geometry-and-trigonometry

[upper karma]

I just don't know why it is like that

q

4263

[rare talon]

Can you translate the question?

[upper karma]

You are standing on a 3,6m high edge with a 6m rope

huw much closer does the boat come to the edge if you pull the rope in 1,5m

I know the solution for the answer

[rare talon]

To be fair, I don't understand the question

[upper karma]

but I don't know why

[rare talon]

What do you mean by "pull the rope" in 1.5 m?

[upper karma]

So a rope is tied to a boat from a dock

the rope is 6m long and the dock is 3,6m tall

[rare talon]

The rope is the solid line

Right?

In your picture

[upper karma]

yes

[rare talon]

Then

?

[upper karma]

how much closer does the boat come to the dock

if you pull it in 1,5m

[rare talon]

Which one is 1.5 m?

[upper karma]

rope

[rare talon]

So the rope is 1.5m shorter?

At the 2nd state?

[upper karma]

yes you pull it in so it is 4,5m instead

as far as I understand

[rare talon]

Okay so the dashed line is 4.5m

[upper karma]

so I know that you are supposed to use

[rare talon]

You can see that picture

<ABC is 90 degrees

[upper karma]

(sqrt) (6^2)-(3,6^2)-(4,5^2)-(3,6^2)

[rare talon]

Don't use that

I don't want to use that at all

[upper karma]

it tells me to

[rare talon]

Nah

No need, memorizing formulas will make your brain burns

Too much

Okay see my picture

[upper karma]

yes

[rare talon]

What length does the question need you to ask?

The difference between BC and BD right?

[upper karma]

I think so

[rare talon]

Yeah the question is asking how much the boat closer if you pull the rope by 1.5m

[upper karma]

yeah

[rare talon]

So you should find how close the boat to the dock at initial state

Which is BC

Okay?

Can you find BC?

[upper karma]

yes

the length?

[rare talon]

Yeah

[upper karma]

um first it was 6

[rare talon]

6 is the length of the rope

Which is AC in my diagram

[upper karma]

oh

idk then

[rare talon]

https://cdn.discordapp.com/attachments/326138757474680852/373205235902185482/image.jpg

[upper karma]

D?

[rare talon]

Do you notice that <B is right angle?

Open the full picture

It got cut here

Open the link I mean

[upper karma]

oh

yeah

[rare talon]

Okay

From my model you should agree that AB is the height of the dock

[upper karma]

yes

[rare talon]

And AC is the length of the initial rope before you pull it

Now

Can you find BC?

Which is the distance of the boat to the dock in it's initial state

[upper karma]

I can use pythagoras

[rare talon]

Yeah!

Now find it

[upper karma]

can you make up some random numbers

instead of what we had?

[rare talon]

You hate decimals?

[upper karma]

because I know the solution already

[rare talon]

😛

Okay

I can

[upper karma]

I don't want to base it off

me knowing deep inside what the answer is

[rare talon]

Wait a minute

Okay

We have the height of the dock is 15 meters

[upper karma]

okay

[rare talon]

And the length of the rope in initial state is 25 meters

Now if you pull the rope by 8 meters, how much closer the boat will be to the boat?

[upper karma]

so the dotted line is 17m

[rare talon]

Yeah

[upper karma]

in order for me to use pythagoras

I should base it off the rope

[rare talon]

Well

Which triangle do you want to use pythagoras on?

ABC / ABD / ACD ?

[upper karma]

ABC

[rare talon]

Good

We know AB = 15, AC = 25

From the problem information

[upper karma]

yes

[rare talon]

Then?

What should you do?

[upper karma]

sqrt (25^2)-(15^2)

[rare talon]

The result is?

[upper karma]

20

[rare talon]

Okay

Which length has the length of 20?

BC / BD / CD ?

[upper karma]

BD?

[rare talon]

Make yourself consistent (hint : What triangle do you use on pythagoras? )

[upper karma]

oh

[rare talon]

So what is it?

[upper karma]

yeah it's BC then

[rare talon]

Okay

Good

Now

You pull the rope by 8 meters

So we have 17 meters on the dashed line (AD)

What you're gonna do with that?

[upper karma]

oh should I use the new information

for another pythagoras?

[rare talon]

Maybe c:

You know what is the requirement to use pythagoras right?

[upper karma]

90 degrees

[rare talon]

Yeah

Then?

I mean

[upper karma]

idk what the other requirements are

[rare talon]

What should you do?

[upper karma]

oh

wait

[rare talon]

I'll let you think

[upper karma]

is it sqrt(20^2-17^2)

[rare talon]

Which triangle do you use for pythagoras?

Can I know?

Don't guess

[upper karma]

oh yeah wait

I'll name em and look on the paper

[rare talon]

Okay

[upper karma]

I am using CB and AD

which isn't too smart

wait hmm

[rare talon]

No

There is no 90 degrees angle there

I have to deny it sorry

[upper karma]

yeah I know

wait

sqrt (17^2)-(15^2)

[rare talon]

What triangle?

[upper karma]

witch BD and AB

[rare talon]

Okay good

But in the future

Make sure name the triangles

Not the line

Which is triangle ABD

Don't worry I understand what you're doing though

And the result is?

[upper karma]

8?

[rare talon]

Which length?

[upper karma]

the boat shouldn't it be

[rare talon]

No no no

I mean

Which length on my diagram?

[upper karma]

BD

[rare talon]

Okay

So BD = the distance of the boat to the dock after I pulled the rope = 8m

Agree?

[upper karma]

yes

[rare talon]

And BC = the distance of the boat to the dock in initial state = 20m

[upper karma]

yes

[rare talon]

So, how much closer the boat is?

