geometry-and-trigonometry

[acoustic fiber]

😁😁

hardwork seems to be paying off

[upper karma]

Several rectangles are drawn on a plane surface in such a way that their intersecting lines form 18,769 areas not further subdivided. What is the minimum number of rectangles that must be drawn to form the described pattern?

you can do it with less than 100 rectangles

but i cant find the pattern

a little help, guys?

[dark sparrow]

wow oof

[upper karma]

i can do it with 10 times less rectangles than areas

but thats the best i got so far

[oak minnow]

how do I do

[dark sparrow]

what is holding you up here?

we aren't going to just do this for you, yknow

and most importantly, you have once again posted multiple problems with no indication of which one you are stuck on!

[upper karma]

@dark sparrow who are you talking to/

?

[dark sparrow]

to fuhrer1

[upper karma]

can anyone help me with mine?

everytime i post a problem on this server no one helps

[coarse stag]

Fuhrer1, use the law of sines. Then I think it'll be clear.

[patent kite]

Law of cosines

[upper karma]

D

Actually E

But ok

oh shit

[clear haven]

lol

[upper karma]

Yeah, ok let's use the formula

[clear haven]

ok so how ur supposed to do it

is by plugging in 3r where r is supposed to be

[upper karma]

I’m asking

$ C = \pi*r^2 $

[clear haven]

and squaring the entire quantity 3r itself

[somber coyote]

rudy:

[upper karma]

yeah

Did it in my head and for some reason thought of 3*2 instead

Omg

[clear haven]

[upper karma]

So ez

[clear haven]

yus

feel gud

[upper karma]

Why does it seem hard

[clear haven]

dunno

[upper karma]

Because you don't understand the: 3 times larger

[clear haven]

maybe question wos intimidate 👻

[upper karma]

No I mean I’m good at math

But it’s so simple it’s hard

://:

[clear haven]

[upper karma]

Huhhh

two positives make a negative?

Extraneous solution????

extraneous solution != negative

[clear haven]

tf?

is goin on

[vernal void]

[clear haven]

[upper karma]

🅱️🅱️🅱️🅰️

the fish is thonking rn

[vernal void]

[frigid wasp]

You know how sinh, cosh and tanh have definitions involving e^x etc., do "normal" trig functions have that too?

[dark sparrow]

yeah, but you need to go into the complex numbers to get it

$\cos(x) = \frac{e^{ix} + e^{-ix}}{2} \ \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$

[somber coyote]

Ann:

[dark sparrow]

has to do with euler's identity

$e^{ix} = \cos(x) + i \sin(x)$

[somber coyote]

Ann:

[frigid wasp]

But then u end up with sinx defines in terms of sinx

Surely?

[dark sparrow]

mh well the thing is

usually euler's identity isn't considered a definition

rather, the exponential function is defined (via its taylor series), and then cos and sin are defined from it like i wrote above

and euler's identity just follows

[gray saddle]

k3 and h2 both have a radius of 3 what is the radius of h1

[storm hatch]

Is that all of the given information?

[gray saddle]

K3 is touching h1 and h2

And AB

[storm hatch]

Does it tell you that BC = AC, BC = ½ AB or C is the midpoint of AB?

[gray saddle]

Nope

[storm hatch]

Does it mention anything about C at all?

[gray saddle]

No just that h2 is the half circle of cb

[storm hatch]

So you can't exactly establish that BC is the radius of h1.

[gray saddle]

Bc?

[storm hatch]

I think I just suck at Geometry. Could you send the whole question, with the text?

[gray saddle]

K ill translate it

Let C be a point on the line AB. About AB and CB semicircles h1 and h2 are built on the same side. The circle k3 touches h1 from the inside,
h2 from the outside and also the line AB.
If the radii of h2 and k3 are both 3cm , how big is the radius of h1?

[storm hatch]

I'm sorry, man. I don't know.

I know BC = 6 units, but since you can't prove AC = CB, then you can't assume that the radius of h1 = 6 units, unless I'm missing something obvious.

[gray saddle]

Oh im stupid

[frigid wasp]

@gray saddle is that from the ukmt intermediate Olympiad? I had a similar question when I did that

[gray saddle]

I got it from the german math olympiad archive

https://www.mathematik-olympiaden.de/moev/index.php/aufgaben/aufgabenarchiv

[frigid wasp]

ok

try connecting centres?

[gray saddle]

Why? How should this help?

[real mulch]

the center of h1 (O) lies on AB, let's denote the distance BO with x;
then we can write an equation for x using the fact that x is equal to the distance between O and the furthest point of k3 (from O)

@gray saddle

[lime flame]

Disregard stuff I wrote

[plucky marlin]

https://prnt.sc/nbn3ls

i was able to draw a plane that goes thru both BC and EH

@lime flame can u draw planes thru the other lines

[upper karma]

https://cdn.discordapp.com/attachments/462028456323186690/566811937413922826/image0.jpg

does this make sense

[upper karma]

hey i need help with a question. can anyone DM me?

this is the problem i need help with

[upper karma]

nvm i got it right

[dark sparrow]

can you enlighten me as to what "rounded distance" is meant to be, then

bc i've never seen that term before

and can't seem to decipher it in context

[upper karma]

@dark sparrow can u help me

with mine uwu

[dark sparrow]

ok what's holding you up

[upper karma]

literally

everything @dark sparrow

lol

I know its a similar triangle

ok im done

[dark sparrow]

"it"?

what do you mean by "it"?

[upper karma]

BEF and EDC

theyre similar @dark sparrow

[dark sparrow]

BEF and ECD but yes

........ok wait

on second glance this problem appears to be ill-posed.

[upper karma]

@dark sparrow yes right

its worded badly right

[dark sparrow]

the wording's fine, it's just that i feel like it might not be enough info

[upper karma]

oh

hm

if u say that, i believe u

brb insulting this ass😡 😡 😡

[upper karma]

^/s btw

[upper karma]

why is $$\sin{\theta} = \sin{-\pi - \theta}?

