geometry-and-trigonometry

[young hemlock]

ok

[little osprey]

So what would it look like? (the formula)

[young hemlock]

90 by 360

[umbral snow]

You need the perimeter of a whole circle of radius 7, then multiply it by 3/4 since it's a 3/4 circle. That gives the three curvy bits

[little osprey]

Don't give him the answer, try to make him understand

[young hemlock]

yeah i dont mind

i want to learn

[umbral snow]

I wouldn't go as complicated as a radians formula

[little osprey]

Wdym

It doesn't need any radians

[umbral snow]

Just make whole circles, then cut parts of them out by division

[little osprey]

That's not how the perimeter works

it needs the arcs and the sides

Anyways...

@young hemlock You still following?

[young hemlock]

yeah

[little osprey]

So um what do you get for the arc lenght?

lenght*

length*

[young hemlock]

90 by 360?

[little osprey]

yeah and if you keep that in the formula?

[young hemlock]

90 by 360 . times 2 pi 7

[little osprey]

Nice

So what do you get?

like the answer

for that

[young hemlock]

10.99557429

[little osprey]

Nice

And that x 3

Cuz there are 3 arcs

You got that step?

[young hemlock]

ye

[little osprey]

Then add 2x7

And you're set

for perimeter

[young hemlock]

uh

[little osprey]

Yeah?

[young hemlock]

well

the answer is 47.0 cm, 115 cm square

[little osprey]

What did you get from the calculations

[young hemlock]

461 point something

wait hold up

i messed up

[little osprey]

2x7+270xpix7

[young hemlock]

ok

u were right

i did times 14 by mistake

[little osprey]

Then round that by 3 s.f

[young hemlock]

also

why do we add 2 times 7

[little osprey]

Cuz there are two radius sides to add as well

So you see the 3 arcs right?

[young hemlock]

yeah

[little osprey]

do you see the 2 radius's?

[young hemlock]

2 radius sides...

[little osprey]

that are besides the arcs

Do you not see it?

[young hemlock]

the curve?

[little osprey]

Those sides

The curves were calculated already

[young hemlock]

oh

ok

👍

[little osprey]

Awesome

You can do area?

Or...

[young hemlock]

ill try

ok can't figure it out

@little osprey can u help

[little osprey]

k

whats the formula of a circle

area

[young hemlock]

pi r square

[little osprey]

yeah

so if it's a quarter

[young hemlock]

divided by 4

[little osprey]

what do you think it would be

yeah

and then there are 3 of those quarters

so what would that become

[young hemlock]

times 3

OH

[little osprey]

yeah

lol

[young hemlock]

i got it

sry for caps xd

[little osprey]

no worries

Shows you have expressions

[young hemlock]

lmao

[upper karma]

any help much appreciated

Matt watches as a hot air balloon rises and comes to a stop at a height of 130 feet. The angle of elevation of the hot air balloon from where he is currently standing is 30 degrees. How far does Matt need to walk if he wants to stand directly underneath the balloon?

[hallow smelt]

i think its 130sin60/sin30

the 60 and 30 are degrees

[hallow smelt]

does this only make one triangle?

[upper karma]

What’s his error?

[astral hornet]

Do you know how to do an angle bisector construction?

[upper karma]

$cos(5x)-cos(15x)=(3*sqrt(3)*tan(5x))/2$

[somber coyote]

rudy:

[upper karma]

Hmm

[pallid cloud]

@upper karma his mistake is he considered CB=CD, and it is clearly not the case

[pallid cloud]

Metric relations exercise

But if you could send a better image

I can barely see

@lime flame

[lime flame]

oh srry

can't at the momment

ln is an altitude and klm is right

and km is 16

[pallid cloud]

I'll tell you what you should try

[lime flame]

ok

[pallid cloud]

For A see if $LM^{2}=MN×MK$

It's called Euclid's theorem

If it is verified

Then LMK is right at L

And the reciprocal is also true

Anws

For B the same thing

[somber coyote]

Patrick Salhany:

[pallid cloud]

For B the same thing

I guess B is correct

$LM^{2}=12^{2}=144$

[somber coyote]

Patrick Salhany:

[pallid cloud]

$MN×MK=9×16=144$

[somber coyote]

Patrick Salhany:

[pallid cloud]

So B is correct

[dark sparrow]

oh god two col proofs

okay but actually it's not hard to go from AD = BC and AB = CD to ABCD being a parallelogram

draw a diagonal, get two congruent triangles (by SSS), prove some angles equal to one another

well like ok lemme make this more concrete

triangles ABD and CDB are congruent

angle ABD = angle CDB

the triangles are congruent by SSS

[dark sparrow]

what's holding you up?

@young hemlock

[young hemlock]

i dont understad

how to do it

its part of the arcs section questions

[dark sparrow]

well ok so like

you know what a clock looks like, right

[young hemlock]

yes

[dark sparrow]

well

what curve does the tip of the minute hand follow?

