geometry-and-trigonometry

[olive sapphire]

ok

send

haven't done tri in a year

but let's see

[prime vigil]

not trig

geometry

10th grade

[olive sapphire]

ok

send

[prime vigil]

im in 8th grade

and im 13

dont ask

[olive sapphire]

why

wtf

[prime vigil]

idk

anyways

[olive sapphire]

practicing early?

[prime vigil]

no

i take a class on geometry'

in school

they offer that at my school

anyways

what do any of these mean\

[olive sapphire]

think about it

[prime vigil]

what

[olive sapphire]

if you live in abilene can you live in texas

[prime vigil]

yews

yes

abilene is in texas]

[olive sapphire]

so that one is true

[prime vigil]

yes

see i know that stuff but what does converse, contrapositive etc mean

[olive sapphire]

idk about second one

[prime vigil]

im not good with making inferences about latin/greek roots

[olive sapphire]

inverse is if you live in texas do you live in abliene

[prime vigil]

no

[olive sapphire]

:Dewad

d

are u sure

[prime vigil]

i mean you can live in austin

you dont have to live in abilene

abilene is one city

[olive sapphire]

yes

i thought u said no to my explanation

[prime vigil]

so if you live in texas you CAN live in abilene

[olive sapphire]

anyways

so that is false

[prime vigil]

by no i meant false

sorry for confusion

[olive sapphire]

"To form the converse of the conditional statement, interchange the hypothesis and the conclusion. The converse of "If it rains, then they cancel school" is "If they cancel school, then it rains." To form the inverse of the conditional statement, take the negation of both the hypothesis and the conclusion."

https://www.varsitytutors.com/hotmath/hotmath_help/topics/converse-inverse-contrapositive#:~:text=To form the converse of,the hypothesis and the conclusion.

[prime vigil]

thanks

imma go npw

[olive sapphire]

👍

[prime vigil]

bye

[olive sapphire]

bye

[sage pelican]

Anyone know how to solve this?
The square of an angle measure is 15 degrees more than twice its complement. Find the measure of each angle.

[silent plank]

what have you tried so far?

[drowsy cairn]

Do you still need help with this?

[bronze shell]

in "sin^-1"

the -1 just means its inverted right

[random acorn]

sin^-1 is the arcsin function, not 1/sin(x). Basically u input a value between -1 and 1 and it tells you what angle between -90 and 90 degrees gives you that sin value

Or, sin^-1(sin(x))=x

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

[upper karma]

.

[timber tapir]

aye

[timber tapir]

how can i find the cos(2pi) using unit circle?

based on the unit circle at angle pi the coordinates are (-1,0)

so i took the first coordinate which is cos, and multiplied by 2

but that's wrong so what do i do

[surreal trail]

hiya, can someone please help me find the equation for this cotangent function

[upper karma]

that are degrees?

if that are degrees, then, the function is $\frac{1}{tan(x-90\textdegree)}+2$ or in radians $\frac{1}{tan(x-\frac{\pi}{2})}+2$

[somber coyote]

R0bin

[timber tapir]

since sin^2(x) + cos^2(x) = 1, what does sin(x)+cos(x) =?

also 1? because if u sqrt both sides its still 1

im trying to solve the inequality cos(t) + sin(t) > 0

so i've set it = 0 just to see when its = 0

and then solved to get sin(x)/cos(x) = -1 which is tan(x) = -1

[lapis cypress]

hello guys I just wanted to know if we can check that cos(90+x)=-sin(x) quickly through some website?

[upper blaze]

What do you mean through website ? You want to prove the formula or whaT?

[lapis cypress]

i just want to quickly know if the equation is true or not

[upper blaze]

Woho mate 😄 We will go over this but this is something you should never do 😄

[upper blaze]

Fastest intuition way to see if formulas like this one are true is to look at unit circle and if you want proof of it you use formula for cos(x+y) 🙂

[lapis cypress]

this formula?

[upper blaze]

yes

[lapis cypress]

alright thanks I was a little confused because theta and 90 were in the same bracket.

[timber tapir]

yeah… i figured that out hah

[trail mist]

If point P is (0,5) on the terminal side of angle theta, how do i find the lengths of the legs to find the trigonometric functions?

[junior veldt]

Use algebra, replace the width with a variable and find out what the longer side is in relation to the width(hopefully I make sense)( @upper karma )

Also don’t forget there’s 4 sid3s to fence :)

[upper karma]

Hello, there is a question.

If anyone can help me, please DM me.

It's of coordinate geometry.

[valid harbor]

just ask here

[upper karma]

Let A, B be two Points on the plane such that A is different from B, then there is a unique line that goes by them

a plane i mean not the

[dark jacinth]

Hey guys, so I know how rotational matricies work, however, I wasn;t too sure how to apply them to points, could anyone explain that process? It would be greatly appreciated.

[dark sparrow]

make your point into a column vector x and then multiply R * x (where R is your rotation matrix)

this will apply the rotation encoded by R

@dark jacinth

[loud shard]

how do you simplify this?

[charred flicker]

Hello?

IS this free?

I would need help

Anyone here? when is the best time to be here?

[fervent rover]

I think u should just put the prob up

[upper karma]

Yup

[charred flicker]

https://fog.ccsf.edu/~wjeh/math95/circle_approach/trig_exercise_circle.pdf I'm on 3. g.

Is this a 30 degree triangle?

[valid cradle]

to go from radians to degrees, you multiply by 180/pi

so just multiply the given angle (which is in radians) with 180/pi, and you'll get your answer

[charred flicker]

Well I was doing the usual and I got 210 degrees

what do I do next?

[charred flicker]

still here?

I'll be back tomorrow

[hearty maple]

help

[tardy spoke]

hello, i can help you with this problem, please DM

[compact oasis]

Seems to be easy but i am having trouble

[dark sparrow]

they've misspelled nonagon

anyway, are you familiar with complex numbers? @compact oasis

[compact oasis]

Yes, i am familiar with them

[dark sparrow]

consider the nonagon as being in the complex plane, oriented and scaled such that its vertices occupy the 9th roots of unity

and let A be the vertex sitting at 1

let u = e^(2pi/9 i) be the first vertex counterclockwise from 1

then you have $x = |u-1|, y = |u^2-1|$ and $z = |u^3 - 1|$

[somber coyote]

Ann

[compact oasis]

I am anot that confortable with complex numbers as of now, can you give me another approach?

