geometry-and-trigonometry

[keen aspen]

2/cos2-(sin2+2)/(sin4) right

For lhs

[blazing kelp]

yes but we have to follow bodmas right ?

[keen aspen]

?

[blazing kelp]

nvm

[keen aspen]

2cos2/(cos4)-(sin2+2)/sin4

2cos2sin4-sin2-2/(cos4sin4)

[blazing kelp]

how do you get sin2+2 ?

it should be sin4 + 2 ?

[keen aspen]

Combined numerator

No

Actually this is kinda hard

Im at work right now so maybe one of the Helpers can assist you

<@&286206848099549185>

[blazing kelp]

<@&286206848099549185> can anyone help me ?

[timber hinge]

what

smh

=pup is \frac{2}{\cos\left(x\right)^2}-\frac{1}{\sin\left(x\right)^2}+\frac{2}{\sin\left(x\right)^4} equal to \cot\left(x\right)^4-\sin\left(x\right)^4

did we kill the mathbot as well?

rip

They're not the same

[valid agate]

,w is \frac{2}{\cos\left(x\right)^2}-\frac{1}{\sin\left(x\right)^2}+\frac{2}{\sin\left(x\right)^4} equal to \cot\left(x\right)^4-\sin\left(x\right)^4

[somber coyote]

Display results online and refine query

Requested by Puerøsola#0494

[timber hinge]

@blazing kelp You must've copied it wrong

[blazing kelp]

really ?

I guess im getting 4 marks in my comp exam then

ty guys

[timber hinge]

guess so?

[blazing kelp]

was racking my brain whole night

question was wrong all along haha

@timber hinge sorry for troubling you

[brave hill]

https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_20

could someone please explain to me why triangle EMI is a right isosceles triangle

all the solutions kind of assume that so i'm confused

and why is M the midpoint of the diameter arc

of arc EI

[brave hill]

<@&286206848099549185>

[eager pendant]

@brave hill note that AIME is cyclic and AM is bisector of BAC (easy to see if you draw a diagram)

so by properties of cyclic quad, EMI is right isosceles

[shy heath]

can anyone help me out iwth a question pls?

im supposed to graph |z-1| < 1 < |z+1|

idk how

z is a complex number

[upper karma]

magnitude of a complex number is well-defined, though

$|a+bi| = \sqrt{a^2 + b^2}$

[somber coyote]

TendentiousTorturousTopics:

[shy heath]

i did it

Give a geometric proof (i.e. using the vector analogy) that for any two complex numbers
z1 and z2, |(|z1|-|z2|)| =< |z1+-z2| =< |z1| + |z2|

idk how to do geometric proofs

[warm pine]

Try drawing a triangle

[lethal iris]

is circle theorem under this topic?

[gray blaze]

I believe this belongs here, but if it doesn't I can delete and move my question to the appropriate channel.
Given that m1v1+m2v2 = m1v1f+m2v2f and v1+v1f = v2+v2f
How would I expand this for multiple objects?

IE, m1v1+m2v2+m3v3=m1v1f+m2v2f+m3v3f and ?

[brave hill]

https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_13

can someone please explain the '"proper solution" to this problem better

I drew an accurate diagram and my ruler says the centroid square is a little under 2 cm while the entire square was exactly 4 cm

so I got ~ 200 for the area as 14^2

but i feel like i'm missing out on the proper centroid method

or perhaps coordinate bashing with something something homothety at point P which Idk how to do

I mean 200 IS the right answer, i just don't know the PROPER way to do this

[spiral bear]

Drawing and finding the area is not really the best way.
I find it easiest way by coordinate plane its simple and won't have to brainatorm much.
But I cannot relate how would diagonals of formed quadrilateral would be related to sides of given square

It is however clear that no matter where the point P is inside the square the result would be same

[brave hill]

Okayyy so how do I coordinate bash

[manic wharf]

can someone solve this: 49^(x-1 =7√7

[upper karma]

Can someone also solve this

https://gyazo.com/aa3309f53b164d75a4f9c84772d71eb2

cause my dumbass is confused

2.1 m^3 of concrete for the first floor, 1.2 for the second
therefore 65(2.1+1.2) = 65*33/10

Are you answering my question? mobius

yes

Thx

lol np

[manic wharf]

can u answer mine 2 pls

[spiral bear]

@brave hill then shift coordinate axis by x/3 and y/3. You'll get a square. And it's easy afterwards

[upper karma]

@upper karma

How old are you btw?

16

@manic wharf use exponent laws

[brave hill]

Omg I'm 16 ttooooo

[upper karma]

@upper karma You still there?

yes

Ik im dumb for this

afk

one sec

[jaunty plume]

49^(x-1)=7√7
==> 7^(2x-2)=7^(3/2)
==> 2x-2=3/2 @manic wharf

[manic wharf]

thank u

[upper karma]

okay back

[spiral bear]

Zuagaikotsu. got it finally.😅 if you don't like coordinate algebra you can also exploit the fact that vertices of quadrilateral formed would be on the sides of square which made by joining mid points of of the sides on original square.

