geometry-and-trigonometry

[night vigil]

Ok

Thanks!

[silent plank]

if you were to consider HK as your base,
the altitude to that would be MJ

[upper karma]

you could also use Apollonius theorem + pythag

[copper marten]

its asking me for the scale factor to get from the base of the large pyramid to the base of the smaller pyramid at the top, but im not sure how to do it (??)

[trim breach]

The triangles in the top pyramid and the whole pyramid are similar.

You can set up a proportion between similar parts.

[copper marten]

so like this?

144 = height of the whole/biggest pyramid
144-96 = 48

48/144 = 1/3

1/3 is the scale factor (?)
(dilated from the apex)

[lament nova]

Can someone tell me why I'm getting these wrong?

[silent plank]

you'll need to tell us what you actually did to get those values

@lament nova

[finite swan]

You are pretty close

Check again which value is equal to 6.5

[lament nova]

i managed to get it done thx!

[onyx marlin]

would anyone pelase help me

[lament nova]

whats up

[onyx marlin]

can you please show me how to solve the one aboce

[lament nova]

oop

not my thing XD

[earnest sphinx]

What's a recommended software to use to do geometry? For example, for working with parametric curves?

Is MATLAB the best option?

[twin phoenix]

Try using the proprieties:
• cos(2x) = cos²(x) - sin²(x)
• sin(2x) = 2sin(x)cos(x)

And expand the right hand side using tan(x) = sin(x) / cos(x)

[somber coyote]

xi64

[upper karma]

@onyx marlin

there is a typo btw

(a - b)^2 = a^2 - 2ab + b^2

i mistyped + instead of -

[crimson holly]

$4/3 n + 3/4 = 1/2$

[somber coyote]

rayy

[upper karma]

what is the question ?

[frigid flower]

Hello i really can't find any material on my issue because i don't know what this is called. Basically, the exercise goes like this:

AB=5cm
I need to find points (named T1,T2,T3...) that are 2cm away from AB.
Also the angle ATB must be 90°

Help please, i don't even know what method to use to get these points

[trim breach]

I don’t really know what you are describing. It sounds like ATB is a triangle with hypotenuse AB = 2, but I am not sure what you mean by points away from AB…

Do you have a picture?

[inland salmon]

someone help please!

[silent plank]

consider circumferences

[past geyser]

@inland salmon calculate the circumferences of each circle

that is the equivalent of 1 revolution

[vestal canopy]

HELP ME

PLEASE

@everyone

[silent plank]

don't ping everyone

apply secant-secant theorem

[finite jackal]

@delicate lagoon these triangles must be similar, so just select all the ones that show the correct proportions

[upper karma]

HELP

[left plank]

Is octahedron here inscribed in tetrahedron??

My teacher claims its not

[silk patio]

I think it is

[left plank]

There are multiple ways to inscribe octahedron inside tetrahedron right?

[left plank]

🙏 pls

[left plank]

I think it is, but I'd rather get confidence from you before approaching my teacher

[trim breach]

The altitude is perpendicular to the bottom line, which is why it forms a 90-degree angle. Using supplementary angles, and the total degrees in a triangle, you can solve the left triangle. Then, you should be able to solve the rest of easily.

[ruby anvil]

dont know how to do this

[storm portal]

125˚ can't be obtained by summing and subtracting

so

you might have to use a calculator

unless someone else knows

¯_(ツ)_/¯

[ruby anvil]

ok thx :)

[upper karma]

this is for geometry

is this already in use

[tawny cypress]

How do I solve b and c

[trim breach]

B is calculated finding the hypotenuse of triangle ACG.

[shy heath]

you can use pythag to find AC then use pythag to find AG

[finite jackal]

@upper karma angles in a triangle add to 180 degrees and the little squares in the triangles represent 90 degree angles

thats all im going to say

[shy heath]

find AC first
7.2^2 + 9.6^2 = x^2

x^2 = 144

x=12

then same thing for AG

[tawny cypress]

I found b

I got 12.5

[shy heath]

ahh

yeah

thats correct

[tawny cypress]

Now I'm confused w c

What do they mean by floor

[shy heath]

C is finding the angle of GAC

[shy heath]

the angle line AG makes with plane ABCD

[shy heath]

you can use trig to solve for GAC

[tawny cypress]

Ahhh ok I see

Thank you

[shy heath]

np

[tawny cypress]

So tan 3.5/12?

[shy heath]

yeah you can do that

but

you have to use the arctan function or tan^-1

because your solving for an angle

[shy heath]

arctan 3.5/12=x

you can also do

arccos 12/12.5=x

[remote bronze]

is this correct

[trim breach]

It should be considering the volume relation between pyramids and prisms is 1/3. I don’t think it matters that it is a right pentagonal prism, but if someone could confirm, that would be great.

[tiny snow]

Let's say I am given a "box" (not drawn to scale) with the following side lengths. Would there be a way to figure out the metric given this information?

x_1 and x_2 are colinear, and so is y_1 and y_2.

[junior light]

Is the box supposed to have all sides equal?

Or just opposite sides or sth like that?

[tiny snow]

Not necessarily, just opposite sides. In this particular case If we choose x_1 = y_1 = 0 it will collapse to a right angled triangle, with the hypotenuse corresponding to that you would get using Pythagoras. So it's very likely that the metric is the usual metric on R^2.

