geometry-and-trigonometry

[shut swift]

can someone help me here please?

[chrome fiber]

since they're congruent, x^2 - 10 = 5x + 4

solve for x

then use pythagoras theorem in any one of those to find the height

then area is 1/2 * 15 * height

times 2

[shut swift]

thanks

[chrome fiber]

np

[latent canopy]

A radius = 4

B radius = 2

how can we know small circle radius?

step by step

[upper karma]

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

[timber hinge]

@latent canopy still need help?

[latent canopy]

yhea

[languid lodge]

Looks like a problem that I solved from a math competition.

I will have the solution within 30 minutes if that not too late.

[languid lodge]

Oh wait, nvm. I can find the small circle radius if I were given the distance between the bottom centers of circle A and B.

Then I would be able to solve it using systems of equations when connecting the radii to the centers of B to smaller circle and centers of A to smaller circle.

[latent canopy]

I only have that information

Can anyone still help me?

[upper karma]

@latent canopy You might want to give people that super helpful link you showed me yesterday

[serene field]

Label the points of tangency. Find the relation between each point of tangency (will involve each radius), and create the general formula.

[frigid wasp]

@limpid basin ik it was a few days ago but what did u mean by law of cosines?

or about the question i asked in general

[latent canopy]

@serene field how am i suppose to do that?

btw im really confused so...

[umbral snow]

@gilded needle
Doesn't seem like enough info

[gilded needle]

@umbral snow ?

[umbral snow]

You're still missing an important part. I see now that you are breaking a vector into its components, and whenever you do that, the angle opposite the original vector is 90°

As such
Gy = 9.81cos(55)
Gx = 9.81sin(55)

This is a general pattern. Whenever you break a vector into its components:
The opposite side is hsin(θ)
The adjacent is hcos(θ)

[marble perch]

How do we find x and y? And I would!’t know how to do AC

[upper karma]

EDC is a right angle, so that should help you solve for x

[errant ether]

DE bisects AC so AD = DC = AC/2

[marble perch]

yeah

how do I solve: 6y+10=y2 + 3y

[serene field]

Bring every term to one side and apply the quadratic formula?

I assume that's y^2.

[marble perch]

y^2*

yep

[surreal cipher]

Guys, quick question. What regular prism has the best V:SA ratio?

[rain tulip]

icosahedron?

@surreal cipher

icosahedron has lowest SA : V ratio so therefore it must have highest V: SA ratio

[nova pine]

What is the area of an iscosceles triangle with a hypotenus of 20cm?

[meager lantern]

Do you mean like this?

[surreal cipher]

@rain tulip
This is my problem.
I need a regular polygon to use as a face for a prism that extends 22 feet. What regular polygon is the best face choice?

[hazy granite]

@nova pine
x^2 + x^2 = 20^2
2x^2 = 400

2x = 20

x = 10

[grim sorrel]

Yeah $$\sqrt{2x^{2}}=\sqrt{2}x$$ and not 2x

[sick veldt]

Are these two questions of mines correct?

[hard gale]

Yus

[sick veldt]

thanks

[upper karma]

inverse tangent is defined as:

[somber coyote]

Kylixir:

[upper karma]

is there a way to define a variation of arctan such that

[somber coyote]

Kylixir:

[upper karma]

Or should I just define my own function as

[somber coyote]

Kylixir:

[valid agate]

If you really need it I would say define your own function

It's somewhat unusual tho

[upper karma]

$arctan(\frac{Qx}{P})$

[somber coyote]

Kylixir:

[upper karma]

Looks like I don't need to make my own function after all

[valid agate]

Um

Oh

yeah

That works

[fresh axle]

https://i.gyazo.com/49d41054624cc2f28b02cb582aac4f44.png

why is using arctan((-8) / 2) followed by adding 360 not the correct answer?

i'm confused over this

[upper karma]

One significant digit after the decimal point

[fresh axle]

no, the problem is that the correct answer is 436.0

[upper karma]

Like 284.1 (or whatever)

[fresh axle]

the german part at the bottom states: "the right answer is 436.0"

[upper karma]

Well, that can never be right for degrees. And I don't know of another unit that goes that high. Even gradians only go to 400

[fresh axle]

huh, guess i'll have to ask my professor for more clarification

thanks for the help

[upper karma]

Sure

[thorny solar]

Can someone help me with this? My teacher didn't explain it at all

Sorry about that white thing, it's my dogs tail

[upper karma]

