geometry-and-trigonometry

[weak lichen]

can anyone help me?

[azure pebble]

can someone help me with this problem

i got 12 but the answer isnt there

<@&286206848099549185>

[onyx cloud]

how did you get 12

[azure pebble]

im not doing it right man

[onyx cloud]

how did you get 12

[ornate marsh]

are you using the proper formula?

[azure pebble]

i dont think so

i dont know wh yi said 12

i got 6

let me show u what i did

[ornate marsh]

yes

[onyx cloud]

what's the length of the radius then

that image looks right

[azure pebble]

i put 3

then diameter must be 6

but the answer isnt there

[onyx cloud]

ah

[azure pebble]

so i dont know

[onyx cloud]

the image is wrong then

[ornate marsh]

hmm ok

[onyx cloud]

the 3 on the right

[ornate marsh]

lemme get you the formula

[onyx cloud]

it looks like you think the 3 is for the arc

but its only for the height of the point

so it would look more like the following

[azure pebble]

redrew it

[onyx cloud]

yes exactly

so the radius is not 3

the radius

is the red line

right?

[azure pebble]

yes

[onyx cloud]

so if we have that

[ornate marsh]

center: (h, k)
equation: (x−h)2+(y−k)2=r2

[onyx cloud]

what formula can we use to find the red side?

[ornate marsh]

(x−h)^2+(y−k)^2=r^2

my bad

[azure pebble]

yeah the formula squid put up

[onyx cloud]

we can simply use pythagoras theorem, which is essentially what squid has

3^2 + 3^2 = r^2

[azure pebble]

yes

i prefer thinking of it on a more intuitive level

[onyx cloud]

we've gone as intuitive as possible

we can't move forwards without crunching a few numbers

[azure pebble]

yeah

so it would be the square root of 18

[onyx cloud]

can you solve for r?

exactly

that's the radius;

how would you compute the diameter?

[azure pebble]

wouldnt the diameter just be square root of 36?

[onyx cloud]

no

[somber coyote]

maximo

[azure pebble]

ohhh

[ornate marsh]

you must simplify all terms no?

yeha that's what i was about to say

[azure pebble]

so it would be 2 times the square root of 18

cant get simpler than that without the calculator

[onyx cloud]

can you simplify that further?

no

let me rewrite it

[azure pebble]

oh

i think i found a way

[onyx cloud]

lets see it

[ornate marsh]

ok go ahead

[onyx cloud]

nice

[ornate marsh]

maximo is better at this than i am, listen to them 😩

[azure pebble]

and thats the answer isnt it

nah bro u guys are greater together

thanks

[ornate marsh]

yes that would be the answer

[prisma whale]

[azure pebble]

im gonna try a problem ok

[prisma whale]

This server is cool

[azure pebble]

and if i dont get it im gonna ask for help

[ornate marsh]

i'll be online so just ping me, you don't have to wait and ping helpers

if i can't answer then go ahead and wait and ping helpers

[azure pebble]

ok

[ornate marsh]

our discussion is finished in this channel for now, so it's free to use for other questions!

[ornate marsh]

i agree

[upper karma]

Can you help me with this please

[ornate marsh]

lemme take a look

[upper karma]

Ok thanks

[ornate marsh]

hmm ok

show me what you've done so far

or are you not sure how to attempt it

[upper karma]

well

Idk if this is right

So not really sure

[ornate marsh]

ok lemme look through and see

the image is still loading hold on dear

[azure pebble]

omg bro i couldnt get the answer

:/

[ornate marsh]

ok marco11, i'll be with you in a second

so lx, you're solving for P

[upper karma]

Yea

[ornate marsh]

right?

ok

[upper karma]

Mhm

[ornate marsh]

ugh give me a moment my brain isn't working

i'm looking through your work

[upper karma]

Yeah I’m a it confused also

[ornate marsh]

i don't see an error

[upper karma]

Is that how it’s supposed to be solved tho

I got 1/2

And - root3/2

For the answer but

[ornate marsh]

that's how i would have done it

[upper karma]

I thought the answer was supposed to be a degree angle

So that’s why I was a bit confused

[ornate marsh]

yeah i'm working through it right now and i'm not getting a proper answer either

[upper karma]

Mhm

Just a little confused about that

[ornate marsh]

ah that's because it's not supposed to be a degree

you're looking for the terminal side of an angle

that's a side length

[upper karma]

O

Makes sense now

[ornate marsh]

mmhm

[upper karma]

So do you think my answer js right

[ornate marsh]

are you good to go or do you need more help

i think you are right

[upper karma]

I’m good thanks

[ornate marsh]

awesome!

@azure pebble ok ready when you are

[azure pebble]

ok

im trying to solve this

@ornate marsh

[ornate marsh]

ok lemme look

ah ok got it

show me what you've done so far

[azure pebble]

ok

[ornate marsh]

and we'll see

[azure pebble]

i did it on a piece of paper so ill have to draw it on ms paint

[ornate marsh]

can you not take a picture of the paper and send it to me

if not that's alright, take your time

[azure pebble]

are they equal?

