geometry-and-trigonometry

[marble topaz]

So you have learned that much

[upper karma]

Yea haha

[marble topaz]

yeeaaah haha

[upper karma]

Yessirr

[marble topaz]

woww this would be so freakin easy with some quick trig application

[upper karma]

Yea I bet

But idk what to dooo

[marble topaz]

I'm wondering if those larger triangles are in fact similar to the smaller ones at the "top"

If that was the case, all you'd have to do was find out whatever factor the larger triangle was being multiplied by

and then you could do that to find the crossbar

Do you see what I mean?

[upper karma]

Yea, I gotchu

But idk if they are

All I see is one right angle in both of them

Idk how to determine the rest of the angles to see if they’re congruent

[marble topaz]

Let me check the other pages from your textbook

You might want to look at your notes as well. I'm sure you guys have covered diamonds

[upper karma]

Nope

Lmaoo

In science we did tho

Not math lol

With the moths hardness scale or smth

Mohs?

Idk

[marble topaz]

That theorem 7.5 thing in your book looks like it applies

I can't read the rest though

I'm sorry

[upper karma]

Do not be the sorry

We will do this!!

[marble topaz]

Oh fuck

okay

[upper karma]

Stay positivee

Lmaoo

Let’s see

@ everyonee

Work on my problem!!

[marble topaz]

I found those mean and leg theorems you were referring to

[upper karma]

Good job

[marble topaz]

https://www.slideserve.com/keegan/geometric-mean-theorem-i

Just some real simple algebra to solve for whatever you want

Okay yeah the first and second slide are exactly what you need

Now that the application is a bit more clear, do you see what you need to do?

[upper karma]

Mhm, kinda

[marble topaz]

Okay

Give yourself a bit of time to work through it and if you're still stuck I'll try to help some more

In a bit I'll try to work it out

[somber coyote]

Xoc:

[twin prawn]

\theta

[dark sparrow]

$s = \theta r$

[somber coyote]

Ann:

[dark sparrow]

lmao quora

[sudden locust]

Tangents and Circles 🤓 👉 https://youtu.be/shoRxnTtjkA

[quiet rain]

hi little confusion of myself. So we can state that arccos(x)=cos^-1 (x) right ? But isn't the definition of sec (x) = 1/cos(x) = arccos (x) ?
please @ me if you have the answer. thanks

[gritty siren]

it's a little notational issue

here the ¯¹ has nothing to do with the usual a¯¹=1/a

it would be better if people would refrain from writing cos¯¹ at all actually

@quiet rain

[quiet rain]

Thanks sir you really helped me.

[tacit meteor]

Tan

Hey guys

Is it possible to algorithmicly devide a triangle in to 2 right triangles

?

[dark sparrow]

draw an altitude

[tacit meteor]

Do it on a computer

Algorithmicly

[dark sparrow]

...

[tacit meteor]

Are u french

?

[dark sparrow]

no

[tacit meteor]

Is that the ai thing

On youtube

Instagram

?

[dark sparrow]

i speak french (somewhat shittily) but i am not french myself

also,,, what

[tacit meteor]

Ok

[dark sparrow]

what are you talking about

[tacit meteor]

An tuong

Iv heard that name on a ai generating faces

I guess after alan turing

Anyways

[dark sparrow]

uhh

no

Tuong is the name of another honorable

[tacit meteor]

Ok

[dark sparrow]

my name translates to "Ann, discount Tuong"

[tacit meteor]

Actually i have managed to right the algorithm

But when i change the triangle sides it ceases to function

[gritty siren]

Hi, I'm Tuong

[tacit meteor]

Howdy

[hybrid jasper]

Given that cos(θ) = -3/5, determine the following: 

sin(π - θ) 

how would I do this?

[hybrid jasper]

<@&286206848099549185>

[river forge]

i'd recommend drawing a triangle that satisfies your condition cos(theta) =-3/5

[hybrid jasper]

yeah that'd be in the 2nd quadrant

with a hyp of 5

and one of the legs being -3

[river forge]

yes

[hybrid jasper]

I just don't understand (pi - theta)

like I get its 180 - theta

but like I can't visualize where the triangle will end up at

[river forge]

sin(pi-x)=sin(x)

there are a few ways to reach this conclusion

[hybrid jasper]

mhm

what if it was cos or tan

or sec/csc/cot

[river forge]

cos(pi-x)=-cos(x)

the rest can come from those two identities

namely tan and cot are periodic on pi

but you flip for them

since you have a negative x

the best thing tho would be think about these on the unit circle i guess

[hybrid jasper]

yeah that's what I'm tryna do

like I get how to do csc(theta - 2pi), csc(pi + theta), etc

but for pi - theta it just confuses me somewhat

so these work regardless of which quadrant theta is in?

would it just be better to think -theta + pi
if triangle is in Q1, -theta is Q4 + pi is Q2
if triangle is in Q2, -theta is Q3 + pi is Q1
if triangle is in Q3, - theta is Q2 + pi is Q4
if triangle is in Q4, -theta is Q1 + pi is Q3

