geometry-and-trigonometry

[narrow sleet]

@woeful flame -1.1023 not -1.0123

@pallid comet We simply it actually

So we just find an expression of x in terms of n

[woeful flame]

yes i fixed it to -1.1023 + pi

but it still wrong

maybe there is a mistake in the book? o_O?

[narrow sleet]

Try 3 decimal places

[woeful flame]

what is it?

-1.102 ?

[narrow sleet]

I think

[woeful flame]

tried

wrong

[narrow sleet]

Pi - 1.102?

[woeful flame]

still wrong

[narrow sleet]

Then Idk🤷

[woeful flame]

ok thank you so much guys

you both helped me very much

and there is probably a mistake in the book

[narrow sleet]

I hope there is

Wlc

[woeful flame]

it happens i hate that

[vital meteor]

How do you solve those sinosodial equations with two trig functions

[pallid comet]

give an example and i'll see if I know it

@woeful flame well, at least you know just a bit more than you did 19 hours ago lol

[lament bay]

Which axiom is it?

[narrow sleet]

Can we look at the diagram?

[lament bay]

Sure

@narrow sleet

[narrow sleet]

Why AB -AD = AC - AE instead of AB - AE = AC - AD

[lament bay]

oh, my bad

[narrow sleet]

Well I don't know what axiom is that. Probably other people can help

[lament bay]

ok

[lament bay]

<@&286206848099549185>

[waxen gorge]

,rotate 90

[waxen gorge]

Use similarity

There r more specific rules I guess

Think about the side correlations

[lament bay]

idk 😭

[pallid comet]

Can someone prove for me this statement from wikipedia?

"The size of a hyperbolic angle is twice the area of its hyperbolic sector."

[lament bay]

@waxen gorge can you show me the steps

[waxen gorge]

@pallid comet didn't someone already post

[pallid comet]

I posted something resembling that, now i have a different question

[waxen gorge]

Well what's the area of a hyperbolic sector

[pallid comet]

that's a great question.

[waxen gorge]

https://youtu.be/h-gKfQkHQSA

Just watch this tbh

[pallid comet]

I have

[lament bay]

ok leave me hanging i see

[pallid comet]

lol sorry @lament bay I'll go now

[worthy igloo]

not unlike the circular case

where the area of a circular sector bounded by (0,0) - (cos t, sin t) is t/2

[plucky marlin]

why not use some beautiful integrals instead of hyperbolic things

[waxen gorge]

@plucky marlin that's like telling someone to integrate a circle to show (what mniip said)

[plucky marlin]

[real sparrow]

[narrow sleet]

[worthy igloo]

well you know how the area of the unit circle is pi

it should be clear from geometric reasons that the area of a sector of 2pi/k radians is pi/k

from similar geometric reasons it's clear the area of a sector of 2pi n/k radians is pi n/k

and then with a bit of analysis it's clear the same holds for any real number

[timber hinge]

"bit of analysis"

[left folio]

I have to compute the limit of x that approaches 0 of sin(x)/cos(x)

The answer would be 0 but can someone explain to me why that is, given they change at the same rate as x changes?

I understood it, sorry

[supple bramble]

How do you prove that the hypotenuse is shorter than the sum of the sides in a right triangle?

[narrow sleet]

2ab> 0
a^2 + 2ab + b^2 > a^2 + b^2
(a + b)^2 > c^2
a + b > c

[plucky marlin]

: O

[narrow sleet]

@supple bramble

[supple bramble]

What does a^ mean?

Is it just a^2?

and you missed the 2?

[soft kraken]

Yes.

[supple bramble]

I don't know how that proof is even possible

but I love it

[plucky marlin]

ikr

its goddamn beautiful

[narrow sleet]

I mistype it. Sry

[supple bramble]

Have you seen it before or did you come up with it on the spot?

[soft kraken]

That's a common proof.

[narrow sleet]

I thought about it before

And alao it's a common proof

[soft kraken]

This is like basically a corollary of the triangle inequality theorem

[narrow sleet]

^

[woeful flame]

hi

in this quesion there is triangle
ABC
angle a against alfa angle b against beta and angle c against gama
i need to use the Law of Sines
but in question 1 im getting two results
44.778 that is correct by the book
but im also getting alfa=135.22
(180-44.778)
i dont understand why it isnt a result aswell?

that is the Law of Sines

[empty moss]

@woeful flame It has to do with the order. There is a specific order for different triangles when using the law of sines. The answer of 135.22 degrees is wrong due to the fact you used the wrong order.

[woeful flame]

@empty moss what do you mean

?

look at the picture sir

there are two results also in the picture

what do you mean by the "specific order"?

i never heared about somthing like that

i know that if i got the smaller angle in the data, i could have 0\1\2 results

and when i get the bigger angle in the data I could have only 1 result

but here a is bigger then b

so angle Alfa must be bigger then 25 degree

and also 44.778 and also 135.22 looks correct

[empty moss]

44.778 and 135.22 are the angles of A2 correct?

or are you talking about A1

[woeful flame]

A1 = 135.22

and A2 = 44.778

[empty moss]

ah ok

[woeful flame]

ohhhh

but they are asking for angle alfa

[empty moss]

^

[woeful flame]

so Alfa is A2

right?

