geometry-and-trigonometry

[high wren]

yeah

[dark sparrow]

with that in mind, having solved $\csc(z) = \sqrt{2}$ for $z$, are you able to solve $\csc(x/4) = \sqrt{2}$ for $x$?

[somber coyote]

Ann:

[high wren]

well im not sure

does it have to do with something multiplying the sqrt 2 with the 4

oh wait

do you plugin pi/4 and 3pi/4 into the x?

_>

[dark sparrow]

no you're overthinking it again

[high wren]

i still have no clue then

[dark sparrow]

$\frac{x}{4} = \frac{\pi}{4} + 2\pi n, \frac{3\pi}{4} + 2\pi n$

[somber coyote]

Ann:

[high wren]

is that the answer?

[dark sparrow]

not quite.

you still need to multiply both sides by 4 to get x

[high wren]

so

pi + 8piN, 3pi + 8piN

?

[dark sparrow]

there

[high wren]

???

[dark sparrow]

yeah there you have it

[wind heart]

Uh I need help

No matter what I do

I get 15 for x

Oh wait nvm I think I got it

[unborn jacinth]

how many possible ellipses exist

with a given center

and foci that are x units from the center

[dark sparrow]

infinitely many; the major axis can be as big as you want so long as it's bigger than the distance between the foci.

[unborn jacinth]

o wait @dark sparrow I forgot this part: the major axis is of length 10

[dark sparrow]

ah

ok, followup

are the foci themselves fixed? or is only their distance from the center given

and are we counting two ellipses differing by a rotation as distinct or not

[unborn jacinth]

1)only distance is given
2)rotations not included

[dark sparrow]

ok alright

then yeah there's only one possible ellipse

[unborn jacinth]

ok

[drifting parrot]

I need help finding the standard form of this parabola

I’m not sure if I did this right

[twilit zenith]

I think it's correct except for the fact that you seemed to have added two different numbers to either side.

Notice that we add the (-b/2) term only if we're turning a term of the form x² + bx into the form (x + a)² ,which isn't the case with the right hand-side of the equation.

You only needed to add (-10/2)² to either side, and the equality would have remained.

[drifting parrot]

ok, i'll try that

I'm still confused as to why I have to put (-10/2) on both sides.@twilit zenith

[twilit zenith]

As a general rule, in order to maintain equality, you must add the same term to both sides. If you add, for example, +10 to one side and +20 to another, then you've lost the equality.

As to what specific number you have to add to both sides, that does depend on the b, and you've correctly identified that the value you need to add is 25.

[drifting parrot]

ok

Thank you for the help!

[drifting parrot]

Can someone help me with this problem?

[surreal bolt]

okay. what do you know about parabolas?

[drifting parrot]

I know that they have a focus, directrix, and vertex

I’ve had problems like this before, but the zeros make it difficult

[surreal bolt]

so how do you find "p"?

[drifting parrot]

I’m not sure how to find p without a given value

[surreal bolt]

iirc p is the distance between the focus and vertex

well +/- p

[drifting parrot]

So do you subtract the vertex value minus the focus?

[surreal bolt]

well, to be complete I have to say ... ya gotta find what direction the parabola is opening to

vertex is (7, 0) and focus is (0, 0) so which way does the parabola open towards (left, right, up, down)?

[drifting parrot]

I know that if p is greater than zero then the parabola will open downwards and if p is less than zero then the parabola will open downward, but how do you figure out whether or not it opens left or right?

[surreal bolt]

sorry D/Ced

Parabolas always open from the vertex to the focus.

[celest sky]

I am dumb

[surreal bolt]

we all are ...

could you make a more enlightened statement?

[celest sky]

My math teacher got fired at the start of the year so no one has taught me this yet

[surreal bolt]

yeah. There's a lot of people in that boat

[drifting parrot]

Soo, parabolas never open to the left or right?

[surreal bolt]

um, this one does. The Vertex is to the right of the focus.

[celest sky]

My teacher got fired for talking trash about Mexicans and illegal immigrants and “how great trump is”

[surreal bolt]

I'm glad you think that's important to getting help on this server.

[celest sky]

And the school can’t find anyone that can teach us

[surreal bolt]

yeah ... are you doing it all online?

[celest sky]

So we get handed work to do but we have never been taught this

Yep

But all videos are restricted on our chrome books

So I can’t get help

[surreal bolt]

@drifting parrot still need this answer?

[drifting parrot]

Yes

I actually have a question

[surreal bolt]

um, what grade are you in, @celest sky

[drifting parrot]

Soo, from what I looked up, if the equation starts with y^2, then you are dealing with a vertical parabola? Am I correct?

[surreal bolt]

y = ax^2 + bx + c is vertical.

opens up or down depending on the sign of a

[drifting parrot]

It has nothing to do with what the equation starts with?

[surreal bolt]

So if y = ax^2 that's opens up or down ... if x = ay^2 that opens left or right.

[celest sky]

@surreal bolt sophomore is highschool rn

[surreal bolt]

you should complain to the school about not being able to see the video. That's crazy.

[drifting parrot]

I'm confused. There's no pattern

[celest sky]

I know algebra, trig, some calculus and statistics, but only done geometry

[surreal bolt]

um, what kind of pattern were you looking for?

again, I can't help you view your videos ...

[celest sky]

Well the school knows and they think we will be using YouTube and other websites for fun instead of using it for educational purposes

[surreal bolt]

yeah ... so your teacher doesn't know your videos are blocked?

or ask the head of your math department about it

@celest sky seriously. If you haven't at least tried that, well anyway you should try it.

[drifting parrot]

I was looking for a pattern like “ if the equation starts with x^2 then it is a vertical parabola and if the equation starts with y^2 then it’s a horizontal parabola”

[surreal bolt]

um, yeah if (x - h)^2 + k = ay that opens upwards or downwards.

ahh I flubbed the parent equation

(x - h)^2 = a(y - k)

[drifting parrot]

I meant if the equation is given in something other than standard form like this.

[surreal bolt]

It boils down to ... if there's an x squared term and no y squared term, it's up or down. and if there's a y squared term with no x-squared term it's a left or right.

assuming it's a parabola at all

[drifting parrot]

Yes

[upper karma]

I sense a conics conversation...

