geometry-and-trigonometry

[snow crystal]

But I will dm you

[smoky jetty]

thats fine

do u do olympiads?

[snow crystal]

Right now I'm preparing for one, yeah

[smoky jetty]

oohh

national?

[snow crystal]

I have like 3 months left

[snow crystal]

Yes

[smoky jetty]

thats ncie

try to join the math olympiad server linked in competition math channel

[snow crystal]

Also that exercise got my kind of confused cuz back when I was doing geometry in my school we would solve such problems doing the exact same thing I did (without giving the approximate result but the actual number even if it wasn't rational)

[pseudo viper]

lol i definetly had no chance at solving that

[silk dove]

@snow crystal i thiiink i see the mistake?

if there is

kinda forgot trig rules but one sec

[snow crystal]

huh

[silk dove]

shouldn't this be tan(a) = 4/8?

instead of sin?

[snow crystal]

no, why

I'm using law of sines

for the triangle DEF

[silk dove]

but consider this for any right triangle. if a = 30 deg, then that would mean AD = 8 which is not true

[snow crystal]

wait I think I see the mistake

triangle DEF is not inscribed in any circle

So I cannot use law of sines like that

that makes sense now

[silk dove]

also

[snow crystal]

so I get that tan alpha = 1/2

[silk dove]

law of sine is kinda weird when not all of the sides are acute

[snow crystal]

you mean angles

[silk dove]

angles yeah my bad

[snow crystal]

I mean in this case it doesn't make any difference

[silk dove]

we love that 26.57 degree angle 😭

[snow crystal]

that hurts so bad

and then you have to do all the calculations

naaah 💀

[silk dove]

ill try tio like approximate some the solution bc thas just what we do

[silk dove]

i got 1.252 which is like same thing that the other guy got, GGGYeah

[snow crystal]

cool

[silk dove]

would love to show you my solution if you like reading messy, sloppy, bad-pencil, cumbled-paper, incoherent, non-standard writing

[snow crystal]

I like challenges

you can post it here

yeah I did all the calculations and got ~ 1,25 as well

[silk dove]

Too embarassing may I dm you

[snow crystal]

sure

[smoky jetty]

thats nice of yall

[smoky jetty]

how long did it took u get?

i realized earlier than u could straight up go to SSS law of sine case to ease the calculation a bit

[pseudo viper]

what's there to learn trig wise after learning the core trig functions & how to use them to find sides / angles + law of sine / cosine?

[weary hound]

I guess this is because of the property that sin(x)= sin(π-x) so you have to find if the angles are acute or not

[snow crystal]

sketching graphs of trig functions, solving trig equations and inequalities

(for example)

[smoky jetty]

yeah, or maybe if u have a trig book, dig into those lessons

[silk dove]

yea

[pseudo viper]

hmmmm
I will look into that

trig book might be useful

only rly learning trig rn for game dev stuff

[smoky jetty]

ooh interesting

[snow crystal]

wow I'd never think that would actually come inhandy in life

[smoky jetty]

Joe casted a shadow extending 2 met-

[tardy plover]

is that a homework question?

[snow crystal]

LMAOO

[weary hound]

Avg homework question stereotype where you have to find how close Joe is to the lämp and stuffs

Honestly I wish they could provide an example of how math is used irl

But yeah that's why universities exist probably

[tardy plover]

people use it to calculate the height of buildings by looking at the building's shadow

no one does that, but that's what its used for apparently

[vivid plinth]

what game are u making?

[pseudo viper]

not making any game rn at this time
but realised i need to scratch up on my math skills

[pseudo viper]

this is why i struggled a lot in school

• they just sucked at teaching people that have any sort of different thought process

[weary hound]

What do you expect, school is standardized, also has to differentiate student with grades, and it's very slow to change the legislation or something

[pseudo viper]

i understand why it is the way it is
doesn't change the fact that it sets up a lot of students for failure

[smoky jetty]

what grade are u in rn?

[weary hound]

University freshman

I'm here because I'm mostly bored

[smoky jetty]

oh lol

[cyan crow]

What are euclids postulates

[grave pond]

That's something Google is good at answering.

[fiery oriole]

is sin^2 x the same as (sin x)^2 or sin(x^2)?

[snow crystal]

sin^2 x = (sinx)^2

[velvet lark]

This question came up in my precalc course (not on a quiz or anything graded, but as a practice question), and im completely at a loss when it comes to solving it. can someone help?

[smoky jetty]

the tan function gives u the ratio of opposite side and adjacent side. Now, it asks for the value of the ratio of secant function in the third quadrant. What do u know about the sides in the third quadrant?

[topaz panther]

A frame in the shape of an equilateral triangle encloses three circular discs of radius length 5 cm so that the discs touch each other. Find:

a. the perimeter of the smallest frame which can enclose the discs
b. the area enclosed between the discs.

I got the perimeter but I am completely lost on what to do for area

[smoky jetty]

what does it mean by the smallest frame?

(for b) If the area asks for this area, then I think u can find the area of the equilateral triangle such that adding the area of those sectors with degree measure of 60 each, and that area x is equal to the area of the scaled-down triangle

u could refer to the radii as lengths of the triangle

[topaz panther]

thank youu

[velvet lark]

Would this be how you would solve it? Or did I make a mistake

[warm warren]

i think i found the mistake, the theorem says that the sin formula can be used for triangle which belongs to circle but the one you did that for(4/sin(a)=8) is wrong bc its half of it

[snow crystal]

yeah I already found that yesterday

[warm warren]

ooh then sorry i just saw it

isn't tho? i think its a simple question which requires trained eye to these kind of problems so you can pave your way and draw extra lines

ooh i love these questions, my favourite one is this

[smoky jetty]

seems about right

[velvet lark]

ok thank you!

[sharp lodge]

<@&286206848099549185>

[snow crystal]

what have you tried

@sharp lodge

[novel scaffold]

if (sinx)^n = sin^n(x) then why isn't (sinx)^-n = sin^-n(x)

[snow crystal]

because (sinx)^-n = (1/sinx)^n

[novel scaffold]

yeah i know that

but why

[snow crystal]

because if you raise a number to a negative power you get its reciprocal

[novel scaffold]

hmm

but then technically (sinx)^-n = sin^-n(x) is correct too

you could just evaluate sinx first and then reciprocal

[snow crystal]

but sinx is a single term

you can't break it into pieces like that

[novel scaffold]

what

then isn't sin^n(x) pieces too?

