geometry-and-trigonometry

[spark stag]

again, this is solely to help visualize where sine/cosine graphs come from

if you just want to visualize "the circle", then the visualization at 11:38 is much better.

but changing height is done to capture changes in theta

so that when they rotate the thing to make it look like a sine/cosine graph

the height now serves as the theta axis

(also lmao, does that video use sonic music at 12:30)

[upper karma]

@onyx sigil So... at http://demonstrations.wolfram.com/SineAndCosineIn3D/, does the cuboid represent an actual object in real life?

[spark stag]

sure

electromagnetic radiation

e.g. light, radio waves, etc

but for the most part, that's just a tool to visualize sine/cosine graphs

and how a change in an angle's value corresponds to the change of the value of sine/cosine on a graph

[upper karma]

I just need to know... the unit circle is supposed to be visualized as a flat circle on a table or a flat circle in the air being held up?

[spark stag]

the former

flat circle

[upper karma]

on a table

[spark stag]

sure

on a plane

that plane can be a circle, or anything else flat

again, the point of the 3d visualization was a (convoluted) way to demonstrate

how the "cosine is the x value, sine is the y value" intuition

translates to the graph of cosine/sine

they graph the "distance around a circle", roughly speaking

[which is why they're periodic with period 2pi]

you don't need to worry about that specific visualization

as long as you understand how sine/cosine graphs are generated

[upper karma]

I don't understand anything about trigonometry, even from the basic triangle; sure, I know how to do formulas and whatnot but I want to be able to visualize why I'm doing it. What does it all mean. I'm simply memorizing things my teacher tells me. No explanations were given as to what this all means.

[small raptor]

Have you been shown the unit circle?

A lot of the Trig concepts from from the unit circle

[upper karma]

Yes, but I have no idea why we use the unit circle. How does the unit circle help with measuring triangles? Is it just to measure the angles?

[small raptor]

Think less about it as a tool, and more of about it as a concept. Within angles, one of the measures are degrees of rotation. When you hit 360, you have come to a literal full circle. So, yes, the unit circle is what allows you to identify the angles, but this is where sin, cos, and tan gets their identities/properties.

the points of a triangle can also be seen as points on the unit circle, this allowing you to identify angle, and even a unit length of the sides.

https://www.mathwarehouse.com/trigonometry/sin-cos-tan-all-graph/animated-gif/all-trig-graphs-gif_orig_large_optimized.gif - you can actually see the triangles form as the point rotates around the circle

[upper karma]

So the unit circle can be used to measure real life objects and for things as small as radio waves or whatever?

[small raptor]

Technically, yes. Every time you use a trig function, you are utilizing the conceptual property of the unit circle

[upper karma]

my brain has never seen such math

[molten oracle]

Sup guys, need some help figuring out the best calculations to use...

The situation is quite complex, with multiple objects receiving usage of the desired solution for themselves and multiple component parts per frame, in a computer simulation.

Simplified... Take a sphere, apply an arbitrary force to it, in the form of a vector. The force data contains only the magnitude, direction, and point of contact. The sphere itself possesses a mass, and in this case we are not computing those calculations. The centre of mass is done, we just need to find the calculation that can process the force, coming from any arbitrary direction, into its tangential component, and its non-tangential component, enabling the force to be split into the translational component and rotational component.

Short and sweet... A force hits the sphere at a contact point which is 45 degrees off the tangential plane, how do i split it into the tangential and non tangential parts?

@me if anyone responds :)

[sly marlin]

@molten oracle
The net force gives the direction of the translational component.

The torque the force exerts wrt to the centre of mass gives the rotational component

[molten oracle]

Im programming an arcade simulation, i specifically need to calculate that tangential force.

@sly marlin

[sly marlin]

So, you need to calculate the tangential force for reasons other than calculating the rotational component?

[molten oracle]

It is for doing that...

But not in this way. It is to do with a force accumlator setup in a physics system, if i can seperate the force into tangential and non tangential components at the point they are read from the accumulator, it makes the total accumulations much easier.

If i could find an easy way to simply seperate the vector into the component parrallel to the radius vector between the centre of gravity and the contact point that would work too, and the remainder, that would work too.

@sly marlin (Ill @you when i reply, let me know if that is annoying tho :) )

(Also, im happy to put together the equations and stuff, what im after is the technique usually used - this must be a common problem.)

[sly marlin]

That's okay since I'll notice it

I jump around a lot in this server

So, you are given a sphere

And you have the position the force acts on the sphere

[molten oracle]

Yes

and the magnitude/direction of the force

[sly marlin]

and the radius vector is just the vector from the position of the force to the centre of the sphere

[molten oracle]

Yes

[sly marlin]

then afterwards, all you need to do is some dot products

Let R be the unit radius vector and F be the force

[molten oracle]

slow down a sec...

[sly marlin]

Then the component parallel to the radius is (F dot R)R

[molten oracle]

I am not in need of the momentum calculations

i have all that done hehe

[sly marlin]

The tangential component would be F-(F dot R)R

[molten oracle]

Oh, i see

my apologies for jumping ahead hehe

So, F is a vector and R is a scalar?

[sly marlin]

No, both F and R are vectors

R is especially a unit vector

[molten oracle]

Ah, i see, the radius vector

[sly marlin]

if you can't get a unit vector easily, then (F dot R)R/(R dot R) would do

as well as F-(F dot R)R/(R dot R)

[molten oracle]

Ok, why specifically must it be a unit vector?

[sly marlin]

Well, it doesn't

Your formulas would become this otherwise

[molten oracle]

but if it is, you can simplify out some of the variables?

[sly marlin]

Yes

(F dot R)R and F-(F dot R)R, since R dot R=1

[molten oracle]

Ok, thats wonderfully helpful, thankyou :)

Would you like to hear what its for?

Im designing a space game, an mmo, which will be running an entity component system simulation on the server, and copying its data using monobehaviours on the clients, but, the game has a very cool feature, already put together, which is a ship editor. Part of what it does is balance the ship, offsetting the position of the origin to make an object whose balance is as perfect as possible given the shape. The intention therefore is to make a physics system which can be constrained through methods understandable to the layman, thereby allowing ship design to be tuned through planning how you think the parts will affect the ship's handling. During the balancing, it results in an object which can be treated like a sphere, with its simplified arcade physics causing it to act even more like a sphere. So, when a contact force is received at any point on the object, it will then be split, as mentioned here, and then accumulated to two seperate accumulators for translation and rotation, then as they are added they become two single forces. Those final forces are what gets applied to the object. It means i can divorce management of the force sources from the force management and application... Anything can add a force to the accumulator, and easily so.