[empty moss]

yes

[woeful flame]

dayum

waste almost 2 hours for that

thank you very much dude

[empty moss]

glad you got it

[woeful flame]

wait but there are still

two results

@empty moss no one said that Alfa must be A2 and not A1

[empty moss]

first you gotta find the C value and the A2 value in the triangle of A2BC

Then, using that information, you can solve for the rest of the angles

there are 3 triangles,. you have to solve for all of them

theres 3 boxes due to the fact that there are 3 total triangles

[woeful flame]

no

im talking just about question 1

about angle Alfa

[empty moss]

ah

[woeful flame]

I dont get why there is only 1 result

and not two

can you explain to me why?

[empty moss]

for the triangle of A1BC?

When solving for the length of C and the angle of gamma, did you check your answers?

the angle of gamma cannot be less than the angle of Alfa is side A is bigger than side C

[woeful flame]

you dont get me sir

for angle-Alfa = 44.778 there is one result for C

and for angle-Alfa = 135.22 there is different result for C

[empty moss]

Ok

[woeful flame]

like 2 cases

but both of them are correct

[empty moss]

when alfa is 44.778, is the length of A greater than C?

[woeful flame]

i dont understand

[empty moss]

Ok

you know Pythagorean theorem. The length of the side across from the right angle is always the biggest right?

[woeful flame]

wait what?

i dont understand this words

i probably know that but just dont understand your languege

[empty moss]

Ok I will find picture

This

[woeful flame]

leg^2 + leg^2 = hypotenuse^2

?

[empty moss]

yes

[woeful flame]

Pitagoras?

[empty moss]

the Right angle, or the red angle

is the biggest in this triangle right

[woeful flame]

90 degree must be the biggest angle in triangle

[empty moss]

yes

[woeful flame]

thats what you mean?

[empty moss]

so the side across from the biggest angle must be the largest

[woeful flame]

im just getting 2 results

[empty moss]

otherwise it isnt triangle

[woeful flame]

and dont understand why the second result is wrong

can you tell me why?

gama = 110.222

and c = 13.322

but when Alpha = 135.22

gama = 19.78

and c= 4.804

like there are 2 cases

if Alpha = 44.778
so: Gama = 110.222 and c = 13.322

if Alpha = 135.22
so: gama = 19.78 and c = 4.804
.
(2 cases)
.

why the second result is wrong?

i dont get it

that is my question

[empty moss]

I am unable to answer this question, because I don't completely understand why it's wrong either

[woeful flame]

ok sir thanks anyway

[upper karma]

Hi

Is sin60.sin10-cos60.cos10 equal to -cos(60+10) or cos(60+10) ?

[lament bay]

https://cdn.discordapp.com/attachments/531171430323322882/531171898223099934/image0.jpg

not sure what to do

[dapper patio]

CF and EF are parallel, likewise DE and CF for it to be a parallelogram

angle DCF = angle DEF

idk really something maybe to help

[lament bay]

<@&286206848099549185>

[upper karma]

@lament bay hey

[lament bay]

I’m struggling 😔

[upper karma]

So the basic definition of a parallelogram is a four-sided rectangular polygon with opposite sides that are parallel

Assuming our triangle is an equilateral, we can use properties of angles to prove that CDEF is a parallelogram

So we need to prove that DE and CF are parallel, as well as CD and EF are parallel

[lament bay]

Ok

[upper karma]

One sec

We can use angles to prove this, but do you know what the internal angle of an equilateral triangle is?

[lament bay]

Uh not really

[keen tangle]

That's not an equilateral triangle though

[upper karma]

Oh wait, I thought the problem said that

No worries, instead of 60 degrees we can label angles with variables instead

[keen tangle]

I'm assuming you wanna prove by similarity, then yeah

[upper karma]

@lament bay you still there?

[lament bay]

Yes

@upper karma

[upper karma]

I know what approach we can take but I'm having a problem proving that the midpoints of AC and AB are on the same level

This is a skewed version I've drawn so we don't assume it's equilateral

[lament bay]

Ok

[keen tangle]

ik I'm late but you can solve it with similar triangles @lament bay ?

[upper karma]

Oh.... I completely forgot about that...

I guess we can use that intuition to prove that CDE = 180 - DCF

[serene field]

🤔

Triangle Midsegment Theorem?

[keen tangle]

I mean, that literally gives you the answer straight away

And also how would you prove that without similarity?

[lament bay]

Gonza

Do you know indirect proofs?

[keen tangle]

Yea ofc

[lament bay]

https://cdn.discordapp.com/attachments/531195604487897116/531239389825859584/image0.jpg

I think I know how to do the first one

but I dont know the specific theorem

[keen tangle]

What specific theorem?

[lament bay]

I need to just prove that since the medians are not congruent there is no way for the sides to be congruent by doing some sort of calculation or theorem that creates a falsehood

[keen tangle]

Well you can do as they said

Assume two sides are congruent and prove that medians are congruent

That's what indirect proof means

Also (tbh most commonly) known as proof by contradiction

[lament bay]

uh huh

https://cdn.discordapp.com/attachments/531184651914706954/531209138764513281/image0.jpg

(-1.5,2) (-3.5, -1.5) (-2,-1.5)

Are my answers right?

