}{65}$$

[somber coyote]

Gonzo17:

[keen tangle]

or as you said

[supple haven]

okay I think I understand it now

thanks for the help fam

[supple haven]

$$\frac{\frac{17}{15}\frac{{-17}{8}}

[somber coyote]

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

[supple haven]

fuk

can someone help me out with this complex fraction?

17/15/-17/8

[somber coyote]

Yue:

[serene field]

This?

[supple haven]

yeah

thanks im not very good with the commands

[serene field]

Flip one of them and multiply across the numerator and denominator

And simplify as needed

[supple haven]

ah thanks

[supple haven]

sin 3pi/4

is 1

so would csc 3pi/4 be

-1?

<@&286206848099549185>

[jaunty horizon]

Definitely not

[supple haven]

oh in that case, how can I find the csc, sec, cot for problems like that?

@jaunty horizon um could you help me out?

[jaunty horizon]

You put it into a calculator

[supple haven]

there must be a way to find it withuot one

i can find sin cos and tan without a calculator

[jaunty horizon]

impractical

[supple haven]

not really

its really easy to get sin cos and tan without a calculator

so long as you utilize your knowledge of the unit circle

and my teacher doesnt allow calculators anyway so its not like I could use them even if I preferred to do so

[somber coyote]

Yue:

Yue:

Yue:

[full zenith]

Do you know the relationship between tan, csc, cot, etc and sin & cos? If you do, you can use your knowledge of the basic unit circle angles to find the values of those trig functions as well 😃 @supple haven

[supple haven]

I think I was able to figure it out

but thank you very much 😄

[full zenith]

Glad to hear that 😃 @supple haven

[scenic nacelle]

Anyone willing to check my answers? I’m really new to this and I’m not sure if I’m doing it right

[upper karma]

what

i need help

[jaunty plume]

write an equation

[upper karma]

hmm

?

[jaunty plume]

okay

start with a picture then

[upper karma]

hmmmm

[upper karma]

[jaunty plume]

that's a right triangle

ya?

[upper karma]

ya

"twice the difference"

what

[jaunty plume]

try to write the question in terms of x and y

well

the difference is one minus the other yes?

so just 2x that

[upper karma]

2x - y

• 12?

math can be confusing

hmmm

brb

[jaunty plume]

should prob be absolute value of the difference

[upper karma]

hmm

x = 2y - 24

[jaunty plume]

or x = 24 - 2y

[upper karma]

oh

[jaunty plume]

and what else do you know about x and y

[upper karma]

my brain just stops when it comes to math

why

[jaunty plume]

¯_(ツ)_/¯

what do you know about the angles in a triangle

[upper karma]

one of them is 90

[jaunty plume]

yea

[upper karma]

hmmmm

[jaunty plume]

and

they sum to

[upper karma]

180

[jaunty plume]

so

what's x+y

[upper karma]

90

[jaunty plume]

now you got yourself 2 equations to solve

[upper karma]

so

y

90 - 12?

idk

hmm

[jaunty plume]

wut

[upper karma]

nah

[jaunty plume]

x+y=90 is first equation

[upper karma]

ima get it

[jaunty plume]

what's the second

[upper karma]

[jaunty plume]

https://discordapp.com/channels/268882317391429632/326138757474680852/519014096209707008

[upper karma]

god i am bad

[jaunty plume]

what don't you get

[upper karma]

idk man

[jaunty plume]

lol

you have x+y=90 right

[upper karma]

yep

[jaunty plume]

and the other one is the sentence thing

x = 2(y-12) or x = 2(12-y)

right?