I could have solved the problem with trig... But i didnt see that solution as being efficient. And... failing on research :)

Thanks @sly marlin

[sly marlin]

cool okay

[molten oracle]

Is there any particular reason its written with the scalar variable after? F-(F.R)R instead of F-R(F.R)?

Also, i guess that would suggest to calculate both of the components, you simply have
F-(F.R)R = Tangential
(F.R)R = Component parrallel to radius vector

[sly marlin]

@molten oracle No reason at all

[molten oracle]

Cool

[sly marlin]

You can write it F-(F.R)R or F-R(F.R)

[molten oracle]

I was just wondering if there was a convention :)

[sly marlin]

Well, I typically would put the scalar in front

Which is (F dot R)

[molten oracle]

So is (F.R)R is correct for the radius parallel force component?

[sly marlin]

The component of force parallel to the radius

[molten oracle]

Im just deriving here - thats the part you remove from the force to get the tangent hehe

[sly marlin]

yes

Because F dot R is the length of the projection of F on R

and we want to multiply that by R to get the vector we get when we project F on R

[molten oracle]

Is there a particular topic of maths i could look up for this? I keep failing at finding anything more complex than basic stuff... I guess looking at vector projection?

I just dont know the right terms to search hehe

[sly marlin]

Hmm

[molten oracle]

I've got all the basics already researched and taken in, vectors in 3d, rotating vectors etc, dot products and cross products, now, im mostly looking through for slightly more complex stuff.

[sly marlin]

This is pretty basic stuff

properties of dot product, properties of cross product

[molten oracle]

Hehe yeah, hence why im keen to read up - i did some stuff more advanced than this, but this was like a real hard topic to find anything for

Ok, cool, ill look at those, thanks again :)

[upper karma]

@small raptor I finally understand everything after watching 28 videos.

[molten oracle]

hehehehe

Videoz iz gud, seez?

respectfully salaams every youtuber who ever made a tutorial video to upload

[vague mantle]

How much is the sine of 30 degrees?

[dark sparrow]

,w sin(30 degrees)

[somber coyote]

Display results online and refine query

Requested by Ann#0413

[vague mantle]

thanks, but this is just the beginning & I have no clue about anything... praying to at least pass math class

[upper karma]

Someone help me with Geometry

[silent plank]

do you have a specific problem you're struggling with?

[upper karma]

I’m trying to study

I don’t have any problem

[night drift]

Does anyone know why there are cofunctions of sin, cos, and tan? Example: sec, csc, cot

From my understanding it’s literally the inverse of whatever it’s co function is.

It’s confusing to me because is tan-1 not the same as cot?

[upper moat]

The difference is that csc, sec, and cot are the reciprocals of sin, cos, and tan, whereas sin-1, cos-1, and tan-1 are the inverses

cscx = 1/sinx, secx = 1/cosx, cotx = 1/tanx

[silent plank]

arcsin, arccos, arctan are inverse functions
csc, sec, cot are multiplicative inverses

[upper moat]

So what you end up with is cscx * sinx = secx * cosx = cotx * tanx = 1, and sin-1(sinx) = cos-1(cosx) = tan-1(tanx) = x (barring any undefined values)

[upper karma]

dm me if u know how to do this pls

[silent plank]

what ahve you tried

[upper karma]

wym

[silent plank]

where did you get stuck?

[upper karma]

okay so what happened was i was absent one day when i had a quiz and my teacher made me take the quiz instead of learning the lesson so idk how to do this

[silent plank]

i'll split this into 3 smaller questions
do you know how to find the gradient of a line?
do you know how to determine if lines are parallel from their gradient?
do you know how to determine if lines are perpendicular from their gradient?

[upper karma]

we never learned gradient

Gradient = Slope.

oh

[silent plank]

that should've been one of the first things you learned

[upper karma]

then yea ik slope

Ramonov, schools don't call it gradience until like Precalc+.

Or at least from where I'm at.

yea

[silent plank]

ah ic. i was taught both

[upper karma]

oo

[silent plank]

so, can you determine those properties from their slope?

[upper karma]

yes

[upper karma]

ok

[upper karma]

Can someone help me

[dark sparrow]

with what

[upper karma]

4

@dark sparrow

[dark sparrow]

#4?

[upper karma]

Yeah

[dark sparrow]

but you've already written all your answers for #4

[upper karma]

Yeah I know but I don’t remember how I got that

Because when I sin2

I get a different number

[dark sparrow]

are you using a calculator

[upper karma]

Those are my notes I’m doing a problem rn hold on

Yes

[dark sparrow]

are you getting something close to 0

[dark sparrow]

is your calculator in degree mode

[upper karma]

Yes

[dark sparrow]

switch it to radian mode

and try putting sin(2) in now

[upper karma]

Okay I got it thank you

This would just be 30 degrees right

On the unit circle

[dark sparrow]

not 30 degrees, no

but you don't need theta itself

you're practically GIVEN cos(θ) and sin(θ), and everything else can and should be calculated from those.

[upper karma]

Ok

So to find tan

I just do opp over adj

Or just look on the unit circle because it looks to be 30 degrees on the circle

Meaning the tan would be sqrt 3 / 3

[dark sparrow]

no!

@upper karma theta is NOT 30 degrees and its tan is NOT sqrt(3)/3.

tell me

according to that picture

what is sin(θ)

[hybrid walrus]

Anyone know how to do this

[upper karma]

[sudden locust]

Brand new video AND mini-series on my channel! This series will be all about The Vertical Angles Theorem. In Part 1 I will prove why Vertical angles are equal. Any feedback would be greatly apprecieted, I want nothing more than improving. 🥰 https://youtu.be/YDbfSTHXzLE

[upper karma]

Someone send me Geometry state practice test with example cuz idk anything about geometry I took it already but I need to practice. If you got a link for it please mention me and send me that link.

[sudden locust]

@bold terrace Thanks alot for the feedback.. Im thinking I rather go TOO in depth than TOO fast tho... Maybe I should tell the audience that its possible to speed up the video by 1.25x or 1.5x..?

[quiet mason]

nope

that would be lame

just remake

i mean if you already have all thr work from the last vid arranged to be used

[sudden locust]

👌

[upper karma]

Someone help me?

[silent plank]

what have you tried?

[upper karma]

??

