geometry-and-trigonometry

[graceful hull]

So I’m drawing sort of triangle figures

But I always forget the arrows

Where do they go?

[hazy granite]

so the dot is 0 right

relative to the dot, place the first sum

[graceful hull]

6 across from the dot?

[hazy granite]

no actually A

the first part of the sum

[graceful hull]

ive done the sums

read the vectors above the 3 squares

[hazy granite]

the sum is the final result

im pretty sure you have to have to draw arrows which move from the dot to the first vector to the second

[graceful hull]

ah ok

ill do 1

then check with you

How’s this looking?

[hazy granite]

no you have to draw A

[graceful hull]

What do you mean?

[hazy granite]

which is 1x 3y

[graceful hull]

Uh

So erase that?

And do 1 across 3 up?

[hazy granite]

yeah

[graceful hull]

Draw a vector diagram to show each situation including the resultant vector

oke

[hazy granite]

yeah draw the resultant

keep the resultant vector

just remove the lines forming a right angle

[graceful hull]

Wait

For A

It says to draw A + B

[hazy granite]

yeah

draw point A first

now take point A as 0

now draw point B relative to point A

[graceful hull]

Ok let me rub this out

[hazy granite]

now draw a line between the dot and point B

that's the resultant vector

[graceful hull]

Blurry but like this?

[hazy granite]

yeah keep the ones with arrows

remove the right angle

[graceful hull]

Sure

[hazy granite]

now draw point A

[graceful hull]

1 across from the dot?

And 3 up?

[hazy granite]

yes

[graceful hull]

like so?

[hazy granite]

yeah that's correct

now taking point A as 0,draw point B

5 right and 2 up from point A

which leads to the resultant line

[graceful hull]

So thats all done?

[hazy granite]

yeah

[graceful hull]

So b+d

Same process?

[hazy granite]

yeah

[graceful hull]

I’m confused

I can’t go 5 up for D

I mean*

B

Can’t go 5 up on B

Nvm

It’s 5 across

XD

Don’t understand what I’m doing

Don’t understand what I’m doing

[hazy granite]

use pythagoras theorem

think of it as a right angle

[graceful hull]

No idea what that is

Ok

[hazy granite]

a^2 + b^2 = c^2

thus sqrt(a^2+b^2) = c

[graceful hull]

I dont understand any of that

is it 10?

[hazy granite]

yeah

[graceful hull]

awesome

Geometric reasoning kills me inside

[hazy granite]

there are just a couple of reasons for this one

[graceful hull]

Ive forgotten all of them

[hazy granite]

the main figure with 2 parallel lines and 1 other line

[graceful hull]

2nd one on the top?

[hazy granite]

6 reasons for that

[graceful hull]

or the bottom 2nd

[hazy granite]

both of them are the same

[graceful hull]

ok

I dont know any of the reasoning

[hazy granite]

vertically opposite - 2 lines forming a cross, the angle in the opposite side is equal

[graceful hull]

For those 2?

[hazy granite]

no im stating the reasons

[graceful hull]

Ah

Angles on a straight line? @hazy granite

[hazy granite]

heres the reason

i dont think its clear

[graceful hull]

dont get it

[hazy granite]

they are all related to x

its reasons

[graceful hull]

Adjacent angle?

[hazy granite]

2 of them

[graceful hull]

Can you give me a clue or something

I dont get it sorry

[hazy granite]

the 3rd one is an adjacent angle

on a straight line

now in a diagram like this

everything is either supplementary or equal to

[graceful hull]

Corresponding angles on equal lines?

For 2

[hazy granite]

eya

[graceful hull]

?

Help me out please

I dont get any of this

[uncut portal]

Guys, i have a question, cos^2x = cosx * cosx?

Or this don't work that way?

[serene field]

Double angle identity.

cosx * cosx is = to cos^2 (x)

Wait.

I'm high or something jfc

Yes, cosine squared of x is equal to cos(x) * cos(x).

[uncut portal]

Are you sure?

[serene field]

Yep.

[uncut portal]

I don't see the red line.

Maybe you just high.

[serene field]

It overlaps!

[uncut portal]

Oh

Then i high.

[serene field]

We're all high, when it comes to math.

[uncut portal]

Well if cos^2x = cosx * cosx, then 1/cos^2x = cos^-2x?

I don't know what is sec is so

Oh

Thanks.

[meager lantern]

^ cheap trick

[uncut portal]

Is it something bad?

Or i will raise your profile?

[meager lantern]

It's gonna give him rep points with Tatsumaki bot

[uncut portal]

How this point important?

How much?

Omar

Your answer?

[meager lantern]

Hmm

His methods are cheap, but if he helped you you can rep him

[uncut portal]

t!rep salty boi

[loud cedar]

🆙 | Tokyo31 has given @upper karma a reputation point!

[meager lantern]

cheap af

[uncut portal]

He has a dog?

[meager lantern]

no honor

We should probably move to #chill

come

[stone pike]

Howdy

I'm trying to convert a triangle function into some code to draw it on screen.

wikipedia says p is "period" and a is amplitude. I'm setting the amplitude to 1, so that simplifies it a bit.