[river forge]

i mean

case bashing probably isn't ideal

i sorta think about it as moving along the unit circle

[hybrid jasper]

thats what I do for the other scenarios

idk why im not grasping this

[river forge]

i guess think about it like flipping the unit circle over

the y axis

so instead of 0 being at 0 and going counter clockwise, we start at pi and go clockwise

this is equivalent to flipping any angle theta over the y axis as well

so Q1 becomes Q2 and vice versa, and Q4 becomes Q3

but sin in Q1 and Q2 is the same sign, and same for Q3 and Q4 but cos has the opposite

[hybrid jasper]

mhm

@river forge so for e. did I just get lucky since that triangle was supposed to be in the first quadrant

[wind heart]

When writing proofs

Like

If I state “<ROB and <TOV are right angles”, or like is <TVY congruent to < TYV or, segment TUC congruent to segment KSZ, do they all have to match up?

Like for the first part, does R have to match up with T?

For the second part, does V have to match up with Y?

[river forge]

i mean

[wind heart]

Or does that mean when triangles are congruent?

[river forge]

so tan(theta)=sin(theta)/cos(theta), so tan(pi-theta)=sin(pi-theta)/cos(pi-theta)=sin(theta)/-cos(theta)

[hybrid jasper]

yeah but lets say I had to draw the triangle in addition to evaluating

it wouldn't be in Q3 right

[river forge]

it would be in Q1 yes

[hybrid jasper]

yeah ok I get it now

-theta + pi makes it easy to understand

[river forge]

again because the mapping described sends Q2->Q1

[hybrid jasper]

like this

ty for the help

[wind heart]

<@&286206848099549185>

Also, would this be correct?

[silent plank]

you should've continued to use "congruent" in the last line

[rich wolf]

Car

=

Box

[wind heart]

Oh ok

Does the ordering for congruency matter though?

Like , here’s an example

Would R have to match up with S for the first one?

Like, would R and S have to be equal to each other

Or does it not matter?

[rich wolf]

Yes the letters have to represent the same parts of the triangle, or angle, or figure

[wind heart]

Does that apply to triangles too?

Like triangle congruency

[rich wolf]

Yeah

[wind heart]

Like

Here’s why I’m confused

On a YT video, it says ABC is congruent to ECD

but like

I don’t think that like

B and C are congruent

[rich wolf]

Yeah the video didnt say it right

ABC=EDC

[wind heart]

Oh ok

[rich wolf]

It doesnt really matter that much tbh but its the type of thing your teacher will call you out on

[wind heart]

So also, when there’s two congruent right angles on the same side

Letter*

would you just name the entire triangle

Instead of angle c is congruent to angle c

Lastly, would this be correct?

Oh I forgot to add the triangles for 4

But like

[silent plank]

if its ambiguous, label it properly

[wind heart]

Imagine there being triangles before it

[dark sparrow]

what are you asked to do

and what is giving you trouble

[pallid knot]

find k

and i have genuinely no idea how to do it

i found the length of OM but that's it

[narrow sleet]

Hint: OA is ALSO the radius of the circle

[pallid knot]

OH

thanks

:)

[civic nexus]

i need help with this question

show that 2cos(pi/4 + x)*cos( pi/4 -x ) = cos2x

[narrow sleet]

Use cos (A + B) = cosAcosB - sinAsinB and cos (A - B) = cosAcosB + sinAsinB

[civic nexus]

thank you!

[upper karma]

my turn for help!!!!!!!!!!!!!!!!!!!!!!!!!!!

Write a paragraph proof of the 30,60,90 Triangle Theorem.

Given --> Triangle JKL is a 30,60,90 triangle.
Prove--> The hypotenuse is twice as long as the shorter leg and the longer leg is sqrt3 times as long as the shorter leg.

Plan for Proof: Construct Triangle JML congruent to Triangle JKL. Then prove that Tringle JKM is equilateral. Express the lengths of segment JK and JL in terms of x.

[narrow sleet]

JKM is an equilateral triangle

[upper karma]

but we have to prove that

[narrow sleet]

Since JLM is congurent to JKL, < JML = 60°

And < LJM = 30°

[upper karma]

but you have to prove how the triangles are congruent

it is not given that those two triangles are congruent

it is only given that jkl is a 30, 60, 90 triangle

[narrow sleet]

JL = JL (Reflexive)

KL = LM = x

It's KLM given as a straight line?

Wait

[upper karma]

we dunno

but there's a right angle

so i assume so

because the other angle must be

90 degrees

[narrow sleet]

Didn't they tell you contruct the triangle JLM, not proving it?

No, u can't do that

[upper karma]

this is the image they gave in the book

but its a 180 degree angle line

its going to be straight

i just redrew the image because

my mom took away my phone

so i cant take a picture

[narrow sleet]

So they did mention it's a 180 degree angle line?

[upper karma]

no but look

there's a 90 degree angle for angle klj

so the angle opposite of it, jlm, is going to be 90 degrees too

they are perpendicular lines, segment km and jl

so they must forn 4 90 degree angles

[narrow sleet]

Ok

That's why we need the full question to understand it

But assuming it did saying KM and JL are perpendicular

[upper karma]

i gave you everything

it didn't

but you can see they are

[narrow sleet]

Then < JLM = 90°

[upper karma]

because angle klj is 90 degrees

[narrow sleet]

What happened if it's 91°?