I can see the answer but I wanna know how to solve it and get a answer

[silent plank]

have you thought about how to approach the problem?

[upper karma]

Yo bro I’m learning this by myself s

I don’t have any classes.

[silent plank]

you should still have at least some knowledge about what you need to do

like using a very famous theorem

[lone swift]

Hey can somebody familiar with Desmos can tell me what theta actually represents in Desmos when graphing?

originally I thought Theta in Desmos was just a variable representing the values from 0 to 2pi

but thats definitely wrong

[zenith ember]

Theta is usually an angle

Measured in radians, going counterclockwise from the positive x axis.

It is widely used in 2D graphing in Polar Coordinates

[lone swift]

Hmm

Is it accurate to say that theta is just a variable that holds every value from 0 to 2pi ? since 0 to 2pi is the full range of angles in radians?

[zenith ember]

Not really

Angles above 2Pi are possible.

And angles less than 0 are also possible.

They map to angles between 0 and 2 pi.

[upper karma]

@silent plank I’m slow so idk man I just wanna do that and I have to do Exam in 2 months

[silent plank]

what theorems do you know about right angled triangles?

[umbral snow]

It's in the middle lol

So take the average of π/2 and π

That's 3π/4

Average of a and b is (a + b)/2

So (π/2 + π)/2

= π/4 + π/2
= π/4 + 2π/4
= 3π/4

[silent plank]

lcm of 2 and 4 is 4

different order of doing things

(π/2 + π)/2
=(π/2 + 2π/2)/2
=(3π/2)/2
=(3π/4)

[quick sage]

how do you calculate this?

I can't remember the formula

is it just 1 regardless of theta?

[proven plover]

Yes its 1

[quick sage]

alright cool

[idle bloom]

hell you don't even need theta in there

you have literally anything in there

a function

multiple numbers

doesn't matter

as long as the argument of sin^2 and cos^2 are the same

it still equal 1

which is really cool

[somber coyote]

RokettoJanpu:

$\sin^2(\text{blah})+\cos^2(\text{blah}) = 1$

[quick sage]

thanks a lot

oh

if it's in the second quadrant it's negative correct?

[proven plover]

Negative what

[silent plank]

what's negative

[proven plover]

cosine is x and sine is y

[quick sage]

sin^2 + cos^2 =-1 in the second quadrant

[silent plank]

no

[proven plover]

that will never happen

[quick sage]

it's just always 1?

[silent plank]

yes

[proven plover]

Yes its a very important identity that you can derive other identities with

[quick sage]

When I want to calculate this do I calculate it as (-6/sqrt(13))(sqrt(13)/7) or do I need to change something in the postives/negatives?

I'm not sure if it being in the second quadrant is just stating that it's -6

or if i need to change something when inputing the values into cot

[proven plover]

Would you know how to draw that triangle?

[quick sage]

yeah

like this correct?

[proven plover]

No

[quick sage]

hm?

[proven plover]

oh wait

nvm

lol

ok so do you know the identities for cotangent and sine for a triangle?

[quick sage]

yeah sin is opp/hyp and cot is adj/opp

[proven plover]

well figure them out and just multiply

[quick sage]

so I do not need to change any of the values to negative?

like

(-1)cot

since it's in the second quadrant

[proven plover]

well the final answer is cotangent * sine, what are the signs of the trig functions in the second quadrant

[quick sage]

negative for cot and positive for sin

[proven plover]

so your answer should be what sign?

[quick sage]

negative?

[proven plover]

Yes

[quick sage]

okay

so

Lemme get this straight

(-6/sqrt(13))(sqrt(13)/7)

this is cot*sin

is it like this |(-6/sqrt(13))(sqrt(13)/7)|*(-1)?

[proven plover]

Why would you multiply by -1

[quick sage]

because it's negative

It's in absolute value

[proven plover]

Its already negative, I just told you that thing so you could make sure your final answer was the right sign

[quick sage]

alright I get it

I think I'm just over complicating it

[proven plover]

But yeah the first thing you said is the right answer

[quick sage]

I see alright thanks

[upper karma]

How do I find inverse sin of radical 3/2 in radians?

Here is what I did so far: On my calculator I found the inverse of radical 3/2 and got 1.047197551. Sin is positive in the first and second quadrants. But, my teacher said we had to subtract 1.047197551 from pi.

Why do I have to subtract from pi? Why can't I use the first quadrant?

My question may not make sense because this is the first time learning this and I just am so confused.

[silent plank]

write it as sqrt(3)/2 instead of radical

on the domain of 0 <= theta <= 2pi
there are two solutions and you would need to provide both

sqrt(3)/2 is also the ratio in a special triangle.

and the answer should be left in exact form unless told otherwise

[upper karma]

I'm still very confused. My teacher said to use the value in quadrant two or something like that, so we subtracted from pi.

I just need to know why the second quadrant specifically.

I know how to do the problem.

[silent plank]

what was the original problem?
did it ask for quad2 specifically?

did it state a domain

[upper karma]

@silent plank

I know my answers are probably correct — we did this in class.

[silent plank]

from what i'm seeing, assume they want the domain of 0 <= theta <= 2pi and list both solutions

do you know the special angle for
arcsin(sqrt(3)/2)?

[upper karma]

No

What does <= mean?

[silent plank]

less than or equal to

[upper karma]

That’s what I thought, okay

[silent plank]

the special triangles/angles will come up a lot and you should definitely learn them

[upper karma]

The thing is, I’m in calculus and we get a reference sheet for all of that.

[proven plover]

Memorize from 0 to pi/2

[silent plank]

well you only really need the first quadrant and use ASTC to figure out the rest when needed

[proven plover]

Wait why are you doing those types of problems in Calculus

[silent plank]

its basic trig

[proven plover]

You sure its not precal?

[upper karma]

It’s a preliminary chapter

The thing is... the math department at my school asked everyone if they wanted to go into calc right from trigonometry — I’m in a class with a bunch of honors kids and don’t know ANY of this stuff.

[silent plank]

familiarise yourself with the unit circle posted above.
1st quadrant (which includes the special angles/triangles) and
when trig functions are positive/negative from ASTC

[upper karma]

@silent plank Yes I know all of that

I just don’t know why we’re choosing the second quadrant over the first

[silent plank]

i've said many times you need to put both

[upper karma]

She never said to

We did examples in class and never put two answers.

[silent plank]

was there something else on the page?

its cut off

[upper karma]

Nope

Those are just examples of the same type

Probably all wrong

[silent plank]

whos writing is that?