<@&286206848099549185>

[vital frost]

Need help with 14-17

[serene field]

Recall your trig functions, specifically SOHCAHTOA

[frigid wasp]

This is prob a stupid question but is there a solution to sin(x)=i?

Sorry if this is the wrong channel

[clear haven]

maybe square both sides idk

lol

no clue

am being dumb rn i am

[umbral snow]

@frigid wasp
Not a stupid question, there are solutions to that yes. If you go on to take complex analysis you'll learn more about solving it

[waxen gorge]

@frigid wasp you want to solve sin(x) = i?

Well we can use the complex definition of sin

Do you know what it is?

$ \sin{z} = \frac{e^{iz} - e^{-iz}}{2i} $

[somber coyote]

Pleb':

[waxen gorge]

So we can set that equal to i

We get: $ e^{iz} - e^{-iz} = -2 $

[somber coyote]

Pleb':

[waxen gorge]

You can go solve that using several techniques

In the end however you get

Wolframalpha much owo

Hm it seems there are better forms

[devout moon]

yo guys

does anyone know how 3D vector directions work? the one with three angles

[narrow sleet]

What? Example?

[devout moon]

so i read that you can represent diection in the 3d planes

through three numbers, each denoting a rotation in each axis

like[90,50,30]

[real sparrow]

Use the direction ratios I guess ?

[frigid wasp]

Thanks @waxen gorge

[shadow anvil]

@devout moon you mean polar vectors?

Because only two of the numbers are angles and one is the magnitude of the vector

[primal oyster]

Help me solve the thing, for my math homework. Thanks <@&286206848099549185>

[astral hornet]

I guess Try to write some angle sum relationships and use substitution

[astral hornet]

Also, the line from right angle perpendicular to hypotenuse forms similar triangles

[upper karma]

Any help with this? Can't see a linear algera topic: https://math.stackexchange.com/questions/3063813/adjoint-operator-g-ffff

[upper karma]

How would I solv 3cos^2 x=7sinxcosx for 0 <= x <= 360

[dire rampart]

try dividing both sides by cos^2x

[upper karma]

Thought so

Been off school for 2 weeks my head doesn't work u know

:p

so

3=7sinxcosx / cos^2x

then is it something to do with sinx/cosx = tan

[dire rampart]

yup

[upper karma]

7tanx cosx/cosx

so 3=7tanx

tanx=3/7

?

[dire rampart]

yes

[upper karma]

tyty

And then for the range

Find the first value

and add 180

[dire rampart]

mhm

[orchid solstice]

anyone know how to do question 3

[astral hornet]

use trig ratios to fill in other sides and angles

[tawdry pivot]

@orchid solstice use sine rule

[gloomy slate]

Hi, anyone here? I got a question if you guys dont mind

[narrow sleet]

[gloomy slate]

On this bottom part of the example, why is it when he square roots the sin that it ends up with a + - before the answer? I havent taken a math class in almost a year and kinda forgot

[narrow sleet]

Ex: x^2= 4

x = 2 and -2

So x = +/- 2

Same ide with that

[gloomy slate]

oh that makes sense thanks

I got one more question if you dont mind

In this question it says that the theta is in quadrant 3, I got 1.05 after using the correct identity, the answer IS 1.05 but shouldn't it be -1.05 because it is in quadrant 3?

[hard gale]

should be 1.05

in q3, sin and cos are both negative

[gloomy slate]

is that something that should just be memorized or?

[hard gale]

it's cool for sanity checks like that, but it's not super essential to remember it

so how did you get cot ?

[gloomy slate]

I plugged -1.45 into this identity

then did the algebra

I had a problem earlier that said to make sure that your answer is negative because it was in quadrant 3, so I assumed it applied to all signs

But I guess its only for sin and cos

[hard gale]

yep

[gloomy slate]

Alright cool thanks

[hard gale]

👌

[marble perch]

Aaaaaa helppp

I have to solve for X

this is an interior polygon

[upper karma]

here im pretty sure you add up the number of sides to find the sum of the interior angles @marble perch

so

$$sum=180(n-2)^\circ$$ where n is the number of sides

[somber coyote]

Jaboi:

[marble perch]

ty

[upper karma]

@marble perch hey, did you find the answer?

[marble perch]

yes

[upper karma]

[upper karma]

this makes no sense to me

shouldn't first term be y = x + 4?

no nvm

I see now

what do you need help with luv @upper karma

Well you see, for y = 10 - x

at x = 6 (R)

it is at the top of the right corner

so y = 10 - 6 = 4

So you know the height is 4

And you know that to find an area you just need height * base (length)

You know that at y = 4 is the top

so for y = 5 + x = 4 you can easily see that x = - 1

and you are already given outer length 6

how far is 6 from - 1?

Good, so O = -1

And now you know that length is 7

and height is 4

so 28 units^2 to be exact, but did you understand why?

did you follow?

i am not confident in my explainations because I explain it how I would understand

i should stop helping

[tawdry pivot]

For a convex polygon with equal sides, prove that the sum of all perpendiculars, drawn from a point P inside the polygon, stays constant.

I've done it for a triangle, no idea how to proceed.