[dark sparrow]

...oof

any other approach would involve very painful trigonometric bashing

[compact oasis]

Oh, alright i will try to learn more about complex numbers first today

[nocturne remnant]

I got a solution that involves not *much trig bashing

The trig could probably be averted completely with a few more Ptolemy’s

[tough nest]

solved without advanced trig (does law of cosine counts idk)
consider triangle ABD with law of cosine:
z^2 = x^2 + y^2 - 2xy cos(ABD)
consider isosceles triangle BCD, since it's a nonagon, BCD = 140 degrees, and therefore CBD = 20 degrees
and making angle ABD = angle ABC - angle CBD = 140 - 20 = 120 degrees
therefore, z^2 = x^2 + y^2 - 2xy * (-1/2) = x^2 + y^2 + xy

[compact oasis]

Wow, thank you so much

[tough nest]

[compact oasis]

😃

[sharp depot]

Does anyone know if $\frac\pi 2+2\pi\mathbb{Z}$ is a valid expression for saying: $\frac\pi 2+2k\pi$ where $k$ is every number in $\mathbb{Z}$?

[somber coyote]

NoRysq

[sharp depot]

I hope this fits here #category-theory was not the right place apparently

nvm solved

[worthy gust]

need help with this geo question

[charred flicker]

I need help anyone available ? good afternoon

[slender hazel]

hello hello

i’m taking trigonometry currently and i know how to do this problem but could someone explain why it’s necessary to use the 30-60-90 triangle rather than using the distance formula with the coordinates that lies in the quadrants? like (1,1)

[raw tiger]

I would just use the definition of cot and use the unit circle

[dark sparrow]

#bots #latex-help #latex-testing

[hearty maple]

pls help me here

[white swift]

Mmm tfw when you cant solve an 8grader problem 😭

There’s a given triangle ABC with <C 80, angle bisector CL crosses the perpendicular bisector of AB at point Q. What is <AQB?

This I believe

No im talking about me as a postgrad engineer I solved your question in one of the Q channels

[white swift]

<@&286206848099549185> bless you

[nocturne remnant]

Q should be outside the triangle lol

[upper karma]

70=10c
c=7
70=6d-2
6d=72
d=12
@upper karma

idk im bad at math idfk if this is right lol

[barren meadow]

How does this turn into this?

[nocturne remnant]

$\sin^2{x}+\cos^2{x}=1$

[somber coyote]

elon mass

[nocturne remnant]

this allows for some simplification on the top

[smoky talon]

sum1 pls help me

does y = 122

x = 65

does z = 57

sum1 pls help

immediatellylylyllylylyl

<@&286206848099549185>

[stable maple]

ans coming same as yours

[rapid bone]

@smoky talon Yes, x = 65 y = 122 z=57

good job!

[upper karma]

please <@&286206848099549185>

[lime crown]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[dark sparrow]

@upper karma do you still need help with this?

[upper karma]

YES

<@&286206848099549185>

[dark sparrow]

have you made any progress?

[upper karma]

yes i have

[dark sparrow]

please share

[upper karma]

i try to factor => doesnt work

i try to blur away the -1/2 thing but it end up to complicated for me

[dark sparrow]

are you familiar with complex numbers at all

[upper karma]

u gonna use the euler ?

[dark sparrow]

yes

[upper karma]

ok

do it

[dark sparrow]

2cos(t) = e^(it) + e^(-it)

[upper karma]

ok let not use it

the e t hing

i dont know about that

[dark sparrow]

bruh

[upper karma]

can u use only caculation of trigonometry

like cosx + cosy cosxcosy

[dark sparrow]

idk if i can turn this into pure trig

[upper karma]

what does the e thing do ?

[dark sparrow]

i thought you knew about euler's identity

e^(it) = cos(t) + i sin(t)

[upper karma]

i knew it because of google , but nobody teaches me that

[dark sparrow]

...okay then forget about it for now

hold on

[upper karma]

and i dont know what does the constant do ?

[upper karma]

hold on what

[dark sparrow]

hold on as in give me some time

[upper karma]

ok

[dark sparrow]

well shit i can't figure out how to do it without euler

[upper karma]

so can u do the euler for me

and i will figure it out

my teacher throw that shit to me without teaching me the euler

[dark sparrow]

er

i forgot to say k in Z but k should be a non-multiple of 11

[upper karma]

e^(i pi) = -1 ?

i only know that

[dark sparrow]

well damn

[upper karma]

what ?

[dark sparrow]

i don't have anything else to say

[upper karma]

explain to me the last 3 lines

i understand all the above

[dark sparrow]

the last three lines are just recognizing a GP and dividing out by things that are known not to be zero

[upper karma]

from line 2 to line 3 please

wheres the -1 go?

[dark sparrow]

add 1 to both sides

recognize 1 as e^(0ix) [a.k.a. e^0]

[upper karma]

ok i see

ok thank u

[eternal plover]

Why does sine of theta specifically equal opposite/hypotenuse

[charred flicker]

for 3g I'm getting the opposite wrong
I don't know why. https://fog.ccsf.edu/~wjeh/math95/circle_approach/trig_answer_circle.pdf

<@&286206848099549185>

[exotic skiff]

Can someone help me with geometry?

I need an A to get first seating in my highschool magnet, but I have an E right now.

[storm portal]

That is how it's defined

[eternal plover]

Why though?

[raw tiger]

Long ago mathematics needed something to describe relations between sides and angles of a right triangle. Sine is just one of the 6 possible combinations to describe an angle in relation to the sides of a triangle

[eternal plover]

that's not the answer I'm looking for tbh, it had something to do with like chords and saying sine = half chord something

It does make sense if you know about the unit circle

3-4-5 triangle

cos = 0.6, sin = 0.8

and thats the coordinates

[raw tiger]

It is especially helpful when dealing with polar coordinates

[eternal plover]

What im asking is how they actually founded sine of theta

why did it mathematically make sense to say sin of theta = o/h

[raw tiger]

Like many mathmatics operations we just decided a name for it. All the trig functions represent is just ratios between sides.

[sharp coral]

i just dont understand math 🥲

[still mica]

what am I missing here? am i picturing it wrong?

[upper karma]

Try using the law of sines

Also, I think you got one of the angles inccorrect

The one in the top right should be 5 degrees (I'm pretty sure) because 180-31 = 149, and 180-(149+26) = 5.

[quartz cedar]

Can someone please help me on this Trigonometry problem?
cot t = -1.2345, sect < 0. Find t

[upper karma]

<@&286206848099549185>

[south onyx]

enter the coordinates of the vertices of (Rx-axis ∘ r(180°,O))

[raw tiger]

what does equidistant mean?