I'm using this triangle wave in an Oscillator whose internal counter operates within the range of (-pi, pi)

I'm trying to figure out what p represents, if my oscillator is defined as operating at 0.5hz in a 60FPS environment.

is p 60?

[shut marten]

Can someone help me out with an ellipse problem?

I can't figure out how to calculate the length of the latus rectum given the equation of directrix and the center.

<@&286206848099549185>

[steady sleet]

#❓how-to-get-help, rule #4: If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once.
Please read the rules: #❓how-to-get-help!

[soft kraken]

^ The above guy uses self bots it seems.

[meager acorn]

so far ive gotten

sin(pi/2+x)
cos(pi/2+x)

whats the next step >?

[limpid basin]

sin(pi/2 + x) = cos x

cos(pi/2 + x) = - sin x

-cosx/sinx = -cotx

[meager acorn]

wait

@limpid basin

so i dont have to use sin (a+b) formula?

[umbral snow]

You don't have to but that works as well

There's a tan(a + b) formula

But Plum's suggestion is easier

[meager acorn]

so if i use that plums formula

i be good on test

kaynex

sin^2x = 1-cos^2x
cos^2x = 1-sin^2x?

[umbral snow]

Yus

[meager acorn]

so for this example is there a short cut formula?

like number 11

sin[pi/2+x] = cos

5pi/2 = 90deg , 0 rad

cos = 1/sinx

[upper karma]

Er, that last one's not true

[meager acorn]

do i have to use product to sum identties?

[upper karma]

(thinking)

Oh, cos(x) = sin(pi/2 - x) (the "co" identity") may help doublecheck I'm right and try it

[meager acorn]

cosx cos (5pi/2) + sinx sin (5pi/2)

cos (5pi/2) = 0 , sin (5pi/2) = 1
if you draw unit circle it should be 90 deg

[upper karma]

Wait, what?

I'm not sure how you got
cosx cos (5pi/2) + sinx sin (5pi/2)

[meager acorn]

cos(a+b)= cosAcosB-sinAsinB

A= x
B = 5pi/2

cosx cos (5pi/2) - sinx sin (5pi/2)

cox(x)(0)-sin(x)sin(1)

-sinx

@umbral snow

is this correct?

usually i know a cos (u+,-v)

if you have + u change the the identies to -

[umbral snow]

That identity is wrong

[meager acorn]

????

[umbral snow]

No it's not

[tropic shard]

cos(x+½5π)=cos(x+½3π)=cos(x+½π)

[meager acorn]

wait

1/2 5pi?

5pi/2

[tropic shard]

Same thing lol

I was saying (½)(5π)

And we know that for every 2π rotation, we're back to where we've started, ∴ we can reduce by 2π's.

[meager acorn]

so i did wrong?

entire thing

[tropic shard]

Seems so.. Say x=0, then cos(½π)=1, which is sin(0)

[meager acorn]

but whati did

cos(x)cos(5pi/2) - sin(x)sin(5pi/2)

i changed

cos 5pi/2 to 0

sin 5pi/2 to 1

[upper karma]

@tropic shard cosine doesn't repeat every pi, its every 2pi

[tropic shard]

That's what I said?

[meager acorn]

scosx x 0 -sin x 1 =-sin x

[upper karma]

Castle: cos(x+½5π)=cos(x+½3π)=cos(x+½π) -- I'm pretty sure 5pi/2 and 3pi/2 are only pi apart

[tropic shard]

"So we can reduce by 2π's" ... "for every 2π rotation"

Oh

[meager acorn]

i m not solving

i m trying to identity

[tropic shard]

Lol I see what you mean.

[meager acorn]

😄

[tropic shard]

Well basically cos(x+½5π) = cos(x+½π) = (think of the unit circle)

[meager acorn]

thats why i said.. because

5pi/2

5/2 =2.5

2pi .5

90deg

if u look in unitcircle 0,1

[tropic shard]

Yes, that's right so far

[meager acorn]

now

we have cosx

(cosx) x 0 =0

-sinX x 1

[tropic shard]

Why'd you multiplied by 0?

[meager acorn]

because

COS (A+B)

COS (u+.-v)

if +

[tropic shard]

Oh

Yea

[meager acorn]

cos u cos v-sin u sin b

v

the sign changes

so did i do it right?

i m so lost on this identities

[tropic shard]

Cos(x)Cos(90°)-Sin(x)Sin(90°)

[meager acorn]

i m trying to solve just looking at identies formulas

[tropic shard]

That's correct as well, just instead of (Sin b), it'd be (Sin v) in your identity.

[meager acorn]

t!rep @tropic shard

[loud cedar]

🆙 | Yupki, you can award more reputation in 1 hours, 28 minutes and 23 seconds.

[tropic shard]

Oh, thanks

[meager acorn]

forgot to rep you yesterday too

[tropic shard]

That's okay xD

[meager acorn]

this problem

-_-

[upper karma]

Do you know Euler's formula? If not, I probably can't help, although I could suggest a few things to try

[meager acorn]

i m stuck

LOL

[tropic shard]

Okay this one has a few identities to remember

[meager acorn]

let me take a pic

[tropic shard]

cos(3x)=cos(x+2x), right?