[surreal bolt]

sigh ... am I allowed to make a ... coney joke ...

[upper karma]

do it

[drifting parrot]

Oh no

[surreal bolt]

Ai just did 😛

[upper karma]

Wow how did I not get that

[surreal bolt]

math dad jokes ... how I've fallen

[upper karma]

oh Jebus

[surreal bolt]

anyway @drifting parrot I'll just show you the process for the specific problem you had.

[drifting parrot]

Ok

[surreal bolt]

I'm looking at graph paper and plot (7,0) and (0,0)

Parabolas always open from the vertex to the focus. So, in this case, from (7,0) to (0, 0) ...

|p| is the distance between the vertex and focus so |p| = 7.

[drifting parrot]

Did you use the distance formula for that?

[surreal bolt]

you can

but you might notice one is exactly 7 units to the left/right of the other

they're both on the x-axis.

So yeah from all this info, x = a(y - k)^2 + h is the equation we need but we have to fill in all the variables except x and y.

[drifting parrot]

Ok

[surreal bolt]

1/a = 4p and |p| = 7 so a = +/- (1/28)

[upper karma]

That’s awesome

[surreal bolt]

um, are you sure the 7 goes on the left?

[drifting parrot]

Hold on, I think I did that wrong

[surreal bolt]

I dislike conics but at least you're not asking me about one with a line of symmetry rotated 30 degrees from vertical ...

[drifting parrot]

Oof

I don’t like them either, I’ve always had a hard time with them

[surreal bolt]

um ... isn't 7 usually to the right of the y axis ...

[drifting parrot]

Ugh! I can’t believe I did that 🤦🏾‍♀️

[surreal bolt]

it's okay. I jumped out of a moving car on the highway

funsies

[drifting parrot]

😶

[upper karma]

why u no like conics? They’re useful if you want to set a fleet of Roman-empire-age Roman ships on fire

with a parabolic mirror

and the sun

[surreal bolt]

oh no worries. My teeth are still all there. My knee hurts every once in a while though!

[drifting parrot]

Yikes

[surreal bolt]

it's cool.

[surreal bolt]

yeah so it opens left.

[upper karma]

|| Greek inventor Archimedes is said to have used [parabolic] mirrors to burn ships of an attacking Roman fleet ||

[surreal bolt]

x = a(y - k)^2 + h

[drifting parrot]

Why is 1/a equal to 4p?

[surreal bolt]

a is negative

yeah I never liked using that conversion either.

there's something else I like better.

[drifting parrot]

This?

[celest sky]

@surreal bolt the problem is that I don’t have a teacher

[surreal bolt]

well you also don't have ideos

[celest sky]

The school throws work at us according to what we should know with the curriculum

[surreal bolt]

lol kinda double bad

look man, I feel for you cause it took me 10 years to get a MS

but I can't help you with every problem you have.

[celest sky]

It’s fine

At least my work isn’t graded

[elfin ingot]

yo can anyone help me with a problem>

?

[surreal bolt]

sigh ... it's going to be a mess for a while down the line.

[celest sky]

My school district removed grades due to corona virus and gave us all credit except for the kids failing. The teachers give work out for the kids failing so they can get credit for the year but my mom still wants me to do the work

[surreal bolt]

There's this nice rectangle that occurs when you have a parabola.

The distance from the vertex to focus is p. the vertical distance from the focus to the parabola is 2p on either side.

so we have (7,0) is the vertex.

(0, 0) is the focus.

(0, 14) and (0, -14) are on the parabola also.

that's easier for me to remember than 1/a = 4p

doh

lol

[drifting parrot]

that's sounds a lot harder

[elfin ingot]

please help

[surreal bolt]

it's a box with 4x the length as width or 4x the width as length depending on which way the parabola opens.

[surreal bolt]

-28 actually

|p| = 7 means p is plus or minus 7.

[drifting parrot]

So because a is negative, I pick negative 28?

[surreal bolt]

the way I see it is, if it opens down or left, the a is negative

[drifting parrot]

Ok

[surreal bolt]

sigh ... sorry it took so long.

do you have the answer key?

[drifting parrot]

It’s fine. It takes me awhile to understand things.

Number 15

[surreal bolt]

I still like the "find a point on the parabola" method. (0, 14) and (0, -14) should be solutions.

and they are.

the whole "a" thing is just fodder to forget another equation.

I'm about to go ...

[drifting parrot]

Ok

Thanks for trying to help me

[surreal bolt]

but yeah. opens left. vertex (7, 0) and (0, 14) is on the parabola.

(y - 0)^2 = (T)(x - 7)

(14)^2 = (T)(-7) --> T = -28

[drifting parrot]

What is the T for?

[surreal bolt]

whatever should be there!

(y - k)^2 = (something)(x - h)

you don't need to remember which side a is on or the relation between a and p.

[drifting parrot]

Ok

[surreal bolt]

sorry I'm a bit unorthodox here as I find goofy ways around memorization.

[drifting parrot]

Lol

[surreal bolt]

alright. It's 1 PM and I am waiting for a phone call and couldn't reallyl sleep last night.

time to leave lol

[drifting parrot]

Alright. Thanks again!

[surreal bolt]

thanks for letting me try 🙂

[drifting parrot]

Ofc!

[glad gorge]

Does anyone know what formulas I should use for this?

[upper karma]

Theorems*

[glad gorge]

Oh yeah

[upper karma]

Most if it is common sense and that on the center the angles sum up to 360

[glad gorge]

Yea

[upper karma]

I dont think i miss anyone more

[glad gorge]

But I’m not sure how to use the outer side measurement to find the other angles inside

[upper karma]

The 130 one?

[glad gorge]

No like

Using 48 how am I suppose to find x

[upper karma]

Oh

Its called central angle theorem

[glad gorge]

Oh

[upper karma]

Wait lemme post it here

Its really dumb and simple but idk

[glad gorge]

K

Ohhh

I got itttt

Ty

[upper karma]

Ok good

Np!

@elfin ingot

according to the "inscribed angle" theorem it should be half that (212/2 = 106), but you don't have that as an answer, so...