I would not like to employ that trick though, and I'd rather figure out how to get a metric out of that. Even if it means more conditions have to be satisfied. I'm working on a paper and I will be getting more exotic boxes than that one.

[junior light]

Aaah

This also might be a better fit for #point-set-topology tbh

This isn't exactly at the pre-uni level haha

[tiny snow]

Ah, true 🙂 I'll give it a try there, thanks.

[junior light]

No worries. I'll be trying it too, although I don't know of any metric besides L1, L2 and L∞ 😛

And the discrete metric obviously

@tiny snow I guess the fact that $\sqrt{x_1^2+y_1^2}=\sqrt{x_2^2+y_2^2}$ is independent of our choice of metric since these are just real numbers, right? This gives us $$(x_1-x_2)(x_1+x_2)=(y_2-y_1)(y_2+y_1)$$as a constraint. Would this only be consistent with the $l_2$ metric?

[somber coyote]

Manan.

[junior light]

Since we want $d(y_2,y_1)=d(x_2,x_1)$

[somber coyote]

Manan.

[junior light]

This problem does seem beyond my depth

[tiny snow]

No, that box actually looks like this (roughly sketched)

[junior light]

Aahhhhhh

So uh, you want to find a metric which ensures it retains a convex quadrilateral shape?

(given that the dimensions are as mentioned in your last diagram)

[tiny snow]

My goal is to figure out that given those side lengths I must be in the real plane, and not, for example a hyperbolic plane.

[junior light]

Hmmmmmm

Okay, so we're confined to R^2 and metrics on R^2, which ensures the above figure stays a convex quadrilateral

Let's just work with the assumption that none of the four values is 0 to avoid collapsing-into-triangle kind of degeneracy

Also, x_1 and all that don't have any relation to coordinates, right? They're just real values?

[tiny snow]

They're just real values.

[junior light]

Okay

[tiny snow]

I think there's some constraint missing in order to get a handle on this.

[junior light]

Honestly I think so too

I'd be amazed if this problem could be solved with only this much information

I'm still trying to wrap my head around the problem itself lmao

I also think the values of x_1, etc. would impact the choice of metric

[tiny snow]

I can choose x_1 and x_2 freely as long as they are colinear, and same for y_1 and y_2.

So, making those two sides equal in length is possible.

[junior light]

What would it mean for x_1 and x_2 to be collinear? They're just real values, right?

[tiny snow]

I mean, geometrically they're along the same straight line on the space.

[junior light]

Is that a constraint on the coordinates of the box?

Any two points on the plane are collinear anyway

You can always get a line through them

[tiny snow]

Yes, sorry, these are on straight lines. I don't want to rule out that there could be other lines through two arbitrary points. Think great circles on a sphere for example.

[junior light]

Hmmmmmm

[tiny snow]

Maybe I should rephrase the question to how much information can you get about a space, if you can construct boxes with various side lengths at various places in the space 🙂

[junior light]

Sure, that might help!

I'll try thinking more about this problem, limited as my understanding of both geometry and analysis may be. 😛

[tiny snow]

Thanks for the help so far!

[junior light]

I wasn't able to add anything meaningful. Let me know if you have any updates!

[barren patio]

Truth or deception?

[humble pulsar]

I mean given that's not pi, deception

However the problem of squaring the circle under using a straight edge and compass is known to be impossible

[tawny cypress]

ohh yeah that's what i meant tyvm again

[barren patio]

I solved that problem too.

[humble pulsar]

Ok.. how?

cause given you posted something with an incorrect value of pi, I doubt you did

[barren patio]

Did you check the squaring of the circle with that value 3.1630625 on calculator?

[humble pulsar]

why would I use that value?

It's not pi, so using that value isnt the area of a circle w/ radius 1

[barren patio]

wait, I will send you.

[humble pulsar]

I asked for your proof of squaring the circle under the typical restirctions

not a radian calc

[barren patio]

I'm sending you a PDF

[humble pulsar]

cool, can you stop fucking pinging me

[barren patio]

ok, sorry

[humble pulsar]

You did not solve a problem from ancient Greece if you used a value of pi that isnt pi

[barren patio]

I'm talking to you, so it's my habit. Sorry again for that...

Why are you immediately convicting and you haven't even looked at the evidence yet?

[humble pulsar]

cause what you have posted is grounds enough to null your proof

[barren patio]

moshill1 Let me know when you read.

[humble pulsar]

so yeah, 2nd line already missed the point of squaring the circle

which is that you cannot create a square using a straightedge and compass that has the same area as a circle

Squaring the circle does not say it's impossible to make a square with the same area period

[barren patio]

Do you have a better solution?

[humble pulsar]

cause that's trivially define the side length as $\sqrt{\pi}$ and pack up

[somber coyote]

moshill1

[humble pulsar]

No, since it's known impossible

[barren patio]

lol

[humble pulsar]

no amount of precision in your straightedge, even if it's graduated, will give you square root of pi

since sqrt(pi) is irrational, and thus always more decimals

[barren patio]

🙂 Ok 🙂

I don't want to argue or convince who is right. I just want to share what is reasonable and logical. He who has eyes, let him see, and he who has ears, let him hear. With love.

[humble pulsar]

Then share something reasonable and logical

[barren patio]

I've already done that.

[torpid hearth]

@barren patio pi is irrational?

[humble pulsar]

You did not, since the 2nd point of that article asserts that pi is rational, yet it is not proven that it's rational

whereas pi has been proven to be irrational, transcendental at that

[barren patio]

@torpid hearth Pi is a rational. Check countless times in countless ways.