[upper karma]

it's not 91

because thats a right angle

so the other angle must be 90 degrees to equal 180

and if the triangle is equilateral and split into 2

it would be a 180 degree line

[narrow sleet]

IF!

[upper karma]

for segment km

BUT IN THE-

[narrow sleet]

Which we need to prove that

[upper karma]

ack, ima let you speak, i feel so rude lmaoo

[narrow sleet]

It's ok

The thing is if they says either KLM is a straight line or JL and KM are perpendicular, then we can't prove it is 90°

Otherwise, there is no indication it is 90°. Not just because the angle beside it is 90°, then it automatically become 90°

Or unless the diagram did show it's 90°

Or like I say, they tell you to construct, not to prove.

@upper karma So anything?

[upper karma]

nah fam

im just

trying to draw fractals on my page LOL

[narrow sleet]

Lol

[upper karma]

yesyes, but we continue!

[narrow sleet]

It's late here

I better go to sleep

Maybe you can have someone else to help u

[upper karma]

:(((((((((((((((((((((((((

thanks for trying though

i truly appreciate it

goodnight!

[narrow sleet]

👍

[upper karma]

wait, 15 minute over

i can ping them...

nah, im scared

too many smart people=

scary

[narrow sleet]

All I can say is that the hypothenuse is 2x because equilateral triangle has the same length fpr each side and JL is sqrt(3) x because of phytagoras theorem

Good luck

[upper karma]

ohh

TYSM

and you too

good luck sleeping!!!!!!!!

[narrow sleet]

😂

[hybrid jasper]

,w evaluate cot^-1(tan(5pi/3)

[somber coyote]

Display results online and refine query

Requested by Pingu#4102

Pingu:

[hybrid jasper]

why does symbolab say 5pi/6

[narrow sleet]

arctan (- 1/ sqrt(3)) = pi - arctan ( 1/ sqrt(3))

[hybrid jasper]

what

[narrow sleet]

arctan ( - 1/ sqrt(3))

The inside is negative

Since it's arctan

[hybrid jasper]

I got -pi/6 as my answer

[narrow sleet]

And tan is negative in Q2 and Q4

[thorn talon]

they define the arccots differently

it seems

[narrow sleet]

Oh

[thorn talon]

wolfram alpha and symbolab picked different ranges

[narrow sleet]

• pi/6 is also correct

[hybrid jasper]

5pi/6 is too?

how so

[thorn talon]

WA has arccot with range -pi/2 to pi/2

whereas Symbolab seems to have 0 to pi?

[narrow sleet]

Ah

[hybrid jasper]

5pi/6 works too?

[narrow sleet]

Yeap

[thorn talon]

consult whatever definition your textbook or class is using

[narrow sleet]

Or look at the range they gave you in the question

[hybrid jasper]

I got no range

[narrow sleet]

Yeah then like qhat @thorn talon said

[hybrid jasper]

arcctan is restricted to quadrants 1 and 4 right

[narrow sleet]

Depends

Some teacher says is q1 and q2

Some say q1 and q4

[hybrid jasper]

my teacher said arcsin and arcctan were both 1 and 4

but for arccsin how could the domain be [-1, 1] if its only 1 and 4

[narrow sleet]

Arcsin (1) = pi/2

Arcsin (-1) = - pi/2

[hybrid jasper]

oh wait nvm I was thinking of cos coords lol

[silent plank]

range of arctan is generally accepted as
(-pi/2, pi/2)

[thorn talon]

https://www.intmath.com/blog/mathematics/which-is-the-correct-graph-of-arccot-x-6009

kind of interesting

[silent plank]

yeh. the continuity argument.

[robust socket]

$\tan(x)<\sqrt{2}-1$

[somber coyote]

Umma.Gumma:

[robust socket]

I would appreciate some guidance on this

[elder surge]

wdym

[robust socket]

I'd like to solve it without recurring to graphs

[elder surge]

well

you first need to find where tan(x)=sqrt(2)-1

so

$\tan{x}=\sqrt{2}-1$

[somber coyote]

Angelium:

[elder surge]

$x=arctan(\sqrt{2}-1)$

[somber coyote]

Angelium:

[elder surge]

right?

[robust socket]

yep

[elder surge]

so then you know that tan(x) is a periodic function

right?

[robust socket]

$x < \arctan(\sqrt{2}-1) + k\pi$

[somber coyote]

Umma.Gumma:

[robust socket]

is that it?

[elder surge]

what is k?

in this case?

[robust socket]

any number belonging to Z

[elder surge]

yep

then you can write that as

$x < \arctan(\sqrt{2}-1) + k\pi$, where k is an element of $\mathbb{Z}$

wait

uh

[robust socket]

$\mathbb{Z}$

[somber coyote]

Umma.Gumma:

[elder surge]

ye that

[somber coyote]

Angelium:

[robust socket]

ok, thank you for your help

[elder surge]

no problem

this is actually my first time helping someone

here

which is nice

[robust socket]

:)

[elder surge]

:)

need some help somewhere else still?