[upper karma]

square

[tawdry pivot]

Just keep trying more shapes, like square rhombus, pentagon etc... until something clicks?

[upper karma]

No

Or you could do proof by brute forcing all convex polygons but that would take a while

[tawdry pivot]

there's an infinite amount of them?

[upper karma]

There may be, but you are supposed to prove that the sum of all perpendiculars stay constant

Just like the sum of all angles in euclidean triangles are 180

it is constant

[tawdry pivot]

Looked at the answer and all you do is draw lines from P to the corners of the polygon, and sum up all the triangles. rip me I thought this would have a hard solution.

[upper karma]

i kind of gave you that hint

[frozen saffron]

Can u guys help me

Show me how it's done and I'll just do the rest

[upper karma]

I need help with one geometry question

can anyone help me

[astral hornet]

since t bisects ge and o bisects gm, that makes to midline

and by midline theorem, it'll be parallel to em

you can find gto as linear pair with eto

and tog with parallel + transversal angle relationships to emo

then angle sum theorem to find g, then e

[upper karma]

can anybody help me with a proof question on my homework?

[thorn talon]

what's the question?

[upper karma]

sorry that’s all blurry

[thorn talon]

that seems annoyingly methodical

but ok

[upper karma]

lmao

[thorn talon]

well

you know two pairs of angles are equal

[upper karma]

yes

[thorn talon]

so then, the two triangles have to be similar right?

[upper karma]

mmhm

[thorn talon]

because if two are equal

then the third angle has to be equal

so those two triangles are similar because they are equiangular

[upper karma]

oh

[thorn talon]

and then you can basically repeat the same process for the other triangles

[upper karma]

hmm

this is dumb

[thorn talon]

it is

i don't see how this homework is supposed to be useful for geometry proofs tbh

[upper karma]

🤷

can u also help me w one with actual numbers

[thorn talon]

ok

[upper karma]

the length of the hypotenuse of a 30-60-90 triangle is 17 yards. what is the length of the other two sides?

[thorn talon]

one sec

something like that

just spam out your trig ratios basically

[upper karma]

we haven’t learned trig yet woops

[thorn talon]

rip

ok other way

hmm

do you know where this triangle is derived from?

[upper karma]

wym

[thorn talon]

this special triangle

do you know that it is made from cutting an equilateral triangle?

[upper karma]

oh yeah

[thorn talon]

well, we cut an equilateral triangle in half

the hypotenuse is from the original

the bottom side however

is cut in half

the equilateral triangle originally had all sides with 17 yards

now we cut it in half

so the bottom side should be 17/2 yards right?

[upper karma]

oh yeah

8.5

[thorn talon]

and then pythagorean theorem for the left side

yeah

[upper karma]

so 8.5 and 8.5 square root of 3?

[thorn talon]

yeah

[upper karma]

yey

ty

[thorn talon]

np

[upper karma]

i have one more question hehe but i don’t wanna seem annoying

[thorn talon]

it's fine

post it

[thorn talon]

you really haven't done trig?

[upper karma]

nope

[thorn talon]

ok that's fine

[upper karma]

florida has really bad education

[thorn talon]

do you know where this special triangle comes from?

[upper karma]

um

a square cut in half?

[thorn talon]

yep exactly

so then the side lengths must be equal right?

[upper karma]

oh yeah

[thorn talon]

essentially x^2 + x^2 = (18sqrt2)^2

solve for x to get the two side lengths

[upper karma]

and then i just have to add right

[thorn talon]

yep

exactly

[upper karma]

i got 648 for 18sqrt2^2 am i dumb

oh wait nvm

[thorn talon]

nah

that's fine

it's correct

so 2x^2 = 648

and solve further

[upper karma]

div 2 equals 324

[thorn talon]

yep

[upper karma]

324 rooted is uh

18

[thorn talon]

yep

so 18 is the side length

[upper karma]

wait which one

AR?

[thorn talon]

IT = TG = AR = 18

[upper karma]

oh ok

[thorn talon]

IT and TG equal cause special triangle

and equal to AR cause the sides match up

[upper karma]

so

104 is the perimeter

[thorn talon]

hmm

don't think so

[upper karma]

oh

i did 18+18+(43-18)+43

[thorn talon]

that is mostly correct

but you shouldn't have two 18's at the start

TG is 18 and that's fine

IT is not part of the perimeter though

you need IG which is 18sqrt2 instead

[upper karma]

oh yeah i’m dumb

[thorn talon]

it's fine

mistakes happen

[upper karma]

so just 86

[thorn talon]

then you add the hypotenuse of the triangle

should be 86 + 18sqrt2

[upper karma]

25.5

111.5

[thorn talon]

yeah

if you need to round to 1 decimal place

[upper karma]

yeah it said round to the nearest hundredth

[thorn talon]

o

need to go one decimal place more then

[upper karma]

then 111.56

[thorn talon]

your first decimal place shouldn't round

your next one is the only one that should

[upper karma]

woops 111.46

[thorn talon]

yep

that's good

[upper karma]

yay

thanks so much

sorry for being dumb

[thorn talon]

Np

It's fine

Part of the learning process

[acoustic fiber]

need help, 10(a) I got the answer for b, it's 9cm squared

[plucky marlin]

https://prnt.sc/m4hvgn

9+7+6+6 = 28

[acoustic fiber]

how did you calculated 7 cm^2?

