geometry-and-trigonometry

[analog quarry]

okay wonderful

don't use that

the question asks you to leave it in terms of pi

so you can actually just leave it as π and treat it a bit like an algebraic term (like x or y for example)

and your final answer will be something times π

[edgy apex]

yea

Gotcha

[analog quarry]

okay so the question gives you the size of the radius, 7m

see if you can find the area of each circle using πr^2

leave it in terms of pi

so it'll be a number times pi

also just so you're aware, both circles have the same radius because that's how a cylinder is defined

[edgy apex]

Skurt I love you bro

[analog quarry]

haha np

[edgy apex]

just found these in my folder

[analog quarry]

they'll help lots

[edgy apex]

yes

[analog quarry]

we were about to arrive at the formula for SA of a cylinder anyway

it's 2πrh + 2πr^2

the 2πr^2 comes from πr^2, the area of each circle, times 2 because there are 2 circles

2πrh is the size of the rectangle that wraps around because it's base x height, where the base is the circumference of the circle

so you substitute 2πr which is the circumference into B x H and you get 2πr x H which is 2πrh

add them up and you get the surface area :)

[edgy apex]

Betttt

[torn gazelle]

@analog quarry

[analog quarry]

wassap

[torn gazelle]

could you help me rq

so for a, b^2 would represent the base squared

[analog quarry]

that is tremendously hard to read

[torn gazelle]

ima take a smaller pic

[analog quarry]

okay thank you

[edgy apex]

LOL

[torn gazelle]

Is a correct?

[analog quarry]

lemme take a look here

kind of

you're on the right lines

i think you had the right answer in mind but worded it wrong

b^2 represents the area of the base

do you understand why?

[torn gazelle]

oh yes

[analog quarry]

also c is incorrect

[torn gazelle]

that makes so much more sense then how i did it

[torn gazelle]

what would that be then?

[analog quarry]

let's go over the method here

the question's telling us that a = 0

so we're gonna substitute 0 wherever a appears

1/3 * (0^2 + 0b + b^2) * h

you simplified that to get 1/3 * (b + b^2) * h

can you see the error there

[torn gazelle]

yes i do

[analog quarry]

what would the corrected answer be

[torn gazelle]

1/3(0^2+0b=b^2)h

[analog quarry]

that's correct but it can be simplified, which is probably what the mark scheme wants for that question

lets go over each term

1/3 is a constant so you can leave it

what's 0 squared?

[torn gazelle]

1/3(0^3+b^3)h

[analog quarry]

not a trick question

[torn gazelle]

that simplified correct?'

[analog quarry]

no

[torn gazelle]

0

[analog quarry]

yeah, and if a term that's just 0 appears, you don't have to write it down

so you can eliminate 0^2 entirely

[torn gazelle]

so its 1/3(b^3)h

[analog quarry]

not exactly

lets look at the next term

what's 0 times b

[torn gazelle]

b

[analog quarry]

again, not a trick question

[torn gazelle]

lol

[analog quarry]

not quite

1 times b is b

0 times b is just 0

[torn gazelle]

oh

[analog quarry]

ANYTHING times 0 becomes 0, doesnt matter what it is

are you happy with that idea

[torn gazelle]

rookie mistake

[analog quarry]

haha it happens to the best of us

[torn gazelle]

so the final formula would be 1/3(0+0+b^2)h

[analog quarry]

yeah, and you can eliminate any terms that are just 0

[torn gazelle]

then we keave the zeros out

[analog quarry]

so how does it simplify

[torn gazelle]

1/3(b^2)h

[analog quarry]

yes that's exactly right

[torn gazelle]

and thats the formula for a pyramid?

[analog quarry]

for this type of pyramid yes

in part d you should specify what kind of pyramid it is

i believe it tells you in the question

pyramids are categorised by the shape of their base

[torn gazelle]

square pyramid

[analog quarry]

i think you'd normally say square-based instead of just square

but yes that's right

[torn gazelle]

oh

also, for b we didnt asnwer what a^2 equaled

[analog quarry]

okay so take a look at your answer for what b^2 represents

[torn gazelle]

theres also a problem d and f

[analog quarry]

we said it was the area of the base, right?

[torn gazelle]

the area of the base

[analog quarry]

because it's a square, b is the side length, and the area of a square is the side length squared

the top of the frustrum is also a square and its side length is a

so what do you think a^2 represents

[torn gazelle]

so the area of the top

[analog quarry]

yes that's right

[torn gazelle]

ok

e says "what does the volume expression become if a=b?

then f says, for what type of geometric solid does your answer to part e give the volume

[analog quarry]

okay so

lets take the original expression for volume then

1/3 * (a^2 + ab + b^2) * h

now wherever we see b, lets replace it with a since a = b

so we get 1/3 * (a^2 + aa + a^2) * h

which immediately becomes
1/3 * (a^2 + a^2 + a^2) * h
because a * a = a^2

can you find a way to simplify it even further

[torn gazelle]

do you mean b?

it said if a=b?

[analog quarry]

it's the same thing

[torn gazelle]

ok

[analog quarry]

you can use either variable

[torn gazelle]

so it would be

[analog quarry]

1/3 * (b^2 + b^2 + b^2) * h is also correct

[torn gazelle]

1/3(b^6)h?

[analog quarry]

not quite

when adding up square numbers, you dont add the indices

you only add the base

think of b^2 as some new variable, x

what would x + x + x be

[torn gazelle]

oh

so it would be

1/3(6b)h?

[analog quarry]

not quite

[torn gazelle]

no

thats not right

[analog quarry]

you leave the indice alone completely

[torn gazelle]

wait im confused

oh wait

i know it

[analog quarry]

go on

[torn gazelle]

1/3(b^3)h

[analog quarry]

still not there yet

it's much simpler than you think

[torn gazelle]

its just Bh

[analog quarry]

you're adding a number to itself 3 times

what does that sound like

super super basic and fundamental piece of math

[torn gazelle]

what does adding it sound like?