[snow crystal]

wait , you mean this? $\sin^{-n}\left(x\right)$

[somber coyote]

iammax420

[novel scaffold]

yes

why isn't (sinx)^-n equal to that

just like (sinx)² = sin²x

[snow crystal]

but sin^-n (x) means the same as 1/sin^n (x)

they may look different but it's exactly the same thing

[novel scaffold]

someone said u can't write it like that

[snow crystal]

welp they were wrong

who told you that

a teacher?

[novel scaffold]

a friend

he asked teacher, teacher said it's some binomial stuff

idk

welp thanks

[snow crystal]

take arcsin for example

this is the inverse function to y = sin x

and arcsin is basically sin^-1 x

which is (sin x)^-1

[novel scaffold]

hm yeah

[novel scaffold]

wait how

that's 1/sinx

which is different than arcsin

[snow crystal]

yep

[novel scaffold]

and 1/sinx isn't the same as arcsinx

[snow crystal]

oh wait lol

[novel scaffold]

so how is (sinx)^-1 = sin^-1x

i mixed up things

[snow crystal]

turns out 1/sin x=csc x

[novel scaffold]

eeee

[snow crystal]

ok my bad apparently

so turns out sin^-1(x) is not equal to (sin x)^-1

interesting

[novel scaffold]

yes

it's not

and that's what i asked, why is it not

[snow crystal]

ohhh wait that makes sense

so basically sinx=b/a

(sinx)^-1 = 1/sinx = a/b

which is csc x

same thing with other trig functions

[novel scaffold]

yeah ik that

[novel scaffold]

but what's the explanation behind this

[snow crystal]

ok I think I know now

sin^-1 x is the inverse of the result of sin x

whereas (sin x)^-1 is the inverse of sinx

[novel scaffold]

u said opposite

(sinx)^-1 is inverse of result of sinx which is basically 1/sinx

that's all sorted out

what I'm asking is

why isn't (sinx)^-n = sin^-n(x) just like (sinx)^n = sin^n(x)

(sinx)² = sin²x but this doesn't work when u replace 2 with -2

why is that

what's the reason

[snow crystal]

hmm

[royal scarab]

actually the formula has a condition that says n must be a Natural Number. That's why (sin x)^n = sin^n(x) is only applicable for n = 1, 2, 3, 4.........

Hope it helped your doubt

[snow crystal]

but why is that

where does that come from

[royal scarab]

Idk the exact reason of this rule. But (sinx)^-2 is just simply written (cosec x)^2 or cosec^2(x) and not sin^-2(x)

the reciprocal of sine function has a separate denotion "cosec". That's why we don't need to write sin^-2(x) separately

[snow crystal]

that has to do something with arcsin x

probably

[royal scarab]

actually it has nothing to do with arcsin x or sin inverse. cosec and sin are inter-related but arcsin is not related like that

arcsin is the inverse of the sine function. The -1 on sin in arcsin or sin inverse denotes the inverse the function. For example f^-1(x) is the inverse of f(x) not reciprocal

[snow crystal]

ok but it is related to the question that they asked

because sin^-1 x = arcsin x

and that got him confused

[royal scarab]

haha the confusion is too common. I also had the same two years ago

[snow crystal]

tbf I haven't even studied this "inverse trig functions" thing so I shouldn't have said anything lol

[snow crystal]

also I'll ask my teacher about it

[royal scarab]

i see. inverse trig function is not that hard by the concept tho. Questions can be

[royal scarab]

good! if possible let me know too what he thinks about it

[snow crystal]

I will

[novel scaffold]

nah not that

arcsin is fine

its inverse of sin

[novel scaffold]

i want the reasoning behind this

[novel scaffold]

but why

whats the reason that (sinx)² = sin²x is only valid for natural numbers

why cant u do the same with negative integers

[royal scarab]

i can't provide exact mathematical or logical answer to this question. you can take it like a tradition/manner/rule to write or denote like this

[novel scaffold]

eeee

im doing that already from 2 years

[snow crystal]

As I mentioned before

I will let you guys know as soon as I find out

[novel scaffold]

okie

[snow crystal]

Funny thing is that there is plenty of "honorable people" or sum other graduates on this server that can easily answer this question but for whatever reason they won't

That would save us some time

[novel scaffold]

lel

maybe no one saw it

i cant find anything online either

[snow crystal]

Neither can I

[silk dove]

where did u even hear that

this is the first time im hearing that

actually kinda makes sense cause of sin^-1(x)

if i had to guess i would say that its just a convention

damn not even desmos lets me try

[wind coral]

If you want sin^a(x) in desmos try sqrt(sin(x))^2a. They are similar except for the fact desmos won't show an accurate graph.

[pseudo nebula]

Notation for sin makes no sense

[silk dove]

I was just playing around with the notation but thank you nevertheless

[pseudo nebula]

I assumed so too

the notation is just weird

[wind coral]

Np

[vivid plinth]

I always just read sin²x as (sin x)²

I don't encounter cases with the notation sin^a(x) other than sin²(x) often, so I never rly thought abt it

In the case of $\sin^{-1}(x)$ I just naturally assume arcsin

[somber coyote]

utopian_vision

[south helm]

There is no reason. This is a question of notation, not on the properties of the sine function. You can do this, and nobody is going to scold you, as long as it is clear from context what you mean. The problem with this method of notation is that sin^(-1)(x) is already used for something: arcsin, the inverse function. And, in general, f^(-1)(x) is usually used for the inverse function, while f(x)^(-1) is the reciprocal. But arcsin(x) is already an alternate notation, so you are free to use sin^(-1)(x) as notation for 1/sin(x) all you want, as long as it is clear to the reader that is the notation you are going with. But, at the end of the day, you can avoid all of this by simply using sin(x)^a. That is absolutely unambiguous. Yes, it means you have to use parentheses, because sin x^a is also ambiguous (it could be sin(x^a) or sin(x)^a), but it avoids all of these issues, nonetheless.

[south helm]

Who told you this, or where did you read that? How is that even a "formula"?

[novel scaffold]

hmm

[royal scarab]

My maths teacher told me this 3 years ago that n must be a Natural number. But I read your explanation too. Your's is more clear. It's just a matter of convenience as long as the reader can understand what I am trying to convey

[novel scaffold]

makes sense

[novel scaffold]

they probably told that as a convention

cuz you dont usually do (sinx)^-n = sin^-n(x)

[royal scarab]

yeah, you are right. It's all just convention

[hollow dust]

or sin(x)^a

[deft drum]

If we have 3 position vector points on the plane, we can find two displacement vectors. Then we find normal vector $\vec N$, and to find equation of plane $\vec N \cdot (\vec r - \vec r_1) =0$ is done. I have problem understanding why we have to find the normal vector to plane.