[robust socket]

not at the moment but I'll ask for sure shall the need arise

thanks

[elder surge]

thank you too

[robust socket]

actually, checking the solutions it seems that the inequality is verified for completely different values

i.e. $-\frac{\pi}{2}+k\pi<x<\frac{\pi}{8}+k\pi$

[somber coyote]

Umma.Gumma:

[elder surge]

i mean

wait a minute

$arctan(\sqrt{2}-1)=\frac{\pi}{8}$

[somber coyote]

Angelium:

[robust socket]

how did you check this?

nevermind got it

[elder surge]

wolframalpha

[robust socket]

I wonder where the \pi/2 comes from

[edgy raptor]

I know this is no geometry but maybe someone can help me finding the theta and the vector Fab

[quiet mason]

translate bro

[steep temple]

is this diagram geometrically accurate

what would GAE or BCF be?

can it be simplified?

[quiet mason]

what are your known angles

like did the question provide x or y

[steep temple]

its not a question

i just want a diagram i can always run to when i need help

the segment in the question i was doing is not even oriented like that

but i imagine the orientation of the segment is irrelevant

so; is the diagram accurate?

[edgy raptor]

Well the mass on the weight is 10kg and i need to find the vector Fab and the angle

[quiet mason]

have you learnt of free body diagrams

if no,learn and try to apply one here

[edgy raptor]

I just found the solution

Im pretty stupid hahah

I fucked up the whole time with finding that angle

[steep temple]

https://discordapp.com/channels/268882317391429632/326138757474680852/665953393151574037

[upper karma]

--

[gritty sail]

how to solve

-7(cos(pi/7)+cos(3pi/7)+cos(5pi/7))

??

[turbid rain]

is separated axis theorem best way to determine object collisions?

[weak shoal]

???????????????

[gritty sail]

Nvm for my q

[upper karma]

Hello, I need help.

Related to a problem-.

I'm trying to search for a coherent solution for 3 days, still I dunno how to answer it correctly

I mean, the procedures.

This is the question:
The figure shows a sketch of the location of 4 police emergency posts at points A, B, D and C; in addition to the central station located at point E. If stations A, B and C are at the same distance from each other, as well as B, D and E, in addition to CD = 20 km, find the distance from station A to central station E

So I made this, but still

People told me "The answer is 20"
But why?

I don't want a direct answer I want the procedures to learn better

[frank sail]

How'd you get the angels?

Angles*

I mean if you have the angles you'd be able to calculate the sides with cos/sin

[upper karma]

@frank sail

[frank sail]

No?

[upper karma]

are you good with algebra 1

[frank sail]

I wouldn't consider myself good at algebra

Why

[upper karma]

i had a question

[frank sail]

I study maths in argentina

I'm not quite sure what alg1 contains

But I can give it a try

You are better off asking one of the <@&286206848099549185>

[abstract arch]

Please don't ping before actually having asked the question.

[upper karma]

I don't fully understand this. I just need 2 sentences answering the question...

[upper karma]

@frank sail The angles are simple to get

Remember the fact
"if 3 sides of a triangle are the same, then all angles are 60°"

[frank sail]

Aaahh I had forgotten than

Ty

[brave zephyr]

Can someone walk me through/help me on this problem. I’ve been stuck on it for like the past 10 minutes.

[thorn talon]

You should be able to draw some triangles

I see some ratios

I don't know if the ratios are correct

But it's the right idea

[wind heart]

https://media.discordapp.net/attachments/553259767712186401/666423274858741775/20200113_182731.jpg

Can somebody help me on this stuff?

I was kind of absent today and my friend gave me the WSH but I have no idea what any of it means

[wind heart]

<@&286206848099549185>

[upper karma]

@wind heart which problem

@wind heart 3 and 9 are incorrect

[wind heart]

I guess the first four if it dosent trouble you*

Like, I don’t know how to do them

Is the thing

[upper karma]

Oh wait 4 is wrong too but like kinda not

What can you tell me about equilateral triangles

[wind heart]

Every side is congruent to each other?

[upper karma]

And...?

[wind heart]

And angle

[upper karma]

Correct

What's the sum of interior angles for a triangle

[wind heart]

180

[upper karma]

Ok for number 4 is it an equilateral triangle

[wind heart]

Yes

[upper karma]

So the angles are...?

[wind heart]

60 each

Which would be

180

[upper karma]

180 total but they are asking for one of the angles

[wind heart]

Oh

[upper karma]

Make sure you write the degree symbol

Not a percentage

For number 3

Is that an equilateral triangle

[wind heart]

Iscocele

Isosceles

[upper karma]

What's the other angle

The angle that's not written in

[wind heart]

Uhhhh

Oh wait

It’s also

Equal are risk

Equalateral*

[upper karma]

Yes it's an equilateral

[wind heart]

Sorry autocorrect

[upper karma]

So what would the side be

Np

[wind heart]

Well <H would be 60

But the sides

I’m not sure

Because one of the sides is 12

Would that also mean the other sides are also 12?

