geometry-and-trigonometry

[upper karma]

@signal hornet Can't tell what you are asking. But the second picture you sent is correct.

[signal hornet]

weird how WA is wrong then

[noble heath]

Wolfram alpha interpreted it as asking to convert (-2, 3pi/4) from cartesian to polar

[wooden swift]

I’m done with trig and geometry but forgot how to find angle A and C, can someone help

[signal hornet]

Good thing you managed to get the air pods in the shot, lol

[signal hornet]

@noble heath any idea how I can get WA to draw it properly?

[noble heath]

,w convert (-2, 3pi/4) from polar to rectangular

ok then

[somber coyote]

Display results online and refine query

Requested by Icy001#6642

[noble heath]

Jeez I can't figure out how to do it

Therefore, it's impossible

[signal hornet]

I've not been having luck with WA lately

it's not that smart

[signal hornet]

how can I conver the polar coord (3, pi/2) to rectangular?

[weary drift]

polar $(r,\theta)$ to rectangular $(x,y)$:\\$x=r\cos(\theta),y=r\sin(\theta)$

[somber coyote]

RokettoJanpu:

[signal hornet]

oh, sorry

I meant to ask which function I should use on the ti nspire

there is 2 of them

and I can't get either of them to work

I know how to do it by hand, I just want to check my work on the calc

[weary drift]

,calc cos(pi)

[somber coyote]

Result:

-1

[signal hornet]

Polar to Rectangular, x and Polar to Rectangular, y

why not just say zero WA

[weary drift]

imprecision ig

[signal hornet]

why does the calc say pi instead of just 0?

[dark sparrow]

bc you're converting the other way around

rect to polar

[signal hornet]

The input is rect though

and I want polar

It should just say [1 0]

[dark sparrow]

(-1, 0) in rectangular does not correspond to [1, 0] in polar.

[1, 0] in polar is (+1, 0) in rectangular.

[signal hornet]

what?

[dark sparrow]

imagine trusting symbolab

(-1, 0) is literally on the boundary of quadrants 2 and 3, yet shitterlab fails to recognize that

[signal hornet]

wait, symbolab is actually wrong

[dark sparrow]

yes it is

[signal hornet]

who can I trust?

WA?

my calc?

[dark sparrow]

WA more so than symbolab by a mile

your calculator is correct in its conversion

[signal hornet]

Let me show you my method on paper, because that also yielded (1, 0)

[dark sparrow]

just convert [1, 0] from polar back to rect!!! you won't get (-1, 0)!!! does that not make any sense?????

[signal hornet]

I believe you, that is not the issue

[dark sparrow]

ok show what you did on paper so i can see where you fucked up

[signal hornet]

Let me double check first

in case It is some basic fuck up

[dark sparrow]

θ = tan^-1(0/-1)

ding ding ding

tan(θ) = 0 sure but this doesn't mean θ itself is 0

[signal hornet]

i just typed arctan(0) into my calc and got 0

so theta should be 0

But at the same time I know that tan(pi) = 0

so how do I know which thetangle (either pi or 0) is the right one?

[silent plank]

from the position of (-1,0)

[signal hornet]

Yes, but is that the way I am supposed to do it?

[silent plank]

yes

[signal hornet]

Alright then

I am still in awe that symbolab is no good

[dark sparrow]

@signal hornet NO!!! ARCTAN IS NOT THE TRUE INVERSE OF TAN!!!!!!!!!!!!!!

[noble heath]

$\arctan$ is the true inverse of $\tan|_{(-\pi/2,\pi/2)}$

[somber coyote]

Icy001:

[signal hornet]

@dark sparrow well that complicates things

[upper karma]

https://cdn.discordapp.com/attachments/649919947937677312/653862009053642752/image0.jpg

help

[open zealot]

I'm studying for my GRE exam and running through some sample GRE trig questions. I wanted confirmation on my logic of this question, could anyone provide it?

]

Given the naive idea that the triangle is equilateral, and the perimiter being 6 implies the side lengths are 2. Therefore the radius is 2, in which case the circumference of the circle is greater than 12. However I do not feel it is safe to assume the triangle is equilateral.

what is the correct way to approach this?

[inner sandal]

OR and OS are equal since they are the radius, which implies ROS is at least isosceles

Use your knowledge on isosceles triangles (and 180 degrees in triangle) to prove it's equilateral

Once you have proved it is equilateral you can then go on to prove the circumference is greater than 12

[open zealot]

If two sides of a triangle are congruent, so are the angles opposite of those sides. Therefore angle RSO and angle ORS must be congruent. They must be congruent and add up to 120 degrees, therefore they are both 60 degrees, therefore the triangle is equilateral. Is this the right logic?

[inner sandal]

hmm

[...] must be congruent and add up to 120 degrees

being congruent doesnt mean they add up to 120

you know that say the angles are R, O and S

R + O + S = 180

R = S = x
O = 60

2x + 60 = 180 => x = 60

R = S = x
this is from the fact they are congruent

[weak shoal]

Lmao, if you just calculate the length RS in terms of the radius, you will realize that it’s the same as the radius

So it’s equilateral

Anyways, you also know, for a fact, that it is isosceles so RSO = ORS

[open zealot]

That makes sense. I understand the initial logic. I am not sure what you meant by calculating the length of RS in terms of the radius, and I'm curious about that. Could you expand?

[inner sandal]

just calculate RS in terms of r

why

there is no need to do that

[weak shoal]

? Why not? It’s a valid way of getting to the solution.