I mean, it doesn't look like a 212 angle

it looks a little wider than 90 degrees, so it's probably 106

anyway, I'm off to bed

[normal solar]

What are some of the things I NEED to know to move to calculus?

[summer spire]

a strong understanding of algebra, at least

[novel flax]

algebra trigonometry and precalculus

[acoustic jungle]

knowing all the minecraft recipes by heart.

[upper karma]

Lmao

[royal cipher]

if u were to draw 4sin(t) as a circle, would the radius be 4?

or not since u dont have the cos formula

[acoustic jungle]

you're scaling everything up by 4?

[upper karma]

Help

[acoustic jungle]

it's on the video

watch the video

[versed river]

you can use the fact that opposite angles created by intersecting lines are equal

aand that triangles angles add to 180

[acoustic jungle]

yeah that isn't going to help you find x.

[silent plank]

^

[acoustic jungle]

you will have to watch the video.

[silent plank]

question is evil and requires you to know what to construct and/or maybe use a decent amount of trig

[eager pendant]

either you construct something or you just spam law of sines to verify an angle works

[upper karma]

Cool ok

[upper karma]

note: precalculus is just glorified algebra II. Move there if your algebra or trig course didn’t give you a concrete understanding of algebra or trig.

[surreal ingot]

Is integer without an undefined function and doesn't have a limit. It's just normal graph. Like X+1

[upper karma]

whered that come from?

[surreal ingot]

You're talking to me?

[upper karma]

Yes

I don’t understand the relevance of that message

[surreal ingot]

Oh

Sorry I was trying to understand what integer mean lol

The different between undefined functions and doesn't exist limit

[upper karma]

uhh

ok then

Good luck.

[surreal ingot]

Like different between infinity, undefined function and doesn't exist limit

Uhhh

[novel flax]

?

[surreal ingot]

I thought you were going to help.

Xd

[upper karma]

I mean I can

I’m just having a hard time understanding the question

[surreal ingot]

Oh

[upper karma]

Something about functions and where limits exist?

[surreal ingot]

I'm trying to figure out what is different between integer, undefined and doesn't exist limit in a graph.

[upper karma]

Hmmmm

[surreal ingot]

Chapter is finding limits using tables and graphs

[upper karma]

Well

If you want to find a limit

First find a discontinuity in your graph.

Then test numbers that approach your discontinuity...

(Right now I’m not the best person to be asking, but I’d suggest you look for some answers to your questions in places such as #❓how-to-get-help if you have specific questions.)

[surreal ingot]

I mean

It's trigonometry

[upper karma]

ok...

[surreal ingot]

Should I ping helpers?

[upper karma]

if you have a specific question yes

[novel flax]

what's the question

[surreal ingot]

How do you determine undefined, doesn't exist limit and integer on a graph

[novel flax]

The function isn't defined at the limiting value but that doesn't mean the limit doesn't exist!

[silent plank]

also integer is improperly used here. in this context it should be a constant

[surreal ingot]

Like x+1

Consider as constant?

[silent plank]

no as in what the limit is

as x approaches something

[surreal ingot]

Ohh I see

[silent plank]

it would be much better to explain if you have a specific example

also this isn't geo/trig

[surreal ingot]

As x is approaching 1, f(x) is approaching 2.

I mean it's consider as geo/trig because it's chapter 11 finding limits using tables and graphs

[upper karma]

Chapter 11 in what?

[surreal ingot]

It's introduction toward to Calculus

[silent plank]

so pre-calc

[upper karma]

Well that would be precalc

Or calc... I guess it’s acceptable...

[surreal ingot]

Idk. It mixed precalculus stuff and trigonometry stuff.

[upper karma]

Precalc isn’t really its own course, granted.

Math committees and what have you mashed together algebra II, trig, and a bit of limits and continuity into a new course, deemed “pre-calculus.”

[surreal ingot]

I mean precalculus combined algebra1 and 2

[upper karma]

I despise the creation but it is a good review and Segway into the calculus.

[surreal ingot]

It looks a bit easier than my previous chapter.

Lol

Where we do sinx and cosx

[upper karma]

that’d be trig??? Bit vague of a statement.

[surreal ingot]

I mean I have the precalculus 5e Blitzer

If you want to check it out lol

[upper karma]

uhh

Ok

I might, but I have my own studying tool for pre calc.

Anyway

[surreal ingot]

Ok. I probably do my quiz now lol.

Since I'm ready

[upper karma]

Ok.

[surreal ingot]

See you later

[upper karma]

Cya

[olive scarab]

I've been stucked with this problem for a while:
give an ellipse (E) $ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1 $ M $\in$ (E) (M is mobilize) 2 focis F1(c;0) and F2(-c;0) . Prove that the incenters (or the center of inscribed circle) MF1F2 is mobilize on an ellipse, find the equation of that ellipse.
sorry for bad english

[somber coyote]

.,.:

[upper karma]

At t = 0 a car is due 2 km north of you heading west moving 13 km/s after 1.5 s. What is the angle in radians?

is this question written wrong?

I have no clue what it is trying to say

[acoustic jungle]

you shouldn't have a period after 1.5 s

I believe the question is asking, "after 1.5 seconds, what is the angle in radians?"

@upper karma this is probably what the question is asking.

[upper karma]

Wait give me a second to understand

Ah ok

@acoustic jungle when I calculate for theta I get 0.3 something

But that's not an answer for the question

[languid finch]

hey can someone help with geometry?

<@&286206848099549185>

[upper karma]

this is the right channel, yes

what's the problem?

[languid finch]

im very confused

if someone can help thatd be epic

[upper karma]

I'm gonna assume "R" is a point

similar to how "P" is a point

and they give you a value for the y axis for this "R" point

[languid finch]

-2?

[upper karma]

yes

do you see where "P" is on the grid?

[languid finch]

-5, 2 right

[upper karma]

since they don't give you any values for the x axis...

that means it's either on top of point "P" or under it

and since they give you a negative value for the y axis...

where do you think point "R" should be on the grid?

relative to point "P" 's location

[languid finch]

-5, -2?

[upper karma]

2 units down from (-5, -2) is... ?

[languid finch]

-5, -4?

