geometry-and-trigonometry

[upper karma]

2m is the radius of the circle?

[royal edge]

No, it is part of one corner. Once I finish washing the dishes I’ll send the tangent points

[upper karma]

Aight

Wait that's easy

[royal edge]

The dishes? Yeah I agree

[upper karma]

Show your diagram

So I know how your points are called

[royal edge]

I circled the points of tangency, labeled congruent segments

I just don’t know how to find x neither the segments connected to point O

[upper karma]

You are done

[royal edge]

OH

I see it now!

[upper karma]

x=10-2

[royal edge]

X=8

[upper karma]

Yess

[royal edge]

Thanks man

[upper karma]

I didn't do anything :)

[livid sequoia]

is this right?

[silent plank]

show work

justify your choices

[livid sequoia]

how do I find the area of the triangle?

[silent plank]

apply the appropriate formula

[livid sequoia]

1/2bh

so its A and B

[silent plank]

yes

[livid sequoia]

I think one of them is wrong but I dont know which

I also put G

[dark sparrow]

G is wrong

and also it's wrong that you didn't select C

[livid sequoia]

So its C,D and E?

[arctic vortex]

does this look good

[upper karma]

Just expand all of them

[silent plank]

if you were able to get D, the full expansion should be trivial

[upper karma]

If you expand then you cannot have any doubts

[livid sequoia]

Im completly sure its C D and E

[silent plank]

good

[livid sequoia]

this is the last one can anyone explain what a rigid motion is?

oh im sorry vocal

[arctic vortex]

what

i know what a rigid motion is

its when the object moves

but the size and shape are the same

and why are u saying sorry

wait no

the size changes

right

[livid sequoia]

isnt it when it rotates or something?

[silent plank]

rigid motion consist of stuff like translation, rotation, reflections

[upper karma]

So no dilation?

[silent plank]

nope

i.e. this is a question about congruency

[livid sequoia]

I found out that C was wrong

[silent plank]

are you sure?

[livid sequoia]

o wait

35+80+65=180 so its congruent

[silent plank]

you should've selected A instead of B

[livid sequoia]

are they all being translated?

[upper karma]

I was still looking at the previous question

[livid sequoia]

thanks guys

[upper karma]

Damn

[eternal crag]

np

[upper karma]

Thanks Pub

[azure reef]

np

[upper karma]

Thanks

np

[azure reef]

thanks

np

[upper karma]

You are welcome.

[meager rapids]

@ me

[pallid sierra]

no

Sure

Nvm

Good job

[sinful junco]

Can someone help me with this problem, please?

[silent plank]

,w area octogon with side 16

[somber coyote]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by ramonov#0921

[upper karma]

@meager rapids

[meager rapids]

So am I wrong or right 1236 is different to my answe

[pallid sierra]

Close enough

I don’t know if you are rounding

But since it’s close I assume you are understanding

[meager rapids]

Ok

[silent plank]

you should avoid the temptation to round things

[meager rapids]

Ok

[silent plank]

and should start that practice as early as possible

[rich wolf]

Rounding is ok depending on context

[meager rapids]

There isn’t any context just says fill in boxes I usually just round to leave it cleaner

[rich wolf]

Is it marking the answers as correct

[pallid sierra]

It’s a google doc

[meager rapids]

No just some practice

[silent plank]

,w calc 8+8sqrt(2)

[somber coyote]

Results provided by WolframAlpha

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Requested by ramonov#0921

[pallid sierra]

Cool

I see!

[silent plank]

that is a lot cleaner

and more correct

[pallid sierra]

You are a genius

[somber coyote]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by eZkyol#3696

[sinful junco]

Can I get some help on this problem, please?

😭

Why no one help me?

I've been asking this at 5 different servers already

[upper karma]

I'm trying

[silent plank]

the amount of help likely to be given is inversely proportional to the amount of spam

[misty plume]

Anyone can help with the squared part 3?

It seems intuitively correct but can’t seem to work to it

Equating the lines don’t seem to help since too many unknown

[upper karma]

Hmmm

can zoom ib

n

i can't read it

Yes u can

[sleek thistle]

i put x=8 and it said it was wrong?

isn’t it supposed to be 12/9=2x-8/6?

[drifting drift]

looks right to me

8:12 6:9

[rotund hull]

8 should be right either way you look at it I believe

[meager rapids]

But this one gives me a number inside shape

[white pike]

            double projectionDuration = ((projectionVelocity * 2.0) * Math.Sin(projectionAngle)) / 9.80665;
            double projectionHeight = ((projectionVelocity * projectionVelocity) / (9.80665 * 2.0)) * (Math.Sin(projectionAngle) * Math.Sin(projectionAngle));```
Trying to calculate a projection but it does not seem to be correct, anybody knows what is wrong with the formulas?

[somber coyote]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Dripguy#5995

[meager rapids]

@ me

[upper karma]

@meager rapids ok

[ancient spire]

What are the base angles of a trapezoid?

are all angles the base angles?