[somber coyote]

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

[upper karma]

ughh

[dark sparrow]

$\sin(-\pi-\theta) = \sin(2\pi - \pi - \theta) = \sin(\pi - \theta) = \sin(\theta)$

[somber coyote]

Ann:

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

[dark sparrow]

i mean

what i did is $2\pi + (-\pi - \theta)$

[somber coyote]

Ann:

[upper karma]

oh wait i get it now since $sin(\pi - \theta) = sin(\theta)$

[somber coyote]

taven:

[upper karma]

thanks

[dark sparrow]

we can do this here if you want

but what's holding you up here?

[fair vine]

I found AB, OM, OP, OM, AP but I cant find any way to get AN or anything near N

[dark sparrow]

ok well

APN is a straight line

so you know AN = k*AP for some constant k

[fair vine]

Ok yes

[dark sparrow]

and you also know that ON = m*OB for some constant m

[fair vine]

Yes

[dark sparrow]

try going off of that

[fair vine]

Ok

ill try, if i cant solve it can i @ you?

[dark sparrow]

sure

[upper karma]

is cos(2x) = cos^x - sin^2x?

sorry for such a dumb question

[fair vine]

@dark sparrow 😅

i tried... nothing good

[dark sparrow]

what did you try

[fair vine]

So AP is
2/7 b - 5/7 a

AN is

$$\frac{\frac{2}{7}b-\frac{5}{7}a}{m}

[somber coyote]

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

[fair vine]

$$\frac{2b-5a}{7m}

[somber coyote]

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

[fair vine]

then i found that ON = b/k

and if we go the other way using ON

AN = b/k - a

$$\frac{2b-5a}{7m}=b-ka

[somber coyote]

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

[fair vine]

this is probably wrong idk what to do from here :?

[dark sparrow]

okay so like

$AN = -\frac57k a + \frac27k b \ ON = \left( 1 - \frac57k\right)a + \frac27 k b = mb$

[somber coyote]

Ann:

[dark sparrow]

since a and b aren't parallel, the only way the last equality can be true is if $1 - \frac57k = 0$

[somber coyote]

Ann:

[fair vine]

so -5/7k = -1

k = 7/5

[upper karma]

uwu what's the problem you're working on?

[fair vine]

Scroll up i made a copy of it

[upper karma]

https://cdn.discordapp.com/attachments/326138757474680852/566974818231255080/Untitled.png

that one?

[fair vine]

Yes

[dark sparrow]

yes k = 7/5

do now you know m = 2/5

[fair vine]

Yes

All i need is K though

[dark sparrow]

no, you need m

ON = m*OB, remember?

[fair vine]

Yes

[upper karma]

i cant see

anything

[bleak rivet]

Im trying this question here but im stuck part way though

I started by using the vertex form: Y = a (x - h)^2 + k

Then I subbed in the numbers:
300 = a(2100)^2

and am stuck with 1 / 14700 = a

Answer is allegedly:

so I figured I was on the right track

[umbral snow]

You have y = (1/14700) x²

Same thing

[bleak rivet]

so whats the process of rearranging to that?

[umbral snow]

Multiply by 14700

[bleak rivet]

ahh i see

Thanks

[solar shale]

@me

How do I solve this, that’s the correct answer btw

[solar shale]

<@&286206848099549185>

[upper karma]

Area of.....?

Yeah idk what it is asking

Take another pic

[solar shale]

I sent the wrong one

If you don’t mind just write out all the steps I’m kind of busy so I can check out later

[upper karma]

You want to know how to find arc length

?

[upper karma]

owo

thats cool @upper karma

Some of my classmates think I'm weird keepiong whiteboard markers and shit on me. They call me the second professor when we do study sessions and I do powerpoints and use the whiteboard a lot

[acoustic fiber]

What exactly are sin theta and cos theta? can anyone explain me in simple terms?

[proven dome]

um, they’re ratios of the sides of a right triangle @acoustic fiber

so sin θ is the ratio opp/hyp

and cos θ is adj/hyp

[dark sparrow]

that works for θ between 0 and 90° ofc

[acoustic fiber]

cool man thanks

[oak minnow]

How do i do this one

using the sine rule

[dark sparrow]

you don't need the sine rule here.

or are you explicitly asked to use it somehow?

@oak minnow

hello?

[hallow smelt]

how come they want me to use the law of cosines and the quadratic instead of just using the law of sines?

[plucky marlin]

yup sin rule

[dark sparrow]

@hallow smelt either way is possible

[hallow smelt]

but if i use the law of sines i get possible triangles if i use the law of cosines no real answers

im confued

[plucky marlin]

y wud someone use cosine rule

[dark sparrow]

@hallow smelt show work?

[hallow smelt]

ok

I have a page of these

where they want to see the quadratic

wait guess the law of sines doesnt work for that triangle I thought it did maybe it was another

[plucky marlin]

ofc law of sines work

idk how law of cosines will work on that

[hallow smelt]

I hate the quadratic... even my chem prof hates it. No pH or pKa etc where we have to use it

[plucky marlin]

,w solve for x in 50^2 = 65^2 + x^2 - 265x*cos(55)

[somber coyote]

Display results online and refine query

Requested by soap#7503

[plucky marlin]

well fak

[dark sparrow]

[plucky marlin]

well even sine rule gives you sin(B) = 1.06489 > 1

ayy voila

no real B can satisfy this relation

🍻

[hallow smelt]

do you guys know how to use the solver function on a ti-84

to solve the quadratics?

[plucky marlin]

https://www.youtube.com/watch?v=4lyWrli7oQ0

[upper karma]

welcome to the next generation of laziness

[plucky marlin]

[hallow smelt]

how do they just know thats a 135 degree and 45 degree angle?

[plucky marlin]

"north east"

[hallow smelt]

so two directions means 45 degrees?

[plucky marlin]

well yeah

[dark sparrow]

NE is at 45° angles to north and east @hallow smelt

[gray saddle]

I've solved it it's 3+2√3 =~ 6.46

Was way easier than i thought

[dark sparrow]

attach this diagram to a mirror image of itself so that it becomes 3 circles inscribed in a larger circle

[gray saddle]

That's what I did

[vital frost]

Would like help

[upbeat shore]

Basically, just get the volume of both the roll and the hole inside. Subtract the hole's volume from the roll's and then you have the volume (which I'm assuming is the thickness) of the actual roll. Volume for cylinders is V=Bh
(B=area of the base)

Someone please correct me if I'm wrong. Also sorry for being bad at explaining it as I'm usually the one getting helped.