[zinc grail]

hello ppl

I understnad that this would be a semi circle then I get stuck

[cold plaza]

Use the formula for the perimeter of a circle in terms of radius

2pir

In this case only half of that is equal to 8 pi

So we have pi * r = 8 pi

r = 8 then

[pure kraken]

Is ABC always equilateral?

[thorn talon]

assuming that is a straight line from A to bottom right

and B to bottom left

should be

wait

[pure kraken]

If so, how can I prove it?

[thorn talon]

i made too many assumptions

idk if it's true or not

[jaunty plume]

it is true

like you said assuming the straight lines

angle ACB will be 60

and side AB is parallel to the base

[thorn talon]

is that true though?

that was the other assumption i was gonna make

which makes everything easy

[jaunty plume]

is what true

[thorn talon]

but idk if it's a valid assumption

AB parallel to base

[jaunty plume]

yes because AB = AC

sorry

AC = BC

[thorn talon]

how can we prove that?

[jaunty plume]

let me add more lables

labels

CD=CE

and AD=BE

so AC=BC

[pure kraken]

I see thanks

[jaunty plume]

obviously not true if DE≤DC though

since then ABC doesn't exist

[shrewd lintel]

does PF1 and PF2 have a name?

[plucky marlin]

blue line

[shrewd lintel]

yes

oh

no cheeky answers plz

[plucky marlin]

straight blue line

[shrewd lintel]

💤

[dire rampart]

not sure it has a name its just the distance from point P to one of the focii

[shrewd lintel]

hmm okay thanks

[supple haven]

Wtf would sin^2 - cos^2 be?

[keen aspen]

-cos(2x)

[limpid basin]

Focal length

Huh or nvm

[keen aspen]

With lens?

[limpid basin]

I thought on an ellipse it was also called that

been calling it wrong then oops

[keen aspen]

Oh lol close, if the lengths of the major and minor are in terms of sine and cosine :v

Also if it were hyperbolic it would be -1

[clear haven]

no

not focal length

there is no particular name for the blue lines

[upper karma]

Use the Pythagorean Theorem to find the distance from B to the center and the distance from the center to D.

[vital frost]

so i split the kite in half @upper karma

[upper karma]

vertically, yes

[storm yarrow]

Hey guys need help with basic hs trig

That is what my instructions are

I inserted the exact same thing into my calculator but got a different answer, a bit confused as to what i did wrong

[clear haven]

r u in degrees or radians

[storm yarrow]

Not sure actually

How do i check?

[clear haven]

go to mode

[storm yarrow]

Im in degree

[clear haven]

hmmmMMMMMmmmm

[upper karma]

LOOOOL

[storm yarrow]

Sooo, i go to radians then?

[clear haven]

ono

[upper karma]

NO wait

You are in rad

[clear haven]

he's in rad

[upper karma]

I just tried on my ti 84

[clear haven]

what

oh

lol

then he needs to select degree

kek

she

idk

they

yes

[upper karma]

You need to go to degree man

[clear haven]

yeah degree needs to be black

[storm yarrow]

Ohhhhhh

Okay yeah in degree now, will try it out again

Yeah i got the correct answer now, thanks

[clear haven]

[vital frost]

@upper karma ok I got the answers, now for this

[upper karma]

consecutive, opposite, diagonals, bisects, perpendicular, right, congruent

make sure you know why

[vital frost]

@upper karma

[upper karma]

,rotate 90

[upper karma]

which part?

[vital frost]

all

[upper karma]

Well, you should know basic coordinates

If I move a units to the right from the origin, where do I end up?

The area can be found by using the fact that OR bisects PS and finding the areas of all of the triangles

you can also change the kite into a rectangle by moving its lower half and fitting it

[vital frost]

hmmm

[brave hill]

https://artofproblemsolving.com/wiki/index.php/2018_AMC_10A_Problems/Problem_21

can someone please explain to me how they're able to take the square root of 2a - 1

i understand how to factor the expression into the form they did but I don't understand why the other two intersection points must be +- root 2a-1

[keen aspen]

x^2-(2a-1)=0

to find its corresponding 0 you move it over and take the sqrt

x^2=2a-1

x=+-sqrt(2a-1)

[brave hill]

OOOOOOOOOOOOOOOO

dammit it feel so dumbbb

lmaooo

[vital frost]

@upper karma alright i got the others, i dont understand c and d

[upper karma]

I believe such a request is bannable. #rules

I know

just read it

@vital frost you can find the area by adding the areas of the triangles

@vital frost 2c is the length of PS

[upper karma]

for this one

i know that a is 110

cuz angle subtended by same arc

but how do i find b

OH FUCK

nvm

its 220

angle at a centre

[jaunty plume]

inscribed angle theorem

[upper karma]

theorem

[jaunty plume]

ye

[jaunty plume]

tan(theta)=x so x is opposite and 1 is adjacent

that makes sqrt(x^2+1) the hypotenuse

@wild inlet

[jaunty plume]

np

[upper karma]

Hello, I need some help with this chunky one. To be honest, I have solved way more complex questions, and yet I have been staring at this for 10 minutes.