[south onyx]

which one do i do first/ enter the coordinates of the vertices of (Rx-axis ∘ r(180°,O))

[dark sparrow]

are you asking where the name sine came from

cause it's a pretty long story

but the TL;DR of it is that it ultimately came from the Sanskrit word for bow string but it went through a bunch of broken telephone involving Arabic and Latin

[earnest basin]

Hello

Is anyone free today

[dark sparrow]

no, clearly nobody on this server of thousands has even a millisecond of time to spare on this day in particular (/sarcasm)

[upper karma]

Can u tell me some of the theorem that looks like the Pascal and the Newton ones ?

[eternal plover]

I'm just asking why does sine equal opposite/hypotenuse and it had to do with chords in a angle or something
I got my sources from this and other quora stuff: https://www.quora.com/What-does-sine-mean-and-how-do-you-prove-that-sine-is-equal-to-the-opposite-side-over-the-hypotenuse

[dark sparrow]

sin(θ) = opp/hyp is the definition. there's nothing else to it.

[eternal plover]

Welp ig thats my answer

thank you very much ma'am

[oblique tiger]

Y’all get this Bcs i fine

Dont

[humble pulsar]

ok...?

[oblique tiger]

It’s loading

Anyone ?

[stable maple]

Which question?

[oblique tiger]

3

3

[stable maple]

k wait let me try

[oblique tiger]

Okay

[silent plank]

what have you tried so far?

[oblique tiger]

I barely know this and he expected us to do it, we had a sub and I try another paper nope I don’t get it

[silent plank]

do you know the slope formula?

[oblique tiger]

Yes I know the formula

[silent plank]

can you try applying it

[oblique tiger]

Okat

[silent plank]

also your answer for Q2 is wrong

[oblique tiger]

It is

How

[stable maple]

yes just saw that

[oblique tiger]

Oh

[floral fjord]

help please?

[somber coyote]

@tawdry pivot

@tawdry pivot

[tawdry pivot]

remember to argue why we take the principal square root

@floral fjord

[oblique tiger]

Yall know 5

[upbeat sundial]

Idk if this is in your course or if it works but I'd apply distance formula

This is what I did.

First step is distance formula, then you just apply the points given to the placements and solve

I mean I am an idiot 🤷

But that's how I'd do it

[floral fjord]

tell me the answer if u know

[verbal night]

Does anyone know how to do this?

[upbeat sundial]

I never learned this but honestly I'd just try and graph it out

Maybe you tried that already

But that's all I can think of

[verbal night]

I am a bad drawer so it just came out as a mess

[upbeat sundial]

I'd just draw the origin

Then draw x and y axis with a thick line

Then put in points for the x and y values

Use a ruler

Might help, might not

[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!

[pure knoll]

wait wrong chat sorry

[calm imp]

How do I do question C

[stable maple]

do we have to find angle f and g?

[calm imp]

yes

[stable maple]

k

[calm imp]

do uk how to do it?

[stable maple]

I am trying

[calm imp]

ok

[stable maple]

is any line given parallel or something

[calm imp]

no

[stable maple]

k

is the answer 65,37?

[stable maple]

?

[calm imp]

idk

i came here for help

i think

[stable maple]

I dk if this the right ans but I can tell you how I did it

[calm imp]

can u explain how u got it

[stable maple]

yes

F+G=102 exterior angle property and angle 65 and angle opposite to it will also be equal to 102 so let the opposite angle be y so from equation 65+y=102 we will get y=37 and as F+G=102 and angle 65 + angle 37 =102 I put them in equal and compared the equation so f=65 and g=37

[stable maple]

Did you understand it?

[calm imp]

no💀

[upper karma]

Is there a fast way to get the quadratic form of an ellipsoid given the rotation matrix, length of each of the three-axis, and the location of the ellipsoid's center in some world coordinate frame?

The straightforward approach seems to be applying the homogeneous transformation matrix (new_h = [R t; 0 0 0 1] * old_h) on some general point on the surface of the ellipsoid and then plug the resulting x, y, z (new_h = [x; y; z; 1]) coordinates into the general equation of an ellipsoid and expand that to get a quadratic form (writing x2/a2 + y2/b2 + z2/c2 = 1 in terms of the old_h, rotation matrix, and translation matrix parameters).

But is there a faster way because the calculations are tedious and I was hoping there is some generic quadratic equation of an ellipsoid but I can't seem to find it on the internet.

[sleek scarab]

I have a problem: a circle is inscribed in an isocele trapezoid with bases equalling 40cm and 10cm. Find the area of the triangle created by the touching points of the circle and the trapezoid's sides, and the touching point with the smaller base.

i have no idea how to do it

[arctic prawn]

sry didnt mean to be rude..

[brazen sluice]

@sleek scarab how much have you solved until now?

[sleek scarab]

this much

ignore the serbian lol

[brazen sluice]

Alright

Now that you know h you know R(the radius of the circle) and that is h/2 = 10. After that draw to lines that are perpendicular to the AC and BD - those are radiuses. If you draw another radius perpendicular to CD (that would be OF) you will end up with two new shapes - CHOF and DGOF

[sleek scarab]

I follow

[brazen sluice]

These shapes both have the same angles - two are 90 degrees, one we will denote as x and it will be the angle formed at the lower base when AC or BD meet with CD, and the other will be 180 - x. Those are HOF and GOF. Then you will end up knowing the angle HOG, which is 360-2×(180-x), thus HOG is 2×x

We can easily find the value of cos(2×x), since we can use either of the right angled triangles on the left and right. We can use the formula cos(2×x) = cos^2(x) + sin^2(x). By using basic trigonometry we can determine that cos(x) = 3/5 and sin(x) = 4/5

We can then use this formula : HG^2 = OH^2 + OG^2 - 2×OH×OG×cos(2×x)

This leaves us with HG = 16

Then I used the formula 2×R = HG / sin of HEG. That is 20 = 16 / sin of HEG. That means the sin equals 4/5

I can then find out the value of that angle's cos with the formula 1 = sin^2 + cos^2, and that is -3/5 because the triangle is clearly obtuse, since the center of the circle is outside that triangle.

I can then find out what the values of HE and GE are by using the same formula I used to find HG. This leaves with HE=GE=4√5. We can then use this formula for the area: S = (HE×GE×sinHEG)/2

And then the final answer will be 32

Any questions?

[slate steppe]

Could someone help me? I’m suppose to find x and y and the red lines are parallel and the blue lines are parallel as well

[stable maple]

as red lines are parallel so 3x=105(Alternate interior angle) and in the triangle in which two angles are 2y and x given the third angle in the triangle will be equal to 105(Alternate Interior angle) so u know already know the value of x from equation 3x=105 so apply angle sum property and you will get value of y also

[slate steppe]

Thanks

[sleek scarab]

no. Thank you so much tho

it was as clear as my tears

[brazen sluice]

@sleek scarab lol

Btw I am Bulgarian

So greetings, neighbor

[ocean bison]

Hello, I was wondering why it became sin(60)? He said that 120 degrees is in quadrant 2, and it forms a reference angle of 60 degrees. He said that the reference angle is an angle between the hypotenuse of a triangle and the x-axis. He also said that 120 degrees isn’t in the 30-60-90 triangle. But afterwards, he formed that triangle here.

