geometry-and-trigonometry

[dark sparrow]

oh, it asks you to find angle A itself as well?

[mortal stirrup]

yeah

[dark sparrow]

fair enough

well now you just need to find its tan

[mortal stirrup]

ok

[dark sparrow]

you found its sin and cos correctly

[mortal stirrup]

hmmmm

tan A =8/15

[dark sparrow]

yup

[mortal stirrup]

got 28 too

[plucky valve]

Help

[dark sparrow]

don't double-post.

[sick veldt]

Is this correct?

<@&286206848099549185>

[twin prawn]

Yes

[sick veldt]

Okay can you confirm two more?

sorry

forgot I can just ask

I just need these two checked

hello

[twin prawn]

Last one doesn't seem right

[sick veldt]

ahh yeah

that one was tricky

I'm not too sure about it

I think its B

maybe

I don't see how A and D are gonna be correct

:/

[twin prawn]

Why would A not be correct?

[sick veldt]

cause I don't see why we would do half of 72 for

[twin prawn]

You should calculate the angles and see for yourself

[supple haven]

can someone help me with this?

<@&286206848099549185>

[lime flame]

Can someone help me with a problem

[shadow anvil]

[hard gale]

@lime flame pong

[lime flame]

nvm i solved it

[upper karma]

anyone know how to do this?

[tacit swan]

I don't understand how the answer is 4/5

[swift lava]

$\cos{(90-\theta)}=\sin{\theta}$

[somber coyote]

Eddie:

[swift lava]

$\cos{(90-\theta)}=\cos{90}cos{\theta}+sin{90}sin{\theta}$

[somber coyote]

Eddie:

[swift lava]

$\cos{90} = 0$

[somber coyote]

Eddie:

[swift lava]

$\sin{90}=1$

[somber coyote]

Eddie:

[swift lava]

so plug those in and you get $\cos{(90-\theta)}=\sin{\theta}$

[somber coyote]

Eddie:

[tacit swan]

I think I get it

So you're saying, that cos (90-whatever) equals the other complemntary angle, which is sin

[swift lava]

Right

[tacit swan]

so if sin theta is 45

cos theta would be 45

[swift lava]

Exactly

[tacit swan]

I meant as in the angles but yeah

[wind trail]

👀hi

[split raven]

Could someone please explain to me as to why cos(-A) = cos A?

[dark sparrow]

cos(A) is the x-coordinate of the point on the unit circle that corresponds to A

[split raven]

Ohh

Thank you @dark sparrow

[little fulcrum]

How would you find the length of y if this is all you have set? No angles either

I was told y should equal 8.25

[fleet wolf]

similiar triangles

assuming that y is parallel to x

[little fulcrum]

So that would be

3/7 = 6/X => which would mean that X = 2.571

and then

2.571/Y = 7/8 ?

[dark sparrow]

3/7 = 6/X does not become X = 2.571

[little fulcrum]

14* sry

So once I got that X = 14, I am unable to do the same with the Y like this

14/Y = 8/5 that would get me 8.75 which doesn't equal 8.25

[dark sparrow]

who told you y = 8.25

[little fulcrum]

Apparently it's 8.75, I write like a moron, thank you tho!

[vernal tusk]

is that even a real problem? because you have to assume parallelism

i thoughr you can never assume that

[wind trail]

[upper karma]

https://i.imgur.com/rehi1uT.png

How are they arriving at theta = 30 degrees?

I am attempting to resolve the complex fraction portion and am not sure what they're doing to cancel out pi...

[umbral snow]

π/6 ÷ π
= π/6 × 1/π
= 1/6
@upper karma

[upper karma]

Thank youuuuuuu.

[winged saddle]

hi

can someone help me with a question

i am being dumb right now imbrain dead

[upper karma]

[winged saddle]

no understand how the fox's hand can turn from left to right?

[upper karma]

Literally add the numbers ._.

[winged saddle]

oh

wait

but can u explain whats going on tho

cuz im not reading the question right or something

[upper karma]

You have to be 13 to use Discord

[winged saddle]

oh crap

ok i understand how you add the numbers just how does the fox's hand change??

dafuq

[wind trail]

Hi guys I’m looking for a good trig book so I can learn on my own ?

I prefer physical

Any recommendations please?

[hallow smelt]

I need help graphing sin

[winged saddle]

go on desmos

and write sin(x)

[hallow smelt]

no that doesnt help...

[winged saddle]

ok

[hallow smelt]

I want to be able to sketch it

[winged saddle]

oh

[hallow smelt]

demos is exact

[winged saddle]

well u know u can plot points for special angle values

[hallow smelt]

-3/2 sin(0.2T+1.4)

the shift is not in pi

this makes my life bad

[winged saddle]

and u can think about the unit circle when you do that because the angle's y coordinate is the sin

[hallow smelt]

I know what Omega is / Frequency etc.

