geometry-and-trigonometry

[dark sparrow]

it'll get you the perimeter

notice, however, that you can carve this room up into two squares and a rectangle

[robust breach]

area is the inside right.

[dark sparrow]

...yes

[robust breach]

how the heck do we count the inside of a square?- im sorry im stupid and I am extremely bad at math...

[dark sparrow]

a square is just about the easiest shape to compute the area of...

all area is based off of squares and rectangles

if the side length of a square is $s$, then its area is $s \cdot s$, or $s^2$.

[somber coyote]

Ann:

[dark sparrow]

this is the reason we call things raised to the second power "squared"!

[robust breach]

wha-

I have no idea what the heck you just said... like I said I can not do math.

[dark sparrow]

mmh

...

the person left.

[supple abyss]

[dire rampart]

@GamerDood#9867

rip

[plucky marlin]

🇫

[violet cave]

https://gyazo.com/eb44a728dbc5003376b0aecdda2a4f38

How can i calculate the area of this

[dark sparrow]

blue triangle minus purple triangle

[violet cave]

ah thanks

[dark sparrow]

subtractive area calculations like this are pretty useful a lot of the time

[upper karma]

whats the difference between a circle and its circumference

there isnt one right

a circle is the locus some distance from a set point, not the figure bounded by it becuase thats a disc?

[steady sleet]

A circle is a shape

the circumference is the length along the outside of the circle

[upper karma]

a circle is formally a line though

[pallid cloud]

More like a line is a circle

[upper karma]

?

[steady sleet]

I'd rather say a circle is a collection of points

[pallid cloud]

You can consider a line as a circle with radius infinty

[upper karma]

well yeah thats what i said before

a locus some distance from a set point

and technically a line segment or well in this case a curve is also a collection of points
my question was is there an actual distinction between a circle and circumference

[pallid cloud]

The locus of the points with a constant distance to a given point

Well yeah

The circle is a geometrical object

Its circumfrence is a number

Perimeter

[upper karma]

oh alright so the circumference is the length itself

[pallid cloud]

What's the difference between surface and area

Same thing

Area=measure of surface

[steady sleet]

I think you are confusing objects with their measurements

[pallid cloud]

Surface is geometric object
Area is measure of the surface
Circle is geometric object
Circumference is measure or the circle

[scarlet path]

I need help

[pure shore]

any help

[devout shell]

well use the fact they are similar triangles to help you

[upper karma]

Hello! Can I get a little help? Nothing too hard

<@&286206848099549185>

[vagrant elk]

smh

don't p i n g

[devout shell]

you didn't even post a question...

lol

[vagrant elk]

^

[upper karma]

Sorry D: I'm trying to find the rules but is kinda hard to find

[vagrant elk]

no worries, just put your question here (or preferably in a question channel)

[devout shell]

well nothing will undo the ping, so just ask it now lol

[upper karma]

Okay ':D I'm sorry. I'm doing example B (it already done but I want to know how do they did it) and I want to know why the 16 went over -56

[devout shell]

how do you isolate cos(R) then?

[upper karma]

Yes

[devout shell]

$-56\cos(R)=16$

[somber coyote]

⚡Amphy⚡:

[devout shell]

Isolate cos(R) for me

I really you know how to...because how did you even get this far in geometry then without knowing algebra basics...

[upper karma]

Long story...

[devout shell]

isolate cos(R) please

[upper karma]

Like this?

[devout shell]

perfect

that is exactly what you are supposed to do

[upper karma]

Thank you very much! If I want to ask more questions, which channel should I use?

[devout shell]

you can keep using this one then, since you already occupied it lol

[upper karma]

:o okay

It is okay if you correct my answers or I can only ask questions?

[devout shell]

[upper karma]

Is this okay?

[devout shell]

well no original triangle to reference

[upper karma]

Sorry!!

I can show you the one on my notebook if it's better

But that's more exact

[devout shell]

you are incorrect

you messed up here

580 - (576 * cos(21))

$580-(576\cdot \cos(21)) \neq 4\cdot\cos(21)$

[somber coyote]

⚡Amphy⚡:

[upper karma]

Then square root?

[devout shell]

yes, but you have to realize why you were not allowed to do what you initially did

[upper karma]

Because 576 is connected to the cos21?

[devout shell]

recall order of operations

PEMDAS

multiplication comes before addition or subtraction

[upper karma]

Trueee

So dumb

[devout shell]

so you have to multiply 576cos(21) first

then you can do the subtraction

then you take the square root of everything

[upper karma]

Yes yes

[devout shell]

[upper karma]

Now comes the hard ones, the ones the ones that got my class all messed up

[devout shell]

lol hardly a challenge for us though

this will be easy to work through, but I'm not going to be the one doing most of the work, I just give pointers lol

[upper karma]

It's okay! You are giving the support I needed

I think it's right!!!

[devout shell]

give me a minute lots of stuff happening at once

[upper karma]

Take your time :0

[devout shell]

perfect

it is right

[upper karma]

:D great!!

The last one!

[devout shell]

correct

[upper karma]

Amazing!!!

Thank you so much Amphy!!

[devout shell]

[upper karma]

Have a nice night :)

[upper karma]

how would i calculate the angle of a point in a plane compared to another point in a different coordinate in the same plane? like A(2;4) and B(3;7)

[devout shell]

you mean like if there was a vector from (0,0) to A and a vector from (0,0) to B, what is the angle between them?

[upper karma]

a line between A and B and the angle of that line

[devout shell]

can you show picture of what you mean?

[upper karma]

is there a tool for drawing a plane and points on it?

or do i have to go on ms paint

[solemn holly]

What have i done wrong here?

[devout shell]

use desmos lol

[solemn holly]

The three circles have the same radius "r", the lines are paralel and circles are tangent

[devout shell]

geogebra is good too

[upper karma]

thanks

ok, imagine a line between those two points. i want to know how to calculate that line's angle based on A looking at B

[devout shell]

angle that line makes with the x axis then right?