Ah yes, read my comment as well

The one below that remark

[open zealot]

angle RSO equals angle ORS, yes

[weak shoal]

Ye

And their sum is 120

[open zealot]

right. 2x = 120, therefore x = 60.

it makes sense now. I didn't note that since OR and OS are both the radius they are equal, my mind slipped. the rest makes sense.

[weak shoal]

Mmmh

Multiple ways to get to the solution, buddy

[open zealot]

I appreciate the explanations 🙂

There was a similarly naive question which I didn't feel like I approached correctly. Let me see if I can recreate it. Sorry if these questions are super simple, I haven't touched trig in too long.

Something along this lines, given that PS = SR, what can be said about the relationship between angle X and angle Y?

Again, I have little confidence in my observations but I think angle QRS would be congruent to angle QPS.

[inner sandal]

that is not true

[open zealot]

What observations are true about the triangle?

[hybrid jasper]

let’s say I was given theta = -60 to find all six trig functions so I would think 360 - (-60) = 420 and do 420 - 360 to get the reference angle for the triangle right

or does that only work for that angle

[open zealot]

I'm not sure.

[inner sandal]

@hybrid jasper wdym all six trig functions

what are you tryna do

[hybrid jasper]

sin cos tan csc sec cot

[inner sandal]

so like sin(-60), cos(-60), tan(-60) ...

[hybrid jasper]

yea

[inner sandal]

ok well first

sin(-x) = -sin(x)

cos(-x) = cos(x)

tan(-x) = -tan(x)

[hybrid jasper]

I got a whole problem wrong because I thought 360 - 60 = 300 and 270 + 30 = 300 and used 30 for ref angle

[inner sandal]

and you should know 60 degrees right

360 - theta doesnt work for sine

that is for cosine

[hybrid jasper]

and tan?

or no

[inner sandal]

nope

sin(180-x) = sin(x)
cos(360-x) = cos(x)
tan(x+180) = tan(x)

sin(360-x) = sin(180-x + 180) =\= sin(x)

[hybrid jasper]

so whats the best way to find a ref angle

[inner sandal]

with the things i sent

[hybrid jasper]

wtf

[silent plank]

do you know the definition of the reference angle?

[inner sandal]

sin(x)=-sin(-x)=sin(180-x)
cos(x)=cos(-x)=cos(360-x)
tan(x)=-tan(-x)=tan(x+180)

[hybrid jasper]

not sure how to explain it

but I know it visually

[silent plank]

smallest angle between the arm and the x-axis?

[hybrid jasper]

yeah I guess that works

[silent plank]

so the reference angle for -60° would just be 60°

find the trig ratio for that and adjust signs appropriate ly

[hybrid jasper]

yeah I get that but like in my head I'd think 360 - 60

because from that ref angle I have to construct a special right triangle and evaluate values from there

[silent plank]

the reference angle should be between [0,90°]

[hybrid jasper]

yeah

so I thought 270 + x = 300

x = 30

and used 30

I get why its 60 but I'm not visually seeing it

[silent plank]

which you can get from 0- (-60°)

[hybrid jasper]

well yeah

[silent plank]

why are you using 270?
that's the y-axis

-60° is 60° clockwise from the positive x-axis

[hybrid jasper]

yeah

[silent plank]

so trivially, the reference angle is 60°

[hybrid jasper]

so let's say I had -120

I know thats in the third quadrant

360 - 120 = 240

and 180 + x = 240

so 60 would be that ref angle too

that's how I think for ref angles so it made sense using 270

[silent plank]

yes

not the part about 270

[hybrid jasper]

so I can't use 270 or 90

[silent plank]

you COULD do
180 + (-120)

[hybrid jasper]

yeah that makes sense too

[silent plank]

equating those 2 will get you
360 - 120 = x + 180
x = 180 - 120

you should know the ratio for that angle,
then adjust the sign for the quadrant

[hybrid jasper]

mhm

[dire obsidian]

What is the value of cos13pi

[simple scaffold]

-1

[dire obsidian]

I am told the point is-

ok

but how

ik its (-1,0)

but where

do

u get that

[simple scaffold]

That's how math works

[dire obsidian]

er mkay.

[simple scaffold]

I really don't know why that is what it is

We learnt it in 12th grade, but that's been eons

[vast dune]

cos 13pi = cos pi

which if u look at the unit circle end ups being at the point c(-1, 0)

[dire obsidian]

ok

what about co8pi

cos8pi

how come thats 1

@vast dune

[vast dune]

use equal angle indenitity

[dire obsidian]

didnt leanr that

learn*

[vast dune]

cos (8pi - 4(2pi))

equals to cos 0

[dire obsidian]

lol yeah

i havent learned that yet

im in algbera 2

algebra 2

[vast dune]

its just logic

[dire obsidian]

this is our intro to trig + precalc

it doesnt make sense to me

[cosmic rampart]

Okay

[vast dune]

400 deg angle is same as 40 deg angle

[cosmic rampart]

The angle goes around the circle, until it reaches a full 360 degrees or 2pi in radians

[dire obsidian]

uhuh

[cosmic rampart]

And then the cycle repeats, but the angle doesn’t

[dire obsidian]

mhm

so it lands back at

(1,0)

[cosmic rampart]

Yes

[dire obsidian]

ok

so what is it with the cos13pi then

i have to do

3 cycles?

• 1 extra

so thats in quad 2

or

-1,0?