[upper karma]

BD = 8

[rare talon]

It is right, but it's not the answer I'm lookng for

[upper karma]

oh

25-8

no

[rare talon]

What is the question asking?

[upper karma]

wait

[rare talon]

Don't guess

[upper karma]

oh yeah

how much does the boat come

[rare talon]

Closer

[upper karma]

25-8 then isn't it

[rare talon]

?

[upper karma]

yes

[rare talon]

25 is the length of the rope

[upper karma]

oh

[rare talon]

It has nothing to do with how much does the boat come closer

[upper karma]

you mean after they pulled it in?

17-8

[rare talon]

Uhhh

Not that

Okay

Let me rephrase

We know that:

1. The distance of the boat to the dock in initial state = 20m = BC , as we counted before
2. the distance of the boat to the dock after I pulled the rope = 8m = BD, as we counted before

So

How much the boat go closer to the dock from the initial state to the state where I pull the rope?

Use common sense in this

[upper karma]

12?

[rare talon]

Yeah

If you can give me the reasoning, it would be good

(I don't expect 20-8 as your answer)

[upper karma]

I honestly don't know

I rather prefer the formula in the book

[rare talon]

Hmmm

How to explain this

Okay you know that before I pull the rope

The distance from the boat to the dock is 20 meters right?

[upper karma]

yes

[rare talon]

And after I pull the rope, now obviously the boat is closer, and we know that the distance from the boat to the dock now is 8 meters

So it should make sense that the boat has COME CLOSER to the dock by (20-8 = 12) meters

[upper karma]

yes

[rare talon]

Since I pull the rope

Yeah

Then it's the answer

[upper karma]

how did I come up with 8 again

[rare talon]

Didn't you count?

The length of BD before?

By pythagoras on triangle ABD?

(you say it from line BD and AB)

[upper karma]

oh yeah

17 n 15

[rare talon]

Yeah

😄

Now do you get it?

I can think another way explaining that if you're still confused why we take 20 - 8 as the answer

With another analogies

[upper karma]

the dock was 15m right?

[rare talon]

It is 15 meters tall

But it doesn't matter, since no matter how tall the dock is, it won't affect the distance from the boat to the dock

(hey dock can't move itself, in fact we move the boat by using rope!)

So only we can control the boat to be further or closer to the dock

By pulling the rope

[rare talon]

I hope this make sense now

12m closer

[upper karma]

same thing

I tried to translate it into English

[rare talon]

Oh

I see

😄

[upper karma]

so 12m is correct?

[rare talon]

It is

Well actually

If you want a word from me

I can see your solution looks rather disorganized

It's hard to stay consistent on what you're looking for if you don't specify the triangle you use for pythagoras, writing pythagoras equation if possible (for example : AB^2 + BC^2 = AC^2 on triangle ABC)

So we can find BC = 20

For example that

[upper karma]

could you type a solution on paper

that would look better?

[rare talon]

Wait

I would write it at least like that

I'm adapting to your solution

But I'm writing it clearer

Also I'll go to sleep, so try to mention @ help regarding that

[unique prism]

hi, how do i construct a square when it says to "precipitate a normal" twice?

it says to draw a line AB and by precipitating a normal twice construct a square where all the sides are the same length as the line AB

[drowsy pilot]

so if you have a line

[dark sparrow]

precipitate?

[drowsy pilot]

normal = perpendicular

[dark sparrow]

drop, surely?

[drowsy pilot]

make your first normal, which will be perpendicular to your original line

you'll then be able to place another normal at the opposite end of the line; you should now have 3/4 of a square

the last side comes from finding the normal of the first normal

@unique prism

[unique prism]

ooh alright thanks

[amber tulip]

HELP please

[dark sparrow]

[amber tulip]

I don't know what I'm doing wrong

find x

[dark sparrow]

mhm

[amber tulip]

using trigonometry

[dark sparrow]

mhm

[amber tulip]

sin(41)=12/x

[dark sparrow]

mhm

[amber tulip]

there for x=12/sin(41)

[dark sparrow]

mhm

[amber tulip]

x=-75.651...

[dark sparrow]

is your calculator in radian mode?

== 12/sind(41)

[charred spear]

18.29103704

[amber tulip]

I don't know, I'm using a graphic calculator

[dark sparrow]

try having it do sin(90)

if you get 1, it's in degrees
if you get something weird, it's in radians

[amber tulip]

sin(90)=0.8939966...

[dark sparrow]

yup, it's in radians

[amber tulip]

oh, okey. I don't know what radian means.

[dark sparrow]

it's a unit for measuring angles

2π radians = full circle

it's the default unit in most post-highschool math :p

[amber tulip]

oh, that circle thing to make sin and cos waves. I saw it on numberphile before but I haven't been taught yet

am I able to change the calculator in the "other mode"?

[vale raven]

What's your calculator model?

[dark sparrow]

"degree mode" is the word you're looking for

:p

[amber tulip]

@vale raven it's Casio fx-9750GII

[dark sparrow]

Google "Casio fx-9750GII degree mode"

[vale raven]

Ohh

Alright

Press Shift, then Menu

Then scroll down to the Angle option

Then press F1 for degrees

[amber tulip]

ohhhhh, it worked! thank you! squareroot2 and kaoffie

I'm still not entirely sure what the difference is though. I have to read it up. thanks again

[dark sparrow]

it's like feet vs meters, in a way

but for angles

[vale raven]

Here's a helpful chart that everyone seems to use all the time

the inner ring is degrees - what you're familiar with

the middle ring is radians

[amber tulip]

so 2pi is the length of the whole circle, and points across it is the ratio?

they should of used tau

it'll make the radios bit more cleaner

[topaz valley]

you can use tau

[vale raven]

[topaz valley]

we wont stop you

[vale raven]

we'll just silently judge you for being different

You'll get used to the fact that π is a semicircle :P

[amber tulip]

haha, is there some sort of stigma associated with Tau?