[somber coyote]

tbhaxor

[deft drum]

If we have two vectors $\vec S_1$ and $\vec S_2$ we can consider them as basis vector and span the whole plane. Why do we need normal vector, this part is bugging me.

[somber coyote]

tbhaxor

[deft drum]

What I think is that, its related to how the plane lies in the space. In 2D we describe line by two points specficially touching the x and y axis. Similarity here in 3D, when we have plane there are infinite ways of its orientianation which also changes with perpendicular vector (aka normal vector).

If this is wrong, please correct me

[south helm]

You do not have to find the normal vector. A parameterization works just fine. But it is a natural choice for a single equation form. The short answer is because the equation is defined by N dot R = 0. The presence of a dot product equaling zero should clue you in that there is some sort of orthogonality relationship at play.

The long answer is as follows: We desire that $\vec{N}\cdot \vec{r}=\vec{N}\cdot \vec{r}_1$. Thus, a particular solution for $\vec{r}$ is going to be $\vec{r}=\vec{r}_1$, and the rest can be obtained from a linear combination of vectors in the kernel of the linear transformation $\vfunc{f}{\bR^3}{\bR}{\vec{x}}{\vec{N}\cdot\vec{x}}$. That is, vectors orthogonal to $\vec{N}$. Since we defined $\vec{N}$ as being orthogonal to the $\vec{S}$ vectors, then this equation describes exactly the plane spanned by the $\vec{S}$ vectors, which passes through $\vec{r}_1$.

[somber coyote]

st.jamie.

[upper karma]

can someone help me with these

[grizzled breach]

7. Then OP is congruent with PN.

Oh wait. It is asking for angles haha.

[upper karma]

yea i think sso

[grizzled breach]

Do you see that in 7, segment LM and segment MN, both are congruent and related and both have the M? So using that, then we can also say that the answer for 7 is then : angle LMN is congruent to angle NML.

[upper karma]

ohhh

[grizzled breach]

Do you see it?

[upper karma]

wait so for 8 will it be like

NPO is congruent to OPN?

[grizzled breach]

I wouldn't say you are wrong, but I don't think they are asking for the angle (you are giving angles as answer). The are givving you the angles for you to give them the segments (the inverse of number 7)

Give them the segment of line that are congruent. The connecting point is P which I think you notice already.

[upper karma]

oh so what should i give as an asnwer then?

[grizzled breach]

What is the opposite segment of the angle PNO?

[upper karma]

ONP?

[grizzled breach]

ONP is an angle, not a segment (segment are described by 2 points, angles are described by 3 points)

What is the opposite segment of the angle PNO?

What is the opposite segment of the angle PON?

[upper karma]

PON?

[upper karma]

PNO?

[grizzled breach]

PON is an angle. I am not askin for the angle but for the segment of line opposite to the angle PON.

[upper karma]

ohhh

wait I don't understand

I'm very bad at this topic

[grizzled breach]

What is that?

What you dont understand?

[upper karma]

the oppsite segement stuff

[grizzled breach]

Ok, in the picture, if you see ccarefully, look at the angle MNL. What is the opposite segment of line of that angle?

The answer is LM

Or ML.

LM is a segment of line.

[upper karma]

so this is PO?

[grizzled breach]

Yes.

PO also has to letters only which makes it a segment of line.

[upper karma]

ahhh

[grizzled breach]

You have to be careful and be very strict with the definitions.

[upper karma]

yea i see that now

[upper karma]

so for 8 It's PO congeuent to segment OP?

[grizzled breach]

No. But PO or OP is correct inthe first blank space if they are asking for line segments.

[grizzled breach]

PO or OP is the segment opposite tot the first angle, What is the opposite segment of the angle PON?

[upper karma]

oh

[upper karma]

PN?

[grizzled breach]

Yes.

So the answer would be...

[upper karma]

PO is congruent to PN

[grizzled breach]

Yes, that should be fine but a cleaner answer (a little detail in the order) would be: OP is congruent to PN

[upper karma]

ohhh okay

Thank you.

[snow crystal]

what can you say about the sides lenghts

I'm talking about 11

[upper karma]

um

it's 16:12

[shadow plover]

In 11 look at the sides how do you go from 16 to 12

[upper karma]

what do you mean

[snow crystal]

I mean there not corresponding

you have 8:4

but you also have 16:12

the ratio is not the same

[upper karma]

oh so

what do i do

[snow crystal]

what does this mean to you?

[upper karma]

the ratio's are unequal

[snow crystal]

so are the triangles similar?

[upper karma]

no

[snow crystal]

there ya go

[upper karma]

ohhh

so

for the answer

i just write the ratios aren't the same, therefore the triangles aren't similar

what about 12

[snow crystal]

what can you say about 12

[upper karma]

one triangle is smaller than the other

[snow crystal]

what about the angles

[upper karma]

they arent the same either

[snow crystal]

hmm you sure?

there is one particular angle

[upper karma]

right angle?

[snow crystal]

nope

[upper karma]

ummm

which angle

idk which one it is

[snow crystal]

what do you call these angles in english?

I actually don't know

[upper karma]

um wait lemmie check

[shadow plover]

vertical angles

[upper karma]

vertical?

[snow crystal]

ok

so they're vertical angles

[upper karma]

yeah

[snow crystal]

and also the ratios of corresponding sides are the same

10/24 = 15/36

which means ...

[upper karma]

there similar

[snow crystal]

yep

[upper karma]

ohhh

thanks

[snow crystal]

np

[gaunt lotus]

Opposite

[night idol]

How would I solve for x and y without using trig like without SOHCAHTOA

[silent plank]

pythag

[wanton shell]

@gaunt orbit

ye dheko

[shadow plover]

Use 9^2 + b^2= 15^2 then use that information to solve bc

[gaunt orbit]

see what

[smoky jetty]

might have done a more rigorous approach but, you can find the side of DC using pytha, then find length of DB using pytha again and subtract it to DB from DC. Then, find area of triangle DBA and subtract it from the area of triangle DAC. Finally, the area of triangle ABC is equal to (x*AC)/2

[night idol]

Woahhhhh

That actually makes sense, THANK YOU

[smoky jetty]

np!

[upper karma]

vertically opposite angles

[hardy night]

u can find BC by use 2 similar triangles DBA and FBC. It's faster

[smoky jetty]

thought of that as well, but how could we prove that they are similar?