[young hemlock]

i dont know

[dark sparrow]

imagine the minute hand of a clock

how does it move as time goes on?

[young hemlock]

clockwise

[dark sparrow]

it rotates.

so what curve does its tip follow?

[young hemlock]

um

[dark sparrow]

imagine the hand rotating

[young hemlock]

ok

[dark sparrow]

focus on its tip

how does it move?

[young hemlock]

it moves around

[ember obsidian]

Shape

[dark sparrow]

it moves along a ____

fill in the blank

[young hemlock]

circle\

[dark sparrow]

great

so the tip of the minute hand moves in a circle

[young hemlock]

yes

[dark sparrow]

so

so if you follow the tip of the minute hand over a certain length of time (here 35 minutes)

what curve does it trace out?

[young hemlock]

a circle

[dark sparrow]

not necessarily

it takes an hour to go fully around the circle

so what if we only track its movement for less than an hour

[young hemlock]

so semicircle

[dark sparrow]

no, not necessarily.

that's if it was half an hour.

it traces out a ___

[young hemlock]

I don't know.

[dark sparrow]

does the curve i drew here look familiar?

[young hemlock]

No

[dark sparrow]

there's a special name for this kind of curve

[young hemlock]

What is it called

[dark sparrow]

that's what i should be asking you ;P

[young hemlock]

ok

perimeter

[dark sparrow]

no

you even mentioned the word i'm asking you for

in this very conversation

this curve is called an arc.

[young hemlock]

Ok

[dark sparrow]

honestly if you didn't know what that was then you should go review your basics

bc you will not be able to do this exercise not knowing what an arc is

[young hemlock]

Ok I got the answer

[young hemlock]

can someone help me in this sum

[dark sparrow]

what's holding you up here?

[young hemlock]

well

I'm trying to find the volume

[dark sparrow]

that's what the problem is asking, yes.

[young hemlock]

But I don't think I'm putting the correct values

[dark sparrow]

how are you trying to find the volume?

[young hemlock]

uh

using the formula

pi r square times height

[dark sparrow]

what does that formula apply to?

[heady urchin]

Ann I need help after this guy

[dark sparrow]

ok

[young hemlock]

oh i think i used the weong formula

[dark sparrow]

no formula exists in isolation

[young hemlock]

its for cylinder

but this isnt cylinder

[dark sparrow]

indeed.

[young hemlock]

lemme try it with the other formula

[heady urchin]

Ok ann

[dark sparrow]

"the" other formula?

sorry @heady urchin, we're not done here yet.

@young hemlock what do you mean by the other formula?

[young hemlock]

um

area of cross section times length

@dark sparrow would it be width times depth times height

[dark sparrow]

i don't know, would it?

[young hemlock]

i dont know thats why im asking you

[dark sparrow]

try to answer the question yourself

"width * depth * height" does give you the volume of a shape. what is that shape? does your pool have that shape?

[young hemlock]

cuboid

[dark sparrow]

and is your pool a cuboid?

[young hemlock]

yes i think

[dark sparrow]

are you sure about that?

[young hemlock]

yes

[dark sparrow]

you'd be wrong.

the faces of a cuboid are all rectangles, but two of your pool's faces are trapezoids.

[heady urchin]

Ann is there someone that I can msg I need help ASAP lol

[dark sparrow]

i mean you can PM me i guess

why the urgency though?

@young hemlock are you still here?

[blissful cedar]

The pool isn't at a constant depth also

[dark sparrow]

...

[young hemlock]

yes

im still here

ill move onto the next question since i cant seem to answer this one

[dark sparrow]

as you wish.

[upper karma]

@young hemlock I think I can help

Or have you figured it out already

[young hemlock]

@upper karma I asked someone else and he helped me

No worries 👍

[upper karma]

Okay

[flint compass]

Why do you divide 2pi by the coefficient to get the period of sinusoidal function?

[forest niche]

Help me 😭

[dark sparrow]

what's holding you up here?

[forest niche]

I’m just lost all around

Where do I start?

[dark sparrow]

ask yourself this:

• What is this thing in front of me?
• What am I asked to do with it?
• What can I use to do that?

[forest niche]

It’s a rhombus?

But it’s cut weird?

[dark sparrow]

no, it is not a rhombus. its sides aren't known to all be equal to one another

it is, however, a parallelogram

[forest niche]

OH

Bxh?

[dark sparrow]

you've asked yourself the first question but showed no indication that you have asked yourself the other two

and i'd like you to spell out your answer to the second question before blurting out a formula without even saying what that formula does

(also, don't use x for multiplication, for god's sake!)

[forest niche]

Uh the 11 is the base?

And the 6 is the height

[dark sparrow]

please read what i am saying

[forest niche]

Idk you’re confusing me right now

[dark sparrow]

you've answered "What is this thing in front of me?"