This is pretty much vertical and corresponding angles stuff. Unfortunately, the whole picture is a bit fuzzy. If no one else offers to help in 5 minutes, feel free to PM me

[thorny solar]

Ok

This should be less fuzzy

[upper karma]

Better. Remember vertical angles or PM me if no one else helps

[upper karma]

How does 2sinx - 2sin^3x

Factor to

2sinx(1-sin^2x)

Someone pls

[jaunty plume]

because 2sinx(1-sin^2x) expands to 2sinx - 2sin^3x

[heavy escarp]

Can someone do (i)

[limpid basin]

First let $$x = (1 + i \tan\theta)^n + (1- i\tan\theta)^n$$. Notice that multiplying x by $$\cos^n\theta$$ yields $$x(\cos^n\theta) = (\cos\theta + i\sin\theta)^n + (\cos\theta - i\sin\theta)^n$$

[somber coyote]

Plum:

[limpid basin]

Weird bot

Also you can finish it using de moivres

[steady sleet]

[heavy escarp]

I'm still a bit confused

[limpid basin]

About what

[mellow scroll]

I've been stuck on this problem for the past hour, I feel like the solution is literally in my face but I can't see it.

Could someone give me a little help?

😃

[rain tulip]

Is that from egmo

@mellow scroll

[limpid basin]

Seems like it

[mellow scroll]

Yeah, it is

@rain tulip

[peak talon]

Hey guys, I'm a little confused about trig on the Cartesian plane. When calculating theta, it's always between the x axis and the hypotenuse right? But in my book it says something about "180-theta" when it's in the second quadrant. What does that mean / when do you use it?

[mellow scroll]

Could you take a picture @peak talon ?

[peak talon]

Uhh my phones camera is broken but I'll try find an example online and screenshot

[mellow scroll]

Alright

And by calculating theta, I presume it is to solve for the possible values of theta given that some trigonometric function, when applied to theta, equals <insert arbitrary ratio>.

[mellow scroll]

AH yes

[peak talon]

It's just confusing when to use theta and when to use 180-theta or whatever

[mellow scroll]

Alright

The 180 - theta stuff

is to calculate for something known as the "basic angle"

do you know what the basic angle is?

[peak talon]

Nope

[mellow scroll]

Alright

Do you have a calculator atm?

[peak talon]

Yea

[mellow scroll]

Try putting in tan(135 degrees)

[peak talon]

-1

[mellow scroll]

Correct

now, 135 degrees is in the second quadrant right?

[peak talon]

Yeah

[mellow scroll]

Now try tan(45)

[peak talon]

1

[mellow scroll]

What is common between the 2 values?

[peak talon]

They're both 1

Just ones negative

[mellow scroll]

In other words, the have the same magnitude

The basic angle is the acute angle that when put through the same trigonometric function, will spit out a value of the same magnitude as the theta.

[peak talon]

So it's like the interior angle?

[mellow scroll]

Not quite

[peak talon]

Cause what confuses me is that like in my book there's an example thats a triangle in the second quadrant and theta goes from the x axis to the hypotenuse and it says theta is between 90 and 180 but then when you work out theta with shift sin or whatever it returns something like 22 degrees

[mellow scroll]

yrd

yep

[peak talon]

I've never heard of that lol.

I'm in grade 10 if that matters

[mellow scroll]

It basically means the value of your inverse sin (arcsine) will only give you an angle between 90 degrees and -90 degrees.

pi/2 to -pi/2 in radians

?

[peak talon]

Okay I don't wanna get too ahead 😂 I just wanna know, why does it say theta is between 90 and 180 but then when you work out theta it's 22 in my example? When you work out that are you working out the interior angle so to get the whole angle it's 180 - 22?

[mellow scroll]

nope

@peak talon could you type out the question?

or example

[peak talon]

Why is this so confusing to me ffs

Okay so the question sentence is: if 13sinTheta - 5 = 0, and then some questions

[mellow scroll]

Give the range for theta

nvm

you said between 90 and 180

[peak talon]

And in the beginning it says "Theta E[90, 180]"

[mellow scroll]

mb

let me fire up paint

[peak talon]

Actually wait it says Theta E[90, 360]

I might have written it down wrong tho

[charred spear]

Command disabled

The sever owner has disabled that command in this location.