[vast dune]

just do minus 2pi until u cant anymore

with cant anymore I mean before u get a negative angle

[cosmic rampart]

Yes

[vast dune]

cos 13pi = cos pi

[dire obsidian]

so if its cos15pi would it be (0,-1)

[vast dune]

no

that would equal to cos pi

[vast dune]

which has c(-1, 0)

cos can never be 0 when u get a full angle like that

u need a fraction for that

[cosmic rampart]

y=1 at pi+k2pi

[dire obsidian]

er wtf

[cosmic rampart]

Where k is all integers

[dire obsidian]

lol

idek

what

the hell that is

[cosmic rampart]

So look at the graph

[dire obsidian]

i just started unit circle yesterday

[cosmic rampart]

It can be hard to grasp

[dire obsidian]

yeah

[cosmic rampart]

It was for me at least

[dire obsidian]

especially

for a sophomore

lol

[cosmic rampart]

I recommend Kahn academy

[vast dune]

take a piece of paper and draw the unit circle with the angles ur looking for

than look at the cords

[dire obsidian]

his vids are long

id rather watch organic chem tutor

uhuh

[cosmic rampart]

Wait

Can you ask a specific question

That way I know what you want to know

And could probably help you better

[dire obsidian]

i pretty much get it now

i can just draw a graph

and count

cause the quadrantal angles are represented by #/2

so (1,0) is 2pi or 4pi/2

(0,1) is pi/2

(-1,0) is pi or 2pi/2

(0,-1) is 3pi/2

[vast dune]

2pi/2 is the same as pi tho

[dire obsidian]

yes

lol

[vast dune]

also u should write it down the full way probs

pi + k pi with pi element of integers

etc

[gritty sail]

y=rsintheta

x=rcostheta

Substitute

[dire obsidian]

how is this located in the 3rd quadrant

[weak shoal]

What’s $-\frac{8\pi}{3}$?

[somber coyote]

Abhijeet Vats:

[weak shoal]

In degrees?

[dire obsidian]

no

exact value

hats the answer key

thats*

[weak shoal]

@dire obsidian Your question was: How is this located in the 3rd quadrant?

What exactly is ‘this’? Also, I’m asking you to compute the angle in degrees.

[gritty sail]

Isn't polar coord just represent as r = something

You get r = 4csc(x)

Uh by isolating r

Bruh

Isolate r

By itself

[somber coyote]

Jason_Bjorn:

[dire obsidian]

can someone help me visualize why 4pi/5 and -14pi/5 are not coterminal?

[fossil lotus]

Why do you think they should be coterminal?

@dire obsidian

[dire obsidian]

They would share the same quadrant

[fossil lotus]

What smaller angle is -14pi/5 equivalent to

[silent plank]

how are you doing the conversion for -14pi/5

[dire obsidian]

would -14pi/5 be equivalent to

-12pi/5

and -16pi/5?

Adding 2pi and subtracting 2pi?

[fossil lotus]

That's adding and subtracting 2pi/5

[dire obsidian]

o

do i have to subtract by

2pi

and make it a common denominato

so thats

10pi/5?

[fossil lotus]

How did you get that?

[dire obsidian]

cause

-14pi/5 -2pi

wouldnt you have to multiply

-2pi by 5

so that it has the same denominator

[fossil lotus]

Yeah, what does that give you

(I would add in this case rather than subtract, since it's a negative angle)

[dire obsidian]

so

-4pi/5

?

[fossil lotus]

Yes

[dire obsidian]

o

and then the other is

4pi/5

so tahts not -4pi/5

got it

[fossil lotus]

Yep

[dire obsidian]

okay i get it

tysm!

my teacher did not make the 2pi rule clear

so i had to watch vids

[fossil lotus]

Np

[signal hornet]

does this looks right?

[inner sandal]

@signal hornet yep

[signal hornet]

thanks

I hate how there is no way to check my answers on this stuff

[inner sandal]

https://www.desmos.com

@signal hornet

[signal hornet]

that graphs a parabola

[inner sandal]

put in polar equation and put in cartesian equation and they should represent the same curve

[signal hornet]

ahh

thanks

[inner sandal]

np

[signal hornet]

I thought I could multiply both sides by r

but then the left side looks horrible

any ideas?

[cinder goblet]

You square root

I would assume

The only way you can solve for a square variable is square root

@signal hornet

Try it. Cos the angle first. And the square root the answer

[signal hornet]

what do you mean cos the angle?

I am not given theta

I just have to rewrite the equation

[cinder goblet]

Then r=

(Cos 0•)^1/2

That’s the same as square rooting

If you’re solving for r. All you do is put a square root on both sides

[signal hornet]

This is what I did

[cinder goblet]

You don’t r times r

[signal hornet]

well let's see your solution then

[cinder goblet]

The only way you can get a square variable on its on is y square rooting

[signal hornet]

no, you need to rewrite the entire equation with only x and y

no trig, no theta, no r

[cinder goblet]

Can you send pic of the question?

[signal hornet]

sure

[cinder goblet]

So theta is y?

And r is x?