[upper karma]

Mine

[silent plank]

red/pink/purple

all of them?

[upper karma]

Yes

It’s so unorganized because I’ve been trying to write down everything my teacher says during class haha

[silent plank]

based on the level your at, they probably only want the first quadrant for this question
your teacher wanting the 2nd quad was probably for something else

in the second section i can see that they want all solutions in [0,2pi]

[upper karma]

I didn’t learn that yet

That entire section

[silent plank]

what did you mean when you said

Yes I know all of that

[upper karma]

The ASTC part

What is positive and what is negative

[upper karma]

Yes....

That is the part I know

[proven plover]

Well what were you asking for

[worthy root]

In my textbook i saw two equation . dont know if they are correct

$1^{\degree}=180/\pi$

[somber coyote]

Krishna:
Compile Error! Click the reaction for details. (You may edit your message)

[weary drift]

$360^{\circ} = 2\pi$

[somber coyote]

RokettoJanpu:

[weary drift]

then it follows that...

[dark sparrow]

$1^{\circ}$

[somber coyote]

Ann:

[weary drift]

$1^{\circ} = \frac{\pi}{180}$

[somber coyote]

RokettoJanpu:

[dark sparrow]

1° = π/180 radians

[worthy root]

yes exactly

but my textbook has some wrong equation

[upper karma]

How can I find the inverse function of this problem? The correct answers are in green. I know the first answer is .6155, but how do I manage to get 3.7571 for quadrant III? .6155-pi gives me a negative number...

[spark stag]

you should be adding pi, not subtracting

note that angles in quadrant III are a little more than pi

so we add them to pi

rather than subtract pi

[upper karma]

THANK YOU!

Can you tell me what to do in each quadrant, like subtract pi or add pi?

[spark stag]

quadrant I is just the reference angle; we add it to 0, i guess, if you want to think of it like that

for quadrant II, remember that the reference angle is drawn from the x axis, so:

we subtract the referece angle from pi

the entire half-circle is pi, and the reference angle is the bit "taken out of" pi

to produce the blue line here

a similar argument applies to quadrant IV, but now we've went around a full circle, not a half-circle

so it's actually subtracted from 2pi.

[upper karma]

Thank you very much!

[upper karma]

@spark stag Do you know how to do question 2?

Arc tan (-1)

[spark stag]

arctan(-1) is the value theta such that tan(theta) = -1

that means we need $\frac{\text{opposite}}{\text{adjacent}} = -1$

[somber coyote]

Namington:

[spark stag]

if you prefer, you can think in terms of $\frac{\sin(\theta)}{\cos(\theta)} = -1$

[somber coyote]

Namington:

[spark stag]

this means that sine and cosine need to have opposite signs (i.e. quadrant 2 or quadrant 4)

and need to be equal to each other

(i.e. a refernece angle of pi/4)

so, on the unit circle, we have theta = 3pi/4 and 7pi/4

since we take pi - pi/4 and 2pi - pi/4

but wait! the domain of arctan is restricted to (-pi/2, pi/2]

it turns out, 7pi/4 = -pi/4

since clockwise rotations are negative

that fits within the domain, so there's the solution: -pi/4.

the value 3pi/4 doesn't correspond to a solution, as there's no way to make it fit in the domain (-pi/2, pi/2]

even if we make it negative, that's -5pi/4

which is still out of the domain

[upper karma]

I didn’t learn any of that...

In terms of restrictions

That is what I’m doing @spark stag

[spark stag]

i mean, the question asks for arctan(-1)

not arctan(1)

[upper karma]

My teacher said to do that whenever there’s a negative

I tested what I got and did tan (5.4978) and tan (2.3562) and got -1; I don’t know why she gave those answers that I showed you in pink @spark stag because when I test them they didn’t even give me -1

[spark stag]

ah yeah, it appears the answers in pink are incorrect

sorry, should've checked those

bizarre

[upper karma]

So am I right?

I mean, it works....

[spark stag]

seems fine

(except for codomain issues, but i guess your class isnt too picky about that)

[grizzled zinc]

I have no idea what I'm doing wrong for those 3. #6 I was thinking youd just subject but its not even so idk. #8 I keep getting two numbers that aren't evenly dividable and #10 is the same problem

[silent plank]

youd just subject
?

[grizzled zinc]

What

[spark stag]

what do you mean by "just subject"?

do you mean "subtract"? because yes, that's correct

you subtract

[grizzled zinc]

For #6 right?

[spark stag]

yes

[grizzled zinc]

Oh ok it just made me second guess myself because it wasn't even

[spark stag]

i'm not sure why it being even matterrs

[silent plank]

why would it matter whether it was even?

^

[grizzled zinc]

idk I just thought it did

Then for #8 my equation was 9x-39+47=3x+10

But it ended up being 6x=22

[spark stag]

are you sure about that?

[silent plank]

you didn't add the lenghts properly

[spark stag]

DF is the entire length of the line

we should have

$DF = DE + EF$

[somber coyote]

Namington:

[spark stag]

right?

but somehow you have

$DF + DE = EF$

[somber coyote]

Namington:

[grizzled zinc]

Oh

Oh wait okay

I see what I did wrong

So then I made the same mistake for #10

It should be 7x-27=3x-5+x-1

[upper karma]

@spark stag Thank you so much for helping me! I sat in math class for two days being so confused along with other students!

[zinc silo]

Hello friends

ive been working on this problem for the last like hour

and dont know what to do anymore

https://i.imgur.com/r267iDr.png

[median crown]

Uhhhhh

I think I can 200 IQ trick shot this one

[median crown]

idk

I gotta do something

[sudden locust]

New Vertical Angles Theorem video is out! This is part 2 of my mini-series, and we will now start to solve some problems. Do you know your vertical angles or do you need some refreshment? 🙈😊🤓 https://youtu.be/8x5wY9EfBFQ

[fringe dirge]

Your advertising here is getting pretty annoying

[worthy root]

Lmao

[sudden locust]

😪

[quiet mason]

and the emotes

[sudden locust]

@quiet mason So sorry that you feel annoyed, Im pretty sure all my messages will disappear if you block me.. (want to put a sad emote here but I refrain)

[dark sparrow]

they won't disappear

they'll get hidden under "n BLOCKED MESSAGES" tags

[quiet mason]

holy shit

@sudden locust do whateved you want lol

it dosent matter to me

if it dosent affect me

also it dosent matter to me if you get banned

[sudden locust]

I will tone down for sure.. Appreciete your honesty Leonhard 🙏 👌

[upper karma]

sorry to interrupt, but what’s that topic called, where you have one triangle, it gives you 3 sides with variables, you need to find the length/angle of triangle using sine,cos and tangent?