[upper karma]

yeah

[devout shell]

one way is to see that theta is equal to arctan(y/x)

so arctan(4/2) will give the angle

well perhaps no

hmmm

because then arctan(7/3) would be different

I should play with it geometrically then

"If the slope of a line is m then the angle is θ=arctan(m)"

looks like that's your answer, had to look it up lol

[upper karma]

i'm gonna be honest. i'm bad with angles so i don't know how sin, cos, tan and arctan works

neither θ

[devout shell]

theta is common notation for the angle

arctan means tangent inverse, arc[insert trig function] means inverse[of that trig function]

[dark sparrow]

θ is just a letter

a greek letter, as it were

[upper karma]

i need to be able to be good at trigonometry to be able to make 2d games but i just suck at it so much

i'm better with algebra

[devout shell]

you learn slowly

no one gets it right away

[upper karma]

i don't understand cos, tan and sin

how did they appear, how are they calculated

why isn't sin(2) = sin(1) + sin(1)

[devout shell]

more than we can explain over discord lol, you'll take a course on trig though and you'll learn it then

[dark sparrow]

why would sin(2) be equal to 2sin(1)

in all honesty, khan academy has a good trigonometry course

[unkempt palm]

Ok so i rlly need help with some trig h(x) = 3 sin(2x) sin(10x) basically i get the amplitude and the two periods of the functions individually but when multiplying 2 functions how exactly does it work how is the new period determined how is the new amplitude determined etc.

[dark sparrow]

uh

well ok for starters this won't be a pure sinusoid

second you can rewrite it using a product to sum identity

[unkempt palm]

yea i tried graphing it and got a period of pi/2 which i didnt understand how the two periods related to that value

[dark sparrow]

$\sin(u) \sin(v) = \frac12[\cos(u-v) - \cos(u+v)]$

[somber coyote]

Ann:

[dark sparrow]

so you get $h(x) = \frac32[\cos(8x) - \cos(12x)]$

[somber coyote]

Ann:

[unkempt palm]

im a little bit confused about the period

because when graphing i observed a period of pi/2

[dark sparrow]

indeed

[unkempt palm]

and i do understand how 2 and 10 result in a pi/2 period

not*

[dark sparrow]

cos(8x) has a period of pi/4 and cos(12x) has a period of pi/6

the "greatest common integer multiple" of these is pi/2

[unkempt palm]

ok

so the greatest common multiple is the period?

[dark sparrow]

well... generally, yeah.

[unkempt palm]

and whilst still on this how is the amplitude discoverd

is it just 3

[dark sparrow]

well it can't be greater than 3

[unkempt palm]

because when graphing my valuse didnt go from 3 - -3

[dark sparrow]

but to find the exact amplitude i'd differentiate the thing and find its extrema

it suffices to find them on [0, pi/2]

bc periodicity

[unkempt palm]

ok

thx this helps alot

@dark sparrow sorry to bother but i just had one question how did you go from 1/2[cos(u-v) - cos(u+v)] to 3/2[cos(u-v) - cos(u+v)]?

[dark sparrow]

huh what

[unkempt palm]

sorry

i fixed it

[dark sparrow]

your thing had 3 as a coefficient

[unkempt palm]

so the coeffecient becomes the numerator

[dark sparrow]

what

no

jeez

i just went from $3 \sin(10x)\sin(2x)$ to $3 \cdot \frac12 [\cos(8x) - \cos(12x)]$... don't overthink it

[somber coyote]

Ann:

[unkempt palm]

oh i see sorry im very new to this type of trig as i am first semester year 11 thanks so much for the help

[upper karma]

#help-3

[rustic cairn]

Whats the proof for the second one

[dark sparrow]

what

sorry, the picture is cropped too much

[rustic cairn]

H

[ornate shadow]

For SAS:
OM = HG S
<O = <H A
Since the angle and side are connected and we need another connected side for SAS(i.e. angle included between two sides)
HJ=OK S

For ASA:
This time one side should be connected to the two angles. The only other connected angle to the given sides is<M and <O

So,
<M=<O

If by HL you means RHS,
KM=JG H

[upper karma]

Hello, I'm studying trigonometry with unit cirlce and I don't understand why a cosine or a sine can be negative ? Isn't it a lenght divided by another length ? Sorry if it is the wrong place to ask this

[real mulch]

it's signed length

it conveys information about direction

cos(x) and sin(x) are the coordinates of the point on the unit circle that you reach if you turn x radians around the origin in the counter-clockwise direction starting from the point (1, 0)

the x-coordinate of the red point is cos(x),
the y-coordinate of the red point is sin(x)

[upper karma]

Thank you so much for this explanation !

[low herald]

can you guys help me out with this

If tan(x + y) = 33 and tan x = 3, find tan y.

[half gull]

there's a formula for tan(x+y)

you can find it by using formulas for cos(x+y) and sin(x+y)

[low herald]

you can find it by using formulas for cos(x+y) and sin(x+y)

I tried using that but I couldn't get anywhere

[upper karma]

what did you try

[ornate shadow]

tan(x+y)=(tanx +tany)/(1-(tanx)(tany))

Plug in the values of tanx and tan(x+y) to get tany

[devout shell]

Easy way to think of it, area of base * height

area of the base is a quarter circle

Area of a quarter circle: $\frac{\pi \text{r}^{2}}{4}$

[somber coyote]

⚡Amphy⚡:

[devout shell]

Base Area * Height = $\frac{\pi \text{r}^{2}}{4} \text{h}$

[somber coyote]

⚡Amphy⚡:

[devout shell]

height is 10, but also check your area

you used the formula correctly? If you did then you got the right answer

no need to ask me if you followed the formula correctly

your answer was?

correct

for regular solids, to find the volume, just remember that it will be:
(Area of the Base) * (Height)

[vagrant elk]

bruh

[devout shell]

why didn't you just post there though?

