geometry-and-trigonometry

[wary bone]

well thats not real life, else it's close. exactly my problem

[remote heart]

yea

[upper karma]

u guys are Awesome! i actually learn a lot

Thanks

[little osprey]

looks god

[remote heart]

desmos says yes

I'm 'MURRRICANNN by the way, jc

[little osprey]

good job

ah

cool

[remote heart]

yeye

are you?

[little osprey]

Nah im not MURRRRICAN

[remote heart]

o

[little osprey]

@wary bone A pipe?

[remote heart]

I've seen people spell it strangely like "American", are you this?

[little osprey]

Ppfffft

Im no uneducated fool

[remote heart]

ah, good

"American" just isn't natural

[little osprey]

Doesnt sound free enough ngl

[wary bone]

wdym a pipe?

[prime jewel]

I got it, thank you guys.

[remote heart]

nice!

[prime jewel]

I forgot i had to divide.

I can do complex problems but can't understand basic ones! Nice 😄

xD

[little osprey]

like a uh pipe

you know

water pipe?

[remote heart]

but how do you get the surface area?

[wary bone]

yeah but in what case would i only know the cross area but not the radius of that pipe?

[remote heart]

without knowing radius, in that case

[little osprey]

in the case that you work for a company

thats a really hard question

ig you would know the radius as well

[wary bone]

yeah i think about that a week now for my 8th grader and can't find any suitable

[little osprey]

what was the initial thing

[wary bone]

you example is still a good idea, if i think about pneumatic delivery (if thats the right word for shooting letters through pipes)

[remote heart]

it is

[wary bone]

i need a real circle where i know the surface area but not the radius

[little osprey]

Candle maybe

im just throwing ideas

[remote heart]

what about a fan?

I'm not sure

like you know the area that the fan blows, and you know it's circular

[little osprey]

huhhhh

thats actually good

[wary bone]

if i think about a fan for air condition that could work, cause i could start with the air flow

[little osprey]

Theres a room for teachers lounge

[wary bone]

but a lot detail to do for 8th grade. can't beleive there seems no simple one

ah ty

[little osprey]

Maybe the other teachers have a good idea

yeah no worries

[versed iron]

can someone help me with a bunch of these? I'm in grade 9

[compact spire]

https://i.gyazo.com/03f6982a0c0c438c50f01005dd272f9f.png

[prime jewel]

how do you find angle B?

Law of Sines?

[compact spire]

Hmm

[wary bone]

@versed iron in fact you only need to draw a lot of (partial) circles. you know how many edges a hexagon has? try to get that many on the starting circle at equal distance

[versed iron]

@wary bone idrk even where to begin, quarantine school has been tough ngl

[little osprey]

@prime jewel Is that all the info you're given?

[prime jewel]

All the info is in the picture, i can put the problem too

[wary bone]

begin with: what s a regular hexagon?

[prime jewel]

I'm looking for AD

But i need Angle AEC

i think

ABC*

[wary bone]

ABD** 😉

[prime jewel]

I think so

I'm not sure

[remote heart]

@compact spire still need help?

[compact spire]

yes

[remote heart]

okay

[compact spire]

im making slow progress

had to redraw this thing twice now

ill send a pic

[remote heart]

okay

[compact spire]

okay so this is what I have so far

forgot to write sin65 on the bottom

$ x = 300sin(40)/sin(65) $

which after crunching the numbers I got 141.8 degrees

[remote heart]

?

how'd you get degrees?

[compact spire]

well because im using degrees

the 40 degrees and 65 degrees

[remote heart]

yes, but sin(40) and sin(65) are numbers

[compact spire]

wait I wrote 200 and not 300

[somber coyote]

Pony Boy:

[compact spire]

212.7

Oh okay

so its pounds not degrees

[remote heart]

yes

[compact spire]

so tension x = 212.7 lbs

and tension y = 319.7 lbs

$ y/sin(75) = 300/sin(65) = y =300sin(75)/sin(65) $

[somber coyote]

Pony Boy:

[compact spire]

no

x = 300sin(40) / sin(65)

so the 300 is only applied to the sin(40)

[remote heart]

but since it's being multiplied you can divide sin(65) by sin(40)

and multiply that by 300

[compact spire]

Okay

[remote heart]

but that equals ~270.35...

[mystic shadow]

Just a simple geometry question

in a sphere, what portion of it corresponds to .2 steradians

[upper karma]

@prime jewel are AB and DC parallel?

[little osprey]

looks like vectors tbh

which idk about

[upper karma]

there's this property that if AB and DC are parallel, the angles between them are also congruent

and all the angles from a triangle need to add up to 180 degrees

it's really easy to find angle ABC, imo

and the "magnitude" is just a fancy word for size of the edges

then, once you know two angles, you can use the law of sines to get the length (or "magnitude") of the other edges

also, yes, if we're talking about vectors, those edges are parallel

because we're talking about a cartesian coordinate system (x, y)

basically a grid

@prime jewel do you still need help with this?

[tacit quail]

I'm working on a robot arm that can draw things. Is there anyone that can help me with the inverse kinematics?