Well you can say cos(x+2x)=cos(x)·cos(2x)-sin(x)·sin(2x) — (double angle of cosine)

[meager acorn]

cos^3x-sin^2cosx - 2sin^2x cosx sinx

i got cos^3x-3sin^2x

but what do i do with cos x sinx

??

[tropic shard]

You distributed wrong on one of those I think

[meager acorn]

2sin^x cosx sinx

isnt it?

[tropic shard]

That'd just be 2sin²(x)cos(x)

[meager acorn]

wait what

sinx cosx?

[tropic shard]

No, since you only distribute once because the 2sin(x) is multiplying with cos(x).

So you'd just get 2sin(x)cos(x)sin(x), which is 2sin²(x)cos(x)

[meager acorn]

wtf

[tropic shard]

So, Cos³(x)-sin²(x)cos(x)-2sin²(x)cos(x) = cos³(x)-3sin²(x)

[meager acorn]

....

wait...

[tropic shard]

If it's A(BC) it becomes ABC, if it's A(B+C), it becomes AB+AC

[meager acorn]

i m lost

lol

one sec

cos^3x-3sin^x

but where did cos x go

[tropic shard]

cos³(x)-sin²(x)cos(x) is what I see for the left

So the cos(x) is still there

[meager acorn]

let me do the distrube again

i feel like i m overthinking this

[tropic shard]

lol

I hate it when I overthink things >.<

[meager acorn]

[ 2sin x cos x] sin X

u get

2sin^2x cos x sin x

[tropic shard]

Nope

[meager acorn]

wait

WAIT

....

tell me

if i have 2sin x + cos x

2sin x cos x

is 1 thing

[tropic shard]

Yes, 2sinxcosx is one thing

[meager acorn]

if therFUOCSKLFJASLKFJASDLKFJASDLKFJASDLKFASDF

FASLDKFJLKASDFJLKASDFJLKASDFJLKSADJFLKASDFAS

[tropic shard]

lol

[meager acorn]

💥 🤦

the other side can be distrube 2 times because there is - sign

[tropic shard]

Yes.

[meager acorn]

SO

cos^3x -3sin^x cos x

[tropic shard]

cos³x-3sin²xcosx

You were missing the 2 on the exponent for sine lol.

[meager acorn]

cos^3x -2sin^x cos x

nono

oh my god

cos^3x -3sin^x cos x

thank you again castle

[tropic shard]

np!

[meager acorn]

i hope least i dont fail test

[tropic shard]

Good luck!

[meager acorn]

tomorrow

i need cool off my head

[tropic shard]

Yes, don't overload yourself, I've seen you do a lot of problems lol.

[meager acorn]

thank you

i need just cool off my head and be back in 1hr

[tropic shard]

Good plan.

[meager acorn]

going to walk or jog with puppy

[tropic shard]

Aw, cute

[subtle cedar]

help please

[astral hornet]

when an altitude is drawn from the right angle to the hypotenuse perpendicularly, the original triangle is split into 2 similar triangles, with both of those similar triangles similar to the original

using that, you can write proportions to solve for x

[subtle cedar]

how would i go about that

[astral hornet]

if you need help visualizing the similarity, draw the 3 similar triangles in the same orientation with known specs labled

[subtle cedar]

i think it’s 13.9

[umbral snow]

Let θ be the upper corner angle.
cosθ = x / 16 (by the total triangle)
cosθ = 12 / x (by the upper triangle)

x/16 = 12/x
x² = 192
x = 13.9
@subtle cedar

[wild hamlet]

Ye?

What do you not understand about it?

[meager acorn]

i found my opp

[wild hamlet]

Which is...?

[meager acorn]

the answer says - 119/169

12 opp

[wild hamlet]

so, cos(theta)cos(theta) - sin(theta)sin(theta)

is cos(2theta)

[meager acorn]

did u use sum and difference identites

[wild hamlet]

yes

[meager acorn]

since find cos 2 theta

but

something that i dont understand

cos2 theta = double anggle or half angle right?

[wild hamlet]

double angle

[meager acorn]

cos^2 theta - sin^2 theta?

[wild hamlet]

yes

[meager acorn]

one sec

now how do use cos^2 theta - sin^2 theta

[wild hamlet]

well, we know what cos(theta) and sin(theta) are, right?

[meager acorn]

no

-5/13 - 5/13?

[wild hamlet]

-5/13 is cos, 12/13 is sin

[meager acorn]

oh wait

i red it wrong

cos^a - sin ^2 a

cos =-5/13
sin = 12/13

-5/13 - (12/13)

[wild hamlet]

both are squared though

[meager acorn]

oh...

(-5/13)^2 - (12/13)^2

[wild hamlet]

yep

[meager acorn]

-119/169

nice

[past siren]

How would I know which angle to go by to find tan of XZY?

I was going by the 90 degree angle but that wasn’t any help

[upper karma]

If no one else offers to help, feel free to PM me in 5 minutes

[past siren]

Kk

[meager acorn]

tan = opp/adj

7/8

SOH = opp/hyp
CAH = adj/hyp
TOA = opp/adj

[hazy granite]

is the formula for finding the diagonal of a cuboid a^2 + b^2 + c^2 = d^2?

[limpid basin]

A space diagonal

[sand geode]

What's a way to use matrices for transformations from any given point other than the origin or a line other than the x & y axis? (rotations, enlargements, reflections)

[umbral snow]

@sand geode
Do you have a specific example in mind?