[upper karma]

https://gyazo.com/b0255c98d6bb23c7195fa86e8b44310b

will love who ever can help me out

[devout shell]

are you familiar with the definitions of the six trig functions?

you will also need to solve for c first, how would you go about doing that then?

[upper karma]

A^2+B^2=c^2

i did that

and got squareroot(58)

[devout shell]

that is incorrect

a=7 and b=10

check your squaring and addition again

[upper karma]

149

then simplify that right?

[devout shell]

now take its square root and you have the length of c

[upper karma]

12.2

is it 2 square root(74.5)

[devout shell]

you could just leave it as sqrt(149) if you wanted

[upper karma]

ok

[coral terrace]

can anyone assist?

[upper karma]

so sin= 7/squareroot(149

[devout shell]

yes, just follow the definitions of the trig functions and you will get it right

[uncut pagoda]

Can anyone help me with this?

https://cdn.discordapp.com/attachments/325466388560871425/576197524583481344/20190509_190251.jpg

[devout shell]

what is the aim of the problem?

[uncut pagoda]

Trying to find the base and leg
Or two legs if that makes sense

[devout shell]

are you familiar with sine and cosine?

[uncut pagoda]

@upper karma

[upper karma]

Yes

[devout shell]

you had question?

[upper karma]

Yeah

[devout shell]

[upper karma]

https://gyazo.com/11309da90947baeef9f63d10fceb2b2c

sooo

i understand to do 6^2+ b^2 = c^2

but like how do i do that without have b or c

[devout shell]

use the trig ratios to help you solve

you have an angle and an opposite side, what could you use to find c?

[upper karma]

B and A but how

[devout shell]

you can use a trig ratio, as I had already stated

[upper karma]

I see what you want me to do but i dont know how to do it

[devout shell]

type it out then

what do you think I want you to do?

[upper karma]

65/25

[devout shell]

what does that even represent?

[upper karma]

Opposite and hypotenuse

[devout shell]

Let's just start from the definition then.\ $\sin(A)=\frac{\text{opp.}}{\text{hyp.}}$

[somber coyote]

⚡Amphy⚡:

[devout shell]

right? you know that is true

[upper karma]

yeah

[devout shell]

so we have an angle, A

what is that angle?

[upper karma]

65

[devout shell]

and what is the opposite side?

[upper karma]

25

[devout shell]

do what do we want to solve for?

[upper karma]

2.6

[devout shell]

I didn't say anything about numbers, I said what are we trying to solve for?

[upper karma]

so that will be c

[devout shell]

$\sin(65)=\frac{25}{c}$

[somber coyote]

⚡Amphy⚡:

[devout shell]

solve this equation for c, and that will be the answer

[upper karma]

27.5

[devout shell]

correct

so now you can get b

@upper karma you had a question about the area of a regular polygon?

[upper karma]

I got an email from my math teacher that said my previous answer which the paper said was incorrect is actually correct, so I'm sorry for bothering you

if I ever have further questions I'll ask

[devout shell]

ok then lol

[upper karma]

I am having trouble getting b is this right so far 36+b^2=756.25

[devout shell]

you can also use a trig ratio as opposed to solving it with pythgorean thm

[upper karma]

ok so i would do opposite/adjacent

[devout shell]

what is that trig ratio called?

[upper karma]

opposite/adjacent

[devout shell]

it has a name

[upper karma]

tan

[devout shell]

tan(x)

[upper karma]

ok

[devout shell]

you take the tangent of something, tan on its own doesn't mean anything

it's a bad lazy habit to just write sin, cos, tan

[upper karma]

oh right ok

[devout shell]

because they are functions, not just words

so you should get used to writing sin(x), cos(x), and tan(x)

[upper karma]

ok.... will it be then tan(x) = 6/b

[devout shell]

x is what though?

[upper karma]

25?

[devout shell]

but that is correct, just put in the correct number for x

x is the angle

wait your answer for c is wrong

I was just going off the numbers you gave me, but like since when was 25 degrees the length of the opposite side?

[upper karma]

65 is the opposite

[devout shell]

opposite means opposite side...

the only place an angle will appear is in the x for the trig ratio, the rest of the number are lengths

[upper karma]

ok so

the opposite of c is a

[devout shell]

I don't even know what you are trying to say there

[upper karma]

idk what im saying honestly im lost

once u said c is wrong u lost me

[devout shell]

because you used angle measures in place of length measurements

you will have to review the trig ratios then

[upper karma]

yeah i see that now but idk what to do from there

[forest thunder]

are you guys talking about SOH CAH TOA

[devout shell]

yes, but we are making our way through it

[forest thunder]

Sin == opposite/hypoteneuse (english??)

ok just curious

[devout shell]

correct

[upper karma]

I want to get c so i have to do ........ sin?

[devout shell]

so how about you repost the screenshot and tell me all angles measures you have

and tell me what lengths we are given

[upper karma]

https://gyazo.com/11309da90947baeef9f63d10fceb2b2c

all angles are shown there

a=6, A=65 and B= 25

[devout shell]

ok, so we want c

[upper karma]

yes

[devout shell]

and we have a

so write out how we would solve for c, fill in the numbers

[upper karma]

would it be 6/c

[devout shell]

6/c equals what?

you have to say what it is equal to

that is very important

because 6/c is just some ratio, it has no meaning to me

[upper karma]

sin(x)= 6/c

[devout shell]

what is x then?

[upper karma]

x would be 65

[devout shell]

ok so write that all out and then solve

[upper karma]

c = 6.6

[devout shell]

yes

now you can solve for b however you want

[upper karma]

so b is 7.56

[devout shell]

show me how you got that answer then

[upper karma]

6^2+b^2=6.6^2

36+b^2= 43.56

b^2=7.56

[devout shell]

b^2=7.56

you need to take the square root

because you have b^2, not b, and you want b

[upper karma]

how do i take the square root out?