[upper karma]

yea

[jaunty plume]

so what you need to do is solve for x and y for the first one

( x = 2(y-12) )

then solve for x and y for the second one

( x = 2(12-y) )

one of the answers will make sense

the other won't

[upper karma]

what

[jaunty plume]

what

ok i'll show you the first one

[upper karma]

ok

[jaunty plume]

you have x + y = 90

that's the same as x = y - 90

[upper karma]

ok

[jaunty plume]

but you also have x = 2(y - 12)

so you can do

y - 90 = 2(y - 12)

right

that gets rid of the x so you can solve the equation

y-90 = 2y-24

[upper karma]

my head hurts

oof

[jaunty plume]

subtract y on both sides

-90 = y-24

add 24 on both sides

-66 = y

negative angle doesn't make sense (at least in this context) so you can disregard that

that means the solution will be the answer to the second set of equations

[upper karma]

why am i in this class

[jaunty plume]

x+y=90 and x = 2(12-y)

[upper karma]

ill just ask my teacher

thanks for trying

[jaunty plume]

have you learned about simultaneous equations

[upper karma]

god

no

im in 8th

[jaunty plume]

what about substitution

[upper karma]

yeah

[jaunty plume]

alright you can do it with substitution too

x = 90-y right?

[upper karma]

i dont think this was in the lesson but ok

[jaunty plume]

hm

[upper karma]

this was a while back too

like

weeks ago

[jaunty plume]

how did you do it

[upper karma]

idk

[jaunty plume]

just assigning values to x?

ie x = 2 or something

[upper karma]

ill just ask my teacher

[jaunty plume]

oki

[upper karma]

😑

[jaunty plume]

[upper karma]

that made my brain hurt

math is my kryptonite

[jaunty plume]

easiest way i can explain it

¯_(ツ)_/¯

[sick veldt]

@hard gale

you miss me

hoi hoi hoi

[dire rampart]

yup looks good

[supple haven]

1/? = 4

how do I figure that out

[dire rampart]

set a random variable as what you want

so $\frac{1}{x}=4$

[somber coyote]

lemon catto:

[dire rampart]

rearrange for x

[glad ocean]

so 4x = 1 by multiplying both sides by x

divide both sides by 4 and x = 1/4

so 1 / (1/4) = 4

because division is the inverse operation of multiplication, the statement is synonymous with 1 * 4 = 4

but you should put this in algebra next time because fundamentally it seems far closer to algebra than geometry or trigonometry

so #prealg-and-algebra

[supple haven]

Csctheta =4

So how would i find sin?

[sick veldt]

@hard gale hiya pal

heheheheheh

You know the drill

😛

[hard gale]

🍻

[supple haven]

?

[upper karma]

Kira, draw your reference triangle

[sick veldt]

wx = 0, xy = -3, yz = 0, and zw = 0

so it is an isosceles?

Hmm maybe it isn't

[upper karma]

I'd draw a reference figure just to make it easier

[sick veldt]

nvm got 100% on test

[lament bay]

can you guys check my answers

<@&286206848099549185>

[paper glacier]

you didn't do c? 🤔

also what's the S in "A__S__A" ?

you could also prove, for ex 2, that DE is parallel to CB

[lament bay]

I didn’t know how to do C

[little osprey]

I mean you can calculate angle ABC and angle ACB right?

[lament bay]

Yea

[little osprey]

Then since BD bisects angle ABC, angle DBC should be half of angle ABC

@paper glacier Btw it's Side.

[lament bay]

What should the order be?

[little osprey]

Wdym order?

[paper glacier]

ah i see

[lament bay]

Isn’t there a specific order in steps I’m supposed to write

In

[little osprey]

Find out angle ABC and angle ACB

[paper glacier]

well you find ABC... then DBC

[little osprey]

Then say since BD bisects angle ABC, angle DBC should be half of angle ABC

[paper glacier]

and say idem for ACB and DCB

[little osprey]

Then do the same for ange ACB

Frogu can you help him/her?

I gtg.

[paper glacier]

[little osprey]

Don't worry

[paper glacier]

sure :p

[little osprey]

You'll be fine

Yeah, thanks!

[paper glacier]

[lament bay]

So

How would I write it formally?