[upper karma]

the coordinates for point "P" are (-5, -2), correct?

no

wait

[languid finch]

-5, 2

[upper karma]

the coordinates for point "P" are (-5, 2), correct?

[languid finch]

ye

[upper karma]

right

so you need to move down from there, by two units

where do you end up?

(-5, 0)

[languid finch]

o

[upper karma]

yeah, because relative to point "P", point "R" is -2 on the y axis

so you go to point "P", and just move down from there by 2 units

[languid finch]

ok thanks

so -5, 0 would be the answer?

[upper karma]

yes

[acoustic jungle]

@upper karma well what did you do?

is the answer to the question 1.47

[upper karma]

Ooh

I figured it out

I just wasnt uing inverse tan

Mb

That was a goof

[acoustic jungle]

nice

[upper karma]

Yeah the answer was 1.4

1.4 what?

radians?

[acoustic jungle]

1o0 wHaT? eLePhAnTs?

[upper karma]

That just made my day.

[languid finch]

is this correct?

<@&286206848099549185>

[acoustic jungle]

not 15 minutes.

[languid finch]

??

[pallid cloud]

@olive scarab I think I solved your problem

But I solved it in a very unorthodox way

I am sure they don't want the solution like that

But I'll send my solution anyways, and explain what I did

[languid finch]

can anyone help wid mine

[pallid cloud]

The idea is to say that there exists a unique ellipse having its vertices the vertices of a rhombus

And I used a metric relation in a triangle, s=p.r, where s is the area, p the half-perimeter and r the inradius

I took 6 particular positions for the point M on the ellipse: B, B', K, K', K'', K'''

B and B' secondary vertices

K, K', K'' and K''' the points of intersection of the perpendicular to the main axis at each foci with (E)

Primary and secondary axes are axes of symmetry of the figure

You can then conclude that (BK) and (BK') on one hand and (BK'') and (BK''') on the other hand intersect at two points I and J that will make BIB'J a rhombus

I am sorry not B and B' but L and L' new points with coordinates determined

I am not 100% sure of my solution

But I showed you my work
So if there is something wrong someone tell me please

When I said B B' K K' K'' K''' I meant new points....

With coordinates determined (wrote them in my solution)

[devout harbor]

@languid finch so the point (5,2) becomes (-2,6) or is it the other way around

[pallid cloud]

@languid finch when composing translations, you add the vectors of translations, so add the coordinates of the given vectors

[devout harbor]

im confusd myself, but looking at the coordinates, -2 becomes 5 by adding 7 and 6 becomes 2 by subtracting 4, so b looks legit

[languid finch]

anyone know the answers to this one:

[pallid cloud]

$t_{\overrightarrow{u}} \circ t_{\overrightarrow{v}}=t_{\overrightarrow{u} + \overrightarrow{v}}$

[somber coyote]

Patrick Salhany:

[pallid cloud]

@languid finch use your eyes

Look at the diagram

You have the axis of symmetry

And the coordinates of the point

So determine the coordinates of the new point

e.g. (x;y) becomes (y;x) for the reflection with axis y=x

[languid finch]

would thisbe right

@pallid cloud

[pallid cloud]

A''(-6;-4)

Now find A''' and A''''

[upper karma]

@upper karma @acoustic jungle

or maybe I goofed it

I dunno, I'm really tired

I'm going to bed

leave a note if it's wrong

[languid finch]

@pallid cloud would A''' be (-6, 4)

[pallid cloud]

No

When reflecting with respect to y=x, a point that has coordinates (x;y) becomes (y;x)

A'' is (-6;-4)

So A''' is then (-4;-6)

[upper karma]

@upper karma why is it 2000

the car is traveling 13/kms

[acoustic jungle]

@upper karma it is wrong.

don't worry about it.

cheese pizza.

[drifting parrot]

Does anyone know how to find the endpoints of minor axis?

This is an ellipse btw

[upper karma]

hey, i had this question come up, i can't figure out how to find the diagonal without 3 givens. does anyone have an idea? https://cdn.discordapp.com/attachments/139130894400225280/704832971139448872/unknown.png

[pallid cloud]

Well you can't

You have to have the height

[upper karma]

yeah that's what i thought

[pallid cloud]

But they said rectangular prism

Does that mean that the height is equal to tge width?

[hollow finch]

The height should be 10 right?

[upper karma]

is that a rule with rectangular prisms?

[pallid cloud]

Sorry my vicabulary in geometry is french, so english terms might be confusing to me

So for rectangular prisms height=width?
If yes, then there you go

[upper karma]

i learned math in french too i get it

[pallid cloud]

@drifting parrot endpoints on minor axis you mean secondary vertices?

[drifting parrot]

Is that the same thing?

[pallid cloud]

It should be I guess

They're still considered vertices

So determine them

[drifting parrot]

ok

Is there a formula for the endpoints?

[pallid cloud]

Well for an ellipse x^2/a^2 + y^2/b^2 = 1 with a>b

Vertices:
Major axis: A(a;0) and A'(-a;0)
Minor axis: B(0;b) and B'(0;-b)

When you have (x-s)^2/a^2 + (y-t)^2/b^2 = 1

You just do a translation

Like what you did for major axis vertices

Adding 3 for x and 1 for y

@drifting parrot

Got it?

[drifting parrot]

no

maybe i'll get it if I reread

[pallid cloud]

Yes

I mean if you know how to determine the major axis vertices
Same thing for minor axis vertices

Anws if you don't get it ping me

[modest spear]

i know csc is hyp/opp right

i just get confused with these type of things

[drifting parrot]

@pallid cloud Yeah, I still don't get it

[devout harbor]

@modest spear Yes

[pallid cloud]

,rotate 90

[modest spear]

@devout harbor so what else?

[pallid cloud]

,rotate 180

[devout harbor]

@modest spear So you have cosec31 = d/3

[pallid cloud]

@drifting parrot

[modest spear]

oh

[devout harbor]

Now isolate d

[modest spear]

so it'd be

3 cosec 31

[devout harbor]

Yes.

[modest spear]

Ah

thank you sir

[drifting parrot]

@pallid cloud I was reading what you just sent

where did the zero come from?