[forest niche]

Ok

[dark sparrow]

what i'm now asking you to do is to answer the next question on that little list i sent

[forest niche]

I need to find the area

[dark sparrow]

okay, good

and now the third question

"What can I use to do that?"

[forest niche]

And use base times height?

Because it’s a parallelogram

[dark sparrow]

sure, you can use the formula that states the area of a parallelogram is equal to its base times its height.

[forest niche]

Okay, thank you!

[dark sparrow]

you're welcome

there are two things you need to find out no matter what problem you're doing

what to do, and how to do it

in this problem, the former was harder than the latter

[forest niche]

Ok I appreciate it

[fiery iron]

If area of polygon P1 constructed from center of circles C1 is greater than that of polygon P2 constructed from another set of circles C2, and each circles is having same radius, does it guarantees that area( C1 ) > area( C2) if area(P1) > area (P2)

https://i.postimg.cc/ZRYLMnv9/area-of-circles.png

[upper karma]

how to solve the second question?

[tropic shard]

When in doubt, quadratic equation.

[upper karma]

@tropic shard elaborate on that pls ;-;

[valid agate]

@paper verge Please don't ghost ping

[upper karma]

So that's who did that

[valid agate]

I've got a pounce on mod pings

[upper karma]

<@&268886789983436800> maybe you guys can answer my question? i really need help rn

[native night]

don't ping mods for hw help

[upper karma]

oh sorry i didn't know that

[tropic shard]

To solve for x for x²-7x-8=0, you'd use the quadratic equation for A=1, B=-7, and C=-8

[upper karma]

@tropic shard sorry thats the third question

i said second

[dire rampart]

[faint orchid]

Can someone help me figure out this identity? I think it has something to do with Euler's Identity but not sure

could be overthinking it

Ignore the 2 and 1 on the far left, those are just initial conditions for a discrete time system

Nevermind, I think I understand. Its just basic unit circle manipulation

Looks like the combination of both the sin and cosine terms yield the -pi/6

[oak minnow]

Can someone help me with trig

like the basic one

how do u do question b

[dark sparrow]

you don't need trig for this

[oak minnow]

waddaya mean

im stuck

[dark sparrow]

you're overthinking this, surely.

[oak minnow]

I used the cosine rule

[dark sparrow]

oh my god what

look

this is a right triangle.

[oak minnow]

nvm

tangent

[dark sparrow]

no, even simpler

you know two of its sides and want the third.

[oak minnow]

yea

[dark sparrow]

there's a theorem which relates the three sides of a right triangle

very famous

and you definitely know what it is

[oak minnow]

ye

[dark sparrow]

yeah so what is that theorem

[oak minnow]

pt

[dark sparrow]

pt?

[oak minnow]

its a shorter way of saying it

pythagoras theorem

[dark sparrow]

yes

so

[oak minnow]

or however you spell it

[faint orchid]

If you know the theorem why aren't you able to deduce the answer from the theorem?

[somber coyote]

warmpoptart:

[oak minnow]

yeah

[dark sparrow]

yeah so

what's the problem

[faint orchid]

You have A and C

[oak minnow]

like the answer its giving me in in a square root

.help

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[faint orchid]

The answer is allowed to be in square root form, not every triangle has perfect sides with integers.

in this case you should have the formula to find the side B

[oak minnow]

I dont understand them though

[dark sparrow]

you don't understand what

[oak minnow]

I'd like it in the sin, cos, tan form

√35 m

[faint orchid]

$b = \sqrt{6^2 - 1^2}$

[dark sparrow]

6^2 - 1^2 @faint orchid

[oak minnow]

how'd u type that out so fast

jesus

[dark sparrow]

the answer is sqrt(35) m

[faint orchid]

Ah yeah my bad

[dark sparrow]

what more do you want

[somber coyote]

warmpoptart:

[faint orchid]

What do you mean by you want it in 'sine, cosine, tan' form?

it's asking you for the length of the edge.

[somber coyote]

warmpoptart:

[oak minnow]

ah

how tf did u do that

I need to learn those commands

I know but

[faint orchid]

Check the resources tab for a LaTeX cheat sheet

[oak minnow]

wait

[faint orchid]

#resources

[oak minnow]

6^2+1^2

is 37?

[faint orchid]

its minus, I made a mistake and corrected.

[oak minnow]

not oh

[faint orchid]

6^2 - 1^2

[oak minnow]

so we just put the question the form of a^2+b^2=c^2

?

[dark sparrow]

what

[oak minnow]

nvm

so for this one, question b, how do you find the height of it?

would it be 1/2ah

wait

[dark sparrow]

again

this is a right triangle

and you know two of its sides

and you want the third...

