geometry-and-trigonometry

[tribal sequoia]

which question

[frail ridge]

Hello geometry

I have a pretty difficult problem

So I have a figure that shows three parallel lines cut by a traversal

All the angles have linear binomial measures

To get the value of the variable, I need to find a way to make angle 3 equal to angle 2

Since angle 4 has y as its variable unlike angle 2 and 3 where its x

[smoky jetty]

the only way I could think of is if angle 3, 2, and 4 and their respective supplementary angles (summing up to 180 degrees) is bisected by the transversal of the 3 parallel lines

giving you x=y=180/2 ??

[upper karma]

Help

[exotic yarrow]

Adjacent angles in a parallelogram are supplementary, so ||x+(x+30)=180||

[upper karma]

So the answer would be 150.-.

[exotic yarrow]

That's the value of 2x, not x.

[upper karma]

75

That makes a lot of since thx for helping

[exotic yarrow]

👍

[fiery oriole]

a = c and d = b based on it being a parallelogram

oh hold on it was answered nvm

[upper karma]

Thx for the other hint though

[fiery summit]

helo

i need help with my geometry homework pslspsppslspsls

[trim dirge]

for this problem would I have to divide 630 by tan(50) or multiply 630 by tan(50)

[floral shale]

Drawing the scenario will help

[upper karma]

Hi guys!! I need some help with my geometry review packet, right now we are on a unit about area, volume, and surface area! If someone could take a look at these problems and show me how to do it step by step that would be super helpful! No worries if your not up to doing the entire packet, anything would be great! Some of the problems have some of my work but I don't think I did any of it right. Thank you!

[storm portal]

I’d start by asking and sending 1 Q at a time

[upper karma]

Sorry about that! These are the two questions I’m having the most difficulty with

[storm portal]

Circle question first

You have the incorrect center

But radius is good

[upper karma]

-5, -3

• is negative points, - is positive points

Second one, use 180 and subtract whatever arcs you need to find

KJL is major arc

Find angle G

360- angle G = KJL

180 - 72 - 30 = JK

[upper karma]

27 and 29

[tribal sequoia]

Use trig ratios to solve

[tender urchin]

uhm

[pliant nest]

Hey, how does one calculate the Volume of a Triangular Prism when the base has no defined height?

How’d I be able to calculate the area of the base (triangle) without knowing the height, only the length of the three sides and that it has a 90 degree angle?

Like in this example below.

Like damn, I know I shouldn’t be struggling since calculating Volume is so easy and I have 0 problems whatsoever with the other problems

[dark sparrow]

the base is a right triangle tho

so you can take its legs as the base and height...

or yknow

see the right triangle as half (in the literal sense) of a rectangle @pliant nest

[upper karma]

Help

[plain burrow]

two add up to 180

[upper karma]

89

Welp I had the wrong answer

[upper karma]

I suck at geometry

[upper karma]

test

[blazing basalt]

does anyone know a website i can use to plot a bunch of circles of different sizes?

i have the x, y and radius values

[upper karma]

No 💀💀

[blazing basalt]

😭

[pliant nest]

Okay, so basically:

B = 4 x 3
B = 12 mi squared

V = 12 x 5
V = 60 mi cubed

or half of that which would be 30 mi cubed?

But uh…the answer is somehow supposed to be 10 mi cubed.

[upper karma]

Thx

[dark sparrow]

the volume of a pyramid is one-third of the product of its base area and height.

[stuck oak]

Why is the restriction for arctan or any inverse trig function based around the origin or 0

For example, arctan domain is from -pi/2 to pi/2

Why don’t we go from 0 to pi instead

[silent plank]

For example, arctan domain is from -pi/2 to pi/2
did you mean range?

[stuck oak]

I do realize that the the horizontal line y=0 would pass 2 points but still

[stuck oak]

Umm

I thought it’s now domain

Ok nvm

I meant range

I mistyped that

[silent plank]

convenience, continuity

[stuck oak]

I sorta understand convenience but continuity would still work in my case no? 0<y<pi

[silent plank]

tan(pi/2) is undefined

if you tried going from 0 to pi instead of
-pi/2 to pi/2

[tardy yacht]

i dont understand solving for trigonometric equations, especially in the negative domain. like why is there a billion different ways to figure out different questions. how am i supposed to get it right each time. it just doesnt feel like theres a single process and when i figure out a process for one question its all stuffed up for another question

[silent plank]

show specific examples

there's no one size fits all formula/process

understand the basic principles and apply those to your problems

[silent plank]

show your attempt

[tardy yacht]

My working is very messy and all over the place and some incompleted, I apologise for the inconsistency. Many attempts have been put in the bin. The last one is a successful attempt. I’ve been focusing on tantheta=-1

[silent plank]

which question are you focusing on atm

[tardy yacht]

g) tantheta=-1

[silent plank]

can you post the attempt for that only separately

[tardy yacht]

the attempt has been posted in the first picture for tanx=-1.

[silent plank]

not sure what you're doing with neg domain

[tardy yacht]

putting negatives in front of it

[silent plank]

the angles for every quadrant you set up should pretty much be the same

[tardy yacht]

I don't understand

[silent plank]

with what you had with ASTC, with A in the top right

and none of this stuff

[tardy yacht]

tan is negative in the second and fourth quad

[silent plank]

yes

[tardy yacht]

and then when you want to find a negative angle you put a negative in front of each thingy and reverse the

i dont even understand what im saying

places where tan is negative

[silent plank]

theta = arctan(-1) = -arctan(1) = -45°

[tardy yacht]

correct

that's straightforward

[silent plank]

which would be represented

in that location

[tardy yacht]

yes

315°

[silent plank]

well arctan(-1) gets you -45° directly already

[tardy yacht]

wait i remember my teacher said to just change it to positive because negatives are too difficult or annoying to deal with

[silent plank]

and to get other solutions you can add integer multiples of 180°

[tardy yacht]

I get that

But how will I know that's all the possible answers?

and I tried that on another question with a negative angle and it was wrong. I think it was 4e, tant=-1/sqrt3 which is -30

and that's not true I just contradicted myself.

I'm not even sure what I'm doing wrong

[silent plank]

-30
yeh, that's one of the solutions,
and +180° to get the other

[tardy yacht]

The pieces seem to be coming together but I'm sure it will come crashing down.

Like for example this, it was provided by my teacher in an example but I don't understand it

[silent plank]

what about that don't you understand

[tardy yacht]

the negative angles

[silent plank]

think of it more as the rotation from the positive x-axis

i'll make some adjusments

[tardy yacht]

the 4th quadrant is initially 360-t but I would think to put a negative in front to make it -(360-t) = -360+t?