[pallid cloud]

That is a property

In the natural system of the ellise, where the x axis is the major axis and the y axis is th minor axis, the major vertices have coordinates (a;0) and (-a;0) because they're on the x axis

And the minor vertices have coordinates (0:b) and (0;-b) because they're on the y axis

By natural system I mean when the equation if the ellipse would be just $\frac{x^2}{25}+\frac{y^2}{9}=1$

[somber coyote]

Patrick Salhany:

[pallid cloud]

But here we have a transaltion, like we grabbed the ellipse and moved it to be centered aroud (3;1)

So its equation became $\frac{(x-3)^2}{25}+\frac{(y-1)^2}{9}=1$

[somber coyote]

Patrick Salhany:

[pallid cloud]

So logically the vertices have to move

So we add 3 on the x coordinate of each vertex and 1 on their y coordinates

Get it?

[drifting parrot]

not yet. Could you explain, in detail, how you solved this?

[pallid cloud]

Well all I did is explained in details

Re read please

If you didn't get it

That means you probably didn't focus on the course concerning the properties of ellipses and their equations and formulas and coordinates of their vertices ...
Please re read
And revise the course
So you get familiar with the properties
I hate to say it, but some formulas have to be memorized

[drifting parrot]

I can't revise the course. i wasn't the one who created it

[pallid cloud]

You don't have notes?

[drifting parrot]

I do, but the required reading doesn't explain the endpoints of minor axis

[pallid cloud]

I told you

They are the same as the vertices on major axis

Like you did for the vertices on major axis

You do for the endpoints on minor axis

But instead of working with x coordinate you work with y coordinate

And I explained all above in the chat in details and how I got the calculations

[drifting parrot]

I'm sorry. I've had this stuff before and had a hard time understanding it

[olive scarab]

@pallid cloud your solution was pretty strange with these kind of problem, but I'll try it, thank you

[pallid cloud]

I know

I even indirectly used Pascal's theorem in projective geometry to justify the fact that there exist a circumellipse to a hexagon having the intersection points of opposite sides on the same line

[scarlet hare]

Anyone know how I would solve this?

[olive scarab]

hm...but I think we need to prove it in every case somehow

[upper karma]

How would I go about doing this?

my first strategy was to look for common factors to make it into a pythagorean triple

[burnt shale]

I can't seem to find the radical version

[hollow finch]

my first strategy was to look for common factors to make it into a pythagorean triple
@upper karma A hypotenuse length is longer than any leg on the triangle

[acoustic jungle]

use tan(x+y)=(tan x + tan y)/(1-tanxtany)

[hollow finch]

So just find the 3rd length and see whether it is the longest or not

$\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}=\tan(x+y)$

[somber coyote]

Lσνιηg✧Sσνєяєιgη:

[hollow finch]

Ya use this

[burnt shale]

oh, didnt know it was that simple

[acoustic jungle]

use the formula to find tan (2x)

@burnt shale

[deep trail]

help

https://gyazo.com/49dbde647e2095dbd27d69709a0e0fff

[umbral snow]

Which trig function are you looking to use?

[deep trail]

sin

[umbral snow]

Exactly!
sin(36) = x/10
Is the form you're looking for

[deep trail]

im still confused

is it just 36 = x/10?

wait nvm

[umbral snow]

See how that form happens?

[deep trail]

yes

[upper karma]

👍

[umbral snow]

Need any more help with the question?

[deep trail]

no i dont think so but just to be clear the asnwer would be sin(36) * 10 right?

[umbral snow]

Yussir that's it

[deep trail]

ty

[umbral snow]

Feel free to ask if you have anything else

[deep trail]

@umbral snow https://gyazo.com/7e9e0a15e896bb439dcbf166ee2de2cd helppp

[umbral snow]

The exact same question, still even using sin

But you'll want to draw a picture to get a sense

[deep trail]

the question just doesnt make sense. if the ship is on the ocean floor why is the point 60 meters below the ocean floor. is it in the ground?

[umbral snow]

The wreck is on the ocean floor

[deep trail]

oh i think i get it

so itd be like this

https://gyazo.com/5c50a0fa9ce25aec751f3278c460fa9e

so its tan(12.5) = x/60 i think

[umbral snow]

Careful, angle of depression measures looking down from the horizontal

You drew it looking up from the vertical

[upper karma]

how is a semi-circle translated to the positive/negative direction of the x-axis, and the positive/negative direction of the y-axis? does anyone know? Please @ me

<@&286206848099549185> ping me, ty

[cedar harbor]

Would anybody know how to solve these?

[novel flax]

yes

[rich wolf]

@upper karma think about it

the equation for a circle (radius 1, at origin) is x^2+y^2=1

if you try solving for y (or x, doesn't matter really) you get y^2=1-x^2

and when you take the square root it has to be plus or minus

so it's represented by two different functions

each composing 1/2 of the circle

[upper karma]

@rich wolf sorry i don't understand your explanation

Never mind i figured it out

[bitter ferry]

hey can anyone here help with trig?

[dark sparrow]

[bitter ferry]

"Consider a right angle triangle with the ratio sin 35 = ab. Draw a diagram and fill in as much information as you can about this triangle."

teacher never properly gave us anything so im kinda confused

[dark sparrow]

are you familiar at all with the basics of trig

perhaps the mnemonic SOH CAH TOA may ring a bell

[bitter ferry]

like the sohcahtoa?

yeah

[dark sparrow]

yeah so

can you draw me a right triangle where one of the angles is 35 degrees

(also are you sure you copied the problem correctly? bc i'm like 90% sure "ab" is supposed to be a/b but the slash got lost)

[bitter ferry]

it was copied from the doc

its alright if it isn't exact right?

[dark sparrow]

okay i guess this will do

i don't think you're expected to make a to-scale diagram

anyway

the somewhat unclear wording of the problem is making it hard for me to explain clearly how to continue

[bitter ferry]

yeah i know the teacher has his history

oh ive got it

the a/b didnt properly paste

[dark sparrow]

that's

...

ok

great

i asked about this

but ok at least we've got it figured out

[bitter ferry]

lemme actually screenshot it for you

should've done this earlier sorry

[dark sparrow]

uh huh

yeah

i suppose the intended course of action is to label one of the sides as a and another as b in a way that makes sin(35°) = a/b

[bitter ferry]

something like this?