[ornate marsh]

the problem is not asking for one equation

give me a moment to draw it out

ok

does this make sense

@azure pebble

[azure pebble]

yes

[ornate marsh]

ok awesome

[azure pebble]

i dont see any equality between them

misinterpreted it

[ornate marsh]

that's because there is none

yes

[azure pebble]

if theyre added together they are equal to 180 degrees im sure

[ornate marsh]

yes!

can you go ahead and do that

[azure pebble]

ok

@ornate marsh

[ornate marsh]

um yes

[azure pebble]

ok

[ornate marsh]

ok lemme see the problem again one moment

do you have answer choices

i would like to fact check before telling you anything

[ornate marsh]

ok thank you

alright give me one moment to solve all the way through

ok go ahead and solve for x @azure pebble

[azure pebble]

ok

@ornate marsh

is this how u factor it

[ornate marsh]

yes

wait no

check your signs

[azure pebble]

-12 + 4 = -8

-12 * 4 = -48

[ornate marsh]

wait no you're right my bad

gimme a moment

ok so now that you have solved for x, do you know how to set up the equation for angle 3

[azure pebble]

no

[ornate marsh]

@azure pebble sorry that this is taking forever my online sketchbook does not want to cooperate

[azure pebble]

all good man

just use ms paint

lol

[ornate marsh]

okay, do you know the vertical angle theorem?

[azure pebble]

yes

wait

so that means x is either 12 or 4 right?

[ornate marsh]

x is either 12 or -4

[azure pebble]

yeah

pretty sure it can only be 12

[ornate marsh]

no, it can be both

[azure pebble]

ohh true

[ornate marsh]

negatives do not matter since we are not talking about angles

ok so give me the vertical angle theorem

[azure pebble]

the vertical angle theorem states that vertical angles are always equal

so therefore

since x is either 12 or -4

[ornate marsh]

no do not worry about x right now

focus on the angles

[azure pebble]

but shouldnt we plug x in

[ornate marsh]

yes, but in order to do so, you need to know which equation

yes perfect

now plug x in

[azure pebble]

the answer is 24

@ornate marsh

[ornate marsh]

ugh yes

sorry my computer is acting up

here is the entire thing for reference

[azure pebble]

nice

all of these questions are alot of work lmao

[ornate marsh]

yes

geometry takes a lot of big brain power and many braincells

many functioning braincells

[azure pebble]

nah bro the whole year has been pretty ez

but these questions here

take multiple skills

[ornate marsh]

well i'm glad i could be some form of assistance!

[azure pebble]

yeah man

glad to have you

[ornate marsh]

mmhm np

[azure pebble]

what does the first one mean

[ornate marsh]

like the first answer option?

[azure pebble]

i know what a reflection is

but

idk what T4,0 is

i think T means translation

[ornate marsh]

oh yeah

t means translation

so a reflection, and then a translation of 4 units across the x axis

[azure pebble]

-4 units

and that doesnt look correct lmao

[ornate marsh]

yeah

you're going 4 units backwards

[azure pebble]

B is on the opposite side

heres how im picturing this reflection

[ornate marsh]

yeah

you got the right answer selected

[ornate marsh]

ok go on

[ornate marsh]

yes

and then you would move that 4 units to the left for the first one

[azure pebble]

yeah

and it wouldnt work

[ornate marsh]

correct

[azure pebble]

give me 2 minutes

im gonna do another one

[ornate marsh]

ok

yup, just ping if needed!

[azure pebble]

im dedicated to finishing 95% or all of this homework today

it's due in 3 days and i hate keeping it on my to-do list

[ornate marsh]

i'm here until 12 AM EST, and it is currently 9:52 PM

[azure pebble]

same bro

[ornate marsh]

so i'm here for a while

[azure pebble]

lets go

[ornate marsh]

haha yay!

[azure pebble]

ok i got this question right

[ornate marsh]

ok good

just ping me when you need help or even just a fact-check

[azure pebble]

ok

@ornate marsh

is this correct

[ornate marsh]

no

[azure pebble]

its a cyclic quad tho

the opposite angles are supplementary

[ornate marsh]

supplementary angles are angles that add up to 180 degrees

[azure pebble]

yeah

thats a property of a cyclic quad

[ornate marsh]

wait i read that totally wrong

hold on

no wait i read that right

[ornate marsh]

yeah okay

yes that is true

good job

[azure pebble]

bet

ty

[ornate marsh]

the picture was not clear for me

so i didn't see the angle properly, but in the one you just sent, it is correct

[azure pebble]

got 4 questions left

yessir

is this correct @ornate marsh

[ornate marsh]

pic is loading hold on

ok i can't see the answer choices because my compter has decided to commit die, but you shouldv'e gotten SSS

side side side

because that's what i got

@azure pebble is that what you got

[azure pebble]

yes

i got that

[ornate marsh]

ok awesome!

[azure pebble]

ok bro

2 more questions

[ornate marsh]

ok ok take your time

[azure pebble]

@ornate marsh

is this enough information?

[ornate marsh]

i don't know how to solve that

[azure pebble]

@onyx cloud

[ornate marsh]

i wouldn't even know where to start

sorry about that my guy

[azure pebble]

all good bro

[ornate marsh]

maybe the next one if you need help but yeah

[azure pebble]

nobody knows everything lmao

u did ur best

[onyx cloud]

mns and rnt?

what does that mean lol

[azure pebble]

FR

like what

[onyx cloud]

im inclined to believe that they are similar triangles

in which case the answer would be 10

[azure pebble]

same

[onyx cloud]

oh i see

they are saying mns and rnt are straight lines

which makes this an image with 2 similar triangles

so yeah, in short, 10

[azure pebble]

i got 1 question wrong

this one

[onyx cloud]

that's for a parallelogram

rhombus has diagonals perpendicular to each other

and bisect each other as well

[azure pebble]

so b

[onyx cloud]

yes

wait

i think they don't bisect each other

[azure pebble]

ok thanks @onyx cloud @ornate marsh

what

[onyx cloud]

but they bisect the angles

[ornate marsh]

yeah no problem!!