[analog quarry]

adding something to itself a number of times

[torn gazelle]

then you can put it to a certain power

[analog quarry]

that's one step beyond what you need

what you're doing is just multiplying it, right?

[torn gazelle]

yes

[analog quarry]

if you add a number to itself 3 times, that's the same as multiplying it by 3

[torn gazelle]

oh lol

1/3(3b)h

[analog quarry]

almost

nearly there

you're adding b^2 to itself, not b

[torn gazelle]

3b^2

[analog quarry]

yes

so it's 1/3(3b^2)h

[torn gazelle]

and what formula is that?

[analog quarry]

well it can be further simplified, can't it

[torn gazelle]

wait wait wait

[analog quarry]

what's 1/3 times 3b^2

[torn gazelle]

1/3 of 3 is just 1

[analog quarry]

yes exactly

so 1/3(3b^2) just becomes b^2

[torn gazelle]

so its just b^2h

[analog quarry]

yes

well done

[torn gazelle]

and thats a formula for?

[analog quarry]

that's the formula for a square-based cuboid

or a square-based prism

whichever you prefer

[torn gazelle]

oh cool

[analog quarry]

it's NOT the general formula for a cuboid because not all cuboids have a pair of square faces

[torn gazelle]

thanmk you so much for the help

[analog quarry]

no problem man

[torn gazelle]

i have a couple of basic questions to ask

just to clarify

[analog quarry]

ask away

[torn gazelle]

in this one

AB=90 using pythagareons thrm

the for b, how would i go about getting the angle measure for BAO

[analog quarry]

careful

in the pythagorean theorem, a^2 + b^2 = c^2

not c

you've done c = a^2 + b^2

do you spot the error there

[torn gazelle]

yes

so i need to take the sqr root of 90

[analog quarry]

that's right

and that's the length of AB

[torn gazelle]

which is

[analog quarry]

just leave it as root 90

always best to leave them in that form

otherwise you can lose precision when multiplying those long decimals

[torn gazelle]

ok

then we can find the angle?

[analog quarry]

you actually don't need the hypotenuse to find the angle

i assume you've covered your basic trigonometry rules

SOH CAH TOA

[torn gazelle]

yes

we could use tangent

[analog quarry]

yes you can

tan x = opposite / adjacent

[torn gazelle]

so what would i put in the calc?

[analog quarry]

you know about using the tan^-1 function to get an angle from a tan function?

it's the inverse tangent function

[torn gazelle]

yes

[analog quarry]

good good

[torn gazelle]

my phone calculator doesnt have it though

[analog quarry]

that's unfortunate

use a scientific one online

[torn gazelle]

oh shoot

[analog quarry]

make sure it's using degrees and not radians

[torn gazelle]

ok

so sqr root of 90

then

inverse of tangent

[analog quarry]

yes so

doing opposite / adjacent, or 3 / 9, will get you the tangent of the angle

so to get the angle itself, you have to do tan^-1 ( 3/9 )

do you understand why?

[torn gazelle]

yes

[analog quarry]

do that and you should get the answer

[torn gazelle]

so its like

83.98

which is 84

[analog quarry]

yarp

well done

oh actually be careful

[torn gazelle]

of what?

[analog quarry]

i dont know what you did on your calculator

but mine gave 18 degrees

[torn gazelle]

oh

i put in rad90 inverse

so i did it wrong

[analog quarry]

that'll do it lol

[torn gazelle]

would it be 18 whole>

[analog quarry]

yes

it says to the nearest degree

so you round to the nearest integer

[torn gazelle]

ok

then for c i got 84.85cm^3

using 1/3 times pir^2 times h

@analog quarry

[edgy apex]

who can help

#9

[lime crag]

@edgy apex what is the surface area forumla?

[edgy apex]

here

@lime crag

[lime crag]

What kind of shape is it?

[edgy apex]

Pyramid i tjink

i’m on this one now

#8

[lime crag]

Same thing

What shape?

[edgy apex]

Cylinder

but hexagon

[lime crag]

prism

Use the prism formula

[austere mortar]

can I try and explain how the formulas roughly work? Becouse I think that might be way more valuable then just helping you with specific problems

[edgy apex]

ok

[austere mortar]

then gimme a moment please, Ill try to explain it on the basis of that hexagon prism thing and in advance sorry if my english will be a bit wonky at times

alright

every 3d shape is made out of 2d shapes and by calculating the areas of these shapes we can calculate the area of a 3d shape

for example a cube - if we want to calculate its surface area we'll first have to calculate the area of the squares that make it

so here it'd be a^2 and since a cube is made out of 6 of such squares we just multiply that area by 6 - 6*2^2

now a hexagon prism. Its made from six rectangles and two hexagons wich form its bases

now these hexagons are made from six triangles that if theyre even you can caltulate their are with this

where this formula is derived from just normaly calculating the area of a triangle by multiplying its base and its height and dividing it by 2

[edgy apex]

i’m doomed bro

[austere mortar]

why so?

[austere mortar]

but you can get through it, trust me, you just gotta give it a little time and try to understand it

[edgy apex]

i love you bro

[austere mortar]

but give it your best. I genuinely understand what youre going through since at the start I also was pretty bad at geometry but giving it time and trying to understand really helped

[austere mortar]

so, generally speaking, how would you calculate the area of this?

[edgy apex]

wowww

i got these

[austere mortar]

but try to understand it, itll be easier for you to remember and will build a better understanding of geometry atleast if you try to understand how to calculate areas of more rectangular and triangular figures since cones and stuff with circles could be a bit harder for starting

[torn gazelle]

Could you help me real quick

[austere mortar]

I can try

[torn gazelle]

Let me take a picture

[austere mortar]

ooo, scales

with the full thing?

[torn gazelle]

Yes

Let me take the other picture

@austere mortar

[edgy apex]

ok

[austere mortar]

I can try and help with the first one since imma need to go in a moment. But lets start with the first question

[torn gazelle]

Ok

[austere mortar]

whats the scale of the monsters height compared to each other?