[dire rampart]

you want to find the area of it?

[upper karma]

Yes, without rounding sides apparently

[dire rampart]

you can uh split it up into diff shapes

triangles and rectangles

for eg drawing a line from the point 3,3 to the point 2,3 forms a triangle

[upper karma]

True, I got you there

[jaunty plume]

1,2 to 3,-3

then you have a rectangle and a trapezium

[upper karma]

1,2?

[jaunty plume]

(1,2)

(3,-3)

oh right

(2,1) i meant

😦

[upper karma]

remember the formula for the area of a parallelpiped 😉

[lilac schooner]

Woke up in the middle of the night im too tired can somebody confirm or deny ping my pc account so I can see (@thatNickdude#3799)

[winged saddle]

use pick's theorem

wat

wojfoiewjfweijfa

[proper citrus]

Uhh guys

Is the sum of an exterior angle and an interior angle of a polygon not equal to 360?

[glad ocean]

it's equal to 180°

An interior and exterior angle are supplementary angles

@proper citrus

they form a straight angle

It’s 180

[proper citrus]

What is an exterior angle anyway?

"the angle between a side of a rectilinear figure and an adjacent side extended outward."

I see.

[jaunty plume]

[light notch]

Quick question, how can I find out the value of r in an equation like this?

Because whatever I do, I end up dividing r by r and having no variables

It must be possible, since Wolfram|Alpha is able to give me the answer

[gritty siren]

Well you can rearrange the thing into
r=(sin 25°)(13.3+r)
maybe it's easier to work with now

[light notch]

I tried that too, but I'll try again

I'm just going in circles

[gritty siren]

r = (sin 25°)×(13.3+r)
r = (sin 25°)×13.3 + (sin 25°)×r
r - (sin 25°)×r = (sin 25°)×13.3
how about now?

[light notch]

The only way I can find is to actually calculate (sin 25°)xr, which is around 0.423r, and combine r - 0.423r to (1-0.423)r = ...

That does work

But hmm, that's something I've never had to do before

[hard gale]

(you don't even have to get a decimal value for sin 25 to combine)

[light notch]

Oh?

[hard gale]

$$r - \sin(25)r = 13.3\sin(25) \iff r(1-\sin(25)) = 13.3\sin(25)$$

[somber coyote]

emeric75:

[hard gale]

i hope you can do the division yourself

[light notch]

Yeah xD Thanks

[hard gale]

[light notch]

[astral hornet]

You can use R (reflexive) and the given congruent angles for AA sim

Rus to show 1:2 ratio

Something like that

[storm yarrow]

Bit confused as to how I will start this, havent done math in a really long time

[astral hornet]

Trig ratios

[steel hawk]

DE = hypotenuse EC= opposite (from the angle) sinx = opposite/hypotenuse

@storm yarrow

[storm yarrow]

Thats what i ended up getting

[steel hawk]

Yup 9.7 is correct @storm yarrow

[upper karma]

Hey

What do I do when I'm trying to see if a coordinate figure has perpendicular angles when one of the lines is vertical

https://gyazo.com/1f59a4f24227a1be85fe12675c1d92d3

as in this case

[steel hawk]

When two lines are perpendicular, what does that mean about their gradients. @upper karma

[upper karma]

they multiply to -1

[steel hawk]

Yes, so do you know the coordinates of these lines or their equations?

[upper karma]

Okay look, you can't do that because of htis:

Slope of the first line: -2/0

second: 0

this is implying that -2/0 * 0 = -1

[steel hawk]

That is very true xD I’m actually an idiot sometimes

[upper karma]

i think i know a way to do it maybe

nvm

[steel hawk]

What was your thought?

[upper karma]

well i was just gonna do it without computatioin but then you get that 0/-2 = 0/-1

WAIT

THATS RIGHT

ok im not stupid

[steel hawk]

Yeah that sounds about right

[storm yarrow]

How would i start this off?

[steel hawk]

CB = hypotenuse AB = opposite (to angle x) therefore sinx = AB/CB

[storm yarrow]

Would i divide AB/CB and inverse cosine to get the answer?

Thats what i got

[steel hawk]

No because cosx is adjacent/hypotenuse and the adjacent in this case would be AC

Now do arcsin of 58.46 to get x

[storm yarrow]

Arcsin?