[tardy yacht]

Thank you

[pliant nest]

Huh? Oh shit, it’s a Triangular Pyramid 😹, I forgot the fact it’s 1/3rd of a Triangular Prism

Like how a Cone is 1/3rd of a Cylinder

My bad, thanks G

[silent plank]

the 4th quadrant is initially 360-t
all the signs are already accounted for

[pliant nest]

Wdym? I fked it up lol

[silent plank]

if you're using a reference/related acute angle,
you apply the - to that to get a position in Q4

e.g. with
tan(t) = -1
the reference angle would be arctan|-1| = arctan(1) = 45°
and you apply the - to the 45°, to get the result of -45° for Q4

alternatively you could subtract 45° from 360° to get 315°
which is co-terminal with -45°

however 315° isn't in the desired interval

[upper karma]

how do i remember all trig functions behaviours?

[tardy yacht]

wait back up on the absolute value with arctan why are you doing that

[silent plank]

for the reference angle / related acute angle you take the magnitude to work with positive values as per your teacher's recommendation

and then apply signs and/or quadrant shift afterwards
depending on what you started with

[tardy yacht]

but wouldn't that get confusing

[silent plank]

not really

[pliant nest]

So essentially this is:

B = (4)(3)/2
B = 12/2
B = 6 mi squared.

V = (6)(5)/3
V = 30/3
V = 10 mi cubed.

Got it, thank’s Ann. 💗

I can’t believe how I didn’t realize that it’s a Triangular Pyramid and not Triangular Prism 😅

[tardy yacht]

i follow a little bit but not enough to put it into practice

[silent plank]

do you have any issue for non-negative values?

[urban raft]

can i get a function to plot a spring like structure in three dimensional space ,

[tardy yacht]

no

[silent plank]

the k above is the reference / related acute angle,
i.e. the acute angle between the terminal side and the x-axis

inverse-trig-function |your value|
gives you that angle

and getting the angles in your desired quadrant is pretty much
applying angle addition/subtract
and 360° in a full revolution

[upper karma]

how do I be doing these ones?

i can't mathing

[junior pine]

Giys

I need help😭

I actually am so lost haha

[grave pond]

You will continue to be lost unless you actually explain what you're lost with.

[junior pine]

OH SORRY

HAHA

ok so basically

I think I understand trig but thing is

How Dyu know what t label it

Yk how ders sin cos tan

Like da frik is dis wan

Cus icl all of them look disame

Da only topics I don’t get is trig n geo🥲

[vocal cloud]

do you need to find theta?

[junior pine]

Huh

Theta?

[vocal cloud]

this guy

[vocal cloud]

its the angle you have marked witha 0 with a dash through it

[junior pine]

Oh hahyah

[vocal cloud]

its called theta

[junior pine]

I need t find day

Ohh

Oki

[vocal cloud]

you need to find that correct?

[junior pine]

Idk what t label it cus all looks disame lel

Yup I do

[vocal cloud]

Ok

you have all 3 lengths of the sides correct?

[junior pine]

Like u can’t tell me dis looks all disame

[vocal cloud]

it does but you get used to it

[junior pine]

🥲

[vocal cloud]

Do you have the length of the 3 sides?

[junior pine]

I’m abt t have a breakdown over maths

Uh

Do I

UHM

I think

[vocal cloud]

the hypotenuse, opposite and adjacent

[junior pine]

OH

YAHYAH

I do

[vocal cloud]

ok good

Now

lets use

trigonometry to find out angle

(il call theta x cos idk how to put it on keyboard)

[junior pine]

Oki

[vocal cloud]

We have all 3 so we can use any of the sin,cos or tan

I will use tan

TAN(X)= Opposite/adjacent

correct?

[junior pine]

How Dyu even know t use tan

[vocal cloud]

You can use any of them in this case

[vocal cloud]

Look at the information theyve given you

[junior pine]

Ohhh

Aha 🥲

[vocal cloud]

Do you have the length of the sides of the hypotenuse, opposite or adjacent

[junior pine]

Idk do I💀

[vocal cloud]

do you have a number labeled next to them? or in the book does it?

[junior pine]

I do

[vocal cloud]

Then you do

[junior pine]

Oooooh

Oki

[vocal cloud]

and now

angle x

what is opposite it?

[junior pine]

Angle x

Eh uh

Nah nah my poor brain

[junior pine]

Wait

[vocal cloud]

just reposting

[junior pine]

Ph yes lol thanks uh

Idk💀

[vocal cloud]

ok

imagine you are standing at

angle x

Ok?

[junior pine]

Yeah

Ok

[vocal cloud]

and lets say you are under a building

and its the same shape as that triangle the area you are standing in

What are you seeing is opposite or facing you

[junior pine]

Lord my brain ok

Let me process dis

[vocal cloud]

i am horribel at explaining lol

[junior pine]

LMAO NO

I JUST DONT VET TRIG DATS ALL

[vocal cloud]

ah i see

[junior pine]

see I’m in foundation level for a reason💀

Foundation level is baby maths

Idk why they have trig in der

[vocal cloud]

foundation level?

[junior pine]

Yeah

[vocal cloud]

what country are you in

[junior pine]

Ireland

[vocal cloud]

YOOOOOOOOO

SAME

Dublin?

anyway how tf do i explain this

[junior pine]

WAIT

DUBLIN

WHAUHA

NAHHH

WHAT A COICODENE

[vocal cloud]

yeah ikr

[junior pine]

nah my English bro

I’m doing leaving cert dat why

[vocal cloud]

i got jc in a5 days

[junior pine]

And I’m cramming

BRO GOT JC

[vocal cloud]

LC?!!?!?

LC?!?!?!

[junior pine]

UR LUCKY

FFS

[vocal cloud]

MY GUY

[junior pine]

😭😭

[vocal cloud]

TRIG IS SO EASY

[junior pine]

LMAO

NAH SHUTUP MAN

[vocal cloud]

alr

[junior pine]

MATHS IS HARD FOR MY NON EXISTENT BRAIN

[vocal cloud]

fair enough i understand that

[grave pond]

Indoor voices, please.

[vocal cloud]

ok

sorry

[junior pine]

Damn I was gonna send a request t u

Lord maths is making me cry

[vocal cloud]

idk how to explain this

[junior pine]

See dis is why I wud never graduate as a maths degree

[vocal cloud]

what side are you looking at

from angle x

[junior pine]

Like

Wait

[vocal cloud]

dont think its trig

just imagine you are standing somewher

[junior pine]

Der

[vocal cloud]

yes

[junior pine]

Now what am I gonna look at here💀

[vocal cloud]

what side are you looking at

[junior pine]

Like where I’m standing at x or?

[vocal cloud]

imagine ur under a building

at x

what do you see

[junior pine]

https://tenor.com/view/crying-meme-black-guy-cries-sad-man-thank-god-for-my-reefer-hood-news-gif-24902056

[vocal cloud]

is it the opposite

the adjacent

[junior pine]

Ehm

[vocal cloud]

Is it opposite yours eye or is it next to it?