I have 2 arms of length L1 and L2. Given (x,y), I need to find q1 and q2 so the pen mounted on this arm reaches this point.

[upper karma]

law of sines

what's the length of L1 and V?

@tacit quail then, to get the angle you need the arcsine

https://www.khanacademy.org/math/trigonometry/trig-equations-and-identities/inverse-trig-functions/v/inverse-trig-functions-arcsin?modal=1

[upper karma]

is 4sinx • cosx - 2sinx = -2sin^2x?

[dark sparrow]

no

[upper karma]

oh, could u tell me then?

what it equals to ;_;

[dark sparrow]

$4\sin(x)\cos(x) - 2 \sin(x) = 4 \sin(x)\cos(x) - 2\sin(x)$

[somber coyote]

Ann:

[dark sparrow]

if you wanted to SIMPLIFY it rather than merely get an example of something it is equal to, then i can offer the simplification $$4\sin(x)\cos(x) - 2 \sin(x) = 2\sin(x)(2\cos(x) - 1).$$

[somber coyote]

Ann:

[upper karma]

ahh, sorry i should have specified. thank you

If I do tan^-1 (0) in a calculator i get zero! Why?

[dark sparrow]

tan^-1 is bad notation

tan^-1, better known as arctan, returns values between -90° and 90°.

[steep marsh]

So if anyone has the time can someone explain how am I supposed to calculate something like cos(A) = 5
I haven't properly learned this but got a test this morning and I can't figure it out

[dark sparrow]

cos(A) = 5?

[steep marsh]

Just a random one

[dark sparrow]

there are no real solutions to cos(A) = 5, just so you know

[steep marsh]

The more ya know

I just need the concept of how I would solve it explained, Can't find anywhere online that I understand it

[upper karma]

@steep marsh https://www.khanacademy.org/math/trigonometry

[wheat river]

Unless you have very specific values, cos(x) = something can rarely be solved in any nice way. For example, I don't think cos(x) = 1/7 has a solution which can be described in any "nice" way. That's why people will just write that the solution to that equation is cos^{-1}(1/7).

but yeah I i magine that khanacademy link will teach you something good about the unit circle; that's usually the best way to calculate sine/cosine-things

[gilded acorn]

I have no idea what I did wrong here

[acoustic jungle]

@gilded acorn negatives

[gilded acorn]

And how do I do this if there are no right angles?

[acoustic jungle]

There are.

[tawny linden]

can someone explain this my teacher refuses to teach us and i dont understand how proportions work in these triangles

[hollow tiger]

the way I do this is I look at what elements of the triangles are compared

like in the example number 2, a and c are hypothenuses in triangles afd and abc respectively

and f is the smaller cathet in the triangle afd

so the question mark should have the smaller cathet in the triangle abc, which is b

[acoustic jungle]

Why are there no right angle symbols

but yeah you would look at similar triangles

If you don't know what I mean, the answer for the first one would be a/c=f/b

a/c is also equal to d/a

[pliant matrix]

!clean

[upper karma]

hey guys

i need help with circles

3 questions i dont get cuz i didnt attend class cause my computer was bugging out

[lucid nacelle]

You can ask

[upper karma]

yea i was taking pics

[acoustic jungle]

First one ON = MO because a theorem - I apologize I don't know the name of that. Check the organic chem tutor on circle theorems, it should be in the first 10 minutes of the video

second one for a) CB=4 and you would use pythagorean

for b and c use trig

Third one from another theorem a perpendicular bisector always passes through the center of the chord

it's an if and only if

so if it's a bisector and passes through the center, it's 90 degrees

[upper karma]

thank you so much

i checked with my friends and your answers were right

thank you @acoustic jungle

[lime storm]

i’m having some confusion with this problem

[dark sparrow]

no wonder

it's impossible as drawn

line WL definitely does not pass through the center

hang on wait

i spoke too soon

this is possible

[lime storm]

I emailed my teacher and she has not gotten back to me so im stuck

[dark sparrow]

angle WLK can be found as angle MLK minus angle MLW

MLK is 90°, MLW is half of arc MW

by the inscribed angle theorem

[lime storm]

so is the Angle WLK 55°?

[acoustic jungle]

You are right.

[lime storm]

Thanks!

[willow tendon]

I need some help with probability

[silent plank]

the probability of your probability question getting answered in the geo-trig channel is very close to 0.

[willow tendon]

😂😂😂

So what channel would I go to

[silent plank]

one of the unoccupied greek letter channels

[willow tendon]

Ah ok thx

[upper karma]

the probability of your probability question getting answered in the geo-trig channel is very close to 0.
@silent plank savage

Lmao

[upper karma]

<@&286206848099549185> how do prove a shape is a square

i need help

[knotty merlin]

also

stop ping helpers

[upper karma]

oh ol

[knotty merlin]

you need to wait 15 mins

after askin the question

[upper karma]

ok

[knotty merlin]

to ping helpers

[summer spire]

what is the definition of a square?

[upper karma]

4 equal sides and 4 90 degree angles

[summer spire]

for problems like this, I usually start by gaining as much extra info as possible and hope the answer falls from those observations

are there other statements you can prove about this shape right off the bat with the information given?