I need to know how to graph that without it being a mess

[winged saddle]

daheck is that

anyway

what i would do is for example, when the angle is pi/4, (45°), i would construt a triangle that is 45 45 90 triangle and i would figure out that if it is inscribed in the unit circle the y coordinate is sqrt2/2 and so the value on the graph y = sin(x) would be sqrt2/2

and just keep plotting points like that

[hallow smelt]

2pi/omega is period omega/pi is frequency , shift is 1/omega*1.4 amp 3.2 thats what i mean. I get all that

[winged saddle]

remember that a full rotation around the unit circle is 2pi, so the period of the function is 2pi, meaning htat if you have an angle theta, then sin theta= sin(theta+2pi)

ok

oh

then just sketch it lol

[hallow smelt]

try to sketch it yourself

[winged saddle]

okie

[hallow smelt]

thanks

[winged saddle]

lol just like write a graph and then put poiints ;-;

[hallow smelt]

-_-

[winged saddle]

bruh nits so inprecise tho

;-;

[hallow smelt]

that is what im talking about

[winged saddle]

oh

[hallow smelt]

very hard to do spacing

[winged saddle]

bruh then i dont know

sorry

i am very useless

[hallow smelt]

they are trying to bread crumb physics to us

and its not fun

[winged saddle]

.-.

laol

[inner lantern]

how can you do this?

find OD,i mean the radius

[midnight socket]

@hallow smelt You still here by chance?

[stiff topaz]

@inner lantern this might help you a bit

[inner lantern]

oh my god

youre the best

[midnight socket]

:pog:

[stiff topaz]

I would start by drawing a line from O to T

[inner lantern]

yes

[stiff topaz]

and work your way from there

[inner lantern]

unit circle

<3

wait

but i do it ABO

the trigonometry way?

[stiff topaz]

I mean

[inner lantern]

its the same, right?

[stiff topaz]

first of all

since T is on the circle

that means OD = OT

that translates to BC on that image I sent

the radius

so the orange line on the image

is TC in your problem

I think

do you have an answer key to check your work with? I don't want to accidentally help you incorrectly

[inner lantern]

nope

[stiff topaz]

oh

rip

[inner lantern]

but will this also work?

[stiff topaz]

it would

[stiff topaz]

but you don't know OB

[inner lantern]

i donnt even know the angle

lol

[stiff topaz]

I'm just wondering

what has your teacher taught how to do?

[hallow smelt]

@midnight socket whats up

[midnight socket]

You still stuck on that q?

[stiff topaz]

but using the method I would use

[hallow smelt]

i didnt do it good, but i ended up doing a spacing of just 7

and just estimating it

[inner lantern]

i learn it from khanacademy

[stiff topaz]

you know that tan(theta) = 6 because TC = 6

[midnight socket]

go to #help-3

[stiff topaz]

so you could solve for theta

oh

okay then

anyway you would solve for theta using tan(theta)=6

then you can use the triangle ABO that you made

actually

no

you couldn't

those are separate angles

sorry

[inner lantern]

yes

and i only know 90 degrees exists

[hallow smelt]

-7+7.8then just estimating an extra .8+.8 in-between each increment of 7 @midnight socket did you have a better way?

[stiff topaz]

you could find an angle for TOC

[midnight socket]

You could just find the exact, I'll show you hot to di it.

@hallow smelt

just go to #help-3

[stiff topaz]

you would then know TOA as well because they're complementary

[inner lantern]

so you want me to work TOC out, right?

[stiff topaz]

yes

the angle always helps

start there

meanwhile I'll try to work this out myself

btw anyone else feel free to correct me if I'm helping them incorrectly

by the way @inner lantern does that say BT = 4?

[inner lantern]

it didnt say that

WAIT

[stiff topaz]

what's that mark next to BT then

[inner lantern]

nope

i dont think so

xd

[stiff topaz]

what's the mark next to BT

is that just "y"?

[inner lantern]

no

its not the printer ink tho

[stiff topaz]

then what is it?

that might be necessary for the problem

[inner lantern]

i guess its nothing

[stiff topaz]

oh

ok

[inner lantern]

wow

wow

i thought BC was 6

but it is TC that is 6

🤦 me

[stiff topaz]

that's why you can use tan(theta)=6

because TC is the representation of tan(x)

[inner lantern]

okay sir

have you got the answer tho

i really stuck

[stiff topaz]

I'm busy with something else right now so I'll try to help you in a bit

sorry

[inner lantern]

BT = 4 @stiff topaz

good news

[stiff topaz]

oh good

that means AT is a chord of the circle

a segment such that both endpoints are points on the circle

[inner lantern]

the answer is 8

[stiff topaz]

you can also use the fact that ABT is an isoceles triangle

oh you got it?

[inner lantern]

how would you even solve for angle B

that is impossible

but we do know the side though except the hypotenuse

weird

[stiff topaz]

you know lots of information

[inner lantern]

yeah beacause i tried that

xd

[stiff topaz]

it all comes down to the chord

that's how you'll work toward the radius

it's also a part of the isoceles triangle

[inner lantern]

YESSS

but we still dont know what AT is

[stiff topaz]

i would try to define a system of equations

[inner lantern]

we need to find the length of the chord though

[stiff topaz]

yeah

[inner lantern]

yeah, insufficient information

the problem is that

B is not a right angle

[stiff topaz]

do you know the formula for length of a chord

[inner lantern]

no

[stiff topaz]

it's 2r*sin(theta/2)

where r is the radios

radius*

and theta is the angle between them

we know theta

[inner lantern]

where are you defining it as theta

B?

