geometry-and-trigonometry

[honest pumice]

a can can be 0 to pi/4 on the positive x and y axis

and many other values

but the unit circle has a circumference of 2pi

and can keep measuring a with more than 2pi

or less than 0

by just adding 2pi + the extra angle you want

you should really study the unit circle

it's essential to know it

[valid magnet]

yes

[honest pumice]

can't really explain it with words here in a chat

[valid magnet]

i'm trying to figure out how to calculate the hypotenuse of a right triangle w/o doing sqrt(x^2 + y^2)

[honest pumice]

https://www.youtube.com/watch?v=1m9p9iubMLU

[valid magnet]

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

[pearl finch]

can someone show me how this is wrong? tag me if u have a reasonable explanation plz

[pure cape]

remember, they are asking for the slope of segment WZ not segment YZ

segment WZ is perpendicular to YZ at Z

[pearl finch]

@pure capedude how do u know that

without drawing anything up?

[pure cape]

that's because WXYZ is a rectangle

[pearl finch]

so a rectangle has specific rules is what you're saying?

[pure cape]

what do you mean by specific rules?

[pearl finch]

like, specific rules when it comes to questions like this

[pure cape]

if a rectangle is named WXYZ, the vertex has to follow that order

[pearl finch]

so something like this

im so confused lol

something's just not clicking

[pure cape]

for example this rectangle is not WXYZ

this rectangle is WYXZ

it has to follow an order, thats just how they are named

[pearl finch]

oh

[pure cape]

this is a WXYZ rectangle

[pure cape]

so here, you can see that WZ and YZ are perpendicular

and the question asked for the slope of the segment WZ, and not YZ

but you answered the slope for segment YZ

[pearl finch]

ohh i get what ur saying

it always has to have a certain order

[pure cape]

yes

you can read it from any vertex you like

[pure cape]

but it has to either go in anti-clockwise or clockwise order

[pearl finch]

yeah lol

[pure cape]

for example that rectangle could also be called XWZY

or ZWXY

it doesnt matter, you just need to get the order right

[pearl finch]

yeah but it's always following that order

yep

and since it's perpendicular to the slope we already know

it'd have to be 3/2

[pure cape]

yes

[pearl finch]

tysm'

[trim laurel]

i put the points but how do i get the angle?

[dark sparrow]

you don't need the angle itself tho

[trim laurel]

yea ik but i need to get it

so then i can get the ratios

i dont understand how to get it tf

[dark sparrow]

no you don't need theta itself

[trim laurel]

what do i need

[dark sparrow]

you need the point where the ray OP intersects the unit circle

its coords will be (cos(θ), sin(θ))

you can also find this by dividing each coordinate of P by the length of OP

[trim laurel]

what?

[dark sparrow]

O stands for origin of course

[trim laurel]

we never did that before

do i even need theta

if i have a, b, and c

[dark sparrow]

this is what i was talking about

[trim laurel]

we never did that tho

this is hs

it says to get the ratios

[dark sparrow]

but you can also get yourself a right triangle instead. just be mindful of the signs

[trim laurel]

it is a right triangle already though no?

[trim laurel]

yea

thats what i got

[dark sparrow]

yeah ok so you have two of its sides given to you

[trim laurel]

9 is adjacent

then find hyp

[dark sparrow]

and the third is the hypotenuse, which can be worked out using the pythagorean theorem

[trim laurel]

i got square root of 63

[dark sparrow]

sqrt(63)? that looks sus

[trim laurel]

which is 7.9 but i wrote it as 3 times the square root of 7

[dark sparrow]

show work?

[trim laurel]

okay

[dark sparrow]

i think you fucked up your arithmetic somewhere

[finite edge]

-12 times -12 is not -144

[dark sparrow]

^

[trim laurel]

ohhh i fucked that up

okay thanks

[dark sparrow]

i mean you should've gotten a red flag when you tried to shove a negative number (-63) under the root

instead of just sweeping that minus under the rug

[trim laurel]

helped more in 2 minutes than anyone else

so hyp is 15

[dark sparrow]

yes the hypotenuse is 15

now you have enough info to get all three trig ratios of theta

[trim laurel]

appreciate it

got it

i have 2 more questions that im reallllly confused with

if ur willing to help

[dark sparrow]

is #4 the same angle θ that we worked with in the previous problem

[trim laurel]

i dont think soi

so

i thought it was

[dark sparrow]

no i mean like

[dark sparrow]

is there a different point or something

[trim laurel]

maybe this will help

[dark sparrow]

for problem 4

[trim laurel]

yea i think its completely different

[dark sparrow]

okay then youll have to give the coordinates of the new point

[trim laurel]

so that means it is connected to 4

because how else would we get the coordinates

[dark sparrow]

...

???

[trim laurel]

im saying like

it says of the right triangle

oh im an idiot

[dark sparrow]

in problem 3 we worked with the angle whose terminal ray contains the point P (9, -12)
in problem 4 are we still looking at the same angle or not

[trim laurel]

no

[trim laurel]

.

do we use the 45 degree triangle for this?

[dark sparrow]

what????

hold on

what angle is even being talked about in #4?

is #4 a continuation of #3 or not???