[silent plank]

AA

[smoky jetty]

but wouldnt it require to do the sohcahtoa thing?

[silent plank]

no

[smoky jetty]

how come

[silent plank]

AA is sufficient to justify similarity

vert angles + right angles

[hardy night]

it's too easy. You cann't prove it ??

[hardy night]

yeah, that's right

[smoky jetty]

oh right, lol my bad for not seeing it

yeah that's surely a better way

[tardy swift]

trig is hard

[pseudo viper]

would cross product fall under geometry?

[weary hound]

Perhaps but isn't that more of vectors

[pseudo viper]

hm true

[west lintel]

does anyone know where I can get a ti 84+ for the cheapest price possible or a ti inspire for the lowest price

[upper karma]

how would i do these

[gilded hatch]

Which one?

Use for the tasks 14 and 15 the intercept theorem 🙂

[upper karma]

idk that

[upper karma]

13.14.15

all e of them

3*

[gilded hatch]

Hmm I see

For 13 you can use the pythagorean theorem

Do you know that?

$a^2+b^2=c^2$

[somber coyote]

gilgamesh5000

[upper karma]

yes ik that

but idk what to subsitie for a , b and c

[gilded hatch]

Solve the first equation after that the second...

[upper karma]

ohhh

so i do 15^2=h^2+9^

and then i solve the other

[gilded hatch]

ye

[upper karma]

okay

give me a min

for the first one

h=12

[gilded hatch]

Yes

[upper karma]

for the 2nd one

do i solve for h or x?

[gilded hatch]

You have the value for h 😁 it's 12

[upper karma]

ohhh

[gilded hatch]

You can only solve for x😁

[upper karma]

that makes more sense

so that's the answer for 13?

[gilded hatch]

$20^2 = 12^2 + x^2$ solve it

[somber coyote]

gilgamesh5000

[upper karma]

okay

x=16

[gilded hatch]

Nah

Wrong

[upper karma]

how

[gilded hatch]

$20^2 = 12^2 + x^2 \mid -12^2 \ 20^2 - 12^ 2 = x^2 \
400 - 144 = x^2$

[somber coyote]

gilgamesh5000

[gilded hatch]

Now you can calculate

[upper karma]

yea so 400-144=256

and the 256=x^2, so u square root 256 to remove the ^2 from the x

and you get 16

[gilded hatch]

Oh yes

My bad😅

[upper karma]

it's okay lol

so that's my final answer right?

[gilded hatch]

Yes

[upper karma]

ahhh

[gilded hatch]

x=16 is your final answer

[upper karma]

yes

[upper karma]

do i do the same with 14 and 15?

[gilded hatch]

With intercept theorem

[upper karma]

what's that

[upper karma]

do you still need help?

[smoky jetty]

it'd only work if they're right triangles right? tho not indicated that theyare

[lapis torrent]

Someone please suggest me some good geometry problems.

[smoky jetty]

look up AOPs problems

[upper karma]

no i finally understood it lol

[sharp lodge]

<@&286206848099549185>

[last quiver]

you can use thales theorem for part a

[sharp lodge]

thalas..?

[languid pawn]

If three points A, B, and C lie on the circumference of a circle, whereby the line AC is the diameter of the circle, then the angle ∠ABC is a right angle (90°).

[sharp lodge]

oh

what abt part b?

<@&286206848099549185>

[magic otter]

take coordinates (rcosQ , rsinQ).

Parametric coordinates , got it ? or i tell ?

[sharp lodge]

i dont knw all that yet

im in highschool..

[last quiver]

it can be done with Pythagoras theorem as well

[sharp lodge]

how

we only knw one side

[last quiver]

Considering O as origin , can you figure out the angle BOC ?

BC = 3cm

Radius = 3cm

OB = 3cm

What triangle is this ?

[sharp lodge]

a right angled.. according to part a

NOO its not right angled

idk whut it is

this is why i hate math

[last quiver]

Ahh sorry I didn't specify properly, what triangle is BOC

[sharp lodge]

but isnt the question asking for abc..?

[cerulean robin]

I know you are asking for help, but don't spam @ Helpers (sorry helpers discord automatically joined the mention).
Spam in this server is usually defined as more than once per 15 minutes, refer to #info - rules

[sharp lodge]

oh srry i didnt knw that

[last quiver]

Yes but it's a step to reach the answer

[sharp lodge]

ohh

[cerulean robin]

no problem

[sharp lodge]

@last quiver i dont get u...

why would we need boc for the asnwer?

[magic otter]

it can be done using straight lines

[sharp lodge]

this is so confusing 😭

[last quiver]

We need to figure out the coordinates of point C and in order to do that first we try to figure out that the triangle BOC is equilateral triangle (and argue that the x coordinate of C lies midway on x-axis) and then apply Pythagoras theorem to figure out the y-coordinate of point C

[sharp lodge]

ohh

[last quiver]

Although I'm not sure if there's a shorter (and easier) way to do this.

[last quiver]

So can you figure out what (type of) triangle BOC is (given that BC = BO = OC = 3cm) ?

[sharp lodge]

but the problem is we dont knw what type of triangle it is

[magic otter]

sending solution in 5 min . please wait .

[last quiver]

hmm it's a equilateral triangle since all sides are equal

[sharp lodge]

how do uk all sides are equal tho..?

[last quiver]

It's given BC = BO = OC = 3cm

[sharp lodge]

theey only gave bc

[last quiver]

Yes and the radius is 3cm which gives the length of other two sides (which are radii themselves)

[sharp lodge]

OHH

[last quiver]

So can you now figure out what's the x-coordinate of point C ?

[sharp lodge]

is it 1.5?

[last quiver]

Yes

[sharp lodge]

so AB= 6 and bc = 3

we can use phytogoras theroem ..?

oh nvm we dont need to find AC

[sharp lodge]

<@&286206848099549185>

[lusty acorn]

sure

1230 A and B are 2.5 km closer compared to midday

[lapis torrent]

If c is not point a or b , then acb = 90

[cerulean robin]

This is closer to correct usage, but we request that you wait 15 minutes after posting your question at least before you DM someone.

[sharp lodge]

it already was 15 mins bro

[lapis torrent]

Bro the bot is cracked

[runic crag]

15 minutes after posting your question.

[runic crag]

What part do you not understand?

[sharp lodge]

OH

[sharp lodge]

c

[runic crag]

Oh. Just convert m/s to km/h and calculate both of their distances, then find the distance between them.