[vale raven]

Not really, but kinda, but not really

[amber tulip]

I know, there's no tau symbol on my calculator anyway

[vale raven]

My calculator lets you type τ = 2⋅π

then you can just use τ

but there's no τ button so it's not practical

[amber tulip]

which calculator are you using?

[vale raven]

TI nspire CX

Well

τ is relatively okay if you compare it to the dozenal people

[amber tulip]

wayy out of my budget. you could set it to the litter T for tau, if you want to get out of your way in using tau

[vale raven]

You could go for the extreme and support both tau and dozenal

then you'd have τ = 6.349416967B635108B

[amber tulip]

dozenal = base12?

[vale raven]

Yea

[topaz valley]

you could go real crazy and start using the medieval imperial system

[amber tulip]

yuck

is there any point for base12 number system except for easier divisions and ratios?
relearning how I count is really gonna throw me off.

actually the same thing could be applied for tau...

[vale raven]

Yea

No one's stopping you from measuring time in square root pound force per stone acre

which is about 84 milliseconds

[dark sparrow]

microfortnights are clearly the most sensible unit

[amber tulip]

according to wiki the speed of light is 1.8026 megamicrofortnight

[vale raven]

Nanomillennia?

about 30 seconds

1.8026 megafurlongs per microfortnight

[amber tulip]

what about 'gigaattodecade'

[dark sparrow]

according to wiki the speed of light is 1.8026 megamicrofortnight
what?

[amber tulip]

1Gad=0.3154s

yeah, I didn't get it as well

[vale raven]

I mean

[dark sparrow]

btw, you don't normally attack SI prefixes like that

[amber tulip]

I know, I just tried to make it more absurd

[dark sparrow]

giga atto is the same as nano

[amber tulip]

better one is nanodecade

oh, you've got it

[upper karma]

What kind of triangle is this again?

[dark sparrow]

define "kind"?

if you're classifying by angles, it's obtuse
if you're classifying by sides, it looks scalene

[upper karma]

Thanks.

[vestal pine]

hello!

if i write cos^3(x) is the same as cos(3x) ?

i mean, no it isn't

but i'm confused by what's happening here

https://www.khanacademy.org/math/precalculus/imaginary-and-complex-numbers/multiplying-and-dividing-complex-numbers-in-polar-form/a/visualizing-complex-powers

i don't understand how we can get the cos(60+60+60)

[dark sparrow]

oh

(cos(x) + i sin(x))^n = cos(nx) + i sin(nx)

[vestal pine]

if i do the math i should get the same result? or is there some other intuition inbetween?

[dark sparrow]

there is an intuition behind the formula i just posted

multiplying by a complex number is a rotation

optionally combined with a scaling

[vestal pine]

mmm i think it makes sense, because i've seen that taking the powers is like rotating each time by the same angle.

so that would give me the rectangular form with that angle

thanks!

[surreal bolt]

It is based on the sum of angles identity. Just n times for the same angle.

[prime urchin]

I require assistance with solving 2 questions

[keen aspen]

2x+90=5x+21

180=47+2x+13+c

3x=5y-23

[prime urchin]

what? @keen aspen

[keen aspen]

For the second one

Well question 5

[prime urchin]

oh

for question 5 I got X=24 and Y=19

[keen aspen]

Correct

[prime urchin]

alright thanks

this would be my last question

[keen aspen]

Look at angles 6 12 and 15

They make up a trianglw

Triangle*

[prime urchin]

yeah

[keen aspen]

Meaning all angles add up to 180

[prime urchin]

right

[keen aspen]

Let's say x = measure angle 6

12*

[prime urchin]

yea

[keen aspen]

92+x+41=180

Solve for x

And that'll give you the value for angle 12

[prime urchin]

47

[keen aspen]

Ok

So now that you have 12 you can find 11

You see how angle 11 and 12 are congruent to angle 2

[prime urchin]

alright

[keen aspen]

Because they are corresponding

So since 2 is 103

103=x+47

x is the angle value for 11

[prime urchin]

so 56

[keen aspen]

Ok

So those are your answers

47 and 56

[prime urchin]

alright, thanks

[keen aspen]

Np

[steady widget]

not sure how to solve this, too many unknown's to use heron's formula (given my luck, I'm probably missing something blatantly obvious)

[fathom wraith]

the area must be 32 I think

[slender socket]

Am I being a muppet but is it not 32cm^2

Yeah

[steady widget]

how?

[slender socket]

If you look in each of those boxes

Half the box is shaded

Half is unshaded

[fathom wraith]

the area does not depent on where the right vertex is, so I took all of em to A

[slender socket]

That’s just the nature of triangles

[fathom wraith]

but that's not very rigorous (I'm talking about my method)

it needs a bit more explaining

[steady widget]

so ur saying the bases of the three unshaded triangles are 4, 3, and 1?

[slender socket]

No

You don’t need to know the bases

They’re unknown

[steady widget]

oooohhh

ok

again, blatantly obvious overlook

thanks

[slender socket]

No problem, it’s a tricky looking question

[steady widget]

i was trying to use heron's formula lol

[fathom wraith]

you couldn't have done it that way

there are way to many triangles and variables

[steady widget]

yea

Way too many

[upper karma]

yo guys

can you prove lines are parallel without any angle measures

with converse theorems?