[forest niche]

Help!

Find area?

[umbral snow]

Break everything into rectangles
Find the lengths and heights of those rectangles
Use area formula

[forest niche]

I know but the bottom rectangle doesn’t make sense

How do I find the length?

Or width*

[vital frost]

uh

[oak minnow]

Can someone link me a good tutorial that explains and shows how to find the points and magnitude using the geometry?

[dark sparrow]

,w define vague

[somber coyote]

Display results online and refine query

Requested by Ann#8192

[oak minnow]

,w define bitchy

[somber coyote]

Display results online and refine query

Requested by Fuhrer1#6111

[supple abyss]

[dire rampart]

lmao

that was indeed extremely vague tho

points and magnitudes of what?

[oak minnow]

You know how some questions have "Find ABC, ACB and BC"

Thats what I mean

I don't know what their called properly

Questions like those

I need to practice them

How do I find AY

I found AX

[dark sparrow]

give that unnamed intersection point a name

say, Z

and consider triangle ABZ

there are some angles you can compute here w/o much trouble

[oak minnow]

What do you mean by intersection?

[dark sparrow]

i mean exactly what i said

look at that diagram

[oak minnow]

Oh so you mean the center?

[dark sparrow]

it'd be a bit of a misnomer to call that point the "center" of anything, since there is no shape it could be the actual center of

but i am referring to the intersection point of AY and BX

[oak minnow]

Yeah, thats what I was also referring to

So do I solve for A and B to get the intersection Z

[dark sparrow]

that last sentence of yours is meaningless. try again.

[oak minnow]

Ok... so I presume not

[dark sparrow]

what do you mean, "solve for A and B"?

A and B are points.

[oak minnow]

those angles

[dark sparrow]

1. each of these vertices has more than one angle at it.
2. you are already given all the relevant angles both at A and at B.

like, they're right there in the diagram

[oak minnow]

Welp, you confused me

[dark sparrow]

alright, let's back up

https://cdn.discordapp.com/attachments/326138757474680852/567599345734516737/Screenshot_2019-04-16_at_16.36.27.png

here's the diagram, so that i don't have to scroll up for it

[oak minnow]

good

[dark sparrow]

you say you found AX, right?

[oak minnow]

aye

[dark sparrow]

how did you find it?

don't tell me the actual value of AX, for i do not care about it. i care only about how you got there.

[oak minnow]

We know that A is 32

[dark sparrow]

no

"A is 32" is meaningless

[oak minnow]

ok then how

[dark sparrow]

might you have meant that angle ZAB is 32°?

carry on. you were describing the work you did to find AX.

[oak minnow]

Welp you confused me again by saying it was wrong,

so I need to find another way

[dark sparrow]

no

i did not say that what you said is wrong.

i merely said that what you said is meaningless (as written).

[oak minnow]

meaningless

[dark sparrow]

do not put words in my mouth.

[oak minnow]

meaningless means "false"

well

its a synonym for it,

[dark sparrow]

no

it is not

"meaningless" and "false" are definitely NOT synonyms.

here is an example of a false sentence: "Two plus two equals five."
and here is an example of a meaningless sentence: "Godfg wkejrl fhadfu sjsfj rgwkl."

[oak minnow]

oof, but why are you using dots everytime

[dark sparrow]

what?

[oak minnow]

full stops

in every sentence

it makes it look like that you're some kind of grammar nazi (hence my name)

[dark sparrow]

i'm not a grammar nazi

[oak minnow]

coincidentally

[dark sparrow]

not being a grammar nazi does not mean abolishing punctuation and spelling in their entirety, though

[oak minnow]

yes but people know the meaning of what they say anyways

[dark sparrow]

anyway, i don't know why you're bringing this up now

[oak minnow]

waddaya mean

Oh

[dark sparrow]

i don't know why you're bringing up my typing style

especially when it has nothing to do with the geometry problem at hand or my clarification to you that "false" is not at all the same as "meaningless", and my yet-unexpressed astonishment at how one could possibly conflate the two

[oak minnow]

uhh forget it,

[dark sparrow]

let me just get the diagram again...

https://cdn.discordapp.com/attachments/326138757474680852/567599345734516737/Screenshot_2019-04-16_at_16.36.27.png

now describe to me, once more: how did you get the length of AX?

[oak minnow]

first off i'm not sure if the answer I got is correct but here is what I did:

to find AX I did

50sin(32°)/sin(88°)

[dark sparrow]

and what is your logic behind that?

[oak minnow]

A is 32° and X is 88°?

[dark sparrow]

...

[oak minnow]

alright, i've had enough,

I'm going to ask someone else

[dark sparrow]

as you wish

[oak minnow]

ty for your much valuable time madam

[dire rampart]

[upper karma]

lol

[sleek pendant]

hey can you send me a really dificult geometry problem

[dire rampart]

what kind of geometry

[vague pagoda]

I have a exam on geometry next Thursday

[sleek pendant]

sth with shapes

[dark sparrow]

that covers basically all of geometry tho

[sleek pendant]

well even better , please sb send me a really hard problem , i want a problem that my teahcer wont be able to solve , because she always says she is too good and stuff like that

[upper karma]

Can someone point me in the direction I should start with?

Is it just double trig?

[dire rampart]

@sleek pendant here's a fun area problem

[sleek pendant]

nice thanks a lot

[proven dome]

@upper karma split the figure into three shapes first, then find the other side length of the triangle with the side AB

[solar shale]

For Area of Circle it says = 64 pi I’m curious as where the 64 comes from?

[vague pagoda]

Area of a circle is not 64pi??

[solar shale]

Well look in pic it says Acitcle=64 pi

I just want to know where the 64 came from only way I got 64 was by adding 16+16+16+16

[vague pagoda]

its pi * (r)^2

r = radius

[solar shale]

Wait but it doesn’t say on the pic and the answer is right?