Since they've been moving from 12:00 to 12:30, time elapsed is 1/2 hour.

If you've still got doubts, ping me.

[sharp lodge]

okay

[runic crag]

Have you got doubts?

[sharp lodge]

how to convert into km/h ....i forgot..

[runic crag]

,tex $\frac{xm}{1s}=\frac{\frac{x}{1000}km}{\frac{1}{3600}h}=3.6\cdot\frac{xkm}{1h}$

@sharp lodge

[sharp lodge]

okayy

[somber coyote]

rageiplier

[marsh pagoda]

The angle between h and x is not said to be 90°

[upper karma]

Gang what do l have to do here to find the diameter

[silent plank]

<@&268886789983436800>

[wind coral]

<@&268886789983436800>

[grave pond]

Oh, they spammed in multiple channels and then left.

[smoky jetty]

a fun geometry problem I found online

solved it in 15 mins, actually easier than I expected. I bet yall could get it faster

[smoky jetty]

are 1,6 and 0,6 the midpoints?

[foggy parcel]

oo the shit's crazy af

goin to grab some papers

[carmine crag]

GH is of 1 unit

draw a line parallel to gh from b it will be of same length

consider the point where the line from b cuts GC E

wait whats the length of GC and HB

[flat kelp]

6 units?

Wait what

[floral shale]

How much you wanna bet it's just 48

[foggy parcel]

nah basically that's the one lol

@smoky jetty

[rustic yew]

how do u do this?

im so confused

[fluid stream]

What do you know about the angles of a parallelogram

[rustic yew]

umm opposite angles are congruent

so angle a = angle c

[fluid stream]

Ok

Anything else?

[rustic yew]

uh angle a + angle b = 180 degrees

[fluid stream]

Ok good

Now you have two equations and two variables

[rustic yew]

ye

so 100 + 13x = 180 right @fluid stream

[fluid stream]

Yes that's what you get from one, now you can solve for x

And then you can write the other one and solve for y

[keen gorge]

confused

[smoky jetty]

which one?

[timber cargo]

I got 37.12594146315

[upper karma]

GC length is 1,6 and BH length is 0,6

[clear remnant]

https://youtu.be/pUSI91Mo7LU

[upper karma]

Adjacent Angles Of Parallelogram Is Always 180 Degrees

[sturdy dawn]

hi can anyone tell whats an peremittre triangle??

plz pretty urgent

[errant lake]

cant u just Google it

I Need help with this problem

this is what I’ve done so far

And btw there’s no trig involved cuz the lesson is geometrical transformations like rotation translation dilation and stuff like that

[grave pond]

I don't see how you can hope to end up with degrees without doing some trig along the way.

[upper karma]

💀

[spring oxide]

First thing what number is larger ET or TF?

[smoky jetty]

not quite, but can you show your sol?

[timber cargo]

Yesterday, I realized that I made a mistake on that problem

Quite early into it

but I'll redo it

[smoky jetty]

ight gl!

[upper karma]

how do i do this? Also option D is AAS

is the answer B?

[dull light]

no

uh look at the side ratio

[upper karma]

i am

[dull light]

then why is it AA

you only have one pair of equal angles

[upper karma]

ahhh

wait what is it then

[dull light]

SAS

two pairs of side ratios

[upper karma]

okay that makes more sense

how do i do this

[smoky jetty]

what do u know about similar figures' properties?

specifically, what is the relationship between fig 1's angles and sides to fig 2's?

[upper karma]

all the angles are congruent

like all corresponding angles

[smoky jetty]

how about the sides?

[upper karma]

uhhh

yea idk

[smoky jetty]

the corresponding sides are proportional

u can think of it like this, the figures are like the images u scale up or down in MS Word

[upper karma]

yeah okay

so i make an equation

[smoky jetty]

yeah

[upper karma]

ahh okay

idk how to make the equation

[smoky jetty]

.

[upper karma]

yea ik that

OH WAIT

wait nvm the equaton i made doesnt make sense

[smoky jetty]

now u could set ratios that would make them in proportion

and frmo the given figure, u can have more than one approach, as long as they are in proportion

[upper karma]

so like x=16, 38=19 right?

[smoky jetty]

clearly, 38 is not 19 isnt it?

[upper karma]

yea

[smoky jetty]

right, so what ratios of corresponding sides do u think could u set in proportion?

meaning the value of their ratios would give the same answer/quotient

[upper karma]

um idk

im lost

[smoky jetty]

i.e what sides from the larger figure is comparable to the smaller figure?

[upper karma]

x and 16?

[smoky jetty]

there

and what other sides are comparable

[upper karma]

38 and 19

[smoky jetty]

right

and a way we could compare things is by setting a ratio

[upper karma]

yea

[smoky jetty]

and when ratios are in proportion, it means that the two set of things that we compare are equal. Simply an equation of ratios

[upper karma]

the answer is B

[upper karma]

yea

[smoky jetty]

yup, and could u explain how u were able to get it?

[upper karma]

I guessed cuz the time was gonna end

[smoky jetty]

ohh, lmao i didnt expect that

[upper karma]

hahaha I was lucky

I'm gonna study my mistakes tho

[smoky jetty]

but i'd suggest at least reading more about it on the internet or wherever u could, coz u seem to be confused on similar triangles

yeah that'd help surely

[upper karma]

yea i don't understand them at all

i got all the similar triangle questions wrong

[smoky jetty]

u can look up khan academy and browse videos all about similar triangles, they're free online

[smoky jetty]

did u listen in class? or there was some trouble u encountered while it was being discussed

[upper karma]

it's an online course

so there is no class

just work

[smoky jetty]

oh

[upper karma]

but the hw she gave us was so easy

and this assinment was sm diff than the hw's she gave

[smoky jetty]

https://www.khanacademy.org/math/geometry/hs-geo-similarity/hs-geo-triangle-similarity-intro/v/similar-triangle-basics

[smoky jetty]

welp, might be a call to look into it deeper

[upper karma]

yea i will

[upper karma]

option D is ASA

i think the answer is C

so it's c?

and this is C

[elfin osprey]

can anyone help me with a two column proof problem?

[trail trellis]

I struggled with these too lol

[elfin osprey]

what should i do lmao theyre so hard

[trail trellis]

I learned in a different language so I don't know but

They're hard right lol

[elfin osprey]

yeah

[trail trellis]

Did you learn theorems for it

[elfin osprey]

a few but not enough for me to do this problem correctly

i'd know how to put the theorems down once i heard them but i dont know what to look for yk?

idk the first step

and i dont know theorems well enough to do it in reverse

[trail trellis]

Ohh ok

Do you know the SAS theorem thing

[elfin osprey]

yeah side angle side, you can tell if two triangles are congruent if they share a side, an angle, and another side. right?