[junior pine]

Yes…no?…idk🥲

[vocal cloud]

oh dear

[junior pine]

Allow man

[vocal cloud]

i am

[junior pine]

I’m in foundation for a reason

Ur in higher level aren’t u

[gaunt lotus]

Hint it starts with o

[vocal cloud]

yep for jc

[junior pine]

Ffs higher level ppl

[junior pine]

Huh

[vocal cloud]

meet my lord and saviour

https://www.youtube.com/watch?v=PUB0TaZ7bhA&ab_channel=TheOrganicChemistryTutor

[junior pine]

LMAO

[vocal cloud]

the organic chemistry tutor

[junior pine]

😭😭

[vocal cloud]

he saved me

[junior pine]

Imagine I’d die laughing if we’re neighbours

[vocal cloud]

from going to ordinary

[vocal cloud]

hehe

[junior pine]

Oh lord

Dublin 6💀

[vocal cloud]

ah

city centre?

[junior pine]

Yeah sure

[vocal cloud]

try understand the video i sent

[junior pine]

Okie

[vocal cloud]

good luck for LC mate

[junior pine]

I’m dying I’m telling u

I’m cramming

[vocal cloud]

do you do geo or history for lc

[junior pine]

History

[vocal cloud]

is it hard?

[junior pine]

Yes💀

[vocal cloud]

ok

[junior pine]

A lot of ppl regret choosing it

[vocal cloud]

i gotta do my irish but cya

[junior pine]

LMAO

[vocal cloud]

the video imo explains it really well btw

its helped alot of my friends get 90s in their trig test

watch it .5 speed and stuff

[smoky jetty]

just so u know, size does not matter for trig ratios of right triangle as the scaled-down or up version would mean that u could simplify the terms in the ratio by their gcf and giving u the same result compared to the original size.

[junior pine]

Ik that

I have it in my notes

☺️

[smoky jetty]

thats neat

are u clear on the ratios part now?

[smoky jetty]

get the ratio equal to whatever function that has the segment NO in it

could be the ratio of secant 75 deg

[junior pine]

Who I talking

[upper karma]

what you guys think of these flyers here for my trig article?

[light aurora]

<@&268886789983436800>

[slow spruce]

What are you trying to show? Formatting needs work; text overlapping your images, tl;dr, font too small to read, etc.

[upper karma]

then read this

[upper karma]

this is a better way to do trigonometry in every case basically and it simplifies the effort needed

im not much of a programmer but this is my preemptive benchmark results on a CPU by google bard in multistep vectors and gfx

[slow spruce]

Do you know about latex/ beamer? The ^-1 everywhere are an eyesore

[upper karma]

ya sorry bud its done in word with latex but i didnt realize i coulda superscripted that. i will fix it next time i update it i hope it doesn't make it impossible to understand lmao

[slow spruce]

No it doesn't

[upper karma]

i swear after you study it all a while it makes yr mind feel more efficient like you just took a selenium pills or smth; like youll look like my pfp lmao

[slow spruce]

Is there a reason you don't use radians?

[upper karma]

ehhhmm... well actually the calculation comes in radians, did i neglect to put theta in deg = to the radians? i should thx

right yea i did... well i figured it doesn't quite hurt since deg is a radians alternative and the formula to convert units is elsewhere. like to say 3kg != about 6lb is kinda wrong anyways, will fix next edition im savin this chat

ive been working on this for over two years basically, since 2019 and only started with publishable material like 2 years ago and overdosed from frustration once or twice even flunked a semester; but it turns out this is probably 6x better for a physics or geo student or computer for trig and is probably the best way as ive explored other theorems for a long time

but i found the formula myself without knowing arcsin and while the proof to derive it is only a few major steps and not common knowledge to most undergrads; or even obvious to my other profs it is about as diff as the diff series for pi credited to diff mathematicians like the binomial pair series by ramanujan or leibnitz

so figuratively plus the benefits its publicationworthy

[flint steeple]

question about this

i have issues with estimating angles based on images

how do i do this? (self-learning because im bored)

[exotic yarrow]

It might to help to convert to degrees

1/2 of a radian is quite a lot of wiggle room

You can ||assume it's a straight angle and still be within the margin of error||

[flint steeple]

oh interesting, ive been using a method where estimate using radians but i keep getting the answer wrong

[exotic yarrow]

tbf radians is probably new to you, it takes time to get used to thinking in radians intuitively

but once you do, you never voluntarily turn back to degrees lol

[upper karma]

I found the trick to all these kinda questions, it's all about equivalent equations via multplication basically. https://thirdspacelearning.com/wp-content/uploads/2022/04/speed-distance-time-image-2.png (example of what i mean from basic physics)

just remember these concepts and it's free lol

Like this:

AB=Hyp
Opp/Hyp=0.82 (at 55 degrees)
Opp=5
5/Hyp=0.82 (at 55 degrees)
5/0.82=Hyp (at 55 degrees)
5/0.82≈6.1
Hyp≈6.1
AB≈6.1

this should hopefully be the right answer

[tardy yacht]

how can you get the calculator to give things in terms of pi

like if a value is given as 1.5707.... which is pi/2, how do i make the calculator show its pi/2 without having to figure it out myself

or is that not possible

[silent plank]

depends on the calculator

if your question involves special angles / ratios, you shouldn't really need the calc for it anyway

[tardy yacht]

its asking for answers in terms of radians. its more of a little hiccup to deal with it but i found you can just divide it by pi, then whatever number you get you multiply it by pi and thats the exact answer

like 1.047195.... /pi=0.3333 therefore 1.047195......=0.333pi=pi/3

[silent plank]

if your question involves special angles / ratios, you shouldn't really need the calc for it anyway

[tardy yacht]

but

if they want the answer in radians

actually

you're right

[wind sedge]

Ayeee

[vocal cloud]

wat

[wind sedge]

Nothin

[wild rivet]

it is

[odd geyser]

Hi! Given 17 cubes, the sides of which are 1 centimeter long, how would one find the smallest possible hollow cube that can fit those 17 cubes within it? Thanks!

(Please ping me if you think of anything!)

[grave pond]

That sounds hard. The analogous problem of finding the smallest square that will fit 17 identical squares is open, for example. On the other hand, with cubes there's an easy upper bound of 3 (since 3³>17), which doesn't leave a lot of room for being creative with tilting some of the cubes...

[odd geyser]

it does sound quite hard!

i was out of ideas so i decided to consult a math space (this one)

[grave pond]

Do you have a hydraulic press available? (Cf https://xkcd.com/2740/)

[odd geyser]

cant not bring up this one while were talking xkcd and unsolved math problems https://xkcd.com/2529/

[smoky jetty]

yeah seems about right

[dire hull]

I am very troubled by a question

[dire hull]

So I did infact do the cosine law

However I’m troubled by making the triangle itself

Here is my work in part B

But I’m still getting the triangle drawing wrong

Like my drawing makes no sense

[exotic yarrow]

Because you need to redraw it with angle A being obtuse

[dire hull]

Oh, okay wait lemme try doing that rn

That’s the way i drew that

But in order for me to do my cosine law I need to get my letters to certain places

My work in part B doesn’t match my drawing in part a

[exotic yarrow]

then rewrite it so it does

||a=sqrt(b^2+c^2-2bc cos A)||

[dire hull]

OH

Omg I’m acc so stupid 🤦🏽‍♀️

Sorry and thank you 😭

I was stuck on this for a solid 30 mins and then did some other questions 💀

But 21.3 seems odd because all the other two numbers are whole numbers that aren’t decimals

[steady estuary]

For b, wouldn't it be because it lies in the 3rd quadrant where its x and ys would be negative?