[torn gazelle]

16:40

[austere mortar]

or 40:16

[torn gazelle]

So it would be 2.5 times the size

[austere mortar]

and since youre trying to calculate how bigger the larger monster is you can just calculate it as k = 40:16, becouse (smaller monsters height)*k will give you the larger monsters height where k is just the scale

[torn gazelle]

Ok

[austere mortar]

yep

[torn gazelle]

So 2.5 is the answer for problem A?

[austere mortar]

and now if we're calculating surface area we just square the scale, so k^2 will be equal to what?

[torn gazelle]

2.5^2 is 6.25

But is 2.5 the answer for A?

[austere mortar]

yeah, but its best to write it fully as k = 40:16 =2.5, its just more clear

[torn gazelle]

Ok

[austere mortar]

and you already have the answer for B

[torn gazelle]

That’s all it is?😂😂

And then the volume would be 2.5^3

[austere mortar]

exactly

and we're done ^^

[torn gazelle]

Ok

But then…

[austere mortar]

yes?

[torn gazelle]

^

We have d,e, and f

[torn gazelle]

I appreciate your help, if we could finish these 3 that would help soooo much

[austere mortar]

then to start - got any hunch on what we could do based on question B?

[torn gazelle]

6.25 times 60ft?

Then for e, we do 400 times 15.625?

[upper karma]

can anyone help

look in algebra one

[austere mortar]

yes

[torn gazelle]

So d is 375ft

e is 6250 pounds

Then how do we find f?

[austere mortar]

heres a kind of way to understand it where t represents the value youre upscaling/downscaling where its square(idk how x^n is called in english, sorry) is already usually calculated such as in 60 ft^2 becouse it comes from (√60 ft)^2

[torn gazelle]

Ok

So am I right?

[austere mortar]

not currently, sorry

[austere mortar]

gimme a moment to calculate it please

[torn gazelle]

Ok

[austere mortar]

all correct

[torn gazelle]

Ok cool, how would I find the length of the arms

[austere mortar]

F is asking just about feet, in other words a measurement thats just in one axis or rather one dimension so howd you upscale it?

[torn gazelle]

Wouldn’t I find the difference between 375 and 6250 and then multiple that by 10 ft

That might be the totally wrong idea

[austere mortar]

you already know the scale, so its not needed, rather its a question of weather its just plain k[thats k^1]; k^2 or k;^3

becouse its just asking for their lenght

[torn gazelle]

I’m so confused😂

It says that the arms are 10 feet long.

Then it says how long should it’s arms be?

[austere mortar]

if the scale is 2.5 then we multiply it by 2.5 becouse thats are scale with wich we need to upscale it

[torn gazelle]

Ok

So 2.5 times 10?

[austere mortar]

yeah

[torn gazelle]

The arms should be 25

That’s actually so easy witw

I was expecting it to be impossible 😂😂

[austere mortar]

the only thing that you have to watch out is weather youre upscaling or downscaling something, becouse the scale is relative since it can be a/b or b/a depending on what youre trying to calculate

[torn gazelle]

So am I wrong?

It says we made the smaller one, with the arms being 10ft so we need to upscale it

Which is 25

[austere mortar]

no, not at all. Just if theyd require you to calculate the measurements from the big gorilla to the small one then youd downscale it

[austere mortar]

yeah, Im just giving you a little heads up for potential future exercises to solve that'd involve scales

[torn gazelle]

Ok, thank you so much!

Very helpful, hope you have a great day!

[austere mortar]

no problem and I hope you have an amazing day too, now I gtg so bye ^^

[torn gazelle]

alright, see ya!

[weak flare]

How do I do this...

<@&286206848099549185>

[visual marten]

What is arc " BA

[weak flare]

It's not given

[visual marten]

You can figure it out

[weak flare]

152/2 = 76

[visual marten]

I mean sure

[weak flare]

Is that not how it's done...

[visual marten]

That is the right answer somehow

[weak flare]

It's how I was taught lol

76/2= 38

So D=38..?

Im not so bright when it comes to geometry

[visual marten]

I mean

Yes

somehow

You are correct

Goodjob

[weak flare]

Thanks

[visual marten]

[weak flare]

Wb finding pq

How do I do that...

[visual marten]

,rotate

[visual marten]

🤷‍♂️ idk

[toxic estuary]

do people actually help here

or do people just send for help

[slim dragon]

Is F the center?

[weak flare]

No

[toxic estuary]

i’m sped asf

pls help😭

[slim dragon]

Is G the center?

[slim dragon]

Then idk

[weak flare]

Solved it.

[slim dragon]

How did u get BE to be 46

[slim dragon]

If so the 60 degrees is 1/6 of the circle

So u can set up a proportion

(60/360)=x/2pi*r

x being the length of the arc

[toxic estuary]

that’s so confusing

[nocturne remnant]

i believe there is not enough information - as point E varies on the circumference, angle CFE changes while the givens stay the same. unless some other information is given

[toxic estuary]

so what’s the answer

2x?

how do you do the pi sign

like 2pi

is that the answer

[toxic estuary]

is it A or D

[timber cargo]

length of sector = rtheta

[steady vault]

No isn't it 2pir X angle/theta

[faint basin]

Guys pls help

[weak flare]

Vertical angles, so the opposite side is the same

[zenith field]

did you try using SohCahToa?

Sine=opp/hip Cosine=adj/hip Tangent=opposite/adjacent

[craggy briar]

Guess you could use F- and Z-angles and/or prove similarity.