[trim laurel]

no

i dont think it is, it wouldve said

[dark sparrow]

okay then WHAT ARE THE DATA FOR PROBLEM 4

theres no data

no data no answer

[trim laurel]

lmaooo thats what im saying but i think he told us that whenever we see "right triangle"

we use special triangles

[dark sparrow]

who's "he"

[trim laurel]

my teacher

[dark sparrow]

and who is he to claim that right triangles with angles other than 30, 45 or 60 degrees just don't exist

[trim laurel]

hes not saying that

[dark sparrow]

either your teacher has done a poor job of writing down the problem(s) in a clear and solvable manner

or you're misleading me into thinking there is no data given for problem 4

[trim laurel]

hes saying that whenever we do get one of those angles, we use the hyp adj and opp angles he gave us

[dark sparrow]

okay but we don't have an angle at all

let alone any of those three

[trim laurel]

misleading? but how would I know because it doesnt say anything about 4

i mean 3

[dark sparrow]

fucking hell

this is a communication breakdown this is a communication breakdown this is a communication breakdown

[trim laurel]

ill look in my notes for something related to this

[dark sparrow]

your notes are useless

this isnt a notes issue

this isnt a knowledge issue

this is a reading comprehension issue

[trim laurel]

yea i know

because the question doesnt say if its connected to the angle we got in 3 or not

[finite edge]

hey ann the diagram says use it for the next 2 questions

so 4 has the same point P

[trim laurel]

ohhh

[dark sparrow]

SEE THATS WHAT IM TALKING ABOUT

[trim laurel]

it does

[finite edge]

^

[dark sparrow]

but moe was insisting

[trim laurel]

i didnt see that lmao

[dark sparrow]

that that wasnt the case

for some reason

[trim laurel]

LMAOOOO

[dark sparrow]

ARGH

okay

[trim laurel]

i wasnt insisting i said i think so looool

my bad fr

[dark sparrow]

so we still have our triangle with sides 9, 12 and 15

[trim laurel]

-12*

yea

[dark sparrow]

we're looking at the reference angle

which is in the first quad by defn

so that side is just 12

[trim laurel]

why first?

[dark sparrow]

reference angle

[trim laurel]

opp is negative

[dark sparrow]

reference angle

the reference angle of θ is the angle in the first quad whose sin and cos are the same as the original up to sign

[trim laurel]

i still have no idea what you mean

wouldnt it be in quad 4

[dark sparrow]

theta itself is in quad 4

but im talking about

the

reference angle

[trim laurel]

yes ik

[dark sparrow]

and if you didnt know what that means you could have and should have asked me outright

and made both of our lives easier

[trim laurel]

ik what it is

[dark sparrow]

great

so you know that the reference angle always lies in the first quad

by its very definition

[trim laurel]

WHAT

bro WHAT

[dark sparrow]

don't call me bro...

[trim laurel]

so ur saying that there cant be a reference angle in quad 2 3 or 4?

[dark sparrow]

reference angles are acute

by definition

https://www.purplemath.com/modules/radians3.htm

no matter which quad the original angle is in, the reference angle will be acute

[trim laurel]

aight

i just thought that since we're in quad 4 i thought the reference angle would be also

[dark sparrow]

no the reference angle is acute

no matter what quadrant the original is in

the reference angle is acute

as ive said like a dozen times already

[trim laurel]

so to get an angle i can use cos-1(3/5)?

[dark sparrow]

cos^-1(3/5) yes

if you wanted you could also do sin^-1(4/5) or tan^-1(4/3)

these all give the same result

[trim laurel]

and that gives me the reference angle

so ref angle is 53 degrees

and the other angle is 127?

[dark sparrow]

,calc acos(3/5) * 180/pi

[somber coyote]

Result:

53.130102354156

[dark sparrow]

okay yes 53°

but why would the other angle be that large all of a sudden

its also supposed to be acute

which 127° decidedly isnt

[trim laurel]

oh

so how do i get the other angle

[dark sparrow]

how did you get this 127° angle?

[trim laurel]

oh is it 180-53/2

[dark sparrow]

no? why half

did you subtract 53 from 180?

[trim laurel]

yea

[dark sparrow]

and who will account for the right angle that's already present?

and which contributes 90 degrees to the angle sum?

which you just went and lumped in with the other acute angle...

[trim laurel]

is the other angle 37 degrees

180-90-53

[dark sparrow]

yes it is

[trim laurel]

thank u

helped a lot i appreciate it

what kind of triangle do i use for this

if it says how long, would it mean the adjacent side?

[dark sparrow]

make a diagram

wait hold on

"at an angle of 15°"?

that's ambiguous

[silent plank]

the wording of the problem is so bad and vague

[dark sparrow]

theres like a dozen different places the 15° angle could go honestly

[trim laurel]

the angle is for the side 30m

thats what he means

yea ik he words all his shitty questions like this

[dark sparrow]

"the angle is for the side 30m" is not any better

[trim laurel]

yes it is

[dark sparrow]

can you draw a diagram that shows your interpretation

[trim laurel]

we use the sine law for this

[dark sparrow]

because i'll be honest

i have no idea what's going on

and i need you to explain yourself to me

with a diagram

[trim laurel]

im trying to figure out what triangle to use for this

[dark sparrow]

why is "make a diagram" not the first priority in your process for solving geometry problems???

can you draw a diagram

[trim laurel]

because ive never made one for trig lol

yes i will

[dark sparrow]

okay make your diagram and ping me when it's done and show it to me

[trim laurel]

nah i cant even understand it

[silent plank]

is probably the most reasonable interpretation

[trim laurel]

no but the 15 should be on the other side

because it says 30m at an angle of 15 degrees

[silent plank]

that statement implies that the 30m side forms an angle of 15degrees with another side

here, its the length of the pond AB

[trim laurel]

yea im confused bro

i wouldve put 15 at the other side because its referring to 30m

[dark sparrow]

the question itself is confusing

[silent plank]

ignoring the wording of the question,
would you be able to calculate the length of the red line AB?