[hard gale]

width of 7cm, height of 1cm

[acoustic fiber]

thanks @hard gale and @plucky marlin

[plucky marlin]

anytime fren

[hard gale]

[plucky marlin]

[acoustic fiber]

[clear haven]

[tawdry pivot]

[gritty siren]

[dire rampart]

[upper karma]

[umbral snow]

[limpid basin]

[sly charm]

anyone wanna do some trig w/ me

[wild hamlet]

[sly charm]

ok 😃

\csc \left( \frac { 5 } { 6 } \pi \right) \cos \left( \frac { 1 } { 6 } \pi \right)

whats the latex command

tex=\csc \left( \frac { 5 } { 6 } \pi \right) \cos \left( \frac { 1 } { 6 } \pi \right)

[wild hamlet]

$\csc \left( \frac { 5 } { 6 } \pi \right) \cos \left( \frac { 1 } { 6 } \pi \right)$

[somber coyote]

Yeat:

[sly charm]

ye

wanna walk me through this guy, not just the answer? 😃

cos idc about the answer

[wild hamlet]

you find the values for each then multiply them together

you can use a unit circle

[loud cedar]

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

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

[wild hamlet]

https://upload.wikimedia.org/wikipedia/commons/4/4c/Unit_circle_angles_color.svg

csc(x) = 1/sin(x)

[sly charm]

can you explain this summation identity

sin(x+y)=sin(x)+cos(y)+cos(x)+sin(y)

[wild hamlet]

$sin(x+y)=sin(x)cos(y)+cos(x)sin(y)$

[somber coyote]

Yeat:

[wild hamlet]

There's a geometric image that proves this. lemme find it

http://daylateanddollarshort.com/bloog/wp-content/uploads/2013/03/angaddsub.jpg

[sly charm]

1 is 5pi/6 in this case?

the radius?

hypotenuse

[wild hamlet]

the radians, not radius

1 is always the hypotenuse

[sly charm]

and thats at angle 5pi/6

[wild hamlet]

yes

[sly charm]

how do i find the roots for a radian

for example 2 pi/3

what is sin cos and the cooresponding root, and why?

[upper karma]

@sly charm still need the answer?

[haughty breach]

hello, how would i solve for the lengths for number 17?

[plucky marlin]

spam trig

[empty moss]

Use trig functions such as tan and cos or tan and sin in order to find the missing values

otherwise known as "spam trig"

[haughty breach]

how do i know which one is adjacent

[plucky marlin]

that depends on the angle ur working with

[empty moss]

what does adjacent mean?

like definition

[haughty breach]

next to

[empty moss]

ok

with any angle (other than 90)

[haughty breach]

I see I see

[empty moss]

Good job

[plucky marlin]

there's no fixed adjacent side bruv

it depends on the angle ur working with

[empty moss]

^ (You always (only like 1 case in which there isn't) have one that isn't)

[haughty breach]

Ok imma work w the 62 degrees

[plucky marlin]

so then adjacent would be BC

[haughty breach]

opposite AC

[plucky marlin]

yeah

[haughty breach]

hypotenuse BA

[empty moss]

yes

[haughty breach]

Oki

[empty moss]

the hypotenuse stays constant

so you always know adjacent and opposite

[haughty breach]

hypothenus is always across the 90 angle correct?

[empty moss]

yes

[haughty breach]

so if i wanna find BA

i would have to use csc?

[empty moss]

sin^-1

different things

csc is different from sin^-1

theres (should be) a button on your calculator for sin^-1

[plucky marlin]

did u just assume what button his calculator has

[empty moss]

I put should be

ok

[empty moss]

yes that one

nice job

well actually you don't need that one

[plucky marlin]

@empty moss

[empty moss]

r\woosh

You don't need sin^-1

[haughty breach]

okay

[empty moss]

you do sin(62) = opposite/hypotenuse

then you input values you know

[haughty breach]

got it

[empty moss]

Explain joke plz @plucky marlin

[haughty breach]

i got 24.91

@empty moss

[empty moss]

ok

lemme check

=pup 22.6/sin(62)

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@empty moss

[empty moss]

howd u get that?

[haughty breach]

i forgot .6

whoops

[empty moss]

ahhh

[haughty breach]

thank you!!

[empty moss]

Enjoy the trig

Your Welcome

[haughty breach]

[upper karma]

can anyone help me with this

question 36

<@&286206848099549185>

[astral hornet]

use point slope form to find equations of the orthocenters

then use those equations to find the 3rd point

[upper karma]

dang it

it's a simple question

and yet I didn't know how to do it FUCL

FUCK

[magic chasm]

Not sure I should ask here but I got this function in a trig class: pi * x/(1+x^2)

The range should be R

But in the correction the prof wrote R \ { -1,1}

[unreal flame]

need help
I need to write "m" with using of a, b, c, and gama(y)
data: BD = DA
i think i should do it with cosinus sentence
but i dont know what to do

I used the Law of Cosines

and got

c^2 = b^2 + a^2 - 2ab*cos(y)

but i need to use "m" also
and I dont see how can i do that

[torpid horizon]

@upper karma

do you really have the balls to take out ur phone and take a pic of your math test?