[tropic stirrup]

coordinates system

[dark sparrow]

@upper karma what are you given?

[upper karma]

That two angles are supplementary @dark sparrow

[dark sparrow]

uh

what two angles

care to share a pic?

[upper karma]

Yeah sure, I'll draw it up when I'mm out of class

[main sluice]

@tropic stirrup complex bash

😄

[upper karma]

@dark sparrow Here's a quick diagram I made

[dark sparrow]

yeah nop, nothing's restraining that ray

it can point wherever

[upper karma]

But couldn't you say that the lines are parallel due to the corresponding angles theorem converse?

[dark sparrow]

what lines?

XY and AB?

there's no angle here that's related to AB

you have no information about AB

[upper karma]

But can't you deduce that angle 1 is corresponding to the angle next to letter A? @dark sparrow

Also on another question, are there any angle bisectors in this question?

My thought is no, but is it possible for angle B to bisect it?

[dark sparrow]

But can't you deduce that angle 1 is corresponding to the angle next to letter A?
no. you know nothing about ray AB

you can position it however you like

regarding the other thing

[upper karma]

huh

[upper karma]

yeah i figured the other thing

[dark sparrow]

OQ bisects POR and OS bisects ROT

and OR also bisects QOT

[upper karma]

i did a dumb mistake and thought the bisector was refering to the entire line!

ok thank you

but @dark sparrow because we know that the ray AB is actually placed there in the question

doesn't that mean that the angle would have to be corresponding with the angle on top?

[dark sparrow]

no!

it's not specified to us how exactly the ray is placed!

just because it appears parallel to XY doesn't mean it's actually parallel!

[upper karma]

But it's a proof 😮

why would they give us something that cant be proven 😭

[dark sparrow]

can you show the diagram in your book?

[upper karma]

It wasn't in a book

It was actually a test question that i'm having extremely bad after thoughts about xD

@dark sparrow but?:

Could you do that

oh wait

no youre right

you cant do any of this

because we dont know anything about ray ab

so we'd just be taking the assumption that they are equal to each oteher

and we dont know that they are equal to each other

right @dark sparrow

[dark sparrow]

yes

[leaden kindle]

How would i find the length of DE in this triangle

I know ABC ~ DEC ofc

[keen aspen]

Since AC is equally divided into two the DE is exactly 1/2 of AB

So DE = 2.5

[keen aspen]

Definition of a midsegment I think lol

[upper karma]

@dark sparrow Actually, you could technically do that question right. One of the givens is that line z and line w are parallel, so you could use the alternate interior angles to know that angle 2 is in the interior part on top of angle 1.

And if you use the converse of the corresponding angles theorem then you can say that angle 1 is 180-angle2 and because of the consecutive interior angles theorem, the angle 1 on line ab is 180-angle2 as well?

[dark sparrow]

uh wat?

can you repost your picture?

[upper karma]

this maybe ?

yes its that diagram

👌

?

[dark sparrow]

...one sec

[upper karma]

oh yeah np

[dark sparrow]

you're trying to say something about angles 3 and 4

[upper karma]

3 and 5

[dark sparrow]

you know nothing about 5

it can literally be whatever

[upper karma]

but couldn't you derive an equation

[dark sparrow]

from where?

[upper karma]

Because you know W and Z are parallel

[dark sparrow]

so what?

z has nothing to do with 3 and 5

[upper karma]

And you know angle 1 and 2 are supplementary

[dark sparrow]

yes, so what?

[upper karma]

well angle 2 is equal to angle 3

[dark sparrow]

so what?

[upper karma]

so angle 5 must be 180-angle 3

[dark sparrow]

you'd need XY and AB to be parallel to be able to make any statement about 3 and 5!

no!

no!!!

[upper karma]

oh

oh

oh

wait

but we know w and z are parallel

[dark sparrow]

so WHAT??

you could remove z from that diagram entirely

1, 3 and 5 would stay there

1 and 3 would obviously still be supplementary

[upper karma]

and 1 and 5 would be corresponding angles

oh crap i said something wrong again didnt i

[dark sparrow]

i mean i guess technically? as much as you can call the angles positioned like this for not-necessarily-parallel lines 'corresponding'

i'll repeat

you know nothing about angle 5

[upper karma]

xy would have to be parallel to ab to say that, right, because there is no way to say that they are supplements or that the angle formed is the same - the only way that happens would be if it was already parallel?

[dark sparrow]

...yes

[upper karma]

Hm,

That makes sense!

Thank you @dark sparrow for putting up with my slowness lol

[unique prism]

"what can we say do the points on the perpendicular bisector have in common?"

im bad at translating sorry

but this is for geogebra, idk if anyone knows what that is

idk the answer

i have until friday

[mossy folio]

@unique prism Any gives?

Is that line cutting the circle in half?

Is AC and BC equal

[unique prism]

yes and yes

[mossy folio]

Any given angles?

B/C

That triangle is isosceles

With two acutes and an obtuse

[unique prism]

i made the figure myself from instructions

im not familiar with english terms brb

[mossy folio]

Where are you from?