[upper karma]

So what does the sine function do

How does the calculator calculate it

[dark sparrow]

those are two different questions

there are ways to program the calculation of the sine function but none of them are particularly enlightening if you want trigonometric intuition

so what do you want exactly

[upper karma]

ann

[dark sparrow]

yes?

@upper karma what did you want to ask me

[upper karma]

ann

can you help me

suppose we have this

using only the picture

how can you determine the terminal point of Aj

assuming unit circle

this means no algebra at all, no pen and paper

@dark sparrow

[dark sparrow]

uh

so i am to assume that the black curve that looks like a circle actually is a circle

that the tails of both of these vector-representing arrows are the center of said circle

[upper karma]

assuming unit circle
@upper karma

[dark sparrow]

that j stands for the second vector in the standard basis of R^2

[upper karma]

yes

[dark sparrow]

ok well alright then the blue vector makes an angle of π/2 + θ with the positive x-axis so its coords are [cos(π/2 + θ), sin(π/2 + θ)]

[upper karma]

oh look

algebra

[dark sparrow]

(cartesian coords)

...

jeez dude idk your constraint is weird

[upper karma]

ann when i ask for no visual solutions:

[dark sparrow]

no algebra at all, no pen and paper

[upper karma]

ann when i asked for no algebra solutions:

yes

because i was told that

if i rotated my head

i could see it instantly

but i don't

and i'm wondering why

[dark sparrow]

yeah if you rotate it so that j points to the right

[upper karma]

ok?

[dark sparrow]

idk tho it looks like there's some accompanying text to the picture that you're just choosing to leave off

and leave me in the dark

[upper karma]

...

it literally says you may have to tilt your head to see it clearly

and the coordinates of the point

ugh

[dark sparrow]

"Aj has coordinates [-sin(θ); cos(θ)]. You may have to tilt your head to see it clearly."

is this verbatim what the text says

y/n

[upper karma]

something like that

it's not relevant, the relevant part is the head tilting part

i don't fucking see it

[dark sparrow]

here if the picture was this

would it be clear that the blue vector is [cos(θ); sin(θ)]

[upper karma]

ofc

[dark sparrow]

or is that too much algebruh for your highness

[upper karma]

..

[dark sparrow]

anyway this is a rotation of the original picture by a quarter-turn

[upper karma]

yes

[dark sparrow]

or idk you could like

imagine the components like this

and then rotate THAT back

[brisk ginkgo]

Hello is this taken over?

[upper karma]

yes

the army is in place

[dark sparrow]

poly are we done here for now

bc if so bby can ask their question

[upper karma]

i guess

this is not doublechecking your work

you are just pasting some randomly selected answers

there is no work to check

at all

[brisk ginkgo]

I mean if I got wrong explain what I did wrong

My answers aren’t randomly selected lol

[upper karma]

how can we explain what you did wrong

when we don't know your work

"x + 7 = 10"

"x = 9"

can someone check my work and tell me what i did wrong

the only thing we can tell you is "your answer is wrong"

would that be good

[brisk ginkgo]

Yeah

[upper karma]

cool

[brisk ginkgo]

Don’t give me the correct answer lol @upper karma

[upper karma]

use a CAS to check your answers

[brisk ginkgo]

Because this my practice thing I need to understand

What is CAS

[upper karma]

https://www.sagemath.org/

https://www.wolfram.com/mathematica/

https://www.maplesoft.com/

any one of those 3.

[brisk ginkgo]

@upper karma how does sage work

[upper karma]

you install it

[brisk ginkgo]

I have to download something

[upper karma]

give it variables

and then let the computer solve it for you

[brisk ginkgo]

I’m on my phone

Too much I’m a process

[upper karma]

time to buy a computer then

[silent plank]

1. this isn't geotrig
2. show work means show work.

[brisk ginkgo]

I use my school computer I can’t download those stuff

Oh right yeah why am I in in the geometry channel

[upper karma]

but ramonov's point 2. is what the real problem is

[brisk ginkgo]

I thought the other channels were used

[upper karma]

you just give us some random answers

and you expect us to tell you what is wrong

why would we do that

there's nothing to say even

[brisk ginkgo]

Bruh then I have to pull up my notebook oh god

[silent plank]

so we don't actually have to redo the problem from scratch.

[upper karma]

oh god oh no not your NOTEBOOK

[brisk ginkgo]

It’s like downstairs

And sorry for posting it in the geometry channel.

[upper karma]

so this confirms what i said

[silent plank]

and getting the correct final result doesn't mean you did it properly

[upper karma]

you just select some random answers

and expect us to tell you which are right

because you have no work to show

[brisk ginkgo]

It’s not selected I’m positively sure it’s correct I just want to double check I do have my work

[upper karma]

lol

"it's not randomly selected, i just need to get my notebook which i left downstairs and it is impossible for me to get because downstairs is approximately as far as mars"

[drifting drift]

I completely forgot how to do this. Can anyone jog my memory?

wait is the diagonal 10

[dark sparrow]

maybe

[drifting drift]

lol

[dark sparrow]

yes the diagonal is the diameter here

[upper karma]

@drifting drift just think about it

what is the definition of a diameter

[drifting drift]

nah I got it right after I posted the question

wait but if I solve with pythag to get 2sqrt(-x^2+25), that doesn't make sense because it implies that as x gets bigger, the side cannot exist

oh wait of course that makes sense

the radius is limited

it isn't part of the question but I am curious, is there a way to solve for x?