[azure pebble]

they would have to bisect each other to form the angles though

[onyx cloud]

take this image for example

they don't bisect the actual diagonals, but the angles

actually i think they might need to bisect each other to be classified as a rhombus

i have no idea honestly, but the best answer to that question seemed to be B

[finite sky]

answer is 8???

similar triangles???

@azure pebble

nvm i mean 10

[covert ridge]

how do i solve this - "In a windstorm, a tower was bent at a point one fourth of the distance from the bottom. If the top of the tower now rests at a point 60 feet from the base, how tall, in feet, was the tower?"

[upper karma]

u can calc height of tower as 25 tan47 degrees

and then angle from y is arctan (height/15)

[oak lotus]

I need some help at exercise 4 c)

I need to calculate that sum

What should I use gauss one?

Tag me when if anyone can help thanks in advance!

[dark sparrow]

tan(x) = -tan(180°-x)

@oak lotus

[oak lotus]

And how it should help me?

Like, I don't get it

[dark sparrow]

a number plus its own inverse is 0

tan(1°) + tan(179°) = 0

tan(2°) + tan(178°) = 0

etc

[oak lotus]

OHHHH

So everything is zero right?

[dark sparrow]

yes

[oak lotus]

So tan of 1 is the inverse of tan 179? Right but how? Like can you explain it to me please?

[dark sparrow]

inverse as in additive inverse. i didn't want to say negative

[dark sparrow]

see this with x = 1°

maybe it'd be better to say opposite

[oak lotus]

So, like tan of 1° is the opposite of tan of 179°

[dark sparrow]

yes that's what i said

[oak lotus]

But why?

Like, don't get me wrong I am not making fun of you or something

But I really don't get it why it's the opposite.

[dark sparrow]

so you want me to explain why tan(x) = -tan(180°-x)?

[oak lotus]

Yes, if you can and want to.

[dark sparrow]

$\tan(180\dg - x) = \frac{\sin(180\dg-x)}{\cos(180\dg-x)} = \frac{\sin(x)}{-\cos(x)} = -\tan(x)$

[somber coyote]

Ann

[oak lotus]

THANKS

I got it

[limber night]

I need some help plssss

[dark sparrow]

there is a nice formula that gives you the sum of the angles in a polygon with n sides

[upper karma]

how do you go about solving geometry problems that ask you to prove that a line AB passes through a certain point if B is varied across and arc or whatever?

[silk patio]

Too broad of a question

There are lots of ways and it depends on the problem

Do you have a specific problem?

[upper karma]

yezz

[silk patio]

Identify what the fixed point is, in this case it’s the orthocentre of the triangle

[shut venture]

I have a stupid question i always confuse myself when in an exercise some weird external or internal angle let me screenshot a pic.

weird cant seem to find it

[kindred peak]

I’ve done some work already for this problem I’m trying to do

But my question now is how to find the arch?

[upper karma]

Arch length is s=r(theta)

There's an equation for this type of mechanical problem

I could give it to you but idk if you want to figure it out with the equation for cross belt drives

[kindred peak]

Nah cause It’s probably way over my grade level 😂

What is theta?

[kindred peak]

Ohh

But that’s the thing

I don’t know what the 0 thing is

[upper karma]

For now, you could substitute it for pi + 2x, because its half the circle plus two equal lengths

And pi + 2(other variable) on the other circle

Does that make sense

[kindred peak]

Does the “0” has to be related to whatever the circle shape is?

[upper karma]

0? That's theta, meaning the angle

[sturdy yacht]

when do you sqaure root during the Pythagorean theorem?

[kindred peak]

I understand that, I was just asking if being able to substitute pi + 2x is ok for my problem

Or is “pi + 2x” situational

[upper karma]

Oh, the x variable would be situational, dont mix up variables

[wispy cedar]

can someone help me with 29

[south ginkgo]

Transpose sinx and divide the entire equation by 2

[rancid fulcrum]

how to solve for angle N in a triangle when m = 62, n = 31 and o = 28?

[dark sparrow]

you know all three sides and you are looking for an angle

which law relates three sides and an angle in a triangle?

[rancid fulcrum]

cosine?

[dark sparrow]

correct, the law of cosines.

[rancid fulcrum]

ayee but when I tried to solve for N, it results into a math error

[dark sparrow]

are you using a calculator?

[rancid fulcrum]

yeah

[dark sparrow]

can you sh-

wait

hold on

are you sure you got the side lengths right

62, 31 and 28?

[rancid fulcrum]

mhm

[dark sparrow]

this breaks the triangle inequality

there does not exist a triangle with side lengths 62, 31 and 28.

[rancid fulcrum]

ohhh okay

[dark sparrow]

are you sure you have the side lengths right?

[rancid fulcrum]

yes maam double checked and triple checked

[dark sparrow]

are you sure the problem also does not allow for the possibility of the triangle being impossible?

[rancid fulcrum]

i guess?