(bring up the wording issue with your teacher)

[trim laurel]

yea probably

id use cosine law

no i did it wrong

because i got 22 m

that doesnt make any fucking sense

@dark sparrow did u figure it out?

[dark sparrow]

going by ramonov's picture... let me calculate it rq

yeah 22 is way off. how did you get it?

the answer i got is ||78 m, to the nearest meter||

[trim laurel]

i used cosinse law

[dark sparrow]

show your work.

[trim laurel]

so i did (50^2+30^2)-(2(30)(50)cos(15))

[dark sparrow]

this does not correspond to ramonov's picture

[silent plank]

ur not applying the cosine law properly

[dark sparrow]

note the placement of the 15° angle. it is not between the two known sides.

[trim laurel]

yea its for the unknown side

oh

i mean 30m

[dark sparrow]

this assumes the 15° angle is at B, which it is not

[trim laurel]

so what do i do

[dark sparrow]

apply the cosine law properly.
perhaps you could denote the unknown side with a letter like x, and write out the law of cosines in terms of that

[trim laurel]

okay

[dark sparrow]

in your case you will get a quadratic equation in x

[trim laurel]

im not doing it right

can u please show ur equation

[dark sparrow]

$50^2 = 30^2 + x^2 - 2 \cdot 30 \cdot x \cdot \cos(15)$

[somber coyote]

Ann

[trim laurel]

i dont understand how angle 15 is corresponding to 50m

the angle is for 30m is it not?

because the angle isnt right next to the side, its the opposite of the side

@dark sparrow

[dark sparrow]

??

the 15° angle is between the 30 and x, and opposite to the 50

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

[trim laurel]

oh

I thought that 30 m was at an angle of 15

and ur saying that its between 30 and x which i understand, but if it was 15 degrees would be right next to the 50m

[dark sparrow]

but if it was 15 degrees would be right next to the 50m
then you would have a different diagram

[trim laurel]

Yea that’s what I mean

That’s why I think the diagram he made was wrong

But I was wrong too though

[hot flume]

I may have an idea

I used CH as the height of the triangle and now I have 2 rectangle triangles

Therefore I can use trivial trigonometry

But maybe I'm wrong, what do you think?

[trim laurel]

Well what’s ur answer

[hot flume]

Around 78

[trim laurel]

Also

[hot flume]

Seems legit

[trim laurel]

Why did u put 15 on the 30 m side?

Shouldn’t it be on the 50m

[hot flume]

I saw the draw above

So I put it here

[trim laurel]

If the question said 30m at an angle of 15 degrees

Would it be on the 50 m side?

[hot flume]

U mean, the angle ACB would be 15 °?

Or cba?

[trim laurel]

No

Put 15 degrees at the other corner

That’s what I mean

[hot flume]

O okay

[trim laurel]

Yea would that be right

Because the angles have to be on the opposite side

Idk though because ann said ur way is right

[hot flume]

Yeah but maybe my way doesn't answer the right question

Maybe ur right and this is not the right angle...

I don't really know, did a teacher gave it to you?

If so, u could ask him

[upper karma]

what does this sign mean

[dark sparrow]

"straight line" maybe?

[upper karma]

If u have 2 points, connect them with line segment and draw two circles with center one of them radius line segment length it will intersect in 2 points. if u connect those they will intersect line segment. How to prove that that intersection point is the middle of the 2 original points?

[dark sparrow]

so you have two points A and B, and you have two circles with radius AB, one centered at A and the other at B

and they intersect at two other points C and D

[upper karma]

yea

[dark sparrow]

and youre asking how to prove that CD bisects AB

[upper karma]

no

uh maybe

idk what bisects means

intersects at mid

[dark sparrow]

splits into two equal parts

yes

okay

there are different ways to go about this

the simplest would be to just say the whole picture is symmetric about CD or something

[upper karma]

yea it seems to be

how to formalize that tho

[dark sparrow]

mmm

[tiny snow]

@upper karma If you are allowed to use knowledge about a rhombus, the construction in blue is sufficient. If not, you can create two larger triangles (orange) which you can prove are equilateral, from which the symmetry then follows.

[tired knoll]

howw

[dark sparrow]

not enough info lmao

oh wait

AC = 2

misread as OC

[tired knoll]

whats ab thou

ik the traingle is a right triangle

[dark sparrow]

AB is what you need to find lol

what's "it"

[tired knoll]

yepp ToT

[dark sparrow]

OAB is a right triangle thats correct

OAC however is equilateral

all of its sides have length 2

now what do you know about equilateral triangles

(also, never use the word it in math unless you are absolutely 110% certain that you understand exactly what the word refers to)

[tired knoll]

oo

[tired knoll]

lol what does it mean

[dark sparrow]

??

[tired knoll]

nvm

but how do you find ab from that?

[dark sparrow]

now what do you know about equilateral triangles?

[tired knoll]

they have the same angles and sides

[dark sparrow]

how big is each angle in an equilateral triangle?

[tired knoll]

60

degrees

[dark sparrow]

60 degrees, that's right.

[tired knoll]

oooooo

nvm

thank you god

[dark sparrow]

so in particular, angle AOC (which is also known as AOB) is 60 degrees

[tired knoll]

tyty

[upper karma]

what's the difference between barycenter and circumcenter?

i can see that it does 3 90° angles

[dark sparrow]

eh?

afaict barycenter is another name for centroid in the context of triangles

centroid is the intersection of the medians

and the circumcenter is the intersection of the perp bisectors

[upper karma]

oh

why is a circumcenter commonly shown with a circumscribed circle on the triangle

[dark sparrow]

because that's what the circumcenter is

it's the center of the circumscribed circle lol

[upper karma]

oh

[valid magnet]

https://imgur.com/a/LrO7yA6

https://i.imgur.com/LzYo3Dt.png

[median needle]

help me in this one i am feeling stuck tried various ways to approach the question but still stuck

[lost hound]

im taking a trig final on tuesday if anyone has any practice problems that i could use it would be much appreciated

[zenith garnet]

this is just a trapezoid right?