[silent plank]

there is insufficient info to determine the value of x

[drifting drift]

oh ok

makes sense

[heavy spade]

can someone help with my homework question

determine two angles between 0 and 360 that have a cosecant of -2/root 3

[silent plank]

express it in terms of a trig function youre more familiar with and review special angles and/or unit circle

[small perch]

is this correct?

180 degrees is of course half a circle.
180 degrees = 3.14159... radians.
3.14159... radians = pi

I'm writing notes and I want to make sure I'm understanding it correctly

[gray mason]

pi radians.

(pi rad for short, or just pi when you're using radians)

[small perch]

Yes, I know it's writting that way. I'm trying to make sure I'm understanding it correctly

[gray mason]

180 degrees = pi radians.

[silent plank]

pi (radians) = 180°

technically writing radians isn't necessary

[gray mason]

Thanks for correction Ramonov.

[small perch]

lol yeah i know tha. i guess my actual question is ... have we been working with radians this entire time?

when we learn about pi for the first time are we technically learning radian and they just don't mention it because that will confuse everything

[gray mason]

I think when you're just starting to use radians to measure angles you should call them radians. Drop it later on?

[silent plank]

surprise, when you calculated areas and volumes of circles and spheres, you've been working in radians

[small perch]

@silent plank okay, so I'm not correlating that incorrectly then

[gray mason]

why did NO ONE tell me this?

But yes it makes perfect sense.

[pure pivot]

Can anyone help with this

[gray mason]

What have you tried?

[pure pivot]

I tried making a triangle

H/A

I’m kinda lost atm

[gray mason]

I don't think making a triangle would work, considering you're dealing with an angle greater than 180 degrees.

[pure pivot]

Oh

[gray mason]

angle in quadrant III

[pure pivot]

How should I start the problem then

[gray mason]

What is the secant ratio of an angle?

[pure pivot]

H/A

[gray mason]

or?

[pure pivot]

1/cos

[gray mason]

Can you find cos?

[pure pivot]

-5/6?

[gray mason]

(a better way to think about this: since they ask you to find cot, and cot(x) = cos(x)/sin(x), you need to find cos(x). Where can you find cos(x)?)

Yup.

[pure pivot]

So then I need to find sin?

[gray mason]

Yes, it's in the question.

[pure pivot]

I’m lost 😂

How do I find sin when I don’t have the hypotenuse

[gray mason]

Please go back to basic trigonometry.
Recall that $\sin^2(x) + \cos^2(x) = 1$.

[somber coyote]

AdventurousAndrew:

[upper karma]

Is there a way to calculate the sine function without a calculator

Dosent need to be precise

Is it a Taylor series

[versed river]

Taylor series, you can eyeball it, other estimations, you can get plenty of stuff with exact values

like sin(45 degrees) is precisely 1/sqrt(2)

[upper karma]

low order taylor series is doable by hand

and $\sin(x)\approx x$ for small $x$

[somber coyote]

Botn:

[upper karma]

technically a low order taylor series anyway

[upper karma]

Thanks

[dark sparrow]

the translation is ok

they are called solids of revolution

i'm not sure what the question is asking you tho

is is asking you to describe the shape of each of those solids?

[marble mesa]

If you follow the arrow, what solid of revolution would it be

that's what they ask me

[dark sparrow]

ok so they're asking you to describe the shapes

the shape in (a) will give you a solid composed of two cones sharing an apex, while the shape in (b) will give you a solid composed of two frustums sharing a base (assuming the vertical sides that look perpendicular to the axis of rotation are perpendicular to the axis of rotation)

[marble mesa]

yes i g

Like this?

the (a)

no, right?

[dark sparrow]

yeah

[marble mesa]

yes..?

[dark sparrow]

that's what it looks like

well there's gonna be circular caps at the top and bottom

but this is what its lateral surface looks like

[marble mesa]

wow I don't have that visual perception... Alright, thanks

I thought like it's being turned horizontally

forming an strange cylinder

[dark sparrow]

it's being rotated around the axis

the axis stays put

[marble mesa]

ok, now I see, wow! amazing

Is there a tool or an image where I can visualize the two frustrums sharing base?

I can't manage to see it

[dark sparrow]

lemme try to make a sketch

[marble mesa]

👍

[dark sparrow]

it will be easier if you turn your head so that the axis becomes vertical

shitty sketch but

whatever i guess

[marble mesa]

wow, you even draw great!

thanks

[arctic vortex]

are these 2 ones correct

[dark sparrow]

let's see

[upper karma]

2nd is correct

[dark sparrow]

#1 seems ok too

lemme just doublecheck my own arithmetic

[arctic vortex]

ok

[dark sparrow]

yup checks out

[arctic vortex]

also how do i do this

yup checks out
@dark sparrow thanks

i learned how to quote yesterday

do i find the sums of all the parallelograms

[upper karma]

all central angles are the same

[arctic vortex]

oh

[dark sparrow]

i don't know do you
you might just be overcomplicating this

[arctic vortex]

so its 90?"