[vague pagoda]

?

what is the question

I can’t see

[solar shale]

Find area of shaded region

[vague pagoda]

O

[solar shale]

I mostly understand it just have trouble seeing where the 64 came from

[vague pagoda]

Yeah should be right

[solar shale]

Wait so the 64 comes from where tho?

I assumed 16 is half the circle so 32 is the whole thing and there are 2 circles so 32+32=64 not sure tho

[vague pagoda]

16 is the diameter of the circle

16cm/2=8cm

pi*(8cm)^2

pi*64cm

[solar shale]

Oh ok tnx 👍🏼

[vague pagoda]

No problem

[dark sparrow]

chem?

[blissful cedar]

I need help drawing the hexagons with chlorine attached

aka the benzene rings

[vast garnet]

for an arbitrary equilateral triangle, what's the maximum distance between two points

[surreal bolt]

side of the equilateral triangle?

In triangle ABC, the circle containing arc BC and centered at A can be drawn.

Clearly the triangle is within the circular arc drawn. Thus the points in the triangle must be at most AB or AC (they're equal) away from point A.

The farthest away IS in fact AB away (i.e. point B). Rinse and repeat for the circle center at B, and then at C.

[vast garnet]

Doesn't this show that the points farthest away from the vertex of an equilateral triangle are just the other vertices?

How does this apply for arbitrary points...

[last lichen]

how is cot() = cos()/sin()?

o dw i found out

[cunning cosmos]

hey guys, can someone proof my work and see if its correct

[dark sparrow]

all good @cunning cosmos

[oak minnow]

how do I do c

[dark sparrow]

what have you tried so far and where are you stuck?

what do you think you can use to solve 3(c)?

[oak minnow]

cosine law

?

[dark sparrow]

well why don't you try that then?

[oak minnow]

I have but im getting wrong outcomes,

[dark sparrow]

can you show your work, then?

so that i can see where exactly you went wrong

[oak minnow]

How am I supposed to show you

my phone is getting fixed

all I have is a digital textbook

[dark sparrow]

idk type it out or something?

[oak minnow]

ok but you might not understand it very well

[dark sparrow]

i will ask you to clarify whatever i am unable to parse

[oak minnow]

k

b^2+c^2-a^2/2ab

35^2+46^2-27^2/2(35)(46)

[dark sparrow]

woah ok wait

first off

is that meant to be one big fraction

if so, parentheses

[dark sparrow]

careful there.

the reason you're getting the wrong answer is bc you're mixing up the notations

[oak minnow]

wait what

I thought that was the formula

[dark sparrow]

you want to find angle B, not angle A

so the letters switch up

[oak minnow]

oh yeah

[dark sparrow]

the sides that go in the denominator are those that are adjacent to the angle.
the side that appears with a minus sign in the numerator is the one opposite the angle.

as a sanity check, the formula should remain unchanged if the values of the two sides adjacent to the angle are swapped.

so following the notation set out in the problem

$\cos(\angle ABC) = \frac{c^2 + a^2 - b^2}{2ac}$

[somber coyote]

Ann:

[oak minnow]

ok, so if I wanted to solve for lets say angle ACB will the formula remain unchanged

nvm, i figured it out

[zealous bough]

Anyone know why I got this wrong? I can’t find triangle B

[real mulch]

Is that 7cm given?

Also, you cannot say that the angles on the right are 45 and 45 degrees

Because they arent

The area of a trapezoid is (a+c)*h/2, where a and c are the parallel sides, and h is the height

[zealous bough]

I got triangle A right

I assumed triangle A = triangle B

And found the area

Which I guess is wrong?

7 cm was given

So would I do (4+4) * 7/2? @real mulch

[real mulch]

No, because the bottom side is not 4 units long

[zealous bough]

So how do I find this stuff

Oh it’s 4+3+3?

Because the triangle on the right = 3

[real mulch]

The triangle on the left is half a square, therefore its bottom side is 7 units long

[zealous bough]

So 4+3+7? That’s the total bottom?

[real mulch]

It seems to me, but I'm probably wrong because there are no radicals here, but the statement talks about them...

[zealous bough]

Yeah you don’t need to use radicals that’s only for extra I think

[umbral snow]

@zealous bough
I don't believe you got triangle A right. The base is 7, so the area is 7×7/2

Triangle B has a base of 3, height of 7. It has area 3×7/2. You got the right triangle correct

[vast garnet]

if I show that in an equilateral triangle, the farthest distance from a given point must be a vertex, then can I fairly assume that the maximum distance between two points in said triangle is the length of the side

pls ping me for the answer

[real mulch]

I think so

that's true for any triangle btw

[fierce snow]

Been stuck here for like 30mins

I some how got the answer to be 53° for alpha but it was pure luck and idk how to actually do it

[dire rampart]

only the vertical component of F causes a moment about P

@fierce snow

[fierce snow]

Is the vertical component of F = 20/sqrt(5)

?

@dire rampart

[dire rampart]

how did u get sqrt5

its 20sinalpha

[fierce snow]

isn’t that the perpendicular distance from F to P?

And since M is 20Nm

[dire rampart]

no, perp distance is 2m

[fierce snow]

oh right xD

[dire rampart]

[fierce snow]

The answer is 53° or 180°

I see how to get 180° now but still baffled about the 53

[dire rampart]

[fierce snow]

This is gonna keep me awake 💔😭

[vernal void]

idk

[dense tundra]

could anyone give me a good website on how to find the foci and vertex of an ellipse and hyperbola by just having the equation or just having only the foci or just having only the vertex and ping me please? i have to go to class now

[slate pewter]

howi can something like this

[dark sparrow]

you'll have to translate from turkish 😛

[slate pewter]

the translate = the shape in the right his radius is 12

what is the MN long

pi eqaul to 3

[real mulch]

pi equal to 3
I am convinced

we're moving

finally, at last, they understand

[dark sparrow]

i think the problem is just asking to approximate pi as 3? idk

whatever. doesn't matter

you know how to find the circumference of a circle given its radius, right? @slate pewter