@paper glacier

[paper glacier]

so how did you find ABC?

[lament bay]

What do you mean?

I’m just trying to get the steps on what to do

@paper glacier

[paper glacier]

yeah that's the first step... finding abc... using the fact that a triangle has 180° angle in all

[lament bay]

How would I say it?

Triangle ABC is what?

[paper glacier]

nono... angle abc

[lament bay]

How would I write it with symbols

[paper glacier]

and idk which theorems who guys have used

but ABC = (180 - 80)/2

=50

[lament bay]

ok

What do I do next

[paper glacier]

then, knowing that BD bisects abc, you know that dbc is half of abc

[heavy verge]

By the SSS(Side-Side-Side) postulate of triangles, if 2 angles in a triangle are congruent, such as ∠C and ∠B, then that means the opposite sides to those angles are also congruent, meaning AC=AB (1).

If AD=AE then

AC - AD = DC (2)
AB - AE = EB (3)

But AD=AE and from (1) AC=AB so
From (1), (2) and (3) AC - AD = AB - AE = DC = EB (4)

By the AA (Angle-Angle) postulate, triangles CFD and GBE are similar, meaning that

CF/DC=BG/EB

But from (4) we have that DC = EB so

CF/DC=BG/DC => CF=BG

[upper karma]

yo

I need Mad help

geometry test tomorrow

<@&286206848099549185>

oh no

can u help

no

[paper glacier]

what's the problem?

we can't help if you don't ask

[upper karma]

ok

i have test

on some shit with circles

and formulas

like 2 times angle equas arc minus arc then its wrong sometimes right sometimes makes me mad

[dire rampart]

!15m

[upper karma]

wut

[dire rampart]

smh

you gotta post a question first, then wait 15 mins before pinging helpers

read rules plz

[upper karma]

ok i join 4 minute ago

post the question

ok one sec

it is what is the unit of Arcs where it has no angle and other way aroudn

so can someoen help

lol it says in your book

ok ill take a picture

man this take too long

anyone want to help in DM

this no help

dude you are embarassing

your question is a definition, it says in the book if you bother reading

post a problem instead

ok

whatever fuck this bullshti i leave no one HELP EM

[dire rampart]

[serene field]

[copper valve]

amazing

[glacial cedar]

I need some help on this project im doing

[upper karma]

post it

[upper karma]

neeed help

I am trying to proof this

like a stament and reason chart

@serene field

[torpid horizon]

what are you trying to prove

[upper karma]

Is

Why M is the medican

And

[torpid horizon]

oh

[upper karma]

PQ and Pr how are thye congruent

[torpid horizon]

what are you given?

[upper karma]

is up htere

is the given

[torpid horizon]

just the diagram?

[upper karma]

yes

well

[torpid horizon]

are there any given markings

[upper karma]

well

congruent markins

[torpid horizon]

because it looks like you drew the current ones by yourself

yes

[upper karma]

ye

The box is the given

along the diagram

[torpid horizon]

like

the lower right box?

[upper karma]

yes

[torpid horizon]

thats

not helpful

are the markings actual markings

that are given

[upper karma]

yes

[torpid horizon]

ok

well

it says in the diagram pr is congruent to pq

[upper karma]

Yes

[torpid horizon]

so the reason for that is given

[upper karma]

Yes

[torpid horizon]

draw qm and mr

2 points determine a line

[upper karma]

M is the median

[torpid horizon]

wait

what are you trying to prove

m is the median?

???

[upper karma]

ok

[torpid horizon]

??

[upper karma]

I am trying to prove angle 1 and 2

with Why

[torpid horizon]

ok

you could have said that earlier

[upper karma]

sorry

[torpid horizon]

after qm and mr are drawn

they are congruent

because definition of a midpoint

then

[upper karma]

ok

[torpid horizon]

you say pm is congruent to pm
reflexive property

[upper karma]

could you make a chart of this? After?