[signal hornet]

no

theta is the angle

[cinder goblet]

Omg lol. Just send the whole question

You don’t need to clip it out

[signal hornet]

that is it

[tawdry pivot]

@signal hornet what you did looks good

[signal hornet]

ok, thanks

[tawdry pivot]

tho (x,y)=(0,0) is a problem but I think we just ignore it

[signal hornet]

yeah, no way to get x and y

that would be way too much work

well in this caase, isolating y

[tawdry pivot]

no, I mean that your final equation permits the point (x,y)=(0,0)

even though it's kinda weird for the polar equation

[signal hornet]

huh, not sure

[tawdry pivot]

theta is kinda undefined at the point (x,y)=(0,0). see https://en.wikipedia.org/wiki/Polar_coordinate_system#Converting_between_polar_and_Cartesian_coordinates

[keen rose]

https://media.discordapp.net/attachments/536995777981972491/654256567172202496/353305252633772052.png?width=874&height=656

ok lads

i know the best method to work this out but i keep getting the wrong answer

im supposed to make a smaller circle that passes through P A Q and R

and then solve the two circles simulatneously to find P and Q

the only problem is that i cant find the equation of the smaller circle

😦

<@&286206848099549185>

[open sinew]

@tawdry pivot when r is 0, any value of theta is valid

so the working is correct

unless thats not what your point was about

[mystic pebble]

What is n here?

and how do you reach it

maybe i left out some essneitla info, i drew this off memory from my mock gcse i had this mreoning

[open sinew]

bruh

[upper karma]

[wind heart]

I have a question

What do those little circles represent??

Like, for example, on angle D theres 2 half circles next to it

What are those?

[upper karma]

Any idea...?

Not sure how to start this

[wind heart]

Also, when identifying all congruent and corresponding parts,

For example, from this if AB was congruent to EF, could it also be BA = FE ?

Would that be correct?

[wind heart]

<@&286206848099549185>

Or could BA = EF?

[gritty sail]

The circles mean congruence

That's the same side so yeah

@wind heart

[wind heart]

Oh ok cool then

Thanks

And uh

What do the tick marks mean

[gritty sail]

Congruence

@upper karma just prove that the top triangle is equilateral. You know all the angles, including angle D, and all angles in a quad add up to 360

Wait u don't even need that

All angles in a triangle add up to 180

[torpid torrent]

each of two congruent circles passes through the center of the other, as shown. If the area of each circle is 16pi, what is the area of the triangle whose vertices are the centers of the two circles and one of their points of itnersection

pls halp

i am confused

4 is radius

but i got 8 pi as area

and ik that is wrong

[fossil lotus]

How did you get 8pi? @torpid torrent

[torpid torrent]

bh/2

and 4pi*4pi/2 is 8 pi

[fossil lotus]

4pi*4pi/2 would be 8pi^2

But where did you get 4pi?

[torpid torrent]

so is that right

[fossil lotus]

It's not

[torpid torrent]

i need help

[fossil lotus]

Where did you get 4pi?

[hybrid jasper]

could someone explain this

like how would you know it would be a 30-60-90 triangle

and know to use 2a, a, a sqrt3

[fossil lotus]

What's the definition of secant?

@hybrid jasper

[hybrid jasper]

H/A or 1/cos(theta)

[fossil lotus]

Right

So if secant is H/A, and sec x = 2, then H/A = 2

So the hypotenuse is twice the length of the adjacent side

That's where 2a and a come from

[hybrid jasper]

oh

same thing for the other triangle too right

makes sense

[fossil lotus]

Yeah

[torpid torrent]

@fossil lotus

i got 4 pi

from the area being 16pi

and using the area equation

pir^2

[fossil lotus]

So if the formula for area is pi(r)^2

And r is 4pi

Then the area would be pi(4pi)^2

Right?

[torpid torrent]

so its just 4

the radius

ok

[fossil lotus]

Yeah

[torpid torrent]

so area is just 16 then

[fossil lotus]

How so?

[torpid torrent]

if 4 is radius

then 4*4 is 16

oh wait

/2

that means its 8

[fossil lotus]

What formula are you using for area

[torpid torrent]

bh.2

*bh/2

[fossil lotus]

How do you know that the height is 4

[torpid torrent]

because the distance from the center to the outline of the circle

has to be 4

[fossil lotus]

The distance from the center of one of the circles to its perimeter is 4

[torpid torrent]

yes

it mustbe the radius

and that also must be the height

[fossil lotus]

But the line segment that corresponds to the height of this triangle doesn't touch the center of either circle

[torpid torrent]

but the bottom segment base segment is the base

and then

like one center

to the top

would form a right angle

and the other segment could correspond to height

it doesnt have to necessarily be from the top vertex to the bottom segment as an altitude right

[fossil lotus]

Uh

Would it be possible to draw how you're picturing this

And upload it here, so I can see more clearly how you're approaching this

[fossil lotus]

Ah

[torpid torrent]

act as if theyare intersecting in their centers

[fossil lotus]

That's not the height

[torpid torrent]

hmm

[fossil lotus]

That's higher than the height of the triangle

[torpid torrent]

im confused as to how i would get height then

[fossil lotus]

This is the situation, right?

So first of all

What do you know about the triangle

Obviously the bottom side has length r

Can you figure anything out about the other two sides

[torpid torrent]

hm

we have leg and hyptonuse

can we get height

[fossil lotus]

You know the hypotenuse?

[torpid torrent]

well

it would just be r

[fossil lotus]

That's right

So do you know how to find h?

[torpid torrent]

c^2-b^2= a^2

and a^2 would be height

[fossil lotus]

Right

Well

a would be height

But otherwise yes

[torpid torrent]

well yeah

ok thanks

[fossil lotus]

Np

What did you get?

@torpid torrent

[torpid torrent]

sqrt of 32

for height?

[fossil lotus]

What did you have for b and c?