[junior light]

Yes. How would you prove it, though?

[warm locust]

is there a way to construct the height of a triangle with ruler and compass? (there must be i feel like) i want to prove that, given a triangle, you can construct a square with the same area. my approach was to just construct sqrt(1/2gh)., because i know that you can construct the square root of a constructible number. 1/2 is constructible and g is given, so i just need to show that h can be obtained

[tame panther]

Cant you just measure from one point of the triangle, perpendicular to the opposing side to find the height?

[warm locust]

but can i measure perpendicular ?

well wait

no i think you are right. lmao i

m dumb

[tame panther]

Keep one side of the ruler on the top point, and move the other side of the ruler around until you have a 90 degree angle with the bottom of the triangle

[warm locust]

yea, thanks 🙂

or wait no if i keep the needle of the ruler on the topside of triangle then i can measure the distance until i reach the base, but i can't say for sure that it is a right angle. I can just approximate

hmm well i think this was maybe lost in translataion, compass in my language can just be used to connect two points with a straight line

so i can't measure angles with it

[tame panther]

I got compass mixed up actually

I was thinking you could measure angles with it

I'm trying to think how to measure a right angle with your tools

I have an idea but it is a little cumbersome

Put one side of the compass on the top of the triangle, put the other side where you think the perpendicular point is

Draw a small portion with the compass and it should be tangential to the base

That tangent point is perpendicular

[warm locust]

yup i found another way too

[tame panther]

Great. That also works

That way is more accurate too

[warm locust]

i really like these construction problems 😄

pretty elementary but still satisfying if you get it

[tame panther]

Yes I get it. Math can be very satisfying

I recall reading about an ancient greek mathematician who spent lots of time trying to solve geometry problems with only a ruler and compass. Possibly pythagoras?

[warm locust]

i think many greek mathematicians thougth about these constructions. I mean euclid formulated his entire elements in terms of construction with ruler and compass afaik

[tame panther]

It must be Euclid that I am thinking of

[warm locust]

could be ^^

[tawny dragon]

Are you guys familiar on how to find circle diameters. If so can you help me out with this question?

[mellow gate]

Tan can or cannot involve hypotenuse?

If opposite is hypotenuse?

[tiny snow]

The only angle facing the hypotenuse is the right angle.

[tiny snow]

OD is a radius. MOD is a right angled triangle, with the legs OM and MD known, so you can use Pythagoras.

[vapid stag]

Hello, my professor is showing us the ways to find the asymptotes for sin,cos function through using -2pi < x < 2pi. Is this part of the squeeze theorem? I watched a lot of tutorials on trig and they all showed a different way to solve the problem. Which one would you guys recommend?

[humble pulsar]

sine and cosine dont have asymptotes

[vapid stag]

i meant to find the points to graph it over one interval period.

i'm still confused on whether to use it or not.

[humble pulsar]

Ok so if you're picking points to plot the graph.. where does squeeze law come into play?

[vapid stag]

let me see if i can pull up an example he did

this is what he used. which I'm kind of confused on what it is. and wonder why it's so different than how all the youtube tutorial is lecturing a different way to find the solution.

[humble pulsar]

that's not squeeze law

that's just the domain of 1 period

[vapid stag]

understood

i'm kind of lost in what it's supposed to help me find. do you by chance know the purpose of it?

[humble pulsar]

to find the domain of a period

sine is 2pi-periodic so $\sin(\theta)=0 , D={\theta \in\mathbb{R}|0\leq \theta\leq 2\pi}$

[somber coyote]

moshill1

[dapper bronze]

20 and 21 can anyone help?

[trim breach]

MH and MA are radii of circle M. The tangents meet at point T, so plane MATH is a kite.

What immediately comes to mind is drawing segment HA to form right triangles within the kite and solving that way.

[waxen roost]

need help with this

[trim breach]

Were you given the formula to find volume of a cone in your notes?

[waxen roost]

nvm I solved it

I saw a tutorial on youtibe

can someone tell me how to solve this

[ember canopy]

use the formula of volume for cones

where h=6 and r=8

thats the bigger cone

then find the volume of the smaller cone

and subtract it from the big one

[waxen roost]

how do I find the smaller cone volume without a base?

[final pine]

Whole - the other piece

The whole cone is 6 and 8 and the small cone is 3 and 4

[waxen roost]

I hope I did it correct

@final pine

[golden garnet]

Anyone wanna help me solve this? This is an ixl study guide for my upcoming quiz and was stumped on this question. The assignment does not go towards my grade but I would like to know how to do a question like this.

[blissful willow]

its 125 cm squared

[dark sparrow]

@blissful willow do not give out answers here

[cosmic sundial]

I can help! DM

[long garnet]

i'm confused about this review question for our test, can anybody help?

[faint bramble]

such a good question i can't figure this one out

[slim saddle]

the volume of the barrel is 20pi

[slim saddle]

<@&286206848099549185>

[compact jasper]

Sometimes i see this formula. but what is C? 🙂

[silent plank]

circumference of the circle

[trim breach]

Square prism basically means that the two-dimensional square was given depth, which is your height. So two sides are the same by definition of a square, and the problem tells you what the height should be in terms of a variable.

