geometry-and-trigonometry

[silent plank]

(unless you omitted that part)

[unborn jacinth]

well in the screenshot the base was 4

[silent plank]

oh whoops I used the wrong term

[steel pawn]

ram, does this mean that the integral of 1/x is ln |x| because you have to retain the domain of the initial function for the parent function?

[silent plank]

mindblank

[steel pawn]

like the domains have to exist both ways

[silent plank]

i meant the x term being raised to the power of 2

[unborn jacinth]

o

[silent plank]

yeh integral of 1/x is
ln|X| + C

[unborn jacinth]

im sorry could you repeat that again ramonov

I didn't quite understand

when its |x|

and when its not

[silent plank]

$\log_ba^c = c\log_b |a|$ if $a^c > 0$

[somber coyote]

ramonov:

[steel pawn]

so you have a line right

and you have two parrel lines shooting through

so the angle they were shot from the horizonal stays the same, but there's a different distance down the firing range

[silent plank]

@mossy narwhal your third line is the y-axis

[mossy narwhal]

So what

Cant make all

Equal

Or try each 2😯

[silent plank]

y-axis and line 1 will give a point
y-axis and line 2 will give another point
and seems you already have the point from line1/2

[steel pawn]

but they'll shoot into they line and create two angles for both sides of the target line
the target line has two sides, but for the angle across both of those two sides we know that it's 180 degrees because a line is 180 degrees

[mossy narwhal]

I don’t understand the last sentence @silent plank

[silent plank]

I can find out the intersecting point from 2 linear equations.

[steel pawn]

so both angles on either sides of how you shot the target should add up to 180 degrees. granted, you shot two shots at different positions x distance away. But you shot them at the same angle in reference to the horizontal and the target line. so both sides of the trajectory of how the bullet hit should add to 180 identically for both angles

[silent plank]

i assumed you found the intersections of
y+2x=5 and y+3=2x

[mossy narwhal]

Oh ok wait

I found the first 2 u mentioned

[steel pawn]

so you've got an a and b angle from both shots that are identical. a + b = 180 and c + d = 180. in terms of this, angle 12 and 13 are your a and b and angle 14 and 5 are your angle c and d

[mossy narwhal]

Lemme show u if im correct

[steel pawn]

and we know angle a = c and angle b = d.

[mossy narwhal]

,rotate

[steel pawn]

so if you know angle 5 and 12 you can add them together to get 180 and then solve for x

[silent plank]

seems you mixed up the x and y axis
what is the value of x at the y axis?

[steel pawn]

plug in x for angle 5 and you get angle 3, which is a complimentary angle to angle 5 and so it is the same

[silent plank]

(the y-axis is the vertical line where x= ?)

[steel pawn]

@upper karma

[mossy narwhal]

0

Oh

Yeah

,rotate

[mossy narwhal]

All done?

[silent plank]

,rotate

[mossy narwhal]

Second part find area is easy

[silent plank]

you didn't fix your values for 1 and 2

[mossy narwhal]

Why

I made them equal

The y equation

Found x

[silent plank]

(the y-axis is the vertical line where x= ?)
0
Oh
Yeah

[mossy narwhal]

Ik buttt

Thats not any asymptote (x or y)

[silent plank]

so when finding the intersection, instead of substituting y=0,
you should be substituting x=0

[mossy narwhal]

Its 2 equation u make them equal

[silent plank]

yeh... and the equation of the y-axis is x=0

[mossy narwhal]

Um

Let me circle😂

[silent plank]

i.e. solve
y + 2x = 5
x = 0

[mossy narwhal]

We dont

Substitute the answer of x?

[silent plank]

wdym?

[mossy narwhal]

Look at the third thing i dud

I found

X=2

So I substituted

To find y

,rotate

[mossy narwhal]

For the third one

I made them equal

Find x

[silent plank]

yeh, the 3rd one was fine

[mossy narwhal]

Substitute it with one of them to find y

[silent plank]

the issue is with the first 2

[mossy narwhal]

Oh

I did

Exchange them

[silent plank]

the first line, you set y=0

[mossy narwhal]

U didnt look at that part😂

Just look at thte answer

I understand now

[silent plank]

and its wrong

[mossy narwhal]

Ok x=0

Got it

[silent plank]

(what you did was find the intersection of the line and the x-axis and then reversed the x and y coordinates)

[mossy narwhal]

Yes i got it

I just

Forgot to exchange the way

But i did understand

So i exchanged the answers only

Oh

The answers might be wrong

For not exchanging the way?

[silent plank]

what are your intersection points?

[mossy narwhal]

Ok lemme redo

@silent plank now perfect?

[silent plank]

yeh thats better,
now find the area

[mossy narwhal]

Ok thanks😄

@silent plank im stuck😂

[silent plank]

drawing a rough sketch will help

[mossy narwhal]

I did

Its not right triangle

And i got 1 length only

[steel pawn]

no you have 2 right triangles

[silent plank]

doesn't need to be

[steel pawn]

split the big triangle in 2

[silent plank]

whats the distance from the point (2,1) to the line?

no need to split

[mossy narwhal]

Which line

[silent plank]

uh y-axis

[mossy narwhal]

Oh

Yeah the distance formula

Right?