[upper karma]

Opposite sides are equal and parallel

[summer spire]

how do you know?

[upper karma]

Because in parallelograms opposite sides are congruent/parallel

[summer spire]

what else can you add to the proof pile?

[upper karma]

um

does the diagonals get bisected by the midpoint?

[summer spire]

how would you prove that?

[upper karma]

idk

[summer spire]

also, is the "type of statement" a drop down box? Or do you have to type something in by hand?

[upper karma]

its a drop down

it gives you types to fill in

Would i need to prove the inner triangles to be equal in order to prove the sides to be equal via cpctc?

[summer spire]

worth a shot

[upper karma]

idk how imma get through these problems

there are 3

[chrome ice]

guys

what more proof do i need for this problem?

[summer spire]

@chrome ice can you move to one of the #❓how-to-get-help channels, please?

[chrome ice]

o sry

[summer spire]

@upper karma after thinking about it a bit, the most focused approach would be to prove the requirements of more strict quadrilaterals

so what requirements are different between a parallelogram and a square?

[upper karma]

you can say the opposite sides are parallel

idk what you mean tho

[summer spire]

that's part of the definition of a parallelogram, so that's a given

I mean all squares are rectangles, with the restriction that all sides are the same length

[upper karma]

oh ye

[summer spire]

and in turn all rectangles are parallelograms, with the restriction that all corners are right angles

[upper karma]

ok

so what should i try to prove then

[summer spire]

how do you go from parallelogram to rectangle?

[upper karma]

4 equal angles

[summer spire]

can you prove that?

[upper karma]

i proved that 2 of the angles are congruent

because one was given as right

and opposite angles are congruent in a parallelogram

so how would i do the other 2?

[summer spire]

CBD is a right angle?

[upper karma]

Nah

DAB was given as right

i proved that DCB is also right

idk where to go from there

[summer spire]

sets of parallel lines always preserve the angle between them

so if BA is parallel to CD, and BC is parallel to AD, the angles must be the same between all lines in each set

if what I'm writing makes enough sense

[upper karma]

idk how to put that into my software

[summer spire]

could also take an algebraic approach, if that's an option. All angles in a quadrilateral must add up to 360 degrees, and
you've already proved BAD = BCD = 90, and
from the definition of a parallelogram, opposite angles are equal, so
the two remaining angles must equally share the remaining 180 degrees

[upper karma]

ok

i got it

all 4 are now proven right angles

whats next now?

[summer spire]

now to go from rectangle to square

[golden panther]

YO! how do turn x=3tan(t) , y=2cot(t) into rectangular equation

[upper karma]

so i know the opposite sides are equal

but how do i prove adjacent sides are congruent

[summer spire]

I think I'll have to say that the key lies with point E

[upper karma]

wdym

How do I do the first one

oh

I know the vertical displacement is 4

and the amplitude is 11

Idk how to find the period when its really wack like this

or the phase shift

so find the values of the angles formed by diagonals and then say that sides opposite of congruent angles are congruent?

[summer spire]

I think that would work

[upper karma]

can someone help me with the trig problem?

anyone?

[high zephyr]

ping helpers only after 15 mins

and just ask the question

no need to ask if you can be helped or not

[summer spire]

@high zephyr he already posted the problem a few messages up

[high zephyr]

oh he did?

soz i didnt see that

[upper karma]

so

can i @ someone

[rich wolf]

bro just ask it

don't ask to ask

[upper karma]

https://cdn.discordapp.com/attachments/326138757474680852/700884461428015214/unknown.png

i literally

puit

the

questionm

in

here

[rich wolf]

sure thing buddy

[upper karma]

thank

[rich wolf]

what seems to be the problem here

[upper karma]

Alr

So I don't know to find the phase shift

and

period

I know the amp is 11

and the vertical displacement is 4

[rich wolf]

write what you know so far into the equation

[upper karma]

ok

11sin(x) + 4

and 11cos(x) + 4

[rich wolf]

do you know the formula

[upper karma]

for Sin?

[rich wolf]

for a general wave function

[upper karma]

Isnt it like

Amplitude*Sin(Period[Phase shift +/- Angle]) + Vertical Displacement

[rich wolf]

m + a(b(x-s))

yeah pretty much

[upper karma]

so right now its

4 + 11(b[x-theta])

@summer spire how would i get two of the consecutive angles formed by the diagonals to be equal

im still on the problem

[rich wolf]

so let's just write it as a sin function first

where does sine start?

[upper karma]

so the sine function should be starting at 4

sine0 = 0

add 4

thats where the period starts

but it appears that

sine0=0

[rich wolf]

4 is the y-value

[upper karma]

yea

[rich wolf]

you need the x value of where sine starts

[upper karma]

yea thats the thing

idk

where it starts

becuase

sine0 in that equation is 0

instead of 4

[rich wolf]

let me just tell you one thing

sine starts at the midline and goes up

[upper karma]

yea

exactly

[rich wolf]

at what point on the graph does the function start at the midline and go up?