[dark sparrow]

just a very tiny bit

90 degrees huh

six 90-degree angles fitting around a vertex

[arctic vortex]

o nvm

[dark sparrow]

are you sure theres enough room

[arctic vortex]

45

[versed river]

are you guessing

[arctic vortex]

because its a 45-45-90 triangle

[upper karma]

....

[dark sparrow]

six 45-degree angles

[upper karma]

just do the division

stop guessing

[arctic vortex]

idk how to do it

[dark sparrow]

there are 6 angles around the central vertex

[upper karma]

how about 360/6

[dark sparrow]

they are all the same size

[arctic vortex]

how about 360/6
@upper karma how did u get that

[dark sparrow]

there are 6 angles around the central vertex
they are all the same size

[arctic vortex]

ohhh

[dark sparrow]

they add up to 360 degrees

bc yknow

360 degrees

is a full circuel

circle*

[arctic vortex]

yes

thanks

[mighty condor]

i would say that is an optical illusion, all angels are the same

[dark sparrow]

👼

[versed river]

no the ones in the new testament are different from the ones in the old testament

[mighty condor]

oh sorry my bad

[arctic vortex]

is this 6 because 360/60

[upper karma]

yep

[arctic vortex]

ok

[upper karma]

an hexagon doesnt fit under that cardboard though

[arctic vortex]

this one is 49

so it cant be 6?

or does it not matter

[versed river]

,w solve 2x+15+3x-20+x+7+x+15=360

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Sneaky#0928

[versed river]

49 is right

[arctic vortex]

how did u do that

,w solve 1+1

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by VOCAL#1159

[arctic vortex]

oh wow its a smart bot

[versed river]

,w 23/33

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Sneaky#0928

[versed river]

don't spam it in here though

[arctic vortex]

ok

[dark sparrow]

the bot only queries wolfram|alpha

[arctic vortex]

what method do i use to solve this

i use mathway.com

[dark sparrow]

mathway is kinda meh

[upper karma]

I use paper

[versed river]

what do you know about sum of exterior angles of any polygon

[arctic vortex]

equal 360

[versed river]

ye

[arctic vortex]

49

because 360-311

[versed river]

,calc 360-18-21-32-48-57-65-70

[somber coyote]

Result:

49

[versed river]

ye

[arctic vortex]

ty

[mighty shuttle]

@dark sparrow HM?

[dark sparrow]

???

what

[cosmic pebble]

#questions

For a list of rules about asking on this server, please see #❓how-to-get-help.

[mighty shuttle]

@dark sparrow I was referring to your comment about the bot . That comment was vague

Mb

I mean your comments were a bit nebulous

[broken marten]

Can anyone help me with 5

[glacial haven]

What’s giving you trouble

[upper karma]

, rotate

[broken marten]

#5

i dont understand how to do it

i used to know how but i just dont remember anymore

@glacial haven

[glacial haven]

ok would you use cos sin or tan?

[urban egret]

@broken marten hi did you figure it out yet? As Nelson asked which angle did you want to use

[upper karma]

k

[broken marten]

oh yes sorry

i didnt realize you responded and i asked my friends to help

@glacial haven thank you, but i figured it out

[drifting sorrel]

I am pretty lost on where to start with this problem. I know it has something to do with vectors, but I really don't understand how vectors relate to the other things we've been doing with sin/cos/tan. Can you point me to some resources that might be able to explain the concept used for this problem? Or the name of the concept so I can look it up?

[hard geyser]

@drifting sorrel in general, with this kind of problem, it is helpful to decompose each force in perpendicular(x,y) components

here, vectors only represent the direction and the amount each force acts upon

[drifting sorrel]

I was trying to do that, but I've only been given one angle (45 degrees) so I don't know how to find the "y" portion of the vectors. Is that what you mean when you say decomposing it?

[hard geyser]

yes

are you supposed to use calculators?
or do they not let you?

[drifting sorrel]

They allow us a TI-30XS

Are they expecting me to assume that these lines are exactly the same distance apart when decomposing? So I could form a 90 degree triangle by drawing a line down the middle and have 22.5 degrees on either side?

[hard geyser]

@drifting sorrel wym same distance apart?

you could do that, yes

that is the first approach that comes to mind

[drifting sorrel]

Like so

And then I could add the force of both on the middle line to find out the force going to the right?

[hard geyser]

dont forget the two other components

like you described

if you add all of them, youll have two components at the end

then you just need to make them into a single one

[drifting sorrel]

OH I think I see. So I've found the sum of forces in the x direction

and there is also a y

[hard geyser]

yes

sum the ys

[drifting sorrel]

I got approximately 4478.8 pounds

Woops

4476.8

[hard geyser]

lemme see

seems about right

[drifting sorrel]

Woot woot!

