geometry-and-trigonometry

[vital frost]

LW and CD I'm a little shaky about

[keen aspen]

Not sure how to calculate CD if you dont have an angle value

Also you can use Law of Cosines to find LW

[vital frost]

Yeah and did u get 4.3?

[keen aspen]

LW^2=5.1^2+5.1^2-2(5.1)^2cos(50)

Or better yet law of sines because of isoceles

== sind(65)

[charred spear]

sin(13×π/36) = 0.90630778703665

[vital frost]

Ye I used sine

[keen aspen]

==sind(65)/5.1

[charred spear]

10×sin(13×π/36)/51 = 0.177707409222873

[keen aspen]

==sind(50)/(0.177707409222)

[charred spear]

500‚000‚000‚000×sin(5×π/18)/88‚853‚704‚611 = 4.3107062697763

[keen aspen]

Yeah 4.31

[vital frost]

ye

So back to the other problem

Problems*

How do u get that

A,B, and C

[keen aspen]

Look at the triangle with 50° in correlation to LW

[vital frost]

Which im not sure what to do

[keen aspen]

What's the max value for LW if you were to make the angle as big as possible

[vital frost]

🤷

[chrome fiber]

cant you just use the triangle inequality

and the reverse inequality

to find the range

[vital frost]

I'm pretty sure

[chrome fiber]

instead of whatever ur doing, pjs

[vital frost]

@chrome fiber yeah I asked other people and they said they used sine

Which we have not learned about

[chrome fiber]

yeah no

[vital frost]

Yet

[chrome fiber]

screw that

you dont need it if you havent learned it

[vital frost]

yeah so what do I do to find it without sine

[chrome fiber]

for a triangle with sides $a, b, c$, the range of any side is given by $|b - c| < a < b + c$

[somber coyote]

june:

[chrome fiber]

triangle inequality basically says that the length of any side of a triangle is always going to be less than the sum of the other two sides

[vital frost]

So...

yeah basically my geometry teacher taught this for one day

[keen aspen]

Bruh that's legit where I was getting at

0<LW<10.2

[chrome fiber]

reverse inequality says that the length of any side of a triangle is always going to be greater than the absolute value of the difference of the other two sides

idk pjs, you were confusing the poor guy

[vital frost]

And she just went over it fast because it was going to be thanksgiving break

[chrome fiber]

so suppose you have a triangle of sides a, b, c

take side a

triangle inequality says its going to be less than sum of the other two sides

[vital frost]

So I'm shaky on how to get CD, DL, and LW

[chrome fiber]

which is b + c here

so you have a < b + c from that

reverse one says

any side will be greater than absolute value of the difference of other two sides

so take a

absolute value of difference of other two sides is |b - c| or |c - b| (order doesnt matter here)

so a > |b - c|

you can combine those two inequalities

$|b - c| < a < b + c$

[somber coyote]

june:

[vital frost]

uh brain is hurting

[keen aspen]

Notice that if you were to make that angle of 50° all the way out to 180°(but not yet 180 due to it still being a triangle), it would approach a line segment of the two sides added together

[vital frost]

So how do you get CD, DL, and LW

Without sine

[keen aspen]

And vice versa with making your angle approach 0 will allow LW to be the two side lengths subtracted

[jade solar]

So, In ACD 4.2 - 2.5 < CD < 4.2+2.5

[keen aspen]

https://www.khanacademy.org/math/geometry/hs-geo-foundations/hs-geo-polygons/v/triangle-inqequality-theorem

Here's a good representation

[vital frost]

The answer would be nice

[keen aspen]

0<LW<10.2

[vital frost]

Is that for the ABC part?

[keen aspen]

Well C as a compound inequality

[vital frost]

Ok

So a, and b?

[keen aspen]

Upper bound for length of LW is 10.2 so LW<10.2 for a

[vital frost]

Ok

[keen aspen]

And LW>0 for lower bound

[vital frost]

So how would you find CD and LW without Law of sine and cosine, and what are the answers @keen aspen

[keen aspen]

You cant really

Unless the triangles were right

You could use pythagorean theorem but nope

[vital frost]

so u do nothing?

[keen aspen]

Well you can find the length but not without trig

[vital frost]

Yeah how do u find that

Or is sine the only way?

We got another one to do also

But it doesn't have more steps to it

It's just solve

[keen aspen]

It's easy to find PR and RJ

The other two not so much

[vital frost]

PR also 1.9

[keen aspen]

Yes

[vital frost]

I'm guessing RJ 1.7

JM and MV not sure

@chrome fiber could probably help on this one?

[keen aspen]

Yeah maybe June can find the lengths, idk if its possible

[chrome fiber]

which ones?