[devout shell]

what?

something is being squared right now, so to taking its square root will undo that squaring

[upper karma]

k im back

so b=3.78

right

[devout shell]

I guess, I can't do square roots in my head, but that sounds correct

[upper karma]

c is right but b isn't

[devout shell]

so then you messed up a bit somewhere or there is round off error

you can just use tan(x) to solve for b instead if you wanted

that should be more accurate

wait...how did you get 3.78?

I should have realized that was way off, but oh well

square root does not mean divide by 2 if that was what you did

[upper karma]

i just divided by half and crossed my fingers couldn't figure it

[devout shell]

no...

you know what a square root is?

[upper karma]

yes but like i don't understand how to get rid of it

[devout shell]

you were attempting to get rid of a square, not a square root

to get rid of a square, you take the square root

[upper karma]

ohhhhhhhhhh

2.74

[devout shell]

that sounds more correct

should be in that range of numbers

[upper karma]

ok thanks for helping and dealing with my stupidity

[devout shell]

it's all part of the learning process, that's all, nothing stupid about it

[solar shale]

Answer is right but how would ik it’s a right angle? @me

[solar shale]

<@&286206848099549185>

[dark sparrow]

@solar shale is that line labeled as "15" meant to be tangent to the circle

[solar shale]

So if there is a tangent then it means 90 degree angle

[dark sparrow]

a tangent is perpendicular to the radius drawn to the point of tangency.

[solar shale]

Ah ok

[granite canyon]

<@&286206848099549185>

;3;

can anyone guide me through this

[dark sparrow]

(in the future, post your q, then ping helpers if you don't get any replies within 15 mins)

[granite canyon]

oh

srry

[dark sparrow]

well what have you tried so far

[granite canyon]

uh

I know that you have to divide 360 by 4?

[dark sparrow]

no, you never "have to" do anything

[granite canyon]

o

[dark sparrow]

don't think in terms of "do i have to do this"

[granite canyon]

ok

[dark sparrow]

think in terms of "if i do this, what'll that give me?"

[granite canyon]

that makes more sense

uhh

[hallow smelt]

so my original equation is r=2sin(x)

I get how we wind up x^2+y^2-2y=0

I don't get how they complete the square of that to graph it? how can you complete the square with three terms

[dark sparrow]

complete the square on y^2 - 2y

x^2 is already a completed square

[hallow smelt]

negative b over 2 a right? 2/2

x is already a completed square? because its the only x term?

[dark sparrow]

no

x**^2** is already a completed square

because there is no x^1 term

[hallow smelt]

oh ok

cool I see it now

my teacher wrote it like this though

(x+0)^2 + (y-1)^2 = 1

is there any reason to write x like that?

is x^2 wrong?

[dark sparrow]

maybe to accentuate that 0 is the x coord of the center, but not for any other purpose

[hallow smelt]

they want me to go from rectangular coords to polar. x^2 + y^2 = 9 I know x^2 + y^2 = r^2 so r = 3

how do I get the angle though?

[dark sparrow]

r = 3

that's it

r = 3 describes the circle of radius 3 centered at the origin

[cloud merlin]

https://cdn.discordapp.com/attachments/326138757474680852/576251360224346142/Screenshot_2019-05-09_at_11.30.32_PM.png

im pretty sure its gona be like square root of 3 to the power of 2

[devout shell]

if only it explicitly said it was a square, then that would make things easy lol

[spark stag]

@cloud merlin not quite

[cloud merlin]

it is a square

[spark stag]

does this help?

[devout shell]

there shouldn't be any square roots in the final answer lol

[spark stag]

(sorry for the terrible drawing)

and yeah its regular and 4-sided, so its a square.

[devout shell]

the answer for this is pretty easy to find lol

[spark stag]

saying its "pretty easy" isnt exactly helping

although man, TAing would be a lot easier if so

[devout shell]

thought he figured it out already

I'm thinking he knows what to do at this point though

[cloud merlin]

look ill draw it out

[devout shell]

did you get the answer though?
or you still thinking that it's going to be 3 m^2?

[cloud merlin]

nvm

its square root of 18

[spark stag]

erm

the side length is the square root of 18

whats the area?

[cloud merlin]

square root of 18 to the power of 2

[spark stag]

can you simplify that?

[cloud merlin]

here let me show you

[spark stag]

$\sqrt{18}^2$

[somber coyote]

Namington:

[spark stag]

i just want you to rewrite that in simpler form

[spark stag]

yes, your work is correct

but your answer can be simpified

if we take the square root of something, and then square it, what are we left with?

[cloud merlin]

i gtg ill be back later

[spark stag]

.

$\sqrt{18}^2 = (18^{\frac{1}{2}})^2 = 18$

[somber coyote]

Namington:

[spark stag]

and then you need a unit of m^2

so, the area is 18m^2

that rule should intuitively make sense too: if x*x = some number

then the square root of that number is x

and x^2 = x*x = that original number

in other words, $\sqrt{18}^2 = \sqrt{18}\times\sqrt{18} = 18$

[somber coyote]

Namington:

[spark stag]

since sqrt(18) is the number that, when multiplied by itself, gives 18.

[sleek knot]

To find the height of a pole, a surveyor uses a theodolite. The theodolite is 4 ft tall. The surveyor stands 140 feet away from the base of the pole. Then she measures the angle of elevation to the top of the pole to be 38o. What is the height of the pole? Round the answer to the nearest foot.

can somebody help me

[devout shell]

well the height of the pole will be given by x+4, but then the question is what is x?

assumption: that is a right triangle

not explicitly given that it is a right angle but looks close enough so that works for me

anyways, we have an angle and a side given to us, and we want to find another side, what trig ratio will help us out?

[sleek knot]

yeah im just unsure how to start the problem off

trig ratio?

[devout shell]

trigonometric ratio, (sine, cosine, tangent, etc...)

[sleek knot]

oh those three

we usually do sin in class

isnt like sin 44 = 4/140

[devout shell]

where did you get that picture though? says angle of elevation is 38 but then there is 44?

then it seems like you added in 38 on your own?