[thorny fossil]

Yes, or alternatively one can use tan to show it

[tawny grove]

what are the uses of sine cosine and tanget ratios that are greater than 180

the sine cosine an tanget are just ratios of the sides of triangles

but how can the ratio be negative

[feral canyon]

HELP

[grave pond]

What the cosine and sine really are is coordinates of points on the unit circle, and the circle has points on both sides of the coordinate axes.
The interpretation as rations of sides lengths in a right triangle works only for angles between 0° and 90°.

[upper karma]

what is the fomula

formula*

for base area

in prisms

[grave pond]

There's no special formula that you use for prisms specifically. The base of a prism is (by definition) some plane polygon, and you compute it area in the same way as you would compute the area of the same polygon if there wasn't a prism built on it.

[surreal plover]

How do you find the area of the base of a prism

[grave pond]

That's what I answered in the post immediately above yours.

[surreal plover]

Well I mean more like for surface area

[grave pond]

You find the area of a polygon in the same way no matter what you're going to use that area for after you've found it.

[surreal plover]

Oh

I’m just awful at surface area then

[grave pond]

If you have a particular polygon you need to find an area of, we can discuss that, but simply saying that it's the base of a prism is not going to help us narrow in on an area calculation that works for it.

[surreal plover]

Oh ok

Let me find the problem

I’m supposed to find the surface area

But I can’t figure it out

[steady estuary]

Wouldn't it be 90?

[grave pond]

Okay. So the base here is presumably a regular hexagon with side length 2, and you want to find its area.

(This is a pyramid, not a prism, but fortunately the area of the base doesn't depend on what we're building on top of it).

The simplest way to find the area of a regular polygon is to draw lines from each corner to the center, dividing it into congruent triangles, one for each side.

You may alreay know that in case of a hexagon, these triangles end up being equilateral.

Can you find the areal of an equilateral triangle with side length 2?

[grave pond]

90° is one of them, yes.

[surreal plover]

Maybe

I’m completely lost

[grave pond]

The case for an equilateral triangle can be done with just Pythagoras, but is perhaps more instructive to use some trigonometry -- do you have trig available?

[surreal plover]

I do

[grave pond]

Then we should perhaps look at the general case of a regular n-gon with known side length.
Draw lines from each corner to the center, dividing it into n isosceles triangles. Then more lines from the center to the midpoint of each side splits those into 2n right triangles that you can do trigonometry on. (And once you have the side lengths of a right triangle, its area should be easy).

[surreal plover]

Oh that makes sense now

I was thinking of it wrongly then

[maiden brook]

technically it does but in this case we are assuming that perpendicular intersects the "center" of the hexagon

[grave pond]

It doesn't even technically. The area of a given polygon is the same no matter whether you're building a prism on top of it, or a pyramid, or nothing at all.

(The area of the other sides of the pyramid will depend on where we put its apex, sure).

[maiden brook]

yeah and for the purposes of the problem u have to assume it’s at the “center” (is there a better word for this?) of the hexagon?

[maiden brook]

sorry I misinterpreted what u had said

[inland grove]

What's a pythagoream triple?

[weak pawn]

just a set of integers that fits into the pythagorean theorem; e.g., 3, 4 and 5 or 5, 12 and 13.

[quasi perch]

Can anyone please explain why point B is

x = $q_3\cdot sin(q_2) + L\cdot cos(q_2) \\
z = q_1 + q3\cdot cos(q_2) - L\cdot sin(q_2)$

[somber coyote]

jordinho

[grave pond]

It isn't, as far as I can see -- you need to add L to the expression for x.

[upper karma]

https://media.discordapp.net/attachments/1115279150568190002/1115279689875984494/1092420235748982964.png

[last ocean]

Forgot formula of how to find length of bisector line.

Does anyone remember

[silent plank]

more context?

what's the original problem

[short dagger]

Sorry for the language of the question. In short, the radius of the pizzas is 10 cm and it asks for the area of the KLMN square.

[smoky jetty]

could be referring to the special triangles

[upper karma]

is this true or false for all right triangles

[grave pond]

Multiply by c on both sides and compute the area of the triangle in two different ways.

[upper karma]

20*20=400cm^2

tray must be 90 degree

so it is either rectangle

or square

so it's l*b or a^2

and the sides got diameter of pizza

so square

20^2

[upper karma]

oh so it's gotta be true

cus hc/2 = ab/2

and that's just two different ways of showing the area of the triangle

pretty cool then

[grave pond]

Eyup.

[upper karma]

thanx

[grave pond]

You can also do it with similar triangles, but it's easier to get lost in the algebra that way.

[short dagger]

answer key says 800

[upper karma]

yeah i agree

[grave pond]

One easily sees on the diagram that the sides are longer than the diameter of the pizza. If they had the same length, the two half-pizzas that meet in a corner would overlap.

[dire hull]

I wanted to know if what I was doing was right and whether the answer to this was reasonable or not

And whether we cancel out the squared and the square root for 30k^2 under sqrt

[fair vapor]

yo

i have a question

[timber cargo]

Don't we all

[fair vapor]

indeed

[lethal nacelle]

can someone solve

i wanna check my answer

https://cdn.discordapp.com/attachments/1115479973835325491/1115480895323910204/IMG_1686.jpg

and this

[lethal nacelle]

find ab

[versed glacier]

[10 * sin(15)]/sin(135) = AB

[ember stratus]

in this video https://www.youtube.com/watch?v=yRhYsq0qOWw

how did they find 0:26 2 sin x/2

[smoky jetty]

tbf that triangle is hella inaccurate in regards to angle B

[prime musk]

heyy

anyone online?

[zealous pike]

Have someone read this book? I want to know what background in math I need to read it

[upper karma]

can someone show me how to solve these?

[exotic yarrow]

Г) Convert the argument to a power of 10
д) Rewrite the exponent as a single log with a coefficient of 1, then apply the definition of a log (as the inverse of the exponential)

In the future, questions like this belong better in the #prealg-and-algebra channel than here.

[exotic yarrow]

Direct formula substitution

[upper karma]

🙏🙏

[silver stream]

You are given a leg of value 10, specifically adjacent to the angle of 52°, which trigonometric function deals with adjacent sides? Cosine. To solve for the hypotenuse we can think back to the different definitions for trigonometric functions, cosine is defined as cos(θ) = adjacent/hypotenuse. Using this information you can create the equation cos(52°) = 10/x (let's set the hypotenuse equal to x), from here just solve for x: (x) cos(52°) = 10/x (x) **--> **xcos(52°) = 10 (divide by cos(52°) --> x = 10/cos(52°) (you are given a approximate value of cos(52°), let's use that) --> x = 10/0.616 **--> ** x ≈ 16.23. This means that AB ≈ 16.23

Note that the question gave you an approximate value of cos(52°) that is rounded so hopefully that is the decimal they are looking for.