[torpid hearth]

https://www.youtube.com/watch?v=PgKmstECld0

[barren patio]

There is a lot of garbage on Wikipedia and on the Internet. The truth is barely noticeable.

[torpid hearth]

I can't tell if this is a troll or not

[humble pulsar]

It's clearly a troll

[wanton storm]

Now prove it's transcendental

[humble pulsar]

https://en.wikipedia.org/wiki/Proof_that_π_is_irrational

6 proofs by different mathematicians that $\pi\notin\mathbb{Q}$

[somber coyote]

moshill1

[torpid hearth]

this is so sad

who takes the time to make a fake research paper

[barren patio]

If I’m a troll, I’m sorry I came to this channel of my own free will. 😦 It is easiest to condemn. I was born in the country where Nikola Tesla was born. As a token of love, I will greet you.

[humble pulsar]

and they gone

[wintry tundra]

Lol

[eager rock]

need help

[wintry tundra]

With what specifically

[eager rock]

the whole thing 😭

[wintry tundra]

That's a lot of work

I mean I can just generally teach you how to do proportional triangles

[eager rock]

sure please

i keep getting decimals in my answers when im solving x

and idk if thats right

for number 2 i got 12.6 for GH and 9.6 for Kj @wintry tundra

[wintry tundra]

If its only tenths then probably right

As long as ur taking ratios for lengths of similar sides

Which ima guess u are

Also don't expect to get all the answers bc here we only help you get to the answers

So its mainly u that gets the answers in the end

[near imp]

I need help

[upper karma]

can someone explain this to me? my english is really bad and im not very good at word problems

[upper karma]

ohh i see it now, thank you!

[tawny cypress]

How do u solve d

[tired pine]

It’s a square pyramid

You have the diagonal of the square from (c), and you’re solving for the side of the square

Do you know special right triangles?

actually wait you don’t even need that

do you know pythagorean theorem

[proven breach]

Anyone know how to do this?

[tired pine]

Do you know the double angle identity for cosine?

[proven breach]

This?

@tired pine

[tired pine]

Yep

[proven breach]

what do i do with it

[tired pine]

Which one can you do with the info you have

[proven breach]

the second one

[tired pine]

You have cosine

yes

plug in

[proven breach]

plug in where

[tired pine]

You can use $\cos(2A)=2\cos^2(A)-1$

[somber coyote]

CST

[tired pine]

Since you have $\cos(A)=\frac{\sqrt{3}}{6}$

[somber coyote]

CST

[proven breach]

how i get cos^2 in my calculator

[tired pine]

you don’t need the calculator (cos)^2

[proven breach]

ok

[tawny cypress]

Omg I'm literally not thinking damn I feel dumb but thank you I get it now

[proven breach]

what do i do

[tired pine]

in this case

You have cos already

so

[proven breach]

im confused

[tired pine]

square the value you have, then multiply it by 2, then subtract 1

[proven breach]

thanks, got it

[worthy lichen]

oh wait nvm this isnt the help channel

[valid sinew]

This is the help channel

[worthy lichen]

oh, i thought it was for discussion

i got the help i wanted, but thanks anyway

[inland dagger]

can someone help me with this? https://images-ext-2.discordapp.net/external/pSArO3yj9X8X3JLyVqbGmG6pGLUI-oEJAdFh8Wpo1C4/https/i.imgur.com/WVsB7Y3.png?width=821&height=217

[junior light]

@inland dagger What have you tried so far?

[pure bronze]

@inland dagger Do you still need help with that question?

[copper marten]

49

@inland dagger
find scale factor:
ST and SQ are corresponding, so we can divide those to find the scale factor
42/24 = 1.75

TQ is corresponding to QR and we know what TQ is which is 28
28 x 1.75 = 49
so, QR = 49

[azure basalt]

Hey guys, this may sound as a stupid question, but how can sine/cos have inputs > 180 degrees. from what I now those functions are just the ratio of triangle sides. triangle interior angles add up to 180. so there is no triangle with angle > 180. what am I missing? 😢

[trim breach]

Those functions are based on the unit circle, but can be applied to triangles.

[azure basalt]

so later on I will learn the relation of sine/cos etc. in relation to the unit circle?

I am already solving unit circle problems now without a problem, It's just that I don't really grasp the full picture/relations

[trim breach]

Yeah. Essentially, anything greater than 2π radians (360 degrees), the angle in those trig functions yields the same value as that angle minus 360 degrees.

So you wouldn’t really have a 400 degree angle in your triangle, but you could have a 40 degree angle since it is the same value within the trig function.

[proven breach]

Does anyone know what this would be?

@tired pine

[wintry tundra]

If I'm not wrong arccos is the inverse of cos

And $f^{-1}(f(x)) = x$

[somber coyote]

hiidostuff

[wintry tundra]

Also lol 69 nice

[proven breach]

So it’s pi over 5? @wintry tundra

[wintry tundra]

Mhm

[proven breach]

Thanks

[wintry tundra]

Np

[proven breach]

Does anyone know this one?

@wintry tundra

[humble pulsar]

tan is odd, so $-\tan{x}=\tan(-x)$

[somber coyote]

moshill1

[proven breach]

@humble pulsar so ya what’s the answer

[humble pulsar]

find out when you use the hint I gave you

Im not giving you the answer

[proven breach]

2 square root 5?