[upper karma]

Precisely

[wind heart]

So would FH be 36 or 12?

[upper karma]

Is this ur friends hw or yours because 1 and 2 are correct

[wind heart]

Because I don’t know if it’s asking for FGH or FH

it’s my friends

[upper karma]

Oh

[wind heart]

I was absent so I asked for it

[upper karma]

Oh ic

[wind heart]

But he gave us no notes for it so it’s hard for me to understand it

Which is why I’m asking

[upper karma]

So what problem do you have with the first one

[wind heart]

So like

[upper karma]

AB with the line on top stands for the side from point A to B. Same with BC.

So if you have two sides that are congruent what do you have

[wind heart]

congruent segments?

[upper karma]

What type of triangle

For example the previous examples we worked on equilateral what kind of triangle is when 2 sides are congruent

[wind heart]

Isosceles

[upper karma]

Correct

The question asks what angles are equal in an isoceles triangle

Can you figure them out?

[wind heart]

Not really

It say <BAC congruent to <BCA

But does that mean that like

E is inside the angles?

[upper karma]

What

[wind heart]

Like

There’s an E

[upper karma]

E is just a point of a triangle within the triangle

They are two separate triangles

[wind heart]

oh

[upper karma]

But overlapped

Yes E has nothing to do with this

So what angles would be congruent to each other

[wind heart]

Would CAB Comfruent to BAC be wrong?

[upper karma]

Yes

Because they connote the same angle

<CAB and <BAC means the angle "in between" line CA and AB

Or BA and AC

With the A being the vertex of where the angle is at

Does that make sense or nah

[wind heart]

Uhhhhh

Not really sorry

[upper karma]

That's fine

I'll brb lemme make a drawing

[wind heart]

Ok

Thanks

[upper karma]

can anyone help me do a proof im very confused

@wind heart

Note that the top can also be written as <CBA and the bottom can also be written as <BCA

[wind heart]

Wait so those two are congruent?

[upper karma]

Nope

It's never defined

So don't assume that

Given: AB||CD CA||DB
Prove: AB Congruent to CD

https://cdn.discordapp.com/attachments/664989932422561802/666429917491232808/unknown.png

any1 know how to do this
?

@upper karma another channel pls

[wind heart]

Oh it’s because there’s no congruent tick marks on the sides?

[upper karma]

Yes

Ok back to the problem

#2

What angles are congruent

[wind heart]

Would EC and EA be fine

Rather than EC and AE

If so I understand it

If not then I’m confused

Oh and for angles it’s EAC and ECA

[upper karma]

Yes

It's given that EC and AE are congruent

But those connotes sides

But yes that's correct

Anything else?

[wind heart]

Uhhhh one more

4

5*

[upper karma]

Ok in 5's triangle what can you tell me

What kind of triangle is it

[wind heart]

I can tell you that uhhhhh <S is congruent it <T

[upper karma]

Which makes it a...?

[wind heart]

Isosceles

[upper karma]

And what else can you tell me abt isoceles triangles

[wind heart]

Um idk

[upper karma]

Congruent angles and..?

[wind heart]

sides

[upper karma]

Yes

So what sides are congruent to each other

[wind heart]

RT and SR?

[upper karma]

Yep

RT=SR

What is RT and what is SR

[wind heart]

RT is 9

so you set SR to 9?

And then find z

[upper karma]

Yes

Exactly

Everything cool?

[wind heart]

Ummm

If it dosent trouble you

Can we go over 6?

(Last one for sure)

But like 6 is just weird

[upper karma]

Sure send it

It's not weird if you go back to the basics

Just think abt what you know

[wind heart]

So it’s isosceles so far

[upper karma]

Okay what can you tell me about that triangle

Okay

[wind heart]

Congruent sides and angle

[upper karma]

Okay

So what do you do next to find x

[wind heart]

So W would also be 62

So it would be

124

[upper karma]

What

[wind heart]

124 = 4x - 2?

[upper karma]

No

[wind heart]

o

[upper karma]

Do you know what corresponding angles/sides are

[wind heart]

WY and XY?

[upper karma]

What's that

[wind heart]

wait

XW and WY

[upper karma]

What's that

[wind heart]

Idk

the sides

[upper karma]

What abt those sides

[wind heart]

they are congruent segments

[upper karma]

Okay

So what angles are congruent

[wind heart]

X and Y

[upper karma]

Yes

So how would you go about finding x

[wind heart]

Idk tbh , I would still do the previous thing I came up with

[upper karma]

Look at the previous exercise

We did the exact same thing just backwards

[wind heart]

4x - 2 = 16?

[upper karma]

Where'd you get 16

[wind heart]

From the segments

[upper karma]

?

[wind heart]

The two 8’s

[upper karma]

Why would an angle ever equal to a side?

[wind heart]

Idk

Would it be 62 = 4x -2?