[peak talon]

Yeah I know it's in the second quadrant. But what confuses me is that in the drawing it says theta goes from x axis to hypotenuse. But then if you work it out its 22👀

Yea it is

Shift sin

I sin^-1 both sides

Which gave 22 or something degrees

[mellow scroll]

^

[peak talon]

How does that work?

You've just hit the nail, that's what confuses me

Why does the exterior angle (theta) have the same ratios as the interior one?

So when you're working with sides you use theta but if they ask you for the size of theta then you work out the angle using shift whatever and then - 180 or however much you need to get it?

Yeah that's what I meant

But for the second quadrant it's 180 -

[mellow scroll]

@peak talon Do you know what theta looks like?

The angle its referring to?

[peak talon]

In the example theta goes from the x axis to the hypotenuse

I wish my phones camera worked so I could show

It just confuses me how theta is clearly more than 90 degrees in the drawing but when you work it out its "Theta = 22"

[mellow scroll]

'

basic angle also known as reference angle

since the triangles are just reflections, the basic angles are equal

In this case, the arcsine will give you alpha (basic angle), so you have to use that to find theta

since theta was specified in the question to be in the second quadrant.

[peak talon]

Ahh

I'm getting it now

[mellow scroll]

If the question gave another range

😃

for example

0 <= theta <= 360

then you would have to give both arcsine(theta) and 180 - arcsine(theta) as answers. Since both are possible values

[peak talon]

So I can't go from "sinTheta = 5/13" to "Theta = sin^-1 * 5/13" cause that doesn't give theta it gives the basic angle?

[mellow scroll]

kek

if a question asked that

I would give the general forms for theta

(arcsine(theta) + 360n) and (180 - arcsine(theta) + 360)

n E z

yea

you're right

we have to use Tau Radians

hueheuheuhue

[inland karma]

someone help me

[somber coyote]

omar23:

[astral hornet]

which problem?

[meager acorn]

i have a few way to simplify but i dont know what i should do next

sin^4x-cos^4x+cos^2x
1-cos^2x+1-cos^2x-cos^4x+cos^2x is this how you simplify or do you make everything to sin?

[wild hamlet]

I guess thats fine for simplifying

[meager acorn]

but

in the answer it says

Sin^2x

i dont know how

[wild hamlet]

sin(x)^4 + sin(x)^2

[meager acorn]

wait

i am lost

what happens to cos

[wild hamlet]

2sin(x)^4 - cos(x)^4 + cos(x)^2 = 2sin(x)^4 - cos(x)^2 * (cos(x)^2 - 1)

cos(x)^2 = 1 - sin(x)^2

cos(x)^2 - 1 = -sin(x)^2

sub it in

[errant ether]

no, the sin^4(x) cancels out

[wild hamlet]

you forgot the 2

[teal dragon]

there is no 2

2 is the question number

[wild hamlet]

problem 2 then

[meager acorn]

bit confused

[wild hamlet]

anyway

[teal dragon]

it does cancel out

cause there is no coefficient 2

[meager acorn]

so sin^4x-cos^4x = 1

isnt it

because sin^2-cos^2 = 1

[errant ether]

sin^4(x) - cos^4(x) + cos^2(x) = sin^4(x) - (1 - 2sin^2(x) + sin^4(x)) + cos^2(x)

[meager acorn]

if you have ^4

double 1?

[teal dragon]

when you bring a binomial to a power

you cant distribute the power

have you learned foil?

[wild hamlet]

sin(x)^2 + cos(x)^2 = 1, not subtract

[meager acorn]

oh yea

my bad

just realized its +

not-

[teal dragon]

it wouldnt be 1 either way

[meager acorn]

is this mean i have to make everything to cos?

sin^2x = 1-cos^2x

[tawdry spoke]

Oo trig identities

[wild hamlet]

no, make everything to sin

[meager acorn]

whats the identies for cos to sin

[tawdry spoke]

Wats the prolem

[wild hamlet]

cos(x)^2 = 1 - sin(x)^2

[teal dragon]

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

[charred spear]

Command disabled

The sever owner has disabled that command in this location.