[oak minnow]

i already did that one

but

[dark sparrow]

but what

[oak minnow]

i have another question

this one I cant figure out

[dark sparrow]

make a picture!

[oak minnow]

I did

its not very good

help

[dark sparrow]

now it's trig time

this is a right triangle

you know a side and an angle

[oak minnow]

yes

how do i find the third

[dark sparrow]

i mean there are two sides to be found but those are different subproblems

are you able to find AB?

[oak minnow]

nah

wait

wouldn't it be 3^2+3^2

sqrt of that

[dark sparrow]

why?

you don't know if AC = 3

[oak minnow]

how do i find AB

help

plz

[dark sparrow]

well you said you were doing basic trig, didn't you?

[oak minnow]

could u please just explain it out

[dark sparrow]

sorry, i was having lunch.

@oak minnow still here?

[oak minnow]

yes

I was waiting for a reply

[dark sparrow]

k

so

how familiar are you with sin, cos and tan

[oak minnow]

a fair bit

[dark sparrow]

well

do you know what they are

[oak minnow]

what

yes

[dark sparrow]

well then

what are they

if you answer this question, it may help you solve the problem

[oak minnow]

what do you mean

sine, cosine and tangent

[dark sparrow]

you said you knew what these are

state their definitions to me

[oak minnow]

oh

for the question

[dark sparrow]

no, forget about the question for a moment

tell me what sin, cos and tan are

[oak minnow]

i just use the sohcahtoa thing

[dark sparrow]

i mean. okay, yeah, that's a mnemonic to help remember the definition

ok

alright

so back to the problem

[oak minnow]

ya

[dark sparrow]

you know BC and you want AB, and the one angle you know is angle A

which of sine, cosine and tangent is it possible to use here?

[oak minnow]

erm

cosine

[dark sparrow]

why?

(i'm going to ask you "why" a bunch. you gotta be able to explain what you do.)

so, why cosine?

[oak minnow]

because we need to find

idk

i give up

im not good at trig

[dark sparrow]

okay look

this is just logic

common sense

don't give up

you were mere inches away

[oak minnow]

ok since we have both sides BC we need to find AB

and to do that

we need to find the sides

[dark sparrow]

no, no

look

there's two sides you care about

you care about BC because it's the only side you know

and you care about AB because you want to find it

does that make sense?

don't overthink it.

[oak minnow]

yes ok but

how do I find the other side

[dark sparrow]

well

wait

[oak minnow]

the length of

[dark sparrow]

what other side

[oak minnow]

the ladder

[dark sparrow]

well

this is where trig comes in

[dark sparrow]

so you know the mnemonic "SOH CAH TOA"

[oak minnow]

eah

[dark sparrow]

what does it stand for?

[oak minnow]

ok

sine = opposite/hypotenuse, cosine = adjacent/hypotenuse , tangent= opposite/adjacent

[dark sparrow]

okay good

so

back to this diagram

is BC the opposite, the adjacent, or the hypotenuse?

[oak minnow]

adjacent

[dark sparrow]

are you sure?

[oak minnow]

i think so because its adjacent the opposite?

[dark sparrow]

adjacent means "next to"

next to the angle, in this case

BC is the opposite

[oak minnow]

oh

yeah

cause of the angle

[dark sparrow]

yes bc the angle is at the center of all this

so now

is AB the opposite, the adjacent, or the hypotenuse?

[oak minnow]

hyp

[dark sparrow]

okay

so you have the opposite and the hypotenuse.

[oak minnow]

sine

[dark sparrow]

great

are you able to write this out?

[oak minnow]

so would it be

3/sine26

6.84

[dark sparrow]

it = what?

,w 3/sin(26°)

[somber coyote]

Display results online and refine query

Requested by Ann#8192

[dark sparrow]

...yes, AB = 6.844 m

[oak minnow]

how you have u done it

[dark sparrow]

that's exactly how i would've done it. this is one of those problems where there really is only one solution

[oak minnow]

thanks ann

[upper karma]

So I'm really struggling with the Trigonometric function like... i know how sin, tan, and cos works, but when i typed it in the calculator, it never came out right

[dark sparrow]

is your calculator in degree mode or radian mode

[upper karma]

Rad

[dark sparrow]

put it into degree mode.

[upper karma]

It's on

I'm not sure how to find the c and b if I only have the 67.2

[dark sparrow]

...wait ok what

lemme read the problem again

are you given only a = 67.2?!

[upper karma]

Yup

The C is just 90°

[dark sparrow]

together with the right angle, that means you only have two elements in the triangle

you need three to solve it

[upper karma]

Unless if i just click "there is not enough information to solve it"

[dark sparrow]

there is not enough info.

[upper karma]

Ah i figured

I have another question that's more simpler, but i keep getting it wromg again

If i typed it out, the number somehow got way too small

And I'm not sure what i did wrong
I put TAN 32.7/18.6

[dark sparrow]

you do not want tan(32.7/18.6)

what does inside the tan would be an angle

you want the inverse tan.