[unique prism]

nordic country

[mossy folio]

oh ok

So like

What I can tell you are this

You have a line that splits the circle in half

You have a chord drawn within the circle

And AC and BC are equivalent

Thus you can assume angle A and Angle B are the same and are acute

Thus this is an isosceles triangle

[unique prism]

ooh okay

[mossy folio]

What level is this for

HS or College

[unique prism]

i think in that school system its not even hs

[mossy folio]

ok

[unique prism]

not sure tho

[mossy folio]

alright

[unique prism]

anyway, do you know the answer to "what can we say do the points on the perpendicular bisector have in common?"

or did i missunderstand something

:/

[mossy folio]

hm

I mean

if you tilt the circle

you can notice

the cord is 10 units long

and the height of the triangle is 3 units

[unique prism]

like this? Size wasnt really mentioned in the instructions though

i could try to translate it all if you want

its pretty short

[mossy folio]

oh

it wasnt

you can say that the perpendicular bisector creates two right triangles

that are both equal to eachother

[unique prism]

hm but that doesnt exactly answer the question i think

sorry for bothering you with this

[mossy folio]

what can we say do the points on the perpendicular bisector have in common

Is this the exact wording?

[unique prism]

no not exact, its translated by me so its not perfect :/

[mossy folio]

oh

[unique prism]

"What can we say is common to all points on this midnormal?" - google translate

[mossy folio]

that the triangles share the same points on the midnormal for a certain range of y values

[unique prism]

hmm

sorry lack of sleep makes it hard to process that sentence

[mossy folio]

oof

[unique prism]

a. Draw a segment AB with [function]. Draw a midnormal (?) on the segment. Put a point C on the midnormal, a bit over AB. Use [function] and draw a circle with the center in C and with A on the circle pherifery (?).
b. [question]?
c. Draw the segments AC and BC. [question]?
d. Use [function] and move C. [question]?
e. What can we say is common to all points on this midnormal? (/the question i cant figure out the answer of)

[mossy folio]

idk

I was never strong at geometry

[unique prism]

hm okay thats fine

thanks for helping though

and that was the instruction btw

[sullen flint]

How to prove that triangles ABC and BCD are equal ?

is it like AC || BD
∠A = ∠BDC
∠C = ∠CBD
△ABC = △BDC

[dark sparrow]

what are you given?

[sullen flint]

nothing just question and this figure/shape

[dark sparrow]

uhh

is there anything written beside it?

because in the absence of any information, this is just four random points connected with lines

with nothing deducible from that

[sullen flint]

but it says that ABC and BCD are triangles
i said it above image
"How to prove that triangles ABC and BCD are equal ?"

[dark sparrow]

can you post a picture of the whole worksheet/page you got this from?

[sullen flint]

its in another language hold on ill translate every word

Given steady(right angle) trapezoid/trapeze ABCD. Prove that triangles ABC and BCD are equal.

[dark sparrow]

okay, so you're given that ABCD is a right-angled trapezoid

and... what else?

[sullen flint]

thats it

[dark sparrow]

do you want to prove that those triangles are equal, as in congruent

or just that they have equal area?

[sullen flint]

yes equal area i think

[dark sparrow]

they both have height equal to AB and they both have BC as their base

[sullen flint]

I only know how to do triangle within the triangle but how to two triangles in right-angled trapezoid

[dark sparrow]

okay first off

in the thing you just posted

those two triangles aren't equal

they're similar

and second, the triangles in your question are not equal, nor are they similar

they have the same area

[sullen flint]

proportionate ?

hm

[dark sparrow]

no!

[sullen flint]

oh wait in that right-angled trapezoid question it says:Prove, that triangle ABC and BCD areas are equal.

sorry i missed word areas earlier

😭

[keen aspen]

Cavalieri's Principle I would say lol if you moved AD over to the left a bit and make it look like a normal trapezoid you will see visually that their the same based on their diagonals

I'm not good with proofs though lol

Using logic that's what I would say

[manic glen]

area ABC = 0.5⋅AB⋅BC
area BCD = 0.5⋅AB⋅AD-0.5⋅AB⋅(AD-BC)

let's start with that and then see how things cancel out, in other words the question is:
prove that 0.5⋅AB⋅BC == 0.5⋅AB⋅AD-0.5⋅AB⋅(AD-BC)
so then you expand 0.5⋅AB⋅(AD-BC):
0.5⋅AB⋅BC == 0.5⋅AB⋅AD-0.5⋅AB⋅AD + 0.5⋅AB⋅BC
then cancel out:
0.5⋅AB⋅BC == 0.5⋅AB⋅BC

some further context:
AD and BC are parallel because A and B are right angles, according to the definition of a right-angled trapezoid
create a triangle BDD' where D' is a (right-angled) projection onto the extension of BC, since BC and AD are parallel, and A and B are right angles, ABDD' forms a rectangle.
DD' is of length AB because ABDD' is a rectangle
BD' is of length AD because ABDD' is a rectangle
CD' is of length AD'-BC
the area of BDD' is 0.5⋅AB⋅AD
the area of CDD' is 0.5⋅AB⋅(AD-BC)
the area of BCD is the area of BDD'-the area of CDD' or (0.5⋅AB⋅AD) - (0.5⋅AB⋅(AD-BC))

[tropic stirrup]

Uh

Why do you have to be so complex

triangle ABC and trangle BCD has the same base BC and their height is all AB
And thus their areas are equal

Done

[manic glen]

oh right xD

well... guess that's why i always got F's for proofs 😛 (at least i actually got an answer for a change)

[sullen flint]

@tropic stirrup but if you had to prove that ABC and BCD like here

[tropic stirrup]

Where is BCD

What are you trying to prove

[sullen flint]

no i mean earlier

that trapezoid

[tropic stirrup]

Oh

[sullen flint]

how to write proof

[tropic stirrup]

I'm confused as to what you're asking me
Can you repeat what you're saying? ;;>w>

[sullen flint]

i have to write proof why in trapezoid ABC and BCD areas are equal but i have to write proof like shown in that triangle one

i dont know how to explain

[tropic stirrup]