[keen aspen]

Oh wait maybe I'm stupid

[vital frost]

Recent Post for JM and JV

[chrome fiber]

its asking for a range

[vital frost]

Yes

[chrome fiber]

just use what i told you

[keen aspen]

KMJ could be 28° maybe?

[chrome fiber]

$|b - c| < a < b + c$

[somber coyote]

june:

[chrome fiber]

for both of those triangles

[keen aspen]

Oof range lol nvm okay yeah

[vital frost]

That's what I'm supposed to use for the 1st too lmaoo

Answers would be nice like I said

[keen aspen]

Wow I need to read questions better

[vital frost]

Because.my geometry barely went over this

And it's due tues

[keen aspen]

Look at PR

[vital frost]

Uh huh

[keen aspen]

The shortest length would be 1.9-1.1

And the biggest length would be 1.9+1.1

[vital frost]

So 0.8 and 3.0

Ok

[keen aspen]

Okay so do the same for the other triangles

Calculate the ranges

[vital frost]

So MV the longest ranglr would be 4.8 and the shortest would be 0.6

@keen aspen yes?

For rj idk what to do

[keen aspen]

Correct

The side lengths are 3.6 and 3.6

So 0 and 7.2

[vital frost]

Ok

[keen aspen]

And now JM

[vital frost]

And JM?

[keen aspen]

2.1-1.7

2.1+1.7

[vital frost]

0.4 and 3.8

That's all u had to do?

[keen aspen]

Now add up all the lowest bounds

And add up all the highest bounds and thatll be your range

[vital frost]

Ok

Wow

I made this so complicated then it should be smh

[keen aspen]

Lol

[vital frost]

I really appreciate y'all helps @keen aspen @chrome fiber

[chrome fiber]

np

[vital frost]

life savers lmao

[jade solar]

@clear haven I ended up doing something like this

Does your method with vectors do something like this?

[left folio]

"Show that x^2 +y^2 = r^2 is orthogonal to y = mx. Conclude that the family of circles centered at the origin is an orthogonal trajectory of the family of lines that pass through the origin. Note that there is a technical issue when m = 0. The circles fail to be differentiable when they cross the x-axis. However, the circles are orthogonal to the x-axis. Explain why. Likewise, the vertical line through the origin requires a separate argument."
Do i just plug in some numbers for x and y and graph it or is there a formula i don't know of?

[left folio]

Please some help

[patent robin]

wait wat

so what's the question

do they want you to prove that the circles are orthogonal to x axis

[chrome fiber]

@left folio you have to show that the slope of x^2 + y^2 = r^2 times the slop of y = mx is -1

[left folio]

Since y'=-x/y i understand that but how

[chrome fiber]

an intuitive approach would be to think of all y = mx as all every straight line passing through the center and since the circle is centered at the origin, every single line must be a radius to the circle

which is always tangent to any point on the circumference

from (I)

$y' = -\frac{x}{y}$

[somber coyote]

june:

[chrome fiber]

from (II)

$y' = m = \frac{y}{x}$

[somber coyote]

june:

[chrome fiber]

what's the product of those two?

[left folio]

Sorry but the product of what exactly

[chrome fiber]

the two slopes

[sick veldt]

Is 16 correct?

I did

12^2 - 20^2

= 256

squared down

= 16

[left folio]

@chrome fiber I see that m is the slope in the equation y=mx but i don't see any other slope in the equation y'=-x/y

Sorry if i sound stupid but i don't know

[chrome fiber]

@sick veldt use another question channel below

@left folio well yeah

so, the slope for the circle is y' = -x/y, right?

[left folio]

Yeah

[chrome fiber]

and the slope for the line is m

but you also have this

y = mx

[left folio]

Do i count them both as functions?

[chrome fiber]

so m = y/x

[left folio]

And multiply using the function's rules and so on?

[chrome fiber]

not the circle

[left folio]

ok

[chrome fiber]

but anyway

so for

slope circle = -x/y

slope of line = m = y/x

you need to prove that slope of circle times slope of line = -1

for orthogonality

[left folio]

Mmmh i understand

[vagrant bolt]

does anyone here know i

anyone here know to anwser this

[last pollen]

tan (90-x) =sin(90-x)/cos(90-x)

sin(90-x) = cos(x)

(I think)

and so on

[vagrant bolt]

in terms of tan x

[last pollen]

cos(90-x) = sin(x)

so it's going to be 1/tan(x)

and so on and so forth

[vagrant bolt]

how does cos(90-x) = sin(x) go to 1/tan(x)

[last pollen]

that's not what I meant

tan(90-x) = cos(x)/sin(x) = 1/tan(x)

[vagrant bolt]

thanks what about the 4d)

[last pollen]

well you know sin and cosine are periodic functions

so you can subtract 360 from the functions argument

[vagrant bolt]

is the arg the brackets rihgt

[last pollen]

the argument is what's between parentheses

[vagrant bolt]

so sin(540-x)/cos(540-x) --> sin(180-x)/cos(180-x)

[last pollen]

right

[vagrant bolt]

what now

[last pollen]

you should be able to figure out yourself

👀

[vagrant bolt]

ill try but im pretty retarded on this topic

[last pollen]

lol

well it's just about applying formulas

do you have some formula in your notes

that could help?