[dark sparrow]

no, that side labeling will not result in sin(35°) = a/b

[bitter ferry]

am i meant to label the vertical side as well?

[dark sparrow]

no, you've just labeled the wrong sides

i mean

you'll end up labeling all three

[bitter ferry]

oh

[dark sparrow]

it's just that with the way you've labeled the sides now, you get a/b = sec(35°).

[bitter ferry]

so the b would be the vertical side

[dark sparrow]

no

[bitter ferry]

the a would be the one on the bottom

[dark sparrow]

no

that would give you tan(35°) = b/a

not what you want

[bitter ferry]

oh

and im trying to find a/b= sin(35°)?

[dark sparrow]

you're trying to find which side to label a and which to label b in such a way that makes a/b = sin(35°)

which should be pretty obvious if you know
(a) the mnemonic SOH CAH TOA
(b) what stands for what in said mnemonic

[bitter ferry]

he never told us about mnemonics

but the recent work we've done is working with theodolites

[dark sparrow]

you told me you were familiar with "SOH CAH TOA"

[bitter ferry]

yeah and how they are placed on a right triangle

hyp opp and adj works with this problem right?

is this how its meant to be?

[dark sparrow]

no

if sin(35°) = a/b, is a meant to be the hypotenuse, the opposite, or the adjacent?

[bitter ferry]

i guess the a is the opposite and b is the hypotenuse

[dark sparrow]

now look at your last picture and see that it's got the labels the wrong way around

[bitter ferry]

yeah

sorry bout that

[upper karma]

Anyone know about Awesome math program?

[upper karma]

Yes it’s awesome right

An awesome program 👏

Do you know about it?

[gray marten]

Do it know about you?

You do it about know?

Try #competition-math

[upper karma]

I m selected in it

to attend this camp

[fringe dirge]

I attended a long time ago

I could probably answer questions you have, but I've forgotten a lot of things since I went I think 7 years ago now

I also went to the Berkeley campus, so that might make a difference too

@upper karma

[quiet mason]

wow titu andrescuu is the director

[brazen depot]

in a "solve the triangle" situation where the givens are all 3 sides, i solved angle a using law of cosines and i get 22.08 degrees. Then, i used the sin(angle a )/side a=sin(angle b)/side b to get angle b, which was 25.02 degrees. However, my math teacher used sin(angle a)/side a=sin(angle c)/side c to get Angle C, which was 41.1 degrees. I tried solving her way and got 41.1 degrees for angle c too, and I triplechecked my method of solving for angle B. With her method, angle B would be around 116 degrees while with my method, angle C would be around 132.9 degrees. However, if I went along with trying to find Angle b first, why would I be wrong? Or would I be right? I know there can only be 1 triangle too.

(picture for context)

[upper karma]

your formula for law of cosines seems wrong

try it now

https://www.mathsisfun.com/algebra/trig-cosine-law.html

[brazen depot]

@upper karma Ohhh I just used a diff formula and Angle A is right. I'm just getting Angle B wrong (using law of sines)

I checked again, and I multiplied Angle B wrong. But when I multiplied it correctly, I got 66.3 degrees which is not 116.8 degrees 😅

[silent plank]

recheck your calculations for your application of the sin law.
it is also recommended that you apply the cosine law where possible instead of sine to avoid dealing with the ambiguous case

[brazen depot]

ambiguous case?

ok ill do the cosine law

[silent plank]

since there are two solutions to
sin(theta) = c; for 0<c<1 , 0 < theta < 180°

[brazen depot]

Ohhh

ty! ill try to do the cosine laws when possible then

Thank you!!!

[silent plank]

the algebra was also a bit dodgy
even applying the sine law you should've reached
sin(B) approx 0.8928

[brazen depot]

yes i realized i multiplied incorrectly (accidentally pressed only 9 instead of 19 in my calculator) and got 66.3 degrees

[silent plank]

seems there's another calculation error

[upper karma]

@silent plank but aren't there two solutions to cos(theta), as well?

[silent plank]

you should be getting
B ~ 63.2° or
B ~ 180° - 63.2° = 116.8°
and choose the appropriate one if possible.

not for 0 < theta < 180

[upper karma]

right

yeah

since it's a triangle

all angles must add up to 180

[brazen depot]

but why would u subtract 180 with 63.2?

[upper karma]

if it's an obtuse angle (>90 deg) you subtract it from 180

[silent plank]

to get the other solution for sin in that interval

sin(63.2°) = sin(180°-63.2°)

only using information from 2 sides and angles like that won't necessarily tell you whether the triangle is acute or obtuse

[upper karma]

because sin can have two answers, looking at the picture I posted above: https://discordapp.com/channels/268882317391429632/326138757474680852/704997866455826453

that's why sin(theta) = sin(180-theta)

[brazen depot]

I see, so becz side b is the biggest side, then angle b should be the biggest angle kind of thing

and that's why you choose 116.8 degrees

Thank you so much guys!! This makes a lot more sense now

[silent plank]

you should encounter questions that deal specifically with this later

[tardy junco]

Is arctan(tan(π+π/4)) equal to π/4 or undefined?

[upper karma]

well, what is the result of arctan(tan(x)) ?

[silent plank]

it would be pi/4

[upper karma]

how do you figure that?

what is the result of arctan(tan(0.3)), for example?

ah, ok

yeah

no

pi + pi/4 = 2pi/4 = pi/2

not "pi/4"

[silent plank]

OOPS: order of operations

[upper karma]

😦

[versed river]

pi+pi/4=4pi/4+1pi/4=5pi/4

[silent plank]

arctan(tan(x)) = x for -pi/2 < x < pi/2

arctan(tan(π+π/4)) = arctan(tan(π/4)) = π/4

[tardy junco]

@silent plank here is my reasoning.

F-1( F(x) ) = x is definition of inverse function.

And for inverse functions the range of F-1 must be the domain of F and the domain of F must be the range of F-1.