[foggy talon]

Hello can someone tell me how to solve this

[dreamy ridge]

The two expressions are equal

Bc thats the property of rectangles

[valid magnet]

(4c) - 25 = (6c) - 59

((59 - 25) / (6-4)) = 17

we notice that the differnce between 59 and 25 are 2c

2c = 34

59 - 25 = 34

(59 - 25) / 2 = c

@foggy talon

when you look at the 25 and to get to the 59, you add 2c to 4c

[foggy talon]

so 6c?

4c+2c=

6c

Thank you @valid magnet

[valid magnet]

4 * c + 2 * c

not sure if it's equal to 6 * c

4 * 5 + 2 * 5 = 30

6 * 5 = 30

[valid magnet]

(4 * c) * (2 * c) = ?

[quartz cedar]

Would anyone recommend any books that could help you succeed in a trigonometry class?

[valid magnet]

i might be able to find you some free books related to trigonometry

[quartz cedar]

That would be helpful. Thank you.

[tall lava]

Anyone down to teach me some trigonometry during my 4 hour car ride? 😂

[trim breach]

I can’t guarantee you four hours of help, but if you have questions definitely ask them here.

[tall lava]

Driving atm. I’m in 8kb channel, though. I just barely started learning and I don’t want to waste four hours doing nothing but drive. 😂 If you or anyone who wants to talk with a mic can just start from the beginning , that’d be helpful.

[foggy talon]

8c?

[valid magnet]

not 8(c^2) ?

[foggy talon]

Then I was mistaken

[quartz cedar]

soh cah toa = sho cha tao

[simple bear]

How to verify this identity using de Moivre's Theorem?

[zinc cradle]

Hello all. Anyone good with some shapes

[coarse shoal]

could someone help please?

[tender lion]

i need to place these in a trigonometry circle

[trim breach]

I would recommend using trigonometric ratios to find x. The first triangle you have to actually do the algebra, but the second triangle can be solved easily if you recognize a pattern.

[upper karma]

For finding a length of an arc in geometry can it end as a decimal pi or just a decimal. Or does it have to be a whole number pi?

[rancid sigil]

@abstract barn put it here

[abstract barn]

k

[rancid sigil]

<@&286206848099549185>

Please help him

[steady shoal]

Why aren’t you helping him?

[abstract barn]

They dont understand geometry

SO they recommended I join this server

[steady shoal]

How old are you ?

[abstract barn]

16

[rancid sigil]

I aint smart, I don't do the math for this server

[steady shoal]

Do you know what the interior angles of a rectangle are ?

[abstract barn]

Yes

Ive repeateadly tried to solve, and I keep getting 14

and it just doesnt add abck up

*back

[steady shoal]

Can you share what you did

It’ll be easier if you use pen paper rather type

[abstract barn]

SO basically I used the given numbers and variables (3x-30) (2x+10) and multiplied by 4
Then I did 360-120(sum of the -20)
240+40(sum of the +10)
280/20=14

then I plugged 14 in and it just wouldnt add back up to 90

[steady shoal]

Why did you multiply (3x-20) and (2x+10)

[abstract barn]

didnt multiply them by each other but by 4

[steady shoal]

Why

[abstract barn]

I guess I tried to take a shortcut and screwed up tbh

[steady shoal]

No you possibly took the longest path

[abstract barn]

oh

[steady shoal]

What’s the interior angle of a rectangle

[abstract barn]

360

in total

[steady shoal]

And at the corners ?

[abstract barn]

90\

[steady shoal]

Use that instead

[abstract barn]

what do you mean by that

[steady shoal]

You’ve been given an angle which must equal 90

[abstract barn]

Im not sure exactly how to set it up

[steady shoal]

Angle C is 90

You’ve been given angle c in terms of x

(3x-30)+(2x+10)=90

[abstract barn]

ok so its 22

Thank you so much

[grim beacon]

nice

[upper karma]

This is simply melanous theorem , do u know it, or do i show u a proof?

There r plently of ways of doing this tho

[waxen glen]

idk what is it

let me research a bit please

$$\frac{AD}{BD}\frac{DF}{FE}\frac{BF}{FC}=1$$

[somber coyote]

Hyl1s

[waxen glen]

@upper karma like this?

[upper karma]

Gimme a second pls

Lemme work this correctly

[upper karma]

Ok a good way to solve this is areas, join be and af

Then let area of the big triangle abe be A

then area of adf = 5/9 area of ade = 5/9 ( 2/5A)

= 2/9 A

similarly area of bdf is A/3, and of bfe = 4A/15

Is this good?

[rich wolf]

,w 3/323 - 1/121

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by AMD#4434

[upper karma]

how do I do this

[grim schooner]

use the two-tangent theorem

[grim schooner]

two segments tangent to a circle and that meet at an endpoint are both equal

so 28=3x+4

24=3x

x=8

[waxen glen]

i didint understand it

is there any video explains this concept

@tiny snow

[tiny snow]

Not that I am aware of... It's only arithmetics and similar triangles, with one additional parallel line drawn.

[topaz nimbus]

do it by the similarly criteria

like BPT

basic proportionality theorem

[upper karma]

Guys.

It's simple:

I've found these three value scopes.

Should be there more?

Should be there the fourth one?

[sturdy iris]

I think you're fine

[zinc cradle]

If row 1 is length of outside line, then row 2 is the next inner line, what is the ratio of row1/row2 for an octagon

The numbers on the sketch are from ruler observation, not maths

[upper karma]

First to find the length of an n sided polygon in circle

Can you do that?