Find that

Then i found h

[silent plank]

don't really need the full distance formula for that

[steel pawn]

the y axis is the line x = 0 right

[mossy narwhal]

Then idk T - T

[steel pawn]

and you have a point with an x value of 2

what is the change in x

[silent plank]

^

[mossy narwhal]

2

[silent plank]

yep

[steel pawn]

multibene

[mossy narwhal]

Man Im dumb

[steel pawn]

nononono

[mossy narwhal]

😂

Its true

Ok

[steel pawn]

the more you do stuff like this the more it makes sense to approach it like this

[mossy narwhal]

A= 1/2 x 8 x 2

[silent plank]

yep

[mossy narwhal]

😄thanks a bunch ❤️

[silent plank]

also use \* for multiplication here

[mossy narwhal]

Oh ok

If I remember

@steel pawn @silent plank wheres my welcome btw

[steel pawn]

wait

oh my god

[mossy narwhal]

🤔

[steel pawn]

I'm retarded

[mossy narwhal]

No worries

U r not alone

If thats what u think about urself

What about me T - T

[steel pawn]

I can solve for the distance between the earth and the moon but I didn't realize the area of a triangle for any triangle was just base time height * 1/2

I thought that only applied to right triangles

fuck

[mossy narwhal]

Lol

Glad to be of help

[unborn jacinth]

my desmos is stuck like this

how do I fix it

its stuck zoomed out

I tried clearing cache

nvm

I got it

[mossy narwhal]

,rotate

[vagrant steeple]

what do i need to know to solve
arcsin(sin50/9pi)

[dark sparrow]

uh

is this $\arcsin\left(\frac{\sin(50)}{9\pi}\right)$? or $\arcsin\left(\frac{\sin(50)}{9}\pi\right)$? or $\arcsin\left(\sin\left(\frac{50}{9\pi}\right)\right)$? or, god forbid, $\arcsin\left(\sin\left(\frac{50}{9}\pi\right)\right)$?

[somber coyote]

Ann:

[vagrant steeple]

the third one but pi is on top

[dark sparrow]

$\arcsin\left(\sin\left(\frac{50\pi}{9}\right)\right)$?

[somber coyote]

Ann:

[vagrant steeple]

yup

[supple abyss]

lol

[dark sparrow]

ok so to answer your question of "what you need to know": you need to know the definition of arcsin and some of the properties of the sine function

[vagrant steeple]

so special right triangles ?

[dark sparrow]

no.

(it is to be noted, by the way, that arcsin is NOT simply "the inverse function of sin" as sin does not have a true inverse.)

[somber coyote]

Display results online and refine query

Requested by ShriHari#1305

[hallow rose]

,w calculate above

[soft gulch]

The height would be AC if you were take BC as base

What if you take AB as base

Would the height line bisect the angles of the vertices?

That was my question

I have solved it tho

[sudden locust]

Did you know that the Pythagorean Theorem applies even for half-circles? In this video, I explain the logic behind it!

https://youtu.be/Ir3MUW_VkgQ

Hope you are having a great Wednesday so far! 😊

[mighty narwhal]

Did you know that the pythagorean theorem applies for any blobs that are similiar?

[sudden locust]

As long as the figure is in 2D it applies! Its really cool imo! @mighty narwhal

[upper karma]

@sudden locust Great video! Keep it up!! :)

[sudden locust]

@upper karma Hey, thank you so much, I really do appreciete your comment! 🥰👍

Most of the time I get negative feedback in here, so kind of considering to stop posting because some see it as annoying, knowing some people appreciete it means more than you think!

[torpid torrent]

you guys know any cool application videos of rigid motions
like any type ofsub 2 min video
on rigid motions
thats rly cool
like on its application in real life
im tryna find one thats interesting but im having a bit of troubel

[slender frost]

is anyone able to like, dm me and help me out? I have a big test coming up tomorrow and don't have a single clue how to do any of the material

[upper karma]

@slender frost what math class is it? If it is pre-algebra sure if it is algebra 1 sure if it is geometry what topics will be covered on the exam and if it is Algebra 2 what topics if pre-cal or cal 1 or above srry

[slender frost]

it's trig

[dawn gorge]

Indeed help

I need help

If the equation is mad = bad and Andy = mad then the equation is also Andy = Bad am I right?

[sly marlin]

@dawn gorge depends, because we need to apply transitivity and case insensitivity.

[chrome fiber]

🅱&

[zenith ember]

In the future, please do not delete the ping message after the fact. It doesn't make the ping go away, it just makes us have to look harder to figure out what happened.

[upper karma]

oh sorry

[bleak lintel]

I don't know if this fits in here but it's with vectors. I'll translate what the excercise tells me to do.

The point a is defined by the vector oa = Ex - 2Ez, the point b by vector ob = 2Ey + Ez. A point C lays on the line AB in between A and B with ||CA|| = 2||BC||. Determine the angle côb.

I've tried this and with some fiddling I got the answer but I'm curious on how to get a formula to determine the point C as I didn't do that. I just went with converting ||CA|| = 2||BC|| into something that becomes c = (a+2b)/3 which then leads to x = 1/3, y = 4/3 and z being -1/3. This on the otherhand isn't correct apparently as I have to use z = 0 to get a correct calculation of the angle using the dot product.

Can someone help me with this? 🙂

^the original one, in Dutch

I might be asleep whenever someone answers so please tag me if you got a solution-

the solution for the angle is cos^-1(8/sqrt(85)) whihch is around 0.52 radians

^the answer the course/professor gave me.