[humble pulsar]

why is it that?

[proven breach]

Cuz that’s what I got

Idk

[humble pulsar]

by doing what?

[proven breach]

Idek

[humble pulsar]

Ok then idek if it's right

[proven breach]

Why so mean

[humble pulsar]

It's not mean, trying to see if you actually solved it or blindly guessed

Besides, the final answer isnt important 99% of the time

[proven breach]

I guessed idk how to do it

[humble pulsar]

what's the relationship between cot and tan?

[proven breach]

I forgot

[humble pulsar]

Ok so look at your notes, or the trig cheatsheet that's pinned in the channel

[proven breach]

Not doing all that, just gonna guess

[humble pulsar]

Ok

[proven breach]

😔

Just tell me if I was right

@humble pulsar pls

[humble pulsar]

Yoiu have a 25% chance of being right

[proven breach]

😑

[humble pulsar]

Look, if you wanted to learn how to get the answer, I was more than willing to explain the couple steps it took

[proven breach]

U wouldn’t explain

I asked u to

[humble pulsar]

However your attitude of "just gonna guess" changed my mind

so just guess

[humble pulsar]

I gave you a nudge with how to get tan(-x), then told you you need a relationship b/w tan and cot, also told you the relationship was pinned in the cheatsheet. I did help you, and did try to explain it

I am not going to spoon-feed someone answers if that is what you expect help and explanation(s) to be

[proven breach]

Damn u one of those smh

[humble pulsar]

Yes, I'm one of those people that actual helps

[proven breach]

@humble pulsar well thanks anyway

[wintry tundra]

Ahh ur one of those smh, those who want the answers spoon fed to them

[proven breach]

no im not

[azure basalt]

Thanks bro! 😃 Appreciate the info!

[onyx viper]

hey any tips on this

[wintry tundra]

Well idk why u boxed only sin(x) = A

When the equation is actually arcsin(x) = A

Where arcsin is the inverse function of sin

[onyx viper]

yea that was a stupid mistake

[wintry tundra]

So did I help orr

Idk what ur confused about

[onyx viper]

not sure how to find the arccos and if it can be a real value

[wintry tundra]

Hmm

Well sadly the arc mistake is really the only thing I can help with

Sorry

But maybe someone else could help out

[onyx viper]

it is ok just not sure where even to begin tbh

[desert copper]

help me guys

[shrewd tendon]

😦

i got a 70 on my geometry class

circles were my enemies

[desert copper]

sad

[shrewd tendon]

yea

[shrewd tendon]

intesesting name tho

[desert copper]

ahhaa thanks

[shrewd tendon]

can u help me with this by any chancve

[desert copper]

sorry i can't

[shrewd tendon]

rip

😦

[desert copper]

that's hard for me

[shrewd tendon]

yea math has never been my thing

i was the smartest in my class until 3rd grade lol

[desert copper]

i was until 7th grade

hahaha

[shrewd tendon]

damn

[desert copper]

ii think my besties influence me

[shrewd tendon]

well good luck m8

i never had friends lol

i just became ass after 3rd grade lol

[desert copper]

now what grade are u

?

[shrewd tendon]

junior in hs

its been a while lol

[desert copper]

hahah how old are u

[desert copper]

me too i will be 11th grade next year

[shrewd tendon]

nice

[shrewd tendon]

i just turned 17 yesterday lol

[desert copper]

happ bday

happy*

to u

XD

[shrewd tendon]

lol tks :)\

🙂

well im off to find the answer

to my question

[desert copper]

[shrewd tendon]

https://tenor.com/view/chuck-e-cheese-fortnite-dance-orange-justice-gif-15868872

[wintry tundra]

what is it that was confusing about circles to you? i could help you with that

[shrewd tendon]

lol the whole thing but lukuly i passed that unit

i really appreciate the help tho brodie

[wintry tundra]

np

i mean i guess one thing to remember with circles tho is that most things you are trying to find in geometry with circles relate to how much of the angle is to 360

[glacial dawn]

Amazing pfp hiido

[wintry tundra]

thank you stephen

[shrewd tendon]

whoa picture is amazing lol

usually i struggled with the tangents cosign and the other one

i would struggle to identify each side

[wintry tundra]

but if you have an angle whos joining point is on the middle of the circle, the curve at the end, like the crust of the pizza slice, it will be the radius times the angle

[shrewd tendon]

and how to solve

[wintry tundra]

tangent cosine and sine?

ahh i see

just remember that they are ratios on a triangle

like comparing sides

the reason we even bothered to have sine cosine and tangent is because you need to find weird ratios of triangles all the time

if ur in engineering or physics that is

but for my robotics team i am mainly the mathematician and i have to work with circles a lot for wheels

and holes being cut in parts of the bot

which are usually circular holes

but good luck with geometry and algebra 2 next year is what i guess u would take

[slow fossil]

hello, I need help on something guys

https://cdn.discordapp.com/attachments/409596215697866773/832504463368912956/unknown.png

[fast pulsar]

Here, I have this drawing for you at least

[slow fossil]

thanks

<@&286206848099549185>

<@&286206848099549185>

[pseudo pumice]

Stop double pinging, please.