[torpid torrent]

4 and 4

[fossil lotus]

Ah

[torpid torrent]

16 and 16 is 32

so 32 = b^2

[fossil lotus]

The pythagorean theorem only works for right triangles

[torpid torrent]

oh no

so itd be 4 and 16

[fossil lotus]

Yeah

[torpid torrent]

so sqrt of 20

• 4

[fossil lotus]

Look at the formula again

[torpid torrent]

20 = b^2

sqrt(20) = height

[fossil lotus]

What was the formula

The version of the pythagorean theorem you're using

[torpid torrent]

a^2+b^2=c^2

ohh

wait

i added sry

so its just sqrt of 12??????

[fossil lotus]

Yes

[torpid torrent]

*4

/2

2*sqrt of 12

[fossil lotus]

Yes

Usually you would rewrite sqrt(12)

[torpid torrent]

so can u simplify that at all

[fossil lotus]

Usually you want to simplify radicals so that the number under the radical is as small as possible

E.g. if it were sqrt(18) then you would rewrite it as 3sqrt(2)

[torpid torrent]

hm

[fossil lotus]

(since sqrt(18) = sqrt(9*2) = sqrt(9)*sqrt(2) = 3sqrt(2))

[torpid torrent]

so

sqrt of 12 is just

sqrt(3*4)

or sqrt(4)*sqrt(3)

or 2*sqrt(3)

*2

so 4*sqrt(3)

[fossil lotus]

Uh

Your second-to-last step was right

[torpid torrent]

wym'

42sqrt(3)

all divided by 2

oh so justs

2(sqrt)(3)

[fossil lotus]

Oh wait

Sorry I didn't realize you were also calculating the area

I thought you were just simplifying sqrt(12)

[torpid torrent]

so 4*sqrt(3)

[fossil lotus]

Yeah, 4sqrt(3) is right for the area

[torpid torrent]

ok thank u

[fossil lotus]

Np

[torpid torrent]

thats the final answer then

[fossil lotus]

Yep

[upper karma]

Does anyone know how to prove this, this should probably be easy to prove but I'm a little stumbled

[silent plank]

let D be the perpendicular intersection point

what can you tell me about AD and CD?

[upper karma]

AD and CD are equal

[silent plank]

does that give you any ideas?

[upper karma]

AD = 1, CD = 1, AC = sqrt 2

[silent plank]

uh, they are in that ratio but those are not their lengths

what information is needed to find AB?

(AB is the sum of the lengths of what?)

[upper karma]

AD and BD

[silent plank]

and as you said earlier, AD=CD

are you able to find CD and BD (using trig)?

[upper karma]

Yes

So then lets see

sin30=1/sqrt 3

[silent plank]

no

[upper karma]

Wait

sin30=1/x 😅

[silent plank]

still no

the side opposite the 30° is not 1

the side is represented as CD

[upper karma]

But doesn't AD=CD

[silent plank]

yes, but AD isn't known to be 1 either

the idea was
AB = AD + BD
AB = CD + BD

[upper karma]

Yeah I get that.

[silent plank]

so from trig,
sin(30°) = Opp/Hyp = CD/x right?

[upper karma]

Yes

[silent plank]

what's the numerical value of sin(30°)

[upper karma]

1/2

[silent plank]

hence the length of CD is: ?

[upper karma]

1/2 x

[silent plank]

similar idea for BD

[upper karma]

so sin(60) = BO/x

BD*

[silent plank]

sure

[upper karma]

BD = (sqrt3)/2 x

OHH I GET IT NOW

[silent plank]

though you could've used cos(30°)

[upper karma]

THANK YOU SO MUCH

I don't even know how I got stuck on the question but oh well, not my day today haha

[silent plank]

np

[sudden locust]

The relationship between the area of a circle and a square https://youtu.be/OP7-agwP2Co

[weary drift]

Infinitely many lines pass through (a,b)

[upper karma]

^

[idle bloom]

You need a minimum of two points to define a line

[weary drift]

Provide another constraint and we’ll see if there exists a unique line satisfying the constraints

[upper karma]

If you have two sets of coordinates to add on, you could use the formula
y-y1=m(x-x1)

with just coordinates (a,b), you can’t get a gradient out of that

[idle bloom]

Oh well think about why it always passes through

[upper karma]

Was there any values for a and b

[idle bloom]

^

And that actually answers your question

You can pick any a and b right?

That means you have infinitely many slopes

Through one point

So bam

You can't define a line with just one point

Unless you have other restrictions

I know but I mean it answers this question

" If i wanted to find the formula of a line y such that y is linear and always passes through coordinates (a,b), how would i go about doing it? "

It's aight

[upper karma]

I was confused at first in terms of what you were trying to prove

Hmm better phrased

Still infinitely many lines can pass through (-1,0) for all a and b

Wait what, I’m so confused

To summarise: You need two points to define a line

For example, there’s a line that can pass through (-1,0) and (0,0) or (0,1) or (1,0) or even (0.00001, 0)... and so on

[wind heart]

How is 10) ASA and not AAS?

How is 7) not congruent and not AAA

[shell raft]

Shini, can you tell me what congruent means?

https://www.onlinemathlearning.com/congruent-triangles.html#sas

[upper karma]

Didn’t even know there was an ASA, in the Australian curriculum, we are only taught that there is SSS, SAS, AAS, and RHS

[crisp mulch]

@upper karma I think ASA is the same as AAS

I live in Australia too

😅 😅

[upper karma]

Oh nice

[crisp mulch]

NSW

[upper karma]

Same

[crisp mulch]

Graduated highschool?