[vagrant bolt]

is 4d tan(-x)

the book says -tanx but im assumin tan(-x) is the same thing

yea thanks a lot i finally understand this shit

tbh the topic is acctually fine its just those questiosn which confused me the fuck out

[jade solar]

As a visual person I find answering questions like this easier when I look at the graphs and try to figure them out

[vagrant bolt]

yea i plot them on desmos

@last pollen the -360 thing for the last one, does that work for the

would that work for them all

[last pollen]

yes

all of them are periodic functions

and yes tan is an odd function

so tan(-x) =-tan(x)

this comes from tan being the product of two functions

one is even the other is odd

so you get an odd function

sin(x) is the odd one

cos is even

[vagrant bolt]

ok so the first one is 1/tanx , but if i minus 360 from sin(90-x)/cos(90-x) --> sin(-270-x)/cos(-270-x)

and then do tan(-270-x) --> -180 --> tan(-450-x) --> ?

@last pollen

[last pollen]

that doesn't help much

it's not wrong, but won't help you with this excercise

you should follow blah's advice and make a drawing

[somber crane]

I see you cardi 👀

[clear haven]

@jade solar my method with vectors is slightly related but is very very different

just dw about it haha

[vagrant bolt]

how is tan(270-x) --> 1/tanx and tan(360-x) --> -tanx!

[lament bay]

did I do this right?

[glad ocean]

mostly but right AC = AC

because cf is an extension of ac

but AC is the actual sides of the triangle

@lament bay

[lament bay]

Oh my bad thanks!

[glad ocean]

and DAC and BAC should be congruent by substitution

[lament bay]

Really?

[glad ocean]

yes

[lament bay]

How and when should I use substitution property?

[glad ocean]

ok so (this might take some time to type):

After vertical angles

<BAC ~= <CAD

~=*

by that I mean congruent

[lament bay]

Can you help me on #2?

[glad ocean]

ok

so everything given is given

[lament bay]

Yes

So is it ASA?

[glad ocean]

EF = EF because it is reflexive

yes

[lament bay]

So if shapes connect within each other via lines? It always means reflexive ?

[glad ocean]

yes for the segment that is on both triangles

[lament bay]

So it’s like that?

[vital frost]

Wait @keen aspen how do u her CD DL and LW again?

[glad ocean]

@lament bay after Ef = Ef:

De + ef = cf + ef

[vital frost]

Add what and what

[lament bay]

Segment addition postulate?

[glad ocean]

yes

Then DF = CE

by substitution

[lament bay]

I still don’t get when you use substitution

[glad ocean]

let's say for example:

[keen aspen]

@vital frost your range

[vital frost]

Yes

[keen aspen]

LW is from 0 to 10.2

[vital frost]

But what numbers

Oh ok

DL is?

[glad ocean]

@lament bay 1 + 2 = 2 + 1

3 = 3

by substitution

[lament bay]

How do I know on a triangle tho?

[glad ocean]

Use it after you use addition property to combine shorter segments into a longer segments

[lament bay]

So it almost always comes after segment addition postulate?

[vital frost]

So DL and CD @keen aspen ?

[glad ocean]

https://www.dummies.com/education/math/geometry/the-transitive-and-substitution-properties/

[vital frost]

Would it be 4.8+3.1 and 4.8 - 3.1

[glad ocean]

@lament bay use the article above

[keen aspen]

Yep

[lament bay]

Ok give me a sec

[vital frost]

And for CD...

4.2 and 2.5? @keen aspen

[keen aspen]

Yep!

[glad ocean]

brb

[vital frost]

Ok cool, thanks

[lament bay]

@glad ocean I’m really bad at this 😦

[reef inlet]

are you posting this from a class...

lol

[lament bay]

Yea I’m doing it in my chemistry class

We’re having free time atm

[glad ocean]

nah ur not bad

Everyone learns at their own pace

it's really a matter of studying

[lament bay]

I always never know the next step after given lol

Like I don’t see the angles or theorems I guess haha

@glad ocean so what should I do for the next step?

[alpine solstice]

I don't think you technically need the rest of that to prove AB = CD

B and D are both given to be right angles, so A and C must be right angles as well

Which makes this a rectangle, and we know opposite sides on a rectangle are equal in length

Or, since BF and ED are equal, that means the hypotenuse for the 2 right triangles are the same, which means the lengths are going to be the same as well

[nocturne gate]

This

Answer

Now

Please

[keen aspen]

Exterior angle theorem

The sum of the two opposite angles for 1

[nocturne gate]

wh

[keen aspen]

80+60

= Anglle 1

[nocturne gate]

So 140

[keen aspen]

Yes

[nocturne gate]

thank you, I’m helping my friend out with their math work.