[craggy briar]

60/360 = 1/6. circumference = 2 x pi x 6. HJ is 1/6 x (12pi)

[rain rover]

can you round up a decimal using sin or cos rule?

or does it change the entire answer

[analog quarry]

can anyone spot my error? i’ve looked over my workings 5 times and still haven’t seen what i did wrong

[golden idol]

i think in step 2 you forgot the coefficient of cosxsin60degree

which should be -2 but not -1

[golden idol]

but your teacher actually gave it a tick lol

[timber cargo]

What did you just type

[analog quarry]

ah that’ll be why i didn’t spot it then lul

[analog quarry]

tried it again with this in mind and got to the right answer of tanx = 3sqrt3

ty for the help

[golden idol]

no problem

[analog quarry]

<@&268886789983436800>

[golden idol]

what happened

[desert eagle]

"hey free nitro"

[median dragon]

i need help. i know it’s pretty basic but i already forgot how to do it. 😭😭 please someone explain it to me

[hazy quartz]

pythag theory

a square plus b square

then square root your answer

[craggy briar]

10^2 + 10^2 = x^2
x^2 = 200
x = root of 200 is about 14-14.2?

[gusty kestrel]

Yes

[rough sluice]

Does anyone know how to solve this?

[storm portal]

!show

[lime crown]

Show your work, and if possible, explain where you are stuck.

[rough sluice]

I mean idk how to find x and y

[trim glen]

are there like helping vcs

im really struggling with ix;s rn

[eager ivy]

hello

i can help

if still needed

@rough sluice

[golden idol]

pls

i want to know it too

[eager ivy]

okay

the other person can just read these ig

so

lets do x first

do you know the inscribed angle theorem?

[golden idol]

x is easy i think everyone can do it

[eager ivy]

oh

okay

good

it is 105 right?

[golden idol]

wa

i got 115

[eager ivy]

oh rihgt

sorry, calculation error

your right

[golden idol]

Ok

[eager ivy]

so y then

[golden idol]

yes

[eager ivy]

so there is a formula for these problems

whnever you have two intersecting secants

that intersect outside the circle

then

[golden idol]

im not sure if youre talking about tangents

[eager ivy]

no, im not

[golden idol]

Ok

[eager ivy]

then i believe that the arc on the outside of the points of intersection minus the arc on the iside

then divide everything by two

and you get the angle between the secants

ita really hard to explain so i get if ur confused

ill try to fins a picture

[golden idol]

yeah indeed im confusing

[eager ivy]

this should help

the coloration got wierd on my end, idk about u

its still legible tho

[golden idol]

wow

[eager ivy]

the one to the right is what im reffering to

[golden idol]

yeah i got it

learn a new rule today

[eager ivy]

nice

good job

and good luck

bye

[golden idol]

thx

[eager ivy]

hopefully the other person sees theese

[golden idol]

Yeah

[eager ivy]

ok bye now

[golden idol]

bye

[eager ivy]

bye

[rough sluice]

Tysm!! @eager ivy

[random garden]

Hi, is there a name for the shape of visibility horizon of non-uniform scaled sphere?

How would I even begin to compute the horizon?

[unborn cloak]

what is 2+the area of my dik

[visual marten]

0.5 cm

[rancid topaz]

lol he has -1.5 inches

of area

[visual marten]

Yeah he has an innie

[rancid topaz]

a pusshe

lol

[mint garden]

2+i

[rancid topaz]

brooooooooooooo

roasted

[mint garden]

Or $2+x, x \notin \mathbb{R}$

[somber coyote]

Kai The Doge

[unborn cloak]

NAH MORE LIKE 1+200 NANOMETER

[dark sparrow]

#chill exists, y'all.

[mint garden]

Sorry...

[light aurora]

pop pop pop pop pop pop pop pop pop pop pop pop

[dark sparrow]

same applies to you

[light aurora]

i thought i was in chill lmao my bad🥺

[eager ivy]

no problem

[livid sparrow]

how do i find the arc length of a sector with the radius given to me and a missing angle
and the area also given to e
me

[timber cargo]

s=length of radius times theta
and area of sector = (1/2) times theta times the square of the radius

You can operate from there

[wind fox]

How do i get length DE

[slim dragon]

Is there anything else thats given

[eager ivy]

ParfoPizza has a point, if angle C is right, then we could use pythag and life would be easy

[bitter nymph]

If you know the ratios of the triangle yes

[floral shale]

Just use properties of similar triangles

You have enough info

Well

Actually you dont

[zenith pagoda]

math is easy once you get it ;p

[fiery oriole]

cool bot

$2+5

bruh

[worn arch]

$2+5$

[somber coyote]

bacchess

[worn arch]

:)

[fiery oriole]

ohh

$sin(30) = /frac{1}{2}$

[somber coyote]

Gyro

$sin(30) = /frac{1}{2}$

[floral shale]

\

$\sin(30^{\circ}) = \frac12$

[fiery oriole]

thanks

[somber coyote]

Umbraleviathan

[fiery oriole]

wait but couldn’t it just detect it as fraction with 12 and not 1/2

[floral shale]

Without the {} it'll take the preceding character as an input

It's not required for a single character argument

If you wanna do something like 12/24

$\frac{12}{24} = \frac12$

[somber coyote]

Umbraleviathan

[fiery oriole]

👍

[void vault]

i want to have a question to solve

please

[sterile radish]

can anyone please send me coordinate geometry important questions class 9

[plucky geyser]

pls help me

[slim dragon]

Have u heard of pythagorean thrm

a^2+b^2=c^2

a and b are the legs of the triangle

c is the hypotenuse

@plucky geyser

[plucky geyser]

yeah

I got help

thanks anyway tho

[jovial pier]

Helo people

May I ask some advices in studying trigonometry? (3rd Year High School)

[wraith scaffold]

You just need to do a lot of excersises after you have understood the concepts. High School´s trigonometry is not taught as deep so you do not have to worry much, just treat it as other areas like equations.

i am currently 4th year high school

[upper karma]

is this volume for a frustum only true when the bases are rectangles or squares?

[cloud canopy]

hello, i might be turning crazy, but if a problem gives u that a is bigger than b, a minus b is bigger than zero right?

[raven sable]

yes

[austere mortar]

what recourses would be good to learn all of basic trig from the ground up on a highschool level?