Hopefully I did that right 💀

[upper karma]

Thx

[silver stream]

np

[weak pawn]

How would you do 2a?

[weak pawn]

Someone please 😭😭🙏🙏

[upper karma]

is the amplitude pi/6? @weak pawn

[weak pawn]

No clue 💀

I'm trying to relearn trig and someone sent this question, I have no clue how to do it 😭😭

[upper karma]

might try to take advantage of the fact that (1+irt(3) is just (rt(3)-i) rotated 90 degrees in the positive

finding the modulus might be a pain but amplitude can by obtained by mere observation if you identify this fact

or hey even better

use the euler form

the modulus is 4 (?)

please confirm the answers

@weak pawn

[weak pawn]

I don't know the answers

[upper karma]

hmm

[upper karma]

ask the person who sent you this question ey?

[meager citrus]

Anyone know how to do 7E)

[lucid vale]

Did i do this right and how do i calculate the alpha values its like midnight and i have a test tomorrow plz help

[exotic yarrow]

Factor the numerator and denominator

[weak pawn]

they still havent responded with the answers 😭

[meager citrus]

Which side?

[exotic yarrow]

Which side can be factored into linear factors that cancel?

[torn heath]

You could rewrite 1 + i sqrt(3) = i (sqrt(3) - i) and then divide numerator and denominator by (sqrt(3) - i)^15

[meager citrus]

Right?

[weak pawn]

Wouldnt that mean it would have to be -1 in the numerator originally?

[torn heath]

Do you mean that the numerator should have been (-1 + i sqrt(3))^17?

[weak pawn]

yeah

ive forgotten all imaginary number rules so 💀 that might be it

[torn heath]

Well, no, because sqrt(3) - i times i equals 1 + i sqrt(3). Remember that i^2 = -1, so
(sqrt(3) - i)i = i sqrt(3) - i^2 = i sqrt(3) + 1

[weak pawn]

OHH okay yes

thank you 🙏🙏

other than that, wouldnt that interfere with the exponents too?

like ((i)(sqrt(3)-i))^17

[torn heath]

Yes

[weak pawn]

so then how could you divide it

[torn heath]

(i(sqrt(3) - i))^17 = i^17 (sqrt(3) - i)^17

You can divide

[weak pawn]

ohhhhh 😭

okay yes thank you

youre a genius

[weak pawn]

okay i just wanna make sure i did it right, i got
2sqrt(3) + 10i

[meager citrus]

Ohhhhhhhh would the top be (sinx-3)(sinx-3x) and bottom (sinx-3)(sinx+3)

Been a while since I did perfect and difference of squares

Is that right?

[exotic yarrow]

I think you meant (sin x-3)(sin x-3) for the numerator

but yes

[meager citrus]

Ye that’s what I said

[exotic yarrow]

Ohhhhhhhh would the top be (sinx-3)(sinx-3__x__) and bottom (sinx-3)(sinx+3)

[meager citrus]

The x😂

Sorry

[exotic yarrow]

all good mate

👍

[meager citrus]

Typing is a pain

[meager citrus]

Can you help with D)

Had trouble with that one too

[exotic yarrow]

Send what you did so far with it

[meager citrus]

Nothing. I’m not sure how to start

[exotic yarrow]

pythagorean identity

note ||cos^3 x = cos^2 x * cos x||

[torn heath]

Could you show how you obtained that in detail? I think the correct result is 2sqrt(3) + 4

[meager citrus]

On the right side since sin^2x= 1-cos^2x

?

[exotic yarrow]

🤷‍♂️

I'll leave that for you to figure out

[meager citrus]

I can rearrange from here right?

[exotic yarrow]

I can't rlly make out what you're doing because of all the scribbles, but the logic seems fine

[meager citrus]

Sorry, I converted sin^2x to its equivalent, same with tanx, then I found an LCD of cosx between all 3 and got the final line

[weak pawn]

(i^17)(sqrt(3) - i)^2
(i)(sqrt(3) - i)^2
(i)(9 - isqrt(3) - isqrt(3) + 1)
(i)(9 - 2isqrt(3) + 1)
10i + 2sqrt(3)

[lament moon]

i need 2 solutions for doing this, help pls

[meager citrus]

Anyone know what to do from here?

[torn heath]

The third line should be (i)(3 - isqrt(3) - isqrt(3) + 1) because sqrt(3)^2 equals 3, not 9

[weak pawn]

oh yes, dont know how i missed that 🤦 thank you

[abstract kettle]

Plz help with this and provide the proof

[grizzled ridge]

I am in Geometry Honors and was wondering what the forumla for this is? I remember something about calculating chords using a tangent line but I can't find the formula on google. Thanks!

[royal cedar]

it's called power of a point

[twilit plover]

whats the question exactly?

this is what I got on simplification

[pliant pumice]

Hello! If sum of each pair of opposite angles in quadrilateral is equal 180 degrees, does it mean that the quadrilateral is cyclic? If yes, how can I prove it?

[weak pawn]

To find the height you'd use tangent

Since you already know the value of tan(35), youd set it equal to x/100

Then youd multiply 100 to the other side to isolate the X

Getting .7002*100, which equals 70.02. Then you'd add 5 to get the height of 75.02ft

[upper karma]

Thx

[junior pine]

Where’s that Irish guy

Who lives in

Yeah

I wanna ask for notes off him lol

Cus I need trig n geo notes 😔

[upper karma]

Same

[junior pine]

U got some?

[upper karma]

I have sin = opposite over hypotenuse and cos = adjacent over hypotenuse and tan = opposite over adjacent

That’s it though💀💀

[junior pine]

Oh wait that’s good cus idk how t use dat lel

[upper karma]

I need help with this

[junior pine]

Send it thru my tingy uh dms ting

[upper karma]

Well your welcome lmao

[upper karma]

alternate int., corresponding, transitive
in that order

Thx

[proven tusk]

anyone know how to solve this?

[maiden brook]

hint: find the interior angle measures and extend each side

[empty edge]

yes

[cursive sluice]

whats a good resource for geometry like pauls online notes, I want to learn these topics

[buoyant nest]

How would you do this question

[upper karma]

I think there are no solutions between -pi to 0 ( as sin is negative in those regions)

you do have 2 solutions in 0 to 2pi, one in the first and the other in the second quadrant

which can simply be found by satisfying sin(x)=1/sqrt(2)

[buoyant nest]

Apparently you’re ment to use the negative trig identities for this question

[upper karma]

yes I'm aware

but I don't think that changes the solutions

does it?

[buoyant nest]

Maybe? It might change the -square root 2 to a positive

[upper karma]

you mean convert it into 2sin(-x) + sqrt(2) =0?