#just-trigonometry

[distant bear]

why am i bad at math

[umbral snow]

Because it's not internalizing for you - you likely see the symbols and aren't understanding "why" we use them the way we do.

[distant bear]

bruh

[umbral snow]

And that's fair, many teachers aren't teaching that side

[distant bear]

my teacher doesnt teach the steps that throughly

[umbral snow]

A bit blurry, can that be better?

[distant bear]

I got a 59 in geometry

One sec

[umbral snow]

I don't think this is enough info to solve for x

Imagine that 21 line wasn't there, then it would surely not be enough info. Then, you can always add the 21 line in no matter the triangle

Troll question?

[upper karma]

Aren't you suppose to solve for the first triangle, then by similar triangles, use that to solve for the 2nd triangle?

Though I am not sure if this is enough information

[coarse bronze]

it is

you have all 3 angles of the big triangle

[upper karma]

is pythagorus triganomitry?

[coarse bronze]

the pythagorean theorem is just geometry

if thats what you mean

[upper karma]

ah

okay

thx

[lavish void]

Real talk anyone remember how to find the axis of symmetry given a focus and directrix

[primal lynx]

Yes

Prof google will send you to a TA, the organic chemistry tutor

[astral torrent]

Hi

[reef flower]

Hey, I’ve been at this for a while and am a bit stuck. Question asks to find the exact value for: arctan(tan(2pi/3)), I know tan(2pi/3) is -sqrt(3) but I’m not sure where to go from there, should it be obvious to me? Thanks in advance.

[spark stag]

this should be fairly obvious from the definition of arctan, yes

recall that arctan "undoes" tan.

[next saffron]

its like squaring a square root

[spark stag]

the catch is that arctan's range is restricted

specifically, you have to make $\frac{2\pi}{3}$ fit in the interval $(-\frac{\pi}{2}, \frac{\pi}{2}]$

[somber coyote]

Namington:

[spark stag]

(which can be done by thinking "clockwise")

[reef flower]

Thanks, I forgot about the domain of arctan. The way I got there was making tan(x)=-sqrt(3) and to make it easier I found sin(x)=-sqrt(3)/2 and used my knowledge of the unit circle from there to get x=5pi/3 but since its not on the domain I just took -Pi/3 instead. Is there an easier way of doing this (should arctan(-sqrt(3)) obviously be -Pi/3 to me) or is this approach reasonable?

[spark stag]

arctan is the inverse function of tan (if we ignore domain-range issues)

so $\arctan(\tan(x)) = x$, $\arcsin(\sin(x)) = x$, $\arccos(\cos(x)) = x$, $f^{-1}(f(x)) = x$, etc.

[somber coyote]

Namington:

[spark stag]

again, the only catch is that arctan's range isnt quite the same as tan's domain

so we have to "change" the value of x to fit in arctan's range

but besides that, yeah, it's just a property of inverse functions.

[dark sparrow]

arctan isn't the inverse of tan; tan doesn't have an inverse

[spark stag]

yes, hence why i mentioned that we're ignoring some issues

tan isnt invertible; arctan only has a codomain on some of tan's domain

but either way, arctan is defined specifically so that its relatively "easy" to make "look like" an inverse, by just accounting for domain/range issues

(like we did when "converting from" 2pi/3 to -pi/3)

[distant bear]

Idk

[dark sparrow]

:?

[worthy root]

Ann is this correct ?

Ex.4

@dark sparrow ?

[dark sparrow]

i'll get back to you in like 10m

[worthy root]

Okay

[dark sparrow]

yes this is correct

[worthy root]

[dark sparrow]

what

why the sad

[worthy root]

How?

Lol i am not sad in real.

[dark sparrow]

all the work has been laid out

[worthy root]

Woops

[dark sparrow]

what part of it do you not understand

[worthy root]

Di didnt see the diagram

I*

Wait

I still dont get

[dark sparrow]

what do you not get

[worthy root]

Can you rrphrase the question of i dont get it

[dark sparrow]

you have a circle, and a sector within that circle

the sector's perimeter is known to be equal to half the circle's circumference.

you're asked to find the sector's angle.

[worthy root]

If the sectors perimeter is half the 2pir then the anhle it forms with centre is 180°

[dark sparrow]

no.

it's not.

[worthy root]

Why not?

[dark sparrow]

the perimeter is not just the arc. it includes the two radii as well.

[worthy root]

Woops

I forgot the definition of arc sorry

Of sector*

[upper karma]

what is this type of math called

geometric model I think, but I'm trying to find videos about it khan and I can't

I have to find the area of a rectangle on the grid within the line of the equation

[upper karma]

I can't take any pictures

[weary drift]

@upper karma can you express the sides of the rectangle in terms of y or x?

[upper karma]

yes

[zenith ember]

You said, in #discussion that you had a point and a line.

What is the point?

[upper karma]

the point is P

[zenith ember]

Ok...

Please post an image or something of the whole question

Because as it stands, this is not a complete question.

[upper karma]

I completed the question already

I just need to know what kind of math this is called so I can understand it bette

[zenith ember]

Since we don't know what the question is, we cannot answer.

Is it an optimization problem? That'd be calculus. It might be geometry. Or trigonometry.

[upper karma]

"Draw a rectangle in the first quadrant as follows: Two sides lie on the coordinate axes, and the upper right vertex is a point P on the graph of y=12 - 2x, as shown in the figure. The shape of the rectangle depends on which point on the line y=12 - 2x is used as a vertex.

a) Find the area of the rectangle as a function of the x-coordinate of P. Call the function A(x)

b) What is the domain of the function A(x) ?

c) Graph the function A(x) on its domain.

d) What is the maximum value of A(x) ?

there is one point on the line where y=4 and x=4

The domain is (0,6) and the area of the rectangle is 18.

the point on the graph is (6,3)

[acoustic peak]

I mean, just sounds like geometry to me.

[worthy root]

OAb is sector of a circle having centra at O and radius 12cm.If m(angle AOB)=45degree, find the difference between the are of sector OAB and sector AOB

Can anyone draw this with both sectors

[upper karma]

I cannot

[worthy root]

I cant see what is sector AOB

and what is sector OAB

[upper karma]

I haven't done this before, maybe someone else can help.