[obtuse patio]

khan academy

[austere mortar]

alright, thank you ^^

[dry current]

hi im stuck on this question.
how would I square root 56.25?

I did 12.5 * 4.5 = x * x

and got 56.25 = x^2

[safe tundra]

Calculator ? Or change into fraction and get that it is 7.5

[golden idol]

,query sqrt56.25

[somber coyote]

Results provided by WolframAlpha

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

Requested by MochaOhwelp#8034

[golden idol]

[balmy wagon]

i have a small question

im having a little bit of trouble understanding sin^2(x)+cos^2(x)=1

why can't you just square root both sides of the equation from the very start, making it sin(x)+cos(x)=1?

[worthy matrix]

well, you don't get sin^2(x) + cos^2(x) when you square sin(x) + cos(x)

like how (x+y)^2 isn't x^2 + y^2

[balmy wagon]

oh

ok thx!

[fair vapor]

rule of sine

sin B/ b =sin A/a

sin 30/ a = sin x/ 12

[fair vapor]

sin^2(x)+cos^(x)=1

[slim dragon]

<@&286206848099549185>

[upper karma]

apply pythagourous theorem

[heady warren]

You could try out Khan Academy if you haven't already, and maybe places like Alison courses and CK-12. Khan Academy provides lessons and practices, and CK-12 has a lot of textbooks available as well as practices with many chapters.

@heady warren

[torn terrace]

Mate this is probability not geometry and trigonometry

The answer is the reciprocal of the probability of any given combination

So 1/total number of marbles * 1/total number of marbles after removing one marble

The reciprocal of that

[covert frigate]

He's using everything to solve that question.
Give that guy an apple

[subtle karma]

can someone explain how to solve this?

so i can do the rest myself

[exotic yarrow]

Left the server ^

[ebon tinsel]

Ray bc is tangent to the circle. At the point of intersection of u were to draw a line such that is passed through the point of intersection and the radius of the circle then the angle between the ray and this new line is 90 degrees. As this is the case in this quesiton. And we know that a triangle contains 180 degrees then we can deduce the third angle as we know two of them already

[shut sluice]

hi yall i need help w geometry

Choose a room of your choice. Render this room with furniture at a scale of 1:50. Show with the help of calculations how you arrived at this representation.

[slim dragon]

Probability is a chapter in my geometry class

[torn terrace]

that's kind of odd, not a class I'd expect to see it covered.

[slim dragon]

Thats what i was saying

What class is it normally taught in besides yk

Stats

[tight solar]

algebra 2 in my county, maybe even state

[slim dragon]

O i think where im from its taught in alg 2 geometry and pre calc

Idk why

[meager citrus]

How do I know what’s on the x-axis

[thick kraken]

What is the difference between geometry and trigonometry?

[gray moss]

Geometry is dealing with all kinds of shapes and trigonometry is dealing with special triangles which have one angle 90° ig

[thick kraken]

Ah.

[dusky orchid]

can you guys do the question--- If x is obtuse and sinx=5/13, find the other primary trig ratios. Then evaluate for x.

[fiery oriole]

trigonometry uses specific functions, most notably sin, cos and tan and also uses a new unit called radians as well as degrees

[tawny totem]

ughh i cant keep all formulas in my mind of trigonometry they all are a mess

[dire hazel]

god im washed up on geometry

gotta do some for some math placement exam and im legit dying on whats prolly the easiest thing ever

[plain dagger]

same

[dire hazel]

Yeah it was easy

[light stream]

Hi, I just wanted some kind soul to confirm the validity and formatting of my trig answer.

Refer to the graph of 𝑦 = cos 𝑥 to find all the values of 𝑥 that satisfy the
equation cos 𝑥 = √3/2

referring to the unit circle, I came up with this answer: x=pi/6 + 2pi(n) and x=11pi/6 + 2pi(n)

[frosty rose]

helloss, im looking for anyone that can help me in how to draw a line in an orthonormal system just through the equation itself without any reference to a point that is on the line (if it is possible to do so). Thanks in advance !

[tawdry condor]

and circles sometimes

[empty cliff]

how do i remember cos tan sin

[vivid plinth]

You mean the ratios?

[worthy matrix]

SOHCAHTOA

[light stream]

Does anybody know if trig identities are typically provided on your formula sheet on exams or do you have to memorize them?

[dark sparrow]

this sort of humor belongs in #chill

[low oar]

as a mod I am seconding this

[vast saffron]

How to eliminate theta and r to get x, y

Polar to Cartesian

[upper karma]

yep this is correct

[upper karma]

provided (at least where i am)

[upper karma]

i have a question about geometry but i couldn't translate the question (native to english) ;;_;;

i guess i did

In Figure 1, we have a piece of paper in the form of the ABC equilateral triangle. The BDC isosceles triangle is cut from this piece of paper.When then glued to collide [ AC ] to [ BC ], figure 2 is also obtained that the image is obtained. Accordingly, how many degrees is x?

A-) 30 B- )45 C-)60 D-) 75 E-) 80

True answer is C (60)

[frosty rose]

helloss, im looking for anyone that can help me in how to draw a line in an orthonormal system just through the equation itself without any reference to a point that is on the line (if it is possible to do so). Thanks in advance !

[upper karma]

help

[gray moss]

Umm yes

[gaunt nexus]

Can someone suggest some books for trigonometry?

[silent plank]

generic online resource,
not really worth having a dedicated book for trig

[slim dragon]

The way i did it in my out of district course is that i had to memorize them

[frosty rose]

helloss, im looking for anyone that can help me in how to draw a line in an orthonormal system just through the equation itself without any reference to a point that is on the line (if it is possible to do so). Thanks in advance !

[tacit hearth]

hello!

[distant oak]

hii

Someone who speaks Spanish to explain to me the congruence of triangles please?

[lucid anvil]

What is the best way to re-learn geometry as an adult?

Are there any particularly good books that focus on intuitive teaching methods? Or are there any particular geometry books you recommend?