[upper karma]

Nope

Prelim HSC starting next year

[crisp mulch]

Ah so yr 10

[upper karma]

Rn yes

[crisp mulch]

Are U doing well in schoo

School?

[upper karma]

I'm doing aight

[crisp mulch]

DW about yr 10 it's legitimately a joke

Compared to yr 11 and 12

[upper karma]

I'm doing Extension 1 next year lmfao

[crisp mulch]

Haha are U planning on doing extension

2

For yr 12

I'm in yr 12 rn

[upper karma]

No, not worth it

[crisp mulch]

Why not?

[upper karma]

Add me and take this to dms, probably not best to talk about it in this chat haha

[crisp mulch]

Ahaha alright

@wind heart also it's not congruent cause U need a side with AA

Other wise it's just similar

[upper karma]

how do you factor

its so confusing

[fossil lotus]

Is there a specific problem you're having trouble with?

@upper karma

[upper karma]

no i just dont get decompisition

@fossil lotus

[fossil lotus]

Huh, never heard of factoring by decomposition before

But I'm looking it up and I get how it works

[upper karma]

https://youtu.be/6VPDFUm73uA?t=426

@fossil lotus

idk if its a terrible method though for example how would you factor 2x^2 +5x-12

something that has A as something that isnt x

idk how to explain it lmao

like

2x^2

5x^2

etc

[fossil lotus]

Where the first term isn't just x^2?

[upper karma]

yes

[fossil lotus]

Isn't the example in the video like that?

[upper karma]

yeah

i just dont understand

is their a simpler way to do it

[fossil lotus]

The way I learned to do it is just by trial and error

But I think this way might actually be easier than that if you know how to do it

[upper karma]

ok

[fossil lotus]

Do you want me to explain how you would factor 2x^2+5x-12 by decomposition?

[sacred cradle]

yo how would you graph r=2cos7theta on a polar graph

[limber anchor]

Can anyone explain to me how to solve this problem? I’m supposed to be using the 45-45-90 Triangle Theorem and the Pythagorean theorem but I’m just not sure how to get the answer. Also apologies for the camera quality

[cerulean shuttle]

Pythagorean is the way to go

Add the squares of the side lengths and take the square root

Btw one of the axioms of geometry is that the shortest distance between two points is a straight line, so you can tell that the long dashed path is quicker right off the bat.

[limber anchor]

Oh thank you I didn’t know that

My only problem is that it asks for the distances of each path

Man this takes a minute

Wait how do I get the hypotenuse of the long dashed path?

Well actually how do I get the distance?

[thorny charm]

the fact that the lines youre given dont quite form a triangle doesnt matter
you have horizontal parts and vertical parts you can piece together

[signal hornet]

This took me a bit minute, but I got it. Would you do it the same way?

[dire rampart]

Uh

Honestly

It might be quicker to expand cos and sin using double angle

And dividing appropriately

Your working is really hard to follow btw

[signal hornet]

YOu mean re-write tan as sin/cos?

[dire rampart]

Na

[signal hornet]

It's only one step per line

[dire rampart]

I meant writing cosx as 1-2sin^2(x/2)

And sinx as 2sin(x/2)cos(x/2)

In fact

Doing that makes alot of things cancel

And you get the result almost immediately

[signal hornet]

I can't visualize what you typed

is this a common property? cosx as 1-2sin^2(x/2)

[dire rampart]

Yea that's the double angle formula for cos

[signal hornet]

that for cos2x not cosx

[dire rampart]

$\cos(2\theta)=1-2\sin^2(\theta)$

[somber coyote]

the one n only:

[dire rampart]

Yes but theta = 2 times theta/2

[signal hornet]

I knoow that but there is no point in the initial question where there is cos2x

[dire rampart]

Sure there is

[signal hornet]

oh

[dire rampart]

Take a look at this

[signal hornet]

so you rewrite cosx as cos2*0.5x?

[dire rampart]

Exactly

[signal hornet]

hmm

how did you know to try that?

[dire rampart]

Becuz u have theta/2 on the rhs

So naturally youd want similar terms on the left

[signal hornet]

but why with the numerator, why not do that for the bottom?

[dire rampart]

U do it for the bottom as well

Using sin(2x)=2sin(0.5x)cos(0.5x)

Then stuff cancels

[signal hornet]

ok

I will try that method

like this @dire rampart ?

[dire rampart]

Yea

[signal hornet]

nice, thanks

[dire rampart]

Np

I would be more careful with the layout tho

[signal hornet]

how so?

[dire rampart]

Wouldn't put equal sign on each side

You're kinda assuming what you want to prove

[signal hornet]

what do you mean? I am given the first line

I know they are equal

[dire rampart]

So all of that right hand side isnt necessary

No you're meant to show that they're equal

2 different things

[signal hornet]

also, what do you mean "each side". I only have the equals sign in the middle

not on any sides

[inner sandal]

you're writing it out like an equation

something equals something
(rearranged something) equals something

etc

to prove identities consider one side of what you want to prove, and then use algebra to get the other side of the equation (equivalence, not equality)

[signal hornet]

I think what I am doing is fine for my trig class

the proff never mentioned it being a problem

[inner sandal]

it isnt

[dire rampart]

It's a minor thing tbh

[signal hornet]

How am I supposed to write it?

[dire rampart]

If it works in your class dont worry about it

I'd get rid of all the equal signs except the very last one

[inner sandal]

so you'd just write (1-cosx)/sinx on its own for the first line

[signal hornet]

the first line is the problem statement

you mean my first line?

so the second line?

