geometry-and-trigonometry

[upper karma]

?

[dark sparrow]

Eo?

[upper karma]

so*

[dark sparrow]

no, the circumference is not 64.

[upper karma]

How

[dark sparrow]

you said it right here.

it's C = 2πr, not just C = 2r.

[upper karma]

200.96?

[dark sparrow]

i hate to repeat myself.

[upper karma]

201?

…….

[dark sparrow]

the problem says to leave your answer in terms of pi.

so don't do any of this calculator or rounding bullshit.

[upper karma]

yeah ik

[dark sparrow]

you could have, and should have, said "The circumference is 64pi."

[upper karma]

But i was still right either

Way

[dark sparrow]

no, you were not.

[upper karma]

yh its still 64?

[dark sparrow]

no, it is not.

it's not just 64.

[upper karma]

…..

[dark sparrow]

it's 64pi.

pi isn't a trendy symbol for you to stick next to numbers or ignore at will.

[upper karma]

Ok can we continue

Pls

[dark sparrow]

you seem to be under the impression that "64" and "64pi" are one and the same, when they aren't.

[upper karma]

okay

[dark sparrow]

and i would like for that not to be the case.

do you understand why you were not "still right" just because your answer was part of the actual answer [to my question]?

[upper karma]

mhm

[dark sparrow]

okay, so the circumference is 64pi. as we just established.

and now refer to this again.

what's (1/8) * 64pi?

[upper karma]

Uh

8pi?

[dark sparrow]

great.

the concept of "if the person helping you is asking you about something, then maybe it's something relevant to the problem & hence worth finding / thinking about". was this foreign to you?

[inner barn]

Hey, I'm helping my gf with her math since she hasn't done it In a couple years

I'm a bit confused on trig, the question asks for what x is equal to

Would I just write "x=9×(x-3)"

[humble pulsar]

pythagorean theorem..

then solve the quadratic

[upper karma]

hey, i was struggling with solving the missing variable for these types of problems, how would i solve this equation?

do you know of any useful circle theorems that we can use here? if not, try looking up in your notes.

@upper karma

hi and I do not, I can look though.

yeah no, i don't see any in the notes that I have

do you mean this? @upper karma

yes, precisely.

$$(A+B)\cdot B=(C+D)\cdot D$$

[somber coyote]

Al𝟛dium

[upper karma]

couldn't find any better pic, and can't make a drawing rn so yeah, we'll have to stick to that.

okay

but yeah that's pretty much the theorem, are you able to apply it to our problem?

yes, I think I applied B : 9, D : 8, C : 10, A : x correct?

yes.

are you able to form an eqn and solve it from here?

yes, I am

thank you so much.

you're welcome.

[wintry tundra]

is there a video for a proof of this? im curious how one would go about solving this

[upper karma]

consider drawing out triangles PAD and PBC, you can maybe notice something that these triangles have in relationship with each other.

there's a youtube video if you are interested: ||https://youtu.be/wqGmSTPbojs||

[wintry tundra]

really nice proof

simple too

[trim breach]

Question 5 sounds like you need to find the total surface area of that cylinder, then subtract the area from the top and bottom parts from it.

Question 6 wants you to set up a proportion between the surface areas and values of the two cans. So you would need to calculate the surface area for the dimensions of the new can to get a solvable proportion.

[tame quail]

what are the steps i should do for this

@mild cargo

[upper karma]

Could someone help me with this

[wintry tundra]

An arc length is equal to the angle making it

In this case at least

[upper karma]

no.

arc length≠arc measure, you probably meant to say arc measure. arc length is well known as rθ, where theta is the angle from the center in radians and r the radius.

[wintry tundra]

I did

My bad

[marsh kiln]

I just have a quick question. Given a random triangle with one altitude perpendicular to one side how would you inscribe a right triangle with one of its vertex collinear to that altitude and the other vertices lying on the same point as the vertices on the circumscribed triangle? If possible I would like to know how to solve this through a problem and not with a construction method.

[glacial bridge]

E and F how to do please someone

@upper karma yo

help?

[upper karma]

how about drawing a diagram

[glacial bridge]

o i forgot to include its to find the area of a rectangle

@upper karma hol up

[dreamy ridge]

Oh so the horizontal side is 12cos60 and the vertical side is 12sin60 using basic trig definitions

Sin60 = 1/2 x root 3

Cos60 = 1/2

If you don’t remember how to get the exact values of sin, cos and tan 30 and 60, remember it’s the equilateral triangle with side length 2, with a bisecting line straight through the middle

[glacial bridge]

@dreamy ridge u answering me bro?

[dreamy ridge]

Yea

[glacial bridge]

Im not going to lie, can you explain it in geometry terms cause I dont understand sin and cos

please

[dreamy ridge]

No problem but Wdym by ‘geometry terms’ exactly

[glacial bridge]

just explain it without cos and sin

[dreamy ridge]

So have you not met sin and cos x yet?

[glacial bridge]

Yeah

thats it

[dreamy ridge]

I don’t really know how to explain it without trig functions, because the reason the angle is given is so that can help find the sides using sin and cos

Otherwise there’d be no reason for the angle to be given

I think trig is built into the question

Does anyone else know if you can around this without using trig?

Sorry I couldn’t help you man

[jade badger]

wtf does this even eman

mean

[glacial bridge]

one question though so what would 12cos60 and 12sin 60 come up to?

[dreamy ridge]

12cos 60 will go to 6 as cos 60 =1/2

And 12sin60 = 6 root 3

As sin60 = (root 3) / 2

[livid moss]

@jade badger Let's say you have a cube of sidelength 5, you work out the surface area. Then you triple the sidelengths to 15. Then work out the sidelength. What is ratio between the old and the new sidelength? Now it turns out this ratio is independent of the original sidelength. So it doesn't have to be 5. That's why they don't specify a specific sidelength.

[dreamy ridge]

Ah ok, the length of sides will be 3 times longer, so the combined area will be 3^2 times longer as the ratio is being applied in 2 dimensions if that makes sense

[glacial bridge]

can i ask whats the formula for cos and sin

[jade badger]

tank q

[jade badger]

tank q

[dreamy ridge]

So what sin and cos actually means

Is just the ratio of sides of a right angles triangle

So have you heard the words hypotenuse, opposite and adjacent before?