[upper karma]

Go on khan academy and do the geometry unit it helped me a lot.

It isn’t a book but I don’t think a book is that good for learning math

[spiral ruin]

Is this the right place to talk about the computation of pi in a non-euclidean geometry?

[smoky jetty]

I'd like to share a sort of math argument (tho the structure is far from formal) I did for like an hour showing why the given problem in our test had the wrong measures, based on diff theorems. Got concerned about it coz the problem was worth ~27% of the test and who knows if my classmates got wrong answers because of the inaccuracies.
Not really homework, but I just want to ask if it could be verified, coz i allotted much effort to it, lol

[woven bay]

Show that a triangle is equilateral only if its angles verify the relation: sin A/2 * sin B/2 * sin C/2 = 1/8. If anyone could give me some hints it'll be great.

[patent apex]

well since it is equilateral then A=B=C

[candid crescent]

and if its equilateral, angle A=angle B= angle C

[patent apex]

then (sin A) ^3=1/8

[candid crescent]

all of them add up to 180 degrees

[patent apex]

wait

what

never mind

[upper karma]

OMG ur notebook seems to cool

[smoky jetty]

ohh lol thx

[nocturne remnant]

Did you mean “if and only if”?

[ebon depot]

can any one explain con sin. I am doing an AI course and I dont understand

[woven bay]

yes

[worthy matrix]

sorry?

[dark sparrow]

"con sin"?

[nocturne remnant]

Wis con sin or something lmao

[worthy matrix]

real

[nocturne remnant]

The three trig functions wis con sin

[grave pond]

con tah sia

[worthy matrix]

do you mean cos and sin?

[plush cosmos]

Find tg(45-a) if sina and a are the second quadrant angles

help?

[ebon depot]

Yes

[dark sparrow]

have you done trigonometry before

like in any capacity

you might want to look up resources on intro to trigonometry and/or SOHCAHTOA and/or right triangle trigonometry

[patent apex]

sum it up pretty well

with this you can find the size of parts of a right angled tringle

for example

if theta = 50°

hypotenuse is 10

then opposite side is 10(sin 50°)

think of sine and cosine as operations like factorials

[river vapor]

how do i mug up all trigonometric formulas

[drowsy vector]

Like to learn all the basic trigonometric functions including the reciprocal identities or the laws of cosine etc?

[river vapor]

i mean double angle , half angle sin 3a , multiple angle , sum to product

i have to mug up for jee (entrance exam)

[drowsy vector]

The 3 basic trigonometric functions: sine, cosine, tangent. They all give the ratio in length with the respected sides they work with, e.g with the phrase SOH CAH TOA,
we know SOH means sin(x) = o/h, which means the ratio of the opposite over the hypotenuse side, e.g
sin(x) = o/h
sin(x) = 3/5
sin(x) = 0.6

if we turn the decimal to percentage we get 60%, meaning the hypotenuse is 60% larger in length then the opposite side, and if we use inverse sine with this ratio we will get theta
asin(0.6) = x

we get approximately 36 degrees

[drowsy vector]

I dont follow

[tawdry mango]

Just use e^ix = cos x + i sin x

[river vapor]

can i derive an formula with eular form?

[tawdry mango]

You can derive all of them

[river vapor]

although Idk andvanced calculus yet although ug calculus is in my syllabus

[slim tundra]

Hi. I was wondering about a problem for some time and was unable to solve it. Would love to hear the solution from you guys.

We're given a circle, it's diameter, the center of the circle, a random point on the circle and a ruler (only for connecting two dots). Draw a perpendicular line from that random point on the circle to the diameter.

[river vapor]

can you link me with a source , like how do i derive them

[tawdry mango]

$e^{i(x+y)} = e^{ix} e^{iy} = (\cos x + i \sin x) (\cos y + i \sin y) = \cos(x+y) + i \sin(x+y)$

[somber coyote]

DraK(night)

[tawdry mango]

Just manipulate and simplify

[river vapor]

thankyou found some videos on yt related to it

[plush cosmos]

can anyone help me with geometry?

In the right triangle ABC, the angle ACB=90 degrees, the angle CAB=60 degrees, the hypotenuse AB=a. On the continuation of the height CK passed to the hypotenuse, point M is taken, which is equidistant from the lines CA and AB. Find the area of the triangle AMC

[silent hatch]

Can anyone help me solve this?

[ebon tinsel]

As suggested it is noted that the perimeter is 16 cm therefore AB + CD = 8. From this we note that 1/2 rad * (OB + (OB +4) ) = 8

OB can be found through this

Which can then be used to sock everything else

[silent hatch]

alright tysm

[gentle steppe]

can someone explain to me how 1/3pi r^2(r/2) can be simplified into pi r^3/6

given this prob

V

[dark sparrow]

$\frac{1}{3} \cdot \pi \cdot r^2 \cdot r \cdot \frac{1}{2} = \paren{\frac{1}{3} \cdot \frac{1}{2}} \cdot \pi \cdot (r^2 \cdot r)$

[somber coyote]

Ann (glomed)

[dark sparrow]

@gentle steppe

[gentle steppe]

tysm!

i didnt think of r/2 as r*1/2

[plush cosmos]

In the right triangle ABC, the angle ACB=90 degrees, the angle CAB=60 degrees, the hypotenuse AB=a. On the continuation of the height CK passed to the hypotenuse, point M is taken, which is equidistant from the lines CA and AB. Find the area of the triangle AMC

[plush cosmos]

<@&286206848099549185>

[grave bronze]

1 sec I’ll try

[agile laurel]

Try drawing it

[plush cosmos]

i could not draw it correctly

[echo wasp]

💀

[grave bronze]

I have a figure

[plush cosmos]

I let mistake

[grave bronze]

Replace K with M

A Hint AK= CK^2 / KB

@plush cosmos

[plush cosmos]

i will try

[echo wasp]

im actually high

its liek 5 am

good bye

im giving such bad help

[grave bronze]

💀

[grave bronze]