Right side-

left side is me trying lmao

[brave storm]

How would I test using parametric representations of 3D lines that two 3D lines are the same?

say I have two lines A and B

In the form: point + scalar*vector

[fallen ivy]

I think you could just subtract them

And show that's the zero function

[upper karma]

🤔🤔

[tender iron]

I have a question for homework, and I really don't understand it. I'm looking over it and there isn't enough given information to solve.

[tawdry pivot]

ye

[umbral snow]

Yes, I can't relate the two angles in any way

[tender iron]

There isn't enough given information for this too right?

[queen python]

@tender iron is there a measure of <abc?

[tender iron]

No

Everything in the diagram is the given information.

[queen python]

maybe abd just = -10x+58

[tender iron]

xD I tried that

And its wrong

[queen python]

was it wrong

hmmm

[tender iron]

I tried that when I realized you can't solve it

[queen python]

is there no answer?

it looks like its from khan academy

unless im not getting something i think its impossible

abc = -4x+ 99

wait

what

@tender iron did you try that

solve for x

[tender iron]

Yea

[queen python]

did it work?

[tender iron]

It's wrong

Also

it isnt from khan academy

[queen python]

hmmm

[tender iron]

It's from Big Math Ideas

[queen python]

dont they have answers

[tender iron]

I don't think so

If there are I can't find them

I tried searching this up and nothing popped up

[tawdry pivot]

~~use protractor ~~

[queen python]

yeah its not possible with the -4x+99

[tender iron]

xD

Protractor lmao

[queen python]

yeah doesnt look possible

[tender iron]

I'm going to just ask my teacher tomorrow and see what she says because personally I don't see it solvable

[queen python]

this might just be something really simple and i just forgot

or its not possible

who knows

[tender iron]

I came to this discord as last resort because I wanted to see if I can solve it

[queen python]

feel free to ask anytime

im on most of the day so if you need help feel free to ping me

Hopefully i can help

😥

[silent plank]

so there's nothing else on the page...?

[tender iron]

no

[queen python]

ramon do you see a way?

[tender iron]

That is all there is to the two problems

[queen python]

did you try reloading the page

[silent plank]

I mean the angles aren't even ideally labelled

[queen python]

it could just be lag

and somethings not loading

[tender iron]

I have xD

[queen python]

yeah i dont think its possible then

[tender iron]

My classmates all have the same questions too so I don't think its a coincidence

[queen python]

i also hate angles labeled with -something

its not because there hard i just dont like messing with -'s

[bleak lintel]

no answers on my questions from yesterday?

sad

Guess we'll never know ;-;

[vestal oak]

Can anyone help me with this problem

#10

[quaint cipher]

You're supposed to find what?

@vestal oak

[vestal oak]

Find all 6 trig functions

Sin, cos, tan, cot, csc, sec

I am just confused on how to draw a reference triangle for an angle in Quadrant 3

NVM I think I figured it out

[robust socket]

,, sin{2\alpha+\alpha}=sin{2\alpha} cos{\alpha}+cos{2\alpha} sin{\alpha}

[somber coyote]

Umma.Gumma:

[robust socket]

not understanding how to get to cos2\alphasin\alpha

ie the second addend

[slender nacelle]

But you know $\sin2\alpha \cos\alpha$?

[somber coyote]

oscillatingEquilibrium:

[robust socket]

yes that's the formula for sin2alpha

[slender nacelle]

Not really.

[robust socket]

right, misplaced that 2

[slender nacelle]

Yes.

[robust socket]

ok, got to check again then

ty

[slender nacelle]

But that formula comes from the fact that $\sin(A+B) = \sin A \cos B + \cos A \sin B$

[somber coyote]

oscillatingEquilibrium:

[slender nacelle]

That is more fundamental.

[full gorge]

how can i remember the trigonometric circle and the values?

like whenever the professor has to find arctan of an angle she does it spot on

[dark sparrow]

do you mean simple angles like 0, pi/6, pi/4 etc

it's mostly a matter of practice to commit those to memory

[full gorge]

no like

7pi/2

7pi/6

or whatever kind of angle tbh

[cloud meteor]

Hey I'm having a bit of a hard time with this one

a rectangle with area 8 shares 2 sides with a regular polygon of area 80

how many sides does the polygon have

this is a NMO problem

[silent plank]

for a rectangle to be able to share 2 sides with a regular polygon, the polygon will need to have an even number of sides.

for simplicity, let it share the top/bottom edge

draw some triangles, determine the formula for the area of a regular n-sided polygon

[robust socket]

I've just stumbled upon a solved exercise which contains this:

,, 2\sin{2\alpha}\cos{(-\alpha)}=2\sin{2\alpha}\cos{\alpha}

[somber coyote]

Umma.Gumma:

[robust socket]

how is that cos changed from -\alpha to \alpha?

[weary drift]

cosine is an even function

[robust socket]

damn, forgot that. Thanks a lot

[weary drift]

np

[upper karma]

sied length of square= 2cm. Find the area of the shaded area

can someone help me

Vvithout calculus

[silent plank]

are those 2 vertices the centres of the 2 circles?

[upper karma]

yes

the vertices at the bottom left and right ends

therefore the side length is the radius

[silent plank]

yep.
construct lines from the two vertices to the intersection of the 2 circles

[upper karma]

ok a triangle vvith a base of 2

[silent plank]

do you know how to calculate the area of sectors? and triangles?