[torpid horizon]

then triangle qpm is congruent to rpm

no

i gave you reasons and the statements

im not going to do it for you im going to help you do it

[upper karma]

K hold up

[torpid horizon]

then using cpctc

triangle 1 and 2 are congruent

[upper karma]

oooh

can you repeat what you said

[torpid horizon]

go read back through it

its in chronological order

[upper karma]

Thanks @torpid horizon

[slender magnet]

can someone help me with some geometry problems?

[somber coral]

quick question, if anyone is willing to help, need help finding restrictions for trig identities. I know that sinx, cosx and tanx would have to = 0 to find the restrictions but how would you determine that?

[slender magnet]

sorry if some are sideways dont know why that happened

if you can ping me and tell me how to do it with the end results i would be very grateful

[somber coyote]

Yue:

[serene field]

Find the volume of the cone and silo directly.

Consider your trig functions. Remember SOHCAHTOA.

^ for the next question.

[meager lantern]

1. Formula for sum of angles is (n-2)*180 for n = number of sides
   = (9-2)180 = 7180 = 1260

[serene field]

Recall the lateral area, surface area, and volume formulas for a sphere and cube.

[meager lantern]

Oh he beat me to it

[serene field]

[meager lantern]

The rest of the problems are fairly striaghtforward, just recall all the formulas

[somber coral]

ping me if you know, thanks.

[slender magnet]

thanks
dovyou know how to do the rest?

[meager lantern]

Which rest?

The ones I'm speaking about?

[slender magnet]

these two

[serene field]

Recall definitions of trig functions.

SOHCAHTOA.

[somber coral]

do you guys know how to find when sinx,cosx, and tanx will equal to 0?

[serene field]

I like to think of sinx as the y-value on a unit circle.

Similarly, the cosx function yields the x-value.

[upper karma]

I'd call it sin(theta) and cos(theta) just to avoid confusion with x on the x axis

[serene field]

Probably works better, yeah.

tan(theta) is the value received when you extend your angle out to meet the line x = 1.

[somber coral]

then would tan(theta) ever = 0?

[serene field]

Okay, I might've mixed up tangent with something completely different.

If you think about it algebraically, tan(theta) is made up of two other trig functions.

[somber coyote]

Yue:

[somber coral]

yea sin(theta)/cos(theta)

[serene field]

As cos(theta) approaches 0, what happens to the output of the function?

[somber coral]

undefined

[serene field]

And where does cos(theta) approach 0?

[somber coral]

when x = 0

(theta)

90 degrees

and 270 ?

[upper karma]

tan(theta) is y/x

[serene field]

Correct, so we know that tangent will be undefined for values where cos(theta) = 0, which is at those two angles and their coterminal angles.

Similarly, what happens as sin(theta) approaches 0?

[somber coral]

wait what wdym as sin(theta) approaces 0

[serene field]

What happens to the output of sin(t)/cos(t)?

[somber coral]

oh.

it would equal to 0

[serene field]

And since sin(t)/cos(t) = tan(t)

[somber coral]

so tan(t) would = 0 when sin(t) is 0.

[serene field]

That would also = 0 at places where sin(t) = 0.

👍

[somber coral]

thanks alot

[serene field]

No problem, feel free to ask more questions as you come across them. 👍

[deep sedge]

how do u plug #2 in the calculator

[hard gale]

1' (ie 1 minute of arc) = 1/60 degrees

[deep sedge]

so like
as a decimal how do i do it

[hard gale]

so 32°57' = (32 + 57/60)°

Just write 57/60 in your calc tbh

[deep sedge]

ohh ok
and then id do like
cos(32+57/60) = 14.1/x then div 14.1 by both sides ?