[sand geode]

Say I wanted to complete this enlargement using a matrix? I know how to do them from the origin, but not from any given point.

[leaden swift]

I need a little help

[serene field]

[real mulch]

@leaden swift

[leaden swift]

@real mulch .

[real mulch]

[upper karma]

Hello?

i have multiple questions

I’m in 8th, but in geometry

[upper karma]

I’m confused on these

I need a kickstart

(Why am I in this class)

[native night]

@upper karma for 14 note bottom one is isocoles

[long sapphire]

^

it's also how i got x in the diagram

[native night]

isosceles

[long sapphire]

due to it being isosceles (?), two of the angles are the same

the three angles of a triangle add up to 180, therefore 180 = 140 + 2x

[upper karma]

That was fast

[long sapphire]

if you're wondering how i got 140

don't mind the left part

[upper karma]

Thank you

[long sapphire]

15

the lower triangle is also isosceles, two of its sides are of length 40

[upper karma]

I just wanted help, not the answers but ok

idk how.

BUT THANK YOU

Smart

[long sapphire]

so we found the angles of the lower triangles via 180 = 120 + 2x (we're using the fact that it's an isosceles triangle)

[upper karma]

Hmm

[long sapphire]

i don't really need to go over y since you got that yourself

[upper karma]

Yeah

[long sapphire]

if you're wondering how i got 120 again, refer back to those two funky lines that i sent

[upper karma]

Ok

idk why I’m in this class, I get a brainfart every question

[long sapphire]

since the right triangle is equilateral and all angles are the same, we have 180 = 3z => 60 = z; we then got 120 from 180 - z

[upper karma]

its double accelerated or something

[long sapphire]

this is really testing my paint skills

[upper karma]

Lol

Thanks anyways

[past siren]

How can I solve these type of questions?

I have geometry knowledge to an extent

[sharp fox]

Well, you know that the angle C is the same for the big outer triangle and small inner one. Also one of the other angles with each of the triangles is the same value at 28°

So it looks like you can find similar triangles, to me at least

@past siren

[vast flame]

just use trigonometry

it's simple deduction

[past siren]

Problem is I’m not really good with geometry since my geo teacher last year was bad. As in her class was simple to get by

[tropic shard]

Well you see ΔABC is similar as ΔBDC, because they're layed side on side, both with ∠28°, so if the question is AB/(BC) on the ΔABC, follow the same path on ΔBDC.

[meager acorn]

t!rep @tropic shard

[loud cedar]

🆙 | Yupki has given @tropic shard a reputation point!

[meager acorn]

[heavy escarp]

How did they go from sin to cos

[serene field]

Sine product rule

sin(a)sin(b) = (1/2)(cos(a - b) - (cos(a+b)))

[bleak rivet]

Hi im interested in translating a full triangle into points. For instance if I have a triangle that has all angles the size of 60. How could I translate that tp points

[left folio]

Anyone knows a lissajous parameter site? Can't find any and that's weird

[sullen flint]

if for example:
sin135 = sin(90+45) = cos45 = sqrt2/2
then how to do:
sin(30+5/3)

[last pollen]

Ehm you wouldn't use the same rule

[sullen flint]

yes i cant because 30 is not 'included' in reduction formula or something and 5/3 is a mess

[last pollen]

https://www2.clarku.edu/faculty/djoyce/trig/sumformulas.jpg

maybe you can use that

first one

[sullen flint]

i cant take cos, sin from 5/3

[last pollen]

or make an approximation with the derivative

sin(x+h) ≈sin(x)+h*cos(x)

you know sin (30) and cos(30)

but you have to use values in radians for it to work

you could actually take as many terms from the Taylor series as you want I believe

[sullen flint]

maybe i didnt understand the problem
it says:
sin(30+x), when sin x = 5/3 and when x is in second quadrant

[last pollen]

lol

yeah that's something else

well then you can use the formulas I gave you

sin (Alfa+beta)

[sullen flint]

what to do with 5/3, i used the formula and im left with 1/2 * cosx + sqrt3/2 * sinx

[last pollen]

sin(x), what could it be?

[sullen flint]

hm?

[last pollen]

have any idea what could be the value of sin(x) ?

[sullen flint]

something between 90 and 180

i dont know

[last pollen]

you sure?

[sullen flint]

maybe

[last pollen]

sin(x) is between -1 and 1

for any real x

[sullen flint]

oh yea

and

[last pollen]

sin(x) =?

[sullen flint]

(-1; 1)

[last pollen]

your problem is a bit weird

because it says from the onset

that sin(x) =5/3

but that's not possible

[sullen flint]

oh shit wait sorry its 3/5

[last pollen]

well

[sullen flint]

0.6

[last pollen]

then

then we already know sin(x)

we've known it all along

[sullen flint]

what is it

[last pollen]

you've just told me

[sullen flint]

but i cant take sin from 3/5

hold up

[last pollen]

that's not what you have to do

you don't have to know sin(3/5)

[sullen flint]

oh..

[last pollen]

you know that sin(x)=3/5

so you just plug that into your equation

[sullen flint]

oh yea that makes sense

thank you

[tender wraith]

can someone help me with this

[upper karma]

Wouldn't the cotangent just be the reciprocal of the tangent?