[upper karma]

Ohh

60.4

Hm i typed it out and it wasn't right

[dark sparrow]

okay, sorry, my vision is a bit blurry today

you were finding the wrong angle

[upper karma]

Oh oof

[oak minnow]

how would I draw the picture

<@&286206848099549185>

help pls

[keen aspen]

Draw a right triangle with theta being 16 and the height being 400

[oak minnow]

yeah lol thanks man

also for this, i've found XY but how to I find the other values of the pro numerals for the triangle

[upper karma]

trigonometry?

[dark sparrow]

@oak minnow how did you find XY?

(don't tell me the value, bc i don't care about it. just tell me how you found it)

[oak minnow]

dont worry, i did't read the question properly

so what I did was

use the sine rule

[dark sparrow]

uh huh

law of sines

[oak minnow]

10/sine50/sine70

if u know what i mean

[dark sparrow]

...okay, you should never write "a/b/c" like that

did you mean (10/sin(50°))/sin(70°)?

or did you mean 10/(sin(50°)/sin(70°))?

[oak minnow]

something like that

[dark sparrow]

then why did you put that first slash?

[oak minnow]

idk

[dark sparrow]

$\frac{10 \sin(50^\circ)}{\sin(70^\circ)}$

oof

[oak minnow]

yeah

i still dont understand the synbols

$

[somber coyote]

Ann:

[oak minnow]

$

[dark sparrow]

oh $ just tells the bot i want to render a formula

[oak minnow]

, w $\frac{10 \sin(50^\circ)}{\sin(70^\circ)}$

[somber coyote]

Display results online and refine query

Requested by Fuhrer1#6111

[oak minnow]

, ask $\frac{10 \sin(50^\circ)}{\sin(70^\circ)}$

[dark sparrow]

yeah WA gets confused lol

[somber coyote]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Display results online and refine query

Requested by Fuhrer1#6111

[dark sparrow]

,w (10 sin(50°))/sin(70°)

[somber coyote]

Display results online and refine query

Requested by Ann#8192

[oak minnow]

yeah

thats the one

are u a maths teacher

idk what im doing wrong here

(12 sin(38°))/sin(90°)

, w (12 sin(38°))/sin(90°)

[somber coyote]

Display results online and refine query

Requested by Fuhrer1#6111

[oak minnow]

the answer is meant to be 9.46

[dark sparrow]

first off this is a right triangle & you would've done better to recognize that

and second you used the law of sines correctly but found the wrong side

@oak minnow

[oak minnow]

yes

[dark sparrow]

...??? what are you saying "yes" to?

[oak minnow]

the second point you made

so tell me which side i need tofind

[dark sparrow]

if you look at the picture

the side to be found is literally marked

yknow, you won't always be able to get others to read shit for you

[oak minnow]

woah ok

cant i just put in the question in WA and get an answer

but idk how to do that

[dark sparrow]

no, you cannot

consider thinking

you are definitely able to do that question, if only you apply some mental effort

[oak minnow]

i am but i still dont understand :((

[dark sparrow]

this is a right triangle

do you understand that?

@oak minnow

[oak minnow]

Yes

I drew a picture of it but I drew it tilted

as a right triangle

with the points

[dark sparrow]

well

okay

this is a right triangle

so the SOH CAH TOA mnemonic applies

[oak minnow]

sine rule

[dark sparrow]

?

[oak minnow]

but we don't know the opposite ?

[dark sparrow]

you care about two sides: XY and YZ

what are they, in relation to angle Y?

[oak minnow]

right angles

[dark sparrow]

that makes zero sense

how can a side be an angle

please read what i am saying

[oak minnow]

yeah I'm

[dark sparrow]

i mean first off this is the geometry channel

and second i'm not exactly sure whether or not you're shitposting rn

and third, you should get a pingable nickname

???????

[keen aspen]

@upper karma can you not

[dark sparrow]

<@&268886789983436800>

[paper mauve]

@upper karma is everything okay? is there something I can assist you with?

[twin heron]

@upper karma stop

[keen aspen]

Just bean him

[twin heron]

fair

[paper mauve]

im kinda interested to see where this leads tho

[dark sparrow]

beaned

[twin heron]

[fair raven]

[dire rampart]

[oak minnow]

help

[oak minnow]

ok, or not

[halcyon coyote]

[upper karma]

would 1 and 6 be complimentary

No, complementary angles equal 90. 1 and 6 are more than 90 degrees

[empty trout]

2 and 6 would be complimentary

[upper karma]

so 1 and 6 is obtuse

yeah

A: angle 1 and angle 4 Strike Reset
B: angle 1 and angle 6 Strike Reset
C: angle 3 and angle 5 Strike Reset
D: angle 3 and angle 6

I mean out of these

its not a

not 3 and 6

either 3 and 5 or one and 6

3 and 5

[empty trout]

since 1 and 4 are vertically opposite angles, that means that 4 is 90 degrees, making 3 and 5 complimentary

[upper karma]

3 and 5 are also acute

so you can assume they add up to 90 deg

ohhh so they are 90 degrees and acute because both angles are touching each other

they are touching each other and their angles form a complementary angle

alright

[upper karma]

Beth leaned a ladder against her house, as shown below.