[slow fossil]

OK

[torpid hearth]

Did you get this question answered?

oops

@slow fossil

alright I solved it





Imma go to sleep so i'll just post the answers

sorry it's a bit messy lmao

[slow fossil]

Thanks @torpid hearth

good night

[gilded fjord]

hi, can someone help me figure this out? i've been trying to solve it but i'm so confused

[crimson smelt]

part a

49.88+61

[compact jasper]

if i trying to do tan (x) = sin(x) / cos(z) its cant be? The one and only case is tan(x) = sin(x)/cos(x) ? 🙂

[loud wolf]

12x^2+11x-5

can anyone help me factorize this?

[pure bronze]

it has to be the same angle or value

[trim breach]

That doesn’t factor nicely. Can you use quadratic formula for what you are doing?

[pure bronze]

Well it actually does if you use a different method other than the quadratic formula

[vague helm]

Need some help with this problem

<@&286206848099549185>

[torpid hearth]

what is the ratio of cosine? (Hint: SOH, CAH, TOA)

@vague helm

[vague helm]

Um I’m very new to this and don’t really now how to do it at all

[torpid hearth]

here are the 3 trig ratios

[vague helm]

So what way would I answer the problem in?

[somber coyote]

keto11

[torpid hearth]

a ratio is division between two numbers so...

[somber coyote]

keto11

[vague helm]

Ok so how would I start putting the numbers in?

[torpid hearth]

the first thing you want to do is actually draw the triangle, it's a lot easier to see what's going on

@vague helm are you able to upload images to the discord?

[vague helm]

Sorry I’ll brb

And yes I will do that

Helping my dad rn

[torpid hearth]

kk ping me when your back

[remote bronze]

i need help with this

[vague helm]

@torpid hearth ok I’m back

[torpid hearth]

kk

try drawing a triangle

[vague helm]

@torpid hearth done

[torpid hearth]

okay good

now you have to identify which lengths are the hypotenuse, adjacent, and opposite

@vague helm

[vague helm]

So NP is the Hypotenuse, and idk the other ones

@torpid hearth

[torpid hearth]

are you saying this side is the hypotenuse?

@vague helm

[vague helm]

Yeah

Like the hypotenuse leg or something like that

[torpid hearth]

what's the definition of a hypotenuse?

[vague helm]

Oh shit so ON

[torpid hearth]

yep

there'a a really easy way of identifying the sides

[vague helm]

And the opposite is NP

And adjacent is PO

I think

[torpid hearth]

which angle are we looking at?

[vague helm]

Wait what?

[torpid hearth]

We're trying to find the cosine of angle N

[vague helm]

Oh ok so what will the equation look like?

[torpid hearth]

well you have to identify the sides...

you already have 1, the hypotenuse is NO

do you know how to identify the sides?

[vague helm]

So cos(90) 77/85?

[torpid hearth]

90 is the right angle not angle N

[vague helm]

So would I replace 90 with X?

[torpid hearth]

cos(N) = ?

[vague helm]

So cos(N) = 36/85?

[torpid hearth]

yep

and 35/85 is the ratio

[torpid hearth]

from back up here

[vague helm]

So I can use a calculator to answer the question so how would I put this into a calculator?

[torpid hearth]

stop rushing for the calculator

you already have the answer

What is the question asking you to solve for?

[vague helm]

Oh

So the answers 36/85

[torpid hearth]

yep

the only reason you would need a calculator was if they asked you to find the value of N

then you would have to do the inverse cosine

[vague helm]

Yeah and I’m gonna need that soon

I’m gonna finish this one. Then next thing that I don’t know I’ll ask I guess

[torpid hearth]

kk

[vague helm]

Ok I need to find the value now

@torpid hearth

[torpid hearth]

the first step is again to draw the triangle

@vague helm

[torpid hearth]

now you wanna draw the angle that you're solving for just like I did here

[vague helm]

Ok I added the angle

Now what?

@torpid hearth

[torpid hearth]

What's the ratio for sin of an angle?

[vague helm]

It’s 21=20/29

I think

[torpid hearth]

it's these ratios

[vague helm]

So it’s 20/29?

[torpid hearth]

lemme check

yep

now create an equation with it

[vague helm]

So just divide 20/29?

Cus I got 0.69

[torpid hearth]

no you want to create an equation

something = something else

[vague helm]

Oh so 21=20/29?

[torpid hearth]

what's 21?

[vague helm]

The adjacent side

Oh shit nvm

[upper karma]

in addition, be aware of what you are saying.. 21=20/29 is blatantly untrue, this is the same nonsense as saying 0=1.

[vague helm]

P=20/29

[torpid hearth]

these are the equations that you're looking for

[somber coyote]

keto11

[vague helm]

Oh so sin(P)=20/29

[torpid hearth]

yes

you want to memorize these things by heart

[vague helm]

Ok understood

So how would I go about answering now?

[torpid hearth]

how do you isolate P?

[vague helm]

Um idk

[torpid hearth]

if in the last problem to find the value of angle N, you had to take inverse cosine, then to find the value of P you need to take...?

[vague helm]

Just normal cosine?

So 29/20 will equal my answer?

[somber coyote]

keto11

[vague helm]

So what will be A and what will be B?

[somber coyote]

keto11

[torpid hearth]

here

[vague helm]

Oh so the answer will be 0.69?

[torpid hearth]

no

your answer is a off

wait

,calc 20/29

[somber coyote]

Result:

0.68965517241379

[torpid hearth]

o frick me

[torpid hearth]

what pattern do you see here?

[vague helm]

For like the template of the problem?

[vague helm]

Oh so cos wouldn’t have the negative?or that’s it’s not inverse?