I just didn’t understand it cause it’s work for a couple grades above me

[keen aspen]

And you decided to agree to helping him?

[nocturne gate]

well like I don’t want her to get bad grades cause they’re cool as heck, they asked the server I’m in and I decided to step in and help.

[eternal root]

Help

Please🙃

[chrome fiber]

what have you tried?

[keen aspen]

Don't fall for it june

It's a meme question

[hazy granite]

meme question?

[keen aspen]

Yeah

You can't solve for x

[hazy granite]

oh

hm

[keen aspen]

Well you can, but you have to use protractors and stuff

[hazy granite]

you cant find the value of the angle opposite of x

[hazy granite]

uh

its quite blurry

cant see the text

[keen aspen]

@umbral snow Try it!

[hazy granite]

actually maybe you could make a simultaneous equation

unless the one im thinking of are equal

okay actually neverind

40+x+50+y+(20+60+50+30)=360

x+y+70=180

welp nevermind

these 2 are equal

[left folio]

@chrome fiber " For k ̸= 0 and c ̸= 0 show that y^2 −x^2 = k is orthogonal to yx = c. In the case where k and c are both zero, the curves intersect at the origin. Are the curves y^2 −x^2 = 0 and yx = 0 orthogonal to each other?" Sorry for the @ but i'd like to know if i can just say that they're not orthogonal because their product can't be -1 or what else?

[chrome fiber]

If the product of their slopes is not -1, then they’re not orthogonal

Assuming ofc that both slopes are defined

[left folio]

Thanks again, sorry for the @ again 😃

[versed wagon]

someone please help me, i have no idea how to do this

A 8,-6 and B 5, -2 are 2 points in what retion does the line 6x - 8y - 71 = 0 divide [ab]

plsss :(

[sick veldt]

Is my answer correct

[hard gale]

yes iirc what a centroid is

[sick veldt]

Okay how about this one

just do 32*3

96 = AD?

[hard gale]

yes 96$

[sick veldt]

This one is 52?

104/2

52 = NO

[hard gale]

yes maybe consider trusting yourself also

[sick veldt]

I wanna get an exact 100 😛

this test boosts my grade to an A

[hard gale]

k

[sick veldt]

hello?

its only this and one more question left

[hard gale]

yas

[sick veldt]

oops

[hard gale]

ye

[sick veldt]

kk ty

[stable tusk]

hey guys

im given cos x = 1/3 in quadrant 4
how can i find sin (x + pi/3)

hello?

[jade solar]

Use the sin(a+b) formula, find sin x using sin^2 + cos^2 =1?

[lament bay]

can someone help me with some proofs?

<@&286206848099549185>

[keen aspen]

Can I use trig

[lament bay]

I’m not sure

I think I just need to use theorems

[keen aspen]

B is bisected from BF, that means if you were to extend BC and AD into transversals, you can say that EBF and AFB are congruent from alternate interior angles

[lament bay]

How would I write that with symbols?

What do you mean by transversal?

[keen aspen]

Eh, I don't feel like explaining perhaps someone else can help

I am going to sleep

[lament bay]

ok

can someone show me the next step

<@&286206848099549185>

[stable tusk]

sinx=0.40

i get 0.41 for the first

how can i find the other angle

apparently the answer is 2.73 but how do i do it

[dire rampart]

$\sin(x)=\sin(\pi-x)$

[somber coyote]

lemon catto:

[stable tusk]

How do i make this 1-2sin(x/2) equal to 1-sinx

[last pollen]

you can't

are you sure the equation is right?

[torn linden]

Someone help me on 5-7

[signal hemlock]

sure @torn linden

so do you know what the angle of elevation as opposed to angle of depression means?

i believe the wording is what causes the problem for you right?

@torn linden

[torn linden]

No no

So I think #5 is correct

I know that I'm using the correct function on #6, but whenever I set it up I get 0.65

Which is wrong

I'm also confused on #7

[signal hemlock]

yeah on 7

Which part are you stuck on?

What have you drawn so far

[torn linden]

Wait lets start with 6

[signal hemlock]

I gotta go bro

pick 1

[torn linden]

Ok 7

[signal hemlock]

yeah

What do you have so far on it?

[torn linden]

So far I have this?

[signal hemlock]

draw 2 buildings

and label what you got and share here

this is what you got?

https://i.imgur.com/AfXuuiU.png

[torn linden]

Yeah

[signal hemlock]

and you know what angle of declination is

depressiom

[torn linden]

Like that

Yeah

[signal hemlock]

so you have this?

https://i.imgur.com/VrpbZnY.png

[torn linden]

Yes

[signal hemlock]

green is angle of depression 15 deg

yeha and whats the problem?