[trail trellis]

Yes

I think you would use these so first say that Angle RYM and Angle TYN are vertical angles i guess

And Angle MYT and RYN are vertical too

I would start from that

[elfin osprey]

ok let me write that down

[trail trellis]

Oh wait sry im wrong i think

[elfin osprey]

yeah im trying to figure this out it doenst sound right lol

i need to prove that rym and ryn are congruent

[trail trellis]

yeah

Huhhh you could prove that Triangle MYT and Triangle NYT are congruent

[elfin osprey]

they share a side and two angles

ASA

[trail trellis]

yes

[elfin osprey]

since theyre congruenbt

we know that lines ny and my are congruent

[trail trellis]

YES

that would be one side

[elfin osprey]

and then triangle rny and rmy share TWO SIDES because ry!

[trail trellis]

yes

[elfin osprey]

lemme write this down un momento

[trail trellis]

one angle

okay

[elfin osprey]

wait i think thats it

[trail trellis]

Angle MYT plus Angle RYM would be 180 degrees right

[trail trellis]

hm

rlly

[elfin osprey]

i think thats all i needed to prove

i need to prove rym and ryn are congruent

[trail trellis]

Oh?

[elfin osprey]

they share two sides

oh wait i need to prove the angle

[trail trellis]

But it says Triangle not angle

yea

[elfin osprey]

yeah i meant triangle

[trail trellis]

i think we had a way to prove from this

[elfin osprey]

yeah it would be 180

so we can FIND THE ANGLE

ok so myt plus rym equals 180

[peak siren]

da fuq is an angle

[elfin osprey]

which means

angle rym and ryn are congruent

and thats

two sides and an angle?

my=ny, ry, and rym/ryn

SAS?

[trail trellis]

Yeah i think how to solve this is right

[elfin osprey]

we got it

two column proof done

tysm

only 3 more! :))

[trail trellis]

Yay

[trail trellis]

Np!

[trail trellis]

lol

We already have one side done tho

IP and TP

[elfin osprey]

and if bmi and kmt are congruent,

then thats

ok so ipk and tpb share ip and tp

tpb contains tp

and ipk contains ip

and theyre congruent

so thats a side

[trail trellis]

Wait Angle B and Angle K is

[elfin osprey]

oh yeah cuz those are congruenbt

[trail trellis]

Yeah !

You would have to find the theorems tho lol

[elfin osprey]

i can figure those out with google those arent the issue lol

[trail trellis]

Oh okok lol

We can't do ASS so we would need one more angle

so that we could do AAS

[elfin osprey]

side pk and bt are congruent

right

couldnt it be sas? side 1: ip and tp, angle b and k and side pk and bt

[trail trellis]

Why pk and bt?

[elfin osprey]

bmi and kmt are congruent

and ip and pt are

so adding the same length to the same lenght

wqait

i said the wrong thing

pk and bP

[trail trellis]

Yes that would be right, but i think that would be ASS

[elfin osprey]

i dont get the difference lol? if you have two sides and an angle cant that just be sas? like what determines that

anyway

heres the next one

yo.

this is easy

wait

triangle must add up to 180

[trail trellis]

If the side is next to the side it would be ss if there is an angle next to the side it is sa

[trail trellis]

yes

[elfin osprey]

ok lets do this easy one and go back to that

[trail trellis]

okkk

[elfin osprey]

if one side is 60

then they all do

because its isosceles

[trail trellis]

oh yeah

this one would might be tricky to find the theorems tho

[elfin osprey]

that should be fine

[trail trellis]

ok!

this is geometry right

btw

[elfin osprey]

yeah

geometry proofs

[trail trellis]

am i allowed to ask what grade u do geometry

[elfin osprey]

yeah so normally it would be 10th or 11th at my school, but im in a faster and higher course so for me im taking it my freshman year, but for most else they take it 10th or 11th

[trail trellis]

wow

[elfin osprey]

a lot of people will be doing algebra in freshman yeaer

while i did that this past year

[trail trellis]

so u in 9th?

[elfin osprey]

yeah

[trail trellis]

ME TOO

[elfin osprey]

YO

[trail trellis]

but im in algebra2 lol

[elfin osprey]

yeah im taking that 10th lol

[trail trellis]

lol

everybody in my grade is in geometry or algebra1 lol

[elfin osprey]

makes sense lmao

everyone in my grade is either algebra which i took last year or algebra 2 which i take next year

cuz im doing honors geometry freshman year

[trail trellis]

lol

ohhh

[elfin osprey]

in the middle

[trail trellis]

nicee

im lonely tho lol

[elfin osprey]

wdym

[trail trellis]

cuz im the only one in algebra2 in my grade

[elfin osprey]

the only one

like its

you and a teacher

thats it

[trail trellis]

and the older people

but all boys

[elfin osprey]

oh no thaat must suck

in my school

[trail trellis]

YES

[elfin osprey]

the good seventh graders

take algebra t/a with the older kids

[trail trellis]

ohhh

[elfin osprey]

and move onot honors geometry 8th

and then

algebra 2 ninth

[trail trellis]

thats nice

[elfin osprey]

but i did algebra t/a eight grade, which means i go to geometry, not algebra two. so basicallty the seventh graders who were in my class this year are a year ahead of all of us

and two years ahead than the normal math people

[trail trellis]

ohhhh

good system

[elfin osprey]

yeah im rlly scared tho. only because of these damn proofs theyre so hard

[trail trellis]

IKR PROOFS ARE SO HARD

[elfin osprey]

YEAH

ok so theres only a few mroe quesitons i think

[trail trellis]

okkk

[elfin osprey]

wait this is easy too

[trail trellis]

lol

[elfin osprey]

it divides x into two equal segments

and its equilateral

[trail trellis]

yeah

its difficult when u have to use the theorems tho haha

Especially in tests

[elfin osprey]

xjk and xjk triangles and angles are congrent

lamo

jesus

wait

xjk

and

xjf

there

[trail trellis]

lol

Yes then KJ and JF is

congrent

[elfin osprey]

lmao

[trail trellis]

so J is the midpoint?