So if we use that logic, since π+π/4 is not in the range of arctan, this is not defined

[silent plank]

just skimming but that would just mean that arctan(tan(π+π/4)) isn't π+π/4

[upper karma]

f(x)
f^-1(x)

[tardy junco]

No what I meant to say is than tan(π+π/4) cannot be evaluated because it's not in the domain of this 'special' tan

[silent plank]

using those two statements above comes to the conclusion that it would be pi/4

[tardy junco]

I mean domain

[silent plank]

tan(pi + pi/4) = tan(pi/4) = 1 is still in the domain of arctan function

[upper karma]

pi+pi/4=4pi/4+1pi/4=5pi/4
@versed river
this looks ok to me

[versed river]

its correct

[silent plank]

however π + π/4 is outsider the range of the arctan function
and arctan(tan(π+π/4)) \neq π+π/4

[tardy junco]

But see this is tan which has all real numbers as its domain.

What I'm trying to ask is whether the tan which is the inverse of arctan is a different kind of tan with restricted domain

Because the inverse of arctan must have domain equal to range of arctan

[upper karma]

the inverse of arctan?

[silent plank]

not sure what you're getting at...

[upper karma]

the graph of the tangent function looks like this:

The domain of the function y=tan(x) ) is all real numbers except the values where cos(x) is equal to 0 , that is, the values π2+πn for all integers n . The range of the tangent function is all real numbers.

https://www.varsitytutors.com/hotmath/hotmath_help/topics/trigonometric-functions

[silent plank]

wdym by different type of tan

[upper karma]

how do you calculate the "inverse of arctan"?

because arctan IS the inverse of tan

inverse of an inverse?

[tardy junco]

Okay wait.

Arctan(tanx) is equal to x when x belongs to the domain of tanx and hence the domain of arctanx.

π+π/4 isn't in the range of arctanx and hence arctan(tanx) cannot be evaluated

Is what I am asking

[silent plank]

Arctan(tanx) is equal to x when x belongs to the domain of tanx
and hence the domain of arctanx.
such that tan(x) belongs in the domain of the arctan function

[upper karma]

what makes you think it can't be evaluated?

[silent plank]

tan(pi + pi/4) = tan(pi/4) = 1 is in the domain of arctan function

[tardy junco]

@upper karma because I thought that when tan is inside arctan as arctan(tanx) then the domain fo tanx would be -π/2 to π/2

[upper karma]

arctan(tan(x)) = x

[silent plank]

the domain of tan(x) is R \ kpi + pi/2 (where k is an integer)

[upper karma]

arctan(tan(x)) = x

[versed river]

That's only true for values within a certain domain

[silent plank]

arctan(tan(x)) = x for -π/2 < x < π/2
arctan(tan(π+π/4)) = arctan(tan(π/4)) = π/4

[tardy junco]

F-1( F(x) ) = x is definition of inverse function.

And for inverse functions the range of F-1 must be the domain of F and the domain of F must be the range of F-1.

Correct?

[silent plank]

the domain of
arctan(tan(x)) would be all reals except for kpi/2
the range of
arctan(tan(x)) would be (-pi/2, pi/2)

tan and arctan aren't true inverses.
pi + pi/4 isn't in the range of arctan and as i stated at the very start, all you could really conclude from that was
arctan(tan(pi + pi/4)) \neq pi + pi/4

[tardy junco]

So tan has a different domain than the range of arctan. That is fine?

I used to think it would have restricted domain

When it was inside arctan

To apply arctan tanx = x

[silent plank]

to be able to apply the definition, you could apply period shifts

i.e. tan(pi + pi/4) = tan(pi/4)

[tardy junco]

Hmmm. Okay thanks.

[silent plank]

,w arctan(tan(pi + pi/4))

[somber coyote]

Display results online and refine query

Requested by ramonov#0921

Display results online and refine query

Requested by ramonov#0921

[drifting parrot]

@pallid cloud I was able to figure my problem out

[pallid cloud]

Great @drifting parrot

[granite elk]

I have a vectors question

Really simple stuff but i just cant do it

How do i show that they have a common point

and that its a bisection

[tardy junco]

@silent plank wait so arctan ( tanx ) = x, holds for any real value of x?

[silent plank]

no

holds for -pi/2 < x < pi/2

[tardy junco]

But see for any real value of x, tanx gives a value that is inside the domain of arctanx

[silent plank]

(except for npi + pi/2)

[tardy junco]

Yeah that

So x belongs to R - {nπ/2}?

[silent plank]

that would be the domain of tan(x)

[tardy junco]

Yeah but for every number in domain of tan(x), tan(x) takes a value that is in the domain of arctan(x)

So arctan(tan(x)) = x holds for X belongs to R -{nπ/2}?

Right?

[silent plank]

no

since the range of arctan is (-pi/2, pi/2)

it won't output anything outside of that

[tardy junco]

But we just discussed earlier that arctan tan(π+π/4) ) is equal to π/4

[silent plank]

hence, why arctan(tan(x)) = x is only true for -pi/2 < x < pi/2

yes

[tardy junco]

So we can use that logic for any number

And find arctan(tanx)

Even if that number does not belong to -π/2 to π/2

[silent plank]

yes

[tardy junco]

So then arctan(tanx) equals x for any x that belongs to domain of tanx

[silent plank]

no

[tardy junco]

What why

[silent plank]

we established that:
arctan tan(π+π/4) ) is equal to π/4 and not pi + pi/4 right?

[tardy junco]

Oh

So arctan(tanx) is defined for any x belonging to domain of tanx but it is not equal to x necessarily

[silent plank]

apply appropriate shifts such that the angle is within that interval

and then apply the inverse definition over the restricted domain

yes.

[tardy junco]

Okay thanks a lot

And by the way I think tanx is undefined for nπ + π/2 not nπ/2

😅

[silent plank]

fk typo

somewhere

yeh, my mistake

[tardy junco]

It's fine

Thanks a lot

👍

[granite elk]

Chanel cleaR?

anyone knows how to answer this?

[upper karma]

uhh

Definitively, to bisect a line with a Bisector means to divide it into two equal parts. The bisector doesn’t have to divide into 2 equal parts unless they are bisecting each other, right?