[torn linden]

Ohhh

So I find the opposite side to 15 degrees

[signal hemlock]

https://i.imgur.com/4meUhPi.png

solve for x

done

[torn linden]

Right

then 207-x

that length of small building

[signal hemlock]

yeah

[torn linden]

ok can we do 6?

[signal hemlock]

i cant see waht info it has

im on phone

you going to have to type it

[torn linden]

Ok

Find the value of x

Where depression is 18 degress

[signal hemlock]

give me a figure

or something

[torn linden]

ok

[signal hemlock]

zommed in preferable

math on phone is hard ok

im on buss atm

[torn linden]

lol

[signal hemlock]

ok u have bottom part 2km?

and other side x

[torn linden]

yes

[signal hemlock]

and angle?

[torn linden]

depression is 18

internal top one is 72

[signal hemlock]

and you want to solve for x?

[torn linden]

yes

[signal hemlock]

wait waht

do you know how to solve for x?

[torn linden]

yes

[signal hemlock]

then what is the question?

[torn linden]

but when I solve I get 0.65

[signal hemlock]

the angle inside the triangle

you got from 90-depressionangle?

[torn linden]

yes

[signal hemlock]

and that angle is 72 deg?

[torn linden]

yes

[signal hemlock]

and which trigonometric expression combinds opposite and nearby?

[torn linden]

tan

[signal hemlock]

yeah

then wtf?

[torn linden]

so its tan(72) = 2/x?

[signal hemlock]

2km it says?

thats 2000 meters

[torn linden]

Ohhh

lol

Im dumbo

[signal hemlock]

you need to read up on SI prefixes

[torn linden]

Yeah

[signal hemlock]

so now you got tan 72 deg = 2000/x

and now you can rearrange

to solve for x

[torn linden]

ok

[signal hemlock]

ok?

[torn linden]

yes thanks

[signal hemlock]

ok np

my laptop died but im plugging in powerbank now

[torn linden]

lol

[loud cedar]

mitz#9391

Avatar URL Link

[signal hemlock]

am i being hacked again

who did this

[torn linden]

lol

hey @signal hemlock would I use tangent for #7?

[signal hemlock]

well

you dont have to involve the hypotenuse

which means to fastest way is to use tan

the hypotenuse is just a vector in this case

of the shortest distance

[torn linden]

So it is tan right?

[signal hemlock]

well you have nearby

thats the distance between the buildings

nearby = that length u got (forgot what it was)

and x is the delta in height between the buildings

[torn linden]

51

Ok

[signal hemlock]

which means

tan the angle of depression

= x/distance between buildings

then solve for x

[torn linden]

ok thanks

[meager lantern]

9 Lie (?)

[keen aspen]

Proportion

8/20=6/(6+x)

[meager lantern]

ah It's 9

I doubted my answer for a second

You can actually take the ratio between the two segments and not the full sides if I recall correctly

[upper karma]

this is equivalent triangles right

[somber coyote]

omar23:

[meager lantern]

If you would like, you can solve it by cross multiplication after simplifying 12/8 to 3/2:

[upper karma]

wouldnt that be assuming the hypotenuse is the same as that base

[quaint ruin]

Oh thanks

[upper karma]

still is 9 though apparently

[somber coyote]

omar23:

[meager lantern]

Viola!

But you have a good point

When I was his age I was taught that this is a stupid idea, yet it works in some cases.

or so I remember

[upper karma]

20/(6+x)=12/x

[meager lantern]

I initially deleted my answer of 9 because I did this and I remember it might be stupid

[quaint ruin]

Thank you so much!

[meager lantern]

That's the right way @upper karma

But my way works too

[quaint ruin]

oof

[meager lantern]

[quaint ruin]

I have a whole packet to do

Full of stupid shit we didn’t learn

[meager lantern]

What year are you in

[quaint ruin]

9th

[meager lantern]

I remember this was like year 9 stuff

yep

[quaint ruin]

lol

I’m taking stats next semeter

semester*

I wonder why I don’t take algebra 2

🤔

[meager lantern]

You didn't take algebra 2?

[quaint ruin]

nop

[meager lantern]

Hmm

come #chill

[meager lantern]

ABD

[quaint ruin]

Tf

[meager lantern]

wait

[quaint ruin]

But ABD is only half

[meager lantern]

I think that's a lie

[quaint ruin]

Lol

[meager lantern]

My tactic to dealing with those

is to try to align them in my head

[quaint ruin]

Lol

[meager lantern]

ADB?