@glacial bridge

[glacial bridge]

yeah

[dreamy ridge]

So sin (x) where x is an angle is just the ratio of the opposite over the adjacent

Cos(x) where x is an angle is just the ratio of the adjacent over the hypotenuse

And tan (x) = sin(x)/ cos(x)

Which is the opposite over adjacent

So how we could use this is

The fact that sin 60 ( for that question) is the opposite over 12 ( as 12 is the hypotenuse)

Multiplying both sides by 12, 12sin60 = opposite

Now 12sin60 is just a number in your calc

So just type that in

And you get a vertical length

Does that make a tiny bit of sense?

[glacial bridge]

i understand so sin would find the opposite

which is the length

i mean vertical

[dreamy ridge]

So for the width ( horizontally), you’d use cos

[dreamy ridge]

Yep

[glacial bridge]

why did i get a - number bro

[dreamy ridge]

What exactly did you type in your calc?

Ah ok

Is it in radians?

Should be in degrees

Basically there is another type of angle measuring unit called radians but that’s irrelevant right now

Just make sure your calc is in degrees otherwise the answer will be wrong

[glacial bridge]

ah okay

12sin60 i got 10.3923048454

[dreamy ridge]

Yep

So the area is just those lengths multiplied together

And that’s it

Rest of the questions are identical, just think about the side you’re finding and whether it’s cos and sin and you should be all set

[glacial bridge]

idk for the sin one the number suppose to be that big tho u think it should be rounded?

[dreamy ridge]

I would leave it in exact form

In terms of surds ngl

I never use decimals

I mean don’t round it then work out area

Work out area then round it

If you wanna still use decimals

But the Google calc doesn’t give you surds

It just gives you a decimal

In surd form, sin60 = (square root of 3)/2

So 12 sin60 = 6 root 3

So area = 36 root 3

Which is approx 62.4 to 3sf

[glacial bridge]

@dreamy ridge u got a link to a surd form calculator

[dreamy ridge]

I only have a physical calculator

It’s worth buying one, it’ll make your life a lot easier

[glacial bridge]

damn ion got one rn u saying i cant convert to surds online?

[dreamy ridge]

I’m not saying you can’t, I have no idea if you can bc I’ve never done that

But I’m looking rn and it doesn’t seem like you can

As I said, I think it’s worth buying one bc it’s really convenient, unless you’re not allowed them for exams and stuff bc otherwise I wouldn’t get too dependent on em

But you can’t do this stuff without a calc unless you’ve been taught exact values of sin, cos and tan

[glacial bridge]

ig ill just do the pythagorem theorem

[dreamy ridge]

Yeah use phythagoras, so it’s 144- 36

Is the other side squared

108 = other side squared

Root 108 = other side squared

Split into surds and then you’ll see you still end up with 6 root 3 so it still works

Without trig

But you needed to use that one side using trig to get the other

[glacial bridge]

yeah i did the 2nd one i think i got it right

i got 6 as the vertical and 4 root 5 as the horizontal

[jade badger]

help

pl

s

[wintry tundra]

Arc length = the radius times the angle in radians

1 degree =$\frac{\pi}{180}$ radians

[jade badger]

$

[somber coyote]

visual of Petter's ascendance

[jade badger]

ok

[jade badger]

<@&286206848099549185>

[untold peak]

Hey guys I just wanted to know if you could tell me the formula for solving this.

I can't remember it for the life of me

[dreamy ridge]

Solving what?

To get BA

Then you’d use the cosine rule

So BA^2 = 27^2 +21^2 -2(27)(21)cos 33 deg

[untold peak]

Sorry I should've specified

Solving the whole triangle

[dreamy ridge]

No problem, was that it?

Wdym by solving the whole triangle, as in getting all angles

And sides

I’ve never heard that before lol

[untold peak]

Yeah like

Getting all angles and sides

[dreamy ridge]

Ah ok

[untold peak]

Mainly looking how to get side c

[dreamy ridge]

So now we have BA

We can use either use the cosine rule

Or sine rule

To get CBA or CAB

[untold peak]

the assignment is sine and cosine law

[dreamy ridge]

And the remaining angle will be 180- the others

So yeah it’s up to you, you could use the cosine rule rearranged to find one of the angles or you could use the sine rule

I prefer the cosine rule if I can, so for eg Cos CBA = (value for BA^2 +21^2 - 27^2)/ 2(Value for BA)(21)

So you can avoid the ambiguous case of the sine rule

[jade badger]

@dreamy ridge can u help

[untold peak]

Okay so I'm still very confused

How do I find angle A using sine or cosine on this?

[pure bronze]

Use the law of cosines

[elfin frigate]

can someone read this

[trim breach]

,,a^2 = b^2 + c^2 – (2bc)cos(A)

[somber coyote]

Lidoh

[trim breach]

A is your known angle.

a is the side you want to find (which is not adjacent to angle A).

b and c are the two sides adjacent to the angle you are using.

If it helps, rename the triangle using this information.

[trim breach]

You have a 90-degree angle and a 45-degree angle in that triangle. Find the last angle. You should notice something that will help you get the other leg of the triangle, and once you have two legs, you can solve for the hypotenuse using Pythagorean Theorem.

[upper karma]

How do you calculate the arc length and the area of a sector?

I’m a little confused.

For example: I got 45/2 * pi for the arc length.

[silent plank]

how did you get 45pi/2

[upper karma]

for the arc length?

[silent plank]

yes

[upper karma]

pi * 2r

30 pi

multiply by 270/360

90 pi/4

simplify

45 pi /2

Did i do something wrong

[silent plank]

for area you can apply a similar idea

[upper karma]

For that I got 168.75 pi

because fraction form would not be productive

[silent plank]

wdym would not be productive?

[upper karma]

675 pi/4

bulky

I mean

[silent plank]

i guess 168.75pi isn't too bad since its still exact

don't forget your units

[upper karma]

alright, so everything is good?

[silent plank]

yes, where's the confusion?

[upper karma]

just needed to confirm

thank you!