[teal dragon]

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

[somber coyote]

fishcat:

[tawdry spoke]

Lol

[meager acorn]

is that ok to use

my teacher didnt tell me about that one

[teal dragon]

well, you know that $cos^2 x + sin^2 x = 1$

[somber coyote]

fishcat:

[meager acorn]

yes

[teal dragon]

subtract sin^2 on both sides

[meager acorn]

so its same thing than

[teal dragon]

yep

[meager acorn]

sin^2x = 1-cos^2x

cos^2x = 1-sin^2x

[teal dragon]

correct

[meager acorn]

aweseome

taking that on my note

i couldnt think of any

[tawdry spoke]

The answer is sinx^2

[meager acorn]

that opens the door to next equation

yes that whats the practice exam answer says

thank you everyone!

let me get this straight

sin^4x-cos^4x+cos^2

sin^4x-cos^4x+1-sin^2?

[wild hamlet]

yes

[meager acorn]

than what happens to

-cos^4x

[wild hamlet]

cos(x)^4 = (cos(x)^2)^2

sub in 1 - sin(x)^2 and foil

[meager acorn]

hm....

[chrome fiber]

you can just use difference of squares for the first two terms

$\sin^4x - \cos^4x = (\sin^2x)^2 - (\cos^2x)^2\(\sin^2x + \cos^2x)(\sin^2x - \cos^2x)$

$$\sin^4x - \cos^4x = (\sin^2x)^2 - (\cos^2x)^2$$$$(\sin^2x + \cos^2x)(\sin^2x - \cos^2x)$$

[somber coyote]

june:

[meager acorn]

I m so LOST

;;;;;;;;;;;;;;

[upper karma]

which part doesnt make sense to you?

[meager acorn]

can i show you

what i have gotten

[upper karma]

yeah sure

Use june's example as your reference

[upper karma]

bc i think its right

[meager acorn]

let me get mic

[upper karma]

lol its ok

[meager acorn]

so do i foil that [1-sin^2x]^2

[tropic shard]

Yes.

[meager acorn]

if you foil than is it going to be

[1-sin^2x] x [1-sin^2x]?

or [1-sin^2x] x [1+sin^2x]

[tropic shard]

The same sign,

The second line you wrote was the conjugate of [1-sin²(x)] which is something else entirely

So you'd have to do [1-sin²(x)]·[1-sin²(x)] to get,

I'll let you do the honors

[meager acorn]

doing it

1+2-sin^2x+sin^4x?

did i do it wrong

[tropic shard]

Well it's not 2-sin, it'd be -2sin lol

Otherwise, that's correct foiling

[meager acorn]

wait -2sin?

o,o

[tropic shard]

I'd reorder it by powers

oh yea, it'd be -2sin²(x)

[meager acorn]

1-2sinx^2+sinx^4?

[tropic shard]

(a+b)²=a²+2ab+b²

Yes,

So I'd reorder it to sin⁴(x)-2·sin²(x)+1, then plug that in to the original problem for cos⁴(x)

[meager acorn]

let me re-order or write it one moment

tell me if i wrote this wrong

sin^4x-sin^2x+sin^4x

[tropic shard]

Shouldn't be two sin⁴(x)'s

[meager acorn]

i had sin^4x originaly

the question i had

[tropic shard]

Oh I see

[tropic shard]

But you plugged in the thing we did for cos⁴(x),

You also need to distribute the negative to all those terms

So you'd have one positive sin⁴(x), and one negative sin⁴(x), and so on.

[meager acorn]

frustrated right now

i understand till
sin^4x-[cos^2x]^2+1-sin^2x

after we foil

Sin^4x-2sin^2x+sin^4x

what am i misssing?

[tropic shard]

Okay, write for me what we had for cos⁴(x)=?

We got it so that cos⁴(x)=sin⁴(x)-2sin²(x)+1, or no?

[meager acorn]

i got it @tropic shard

[tropic shard]

Nice

[meager acorn]

🤦

[tropic shard]

I got lost on the second line of #2, but I agree with your final answer, good job!

[meager acorn]

it says solve the trig equation

can someone tell me where did i mess up

[static river]

@meager acorn its 1-(cos(x))^2 = sin(x)^2, not 1-cosx = sinx

[meager acorn]

how do you solve this then

sinx/cosx+ 1-cosx/cosx

[static river]

i wanna say square both sides but then you'd get extraneous solutions

i dont think im comfortable enough with my trig identities to help you solve it, i can only point out the mistakes i see

[tropic shard]

sinx/cosx=1-1/cosx
sinx/cosx+1/cosx=1
secx(sinx+1)=1
sinx+1=cosx

Uhh lemme refine that

[teal dragon]

sinx+1 is not equal to cos x

[tropic shard]

It's asking to solve

It's true when x=0, or π

[teal dragon]

whats it solving for

[tropic shard]

When tan(x)=1-sec(x)

[teal dragon]

OH

i misread

[tropic shard]

xD

[teal dragon]

i thought it was just stating the identity tan(x)=sec(x)-1

[tropic shard]

It threw me off too, until I realized halfway.