What do you think

[quaint ruin]

Uh

I think BCA

But i think I’m wrong

[meager lantern]

Oh wait we're not bound by the smaller triangle

[quaint ruin]

Nope

[meager lantern]

I support BDC

[quaint ruin]

LOL

[meager lantern]

Which isn't even an answer choice

LOL

pair of idiots working on something so simple

[quaint ruin]

Yeppp

I wonder how I even got into this class

Everyone in my class has taken geometry A

I’m stuck learning geometry b without learning any of part a

[torpid horizon]

pretty sure its adb

only one with the second angle being a right angle

[quaint ruin]

Ughh

ur maybe right

[meager lantern]

I support that idea

I was looking for the same thing

instead of my useless mental alignment

[torpid horizon]

you just have to see which letters correlate with which one

[meager lantern]

I'm too old for this

at a point where I know neither calculus or algebra

[quaint ruin]

Lol

[supple haven]

Can someone show me how to prove this please?

#63

<@&286206848099549185>

[upper karma]

distributive property

fundamental to your success in math

(a+b)(a-b) = a^2-b^2

@supple haven

[supple haven]

I did that but I ended up getting

cos - cos^2 = sin^2

[upper karma]

no

you shold get 1-cos^2

distribute the 1

1*1 is 1
-cos

then distribute cos

+cos - cos^2

the + and - cos cancel

1-cos^2

hence (a+b)(a-b)=a^2+b^2

[supple haven]

but where did the 1 come from? thats whats confusing me

@upper karma

[upper karma]

thats the distributive property

foiling

=pup foil method

[charred spear]

Wolfram|Alpha didn't send a result back.
Maybe your query was malformed?

[upper karma]

t!yt How to foil binomials

[loud cedar]

📽 | ** https://www.youtube.com/watch?v=fXkRDrpFsNo **

[upper karma]

this guy

watch that you'll be math pro

@supple haven

[quaint ruin]

help

[torpid horizon]

just saying but its difference of squares

for the + and - binomial

[languid lodge]

@quaint ruin What does HL mean?

[torpid horizon]

and @quaint ruin

its C

[quaint ruin]

HL means hypotenuse leg

[torpid horizon]

no not c

sorry

its A

[quaint ruin]

It’s HL

[torpid horizon]

i misread the leetters

[quaint ruin]

Can’t decide between a or b

Lol

[torpid horizon]

side JL and AC

congruent

so its jlk

[quaint ruin]

But

It’s b

Right?

[torpid horizon]

wait

im sped

no

its jkl

[quaint ruin]

bcuz it correspond

[torpid horizon]

i looked at the diagram again

yea ur right

[quaint ruin]

lmaooo

[supple haven]

ohhhhhhh crap I see what I did wrong now

[torpid horizon]

i keep looking at the picture wrong

[supple haven]

I forgot to include the 1 like a big doof

[languid lodge]

@quaint ruin Isn't a and b the same thing?

[upper karma]

learning @supple haven

[quaint ruin]

no

bcuz it has to correspond with ABcy

ABC

To be similar

Help with this

I forgot

how to do this

[torpid horizon]

i believe you use pythagorean theorem for that one

[quaint ruin]

I think it involves

Yea that thing

but

[torpid horizon]

ok i need to domy hw

[quaint ruin]

U know what I do

?

3.6 divided by 1.6

And then use that proportion

[lament lantern]

Similar triangles I think

[quaint ruin]

I got 8

[languid lodge]

Yes it is B

[torpid horizon]

i read ur picture again and i dont think you use pythagorean theorem

idk how u do that

[quaint ruin]

I did 3.5 divers by 1.6

Then did 17.5 divided by 2.2

[languid lodge]

Use the ratios for both triangles.

[quaint ruin]

LOL

yep

[torpid horizon]

o

right

proportions

[supple haven]

multiply by 5

the dumb way (my way) was to add 3.5 until I got to 17.5 then count the amount of times that took

so that way I found out 3.6 times 5 = 17.5

so if thats the case then

wait I might be dumb

[quaint ruin]

How u do dis

[quaint ruin]

And this

[upper karma]

looks like law of sines/cosines

[quaint ruin]

oof

[upper karma]

ez pz

[torpid horizon]

just set them equal

vertical angles are congruent

[quaint ruin]

Ye

There is a y

[torpid horizon]

system of equations

for the 2

after u set equal either substitute or eliminate

[quaint ruin]

oh shit

[torpid horizon]

i think

[quaint ruin]

ughh

[torpid horizon]

you could also guess and check

since they should equal eachother if they are the correct x and y's

[supple haven]

what are the sidelengths for a 45 45 90 triangle?

I know that two of them are rad 2

[upper karma]

isosales ?