[slate pewter]

yes you multıplı the radıus by 2 and after that by pi

one moment i need to do a think

wait me a little bit

back

Right i know @dark sparrow

[dark sparrow]

okay yes alright

MN isn't a full circle but it is an arc of a circle

and its angle is 110°

so it is equal to 110/360 of a full circle

[slate pewter]

yes

one moment i did think a way to solve it

[slate pewter]

i think its like this

@dark sparrow

[dark sparrow]

well, 22 is the correct answer, even though i can't understand your work whatsoever

[real mulch]

I think (s)he's saying that 110/360 = x/72

[slate pewter]

yes multiply the 110*72 = 7920

and divide both sides by 360

for making 360 decepere

and you get x in the left by it self and 22 in the right

this is how i d id d it

[dark sparrow]

is that a long way of saying you calculated $\frac{110}{360} \cdot 72$

[somber coyote]

Ann:

[slate pewter]

i dont know (:

hhhhhhhhhhhh

but the way i did it is better

for me

becuse to divide 110/360=0.305555556

0.305555556*72=22

so i dont know the way i did it is better to my

but thanks

[real mulch]

you did the same thing

[slate pewter]

the shape in the side O1 and O2 the center point there are tow circle given [O1,A] = 4cm and [O1,A] [O1,B ] is right angle of what we see what is the space in the wihte area

pi = 3

i did think about it a little bit but there are a thing wich its not working i know how to find area of a circle

and they given you the radius = to

4

so 44=16 16pi = 48

this is all the space

its right angle so its 1/4

48/4 = 12

the next circle

[dark sparrow]

when you write your work like this, it sure is easy to get completely lost!

[slate pewter]

but

oh for some reasson the messege depered

the next circle is the diameter of 4 so 4/2 = 2 * 2 = 4 4 * pi = 12 but its the half so / 2 and its 6

but

6 + 12 is 18

there no answer 18

so i am confused

[dark sparrow]

as am i, since i am completely unable to follow your work

the language barrier does not help this

[slate pewter]

one moment i will translate it

[dark sparrow]

i was referring to what you wrote here. it's in english, but it's hard to understand.

[slate pewter]

your right my english is not perfect

hmmm

what ever forget i dont care i need to go

thanks you for evry thing

see ya

[upper karma]

@slate pewter, I recommend you to use the @somber coyote bot!

[somber coyote]

Ita:

[upper karma]

I dont know what you are trying to say, but it can help you a little bit

[slate pewter]

ْ]

XD

[vernal void]

[plucky marlin]

[dreamy hollow]

Given two parallel segments with same x-intercept (they share same line equation), how to tell if and how much they overlap? Thanks

[devout shell]

They infinitely overlap each other?

[dreamy hollow]

Maybe overlap's not the right term? Basically the "points" on the line the segments "share"

[devout shell]

Share all points is what it seems to be

Like same x intercept and they are parallel, so I at least think they have to be the same line

[dreamy hollow]

Oh, I should have described the two segments as "collinear" it appears, but need to know if either segments' end points "includes" the others (overlap i guess)

[dreamy hollow]

looks like this may have the answer: https://stackoverflow.com/questions/22456517/algorithm-for-finding-the-segment-overlapping-two-collinear-segments

[upper karma]

Anyone know of a Geometry book?

That goes from a to z

[hallow smelt]

Trying to prove that Sin(x) / Sin(x)+Cos(x) = Tan(x) / 1 + Tan(x)

[gritty siren]

$\frac{\sin(x)}{\sin(x)+\cos(x)} =\frac{\tan(x)}{1+\tan(x)}$

[somber coyote]

Tuong:

[gritty siren]

this?

[hallow smelt]

ya

[gritty siren]

just divide numerator and denominator by cos(x)

[dark sparrow]

also use parentheses next time

[dire rampart]

[cold gorge]

why when trying to get the value of the hypotenuse we do a^2 - B^2 = c^2 but the actual thing is a^2 + B^2 = c^2

[upper karma]

what @cold gorge

The hypotenuse (c) is $c = \sqrt{a^2+b^2}$

[somber coyote]

rudy:

[dreamy hollow]

How would I calculate the minimum rotation (2D) needed to make two vectors parallel? E.g v(0, 1) and v(0, -1) are parallel and the right answer would be 0, not PI. Thanks

[real mulch]

you could use their dot product

arccos(dot_product) is the angle between the two vectors

[dreamy hollow]

Thanks, but doesn't that give Pi for the example vectors I gave? The maximum value for my reqs would be 1/2 Pi, since if an angle > 1/2 Pi could be measure between two vectors, parallelity could be achieved simply by rotating in the opposite direction

[upper karma]

Hi

https://cdn.discordapp.com/attachments/479475790069628949/569658651682406400/test.jpeg

Would <Z be 40 or 70?

OH

<Z would be 30+40 = 70?

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

[vagrant elk]

have you tried using t r i a n g l es

[upper karma]

I am not sure, what they're asking for here, as I have been working with radian and degree measurements up to this point.

[vagrant elk]

doesn't matter if it's degrees or radians, you get the same value (so long as you're consistent and using the thing correctly

[upper karma]

TY

@vagrant elk u mind checking mine out

just in case

uwu

[vagrant elk]

uwu

Z should be 70 i'm p sure (my arithmetic is shit though lmfao)

[upper karma]

uwu kk

ily

[oak minnow]

someone help me w this, how do i find

and how do i label the the parallelogram with abcd

[devout shell]

I need to second to refresh on how this is done lol

[devout shell]

need to look up theroem for this lol, will be back in a minute or so

[upper karma]

Can anyone help out w/ X?

😕

[devout shell]

I really need to review this quickly lol

[upper karma]

$ c^2 = a^2 + b^2 - 2ab\cos ( C) $

[somber coyote]

rudy:

[devout shell]

darn, even this person knows how to latex

I don't know a thing

but working on X for you

[upper karma]

lol thx c:

Law of cosines to find side

C is an angle, a b are sides

[devout shell]

@oak minnow are we allowed to use trig to get the answer? In all fairness I'm pretty sure we can get the answer somehow else, but I can't think how right now

[upper karma]

You can only solve it with trig

:p

Oh

[devout shell]

your problem or his?