[upper karma]

the sinusodial axis is 4

I dont know the x value

I can't find it

i just literally dont know it

[rich wolf]

just draw the points on the graph

where the sine function should start

[upper karma]

I'll just say 1.5

[rich wolf]

they actually did give you the x coordinate, it's a point on the graph lol

look at 10.5

[upper karma]

well thats when sin90=1

[rich wolf]

(10.5, 4)

[upper karma]

so sin360=10.5

[rich wolf]

there's your starting point

[upper karma]

so

its

11sin(x-10.5)+4

[rich wolf]

4 + 11sin(b(x-10.5))

[upper karma]

ya

[rich wolf]

you still need to find b

[upper karma]

mhm

[summer spire]

@upper karma the most concice way I can think of is to use properties of the altitude of an isosceles right triangle, but I'm struggling a bit to come up with a clear explanation

[upper karma]

ok

uh

and how exactly will i be able to find the period

Ik the period is 2pi/B

but there really isnt a critical point that will help me w that

[rich wolf]

look at the graph

how long does it take for the function to repeat itself?

@upper karma still having trouble with that?

[upper karma]

idk

yeah

been on it for like 2 hr

idk bruh

its a little bit before 10.5

for the fucntion to repeat

so i have no clue

[upper karma]

<@&286206848099549185>

aight cool

[high zephyr]

just be patient, someone wll def help

[upper karma]

there was that cool 15m ruke'

[storm mortar]

Can anyone help me with some Trigonometry homework? <@&286206848099549185>

[high zephyr]

first, just ask!

secondly, only ping helpers after 15mins from posting a question

[storm mortar]

Right

So this is a question that I'm stuck on

I'm not sure how to solve it at all and I'm struggling to learn math online

[high zephyr]

is this an online test by any chance?

or is this hw?

[storm mortar]

This is online hw through pearson since the school shut down due to COVID 19

Our tests are on paper through pdf

[high zephyr]

ok good, so not a test

[storm mortar]

Nah

[high zephyr]

well first, you can draw the triangles to help u out

i havent read the question, just skimmed through it

[storm mortar]

So for Sin(s) it would be 5 as Y and 13 for Hypotenuse

[upper karma]

bro

i stg

<@&286206848099549185>

please

help

?

https://cdn.discordapp.com/attachments/326138757474680852/700884461428015214/unknown.png

How

Don’t know

do i find the period for this

bruh

I’m only in geometry

Sorry my man

I stg

fucking corona

Alright

Good luck

ya thanks

[storm mortar]

Bro

If this dumb virus wasnt here I'd be thriving in class right now, its so hard to learn online

[eternal crag]

personally it has been a positive impact on my learning lmao

[upper karma]

@storm mortar

bro

i agree

[gray marten]

hey

so

whups

wrong chanel

[bitter wedge]

@storm mortar, better than doing 100% MathXL from the start.

Based on that problem, it looks like you guys are a tab bit behind.

The next unit is easier, so at least there's that.

Still need help with the problem?

I think I can help since I just took that test not too long ago.

[storm mortar]

This is the one im on

[bitter wedge]

Yes, I know.

The same on you posted above.

[storm mortar]

o

[bitter wedge]

Let me go clean myself up for a second.

[storm mortar]

Wait I think I got it

Nope I don't got it

[bitter wedge]

Alright, let me try to work it out.

Deleted that because it sounded rude.

||And in case I don't know it either||

[storm mortar]

lol it's all good

[bitter wedge]

Did you receive that formula sheet?

[storm mortar]

No we haven't

[bitter wedge]

Are you allowed to use it?

[storm mortar]

Yeah it's just hw

[bitter wedge]

Alright

Cause this is important:

So replace s and t with a and b (respectively)

So for the first problem, you'd use the third formula on this sheet to solve it.

So you'd figure out (cos(5/13)(cos(-3/5)-(sin(5/13)(sin(-3/5).

Tbh, it's really late, I'm really tired, and I don't feel like figuring what each of those values are, but when you do, you just do basic multiplication and subtraction (edited that).

[storm mortar]

Okay thank you

[bitter wedge]

Np

Not sure how much I helped tbh, but I hope it gave you a good start

I feel like this is a lot easier than I'm making it, but why is Angle C = 45?
~~https://cdn.discordapp.com/attachments/326138757474680852/700931536056352878/unknown.png~~

Bumping this post down.

<@&286206848099549185>

[viscid ginkgo]

I'm having trouble with if tan (A+B) = 7 and tabB = 1/43, find tan A

the formula is hard to use

tanA + 1/43 / 1 - tanA /43

[onyx cloud]

@bitter wedge diamonds have 90 degree angles

and the home plate is directly below the pitching spot

which means that it bisects the 90 degree angle

90/2 = 45

[bitter wedge]

ah

I feel like this is a lot easier than I'm making it...

Figured

Thanks for your help

[onyx cloud]

yup

@viscid ginkgo

idk if that helps

[viscid ginkgo]

I'm struggling so hard

I can't understand how to solve it

[bitter wedge]

Order of operations would mess me up when I tried to punch this bad boy into the calculator. Maybe try approaching it in parts.