[zinc cradle]

Correct, if diameter is known, how to calculate the length of each side of polygon

I don't know that equation

[upper karma]

Ok , look at the triangle formed by the centre of circle and any two adjacent points

And split it into 2 right triangle

[zinc cradle]

It is 45degree of arc and the total diameter we will say is 100

[upper karma]

45 degree for any n sided polygon?

[zinc cradle]

In the picture, the first inscribed octagon each member will touch every 45 deg on the circle

This will be at full diameter of 100

[upper karma]

Yes, for octagon n=8 and arc=45degree
For any n, the arc angle will be total angle÷number of angles=360/n

K?

[zinc cradle]

Yes that is easy, but what is equation for side length

[upper karma]

Ye

Look at the half of that

Makes a right triangle

[zinc cradle]

Ahhh.. I think I got it, I look at it as a circle, not a square.

If I imagine the octagon in a square it's much easier

[upper karma]

Okay, that does work for this case, tho equation for length for any n side would be a=dsin(180/n)

d is diameter

Ok?

[trim laurel]

how do i get standard angle from referenec angle

my ref angle is 238, what is it in standard position? please help

[main lintel]

Not sure what is standard angle nor referenc angle? Any definitions in your book?

I think both of these mean the same, and equal 238. But am not sure

[main lintel]

Surface area is the sum of areas of all sides, right?

[sick torrent]

yes

[main lintel]

So what is the problem?

[sick torrent]

I dont know the problem

[main lintel]

By problem I mean, why cant you solve it?

What causes you the trouble

[sick torrent]

yes how to solve it

[main lintel]

You know what is surface area? So where did you get stuck?

[sick torrent]

do i just multiply

all

[main lintel]

No, you add (sum) all surfaces.

[main lintel]

we said this

[sick torrent]

what is the equation?

[main lintel]

You just need to sum each and every side on the image.

You said this is clear to you

[main lintel]

.

[sick torrent]

so 16 + 12 + 9 + 20 +

[sick torrent]

.

[main lintel]

There are 5 sides

bottom, top, and 3 "lateral"

Wait, you are confused a lot.

One thing is "side" the other is "edge"

Edges are the lines. Sides are surfaces.

---

This is one side:

This is not a side but an edge:

[sick torrent]

oh

[main lintel]

There are 5 sides. And there are 9 edges.

For Surface Area you need to sum the areas of all 5 sides (surfaces)

[sick torrent]

624

[main lintel]

How you got that number?

Bottom is: 16*12/2 = 96, top is the same.

[sick torrent]

than midle is 16 times 9

[main lintel]

"lateral" ones are: 16*9, yes.

Then 12*9 and 20*9

[sick torrent]

ok

[main lintel]

Yea, 624.

[trim laurel]

someone help pls

is this 344?

[main lintel]

How you got that?

Yea, I also got 344

[trim laurel]

i used tan-1(2/7) and got 16

rounded to 16

then it said larger than 180 but smaller than 360 so I used quad 4

360-16= 344

[clever ginkgo]

yo

<@&286206848099549185>

I know the ansawer is 6pi

but idk how

they got it

I dont understand how to do this

[latent ore]

Can you see that the shaded area is 3/8 of the circles area?

[clever ginkgo]

yea

[latent ore]

Right well

if the diameter is 8? What's the area of the whole circle?

$A=r^2\pi$

[somber coyote]

.itsjustnai

[clever ginkgo]

so is it

16pi?

[latent ore]

yep

so 3/8 of 16pi is... ?

[clever ginkgo]

hmm

6

:0

[latent ore]

6pi

[clever ginkgo]

Thank youSO MUCH!

[latent ore]

np

[clever ginkgo]

@latent ore wait but how did you get 3/8

like did you just wing it

I just guessed

[latent ore]

I winged it.

@clever ginkgo

[clever ginkgo]

OOOOOOOOOOOO

[latent ore]

I mean

You can also look at it this way

it's obvious the top part is 1/2 of the circle area

and the 45 degrees is 45/360 the circle area, which is 1/8

so 1/2+1/8=4/8+1/8=5/8

so the white area is 5/8

and the shaded one is 1-5/8 which is 3/8

if you're wondering why it's obvious the top part 1/2 of the circle area it's because RP is the diameter

@clever ginkgo

[clever ginkgo]

thank you

my hero

[heavy tinsel]

can someone help me with geometry

[humble pulsar]

Have you tried asking the question? That's usually a good place to start

[heavy tinsel]

Im not sure what the formulas are

[trim breach]

Are you confused about the relationship between scale factor and surface area/volume?

[heavy tinsel]

Im confused on what the formulas are for each one

[trim breach]

It is not necessarily a formula…it is using ratios between the two prisms.

The scale factor is simply the ratio of

[humble pulsar]

\frac{}{}

[trim breach]

^

,,\frac{x}{y}

[somber coyote]

Lidoh

[trim breach]

The ratio between the two prisms surface area is

,,\frac{x^2}{y^2}

[somber coyote]

Lidoh

[trim breach]

You can set up a proportion since you know at least three of these values since they are given in the problem.

[pure cape]

is there a method to solving open form trigonometric equation? For example like x=sin(x)

i have been thinking and it seems like it could only be solved graphically or numerically

[dark sparrow]

open form?