[steel hawk]

Arcsin means sin^-1 so inverse

And sorry you already solved it

But 58.46 isn’t equal to sinx it’s just equal to x

So angle x = 58.64

[storm yarrow]

Ohh okay, my bad, when i said inverse cosine i think i meant arcsin

[steel hawk]

I think you did 😃

All good though

[glad ocean]

x = arcsin(15 / 17.6)

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[glad ocean]

,calc

[somber coyote]

Please give me something to evaluate. See help for usage details.

[glad ocean]

,calc arcsin(15/17.6)

[somber coyote]

The following error occured while calculating:
Error: (intermediate value)(intermediate value)(intermediate value) is not a function

[glad ocean]

,calc arcsin

[somber coyote]

The following error occured while calculating:
Error: Undefined symbol arcsin

[glad ocean]

fuu

[steel hawk]

Oof

[manic wharf]

can someone solve this identity

csc^(2)x+sec^(2)x=csc^(2)x sec^(2)x

nvm

[glad ocean]

Um

2 + 2 = 2 x 2

x = sqrt(2)

I think

[trail minnow]

how do I do steps 4-7?

I don't understand how I'm supposed to subtend those two angles without constructing lines
and if I construct those lines I'm not sure how I'm supposed to get those two perpendicular lines

[astral hornet]

<@&286206848099549185>

[trail minnow]

here is a shitty mspaint drawing conveying what I want to understand and also my agony

[jaunty horizon]

@astral hornet

[manic wharf]

Can someone solve for 9a pls

[fair lily]

@manic wharf ok. So I guess you know $\sin{2x}=2\sin{x}\cos{x}$?

[somber coyote]

inubakari:

[neon stag]

How do I find y?

[dire rampart]

y and 127 lie on the same line

so they must add upp to 180 degrees

[glad ocean]

y = 53°

x = 44°

[upper karma]

Need help

I need help with 2 pages but lets start with question 5. I am not sure how to do it.

Probably looks better this way.

@somber coyote

@astral hornet

[astral hornet]

there's a relationship between inscribed angles and intercepted arcs

intercepted arc's measure is double the inscribed angle's measure

[upper karma]

Can you show me how to do it? @astral hornet

Then I will do the next problem on my own.

[astral hornet]

measure of arc = 2 * measure of angle

arc cd is intercepted by a 36 deg angle

so mCD = 2 * 36 = 72

simple right

[upper karma]

yes

so 72 is the missing angles for CD?

right?

@astral hornet

[astral hornet]

arc cd; yes

[upper karma]

ok

can you draw it out

its hard for me to think ?

@astral hornet

[astral hornet]

its already drawn on your worksheet 🤦

[upper karma]

yeah I mean how you work it out

well can you write your work on papaer will be easy to see

@astral hornet

[astral hornet]

it's just a simple rule: if an inscribed angle intercepts an arc, the intercepted arc's length is 2x the inscribed angle's measure

[upper karma]

ok

[astral hornet]

you just look at your worksheet, know the rule so that you would do 36*2 then you have answer 🤦

[upper karma]

36 like 36 x 36 and x 2?

[astral hornet]

no

in the pic, you can clearly see that there's an inscribed angle of 36 intercepting arc cd

so if you're solving for cd, use the rule: the inscribed measure is 36, so intercepted arc must be 36*2

[upper karma]

ok

like 36 x 2 = 72

[astral hornet]

yes

only being honest if you're struggling to accept a rule like that, gl on geometry 😦

[upper karma]

so you divide 72/360

you get 1/5

what then?

[astral hornet]

since the arc is 72 deg, its length would be 72/360 of circumference

[upper karma]

ok

1/5

idk how to do the formula

[astral hornet]

what formula?

circumference?

[upper karma]

idk

I think its pose to be a decimal answer?

idk

[astral hornet]

it's not decimal if you write it in terms of pi

[upper karma]

ok

Cat?

would it be this way?

@astral hornet

circumfrence = 2 x pi x r
you plug in r with 25
and now you get 50 x pi

[astral hornet]

yes

and the arc would be 72/360 of that circumference

[upper karma]

what about 50?

though

pi

[astral hornet]

50pi is circumference

[upper karma]

ok

so is that the answer?

of CD angle?

[astral hornet]

what cd angle?

your question is asking for the LENGTH OF CD

[upper karma]

yeah

idk what the asnwer htough

[astral hornet]

🤦

you already found that arc cd is 72 deg, and you know a circle is 360 deg

so arc cd is 72/360 of the circumference

which means you calculate circumference * 72/360

if you need like 75+ messages to explain one angle-circle relationship, i'm legitimately concerned for you

[upper karma]

ok got it

I got 10 pi as my asnwer is that correct?

@astral hornet

[astral hornet]

finally yes

[upper karma]

ok x3

done with it

Can you help me with the next problem?

so

cat

@astral hornet

here the next problem

[astral hornet]

congruent chords intercept congruent arcs

[upper karma]

ok

is it 110

[astral hornet]

the congruent ones are

then remaining is arc at

[upper karma]

ok?