[rough roost]

help

[pure bronze]

Find the slope of the function you’re trying to build

[somber coyote]

Hunnydrips

[pure bronze]

And then build a linear function using the particularly well-known formula

[somber coyote]

Hunnydrips

[pure bronze]

And also, this is NOT geometry, please move to an appropriate channel

[dark sparrow]

$y - y_1 = m(x - x_1)$

[somber coyote]

Ann

[upper karma]

Could someone plz help me with the following question? "In circle O, diameter AOB is drawn and point C is located on the circle such that CA = 12 and CB = 16. What is the length of the radius of circle O?"

okay.

what have you tried so far?

Well ive tried drawing it out on paper and im still a bit confused

can i see the drawing?

Sure

that's good.

do you maybe notice any useful circle theorems that we can use here?

try looking through your notes to see if you notice one.

Well inscribed angles are half the of the arc, but idk if that’s very useful here. I don’t notice anymore in my notes

okay that's not the one i was looking for but it's good as well.

are you able to find the angle ACB, considering the theorem you just mentioned among with what the measure of the arc is?

give it an attempt.

Alright, not very good at trig but I’ll try. Give me a sec

this doesn't really contain trig, just consider the theorem you just stated and in order to find ACB, try to see if you notice what the corresponding arc's measure is.

well to find the angle of acb, i thought you need to use tan-1, cos-1, or sin-1

we can't use any of the trig ratios as we don't know if this is a right triangle.

unless you know it's a right triangle you can't use them.

[upper karma]

consider what i said here instead. try to think on what the measure of that arc could be. give it some thoughts.

ok, so angle ACB inscribes the circle of 180 degrees i think because of the diameter so the angle ACB would be 90 degrees?

yes.

good job

Then we can use the Pythagoras therom from there giving you a radius of 10

yes. precisely, good job

Alright, thank you.

[mental crystal]

What does it mean if lines/rays are distinct ?

Google translate does not do it for me

I'm not native english

[tiny snow]

Most likely they want to indicate that the lines or rays do not coincide.

[upper karma]

Could someone help me please

[tiny snow]

Just post your question

[tiny snow]

@upper karma You know the total length of AB (which you have 100% chance of hitting) and you know the lengths of CD and DE, so you can figure out how large a part they are of AB, and thereby the chance of hitting any of those two segments.

[upper karma]

so 11+3 over all all the numbers combined?

[tiny snow]

Yes.

[upper karma]

so its A

40%

?

@tiny snow

[tiny snow]

Correct

@upper karma Next time, post questions like these in one of the #questions channels. This is not geometry or trig...

[upper karma]

ok my bad

[tiny snow]

No worries

[mental loom]

hi, did i do this correctly?

i used law of sines

i dont know if i applied it correctly or not though

[past geyser]

@mental loom wdym

show your work

and we can correct it

but yes, using sine law is the best for this case

[mental loom]

106+30=136. 180-136 = 44, so B = 44
a/A = 24/30, which is 4/5,
b/B = 4/5 and 4/5 of 44 is 35.2 (that is rounded up), so b = 35.2
c/C = 4/5, and 4/5 of 106 is 84.8 (rounded up also) so c = 84.8
so
B = 44 degrees, b = 35.2, c = 84.8

[past geyser]

👀

wtf

this solution is messy and unorganized

show it in full

with the trig ratios

i don't know what you mean by a/A

[mental loom]

a is the side length and A is the angle measure

[eager star]

Help

For the top problem

[inland dock]

73° and 73°

You can do like (360°-(115°+31°+68°))/2 for GIJ angle
And the same for the GHJ

[gilded vault]

who can do trig

useless

[inland dock]

Send trig here , because of the name

May be someone would take it

[trim breach]

In the Law of Sines, you take the sine of the angle measures. I am not sure if that what you did, but your explanation makes it sound like you didn’t.

,,\frac{a}{sin(A)} = \frac{b}{sin(B)}

[somber coyote]

Lidoh

[teal kayak]

How do I find x?

[dreamy ridge]

So we have the hypotenuse and we want to find the adjacent so it’s gonna involve cos

Cos 47 = x/33

So 33cos47=x

[eager star]

thank you so much bro

And for this

would i do 7x+21=(47/2)

?

[humble pulsar]

2 angles in the major and minor of the arc are supplementary

so 7x+21+47=180

[eager star]

oh

so would x be 16?

[humble pulsar]

They assumed the RTP statement was true from the getgo

From the very start it's wrong

Unclear imo

ask your teacher for clarification then

[dreamy ridge]

See what I mean @bleak shell, the problem is the question itself

[narrow seal]

is it c?

[upper karma]

@narrow seal yes.

[storm portal]

you can check using desmos

[narrow plinth]

how would i do 7.10

[storm portal]

You're gonna have to translate that

[narrow plinth]

i ment 7.9

basically it just said solve the inequality for x

[upper karma]

okay.

what have you tried so far?

[narrow plinth]

ive tried

for it to be bigger than 0 either both of the parenthasis have to be bigger than 0 or both less than 0

and for each of them

i've got some funky and crazy numbers which im guessing their not cirrect

Idk what to do now

[icy arrow]

bro wtf is that

what class are you learning that in

am i safe

[narrow plinth]

prolly

[icy arrow]

dayum

[narrow plinth]

4th grade high school

[icy arrow]

is that uuh soh ca toa or something

or sin cos tan

wait

im confused

[icy arrow]

wait what

do you know what class thast is in british system

[narrow plinth]

idk it's the last class of high school like the 4th year

i'm 18

[icy arrow]

ooh my brther class

igcse 2

i think

year 12 i think

[narrow plinth]

i'm just completly lost in this math problem

[icy arrow]

my brain hurted agen

[upper karma]

i have to go, but you can try with weirerstrass substitution on the second one if you are stuck.

tomorrow i will check it surely, if you still haven't got helped.

@narrow plinth

[narrow plinth]

thx

❤️

[upper karma]

https://en.m.wikipedia.org/wiki/Weierstrass_substitution

[narrow plinth]

I mean we're supposed to do this without calculus

[wintry tundra]

That is trigonometry/precalc

Learning about how trig functions work and why/how they are used

I use trig quite a bit on my high school robotics team

[fickle badger]

anyone wanna help me with my hw

<@&286206848099549185> does anyone wanna help me with my hw

[onyx cloud]

#❓how-to-get-help

@fickle badger ^

[dark sparrow]

is this the discord crashing gif

[pure cape]

ig this is where reduced motion comes in handy

[upper karma]

weirerstrass substitution is used in calculus but it's essentially a simple trigonometry substitution.

there's no calculus involved in it.