[upper karma]

no i don't

[dark sparrow]

then you will not be able to calculate this area.

[upper karma]

vvhat is the formula for the area of a sector?

i knovv triangle

[dark sparrow]

(...how come you're typing vv for w)

[upper karma]

my double u key broke

[silent plank]

https://lmgtfy.com/?q=area+of+a+sector&s=g

[dark sparrow]

^

[upper karma]

lol thankyou

ok i get it

and then vvhat?

@silent plank

[silent plank]

think about adding and/or subtracting the areas of sectors/triangles to get your desired ara

[upper karma]

hovv do you get the angle to calculate the sector

[silent plank]

what type of triangle do you have?

[upper karma]

idk

im guessing equilateral

but vvhy

[silent plank]

properties of a circle?

[upper karma]

vvhich property?

[silent plank]

radii...

[upper karma]

ok

so you repeat the sectors tvvice and because they overlap you subtract the area of the triangle? or am i vvrong?

@silent plank

[mighty narwhal]

Yes, that sounds correct

[silent plank]

^

[upper karma]

Thankyou

is it 30 or 60?

[silent plank]

uh what?

[upper karma]

formula for sector is 1/2 x radius^2 x angle of sector

a sector looks like this

[silent plank]

yeh

[upper karma]

hovv can you get a sector from this

VVheres the angle?

[mighty narwhal]

draw a line from the intersection of the two circles from one of the bottom points

[silent plank]

did you draw your equilateral triangle into the diagram?

you should clearly see your angle(s)

[upper karma]

yeah 60, 60, 60

[mighty narwhal]

this terrible picture hopefully helps

[upper karma]

OH

jesus hovv did i not see that

Ok its all clear novv thankyou

[mighty narwhal]

The angle is 60°, you can compute the area of the sector
and as you already said, two times the area of the sector - the triangle's area is the red thingy

[upper karma]

Do you knovv vvhat subtopic of geometry this is?

I need to practice it

[silent plank]

geometry

[upper karma]

K nvm ill just revise all of geometry

[mossy narwhal]

😂

@silent plank hi 👋

Am I not allowed

To ping u for unnecessarily stuff?

[silent plank]

pinging someone to ask a non-math question ...

[upper karma]

arctan(1) + arctan(2) + arctan(3) = ?

[mighty narwhal]

Lets say that it is x
arctan(1)+arctan(2)+arctan(3)=x

take the tan of both sides

tan(arctan(1)+arctan(2)+arctan(3)=tan(x)

Do you know the trig identity for tan(a+b)?

@upper karma

[upper karma]

Yeah.

(tan a + tan b)/(1 - tanatan b)

[umbral snow]

I've seen a pretty good approach for this before, I think

It was geometric

[upper karma]

@umbral snow pls show

I got this bitch wrong in a competition today

sad day

[umbral snow]

Sed

https://math.stackexchange.com/questions/272208/proving-arctan1-arctan2-arctan3-pi

I can't recall it well enough to do it on the spot, but that second one is the one that came up in my head

Think Numberphile did a video on it?

Though that complex number solution is bae

[mighty narwhal]

So basically you have tan(arctan(1)+arctan(2)+arctan(3))=tan(x), and you can use tan(a+b) to expand the LHS

and you will get [a simple number]=tan(x)

[upper karma]

This is actually the nicest one

[mighty narwhal]

Oh wow

[umbral snow]

Competition people are clever

[mighty narwhal]

Thats nice

[late marten]

how do i set this up

[umbral snow]

In general, a very nice way to add arctan angles

How many years passed @late marten?

[late marten]

4

[umbral snow]

So it doubles 4 times

[late marten]

yeah

[umbral snow]

How might one "undo" that action?

[late marten]

?

[umbral snow]

Actually, let's just go algebraically.

[late marten]

yeah

[umbral snow]

Double of x is 2x

[late marten]

yep

[umbral snow]

Doubling it again is
2*2x

[late marten]

ok

[umbral snow]

Or (2)²x

[late marten]

i see

just do 222*2 = 16

then 480/16

[umbral snow]

Doubling it again is (2)³x.
I'm sure you can guess what doubling AGAIN is

[late marten]

yessir

[umbral snow]

So you have
x(2)⁴ = 480

Ya ya, and x = 480/16

[late marten]

then divide by 16

[upper karma]

@umbral snow

I have no idea how this works actually

arctan(-10) != pi

[umbral snow]

arg means argument
That is, the angle of a complex number

[upper karma]

oh.

[umbral snow]

A positive real has arg 0
A negative real has arg π

[full gorge]

angle = arctan y/x

arctan 1/1 + arctan 2/1 + arctan 3/1

[upper karma]

@umbral snow thats beautiful’

Holy fuck!

[umbral snow]

Yeah that's an incredible proof, never would have thought of it myself

And a neat way to add arctans

[sly marlin]

I think we still need a small caveat, that 0 < arctan 1, arctan 2, arctan 3 < pi/2

then we can find the range where the answer should be

[forest lava]

hello

i am but a noob

a noob with a question

[stiff haven]

go on

[forest lava]

uh can i dm you

need 2 send a picture

[stiff haven]

eh sure

[hard gale]

you can send them here no problem

[upper karma]

@sly marlin well

[sly marlin]

yeah?