[hard gale]

"cos(32+57/60) = 14.1/x" yup

Then $$x=\frac{14.1}{\cos(32+\frac{57}{60})}$$

[somber coyote]

emeric75:

[deep sedge]

tyy lol i had the numerator and denominator switched up xd

[hard gale]

[heavy verge]

Since these are unit circles, I know the areas of each of the circles A, B, C, D and E, i.e. a=πr^2, since r=1 then a=π(1)^2 = π. But how to calculate the area in Circle E subtracting the overlapping areas with the rest of the circles?

[dire rampart]

hint: try forming a square by joining up the centres of the circle

[heavy verge]

Oh right! so the area of such square minus the area of circle E right? Ok I did some reasoning based on that, let's see if it's right:

Just realized I had a typo in the second time I wrote Area of A - overlap, pi should be in the numerator

@dire rampart Does that make sense to you?

[dire rampart]

dont really know whats happening here, but i wouldve just done area of A+B+C+D+(Area of square-area of 4 sectors)

[slender magnet]

if someone would tell me the answers and how to get them I would be very thankful

[umbral snow]

Different formula for each shape. You may have a book that tells you these formulas, or you can google them

For surface area, it's enough to find the area of each face. You may need Pythagorean theorem

[heavy verge]

@dire rampart sorry I just realized the mess I made trying to put everything together in that image (I even failed to paste the final answer). However doing A+B+C+D+(Area of square-area of 4 sectors) = 4π+(2^2 -4*((1/4)*π)) = 4π+(4-π) = 3π + 4. Which is the same result I got. I just reasoned it finding the area of the sector which is 3/4 of the area A and used the area of a square with side lengths of 1 (the same length as the radii of each circle): (3/4)A+(1^2)-(1/4)E = (3π)/4+1-(π/4) = π/2+1. That gave me the area A without overlaps. Then multiplied that by 4 and added the area E: 4(π/2+1) + π = 2π+4+π = 3π + 4. But your way is more straightforward.

[sick veldt]

It's simple, but I'm paranoid I'd miss somethin

[hard gale]

is gud

[sick veldt]

I assume this an a right triangle soooo

3x + 5x + 2x = 90

combine

divide

2(9) = 27

oops

[heavy verge]

If line m and n are parallel then (3x+5) and 65 are corresponding angles and therefore congruent. So (3x+5) = 65

[hard gale]

waht's the sum of angles in a triangle?

[sick veldt]

90

[hard gale]

it's 3x+5 degrees

not (3x+5)²

[heavy verge]

o ok sorry

corrected

[sick veldt]

wait im confused so the first ones 20

cause Bug's explanation messed me up a bit

[hard gale]

yas

[sick veldt]

ok

and number 2

[hard gale]

"waht's the sum of angles in any triangle?"

who knows if it's a right triangle?

[sick veldt]

180..?

[hard gale]

yea

[sick veldt]

so do it with the x values = 180?

rather than 90

[hard gale]

so 3x+5x+2x=180

yas

[sick veldt]

x = 18

[heavy verge]

@sick veldt your answer to the first one is correct x=20.

[sick veldt]

kk ty

[heavy verge]

@sick veldt

[sick veldt]

ik

I looked at the first one as a corresponding

and knew they had to be equal

[heavy verge]

Congruent means two angles have the same measure

[sick veldt]

yeah

[heavy verge]

cool

[sick veldt]

I think the two I sent were right but idk

[hard gale]

1st one yeah

the second one tho.....

what's the measure of <ABC ?

[sick veldt]

180?

[hard gale]

i mean the measure of angle ABC

not the sum of the measures of all the angles in the triangle

[sick veldt]

idk

[hard gale]

the triangle is isoceles mate

[sick veldt]

right

[hard gale]

<ABC = <ACB

ie measure of ABC is also 40

[sick veldt]

y = 40

[hard gale]

yeah 100° for the angle at the top

[woeful flame]

hi why my answer isnt correct?