[dark sparrow]

can you show the entire problem text

all instructions

exactly as they are written

[rancid fulcrum]

number 6 only

[dark sparrow]

can you talk to your teacher

email them and say triangle MNO is impossible as 31+28 = 59 < 62, which violates the triangle inequality

[rancid fulcrum]

aytt sure

for now, ill just put my answer as na then

thank you fellow student

[peak mesa]

how can i calculate the circumference of a circle that has a width and a height different (data that is known)

like this

[earnest echo]

First off all, that's not a circle

It's an ellipse

And we don't have closed form expression for it's circumference

[humble pulsar]

There are approximate equations for the perimeter of an ellipse based on the lenghts of the major/minor axes

[surreal bolt]

You can estimate it arbitrarily well with various numerical methods, but yeah. There's no closed form.

[ionic pendant]

@inland quarry

Ask here

Ur transformation question

[inland quarry]

Oh ok

ok so i know the transformations I just don't know how to do them like in my head

once it says reflection

I can reflect it an my brain can process what it would look like but if it says rotation or another reflection or something then I just give up

could someone pleas tell me if they have a trick or something to do this kind of problem?

(I have a final in like 1 hr btw so any quick responses would be great)

@ionic pendant ?

<@&286206848099549185> ?

[ionic pendant]

E

[inland quarry]

I mean there's more than one answer and I have the answer key but I just want to know how you do it in your brain

sorry if that's confusing

like how do you mentally flip it and rotate

and stuff

I feel so stupid rn

[inland quarry]

@sharp plume ?

[livid moss]

Don't ping random people

[sharp plume]

^

[livid moss]

While you wait for someone to help you, you can go read #❓how-to-get-help

[inland quarry]

I have a final in like 10 mins

I won't need help on this question afterwards

[silk wasp]

@inland quarry something that's helped me is by keeping track of two vertices of the shape, the rest will follow

[sharp plume]

you should be completely ready 1 day before the finals tho

[inland quarry]

hmm ok I see. So like I would maybe track E and I in this case?

[sharp plume]

doing exercises 10 mins before a final won't help much

imo

[inland quarry]

its just this prob

[sharp plume]

(doing exercises that u don't know how to solve, revising exercises is alright)

[inland quarry]

I have everything else done solid

[livid moss]

Also, disappointed that you won't need help after the final, which means you don't really care about how to solve the problem because then you would still care after the final :(

[inland quarry]

yes sorry but I did geometry from like last year to now

I started preparing super early

[silk wasp]

Yes, I and E are being used the most so it should help

[inland quarry]

and I like geo for the most part

but transformations just aren't my thing lol

[inland quarry]

ok tysm!

[silk wasp]

Good luck on your final!

[loud flare]

Help plz

[upper karma]

is it possible that a vector can have more than one set of components?

[humble pulsar]

what do you mean?

[bleak shell]

Hi guys, can someone help me with this question

[deft dagger]

https://gyazo.com/3a97db2ea4413ad919a4ef1604a767da

[trim breach]

45-45-90 triangles are a special case of triangle that is best to memorize. You can use the Converse Isosceles Triangle Theorem to understand why the legs are congruent, and then Pythagorean Theorem to get the hypotenuse (AB in this case). However, like I said, you should memorize it after you understand it.

[upper karma]

like is one resultant vector able to be broken down into numerous sets of components?

[humble pulsar]

Yes

[upper karma]

but how? and why?

I thought a vector can only have one set of components

[humble pulsar]

how: different basis
why: different basis

[upper karma]

like how can there be multiple groups/sets of components that each yield the same resultant vector

[humble pulsar]

if you have a R2 vector like [x,y], you can write it as x[1,0]+y[0,1]

[1,0] and [0,1] are a basis of R2

[upper karma]

but then won't the magnitude or direction change ?

[humble pulsar]

No

it's like saying Dog in English vs French

[upper karma]

also we haven't looked at vectors in terms of the coordinate systems

[humble pulsar]

ok so arrow in space vector?

[upper karma]

uhh

i think so

actually yes

[humble pulsar]

Ok so draw the same resultant vector twice (measure with ruler and protractor if you want)

then you can obviously draw 2 different pairs of vectors that add up to the resultant

[upper karma]

this is the part I dont understand

because then, won't it change direction ?

[humble pulsar]

Actually take a pencil and paper and do it

[upper karma]

oh wait could I use parallelogram law for one set

and then triangle law for the second?

yeah im doing it rn'

[humble pulsar]

So pick a "normal" component breakdown of one (the 2 vectors you add are orthogonal)

then do some wacky pair of vectors for the other

since you started with the exact same vector for both, the magnitude and direction remain the same

[upper karma]

but how would I do this ?

[humble pulsar]

Draw a different pair of vectors to the 1st pair you did

same resultant vector, different vectors being added

[upper karma]

but aren't those technically the same components

[humble pulsar]

no?

clearly not

[azure pebble]

ok i'll wait until u guys are done

[humble pulsar]

the 1st vectors are clearly different vectors

[upper karma]

here lemme show you how i drew it

[humble pulsar]

yeah those are different vectors

[upper karma]

but ultimately, are these not the same components ?

[humble pulsar]

they arent

[upper karma]

they are just translated though

[humble pulsar]

one is straight down and one is down and to the right

so they're different

[upper karma]

the components themselves are the same, but the formation is different

[humble pulsar]

this is why we tend to not do geometric vectors in a purely math sense, we use vector spaces

[upper karma]

so does this count as a new set of components ?

[humble pulsar]

yes it's a different set of component vectors

[upper karma]

since the direction of the components is slightly different?