[narrow plinth]

Ok where do i do that substitution

[upper karma]

you can consider using it on both cos(2x) and sin(x), will eventually lead to an inequality with abs signs.

this is just one way to do it of course

[icy arrow]

i just woke up please, no

[gloomy mural]

hi guys, I am having a brain fart rn and can’t solve this, all I found is that EB is sqrt of 80 and that’s it 🤷‍♂️, any help is appreciated

[pure bronze]

Similar triangles

Notice the angles really carefully

[gloomy mural]

oh ye lol like triangle ADC is similar to BCE?

[pure bronze]

ADE

Yes

[gloomy mural]

yeah thx, miss-typed

[pure bronze]

No sweat

Try to find line-segment DE using that

The three triangles are similar, that’s another thing to look at

[gloomy mural]

hmmm like what?

[pure bronze]

What do you mean?

[gloomy mural]

oh wait srry, read the msg incorrectly

Damm I should sleep more or sth

[pure bronze]

What do you need brother

If you have side-length EB

Then you can use the proportionality between the triangles and find AE

[gloomy mural]

nah nah I am good, thx for the little push, triangle similarity is ez just totally forgot about it

[pure bronze]

It’s fine, happens

[wintry tundra]

what

i was just telling you that thats what the class is

[tawdry saffron]

-1+2cos(2x+pi/3)+5sin2x

I have to write this in form of p+qcos(r(x-s))

i tried simplifying but i couldnt get that form

[wintry tundra]

you can write sin as cos with phase shift transformations

[exotic heron]

is it normal that with an initial velocity of 15m/s^2 to get two component vectors that don't add to 15?

[twilit warren]

additive squares is 225 so its still 15

so yeah thats fine

[exotic heron]

additive squares? excuse my ignorance

[twilit warren]

oh sorry

11.49^2 + 9.64^2

thats 225

sqrt(225) = 15

so its fine

[exotic heron]

beautiful thank you v much

@twilit warren , if you can be so kind to help with this last equation. How would I go about solving for t? just a nudge in the right direction I should be able to get it. I just don't ever remember solving for a variable that appears twice like this.

[twilit warren]

oh this is quadratic equation

at^2 + bt = 0

[exotic heron]

ah. thank you I'll look up the quadratic solving method.

[twilit warren]

its just

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

[somber coyote]

aditya

[twilit warren]

but thats for at^2 + bt + c = 0

c = 0 so that falls out

[exotic heron]

oh man this is a wake up call i need to refresh my precal much more.

[twilit warren]

and you already have a and b as coefficients, im just too lazy to rewrite that

also you could solve by graphing if you want an estimation and are allowed that method

[exotic heron]

algebraic only unfortunately

[twilit warren]

yeah that makes sense

[twilit warren]

oh lol btw here's an alternative

you have an equation of the form

$at^2 + bt = 0$

[somber coyote]

aditya

[twilit warren]

right?

just factor a t out

$at^2 + bt = t(at + b) = 0$

[somber coyote]

aditya

[twilit warren]

solve at + b = 0
and then also t = 0

drops it to solving a linear equation

[exotic heron]

thank you, i appreciate the extra explanation

[twilit warren]

np!

[exotic heron]

Would it solve to approx. 1.97?

[twilit warren]

yep, good job @exotic heron

[exotic heron]

thanks to you!

[keen lark]

could i get some help on this please? i've tried going about this several ways and still can't get it

[icy dome]

Is there any equation for these curves?

[dark sparrow]

what even are these curves?

[wise pawn]

https://www.desmos.com/calculator/rfxeosrojj

something like this

[upper karma]

looks like a sine wrapped around a circle

[icy dome]

@wise pawn thanks man.

@upper karma well, this is locus of an electron around a nucleus. So, ya , sine waves wrapped around a circle

[upper karma]

so just use polar coordinates with radius being a sine in terms of the angle

[silk patio]

That debroccoli guy was wrong tho

[sonic trail]

It'll form an ellipse my man. e²=(1-b²/a²). Just try that. Here a = semi major axis and b= semi minor.

[hardy dagger]

@sonic trail are you a PhD?

Or Masters?

[sonic trail]

I am a school student why?

[hardy dagger]

Oh from Pakistan?

Just asking....reply if you want

[sonic trail]

No no. I'm indian.😁

[hardy dagger]

Oh ok I'm indian too

Just started class 11

[sonic trail]

Heyyy. Delhi ncr. I'm a dropper.

[hardy dagger]

Boards cancelled 😋

[sonic trail]

Yeah. Lucky u. Don't slack off tho. Stay at it.

[hardy dagger]

@sonic trail what do you mean by dropper? Could you pls explain the term...

[sonic trail]

I've taken a year gap after 12

To prepare for JEE

[hardy dagger]

@sonic trail yeah yeah thats for sure...other mummy's slipper is waiting

[sonic trail]

Hehehe. Yeah! The real motivation behind studying. That chappal😆

[hardy dagger]

@sonic trail yeah I'll also take a drop....now I'm preparing for getting a scholarship in FIITJEE

@sonic trail yeah 😅

[sonic trail]

Ohh yeah! Do those aptitude type ques. You can score well using that. And try going thru prev years papers + a little bit of 11nth ncert

I was in fitjee too

Hauz khas delhi.

Where are u?

[hardy dagger]

Ok brother....thanks for taking out some time to talk to me 😊😊

I'm fro west bengak

Bengal*

[sonic trail]

Oiii. Khoob bhalo

[hardy dagger]

Bah dhonnobad😆😁

Bye 😊

[sonic trail]

Ba bye

[glacial bridge]

https://cdn.discordapp.com/attachments/269573202018041856/838464240607166464/Screen_Shot_2021-05-02_at_1.17.12_PM.png

[silk patio]

x=-8

[glacial bridge]

fr? how

[silk patio]

By inspection

Do you know how to solve these questions

[glacial bridge]

This one no

[silk patio]

Google “sum of exterior angles”

[fiery flicker]

I come here to learn Pre AP Geometry

Cuz my teacher doesnt teach

I know a bit but i need to pass the class

[old obsidian]

Hello guys , I am in 9th grade and I recently learned about trigonometry in the right-angle triangle , about reduction to the first dial and the Theorems of sinus and cosinus and I didn't understand a thing , could you guys give me some help please ? ( I'm sorry if my grammar is a bit rusty )

[fiery flicker]

Im in 9th too

[old obsidian]

nice

[lofty marten]

special right triangles?