[boreal adder]

This is my math warmup, I forgot how to find the answer

[upper karma]

Pythagorean theorem

[fleet wolf]

if a radius is perpendicular to a chord then it must bisect that chord

[upper karma]

@boreal adder
$x^2 + 16.2^2 = 21.1^2$

Pythagora

[somber coyote]

DSpider:

[upper karma]

revised formula, sorry 🙂

the largest longest side is the hypothenuse

[high zephyr]

longest*

[acoustic jungle]

( ͡° ͜ʖ ͡°)

[rich wolf]

,w sqrt(21.1^2-16.2^2)

[somber coyote]

Display results online and refine query

Requested by AMD#4434

[drifting parrot]

can someone help me with a problem?

I am

Give me a minute

This is a trigonometric function that has multiple angles

[high zephyr]

you can turn secant to its reciprocal trig function

[drifting parrot]

I know that. But i'm not sure where sqrt2) is on the unit circle

[high zephyr]

no, if you turn it to cos 3x, it's not sqrt2 anymore

so cos3x = 1/sqrt2

[drifting parrot]

why wouldn't it be cos3x = sqrt(2)/3?

[high zephyr]

bc the reciprocal of sqrt2 is 1/sqrt2

[drifting parrot]

ok

[high zephyr]

you reciprocated one side and u have to do the same for the other

the boundaries of x are not given in the question

[drifting parrot]

Like this?

My camera won’t focus

[high zephyr]

nono, cos3x

you reciprocated sec3x

[drifting parrot]

oooh ok

[high zephyr]

yup

btw, what're the boundaries for the angle x? it should be given

[drifting parrot]

It wasn’t given

Not sure why

[high zephyr]

alrighty

so what boundaries of x are u planning to do

[drifting parrot]

This problem only needs three solutions if that’s what you mean

[high zephyr]

oh i meant the range of x values

alrighty if the problem only needs 3 solutions im assuming the boundaries are 0<x<pi

@ me when ur back

[drifting parrot]

Sorry about that. My computer shut down @high zephyr

I’m not sure. My professor didn’t ask me to set a range

[high zephyr]

wait..u have a professor?

also no worries

we can assume the boundaries are 0<x<pi

[drifting parrot]

I do

[high zephyr]

so u have $\cos(3x)=\frac{\sqrt2}{2}$

[somber coyote]

nighty:

[drifting parrot]

yes

[high zephyr]

where x is 0<x<pi, let's make that assumption

[drifting parrot]

that would be the first and second quadrant?

[high zephyr]

uh huh

but now u have to deal with the 3x

so u can make 3x equals to t or smth like that

and solve for t within the range of x values assumed

[high zephyr]

im not sure what u did there

why does 3x stays the same when u divide it by 3

[drifting parrot]

oh wait

I'll fix that

[high zephyr]

take ur time

[high zephyr]

that doesnt seem correct

the pi/12 is correct tho

try working it out without using general solutions

it's not tedious

[drifting parrot]

general solutions?

[high zephyr]

as in the formula for reoccurring angles

try solving this $\cos t=\frac{\sqrt2}{2}$

uh

[somber coyote]

nighty:

[drifting parrot]

$pi/4 and 7pi/4$

[somber coyote]

CoraXSkull:

[high zephyr]

where 0<t<3pi

one more

the reason why it's 0<t<3pi is bc 0<x<pi when t=3x

[drifting parrot]

Is that a different way to solve the problem?

[high zephyr]

since the question is equivalent to cos3x = sqrt2/2

to make it easier, you can let t=3x

and then solve for cos t = sqrt2/2

later, u can sub back the 3x afterwards

[drifting parrot]

I don't understand

[high zephyr]

try solving this $\cos t=\frac{\sqrt2}{2}$

[somber coyote]

nighty:

[high zephyr]

try doing that first, for 0<t<3pi

[drifting parrot]

I thought the answer was pi/4

[high zephyr]

there are two more solutions

for t

you'll know what's going on after u worked through this

[drifting parrot]

pi/12, 7pi/12, 9pi/12, 17pi/12, 29pi/12, and 37pi/12

[high zephyr]

nono, for t

try solving for t first

but u got the answer correct anyways

only the first three

cuz we are assuming 0<x<pi

[drifting parrot]

I don't know how to solve for t

[high zephyr]

$\cos x=\frac{\sqrt2}{2}$

[somber coyote]

nighty:

[high zephyr]

are u saying u cant solve for this?

the t is just a replacement of x

[drifting parrot]

I don't understand why I don't get this

[high zephyr]

i think it's best for u to work it out step by step instead of jumping to a solution

$\cos t=\frac{\sqrt2}{2} for 0<t<3\pi$

bear with me for this

[somber coyote]

nighty:

[high zephyr]

try doing that first, after u go through each step, you'll get it

[drifting parrot]

I go around the unit circle three times?

[high zephyr]

nono, going around the unit circle three times is 6pi

[drifting parrot]

I actually solved the problem another way, that's where the solutions came from

once is 2pi, so would going around 2 1/2 times be 3pi?