Nah broo

[plush cosmos]

this from which triangle

[grave bronze]

Triangle ABC
U shd deduce this

As CK^2 =AK * BK

[plush cosmos]

yes i cant continue than

[grave bronze]

Get all 3 30,60,90 triangles

And do basic trigonometry

[echo wasp]

so i was right

bruh

i actually think what ever i do is throwing cuz

5 am comsci is stupid

[grave bronze]

When did anyone say you are wrong

[echo wasp]

i just feel like im wrong

[grave bronze]

I can’t understand whatever that is

[echo wasp]

its liek a trig identity

[grave bronze]

Ohh

[echo wasp]

and teacher took off marks bc it said i cannot divide by 0

even thought 1/tanx exist on teh other side

it was liek 2 years ago

but yea

gave me the anxiety and now im scared of everything in math

[grave bronze]

||Yeah I don’t have trig so how’d I know haha|| damn bro u go

Got it bad for u

[echo wasp]

tirg is easy

just teachers at my hs stupid

cannot consider basic logic

[grave bronze]

Y’all’s in Highschool!?

[echo wasp]

hs yes

[grave bronze]

Which grade.

[echo wasp]

like 11

12?

ish

idk

doing number theory on my own cuz i don't listen in class

[grave bronze]

How u don’t know your own class

@plush cosmos got it?.

[plush cosmos]

@grave bronze I can not find KM

[echo wasp]

i don't listen lmao

probalby grade 11

idk

[grave bronze]

What do u need KM for

[echo wasp]

m the interception?

[grave bronze]

No ‘m’ equidistant from AC and BC (extended)

[echo wasp]

oh bruh

[plush cosmos]

area of AMC=1/2 AK*CM

[grave bronze]

Ohh ya I screwed up

[echo wasp]

okay im actually gonna hop off bye

[grave bronze]

Bye

[echo wasp]

cya 💤

[grave bronze]

Use this

@plush cosmos

Use triangle A’MC

[agile laurel]

you should take a rest.

[fringe kelp]

hey guys i study in 10th class in india any suggestion for me

for my boards

any indian person

[agile laurel]

For Geometry?

If you want to be really good in the Geometry stuff that Grade 10 offers then I'd recommend "Challenge and Thrill of Pre-College Mathematics" by Prof. Ventktachala et al.

It has other stuff too, like Number Theory and stuff

I haven't completed the book myself, but the amount of Geometry questions there were enormous, and it won't be easy to solve. But those questions are very nice to work out, you'll have fun

@fringe kelp

[agile laurel]

But if you want this, I don't have any advice except to know your curriculum well. Knowing Indian Boards, they will never make your exam challenging to solve, at least the math paper.

[fringe kelp]

you are indian right @agile laurel

[kind cairn]

Given is a trapezoid |ABCD with base AB| and . CD | Circles with diameters |BC and A D | intersect at points P | and Q. The diagonals of a trapezoid intersect at S. Prove that P , Q and S | they lie on one straight line.
You very easily can prove that Pq lie on the radical axis of the trapezoid, but then i cant seem to calculate the power of point S
and if i can i get a very weird conclusion that says that BS=CS
whats not true ofc.
anyone any help?

[agile laurel]

yes.

[muted sparrow]

How does one generalize the notion of hemisphere to n+1 dimensions?

I thought about it, and in 3D we can say H_n = S^2 intersecected with p dot n >= 0, i.e. we intersect it with a half-space

defined by some vector n

On the other hand in 4D I have 2 options

i.e. I can take a single vector, thenthe hemisphere can be described in spherical coordinates in [0,2pi)x[0,pi]x[0,pi/2]

Or I could take two perpendicukar vectors for which I must have p dot n_i>=0

Then this type of hemisphere can be described in terms of [0,2pi)x[0,pi/2]x[0,pi/2]

Any obvious issues with my logic?

I believe the first approach should be "the correct thing" since a hemisphere is supposed to be half a sphere

[fair vapor]

how do you do this

[foggy parcel]

do u know the values that makes sin = √3/2 ?

[grave bronze]

sine 60

[sand jacinth]

Duh

[grave bronze]

Nvm it gone

It’s the same one like in #prealg-and-algebra

[grave pond]

(Please don't delete modpings even after they become moot -- that makes it harder for us).

[median dragon]

can someone tell me why Sec(Theta) and Cot(Theta) lies in Quadrant 1 and 2 when they have the same sign rq

[upper karma]

I don't know if this is exactly what you're looking for, but I'll put it this way:

1. Sine and Cosine are both positive when in first quadrant.
2. Only sine is positive when it is in the second quadrant, while cosine is negative.

Hence, you can see that the secant will be negative in 2nd qdr cause 1/(-cos) will give you a negative value for the secant.

Also remember, the definition of the tangent: sin/cos. If both sine and cosine are positive or both are negative, the tangent will be positive too. This only happens at FIRST and third quadrant. If one of them is negative and the other one isn't, the tangent will be negative too, since you'd be dividing a positive by a negative or vice versa. This only happens at SECOND and fourth quadrants.

So, to sum it up:

In the first quadrant: sine and cosecant (+), cosine and secant (+), tangent and cotangent (+)

In the second quadrant: sine and cosecant (+), cosine and secant (-), hence, tangent and cotangent (-).

In the third and fourth quadrant, the sine will be negative, therefore the secant only can be in the same quadrant as the cotangent when they have the same sign, happening only in 1st or 2nd.

I don't know if I made it clear enough.

If you need a visual explanation or something I'll be glad to do it :))

[dire lance]

Two coast guard stations A and B are located along a straight coastline. Station A is 20 miles duenorth of station B. A supertanker in distress is observed from station A at a bearing of S9E,andfrom station B at a bearing of S37E. Find the distance from the supertanker to the nearest pointon the coastline.

.

.

Can anyone help me with this question?

It’s been 4 hours

I can’t figure it out.