[tender wraith]

yeah

ik that.. but im not sure where i did wrong

like there is a sqrt symbol which i cannot simplify bc there are not any perfect squares

idk

[upper karma]

Well, you flip the fraction and then remove the radical by multiplying by sqrt(138)/sqrt(138), which is 1

[tender wraith]

i did flip it but like i dont think thats it

im not sure how you got that either

[upper karma]

OK, PM me and tell me step by step?

[tender wraith]

ok

[rain tulip]

i procrastinate

@tender wraith

138 is divisible by 46.

[tender wraith]

oh

i didnt know..

[left folio]

" A hyperbola passing through (8,6) consists of all points whose distance from the origin is a constant more than its distance from the point (5,2). Find the slope of the tangent line to the hyperbola at (8,6)"

I know that i have to use the implicit derivative form

[last pollen]

what have you tried yet

[left folio]

I don't know how to tackle this

[upper karma]

Hint: definition of hyperbola

[last pollen]

how do you express the distance between a point (x,y) and some other fixed point (x0,y0)

[left folio]

Pitagora

It's not like i'm dumb i know both of those. I don't get the proceidure

[last pollen]

well, assume y=f(x)

write the definition

take the derivative

das it mane

[clear haven]

main mane

[upper karma]

<@&286206848099549185>

Question 3 please

[waxen gorge]

Hi

Well this one is quite simple

For a

U do

3/2 = tan(3x)

So x = tan^-1(3/2)/3

For b

4/5 = tan(x+45) so x+45 = tan^-1(4/5)

For c, tan(2x) = 7/2 so x = tan^-1(7/2)/2

for d, -tan(x-60) = 1/sqrt3

@upper karma

[upper karma]

a is 2/3 not 3/2

The thing is I can get that far but how do I work out the angle for b

*angles

The theta+45 confuses me

@waxen gorge

[waxen gorge]

@upper karma so

Also yes, a is 2/3 my mistake

For b

x = tan^-1(4/5) - 45

the tan^-1 will give the angle

Now u have to subtract 45 to get the theta

[sterile inlet]

hey i have 2x = arcsin(4/5), how do i go on about solving for x?

[dire rampart]

divide both sides by 2

[sterile inlet]

but what is (arcsin(4/5))/2

[dire rampart]

use a calculator

[sterile inlet]

not allowed :(

[dire rampart]

hm

arcsin4/5 isnt an exact value tho

is that all youre given

[sterile inlet]

well i started with tan(x)+cot(x) = 5/2

now im at 2x = arcsin(4/5)

[dire rampart]

how did u reach that exactly

[sterile inlet]

starts top left

[chrome fiber]

whats wrong with that answer?

[sterile inlet]

well i was told arcsin takes a side ratio

and theres an infinite amount of angles to get that side ratio

so an interval is needed

[waxen gorge]

@sterile inlet arcsin(4/5)/2 u need a calc to evaluate it

[rain tulip]

@sterile inlet wait, can't you just leave it as arcsin(4/5)/2

[magic turtle]

Depends what ur teacher wants

[glad ocean]

tRy ArCsIn(2/5)!1!

I know this sounds dumb but what is arcsin the inverse of?

and the reason I'm not just googling this is because I want a coherent but not overly expansive explanation

[umbral snow]

arcsin undoes sin.

You may know arcsin as sin¯¹. Different notations, same thing

@glad ocean

[glad ocean]

ahh

So basically:

arcsin(x) = 1/sin(x)

[umbral snow]

arcsin(sin(x)) = sin(arcsin(x)) = x

They undo eachother.

[glad ocean]

ohhh

thanks

[umbral snow]

Np. Feel free to ask if you have anything else!

[glad ocean]

not at the moment, but thanks1!!1!

well actually

I do have a question but it relates to statistics (check #probability-statistics)

[hazy granite]

its just ratio

since thye are in ratio

angles would stay same

[upper karma]

You're saying ABN and PQS are similar triangles?

[hazy granite]

should be

i meant abc and pqs

[upper karma]

They sort of look similar, but I don't see why they have to be

[hazy granite]

find some pyramid

actually

its just a triangle closing in

[quick perch]

so they are similar triangles therfore angle PSQ = ACB = 75

[clear haven]

i feel like some information is missing but ok

[upper karma]

Cartesian grid would doubtless work

[clear haven]

yeah if it was in a grid things would be easy to discern

[upper karma]

No, I mean you can create a grid yourself

(or is that what you meant?)

[clear haven]

i mean from the picture idk

computer generated problem would have been nicer

basically yeh

[past siren]

Can anyone explain how to solve for x? I know that adjacent angles sum up to 180. 180-106= 74. After I solved for the adjacent angle, I though 74 was corresponding angle to x but I was wrong

[grim sorrel]

Triangle angle sum

[past siren]

Can you further explain? @grim sorrel

[grim sorrel]

Sum of the angles of a triangle is 180 degrees

always.

so in your case 74+23+(180-x)=180

there it is

[past siren]

Thank you I see now

[clear haven]

👁

[hazy granite]

alternatively you could use the exterior angles rule

[meager acorn]

:/

[hazy granite]

?