[high zephyr]

yup

btw, u might have solved the problem, but u dont have the correct number of solutions

this is why the steps im going say next ensures that u have the correct no. of solutions within the given range of x values

[drifting parrot]

yeah, I realized after looking at the answer key the professor gave up

ok

[high zephyr]

try this then

$\cos t=\frac{\sqrt2}{2} for 0<t<3\pi$

[somber coyote]

nighty:

[drifting parrot]

ok

I'll try it

[high zephyr]

brb gotta eat brekkie

[drifting parrot]

alright

wait, breakfest?

[high zephyr]

aight im back

yes breakfast

cuz timezones

[drifting parrot]

oooh

I'm not sure if this is right, buut

[high zephyr]

nani

....you solve this just like a normal trig question

cos t = (sqrt2)/2

t = arccos ((sqrt2)/2)

@modest spear channel seems taken

[modest spear]

huh

[high zephyr]

try the questions channel

@drifting parrot

u here?

[drifting parrot]

yes

[high zephyr]

take ur time and read my msgs

[drifting parrot]

ok

[high zephyr]

i think you're overcomplicating it

literally just cos t = (sqrt2)/2
t = arccos ((sqrt2)/2)

[drifting parrot]

I always do that 😂

[high zephyr]

uhhh

what's arccos (sqrt2/2)

it's a common angle

[high zephyr]

is that...in radians?

[drifting parrot]

yes

[high zephyr]

convert it to degrees then

radians to degrees conversion: times radian by 180/pi

[drifting parrot]

I got 44.99......

[high zephyr]

cora

cos 45 is (sqrt2)/2

it's a common angle

pls dont forget the basics

[drifting parrot]

there are more angles with that cosine

I haven't gotten a problem that used the restriction 0 < t < 3pi

this is new for me

[high zephyr]

ohok

anyways, u seem to have trouble doing trig with common angles

try solving sin x =1/2

dont mind the restriction

[drifting parrot]

ok

[high zephyr]

$\sin x =\frac{1}{2}$

[somber coyote]

nighty:

[drifting parrot]

I have to use my phone to send the photos

[high zephyr]

yes correct, im hoping u can do these common angles in ur head?

[drifting parrot]

I have a unit circle with me

[high zephyr]

you seem to have trouble with $\cos x= \frac{\sqrt2}{2}$

[somber coyote]

nighty:

[high zephyr]

you know u can just type it out right?

[drifting parrot]

how do you type out the pi?

[high zephyr]

for the solution on paper just type:

x=pi/4 and 7pi/4

[drifting parrot]

$ x= pi/4 and 7pi/4 $

[somber coyote]

CoraXSkull:

[high zephyr]

x=pi/4 and 7pi/4
u can just type it out like what i just did

u dont have to use texit

[drifting parrot]

oooh lol

x = pi/4 and 7pi/4

[high zephyr]

ignore that

anyways yes

now that u've solved for that

[drifting parrot]

I feel stupid rn 😆

[high zephyr]

try $\cos t=\frac{\sqrt2}{2}$

[somber coyote]

nighty:

[drifting parrot]

t = pi/4

[high zephyr]

yes indeed

[drifting parrot]

oooh

last time I did this, I had pi/4 but also 7pi/4

soo the 7pi/4 was wrong

[high zephyr]

nonono

that is correct

both of these solutions satisfy 0<t<2pi

pi/4 and 7pi/4 is within 0<t<2pi

there's an infinite number of solutions here, that's why there are restrictions

[drifting parrot]

soo, I wasn't wrong last time? because i had those two solutions

[high zephyr]

last time as in what

cant remember

btw, for common angles like sin, cos, tan with 30, 45, 60 degrees
try not to rely on the unit circle

[drifting parrot]

nvm I couldn't find it

[high zephyr]

anyways

[high zephyr]

,rotate 90

[drifting parrot]

This is what I had before, but some of the solutions are wrong

[high zephyr]

almost all are wrong

the only correct one here is 3pi/4

[drifting parrot]

ok

[high zephyr]

and also, we're still assuming 0<x<pi

so you eliminate the ones outside the range of x values given

for ur working out, that's not how u use general form of trig for cosx

[drifting parrot]

ok

Soo, I start here

Where do I go from here?

[high zephyr]

yes indeed

you still have one more solution

for t

pi/4, 7pi/4 and....

[drifting parrot]

do you add pi/4 to get the next solution?

[high zephyr]

nono, consider the unit circle

you already got pi/4 and 7pi/4

which is within 2pi

you add 2pi to pi/4

bc like the previous revolution, it has the same base angle

[drifting parrot]

Hold on

I had to solve it

[high zephyr]

eliminate the one outside 3pi

nono, u already solved it

[acoustic jungle]

@high zephyr it should be pi/4 and 9pi/4

[high zephyr]

cos (7pi/4) = sqrt2/2

@acoustic jungle

[acoustic jungle]

oh you mean the 4th quadrant

ok.

[high zephyr]

the angle is between 0 and 3pi

[acoustic jungle]

I thought you were adding 2pi, didn't see.

[high zephyr]

okok

[acoustic jungle]

ok.

[drifting parrot]

soo, are my values right?

[high zephyr]

did u read my msgs

i have a feeling that u missed some

[drifting parrot]

yes

i wanted to cut in without being rude

[high zephyr]

eliminate the one outside 3pi

the values are right yes, but one is outside the range of x values

[drifting parrot]

oooh. the 15pi/4

[high zephyr]

uh huh

so u eliminate that

now u have t= pi/4, 7pi/4 and 9pi/4 correct?