[stiff topaz]

no

theta is angle TOC

[inner lantern]

k

[stiff topaz]

because remember how we used tan(theta) = 6?

you can find the complementary angle to that

and that gives you AOT

[inner lantern]

wait

you want me to solve for theta right

im confused

[stiff topaz]

yes

first you solve for angle AOT

[inner lantern]

but u told me to solve TOC

ok

[stiff topaz]

yeah

[inner lantern]

so AOT first

[stiff topaz]

they're complementary angles

[inner lantern]

so theyre the same

[stiff topaz]

you need to solve TOC

before you can find AOT

[inner lantern]

ok

[stiff topaz]

no they're not the same

[inner lantern]

so which onne you wannt me to solve first

TOC?

i mean theta

[stiff topaz]

yeah that's the only thing you can solve for

you need that before you can do anything else

finding AOT is the next step

[inner lantern]

whats that

for any integer n

[hard gale]

just means you can plug any integer n in the formula, and this will get you a sol to your equation

[stiff topaz]

that's not really relevant to the problem but basically it's saying if you add pi to your answer it doesn't affect the solution

yeah

[hard gale]

ie 1.40 + pi, 1.40 + 2pi, 1.40 + 47358429pi are solutions for example

[inner lantern]

oay..

so whats next, sole for AOT

solve

[stiff topaz]

yes

[inner lantern]

does this mean i have to minus 90 to theta?

[stiff topaz]

yes

but

[inner lantern]

but

[hard gale]

that 1.40etc.. is in radians btw

[stiff topaz]

that's if your answer is in degrees

yeah

if it's in radians

[inner lantern]

oh

o

omg

[stiff topaz]

use pi/2

[inner lantern]

that happened to me once

[stiff topaz]

also your image of the problem

it wasn't to scale right?

[inner lantern]

wdym

im stupid

so do i have to convert it to degree or just apply pi/2?

[stiff topaz]

use pi/2

that's converted for you

[inner lantern]

ohh

[stiff topaz]

it's already in radians

[hard gale]

well if you're asked an answer in degrees .......

[stiff topaz]

yeah

i mean

[inner lantern]

(yeah the answer is in cm)

[hard gale]

angle in cm :/

[stiff topaz]

the final answer is a length of the radius @hard gale

[hard gale]

ok i have no idea of the context of the exercise lel

[inner lantern]

your p/2

multiply rightt

i got 2.20798614 degrees

[stiff topaz]

that seems like a really small number

[inner lantern]

unexpectedly small

so now 90-2.207.... ...?

[hallow smelt]

are you guys still working on the same problem?

[inner lantern]

yessss

im stupid

thats why

[hallow smelt]

LOL

[inner lantern]

yeah 😢

[stiff topaz]

it's okay when it comes to math basically everyone is stupid

[hallow smelt]

were all stupid

[inner lantern]

ok so whats next

90-2.207... ???

right

[stiff topaz]

i think

hold on

I'm using desmos to confirm

because i think we screwed up somewhere

[hard gale]

,calc pi/2 * 1.40

[inner lantern]

i think so

[somber coyote]

Result:

2.1991148575129

[hard gale]

[hard gale]

well you want pi/2 - theta (in radians)

[inner lantern]

,calc pi/2*1.40564764

[somber coyote]

Result:

2.2079861496799

[inner lantern]

ok what do you mean @hard gale

[hard gale]

that's the angle you wanna calculate bro

idk why on earth you multiplied

[inner lantern]

so i need to pull bacc

,calc pi/2-1.40564764

[somber coyote]

Result:

0.1651486867949

[inner lantern]

???????? @hard gale

[hard gale]

that's .16 radians

(so what's the questions in fact? that would help me understand what's going on)

[inner lantern]

the radius

thats what that circle wants (hold on, is this a circle? )

its missing its love

guys i need to sleep now, it is 3am @hard gale @stiff topaz

[stiff topaz]

ok

I'll keep working on this later, I have homework

[inner lantern]

anyway, i will stay

[hard gale]

mk

[stiff topaz]

rip

[inner lantern]

stupid bot

[stiff topaz]

I'll update when i have answers

ok luminuous

I figured out that we've been going about this the wrong way

like I said

I'll solve later

but

I used desmos to check my work

and I discovered that the angle that you got from the inverse tangent

is only true when the radius is 1

since we're solving for the radius

you do indeed need a system of equations

[brazen comet]

can someone help me with this mathcounts problem

[low tusk]

Note that triangle ABP is similar to triangle CDP

[brazen comet]

yes

4/5 aspect

[low tusk]

Ok

Then also note that DP:PB = CP:PA = 5:4

As they are similar triangles so the ratio of their three sides are all the same

[brazen comet]

ye

[low tusk]

Also consider triangle DPA and triangle PAB

[brazen comet]

r they similar?

[low tusk]

No

They are triangles with same height

[brazen comet]

k making sure i wasnt going crazy lol

how?

[low tusk]

As they share the same base line and have a common vertex

[brazen comet]

ok

[low tusk]

So the ratio of their area is just the ratio of their base

ie Area of triangle ADP:Area of triangle APB = DP:PB

Similarly Area of triangle CPB:Area of triangle BPA = CP:PA

[brazen comet]

and they are 4/5 of each other?

so area of adp is 20?