[sick veldt]

I did 4x + 4 = 6x - 14

4x - 6x = -14 - 4

-2x = -18

x = 9

2(9) + 7 = 25

[woeful flame]

thats right

BC = 25

[sick veldt]

soooo

this triangle with the tick marks

shows its equal to 90

soooo

25x - 15 = 90

?

[woeful flame]

2y = (25x-15)

and also

2y = 60

y = 30

60 = 25x-15

75 = 25x

3=x

[sick veldt]

yeah

[woeful flame]

i dont get it

what is your question?

if youre right?

[sick veldt]

yeah

I put the answer in the box

and explained it

[heavy verge]

That triangle is equilateral, because the tick marks in each side of it are telling you that the 3 sides are congruent. In an equilateral triangle each of the angles measures 60 degrees.

[woeful flame]

😮

someone have an idea why my answer here is wrong?

[dire rampart]

[hard gale]

tbh no

[dire rampart]

they seem correct

[hard gale]

maybe they want a particular form you didn't disclose to us

[woeful flame]

what?

my english is not very good im using google translate to read and to write

[hard gale]

is that all there is to the problem?

just the equation

[woeful flame]

please use english so I will be able to understand everything

I shared the question here

with my solutions

but it say that im wrong

I just think maybe someone here can tell me what is wrong with my answer

[hard gale]

the thing is, it's right

[woeful flame]

oh

so maybe a mistake in the book

?

it happens

[hard gale]

yeah

[woeful flame]

ok thnks sir

[hard gale]

i'm just checking with wolfram to be 100% sure

=pup sin(x) + 2cos^2(x) = 1

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@hard gale

[hard gale]

wow thx

[woeful flame]

what?

what is means

im right or wrong?

i forgot to tell that answer must be 0 < x < 360

as my answer

90 , 210, 330

so its wrong or right sir?

[hard gale]

it's right

and don't call me sir

[woeful flame]

why not?

and why do you put line on your meesege?

[hard gale]

i'm not even 18 mate

[sick veldt]

God this ones annoying lol

I think its right though

I used geogebra to graph it

since AB is horizontal, we're gonna use the x point which is 1

and since AC is vertical, we're gonna use the Y axis point which is 3

use the midpoint formula i believe

[woeful flame]

what is wrong here? 😦

[sick veldt]

god semester 2 I gotta get into sin cos tan

[woeful flame]

how old are you?

[sick veldt]

16.....

just a child

soooooo is my answer correct...?

.-.

lunch is in 35 min

wanna finish this practice test soon

[hard gale]

==acosd(3/10)

[charred spear]

180×acos(3/10)/π = 72.5423968762779

[hard gale]

seems right though

+there should be two answers in [0,360]

[woeful flame]

I cant find it

i got only this

ohhhhh

nvm

also 287.458

[hard gale]

yea

[woeful flame]

but it still wrong

[sick veldt]

hey hey is mines right?

[sick veldt]

oof

[hard gale]

yus it is firewolf

[sick veldt]

kk

[woeful flame]

sorry fireworl

why my answer is wrong dayum 😢

[dire rampart]

yes

[woeful flame]

lemon do you have any idea why my answer is wrong?

[sick veldt]

You use the centroid theorem here right?

2/3 of the distance from each vertex to the midpoint of the opposite side

so 21*2/3

[dire rampart]

@woeful flame they are correct

[sick veldt]

god i hate these problems haha

sooooo long to solve

and if you mess up its a pain in the ass

[tiny grail]

It’s easy geo lol

Pretty standard

[sick veldt]

Is my proof right?

the Bold ones are my answers

the non-bold one is the assignment answers

Not sure about number 5-7

anyone please

[upper karma]

I think you mean angle 4 in the last step, and your proof seems a bit redundant, but you might be limited re what theorems you can use. Also, what software are you using to do that?

[sick veldt]

Pages :/

mac user

doc file

[upper karma]

Oh, so this isn't the type of software that checks your steps? I was thinking of something totally different. OK, so pretty much nailed the proof, but I think it could be shorter, depending on what theorems you're allowed to use.