Figure not drawn to scale.

37° Strike Reset
B: 90° Strike Reset
C: 127° Strike Reset
D: 180° Strike Reset

180-53

this 180

-53

127

yep

does anyone know the formula for height of a cone

$V = \dfrac{1}{3}\pi r^2 h \
3V = \pi r^2 h \
\
\dfrac{3V}{\pi r^2} = h \
\therefore h = \dfrac{3V}{\pi r^2}$

.

Enter the answer only. Use only numbers. Do not use letters.

Oliver pours 141.3 cubic centimeters of water into a cone-shaped cup.

When he stops pouring, if the water at the top of the cup forms a circle with a diameter of 6 centimeters, the height of the cone is

so radius is 6 right

[somber coyote]

Dr. Nitro:

[upper karma]

6 divided by 2 is 3

yes

use the formula I gave to solve for height

ok

btw thats the pic

Oliver pours 141.3 cubic centimeters of water into a cone-shaped cup

IDk what the number is used for tho

only problem

I got 1331.72Cm

can you check that to see

sure

alright

R=3

does it want it in terms of pi or an exact answer

if it wants an exact answer, what value of pi is it asking for

that what it says

ok

I got

$47.1 \pi$

[somber coyote]

Dr. Nitro:

[upper karma]

that was the question btw

https://media.discordapp.net/attachments/326138757474680852/565635564171165696/A021BC6_250731.png?width=330&height=413

are you asking what values to substitute in?

Well I know R is 3

but idk H

its asking for the height of the cone

@upper karma

$\text{The Formula for the Height of the Cone is:} \
\
h = \dfrac{3V}{\pi r^2}$

[somber coyote]

Dr. Nitro:

[raw prism]

How I do

[upper karma]

$\tan^2(2x) = \dfrac{1-\cos(4x)}{\cos(4x)+1} \
\cos(2x) \left(\dfrac{1-\cos(4x)}{\cos(4x)+1}+3 \right)+3 = 0$

[somber coyote]

Dr. Nitro:

[upper karma]

$\dfrac{\cos(2x)-\cos(2x)\cos(4x)}{\cos(4x)+1}+3 \cos(2x)+3$

[somber coyote]

Dr. Nitro:

[upper karma]

$\dfrac{\cos(2x) \left(1-\cos(4x) \right)}{\cos(4x)+1}+3 \cos(2x)+3$

[somber coyote]

Dr. Nitro:

[raw prism]

so i can multiply the 2x first and then the cos?

[upper karma]

no, you multiply the 2x with the cos

[pallid cloud]

Abdullahsz1, it's cos(2x) not 2x

It's not space between them

[raw prism]

ok

[pallid cloud]

My first tought was to write tan^2 as sin^2/cos^2

Then multiply by cos2x

[raw prism]

yea thats what i did but no work

[pallid cloud]

Then write sin^2 as 1-cos^2

Then multiply the two members of the equality by cos2x

[raw prism]

also why did you add a 1 on the bottom shouldnt it just be cos^2?

[pallid cloud]

Lemme write on a piece of paper

2 mins

[raw prism]

kk np

[upper karma]

or you can substitute the 2x as a variable and then add in the 2x

[raw prism]

shouldnt the for cos on top and on bottom be to power of 2 since tan is also to power of 2?

[pallid cloud]

Yes it should

[raw prism]

,rotate 90

[pallid cloud]

,rotate 270

[pallid cloud]

This is an equivalence as long as cos(2x) is different then 0

To not divide by 0 somewhere

And to not multiply by 0 to be a double implication

[raw prism]

wait wouldnt it be to power of 3 when you multiply the cos^2 (2x) by cos(2x)

[pallid cloud]

@raw prism where ?

[raw prism]

2nd to 3rd step

[pallid cloud]

Where do you see cos^2 that I multiplied by cos ?