It’s because 180 minus the given angle isn’t it? Since it is in quadrant 2?

[dusky surge]

Yea, I think it used sin(180°-θ)=sinθ

[ocean bison]

Oh okay, thank you

[upper karma]

hello!!

can someone help me

[sleek scarab]

Greetings fellow balkan dude

[still mica]

What am I doing wrong with transformed outputs? If I transform the key points on a cos(x) function I keep getting incorrect outputs.

[molten delta]

I need help with this

Nvm

[balmy python]

Can anyone help with part iv?

[storm heath]

?

[formal tartan]

what r the z1 values tho? do you have the polygon area?

[twin kite]

would it be possible for someone to help explain how to do trigonometric inequality?

I'm not quite sure about which should be > or <

[noble crest]

From which rule is AG:GD = 2:1 ????(in 1)

Question is this

[fiery solstice]

yo can someone help me understand why the equation of a plane parallel to the z-axis will be of the type ax+by+d=0? I think the distance from z will be the same for all points on such a plane but that’s as far as I got

[balmy python]

It’s there…

[stable maple]

I think it will be constant because the radius R and r are of same circle so there value will be equal no matter if the triangle increase or decrease so there ratio will always be equal to 1

[tropic fossil]

Halo can someone help me with some sec 3 trigonometry

Thank youuuu

[tardy spoke]

Hello, kindly DM me for help

[formal tartan]

but show some work of yourself bruh...

[balmy python]

??????? Do you not know how to sketch a simple Argand diagram lol

[upper karma]

whats the answer here? thought it was y1-y0 / x1 - x0

[silent plank]

()

[upper karma]

()

[silent plank]

(y_1-y_0) / (x_1 - x_0)

they're equivalent

[upper karma]

this is equivalent to what?

[silent plank]

the selected option

[upper karma]

iisn't the selected optiion the negative version of slope?

[silent plank]

$\frac{k}{k} = \frac{-k}{-k}$

[upper karma]

wait

oh

[somber coyote]

ℝamonov

[upper karma]

lol they're the same

i triied to disprove you wiith an example but it came out equal

thanks

[ocean bison]

Hello, is the reason why the inverse of tangent 1 is 45 degrees is because, the value, which is 1, is positive. So that’s in quadrant 1. And in 45 degrees, it is (√2/2, √2/2) and when you divide them both you’ll get 1?

[dark sparrow]

that sure is a bunch of words you just said.

the 'reason' why arctan(1) = 45° is because tan(45°) = 1.

[nocturne remnant]

Reason:
A 45-degree right triangle is clearly also isosceles
Done

[ocean bison]

Oh okay 😅 I forgot, but how would I know that arctan(1) = 45 degrees? I understand that tan 45 degrees = 1, but how would I know if I was asked what the arctan(1) is? And I didn’t know it was 45 degrees? I have to look at the unit circle and find two coordinates on quadrant one that when I divide will give me 1?

[dark sparrow]

arctan(1) is the angle whose tan is 1.

if you're looking at the unit circle there is a way better way to visualize tan than just 'dividing y by x'

tan(θ) is the slope of a line that makes an angle of theta with the horizontal

[ocean bison]

Oh okay I see, thank you. There’s this one example that I’m confused at too, it says that arctan -√3 = -60 degrees. I’m confused on how it became -60 degrees?

[dark sparrow]

what do you think it should be instead of -60°?

[ocean bison]

60 degrees?

[dark sparrow]

tan(60°) = +sqrt(3) though, not -sqrt(3).

you want an angle whose tan is -sqrt(3).

[ocean bison]

Oh okay, perhaps 300 degrees, I get -sqrt(3) as well as 330 degrees

[dark sparrow]

330 degrees, great

er

no

tan(330°) = -1/sqrt(3).

300 degrees is right.

and it's exactly one full turn away from -60°.

the output of arctan always lies between -90° and 90°.

[ocean bison]

Oh okay, should I subtract it 300 with 360 to get -60 degrees?

[dark sparrow]

...wording

but yes

[ocean bison]

Oh lol 😅

Thank you very much !

[formal carbon]

can someone explain to me why Pythagorean's theorem can't be simplified to a + b = c by taking the square root of both sides?

[quiet blade]

$\sqrt{a^2 + b^2} \neq \sqrt{a^2} + \sqrt{b^2}$

[somber coyote]

peaceGiant

[formal carbon]

ohhh i see

okay thanks

so would you be able to simplify it if it was like this (a+b)^2 = c^2?

[quiet blade]

Yeah, if that were the theorem you could, since we are assuming they are the lengths of a triangle >0

[formal carbon]

ah okay thx

[spark brook]

You could also prove that by doing sqr(5^2+4^2), which is sqr 41, and sqr(a^2) is always a, so 41 doesn't equal to 5+4, or 9.

[slender wasp]

or sqrt(2) isn't 2

[ocean bison]

Hello, how could I find the value if I’m given sin 9pi/2? I converted it into degrees and got 810 degrees which is big, so I subtracted 360, but got 450.

[pure cape]

dont convert it into degrees, it only makes it harder

notice that 9pi/2 is pi/2 plus itself 9 times

that means you are going 90 degrees in the unit circle 9 times

so start from the angle 0, go 90deg (or one quarter) nine times, the y-coords you end up with is the result

[lone copper]

yeah u can subtract another 360deg again to get its position on the unit circle

[ocean bison]

Ahhh yes because pi/2 is 90 degrees. So if I go 90 degrees in the unit circle 9 times, and you said that I’ll start from angle 0 to 90 degrees. So basically I’ll go back and forth to 0 degrees and 90 degrees back and forth 9 times?

[lone copper]

it’s two full cycles on the unit circle+ 90deg

[ocean bison]

Oh okay but why two full cycles on the unit circle?

[lone copper]

each cycle is 2pi so 4pi would give u two cycles, 9pi/2 is 4pi+π/2 which is two cycles+ 90deg

alternatively as waler said above u can have 9 times of 90 deg, and for one full cycle u have 4 times of 90deg, so 9pi/2 gives u two full cycles+90deg

[ocean bison]

Oh okay so one full cycle is 2pi. I see now! How come that 9pi/2 is 4pi+ pi/2 = 9pi/2? How did the numerator become 9?