That diagram doesn't explain a single bit about proving ABC = BCD

[sullen flint]

no i mean same principle that AB || MN etc.

can you explain that ABC and BCD equals are equal like this ?
AC || BD
∠A = ∠BDC
∠C = ∠CBD
△ABC = △BDC

[tropic stirrup]

I just don't see how in the hell that can be used for proving the formula

And no, △ABC = △BDC is not true

[sullen flint]

~

[tropic stirrup]

△ABC is similar to △BDC

And similar doesn't mean the area is equal

But really.... does ABC look like it's similar to BDC

Just look at it

[sullen flint]

proportionate

[tropic stirrup]

yes

But not equal

so for proving ABC = BCD, similar triangles are useless

at least that's what I think

[sullen flint]

but if i had to prove that they are proportionate a proof like this is correct ?
AC || BD
∠A = ∠BDC
∠C = ∠CBD
△ABC ~ △BDC

[tropic stirrup]

No that is completely false

[sullen flint]

AD || BC ?

[tropic stirrup]

You said AC || BD, not AD || BC

[sullen flint]

yes you said that AC || BD is false

what about AD || BC

[tropic stirrup]

AD || BC in itself is correct

but
∠A = ∠BDC
∠C = ∠CBD
△ABC ~ △BDC
this is just not correct in a single bit

[sullen flint]

so how to write a proof 😭

i have to write proof for trapezoid like in triangle one

[tropic stirrup]

My conclusion: You can't write a proof with what you're trying to do

just prove it like how I did, why do you have to prove something in a certain way

[sullen flint]

okay

[tropic stirrup]

👏🏼

[sullen flint]

ill leave it undone

[tropic stirrup]

🤦🏼

[sullen flint]

😄

[tropic stirrup]

;)

[sullen flint]

its holiday now ill ask teacher in a week

[upper karma]

ok so

how hard is geometry

[dense crater]

Heh

How long is a bit of string

[dark sparrow]

@upper karma nobody can answer that question without context

how much math do you know right now? are you in school, uni, or something else? what kind of geometry are you referring to?

[upper karma]

like, high school geometry

[dark sparrow]

and once again, that really depends on how comfortable you are with what came before that

ie algebra and stuff

but even then, it really depends on what the course is actually like

so my final answer is... you can't find out until you start taking the course

a forewarning, though: geometry will most likely be the first setting in which you'll encounter mathematical proof

which, for some, is a huge hurdle

so don't beat yourself up if proofs take time to get comfortable with

[mossy folio]

side angle side!

side side side

angle angle angle

no angle side side

[topaz valley]

breaks into dance

[mossy folio]

starts a chant

[dark sparrow]

uh

[tropic stirrup]

dances insanely

[umbral rivet]

👀

jumps around

[smoky violet]

Given 4 vertices of a square (as position vectors), what is the center of the square? Is it just the mean average of all 4 positions?

[slender socket]

((x1 + x2 + x2 + x4) / 4 , (y1 + y2 + y3 + y4) / 4 )

@smoky violet

[smoky violet]

ty

Does this work the same in 3 dimensions? Is it always (A+B+C+D) / 4 if A,B,C,D are the corners of a square

@slender socket

[dark sparrow]

i'll tell you more

it works in a space of any dimension

and in fact, the center of a regular plane polygon is always the average of its corners

[smoky violet]

Ooo, thanks!

To prove that statement for any regular plane polygon with n vertices, would you just prove that the distance between each vertex and the mean average of the vertices is constant for every vertex?

[upper karma]

I’m having trouble with the problem in the image indicated by the question mark...what am I doing wrong? Don’t those two numbers have to add to 180?

[dark sparrow]

angle 2 definitely isn't 54° lol

and no, they don't

[upper karma]

So then what is it? Because angle 4 and 8 have relationships on the right.

[dark sparrow]

angle 4 isn't 54° either

i take it that the yellow numbers are the ones you put in yourself?

[upper karma]

Yes

I was given angles 1 and 13

[dark sparrow]

yeah, you have 3, 5, 6, 7, 8, 11, 12, 14, 15 and 16 figured out correctly

2 and 4 are wrong

like, 1 through 4 are supposed to add up to 360°, ya feel?

[upper karma]

Angles 2 and 5 are same side interior and they’re supposed to be supplementary though

I know what you mean by adding up to 360 though...

Would angle 4 be 126 instead?

So angles 4 and 8 would both be 126...

[dark sparrow]

Angles 2 and 5 are same side interior and they’re supposed to be supplementary though
but m and q aren't parallel. that argument simply doesn't apply

yes angles 2 and 4 would be 126°

[upper karma]

Would this be correct, then?

[dark sparrow]

2 and 4 are correct, the other red ones are all wrong

angles 6 and 10 are meant to add up to 180°!!!

[upper karma]

Better?

[dark sparrow]

yes

[upper karma]

Thanks

[tropic island]

The perpendicular bisectors of sides AC and BC of ΔABC intersect side AB at points P and Q respectively, and intersect each other in the exterior (outside) of ΔABC. Find the measure of ∠ACB if m∠CPQ = 78° and m∠CQP = 62°.

[past mantle]

The point of an centroid is $$(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$$

[past mantle]

Is that right?

[tropic stirrup]

not considering the grammar, yes

[upper karma]

hahahahaha

[upper karma]

anyone know how to do this?

[upper karma]

Sorry dude, I'm terrible with proofs.

Can someone help me solve this?

[upper karma]

Actually, nvm.