[supple haven]

but Idk the 3rd

right triangle

[upper karma]

2a^2=c^2

a^2+b^2=c^2

where a=b

[supple haven]

i have the iq of a small child, you'll have to be a bit simpler

i just dont remember the 3rd sidelength

[upper karma]

a=sqrt(2)

b=sqrt(2)

c= ?

the pythagorean theorem states

[supple haven]

oh

yeah

[upper karma]

a^2 + b^2 = c^2

[supple haven]

thanks im dumb

[upper karma]

(sqrt(2))^2 + sqrt(2)^2 = c^2

therefore

2+2=c^2

4=c^2 c=2

confirm

[lament bay]

What would I prove for my next step

[lament bay]

<@&286206848099549185>

[meager lantern]

To prove that both (1) and (2) we have to prove that the angles 1 and 2 are 90 degree angles, since doing so would prove they are perpendicular.

one way I devised to accomplish this is to prove that both triangles are congruent.

1. RC = AT (Given.) By subtracting middle segment TC, we can deduce that RT = CA
2. SR = AB (Given)
3. ST = BC (Given)

From this, we can conclude that for both triangles all the sides are equal, and that they are congruent triangles.

For them to be congruent triangles, all their angles must be equal. Which would mean that 1 = 2, let us label 1 or 2 x.
Since 1 and 2 are supplementary angles, 1 + 2 = 180. Since 1 = 2, We can write this as 2x = 180 or x = 90. Proving that both 1 and 2 are each 90-degree angles. Which would make the sides required to be proven perpendicular.

There you go @lament bay Tell me if you spot a mistake in my work or something you don't understand

[lament bay]

@meager lantern yo thanks so much

[meager lantern]

[wise halo]

Can someone help me with this

[hard gale]

you can factor that

[wise halo]

How

Ohhhh

Like (cosx+sinx)(cosx-sinx)?

[hard gale]

yas

[wise halo]

After that?

[hard gale]

when you're solving for x in x(x-1) = 0 (as an example), what do you do?

[wise halo]

So i will say cosx+sinx=0

Then cosx-sinx=0

?

[hard gale]

yus

[wise halo]

Then?

I need to find all the answers between 0-360

Degrees

[hard gale]

solve idk?

we actually simplified our bad ass equation with squares with 2 more basic trig eq

[wise halo]

Like this?

@hard gale

[hard gale]

if you want yes

[sick veldt]

are my answers right...?

[hard gale]

seems roight yes

[sick veldt]

I think its 8

cause 4*2 = 8

8*0.5 = 4

FD = CB?

the uh midsegment theorem

[hard gale]

yus

[sick veldt]

Not sure about this one

I just graphed the ones that the lines touch on

hello

only this and one more left 😛

[hard gale]

yas is correct

[sick veldt]

Last one

This one I'm lost on

Ik I have to use midpoint formula again

but Idk how to find the x's and y's this way

its DF

and DE

I think

halp

[hard gale]

how do you find the midpoint of a segment?

[sick veldt]

use the triangle midpoint segment theorem ?

[hard gale]

i mean just any segement

how do you find the coordinates of its midpoint?

[sick veldt]

x + x/2 y + y/2?

you add the two endpoints and divide by 2 no

?

[hard gale]

your notations are off but i get you

yas

[sick veldt]

yeah its jsut Idk how I find it

so lemme guess

for DE

would the x be 1 and y be 4?

[hard gale]

that's the coords of D

[sick veldt]

right

then

for F

1,1

add and divide by 2?

[vagrant bolt]

is the anwser just the midpoints of ef and df

[sick veldt]

Its on the question

you add them and divide them cause they are endpoints

and we need the midpoints

what do i do lol

add them or no

is it 1, 2.5???

[vagrant bolt]

fe midpoint= (-1,2) and df = (-1,3.5)

[grim sorrel]

Looks like (-1,2) and (-1,3.5) to me

[vagrant bolt]

midsegment is a line joining 2 midpoints and the endpoints are the coordinates at either end so that must be the anwser

i think

[sick veldt]

idk anymore lol

I thought we had to find DE and DF

[vagrant bolt]

the question said parallel to DE

[sick veldt]

ah okay right

hopefully this is right

[vagrant bolt]

in the midsegment theorem the midsegment is joining the midpoints of the 2 lines and the 3rd line is paralle which they stated was De

its also 1/2 in length

[sick veldt]

yeah you got it write

[upper karma]

guys i need help

inmediatly

i have a window of a x y dimensions (600 x 400 i think)

and i wanna create 3 random points

what are the conditions to generate those points in order that the circle made by those 3 points isnt bigger than my window size?

[cold plaza]

@upper karma isn't the largest circle possible a circle with radius 200?