[haughty breach]

Hello, i need some help: how do i do number 19?

[astral hornet]

given are hypotenuse and opposite leg

which is a sine ratio

to find the angle using the length, use the inverse, arcsine

[haughty breach]

so i do

sin^-1 divided by 39/44 @astral hornet

[astral hornet]

sin^-1 (39/44)

not divide

[haughty breach]

Ohh i see

thank you so much

[astral hornet]

np

[neon herald]

ah yes this reminds me of A^+B^=C^

[lament bay]

https://cdn.discordapp.com/attachments/493943807671271424/533123676275736587/image0.jpg

[supple haven]

how would I solve the sin(arcsec(-7/2))

<@&286206848099549185>

[steady sleet]

#❓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 read the rules: #❓how-to-get-help!

[supple haven]

oh my b

I'll just wait till someone hopefully sees this

very patiently

extremely patiently

[lament bay]

<@&286206848099549185>

[supple haven]

oof my dude you'll get me yelled at if you just @ the helpers

[plucky marlin]

wew

trig problems

[steady sleet]

[plucky marlin]

im gonna try it out just for fun

[supple haven]

good luck amigo

im too daft to figure it out myself so if you solve it please enlighten me

[plucky marlin]

okay

i got $\frac{\sqrt{95}}{12}$

[somber coyote]

soap:

[supple haven]

whew thats one heck of a number

can you tell me how ya managed to get that?

[plucky marlin]

wait i used 12 instead of 2

$i got \frac{3\sqrt{5}}{7}$

[somber coyote]

soap:

[plucky marlin]

so first of all

$arcsec(x) = arccos(\frac{1}{x})$

[somber coyote]

soap:

[plucky marlin]

fucc

i drew it wrong

[supple haven]

oof its rad 45 I thought it was just plain 45

[plucky marlin]

so u just need to solve sin(arcsos(-2/7))

draw some triangles and blah blah

[supple haven]

okay so

I got my triangle here

but im blanking on whats next

[plucky marlin]

show me

[plucky marlin]

https://prnt.sc/m5o4nk

arccos(-2/7) = that theta

[supple haven]

so cos of theta = -2/7

but i dont remember what to do after

[plucky marlin]

sin(arccos(-2/7))

but we just got that arccos(-2/7) = theta

so plug that back in

[supple haven]

plug in theta?

[plucky marlin]

yeah

[supple haven]

sdpfjasdklfjhsadkj

im an idiot

I get it now

[plucky marlin]

i mean replace arccos(-2/7) with theta

[supple haven]

so it'd be sin of rad 45 / 7

[plucky marlin]

no just sin(theta)

which then evaluates to rad45 / 7

[supple haven]

oh

so i simplify that and get the final answer right>

[plucky marlin]

yus

[supple haven]

woo thanks man

im rlly dumb so this helps out

[plucky marlin]

=pup solve sin(arcsec(-7/2))

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@plucky marlin

[plucky marlin]

the bot approved the answer

[supple haven]

woo

[plucky marlin]

🎉

anyways i gtg

cya later

[supple haven]

thanks man

adios

=pup solve arcsec(csc((3pi)/2)

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@supple haven

[supple haven]

aw man i did it really wrong then

can someone help me with that?

[plucky marlin]

im bacc

csc(3pi/2) = 1/sin(3pi/2)

that's -1

so that simplifies nicely to arcsec(-1)

which is arccos(1/-1) = arccos(-1) = pi

also the range of arccos is [0 . pi]

so it cant spit out infinite values

[supple haven]

ooo thanks

[upper karma]

@torpid horizon dafuq

[cobalt sun]

Can anyone help me with this question?

I did this much
plus BAC=BDC

[supple haven]

are you sure that you're in math class

and that you didnt accidentally wander into a class that teaches you how to mathematically summon satan?

[sly charm]

[supple haven]

im pretty sure to solve it you have to prove that angle BME is the same as angle CNE

[astral hornet]

Bae and cae

[upper karma]

in a triangle ABC, how to prove that sum of all of the medians is greater than (AB+BC+CA)/2

[keen tangle]

You can do it with just triangle inequality

Actually what I did works for any three cevians, not just medians

Try proving that if M is a point on AB, then AB+2MC>BC+CA

[upper karma]

😮

[upper karma]

I have atrigonometry question

you can ask a question in a #❓how-to-get-help channel

[upper karma]

What is tan?

Sin/cos

Like tan(5)

Is sin(5)/cos(5)

Ok, thankd

Thankd

Thanks

You’re welcome

[dire rampart]

name checks out

[upper karma]

lol can someone help me w trig homework

[hard gale]

ask @upper karma

jk post it

[vernal tusk]

@cobalt sun did you ever figure out that satan question? i'm only in high school geometry but i have no idea other than to use parallel lines cut by a transversal, and i'm curious how one would solve that

[upper karma]

@upper karma which is your question?

[unreal flame]

hi guys

I tried everything I could think about

anyone have an idea how can I solve that and find x ?

i tried to make it

sin(2x)*cos(pie/2) + cos(2x)*sin(pie/2) = cosx

then get

sin2x0 + cos2x1 = cosx

then get

cos2x = cosx

I tried the answer 0 +2pie*k

[plucky marlin]

maybe like sin(2x + pi/2)= sin(pi/2 - x)

[unreal flame]

what do you mean?