[buoyant nest]

Yeah that

[upper karma]

still doesn't change the solutions as you'll ultimately be solving for x

[buoyant nest]

I see that now

The negative x would just be x because of the negative sin rule

I did the working out and got root 2 / 2 is this correct?

[upper karma]

yes

[buoyant nest]

Thanks

[snow sedge]

Hello, how can i find all triangle data only with Ma, Sa, Sb?

I need at least to get example for this one, bc i have 9 more like that 😅 (and no, it's not for exam, more like homework)

[silent plank]

what are
Ma, Sa, Sb

supposed to be

[snow sedge]

My bad

[snow sedge]

M = median

S = middle line

Idk if they're called like that in english

Smth like this

Thanks in advance

[wind shard]

Rectangles of the same color are equal in area, angles of the same color are equal in size.

Why are these triangle similiar?

[dark sparrow]

wrong server

[ancient ether]

Whos online

I have some questions about cos, tan and sin

just one

So is the first letter for the trig ratio always the side that goes clock wise first? Or does it also need to have an X on that side of the angle or for example 15 degrees that or x on that side for you to start the first letter of your ratio for SOHCAHTOA

Is this the right trig ratio?

The adjacent is on X

Today js my last night to work on this so please I cant wait 4 to 5 hours for a response 😒

[median vapor]

Givens: All rectangles (assuming you meant a^2 and r by (r+s) were equal and that c by x and b^2 were equal) are equal in area, same colored angles are the same size.

line r - line AF = line FE
line s - line EB = line FE

b^2 = cs
b^2/s = c

c = line r + line s - line FE

rc = a^2
c = a^2/r

a^2/r = b^2/s
r(b^2) = s(a^2)

just some info i got...

[ancient ether]

Hi

Is the trig ratio in the picture right

[median vapor]

i was replying to someone else, one moment

[rain steeple]

How do I start?

I always get stuck with a mouthful of sines and cosines 😦

[ancient ether]

Rob?

[median vapor]

the trig ratio is correct if A/H = Adjacent over Hypotenuse. if the little angle you drew is the angle you are taking the cosine of though, you might have messed up where you put the adjacent side for this specific angle.

[ancient ether]

hmmm

Oh no that angle was me being dumb

I realized it has to be opposite of where the number degrees is

[median vapor]

alr, but you did get the ratio for cosine right

[ancient ether]

I do the stuff on the calculator to solve now right?

[median vapor]

sure

[median vapor]

probably the double angle formula (to deal with 1-cos2x, there are also good visuals for this on yt)(en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle_formulae).

[ancient ether]

Ok so how do I know if it is multiplication or divi

Division

[median vapor]

if you're talking about trig ratios, cos = adj/hyp, sin = opp/hyp, tan = opp/adj, so on so forth.

the / or fraction is division, though you could rewrite O/A as O * 1/A

[ancient ether]

Sometimes ive seen people do tan and multiplication and also tan and division so now idk what to use when I get tan

For this im solve for the indicated missing side

[median vapor]

alr

[ancient ether]

i got 6.7

[median vapor]

wait a sec

[ancient ether]

oki

[median vapor]

ok

[ancient ether]

I just now seeing my teacher put x down low divide and x up high to mutiply

[median vapor]

so yea

do that

basically, if the sides are >1 or x>1, then multiply by the appropriate sine or cosine

then the opposite, if x<1, divide the known side by the sine or cosine of a known angle other than the 90 degree one that is inside the same triangle.

[ancient ether]

I was looking for the vid that showed me the equation cus my laptop reset 😅

[ancient ether]

gotcha

7.24

Tan46 times 7

[ancient ether]

Oh wait im a dumbass I was supposed to divide

81.86

Do I round up to 82 degrees?

@median vapor

[median vapor]

um

[ancient ether]

…

[median vapor]

honestly, i'd probably wait for a better explainer from another server member, i'm not rly explaining this that well I think.

[ancient ether]

Oh ok

Do you know my answer is wrong

It is?

[median vapor]

i think so

personally, i wouldn't use tangent to find the adjacent side of a right triangle

[ancient ether]

Sin is probably the way to go

[sand tusk]

tan46 = opp/adj = 7/x
x = 7/tan46

tan46 should be a little bigger than 1 since tan45 is 1 so x is approximately 7 id say

x is around 6.7

[ancient ether]

Ah

[ancient ether]

Well I did get this

[sand tusk]

you dont have the hypotenuse length so using sin wouldnt do much cus sin46 = 7/hypotenuse and you need to find X not hypotenuse

[sand tusk]

Yeah did you multiply tan46 with 7?

[ancient ether]

yes

Is that not how u do it

[sand tusk]

nah you need to divide in this case

tan46 = 7/x
x * tan46 = (7/x) * x [x cancels on out on the RHS]
x * tan46 = 7
(x*tan46)/tan46 = 7/tan46 [tan46 cancels out on the LHS]
x = 7/tan46

[ancient ether]

This was my other problem and I got 7 degrees

[sand tusk]

is that angle 84 or 94 i cant see it clearly

[ancient ether]

Cos54x12

[ancient ether]

54

[sand tusk]

oh 54

well 90 + 54 + x(unknown angle) = 180

[sand tusk]

oh wait u need to find the side length?

[ancient ether]

It says solve for missing side

[sand tusk]

oh my bad, A = cos54*12 and O = sin54*12

[ancient ether]

7 degrees?

[sand tusk]

7 units

side length cant be in degrees

[ancient ether]

wait does that mean that the other problem isnt actually 6.7

[sand tusk]

the other problem is 6.7 units

[sand tusk]

like if in this question the hypotenuse was 12 cm, the side would be 7 cm

and if it was 12 feet, side would be 7 feet

An angle is always measured in degrees and radians but in sides if there is no unit mentions its best if you just write(sidelength) units

[ancient ether]

yeah it has to be rhe same unit

Ok im gonna try these other problems

[sand tusk]

if ou find any difficulty just write out the trig ratios in front of you

if they give you the value of an angle, just write down the sin cos and tan of that angle with respect to the triangle like how you did in the previous questions just now

and see which one can be best applied, slowly you can mentally figure out which one you should apply without having to write it all down repeatedly

[ancient violet]

i have been looking at this 360 degree image and im wondering how could i recreate this distortion with other images. is this related to how earth maps are flattened? the mercator proyection and those things?

[upper karma]

heyo guys

[nocturne marten]

guys i got a exam

about trigonometry

i need help fr

[tame gull]

How to find the perimeter of this triangle?

[viscid gull]

What are the assumptions of the problem?

[fiery oriole]

i’m a dumbass but wouldn’t AC be 7x and same with BC?

[warm plume]

im not sure if i understood correctly, but if you want to recreate this exact distortion maybe look up „fish eye prespective” and something might come up?

[hushed urchin]

im confused about the sides, AB is adjacent, AC is hypotenuse, and BC is opposite so my answer is cos = 12/13 but the correct answer is 5/13 can someone explain

[dark sparrow]

no, AB is not adjacent to theta

look at the triangle, look where AB is, look where theta is

they're across from each other!