I didn't

[oak minnow]

hi

[raw prism]

when you multiply the first part (1-cos^2 (2x)/cos^2 (2x)) by cos(2x)

[pallid cloud]

I didn't

What I did is

$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$

[somber coyote]

Patrick Salhany:

[pallid cloud]

$c\neq 0$

[somber coyote]

Patrick Salhany:

[raw prism]

yea but you still have to multiply them with cos 2x casue they are in the paranthesis

[pallid cloud]

What paranthesis bro

The parenthesis are for cos

$\cos(something)$

[somber coyote]

Patrick Salhany:

[pallid cloud]

Do you want a version without parenthesis for cos, sin and tan
Ok

[raw prism]

yea its cos(2x) times that something right?

in the 2nd to 3rd step you only multiplied the 3 by cos(2x) not the other stuff

[pallid cloud]

I'll send a better pic

1 min

[raw prism]

k sorry im not understanding it properly

[pallid cloud]

,rotate 270

[pallid cloud]

,rotate 270

[raw prism]

oh ok i get it now you multiply before turing the sin^2 into 1-cos^2

[pallid cloud]

This is true as long as $\cos(2x) \neq 0$ \
For the 2 equations to be equivalent

[somber coyote]

Patrick Salhany:

[pallid cloud]

Yes

[raw prism]

i suck at math bigtime so thanks a lot

[pallid cloud]

Basically all of us do

[raw prism]

yea i wanted to do engineering or something like that but prob gonna move to a different field cause math is a biatch

[pallid cloud]

Nah maths is good

Don't look at this way

[raw prism]

yea i know but i cant understand it

[pallid cloud]

It is the easiet thing believe me

[raw prism]

its really fun when i understand it I realized that but idk why i spend so much time practicing it but like in 2 weeks its all gone from my brain

[pallid cloud]

We all forget don't worry

Even the biggest mathematicians

A little bit of practice to get better

And most importantly understand what you're doing

[raw prism]

well thanks i'll keep trying for now

[pallid cloud]

You're not asked to be Gauss
But you're asked to understand and practice

Yes always try

[oak minnow]

I dont understand this

[pallid cloud]

Why

What you don't understand

[oak minnow]

howd they get from .5071 to 30.47

[pallid cloud]

Sine inverse

On calculator

Shift sin

[oak minnow]

o

[pallid cloud]

Or $sin^{-1}$

[somber coyote]

Patrick Salhany:

[oak minnow]

sin^-1()

yeah

ok

[pallid cloud]

Ye

Do you know why we have two possibilities ?

[oak minnow]

so just put $sin^{-1}{0.5071}

$sin^{-1}{0.5071}

$sin^{-1}{0.5071}$

[somber coyote]

Fuhrer1:

[pallid cloud]

Yes

But it gives you one answer I guess on calculator

[oak minnow]

yeah

thats right

[pallid cloud]

We have 2 answers, because $\sin(\alpha)=\sin(180°-\alpha)$

[somber coyote]

Patrick Salhany:

[oak minnow]

yeah

it would be 180-30.47

[pallid cloud]

Yes

[oak minnow]

=149.13142

or something

that would be the slope of the angle ?

[pallid cloud]

?

[oak minnow]

nvm

i didn;t word it correctly

[pallid cloud]

Tell me what you mean

[oak minnow]

wouldn't 149 be the angle

of Z

[pallid cloud]

2 angles are possible
30.47...
And 149.53...

Z could be both

[oak minnow]

i

[pallid cloud]

They draw an image for a better explanation

[oak minnow]

hmm ok, will you be on later on today?

I might need your assistance

like in an hour or less

[pallid cloud]

Actually no, bc it's 1:36 am

[oak minnow]

oh gawd

in America

[pallid cloud]

I want to sleeeepppp
Yeah like 7 hrs less

[oak minnow]

Australia is 8 36 am

ok

[pallid cloud]

No me not America

[oak minnow]

you should go to sleep

europe?

[pallid cloud]

Asia, Lebanon

[oak minnow]

oh

[pallid cloud]

I guess we're gmt+3

[oak minnow]

yeah

we are gmt

[pallid cloud]

If you like, I'll be on like for 8 hrs from now

[oak minnow]

aest

yeah sure

ok gn

yoi should get some sleep

[pallid cloud]

Thanks

[oak minnow]

w

[pallid cloud]

Yes I will, now!

[oak minnow]

erm how do I do this

[umbral snow]

Use sin law to get angle B, then the sum of the angles is 180

[oak minnow]

yes but I keep getting different answers

[valid pike]

Isn't it just 99.25°

[upper karma]

What Baynex said

[steady sleet]

\🅱aynex*

wait

🤢 #geometry-and-trigonometry

[upper karma]

do this

[somber coyote]

Haversine:

[oak minnow]

@valid pike yeah its 98.16

but how did u know

[oak minnow]

can someone help me with dis

<@&286206848099549185>

[dark sparrow]

what is holding you up?

[halcyon coyote]

[oak minnow]

I dont get it

[dark sparrow]

how are you getting different answers?

show your work

[oak minnow]

so

can you just tell me on how to do it rather than asking me to "show work"

[dark sparrow]

no, i want to see what you are doing so that i can see where you messed up

[oak minnow]

ok wait

it will take a while cause i'm gonnatype it up

on WA

$8sin(108){frac}/9

what

wait

i'll just show you on paint

it'll be easier i guess

@dark sparrow

yeah so aha

[dark sparrow]

108°?