[drifting parrot]

yes

[high zephyr]

and earlier on, we substitute 3x as t correct?

t=3x

if u scrolled up far enough

[drifting parrot]

yes

[high zephyr]

since u know the three solutions for t, and u know that t=3x

what are the three solutions for x?

[high zephyr]

i have no idea what u're doing at this pt

it doesnt seem like u're reading my msgs

[drifting parrot]

I am

I just don't seem to be getting it

[high zephyr]

ok what are the three solutions for t?

type it out instead of sending another photo

does this seem correct to you?

t= pi/4, 7pi/4 and 9pi/4

[drifting parrot]

ok

[high zephyr]

you get that those are the 3 sols right?

[drifting parrot]

t= pi/4, 7pi/4, and 9pi/4

yes

[high zephyr]

and earlier on, we subbed 3x as t

so t=3x

and u can find the 3 sols for x

which part do u not understand?

[drifting parrot]

My professor has me solving the problem in a different way. I think that's why I'm confused

[high zephyr]

which part do u not understand?

you understood how to find the solutions for t

[drifting parrot]

I do understand

I'm not sure what's wrong

[high zephyr]

there's nothing wrong

the 3 sols u got for t are correct

im not even sure what's troubling you atm

[drifting parrot]

I think it was because I couldn't use 7pi/4.

My professor wants both 7pi/4 and pi/4

[high zephyr]

I think it was because I couldn't use 7pi/4.
@drifting parrot wdym u couldnt use 7pi/4

[drifting parrot]

Based on the restriction, 7pi/4 wasn't in 3pi

[high zephyr]

it is

so does 9pi/4

7pi/4 and 9pi/4 are both within the range 0<t<3pi

[drifting parrot]

I was able to solve it!

[high zephyr]

that's why i was wondering why u were confused

[drifting parrot]

Yeah

I don’t even know

[high zephyr]

whut

[high zephyr]

,rotate 90

[drifting parrot]

I realized what I did wrong

[high zephyr]

,rotate 180

[drifting parrot]

I checked the answer key and finally got the answers right!

[high zephyr]

for ur answer, is that for t or x

seems wrong

[drifting parrot]

X

[high zephyr]

for x, it should be pi/12, 7pi/12, 3pi/4 for 0<x<pi

[high zephyr]

that's why i said earlier in response to ur working out, the way u used the general sols are is incorrect

,rotate 270

[drifting parrot]

Number 6

[high zephyr]

then that means they already gave u a restriction

the restriction seems like 0<x<2pi

then u already have a restriction in the very first place

[drifting parrot]

if that's the restriction for cosine

[high zephyr]

No

that's the restriction for the angle x

the restriction is the limited range of x values for the angle

[drifting parrot]

ooooh

[high zephyr]

the restriction here seems like 0<x<2pi

[drifting parrot]

I didn't read that right

[high zephyr]

now follow on from the solutions of t

since the restrictions for x is 0<x<2pi

and t=3x, so the restrictions for t is 0<t<6pi

keep going to solve for t in cos t = (sqrt2)/2 within the range

[drifting parrot]

ok

I think I got it from this point

thank you!

[high zephyr]

alrighty

glad u got it

[drifting parrot]

same

it was about time

[high zephyr]

phew this has been almost 3 hrs

[drifting parrot]

lol yeah

it's never taken me that long

[high zephyr]

but u understood everything i said right?

[drifting parrot]

not all of it, but i will figure it out.

Even if it takes me awhile

[high zephyr]

alrighty then

gl with ur maths

[drifting parrot]

thanks! you too

[acoustic jungle]

How do I do this with trig. https://www.youtube.com/watch?v=5vhklRWogzo

The question is on the thumbnail.

How would I do this with trig?

[rich wolf]

Bit of lin alg im guessing

Turn that into a system of eqs

[acoustic jungle]

Could you write the equations?

wait

I just have to define a variable as a number

that was what I was missing

thanks.

because I had some equation like sin20/n=sin80/h; sinx/n=sin80/l;h^2+n^2-2hncos20=l^2

so I just have to define n as something like 1 and solve for x.

[upper karma]

how would you simplify tan x sec x into a single number/function?

i only get sin x/ cos^2 x

[rich wolf]

@upper karma what is secant?

[upper karma]

1/ cos x?

[rich wolf]

Yeah so you cant really simplify that further

[upper karma]

ah

so i leave it as tan x * sec x? or what?

[rich wolf]

Yeah just tan * sec

Thats fine

[upper karma]

ok thank u

[dark sparrow]

imo $\tan(x) \sec(x)$ and $\frac{\sin(x)}{\cos^2(x)}$ are about equal in complexity but i personally prefer the latter

[somber coyote]

Ann:

[gritty sail]

help pls

how did they reach the conclusion of EBA being a 30-60-90 triangle

i get how they got the 90 degrees

but how did htey get the other 2 angles

and then after, for all those sides being equal to 5, is that because you first see AB = DE, then because the other angles are the same, DE = DC = BC??