[sleek knot]

oh i forgot about the 38

[devout shell]

can you give me the correct graphic then

[sleek knot]

thats the picture

[devout shell]

because 44 degrees was never given, but it's there

[sleek knot]

its in the picture

[devout shell]

your statement says 38 degrees

[sleek knot]

yeah in the question does say that

[devout shell]

then the picture puts the 38 on the hypotenuse as if it was a distance measurement

so I'm not very sure what is the correct angle to use

[sleek knot]

lets just use 38

[devout shell]

ok so scrap the 44, and replace it with 38, that will change the answer of course

[sleek knot]

so sin 38 = 4/140 ?

[devout shell]

where is the 4 coming from

we want x

[sleek knot]

4 ft

[devout shell]

x is nowhere in that equation

[sleek knot]

oh

so sin 38 = x/140 ?

[devout shell]

4ft is not even part of the triangle we are looking at

what is the definition of sine?

[sleek knot]

sine of an angle is opposite/hypotenuse

[devout shell]

how is 140 the hypotenuse? your picture clearly shows that 140 is a leg

but you are correct with your definition of sine though

just putting in the wrong thing

currently we don't have the length of the hypotenuse

perhaps we should try a different trig ratio

[sleek knot]

wait 38 would be teh hypotenuse

the*

[devout shell]

it said 38 was an angle measure

[sleek knot]

then how are we suppost to find the hypotenuse?

[devout shell]

38 is impossible anyways given one side is 140

we don't need the hypotenuse, we can always use a different trig ratio, unless the only one you have been taught is sine

[sleek knot]

we have been taught* cos and tan but we today we used sin

[devout shell]

what is the definition of tangent and cosine?

perhaps considering other trig functions will be useful because we see that sine is not going to help

[sleek knot]

cos = adjacent/hypotenuse and tan = opposite/adjecent

[devout shell]

so out of the two only one does not need the hypotenuse

we have one of the legs and we want the other leg, so let's use..?

[sleek knot]

tan

[devout shell]

let's use tan(x) because then we won't need the hypotenuse

so what would the equation be?

[sleek knot]

tan 38 = x/140 ?

[devout shell]

no need to be doubtful when you literally followed the formula, you are going to be correct when you apply the formula properly

[cloud merlin]

are you guys doing https://cdn.discordapp.com/attachments/326138757474680852/576626101519122443/2.PNG

?

[sleek knot]

yes

[cloud merlin]

need help?

[sleek knot]

yes

[devout shell]

solve for x for me please

I've been helping though

[cloud merlin]

sure do you guys understand triganomatry

[devout shell]

yes

[sleek knot]

i dont

[dark sparrow]

dsghklf

[cloud merlin]

ok wait a little

[sleek knot]

not a big math guy

[dark sparrow]

can we please like

[sleek knot]

more into science

[dark sparrow]

not mangle the spelling of "trigonometry" like that

please

[sleek knot]

x=109.37998771

[dark sparrow]

granted i've seen worse but

[devout shell]

you know the definitions of the ratios though, we are just helping you to see which one to use

[sleek knot]

yeah

[devout shell]

ok, so it said round to nearest foot, so please do that

[sleek knot]

109 ft

[devout shell]

perfect, now the final step

the length of the pole is given by what?

look at the figure if you are unsure

[sleek knot]

what do you mean given by like the elavation?

[devout shell]

they wanted the length of the pole, which is on the left side of the figure

[sleek knot]

so x

[devout shell]

is that all?

x is only part of the pole

[cloud merlin]

Probably this

[sleek knot]

but in the problem it doesnt say anything about 44 degress

@ eatsleepanime

[devout shell]

this is a strange picture, you sure this is straight from your book?

[sleek knot]

not from the book she send us a link to a google document

[devout shell]

I'm pretty sure that the length of the pole is x+4

[cloud merlin]

Look at the right

[sleek knot]

4 ft on the right

[cloud merlin]

That acute angle

[devout shell]

but original problem said use 38 degrees

but here's the thing

[devout shell]

you need to add 4ft

[sleek knot]

so 109 + 4 = 113

ft

[cloud merlin]

Wait let me get my calculator

[devout shell]

that is the answer, yes

[cloud merlin]

The answer is

[devout shell]

don't use 44 degrees

[devout shell]

the problem gave a different angle measure

[cloud merlin]

Just change the 44 to any other angle

Also this works with degrees so calculators have to be on degree mode

[devout shell]

let's not make this problem any more confusing than the graphic already is by introducing a formula that uses two trig ratios

[cloud merlin]

Well cos can be easily represented with sin

And so could tan

[sleek knot]

so the answer to that problem is 113 ft correct?

[cloud merlin]

Idk I don’t have my calculator

[devout shell]

given what you told me, yes

again that picture is the worse thing ever, because it has two angle measures and one is not even labeled properly, but we went off of what the problem said

but do you see why the height of the pole was x+4?

and not just x

[sleek knot]

yeah

thank you for the help very much

i understand it much better now

[devout shell]

we need a better picture to properly show you things lol, but you can see that it's not always hard and fast about what ratio to use

you take a look at what you are given and choose the ratio that utilizes the givens to find the unknown

[sleek knot]

i also have 1 more problem I did but I'm unsure if its correct if u could check it

[devout shell]

hopefully that one has a better picture

[cloud merlin]

sin is the best like all the other trig functions dont need to exist

[devout shell]

what?