[oak minnow]

oh w8

[dark sparrow]

where on earth did you get 108° from?

[oak minnow]

so i did something like this

@dark sparrow

[dark sparrow]

...and?

[oak minnow]

so what did i do wrong

besides from everything here

[dark sparrow]

so far, you didn't

[oak minnow]

o rlly

the answer is meant to be 98.16

but idk how they got that

180-33.18

doesn't equal to 98

[dark sparrow]

why are you subtracting 33.18 from 180?

[oak minnow]

because of the sine rule?

[torpid mirage]

So you are trying to find theta?

[oak minnow]

yea

[dark sparrow]

because of the sine rule?
?????

here

you found that angle B = 33.18° @oak minnow

[oak minnow]

angle b is the longest side right?

[dark sparrow]

reread the message you just sent.

it doesn't make any sense. how can an angle be a side?

don't overthink this

you know two of the three angles in your triangle.

[upper karma]

Don’t be so hard on him

[dark sparrow]

how am i being hard

do you know how to find the third angle in a triangle given the other two? @oak minnow

[torpid mirage]

Your value for B is incorrect.

[oak minnow]

the angle - 180

[torpid mirage]

sin (38°) / 8 = sin (B) / 9 (Sine rule)
B = sin^-1 [9 sin (38°) / 8] = 43.838 (Inverse of sine)
Theta = 180° - 38° - 43.838° = 98.162° (Interior angles of a triangle)

[dark sparrow]

@torpid mirage is it?

[torpid mirage]

Isn't it 9/8?

[dark sparrow]

wait yes

yes it should be 9/8 shit

i had that before

[oak minnow]

wait

[torpid mirage]

The angle opposite the side is used for sine rule.

[oak minnow]

why would it be 9/8

[torpid mirage]

9cm is opposite angle B.

[dark sparrow]

sin(38°)/8 = sin(B)/9

isolate sin(B)

multiply both sides by 9

[oak minnow]

ok

[torpid mirage]

Is there any difference in the arcsin and sin^-1 function?

[dark sparrow]

@torpid mirage no, and in fact desmos supports both notations

[oak minnow]

whats

a

[torpid mirage]

The angle. In this case, angle B.

You can use practically any variable.

[oak minnow]

oooo I see

180-38-43.1622...

[dark sparrow]

you should've been able to come up with the second line yourself, but evidently you're shaky on the very basics of geometry

[oak minnow]

true

i cant argue there

[dark sparrow]

well then

go back to the basics

and make sure you have those down cold

[oak minnow]

@dark sparrow

This is right?

[dark sparrow]

yes

[torpid mirage]

Yes! Nice!

[oak minnow]

phew thanks

[oak minnow]

How do I do this?

it doesn't state whether its a right triangle

[dark sparrow]

you cannot assume a triangle is right unless it is known to be right

make diagrams

[oak minnow]

i'll use the law of cosines, maybe that'll work

[dark sparrow]

i mean...

you've given five subproblems but no indication of which one you are stuck on

the law of sines and the law of cosines will both probably be needed to solve this problem, each in its own circumstance

[oak minnow]

yeah...

[torpid mirage]

For all of them, sin (A) / a = sin (B) / b = sin (C) / c. So you need to find a, b, c, A, B and C. In (a), a, B and C are given. A can be found using 180° - B - C. Now that you have 3 angles and 1 side, you can use sine rule to get the other two sides.

Every single question gives you 2 angles and 1 side, which is enough information to get the 3rd angle and get all of the sides using sine rule only.

[acoustic fiber]

The difference in the measures of two complementary angles is 12°. Find the measures of the angles?

[dark sparrow]

do you know what it means for two angles to be complementary?

[acoustic fiber]

yes

[dark sparrow]

what, then

[acoustic fiber]

their sum forms 90°

[dark sparrow]

uh huh

so now

you have two angles; their sum is 90° and their difference is 12°

what must they be

[acoustic fiber]

I could do trial and error method

[dark sparrow]

don't

that's gonna take you too long

[acoustic fiber]

correct

let one angle be x

[dark sparrow]

ok

[acoustic fiber]

then x+x=90

[dark sparrow]

huh?

[acoustic fiber]

sorry

[dark sparrow]

no, this is not correct. in writing that, you just assumed the two angles would be equal

[acoustic fiber]

yeahh correct

ok I will think about it

[dark sparrow]

ping me when you come up with something

[acoustic fiber]

sure

I've no clue, how I did that lol, I was just writing down the points and it just hit me and this is the result

@dark sparrow

[dark sparrow]

i mean this is basically exactly what i expected of you

[acoustic fiber]

did I do something wrong?? @dark sparrow

[dark sparrow]

no

on the contrary

this is exactly right