geometry-and-trigonometry

[lone nova]

I think it has to do with a sine-cosine rule

This I think @dark sparrow https://cdn.discordapp.com/attachments/279435476757250049/556834281050210334/nratbig.png?

[dark sparrow]

i'm familiar with the sine and cosine rules dw

it's just that this problem you posted is a bit tricky since it seems like there's plenty of data but it's hard to get a grasp on

[lone nova]

alright

@dark sparrow someone atually came up with a solution but I do not understand it

2.sin(150-x).sin(65)=((2+2.cos(x))^.5).sin(85-(x/2))

[dark sparrow]

mmh

did they give you that and nothing else? or did they back it up with some work

[lone nova]

that is all

[dark sparrow]

wow ok

i mean

i have no idea how they arrived at that

[lone nova]

same

and ? is 40 apparently

oof

[upper karma]

either im hallucinating or you can do this with just the fact that sum of angles in triangle is 180

[upper karma]

right, hallucinating

lol

[lone nova]

ooooooof

@dark sparrow did you come up with anything?

[dark sparrow]

sorry, i got other things to worry about rn

[lone nova]

it is alright, thanks for trying anyway!

[deft umbra]

Did I do this problem right?

[inner lantern]

oh hi @lone nova

[lone nova]

hi

oof

[jade maple]

whats the formula for the size of internal angles in regular shapes

_ look at this digon

[astral hornet]

180(n-2)/n

[lament bay]

https://cdn.discordapp.com/attachments/493942046076174379/556904359724581053/image0.jpg

how do i do this?

[lament bay]

<@&286206848099549185>

[upper karma]

@lament bay
in triangle ABC, let's say the angle at A is α

now what must the angle at B be?

[lament bay]

i did A

can you help me on B

[upper karma]

look at your solution of a)

swap the letters A and B in it

makes sense?

[lament bay]

uh

not really

[upper karma]

suppose you have a right triangle, with a right angle at C, and from C a height dropping onto point D - and you're choosing which point to name A and which to name B

if you say that the left one is A, you get the situation on the drawing you sent

if you say that the right one is A, you STILL get basically the same situation

all the assumptions: right angle at C, at that D is the foot of the height, STILL WORK

so it makes sense that with your new assignment of letters, point a) would still work: so AD * AB = AC²

but if you look at the picture as if B is on the left and A is on the right, and find out what it means, it means the same as DB * AB = BC² in your old way of naming points

so, b) must be true!

[lament bay]

mhm

CDB ~ ACB?

how do i prove these triangles similar

[upper karma]

look at the angles

they both share the angle at B

they both also have another right angle

so the third angle of each has to be equal, and so they are similar

[lament bay]

ok

what proportion would i do?

[upper karma]

to prove b) or c)?

[lament bay]

b

after i prove triangle are similar

[upper karma]

take DB * AB = BC² which has to be proven

divide it by BC and by AB to get an equivalent equation

[lament bay]

uh

[upper karma]

if DB / BC = BC / AB is true, then b) must be true as well

[lament bay]

Like this? @upper karma

[upper karma]

i think so?
ree proof tables

[lament bay]

what triangles i prove similar for c?

[upper karma]

i think you'll agree that some triangle you choose has to have AD as one of its sides

[lament bay]

ok

[upper karma]

likewise some triangle has to have DB as one of its sides

the three triangles to choose from are ABC, ACD, and BCD

[lament bay]

CAD=bac?

[upper karma]

you won't get DB anywhere then

[lament bay]

idk then

[upper karma]

which triangle has AD as a side?

[lament bay]

cad

[upper karma]

and which has DB?

[lament bay]

cdb

[upper karma]

so you've got just one choice left after all

cad ~ cdb is the thing to prove

how to do it? angles

[lament bay]

uh

cda=90

[upper karma]

just name whichever non-right angle alpha, and from that calculate all the other angles

[lament bay]

uh

can you show me

[upper karma]

pick whichever you like from angles: dac, acd, bcd, cbd

[lament bay]

idk 😦

[upper karma]

it doesn't matter which one you choose

[lament bay]

dac and acd

[upper karma]

ok let's go with dac

now acd = 90 - dac

bcd = 90 - acd = 90 - (90 - dac) = dac

that should be enough to prove that the trianglec acd and bcd are similar

[lament bay]

uh

thats kinda confusing

@upper karma

[upper karma]

what is confusing

[lament bay]

what theorems do i use?

[upper karma]

??

[upper karma]

you understand that the sum of the 'small' angles in a right triangle is 90 degrees?

[lament bay]

uh huh

i need theorems for 2 column table tho

[upper karma]

and this statement is a theorem?

[lament bay]

yes

[upper karma]

so what's the problem

[lament bay]

prove (AD)(DB)=(CD)^2

[upper karma]

yes yes ok, but what's the problem with using "the sum of the 'small' angles in a right triangle is 90 degrees" as a theorem

[lament bay]

uh

i dont think thats a official name

[upper karma]

guess i can't help you with that :/

[lament bay]

hmm

what did you do for the 2nd step?

or are all the 2 steps the sum of the 'small' angles in a right triangle is 90 degrees

[upper karma]

bcd and acd give 90 in total because picture

bcd = 90 - acd then

you can substitute for acd after that

[lament bay]

<ACD=90-<DAC

what step is that

@upper karma

[upper karma]

that's because acd and dac must sum to 90 - look at triangle acd

[lament bay]

what about <BCD=90-<ACD=90-(90-<DAC)=<DAC

@upper karma

[upper karma]

well what about that :/

[lament bay]

what theorem

[upper karma]

i have no idea what theorem or what

[lament bay]

uh

why do you know <BCD=.....