[torpid hearth]

what is the inverse cosine?

[vague helm]

Cos-1

[torpid hearth]

and what have I highlighted here?

[vague helm]

Cos-1

[torpid hearth]

so to cancel out cosine you use inverse cosine, then to cancel out sine you use ?

[vague helm]

Inverse sine?

[torpid hearth]

yeah

[somber coyote]

keto11

[vague helm]

Ok ok understood

[somber coyote]

keto11

[vague helm]

So now I have this problem because I finished the section that had the questions you helped me with

Also understood

So how would I go about answering this?

[torpid hearth]

Identify which sides are opposite, hypotenuse and adjacent to the angle G

[vague helm]

So Hypotenuse is I, Opposite is G, and Adjacent is H

[torpid hearth]

those are angles, sides are line segments

[vague helm]

?

Oh

We only have adjacent which is IG

[torpid hearth]

All of these problems are basically the same, why don't you try to work it out on your own for a bit. Use the stuff that I posted above if you get stuck first

[vague helm]

Oh ok

[torpid hearth]

@vague helm did you figure it out?

[vague helm]

,calc 8.9 tan 48

[somber coyote]

The following error occured while calculating:
Error: Unexpected type of argument in function multiplyScalar (expected: number or Unit or string or boolean or BigNumber or Complex or Fraction, actual: function, index: 1)

[vague helm]

@torpid hearth no, no I don’t

I’ve been trying for an hour now and can’t even get past the first question

[torpid hearth]

okay

@vague helm you still here?

[vague helm]

Yes

Also I have a question

[torpid hearth]

wassap?

[vague helm]

I did the exact same thing and got a diffrent answer

And I don’t know how

[torpid hearth]

it's because you're using desmos

later on in math you'll learn about something called radians it's a unit system for degrees basically

if you want to fix it for desmos you want to do this

[vague helm]

So what do I do with my desmos?

[torpid hearth]

go to desmos and find the little wrench icon

click on it and locate this section

[vague helm]

Yeah I see it and I clicked

[torpid hearth]

radians should be highlited in black, click on degrees and you should get the correct answer

[somber coyote]

keto11

[torpid hearth]

@vague helm did you figure it out?

[vague helm]

Yes it worked

Thank you

[torpid hearth]

epic

do you still want help with your problem?

[warm temple]

Talking about radians

How do I convert decimal radians into pi fraction form?

[pure bronze]

which one?

[zealous forge]

the last 4 pgaes

[pure bronze]

when you have similar triangle, you have a special unique attribute, can you guess what that is?

[zealous forge]

wdym

like if the the two triangles are similar if their angles are congruent ?

[pure bronze]

they have a relation of similarity

[pure bronze]

@zealous forge are you here?

[zealous forge]

yeah

[pure bronze]

all right, so when you have a pair of similar triangles, you have a ratio between them

to calculate that ratio, you divide two proportional sides from each triangle

[zealous forge]

ok

[pure bronze]

if BE = 9
AM = 3
ET = 12

then the ratio must be 1/4

since AM and ET are proportional

AM comes from the triangle RAM

ET comes from the triangle BET

and for that reason, what can be inferred?

[zealous forge]

sorry idk what infer mean

s

[pure bronze]

what can you conclude from the ratio of 1/4

if the smaller triangle is RAM and the larger one is BET

and they have a ratio of 1/4

what can you figure out

[zealous forge]

BE

[pure bronze]

BE is already given to us, it's 9 cm

but what can you see and say about the relation between the two triangles?

one is smaller than the other

but by what?

[zealous forge]

is that its 1/4 smaller

[pure bronze]

yup

[zealous forge]

ok thank you

[pure bronze]

anything else?

[zealous forge]

ok can you help me with the last 4 questions please? i think i can figure out everything else

[pure bronze]

sure

13, 12, 11, 10?

[zealous forge]

sorry for asking alot

[pure bronze]

it's fine, I'm here to help ya

[pure bronze]

?

[zealous forge]

yes

please

[pure bronze]

okay, so 10

what do you think you're supposed to do here?

[zealous forge]

i think we have to find the area

[pure bronze]

even better

we dont even need the area

[zealous forge]

so what do we need

[pure bronze]

perhaps some trig could do the trick 🤔

[zealous forge]

Ok lets untie the trick

nvm that was corny nvm

[pure bronze]

XD

it's fine lad

[zealous forge]

are u Australian ?

if u dont mind me asking

[pure bronze]

nope

[zealous forge]

british

?

[pure bronze]

bullseye

[zealous forge]

Nice thats cool

[pure bronze]

but now

what kind of trig could we use to do this?

(hint: this is not a right triangle)

[zealous forge]

umm 1 sec

pls

[pure bronze]

yeah sure

[zealous forge]

sorry wdym by trig

[pure bronze]

trigonometry

[zealous forge]

like what formula we use

?

[pure bronze]

Oh you tell me 😄

have you learned the law of sines?

[zealous forge]

ok 1 sec

[zealous forge]

yeah]

[pure bronze]

well then?

what do you think?