[low tusk]

Yup

[brazen comet]

and when u do it to the other side the area is 20

25+16+20+20 = 81

and 81 is correct

[low tusk]

You got it

[brazen comet]

mind blown

i'm in algebra 1 learning geometry for mathcounts so i'm not the sharpest at it

[upper karma]

translation:

If the line PT is drawn in the quadrilateral ABEC, which contains the midpoints of BC and FD, we assume p/q = z/y

anyone have a rough idea of what the fuck this theorem is?

lol

[solar shale]

How do Ik when to use a division sign or when not when setting up a cos sin or tan problem Like x=3/tan35 or 3tan35 like how would ik when to set them up / or not

[supple haven]

Im really confused on how im supposed to solve this

@solar shale what do you mean?

[solar shale]

I learned how to do it don’t worry about it @supple haven

[supple haven]

ah okay

gl man

[upper karma]

I don’t understand how to solve 9-18

[astral hornet]

First find the measures of the interior angles

Then, angles with vertex in center have 1:1 ratio with Arc intercepted

[supple haven]

<@&286206848099549185> its been a bit since I asked so I hope im not breaking any rules by pinging y'all

[river forge]

you want to add the 3 vectors

[supple haven]

ah okay

i've only got the magnitude though

[river forge]

so if you set north as your reference line, you can rewrite the problem as what is the sum of vector (35,45), (8,58), and (12, 79), you can rewrite these from (r, theta) form to (x,y) form, remembering that (rcostheta,rsintheta) is the (x,y) component of an vector written as (r,theta)

so, symbolically, with vectors (a,b) (c,d) (e,f), the sum of the vectors is (a * c * e * cosb * cosd * cosf, a * c * e sinb * sind * sinf)

oh shoot

[supple haven]

ahhh thats really smarth

smart*

thank you so much Bullton

[river forge]

30pi/360=x/310

np

i got 81.1

[upper karma]

Thanks.

It was wrong question.

[oblique forge]

I’m very confused on how to graph a sine/cosine equation with a phase shift. I tried looking things up online and it is explained way different than what my teacher taught me.

I don’t understand graphing a sine curve with a phase shift and vertical translation. I get very confused

[wary turtle]

Try making a graph out of the function y = sin(x)

x will take over the values of the trigonometric circle, in radians: 0°, π/2, π, 3π/2 and 2π.

So for example, you'll have x = 0°. Then y = sin(0°), thus y = 0.

Its phase shift is zero, in other words, originates from zero

If I have x = π/2, then y = sin(π/2) which is 1, and so on.

The x axis will have the values of the trig circle, and the y axis will have the output values such as y = 1 which corresponds to x = π/2.

Now you trace lines and get the coordinates, and with them, you draw the sin wave

Same goes for cos and tg.

[umbral snow]

@oblique forge

[tropic shard]

I'd suggest you just go on Desmos and type in y=sin(x-2πa)+b, and wiggle the a and b around to get a feel of what's happening.

[wary turtle]

I think it's better to show simple cases first, so it gets easier to grasp.

[oblique forge]

Okay, thank you! I appreciate it

[tropic shard]

Depends, gotta know the other person's learning style.

[wary turtle]

Knowing it doesn't mean making it hard to understand ¯_(ツ)_/¯

[gloomy zenith]

does anyone wanna help me in a call with some geometry i am willing to pay a small fee

[old seal]

Ans is a but i am not getting it

Where am I wrong?

[supple abyss]

in the interval 0<=x<=2pi it's all values of x except 5pi/4

[narrow sleet]

I think the question should be does not hold, instead of hold

[old seal]

I checked it in mathway app and its also giving answer,5pi/4

[narrow sleet]

x = 5pi/4 + 2n pi is only the equation is equal, not more than

[dark sparrow]

@old seal the problem definitely has a typo. your work is correct

[old seal]

Yes,u guys are right.I just confirmed from my teacher also and he said the same.Thank you so much

[narrow sleet]

👍

[vague pier]

hi can i get some assistance on several problems that Im working on?

[upper karma]

1. Recall the definition of altitude. (5x+8)+(5x+2)+90=180. Find x and plug back in.
2. Find slope of JK. Then use the negative reciporacle of that and the midpoint of JK to find the equation for the perpendicular bisector of JK in point-slope form.

3. Recall the definition of incenter. Because NKX is 90°, we know NP is bisected.
4. Recall the definition of centroid. Recall the definition of median. Something's got a relation.
5. Recall the definition of orthocenter.
6. Recall the definition of circumcenter. Note what's equal to what and use proportions to proceed.

7. The angle is bisected as it forms two congruent triangles by SAS. Set the small angles equal and solve for x and plug back in.
8. Recall properties of medians and their point of concurrency.
9. Recall the definition of altitude. 30+90+ABX=180.
10. Use properties of altitudes as above in 9 and 1.

[vague pier]

sorry im kind of a slow worker,

for 1 I got 42 degrees

for 2, the slope i got was 1. the midpoint i got was (2,4), te point slope form i got was y-4=-1(x-2)

@upper karma

[upper karma]

I'm not checking your work.

[vague pier]

oh alr

[rose goblet]

LMAO

[naive gyro]

I'm sorry about this dumb question, but i couldn't find a solid answer on google. How do i calculate the center point of a n sides regular polygon? (i need a simple way, doing this for a code)

[wild hamlet]

Depends on what variables we have

The simplest way is to take the average of all the points of the polygon which make the edges.