[upper karma]

<BCD+<ACD=90 yes?

[lament bay]

yes i know that

[upper karma]

so, <BCD=90-<ACD

[lament bay]

ok

whats the proportions equation

[upper karma]

there aren't that many equations to try to see which one will give you c

[lament bay]

DB / BC = BC / AB

you gave me this for b

whats it for c

[upper karma]

maybe, just maybe, homework isn't 100% asking math discord on what to do

make the proportion yourself

[lament bay]

idk

[upper karma]

so, find out

try some

[lament bay]

;/

AD/CD=DC/DB

@upper karma like this?

[upper karma]

if it will give you c), yes

[lament bay]

i think it will give c

[normal chasm]

I apologize if this is too obvious, but why do the trig functions look like this?

[astral hornet]

think back to the unit circle

[wintry hatch]

Cosinus and Sinus oscillate because if you look go around the unit circle, you come back at the beginning

[lament bay]

https://cdn.discordapp.com/attachments/493942046076174379/556952779097374722/image0.jpg

[lament bay]

<@&286206848099549185>

[dry totem]

@lament bay, since angle C = 90 degrees, the two triangles created by CD are similar

[lament bay]

Hmm

[upper karma]

I’m not sure if we can assume that CD bisects angle C perfectly

[dry totem]

Hence,

If you can solve that, you'll have your answer.

Those ratios must be the same...

@upper karma . The segment CD is a perpendicular projection.

It creates two similar triangles.

[upper karma]

x =/= x + 7 though

If they were similar they would have the same ratio between side lengths

If CD is a length of 12 that implies that the side that both triangles share is the same, so the triangles are congruent, but that contradicts itself since they clearly aren’t

[dry totem]

The triangles may share one side, but one triangle's length is the other triangle's width...

You'd have to imagine the angles... In this case the angle cab=bcd and the angle cbd=acd

And these triangles share a 90°

[upper karma]

heyyo

so i'm having trouble understanding how to solve this problem
https://cdn.kastatic.org/ka-perseus-graphie/4fef060669261047d838b249057ad15f815e1c0d.png

maximum point is (-2pi, 3) and minimum point is (-pi/2, -4)

as an effect of horizontal streching the cos function of the form

$$h{(x)} = a * cos{(b*x + c)} + d$$

[somber coyote]

taven:

[upper karma]

i get this https://cdn.kastatic.org/ka-perseus-graphie/479e44470a676ac18eb6e3d7884cffd4d04de3ee.png

now the max point is at (0, 3.5)

however, the solution to convert the above graph into the previous one is saying

$$3.5 * \cos{(2/3 (x + 2\pi))} - 0.5$$

[somber coyote]

taven:

[upper karma]

why is it +2pi? shouldn't it be -2pi?

ignore the answer i typed in there

[zenith raptor]

From Kisilev's Planimetry

okay, i have a question about this construction

1. why point E is necessarily equidistant from A and B
2. why DF = FC

to be honest, the entire thing makes next to no sense to me, since there are straight lines connecting the endpoints of the given segments

[dark sparrow]

(1) E is the midpoint of AB by defn
(2) ditto for F and CD

[zenith raptor]

my understanding is that the point is to construct a quadrilateral such that E and F are midpoints of the sides not given

we are given segments congruent to two of its sides (so we are given segments AD and BC) and the line EF which connects the midpoints of AB and DC. right? so, when we translate AD and BC, why is it already the case that E and F are midpoints?

oh, should this be in one of #❓how-to-get-help channels?

[gloomy python]

How do you find scale factors?

[drowsy walrus]

divide

[gloomy python]

Divide what lol

[drowsy walrus]

can you send an example xd

it's kinda like

$\frac{\text{after}}{\text{before}}$

[somber coyote]

xItzDia:

[glad snow]

How to squish a square into a circle using matrices

Does such a thing exist or are matrices only for linear things

Maintaining radial ratios

Ie eigenvectors are the infinite set of vectors origination at origin

[dark sparrow]

matrices represent linear transformations only

[last lichen]

im so confused on how you go from the first line to the second

[dire rampart]

?

cos45 has a value

which is 1/root2

[last lichen]

so u just remember that?

[halcyon coyote]

memorise the unit circle tbh

[last lichen]

unit circle?

first day of trig xD

[halcyon coyote]

oh

but for angles like 30, 45, 60

theres this

which is useful

[last lichen]

oooo

[halcyon coyote]

basically for those angles

[sleek perch]

x D

[last lichen]

so i just memorise that and im gucci

[halcyon coyote]

you have 1/2, 1/2 sqrt(2) and 1/2 sqrt(3)

so its basically like

cos(45) is in the middle -> 1/sqrt(2)

idk this picture is not the best but it does illustrate the idea

[last lichen]

what about like tan and sin

[halcyon coyote]

well this is mostly for sin/cos

[inner lantern]

sin is next to cos

[halcyon coyote]

cos is the x value of a point

sin is the y

[inner lantern]