[inner sandal]

then second line, you'd have = (1-cos(2 *x/2))/sin(2 *x/2)

your first line of work will be the problem yes

you dont start your proof with something thats not the same expression as what you're considering

[signal hornet]

so I just get rid off the equals sign and everything to the right?

that's it?

[inner sandal]

so eventually you'd get to just = tan(x/2) on the last line

[signal hornet]

Oh

[inner sandal]

hence LHS = RHS and identity is true

[signal hornet]

yeah I guess I should have rewritten the left side to tan not the right to sin/cos

[inner sandal]

if you look at your work

you worked from the LHS up until the last step you went from the RHS backwards and met up with your LHS

that's not good

[dire rampart]

so I just get rid off the equals sign and everything to the right?

[inner sandal]

You want to go all the way with one side

[dire rampart]

Yea pretty much

[inner sandal]

yeah and also write equivalent signs on the left

[dire rampart]

And just equate ur final line to tan and you're done

[signal hornet]

like that?

[dire rampart]

Yup

[inner sandal]

yea

[signal hornet]

ok, you could have said it a lot simpler from the start @inner sandal

thanks both of you though

[dire rampart]

Lmao

[inner sandal]

even better: informing the reader you're beginning from the LHS between the first and second line, and then writing = RHS under your current last line

concluding LHS = RHS

[signal hornet]

It's hard to know which double angle formula to pick for cosine

[dire rampart]

Just pick all of them

[signal hornet]

I guess I could try them all

But it'd be nice to get it right the first try

[dire rampart]

Well in that q we just did

We needed to get something that involved tan

And tan has sin on the top

So we picked the sin version for the double angle of cos

Its just about what's best suitable

[signal hornet]

Yes, looking back, it is obvious now

Maybe just doing it more I will learn

[inner sandal]

always look at what you're getting to

if you're ever stuck, you can consider the answer and work backwards to see if you can find what step you're struggling to see

again make sure you are going from one side to the other side for your actual answer

[signal hornet]

yep, I know now

[icy depot]

You can find out the values of angles if we have all 3 sides right?

Which law was it again

Wait I’m dumb it’s the law of cosine

Nvm then

[frosty egret]

https://math.stackexchange.com/questions/3474629/algorithm-to-get-the-list-of-triangles-resulting-from-clipping-a-triangle-agains

Everyone I ask tells me this is some kind of "clever riddle" - in that they have no clue what to do. I'm running out of options. Does anyone know what to do here???

[upper karma]

@frosty egret how do you want to do this?

you say "algorithm" and "list", do you want this coded?

[frosty egret]

it will eventually be code

[upper karma]

do you want it solved "perfectly"? or do you mind slightly inefficient code considering runtime should be extremely tiny anyways

(as long as it's not operating on huge polygons)

[frosty egret]

@upper karma idk why you're considering anything to do with coding.

[upper karma]

because it's a lot easier to quickly code up something works that's not elegant, compared to figuring out the exact method?

just person opinion though; my geometry isn't the strongest 😛

[frosty egret]

@upper karma fair enough. The penultimate goal is to cut out a polygonal 2d hole in 3d terrain and then cut around a corresponding 3d to fill in that hole with this being the less.... search friendly side of the problem.

so any (meaningful, after all floating point has error) overlap or gaping holes that one algorithm produces and the other does not fill will not go unnoticed by pretty much anyone.

[sudden locust]

In this video, I do a step-by-step proof for all the Inscribed Angle Theorem Proofs hope some of you guys will find it valuable! 😃 https://youtu.be/FD50CgAbV-4

[dark sparrow]

why do you always put benedict cumberbatch on your thumbnails

it's honestly gotten quite old

[upper karma]

Why such a complicated reasoning? Easier one to remember is 'angle at centre is twice angle at circumference'

[dark sparrow]

if you want that as a mnemonic then go for it but in math you have to be precise

[upper karma]

Australia's math curriculum is bad

[weak shoal]

Really?

[upper karma]

That's literally how we're taught, short reasoning

[weak shoal]

I've heard it is pretty good from some Australian people in my D&D group.

[upper karma]

Are you american?

[weak shoal]

I'm Singaporean.

[upper karma]

Ah

Well let's see

in Math in the Australian curriculum, you cover a whole lot of things

Unlike america where you have specific classes like 'Algebra 1'

And it's not overly hard until you reach the senior years

Basically, grade 7 to 10 is a joke.

[weak shoal]

Ah well

To be fair, Singapore does cover quite a lot but it's taught badly so there's not much point.

[upper karma]

The only good teacher in Australia is this well known math teacher who also does Youtube called Eddie Woo

[dire rampart]

Ik that guy

Hes a yt legend

[upper karma]

He's gr8

[weak shoal]

I've heard of him. Ah well, I guess the information I got is not all that accurate lol

Cos I got it in the context of being told off for using the word 'axiomatization' when talking about set theory

So, idk

[upper karma]

Oh well

[robust socket]

$\cos(x)-2\sin(x)\cos(x)-\sin(x)+2\sin^2(x)=0$

I could really use a hint about this thingy here

[dark sparrow]

uhhhhhh

are you

sure the last term is 2sin(x^2)

[somber coyote]

Umma.Gumma:

[robust socket]

now it's right :)

[weak shoal]

What have you tried?

[robust socket]

played a bit with that 2sin(x)cos(x), which in sin(2x) form did not lead me anywhere significant

[weak shoal]

Have you tried factorizing sin(x) from sin^2(x) and 2sin(x)cos(x)

?