[naive gyro]

i have only two variables: edges length and number of edfes

[wild hamlet]

Hm

Length and edges

There are n edges, which means the points are stationed around the center 2pi/n radians from each other.

The edge length is L

L/(2sin(2pi/n)) = radius

If you have 3 points, then you can always find the center. Otherwise you can not determine one

3 or more points

[naive gyro]

i'm sorry but i think i didn't understand it properly

i can't interact with other points

so i need a "neutral" way to know where the center is, using only the length and edges

so, what you're saying is something like i have to find the triangle inside the polygon, that way i'll find the center?

[wild hamlet]

Yes

[twin prawn]

Draw an auxiliary line segment from the center of the top circle to the tangent point between the circle and the line

You'll see a pair of similar triangles appear

[old seal]

Hey guys..I have a doubt

Domain of 1/secx = R-(2n+1)*pi/2, right?

[twin prawn]

What's R

[old seal]

Real numbers

[twin prawn]

Wait 1/sec(x) is just cos(x) isn't it

[old seal]

yes...but lets just forget cosx

whats the domain of 1/secx

[twin prawn]

Though I suppose it's undefined when sec(x) is +/- infinity

I.e. the roots of cos(x)

So pi/2 + k*pi where k can be any integer

In conclusion, you're right

It's cos(x) almost everywhere, except it's undefined at those numbers where cos would equal zero

@old seal Well done!

[old seal]

ikr

i am literally baffled for 45mins just for this

so this means cosx is not necessarily equal to 1/secx,right?

not everytime i mean

[twin prawn]

No

[old seal]

so??

please elaborate

[twin prawn]

Your conclusion is correct

The idea is that when cos(x) = 0, then sec(x) is undefined

Which in turn means that 1/sec(x) is undefined

[old seal]

yes

I get it what you're saying

but then if someone asks whats domain of 1/secx,what should i tell them?

[upper karma]

R - points at which the discontinuities occur

[dark sparrow]

R - {π/2 + kπ | k in Z}

[upper karma]

oh you were right, just checked

[dark sparrow]

the function is equal to cos(x) wherever it is defined, fwiw

[old seal]

but as 1/secx = cosx, and domain of cosx is R

how are they equal?

this is where I am stuck

[upper karma]

it's only equal when 1/sec x is defined

[dark sparrow]

1/sec(x) is equal to cos(x) wherever the former is defined

[upper karma]

take a rational function like:
(x-3)(x-4)/(x-3)
this is equal to the function (x-4) at all points except x=3

[old seal]

ya

[upper karma]

same trick here

well, similar trick

[old seal]

ohhhh I get it

Thank youu

everyone

my brain godammit

[upper karma]

lol

[twin prawn]

🍮

[fleet finch]

How does one caracterize the center of the inscribed sphere of a tetrahedron with barycenters ?
I mean without talking of angles'n'stuff

My goal is to generalize in higher dimensions : given a simplex, I need its inscribed hypersphere

[delicate hedge]

Hey folks, new to the channel. I have to solve this question but am a little stumped with how to start

[umbral snow]

That is a tall pole

Using just the bottom triangle, what's the distance from the observer to the mountain?

[delicate hedge]

In this case, its undefined

so could I represent that with a placeholder variable?

[cyan beacon]

So, I don't know if this is the right channel to ask this

By doing P1+(P2-P1)*lambda I find all the points between P2 and P1, with lambda being between 0 and 1

But then I have this formula which is like (P2-P1)*lambda' (derivative of lambda) which should return the speed, am I correct on this?

[umbral snow]

@delicate hedge

Sure, but you can represent it with the given variables

[delicate hedge]

little confused on that one, sorry

[umbral snow]

sin(β) = h/d
d = h/sin(β)

[worthy nimbus]

Can someone help me with that plz

[lime flame]

@worthy nimbus
Hint: Draw a line to make a triangle with the given height. It will give u a 30:60:90 triangle

After that set a side as x

I'm assuming we are solving for area

[worthy nimbus]

Give me a sex plz

Sec*

Done

@lime flame so

Done

[lime flame]

Do you notice anything or have any ideas

bc imma make sure u learn, not gonna spoonfeed lol

You know the formual 1/2ap = Area?

[worthy nimbus]

7×√(6 by √(3

I was getting the radical

From my calculator app

[lime flame]

What?

[worthy nimbus]

Divide 7√(6

By √(3

Since that side is x√(3

Because of the special right triangle

[lime flame]

then your basically saying x equals 1 lol

[worthy nimbus]

Well I'm referring

[lime flame]

Oh ok

[worthy nimbus]

To the rule of 60 30 90

Triangles

[lime flame]

yeah ik lol

So you know the formula 1/2apothemperimeter = area right?

[worthy nimbus]

Yeah

[lime flame]

We are given the apothem which is 7 rad 6 all over 2

and the perimter is 6x

[worthy nimbus]

Wait how

[lime flame]

since one side = x

It's a hexagon

[worthy nimbus]

Ah

Got it

[lime flame]

okay so

If we knew what x whats we would know the area right?