[upper karma]

Okay then tell me what you do know

[wind heart]

If it is then I’m confused

Because the other angle is also 62

So I would think you’d add them up

[upper karma]

Lemme draw something for you

This is an isoceles triangle

[wind heart]

Oh I see wait

<X does is not congruent to <W

So since <X is congruent to <Y

You’d do 62 = 4x - 2

[upper karma]

Yes

[wind heart]

64/4

[upper karma]

16

[wind heart]

16

Thanks

[upper karma]

Yes

[wind heart]

I get it now

[upper karma]

No

Problem

[wind heart]

Well that’s it

Thanks for the help

[upper karma]

The last segment think exactly how you would with the first part

Np bro

[tardy junco]

Sin( pi/2 - x ) and Sin( pi/2 + x)
Dont have a phase difference of pi??

But youre subtracting the phase of first function from pi to get the phase of the second function..

So shouldn't they have a phase difference of pi??

[median crown]

if x=0, are you telling me that pi/2 is +- pi radians from pi/2?

bad bad bad

[tardy junco]

Tilting zero radian mark upwards ?

I didn't understand

[median crown]

aight

now for a phase change of pi to happen, 2n=pi

no?

[tardy junco]

What

[median crown]

you're claiming the phase change=pi no?

[tardy junco]

Idk why it's not pi

It's not pi but I don't know why

[median crown]

look at the figure

lets say x=n

[tardy junco]

K

[median crown]

the middle line is p/2 btw

the bisecting angle

aight so x=n

[tardy junco]

Then n is pi/4?

[median crown]

no

[tardy junco]

Then the bisecting line isn't pi/2?

[median crown]

idk what n is

[tardy junco]

Oos

[median crown]

it could be 69 radians idk

the point is

[tardy junco]

N is pi/2

[median crown]

it could be

but can you garuntee that it is?

[tardy junco]

Do we know what the bisecting line is

[median crown]

yes

pi/2

[tardy junco]

How do we know that if we don't know n

[median crown]

look

[tardy junco]

Oh you mean the vertical line

Makes an angle pi/2

[median crown]

the phase change between pi/2-n and pi/2+n= 2n no?

[tardy junco]

With horizontal line

[median crown]

just do the algebra

[tardy junco]

Okay

[median crown]

subtract pi/2-n -( pi/2+n)

[tardy junco]

You get -2n

[median crown]

yeah

-2n is the phase change

[tardy junco]

Hmm

[median crown]

but does n HAVE to equal pi/2?

is your true for all n?

[tardy junco]

No

Ya

[median crown]

no

n=0

[tardy junco]

No and ya

[median crown]

what

n

o

thats a yes or no question

[tardy junco]

You asked two questions

[median crown]

you just contradicted your own answer

[tardy junco]

Ok I get it. Why is phase difference of -2n = 0

Now

[median crown]

its not

[tardy junco]

[median crown]

the phase change is |2n|

[tardy junco]

But those two functions are the same function so phase difference is zero

[median crown]

,w plot sin(pi-x)

[tardy junco]

So why is 2n = 0

[somber coyote]

Display results online and refine query

Requested by maleb1964#6739

[median crown]

,w plot sin(pi+x)

[somber coyote]

Display results online and refine query

Requested by maleb1964#6739

[median crown]

is that the same graph?

[tardy junco]

Pi/2-x and pi/2 + x

Phase difference is -2x

But are the same function

[median crown]

they aren't the same function

look at the graph

[tardy junco]

Not that one

[median crown]

ooo

whoops

[tardy junco]

Sin(pi/2 - x) and sin(pi/2 + x)

[median crown]

,w plot sin(pi/2-x)

,w plot sin(pi/2+x)

[somber coyote]

Display results online and refine query

Requested by maleb1964#6739

Display results online and refine query

Requested by maleb1964#6739

[tardy junco]

[median crown]

doesn't imply what the phase change is

[tardy junco]

Ya but their phase change must be a multiple of the time period

Why is -2x a multiple of 2pi?

@median crown

X is a variable. How is the phase change a variable for two sine functions

[median crown]

well you have to remember that it's more like moving two functions onto each other

here

just graph sin(-x) and sin(x)

and then make the according changes to each graph

use the identity sin(x)=-sin(x-pi) if you want to do it algebraically

so while they are still equivalent, saying that the phase change = some number probably isn't an accurate statement

becuase it's an infinite amount of them

[tardy junco]

But.. phase difference is zero because they're the exact same function.

Maybe you can say that a function is equivalent to itself in an infinite amount of other phases. But that's just to say that the function is periodic. That doesn't mean the function has infinite phase differences from itself.

Idk

[median crown]

nothing profound here

except algegra

[tardy junco]

But you said the phase difference is -2x. How can phase difference between two sine functions be a variable

@median crown

,w plot sin(pi/2 - x)

[somber coyote]

Display results online and refine query

Requested by Godmother#4032

[tardy junco]

,w plot sin(pi/2 - 3x)

[somber coyote]

Display results online and refine query

Requested by Godmother#4032

[median crown]

it comes down to function manipulation

and my brain is turned off rn

[tardy junco]

How to prove sin(90-x) = sin(90-3x)?