[dark sparrow]

@dire lance do you still need help with this

[echo wasp]

i just pulled an all nighter lmao

[austere mortar]

is this alright or did I just pull invalid proportions from thin air?

[plush cosmos]

t he sides of the isosceles triangle abc are equal to 15, and the base ac is equal to 24. o is the point of intersection of the continuations of the altitudes of the triangle. Find ob

<@&286206848099549185>

[floral shale]

What

[paper heath]

if two triangles are equiangular are there sides are equal to one another

[austere mortar]

you mean if they have the same angles?

[paper heath]

yes but no similar sides

[austere mortar]

if you mean like this, where a trangle has angles equal to 60°, then all of its sides are equal

if like this, where two triangles have the same angles, not necessarily, rather theyre just similar becouse they could be in different sizes

[odd bone]

Can anyone prove the results

[severe orbit]

hello

[austere mortar]

|AD|; |BE| and |CF| are all parallel?

[severe orbit]

<@&286206848099549185>

[austere mortar]

what does that equation sign with ~ above it mean? Ive never encountered it and now Im interested

[severe orbit]

That means congruent

[austere mortar]

ohh, alright

[severe orbit]

can tou help me broC

?

irs my last question

its

[austere mortar]

I can help you with one of those questions for sure so please gimme a moment

[severe orbit]

Ok ty

[austere mortar]

have you tried solving the 5th question yet?

[severe orbit]

Which one the last one?

Got this so far

I know the last 2 are correct

I dont know about the first 2

[austere mortar]

unfortunately I cant really help you with the first two sentances since I have a little trouble with understanding math english bc it isnt my first language, sorry, but the two questions before the last one definitely seem alright

[severe orbit]

Alright

[austere mortar]

I have no idea why theres an in (an + bm) / (m + n), but this is the best I could do by using Thales' proportionality theorem

please ping me with the sollution if youll get it, Im really interested in what I missed

[last ocean]

Need help. please ping me with the sollution

[frosty plume]

i love meth

[frozen ocean]

so do I

[frozen ocean]

Does KM says 8.8 there?

Can't understand your handwriting

[austere mortar]

what does "Sent = ?" mean or rather what is it referring to?

[last ocean]

Just 8

[last ocean]

Area of ENT triangle

Solved it

Need help with this one

[weak musk]

i think youre wrong

cuz BM : MB^

[austere mortar]

wdym?

[coarse fossil]

what is this bro

how do i solve it

[safe tundra]

I think it means mass point

[vivid plinth]

You multiply the points by their respective weights and then find the average.

[raw nimbus]

yo like for a proof
given: opposite sides of a rhombus are parallel
to prove: like lines going through opposite corners of the rhombus are perpendicular (im sorry this is vague idk how to describe proofs)

I can do a drawing if needed

[austere mortar]

I think you'll be able to finish it from here

[austere mortar]

oh wait, |BD| is a median so x doesnt have to be equal to y, then never mind, I messed up, sorry

[last ocean]

I solved that

Its 5:1

@austere mortar

[austere mortar]

may I see how you solved it?

[last ocean]

My writing is a mess but i can explain it

Using area to find ratio of triangle KTD and DTM

Then simplifying it to 5:1

[austere mortar]

I see

[last ocean]

This is more like impossible

It uses same formula but i think i need to construct line

Need help. Ping me if u solved it

[austere mortar]

is |AK| = 2x or 1x? I cant really make it out

[last ocean]

2x

[last ocean]

Solved it

[prime vigil]

for sin(theta)= 1/2 i understand one of the solutions is pi/6 + 2pik for any integer k

but book says theres another solution for quadrant 4 which is 5pi/6 +2pik for any integer k

why is there a 5pi/6?

31

wait x,y = cos, sin

Instead of pi/6 it would be pi/3

[twin aurora]

No you are right, @prime vigil, sin (pi/6) =1/2

But 5pi/6 is another option

[prime vigil]

how tho

[prime vigil]

where do you get 5pi/6 from

[twin aurora]

Do you see how 5pi/6 also has a y value (sin value) of 1/2?

[prime vigil]

Oh wait

yeah

my book said that was in quadrant 4, isnt that quadrant 2?

[twin aurora]

Hmm interesting. But your book also said the solution is 5pi/6

[prime vigil]

Oh wait im an idiot

i looked at cos (1/2) for the one they were showing

not sin(1/2)

[twin aurora]

ha, lol. Happens to all of us!

[prime vigil]

ok makes sense

[onyx narwhal]

Find the radius of this circle, pls help

[coarse fossil]

ok

[coarse fossil]

how do i do this?

[hollow pollen]

got a question, what type of graph goes like this

[coarse fossil]

cube root

[hollow pollen]

thx

[upper karma]

Can someone help me with the second part of this problem

you must use cos^-1 (5/6)

the inverse of cosine

@upper karma

Why do you use that

so we know that from the given sides, cos(x) = 10/12

Im talking about ii

oh

its pythagorean

x^2 + x^2 = 144

[south summit]

you also calculate with trigo

since its isosceles triangle

and D=90

[upper karma]

yea

[south summit]

but pythagorean is easier

[upper karma]

x^2 = 72

so x = 6 root 2

which is around 8.49

@upper karma

the answer is 8.49

Ohhhh

Alright thanks

np

[upper karma]

to me this looks like t is just the distance

so algorithmic way is to use distance formula

[onyx narwhal]

.

[tribal sequoia]

An interesting problem: Solve $$ \tan 3 \theta \tan 2 \theta = 1 $$ for $$ 0 < \theta < \frac\pi2 $$

[somber coyote]

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

[echo wasp]

radians

[echo wasp]

nice username

i agree

[upper karma]

Bro im almost done with geometry a 2 more days and ill be done funny thing is that i started yesterday 💀

[thorny fossil]

Using the complex definition of sine and cosine yields a really nice answer (pi/10)

[upper karma]

help

[tribal sequoia]

Solving it without using the complex definition of sine and cosine also yields two nice answers: π/10 and 3π/10