[azure pebble]

are you guys done

welp both of u guys are offline

can someone help with this problem

@onyx cloudare u available

[upper karma]

@humble pulsar u mind if I dm ?

[azure pebble]

<@&286206848099549185> does anyone know how to solve my problem

nvm i got the answer

can anyone do this problem

[wide thunder]

Do you still need help Marco?

[azure pebble]

yeah but with a different question

this one is confusing me

[wide thunder]

Are you doing a test?

[azure pebble]

no

homework

[trim breach]

Find the ratio of length to width of the original map. This ratio of length to width should be the same for the smaller map too, so check which option maintains this ratio.

[upper karma]

how would I get part d into a single vector ?

[arctic isle]

1/2 FH?

[rapid tartan]

who here is feeling extra nice today and can help me with some geometry homework

[azure pebble]

can you draw it out please

can someone help me

nvm

[edgy scarab]

what is the lateral and surface area of this?

i don’t know the radius

and how do i find measurement

of the hemisphere

[edgy scarab]

anyone?

[dense hound]

anyone know this?

[analog tree]

Yes.

[trim breach]

You can use a trigonometric ratio to get the diameter of the circle and find the radius from there.

[long ferry]

try using sin(a+b) formula

[tropic roost]

Can someone help me with trig

I don't understand but fully

At me if you can help

[umbral sand]

help

[dark sparrow]

DEFQ is a parallelogram, is it not? so it'd only be cyclic if it were a rectangle...

are you sure you got the problem statement down correctly?

also, what's PG doing there? it looks like it has nothing to do w/ the question

@umbral sand

[umbral sand]

i also got defq as a parm

It's from previous year board exams tho

I did write it correctly

[dark sparrow]

yeah well as-is it's asking you to prove the unprovable

so like

yeah this question is fucked up

[umbral sand]

i think so too

[tiny snow]

Either you did not copy it correctly or the problem was not correctly stated. It should read "Prove that DEFG is a cyclic quadrilateral".

[umbral sand]

Printing error ig

[umbral sand]

How'd ypu prove that?

[tiny snow]

You need to show that <GDE + <GFE = 180 = <DGF + <DEF

[mighty igloo]

and there are loads of similar triangels - you might want to use this

[surreal bolt]

Also constructing another line so we have a rectangle might help! 🙂

[dark sparrow]

i would argue that it is absurd to expect students to just know that this was a printing error on the book's end

[dull thorn]

Hello, I have the following question: Given triangle $uvw$, if I have that $a$ is the midpoint of $uv$, and $az$ is the line through $a$ parallel to the angle bisector of $\angle uwv$, how do I show that the length of $uz$ is equal to the length of $zw + wv$?

[somber coyote]

Liria ^(;,;)^

[dull thorn]

I thought that it would have something to do with the incentre because that's the intersection of the angle bisectors

But I don't think that's it

[tiny snow]

It looks like it will only hold if z intersects uw. Move u closer to w here to see that it's not true if z is on uv: https://www.geogebra.org/geometry/gbbdqcdk

@dull thorn ^

[dull thorn]

Ah yea, z has to be on uw

Like z being on uw is part of the question

(for some reason it wouldn't let me use z so I used zzz)

[mental crystal]

How is it possible, I just suck so hard at geometry, I just cannot answer things properly

I read online that geometry requires a gift to understand properly

[dark sparrow]

what are you doing rn in geometry?

there may be multiple reasons why you suck @mental crystal

[upper karma]

you don't need to be born with some kind of "geometry gene" to succeed in high school geometry it's completely possible to succeed in given that you have a solid foundation in algebra and study your in-class notes

[compact jasper]

the circumference of the circle is slightly more than three times its diameter. for unit circle (1.0 ) circumference is Pi . Is it fact or fiction ? 🙂

[upper karma]

this is just a definition of pi

that circumference is pi times diameter

and saying pi slightly great than 3 makes sense, but has much error

[dark sparrow]

@compact jasper the circumference of a unit circle is 2pi, not pi.

[dark sparrow]

<@&268886789983436800>

[weary drift]

👢

[dark sparrow]

why not 🔨 ?

[weary drift]

fewer chars

[edgy scarab]

what happened

[weary drift]

troll

[edgy scarab]

what was he doing

[dark sparrow]

spam

[edgy scarab]

are my messages invisible in this server when it comes to math?

can someone please help me step by step find the radius

[dark sparrow]

they are visible

[echo hamlet]

Can anyone help

I dont understand my homework

[dark sparrow]

@echo hamlet channel busy, please move (probably to one of the #questions-_ channels)

[echo hamlet]

I tried nobody anwers

[dark sparrow]

be patient, nobody has any obligation to make help come immediately

anyway

@edgy scarab i believe that last time you posted this problem we ran up against you not knowing how to find the surface area of a sphere - have you set that straight yet?

oh wait

you said radius

[edgy scarab]

I know how to find lateral area and surface area, but i dont know how to find radius

[dark sparrow]

you have a right triangle there

[edgy scarab]

im putting it here so you dont need to scroll up

the surface area of a hemisphere is 2pi(r) squared

[dark sparrow]

2pi r^2

use this symbol for exponents: ^

[edgy scarab]

Oh yeah

ok

[dark sparrow]

anyway, ok, you're concerned with finding the radius

[edgy scarab]

yes

[dark sparrow]

do you see that you have a right triangle, and that its bottom leg is equal to twice the radius?

[edgy scarab]

yes because 2r = diameter, right?