[lone copper]

4pi=8pi/2

[ocean bison]

How is it still 8pi/2? 😅

[lone copper]

u don’t know 4pi=8pi/2?

[ocean bison]

Nope, but since you said 9pi/2 is 4pi + pi/2. In order to add them, you took the LCM didn’t you? So it became 8pi/2 and pi/2. Then it’ll be 9pi/2?

[lone copper]

yep

[ocean bison]

Oh okay thank you very much for the help! Thank you too Waler!

[lone copper]

lcd to be exact

[ocean bison]

Icd?

[lone copper]

lowest common denominator

[ocean bison]

Oh okay, thanks!

[lone copper]

np

glad to help

[ocean bison]

Oh how about if I have to find tan 7pi. Should I turn 7pi into degrees? But I only get a huge number

What should I do if I get huge numbers and I have to find sin and cos and tan? Should I subtract it to 360 until I get a number that I can find in the unit circle?

[brazen notch]

7pi means 3 whole cycles(6pi) and half a cycle(pi) around the unit circle, now if what u are trying to find is the coordinates of a point in unit circle that is being defined by "tan(7pi)" then it will be the same as "tan(pi)" and thus we can say "tan(7pi)=0/-1", (-1,0) is the coordinate

i just want to answer it, correct me guys if im wrong😅

[ocean bison]

Ah okay, Thank you very much @brazen notch

[brazen notch]

welcomee, glad to be of help

[exotic linden]

help please

[quiet bramble]

Can anyone help me with question 13? @ me

[ember stratus]

Is it possible to prove that BX=CY?

[humble pulsar]

for 13: which of those translations will keep the spacing of the hexagons together and get you between rows?

[distant inlet]

hello?

[low jungle]

its more like recreational math

[upper karma]

Need help with this question
https://cdn.discordapp.com/attachments/893758809988800582/894490825449418792/Screen_Shot_2021-10-04_at_6.25.13_pm.png
7a 7b and 7c

[candid forge]

Maths and chemestry crossover 😳

[nocturne remnant]

chemestry

[gritty onyx]

Please tell hoe to approach

[gritty onyx]

<@&286206848099549185>

[upper karma]

@junior light

reposted in #calculus as well

[junior light]

Thanks xie

[upper blaze]

This one solved?

and what are R1 R2 R3?

[sand zodiac]

help pls, go to #help-7｜zen1thxyz i asked there as well

ping me pls

[valid harbor]

@upper karma what are your ideas

[silk charm]

please stop trolling

[gritty onyx]

I could not solve

[gritty onyx]

R1, R2, r3 are radius of the circles

[balmy python]

How? It’s a problem I made my self and I’m genuinely curious how one would solve it…

[silk charm]

the problem is so poorly formulated that there is no question you're trolling. i'm muting you for a month.

[upper blaze]

Hahaha that question firstly doesnt belong here

[heavy pumice]

A circle of radius 5cm has a chord 4cm from the centre of the circle. The length of the chord is? Use circle geometry.

->Please mention when solved!

[silent plank]

we don't solve things for you here, but we can guide you towards a solution

[dark sparrow]

⬆️ @heavy pumice

[silent plank]

have you drawn a diagram

[heavy pumice]

Bro, I don't want a solution. The answer is 6cm. I need help to understand HOW to do it

[silent plank]

have you drawn a diagram

[winter pumice]

@heavy pumice I just did it on a piece of mail using only the description, and special right triangles. Sketch it out, you'll see it. No fancy circle theorems required.

[gritty onyx]

please anyone help

[loud shard]

can someone please help me with this question?

[valid harbor]

@loud shard Do you know the trig identities?

[loud shard]

a little bit @valid harbor

not very good at them

[valid harbor]

$\cos(a+b) =\ cosa \ cosb - \ sina \sin b$

[somber coyote]

Pealover

[valid harbor]

does it tells something to you

[smoky talon]

x = 5???

x = 30???

x = 40, y = 50???

[dire egret]

yes

[smoky talon]

so everything else is wrong???

y = 77 x = 48???

[dire egret]

yes

[smoky talon]

what abt the other 2 uo top???

step 2???

[dire egret]

i don’t think this is write

right

wouldn’t x be 35 and y be 40

bc same side interior angles add up to 180

[smoky talon]

2x+y = 130 2x-y=30

add the equations up and u get 4x = 180

*160

then

x = 40

so y = 50

[dire egret]

but then 2(40)+(50)+ 70 doesn’t equal 180

dk you mean 2x+y = 110

[smoky talon]

omegullu

im stupid

i added to 200

https://tenor.com/view/brain-trash-spongebob-squidward-dumb-gif-17233216

[dire egret]

lmao it happens

[smoky talon]

do u know how to do 2 column proofs???

[dire egret]

could you say alternate interior angle theory?

for 2

[smoky talon]

no

cause u have to prov the lines to be parallel

and the alternate angle theory

requires the lines to be parallel first

HOLY SHIT

IM BRAINDED

OML

KILL ME

IM SOO STUPID

THATS FOR THE OTHER 2 LINES

omegaulul

@dire egret Find the number of sides a polygon has if the sum of the measures of its interior angles is four times the
sum of the measures of its exterior angles

[dire egret]

let me think

[west perch]

I had a problem which I really don’t understand in physics that uses geometry and trigonometry. This man is standing 3 m from a kid who is 1,5 m from a mirror. The man is 1,5 m and the kid is 0,5 m tall. The eye level of the man is 1,45m and the eye level of the kid is 0,45. What is the minimum length the miroir should be for the kid to see himself completely? What is the minimum length of the miroir for the kid to see the miroir completely?

I drew out the problem but I still don’t get it

I know I have to construct right triangles

but given the eye level of the kid, it would no longer be a right triangle no?

[smoky talon]

u done thinking, i have to sleep soon

[dire egret]

i’m sorry i haven’t learned that stuff yet 😭

[smoky talon]

oof

[dire egret]

why can’t the mirror just be .5 meters tall?

[smoky talon]

the formula for interior angle is 180(n-2)

for sum of interior angles

and exterior add up to 360 always

[dire egret]

is this for the question?

[smoky talon]

ye

[dire egret]

yeah i have never even heard of that

[smoky talon]

so is it 10 sides

???

LETS GO

its 10 sides

boi

i am acting hella stupid today for sum reason

lol

[dire egret]

i knew u could do it

[west perch]

cause they’re asking for the smallest miroir possible ;-;

otherwise the miroir could be infinite measures

[strange river]

Does anyone know how to do this?