How do I do it?

so AB is 110? right? @astral hornet

[astral hornet]

yea ab is

[upper karma]

Ok ^^

[zinc grail]

im so lost on this 😢

[astral hornet]

they're similar triangles of 1:3 ratio

[zinc grail]

why 1:3?

wait wouldnt it be 1:2

because half of the other triangle

so the ED would be half of BC?

[clear haven]

length AC is 3 times as much as AD

[astral hornet]

length of ad is half length of cd means you can fit 2 ad in dc

[zinc grail]

ohhh I see

so the answer for ED would be 1/3 of BC?

[astral hornet]

yup

[upper karma]

need help with this

wait have to eat

[upper karma]

ok

I need help

Solve for the following area. Explain your reasoning give and exact solution. Area of shaded sector =

@astral hornet

[astral hornet]

find the angle of the shaded portion

[upper karma]

ok

[astral hornet]

its that angle out of 360 th of the circle's area

[upper karma]

ok?

So how can I solve it?

[balmy tusk]

is the other angle 240?

[upper karma]

yeah

shaded part is 6

[balmy tusk]

I mean than the angle is just 360 - 240 = 120

so a third

since the total area is 6^2*3.14

[upper karma]

huh

so we plug it in the equation of 2pier

[balmy tusk]

37.68 would be total area

that divided by 3 is 12.56

so area of shaded is roughly 12.56

you understand?

[upper karma]

umm how you gor the area like that

how u divide it by 3?

idk how you got the decimal

@balmy tusk

[balmy tusk]

so the area is, A = r^2*pi

r is 6

pi is roughly 3.14

[upper karma]

ok

[balmy tusk]

so total area is, A = 6^2*3.14

= 37.68

now the unshaded part is 240 degrees

[upper karma]

I got this

[balmy tusk]

a circle always has 360 total

its 6 not 36

[upper karma]

6 x 6 = 36

you square it

2 times

[balmy tusk]

?

where did you get the 2 from?

[upper karma]

wait

[balmy tusk]

its 36*pi then

[upper karma]

pie R Square?

[balmy tusk]

even says it on your sheet..

look at the formula on the paper

r^2*pi

[upper karma]

ok

its this

[balmy tusk]

oh ye typo sry

[upper karma]

I think you did a error

[balmy tusk]

so area is = 113

[upper karma]

ok

[balmy tusk]

unshaede is 240 degrees

circle has 360

difference = 120

[upper karma]

360 - 240 = 120

[balmy tusk]

360/3 = 120

so third

the fraction your paper asks for

[upper karma]

oh

[balmy tusk]

so all you got to do is 113 / 3

so area of shaded is 37

[upper karma]

ooh

so 37 is the answer correct?

[balmy tusk]

37.6 periodicly

[upper karma]

oh

[balmy tusk]

so initite 6

if you need precise

[upper karma]

would you round it

like 37.7

[balmy tusk]

37.7

ye

[upper karma]

ok

ok

Explain your reasoning give and exact solution?

@balmy tusk

[balmy tusk]

?

reasoning should be clear

[upper karma]

ok

Can you help me a last problem?

[balmy tusk]

for exact idk you can always just write (36*pi)/3 since when dealing with pi you really cant get an exact solution

mayybe?

what is it?

[upper karma]

its proof

[balmy tusk]

so you just need to get angle A and C

[lapis night]

does sin^2(x) just mean sin(x)^2 ? doesn't f^2(x) usually mean f(f(x))?

[clear haven]

ok

see

uh

this is called notation abuse

okno

jk

anyway

$sin^{2}(x) = [sin(x)]^2$

[somber coyote]

MetalNinja27:

[clear haven]

i can define the exponentiation/indice operation any way i want if i preface the operation with the defintion

but generally it's the traditional exponent defintion

u'll see that yes sometimes it does mean nesting the function

or that it means to take the ^nth derivative of the function

it just depends

on what context is given

[lapis night]

okay thanks

[eager star]

yooooooo

can yall help me out

with this math problem

In a parallelogram ABCD, the measure of <ABC is 165. Find the measure of <C.

hello?

[lapis night]

15

[thorny fern]

is this allowed or am i only allowed to take out a single 1/2 to make the coefficient 1/32
(am i allowed to factor out 2 halfs from the first line to make it the second line)

[thorny fern]

<@&286206848099549185>

[upper karma]

hi

@thorny fern

[thorny fern]

heyo

[upper karma]

a * b = ab

a * b /2 = ab/2

or

a/2 * b

so you can take 1/2 from one bit

but the issue here is

if you factor 1/2 from 1

what do you get?

[thorny fern]

2

[upper karma]

well

yeah nothing wrong in this

[thorny fern]

so that is a correct factor

bless my tiny heart

[upper karma]

looks good

[thorny fern]

thank you so much

[upper karma]

A point located on a chord of a circle is 8cm from one endpoint of the chord and 7cm from the center of the circle. If the radius of this circle is 13cm long, how long is the chord in cm?