[hushed urchin]

k thx

wait so 12 is considered a hypotenuse?

[dark sparrow]

no

none of this is a matter of "consideration"

the hypotenuse is the side across from the right angle.

the hypotenuse is also the longest side.

[hushed urchin]

yes

[fiery oriole]

seeing as you’re on khan academy, i think it should link a video/article where it explains which side is which

pretty sure that’s near the start of khan academy trig

[hushed urchin]

wut that website isnt khan academy

[ancient violet]

Yea i actually found it. Its called equirectangular proyection

[cyan sandal]

bro why did math start looking like this ☠️

[maiden brook]

bruh

if u think thats bad ...

[slim dragon]

Big ideas math lol

[nocturne heath]

let's say i have two points (0,0), (250, 50) how do i translate the coordinates to (0,0) (141, 210)

we know the slope of the red lines (0.707107, 0.707107) and (0, 1)

[ember stratus]

How did they derive the simultaneous equations of
𝑟3𝑐𝑜𝑠(𝜃3)−𝑟4𝑐𝑜𝑠(𝜃4)=𝑟1−𝑟2𝑐𝑜𝑠(𝜃2)
𝑟3𝑠𝑖𝑛(𝜃3)−𝑟4𝑠𝑖𝑛(𝜃4)=−𝑟2𝑠𝑖𝑛(𝜃2)

[upper karma]

does anyone have a diagram of trig functions and their reciprocals interpreted on a unit circle

like this but without the extra stuff 😭

[frozen ocean]

I've never heard of cvs or crd specificity, if you don't mind telling me what are those meaning

,tex .geom trig def

[somber coyote]

Akira

[frozen ocean]

Not sure what do you meant but here

[upper karma]

literally just 1-sin

i don’t know why it exists

[upper karma]

yeah thanks this is it

crd is probably the length of the chord that subtends the angle theta

[grave pond]

They were somewhat useful for calculations in the days of pencil, paper and precomputed tables. If you had a computation where you had to repeatedly multiply by 1-cos(v) for various angles v, a table of log(1-cos(v)) could be a timesaver -- from there to giving them names is not far.
Of course, 1-cos(v) could also be computed as 2 sin²(v/2), but as Wikipedia says:

Prior to the advent of computers, the elimination of division and multiplication by factors of two proved convenient enough that tables of haversine values and logarithms were included in 19th- and early 20th-century navigation and trigonometric texts.

[dark sparrow]

1. memes go in #chill
2. transphobic memes and mimicries thereof go in the trash

[grave pond]

Right, deleted and @upper karma gets a day in the penalty box.

[dark sparrow]

penalty box?

[past burrow]

hey so im currently on the hunt for geometry courses. Any tips?

[grave pond]

24h mute

[nocturne heath]

I was wondering if someone could help me transform coordinates from the right box to the left box

in the bottom left box, we know 2 variables .. the slope of the red line (0.923077, 0.384615) and the coordinates of the start of the green line (15.3847, 222.117786)

we're trying to produce -200, 100 from the coordinates of the start of hte green line and the slope of the red line

can anyone help me?

[nocturne heath]

i figured it out

Vector Transform_Coordinates(Vector enemy_direction, Vector enemy_start)
{
float rotation_angle = asin(enemy_direction.x / sqrt(pow(enemy_direction.x, 2) + pow(enemy_direction.y, 2)));
return Vector(enemy_start.x * cos(rotation_angle) - enemy_start.y * sin(rotation_angle), enemy_start.x * sin(rotation_angle) + enemy_start.y * cos(rotation_angle), 0);
}

[median vapor]

khan acadamy has at least a High School geometry course so long as you are ok with watching video mainly. There are also open textbooks out there if you prefer reading.

[spiral mesa]

-90degrees means 90degrees clockwise

So, don't you think the point would land on the negative y-axis after the rotation?

[upper karma]

ohh i thought 90 was clockwise

90 degrees*

[spiral mesa]

Generally positive value of an angle denotes counterclockwise rotation

[pliant nest]

I'm stuck on something stupid again 😂

How do I figure out the Surface Area of a Cone when the given values do not include the height?

[teal flower]

Pythagorean theorem

[pliant nest]

What? How would I employ it to figure out the height?

17.9^2 + 8^2 = h?

[heady berry]

h^2 + 8^2 = 17.9^2

h^2 = 17.9^2 - 8^2

[pliant nest]

I thought I was solving for the hypotenuse which I thought was the Height. Makes more sense though

thx

Okay, so now I know the height is 16cm, the radius is 8cm & the length is 17.9cm.

[heady berry]

16 cm?

why?

this number is a little higher than 16 but not equal to 16

[pliant nest]

Yea, I am rounding

I am supposed to round to the nearest tenth

But I still don't know the formula to figure out the surface area of the cone.

I know how to figure out the surface area of the cylinder by unfolding, so getting 2x the area of the circles + the area of the rectangle

But I have no clue about the cone

I mean I did google it and did figure the formula out but it seems so different from how I learned the rest

Apparently I could've just done:
SA = area of circle + pi(8)(17.9)

😭

[heady berry]

real

[pliant nest]

Stuck again, this time with a word problem 😂

"A square pyramid measuring 9 yd along the
base with a slant height of 12.8 yd."

How do I figure out the surface area? I thought I could employ the Pythagorean theorem and do:
a^2 + 9^2 = 12.8^2
a = ~9.1 yd

And I did but I am clueless where to go after that

🤦‍♂️

[silent plank]

you shouldn't be using pythag for the surface area

nor have you applied pythag properly here

[pliant nest]

Damn, guess I should've solved for the hypothenuse

But I am clueless

As to how to solve this square pyramid

[silent plank]

you have the slant height of the pyramid, which is already the height of the triangular faces

[pliant nest]

Idk what to do with it lol

[silent plank]

apply area of a triangle

to get the area of the triangles

[pliant nest]

Sigh if this is all it takes

brb

[silent plank]

and add the area of the square base as well

[pliant nest]

Okay so:

1/2(12.8)(9)
1/2(115.2)
= 57.6^2

A = 9^2
A = 81 yd^2

SA = 81 + 7(57.6)
SA = 311.4

I'm crying rn

ty

[marsh current]

whats the best book I can get which covers significant amount of geometry and trigonometry

[safe tundra]

Egmo

By Evan Chen

[sullen raptor]

Algebra and Trigonometry by James Stewart

[wooden warren]

Hello, I’ve been really thinking about this and I even constructed the suggestion given here, but I have really no idea how to prove this.

Heres my construction and I really do not know how to even get the area of the shape

[grave pond]

Hmm, yes, that's not immediately obvious.

Try comparing the areas of triangles GDE and GAE.