[upper karma]

-pi/2 < Arctan(1) < pi/2

so arctan(1) = pi/4

[forest lava]

why is 4sin(16) negative

[weary drift]

if sin(16) is negative, an angle of 16 is coterminal with some angle in the 3rd or 4th quadrant

[forest lava]

@weary drift bruh

wot

[weary drift]

wassup dude

[forest lava]

what am i doing wrong dude

this should not be a negative angle

i mean

not an angle

length

[weary drift]

you workin' in degrees or radians, dude?

[forest lava]

degrees

[weary drift]

put your calculator in degree mode, dude

[forest lava]

it is i think

im using excel

[sly marlin]

working in grads

[weary drift]

i'll break your knees, element

[sly marlin]

what, by putting the angle at 200?

[forest lava]

@weary drift why it no work 😭

[weary drift]

look up the conversion factor from radian to degrees, dude

[sly marlin]

=sin(radians(16))

[forest lava]

bro

i am not using radians

[sly marlin]

see what the radians function does

[forest lava]

i measured it by hand

using degrees

[sly marlin]

...

[weary drift]

by hand???

[forest lava]

what can i say

😎

no im trying to apply mathematics to some geograpphy fieldwork i did

[sly marlin]

just see what the radians function does in excel

[forest lava]

WHY

[sly marlin]

because excel's sin likes radians

[weary drift]

or else we will twist your ankles a full 400

[sly marlin]

so give it radians

[forest lava]

what

no

im using degrees

[sly marlin]

https://exceljet.net/excel-functions/excel-sin-function

[weary drift]

think of it like this

the radians() function is a translator between you and excel

you prefer to speak in deg, excel prefers to speak in rad

[forest lava]

oh ok

wait

[weary drift]

you put whatever degree measure you want into radians() and excel can then know what you want to say

[forest lava]

hmm

OMG

I LOVE YOU GUYS

TYSM

@sly marlin @weary drift RADIANS ARE THE BEST

[weary drift]

agreed, dude

[flint pelican]

I’m kind of confused

I can tell it has a right angle but

How do you know the small white triangle is 45 45 90

How do you know it’s 45 too?

[rain charm]

@flint pelican The angle will be 45 when the sides lengths are the same. Notice if you rotated the octagon the sides would all match up. So that unknown side of the white triangle will match up with side length x. Hence the sides have the same length x.

[flint pelican]

I mean they look the same so I can see what you mean

But is there a way to know they are exactly the same because what if one is slightly bigger than the other

Yeah I can’t see it idk how you can tell for sure

[sly marlin]

@flint pelican what's the interior angle of a regular octagon?

alternatively you can reflect it over y=x

[rain charm]

@flint pelican You can tell for sure because it’s a regular octagon. So if you rotate it about the center the shape doesn’t change.

Like element said you can also verify by considering the interior angle sum

[flint pelican]

All angles are 135 in an octogon right?

[sly marlin]

In a regular octagon, yes

@flint pelican

[flint pelican]

I just noticed that

Both the x side and the other side of the white triangles

Are heights for the trapezoids inside the octogon

Would that help? Since they are both heights that means they’re equal right?

[sly marlin]

heights of a trapezoid with 2 parallel sides, yeah

[robust socket]

..is this actually correct?

,, \cos^2(120+\alpha) = \cos^2(120)\cos^2(\alpha)-\sin^2(120)\sin^2(\alpha)

[somber coyote]

Umma.Gumma:

[green slate]

Isn't cos²(x + a) = (cos(x + a))²?

[robust socket]

never seen that tbh

[sly marlin]

probably not

[robust socket]

what would be the best course of action there

I'm running out of ideas

[silent plank]

firstly do you understand cos²(x + a) = (cos(x + a))²?

[robust socket]

as I said I never seen that form before but I can take it for granted

I mean, it makes sense to me

[wild hamlet]

Yeah it's true, just a difference of preference

They're the same

I prefer cos^2 form

[silent plank]

cos²(x + a) = (cos(x + a))*(cos(x + a))

[wild hamlet]

Though if you type it in a calculator (ex Desmos) they won't recognize it

[silent plank]

the square is usually written between to function and the angle to reduce disambiguity about whether the angle is squared

thought desmos did recognise it

[wild hamlet]

Desmos does with 2, but not 3

[silent plank]

ah

[wild hamlet]

Then there's the inverse notation which has -1, it sucjs

[silent plank]

yep so just apply the angle sum identity, and then square it

[robust socket]

great let me check

[wild hamlet]

Ye

[robust socket]

sorry it took a while, the whole thing was

,, \cos^2(\alpha)+\cos^2(120+\alpha)+\cos^2(120-\alpha)

[somber coyote]

Umma.Gumma:

[robust socket]

worked like a charm, ty

[winged saddle]

trig is freaking stupid

[boreal light]

Would anybody be able to provide me with a fairly straightforward tutorial on how to solve for solutions in trig functions with variable inputs such as (for example) f(x) = sin10x

[steep temple]

solve what

[boreal light]

Like if you were trying to find critical points, so you took the derivative of the trig function and set it to zero.