(cos,sin)

yes

[halcyon coyote]

and angle 0 is (1,0) then

and you go counterclockwise

[last lichen]

uhhh so confused haha

dw ill find some youtube tutorial explaining this

[inner lantern]

and tan is just sin/cos

[last lichen]

mmk

[sleek perch]

mmk

[inner lantern]

i know you @sleek perch

[sleek perch]

lol

[inner lantern]

pi song stucks in my head

[half saffron]

@last lichen sorry for the late reply but

If you're allowed to use the calculator in the exam

[last lichen]

we arent

thats the problem

[half saffron]

dammit

ah I see

[last lichen]

yeaaa

[half saffron]

cuz if you could you could just square the ans

[last lichen]

hence why im asking

[half saffron]

and then yeah :<

But yes this thing gotta memorize unfrotun atel

[last lichen]

reeeee

[lament bay]

Does any he know how to do this

[zenith raptor]

@lament bay there are three right triangles in the figure, and three unknowns

[lament bay]

Ok

[astral hornet]

Right Triangle with altitude drawn to hyp creates similar triangles

[lusty quest]

guys, pi is half of a circle so 2pi is a whole circle

the circumference formula is 2pi * r

what does radius have to do with this?

ok nvm i figured it out

[wary turtle]

It's 2r

Since pi equals to the circumference/diameter

That is the relation you may be looking for

[lusty quest]

yes, makes sense now

thanks

[wary turtle]

You're welcome

[gloomy python]

can someone help with this question

The distance from atlanta to savannah is 5 3/4 inches on the map. how long would it take to drive from atlanta to savannah if you drove at an average of 65 miles per hour?

[umbral snow]

Need a map → size function

[gloomy python]

ok

[upper karma]

How do I solve this

I've never learned the formula relating the sides of a quad before

[scenic trench]

It’s a ratio

I think you’d need to pair the similar sides and find the ratio.

Then extrapolate x using that ratio and the similar side

[upper karma]

Can you elaborate?

Do you mean I should set up a system of equations?

[dark sparrow]

ML/SR = MJ/SP

corresponding sides are in the same ratio

[upper karma]

ahhh

so much simpler than I thought

so the missing side equals 3.6?

[lusty quest]

you only need 2 sides of each, no?

[upper karma]

I think you can just say 5/x = 4.5

then solve for x

[lusty quest]

4/3 = x/27

[upper karma]

Is my answer correct @dark sparrow

[lusty quest]

2.7*

[upper karma]

you can edit you post by right clicking it

@lusty quest

[lusty quest]

yeah

[upper karma]

kk

[dark sparrow]

you only need 2 sides of each, no?

is that not exactly what i did lmao

anyway yes it's 3.6

[lusty quest]

sorry i didnt look

[last lichen]

How do you convert Tan to Sin?

[dark sparrow]

not possible to do uniquely unless you know the quadrant theta is in (though the two possible values of sin(θ) differ by sign only)

otherwise tho, draw a right triangle

[sharp moon]

halp

pls god somebody im in dire straights

[half saffron]

Use $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$
and $\sin(a+b)=\sin(a)\cos(a)+\sin(b)\cos(b)$

[somber coyote]

Itadakimasu!:

[half saffron]

@sharp moon

[last lichen]

Lol he left

How would I draw that @dark sparrow

[dark sparrow]

here's a right triangle

[last lichen]

Yea

[dark sparrow]

can you label two of its sides in a way that makes tan(θ) = 2?

(don't overthink it)

[last lichen]

2/1?

[dark sparrow]

so what's the hypotenuse now

[last lichen]

Sqr root 5?

[dark sparrow]

yes

so what's sin(θ) now?

[last lichen]

2 over sqr root 5

[dark sparrow]

and there you have it

[last lichen]

Oooooo

Tyvm lol

Brain ain't working haha

[elfin ibex]

∆ABC ~∆DAC (AA similarity postulate)
∆ABC~∆DBA (AA similarity postulate)
Therefore
∆DBA~∆ABC (AA postulate)

If you can't solve it from there, I can post another line of the solution

---

Let x equal DA
x\2 = 8\x
x^2 = 16

x=4

__
Let y = AB
y\2 = 10\y
y^2 = 20
y = √20

---

Let z = AC
8/4 = z/√20
z = 2√20

I'm not confident in my answers though.. I only covered it briefly on khan academy a week ago.

Might be wrong

The square roots can still be simplified

[lament bay]

Why are they similar? @elfin ibex

Which angles are congruent ?

[half saffron]

@lament bay okay I solved your question using a relatively simple method

You need to utilise Pythagorus (or however you spell his name) theorem whenever you can and form a set of equations, from there, perform substitution to find your desired lengths

[tender adder]

How do you "prove" this

excuse the quick drawing

[half saffron]

Use circular properties

Draw out a diagram and label the two angles

[supple abyss]

[solar shale]

How do I solve

@me

[blissful cliff]

Is it not just 360/14, or am i thinking about this wrong?

[solar shale]

Lol I’m not sure

I think your correct just checked

[blissful cliff]

Like for a triangle it would be 120 degrees, and for a square it would be 90 degrees. So the formula is 360/n, where n is the number of sides of the regular polygon

[solar shale]

👍🏼

[last lichen]

Why is it that those two values are used?