[worthy root]

<@&286206848099549185>

[acoustic peak]

Could you provide an image?

[worthy root]

Its not given

[vague mantle]

can somebody help me with trigonometry identities?

better said, with my homework

I'm given cos α = 7/25. I need to find sin, tg & ctg

<@&286206848099549185>

[chrome fiber]

!15m

oop

read #❓how-to-get-help, you can only ping after 15 mins have passed

[vague mantle]

this is my first time pinging

[acoustic peak]

Well, now you know.

Just keep SOHCAHTOA in mind when doing trig, in my opinion.

[vague mantle]

still don't understand what you mean but ok

[fringe dirge]

idk why people use that as an excuse

[acoustic peak]

When posting a question

You're supposed to wait at least 15 minutes before pinging the helpers.

[worthy root]

But thi channel was ocuupied by me

this*

[acoustic peak]

Your latest question was after Damanug, there was a 3 hour window from then to your previous help session, from what I see.

Not that I have anything to do with it.

Just pointing it out.

[distant bear]

is the answer 270 degrees counterclockwise and 90 degrees clockwise, just checking

[coarse bronze]

technically there are infinite answers but yes those are the ones expected

[distant bear]

?

[olive solar]

if you rotate something 360 degrees, then you get the same as when you started, so if you rotate that 90 degrees and then 360 degrees more, you get the same result

which is why you'd technically have infinite answers

[distant bear]

oh ok

[median crown]

[wooden swift]

How do you convert 6 radians to degrees?

[zenith ember]

How many degrees is Pi radians?

[wooden swift]

Um I don’t know

Our teacher didn’t really tell us

[zenith ember]

What did your teacher tell you about radians?

[wooden swift]

That’s the thing

I don’t even know what a radian is

[zenith ember]

It's a measure of angle, like degrees.

In a similar way that feet and meters are both measures of distance.

[marble parcel]

stuck on this problem

[zenith ember]

But if you're learning radians, you should know -something- about them.

[wooden swift]

How many degrees is in a radian?

[zenith ember]

It's much easier to answer the question I proposed earlier.

I'm still trying to get a handle on what you've been taught.

[wooden swift]

Not a lot

We know like the unit circle and then minute second stuff and special triangles

[zenith ember]

Specifically about radians.

You haven't been taught that there are 2Pi radians in a full circle?

[wooden swift]

Just the like pie/180 stuff but I’m iffy on it

[zenith ember]

Yeah.

Ok.

that's what I was getting at earlier.

Pi radians = 180 degrees

That's the conversion factor.

[wooden swift]

So 6 radians would be 180x6

[zenith ember]

Not quite.

I'm not saying 1 radian = 180 degrees.

Pi radians = 180 degrees.

$\frac{6 rad}{1} \cdot \frac{180\degree}{\pi rad}$

[somber coyote]

samanthaCS:
Compile Error! Click the reaction for details. (You may edit your message)

[zenith ember]

Bleh.

That's terrible notation, but I'm hoping it gets the point across.

[wooden swift]

So I’m the calculator do I put 1080/pi

[marble parcel]

<@&286206848099549185>

[queen python]

To get the help of helpers send a problem

[marble parcel]

Scroll up a little

[queen python]

Then if nobody has answered I’m 15 minutes ping ONCE

Oh

[marble parcel]

Yeah

[zenith ember]

Do you understand what "Intersection" means in this context?

[marble parcel]

isnt it where line L intersects ABC

[wooden swift]

Samantha in the calculator do I put 1080/pi

[zenith ember]

It's helpful if you don't use the word when defining the word.

[marble parcel]

ok so like “passes through” figure ABC

[zenith ember]

Dimmed Light: I honestly don't know how much more assistance I can give you without literally doing the problem for you.

I can definitely say 1080 is the wrong numerator.

It may be more useful, toby, to think of these as sets of points.

So the triangle, ABC, is the set of points that make up the outline of the triangle.

[wooden swift]

Can you do a problem like it with a different number of radians? Like an example problem

[zenith ember]

No, I really don't think I can.

[wooden swift]

Why?

[zenith ember]

Again, I do not think I can give you more help without practically doing the problem for you.

[wooden swift]

But can you do it with different number of radians?

[zenith ember]

This is a simple conversion.

I've literally written the calculation you need to do already.

[marble parcel]

Still a bit confused samantha. So for 5a would the intersection be DE?

[zenith ember]

The way I interpret the question, it would be D and E, separately.

ie, the set of points that is common to both the triangle (but not the region inside it), and the line.

[wooden swift]

Is that how?

[zenith ember]

Yes, that's the right answer.

[wooden swift]

Okay thanks

[upper karma]

James runs a bicycle around a circular ring with equation $ x^2 + y^2 - 26x + 16y - 392 = 0$ at precisely $\frac{3\pi}{5} $units per minute. How many hours does it take for Iames to run around the ring 12 times?

[somber coyote]

rudy:

[upper karma]

So, I completed the square and got that the radius was 25.

Since r = 25, C = 50\pi

And so: 50pi/(3pi/5)

= 250pi/3pi = 83.3

So it takes 83.3minutes to run a full circle

Then the answer is:

83.3*12

999.96

Or almost a thousand

Lets approx to a thousand

Then it must take 16hours with around 40minutes to complete the 12 rides, right?

[terse leaf]

"Prove that ab to bc is Phi, so |AB|/|BC| = |AB|+|BC|/|AB|

I tried everything, similar/congruent triangles

The only thing I know for sure is that trainglr AB and top is similar to the big triangle

[wise pawn]

do you have more information

other than being parallel to the base, how high up is it

is it supposed to be cutting through the center of the other side?

[terse leaf]

A & B cuts the traingle in the mid

And all the sides are equal

So |AB| is the mid parralel of the triangle

[sly marlin]

hmm

The triangle is not necessarily equilateral, right?

[terse leaf]

All the side are equal

[sly marlin]

the triangle is equilateral?

[terse leaf]

Yes

[sly marlin]

hmm

Extend AC

then you are more or less done

intersecting chords

@terse leaf

[terse leaf]

@sly marlin Thanks, I'm trying rn

Not coming out with using only intersecting chords

[sly marlin]

solved it?