[pure cape]

im not sure how to say it

but its something like x=sin(x)

[dark sparrow]

transcendental equations

these are hard to solve exactly in most cases

[pure cape]

because im wanting to solve an equation that is almost like that over the variable, but since i cant do it algebraically, ig im stuck

more specfically Asin(bx+c)=x

[dark sparrow]

hm

is there any reason why you need the exact values of the solutions

[pure cape]

so basically thats part of a bigger problem. I need to find functions whose inverses are tangent to the original functions at point (a,b) on the line y=x

basically like e^(x-1) and ln(x) +1 which is the only exponential function i found

so i found the best way to do this is that since the inverse is symmetric to the original function thru y=x, we just need to find a function that is tangent to y=x

so i split it into the basic types of functions, (linear functions are off the table), power functions, which does not exist any that fits into the description, exponential functions (which i did and found only one function which is f(x) = e^(x-1) )

now im working on trigonometric functions which is really scary, i tried playing around, and found that sinx and arcsinx actually works

but the thing is, those are not the only one i think.

so which is why i need to work on it, i start off with the standard form f(x)=Asin(Bx+C), find the tangent point from the equation f(x)=x, and then substitute that x into f'(x)=1 to figure out the coefficients. im stuck at the second step

hmm i guess i think we can assume from the fact that sinx is tangent to x at (0,0). And it seems like only sin(x-a-2kpi)+a where a is in R, k is in N fits into the description

hmm

[hallow wedge]

please help

[tiny snow]

@hallow wedge There is something wrong with the terminology in the question. Altitudes intersect in the orthocenter. The medians of the sides intersect in the centroid.

[upper karma]

Pretty sure this falls under geometry

How do I solve for side length with only the area of the figure

[trim breach]

Is it a regular polygon?

[upper karma]

Yes

[trim breach]

You can think of the polygon in right triangles by drawing apothems and radii.

[upper karma]

That’s the thing it only gives me the area of the complete hexagon, not the apothem, side lengths, radius, or perimeter

I know the central angle is 60, and divided into a right triangle to come to 30 degrees but I have no further information to set up 306090

Each triangle has an area of 81 square root 3

[trim breach]

You have to do a bit of algebra. Might want to take a look at this video since they explain it better than I feel I could.

https://youtu.be/CG8ipDUeyRs

[upper karma]

Okay, thanks!

[pearl finch]

does anyone know how to solve this problem

[dark sparrow]

pretty sure a lot of people do

what's giving you trouble here? @pearl finch

[pearl finch]

bro idk

i just have no idea how

to solve this problem

like ive never seen problems like these before

[dark sparrow]

don't call me bro please

okay, let's see here.

[pearl finch]

ok just please help with this problem

some of these questions take my teacher so long to solve

these are the most difficult problems ive seen all year

[dark sparrow]

got a phone call sorry

[untold cosmos]

can someone help me on my geomtry test?

[dark sparrow]

@pearl finch do you know how to find volumes of cylinders and cones?

[pearl finch]

yes

[dark sparrow]

okay

so let's say you already knew the volumes of both of your shapes. would you be able to find how many times the cone's volume fits in the cylinder's volume?

[pearl finch]

i dont think so

[dark sparrow]

okay let's try a simpler problem. set that one side for a while

let's say you have a cup that holds 0.4L of water and a big jug whose volume is 6L

would you be able to find how many cups it'd take to fill the jug?

[pearl finch]

6/0.4?

[dark sparrow]

so you do know

[pearl finch]

oh wow

[dark sparrow]

same concept here

[pearl finch]

question was much less difficult than i imagined

hold on a minute

[dark sparrow]

except your cup is conical and your jug is cylindrical :p

but that only affects the calculations of their volumes

[pearl finch]

@dark sparrow thanks

is it normal to forget most of the math you've done in the school year if you havent practiced them in a while

[dark sparrow]

not really something specific to math

lack of practice in anything will make you forget it eventually

[pearl finch]

i feel so dumb because i cant remember enough to solve most of the problems :/

[dark sparrow]

sometimes you gotta step back and think

like. here the key was to understand that the volumes of your cone and cylinder were the necessary stepping stone between the givens and the answer

[pearl finch]

i removed all the old papers for the class because there was just nowhere to put them

and they were very messy

[dark sparrow]

that's unfortunate

[pearl finch]

some of the problems took my teacher 15 minutes to do lol

so im just wondering how i can complete it

is this AAS

[dark sparrow]

the distinction between AAS and ASA is nearly pointless as far as triangle congruence tests go but yeah the intended answer is AAS it looks like

[pearl finch]

how about this one

how is 67 wrong

[dark sparrow]

one moment

how'd you get 67?

(180-46)/2?

if so, that gives you angle ADB (or ABD, they are the same)
so it's simply not what was asked for

[pearl finch]

i assumed BAD was equal to ADC

[dark sparrow]

yeah well

that assumption is garbo

[gusty estuary]

Do you know the correct answer?

[pearl finch]

no

[gusty estuary]

okay, this is how i worked it out:

[pearl finch]

ok i see a way how to get the answer

[gusty estuary]

First you take 46 and subtract 180 by it, because its a straight light (AD).

Once you get that answer, divide it by 2 to get the measure of angle B and D.

Once you have the measure of angle D, subtract that number from 180, which is going to equal the other side of D (ADC).

Then you subtract the new number from 180, and divide the new number by two to get your final answer

[pearl finch]

i got 33.5

[gusty estuary]

Yep

[pearl finch]

thanks

[gusty estuary]

Could anyone help me out with this?

i read the question wrong. i might need help with a diff one

[shut venture]

@gusty estuary do you mind tell me what book is that?

[gusty estuary]

@shut venture but it's a question made by my teacher

I'm confused on this one

[shut venture]

Ah ok

The questions seem cool.

[gusty estuary]

yeah he makes storylines and stuff like that

[shut venture]

I need to learn how to do those 😄

[gusty estuary]

haha

[shut venture]

It seems motivating to learn like that

[silent plank]

draw circles of radius 5 centred at those points

[shut venture]

@silent plank why radius 5?