[mellow scroll]

H is the orthocentre of ABC

Having a lot of trouble finding the equal angles

Right now, trying to prove that angle FDB == angle BAC.

[dire rampart]

wtf is that diagram

[mellow scroll]

The six cyclic quadrilaterals in the orthic triangle

Drawn by your's truly

[dire rampart]

you should probably try #geometry-and-manifolds

[mellow scroll]

hmm

Alright @dire rampart but I don't think it's worthy of being under there considering that you can prove via angle chasing but I feel that I'm close to solving it

OH SHIT

I DID IT

[dire rampart]

nice

[mellow scroll]

FAE == FDB

time to do this 2 more times and I can proceed to the next question 😄

[dire rampart]

rip

[mellow scroll]

Hail Euclid

Turns out it was simpler than I thought

Just needed to use 2 angles to find expressions for all the angles I need to prove

[mellow scroll]

@normal bronze

What exactly is equivalent

The perimeter? Area?

I don't really know what is meant by saying two rectangles are equivalent but I have a hunch that you need to formulate an equality and solve for X.

[normal bronze]

The answers are on the btm

Its a multiple choice

[rugged moat]

If it's the perimeter,

$2x-2+4x-7=8x-4+x-2$

[somber coyote]

Duckinator:

[rugged moat]

$-3=3x$

[somber coyote]

Duckinator:

[rugged moat]

$x=-1$

[somber coyote]

Duckinator:

[rugged moat]

=pup Simplify 2x-2+4x-7=8x-4+x-2

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@rugged moat

[rugged moat]

So perimeter is 2(-4)(-11)

88

That doesn't work

And sides cannot have a negative legnth either way

So the area must be equivalent

$(2x-2)(4x-7)=(8x-4)(x-2)$

[somber coyote]

Duckinator:

[rugged moat]

Solve that

Get x

Get perimeter

[rain tulip]

@mellow scroll i don't have enough resolve to do egmo

im too lazy

[mellow scroll]

don't need to @rain tulip

Solved it myself 😄

[rain tulip]

ik

but

like

nvm

[mellow scroll]

Got a question

If all angles of a quadrilateral are acute

nvm

[remote geode]

lmao

[mellow scroll]

I just realised how dumb that qn was

Literally if all angles of a quadrilateral are acute then the quadrilateral is obviously cyclic cause its a fuckin square.

-_-

[remote geode]

rip

[mellow scroll]

indeed

fight me

[remote geode]

LOL

[dire rampart]

@mellow scroll lmao this is too advanced even for advanced geometry smh

who tf calls it that last one

smh

[remote geode]

lmao

[mellow scroll]

Oh shit

You’re right

Kill me pls

[lament bay]

Do you know what properties I’m supposed to use? Such as transitive or reflexive?

<@&286206848099549185>

[novel island]

hypotenuse-leg makes the triangles congruent

[limpid basin]

Who said it was a right angle

Use SSS, you don’t need transitive or reflexive

[lament bay]

I’ve done a few but can you check it for me?

I think I’ve done it right

@limpid basin @novel island

[woeful flame]

Find a pair of points on the unit circle that are symmetrical in relation to the x-axis and the ratio
To the beginning of contractions

someone knows?

[astral hornet]

@lament bay #2 👍

#3 put a reason for acb = dce

and i think ur trying ASA?

[lament bay]

It’s not SAS?

[astral hornet]

oh, for #2 you need vertical angles

[lament bay]

For which step I need vertical angles?

[astral hornet]

before last step

[lament bay]

Oh ok

So #3 should be ASA and not SAS?

[astral hornet]

in #3 it looks like your using the right angles, acb=dce and the contained side ca=cd

[lament bay]

Ok

[woeful flame]

help me please

[lament bay]

Am I missing any other steps in #3

[astral hornet]

did you explain why acb=dce

[woeful flame]

Find a pair of points on the unit circle that are symmetrical in relation to the x-axis and the ratio
To the beginning of contractions

I thought its 0,1 and 0,-1

[lament bay]

Gimme a sec I’m thinking

[woeful flame]

but this is wrong

[lament bay]

How do I prove it’s the same

Like a constant

Like what’s the name of the proof

@astral hornet

[astral hornet]

wdym constant?

[lament bay]

It’s part of acb and ecd

BCD

Is apart of the 2 triangles

[astral hornet]

think about angle addition

[lament bay]

Segment addition postulate?

I forgot what it’s called

[astral hornet]

angle addition postulate

[lament bay]

Hehe thanks

@astral hornet I did #5 but I’m stuck on #4

[astral hornet]

in #4, it's a right triangle, with hypotenuse and a leg given

hy-leg

[lament bay]

Can you explain in steps?

[astral hornet]

you first need to prove it a right triangle

[lament bay]

Messed up on #5, not SAS, ASA*

[astral hornet]

then prove that there's 1 pair of hypotenuse and 1 pair of legs congruent

in #5, remember to declare C = C via reflexive

[lament bay]

Ok 👌

Since it’s perpendicular

It tells you it’s a right angle

[astral hornet]

you have to explain that perpendicular lines form right angles

[lament bay]

How would I do that?

Wouldn’t someone already know if it’s given since it’s perpendicular

[astral hornet]

no

when i learned geometry proofs, perp relatoinships have to be explained as creating right angles in order for there to be right angles

[lament bay]

How would I say it then?