[old obsidian]

yes

[lofty marten]

the special right triangles make it easier to calculate the side of lengths of a triangle if you have the angles for them

[old obsidian]

i see

[lofty marten]

so for a 45 45 90 if you are given X you can get the hypotenuse by just substituting

if it was 3 then the side with angle 90 would be 3sqrt2

for 30 60 90 if given X you can plug it in and get the side lengths of angle 90 and 60

hope that helps

[old obsidian]

yes , thank you

[worn marsh]

How do I solve these

It confuses me so much 😂

[full briar]

What are you trying to solve? The angle?

I think the adjacent angle of that is 75deg

180-105=75 (supplementary angle)

@worn marsh

[worn marsh]

@full briar I got the answer for the first

But not the second

Here’s the question

[full briar]

Uh for the second one the rhombus right? The second angle should be 105 and third angle is 75

[worn marsh]

Ok

Thanks

[full briar]

Np

Just note third angle I use supplementary angle and the second one I think it has to do with the property of the shape (rhombus)

[worn marsh]

Alright thanks

[upper karma]

Hello can someone help me with the answers of my geometry hw

[jaunty sail]

@upper karma do you still need help

[sterile cliff]

I'm confused on how the SAS reflects to ABC's congruence to ADC

[upper karma]

If $BC = DC, \angle ACB = \angle ACD$ and $AC$ is common, then triangle $ABC$ is congruent to triangle $ADC$

[somber coyote]

omeganebula

[upper karma]

As 2 sides and the angle between the two sides are equal

SAS will apply

[urban vortex]

looking for some kind of explanation on how to go from degrees to radians in pi form

google offers the formula degree * pi/180

to get rads

but how do i know for example that -60 = 2pi/3

[humble pulsar]

they arent

$-60^{\circ}$ is co-terminal with $\frac{5\pi}{3}$

[somber coyote]

moshill1

[urban vortex]

the -60 i get from the calc

the book technically says

-sqrt(3)

[humble pulsar]

yes their tans are the same, but they arent the same angle clearly

$\tan{x}=-\sqrt{3}\implies x=\frac{2\pi}{3}+n\pi , n\in\mathbb{Z}$

[somber coyote]

moshill1

[humble pulsar]

when n=1 you get 5pi/3 and when n=-1 you get -pi/3

[urban vortex]

can you explain this calculation

[humble pulsar]

Sure, so ik arctan(-sqrt(3)) = 2pi/3, and ik tan is pi-periodic, so any integer multiple of pi added to 2pi/3 will also be a solution

[urban vortex]

why is it 2pi/3

is the specific question

[humble pulsar]

cause it is..?

I mean you could use double angle with tan(pi/3) to determine it's -sqrt(3)

[urban vortex]

how do i know that

[humble pulsar]

unit circle

[urban vortex]

i mean, how do i conjure these numbers

[humble pulsar]

Unit circle, the graph, just learning trig

[urban vortex]

i have (-sqrt3)(pi)/180

[humble pulsar]

idk why you have that but sure

[urban vortex]

allegedly thats how radians are calculated

[humble pulsar]

-sqrt(3) isnt radians

[urban vortex]

right right

ok well that just throws me a step back again

arctan on the calculator just gives me -60

do i use that to get to pi?

[humble pulsar]

$-60^{\circ}\cdot\frac{\pi}{180^{\circ}}=\frac{-\pi}{3}$

[somber coyote]

moshill1

[urban vortex]

um and from this to 2pi /5 pi?

[humble pulsar]

tangent is pi periodic, so any integer multiple of pi added onto a solution will also be a solution

so in this case we're solving $\tan{x}=-\sqrt{3}$ and we know $x=\frac{-\pi}{3}$ is a solution, so $x=\frac{-\pi}{3}+n\pi , n\in\mathbb{Z}$ are all solutions

[somber coyote]

moshill1

[urban vortex]

so its just convention to use positive values?

[humble pulsar]

Depends on how the principle angle is defined, could be [0,2pi) or (-pi,pi]

[urban vortex]

and is a calculator the only way to calculate the arctan?

is it a particularly horrible calculation?

[humble pulsar]

I mean if you want to buy the old trig table book, go ahead and use that

[urban vortex]

its just that we dont use calculators for literally anything else

this seems odd

[humble pulsar]

Cause you dont need a calculator to do basic operations with. . .

[urban vortex]

tan is considered a basic operation?

[humble pulsar]

No

[urban vortex]

but its just y/x

[humble pulsar]

ok what's tan(37 deg)?

[urban vortex]

is there no equivalent for arctan?

[humble pulsar]

I mean there's the taylor expansion, but that's calculus. Just use the calculator

[urban vortex]

alright

so the full route

to solve this kind of question

is tan theta = b/a

use a calculator to get the degrees out of that

then multiply by pi/180

what do you do if theres parameters/variables or something such

[humble pulsar]

or just put your calculator into radian mode. . .

[urban vortex]

this is not as trivially obvious as you might believe

besides the obvious correlation here i dont really know what radian mode is or how to use it

or how to enable it for that matter

and its only by complete chance that i have a scientific calculator nearby ;p

found a nice chart for the 0 30 45 60 90 degree values

realistically these are the only ones we will encounter in questions

fingers crossed

[frosty sphinx]

im a dumbass

[dark sparrow]

this sure is a five-sided shape

what are you asked to do with it?

@frosty sphinx

[unreal beacon]

Is it accurate to say that a radian is the arc length created by a central angle on the unit circle?

A radian is an angle which produces an arc length of 1 on the unit circle

[sturdy iris]

The latter is a great definition

[unreal beacon]

When we say that the sine function relates an angle to the y component on the unit circle, is the angle in degrees or radians ?