[scenic trench]

Because of this^

@last lichen

[last lichen]

Ah

[upper karma]

Help

With my 2-2

[last lichen]

?

[upper karma]

2 MINUS 2 READ

[last lichen]

No

I wont read

[outer otter]

is sin(x + y) the same as sin(x)sin(y)?

[dark sparrow]

no

[outer otter]

k

[dark sparrow]

if it were the case then sin(anything) would be zero lmao

$\sin(x) = \sin(x+0) \overset{?!}= \sin(x)\sin(0) = \sin(x) \cdot 0 = 0$

[somber coyote]

Ann:

[outer otter]

oh

[spark stag]

That would make trig a lot easier

In fairness

[upper karma]

sin(a+b) = sina.cosb + cosa.sinb

@outer otter

[plucky marlin]

sin(x) = x tbh

[dark sparrow]

[upper karma]

imagine if sin(x+y) = sin x + sin y

[upper karma]

what

imagine

i see who you are
you are that bangladeshi guy who was named Integration By Depression

[steady sleet]

[upper karma]

lol

[plucky solar]

I need to prove this and I'm not sure where to start, any ideas?

[dire rampart]

heres a good starting place

say what S is ?

[plucky solar]

oh sorry

it is the area of a triangle

not a specific type of triangle

well actually, I managed to do it

it always happens when I ask something on here

if i ask something i immediately get the answer myself lol

[lime flame]

Can someone explain to me why the answer is A

I didn't know how to do it so i guessed A

and i got it right lol

[lime flame]

<@&286206848099549185>

[upper karma]

50sin(23)

@lime flame

[lime flame]

[upper karma]

Its derived from this

mg is the magnitude (with a direction) of Fg

Since it's on a slope part of it acts down the slope due to gravity

[lime flame]

Is this physics?

[upper karma]

yeah

[lime flame]

Oh this was on my geometry test so

How can I solve it without basic background knowledge of physics?

[upper karma]

You gotta know the angles are the same

[raw prism]

<@&286206848099549185> Could someone help me understand how to do this

[solar shale]

Do I use a^2 + b^2 = hyp^2

[raw prism]

for BC?

[solar shale]

It just says find missing measure

🤷‍♂️

So is that asking for BC?

[raw prism]

use that and then find the angles using basic trig

[solar shale]

Can I @ you when I’m done

To check

@raw prism

[raw prism]

sure

just show me the working i'll tell u if its right

[solar shale]

Idk what to do

[raw prism]

subtrat 100 from eacj side

[solar shale]

Ik that

But after

[raw prism]

then take sqaure root of each side

[solar shale]

Wdym

[raw prism]

allowed to use calculator?

[solar shale]

Yeah

[raw prism]

take the square root of 125

[solar shale]

Ok it’s 11.18

So that’s the answer?

[raw prism]

thats the lenght of the missing side

[solar shale]

So I’m done correct?

[raw prism]

depends do you still have to find the angles or no?

[solar shale]

Just says find measure

[raw prism]

have u learned trig?

[solar shale]

Yes

[raw prism]

ok then I would assume u gotta fine the angles

[solar shale]

How would I start

I know some parts

[raw prism]

you use the tig rule for right tringles

u know sOH cAH tOA

?

[solar shale]

Yup

[raw prism]

yea so use that on any angle and it should give you the correct angle

[solar shale]

Ok also

For this question I would use the same thing a^2 + b^2 =c^2 correct? And after use the sohcahtoa

@raw prism

[raw prism]

yea

[solar shale]

Ok tnx

👍🏼

[raw prism]

but you dont need the lenght

you just need angles

wait u do nvm

[solar shale]

Ok

[raw prism]

since u need all sin cos and tan

[solar shale]

Yeah

[raw prism]

remember to take the inverse of whatever value u get to find angles

[solar shale]

Aight

[raw prism]

Now <@&286206848099549185> can someone help me with this

[upper karma]

,rotate 270

[raw prism]

woah thats convenient

[upper karma]

Triangle with sides 10,10,12

[raw prism]

yep

[upper karma]

S,s,s triangle

Law of sins or cosines

[raw prism]

I tried using cosine rule but anwser didn't come out right

[upper karma]

What's the law of cosines?

[raw prism]

b^2= a^2+c^2 - 2ac * cos B

yea that

i got 53.1 degrees any chance u could do it and check?

[upper karma]

,w Arccos((144-200)/(-20))

[raw prism]

wait]

shit

[somber coyote]

Display results online and refine query

Requested by Babowwz#1394

[raw prism]

i got it right

[upper karma]

That's radians

[raw prism]

just need to times 2

[upper karma]

[raw prism]

cause i found half on theta

shit im retarded

thank u

[upper karma]

,w 1.689 radiant to degrees

[somber coyote]

Display results online and refine query

Requested by Babowwz#1394

[upper karma]

Oof

@raw prism

You know yours is right?

[raw prism]

yea

checked anwers

its 106

i got 53 cause i found half

how about part b though dont know how to do that one

[upper karma]

[I,j]

[12cos(106),12sin(106)]

@raw prism

[raw prism]

oh i just resolve

ok thanks

[upper karma]

Got answers to check?