Thank you so much, I think I'm actually getting this a little now

[hard geyser]

chill kaepora gaebora

[wicked osprey]

I have a trigonometry question

if tan x = -4/3 where 3pi/2<x<2pi and sin y = -15/7 where pi<y<3pi/2 calculate the exact value of tan(x-y)

[silent plank]

where are you stuck?

also there's something very wrong with that question

[wicked osprey]

I have no clue

[silent plank]

sin y = -15/7
...

[wicked osprey]

i just dont know how to do it

thats what the question says

[silent plank]

do you have a pic?

[wicked osprey]

its 9

[silent plank]

-15/17

[wicked osprey]

oh shot

my bad

Im just so lost at this point

[silent plank]

are you able to determine tan(y)?

[wicked osprey]

like on my own?

no way

[silent plank]

yes

[wicked osprey]

I have not a cluw

clue

[silent plank]

use a right triangle, apply pythagoras and/or appropriate trig identities and/or unit circle

[wicked osprey]

can you show me how to do that please?

[silent plank]

one method would be to:
start by drawing a right triangle where sin(theta) = | -15 / 17 |, where theta is the related acute angle of y

[wicked osprey]

ok

[silent plank]

determine the third side using pythagoras

then determine tan(theta) and hence determine tan(y) based on the location/quadrant of y

[wicked osprey]

ok thank you

I will give it a shot

[silent plank]

and then apply compound angle identities

[wicked osprey]

gonna be honest here

i dont know how to solve for tan theta

[silent plank]

tan(theta) gives the ratio of the opposite/adjacent side (relative to the non-right angle in a right triangle)

[wicked osprey]

so that would be 15/8?

or is it the other way around?

[silent plank]

15/8

15 is length of the side opposite theta

[wicked osprey]

right

so the angle is 1.88?

[silent plank]

you're not suposed to find the angle

[wicked osprey]

...

man im so lost

[silent plank]

and doesn't look like it

you are also told the location of y

[wicked osprey]

I learned jack because grade 11 was online

[silent plank]

would tan(y) be positive or negative

[wicked osprey]

positive for sure

[silent plank]

in (pi,3pi/2), i.e. quadrant 3

yes

[wicked osprey]

lets go

[silent plank]

so currently you have: $\tan(x) = -\frac 43$ and $\tan(y) = \frac{15}{8}$

[somber coyote]

ramonov:

[silent plank]

and do you know the compound angle identity for tan?

[wicked osprey]

um

[upper karma]

how would I do this

[silent plank]

no such triangle exists

[gray mason]

It'd make more sense if ABC is a right triangle.

[silent plank]

applying the information at face value,
<A and <B are between 0 and 90°

given that sin(<A) = cos(<B),
they would be complementary. i.e. <A + <B = 90°

(which means that <C would be 90° and the triangle won't actually be "acute". )
It seems they want you to ignore this and just apply the stuff above and/or didn't consider it themselves)

[gray mason]

Yeah. ^
who wrote this question?

[silent plank]

👻

[austere dragon]

wow thats some trigonometry

ok can someone help me with this basic trapezoid area formula?? i cant seem to understand why it would be the average of the area of the lower and top base

[dark sparrow]

one way to see this is via reassembly, cutting off triangular bits from the lower base and putting them on top to make a rectangle

alternatively, if you know how to find the area of a parallelogram, you can also see it by putting together two copies of a trapezoid to make a parallelogram

if you're willing to wait like half an hour i could get back home and illustrate this for you

if you need me to

[dark sparrow]

psst sneaky
you can hold shift while using the line tool in paint to make it snap to the closest horizontal, vertical or diagonal direction

[versed river]

oh you can

i was having trouble with that

thx

[austere dragon]

but what if the trapezoid isnt isosceles?

[dark sparrow]

you don't need it to be

[austere dragon]

wait

hmm yea i forgot that its upside down

[austere dragon]

where does the formula for the area of a circle (pi*r^2) comes from? is it not explained in basic geometry?

[upper karma]

You can actually unfold the circle into a triangle that has the circumference as base and radius ad height

[austere dragon]

wow wait what

where do you learn that? calculus?

[dark sparrow]

it's not directly accessible via basic geo

you do need calc for it

to do it rigorously, that is

[upper karma]

The "proof" I posted is not really a proof

[austere dragon]

cool

[dark sparrow]

personally i prefer the "cut it up like a pizza and rearrange it into an almost-rectangle" approach

[austere dragon]

wait is calculus geometry??

i thought it was like a curve thing

measure curve

[dark sparrow]

no, calculus is not geometry but it relies on geometry a bunch

in some sense

[austere dragon]

owh

[dark sparrow]

like

youre gonna be using graphs in calculus

one way or the other

[gusty spoke]

Last one? How to solve

[upper karma]

@gusty spoke think of first on where D can be if its equidistance from all vertices

Tip: ||think inside the triangle||

[dark sparrow]

why not think outside the triangle

[gusty spoke]

Perpendicular bisector?

I can umm
Find the perpendicular bisectors of AB and BC
Then equate them together to find the point of intersection that is D

[dark sparrow]

then why don't you do just that

[upper karma]

Lol ^

[gusty spoke]

I was confirming

[upper karma]

Remember $m\cdot m_{\perp}=-1$

[somber coyote]

Al𝟛dium:

[gusty spoke]

Did this

[sly belfry]

hello everyone! I'm having trouble getting the cartesian equation of this hyperplane

I'm used to having the constants simplify, but gamma and phi seem to be undefined...