[upper karma]

well

Oh

ught let me solve his

[devout shell]

sure law of cosines will let us get the answer pretty easily but was thinking that maybe they don't want us to not use trig

[upper karma]

Its impossible to find it w/ trig

[devout shell]

oh, so now this problem is not too bad then

let's see if I can use LaTeX to write the answer properly lol

$\overrightarrow{AC} = \sqrt{16 + 25 - 40\cos(132)}$

[somber coyote]

⚡Amphy⚡:

[devout shell]

$\overrightarrow{BD} = \sqrt{16 + 25 - 40\cos(48)}$

[somber coyote]

⚡Amphy⚡:

[upper karma]

damn

u rlly out here grinding lol

[devout shell]

had to look up the arrow syntax, than failed at writing it a few times and looked at how other people set up the syntax and followed that form lol

[oak minnow]

@devout shell we need to use the cosine rule to find the sides

sorry for late reply

[devout shell]

ok, well the answers are provided above lol

but should I explain them a bit?

[oak minnow]

if you could please explain how to solve it that would be helpful

[devout shell]

give it a few minutes as I type it all out lol, it's mainly the copy and pasting of LaTeX that will take up my time

[oak minnow]

np

whered the 132 come from

[upper karma]

He meant 48 @oak minnow

[oak minnow]

yes but AC is meant to equal 8.23

and when you put in 132 it gers 8.23

[upper karma]

OH

[oak minnow]

BD is equal to 3.77

AC is equal to 8.23

[devout shell]

Please excuse any lack of rigor here. I am also going to ignore the units because I don't want to type that out all the time.

Givens:
$\overrightarrow{AD} = 4$
$\overrightarrow{BC} = 5$
$\measuredangle{ABC} = 48$

From the figure you can see that the opposite sides are going to be the same lengths.

The fact that we have parallel relations is important as we can then find $\measuredangle{ADC}$:

$\measuredangle{ADC} = 180 - 48 = 132$

Now that we have that we can use the law of cosines:

$ c^2 = a^2 + b^2 - 2ab\cos ( C) $

I will find $\overrightarrow{AC}$ and let you find the other diagonal.
We take $\overrightarrow{AC}$ as c, and then a and b will be $\overrightarrow{AB}$ and $\overrightarrow{BC}$. Order won't matter in this case.

Your angle will be angle opposite of $\overrightarrow{AC}$, which is either $\measuredangle{ADC}$ or $\measuredangle{ABC}$, both or which are equal to 132.

Now plug it all into the Law of Cosines to obtain:

$\overrightarrow{AC} = \sqrt{16 + 25 - 40\cos(132)}$

[somber coyote]

⚡Amphy⚡:

[devout shell]

clarification, to make it clear, it won't matter which of the other two legs you take to be a or b. c is the important one that you have to choose correctly

[oak minnow]

d ok

[devout shell]

As I had initially stated,I wasn't being rigorous, so the part that maybe be hard to see is how you get the other values for the angles, in which case you will have to look at the THMs provided in your text book pertaining to parallelograms

[upper karma]

https://cdn.discordapp.com/attachments/326138757474680852/569679952899735601/image0-10.png

Can anyone do X 😦

[devout shell]

Pretty sure you realize that you're missing 90 degrees, but to I have to go back and refresh on my geometry THMs, I know those perpendicular diagonals are important

[upper karma]

I just- I don't understand. Is the answer for X 110?

I don't know tho

:C

[devout shell]

I'm trying to figure it out as fast as you are lol

[upper karma]

can...can...i tag @spark stag

[devout shell]

I don't know who that person is lol

[vestal crow]

Have you learnt about cyclic quadrilaterals?

[upper karma]

No @vestal crow but I like learning things 😉

Namington is a happy go lucky guy @devout shell

[devout shell]

I also want to say that it's 110 as well lol

[vestal crow]

Hmm

Well I don't know how to LaTeX

But I'll try my best

[devout shell]

something to do with similar triangles?

[vestal crow]

There are most likely more simpler ways of doing this

But what I found is that the angles in the segments(the 40 and 50 degree are equal) which says that the quadrliateral is cyclic

Which means the other two angles in the other segment are also equal

And yes

It can be done with similar triangles

I just realized

[upper karma]

I assumed that myself

[devout shell]

that's what I'm thinking seeing as how the two upper triangles have the same angle measures, making them similar by the Angle Angle similarity postulate, so (excuse the worst logic ever) that would force the other two triangles to be similar so they must also have the same angle measures

[upper karma]

but

I assumed that but it was just

an assumption

:c

so I didn't think it'd be wise to solve that way

LOL @devout shell yeah

basically that too

seemed like bad logic imo but if it works

it works

;c

[devout shell]

you can also make proportionality relations between line segments I'm guessing which is a more rigorous way to show that they are indeed similar

but that would take a bit to get the relations correct

just name the middle point A or some other un used letter and work from there

[upper karma]

or just guess tbh

yolo

[devout shell]

I mean the intuition is there (albeit everyone is cringing at it right now)

you can definitely show from similar triangle relations that the left and right triangle will have the same angle measures (eyeballing it also tells you the same thing lol)

[upper karma]

im in a 3rd world country @devout shell

We never took that in class

albeit...I know how to do it

due to the

trigonometric

relations

[devout shell]

ok lol, so I don't need to sit down and work out the relations myself

[upper karma]

no bro

u gotta

cus i forgot

so get like 2 or 3 relations

going

pls

LOL

[devout shell]

googles how to draw pictures in LaTeX, lol that would take too long, but I'm working on a paper right now and was just doing this a break, how much time do you have left before you need to turn it in? I can work out some relations later on tonight, or someone else can pick it up from here if they wish

[upper karma]

oh its not for hw :p

this is for fun

[devout shell]

I am working on a paper lol

so I have to finish it...