[upper karma]

yes it is

[cold plaza]

well i found something of use: u can directly find the equation of the circle (hence it's center and radius) from the determinant of a fancy matrix

https://math.stackexchange.com/questions/213658/get-the-equation-of-a-circle-when-given-3-points

its the second or third answer

[valid abyss]

would anyone want to make some questions for me in preperations for my math test tomorrow? it's about surface area and volume

[serene field]

Have you searched google for practice questions with solutions?

[valid abyss]

I have but they're too easy

this doesnt mean im good

[quaint ruin]

How u do diss

[pearl otter]

anyone know the answer?

[last pollen]

have you tried applying the cos(a+b) Formula?

[pearl otter]

yes i got zero as my answer but im not sure

[last pollen]

well try with a calculator

if you get zero you can be sure

[last pollen]

tbh I think that function may be identically zerp

zero

[pearl otter]

ok thank you very much

[torn linden]

Can someone tell me why sine can never go over 1?

[dense bison]

@torn linden well, sin(ax-b) (where a, b are constants) can't because geometrically, sine is the ratio of opposite to hypotenuse of a triangle

And the leg of a triangle can't be longer than its hypotenuse

[torn linden]

Right

ohhh

ok

[upper karma]

How do I find y?

I'll just take it to a questions channel

[tranquil trout]

is this right

[lament bay]

Any ideas on this? I’m stuck

[lament bay]

<@&286206848099549185>

[upper karma]

Well you know CF = BF, right?

[lament bay]

Yes

Transitive property right?

@upper karma

[lament bay]

<@&286206848099549185>

[silent sparrow]

http://prntscr.com/locbhp

[lament bay]

lmfao

[keen aspen]

Law of Sines

@upper karma Still need help ? :v

[upper karma]

Nah. Thanks though

[lament bay]

Help me coach

[silent sparrow]

hello

http://prntscr.com/locyt9

can someone assist me

i think the first thing you do is add B and C

[keen aspen]

Law of Sines

[silent sparrow]

yes i remember

[keen aspen]

Yeah and subtract that with 180

FindA

[silent sparrow]

ok give me a sec

52.16667

6 is repeating

[keen aspen]

Okay and do sin(theta)/opposite side

Thatll give you the common ratio

[silent sparrow]

sin what

[keen aspen]

sin of that angle

[silent sparrow]

sin(52.16667)

this

[chilly cave]

θ

[keen aspen]

Yes

Divided by 356

[silent sparrow]

lol

ya this is wrong

0.0002

[keen aspen]

sin(52.16667)/356=sin(15.3333)/AB

Solve for AB

[silent sparrow]

where did you get 15.3333

[keen aspen]

Angle C

Opposite of AB

[silent sparrow]

ah ok

where did you get 356

i thought its 60

cause mins

isnt 356 for seconds

or no

[keen aspen]

No

=pup calculate (sind(15.33333)*356)/(sind(52.16667)

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@keen aspen

[keen aspen]

119.2m

[silent sparrow]

oh 356 is BC

im sorry

lol

the answer is 119 ?

ok

ya

[keen aspen]

Yeah

[silent sparrow]

ty

[keen aspen]

Np

[silent sparrow]

its better to use them in decimal form

rather fractions

http://prntscr.com/lod204

same thing add A and C

then take that answer

180- ?

[keen aspen]

Yes

You know the ratio

[silent sparrow]

42/60

?

46/60

cause mins

or na

[keen aspen]

sin(37.7)/200

That's the common ratio

You can set that equal to each other parts to find what you need

[silent sparrow]

how did you get 37.7

[keen aspen]

37+(42/60)

[silent sparrow]

ahh

would i set the other one to the same denamator

23+(46/60)

23.76667

?

[keen aspen]

Yes

[silent sparrow]

divide by 2100

200

?

sin(23.76667)/200

[spare garden]

I need help with my Nicolas Cage pong game, I need help finding an equation to find y (I only know soh-cah-toa)

[keen aspen]

tan

tan(angle)=y/dist

y=tan(angle)*dist

[spare garden]

how do I find tan(angle) * dist?

[keen aspen]

tan(x)=opp/adj

[spare garden]

so tan(angle) = y/dist?

[keen aspen]

Yeah

[spare garden]

so y/dist * dist = y?

[long sapphire]

yes but that's not the point

if you did y/dist * dist = y, that ruins the point; you already know what y is and there's no need for this

what you need is to use tangent and plug in your angle

[spare garden]

but tangent is y/distance

[long sapphire]

yes it is

but in y/dist * dist, there is no point in dividing then multiplying dist since you already have y

what i'm assuming is that you have the angle so to get y, you need to multiple tan(angle) by dist

is this programming? if so then your lang's math lib probably has tan

[sick veldt]

Is my proof correct?

@hard gale :PPPPPP I summon you once again for the third day

haha

[hard gale]

cretycrydtvfuybgnhqi,o;syxtcv

oh fking 2 column

[sick veldt]

LMFAO

yeah proofs