[astral hornet]

draw the diagram and you should be able to form a couple triangles to assist with that

[chrome quarry]

err i'm trying to prove this purely algebraically, but i have no idea what the trick is

same with cos(atan(x))

OH i found it

it's basically from the pythagorean identities, sec²x + tan²x = 1

wew

[opal zenith]

guys how do i attempt this question

[upper karma]

differentiate, then put in t = 5

[opal zenith]

Is this correct?

[upper karma]

looks right to me

[opal zenith]

thanks for helping!

[upper karma]

if AC=22

find AF and FC

i found that

ABO and DCO are similar

AB/AO=DC/OC

5/AO=1/OC

OC=y

AO=5y

6y=22

y=22/6

AO=5*22/6

OC=22/6

[upper karma]

lol

[jaunty plume]

1=2 so UV parallel to ST

because RUS is a straight line

idk the names

sec

corresponding angles

it means 3=4 too

so that proves the smaller triangle and biggee triangle are similar

since they share angles

hmm

wait what's the full question

following cannot be proved?

i see

i can't think of another way to prove it

other than similar triangles

oh right midsegments

RU=1/2 RS

So RV = 1/2 RS and VT = 1/2 RS

so RV = VT

hooray for not learning any of these words

wut

oh right i missed a step

the fact that RU=1/2 RS and that UV parallel to ST means UV is a midsegment

and so on

i guess you would also need to say UV = 1/2 ST first

then finally RV = 1/2 RS

uhh

RT

not RS

mb

that is the midsegment theorem

"A midsegment connecting two sides of a triangle is parallel to the third side and is half as long."

no you don't

you have RU=RS and UV parallel to ST

Oh I see

hmm

But if it's parallel it must join with the midpoint

;^;

yeah idk then

similar triangles makes it easy

[glad ocean]

|| inturtles ||

exturtles

[median tinsel]

can anyone explain this unit circle problem to me

[upper karma]

@median tinsel what's the question

[ivory berry]

it depends on the variables/data you have

if the stuff you're working with is all sides+one angle then cosine rule

if its 2 sides and their opposite angles then sine rule

[glad snow]

How to transform a square space into a circle space

I have a (-10,+10)x(-10,+10) plot I want to plot in a circle while preserving relative distance from origin

Circle of radius 10, inscribed in the square

I have a set of points in the square.

Lets say G is a point in the square. When transformed to circle space using the relation EH/EG =EI/EJ we obtain the point H where H represents the point G in circle space.

[glad snow]

Equations need to be in terms of the x and y components of the vector EG

And not using angles

I have to admit I find handling trigonometry in the different quadrants tricky and really want an elegant solution

Elegance is totally optional but I just like it

[hidden parrot]

Hey uh, can anyone help me with a trig question?

If sinx1 + sinx2 + sinx3 +... + sin xn = 1,
sinx1 + 2sinx2 + 3sinx3 + ... + nsinxn = 100,
where x1,x2,x3,....,xn are all unknown variables, what is the minimum positive integer n needed for the equations to be true?

I alr figured out that the answer is 20 via logical deduction, but is there any rigorous solution that n = 20? Thx!

[onyx sigil]

https://www.youtube.com/watch?v=m1GIFsT5Yng

[glad falcon]

Teach me q10

It's coordinate geometry

[fleet wolf]

for line L to be perpendicular to line K, line L has to have as its slope the negative reciprocal of line K’s slope

so line L has an equation like this y=-1/3x+b

we know line L passes through (6,5) so that is a point on the line

5=-1/2+b from which you can deduce b

@glad falcon

[glad falcon]

Yes thank you man 😀

It would be y=-1/3x+7 right?

@fleet wolf

😀

[fleet wolf]

b=11/2

[glad falcon]

What how

Ohhh

Nooo

-1/3*6 = 6

I mean -2

The rearrange 5+2 = 7

@fleet wolf

[fleet wolf]

oh yes haha im dumb

im not sure how i made that mistake

[sleek moat]

I have a gamedev project I am working on and I need to be able to convert a set of latitude/longitude coordinates to the x/y position on a 2d globe. https://i.imgur.com/2e0THNY.png

[umbral snow]

What information do you store about the orientation of the globe? How does the game know where it is looking?

@sleek moat

[sleek moat]

There is a variable that goes from 0-35, which is the amount of rotation the planet has in the current scene, 360 degrees in increments of 10.

[umbral snow]

North pole is always up, south pole is always down?

[sleek moat]

my current idea is X=cos(long)-something Y=sin(-lat)

Always up, the view is prerendered, its not a 3d model.

[umbral snow]

Close.
x = cos(long - current rotation)

Check to see if that goes the right direction though lel

[sleek moat]

But if Lat is not on the equator, then the X needs to be smaller, or the point is off in space somewhere.