[dark sparrow]

yes, or you could also just see it be composed of two radii in a way that is painfully obvious

so look at this triangle in isolation now

are you able to find the length of the bottom leg?

[edgy scarab]

no?

[dark sparrow]

are you familiar with the basics of trigonometry?

[edgy scarab]

yes, with tanget

would it be 14/tan40

[dark sparrow]

so then why did you tell me just now that you could not find it when you just did

[edgy scarab]

because i thought i didnt know

i look at my notes from yesterday

[dark sparrow]

ah, so it took me prompting you on basic trig for you to realize

ok

you have found the diameter of your shape now

[edgy scarab]

so would i do 14/tan40 in calculator

and divide by 2 to get radius?

[dark sparrow]

i don't know, would you?

[edgy scarab]

Yes

i got 8.3

as the radius

[dark sparrow]

does the problem say to give the answer to some number of decimal places?

[edgy scarab]

yes

tenths

[dark sparrow]

okay, then do intermediate calculations with at least twice that precision, so as to prevent rounding errors

[edgy scarab]

and i am supposed to find the lateral area and surface area of the figure

[dark sparrow]

do intermediate calculations with more precision than the answer calls for (or even just do them in exact form if possible), and round ONLY at the end

and no earlier

[edgy scarab]

ok, thank you so much

[rapid tartan]

anyone free later tonight around 9pm est and can help me with some geometry questions

mostly about volume

[dark sparrow]

why such a specific timeframe?

[cloud karma]

how i do these questions ?

[rapid tartan]

lol that’s when ima be home and at my computer

[trim breach]

The scale factor is the ratio of the new to the original. To write the ratio, just find the two corresponding parts. For example, the pentagons have a (1/7) scale factor because to get a side length of 5 from 35, you multiply by (1/7).

[upper karma]

help plz

[hallow ermine]

I'll assume you already know the numerator. What about the denominator?

[tight peak]

can anyone help me out please?

sorry @hallow ermine

i was interrupting you

[sour jacinth]

What is mDE?

[tight peak]

the measure of the Arc DE?

[agile furnace]

heyy is anyone good with tessellations and transformations??

<@&286206848099549185>

[karmic urchin]

anyone know how to solve any of these... i am so confused T-T

[hallow ermine]

For problem number one, left hand side of the equal sign you should expect to get $\cos^{2}x+\sin^{2}x$ as the numerator. What will its denominator be?

[somber coyote]

crabdancerking

[hallow ermine]

Number one is a powerful number. Use it to your advantage.

[karmic urchin]

would the denominator be cosx?

[hallow ermine]

Right.

[hallow ermine]

You know what $\cos^{2}x+\sin^{2}x$ equals to? What is an identity for $\sec x$?

[dark sparrow]

identity*

[somber coyote]

crabdancerking

[karmic urchin]

secx= 1/cosx???

[hallow ermine]

Boom! Done.

[karmic urchin]

really?

[hallow ermine]

Yep.

[karmic urchin]

i thought it was so much more complicated than that LOL... thank you 🙂

[sand root]

Could someone tell me how I would find the angles to work out the length

[hallow ermine]

Just gotta know your trig identities.

[karmic urchin]

yeah, I remember em! gracias

[tiny snow]

@sand rootFirst determine the inner angle y, then use the law of cosines to get the distance c, and finally law of sines to get x, from which you can calculate the new bearing.

Note the two parallel lines, which you can use to determine y.

[dull thorn]

How would I prove that the distance between the intersection of the cleavers of a triangle and the centroid is 3/2 the distance between the centroid and the incentre?

Like given a triangle u, v, w, how could I express j?

I know that g, the centroid, is going to be (1/3)(u + v + w)

[tiny snow]

@dull thornAre you allowed to use the fact that the Spieker (cleaver) center, the incenter, and the centroid are colinear (without having to prove it)?

[dull thorn]

No, the actual question is asking me to prove that these are collinear

Like the actual question text is

And I've been stuck on it for a bit now

I solved the first part which was showing that the cleaver bisects the perimeter (eventually) but I have no idea what to do here

[tiny snow]

I can't see any immediate way to prove they are colinear. There's probably some elegant way to do so, but all I can see is working with lots of angles to prove it. Once you know they are colinear, the "one-third" part follows immediately from the fact of how a centroid divides the median CD in a 2:1 ratio, and that the orange line through D is parallel to the dashed green line through C. Since the red dotted line must adhere to the same ratio.

You already seem to have a sketch in GeoGebra? If not: https://www.geogebra.org/geometry/znz8qfmm

Maybe something similar to how they prove some other centers lie on the Euler line? https://en.wikipedia.org/wiki/Euler_line

[sand root]

thanks

[sand root]

what should i do to determine angle y tho

[dull thorn]

Wait why does the one-third part follow immediately? A centroid didides the median CD in a 2:1 ratio, but what does that have to do with the red dashed line?

And we do know that uh

[dark sparrow]

what are o and g?

[dull thorn]

g is the centroid and o is the circumcentre

[dark sparrow]

hmm

i mean i wouldnt call it immediate but

oh isnt all that what you and roen have been doing

[tiny snow]

Look at the parallel lines. From that you can deduce that y = 100 + (180 - 160).

[dull thorn]

Pardon?

[tiny snow]

I was answering @sand root

[dull thorn]

Ah, I meant in response to Ann

[tiny snow]

Haha, the confusion is total...