[wooden warren]

Ooo they must be equal? Since they have same base and altitude

OH I GOT IT

thank u so much

AE would be a median and it would halve the triangle’s area

[sturdy jackal]

$(\pi)(r)^3(\frac{4}{3})$

[somber coyote]

dmahonjr

[dark sparrow]

that's quite a lot of unnecessary parentheses

[rugged marsh]

I would suggest $\left(\frac{4}{3}\right)$

[somber coyote]

ForJoke

[sturdy jackal]

thank you

[strange crater]

ok i'm taking a step back from analysis and reviewing some simply geometry and i suck, lol

i'm given that this is an isosceles triangle, and i am to find the white part of the area of the triangle

can i assume that what appears to be the right angle indeed is a right angle?

[rugged marsh]

I would hope so

[strange crater]

otherwise there is two possible solutions, right?

[rugged marsh]

wait you said the triangle is isosceles right?

[strange crater]

yes

oh wait no if what appears to be a right angle is not a right angle, then we can't find the are i'm pretty sure

[rugged marsh]

this can only be isoceles if its a 45-45-90

[strange crater]

why not 45-67.5-67.5 ?

[rugged marsh]

that is true

[strange crater]

ugh i'm just gonna assume whoever drew this up is not a sadist

[rugged marsh]

is this all the info?

[strange crater]

ye

[rugged marsh]

yeah just assume 45-45-90

[strange crater]

no it can still be solved in the other case

[last quiver]

yeah it can be solved in both the cases

[last quiver]

all the side lengths of the triangle are not given ?

[weak pawn]

I don't understand the definition of sine part😭

[last quiver]

yk sin α = opposite/hypotenuse ?

[strange crater]

what do you mean

[last quiver]

I mean are all the side lenthgs given , coz in the diagram it's just given that one side length is equal to 10, rest two are not given

[strange crater]

yes, no other are given, but they are easily found if what appears to be so indeed is a right angle, if it's not there is just one other possibility not too hard to find either

[last quiver]

how can you do it in the case of 45 67.5 67.5 ?

[strange crater]

then the two unknown sides must be equal

[last quiver]

yeah

[strange crater]

so we split the triangle in half along the middle, and we have right angled triangle where we know an angle 45/2 = 22.5 and its opposite side 10/2 = 5, so the hypothenuse (the unknown side) is equal to 5 / tan(22.5)

[last quiver]

yes , that makes sense :)

[weak pawn]

but why the cosB?

oh wait nvm

im literally slow asf

[last quiver]

by definition , cos (β) = adjacent / hypotenuse , in this the hypotenuse is equal to 1. So adjacent = cos (β)

[weak pawn]

yeah i gotchu now 😭 idk what i was thinkin earlier

thank you

[last quiver]

yw :-)

[strange crater]

can someone verify i'm not being stoopid: an alpha-degree slice out of circle of radius r, has area alpha/360 * pi r^2 right?

"alpha-degree" being an awkward way to state that the angle between the two side of the slice inside the circle is alpha

[last quiver]

yep

[frosty dew]

Is trigonometry NYS Algebra 2 equivalent?

[upper karma]

i got the answer but it was the opposite like

2401/625

[upper karma]

this is the answer btw

[vivid plinth]

What was your process

How did you get to that

[upper karma]

alright so i got 2401/625 because

wait

nvm i understood it

you flip the fraction so it becomes 7^2/5^2 then flip it again because its exponent is negative

correct

[frozen ocean]

💀

[upper karma]

how?

[vivid plinth]

<@&268886789983436800>

[cinder oar]

The eqn must be: x+y=10

Let A(c,0)=(x,y)

If y=0, then x=10

[upper karma]

thanks i got it

[jade forum]

can you help with geometry?
so there's a triange
pqr and the pqr angle 120 degrees
how do i get the orthogonal projection of line ?
pq and qr is 4 cm

[left ridge]

i hate trig...uguu....

[smoky jetty]

could u elaborate why that should be the equation?

[frozen ocean]

,tex .original

[somber coyote]

Akira

[sudden pulsar]

bruh same

watch im abiutt fail my final tmr

[trim meadow]

How do I do dis

[olive sluice]

if 3x=cosec theta and 3/x=cot theta find the value of 3(x^2-1/x^2)

[olive sluice]

any help with thus

[nocturne remnant]

https://wikimedia.org/api/rest_v1/media/math/render/svg/569181e0b7d46ae94b93ef730ec2f4374dca2ced

Consider using this

[stiff sigil]

Can anyone help me with class 9 cbse ncert maths chapter 6 and 7?

[frozen ocean]

Do you know how to prove $cosec^2 \theta −cot ^2\theta=1$?

[somber coyote]

Akira

[trail sigil]

$1/sin^2\theta-cos^2\theta/sin^2\theta=1$

[somber coyote]

delightful malody

[trail sigil]

$1-cos^2\theta=sin^2\theta$

[somber coyote]

delightful malody

[dark sparrow]

[frozen ocean]

Ghost ping?

[upper karma]

Hi. I'm having some trouble understanding this problem:
"Let ABC be a triangle with a right angle at B. Let D be any point on AB, and let E be the foot of the perpendicular from D to AC. Prove that angle DBE is equal to angle DCE." Is angle DEC a right angle?

could anyone help? 🙂

[frozen ocean]

Is that what the angle looks like for your question?

[young tiger]

First, draw the figure.

Notice how m∠DAE = m∠BAC by the reflexive property of equality.

Also notice how m∠DEA = m∠ABC since all 90° angles are congruent.

This means that △DAE ~ △CAB by AA~.

You can see that in the figure above, m would be the scale factor for △DAE and △BAC.

Now construct segments BE and DC.

Notice how m∠BAE = m∠CAD by the reflexive property of equality.

Notice how AB/AC = AD/AE.

This means that △ADC ~ △AEB by SAS~.

The corresponding angles of similar triangles are congruent.

∴ m∠DBE = m∠DCE

Also, ∠DEA and ∠DEC are supplementary angles.

So if m∠DEA = 90°, then m∠DEC = 90°.

[smoky jetty]

thats neat

[stark wraith]

How do

Dying

[nocturne remnant]

also use the fact that sin^2 + cos^2 = 1, you get a pair of simul equations which can be solved for sin or cos

[maiden brook]

great diagrams!

[nocturne remnant]

cool but you confused AD and AE with each other

Oh this was a day ago

Whatever

[upper karma]

Does that imply that BEC is 90, therefore BDEC is a cyclic quadrilateral?

[upper karma]

This is what has been torturing me. Cuz if m<DEA = 90, then m<BEC = 90 as well, which would make BDEC a cyclic quadrilateral.

m<BDE = m<BCE ---> m<BCE = 90???

[lethal pier]

why

[atomic reef]

could someone help me with these please

[wind coral]

<@&268886789983436800>

[young tiger]

oh shit yeah sorry. fixed it

[young tiger]

∠BEC cannot be 90°.

Notice how m∠DEC = 90°.

By the angle addition postulate, m∠DEC = m∠BEC + m∠DEB

This means that m∠BEC has to be less than 90°.

[young lintel]

I can't find any use of the name "alternating cube" online. Has anyone seen this or did gpt make it up? And do you have a name for this cube?

[floral shale]

This is why you don't use ChatGPT