[sleek knot]

tan64=x/7
x=14.35212689
x=14
wait i need so round to the nearest tenth so it would be x=14.4

[cloud merlin]

all of them could easily be represented by sine

sin

[devout shell]

that would overcomplicate things

[cloud merlin]

still

[devout shell]

why sine? it could have been equally just cosine then

[cloud merlin]

sin is the easiest to calculate

and easy to understand

[devout shell]

what? um no, all of them are fairly easy to understand

you are correct nickk

I'm not sure about rounding; it's whatever your teacher wants

[sleek knot]

wait i need so round to the nearest tenth so it would be x=14.4

[devout shell]

plus if you get rid of everything except sine, then the unit circle would not be nearly as nice

ok, with rounding, you are still correct

[last flume]

All trigs are easy tbh

[cloud merlin]

we dont need to argue

but sin is the best

[sleek knot]

thank you all very much for the help I understand this very much better

[devout shell]

that is overly complicated

[cloud merlin]

thats what people who dont understand say

[devout shell]

then how does the unit circle look for you

[cloud merlin]

what

[devout shell]

that's why lol, once you realize the implications of this to the unit circle you will stop asserting that sine is the best

[cloud merlin]

it is

its the simplest to understand

well maybe cos is also easy

but sin is just cool

[devout shell]

sure, but it's not the end all be all

[cloud merlin]

what does that mean

[devout shell]

it's not the ultimate weapon

you need other tools

[cloud merlin]

sure

does anyone know how to calculate sin

[devout shell]

Now you ruined the simple identity of: $\sin^{2}(x)+\cos^{2}(x)=1$

[cloud merlin]

like exact value?

[hearty barn]

yees

[somber coyote]

⚡Amphy⚡:

[hearty barn]

sin 1+ sin 1 = sin 2

jkjk

[cloud merlin]

no shit

[sleek knot]

for this problem would i also have to use tan?

[cloud merlin]

forget tan

its stupid

[devout shell]

well what are you given?

[cloud merlin]

overcomplicates things

[devout shell]

stop please

[sleek knot]

the problem is just "Find the value of x, rounded to the nearest tenth."

[devout shell]

we are working on something

[cloud merlin]

even pi can easuly be calculated with sin

[hearty barn]

Two questions at the same time?

@sleek knot ping me in one of the question channels

Make sure its empty

[devout shell]

no man, don't leave me to argue about ratios lol

[sleek knot]

which channel idk

i just realized i came into pre uni and this is highschool stuff

[devout shell]

well looks like he's gone....so let's continue

[cloud merlin]

That’s why sin is best

It’s a sick trig function

[devout shell]

please stop, I have a question I am working on with someone

[cloud merlin]

Well forget about that

[sleek knot]

we can argue after this question because its the last one

[devout shell]

but what side are you looking for in the problem?

and what side is given to you?

[sleek knot]

11 28 and x

idk why theres just a little 38 in there

[devout shell]

tell me what side x corresponds to

is it a leg or a hypotenuse?

[sleek knot]

x is the hyp

[hearty barn]

I'm here

@devout shell whose question you want me to handle?

[devout shell]

I'll let wolf take the lead lol

[sleek knot]

the only thing for hard is, is just to choose the correct treg ratio

but i sort of understand now

[devout shell]

stage is all yours

[hearty barn]

lmao it's fine if you wanna

[devout shell]

no no, lol you're eager for it so be my guest, need to get back to work lol

[hearty barn]

lmao okepoke

https://cdn.discordapp.com/attachments/326138757474680852/576636085493301266/3.PNG

[sleek knot]

alright so would it be tan?

[hearty barn]

Whats 32?

[sleek knot]

not sure its just there

[cloud merlin]

anyone watch anime?

[plucky marlin]

yes

[cloud merlin]

xD

[plucky marlin]

[cloud merlin]

one piece and stience gate are sick

[sleek knot]

also sorry if i distracted you of your work @Amphy⚡#0697

[plucky marlin]

i didnt see one piece tho

[hearty barn]

@cloud merlin Warning, don't spam here

[cloud merlin]

then watch it

spam?

im not

[devout shell]

nah I do this on my own time, you're good

[sleek knot]

thanks for the help again

[plucky marlin]

no problem

[hearty barn]

Let's ignore 32 for now?

[sleek knot]

yeah lets do that

[devout shell]

but anyways someone is attempting to help you, you should move into a help channel and lock it to keep spam away

[hearty barn]

lel true

It's fine lets work here

[sleek knot]

so would i use tan or other treg ratio?

https://cdn.discordapp.com/attachments/326138757474680852/576636085493301266/3.PNG

[hearty barn]

We can either use the ratio 11/x or x/11

Right?

Because we have no idea about the other side

[sleek knot]

yeah x/11 because x is the hyp pretty sure

[hearty barn]

Yee

[sleek knot]

so tan 28 = x/11

[hearty barn]

And the angle provided to us is 28

YES

[sleek knot]

lemme use a calculator real quick

[hearty barn]

==calc tan 28

,w calculate tan 28

[somber coyote]

Display results online and refine query

Requested by SolitaryWolf#6406

[hearty barn]

here you go

[sleek knot]

x=5.84880374

round do the nearest tenth so 5.8

[hearty barn]

COOL

it's done

[sleek knot]

thats all i needed thank you for the help

[plucky marlin]

np

[hearty barn]

Want me to kick your banana, @plucky marlin ?

[plucky marlin]

lol just kidding

[hearty barn]

xD all cool

[sleek knot]

have a great night (or morning wherever yall live at)

[hearty barn]

,w time India

[somber coyote]

Display results online and refine query

Requested by SolitaryWolf#6406

[devout shell]

wait the missing side was a hypotenuse

[hearty barn]

Aye

But I wasn't sure if 32 is thhe side length

[devout shell]

but hypotenuse is longest side and it's 5...? but a leg is 11...so that's an impossible triangle lol

some random ass 32 included because the teacher sucks at making graphics, but you guys solved for the opposite side lol

@sleek knot you found a side that you didn't need to lol. I think the graphic confused him

[hearty barn]

OH

OHHH FUCK

AAHHHH

[devout shell]

here's your chance to make it right lol, go for it lol

[hearty barn]

cos 28 = 11/x
x = 11/cos 28

,w \frac{11}{\cos 28}

[somber coyote]

Display results online and refine query

Requested by SolitaryWolf#6406

[hearty barn]

@sleek knot @sleek knot @sleek knot @sleek knot @sleek knot @sleek knot @sleek knot

I MADE A MISTEKKKKKKKKKKKKK

[sleek knot]

oh

thats no bueno

what is it

so i change it before the teacher sees it

[hearty barn]

It's cos and not tan

[sleek knot]

ok ima do that thanks

[cloud merlin]

It’s just this

Here are 2 answers

With tan it’s this

[devout shell]

why would you do that to yourself?