My brain isnt working rn

[silent plank]

after you made enough arguments, the last line would just be what you intended to prove

[upper karma]

https://cdn.discordapp.com/attachments/893758809988800582/895183221627711498/Screen_Shot_2021-10-06_at_4.38.06_pm.png Can somebody give me the working out and answer for this question

[dark sparrow]

@upper karma we don't give out answers here.

[upper karma]

Ok well what about the working

[dark sparrow]

giving out the working would amount to also giving out the answer

[upper karma]

so what do you guys help with

[dark sparrow]

we help you solve the problem yourself.

you tell us where exactly you're stuck (whether it is "i have no idea how to even begin" or something else), and we give you pointers in the right direction.

[upper karma]

yeah Im done anyway, all good

[upper karma]

Source:

I am correct at the 2nd and 4th problem in Area section?

I am not sure if bh/2 is applicable for all triangles

[lone copper]

it should be applicable for all right angled triangles (or when u can split them into right angled triangles)

[misty idol]

So I have the question "Find all the solutions for the equation tan(x) = -tan(2x)"

I did the following:

[fallen girder]

Looks good to me. Idk if you know unit circle stuff but you can go one step further to give x values that satisfy Tanx=sqrt3

[shut jackal]

How do you make theese 5 into a square?

[fallen girder]

Wtf

[shut jackal]

Nah i have been struggling with this

[silent plank]

those look like loose pieces

so it'd be probably easier for you to manipulate

[fallen girder]

Yep

[misty idol]

hmm

[silk patio]

@lapis pine

[lapis pine]

Still no idea

[mellow anchor]

kinda confused, for the y "coordinates" does the definition still hold for the distance if the x and y values are squared?

[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!

[charred flicker]

<@&286206848099549185>

If cos(t) =
−
5
19
and tan(t) < 0, find sin(t) and cos(-t)

[warm mauve]

pythagorean theorem

cos = x/r

so construct a triangle with x = 5 and hypotenuse = 19

solve for the other leg in refernce to theta

[sand jolt]

I'm having a bit of trouble here. I know that they'll all be parallel because the lines all have the same directional vector t(d,e,f) but I'm not sure how to prove that.

[sand jolt]

<@&286206848099549185>

[sand jolt]

Okay got through the above problem but now I'm on a new one. I would appreciate the help if anyone could provide it.

<@&286206848099549185>

[dark sparrow]

@sand jolt still need help?

[sand jolt]

Yes very much so.

I know how to map a line in the form of (a,b,c) + t(d,e,f) like in the previous problem but I'm not sure what to do when it's just flat on the object plane like that.

[dark sparrow]

wouldn't you just take the plane going through the line in the object plane and the viewing point and then intersect it with the image plane

[sand jolt]

What do you mean?

[dark sparrow]

i mean exactly what i said

[sand jolt]

What plane is going through the line?

It's a flat line on the object plane isn't it?

[dark sparrow]

we have our line in the object plane

consider the plane that goes through this line and the viewing point

this plane intersects the image plane along some line, and that will be our image

[sand jolt]

When you say "the plane" do you mean the image plane or the object plane?

Or is there a third plane?

[dark sparrow]

neither

i say "the plane that goes through this line and the viewing point" and i mean "the plane that goes through this line and the viewing point"

[sand jolt]

Oh wait. I think I get what you mean.

[quiet blade]

What is the maximum number of points that two lines can intersect at?

[lapis pine]

1

[quiet blade]

That means that the maximum occurs when each line intersects every other line at a unique point

[lapis pine]

Yes

[quiet blade]

How many ways are there to choose 2 lines from given 5?

[lapis pine]

7?

[quiet blade]

(We are trying to find all possible combinations, choosing any two lines gives us an intersection)

[lapis pine]

So i should use. PnC?

[quiet blade]

A combination can be used here

We have 5 lines, we need to find all possible ways to choose 2

[lapis pine]

Ohh then I can't solve it. I don't know C yet

Pnc

[crystal sable]

I'm grading a student's work and wanted to know if I should give this a 0 or 1 (no partial credit)

Show that if a circle and a square have the same perimeter, then the circle has the larger area

its not exactly circular reasoning, but I feel like it's bad proof writing to assume the statement is true, then equating it to another true statement

[humble pulsar]

what's the total mark allotment?

[crystal sable]

this question is 1 point

[humble pulsar]

and were they explicitly told in class to not start with the RTP?

[crystal sable]

RTP?

[humble pulsar]

required to prove.

[crystal sable]

they weren't explicitly told

[humble pulsar]

Then I'd personally give full credit cause it's "right" but add a note on proper format.

[crystal sable]

thanks

[humble pulsar]

This comes with the caveat of, since you asked in a HS channel, it is a HS student and thus more likely than not is only starting proofs

[crystal sable]

Oop

this is a college class...

[humble pulsar]

what year?

[crystal sable]

first years

most of them didn't take calc in hs

[humble pulsar]

Ok then yeah, full mark + note would be what I'd do

[blissful haven]

$cos(x)sin(x)+x(-sinx)(sinx)+x(cos^2(x))$

[somber coyote]

JPod

[blissful haven]

How can I simplify this down?

[signal swallow]

multiply the sins in the middle out and see if you see something

[blissful haven]

Yea, but from there

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

[somber coyote]

JPod

[signal swallow]

what can you factor out of the last two terms?

[blissful haven]

oof im dumb

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

[somber coyote]

JPod

[signal swallow]

right, now you probably more commonly see the identity written as $cos^2(x)-sin^2(x)$

[somber coyote]

a disappointing son

[blissful haven]

Gotcha

Thanks for your help

[signal swallow]

👍

[charred flicker]

<@&286206848099549185>

give me step by step please

[somber coyote]

Disorganized

[winter pumice]

hint:
use what you know about sine and cosine values on the unit circle (reference angles)

[charred flicker]

Sin=y/r and cos =x/r

[winter pumice]

why don't you sketch all the triangles on the unit circle that reference the 60 degree triangle, from the x-axis.
There are 3 triangles (4 including the 60 degree) that do this.
Evaluate their sines and cosines.

which angles match cos(60) and sin(60)?

[charred flicker]

60 degree triangles

Cos60=2/1 and sin60=1*sqrt3/1?