So the X coord needs to be calculated based on lat and long, while the Y can come from just lat

[umbral snow]

Sorry! You're right. Time to check my spherical coordinates

[sleek moat]

and since the edge of the image is perfect circle, i think the "-something" needs to include a cos and sin. I've been playing around with it and not getting any closer to a solution.

[umbral snow]

Spherical coordinates are as such,
x = ρcosφcosθ
y = ρcosφsinθ
z = ρsinφ

[marsh beacon]

hey how to find the tan^-1 (1/sqrt(3)) wtih the unit circle?

[umbral snow]

Where φ is latitude and θ is longitude, considering where the camera to be facing as (0,0,0)

But we don't want to calculate that in 3D lol

Actually it's just a forward projection so we can forget about either x or y I'm not sure which yet

[sleek moat]

i think i follow

[umbral snow]

x is the one facing us so that one can just be forgotten about.

x = ρcosφcosθ
y = psinφ
Where p is the radius of your circle
And the very center of where we're looking is lat = long = 0

[sleek moat]

I got it to work, thank you so much.

My graphic for the globe is not a perfect ortho projection, so it gets a little misaligned as it turns, but I should be able to come up with a way to compensate for that.

[umbral snow]

Cool cool! Happy to hear it. Let me know if you need anything else!

[sinful grail]

hey any good channels for learning the basics? im in calc 2 and need a refresher

[upper karma]

Can anyone help me solve this? a²+b² = 120

It's a square

[dire rampart]

wdym solve

you dont have enough information to solve for a and b

[jaunty rover]

yeah

you need 2 values

infinite solutions at this point

not infinite but many*

[spiral bear]

It's a circle

[exotic meteor]

Let's say we have to points which are

$T_1= (0, -1, 2) \ T_2=(3, 3, 2)$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

We can say that their z position is equivalent

I would like to find the line that connect those two points

I am using the formula

$\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

Which at the end turns out to be

$\frac{x}{3} = \frac{y+1}{4} = \frac{z-1}{0}$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

Now I am trying to convert that into parametric form

[gritty siren]

dividing by 0

[exotic meteor]

But the 0 in the fraction is giving me problem

Yeah...

If I define z as t

$\frac{y+1}{4} = \frac{t-1}{0}$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

Doesn't work quite...

What am I supposed to do?

[hard gale]

z is a constant for that line as you mentioned, so it's veri weird to parameterise it

[exotic meteor]

So I just set z as 2

And parameterise it from y

[hard gale]

yea or from x, whatever you want

[exotic meteor]

Alright, thanks

[hard gale]

[exotic meteor]

Are three points always coplanar?

[jaunty plume]

yes

[exotic meteor]

$T_1 \ne T_2 \ne T_3$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

Cool

Thanks

[jaunty plume]

as long as distinct and non collinear

[exotic meteor]

I'm learning about planes right now 😃

Ah yeah, makes sense

You can't draw a plane if they're collinear**

[exotic meteor]

Using 3d graphing is making math just so fun

I love how I can pick 3 random points and use math to find the plane that connect those three points 😄

[clear haven]

yes

it is nice

[clear haven]

wym

it should be the same

maybe ur fractions are being weird?

okay

im surprised u have a csc function on ur calculator

lol

[hard gale]

[thorn talon]

That's not the same thing

[hard gale]

i luv conflicting notations

[thorn talon]

True

It's annoying

Trig notation that is

[hard gale]

taht's why i write arcsin :/

[thorn talon]

Don't worry

[clear haven]

arcsin is best notation

[thorn talon]

It's common error

[clear haven]

but yeah its not the same as csc

[thorn talon]

Cause confusing notation

[clear haven]

yeah

[thorn talon]

The right is the inverse of sin

Not csc

[clear haven]

$\sin^{-1}(x) \neq (\sin(x))^{-1} , \asin(x)$

wth

[hard gale]

{}

[thorn talon]

sin^-1(x) =/= 1/sin(x)

[somber coyote]

MetalNinja27:

[clear haven]

wat

it aint givin me the arcsin

but you know wat i mean

it's dumb

using exponents to notate anything but exponentiation (in cases where it's not repeated) is idiotic

and we should emphasize something else

[thorn talon]

Wrap the whole thing in parenthesis just in case tbh

[clear haven]

quantity sin(x)

the whole thing yes

[hard gale]

^_v

[upper karma]

hi

[upper karma]

can someone help me

im stuck on this question

its

[thorn talon]

?

[upper karma]

Omg it's inverted I'm sorry

its q 3

[thorn talon]

my chrome extension seems to flip it

hmm

so it's fine

well

[upper karma]

okay

yeah so its question 3

im not sure how to do ti

it

[thorn talon]

if m is the slope of the first line

then what must be the slope of the line perpendicular to it?

[upper karma]

um?

0?

[thorn talon]

no

[upper karma]

bc it cant be -1/0

can it?