[upper karma]

whats it aboot

[devout shell]

so that means I can't help you right at this moment, so tell you what, how about you try googling similar triangle stuff and see how far you get

if you need to see why the top and bottom triangles are similar, look up the "angle, angle similarity postulate"

[dark sparrow]

can you write out your answers somewhere separate

bc i can't understand which parts of what you wrote pertain to what problems

[devout shell]

I guess

still confused with what the answer should be because I couldn't read the original question in the photo

yea, so you would say that the ratio of the price to area, in other words the price per unit area, is better with the big rug as opposed to the small one

I mean I'm assuming you took the price of each rug and divided it by it's area and found that the big one had a better price per unit area ratio

[somber coyote]

Deddy:

[dark sparrow]

don't mind it

the bot reacted to the dollars

[devout shell]

lol LaTeX goodness

yes, we are saying that the price per unit area of carpet is better when you buy the big rug

[devout shell]

huh?

I didn't check any math, I assumed you did it correctly lol

[green coral]

uhh can I get help for this ?

[dark sparrow]

oh god, bad tex

[green coral]

$prove α=2β, sinβ=\frac{1}{\sqrt{5}} and cosα=\frac{3}{5}$ and α,β are acute

[somber coyote]

Me:

[dark sparrow]

...wait

what

[devout shell]

use the definitions of sin(x) and cos(x)?

[dark sparrow]

you're asked to prove what?

[green coral]

prove α=2β

[dark sparrow]

prove α = 2β given sin(β) = 1/sqrt(5), cos(α) = 3/5 and α, β acute?

[green coral]

yea

haha I'm terrible at using Tex

[upper karma]

$prove, α = 2β,,sinβ=\frac{1}{\sqrt{5}}, and \cos{α}=\frac{3}{5}$ and α,β are acute

[dark sparrow]

no

[devout shell]

well you can start by drawing a triangle that corresponds to $\cos{\alpha} = \dfrac{3}{5}$

[somber coyote]

⚡Amphy⚡:

[green coral]

I drew my two triangles already

[somber coyote]

rudy:

[upper karma]

Looks better now.

[devout shell]

nice

[dark sparrow]

Prove $\alpha = 2\beta$ given $\sin(\beta) = \frac{1}{\sqrt{5}}, \cos(\alpha) = \frac35$ and $0 < \alpha, \beta < \pi/2$

[upper karma]

😐 ann always flexing but ok

[somber coyote]

Ann:

[dark sparrow]

there we go

[devout shell]

how was this on light theme

[dark sparrow]

it might help to compute cos(2β).

and use the fact that cos is one-to-one on [0, π]

[green coral]

uhh haven't done compute cos..

[dark sparrow]

what

[green coral]

nvm ahah

[dark sparrow]

okay if you don't like the word "compute" you can say "find" instead

[green coral]

oh ok

[dark sparrow]

are you able to do that

[green coral]

gimme a minute

oh wait

I got it now

if you find 2β

you can equate the cos on both sides?

or something?

[dark sparrow]

you don't find 2β, you find cos(2β)

[green coral]

yea i mean cos2β

[dark sparrow]

then say what you mean

anyway, after that you make sure it is equal to cos(α)

[green coral]

cos(2β) = 3/5 , cos(α)=3/5

[dark sparrow]

and then, from cos(α) = cos(2β) and the fact that cos is one-to-one on [0, π], you deduce that α = 2β

[green coral]

yea

can I ask more questions?

[dark sparrow]

of course you can

[green coral]

I have these two

Also just 11b

[dark sparrow]

what the hell is "Using the t results" supposed to mean

[supple abyss]

it sounds like there's an equation underneath it

[green coral]

I honestly don't even know haha

That's why I am stuck on 11

[upper karma]

@devout shell LOL I also wonder how it was light theme

[devout shell]

I liked it actually

[upper karma]

anyone know why BC subtends equal angle at P and Q?

[upper karma]

@green coral for q4 try doing cos(2x+3x)=cos(90)=0, where x=18 and then solving for sin(x)

nevermind that doesnt work too well

[upper karma]

<@&286206848099549185> could someone help me with the my question?

thanks!

[real mulch]

Thales's theorem @upper karma

BP and CQ are both altitudes, therefore BPC = 90° and CQB = 90° by the definition of altitude. It follows from those 90° angles that P and Q are on the circle whose diameter is BC (the inverse of Thales's theorem), thus BCPQ is a cyclic quadrilateral. All four points are on a circle.

And QPB = QCB because of the inscribed angle theorem. Both belong to QB chord.

And OQA + OPA = 180, thus we must have QOP + QAP = 180, so opposite angles sum up to 180°, which means that QOPA is a cyclic quadrilateral.

And we can use the inscribed angle theorem again, so QAO = QPO because both angles belong to chord QO in the circle QOPA.

By the transitive property of equality of real numbers, if a1 = a2 and a2 = a3, then a1 = a3.

So QAR = QCR, which means (by the inverse of the inscribed angle theorem) thay QACR points lie on a circle.

By the inscribed angle theorem again, this means that CRA = CQA (both belong to chord CA), and CQA = 90°, which means that CRA = 90°, therefore AR is perpendicular to CB. QED.

[upper karma]

Oh shoot it took me 5 minutes to get it finally but ur answer was really clear, thanks!

so many cyclic quadrilaterals is confusing my brain alot

[hallow smelt]

i've got a trig question, but it's more of an algebra question

is there a formula for when we are multiplying something by the conjugate and it has 3 terms instead of two

Can you even foil three terms?

[upper karma]

yes

you put parentheses around a pair of expressions to turn 3 terms into 2

what they did here was do:
[(1+cos x) - (sin x)] * [(1+cos x) + (sin x)]

[hallow smelt]

I didn't know you could do that. does it matter what you grouped? also is there a name for that

[formal harness]

Just a general question - I was wondering what trig I should know for calculus - do I need to know all the basic identities and their derivations?