So how can we solve for x, using the 30, 60, 90 triangle

[worthy nimbus]

Find the bottom since its cut in half

Times it by 2

And I can find that by

Dividing 7√(6/2 by √(3

[lime flame]

yeah awesome

Oh that's what u said earlier lol my bad

Anyways since u know what half of x is, mutliply whatever u get by two

and plug in the formula

and u get ur answer

[worthy nimbus]

It's not right :(

[lime flame]

what is that lol

u messed up ur paranthesis

[worthy nimbus]

What should I put in then

[lime flame]

and 7×√(6 by √(3 isn't the answer btw, it's half of x

[worthy nimbus]

Ah

Crapo

[lime flame]

Wait let me approach a different method

using a easier method

The Pythagorean theorem

So a2 + b2 = c2

We know that a = 0.5x

b = Given Height

c = x

[worthy nimbus]

Yee yee

[lime flame]

so 0.25x^2 + 73.5 = x^2

[worthy nimbus]

Is 7√(6/2)/√(3

The correct to plug it in my calculator

I use a method that uses the area of the triangle

To find the area of the polygon

[lime flame]

the correct way to plug it in is ((7√(6)/2)/√(3

[worthy nimbus]

Ok

That worked

!!!!

[lime flame]

ayayya

[worthy nimbus]

With my method

Ahhh

[lime flame]

lol

[worthy nimbus]

This makes me very happy

Thanks

[lime flame]

np

[worthy nimbus]

You're the best teacher

[lime flame]

lol thx

Did you find the area of that triangle and multiply by 3

[worthy nimbus]

No 12

12 triangles in a hexagon

[lime flame]

oh wait oops lol

my bad

ok cool

Glad I could help

[upper karma]

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

Not sure what they're asking for here...

[dim pulsar]

@upper karma there must be some context before that question

i.e. a set up

[upper karma]

Nope. No set up. :/

[dark sparrow]

is nothing else said about y and p?

[upper karma]

Nope. Maybe it is a broken problem.

[vale inlet]

Okay, maybe I'm being dumb here, but what is a "wedge of circles" xD

I've never heard this term before

like, is there a picture of a wedge of 3 circles I can look at?

Is this a wedge of 4 circles (up to homotopy)??
https://upload.wikimedia.org/wikipedia/commons/c/cf/Topological_Rose.png

[dark sparrow]

looks like it.

wedge = wedge sum

the wedge sum of two or more pointed spaces is the space you get by joining them at their basepoint

[vale inlet]

thankies, I posted this in here cause was just wanting to see a picture of a shape xD

sure it's topology, but eh

[foggy beacon]

is the reference angle the same as finding the co terminal?

[pseudo pecan]

In this channel we can speak about analytic geometry?

[wintry hatch]

Probably

[worthy nimbus]

Is that right?

[vital frost]

Could someone help me with these questions

[worthy nimbus]

For the first one

You could use pythagorean theorem

Then subtract it by the radius

@vital frost

[vital frost]

so i did Pythagorean theorem and got 40 @worthy nimbus

[worthy nimbus]

Subtract the radius from 40

[vital frost]

40-32 = 8

so x would be 8?

[worthy nimbus]

Yeah

[vital frost]

ok how about for the other 2 @worthy nimbus

[stark pumice]

I need help with this please. I'm not entirely sure what I'm doing.

[rose goblet]

hmm

that stuffbois

LM is x. x^2 = 6*3

@stark pumice

[stark pumice]

okay thanks

[mystic vortex]

i need help with finding the answer to this problem:

ln(x+2) + ln(x-12) = 1

how do i get x

[upper karma]

@mystic vortex u can

[lime flame]

x equals 12.2

[upper karma]

do ln((x+2)(x-12))

and then turn it into a quadratic

and remove line

lin*

[lime flame]

u gotta apply the log rule

[upper karma]

by doing

x^2-10x-24=e

and so we can also sunstitute e @rose goblet

substitute*

with the number 2. balh blah

and just solve

[rose goblet]

wut

[upper karma]

i meant to tag @mystic vortex

sorry

[rose goblet]

lol

[glad ocean]

hey guys is khan academy a good resource to learn algebra II?

for new york state regents exams in the united states

[rain tulip]

im not understanding a problem i believe it has to do with Pythagorean ID

[olive solar]

khan academy is a decent place for alg 2 i think. paired alongside some studying and practice problesm it's p solid

@rain tulip the problem?

[rain tulip]

Suppose that 90° < u < 180°. Find the sign of each function value.

sin 2u

the u is suppose to be theta

[olive solar]

so the range of 2u would be 180 < 2u < 360, right?

[rain tulip]

im not understanding?

are we just putting in 2u in place of u in the original problem?

[olive solar]

what?

you're looking for the sign of the result of sin2u, right?

[rain tulip]

yeah for each function

[olive solar]

so finding the range of 2u would be pretty important, right?

if you multiply u by 2, what is the possible range of it?

[rain tulip]

2u

[olive solar]

so what is the sign of sine from the range of 180 to 360?

think of it in terms of quadrants

that would br quadrants 3 and 4

[rain tulip]

negative?

[olive solar]

yep

[rain tulip]

thats it? wow i feel dumb

how about cos(-u)?

is it 90<cos(-u)<180 and the answer is the sign of the cos in Q 3 and 4?