[sick veldt]

No

This is the proof given to me

through the assignment

I didn't make it haha

I'm allowed to use any theorems

but u have to use the correct one for the type of problem

so for this one its a triangle theorem or exterior theorem problem

[upper karma]

Oh, I'm an idiot. You want to know if what you wrote down is correct, not the proof itself

I think your last step is "subtract <3 from both sides or something"

[serene field]

m1 + m2 + m3 = 180 (interior angles 180)
m3 + m4 = 180 (exterior angle subtends side)
m1 + m2 + m3 = m3 + m4 (substitution)
m1 + m2 = m4 (subtract m3 from both sides)
Is how I see it. Probably not what your prof wants in terms of theorems, though.

[sick veldt]

Yeah I had substitution before

but idk

[main spire]

Hope I’m not inturrupting anything by posting this!

[upper karma]

Suggestion: assign Cartesian coordinates to everything

[main spire]

[vague pier]

Hello, can someone help me with a geometry problem?

Im stuck q.q

[keen tangle]

go ahead

[vague pier]

#10

[keen tangle]

$y=180-87=93$

[somber coyote]

Gonzo17:

[keen tangle]

because they're supplementary angles

and then the sum of angles in a triangle is 180

[vague pier]

ok, i understand. thank you!

wait actually does this mean z=93 or y=93

[keen tangle]

y and 87 'make' a straight line

so y+87=180

and y=93

[vague pier]

so z=36?

[keen tangle]

we know that angles in a triangle add up to 180

and so $z+y+36=190$

[somber coyote]

Gonzo17:

[keen tangle]

but we already know y=93

so $z+93+36=180$

[somber coyote]

Gonzo17:

[keen tangle]

and so $z=51$

[somber coyote]

Gonzo17:

[vague pier]

OH!

I feel so slow now 😓

[keen tangle]

no worries, everone goes through that 😁

[vague pier]

thank you very much for your help :)

[keen tangle]

np

[lusty kernel]

How do I find the phase shift in a sinusoidal function 😦

[serene field]

My explanation varies a bit between trig and physics. I'll explain trig.

[somber coyote]

Yue:

[serene field]

Where:
|a| = amplitude
2pi/b = period
h = horizontal shift
k = vertical shift

[lusty kernel]

Yes

[serene field]

That should tell you everything you want to know about the phase shift, if you can get an equation into that form.

[lusty kernel]

What if it’s already on a graph

And you need to find the equation

[serene field]

Consider amplitude, period, vertical shift

You can choose either the left or right horizontal shift for your 0.

For example:

We can see that the amplitude is 3.

Furthermore, it's translated 1 unit up.

A bit of estimation and we can see the period is 2.

[lusty kernel]

Wait why is the period 2

[serene field]

Distance from crest to crest, is how I like to remember it

[lusty kernel]

Ok

The vertical shift is max+min/2?

[serene field]

I suppose it could be.

[lusty kernel]

I see, what about the phase shift?

[serene field]

Know what kind of function you're starting with

With sine, your output begins at 0

With cosine, your output begins at 1

You can choose to shift your equation to the right or to the left such that it fits the graph.

[lusty kernel]

Uhhhhhhhhh

You lost me lol

[serene field]

🤔

Maybe another method would work.

Look at the part of the graph that would normally correspond to sin(0)

Or rather, a maximum of sine would work

And the value that corresponds to

In the graph I showed above, the maximum value of sine is 4

Which gives us:

We can just evaluate for values of h that satisfy this equation.

[lusty kernel]

Uhhhhhhhhhhh

How do I evaluate that

[serene field]

Ah, the 8 is from another problem I'm working on.

[somber coyote]

Yue:

[serene field]

Or rather

It'd be an angle

[lusty kernel]

-confused-

[serene field]

So the red graph is what we're looking to find

The blue graph is what we've figured out so far.