But in order to solve for solutions you need to take the inverse

And I need a refresher on how to do so

[silent plank]

maybe start with something simpler like
f(x) = sin(x)

[boreal light]

well, its for a group of specific problems that all have an input like ax

Just hoping for a link or something on taking inverses of trig functions with variables

[silent plank]

well would you know how to find the general solution to something like
sin(x) = 0.5

[boreal light]

I know how inverses work, it's comparing it to the unit circle and finding solutions that I am specifically interested in

I know there is something about finding the right quadrant and there being multiple possibilities based on the interval you're working with

[silent plank]

that doesn't answer my question

[median crown]

Just take all solutions between 0 and 20pi

[boreal light]

Yes i know how to find the general solution

[median crown]

and divide them by 10

Hipparchus of Nicaea hates this one trick

Gives you all solutions between 0 and 2pi btw

[silent plank]

if it was
sin(10x) = 0.5
general solution would be for 10x
and then divide by 10 to get x

[boreal light]

Not quite what I was looking for. I actually found a good link online so all good. Thanks

[median crown]

[keen glen]

Hi guys

[green slate]

Just wanted to quickly check whether this is outdated or not. Taken from Lang's "Basic Mathematics". I'm taught in school that Angle QPM = Angle MPQ and that I should differentiate the two angles by writing Obtuse Angle QPM and Reflex Angle QPM respectively. Which is the right way to do this?

[quiet mason]

write reflex <MPQ for the one >π

and obtuse <MPQ for <π

[green slate]

Ah. Thanks. That was throwing me for a loop.

[obtuse tapir]

here's a challenge for my fellow high schoolers

[silent plank]

,rotate 180

[obtuse tapir]

did anybody solve it yet?

[torpid torrent]

theyre the same area right

bc height not to the top vertex of the triangle

all the time

[undone shuttle]

@obtuse tapir i think that i solved your problem

Wait a second

[torpid torrent]

i didnt switch these two around right

the two angles

i mean

[undone shuttle]

@obtuse tapir

[upper karma]

would 17.5 b correct

[silent plank]

17.5 for what? what's p?

[solar moat]

is this HL

#3

[fathom shard]

I need help proving that angle BCE is 1:4 of angle BOC.

[upper karma]

how owuld i go around

attempting https://cdn.discordapp.com/attachments/269573202018041856/640633956538253322/unknown.png

[silent plank]

if you knew BD would you be able to find AD?

[upper karma]

i would

which equation would i use to identify bd

im just stuck on knowing what to use when

[silent plank]

maybe use info in triangle BCD

[upper karma]

which equation?

[silent plank]

what rules related to triangles do you know?

given 2(3) angles, and a side, what can you use to find the other side(s)?

(to find AD, you would use the cosine rule)

[upper karma]

thanm you

[fathom shard]

Got it. Thanks @silent plank

[silent plank]

huh what did i do?

[tidal rune]

Can someone remind me why theta/2 = 2 alpha?

The vertical dashed line bisects the angle theta

The vertical and horizontal dashed lines are perpendicular.

[slender nacelle]

@tidal rune

[tidal rune]

That's Thale's Thm

Oh wait

I see now

[slender nacelle]

Can you complete this circle and see which angle is $theta/2$

[tidal rune]

Thanks

[somber coyote]

oscillatingEquilibrium:

[slender nacelle]

Great!

[winged saddle]

k i take that back trig is amazing holy fk

also did someone delete my message 😂

[torpid torrent]

am i mixing up obtuse and acute

[hearty wolf]

Can someone help me with these questions

[upper karma]

Can someone help me with this question?

[green slate]

So, sin 2x = 2 sin x cos x, correct? How do you find sin x and cos x individually?

[upper karma]

How do I get to 2sinxcosx from sinx+cosx

[green slate]

Let's see.

We can do (sin x + cos x) - (sin x - cos x) for one. What does that give you?

[upper karma]

2cosx

[green slate]

So we can find cos x then, right?

[upper karma]

So I would have to do 1+sqrt(3)/2-1-sqrt(3)/2 to get 2cosx

[green slate]

Yes.

You can similarly find sin x this way.

[quiet mason]

you can not add and all that

but just square both sides of any of the eqns

so its 1+sin2x

[upper karma]

Cosx = sqrt(3) sinx =1/2 is sin2x=sqrt(3)

@green slate thanks so much!

[green slate]

Oh wow, that's smart. That method is probably way better.

[upper karma]

@quiet mason can you explain to me yr method

How does that work

[green slate]

Try squaring both sides of the first equation.

It should pop out to you.

[upper karma]

That is so smart!

I wish you guys were my math teacher! In Ontario teachers get paid an average of 90k

[green slate]

I really need to get better at spotting these clever tricks.

[silent plank]

that doesn't sound that bad

[hearty wolf]

Help

[silent plank]

which question?
what have you tried?

[hearty wolf]

Stuck on 5

@silent plank

[silent plank]

should be doable by inspection.

(if Boris had at least 2c they would have enough)

[hearty wolf]

So 26 cents

[silent plank]

no. if it costs 26c, they would have enough

[hearty wolf]

Im stuck

Im so confused

[silent plank]

(if Boris had at least 2c they would have enough)

so Boris must have less than 2c

[hearty wolf]

1 cent

[silent plank]

assuming prices are in whole numbers, and Boris isn't trying to buy stuff with no money

what's the price of the chips?

[hearty wolf]

25 cents?

[silent plank]

yeh

you can solve this using inequalities if the numbers aren't this small

[hearty wolf]

But how do i explain it?