[raw prism]

I DO LEMME GET THE ANWSER crap caps

[upper karma]

[raw prism]

anwser is 7.2 N

[upper karma]

,Calc 12cos(53)

[somber coyote]

Result:

-11.019393434545

[upper karma]

,Calc 12cos(53degrees)

[somber coyote]

Result:

7.2217802778246

[upper karma]

@raw prism

Its half

[raw prism]

ye thought sp

so*

[upper karma]

I need to review resultant angles cuz I suspect is half, but mathematically I suspect its not always

[raw prism]

well anyway thanks for guiding me

really appreciate it but since i pretty much suck at math i will prob be back here

[dark sparrow]

are you asking to check whether your answer is correct?

bc it is

[plucky marlin]

WTF

[upper karma]

humongous cheese

[plucky marlin]

did he get banned too ?

[steady sleet]

dwai

[olive solar]

yeah they did

[steady sleet]

[lime flame]

You are given a equilateral triangle inscribed in a circle with the center shown. Find the diameter

<@&286206848099549185> Sorry for ping but I need a solution asap

[upper karma]

lol

find the circumradius of that triangle

and that's quite easy with the law of sines

@lime flame why do you need solution ASAP?

also don't expect people to give you solution

I need 50$ ASAP

If its graded then you're asking us to help you cheat fren

@lime flame

If you put off Hw and about to go to class rip u too

Either way you roasted

[lime flame]

Lol :(

[upper karma]

https://gyazo.com/6befad42b38b61646ec20551fee726c8

how would I go about solving thisa

[dark sparrow]

consider making a picture

[civic monolith]

did i do this right?

[tropic shard]

Nope, when you plugged in 36, you forgot to add 48 to it to give you the side of ΔDFE. (DPQE is a trapezoid).

[cloud meteor]

If I have a triangle with side lenghts 2Sqrt(2) 1 and 3

(right triangle)

and theta is opposed to the 1 and the hypotenuse is 3

2 theta*

[tropic shard]

@cloud meteor What about it?

[cloud meteor]

Nevermind

[past siren]

Does anyone know the relationship for this?

[tropic shard]

Vertical angles*

[dark sparrow]

these are vertical angles

[lofty scaffold]

Arent they vertically opposite angles!

?

And they're equal btw.

[dark sparrow]

no they're just called vertical iirc

[somber coyote]

Benjamaster7:

[mint token]

I have a question. I was able to prove it by labelling ABCD in a valid way, but the official solution labelled it in a different valid order and thus had a different solution

would it be necessary to prove both cases? the cases are almost the same but have different point names

[dark sparrow]

what have you tried so far and where are you stuck?

do you have any work to show

it would greatly help me help you

as you say

...

[dire rampart]

casual moment when ur talking with yourself

[upper karma]

hahahahahahahah

[olive solar]

@dark sparrow you okay?

[dark sparrow]

someone asked for help and then deleted their own messages

[olive solar]

o lol

[cold tangle]

If <A = 30° and BC = 6, how can I determine AB?

Has to be something w/ the trigonometric functions right?

[half saffron]

What do you know about sin cos and tan?

[cold tangle]

I got it but thanks anyway

The answer is 12

[vague pagoda]

trig is cool

[cold tangle]

And how about this one?
AC = BC
AB = 4
Calculate the value of the missing sides

[lusty quest]

still use trig

looks like a 90-45-45 triangle

[plucky marlin]

thats not a 90-45-45 triagnle

[lusty quest]

looks like one

[plucky marlin]

bRuH

"looks" like one

[lusty quest]

howd you figure out the angle?

[plucky marlin]

@cold tangle is it a right triangle ?

if it is then yeah it's a 90-45-45

[cold tangle]

Yea it is

[lusty quest]

since its not that, then how would you fing the other angles?

[plucky marlin]

well if its a right triangle then just use pythagorean theorem

[lusty quest]

but other 2 sides are unknown

[plucky marlin]

so ?

$x^2 + x^2 = 4^2$

[somber coyote]

soap:

[plucky marlin]

ig u can solve it ?

[lusty quest]

so x^2 +x^2 = 4^2

[plucky marlin]

yeah

[lusty quest]

im on phone, typing slow

[plucky marlin]

dwai

[lusty quest]

yes but you still gotta know 2 numbers to solve for the third

[plucky marlin]

??

theres just one varaible

[vague pagoda]

??

[lusty quest]

yes

[plucky marlin]

brUH

r u high ?

[lusty quest]

so is x^2 +x^2 = 16 the final answer then?

[plucky marlin]

$2x^2 = 4$

[somber coyote]

soap:

[plucky marlin]

[lusty quest]

oh shit

nice i didmt think of that

[plucky marlin]

lmao

[vague pagoda]

oof now I get it

[plucky marlin]

anyways

solve for x

take the principal square root

cuz x > 0

[vague pagoda]

so what is the answer

[lusty quest]

sqrt 2

[vague pagoda]

o

[lusty quest]

wait no, he didnt square the 4

[vague pagoda]

so 2?

[lusty quest]

is 2x^2 = 16

yes just 2

[plucky marlin]

forgot the square

my bad

[lusty quest]

bRuH r u high

[plucky marlin]

im HigH oN MetH

math*

[lusty quest]

mAtH

[plucky marlin]

[paper frost]

What is the orientation of a plane in space when two of the three variables x,y, and z are missing from its scalar equation

[plucky marlin]

like z = k ?