[robust socket]

was staring this the moment you wrote

let me check

[weak shoal]

Sure

[robust socket]

sorry I was busy

no, that factorisation is not helping

I get to

$\cos(x)-\sin(x)+\sin(x)(2\sin(x)-2\cos(x))$

[somber coyote]

Umma.Gumma:

[weak shoal]

Notice that you could get some sort of common factor if you were clever about it?

$\sin(x) - \cos(x) = - ( \cos(x) - \sin(x) )$

[somber coyote]

Abhijeet Vats:

[robust socket]

$\cos(x)-\sin(x)-\sin(x)(2\cos(x)+2\sin(x))$

[somber coyote]

Umma.Gumma:

[robust socket]

are you suggesting this?

[weak shoal]

Nope

Look closely

Write it on a piece of paper

[robust socket]

I am afraid I don't see this

[weak shoal]

$2\sin(x) [ \sin(x) - \cos(x) ] - [ \sin(x) - \cos(x) ] = 0$

[somber coyote]

Abhijeet Vats:

[weak shoal]

Do you see it now?

[robust socket]

damn yes

thanks

[weak shoal]

Sure

[prisma burrow]

2cot is equal to 2cos/sin right? Or is it equal to 2cos/2sin

[weak shoal]

The former

[upper karma]

may someone assist me with this? i have never actually learned about any of this stuff. ;w;

<@&286206848099549185>

[fossil lotus]

Do you understand what each of those transformations means?

[upper karma]

isnt transformation when you are moving?

[fossil lotus]

It can be

Not all transformations involve moving though

This problem only mentions rotation and reflection, which are two types of transformations

[upper karma]

reflection is when its reflected across an axis, right? and rotation is basically how much it is rotated?

[fossil lotus]

Right

[frosty egret]

@upper karma tbh though highschool does not need to be as precise since not everyone is going to be a mathematician and therefore short neumonics that make sense to the person using it and are helpful in applied uses of that math is perfectly acceptable. Arguably the point of grade school math teaching is to teach an understanding of that math without having to go into insane levels of detail. After all, much of geometry is just flat out intuitive to our visual minds.

[fossil lotus]

The terms are pretty self-explanatory

[upper karma]

what does it mean when it says like rotation 180 around the origin or 270 clockwise around the origin? also, what does it mean when it says reflection across the line y=x or x=y

thats extremely confusing for me

[fossil lotus]

Is it the 180° part that's confusing you or the "around the origin" part?

[frosty egret]

i find it funny how the clockwise is ommitted for the 180 degrees

[upper karma]

both to be honest 😭

[dark sparrow]

do you know what it means to rotate something by, say, a half turn

or a quarter turn

or some other fraction

[fossil lotus]

In this image, the triangle is rotated around the origin

Either 90° clockwise or 270° counterclockwise

[upper karma]

so is this answer to my question

the top left one??

[dark sparrow]

no

[upper karma]

😭

man

this is too hard

is it the top right one?

or does the side also change when you reflect

[fossil lotus]

Do you understand what reflection across an axis is?

[upper karma]

kinda

[fossil lotus]

This is reflection across the y-axis

[upper karma]

so its not a reflection across the y-axis

[fossil lotus]

That's not the answer no

Do you understand what all the options mean now?

[upper karma]

nvm

i got it

it was a rotation 270

because it would have ended up somewhere else

[fossil lotus]

That's right

[upper karma]

thank youuuu

[cerulean parcel]

My teacher never explained how to solve this

[manic crown]

You try it

[dusty coral]

whats shortest way to solve this

ik how to do it

fastest way to solve for the altitude?

or just BC?

[umbral snow]

You can go pretty fast by finding an angle of the triangle, then cutting it in half and treating the cut as a new hypotenuse

Actually, these are similar triangles aren't they?

Even faster

[dusty coral]

yes they are

4/alt = 5/4?

[umbral snow]

Yeah I think that's right

Hmstv. I want to get the same result using both sides

In that case, it has to be alt = 3×4/5

[frosty egret]

I need serious help
my brain is melted today and I can't figure this out
really dumbass thing for me to not be able to do
I have four points
p0, p1, p2, p3
in 2d space
i want to find the interpolation value from p0 to p1 to wherever the line segment represented by those points intersects the infinite line represented by p2 and p3
if the line does not intersect between or at p0 or p1
or does not intersect at all
then the function can return "-1" for an invalid value
this is the lynchpin i need in order to figure out this stupid clipping bullshirt
well
one half of it at least
my brain is liquid butter right now
help

[dusty coral]

why isnt sin(B) = opp/hypotenuse = b/b

im being dumb rn, ik youre supposed to use law of cosines to find cosB and then find sin B

but i dont know why sinB isnt b/b

[upper karma]

b is the hypotenuse of that triangle?

[dusty coral]

yeah

[upper karma]

how do you know

[dusty coral]

im given that

[upper karma]

um

[dire rampart]

is that meant to be a right angle lol

[upper karma]

^

[dusty coral]

no

just any anglew

[upper karma]

a triangle without a 90 degree angle doesnt have a hypotenuse

[dusty coral]

oh yeah

i was wondering why i was dumb

thanks

[upper karma]

the trig functions are defined (at an elementary geometry level) using right triangles only

thats where law of sines and cosines comes from

forming right triangles with a more general triangle

[signal hornet]

The way I see it, there are two approaches to this problem

1. I can assume r =1