[plucky marlin]

u know cos(x) = sin(pi/2 - x)

[unreal flame]

pi - x

not pi/2 - x

[plucky marlin]

it's pi/2 - x

[unreal flame]

cos(x) = sin(90-x)

[plucky marlin]

yeah

[unreal flame]

90 degrees = pi/2 radians

ok

[plucky marlin]

yeah

[unreal flame]

ok i know that

what about it?

[plucky marlin]

i just guessed

[unreal flame]

I dont think it can help

I'll try to ask again i think

hi guys
I tried everything I could think about
anyone have an idea how can I solve that and find x ?
i tried to make it-
sin(2x)cos(pie/2) + cos(2x)sin(pie/2) = cosx
then got-
sin2x0 + cos2x1 = cosx
then got-
cos2x = cosx
I tried the answer 0 +2pie*k

[wild pasture]

can anyone help me with these

[elder oak]

looks like you can use addition method fairly easily for a at the very least

[rose dragon]

@wild pasture set up as matrices and gauss eliminate

[flat cobalt]

Can someone help me with geometry in DMs?

[plucky marlin]

i can try

[flat cobalt]

It’s kinda advanced

[plucky marlin]

can i see it ?

[flat cobalt]

Yea one sec

[eager pendant]

@flat cobalt still need help?

[frosty egret]

https://math.stackexchange.com/questions/3071765/a-collection-of-lines-drawn-between-points-in-a-regular-13-gon-how-to-determin

Looking for some help with this

[upper karma]

What did this symbol mean?

and this one

?

[dire rampart]

first one looks like the subset symbol

second one means equivalent to

$\equiv$

[somber coyote]

lemon catto:

[upper karma]

How is it different than this : =

[dire rampart]

its pretty much the same thing

[stone birch]

sorry to disturb but can someone explain why sin(x)>1 means that there are no solutions

[dire rampart]

what are the range of values of sin?

@stone birch

[stone birch]

basically the question is asking me to show that, 2sin^2(x) +5cos^2(x) =1 has no real solutions using 3sin^(x)=4

3sin^2(x)*

[cobalt sun]

@vernal tusk which Satan question were you talking about?

No I haven't solved it yet

[glad ocean]

'print("Why?")'

Nvm

Code blocks aren't on mobile

[hard gale]

bootiful

you just have to use some weird ticks

``

[glad ocean]

ik the grave marks

I just can't find it on my keyboard

On my phone

[hard gale]

😩

[twilit solar]

A rhombus has sides 10 cm long and an angle of 30°. Find the distance between a pair of opposite sides.

Anyone know how to do this

[astral hornet]

draw out the shape and label it with known dimensions

solve for other dimensions using rhombus properties

then draw in line for height, and use dimensions to find length

[jade jacinth]

Could anyone help me show that lhs=rhs?

[thorn talon]

Double angle identity?

[umbral snow]

60 + 60 = 120 maybe

[glad ocean]

2 tan 60 = tan 120 * 1 - (tan(60)^2)

[graceful trail]

Can someone help me with the easiest geometry?

[upper karma]

a^2 + b^2 = c^2

[graceful trail]

Triangle ABC is an isosceles triangle. AB = AC. What is the measurement of ACD?

I’m trying to see whether my answer is correct or not

[upper karma]

lol you should just read the definitions in your book

[graceful trail]

Haha lmao

I’m stupid, and I can’t figure this out

Would you mind helping with this thing?

[upper karma]

You just need to read the definition in your book about exterior angle in a triangle

Can you spell it out for me

[graceful trail]

I don’t have a book 😭

[upper karma]

WTF

why not ?

[graceful trail]

We’re not using textbooks

My teacher explains it all

I accidently spaced out when he was explaining this

[upper karma]

What the hell, how does he teach then? Are you in america?

[graceful trail]

Yep

I think the answer is 125 degrees

[upper karma]

Anyway this is the theorem
Theorem: An exterior angle of a triangle is equal to the sum of the opposite interior angles.

[graceful trail]

Is that correct?

[upper karma]

Maybe, you can verify it yourself

Angle ACB is 55 right?

[graceful trail]

What is it asking exactly by saying Measurement of ACD

[upper karma]

That is the interior angle

[fiery badge]

the angle

[upper karma]

The exterior angle is ACD*

[graceful trail]

I have no idea

[fiery badge]

ACD can only be an angle, theres no way to deduce the length

[upper karma]

@fiery badge he means the angle

[fiery badge]

yeah

he has the answer

[upper karma]

Anyway now you konw which is the interior and exterior angle right?

[graceful trail]

Not really, unless there’s a parallel line

lol

[upper karma]

No, exterior angle is ACD which you are trying to figure out

ACB is the interior angle wich is already 55

[fiery badge]

"I think the answer is 125 degrees"

[upper karma]

@fiery badge shut up

[graceful trail]

lol rip

[fiery badge]

Ok dude

[graceful trail]

Well, the opposite sides do add up to 125

And the bottom right corner of the triangle is supplementary to the exterior angle, I am guessing?