[paper frost]

that's what i was thinking but im not really sure

[plucky marlin]

thats gonna be a plane parallel to both x and y axis and intersects the z axis at k

[paper frost]

ohhh i get it, thanks!

[plucky marlin]

nice

[covert orchid]

how is this correspondence of angles called ? i was doing a trigonometry exercise and i just saw that i needed to know this to solve it

i want to know more about it but i dont know its name haha

[raw prism]

How I do this?

lel yea hold up

,rotate 90

[raw prism]

there

[raw prism]

<@&286206848099549185> i'm stuck on the problem above could someone guide me

[umbral snow]

@raw prism
Q2 There?

[raw prism]

yes

[umbral snow]

You want to break every force into its components

Let me draw a pic

[raw prism]

ok

[umbral snow]

A little messy, but that's splitting everything up

Let's start with the blue. See how I got those angles?

The 60°, 60°, 30° there

[raw prism]

yea I understaand that

[umbral snow]

Perfect, let's move onto the red. Do those components make sense?

[raw prism]

I was taught to draw them this way sticking out from each line is that fine too?

[umbral snow]

Yeah! Whichever is fine for you. I like my method as it shows the right triangles, and makes the trig clear. Either way, know why the red values work

[raw prism]

ok

[umbral snow]

Anyway, we want to add all of the components in the x-direction, to find the net force in the x-direction.

Likewise, we want to add all of the components in the y-direction, to find the net force in the y-direction

Net force in x = 10cos(60) - 10cos(60) + 6cos(30)

[raw prism]

ok i got that

[umbral snow]

Fx = 5.196

Net force in y = 10sin(60) + 10sin(60) - 6sin(30)

[raw prism]

14.32

[umbral snow]

Yus.

Therefore, the net force on the origin is (5.196, 14.32)

Now, what force should we place on the origin to cancel that out?

[raw prism]

the same ones in the opposte direction?

[umbral snow]

The net force, in the opposite direction. Throw (-5.196, -14.32) on there, and the forces will balance out

[raw prism]

i can just throw it in?

[umbral snow]

That will be the resultant, as it is the force that keeps anything from moving

[raw prism]

so the y resultant should be my answer for "in direction of PQ"

[umbral snow]

Yeah, I think so

Since PQ is the y direction in my construction

[raw prism]

and for maginitude I just take square of both x and y results, add them and then take square root right?

[umbral snow]

Yus yus

[raw prism]

thanks alot

[umbral snow]

It will be longer than 14.32

[raw prism]

ye got 15.23

@umbral snow one last question for now. how would I know which angle to use the 30 degrees between the line and the y axis or the 60 degrees between the line and the x axis

[tropic shard]

If it ain't Supernatural Season 16, Idk why anyone would be that excited.

[dire rampart]

@tired current plz dont multi post

[unique patio]

hey guys

how do i find the angles

alpha

[unique patio]

<@&286206848099549185>

[umbral snow]

@unique patio
Not enough info there. I'm thinking α is set for you in the question

[unique patio]

its just that

[umbral snow]

That's different. Now I know the body is static

So, the sum of the forces on the seat is 0

Sum of X:
40 = P1sinα + P2sinα

Sum of Y:
50 + 75 = P1cosα + P2cosα

It's easiest to take the moment about the top left.
M = 0 = 0.4P2cos(α)

So there, that's three equations with three unknowns:
sinα(P1 + P2) = 40
cosα(P1 + P2) = 125
0.4P2cosα = 0

@unique patio

[unique patio]

hold up

how is taking the moment about top left

only 0.4P2cos

doesnt the 40n

[umbral snow]

I forgot the other forces

Derp

[unique patio]

and 125

oh lol

i see now though

thanks

[umbral snow]

Okay, if that's enough! Feel free to ask if you have anything else

[rain charm]

I’m given cos(x) = 0.788 how do I find x?

Also in another question I’m given sin(x) = 0.6 how do I find cos(x)

[umbral snow]

cos(x) = 0.788
x = cos¯¹(0.788)

You can move the cos to the other side, but it's cos¯¹ if you do

[rain charm]

Of course thx

I can do my other question now

Very helpful

[umbral snow]

The other one is a bit harder. If you think you got it, good luck!

[oblique tiger]

does anything weird/special/interesting happen when you do arctan2(x,y) instead of arctan2(y,x)?

somehow my functions are working with the former syntax

i know that's technically not correct though

[cloud meteor]

That's weird

[torpid perch]

Hello. A Question to curve Discussion: Is the average of the two extreme points in a x³ function always the turning point?

[umbral snow]

Yes, because the average of the roots of a quadratic is always its vertex

@torpid perch

[half saffron]

But it’s

$x^3$

[somber coyote]

Itadakimasu!:

[low tusk]

I think you’re asking for the average of two roots of the cubic function instead of the two extreme points

Then the answer is clearly no

[solar shale]

I don’t understand where the 16x comes from

[highlighted]

The answer is correct I just don’t understand where it came from

[low tusk]

is that $16 \times 192 \tan 22.5 ^\circ$ ?

[somber coyote]

Dragon:

[solar shale]

No it’s 16x=

Ik that 16x has something to with the perimeter

Can’t figure out how

[low tusk]

You divided the polygon into 16 equal parts then computed the area of one part, so in order to calculate the total area, you need to multiply 16 to it

[solar shale]

Ah I see now tnx so much man 👍🏼

Also

If your not given an apothem and base what would I do

Isn’t it like base times number of sides

Actually nvm i figures it out

[low tusk]

That’s great

[cold tangle]

<B = 60°
AC = 9√3

Find AB

[tropic shard]

What have you tried so far?