[granite elk]

Means both lines intersect at their midpoints

they bisect one another

so they both bisect

[soft gulch]

What have you tried

[acoustic jungle]

Try proving MSNT is a parallelogram or something.

[soft gulch]

@granite elk

I'll give you some hints

BS=TA

MO=CN

[upper karma]

MD= DN and SD=DT ?

No?

[soft gulch]

That's what's needed to be proved

[granite elk]

BS=TA i already know that

but its still confusing

How do i show it though

I know that SC=OT

[soft gulch]

Oppo sides of parallelograms are equal

There are multiple ways to solve this

[upper karma]

@pastel anchor use inscribed angle theorem on A

Notice that CBE=20+DBE

[pastel anchor]

alright i found answer its 30

[upper karma]

Umm

The angle of ?

[pastel anchor]

DBE

[upper karma]

Yes

They are asking you CBE remember

So yeah :)

[pastel anchor]

thanks

[upper karma]

Np

Do you need more help on this one?

[pastel anchor]

yea

[upper karma]

Ok wait

Notice that on a circle, 360° total angle on an arc

If you have 250°

Get arc BA

[pastel anchor]

110

[upper karma]

Yeah

About the next step im not sure, im thinking that x is 110 but im not sure

Wait

I have a question

I've been stock on this question its been 30min

[bronze saddle]

The angle will be 125 degrees

Its half of whats BYA

Its a theorem

[upper karma]

@bronze saddle yeah i just found iy

Determines the exact value of each trigonometric ratio. Show your reasoning (using the remarkable triangles and CAST)

If π is 180

Multiply by 5

And divide by 3

[pastel anchor]

WHATS the theorem?

[upper karma]

Wait

Im pretty sure is this one

[bronze saddle]

I just looked at it and the theorem doesnt apply

Since x is not inside the circle

[upper karma]

Idk if this can help you

Yeah

5π/3 is 300°

Bc of what i said before

Due to the unit circle

Sin 300°= (-√3)/2

[bronze saddle]

Yeah x is 300

[upper karma]

@bronze saddle which problem m8

[bronze saddle]

The zezzez question

First one

[upper karma]

Yeah

But zezexx mmm

Oh yeah

$x=\frac{1}{2}\cdot(BYA-BA)$

[somber coyote]

Al3dium:

[pastel anchor]

where did u get 720 wtf

[upper karma]

Yeah tf

[bronze saddle]

Total angle of a shape which has 4 sides?

[upper karma]

Id say x=70

According to the theorem i posted

[pastel anchor]

a 4 sided angle has 360 no?

[upper karma]

But there is no x

[bronze saddle]

Wait nvm me

Yeah 360

[upper karma]

@upper karma bruh im talking to zezexx

Ohh ok

Your question was already solved above

[bronze saddle]

Whats your questions @upper karma

[upper karma]

Already solveeddd

Determines the exact value of each trigonometric ratio. Show your reasoning (using the remarkable triangles and CAST)

Bruh

Above

I've already solved yours above

@pastel anchor did you understood my theorem

[pastel anchor]

yea i looked at it but the x is the angle of those two opposite tangents

[upper karma]

Yes but they are congruent angles

So x=70

Determines the exact value of each trigonometric ratio. Show your reasoning (using the remarkable triangles and CAST)

Can someone help me for this one. Please ?

@pastel anchor

$\theta=\phi$ and $\alpha=\beta$

[somber coyote]

Al3dium:

[upper karma]

Determines the exact value of each trigonometric ratio. Show your reasoning (using the remarkable triangles and CAST)

?

[novel flax]

$/frac{1}{2}$

\frac{1}{2}$

[upper karma]

What were you trying lmao

[novel flax]

latex

[upper karma]

Ikr

But $\frac{1}{2}$ doesnt have anything to do with jpierre3 problem

[somber coyote]

Al3dium:

[unborn jacinth]

I don't really understand how this works

cos(theta) is x/(the first part of x')

so why are they multiplying by x?

[upper karma]

@unborn jacinth bc they got it from

$cos \theta=\frac{x'}{x}$

[somber coyote]

Al3dium:

[unborn jacinth]

how

[upper karma]

$x'=x\cdot \cos (\theta)$

And then the rest of x'

[somber coyote]

Al3dium:

[unborn jacinth]

yeah but why is cos(theta) = x'/x

[upper karma]

Definition of cosine

Its a triangle

[unborn jacinth]

ok I think im not seeing something here

afaik, cos(theta) is that

bottomblue/topblue

[upper karma]

$\cos(\theta)=\frac{adjacent}{hypothenuse}$

[unborn jacinth]

ohhh but the bigger triangle

[somber coyote]

Al3dium:

[upper karma]

Did you got it? :)

[unborn jacinth]

oh yah

ok thx

[upper karma]

Np :)

S = O/H C = A/H T= O/A

Late

xD

mhm

[acoustic jungle]

x' = xcos(theta)+ysin(theta) from similar triangles

[soft gulch]

Soh cah toa is the weirdest shit

[acoustic jungle]

cos(theta)*x is not x'

[upper karma]

Also from definitions

yea but it is a good memory trick

@acoustic jungle ikr its like a way of visualising

He seems to have got it so

[acoustic jungle]

What?

[soft gulch]

Just visualise it

[upper karma]

Yeah

[soft gulch]

Much easier

[upper karma]

??

[upper karma]

Nvm just next question

[acoustic jungle]

why are you setting these equal

[upper karma]

ahh

[acoustic jungle]

when they aren't

[soft gulch]

The functions

[upper karma]

Ikr it was a way of visualising

[acoustic jungle]

do you know what ikr means

[upper karma]

I know already

Although it doesnt fit in tho

lol

[acoustic jungle]

That's because it means "I know right"

not "I know already"

[unborn jacinth]

lmao

[upper karma]

And then the rest of x'
Look

bepis

Thats what i mean

[soft gulch]

i ned bepis

[upper karma]

same br00ther.

[soft gulch]

big amgrey

[upper karma]

do you know what ikr means
@acoustic jungle oh rlly, idc bc its kinda the same :)

[acoustic jungle]

j.

[upper karma]

.

I helped like 30 ppl today lmao

Has to be my record

In one day

good job

deserve helper tag perhaps