[obsidian wedge]

what's the best software for geometric constructions

[sly marlin]

you can try geogebra

try out a couple until you find one you like

[quiet mason]

here phi is golden ratio? @terse leaf

[sly marlin]

yes

[willow oak]

I’m having some trouble with this trig question. I know that sin(-7pi/8) = -sin(pi/8) but I’m unsure about the second part. I think that because sine is an odd cofunction the negative can transfer out and make -sin (pi/8) which would make the question true. However, I’m not sure if I’m misremembering that rule.

[sly marlin]

yes

sin is an odd function

I don't know where you get cofunction

[willow oak]

I was confused on whether or not I was remembering the rule correctly. So the two functions are equal?

[weary drift]

These aren’t functions, each side is just some value. What we’re saying is sin is an odd trig function, so you can take the negative from -pi/8

And make the right side -sin(pi/8)

[willow oak]

Okay. Thank you very much!

[weary drift]

No prob

[terse leaf]

@sly marlin I solved the golden ratio problem yesterday btw. Thanks for your help

[flint pelican]

The heck

Is this even

[primal lynx]

this is polar coordinates

its J right?

[sweet dragon]

can anyone solve this?

[primal lynx]

,rotate 270

[tender iron]

I need help with (a)

[latent galleon]

Please help w/ question 9

[sly marlin]

okay, what did you try so far?

[latent galleon]

i just tried plugging values in my calculator so nothing really, i don't understand how to get the answer at all

[sly marlin]

Can you find |sin x| and |tan x|?

[latent galleon]

the answer suppose to be sinx= -√8/3 and tanx= -√8 i'm not sure how they got those

[sly marlin]

okay can you find |sin x| and |tan x| (without the sign)

[latent galleon]

no i'm not able to find sinx and tanx

[sly marlin]

$\sin^2x+\cos^2x=1$?

[somber coyote]

Element118:

[sly marlin]

@latent galleon

[latent galleon]

sorry, but how do i use this for the question?

[sly marlin]

you have cos x

calculate sin x

[latent galleon]

it doesn't work when i rearrange it and calculate it

[sly marlin]

what happens?

[latent galleon]

so i calculated (cos(1/3)) ^2 then minus that from 1 and last square rooted the full thing and got ≈ 0.33

[sly marlin]

NO

you have cos x = 1/3

[spark stag]

to be honest

i'm confused why you're using the pythagorean identity for this

rather than just drawing a right triangle

[covert orchid]

Hi guys, i have this exercise, but i have no idea what to do, PQRS is a rectangle and it has an area of 90cm^2, and ABCD is a square and it has an area of 36cm^2. The problem says that i have to find the perimeter of the whole figure.

[spark stag]

i mean, you have to use pythagorus either way

but this seems much more motivated than just identity-crunching

¯_(ツ)_/¯

[silent plank]

well the calculations are exactly the same

[spark stag]

yeah

[silent plank]

you didn't subtract it from 1 before taking the square root.

[sly marlin]

@covert orchid There's a number between P and Q?

[covert orchid]

Nope

[sly marlin]

There's not enough information

[spark stag]

if theres no number between P and Q

this cant be solved

[covert orchid]

oh

[spark stag]

we need at least one side length of the rectangle

[covert orchid]

and i have this parallelogram but i think i need more information too

i tried but

[silent plank]

we se a 0 between the p and q. Was that the question number?

[spark stag]

that parallelogram should be solvable

[covert orchid]

yes ramonov

[spark stag]

via angle-chasing

[sly marlin]

But there are no distances

unless x, y, u, v, are angles

[covert orchid]

yes those are angles

[spark stag]

thats what i assumed

[latent galleon]

omg i got it now, i always forget the about using p.t for these questions thank you

[spark stag]

as a hint, note that the blue lines serve as transversals between parallel red lines

and that the internal angles of a parallelogram add to 360

and further, that consecutive angles are supplementary (add to 180)

you should be able to make some systems of equations based off that

[sly marlin]

x okay, but it really looks underdetermined

[covert orchid]

i tried this

i'm stuck at the 5 and 6 part

[flint pelican]

Sorry guys I was afk

Tbh idek

What this is

I think it’s trig so I’m putting it in here

[jovial axle]

do you what pi radians represents?

[primal lynx]

@flint pelican bro i told you polar coordinates

J

[flint pelican]

What the heck is a polar coordinates

[primal lynx]

look it up

[flint pelican]

Pi radians=180 degrees right

Ok

[primal lynx]

its not hard

yes that true

[flint pelican]

It’s not hard? Say lesss

[primal lynx]

you need to know this later on so its a good idea to get it down

[flint pelican]

Ok so uh just search up on YouTube

[primal lynx]

yeah

[flint pelican]

Like khan or organic or whoever

Ok

[primal lynx]

organic

[flint pelican]

Ok

[primal lynx]

2x speed you know the deal

btw is this from ACT?

[flint pelican]

Yeah it’s act

[primal lynx]

nice i figured

i saw basic linalg below

so i was like hmmm what test does this

btw i never took ACT officially

[flint pelican]

Oh

[primal lynx]

i can just tell

ACT is faster than SAT right

[flint pelican]

You took a practice though

[primal lynx]

so easy questions come by a lot

[flint pelican]

Yeah that’s what people say because you get less time on act

[primal lynx]

i did take a practice in class

not class, detention

well both

[flint pelican]

Yeah ig the math isn’t that bad I mean when I finish a test and look back at the answers with no time pressure I can do majority

[quiet mason]

@worthy igloo hey uhm mind to link me platforms to do haskell on?

seems youve done it and i wanna start it with the best advice

[worthy igloo]

why are you asking this here

[upper karma]

Why

haskell is

DISGUSTING

[quiet mason]

where would i ask

[primal lynx]

wut

[worthy igloo]

that's like your opinion man

[quiet mason]

the FP server?

[worthy igloo]

#computing-software if at all

[primal lynx]

fr

[upper karma]

i think its pretty objective

it was QED

[primal lynx]

@flint pelican does the right side of the image say "bro what" lol

[flint pelican]

@primal lynx yes it does lol

[sly marlin]

Can you put that into a calculator?

do you know how to evaluate it? @tired totem

that wouldn't help with evaluating it

and pathetic at best

Just as well

We don't need a rude person here

<@&268886789983436800>

ah all should be well

[valid agate]

[sly marlin]

yeah, that user has been rather rude

[valid agate]

Well if you're going to be like that

,hackban 362359282622267403 -r Joining and leaving just to be rude

[somber coyote]

Hackbanning...
✅ Successfully hackbanned Big Ounce (id: 362359282622267403) and purged 1 days of messages.