[silent plank]

because the problem states that the killer is within 5 miles of them

[shut venture]

oh right i didnt see that

[weak lichen]

can you help me solve this one?

[pearl finch]

how do i solve this problem

i got it right but idk how to solve that shit

[upper karma]

how'd you get it right

but not know how to solve it

[upper karma]

Know what trig ratio applies here?

[pearl finch]

@upper karma the teacher gave out the answer

[upper karma]

have they not given you the process

bruh moment

[pearl finch]

well they did but it got so confusing

i need it to be re-explained

I know RP is the altitude

but i forgot how to use it to solve for the rest of the triangle

[upper karma]

So PT is 12 less than SP right? just clarifying I know it says it there

[pearl finch]

yes

[upper karma]

alright

[shut venture]

Why not using the straight rectangle metric relationships?

[pearl finch]

whats that

[shut venture]

trying to find in english

https://i.ytimg.com/vi/pkCCAW1QrGw/mqdefault.jpg

[pearl finch]

not more formulas oh my god

[upper karma]

Better image

[shut venture]

thanks

[upper karma]

never heard of these so they can explain

[shut venture]

im not entirely sure its the solution

but ill have to write it

@pearl finch what does it mean

is it a 12?

[pearl finch]

what

bro my teacher never taught us this

plus my brain is shit at memorizing formulas

thats how the brain works iirc

[shut venture]

@frank pasturewa sin(25)=x/7

[upper karma]

ok in this case

e is PT

So just find every equations that includes e

and kinda mess around with em

[frank pasture]

?

[shut venture]

@pearl finch you dont need the relationships

[upper karma]

they mean't to ping @weak lichen

[shut venture]

sorry @frank pasture not that used to discord.

@pearl finch you can do triangle similitarity

[upper karma]

according to this

apply it here

[fluid wasp]

Can you help me find formula calculation ?

[shut venture]

Sorry i cant understand the image

[pearl finch]

Bro continue on that

Please

[shut venture]

what do you want to know

[pearl finch]

I still dont know how to solve this problem

[fluid wasp]

Better ?

i wanna know how i can get X if i know r1, r2, a, b

[shut venture]

that seems like sinusoidal function

@pearl finch use triangle similarity

[pearl finch]

Bro theres an altitude and nothing else, what can i do sir

[shut venture]

lets call SP as x so PT = x-12

so 8/x=x-12/8

[pearl finch]

Why not 8/x-12?

[shut venture]

you can do either way

you just have to use the same order

[pearl finch]

U wont get the same answer if u flip the numbers around

[shut venture]

how come?

[pearl finch]

How would the algebra be the same if 8/x is equal to both x-12/8 to 8/x-12

[shut venture]

you have the flip

the other side too

x/8=8/x-12

[pearl finch]

Oh ok

[pearl finch]

What did you get for this

[shut venture]

the dimensions given in the problem

[pearl finch]

No i meant whats your answer for it

[shut venture]

4

x=16

[pearl finch]

Oh

Nvm

Bruh why cant i get that

Wtf

[shut venture]

what do you mean

[pearl finch]

Do you need quadratic equation

can you show me how you got 4

Can you picture and show me

Yes you need quadratic equation

One of the answers is 16

-12 = 4

[shut venture]

its 16-12=4

the other answer is x=-4

but you can have it negative

[upper karma]

where are the four triangles? I mean I do see only two, one is the tiny
one on the left, and the other one, is the one that is subdivided, while
the rest are rectangles

https://imgur.com/Il97R3h <@&286206848099549185>

[onyx cloud]

those trapezoids to the right of the smallest triangle on the left make up bigger triangles

[upper karma]

I'm sorry I still do see only two and not four

[onyx cloud]

do you see it?

[upper karma]

oh, I'm an idiot, thanks

[agile furnace]

Name the single translation vector that can replace the composition of these three translation vectors: <2, 3>, then <–5, 7>, then <13, 0>.

<@&286206848099549185>

[upper merlin]

what happens when you translate using 2 vectors

the resultant translation vector is just the sum of the two vectors

for example going a miles north and b miles east, then going c miles north and d miles east, is the same as going a+b miles north and c+d miles east

so just add the 3 vectors

[woeful shuttle]

could someone help me with a geometry practice because I am struggling because I suck at circles

[upper karma]

post it

[hardy kindle]

hi can someone help me with my hw?
no. 19 and 20 only :D ty

[upper karma]

So for a, what do you know about angles in a straight line

and what do you know what alternate exterior angles

q. 19

[hardy kindle]

all i know is that its a pair of angles on the outer side of each of those two lines

[upper karma]

ok a pair of what angles

[hardy kindle]

transversal

uh

[upper karma]

yeah ok go on

what about them

[upper karma]

19. a=115 b=115 20)a=65 b=65

[dark sparrow]

@upper karma do not give out answers

[upper karma]

?

[dark sparrow]

!

[upper karma]

wy

[dark sparrow]

#❓how-to-get-help

[slate lodge]

https://cdn.discordapp.com/attachments/269573202018041856/842317026399879178/image0.jpg

[upper karma]

Which ones?

[slate lodge]

sorry nvm i watched a youtube video and got it

[small flume]

what did you get for the first one?

i was thinking about this theorem

[slate lodge]

i did 2x-4=x+1

[slate lodge]

^

[tender storm]

wait is it possible for u to help me in #help-8 ?

i have similar problem with circles

[slate lodge]

yea ill try

[tender storm]

thank you

[gray minnow]

pls explain this relation to me

[dark sparrow]

what about it do you need explained?