[sturdy iris]

It doesn't matter

But if we're talking about the function, it is almost always radians

However, it still makes perfect sense to ask the sine of 270 degrees, for example

[unreal beacon]

Alright. Hmm, if it doesn't matter, why is it that when I put sin(pi) in my calculator (calc. is in degrees) - I get 0.05. And when I put sin(90) (calc. in degrees) - I get 1.

[sturdy iris]

By saying "it doesn't matter" I mean that both choices are possible

For calculators there are usually the degrees and radians modes you are free to switch

[unreal beacon]

Right, I just did, and in radians mode I get that sin(pi) is 0. Why isn't it 1 like sin(90) in degrees mode?

Oh, it's because 90 degrees is pi/2, not pi, right ?

[sturdy iris]

It is

[unreal beacon]

Alright, gotcha 🙂 Thanks

[sturdy iris]

You're welcome

[grand geode]

do functions count as coordinate geometry?

like finding domain and range of a graph

is it possible to find the domain and range with given values like this?

[muted finch]

I am pretty interested in math and failry good at it but with the whole pandemic we have been going very fast with new chapters and I just can't focus. Do any of you know good resources to learn about sinuses and cosinuses?

[humble pulsar]

yes, cause you can determine the radius

[grand geode]

thanks

[scarlet basin]

@muted finch sines and cosines*, you could try khan academy? they usually do a pretty good job of being thorough

[muted finch]

Thanks. One more thing. Do any of you know any videos on yt that will help restore my will to learn math?

[hoary prism]

Yo can anyone help me with this

[vapid stag]

is there a way to drop off my past exam somewhere, so people can tell me what i did wrong?

[trim breach]

Someone here would help you so long as it is a past exam like you said.

[inner ingot]

what line represents the lower edge of the front view?

[copper marten]

does anyone have tips on memorizing theorems and postulates for proofs? or can someone list out the main ones

[upper karma]

There are 2 ways to remember geometric proofs, one is to practice them and learn them, the other way is to understand all the proofs, and try reproducing them on ur own, (make ur own variant)

[main hornet]

Well I think it's a tedious task especially the second way ... making your own variant 🙏...I mean if you are preparing for some exam then definitely you should follow the first way.. PRACTICE..And if you are fascinated by mathematics then you can go further to do some research on it🙏🙏..

[upper karma]

Making your own variant means that u understand the proofs, like u need not remember what triangles were congruent, what lines were parallel, what was cyclic, you try to learn the pattern and recognise things on your own

[hollow shard]

Evening, I would like to share my solution video on a geometry problem. Thanks.
https://bit.ly/3uk8ftE

[neon eagle]

which is better

intuitive goemetry or algorithm geometry

as a person learning goemetry right now i would like to know

[wary granite]

is 41c 33

[true sleet]

does anyone know this

im doing a quiz

on changing dimensions

[upper karma]

Help on tests/quizzes isn't allowed

[true sleet]

oh ok

sorry

[pure bronze]

Slowly

But surely!

[small rivet]

someone can help me?

Its in french..

[peak flower]

What don't you get about that task?

It's just definitions you have to learn.

[small rivet]

I dont know how to find amplitude

W a graphic

[peak flower]

half the distance between the highest point and the lowest point.

[small rivet]

okay but in this its not 4? or its the point after p-pi/2

-pi/2

[shrewd kettle]

I don't understand how I'd use this ratio to solve for Figure A, for the other problems I'd usually have an "X", but without one I got thrown completely off the hook

[dark sparrow]

@shrewd kettle your unknown is the perimeter of A

you could find every single side of A if you want

or you could do the smart thing and realize that the perimeter of A is just the scale factor (7/2) times the perimeter of B

[storm tree]

Can anyone help me with my finals for geometry

I would honestly really appreciate it

[shrewd kettle]

Oh that's smart-

Thanks, I'll be sure to apply that

[dark sparrow]

@storm tree is this a test/quiz/exam thats going on right now

[storm tree]

It’s my finals

But it really easy because my teacher gave up on life

[dark sparrow]

not only do i personally not feel like helping someone cheat on a final, it's also kinda against the rules of the server

[storm tree]

I apologize

[finite swan]

we're not graduating for you

[feral sleet]

i could help a bit, but im becuase i genuienly think this will help by concept. for(1b) to sketch the circle find the point at (4,-3) make a line 5 unite from it and start circling.

[finite swan]

1. they left
2. don't help them cheat

[feral sleet]

Ah

Good

[waxen vault]

Hello

Inverse trigonometry?

[dark sparrow]

inverse trigonometry sure is a thing

as a bonus, this is the right channel to ask questions about it

[waxen vault]

Yep its bonus

[dark sparrow]

$y = \sqrt{2}$? or $y = \sqrt{x}$?

[somber coyote]

Ann

[dark sparrow]

anyway this is the kind of thing that can be solved using calculus, specifically the integral for arc length

yes

this will belong in #calculus tho

but in your particular case you will have $\int_a^b \sqrt{1 + \frac{1}{4x}} \dd{x}$

[somber coyote]

Ann

[dark sparrow]

damn youre right

wait whats the original

[somber coyote]

ryane

[dark sparrow]

let's move to calculus

[hollow shard]

Is this a contest problem?

[night vigil]

Here I am visiting again

so there's this really weird problem

and im ataully stumped

can anyone help me?

ok I guess ill ping?

<@&286206848099549185> can you take a look at this problem?
It really has me stumped

[full briar]

is the area of the shaded area of the circle the lateral area of the cone?

that's my current understanding of the problem

[night vigil]

yah

[full briar]

ok just simply speaking: if we are to calculate the area of the shaded region, it should be area of the circle times the ratio of the shaded region

so

$\pi(56)^2 * \frac{225}{360}$

[somber coyote]

birschlatte35

[full briar]

@night vigil

[night vigil]

alright

so

hmm

yah that could work

oh yah that works

thats such a good formula!

[full briar]

i suppose, im just doing based off what im understanding from the problem

np!

[night vigil]

thanks! 🙂

@full briar Imagine you get promoted to helper lol

[full briar]

haha im not usually that active

[night vigil]

ah well

other helpers arent either

[full briar]

lol

[night vigil]

so really only like half the helpers are

[full briar]

i mean if you really want a helper's attention

you should go ask on one of those questions #s channel

[night vigil]

oh

[remote arrow]

Hello everyone!
What would be a good book to start off with to revisit the basics of trigonometry and geometry?