[upper karma]

Hello, I need help. I have to calculate the length between BD. Calculate both of the degree angles on DBC. And lastly calculate the area in ABCD.

[sly marlin]

@upper karma Any ideas how to start?

[upper karma]

Mhm, I think I should use the phytagorean theorem to calculate the length between BD.

[sly marlin]

but you need right angles

[proud mesa]

Are approximations allowed?

It's easy to solve this by using any of the trigonometric laws.

You could either use law of sines or the law of cosines in order to solve this, I mean.

[zenith ember]

I think Element118 was hoping they'd come up with that themselves

[proud mesa]

And also the fact the area of a triangle is absin(theta)/2

Oops

Sorry

[primal lynx]

i mean whatever right, if he knows then he knows and if he doesnt then he doesnt know any other way so its fine

[wooden swift]

What am I doing wrong?

[zenith ember]

You've found the values where cos(x) = ± 1/2

The problem is asking for cos^2

[primal lynx]

even though to solve you root, you must go back

[dark sparrow]

none of these actually satisfy cos**^2**(s) = 1/2.

[primal lynx]

but i cant see what you put

[wooden swift]

So do I do that but with all the 45 degree angles instead cause that’d be 2/4 right?

[zenith ember]

Yes, that's the right idea

[wooden swift]

Alright thanks

[vague pagoda]

um hey

[zenith ember]

👋

[vague pagoda]

i need to find tan β when Ive been given sin β = 0.23

howd i go about this

[zenith ember]

The most straightforward way is to draw a right triangle

Designate one of the angles beta, and then write the lengths of the opposite and hypotenuse

Use Pythagoras to find the unknown leg

And then write down the tan

[vague pagoda]

hmm how do i find the lengths of opposite and hypotenuse

[zenith ember]

You have sin

[vague pagoda]

yeah but only sin

[zenith ember]

That's all you need

[vague pagoda]

oh

this book is crap it doesnt explain anything

[zenith ember]

Sin gives you the ratio

Tan asks for the ratio

So you can just do everything ratiometrically

[vague pagoda]

wut

[zenith ember]

You know SOH CAH TOA, right?

[vague pagoda]

yes-

[zenith ember]

So sin b = .23 means O/H = 0.23

We don't care about absolute distances, just the ratio, so we can say H = 1 to make this easy

[vague pagoda]

nvm ya i realized

i doodled the triangle

[zenith ember]

👍

[vague pagoda]

smh i should know this ive tested into calc 2 level

[zenith ember]

It's easy to forget things you haven't used in a while.

[vague pagoda]

im only 14 so trig and calc is a fairly new thing for me

[zenith ember]

Ah

In that case, don't beat yourself up. You're still well ahead of the curve.

[vague pagoda]

yeah true tbh

im dumb

ofc sin β / cos β = tan β

[cinder charm]

Well I have to do this

[mighty gull]

which one

[cinder charm]

All of em

[mighty gull]

pick 1

i will help you with that

[cinder charm]

Well on the first picture top left what do I put next to “by” on the end

[mighty gull]

well have you found angle

[cinder charm]

No I’m stuck on that one

[mighty gull]

well writte out the given angle

do you see any similar traingle after that

[cinder charm]

What’s the given angle?

Like I’m super new to this

[mighty gull]

ACB and BCD

[cinder charm]

Where did you get that?

Wait nvm

[mighty gull]

its given in the problem

[cinder charm]

My bad

Okay so would it be ABC is similar to BDC

Because that way they both match up

[mighty gull]

yeah, and the 2/3 ratio for sides

[cinder charm]

Nice

So is this right

[mighty gull]

they want you to say like SAS similarity

[cinder charm]

Oh ok

What about this one

[mighty gull]

well what do you think you should do

[cinder charm]

Match up the numbers that match with each other and cross multiply?

[mighty gull]

umm yeh

[cinder charm]

You sureeee?

[mighty gull]

try it

[cinder charm]

Alright thanks man

I gotta go

[jagged holly]

can someone help me please

i dont get this

[cinder charm]

Sorry bro idk either

[lunar crow]

@jagged holly

is the path, the non pool area?

[forest shoal]

can someone tell me how (sin ρ + µ cos ρ)mg/[(cos ρ - µ sin ρ)]= m (Vmax)²/R

simplifies to g(tan ρ + µ)/(1 - µ tan ρ) = (Vmax)²/R

mention me if anyone responds please

[cerulean parcel]

what am i supposed to do with question C

*line n is y=-2/5x+1

[zenith ember]

Hm

I don't see any obvious relationship.

[silent plank]

b looks wrong

[zenith ember]

Agreed

[cerulean parcel]

oops forgot to add the x

[zenith ember]

That's not the only issue.

[silent plank]

it would also be better if the X didn't look like it was in the denominator for a)

what's the relationship between the slopes of perpendicular lines?

[cerulean parcel]

oh shit mb

forgot its the recipricol not the +/-

probs spelt that wrong

[zenith ember]

It's both

[dreamy sequoia]

Is it possible to say for A line AD with points B and C between them, that AB+BC+CD=AD?

[dark sparrow]

no not always

if B is between C and D then that equality doesn't hold

[dreamy sequoia]

Is it possible to use double substitution in a proof

<————————————————>
A B C D

How do I do latex again

[somber coyote]

zbot473:
Compile Error! Click the reaction for details. (You may edit your message)

[dark sparrow]

oh god. two-col proofs?

[dreamy sequoia]

High school moment

Can you help on this?

[dark sparrow]

no

[dreamy sequoia]

Oof

[worthy root]

2*

I tried a lot help me

[surreal roost]

How do I solve this?

[dark sparrow]

what have you tried so far

and where are you stuck

[worthy root]

Me ? Or distance? @dark sparrow

[dark sparrow]

distance.

[worthy root]

Ok

Anyways i found solution on intwent

Distance can you define linear pair of angles ?

[surreal roost]

Well I've tried 50 because ACE looks to be half of ACD which is one 100, but that's wrong so now I assume its somewhere near the 60 to 80 range but I don't want to randomly guess what the answer is and I assume there an actual thing you do to get the answer.

Also a linear pair of angles add up to 180 right?

[weary drift]

yes to your definition of linear pair

[surreal roost]

wait I see it now

[dark sparrow]

@surreal roost no, you cannot assume one angle is half of another just because it looks that way

don't do that

in fact i'm gonna go so far as to say NEVER do that

[surreal roost]

ace would be 80 right

if acd is 100 to add up to 180 ace has to be 80