[silent plank]

let x be the price

B = x - 24, x>24.
M = x - 2
B + M < x

exactly as I described in the words above

[zinc basalt]

parallelogram OPQR is such that OP = [-7,24] and OR = [-8,-1]. Determine the acute angle between the diagonals and OQ and RP can someone help me solve this please

[hearty wolf]

Help with 4

[zinc basalt]

<@&286206848099549185>

[obtuse tapir]

@undone shuttle smarty pants

@hearty wolf it's forming a triangle

trignometry i think

i don't know how to use trig

or i think i can use heron's formula here

[hazy granite]

can someone help me with this question

it was in my maths exam today

i got the answer 16/65

but my friend is getting 16/25

[dark sparrow]

@hazy granite how did you get 16/65?

[obtuse tapir]

Is this all the information

[dark sparrow]

it's sufficient to find what is asked for, so i am certain that yes

[obtuse tapir]

im pretty dumb

[dark sparrow]

@hazy granite alternatively, how is your friend getting 16/25?

[hazy granite]

i took area of trapez as unknown

area of trapez = large triangle - small triangle

areas

so then

12^2/27^2-12^2=16/65

my friend got 16/25 by doing

12^2/15^2

i think

[obtuse tapir]

wait

how do you know

the height and base are equal

[hazy granite]

its a simplified version of the whole think

these triangles are similar

[obtuse tapir]

ah

[dark sparrow]

@hazy granite you missed parentheses but you are actually correct

16/65 is the right answer and your reasoning is solid

[obtuse tapir]

I have not been taught Similiar triangles theorem yet

can somebody explain

[dark sparrow]

"similar triangles theorem" what

[obtuse tapir]

my bad

i meant theorems

[dark sparrow]

uh

honestly they aren't that important

[fringe dirge]

Yeah I don't think I've ever used them

[quiet mason]

just one question lol

is the only definition of cosh(x)=AM(e^x,e^(-x))

and this is the way the others are defined

?

or can someone link me up with some geometric significance

[dark sparrow]

AM

uh

are you seriously using "arithmetic mean"

instead of just writing it as half the sum as literally anyone else would

[quiet mason]

lol

[fringe dirge]

cosh is the graph of a catenary iirc

[quiet mason]

so like a U Graph

[fathom shard]

Any tips on how to prove that EFGH is a rectangle and not a square?

[silent plank]

,rotate

[fathom shard]

I've managed to prove that it's sides are 90 degrees and EH = FG and EF = HG

Angles *

Given : CH = HD, DG=BG, AE=CE, AF=FB, EF = CB/2.
EH = FG =AD/2.
EF is parallel to HG , EH is parallel to FG

[summer spire]

what? If "EH = HG" then isn't it a square? But wait "EH = EG", so EHG can't be 90 degrees.

[obtuse tapir]

Since FG is not equal to HG

Its a rectangle

[quiet mason]

@fathom shard there are too many mistakes

first learn whats a rectangle

[dusty sleet]

how do i find cos of a

[quiet mason]

A

or alpha

cos(A)=0

cos(alpha) = adjacent/hypotenuse

[dusty sleet]

yes

[quiet mason]

=QF/OC

=OG/Do

DO*

[dusty sleet]

so cos a ?

is

[quiet mason]

???

cos alpha?

or a

[dusty sleet]

yeah

alpha

sorry

[quiet mason]

kürbis

yes

[dusty sleet]

XDD

[quiet mason]

just write what i wrote

i hope you do understand what i wrote

[dusty sleet]

but the answer is OF

[quiet mason]

do you?

[dusty sleet]

yea

[quiet mason]

im pretty sure its OF/OC

your books wrong

maybe its a misprint

[dusty sleet]

oh no i understand what you did you did yea but the question was diffrent

[quiet mason]

ok pls translate question

OH

i see

R=1

or radius =1

[dusty sleet]

Give the segment that corresponds to cos α

[quiet mason]

OC=radius

so

[dusty sleet]

Give the line that corresponds to cos α

[quiet mason]

OF/OC=OF

ok?

[dusty sleet]

yes

[quiet mason]

anything more?

you couldve done this yourself

cmon

[dusty sleet]

mann

its for you to say xd

im struggling trying to understand atm

[quiet mason]

but did you understand what i did tho?

its standard

just OC=1

[dusty sleet]

i understand what you did yes

oc=1?

[quiet mason]

They habe given Radius =1

but OC=radius

[dusty sleet]

yes

[quiet mason]

yes so OC=1

ok!

?

[dusty sleet]

yes

[quiet mason]

yeet

thank you for your help

bye

[dusty sleet]

lmfao

xd

[upper karma]

Can someone help me i have no idea

[silent plank]

do you know your trig ratios?

[upper karma]

I have no idea I suck at trig

[silent plank]

soh cah toa?

[upper karma]

Oh yea I know that

[silent plank]

for 11a,
focus on the triangle on the right.
do you think a trig ratio can be applied here to find x?

[median crown]

soh cah toa is primitive

No pun intended

[covert orchid]

.

[pure rain]

i need some assistance

could someone help me with this

[gritty sail]

Help with what

[spice tendon]

I need help please

[keen aspen]

Post your question

[spice tendon]

In geometry we just did a lesson and I didn’t really understand it but I do understand the vocab with it but can you please tell me how I find these measurements

[keen aspen]

No information on the angles?