[full briar]

You should also prob ask it in #book-recommendations

[copper marten]

how do i find the volume when there's like thickness?
the answer to #3 was 180 if thats helpful in solving it

my initial ideas:
splitting the vase into like 4 parallelogram solids
the top 3 solids are the same and the one on the bottom isn't because there is thickness on the bottom

first 3 solids:
since the thickness is 1.5 on the sides, i subtracted 6-1.5 and 3-1.5 to get 4.5 and 1.5, and that's to get rid of the thickness. and then i did 4.5x1.5x2.5 to get the volume which is 16.875, and the 3 solids are the same so 16.875 x 3 = 50.625

bottom solid:
the sides and the bottom are 1.5 thickness
6-1.5=4.5 3-1.5=1.5 2.5-1.5-1
4.5x1.5x1 = 6.75

volume: 50.625 + 6.75 = 57.375
but on the answer key its 95, so i'm not sure how they got that so pls help if u know how to do it

[sand root]

https://cdn.discordapp.com/attachments/825957964149162004/839736245230895154/20210506_173156.jpg

anyone help

[tiny snow]

@sand root Is it (c) you are stuck on, and what have you tried?

Note that the scribbled right angles on your questionnaire are not necessarily such.

Also, you are to prove that |AB| = |BC| (not that |CD| = |BC| which the markings on those segments suggest).

[sand root]

how could i prove they are the same

[tiny snow]

Well, one way is to prove that the triangles ABO and ACE are similar. You then know that |AE| = 2|AO|, and from similarity |AC| = 2|AB|, from which |BC| = |AB| follows.

[sand root]

trianlgfe abo?

[tiny snow]

If you draw a line from B to O you will get a triangle from the three points ABO.

[sand root]

ah ic

how do u get ae = 2

[tiny snow]

|AE| is twice the length of |AO| (diameter vs radius)

[sand root]

ohh ic

thanks

but i dont get why we did the first step

[tiny snow]

I think from the original description they want you to show that the triangles ADE and ACE are similar, but I found it easier with ABO.

The scale factor, 2 in this case, can only be applied if the triangles are similar.

[sand root]

but couldnt we just have skipped to ae = ao^2

[tiny snow]

Not squared...

[sand root]

x 2

[tiny snow]

Here's what I did: https://www.geogebra.org/geometry/sppt2pce

[sand root]

ahh ic ty

[neon hamlet]

im kinda confused, how would you solve for this?

I know W is 70
and its asking for a quadratic equation based on the "solve for all values" part

[silent plank]

why would you even need a quadratic equation for this

applying properties of radii and isosceles triangles to form a linear equation is enough

[upper karma]

all the lines coming from W are radius

[neon hamlet]

ah cool

yeah I just did 2x+70=180

and I have no idea how to do this, maybe splitting the triangle up? but I don't know if YX = WX

[severe rapids]

Anyone good at proofs

[olive cove]

Angle BAO = angle CBD

[trim breach]

It looks like you can find the diagonal of that square and split the larger triangle that way by adding the radius and diagonal to form a leg. Then you could use Pythagorean Theorem since you would have two sides.

[olive cove]

there is no need to use trig

[severe rapids]

Someone told me too

[olive cove]

Angle BCD = 1/2*angleO = 90

So angle BCD = angle ABO

And 2 similar angles is enough to say 2 triangles are similar

[olive cove]

Oh sorry nvm Lidoh was answering another problem

[severe rapids]

What’s the proof though

[plush remnant]

Can someone help me on this please

[scarlet basin]

uhhh

im struggling to understand how my teacher got this

like...why would finding the square root of 960 and 375 here to get the scale factor and volume ratio not work?

but yet it works here?:

[scarlet basin]

anyone?

[rough stone]

is anyone free in here to answer a question?

@fast pulsar

6e^i(pi/12)
=6cos(pi/12)+6sin(pi/12)i
= (This is where my calculator gives me the wrong values from my hw)

cos = 5.9999 sin=.02741 (not rounded yet)
Answer is
=5.796+1.553i

My question is maybe it is a setting problem in my ti-84 calculator but I don't know what is wrong

The problem is moving from complex to rectangular form

[dark sparrow]

radian vs degree mode?

[rough stone]

degrees

I believe I figured it out

I set my mode to real numbers instead of a+bi

stupid of me

I have gotten the right answer

on my calculator thank you for checking in @dark sparrow

[fast pulsar]

Euler's formula is completely false when in degrees. You MUST be in radians for it to be true

$e^{i90}\neq \cos(90)+i\sin(90)$

[somber coyote]

dackid (jump king +)

[rough stone]

tytyty

[plush remnant]

Can someone help me

16 and 12

Iam just confused on how to solve the formul

Formula

[olive cove]

Use soh cah toa

[trim breach]

What your teacher did was simplify the ratio between the surface areas of the triangular prisms by dividing the numerator and denominator by 15. You can see that taking the square root of the simplified ratio gives a scale factor, and that scale factor is cubed to get the volume ratio.

[empty fulcrum]

how to calculate Fp

[wise pawn]

separate out F_BR and F_CR into their x and y components

[idle bobcat]

hi, can someone explain where this 1/10 came from?

[wise pawn]

the equation on the 3rd line

the right side has x/10

the derivative of that is x'/10

[idle bobcat]

kadfasdhfukjasd

thank you lol

[wise pawn]

you're welcome lol, try #calculus next time

[idle bobcat]

yeah i asked another question there

its been quiet

this question is kinda related to trig so i thought id ask here instead lol

oops i mean't I asked in #precalculus not #calculus

[scarlet basin]

Does anyone know how I'd go about learning the volumes and surface area formulas for 3d shapes? there are so many of them and I have a test tomorrow in which we aren't provided a key for them 😭

[idle bobcat]

which shapes are you struggling with

im not the best at geometry but I found it pretty intuitive to determine the volume and surface area formulas by thinking about the dimensions logically

like for a cylinder, the surface area is just the area of the top and bottom circles

and the circumfrence of the base of the circle multiplied by the height

so the area of a circle is pir^2, and since you have two of those, you multiply it by 2 to get 2pir^2

and for the side of the cylinder you basically just want the circumference of the circle multiplied by the height of the cylinder

so you have 2pirh

and then you just add them up

so you have 2pirh + 2pir^2

and for the volume, you are basically just looking for the area of the base of the circle multiplied by the height of the cylinder

so you have pir^2*h

don't know if this helps but it's how I think about things

[empty fulcrum]

I found all sides of the 2 right triangles but how do I use it to find F_P

[muted finch]

Does anyone know if radiants are actually useful compared to deegres? I know you have a formula that you can use with radiants, but are they good for anything other than that?