[astral hornet]

statement would be that angle___ is a right angle

reason would be perp lines form right angles

[lament bay]

what’s angles tho?

[astral hornet]

k and t

[lament bay]

K=t-right angles?

@astral hornet

[astral hornet]

you dont need to prove them equal

in hy-leg, you only need right triangles, congruent pair of leg and hypotenuse

[lament bay]

I’m not sure what you mean by that?

[limpid basin]

If you have two right triangles, to prove they’re congruent you only need to show that their hypotenuses are congruent along with a pair of sides

[lament bay]

Can you show me in a 2 Column table?

[limpid basin]

What’s the problem?

#5?

[lament bay]

Yes

[astral hornet]

hy-leg was for #4

[limpid basin]

#4 is not a given right triangle

Oh sorry my bad

[astral hornet]

perpendicular lines form right angle, triangle with right angle is right triangle

[limpid basin]

just relooked at it

Admittedly not sure about 5 but I’ll relook at it when I have time

[lament bay]

Do I need to include that in my table?

[limpid basin]

Yeah you need to prove they’re right

[lament bay]

How would I show that?

@limpid basin

[limpid basin]

Perpendicular lines

[lament bay]

Ok

So K=t-perpendicular lines?

@limpid basin

[astral hornet]

no need for k=t in hyleg proof

it's k, t right > triangles are right triangle > hypotenuse congruent > leg congruent > triangles congruent

[lament bay]

Then how would I write it?

[astral hornet]

first, put that k and t are right angles since perp lines make right angles

the triangles are right triangles as there is a right angle

then a pair of legs are congruent, using given info

then a pair of hypotenuses are congruent, using given info

then the triangles are congruent, bc hyleg post

for #5, use what you have, but add that perp lines create right angles, right angles are congruent to right angles, and that c=c because reflexive

[lament bay]

Can you type out the steps?

I’m kind of confused what you are trying to say

I have to write it in this format, I’m kind of confused how you write it in that format

@astral hornet

[astral hornet]

for the pair of legs and hypotenuse use the actual segment's name

[lament bay]

? @astral hornet like that?

[astral hornet]

how did you jump to the triangles congruent in step 5 already?

[lament bay]

Because the legs are congruent after you know they are right angles?

[astral hornet]

the triangles aren't congruent at step 5 yet

[lament bay]

What should I add next then

[astral hornet]

well, the triangles congruent save for last step

next: both triangles are right triangles

[lament bay]

How do I write that in steps?

[astral hornet]

then use segment addition postulate to prove kl=vt

[lament bay]

JKL=BDC-Right Triangle?

[astral hornet]

statement would be triangle jkl and bdc are right triangles

reason: they have a right angle

[lament bay]

Ok

What’s after?

[astral hornet]

prove kl=vt

[lament bay]

Ok

[astral hornet]

think about segment addition postulate

[lament bay]

Ok

[astral hornet]

after that check that you've covered 1. right triangle 2. congruent hypotenuse 3. congruent leg

[lament bay]

Ok

What would I do next

? @astral hornet

[astral hornet]

After you've covered everything you can say they're congruent via hypotenuse leg postulate

[lament bay]

Ok thanks

[astral hornet]

Np

[lament bay]

Took a while eh?

Haha

[lament bay]

I got a quick question, how do I know if it’s SSS, ASA, etc? <@&286206848099549185>

[upper karma]

If you are given 2 sides (lengths) of a triangle and an angle, that's SSA

no angles given, S ide side side: SSS

angle side angle

2 angles 1 side

Is pretty obvious

S = side
A= angle

In any combination to do any problem you must be given 3 values

[keen aspen]

SSA is not a thing

[lament bay]

Why is this triangle SAS?

[keen aspen]

Because you have the vertical angles congruent and your two known side lengths

[lament bay]

Thanks

[teal dragon]

@keen aspen SSA is a thing

I think

wait

yeah it is

you can use law of sines to get another angle and solve for the third using the fact they all add to 180

[upper karma]

https://www.youtube.com/watch?v=h4XprRVDm94

[teal dragon]

o

[keen aspen]

Yeah no its not

When you have two adjacent sides and an outside angle you cannot formulate that the triangles are congruent

[astral hornet]

ssa cannot determine congruency unless the angle is right angle

[teal dragon]

why can't you get another angle with law of sines

[keen aspen]

You dont have enough information to formulate using law of sines

[astral hornet]

ssa doesnt guarantee congruency in all cases

https://www.khanacademy.org/math/geometry/hs-geo-congruence/hs-geo-triangle-congruence/v/more-on-why-ssa-is-not-a-postulate

[teal dragon]

you have one side and the opposite angle, and another side

don't you

[astral hornet]

yes

congruence isn't guaranteed unless there is a right angle (hypotenuse-leg case)

[teal dragon]

i understand that it isn't a congruence, I just want to know why law of sine fails

[astral hornet]

it works depending on which sides were given

[clear haven]

ssa grants similarity i think

[astral hornet]

in a hypotenuse-leg case, yes, else, no guarantee

[steady sleet]

sees many instances of "SA"

[keen aspen]

It doesnt

There are examples where it doesnt guarantee congruency

[teal dragon]

pretty sure the only one that grants similarity is AAA @clear haven

[clear haven]

i see

yah

[left folio]

"Find an equation for the tangent line to x^(2/3) + y^(2/3) = a^(2/3) at a point (x1,y1) on the curve, with x1 ̸= 0 and y1 ̸= 0"; i got the differentiate which is =2/3*cuberoot(x). What do i do know?