[elfin osprey]

yep

all done

[trail trellis]

yay

[elfin osprey]

let me see if i have more tho

[trail trellis]

okkkk

[elfin osprey]

idt i have more proofs but i def got more work

[trail trellis]

okk

[elfin osprey]

heres the thing my tutor

assigned me those proofs

and then gave me some other pdf but he never said what to do in it

[trail trellis]

What lol

[elfin osprey]

lemme text him

[trail trellis]

okk

[elfin osprey]

but also

tysm for helping me so much

like rlly

[trail trellis]

Np!

Any time

[atomic osprey]

If I have a circle lying on a larger sphere say on the southern hemisphere and I know that the circle has circumference 10, how can I figure out the distance from the south pole to a point on the circle? From the circumference I can find that the radius of the circle is 5/pi, but I think I would also need something else to find the distance.

[lilac cave]

Hello

[timber cargo]

This is what I got, approximately: 36.53096278

I found the equations for the quater circles and integrated with respect to each of the functions and bounds

[errant lake]

how bout u actually think and try before u say such statements

u can use 30,60,90 triangles to solve this problem by rotating the figure 90 degrees counter clockwise

which is what I partially did before u said what I was doing what hopeless

[errant lake]

what r u laughing about

[upper glen]

my solution

I'll use Google Translate to explain but I don't know if it will translate correctly

first I used marlen's theorem to find the other segment that also contains 2V3, such a theorem says that the sum of the squares of these segments generated by the opposite vertices are equal, in this case for example I said that V6 for the square plus 3V2 for the square was equal to 2V3 for the square plus x for the square such x that gave 2V3 as well, This theorem holds only for when we have a quadrilateral with all its internal angles worth 90 degrees, a rectangle

after that I noticed several congruences of triangles but the one that will serve me is the one that says that the EFT triangle is congruent to the EHT triangle, because it will tell me that those angles described will be worth 45 degrees each

congruence by SSS

I connected a segment of the vertex T next to the square doing 90 degrees

after that I noticed an isosceles triangle 90,45,45 and calculated its sides, I found that they were worth V3 and then in the other right triangle I used Pythagoras and found that the other side that was missing was 3, and this way we discovered a triangle 90,60,30 and solved the problem

[smoky jetty]

seems pretty cool

[smoky jetty]

yep! thats correct

[indigo escarp]

He’ll

Hello

I need help pls

[timber cargo]

Yeah, how did you approach it? Did you do what I did as well

[smoky jetty]

My approach was this: Get the area of the segment of the quarter circle with radius=8 and subtract the area of segment of the quarter circle with radius=4 from it. Next, get the area of the segment of the quarter circle with radius=12, and subtract the area of the quarter circle with radius = 8 from it. Finally, add the 2 areas we've got, i.e 22.84 unit^2 + 13.69 unit^2. Hence, the shaded area = 36.53 unit^2.

[frozen ocean]

[rustic yew]

how do u do this?

somebody halp

is it 120?

[rustic yew]

im taking geometry in 7th grade, is that normal?

[timber cargo]

Yeah, it was a fun problem

[hoary dove]

hi i need hel

help

like asap

[onyx basin]

do you know the sum of angles inside a triangle?

[hoary dove]

yes 180

[onyx basin]

ye

[hoary dove]

x is 90

[onyx basin]

yep

[hoary dove]

what are the others lol

[onyx basin]

this forms a triangle in of itself

you know x is 90

and you also know that another angle is 70

and that x + y + 70 = 180

so what would y be?

[hoary dove]

so why is 20

and z is 20?

[onyx basin]

yes and yes

[hoary dove]

so y is 20

pog thanks

[elfin osprey]

in my school thats not even possible

[rustic yew]

lol

my county allows u to go 2 years ahead in math after 6th grade and I am taking extra summer school to get another yr ahead

[elfin osprey]

what country might that be lol

[rustic yew]

if ur wondering why im doing this, its because im indian this is very normal

[elfin osprey]

i was just abt to say

i am too

[rustic yew]

this is america

[rustic yew]

then u must be a doctor or IT employee eh?

[elfin osprey]

yeah no i get it im doing the entire honors geometry course right now before school even starts so i can stay ahead

nah nah im a hs

[rustic yew]

ok

[rustic yew]

good luck having headaces

[elfin osprey]

yeah :/ this sucks ass im literally sitting here right now doing points of concurrency

[rustic yew]

very simple, I learnt this in kindergarden

[elfin osprey]

aint no way

[rustic yew]

did u forget, im indian

[elfin osprey]

damn im not that indian

[rustic yew]

I would have been slapped my parents and grandparents if I refused

and my great grandparents would have rised from the grave

[elfin osprey]

jesus

[tribal rover]

Can someone explain this please

[upper karma]

Cos of an angle is always equal to the adjascent side/hypotenuse..

Therefore cos 60°=Adjascent side to the angle 60°(1 unit)/Hypotenuse(2 units)
Which is equal to ½

[ember stratus]

P, Q, R and S lie on the circle O ( r ) and are the points of tangency of O 3 ( r 2 ), O 4 ( r 2 ), O 5 ( r 3 ) and O 6 ( r 3 ), as shown.
O 1 ( r 1 ) and O 2 ( r 1 ) are tangent to each other and are tangent to O 3 ( r 2 ), O 4 ( r2 ), O 5 ( r 3 ) and O 6 ( r 3 ), as shown.
Find r 1 as a function of l and L .

[upper karma]

oh wow im doing geometry over the summer to get ahead too lol

i guess it is a pretty common idea

[median vapor]

Hey so I know that you have the standard trig functions and then hyperbolic trig functions (represented by slapping an h to the end of normal trig functions). Are there any other trig functions that follow the naming convention of slightly modifying the spelling of the base trig ratios?

[upper karma]

this question is the hardest question of this year's hong kong diploma of secondary education exam, involving coordinate geometry and trigonometry

the marking scheme is still classified and therefore hk students cannot get them

how will you answer this question if you were the students sitting in that exam?

[languid olive]

math

[exotic yarrow]

Here’s my first impressions:

(a) Straightforward, observe that Q is the midpoint of PR
(b) (i) Angle between two lines formula
(b) (ii) Straightforward
(b) (iii) Angle bisector theorem then find the incentre. Then, do shoelace formula.