[cloud merlin]

do what?

[devout shell]

making it overly complicated for no reason lol

[cloud merlin]

I do the math the way I understand

I don’t memorize

[clear haven]

uh

dunno what to say but

keep an open mind ? lol

[dark sparrow]

what is even being done there

[cloud merlin]

you guys want me to explain them?

the third one is pretty basic but i can explain the sin and cos

[dark sparrow]

i mean it's not even clear what you're going for there

can you post the original problem you were doing

[cloud merlin]

sure

https://cdn.discordapp.com/attachments/326138757474680852/576636085493301266/3.PNG

[dark sparrow]

$x = \frac{11}{\cos(28^\circ)}$

[somber coyote]

Ann:

[dark sparrow]

which even appeared in your work

[cloud merlin]

Here this is the other one explained

[oak minnow]

help, how do I solve this I dont seem to understand it even though it gives a hint

[umbral snow]

Please see #info on helper pings

[dark sparrow]

have you done what the hint tells you to do?

[oak minnow]

yeah I made a line that goes through B and D

[dark sparrow]

okay

so this gives you two triangles

ABD and CBD

are you able to calculate the area of each?

[oak minnow]

BCD

[dark sparrow]

the vertices of a triangle can be listed in whatever order unless you're making a statement about similarity, which i am not.

that's beside the point, anyway.

[oak minnow]

...

ok

[dark sparrow]

there are two triangles there

are you able to calculate their areas?

[oak minnow]

what if I say no?

[dark sparrow]

then we're going to step back from the problem for a moment

but until you say yes or no, i'm not going anywhere.

[oak minnow]

well im 50/50 with that

[dark sparrow]

what do you mean 50/50

[oak minnow]

my guess would be 1/2basinB

[dark sparrow]

i can tell that you wrote sin*B there. that's inappropriate, because sin B is not the product of B and something called "sin".

(and it should be sin(B) anyway)

but that's beside the point

[oak minnow]

I dont care, I just want to know how to do it

[dark sparrow]

ok, you've blurted out a formula. you should probably give it some context, such as describing what the variables in it represent and what its output is meant to be

[oak minnow]

nvm i figured it out

0.5 * 8 * 9 *sin(A)

[dark sparrow]

that's the area of ABD, yes

are you able to calculate the area of the other triangle?

[oak minnow]

hmm

[dark sparrow]

if you were able to do that for the first triangle, then you should also be able to do it for the second.

[oak minnow]

half * 6 * 6.76 * sin(C)

[dark sparrow]

indeed

so now are you able to calculate the area of the entire shape?

[oak minnow]

how do I know where to put the side letters?

like u know how this for eg

[dark sparrow]

the what letters?

[dark sparrow]

you should not be concerning yourself with sticking to one particular set of labels and attaching them to every triangle you see

[oak minnow]

well thats what ive been doing

[dark sparrow]

you should instead understand visually what elements of your triangle the formula involves

[oak minnow]

i dont know

if it was something like ABD how would I state that

with the small labels

[dark sparrow]

okay so look

the formula you know as 1/2 ab sin(C) involves two sides and the angle between them.

[oak minnow]

yes

[dark sparrow]

the exact names of those sides and the angle DO NOT MATTER

and my point is you should not attach yourself to any particular names

[oak minnow]

so I could in between A and B i could put small a?

[dark sparrow]

??

[oak minnow]

nvm you wont understand what im trying to say

ok so for C) it would be ABD + BCD/CBD

[dark sparrow]

i don't know, would it?

[dark sparrow]

no

2. Requesting or offering the exchange of money for completing homework assignments is a bannable offense.(modifié)

<@&268886789983436800>

your problem, not ours

[upper karma]

Ann is french! TIL

This is why she has the Sophie Germain aura!

[half gull]

I think she isn't even if she is almost fluent in french

[dark sparrow]

you're right, i'm not french

and i'm pretty far from fluent in french too 😛

[radiant parcel]

ann's not french

your french is p decent tho from what i've seen

[dark sparrow]

my french is at the level where i can shitpost to some extent but can't read real french literature for shit

[half gull]

most of french people can't read french litterature when it's old

maybe not "cant read"

but have difficulties to understand a lot of sentences

[dark sparrow]

do you have any you could recommend me that isn't like that 😂

[half gull]

No I don't read a lot of books

And I don't think it exists x)

french writers love to use words that nobody uses in common life

maybe some books of Andrée Chedid because she's a recent writer

[gritty siren]

🤔

[autumn kestrel]

how do you solve this one? I know how to find area of a triangle but I’m not sure how to apply it

[half gull]

whats the value of x in terms of theta?

and the value of y in terms of theta?

[trail minnow]

the area of the triangle is xy/2

just solve for x in terms of theta and y in terms of theta and then plug

[autumn kestrel]

arent x and y in terms of theta just sin theta and cos theta or

ohh nvm i see

[half gull]

its 4sin(theta) and 4cos(theta)

[trail minnow]

yeah divided by two

[swift ice]

not 4sincos

its 8

@autumn kestrel its C not D

[autumn kestrel]

yeah I got it

I just put D because I had to guess

[oak minnow]

would be nice if I can get some advice on tis

[dark sparrow]

which part?

a i, a ii, a iii, or b?

@oak minnow

...hello?

[oak minnow]

a ii, a iii and b

@dark sparrow

[dark sparrow]

ok what's holding you up on a ii

what have you tried and where are you stuck

[oak minnow]

with a ii i cant seem to get it

i used the heron's formula as well

[dark sparrow]

heron's formula?

how are you using heron's on that

[oak minnow]

it gives us 3 sides?