Is there a tangent for this equation, if so how to find that?

[soft kraken]

(y-y1)/(x-x1) = dy/dx at (x1,y1)

@left folio That's how it's done. So make sure you differentiate the function correctly.

[left folio]

@soft kraken But what's x1 and y1? Are they both 0? That wouldn't make much sense

[soft kraken]

They are not zero as per the question

[left folio]

Indeed so how do i find the two

[soft kraken]

You don't. There are infinite possibilities of points on that curve where both x1 and y1 are not equal to zero.

[left folio]

I see well that makes sense i suppose

[soft kraken]

Unless you have additional information provided, you cannot find a particular point.

[left folio]

Do i just laeve it at y'=(y-y1)/(x-x1) then?

[soft kraken]

yep

[left folio]

Since it's an astroid i think there's no tangent where x and y are diverse from 0

But i wanted to prove that mathematically

Well i guess that mathod works too, @soft kraken thanks a lot

[soft kraken]

np

[jade solar]

Hi Everyone. I'm trying to convert a point from cartesian to polar coordinates. The formula for getting the angle is atan(y/x). But how do you differentiate between the second and fourth quadrants and the first and the third quadrants?

[clear haven]

see this is why i dont really like that "formula"

im pretty sure it only gives you the angle between the line to the point and the x axis

do you happen to know how to work with vectors?

bc i do a different method to get the angle but it relies on stuff like that

a bit roundabout but it works

but uh yeah that's probably going to take a while to explain

i think we should focus on the main question

[jade solar]

I'm trying to write some code of which this is a part. So I don't know how well it would translate

[clear haven]

hmmmmmm

brb

real quic

imma make some diagrams too ??

[jade solar]

❤

[clear haven]

so we have a point (x,y) in the 1st ??? quadrant

what we do when we do arctan (y/x) is we're making a triangle like the one in the pic to get the angle from the x axis

and for the r term, you just use pythagoras ofc

but what to do if it's in a different quadrant?

we'll see

:

you compute the angle theta and then add 180 degrees to it bc you do angles starting from the positive x axis all around right

counterclockwise

you get kinda what i mean?

oof gtg

i hope this helps sorta?

[subtle cedar]

can i get some help?

[jade solar]

So I'll have to see the sign of x and y and determine the quadrant independently

[astral hornet]

@subtle cedar angle ehf is inscribed in the circle, and when an angle is inscribed in a circle, the angle is a right angle

[jade solar]

That's angle inscribed in a semicircle but we don't know that FE is the diameter right?

[astral hornet]

oh yeah right is for semis. oops

[subtle cedar]

i know the answer isn’t 152 or 118

can’t seem to find the right one though

[jade solar]

What's the fourth option?

[subtle cedar]

180° 152° 118° 100° 90°

it’s not 152 or 118

it’s defo not 180°

either 90 or 100

[astral hornet]

hfg is a linear pair with efh

then you can find measure of angle g

then angle e

then fhe

[jade solar]

How E?

To find E we need FHE

[astral hornet]

triangle sum theorem

you dont need fhe

to find e

there's a big triangle

[jade solar]

But we don't know GHE right?

GHE = GHF + FHE

[astral hornet]

oh yeah

[subtle cedar]

im so lost omg

[astral hornet]

inscribed angle's measure is 1/2 the length of the arc it intercepts

[jade solar]

Is there any was to use the fact that it is the tangent?

[astral hornet]

use angle efh to find arc eh

nvm

[subtle cedar]

it’s 90°

[jade solar]

How'd you solve it?

[worldly glen]

Angle between two vectors is acos(a:Dot(b)) for unit vectors a, b

but how do I know if that

is clockwise or counterclockwise

since it is going to be the shortest due to cos

and always positive

?

[astral hornet]

@subtle cedar angle g = 1/2 (arc eh - arc fh) (angle outside circle formed by a tangent and a secant = 1/2 difference between major and minor arc)

use angle efh to find arc eh

then use above equation to find arc fh

then, use arc fh to find angle e

then use angle e and efh to find angle fhe

[subtle cedar]

@jade solar i didn’t solve it i just guessed

[worldly glen]

nvm I can check the value of the normal dot with an identity vector to be positive=ctrclockwise negative=clockwise

[vital frost]

Someone can help me out on this?

I got 4.3 for LW

I just don't know how to do part A,B, and C

[keen aspen]

God I'm not too good with triangle inequality

Perhaps <@&286206848099549185> could help

[vital frost]

thank you

[keen aspen]

Actually I think I might know it

If you were to make the angle of 50° approach 0, the length LW would approach 0

If you made it as big as possible (up to 180°) the max length approaches 10.2km

So LW has to be 0<LW<10.2

[vital frost]

Ok before you answer that

We had to answer this problem first

@keen aspen

For DL I got 4.4 because of midsegment of BK