[zealous forge]

ok

sinA\a = sin B/b =sin C/c

wait

x is the hypotenuse

[pure bronze]

yup

[zealous forge]

so its should be side a

[pure bronze]

wait no

[zealous forge]

plus side b

[pure bronze]

mate

[zealous forge]

right

[pure bronze]

this is not a right triangle

there is no hypotenuse

be careful

[zealous forge]

Ok

so

the missing angles is

93

and now i can use the formula because i ahve 3 angles

[pure bronze]

yeah, and what do you do now?

yup

and Voila

[zealous forge]

ok

ok

thank u so much

1 sec

am gonna solve and may u tell me if i got it right

please

[pure bronze]

ye

most definitely

[zealous forge]

ok i got for the first side which is A its equal to 7.4

[pure bronze]

which one

x or the unknown one

[zealous forge]

i used the know one to find the of the other sides

[pure bronze]

uhh, what?

[zealous forge]

i ddi this

[pure bronze]

all right, now find x

[zealous forge]

ok i got 5.845

and a bunch of ther numbers

its not fo a though

nbm

nvm

1 sec

A=7.4

[pure bronze]

okay

[zealous forge]

B=5.8

[pure bronze]

weird

[zealous forge]

Yeah

ok its fine nvm

thanks for the help Hunny

not like that

[wintry tundra]

Wait 93 degree angle?

[zealous forge]

yeah

[wintry tundra]

What shape

[zealous forge]

num 10

is the question

[wintry tundra]

Oh ok

[pure bronze]

np

[wintry tundra]

Did u solve it already?

[zealous forge]

no

i couldnt

[wintry tundra]

Ok

Hmm

9.7 / sin 37 = a / sin 50

Will give u one side

Then u can use law of cosines to find the last side

And there u go

Problem solved

So @zealous forge now u should be able to solve ur issue, if you know law of cosines

[zealous forge]

Goddamn

[pure bronze]

didn't I tell you to use the law of cosines? @zealous forge

[wintry tundra]

Do u know law of cosines?

[zealous forge]

no

[zealous forge]

yeah but idk whats that

[wintry tundra]

$c^2 = a^2 + b^2 - 2ab\cos{(\gamma)}$

[somber coyote]

visual of Petter's ascendance

[pure bronze]

I don't recall myself telling you that

[wintry tundra]

Gamma is the side opposite the one ur trying to find

a and b are the sides know

[zealous forge]

telling me what

[pure bronze]

to use the law of cosines

[wintry tundra]

If you use the law of sines, you can find both sides u need to find x

Gamma is the supplement to the sum of the other angles

So boom

That's what u need

[zealous forge]

so if i find the fist side i use that and the orgiginal lside to find X

[wintry tundra]

Yep

[zealous forge]

ok thanks

[wintry tundra]

Np

If u need help calculating I just mention me

[zealous forge]

idk what that law is thi

[wintry tundra]

Remember x has to be larger than 9.7

[zealous forge]

tho*

[wintry tundra]

I told u that law

[wintry tundra]

This is that law

The cosine is why it's called law of cosines

[zealous forge]

ok so anything larger than that i can be sure that its right

[zealous forge]

oh i didnt know that

[wintry tundra]

Yep

Bc think about it

If there's a large angle then it's opposite side also has to be big

Bc the angle makes a lot of space to connect

And 93 is bigger than 50

[zealous forge]

ok

am gonna sovle and can u pls tell me if its right?

[wintry tundra]

Yeah

[zealous forge]

ok i got 12.347034907

for the first one

[wintry tundra]

Yep

So did i

[zealous forge]

lets goooo

ok i need one more

so in this law

the variables are the angles not the sides right?

[wintry tundra]

Ok remember this

English letters are probably sides and Greek letters are probably angles

This follows the same thing

a and b are your known sides

And the weird curly thing called gamma is the angle opposite of c

c being the side ur trying to find

[zealous forge]

so what do i plug in for c^2,a^2 and b^2

[wintry tundra]

a and b are the sides you know

[zealous forge]

i still dont have 3 sides

[wintry tundra]

c is the side you don't know

Yeah ur trying to find c

The right hand side is solvable itself since a b and gamma are now defined

Doesn't matter what sides you put in for a and b

Gamma angle does tho

Has to be able opposite of c

[zealous forge]

so i can do

c^2=12.347034907^2+9.7-2ab cos

then the sign

[wintry tundra]

Yeah

9.7 squared tho

[zealous forge]

yeah sorry

i was just editing it lo

l

[wintry tundra]

So $c^2 = 12.3^2 + 9.7^2 - 2(12.3)(9.7)\cos{(\gamma)}$

[somber coyote]

visual of Petter's ascendance

[wintry tundra]

Is what u should have

[zealous forge]

ok i understand now

how do i make photos like u

when i write an equatin

equation*

[wintry tundra]

Ah with dollar signs and a bunch of backslashes and such

It's hard to explain

[zealous forge]

yeah no thanks am just gonna draw it and send it here

[wintry tundra]

There's so many math functions and therefore so many things to do with the math thing called texit

[zealous forge]

Huh

i got this

idk how to write the last part

[wintry tundra]

Oh

There's a mistake

[zealous forge]

what is it?

[wintry tundra]

$2ab$ is not equal to $2(ab)$

[somber coyote]

visual of Petter's ascendance

[wintry tundra]

For 2ab you only double one of the variable values

For 2(ab) you double both

Here you have 2ab so only one

[ionic talon]

Would 2(ab) be 2a + 2b?

[zealous forge]

so i olny multiply 12.3 by 2?

not it would be 2xab

i think

[ionic talon]

I don't understand.

Where did the x come from?

[zealous forge]

missing side

[wintry tundra]

x means times

I think

[zealous forge]

wait x in the question?