[thorn talon]

it can't

but

but basically

[upper karma]

also what does = /

mean

[thorn talon]

m can't equal 0

it can equal every other value except 0

[upper karma]

so you cant have a slope of 0

but what happens if its just x=9

or y=1

bc thats just a straight line

parallel to the axis

[thorn talon]

yes

in general

m can equal 0

[upper karma]

that doesnt have a slope?

[thorn talon]

just not in this case

yes

horizontal lines (m = 0)

[upper karma]

yeah

[thorn talon]

vertical lines have undefined slope

[upper karma]

oh yeah

but what do i do for this question

[thorn talon]

do you know the property

m1 * m2 = -1?

[upper karma]

um?

[thorn talon]

essentially

for two lines to be perpendicular

[upper karma]

do you mean the negative respicable?

[thorn talon]

the product of their slopes must be -1

yes

[upper karma]

yeah i know that

[thorn talon]

use it

[upper karma]

yeah but i dont have the y=mc+c equation

[thorn talon]

you have m

[upper karma]

i mean i have the y int

[thorn talon]

it is enough

[upper karma]

oh wait yeah i have a point too

[thorn talon]

if m is the slope of line 1

then the slope of the line perpendicular must be the negative reciprocal then right?

[upper karma]

yeah

[thorn talon]

-1/m?

[upper karma]

but m=0

[thorn talon]

no

it states m doesn't equal 0

[upper karma]

but it says m=0

oh

[thorn talon]

it very specifically states m =/= 0

[upper karma]

does a = with a / between it mean it doesnt equal 0

[thorn talon]

yes

[upper karma]

oh that makes sense

i thought it meant m=0

[thorn talon]

do you know how to go from here?

[upper karma]

not really

[thorn talon]

do you know point slope form?

[upper karma]

like i dont actually have the slope

[thorn talon]

or intercept slope form?

yes you do

it's -1/m

[upper karma]

i know slope interecpt

[thorn talon]

slope intercept is fine

[upper karma]

like y=mx+c

[thorn talon]

so you have y = (-1/m) x + c

right?

[upper karma]

yeah

[thorn talon]

you now need to find c

you have the point (0,3)

[upper karma]

so it would be

[thorn talon]

plug that into x and y

[upper karma]

3=-1/m(0)+c

[thorn talon]

yep

[upper karma]

so c=3

[thorn talon]

so c = 3

and you're done

y = (-1/m)x + 3

[upper karma]

oh so i dont need to work out what m is?

[thorn talon]

you don't

it's variable

[upper karma]

okay thats good

[thorn talon]

there's a more specific name for m in this case

but i don't remember it

[upper karma]

its finer

fine

what about 3b

it says find m

if the line passes through (1,-4)

oh wait i have two points

so its just y2-y1/x2-x1

[thorn talon]

go from the equation you found in part a

[upper karma]

wdym?

[thorn talon]

y = (-1/m)x + 3

[upper karma]

yeah?

[thorn talon]

plug the new point in

and solve for m

[upper karma]

i need to find in

m

okay

how do i solve

-7=-1/m

[thorn talon]

flip both sides

[upper karma]

would i go -7x-1

[thorn talon]

no

[upper karma]

or is that only for dominator

denominator

[thorn talon]

hmm

idk

doesn't seem mathematically valid to me

[upper karma]

yeah

ill go to #prealg-and-algebra

well the answer says 1/7

[thorn talon]

-7=-1/m

flip

-m = -1/7

[upper karma]

wdym flip

[thorn talon]

m = 1/7

take reciprocal of both sides

[upper karma]

what

im confused

please explain

[thorn talon]

when you have a fraction

say 1/2

the reciprocal of it is 2/1

[upper karma]

yeah

[thorn talon]

basically flipping top and bottom

taking reciprocal of both sides of an equation

is a valid manipulation

that preserves equality

generally

[upper karma]

yeah

wait so youre saying

-1=-1

is the same is 1=1

[thorn talon]

ehh

basically

when solving equation

[upper karma]

hmmm

[thorn talon]

you know how you need to do the same thing to both sides?

[upper karma]

yeah

[thorn talon]

just like how you can multiply both sides by the same number

or add the same number to both sides

[upper karma]

yeah

[thorn talon]

you can take the reciprocal of both sides

[upper karma]

okay

so when solving

-7=-1/m

you go

-1/7=-m

[thorn talon]

yep

[upper karma]

and then cancel each other

[thorn talon]

yeah

[upper karma]

so its 1/7=m

[thorn talon]

essentially dividing both sides by -1

[upper karma]

yeah

so it would be

-7x-1/x

which would equal -1/7

im confused

[thorn talon]

hmm

what do you mean?

[upper karma]

how does -7=-1/m

equal -1/7=-m

like did i divide it

or times it

[thorn talon]

flip both sides

alternatively

you can cross multiply i guess

[upper karma]

oh yeahhhh