Is the triangle supposed to be isoceles

[spice tendon]

It doesn’t say but we’re learning about isosceles this

tho

[snow drum]

I think it’s isoceles as LM=NM

[keen aspen]

Yeah

Okay, so then LP and NP are the same length

[snow drum]

Yeet

And just solve

[keen aspen]

So set both expressions equal to each other

[spice tendon]

Ohhh

Ok thanks

[upper karma]

Are the two other angles in a right triangle always 45?

[green slate]

No.

It's only if the two sides adjacent to the right angle are equal in length.

[upper karma]

@green slate how do I do this problem? I know how to find angle 1, but idk how to find angles 2 and 3

[green slate]

Well, in this case, it does seem like they are 45°, only because AC = CE.

[upper karma]

Ohhh

Both the side triangles are congruent

[green slate]

Right.

[upper karma]

Tysm

[green slate]

No worries.

[upper karma]

helppppppppp

[keen aspen]

Okay so with whats given the top two angles are congruent

And by the reflexive property, ZX=ZX

So in order to prove its congruent by AAS, you need which angles to be congruent

Because you already have A and S

[signal hornet]

hi

I need help with this one

wait

[keen aspen]

Lol

Got it?

[signal hornet]

I think so

[keen aspen]

What did you get

[signal hornet]

I was thinking -2 isn't on the unit circle

but its sec

one sec

[keen aspen]

Yeah

[signal hornet]

theta = 4pi, 8pi

but symbolab is giving me a different answer, hmm

where did I go wrong

yeah my answers eval to 1 not to -2

I know 2pi/3 and 4pi/3 make sec be -2

[keen aspen]

Okay so cos(3theta/2)=-1/2, meaning if you did 3theta/2=x, cos(x)=-1/2

So x lies within the 2nd and 3rd quadrant

The terminal angle of x is 60 degrees

If you consider cos(x)=1/2

But since its negative, the angle has to open up in 2nd and 3rd quadrant

So you gotta add and sutract pi/3 or 60 to pi

To get your two answers

Then revert back to theta

[signal hornet]

How come this despire being true doesn't help me with the answer?

I know 2pi/3 and 4pi/3 make sec be -2

[keen aspen]

Well thats true if cos(x)=-1/2

[signal hornet]

is it b/c it's not just theta? but 3theta / 2?

[keen aspen]

But its 3theta/2

So do 3theta/2=2pi/3 and 4pi/3

[signal hornet]

so I should substitute? 3theta/2 to x just to make it simpler for me?

[keen aspen]

yeah

[signal hornet]

I figured out what I did wrong

I messed up the arithematic

Instead of divding by 3 I multiplied by 3

[keen aspen]

lol

[signal hornet]

thank you @keen aspen ❤

[keen aspen]

np

[vague berry]

How do you solve this

Please ping me

Hey

[obtuse tapir]

shouldnt the answer be 30-root3

the statement is false

@vague berry

[vague berry]

What?

Please explain @obtuse tapir

[obtuse tapir]

if you solve

the left hand side

you get 30-root3

which is not equal to 30/16

you could write x/2 in inverse form

and then multiply it would be the same thing

and the rest is cheese

@vague berry

[upper karma]

How would I solve this problem?

I understand that angle 3 is y

And angle 1 is 2 y

And angle 2 is also 2y

But the answer key says 84= measure of angle N plus measure of angle L

How did they get this?

[silent plank]

external angle theorem

[upper karma]

Ohhhh

The whole triangle

Idk why I didn’t see that

Do you have any tips for me to catch that?

Or should I just have that theorem on my checklist whenever looking at a problem

[silent plank]

the theorem is a combination of
angle sum of a triangle and
angle sum on a line

I'm assuming you even used it when finding 1

[upper karma]

How would you solve this problem

Yes I did use it I just did not see that it could be used for the full triangle

[keen aspen]

Find OQP first

If a line bisects an angle, what does that mean?

[upper karma]

It means the angles are equal

[keen aspen]

Yep

So whats OQP in terms of angle 1 and 2

[upper karma]

OQP= 180-46-angle 3 and 4

[keen aspen]

Well actually we'll skip a step and use the external angle theorem

[upper karma]

Ok

So angle 1 and 3 is equal to 46 + 3 and 4

[keen aspen]

Mhm

[upper karma]

1 and 2*

[keen aspen]

Well 1 and 2

Ya

And lets say 1 is x and 3 is y

To make it look better

So 2x=46+2y

[upper karma]

Yea

[keen aspen]

Divide out 2

x=23+y

[upper karma]

Ok

Oh angle 1 = angle m + angle 3

[keen aspen]

Ya

[upper karma]

But then how do we move the 46

Or even find y

[keen aspen]

Well first lets write everything in terms of 1 variable

[upper karma]

Ok

[keen aspen]

x=23+y

[vague berry]

@obtuse tapir

Please explain how you got the solution for

I don’t know what you mean by inverse

[keen aspen]

So, 1 is 23+y , 2 is 23+y , 3 is y and 4 is y

[upper karma]

Ok

So then

Ohh

23+y = y + measure of angle m

Wait I still don’t get it

We dont know M

[keen aspen]

What

[vague berry]

Do you guys know how to solve that above

It’s geometry

[keen aspen]

y+M=y+23

[vague berry]

Right triangles I’m using it to solve for x

[keen aspen]

Solve for x?

Ok

[silent plank]

use an open q channel

[vague berry]

Yes

[keen aspen]

Multiply both sides by x/2

[upper karma]

@keen aspen ye we still have 2 variables

[keen aspen]

And yeah go to open q channel