[wintry tundra]

Only multiply 12.3 yes

[zealous forge]

ok

[wintry tundra]

2(ab) is 2a2b

[ionic talon]

So no addition?

[wintry tundra]

No

[zealous forge]

so the last part would look like -24.6 (9.7)

?

[wintry tundra]

Yes

So $c^2 = (12.3)^2 + (9.7)^2 - (24.6)(9.7)\cos{(\gamma)}$

[somber coyote]

visual of Petter's ascendance

[wintry tundra]

Gamma is 93 degrees or 93pi/180 radians whichever one u like best

Degrees is better here so use degrees

[zealous forge]

so what shall i do from here

[wintry tundra]

Multiply the end part by cos 93

[zealous forge]

so -24.6(9.7) cos 93

[wintry tundra]

Yep

cos 93 should be fairly small

[zealous forge]

-75.7448367566

wtf

[wintry tundra]

Is ur calculator is radians mode?

[zealous forge]

nah am using google as my calulator bro

[wintry tundra]

Hmm

[zealous forge]

https://tenor.com/view/umm-awkward-say-what-what-was-that-gif-5424377

[wintry tundra]

End result should be 12.5

[zealous forge]

ok 1 second

[wintry tundra]

With that end part at least

12.5 isn't the final answer

[zealous forge]

1 second

am getting me calculator to see if its possible to use cos on it

[wintry tundra]

Ok

Use desmos

Desmos is a really good online calculator

So is geogebra

[zealous forge]

ok i will

i got this

[wintry tundra]

Your calculator is set to radians

Set it to degrees

Or change 93 to $\frac{93\pi}{180}$

[somber coyote]

visual of Petter's ascendance

[zealous forge]

yeah

[zealous forge]

ok iwill

omg

its right

12/5

12.5

[wintry tundra]

Yep

[zealous forge]

Thank you so so so much dude

[wintry tundra]

Wait no no

U aren't done yet

[zealous forge]

damn ofc u gotta ruin the moment

[wintry tundra]

Now you have $c^2 = 12.3^2 + 9.7^2 - 12.5$

[somber coyote]

visual of Petter's ascendance

[zealous forge]

so i just solve then what ever i get is c^2 right

[wintry tundra]

Yeah

But ur solving for c

[zealous forge]

which is x

[wintry tundra]

Yeah

[zealous forge]

right then am done

[wintry tundra]

Mhm

[zealous forge]

what now lol

[wintry tundra]

Well solve the right hand side

Then I'll tell u

[zealous forge]

aight bet

233

[wintry tundra]

Alright

So c^2 is equal to that

So what should we do to 233 to make it equal to c

[zealous forge]

divide?

[wintry tundra]

Nope

We can take the square root of it

[zealous forge]

oh

i see

ok lemme see

15.26433752247375

[wintry tundra]

Wait

Is that an answer on your homework

Bc that isn't riggt

Did u solve this? $12.3^2 + 9.7^2 - 12.5$

[somber coyote]

visual of Petter's ascendance

[wintry tundra]

Or evaluate it I should say

[zealous forge]

am back sorry

not that answers isnt on the homeworjk

[wintry tundra]

Yeah so it isn't right

So evaluate what I put above

[zealous forge]

i didnt find my mistake

@wintry tundra are you there

Sir

[wintry tundra]

Yes

Sorry

@zealous forge you actually add 12.5 at the end

My bad

[zealous forge]

Oh

thats whyy

its all good

i hope its right this time

16.0586425329

so 16.10

uhh fianlly

thank you very mcuh Sir

[wintry tundra]

Yep

[zealous forge]

for helping me with this

[wintry tundra]

Np

[zealous forge]

Could i ever ask you for help if i need some in the future

[wintry tundra]

Heck yeah I help pretty often

I'd ask for helper role but I don't wanna get mentioned all the time

So dms would be fine by me

[zealous forge]

alr so your like an undercover helper

[wintry tundra]

Not really undercover

Ima be honest I haven't taken trig yet in school

I just self studied it

A while ago

And I'm not bad at it since I use it in my robotics club and engineering somewhat often

[upper karma]

how do I find this intersection?

the semicircles are congruent

[onyx viper]

are the semi circles at at right angle

[upper karma]

yes

[onyx viper]

I think then they should both intersect at the apex

[upper karma]

it seems so

can anything be derived from that?

[onyx viper]

I honestly have no idea I dont see any given angles or line measurements

[upper karma]

pretend the radius is 1

[onyx viper]

what specificaly are you trying to find

the exact cordinate of the intersection or just information about the semi circles

[upper karma]

the area of the part where they intersect

[onyx viper]

oh ok

I have no idea how to help you out with that I would post that in a questions most of them seem to be open

[upper karma]

thank you

[onyx viper]

anyone able to help solve sin^2(x/2)cos^2(x/2)=(sin^2x)/4

[torpid hearth]

uh

double angle identity

[onyx viper]

thanks feel stupid now

[lofty adder]

If a point is defined as being dimensionless or according to Euclid 'that which has no part', then how can it represent anything geometrically. I think this confusion is arising because we use a dot to represent a point but the dot obviously has a part to it which can be observed on zooming in. Can anyone please help me?

[gritty spindle]

anybody?

pls

i only have till 11.59

[dark sparrow]

do you have a calculator with buttons for inverse trig functions?

[gritty spindle]

Nope

Honestly i have no idea whats going on i just want this 1 problem solved? SO i came to discrod