[median crown]

its not

[tardy junco]

But the graphs are the same

[median crown]

scale

[tardy junco]

Ah ya

So the phase difference isn't -2x it's 0

[median crown]

its a whole bunch of stuff

the reality is that they were always the same function

[tardy junco]

Like 0, 2pi,..?

How does 2x become 2pi

K bye I need to study

[median crown]

i started at the wrong point that's why

when you add pi/2, the function just becomes cosine

so the figure should look like this

now that the function is symmetrical, it doesn't matter if you go left or right aka +-x

[tardy junco]

If phase is n, then n for sin(x + n) would be the angle with X axis not Y axis. The diagram is wrong.

n would be the angle with +X axis
180-n would be the angle with -X axis

Projection on Y axis is symmetric of these angles.

I think that's what you were trying to tell

[median crown]

how is the diagram wrong

bruh

thats literally what it is

symmetrical

that doensn't even make sense

[eternal crag]

okay maybe i'm dumb, what is the ratio of x and y?

[eternal crag]

no one is feeling geometry today?!

[twin prawn]

Use similar right-sided triangles and https://en.wikipedia.org/wiki/Angle_bisector_theorem

@eternal crag

[eternal crag]

hmm alright, i tried but got nowhere, i'll try again later

thanks

[twin prawn]

Np

Answer spoiler: ||it's 4/25||

[modern relic]

i dont get how we know the result above is true. Where do you get the stuff on the left hand side from?

[twin prawn]

Hint: draw the three line segments connecting the center of the in-circle to the vertices of the green triangle

[modern relic]

hmm

i did that and found the area of triangle

but i found the area without using the c variable

I'll try againn

[limber seal]

Can somebody help me to do this?

[vague hare]

Do what exactly lol

[limber seal]

All of this

🙃

[dark sparrow]

no but what are you asked to do with all these things

simplify?

[vague hare]

That's usually the assumption with trig problems

[limber seal]

Oh, sorry, wrote "do" instead of "solve"

[vague hare]

Solve?

For what?

[idle bloom]

probably just simplify

[novel tangle]

help?

[grizzled zinc]

I have 0 clue on what equation I need to do for these.

[covert hazel]

@grizzled zinc what's something you know about the internal angles of all triangles, no matter what size and shape?

[limber anchor]

Could someone walk me through how to solve number 9? I have to justify each step and I’m not really sure how to start

[novel tangle]

help plz

[grizzled zinc]

@covert hazel they equal 180

[covert hazel]

ok. so for #7 for example, you know that 180 = (2x) + (3x+5) + 90. how would you go about solving for x now?

[grizzled zinc]

2x+3x+5+90=180

5x+95=180

[covert hazel]

@limber anchor for your #9, you have that angles B and C are complementary. make sure you understand the definition of that word. then, you also have that the measure of angle B is 5 times the measure of angle C. what can we deduce from this?

[limber anchor]

Well B and C added together make 90 degrees right?

[covert hazel]

right.

that's what complementary means.

[limber anchor]

so could i do like 5B + C = 90?

[covert hazel]

^ not exactly, there's a small but important error there.

B+C=90. B=5C. therefore... ?

what's the substitution there?

[limber anchor]

i feel really dumb but like, should C be on the side with 90?

[covert hazel]

substitute the value we know for B (which is 5C) into the equation B+C=90.

then, we'll have an equation that's only in terms of C's.

[limber anchor]

ohhhh okay i am dumb

so 6C = 90

[covert hazel]

dumb is not the same as "not used to it"

there you go

[limber anchor]

And then I divide 90 by 6

[covert hazel]

and you obtain what value for C?

[limber anchor]

15

So C = 15

[covert hazel]

correct! and that completes the proof that the measure of angle C is 15 degrees.

[limber anchor]

Thank you very much

[covert hazel]

no problem. don't get down on yourself - practice more problems like these until seeing the substitutions comes more naturally to you

[grizzled zinc]

Okay i just got home and the ones i didnt so are 8-10

I just don't know what to do with the numbers

I know that outside angle has to be set equal to 180

[limber anchor]

Will do, now to write all that down

[covert hazel]

@grizzled zinc they're done in the exact same way as #7. you know the internal angles of any triangle add up to 180°

so take the angles that you're given, in terms of X's, add them all up and set them equal to 180.

then you can solve for x in the same way that we did before

[grizzled zinc]

Do i not do 5x+7=180 then?

[covert hazel]

for #8 you mean?

[grizzled zinc]

Yeah

[covert hazel]

well, we have to use another piece of information that we know, which is that a straight angle is 180° as well.

so that internal angle is going to be (180 - (5x+7))°, right?

[grizzled zinc]

yeah

[covert hazel]

ok. well, using that value, you can continue using the same method as before

[grizzled zinc]

wait im confused on the 180-(5x+7) now sorry

[covert hazel]

#9 and #10 can be solved in similar manners, you just have to take other little pieces of what you know about certain kinds of triangles and fit them in

[grizzled zinc]

ok

[brave zephyr]

Holy shit when will we ever use flow chart proofs for congruent triangles

Its so useless

Proofs in general

[covert hazel]

I'm sorry you feel that way

as a math person, I disagree

[grizzled zinc]

I'm sorry but I have no idea still what to do for 8

and I keep getting the wrong answer for 9

I keep getting x=20 instead of x=14

[upper karma]

Can you guys help me with my hw this is the question. Recall the equation for a circle with center ( h , k ) and radius r . At what point in the first quadrant does the line with equation y = 0.5 x + 2 intersect the circle with radius 3 and center (0, 2)?