[winter pumice]

those are not the correct ratios, look at your notes or google them

you need to memorize the sine and cosine values of the "special right triangles" to pass this class, this is just one of the first applications

[winter pumice]

ugh, this isn't true, lol

[neat stag]

why is TO angle bisector and OT perpendicular to PQ

[livid moss]

It's because TP = TQ, so angle TPR = angle TQR, and then you can use similarity

[upper karma]

For some reason I can’t find the area for this quadrilateral

I tried cutting them and finding the triangle first

Please if there is an easier way please show me

[willow flame]

JPod

[stable maple]

draw a line parallel to side 7.5 so it make a parallelogram with sides 5 and 7.5 then find area of triangle formed using heron's formula and find the altitude then apply the formula of area of trapezium

[tiny stump]

wondering if i can get som ehelp with both qs pls

,rotate

[tiny stump]

idk if i drew the diagram for 1 right

and 2 i got no idea where to start

for q1 i thought the stake is 10 mettrs from the fence, so idk if it creates a sector or segment

<@&286206848099549185> .-.

[winter pumice]

@tiny stump the picture isn't quite right but I am also suspicious of the claim in Question 2

[tiny stump]

ahhh

yea q2 i just drew a circle

[winter pumice]

Circle is fine, its the radii you drew from the chord instead of the center that are wrong

Join Mathematics voice channel

[tiny stump]

ok one minute pls

[winter pumice]

Nvrmnd

Driving

[tiny stump]

ahhahaha

b careful

[somber coyote]

hehenrique

hehenrique

hehenrique

[tiny stump]

ahhhh

that makes sense

im still trying to wrap my head around the triangle formula

r^2/2 * sin(pi/2 -x)

nvm

i see it

ab is r^2

i dont quite get where the pir^2/4 came from tho...

the second line

[lethal heath]

one quarter of the area of the circle

[tiny stump]

ahhh yes

ok

ty sm

that helped alot

then u equate them and get x=-cosx

got it

ty

[lethal heath]

nop

edit: spelling, lol

[tiny stump]

njo?

ok

wait

gimme a min

so do we do the first line minus the second line?

wait a min

are u sure u dont equate them

if we dont equate them then what do we do

coz the first line is the area of the segment and the second line is also the area of the segment

then we rearrange to prove x

[lethal heath]

yes. first line equals second

[tiny stump]

my other idea is we use the trig graph?

but idk how to apply it here so i dont think we use it

i noticed i could factorise out the 1/2r^2

but i dont see how that helps

[lethal heath]

yes, you should do this

[somber coyote]

hehenrique

[tiny stump]

so factorised i got
$1/2r^2 [(pi/2 - x) - sin(pi/2 - x)]$

[somber coyote]

Kitsune

[tiny stump]

sin(pi/2 - x) =cosx coz of the graph right?

[somber coyote]

hehenrique

[tiny stump]

compound angle right

OK

that makes sense

coz sin pi/2 =1

and cos pi/2 =0

which equals cos x

ok

so we can simplify the factorised expression to:
$1/2r^2[(pi/2 - x) - cos(x)]$

[somber coyote]

Kitsune

hehenrique

hehenrique

hehenrique

[tiny stump]

YES

I GOT THAT

ty

ty sm

i felt big brain i wont lie

ty

i went the longer way i times everything by 4 then canceled 4

ty

then canceled r^2 and was left with pi on both sides

then i solved form there

[lethal heath]

glad to help. lemme see if i can figure something about question 1

[tiny stump]

ahhahah that would be great

u explained extremely as well, im thankful u didnt give me the straight up answer !

[lethal heath]

ok, in first question the whole goal is to find the area of the segment. I trust you now are able to find its area if you are given the central angle right?

so let's focus in finding the angle

[tiny stump]

ye

[lethal heath]

nvm the writings above

[tiny stump]

just drew that

except without the 20m

[lethal heath]

they are all radii

hence 20m

10\sqr(3) comes from Pythagorean theorem

[tiny stump]

yep

[lethal heath]

now apply cosine law to the triangle with sides 20, 20 and 20sqr(3)

and boom

[tiny stump]

is the area of the 1011?

fu--

i dont think it is

one min

its cosine 1/2 right

which is cos 60

so the angle is 60

[lethal heath]

check if you didnt leave a negative sign behind

[tiny stump]

so its cosA = 20^2 +20^2 - (20root3)^2 all over 2*20^2

simplifies to cosA=800-1200 all over 800

-1/2

[lethal heath]

yep

[tiny stump]

gotcha

[lethal heath]

hence A=120 degrees

[tiny stump]

yessir

then its segment formula form there?

[lethal heath]

yes, the one we used before

sector - triangle

[tiny stump]

i convert 120 to radians correct?

[lethal heath]

yes, 2/3 of pi

[tiny stump]

411?

411.56

57*

i was in the wrong mode

shat

245.67

246 nrst sqr m

[lethal heath]

i hope thats right, i dont intend to do the calculations haha

[tiny stump]

yea im pre sure its right

tysm

:))))

[true hound]

Help pls

[turbid wave]

Anybody

¿

Yelp

[void moat]

can u flip it plz

[true hound]

Like?

[void moat]

on my screen its sideways

flip the picture horizontally

[void moat]

now its in portrait

ok thanks

[true hound]

Np

[void moat]

so

it says that 3costheta = 5sintheta

which means that u replace 5sintheta in the equation with 3costheta

therefore u have

[turbid wave]

.

[void moat]

uhhh how do i get the bot working

[true hound]

..

[void moat]

i am trying to show my work using the bot

[true hound]

Texit?

[void moat]

yes

do u know how

[true hound]

,

,help for usage

[somber coyote]

Command for usage not found!
Use the ,list command without arguments to see a list of commands.

Oct 9, 2021, 7:04 PM

[true hound]

,list

[somber coyote]

My commands!

Use ,ls to obtain a briefer listing, and use ,help <cmd>to view detailed help for a particular command, or ,help to view general help.

If you still have questions, talk to our friendly support team here.

LaTeX Rendering

Render LaTeX code and configure rendering options.
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ tex: Render LaTeX code.
​ ​ ​ ​ ​ ​ ​ ​ ​ ctan: Searches the ctan
​ ​ ​ ​ ​ ​ ​ texdoc: Searches the texdoc
​ ​ ​ ​ ​ ​ autotex: Toggle whether your LaTeX is automatically rendered.
​ ​ ​ ​ ​ preamble: View or modify your LaTeX preamble.
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guildpreamble: View or modify the guild's default LaTeX preamble

Guild Admin

Guild administration
​ ​ ​ ​ ​ ​ ​ ​ config: View and set the guild configuration.
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forgetrolesfor: Forget stored persistent roles for one or all members.

Page 1/4

[true hound]

Ahh@void moat try idk

[void moat]

ok

i will try

[void moat]

i will help u tho dont worry

i will make u understand the problem

[true hound]

Ty

[void moat]

i am solving the problem

then i will go step by step

@true hound the first thing i would do is replace the 5sinthetas with 3costhetas

and remeber that sec^3theta = 1/cos^3theta

[true hound]

Yes

[elfin walrus]

Can you help me with this task, because I can't even remember how to start:
The segment CD is the height to the hypotenuse in the right-angled triangle ABC. The points M and K are the midpoints of AD and CD, respectively. If AC: BC = 7: 9 and CM = 14 cm, then the length of the segment BK is?

ex. 8

[elfin walrus]

<@&286206848099549185>

[fluid wasp]

can u solve?