[modest vine]

I think it is 504 but idk

[dark sparrow]

how did you get 504

maybe you were right

[modest vine]

I multiplied 12 x 8 x 6 and then I multiplied 6 x6 x 2 and then subtracted the results

[noble coral]

hi

I can't solve this question

[dark sparrow]

,calc 12 * 8 * 6 - 6 * 6 * 2

[somber coyote]

Result:

504

[dark sparrow]

okay yeah you're all good then @modest vine

[modest vine]

Thx

[dark sparrow]

@noble coral what's giving you trouble here

[noble coral]

I dont understand it

like I need to find the edge but I only have the volume

@dark sparrow

[dark sparrow]

denote the edge length with a letter (like a, for example)
make a diagram of your prism and find its volume in terms of a

set up and solve an equation

[noble coral]

ok

[dark sparrow]

and there you have it

[noble coral]

wait

uhm

what

[noble coral]

@dark sparrow

then I would have to do

a/2 over tan60

how would that work

[dark sparrow]

do you have your diagram

[noble coral]

on paper

ya

[modest igloo]

How do I get angle 7?

[noble coral]

@dark sparrow

I got 15.468

[dark sparrow]

uh

that sounds wrong

[noble coral]

ok

uh i dont get it

[dark sparrow]

i can't follow your work at all

where did (a/2)/tan(60) come from

[noble coral]

u told me to

make a diagram of your prism and find its volume in terms of a

[dark sparrow]

okay like

two convos in one channel is bad

[noble coral]

u need to find the Base Area first

[dark sparrow]

so @modest vine could you please move to a questions channel

[modest vine]

Ok

[dark sparrow]

@noble coral yes

[upper karma]

Do you need answer now Lon

[noble coral]

(a/2)/tan(60) come from

thats where it came from

[dark sparrow]

(a/2)/tan(60) is not the base area of your prism, i can tell you that much

[noble coral]

why

[dark sparrow]

why don't you tell me why you think it is the base area of your prism

i.e. why you think a/(2 tan(60)) is the area of an equilateral triangle with side a

[noble coral]

no i didnt say (a/2)/tan(60) was . the base area

[dark sparrow]

then what is (a/2)/tan(60)?

[noble coral]

part of it

[dark sparrow]

be more precise

[noble coral]

part of the equation

[dark sparrow]

why don't you write out the expression for the base area

in its entirety

and not just part of it

[noble coral]

what i wrote it on the paper

1/2 (a/2)/tan60 (9a)

[dark sparrow]

oh my god why are people posting over each other left and right

i'm out

[supple wedge]

lmao

[remote heart]

if a channel is occupied please don't post in it

[upper karma]

I was getting ad=1010 am I right

[supple wedge]

@upper karma use other channel, also draw a diagram

[noble coral]

Can someone smart help me

[upper karma]

Yes

Please ask

[noble coral]

what

i cant solve this

i did

[dark sparrow]

can you explain your reasoning for why $\frac{1}{2} \times \frac{a/2}{\tan(60)} \times 9a$ is the base area of your prism

[somber coyote]

Ann:

[noble coral]

Just look at the bottom parentheses

And then after that I multiplied it by 2

[dark sparrow]

what bottom parenthesis?

[noble coral]

What

Can u not see

The bottom parantheses

[dark sparrow]

i see a bunch of symbols strewn around in a messy way

[noble coral]

ok

thanks for the help

[modest igloo]

Is it always sometimes sometimes for 345

[remote heart]

that looks like your homework

[novel flax]

3 and 4 are both always

[modest igloo]

Are u sure?

@remote heart it’s a review packet

[little osprey]

@remote heart Isnt this server for math help in general excluding tests

[remote heart]

oh, I'm sorry

[modest igloo]

So it’s all always? Isn’t 3 never since it has to be 2 of the same angles?

[novel flax]

inscribed angle is 1/2 the measure of the arc it intercepts
tangent-chord angle is 1/2 the measure of the arc it intercepts
so if they intercept the same arc, they have the same measure @modest igloo

[modest igloo]

It says here 2 inscribed or tangent

[novel flax]

what textbook is that

i think i have the same one

[modest igloo]

Geometry for enjoyment and challenge

[novel flax]

ok i have the same one

[modest igloo]

So all of 1-4 are always and 5 is sometimes?

[novel flax]

i think the last one is never @modest igloo

[upper karma]

I understand that my teacher is writing out the asymptotes for the function but I can't clearly see what she wrote, does anyone mind explaining what she wrote

[rich wolf]

Bro just use wolfram alpha itll tell you the domain and range

[little osprey]

lmao

[remote heart]

I can't tell that last bit tho

but yea

crap

I thought that said more

[upper karma]

so she wrote Kpi

and that k is all real numbers

right?

[remote heart]

da

[upper karma]

da?

[remote heart]

sorry

yes

[upper karma]

ok

but

K cant be all real numbers because if K was a decimal number

wouldnt that mean there would be asymptotes everywhere?

[remote heart]

no wait

not all real numbers

[upper karma]

doez that double z mean only whole numbers?

i mean single z*

[remote heart]

big Z is integers

ohhh