[cold tangle]

I got it

But thx anyway

[solar shale]

The answer to this is correct but I was wondering why the polygon was divided into 16 different parts

Feel free to @me

[supple abyss]

because this

[tropic shard]

I'm assuming to make it look cleaner.

Otherwise, it doesn't make sense.

[solar shale]

Oh ok so “make it simpler” basically

[tropic shard]

Yeahh

[solar shale]

Is the 2 in square root because of the square being split in two?

[tropic shard]

Yeah, and Pythagorean Theorem.

@solar shale

[solar shale]

Wait is why is there a square root of 3 in this problem

[solar shale]

?

<@&286206848099549185>

[tropic shard]

There's a formula for the 30°, 60°, 90° triangles but you could still use the Pythagorean theorem to find 4√[3]

[solar shale]

Oh yeah that formula but like how would I know it’s that special right triangle

?

[tropic shard]

Exactly.

Lemme write that as normal text so it doesn't look like spam, lol.

[solar shale]

Because Ik there is a 45-45-90 one to

Ok

[tropic shard]

8²+4²+A²
64=16+A²
48=A²
A=√[48]=√[4²·3]
A=4√[3]

[solar shale]

I get the reasoning but I don’t understand how would I distinguish special rights triangles

[tropic shard]

Just gotta memorize that 45°-45°-90°, and 30°-60°-90° are special right* triangles. When memorizing the unit circle, that's where you'd derive those formulas.

[solar shale]

Ok but like how am I supposed to know if it’s 45-45 or 30-60 if I’m only given a 90 degree angle

[tropic shard]

Well I'm assuming it's for the normal polygons, in which case you could use the formula:

,, \frac{(n-2)*180}{n}

[somber coyote]

𝓗𝑒𝓃𝓇𝓎 𝓒𝒶𝓈𝓉𝓁𝑒:

[tropic shard]

To find the angle of each corner.

[solar shale]

👍🏼 Tnx

[jade oracle]

hi guys, I just joined this server and have some issues with my hw

I think i'm supposed to put it in uh

I dont know the name but it's in the form of x=(2pi/3)+2(pi)k, k is equal to all integers

And that's where my issue arises, I have no clue how to find the rest of the angles

oh shit didn't read the rules

[forest niche]

Can anyone help

[dull ice]

Maybr

Triangles UVT and RSV are exactly the same?

Obviously cuz same center

RT = RS?

[forest niche]

I’m lost what?

[dull ice]

So you have 2 triangles correct?

[forest niche]

Yea

[dull ice]

Is there any more information?

[forest niche]

No just that

[dull ice]

Ok

Well do you know the answer?

Im guessing thats what u need help with

[forest niche]

17

[dull ice]

17?

[forest niche]

Idk how to get it

[slim lynx]

Looking for a physicist

[dull ice]

Degrees?

[forest niche]

Oops

35

Sorry

[dull ice]

35

Well thats the only logical response

[forest niche]

How would I know it’s 35 though

:/

[dull ice]

You cant solve for any other angle

Its impossible to mathematically find it

[forest niche]

Alright then

[dull ice]

If they gave you another angle

Then you would be able to mathematically

[forest niche]

OH i GET IT

😂😂

I think it’s an inscribed angle

[dull ice]

But since you only have 1 angle, it must be the same

Yea prob

[forest niche]

I see it

Thanks lol

[dull ice]

Np

[upper karma]

should it be 35?

oops

[acoustic peak]

The line y=(x/4)-9 is moved two units to the left and five units upwards in the coordination system. Calculate the moved line's equation
I'm really struggling with the Analytical Geometry that we were given today

[upper karma]

take two points of the line

then calculate where they will move

actually now that i think about it

you only need one point

since the slope of the line won't change

[acoustic peak]

Idk why I'm finding this so difficult

[upper karma]

well some point (x,y) becomes (x-2, y+5) right

[acoustic peak]

Yeah

[upper karma]

so take any point that lies on the line

[acoustic peak]

Should I rewrite the equation as y-5 = (x+2)/4 - 9?

[upper karma]

(y+5) = (x-2)/5 - 9 you mean

[acoustic peak]

I don't know why, our books show us that instead of writing +5 if it goes upwards, we're supposed to write -5 to get it back to its original position

y = (x+2)/4 -9 - 5

That works too, right?

[upper karma]

yes

[acoustic peak]

Here's where I'm a little confused

X/4 has a denominator of 4

And everything else has a denominator of 1

Should I multiply everything by 4?

[upper karma]

you could do that

[acoustic peak]

But then I'd get 4y

[upper karma]

then divide by 4 after simplifying the right side

[acoustic peak]

Hm

I'll try

https://gyazo.com/224a19ef25726eb849a4e89df5df882f

Yeah that's not looking right

[solemn harbor]

54/4=13.5

but bit complex way to type that upper equation, isn't (x/4)-9 enough?

[acoustic peak]

Idk, that's just how I'm used to writing it.