[dark sparrow]

are your coordinates $(x,y,z,t)$ rather than $(x_1, x_2, x_3, x_4)$?

[somber coyote]

Ann:

[sly belfry]

does it make a difference?

[dark sparrow]

it doesn't

well

[sly belfry]

(I prefer these -- less work and less confusion)

[dark sparrow]

the equation will look slightly different but only up to that change of letters

ok w/e as you say

yeah y=t is your equation as it seems

[sly belfry]

oh! does it sum up to that?

[dark sparrow]

yes

x and z are unconstrained

[sly belfry]

I see

so free variables

cool!

thank you very much

[gusty spoke]

Why did they give the volume here

Can’t we simply calculate surface area using the formula

3 pi (r)^3

Anyone?

[sly belfry]

maybe to save you some work?

or maybe this is not a sphere?

[gray mason]

I mean, it's Oxford, what do you expect?

[hard geyser]

maybe that is not a cut in the diameter

and it shouldnt be, otherwise the volume would be 128π/3

[gray mason]

hemisphere

[fringe fractal]

Guys I have a question :
Let M be a mid point of hypotenuse AB of a right triangle ABC.A line I passes through M orthogonally to CM and intersects AC in a point K.Find angle BAC if AK:KC = 1:2.
What’s the solution to this problem
I’m stuck

[silent plank]

have you drawn a diagram?

[fringe fractal]

Umm

One sec

[silent plank]

the diagram is incorrectly drawn

[fringe fractal]

Oh really

[silent plank]

your line I needs to be perpendicular to CM at M

i'd recommend labelling the points A,B and K last

[fringe fractal]

I’m quite confused on how to draw it
Can you draw it?

[silent plank]

(since depending on your labelling construction of those properties may not be possible)

[fringe fractal]

Oh ok

One sec

I can’t seem to get it

Can you draw it for me

If you don’t mind

[silent plank]

1 sec

[fringe fractal]

Oh aight

[silent plank]

start with the construction of CM

and then construct line I

and then label the intersection K followed by A and B

(and indicated proportion of lengths)

[fringe fractal]

Btw is angle c divide into 45 and 45

Like 2 equal halves

[silent plank]

no

[fringe fractal]

Oh really

Damn

How would you do this question then

[silent plank]

consider thinking about it for a bit with the correct diagram

[fringe fractal]

Ok

I’m not making any progress

@silent plank

[silent plank]

consider constructing a perpendicular line from M to BC or AC to help determine the relative size of line CM

[fringe fractal]

So by that

You mean a line perpendicular to BC

That’s goes to M

[silent plank]

what you posted and deleted looked alright

[fringe fractal]

Oh really

I thought it misread for a sec

[silent plank]

only really need one of them

[fringe fractal]

Oh fr

Aight

[silent plank]

and then apply properties of medians and/or similar triangles / congruent triangles to determine the relative size of line CM

[fringe fractal]

What do you mean by relative size

[silent plank]

in terms of your existing variables or assigned values

[fringe fractal]

So CM would be

[(2x)^2 + b^2]^0.5

But what is b

[silent plank]

that wasn't what i was leading you to

[fringe fractal]

Oh damn

[silent plank]

let the intersection of M and AC be P

[fringe fractal]

Ok

[silent plank]

if you know your median theorems,
AP = PC (which you can prove from similar triangles AMP and ABC)

[fringe fractal]

Oh he I remember this theorem

Aight

[silent plank]

which you can then use to conclude that triangles CPM and APM are congruent be SAS

and hence CM = AM and
angle KCM = angle BAC

and then apply cosine in 2 ways

[fringe fractal]

In 2 ways

?

Wdym

[silent plank]

angle KCM = angle BAC

consider the cosine of those 2 angles

[fringe fractal]

Oh aight

So

For KCM

Cos (kcm) = I/2X

[silent plank]

no

[fringe fractal]

Oh nvm

I did sin

So CM/2x

[silent plank]

yes

[fringe fractal]

3x/AB

[silent plank]

actually forgot parentheses for CM/(2x)

its also time you assign an arbitrary length to BM, CM and AM

[fringe fractal]

So what should I assign them too

[silent plank]

(you don't really need to introduce another variable for it, you could but the algebra would be slightly more tedious)

[fringe fractal]

So what do you recommend me to do

[silent plank]

i.e. you could just let them be 1

lets also let those angles have the value theta

so you'd have:

$\cos(\theta) = \frac{1}{2x} = \frac{3x}{2}$

[somber coyote]

ramonov:

[fringe fractal]

So what’s next

Oh why’s CM=1 and AB=2

Nvm

[silent plank]

CM = AM
was determined from congruency and AM = MB was given.
we can assign arbitrary lengths since we only care about the ratios

[fringe fractal]

AB = 2CM

Ye I forgot about that

[silent plank]

and reduce it to a problem with fewer variables

[fringe fractal]

Aight

So what’s next

[silent plank]

solve for x

[fringe fractal]

X=1/3

Oh

[silent plank]

not quite

[fringe fractal]

Then we can use cosine inverse?