@coral wagon

[coral wagon]

okay

so

can you write the equation of a circle

given its center and radius

[upper karma]

i belive it's x^2 + (y-2)^2 = 9

i'm not sure tho

[coral wagon]

yes

you are right

so x^2+(y-2)^2=9

and y =1/2 (x)+2

you want to find where they intersect

what do you do

[upper karma]

do you plug in y ?

so you would get x^2+ (1/2x+2-2)^2 = 9

[coral wagon]

oay

okay

[upper karma]

what next

[vagrant steeple]

what does this mean by finding a function of a alone?

[silent plank]

in that context, without phase shifts

[vagrant steeple]

so i'm turning it into
a = ??

i tried applying the sum of tangents identity to tan(45 + a)

[silent plank]

what did you get?

[silent plank]

that should be enough

[vagrant steeple]

so all i have to do for this section is have "a" be the only variable

[silent plank]

a is already the only variable

only "a" should be in the argument of the trig functions

[vagrant steeple]

i guess i meant to say what you just said

thanks

[eager kraken]

How do you know what angle you are finding sin cos and tan for?

Is it the angle that is touching the origin?

like how do I know if theta goes in the bottom left side or the top point of the triangle?

[dark sparrow]

How do you know what angle you are finding sin cos and tan for?
depends on context

[eager kraken]

like if I say what's the sin of (-7,24)

[dark sparrow]

that question doesn't make any sense

[eager kraken]

is it 7/24 or 24/25

[dark sparrow]

you can't take the sine of a point like that

"sin of (-7, 24)" does. not. make. sense.

[eager kraken]

well that point and standard position

[dark sparrow]

show the exact statement of the problem.

you're omitting the very things needed to answer your question.

[eager kraken]

The terminal side of theta in standard position contains the given point (-7,24), find sin cos tan

[dark sparrow]

ok, so standard position. that means that θ has its vertex at the origin, and is measured CCW from the positive x axis.

[eager kraken]

so theta goes on the origin vert?

[dark sparrow]

and is measured CCW from the positive x axis.

read what i'm saying carefully.

[eager kraken]

ok

[dark sparrow]

in math you've got to pay attention to details like this, since omitting even one can make you go astray in ways that are nigh-impossible to detect afterwards

[eager kraken]

I mean I did say standard position and that point

the grounds we walk are not pure

[limber anchor]

Can anyone help me solve the top problem? I know 5 would equal 90 because it’s formed by two perpendicular things

But I’m not sure how I’m supposed to relate that to 4 and 6

[silent plank]

do you know the value of
4 + 5 +6 ?

[limber anchor]

180 right?

wait i think i got it

[prisma spruce]

Can anyone help on this? i dont know how to approach these 2

[late glen]

i need help with square root

how do I solve this?

its 1 feet to 21 inches

and im very confused

pls help

[covert hazel]

what is the original question at the top, of which this is part (c)?

[late glen]

"Give the following ratios in simplified form. (If the given quantities are incommensurable, enter INCOMMENSURABLE.)"

[covert hazel]

that doesn't seem to have anything to do with square roots...?

[late glen]

it doesnt?

well that's the theme of it

i need to simplify square root

they gave me that

[covert hazel]

did you get parts (a) and (b) correct?

[late glen]

yes

[covert hazel]

what were those questions and answers?

[late glen]

there you go

[covert hazel]

ok

[late glen]

its very confusing

at least for me it is

[covert hazel]

these definitely have nothing to do with square roots, but I can help you

[late glen]

ok then

[covert hazel]

what can we say about the units "feet" and "inches"? they're both units of...?

[prisma spruce]

I dont think its geometry too

[late glen]

thats what im supposedly doing in math class "square root"

[prisma spruce]

1 foot is 12 inches

[late glen]

well it is in geometry category apparently

yes ik that

[prisma spruce]

So you calculate 12 to 21

[late glen]

ok?

[covert hazel]

doh, I wanted him to figure that out

[prisma spruce]

Lmao

[late glen]

i already knew that

[prisma spruce]

Maybe you could help me with mine tho

[late glen]

...

[covert hazel]

what's yours?

[prisma spruce]

Im stuck at like the last step for both of them

[covert hazel]

15 is correct as it is

it's any x>sqrt(18)

[late glen]

i dont understand how did you get this?

do*

i pressed practice another one

and it gave me that answer

[covert hazel]

@late glen feet and inches aren't the same

[late glen]

(i pressed the 6 on purpose)

ik that

but how do you solve it?

[covert hazel]

a ratio like that has to be both top and bottom in the same units

[late glen]

so what do i type in?

it says 4/5

so there should be able to do something

[covert hazel]

you're skipping to the end

[late glen]

no?