[solemn harbor]

I mean, that many parenthesis are not necessary

[acoustic peak]

[fossil fern]

can someone explain how to factor this
cos^4 x - sin^4 x

basic

[upper karma]

difference of squares

[solemn harbor]

cos 2x

[fossil fern]

ya I gotta verify that its cos2c

[solemn harbor]

whoops shouldnt probably give answer

[fossil fern]

2x

no I have answer

verifying these fuckers is annoying I hate it

[upper karma]

funeris, try plotting
y = x/4 - 1/4
kek

[solemn harbor]

Remember that (a-b)*(a+b) = a^2 - b^2

See where I am going?

if that is true, then also, (a^2-b^2)*(a^2+b^2) = a^4-b^4

now think what a and b is in your case...

[fossil fern]

now what

[solemn harbor]

Don't you remember? sin^2 x + cos^2 x = 1

😃

So that term is eliminated.... you are left with cos^2 x - sin^2 x

[fossil fern]

ugh thanks

now wat

[solemn harbor]

Hmm sin 4x is not 4sin x cos x

2 sin 2x cos 2x

not sure about the first line...

if the line sin 4x - sin 2x is correct, next line could be 2 sin 2x cos 2x - sin 2x, which you could simplify to sin 2x(2 cos 2x - 1) if i didnt make an error 😃

or to 2 sin 2x(cos 2x - 0.5)

[fossil fern]

yup so I do I get the right side from that bulllllshit

second last line is not =sin2x its -sin2x

[solemn harbor]

Like I said I'm not sure about the first line, I didn't check that

and not sure if you need to prove in the first assignment that cos^2x - sin^2 x is cos 2x, or is that known. it's not hard to prove though, but anyway

[fossil fern]

yes first problem is done.
this one is same idea. I have to prove that the left side of first line equals the right side of the 1st line

[solemn harbor]

But I think that is correct

[fossil fern]

ya the cos2x first problem was known

I dont get where this ^3 comes from on the right side

[solemn harbor]

Wait, hmm

Okay I think that is correct

It's just way to substitute trigonometric functions with another

The ^3 thing I mean. Sometimes you want equations with ^3 or something, sometimes with sin 2x, sin 3x, but no exponent... depends on the situation

[fossil fern]

I just get stuck like I am now. tf do I do after this last line to expand it into the right side so I can say it checks out

[empty trout]

any help?

[supple abyss]

I swear you posted this before

[empty trout]

I just join this server lmao

just pinged me when asnwered pls

[supple abyss]

[empty trout]

thx

[upper karma]

The length is one inch more than the width, which is one inch more than the height. The volume is 160.875 cubic inches.

i need to find the length, width and height

so would that mean
L+1=W
L+2=H

[supple abyss]

no

[upper karma]

oh

[supple abyss]

H+1=W

H+2=L

[upper karma]

ok, so we're
160.875 = L x H x W
plug that stuff in
160.875 = L(L+2)(L+1)
is that the right way?

[supple abyss]

ye

[upper karma]

ok, heres the part that confuses me, that ends up to be
160.875 = L^3 + 3L^2 + L
can i use the quadratic formal on that or does it have to be done as
160.875 = L(L^2+ 3L+ 1)

[dire rampart]

@empty trout did u figure it out

[upper karma]

@upper karma you missed something

should be L^3+3L^2+2L

X=4.5

or L in this case

[upper karma]

I need help with 46 49 57 and 58

<@&286206848099549185>

[olive solar]

wait 15 minutes before pinging helpers

[upper karma]

Oh okay

[olive solar]

do you know half angle formula

[upper karma]

Not well

I have the formula for reference tho

[gloomy python]

need help finding x, y and z

[dull ice]

@gloomy python

Still need help?

[gloomy python]

yes pls

[dull ice]

Ok

[gloomy python]

is it possible for u to pm me?

[dull ice]

Yep

[gloomy python]

thx

[oblique bough]

Hey, I need help with law of sines >.>

just like @ me back if u can help

[dull ice]

@oblique bough

[oblique bough]

Oof thanks

[dull ice]

Lol

[oblique bough]

So I have this

But it doesn’t look right

[dull ice]

Ambiguous cases toom

Too?

[oblique bough]

?

[dull ice]

ASS

[oblique bough]

nOni

[dull ice]

K

U didnt arcsin it

Here is a hint

If u ever get domain error or error of some sort

Then it doesnt exist

In this case

If u ever get sin of an angle equals a value more than one, then there are no triangles, but check your math just in case

[oblique bough]

omg im dumb lmao, i should use arcsin

[dull ice]

Yes

U forgot to have sin still there

Sin C equals 1.4

But of you know the unit circle, Sin of any angle will never be greater than 1

[oblique bough]

oof, ok thanks

[dull ice]

Any more help?

[oblique bough]

lemme try this out first

[dull ice]

K

[oblique bough]

ok, so i tried it and i still got 1.4

[dull ice]

Tell me what you did to get 1.4

[oblique bough]

62/44(sin88)

[dull ice]

Ok

So what does that equal

[oblique bough]

1.4

[dull ice]

But what is 1.4 referring to

What is the whole equation

[oblique bough]

sinC=1.4

[dull ice]

Ok

SinC

Means C=arcsin1.4

[oblique bough]

yea

but it game me an error

[dull ice]

Exactly

Because sin can never equal greater than 1

Than 1

[oblique bough]

then i cant solve the triangle?

[dull ice]

It means