[silent plank]

x isn't 1/3

[fringe fractal]

Wait it isn’t

[silent plank]

no

check your algebra

[fringe fractal]

Don’t you cross multiply

1/2x=3x/2

6x=2

[silent plank]

firstly: parentheses \
secondly: $2x \times 3x \neq 6x$

[somber coyote]

ramonov:

[fringe fractal]

Oh 6x^2

My bad

So x=plus or minus (1/3)^0.5

[silent plank]

since you are dealing with lengths here, you can disregard the negative value

[fringe fractal]

True

[silent plank]

and then sub it back in to either ratio and apply arccos

[upper karma]

how can i calculate the position of these points assuming the distance between each point is equally distributed?

[silent plank]

trig

[upper karma]

nice one

[silent plank]

it would be better if you applied more context

[upper karma]

what other details should i include

[silent plank]

where the circle is centered etc

radius

are you expected to also do stuff for higher numbers

[upper karma]

circle is at 0,0; radius 50px; no

[silent plank]

px?

[upper karma]

pixels

[silent plank]

are you able to determine the angle of each point from the positive x-axis?

[upper karma]

360/segments * n?

if you have three points, then it's 360/3 * point(n)

i think

[silent plank]

do the positions need to match the dots present in your figure?

[upper karma]

yeah

[silent plank]

then you should start with the most easily identifiable point (at 270°)

[upper karma]

im not sure you're understanding

[silent plank]

and subtract or add multiples of 360°/n

[upper karma]

or maybe i'm not communicating my problem well enough

[silent plank]

i.e. for the case of 3 points

your points will be located at
270°
270° - 360°/3 = 150°
270° - 2*(360°/3) = 30°

[upper karma]

right, that would give you the angle, but how do I get an x,y coordinate from an angle

[silent plank]

r*cos(angle) for the x-coordinate

r*sin(angle) for the y-coordinate

[upper karma]

oo

okay let me go try it out

@silent plank it works!!

i was being a dumb dumb the entire time, kept using deg when i was supposed to be using radians

[sleek thistle]

guys is the unit circle applied more to geometry or trigonometry?

[silent plank]

trig

[sleek thistle]

oh okay cool thanks

[upper karma]

hello

[glacial haven]

hello

[sleek thistle]

how do i get x?

[upper karma]

RQ/QP=RS/ST

Thales

[sleek thistle]

oh okay thanks

[sleek thistle]

how would i get it here?

[idle bobcat]

can someone explain this to me?

I'm having trouble seeing how any of those options in a single transformation would get from shape A to B

[tulip ingot]

rotation

think about it

[idle bobcat]

aparantely the answer is rotation, but I'm not seeing how you could have a single rotation and have A turn to B

[tulip ingot]

you take A and rotate it counter clockwise

[idle bobcat]

wouldnt you need to translate it as well

[tulip ingot]

well

it's rotation around a point

and that point happens to be a little farther away from A

[idle bobcat]

o

ok that makes sense

for some reason i thought the point of rotation had to be on one of the edges of shape A

[tulip ingot]

oh nah

[idle bobcat]

idk why I assumed that

[tulip ingot]

lol

[idle bobcat]

thank you

[austere karma]

can anyone explain a simple way to solve this problem?

i solved it after a long time but it involved a lot of radicals and is really complicated, so what would be ur approach on this?

[tulip ingot]

wait what

[austere karma]

?

[tulip ingot]

I can't even understand what it's saying

isn't L just 25? or is that after perspective is applied

in that case that would depend on the FOV and so many other factors

[austere karma]

from what i can see, AB is congruent to BC

these are the measurements i can find

but then anything after this is way too complicated for me to understand

[tulip ingot]

also

if a strip of paper was folded that way

with the line going up right

the end of the strip would be going down right wouldn't it?

[austere karma]

i dont think the bottom part is necessary so i just cut it at that

[tulip ingot]

like that

[austere karma]

what do u mean

[tulip ingot]

like

that's not correct topology wise

and origami wise

if you folded the paper in that direction it wouldn't bend to the left

so it just messes with my head and I can't really visualize the problem

[austere karma]

point A goes to point C when it folds

[tulip ingot]

ohhhh

[austere karma]

so i color-coded the line segments that are congruent

[tulip ingot]

I thought that was the crease what where it was folding

[austere karma]

ah

i got 25 because AB is 25, and then its a right triangle, so i used Pythagorean Theorem to get 7 for that small line segment

[tulip ingot]

yeah I thought you were folding the paper backwards instead of downwards

[austere karma]

ah ok lol

[tulip ingot]

hmm wow

I feel like I'm not given enough info to solve that

but if you were able to do it

[austere karma]

i know its doable, but the way i solved for it required a lot of radicals

and i wouldnt be able to explain what i did, cuz my work was all over the place

[tulip ingot]

doesn't tell you the angle or anything

[austere karma]

no, thats all the problem gives

Corner A of a rectangular piece of paper that has width 24 is folded over so that it coincides with point C on the opposite side, as shown in the diagram. If AB=25, find the length of the fold L.