Maybe there’s a better way, but that’s my first thoughts

Ok I just did it, turns out that O, Q, G, I, H are collinear

Heights are the same because Q is a midpoint, so 11/12 (compare horizontal distance)

[upper karma]

I hate this class

[frozen ocean]

it's my favourite class 💜

[raw cobalt]

could someone help me doe this

[upper karma]

m<GME intercepts the big arc GE which is 164 + 62 = 226. Since m<GME is an inscribed angle, it means that it is half the angle it intercepts. 226 / 2 = 113

[heavy crow]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[upper karma]

Oops, sorry about that

[tame dove]

any one got

a algebra 2 cheat sheet

for a final

aka 2nd semester

[sleek inlet]

HOW DO I DO 17 AND 20

[dark sparrow]

@sleek inlet DO YOU STILL NEED HELP WITH THESE

AND WHY ARE YOU YELLING

IN ANY CASE, THE SOLID IN #17 IS A BOX WITH A PYRAMID STUCK ON TOP OF IT, WHILE THE ONE IN #20 IS A BOX FROM WHICH THE PYRAMID HAS BEEN CUT OUT!

IN THE SECOND CASE THE BASE OF THE PYRAMID IS A SQUARE, AND ITS APEX IS PRESUMED TO LIE ON THE BOTTOM FACE OF THE CUBE!

[thin locust]

ive heard that the reason why you cant prove the pythagorean theorem with a lot of trigonometric stuff is because a lot of trigonometric formulas are based om the pythagorean identity (sin²x + cos²x = 1) which is based on the pythagorean theorem (-> cyclic proof)

but cant you avoid this by proving the pythagorean identity using eulers formula which you can prove with a derivative

or is the proof for the derivative of sin and cos also based on the pythagorean thworem which would make it cyclic again

?

like
sin²x + cos²x = (cosx +isinx)(cosx - isinx) = e^ix * e^-ix = 1

[nocturne remnant]

What if:
Define e^x using power series
Define sin and cos using e^ix
Then everything should be good

[heavy crow]

yeah this works. I did it a few years ago, it's interesting. you can define cos and sin as the real and imaginary parts of e^ix and prove that they describe trigonometry. weirdly, you can also define pi in terms of e as making α the first positive zero of cos. then you can prove that the radius of a circle is 4αr=2pi r which means α=pi/2. something like that I may have the exact details wrong

I actually started by defining $L(x) \defeq\int_0^x\frac{\dd t}{t}$, showing it's inverse $E(x)$ is exponential in nature, defining $e\defeq E(x)\implies E(x)=e^x$ and going from there.

[somber coyote]

nixxy nilpotent (raving lunatic)

[heavy crow]

etc. I did it because I thought it was weird that the Taylor series definitions for sin and cos would describe trigonometry, and i wanted to know why.

[frozen ocean]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[violet kiln]

How can this question have 2 answers

I found that x=30

But in the ms it's written 30 AND 210

So how come 210 is an answer???

This is the ms for the question
How is x=30 and x=210 ???

[lost marlin]

30° is a notable angle and you can see the 210° on a trigonometric circle

Btw, you also could have thought what angle that it sin divided by it cosine is equal to 1/sqrt3. You would find the exact same answer

Draw a line that tangent the 0°. The tangent of each angle will be where its extension touches. For example. Take the 210° and draw a straight line all the way to the right. You’ll see that it will pass right over the 30°, so its tangents are equal

[snow crystal]

because tangent has a period of pi, so it has the same value every 180 degrees

[smoky jetty]

afaik tan of 210 and 30 degrees are equal?

30 degrees is the reference angle of 210 in the 3rd quadrant, right?

[warm bramble]

Basically you have tanx=tan(π/6) so x=kπ+(π/6) with k being an integer but now you need to find k
It also gives you that x is between 0 and 360 or in other words 0 and 2π
With that in mind you have 0<=kπ+(π/6)<=2π
If you keep going to solve for k you get -1/6<=k<=11/6
Now combining that with the fact that k is an integer the only options are k=0 and k=1 so you go to x=kπ+(π/6) and replace k with 0 and 1 to find the 2 results
(εφx is tanx, εφθ is tan(any number) and θ is well, any number)

As a simpler answer you can say that tanx=tan(π/6) and keeping in mind that tanx=tan(x) or tanx(π+x)

The answer kinda depends on the material you've covered as well

[lost marlin]

yes, that's correct

[upper karma]

Thx

[whole venture]

Is there any derivation of the trig functions which doesn't require analysis to understand

[prisma wolf]

i need help

[whole venture]

Don't we all

[frozen ocean]

!help

[lime crown]

Please read #❓how-to-get-help

[prisma wolf]

Anybody got me time with this??

[sacred aspen]

no

poopy

[sullen isle]

Hello all! I'm having trouble seeing why this construction works

Any help would be appreciated!

[polar current]

i’m confused on how to solve for the lowest part of the roof and idk if i did it right

[somber coyote]

jahobinthehomeless

[somber coyote]

jahobinthehomeless

[untold crypt]

We know that the number of degrees of an n sided object is 180*(n-2)

But we also know that a circle has 360 degrees

And a circle has infinitely many sides

As n approaches infinity, [180(n-2)] approaches infinity, so how can a circle have 360 degrees?

Like this is a hectogon, a 100 sided object

It has 17640 degrees

[somber dome]

isnt that a circle?

[lost marlin]

when we talk about 180(n-2), we are talking about the sum of the intern angles, that are formed by the vertices of a convex polygon. The 360° degrees in a circle corresponds to the central angle. they just have different natures

if you divide a circle in 4 identical parts, and take those lines you used to divide and put them all together, you’ll be able to form a square

i’ll try to illustrate

see those lines? the ones that forms 4 90° degree angles?

imagine they’re 4 lines, not 2. they’re all the exact same size. if you rearrange them, you’ll have a perfect square

[warm bramble]

CD is an altitude, that's given
Since it's an altitude on a line, it cuts the 180 degrees in 2, so D (of triangle ADC) = D (of triangle BDC) = 180/2 = 90 degrees so DACD and ACBD are right triangles
We know that sinA=opposite/hypotenuse. The opposite of A is CD or in other words h while the hypotenuse of the triangle ADC is b so sinA=h/b
With the same logic for the triangle BDC sinB=h/a
Using that sinB=h/a you can get a(sinB)=h if you just multiply both sides by a and the same goes for sinA=h/b which, if multiplied by b gets b(sinA)=h
You have that both a(sinB) and b(sinA) are equal to h so a(sinB)=h=b(sinA) so a(sinB)=b(sinA)
For the last one you use what we just proven above [a(sinB)=b(sinA)] and you just devide both sides by a*b which gets
a(sinB)/ab=b(sinA)/ab
On the first part the a is simplified and on the second one b
so sinB/a=sinA/a