[somber coyote]

Yue:

[serene field]

At this point, I'd probably just find the difference between their x values and adjust h to compensate

[lusty kernel]

I understand that

So how would I do it without the red graph there

[serene field]

Given just an equation?

[lusty kernel]

No if just the blue graph was there

[serene field]

So what would you be looking for?

[lusty kernel]

Phase shift

[serene field]

The blue graph is pretty clearly not phase shifted, though?

[lusty kernel]

O I mean red

If the blue wasn’t there

[serene field]

I find amplitude, period, and vertical shift

Then create the extra line

And find h via that line

[lusty kernel]

How would you know to put the line where it’s phase shifted to?

[serene field]

Our second graph will contain all the information we found except the horizontal shift

That horizontal shift can be in either direction, to the right or the left.

Finding the distance between the x-value of their relative maximums should tell you the phase shift (h)

[lusty kernel]

Time to become a stripper

[upper karma]

Lol

Can someone please help me to find x

[torpid horizon]

is that even possible

with two congruent lines

[flint bronze]

(x/(2x+5)) = ((x-2)/(2x-1)) I believe

And solve from there

[sturdy flame]

just break it into two different sized triangles and equate them ^

[quaint ruin]

<@&286206848099549185> pls

[steady sleet]

#❓how-to-get-help, rule #4: If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once.
Please read the rules: #❓how-to-get-help!

[quaint ruin]

@steady sleet its been 15min

[valid agate]

Only ask your question in one channel, then

[quaint ruin]

Ok

Soz

[valid agate]

👍

[glad ocean]

hold up

i think i figured out how to prove the parallel postulate

[upper karma]

I know I had posted this but no one had answer it

[buoyant oasis]

The two triangles are similar. The rule for two triangles to be similar is that the ratio of each corresponding sides must be the same.

[upper karma]

So how do I solve it

[buoyant oasis]

Plug in the values and you should get x.

(since you don't have the values of DE and CB, you can throw it out)

[upper karma]

Thank you

[buoyant oasis]

🆗

[somber wagon]

hi

i'm looking for a good book or resource to learn geometry

[upper karma]

Help please

[serene field]

Hint: Side-Angle-Side

[upper karma]

I had got that but do I put that under reason

And that’s it nothing else

[serene field]

That's the reason, but what does SAS prove?

[upper karma]

That it’s congruent

[serene field]

So congruency would be your statement, and your reason would be SAS. You can then go on to prove the rest of it pretty easily.

[upper karma]

The rest of it?

I just need a private tutor at this point *facepalm

PART A

[errant ether]

Do you know what sin(θ) means?

As in what it represents in terms of a relationship between the sides of a right triangle

[magic turtle]

A circle's diameter is 74 meters. Find it's radius.

BOI

[buoyant oasis]

74/2 = radius of circle

[magic turtle]

Wow that was so easy even a baby should know that

[buoyant oasis]

What point are you trying to make

[magic turtle]

It's a hyperbole

[buoyant oasis]

?

[magic turtle]

I can't believe I have to put Creative Writing into this but

a hyperbole is simply an exaggeration of something

[buoyant oasis]

No I get what a hyperbole is but what are you trying to say

[magic turtle]

I'm trying to say that it has been mentioned so many times since a point of life

[buoyant oasis]

Are you drunk

[magic turtle]

Erm... no?

[buoyant oasis]

Nevermind.

[magic turtle]

okay then

[buoyant oasis]

Given that AC = 42 cm and the area of parallelogram ABCD is 672, find CE!

[magic turtle]

idek

Alternate angles are into this

[buoyant oasis]

<@&286206848099549185>

[gritty flare]

By line segment angles

If that angle is 60

And the other one is 40

[buoyant oasis]

e.e my question is the image before that

[gritty flare]

O

Sry

[magic turtle]

nani