[dark sparrow]

the word "radians" has no t in it

anyway, using radians as a unit can often make calculations simpler for things involving rotation

and there are also reasons why radians are more natural but they require calculus to fully explain

[upper karma]

made a circle using lua in unity and shapes, using sin and cos

[fickle sleet]

Radians give vastly cleaner numbers to work with when dealing with Trig Functions

[upper karma]

I'm not sure how to use them to draw circles

[fickle sleet]

that's actually pretty impressive

[upper karma]

oo thank you

[mellow gate]

How do you find x?

[scenic depot]

I would think that by the Pythagorean theorem

[mellow gate]

Sorry blanked out there xD

[scenic depot]

no problem, it happens 🙂

[upper karma]

find x using Puthagorean Theorem

and find ø

with Cos

[tawny dragon]

I'm unfamiliar with ratios in circles. Can anyone help me out, much appreciated.

[sick torrent]

oh ok

yes

thank you

[true sleet]

does anyone know this

[trim breach]

Do you know the relationship between the scale factor and the surface area?

[true sleet]

Ik how to do surface area

wouldnt it be 51 x 4

[trim breach]

Not quite. It gives you the surface area of the shape.

[true sleet]

would I do 4^2

[trim breach]

It wants to know what happens to the surface area when the dimensions are multiplied by 4.

[true sleet]

it increases

[trim breach]

Yes. Now we need to know what the surface area is after we multiply the dimensions by 4.

[true sleet]

51/2 x 4

[trim breach]

Why did you divide by two?

[true sleet]

idk cause we tryne find the dimesion

do u know the formula

[trim breach]

What is the ratio between the dimensions of the shapes?

Let’s start there.

[true sleet]

2 to 1

[trim breach]

You are multiplying all the dimensions by 4.

So the ratio would be 1:4.

[true sleet]

1:4 x 4/4

[trim breach]

No need to multiply.

[true sleet]

so u divide by 4

[trim breach]

1/4 is the scale factor.

[true sleet]

0.0625

[trim breach]

The relationship between scale factor and surface area is this:

If. x:y is the scale factor x^2:y^2 is the ratio of the surface areas.

[true sleet]

so is that the answer

or would I mutiply that

[trim breach]

No.

Your scale factor is 1:4, or 1/4.

So calculate (1/4)^2.

[true sleet]

i got 0.0625

[trim breach]

Yes, or you could also write is as (1/16).

[true sleet]

ok thx

[trim breach]

So we can set up a proportion with this.

Since we know that the surface area of one shape is 51. But we need to get the SA of the new shape.

[true sleet]

its 816

[trim breach]

Yes!

Good job.

[true sleet]

thx

[thorn tiger]

hey

does anyone have any idea on how to solve this type of question?

i was thinking to do costheta = sqrt3/2

but im not sure

[humble pulsar]

yes, that's the 1st step

[thorn tiger]

then what would we do?

use the unit circle?

[humble pulsar]

Yeah since it's an exact value

[thorn tiger]

the answer is

theta=30degrees and 330 degrees?

[humble pulsar]

yes

[thorn tiger]

thanks

[upper karma]

Can someone help me

[storm portal]

@upper karma

Don't ask to ask, just ask!

[upper karma]

Lol i posted the picture it’s just taking a long time

I have to prove this

[storm portal]

oh F

can't read it

[upper karma]

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

[storm portal]

multiply both sides by cos^2x

[upper karma]

Nvm I got it

[empty fox]

Can anyone tell me meaning of this .

Find two transformations (registers) to align the vector op with the unit vector j of the positive axis of y, where o is the center of the coordinate system and p is a given 3D point. Then implement them by applying them on different primates.

[humble pulsar]

No cheating.

[valid magnet]

hey

https://i.imgur.com/4rEoX6k.png

i started off with the blue rectangle

then i multiplied 500 * (800/500) to get the height of the green box (312.5)

500/312.5 = (800/500)

the hypotenuse if the red square is 800 + 312.5

let's look at the point with the black line above the blue box (length 500) and the line to the right of the blue square (length 800)

that line is 312.5 in length

what is this point called?

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

wait those squares aren't equal are they

no they are

sqrt(800 / 312.5) = (800/500) = (500 / 312.5) = (312.5 / 195.3125) = (195.3125 / 122.070313) = ...

[honest pumice]

evening, quick question, sin(pi - a - b) = sin(a+b) right? when using the identity sin(a-b) = sin(a)cos(b)-cos(a)sin(b)

if i rearange sin(pi-a-b) = sin(pi-(a+b)) i can use sin(a-b) = sin(a)cos(b)-cos(a)sin(b) to make it sin(pi-(a+b)) = sin(pi)cos(a+b) - cos(pi)sin(a+b) giving me a result of sin(a+b)

[valid magnet]

(question: how is sin and cos defined?

how do you derive sin and cosin

[honest pumice]

the unit circle

[valid magnet]

can you define sin and cos for us?

[honest pumice]

it's better if i send an image, kind of hard explaining it with only words

but basically, when you make a circle of radius r=1 , the tip of the radius "lands" on the coordinates x = cos(a) and y=sin(a)

where our a is the angle made from the positive x axis with the hypotenuse

[humble pulsar]

the unit circle is pinned

[valid magnet]

how is a measured?

[honest pumice]

this angle a is also the "semi circumference" the hypotenuse line makes with the positive

before calculus was discovered, with measuring tools

they calculated the area with different methods

[valid magnet]

so a can be anywhere from 0 to pi / 4?

[honest pumice]

slicing the circle like a cake

putting very small squares and rectangles inside it

[wise pawn]

yeah that's a good proof, good job you're right

[honest pumice]

thanks for the help