[upper karma]

https://socratic.org/questions/how-do-you-prove-cotx-1-tanx-tanx-1-cotx-1-tanx-cotx

[tiny vine]

Hey all, I want to learn trigonometry and I have no background on it

Im doing well in algebra but I need a good source to learn trig from scratch for school and self-taught calculus

[livid moss]

Khan academy

[tiny vine]

...

Thats it?

Just mr sal khan himself?

[wise hornet]

it works

gods i'm late

[primal lodge]

I am working on deductive reasoning does anyone know a good site to work on it

in geometry

<@&286206848099549185>

[upper karma]

Is there a method that can give me the volume of overlap between two ellipsoids? So far I have discovered algorithms that can check whether two ellipsoids collide but they don't tell me the exact volume of overlap. There is also the monte carlo approach but it is not an exact solution.

[gaunt vigil]

can anyone help me with this one https://cdn.discordapp.com/attachments/509460208737845248/897272369893220372/unknown.png

[junior veldt]

Do you know what bisectors are?

Do you know what these mean?

[weak jungle]

it means the angles are equal

[junior veldt]

I was asking Sukkix/lh

[junior veldt]

i saw that emoji change /j

[humble minnow]

Hi guys,How do I need inverse trigo function for this question???

I'm not looking for the answer,but how to solve it..Thanks!

[upper karma]

I still can’t find the area 😭

[sudden haven]

There has to be some extra information to “fix” the quadrilateral so that it has a specific area.

Is something other than the 4 sides given?

[upper karma]

@sudden haven I’m only in high school and I still don’t know how to do geometry 🥲

[sudden haven]

You aren’t supposed to do this by hand right?

[upper karma]

Yes I’m not sure how to start

[sudden haven]

Two triangles on either side and trapezium in the middle

[upper karma]

How can I get the length using the degree 79?

[sudden haven]

6sin(79) is the distance of D from AB and 7.5sin(67) is the distance of C from AB

Let foot of perpendicular from C to AB be E and D to AB be F. So, DCEF is a trapezium with height EF=9-6cos(79)-7.5cos(67) and parallel side lengths as 6sin(79) and 7.5sin(67)

[upper karma]

How did you get that ????

Oh wow

[sudden haven]

Basic trigo

[upper karma]

And then do the Pythagorean to find the distances for the triangle

And add them up?

[sudden haven]

Pythagorean isn’t required for area of right angled triangle though

Area of DFA is (6sin(79))(6cos(79))/2

[upper karma]

Can I use that formula for the CEB?

[sudden haven]

Yeah

(7.5sin(67))(7.5cos(67))/2

[upper karma]

I’m just curious I never seen this formula before

Where can I find it?

[sudden haven]

It’s just 1/2*base*height lol

DF=6sin(79), AF=6cos(79), angle DFA=90

[upper karma]

anyone know how to mae this in geogebra

[fallen garden]

It looks like the first part requires an inverse function. You find what A is equal to with inverse cos, then you find the function value of A halved

cos of A/2

Note that it's in quadrant 3, where cos is negative @humble minnow

[storm portal]

You can probably just experiment with lines and such

[upper karma]

A = 2 - 4sinx
B = 3 + cosx

A(highest) x B(lowest)

Help

[sudden haven]

A(highest)= 2+4 and B(lowest)=3-1...

[neat stag]

is NML~PQR same as MNL ~ QPR

[covert walrus]

Anyone that may guide me through this? Idk whether this belongs to multivar calc or geometry xd

[dark sparrow]

<@&268886789983436800> ban please

[upper karma]

Guys, guide me doing the number 6 please

[gentle quarry]

Guys do anyone knows about graphing in math?

[upper karma]

try desmos

if u mean graphing functions online

[gentle quarry]

ok thanks

[winter pumice]

See DM

[somber coyote]

Disorganized

[coarse ruin]

Hi can anyone help me out with question 3i&ii) thanks

[neat stag]

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.

[nocturne remnant]

Apollonius’ Theorem can help you secure the third side length

[south forge]

hey can anyone help with this question i have to find x

i have no idea if im in the right server for this but hopefully i am

[lone copper]

law of cosines

[noble plover]

How to solve this.?

[hasty solstice]

use cosine rule

[sonic fern]

https://youtu.be/PDWPnK6pjrw

[covert walrus]

Are there any identities for (cosx)^3 and/or (sinx)^3? If not standalone, maybe in some type of relationship (similar to (sinx)^2+(cosx)^2=1)?

[potent orbit]

Could someone ln a hand here?

[covert walrus]

Anyone? xd

[potent orbit]

I don't know it, sorry

[covert walrus]

Say I have some substitution for cosx = u
How do I alter u so that I have a value for it when I transform cosx to cos(2x)

[humble pulsar]

just use double angle...

[languid ginkgo]

hi, how do i tell whether or not three coordinates, for example (3, 51), (-9, -324) and (12, 204) will make a linear equation or not?

[south forge]

yeah i got it now thank you though!

[quiet blade]

A lot of ways. Simplest one is to check the slope of lines the points go through.
The slope is (y2 - y1)/(x2 - x1) and see if for point A and B, their slope is the same as B and C.

[kind sedge]

It’s perpendicular to the y axis

And then rotated around the y axis

[uncut hedge]

your segment would be something like this. perpendicular means that the angle between the segment and the y axis is 90 degrees

The question is asking you: after rotating this line around the y axis what shape does this trace out?

[kind sedge]

A circle

I’ll be here all night

[uncut hedge]

yep

[lament vortex]

how do you apply the 6 circular functions (e.g., cos, sin, tan, sec, csc, cot) in real life?

[winter pumice]

That sounds like a question your teacher should have answered, but you said it like a question your teacher should have assigned as homework.

Either way, here you go:
https://www.mathnasium.com/real-life-applications-of-trigonometry#:~:text=Trigonometry is used to set,a point in the sea.

[wanton portal]

Hii

[agile geyser]

help me guys on geometry

i have posted the question in question 2

[wanton portal]

I have answered it there, you can check it

[agile geyser]

can i see ur solutin i just want to see how you did it

solution*

[wanton portal]

Does that match the answer you were comparing it against?

[agile geyser]

no its far

may answer is 0.94

ahahahahahaa

but i doubt my answer