[olive solar]

cos -u would go into -180<-u<-90 correct?

put that on a positive 360 range and it means it is between 180 and 270

which is quadrant 3

[rain tulip]

ahhh ok

thank you @olive solar

i struggled with domain and range in algebra and its hurting me more in trig

[upper karma]

I need help with trig

[dark sparrow]

[upper karma]

Lmao hold up

This what I gotta do

[amber raven]

You have a good teacher

[upper karma]

y'all know any good online courses I can take for trig

[upper karma]

Nope but try this book (recommended in awesome math github page) http://www.mecmath.net/trig/trigbook.pdf

@upper karma

[dreamy hollow]

How would I go about taking a simple polygon and sub-dividing it by 'drawing' lines through it? So I draw one line through let's say it's centroid and if it's convex, I'd have two lists of points (can't find this on google)

[sudden gazelle]

So I'm getting started on doing geometry constructions using the following tools (the most powerful set I can reasonably think of using):

foldable tracing paper (allows straight edge constructions and more)
colored pencils and eraser
scissors and tape
push pins and string (allows compass constructions and more. when combined with the folding paper, allows neusis constructions)
sand (for measuring volumes)
lamp (for lighting/shadows)

I'm almost a complete beginner at geometry, so I'm looking for any resources on related topics (straight edge + compass constructions. origami math. etc). Can anyone help me out?

i also considered the tools of cymatics, and electronics + iron filings or other magnetic materials. but i don't have the lab equipment to study that

[sudden gazelle]

I'm searching for a powerful enough set of tools for (at least) solving arbitrary complex polynomial equations either exactly (if Galois allows it) or approximately, using Newton's method.

[umbral snow]

Sounds like you know your tools already. What are you missing @sudden gazelle?

[sudden gazelle]

i don't want to have to derive everything from the beginning lol. i just ordered 2 books ("Geometric Construction" and "Origamics"), but i'm looking for online resources. i can't find an online textbook for geometric constructions of any sort

lots of disparate knowledge in lots of places though

intuitively, my tools should be enough to do the complex Newton's Method though (which if done with non-standard analysis proves the fundamental theorem of algebra), so my goal is to get all the necessary operations figured out

pointwise calculation of complex polynomials and their nth-derivatives. division. arbitrary nth-roots

if my tools can calculate those, that will be enough to solve any complex polynomial equation either exactly whenever algebra allows it, and sometimes even when it doesn't, or with an approximation via Newton's Method

[sudden gazelle]

err... never mind... they're not. 😦 I need to restrict myself to special cases

[rustic cairn]

How do i do 11 and 12

[twin prawn]

Hint: similar triangles

[lime flame]

How do you know they're similar triangles

[twin prawn]

They're isosceles with the angle at the lens being the same

[lament bay]

Can someone help me with these

[upper karma]

Do you know the law of cosines?

This doesn't make sense to me can someone help?

Sin and cos have the same Max value of 1

ofc

cos and sin are both bounded in R

and they are bounded above by 1, so cos(x) = pi has no real solution

Sin also would be the answer then?

Oh

I was confused as to why isn't sin also an answer I just realised it isn't in the given choices.

lmao

[lament bay]

hmm I don’t know any trigonometry yet @upper karma

[upper karma]

Oh my bad I misread it's just Pythagorean Theorem.

Split the base in half.

Check that triangle inequality is preserved.

[lament bay]

Hmm

[wanton prawn]

I have a question. Can I ask it here? It's for similar triangles

[upper karma]

[abstract yew]

If someone can help me do some simple arc length, sector area, radian and special right triangle problems please pm

[upper karma]

@abstract yew

I will

[upper karma]

hello? @abstract yew

okay let me just show you my solution

you notice that the triangle's angles have 30, 60, and 90 degrees?

right?

[tacit solar]

owo

[upper karma]

now you can use 30 60 90 theorem

oh my god cherry

get out

@abstract yew

[tacit solar]

Who’s Cherry? Ew

[abstract yew]

yes

[upper karma]

whoops

I read your name as cherry instead of candy

sorry

[tacit solar]

Ew

[upper karma]

okay @abstract yew

given this

you can solve it

[abstract yew]

for which problem

[upper karma]

in this case x = 12sqrt3

question 6

the one you gave me

https://cdn.discordapp.com/attachments/556266684957327370/556269362852397072/image1.jpg

you sent me this

[abstract yew]

yes

so

[upper karma]

in this case x = 12sqrt3

so according to 30 60 90 theorem

[abstract yew]

so 12sqrt3sqrt3

[upper karma]

you should be able to find the side lengths

yeah

for the opposite

now find the adjacent

I mean hyp

m'bad

okay did I help?

[abstract yew]

yes i think i get it

[upper karma]

okay just do that for all the rest of the questions

[abstract yew]

thank you

[upper karma]

the next question also has 30, 60, 90 theorem

[abstract yew]

Can anyone help me with this

[spiral pawn]

@abstract yew i can help

[glad ocean]

same

so you know that QR is also 7 because QS is and they're both radii

and tangents of a circle are perpendicular to a radius

so just use Pythagorean theorem

7^2 + 24^2 =

625

so QT is 25 and ST is 18 because 25-7=18

[hasty mesa]

Alright I am given a point and a Cartesian eqn of a line
And I was asked to find perpendicular distance

Can I first use dr's to find DC's and then with DC's can I substitute in
A formula to find it in one step??

[hasty mesa]

<@&286206848099549185>

pls help

[upper karma]

Projections?

or orthgonality?