[dark sparrow]

your selected answer is correct :p

[young hemlock]

Is r=(2,-1,3)+t(2,-4,-2) perpendicular to r=(1,-1,2)+k(3,2,-1)?????

[dark sparrow]

are their direction vectors perpendicular?

[young hemlock]

Their dot product is equal zero.

[dark sparrow]

so

[young hemlock]

They are.

Okay. Time to search for a server that teaches how to proof your teacher wrong.

[dark sparrow]

what does your teacher say?

[young hemlock]

The question is as follows:

Find the equation of the straight line passing through the point (2,-1,3) and intersercts the line r=(1,-1,2)+t(3,2,-1) orthogonally.

[dark sparrow]

the issue might be that your line doesn't intersect the given line

[young hemlock]

Oh.

You're right.

I just noticed.
They intersect and are not just perpendicular.
I feel dumb now.
@dark sparrow Sorry. 😦

@dark sparrow THANK YOU SO MUCH. I FEEL SO MUCH BETTER NOW. LOL.

[dark sparrow]

:+1:

[feral fractal]

When given a side and an angle in a right triangle, why won't I get the right results if I use cos(angle)=adjacent/hypotenusis => hypotenusis=cos(angle)/adjacent?

[dark sparrow]

1. it's hypotenuse, not hypotenusis
2. hyp = adj/cos(θ), not the other way around

$$a = \frac{b}{c} \iff c = \frac{b}{a}$$, not $$c = \frac{a}{b}$$

[feral fractal]

https://awwapp.com/b/u9nurs2xc/

I still don't see a way that the hypotenuse could be 39cm as stated in results.

[dark sparrow]

== 10.5/cosd(15 + 20/60)

[charred spear]

10.88755628

[feral fractal]

and the other one 2.8?

[dark sparrow]

== 10.5 * tand(15 + 20/60)

[charred spear]

2.87904182

[dark sparrow]

2.9, if you're rounding to one decimal place

[feral fractal]

then the solutions must be off

thanks!

[dark sparrow]

are you looking at the solution to the right problem? 👀

[feral fractal]

yup. seems wierd

c = 10.5*(cos15°20')

Isn't this correct?

I'm not sure what have you used up there

[dark sparrow]

=tex c = \frac{10.5}{\cos(15^\circ 20')}

[dark sparrow]

mathbot's sin, cos and tan functions are in radians

to make mathbot calculate them in degrees, you need sind, cosd and tand

[feral fractal]

That would be right. But there is no way that they could be ~40cm when using deg.

[dark sparrow]

== 10.5/sind(15 + 1/3)

[charred spear]

39.7074263

[dark sparrow]

you might have marked the wrong angle

[feral fractal]

No, the book was acually incorrect after asking a friend. Also, my calculator has no ctg funcion, could I manually input that into a variable button in the storage of it?

[dark sparrow]

ctg? cotangent?

[feral fractal]

yes.

[dark sparrow]

cot(x) = 1/tan(x)

so if your calculator can do tan, just do that and take the reciprocal of the result

[feral fractal]

or use tan^-1?

[dark sparrow]

no

that's different

tan^-1 undoes tan

i.e. you give it a number, and it returns the angle whose tan is that number

[feral fractal]

oh, okay. so e.g tg=5 => ctg=1/5=0,2?

[dark sparrow]

yes, if tan(x) = 5, then cot(x) = 1/5

[feral fractal]

Thanks! 😄

[upper karma]

What is the name of a shape which has nine sides? Is it nanagon?

nonagon

Thank you.

[dark sparrow]

9-gon if you're lazy

[tropic stirrup]

9-sided polygon if you wanna be a bit less lazier

[surreal bolt]

nonagon I thought.

[dark sparrow]

yeah, nonagon

[past mantle]

Wish me luck on this geometry quiz today

[dark sparrow]

🍀

[vale raven]

Probably 1 or something

[slender gyro]

Begin by drawing p=p', q=q', r=r'
Obviously the second triangle can be obtained by first moving p' either up or down or not. Then q'either up or down or not. Then r'either up or down not.
WLOG you only have to try 1+1+2+4 alternatives.
The one with least intersections is 3 the one with the most of 6.
Not sure how to formalise but it's rigorous enough for me. Assuming i did all the computing right

[cobalt folio]

So I have an ellipse, with the full formula, so I know eccentricity and stuff. The exercise asks me to find what's the min/max distance between a given vertex (I already know it's position) and a given focal point (I don't know how to calculate it's position?)

We just saw ellipses yesterday so maybe I'm a bit more advanced than I should be but I already managed to find all vertices and the formula so I'd like to know how can I finish the excercise

[slender gyro]

Can you make post a drawing with the eccentricity and some vertex and the focal point?

[cobalt folio]

https://i.imgur.com/dbt76E6.jpg

I think that's it

so I need to find max/min distance between planet G and star T

[slender gyro]

Sorry wasnt here

I don't know much about this but the max distance is a+c and the he minn distance is a-c right?

Since the positions that minimize the distance are (a,0) and (-a,0)

With the start being at (c,0) [oops, minor correction here]

[cobalt folio]

is c the center?

I got c from the eccentricity formula but I'm not sure what it represents

I guess it's the center yeah

[slender gyro]

Sorry c is according to wiki, the linear eccentricity

Ie the position to one of the stars is (c,0)

[cobalt folio]

so T is in (0, -c)

so so maybe the min distance is when G is at (0, -480) default pos, and max distance is (0, 480) ?

I have class in like an hour anyways so I'll just check with the prof. Thanks for your help!