[sand root]

ight thanks

i got 871.77 for c and a bearing of 136

when i do law of sine do i add that answer on to 100

[tiny snow]

This is what I meant by the ratio following from lines meeting in the same point between two parallel lines.

[dull thorn]

Oh I see

[tiny snow]

Unfortunately the incenter is generally not on the Euler line, so you cannot use it here.

Does your book prove that theorem? Perhaps you are supposed to use a similar proof here then?

[tiny snow]

Yes, the new bearing would be 100 + x.

[dull thorn]

I'm guessing I'm supposed to use a similar proof

Which theorem does my book prove?

We have this

[lilac adder]

guys

do u know geometry?

faaast

please i need ur help \

[wise pawn]

what's the rush

[upper karma]

What would the formula be?

<@&286206848099549185>

I am very confused.

[lilac adder]

i want this one

[upper karma]

?

[lilac adder]

pls

can u read it?

Consider the right angle xOy. In Βοχηνουμε το Α and in Ο Ο Γ Γ Α Α = 5 cm. From
a random appearance B of Ox we bring parallel to AG that intersects Oy in D construction
NW = 20 cm. We drive M, N, P, S the materials of the segments AB, NW, DG and AG hahahahaha.
Δ1. Note that the MNRS is a rectangle.

Δ2. Complete the MP problem.

Δ3. Yes add to the result ΣΝ.

thats it

can someone solve it?

[sick oak]

ασε ρε

[silk patio]

What’s the rush

[tiny snow]

Ok, that's probably a good indication of what you are supposed to use. I am not familiar with the equations they set up, so I am afraid I cannot help you there. My idea was prove colinearity by showing that green α = purple α = orange α if all three colored triangles share CH as a side.

[tough willow]

<@&286206848099549185>

[lime crown]

From

#❓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 do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[dark sparrow]

@sick torrent i have told you, i believe on multiple times, that you should not simply post the problem without any indication whatsoever of what you've tried, what you've done, and what you're struggling with.

[sick torrent]

I just want to know is the answer 51?

it is ok tho

[tough willow]

<@&286206848099549185>

I've figured out one of the others but I need help with this one

I posted my question previously around 20 minutes ago

[dark sparrow]

then say it lol

if you have an answer to check, then post it along with the problem instead of waiting for someone to bite

[sick torrent]

ok lol

[honest wigeon]

What kind of compass do you guys like to have?

[steady shoal]

Math or kink lmao

[honest wigeon]

Drawing circles

Although the magnetic one is cool too

[dark sparrow]

idr the last time i used one

[trim breach]

I’ll assume that is a pentagonal prism. In that case, you can think of the volume as the area of the pentagon multiplied by the height.

[honest wigeon]

We’re in the geometry channel and euclid’s axioms tell us we can only construct stuff with straight lines and circles 🤔

[tough willow]

What I'm trying to do now

I get that part

What's the perimeter

[tiny snow]

@honest wigeon You can do such constructions in GeoGebra: https://www.geogebra.org/geometry

[trim breach]

The perimeter of the pentagon?

[honest wigeon]

Although online stuff can be fun to a certain extent, there’s a big difference between sketching on paper, and mashing some buttons digitally

[trim breach]

The problem indicates the pentagon is a regular pentagon, meaning all sides are of equal length.

[tough willow]

Yes

[shut venture]

@mellow pewter you're still trying to solve the circumerence inscribed'

[mellow pewter]

we need to find the radius, area and circumference

[trim breach]

Why do you want the perimeter of the pentagon though?

[shut venture]

the area of the triangle is easy

[mellow pewter]

yep

each side is approx 24.25

[tiny snow]

@honest wigeon The ability to have some points being able to move freely in a construction is, in my opinion, often invaluable to understand a problem better.

[shut venture]

the area is ~254

[mellow pewter]

yeah, i got approx 252.9

cause i rounded to the nearest hundredth

how can we find the radius of circle A

[oak lotus]

Hey, I need some help at exercise 1 b.

I am not sure how to approach it

I tried doing sin (b - c) and I reached nothing out of it

[shut venture]

@mellow pewter you can draw the height

or since its all tangent you can draw the radius frm the center

[mellow pewter]

oh yeah

[shut venture]

you use trig or not

funny who knew i could help someone haha

[upper karma]

need help

[storm portal]

Assuming each outfit consists of exactly 1 pair of pants, 1 pair of socks, and 1 shirt

Actually, do you have any ideas, Tankz?

[upper karma]

i have no idea

[storm portal]

Well - it turns out that in order to find the total number of outfits, you need to multiply the number of (pairs of pants) by the number of (pairs of socks) by the number of shirts

Once you know it it makes sense, but I'm trying to think about why this is the case

Assume we just have two items: shirt + pants

there are 5 shirts and 4 pants

all possible outfits include combining 1 shirt and 1 pair of pants

there are 5 possibilities for a shirt, and 4 possibilities for a pair of pants

so if we name the shirts A, B, C, D, and E, and the pants 1, 2, 3, and 4

all the possible combinations with shirt A include

A1, A2, A3, A4

the same applies with B, C, D, and E

creating this list will allow you to see how multiplying the number of options for one item by the number of options for another item yields the total number of combinations

[upper karma]

it would be 20 with just shirts and pants

so idk what it would be with socks includes

i got 21 outfits but idk if thats right

[shut venture]

shouldnt it be 425?

4x2x5

[upper karma]

is there a caculator i can put this into?