[hasty mesa]

no i have to find the distance

perpendicular dist from point to eqn

[upper karma]

whats the point and whats the line

[hasty mesa]

alpha beta gamma is the point

and the line is
x-x1 / a = y-y1/ b = z-z1 / c

dr of that line is a , b , c

only this info is given

[upper karma]

lemme get my notebook i gotta recall dis

[hasty mesa]

okay will be waiting

[upper karma]

so symmetric equations

<a,b,c> is direction vector

[hasty mesa]

ya

can i find dc from abc and use a formula

[upper karma]

so it's projection

Distance = ||PQ X U|| / ||U||

[hasty mesa]

why do u go to projection?

[upper karma]

stupid spoiler

It's a projection it just is fren

Distance is the norm of (PQ X u ) / norm of u

[hasty mesa]

yo pls

[upper karma]

P is a point on the line

x-x1 / a = y-y1/ b = z-z1 / c

make them parametric and say t=0

no i said projection it's not a projection

idk why i said dat

Distance is norm of PQ cross U

divided by norm of U

[hasty mesa]

can i use that formula

d = sqrt( || some long ||)

[upper karma]

I haven't learned it that way

Lemme check it

Lemme get a reference on this one cuz not 100% sure about that formula

<@&286206848099549185>

[hasty mesa]

ya thats why i asked it here

[upper karma]

You in physics or calc for this one?

(not that it matters, don't worry about dat)

[hasty mesa]

im in my room

[upper karma]

lol

[hasty mesa]

btw ping me when u get the answer

[hasty mesa]

[frail hamlet]

Guys hey, I'm struggling with a simple geometry question. We have a cylinder with height h and radius r inscribed in a cone with height 30 and radius 10. I must prove that h=3(10-r)
Thank à lot

[supple abyss]

consider the cross section

[frail hamlet]

I don't get it

[supple abyss]

do you know what a cross section is

[frail hamlet]

Not really

[frail hamlet]

Thank you

[red leaf]

this adds up to 540 right?

[gloomy python]

Need help with number 15

oh my bad the number didnt show up

the word problem one

<@&286206848099549185>

[upper karma]

,rotate 90

[upper karma]

,rotate 180

[upper karma]

151/16.75

,calc 151/16.75

[somber coyote]

Result:

9.0149253731343

[upper karma]

Repulica is approx 1/9th
the actual is 9x bigger than replica

@gloomy python

is/of = %/100

151/16.75=x/100

16.75x=15100

,calc 15100/16.75

[somber coyote]

Result:

901.49253731343

[upper karma]

901% larger

or 9.01x larger

[gloomy python]

Thank u🙏🏽

[heady urchin]

anyone here who can help me with circles?

[umbral snow]

🤔

That's a circle with a face

[upper karma]

lol

[heady urchin]

Preferably help with all the questions lol

[upper karma]

hey

can someone help me with this question?

how would I simplify this?

[csc(-x) + sin(180 -x)]/[sec(180 + x) + cos(360-x)]

$csc^2(180 + \theta)$

[somber coyote]

Death Is The Only Constant:

[upper karma]

what do I do with that identity

[upper karma]

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

This is probably a dumb question, but right now I have to convert the radians to degrees before I can answer their question, is this a necessary step or should I have memorized these degree measures in radians by now?

[umbral snow]

Forget degree measurements, learn radian. You'll learn the system faster, and much better, learning it away from degrees. Having to convert from radians to degrees is harder than it needs to be

• radian measure is an easier system

[upper karma]

Thank you. 😃

[upper karma]

Hey, I have a question that is way out of my league

I have a problem that I don't really think is solvable, but here goes

I have a quaternion(x/y/z/w) that I need to convert to Euler angles while completely avoiding gimbal lock

I, however, don't need perfect precision - at any one point in the dataset, the axes may avoid singularities by having slight offsets

I'm a complete lemon at math, but from what I understand, doing this in Euler is still impossible even under loose constraints, right?

[mystic bloom]

I don't get FL= GA

Def of congruence?

[zenith raptor]

shouldn't it just follow from (6)?

[dark sparrow]

two col proofs 🤢

[mystic bloom]

ye

is this Hyp acute angle?

[upper karma]

what did i do wrong?

[dark sparrow]

show your work

that way we can see where you screwed up

[upper karma]

i do not have it.

[dark sparrow]

how did you obtain the answer you've written, then?

[upper karma]

metal meath

[dark sparrow]

?

[upper karma]

basically i took a hard guess

i just need to get my tesst corrections in

[dark sparrow]

...

i mean

how can you guess and then ask where you went wrong

[upper karma]

im basically asking for an answer

[dark sparrow]

this isn't how this server works

we don't give out answers

so no

[mystic bloom]

you'll not learn lol

[upper karma]

could someone teach me how to do it?

[unique flower]

So basically I have this qs to find the amplitude and phase shift

I got the amplitude is -1.5 and phase shift is 3pi/4 but amplitude can’t be negative

I did try use -1.5 and 3pi/4 to do the equation and got prove that will be 0

What the real ans then if is not -1.5 and 3pi/4??

This is the working to proof that ans is 1.5 and 3pi/4

[lone nova]

https://eylemmath.weebly.com/uploads/6/7/8/3/67831081/triangle_5_orig.png

can someone explain me what is ? equal to

and why?

step by step please

[dark sparrow]

wow oof this looks tricky

[lone nova]

it is indeed