[slender gyro]

Yep

Sorry for the insistency in replying

[cobalt folio]

I don't think "insistency" is the word you want to use

[slender gyro]

Hmm, my tablet just have me Incisura. At least insistency is closer. I'm not even sure what incisura is.

Anyway inconsistency, sorry

[cobalt folio]

don't worry English is not my main language either

I know "incisura" in Spanish not in English idk what it means

[sterile wyvern]

Whats an isometric figure?

[dark sparrow]

a diagram of a 3D object in which the coordinate axes are projected into lines at 120° angles to each other

@sterile wyvern

[upper karma]

Could someone please help with this. I'm finding it difficult to visualise

[surreal bolt]

hmm a 3d problem!

First photo: how far South is she from the building?

[thorn talon]

draw a diagram

[sterile wyvern]

Thx

[thin hound]

<@&286206848099549185> http://prntscr.com/h80ext

[quasi lion]

=wolf sin(x+pi/2)

[charred spear]

Query made by @quasi lion
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=sin(x%2Bpi%2F2)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[quasi lion]

Does that help you?

[thin hound]

no

[slender gyro]

A triangle with hypotenuse 1 has sides equal to cosine and sine

Just use Pythagoras

[dark sparrow]

@thin hound do you still need assistance?

[limber bluff]

what are integrals doing in geometry

[dark sparrow]

wolf queries tend to return Everything™

[limber bluff]

ahh

[long cedar]

guys i really need some help

how do i do this?

my friends are gonna bully me if i get it wrong again..

[topaz valley]

jesus, ok

i'm guessing the final answer isn't given

what have you tried?

@long cedar

[long cedar]

like nothing cuz i dont know how to start

im a noob in maths 😦

[topaz valley]

do you understand what it's asking?

@long cedar your silence is uninspiring

perhaps you should ask this another time

[long cedar]

oh sorry i was trying to do it.. but anyway thank you for trying to help

[topaz valley]

well i've made some progress

are you familiar with the cosine rule?

[long cedar]

yes

[topaz valley]

cause you can start pretty well with just that

[long cedar]

oh so should i just find the angle BEF first

[topaz valley]

well, to be honest, the best approach when one sees a problem like this is to fill in as many blanks as possible

i started by figuring out the lengths of the triangle DCA

[long cedar]

oh ok

[topaz valley]

from that you can find the angle GDA

then from that, the length GA (in the triangle GAD)

[long cedar]

ohhh

how did you find the triangle GDA?

sorry i dont really understand how to do it but i see what you are doing

[topaz valley]

you can use pythagoras' to find DA

right?

[long cedar]

oh yeah

[topaz valley]

so if you make a triangle out of GDA, then you have two sides, and the angle between them

[long cedar]

oh yeah

what about the side CA?

[topaz valley]

that's part of a circle segment

CA = AB = 3

[long cedar]

oh

true

hahahaha im so dumb

ohh im starting to understand much better now

[topaz valley]

😃 yay

[long cedar]

so since i have the 2 triangles, would i need to make another 2 triangles with GFA and AFB?

oh maybe not AFB

[topaz valley]

i was working at afb just now

[long cedar]

oh ok

[topaz valley]

i mean

do it all

figure out every length you can

whether it's helpful or not

[long cedar]

oh ok

thanks so much

[past mantle]

48/52 on my quiz, at least there were no proofa

I guess that's ok

[thin hound]

Ik how to do it

Wasn't even as hard that guy made it look smh lol

[dark sparrow]

who?

[sullen flint]

Completely clueless about this one.

i do too but why

how

[umbral snow]

The hypotenuse is 8 - h

Apply pythag:
4² + h² = (8 - h)²

[sullen flint]

ty

[rugged cargo]

kaynex how come

[dark sparrow]

how come what

[rugged cargo]

i just dont het 8 - h

get

nvm

didnt read it

[topaz valley]

did anyone get around to solving the problem @dankish Haiyodi#8331 posted?

https://cdn.discordapp.com/attachments/326138757474680852/378101142640525312/Screen_Shot_2017-11-09_at_3.45.17_PM.png

[grand ingot]

honestly just coordbash everything except the sector

[umbral rivet]

What is that shape

[crude kraken]

"Oh yeah I know a bit of geometry" gets slapped by that problem

Angle BEF is 118.875, Angle BFE 34.094, angle EBF is 27.03

Aaannd that's the extent of what I can figure out

oh wait

== 78.28 + 34.094

[charred spear]

112.374

[crude kraken]

Those are the values in the order I solved them in, using law of cosines and sines.

If you have any extra questions about that, go ahead and ask. (Fun problem btw) @long cedar

[tidal notch]

ok so i havea geo test tommorrow

and i will tell you guys if i need any help

[crude kraken]

note: We're not gonna help you cheat.

[topaz valley]

hint: if you want our help, don't tell us you're in a test

[crude kraken]

^

I mean uh

No, don't do that.

[topaz valley]

that said, probably a bad idea to ask for help on something time sensitive here

you're more likely to get a lengthy explanation than a quick answer

[crude kraken]

And be careful, if you send us simple stuff like finding the area of a triangle we may pull out some calculus

[vapid kettle]

well there is a (perhaps dangerous) mantra

"Nothing is against the rules if nobody knows you did it."

and sometimes this mantra is

for lack of a better word

good

[tidal notch]

i know that

why would i do that

[tidal notch]

https://cdn.discordapp.com/attachments/351182527316099072/374751838337105923/JPEG_20171030_214913.jpg

https://cdn.discordapp.com/attachments/351182527316099072/374754437064818688/JPEG_20171030_215954.jpg

can someone help me with these

[thorn talon]

Which one?

@tidal notch

[tidal notch]

oh