[tulip ingot]

just 2 side lengths

[austere karma]

yep

[tulip ingot]

for any trig you would need either another side length or an angle

but we have neither

[austere karma]

i havent learned trig so i think this is just algebra/geometry

for the question above, i got this far

and by the looks of it, the triangle that is sticking out on the left looks almost identical to the 7-24-25 triangle

i dont know how to prove that it is (if it even is in the first place)

[tulip ingot]

wait

is the green side also 7?

ohh yes it is

I think

no it's not shat

wait it is

it's mirrored, but I think this is also a 7-24-25 triangle

[glacial haven]

yes

the green is 7

you can easily do it now

[tulip ingot]

wait

how do you find L tho

lmao

[austere karma]

yeah if the green is 7, then it would be very easy

but how do you know if it is 7

[tulip ingot]

because those triangles are similar

[austere karma]

because if its 7

then you can use Pythagorean Theorem to calculate L

which would be 30

[tulip ingot]

this right angle says that they're similar

oh I see

I forgot you could "move" the side lengths like that

[austere karma]

this right angle says that they're similar
can you explain that a bit more please?

i dont really understand why that is

[tulip ingot]

umm lemme draw it

[austere karma]

alright

[tulip ingot]

if that right angle rotates

then the area that is displaced is the same

[austere karma]

ohhhhh

i see it now

so a 90 degree rotation?

[tulip ingot]

nah

the lines are perpendicular

red and blue I mean

[austere karma]

yeah

like this?

[tulip ingot]

yeah

[austere karma]

basically these 2 right triangles are congruent

because of that right angle

ok i see now, thanks a lot!

[upper karma]

Yes

[ebon oar]

Would you say this proof is clear and correct?

[upper karma]

I think you need to say the you called E the intersection between BC and the parallel to AB

Last line I would say: recalling that <B=<DEC,...

Another small thing, I would say that <DEC = <C because of the isosceles triangle theorem

Just to be very precise

@ebon oar

[ebon oar]

I see. Even if the problem says to draw DE, my proof doesn't say that I've actually done it. Makes sense. I'll use the others as well. Thanks.

[upper karma]

np :+1:

[arctic vortex]

how do i solve this

[upper karma]

take the fraction of the circumference it travels in 5 hours over the whole circumference

[arctic vortex]

oh

ok

[upper karma]

sorry

that's not quite right

all you need is the numerator lol

[arctic vortex]

ok

[upper karma]

i misread the question

[arctic vortex]

50.265

[little osprey]

use arc length formula

[arctic vortex]

ok

[little osprey]

h

oh

wait

[upper karma]

ignore jc denton @arctic vortex

[little osprey]

yeah arc length formula

[silent plank]

uh...this question...

[little osprey]

wait no

yeah ignore me

[silent plank]

how far does which part of the hour hand

[little osprey]

?

[arctic vortex]

i dont know

[upper karma]

"How far" 5pi/6 radians

[little osprey]

No but

the hour hand

is shorter

so the circumference is shorter

[upper karma]

no

[azure reef]

find a proportion

[upper karma]

you're overthinking this so badly

rofl

@azure reef that's what i told him you

person

[arctic vortex]

how

[little osprey]

lmao

[azure reef]

but you do not know length of hour hand

[upper karma]

ok

can people please stop typing

@azure reef @little osprey pls

@arctic vortex if the hand traveled to 6 o clock

from 12

how much of the circumference did it travel

[azure reef]

@arctic vortex expresss lenght of hour hand as function of diameter

and then arc length

[upper karma]

sigh

[azure reef]

polynomial. if u just apply what you want you get minute hand

[little osprey]

How do you calculate the length of the hour hand

[upper karma]

^

dude

you don't need to

you are overthinking this problem so much

these problems are just badly designed

that's all there is to it

[arctic vortex]

180

[azure reef]

if problems are badly designed - designers should go in the place where sun never shines

[little osprey]

Yes

[arctic vortex]

each hand is 30 ?

[little osprey]

The moon

[arctic vortex]

so 150

[upper karma]

probably

anyway

[arctic vortex]

but im confused on whether or not solving for arc length or sector

[upper karma]

15/36 * the circumference you found

that's your answer

so 5/12 * 50.25

,w 5/12 * 50.25

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by polynomial#4757

[upper karma]

that's your answer

[azure reef]

but im confused on whether or not solving for arc length or sector
@arctic vortex sector is not needed since it is area

[arctic vortex]

so area is arcl ength

[upper karma]

?????

[arctic vortex]

and circumference is sector

[upper karma]

no?

[arctic vortex]

oh

whats the difference between arc and sector

[upper karma]

circumference is the perimeter of the circle

[little osprey]

arc is the curvy part

sector is the whole thing

[upper karma]

jc denton please.

"the curvy part".

[little osprey]

Alright

hes yours

[arctic vortex]

lol

[upper karma]

spiral

i told you the answer above

...

[arctic vortex]

it was wrong