[dark sparrow]

tell me, what shapes does heron's formula let you find the area of?

[oak minnow]

perimeter of something

[dark sparrow]

no

read my question

what SHAPES does heron's formula let you find the area of?

[oak minnow]

right angle triangles?

idk

[dark sparrow]

no

how are you using this formula if you don't even know what it applies to????

[oak minnow]

i used it on a right angle triangle and got the right answer

[dark sparrow]

i mean that's overkill

way overkill

[oak minnow]

how?

[dark sparrow]

because there is a MUCH SIMPLER way to find the area of a RIGHT triangle.

what you did is the mathematical equivalent of using a sledgehammer to hammer a nail.

[oak minnow]

i made sure i was on the safe side

[dark sparrow]

and my other point is that heron's formula is a formula that lets you find the area of a TRIANGLE. but QBCR is NOT a triangle.

QBCR is NOT a triangle and heron's formula applies ONLY to triangles.

[oak minnow]

we can turn it into a triangle cant we?

[dark sparrow]

you don't need to turn it into anything

it is a trapezoid

go review your area formulas.

[oak minnow]

ok so theres 0.5 * b * h

0.5 * bc * sin(A)

[dark sparrow]

no

no

no

[oak minnow]

0.5 * ac sin(B)?

[dark sparrow]

you CANNOT just blurt out formulae without saying WHAT THEY APPLY TO.

if you don't know what a formula applies to then you DO NOT KNOW THE FORMULA.

[oak minnow]

those are the formulas, no?

[dark sparrow]

if you don't know what a formula applies to then you DO NOT KNOW THE FORMULA.

if you don't know what a formula applies to then you DO NOT KNOW THE FORMULA.

if you don't know what a formula applies to then you DO NOT KNOW THE FORMULA.

if you don't know what a formula applies to then all you know is a jumble of symbols.

,rotate

[dark sparrow]

you should make this kind of infobox for EVERY SINGLE FORMULA YOU ENCOUNTER.

[oak minnow]

what would apply to a ii

[dark sparrow]

you need to identify the shape that QBCR is before you can say what applies to it.

[oak minnow]

how do I identify it?

[dark sparrow]

you should know all of the various types of triangles and quadrilaterals

[oak minnow]

so you say "you should know" instead of telling me a proper explanation

[dark sparrow]

sorry m8, if you don't know what say a square or a parallelogram or a trapezoid is then there is nothing i can do for you

[oak minnow]

what

i know what those are

[dark sparrow]

then you should be able to identify a square, parallelogram, trapezoid, etc. when you see one

[oak minnow]

;/ ok

could you tell me how to apply the numbers into a formula?

[dark sparrow]

no, because that is something you really should know by this point. if you are unable to plug numbers into a formula then you will not get anywhere at all

[oak minnow]

yeh ok but what is the formula of a trapezoid?

do you know?

[dark sparrow]

"the formula of a trapezoid" makes no sense

[dark sparrow]

the most commonly used formula for THE AREA OF a trapezoid is A = 1/2 (a+b) h, where A is the area, a and b are the bases and h is the height.

[oak minnow]

yeah still dont know how to find the height

[heady juniper]

modulus of ai<=1

ai is real

i got kinda far

but the proof is kinda sketch

i assumed for k

then for k+1

i used compound angle

and absolute value of cosine

would appreciate if someone could provide a more formal proof

[weary vine]

Let P(N) be the next statement: sin(x1+x2+...+xN) = a1 sin x1 + ... + aN sin xN, where every a_i satisfies |a_i|<=1
P(1) is true: sin(x) = 1*sin x
P(2) is true: sin(x1+x2) = sin x1 * cos x2 + cos x1 * sin x2 = a1 sin x1 + a2 sin x2
Suppose that P is true for some m>=2
Then sin(x1+x2+...+xM) = a1 sin x1 + ... + aM sin xM
P(m+1) is the following:
sin(x1 + x2 + ... + x[m+1]) = a1 sin x1 + a2 sin x2 + ... + aM sin xM + a[m+1] sin x[m+1]
But sin(x1+...+xM+x[m+1]) = sin((x1+...+xM) + x[m+1]) = sin(x1+..+xM) cos x[m+1] + cos (x1+...+) sin x[m+1]

sin and cos will always be somewhere inbetween -1 and 1

If you multiply two numbers absolute value of which is less than or equal to one you will end up with the number which abs is <= 1

Blah blah blah if P(m) is true P(m+1) is true

[heady juniper]

thanks dord

i was struggling with justifying my conclusion at the end

would it be better to call cos xK+1 a number Ap?

or would that lose generality?

[weary vine]

And since
P(1) & for every m P(m)->P(m+1) you get for every n from naturals P(n) is true

I'd suggest calling every cosine something like a_i

[heady juniper]

"P(1) & for every m P(m)->P(m+1)"

does this just mean true for 1

and if true for m true for m+1?

then true for 2,3........

[weary vine]

Well, P(1) is true since sin x = 1 * sin x

And since from P(m) follows P(m+1) using math induction you get that for every natural number P(m) is true

[heady juniper]

ok

i see now

thanks for the help dord

[weary vine]

Sure

[dark sparrow]

@heady juniper i can't read your thing well

is it meant to be $\sin\left(\sum_{i=1}^{n} x_i\right) = \sum_{i=1}^n a_i \sin(x_i)$

[somber coyote]

Ann:

[heady juniper]

yes

[dark sparrow]

right

[upper karma]

@dark sparrow is the 🐐

[dark sparrow]

the what?

[heady juniper]

goat?

[idle bloom]

Goat yes

[daring zealot]

Forgive the terrible paint job 😛 How can I calculate all the points on the yellow curve around the circle given the radius and angle? I'm working on a 2D game and need all the yellow pixels stored in an array.

[dark sparrow]

wdym all of them