[shut venture]

make a tree

i think you can have 1 shirt 1 pair of socks 1 pair of pants

and so on

[upper karma]

do i use nCr?

[shut venture]

no

[upper karma]

oh

so just make a tree

[shut venture]

yes

try it

[upper karma]

ok

[shut venture]

you can do nCr

but its from 5 pick 1

[upper karma]

how do i use nCr?

[broken lava]

who can help me with geometry please

[shut venture]

@upper karma thats the most basic i believe since you dont have exceptions

you can draw the dashes

[upper karma]

alright

like this?

[shut venture]

you have the same thing

like 1 sock -> 1 Pant -> 1 shirt

@broken lava i can try

?

[broken lava]

Okay

Do you want me to dm or just do it here? @shut venture

[shut venture]

here

[broken lava]

Okay

wait wrong ones

here

[shut venture]

ill take the chance and learn too 😄

[broken lava]

oh u don’t know how to do it either?

[shut venture]

i will study it thanks

im a newb too

[broken lava]

haha it’s ok

[radiant vapor]

how could I find something like this

if I dont have the side length for BDC

[trim breach]

Trigonometry, then identify similar triangles and use proportions.

[shut venture]

sorry ill be back later im having a severe headache

[celest violet]

The points A (0; 6), B (8;7), C (4;-2), and D are the vertices of the ABCD parallelogram. Find the abscissa of the point D. (2 TASK) HELP WHO CAN DO THIS IN NOTE PLS

<@&286206848099549185>

[clear sun]

i got 12.12 can someone help me check

[dark sparrow]

this is wrong

how did you find it?

[clear sun]

uh idk i forgot

oh wait i did law of cosines i think

no wait

[dark sparrow]

wow that's way overkill

[clear sun]

i did 7√3

[dark sparrow]

but why

your angle is 65°, not 60°

[clear sun]

o

am i suppose to use law of sines or cosines?

[shut venture]

law of cosines is when you have 3 sides

or 2 sides and an angle inbetween

[clear sun]

oh i see

[silent plank]

law of sines and cosines in a right triangle is overkill

[clear sun]

then what should i use?

[storm portal]

Just simple trig

[shut venture]

the tangent?

[upper karma]

use sine

[storm portal]

sin(x) = opp/hyp, cos(x) = adj/hyp, tan(x) = opp/adj,

[upper karma]

sine is easier

[storm portal]

tan

[clear sun]

i got 115

[shut venture]

but he doesnt have

[storm portal]

Can't use sine

[clear sun]

for angle b

[shut venture]

the hyp

[upper karma]

oh my bad

[silent plank]

sine doesn't bgive you what you want here

[shut venture]

its the cotangent

?

[storm portal]

(Reposting the image to decrease the need to scroll)

tan(angle) = (side opposite angle) / (other leg)

[shut venture]

or tangent to give BA

[storm portal]

nope

[shut venture]

sorry

[storm portal]

unfortunately

[shut venture]

i meant cotangent

[upper karma]

tan of c should work

[silent plank]

identify the position of sides relative to the angle you want to use
and use the appropriate trig function

[shut venture]

well

[storm portal]

Why are we taking the tangent of C?

That's the right angle

[shut venture]

you can do tangent(35)...

[storm portal]

*25

and that's angle B

[shut venture]

omg im blind

lol

[clear sun]

oh ok

[storm portal]

@clear sun take 90 away from that result

the sum of all angles in a triangle = 180˚

[clear sun]

so the answer is 25?

[shut venture]

BC is 15

[clear sun]

oh

[shut venture]

?

wait maybe i did it wrong im already confused

[clear sun]

its fine thank you

[shut venture]

its 15

@sick torrent scaling up isnt it only doing 3ft*3?

[silent plank]

15 is not the answer they're looking for

[sick torrent]

i dont know

I dont think so

but I am not sure

[shut venture]

@silent plank 15.01?

maybe its wrong then

[silent plank]

15.01 looks alright

[shut venture]

@sick torrent you scale up

[severe rapids]

Area

[abstract saffron]

is it a trapezoid

[severe rapids]

Yes

[abstract saffron]

ok

[severe rapids]

Find the Area of trapezoid WXYZ

[abstract saffron]

drop a perpendicular from Z down to WX

then use the 30-60-90 triangle to find the height

you can find ZY by doing some easy subtraction

u have 2 bases

height

should be ez

[severe rapids]

I don’t know how to find height still

What do I do after drawing the 30 60 90 triangle

@abstract saffron

[sick torrent]

A rectangle with side length 4ft and 7ft is scaled up by a scale factor of 3. What is the area AND
perimeter of the new rectangle?

[trim breach]

Do you know what a scale factor is?

[shut venture]

Nobody helped on my stupid question :\

[upper karma]

is this website complete? https://www.mathwarehouse.com/trigonometry/ ; I know that there is khan academy, but I don't like it, while books, like openstax tend to be bricks

[rapid tartan]

anyone free to help me with some geometry questions? would be much appreciated

[onyx cloud]

just ask

[rapid tartan]

ask who

[abstract saffron]

hi, sorry for late response, was at soccer practice. in a 30-60-90 right triangle, the sides will be in the ratio x to xrt(3) to 2x

since you know the longest side, the hypotenuse, you can find all the other sides

because u just use the ratio

since hypotenuse (WZ) is equal to 8, the shortest leg is the height, which will be equal to 4.

the other side will be 4rt(3), by the ratio