So, 180 - 55 = 125

[upper karma]

Well yeah, or you could just think "what do i add to 55 to make it 180?"

[graceful trail]

B is the exterior

[upper karma]

The theorem is just that these two angles should add up to 180 degrees

[graceful trail]

Which is 125 so, I guess I am right

[upper karma]

Yeah

[fiery badge]

Why was I told to shut up?

[graceful trail]

No idea

[fiery badge]

He clearly understood that from the beginning

you just over complicated it for no reason

[upper karma]

He didn't

Why would he ask to verify the answer if he understood it?

If he understood it he would be confident that the answer is 125

Not ask if it's right

[fiery badge]

Hes asking because he wasnt sure what the actual question was asking

[upper karma]

smh

[fiery badge]

as he said earlier

[graceful trail]

It’s fine guys, lol. Don’t fight xd

[fiery badge]

People ask questions simply for verification, I do so sometimes

[graceful trail]

I got another problem though, which I am literally having a hard time. My teacher said this problem is easy but hard at the same time

Let me pull it up one sec

[upper karma]

What's the point of that, that just means you don't understand it

[fiery badge]

Or perhaps there is no mark scheme available

[upper karma]

Better to ask for an explaintation instead

You don't need a mark scheme if to verify your answer if you understand it lmao

[fiery badge]

that has to be the dumbest thing Ive heard

what if you fully understand your method, but your method is false?

if someone says no you are wrong then you can ask why and find out where you went wrong

it's not that complicated for his question, Im sure he gets it

[graceful trail]

Here

[fiery badge]

I wouldnt have responded like this if you didnt just tell me to shut up for no reason

[graceful trail]

How do you even solve for x? I got no idea

[fiery badge]

You can deduce some of the angles

[graceful trail]

What does that even mean? 🤔

[upper karma]

@fiery badge what

[fiery badge]

the triangle with the 60 and 50 degree angles

you can find out the other interior angle within that triangle from that information

[graceful trail]

What does deducing angles even suppose to mean lmao

[upper karma]

understanding your method is not the same as understanding your answer, I clearly said that if you understand your answer you should have no problem verifying it, understanding the question or method doesn't necessarily mean you understand the answer

[fiery badge]

what

"if you understand your answer you should have no problem verifying it" and what if the logic youre using to verify it are false?

[graceful trail]

Has anyone solved the problem, yet? 🤔

[upper karma]

?

@fiery badge then you're not following the method right and thus not understand it ?

the methods usually have clearly defined steps and definitions, so it's impossible to misverify the answer if you understood the method

[fiery badge]

but if you told him his answer was right maybe he could try explain it and you could verify his method

saves a lot of time and confusion

[upper karma]

if you misverify it then you clearly didn't understand the method thus why it should be explained instead of just verifying and spoonfeeding the other party

[graceful trail]

[upper karma]

yeah true, but most people here just go with a "ok thanks" then move on and ask similar question

if you wanna be wolfram go ahead

[fiery badge]

No but theres only one way he got to 125

there isnt much depth to this question

if he got 125

he clearly knows how he got there, by applying whatever laws/ rules he had

[upper karma]

exactly, so why should he need verification ???

[fiery badge]

i would side with you perfectly if the question was a little more complex

just so he knows with absolute certainty that he is correct?

[graceful trail]

[fiery badge]

You're telling me you've never had a class and then asked your mate "What did you get?" just for verification?

[upper karma]

then just tell him that theorems are well defined so he don't need nothing else to be absolutely certain

yeah, but only when i didn't understand it, i never ask if I understood it

[umbral snow]

[tawdry pivot]

I just answer questions because they are fun, if the op inquires into my method I proceed accordingly. I may ask them if they understand, just to make sure.

[graceful trail]

[upper karma]

@graceful trail i dont use white discord, can't help you sry

[graceful trail]

What does that mean? 🤔

[upper karma]

cant see it

[graceful trail]

Ah

[upper karma]

screen too dark

[graceful trail]

Okay, hold on

[fiery badge]

i suggest you label that point in the middle

and the point in the far right to make it easier to explain

[upper karma]

Well this is about making bigger triangles

Make a triangle out of BCE and measure E

[graceful trail]

It’s not necessarily have to be solved, my teacher just wanted to try it out

[upper karma]

E = ?

[graceful trail]

Hold on

40?

[upper karma]

why the question mark???????

yes tho

[graceful trail]

Verifying, lol

I thought I made a right triangle

So, I was thinking it could be 30

[upper karma]

k, no more verifying be absolute certain from now on

[graceful trail]

Alright

[upper karma]

now make a triangle out of BFC

[graceful trail]

F is 20

I think I know where this is going, lol.

[upper karma]

You do? Enlighten me

[graceful trail]

Hmm, I’ll try my best lol

Actually nvm, lol

I thought it was 110 for a second

But nvm

[upper karma]

You made a mistake, F is actually 50, the C in BFC is just 50 not 50 + 30 (exterior angle)

[graceful trail]

O

I see

Ok, now what?

I tried using algebra to solve x, but turns out, I was wrong, I think

[upper karma]

Why algebra? You can jst find out all the angles

[graceful trail]

Have you figured the answer though?

[tawdry pivot]

I can't figure this out oof

[graceful trail]

xd