[upper karma]

yes

also should know stuff like product-to-sum

[formal harness]

k thanks - was wondering If I could get off easy

[upper karma]

$\cos mx \cos nx = \frac{1}{2} \br{ \cos((m+n)x) + \cos((m-n)x)}$

stuff like this

oop

[formal harness]

thanks a lot

[oak minnow]

https://cdn.discordapp.com/attachments/547942120367849512/570064987222179856/Screenshot_2019-04-23_at_11.53.53.png

need help with 2b

[plucky marlin]

use sine rule to get BC and then the area is $\frac12 \cdot 7 \cdot BC \cdot \sin(100)$

[somber coyote]

soap:

[oak minnow]

thnks

[plucky marlin]

[oak minnow]

ahh one more thing lol

how do i find BC

[plucky marlin]

sine rule

t!wiki sine rule

[loud cedar]

📖 | ** https://en.wikipedia.org/wiki/Law_of_sines **

[oak minnow]

yeah i dont get what they have said, too complex

[plucky marlin]

if that's your triangle then

[somber coyote]

soap:

[plucky marlin]

u can prove it fairly easily

[oak minnow]

so to find BC need a and c

?

[plucky marlin]

i thonk i made a boo boo

[oak minnow]

h

[plucky marlin]

use cosine rule

[oak minnow]

t!wiki cosine rule

[loud cedar]

📖 | ** https://en.wikipedia.org/wiki/Law_of_cosines **

[plucky marlin]

$7^2 + BC^2 - 2\cdot 7 \cdot BC \cos(100) = 9^2 $

[somber coyote]

soap:

[oak minnow]

yeah thats the one

please tell how to find BC

[plucky marlin]

i just gave you the equation ?

[oak minnow]

k but what is BC

[dark sparrow]

this is a quadratic equation in BC

do you know how to solve quadratic equations?

[oak minnow]

yeah mostly

[dark sparrow]

so what's stopping you from solving this equation for BC, then?

[oak minnow]

maybe because i dont know how to ?

[dark sparrow]

you just said you know how to solve quadratic equations

[oak minnow]

this one

[dark sparrow]

it serves no purpose to lie about knowing what you can and can't do

[dark sparrow]

$BC^2 + (-14\cos(100))BC + (7^2 - 9^2) = 0$

[somber coyote]

Ann:

[fleet lake]

$x=\frac{-b\pm
D}{2a}$

[somber coyote]

Krishna_2492003:

[dark sparrow]

@fleet lake i appreciate your effort, but i'd prefer if i handled this on my own

@oak minnow are you unable to solve this equation, using the quadratic formula you just sent - assuming you know its context?

$BC^2 + (-14\cos(100))BC + (7^2 - 9^2) = 0$

[somber coyote]

Ann:

[oak minnow]

ill be back later, need to do my bio

[dark sparrow]

...as you say

[upper karma]

Did they ever come back?

[winged saddle]

🅱

[dark sparrow]

@upper karma evidently, they did not

[woeful mural]

Someone plz help

I need to solve this for tmr mourning

Been trying to do it all day

Only have 3 more boys left

Hours*

[devout shell]

I’ll have time later if no one responds in a few hours

[woeful mural]

Cool

How do you even try to approach this question?

It’s so complicated

Solved another question! Completely different just want to know if it is right

[devout shell]

I’ll get to that later as well

[woeful mural]

Ok thanks

Want me to dm it to you ?

@devout shell

[devout shell]

I could later, have school still so will get to it when I can

[woeful mural]

Ok thanks

[upper karma]

Hi guys. This is a question from a recent pracrice test I had. I wanted to know if this would be 90-45=45?

Close up of the circle.

The question goes: The angle BAC is incribed in the circle, and BC is the diameter. What is the measure of ABC?

oh wair

angles subtended by the diameter are 90 rite

[devout shell]

Triangle whose hypotenuse is a diameter is a right triangle, correct

[upper karma]

@devout shell ur a high level cutesickle

which is like a popsicle

but made out of cuteness

🤔

yeah

[idle portal]

:o

[woeful mural]

How do I solve the first 2 questions

[harsh urchin]

How do i set up a diagram for a question like this? like do i split the right angle triangle in half? and put some sides to represent the time and then convert the time into distance? https://i.imgur.com/B9z6LE9.png

[rain charm]

@harsh urchin |/imagine the straight line goes from the guy to the plane at the start. Then the slanted line goes from the same guy to the plane 2 minutes later

[zealous egret]

Hi, could someone help me out with this? We are meant to use projections to prove this

[cloud meteor]

hmmmm

[zealous egret]

I've been trying to understand it using this, but I don't fully understand

https://math.stackexchange.com/questions/1109142/proving-that-the-dot-product-is-distributive

[cloud meteor]

can you use rectangles?

@zealous egret

[zealous egret]

just projections

and I mean I'm just revising for an exam on the 29th but I thought it's good for my understanding

[upper karma]

so first, what is the relation between projections and the dot product?

[cloud meteor]

because if you can use rectangles it's the easiest thing ever xD

[zealous egret]

well, I understand that the vector projection of suppose vector a onto b = dot product of a and b hat times the b hat

[zealous egret]

Sorry idk how to type maths here

[upper karma]

project onto a instead

project both b+c, b and c onto a

[cloud meteor]

I don't understand what you mean by projections

like geometrical projections onto a plane?

[upper karma]

on a line

orthogonal projection on a line

[cloud meteor]

a

[upper karma]

the line generated by a

[cloud meteor]

that makes sense

[zealous egret]

sigh my Maths is so weak idk how to word it

xD

[upper karma]

can you draw it at least

[zealous egret]

yeah ofc

[upper karma]

draw a line a

draw vectors b and c

which are projected on the line generated by a

and when b and c are added they are "placed one after the other"

it should be clear from the drawing

[cloud meteor]

I can think of a million ways except projection like

[zealous egret]

so this is the diagram I've drawn

[upper karma]

right

[zealous egret]

on paper

[upper karma]

now relate B_A, C_A and (B+C)_A to the dot product

[zealous egret]

Right so am I meant to think of the magnitude of the projections or the vector projections

[upper karma]

use that (B+C)_A = B_A + C_A to deduce the statement for the dot product

[cloud meteor]

Ooo the triangle rule

vector b + vector c = vector (b+c)

[upper karma]

what is ||B_A|| in terms of the dot product?