[teal dragon]

i always get the order mixed lol

[tropic shard]

Could you translate what I said to Tex? My Unicode way makes it look too messy with Paratheses and Brackets.

[teal dragon]

sure

$\frac{\sin x}{\cos x}=1-\frac{1}{\cos x}$

[somber coyote]

fishcat:

[teal dragon]

$\frac{\sin x}{\cos x}+\frac{1}{\cos x}=1$

[somber coyote]

fishcat:

[teal dragon]

$\sec x (\sin x + 1)=1$

[somber coyote]

fishcat:

[teal dragon]

shh you saw nothing

$\sin x + 1 = \cos x$

[somber coyote]

fishcat:

[tropic shard]

Alrighty thanks!

Y'know if there's a way to make it more obvious the answer would be 0∧π?

[teal dragon]

hmm

yeah i got it

$\sin x + 1 = \sqrt (1-\sin^2 x)$

[somber coyote]

fishcat:

[teal dragon]

$(\sin x + 1)^2 = 1-\sin^2 x$

[somber coyote]

fishcat:

[teal dragon]

$\sin^2 x + 2 \sin x + 1 = 1 - \sin^2 x$

[somber coyote]

fishcat:

[teal dragon]

$2\sin^2 x + 2 \sin x = 0$

$2 \sin x (\sin x - 1) = 0$

[meager acorn]

cotx= -sqrt 3 is 30deg

I M LOST

[teal dragon]

wait i messed up

no i didnt

huh

[tropic shard]

It looked right to me at first

Could we get the whole question, Yupki?

[meager acorn]

one sec

[teal dragon]

wait

it was right lmao

[teal dragon]

i deleted it for nothing

[tropic shard]

Yup we're doing that xD

[meager acorn]

😮

[teal dragon]

lemme tex this

$

[tropic shard]

professional $

[teal dragon]

$\tan^2 x = (1-\sec x)^2$

[somber coyote]

fishcat:

[teal dragon]

then you can turn the left into

$\sec^2 x -1 = 1 - 2 \sec x + \sec^2 x$

[somber coyote]

fishcat:

[teal dragon]

subtract sec^2 on both sides

[tropic shard]

Lost me there

[teal dragon]

$-2 = - 2 \sec x$

[somber coyote]

fishcat:

[tropic shard]

Oh, a lot of identities I see

I'm back.

[teal dragon]

alright uh

did i mess up?

that doesnt look right

[tropic shard]

Nope

[teal dragon]

its true at 0 and pi

[tropic shard]

It's right

[teal dragon]

or is it

no it isnt true at pi

[tropic shard]

oh that's odd...

Yeah.. That's wrong somewhere..

Are we sure tan²(x)=sec²(x)-1?

I believe that is..

[teal dragon]

well i just checked on desmos

and i think it is right

[tropic shard]

Yeah, everything checks out..

[teal dragon]

theres no intersection between the two initial equations at pi

[umbral snow]

Take sin²(x) + cos²(x) = 1
Divide both sides by cos²(x)

tan²(x) + 1 = sec²(x)
So that's the right identity

[tropic shard]

Now I have to see my earlier one as to why I was wrong, smh.

So the answer would be x=arccos(1)?

[teal dragon]

yup

[meager acorn]

the answer says 0

i m so lost at the moment

[tropic shard]

So I was right??

[teal dragon]

well you were right about 0

i dont know how you got pi in there

it seemed right to me

[tropic shard]

Oh, cause I'm dumb for thinking cos(π)=1?

[meager acorn]

trig identies starting to messing with my head

....

[umbral snow]

When is sec(x) = 1?
0, 2π, 4π...

[tropic shard]

So that's a yes.

[umbral snow]

And negative

[teal dragon]

the interval was [0,2pi)

[umbral snow]

So 0 is the only answer they want

[tropic shard]

0 and 2π

[meager acorn]

i m so screwed next week test;;;

[tropic shard]

But did they take 2π for you Yup?

[meager acorn]

i dont think so

[tropic shard]

Idk why I saw that paratheses as a bracket, sorry..

[meager acorn]

where do i start?

do i use product to sum and sum to product identies?

[rain tulip]

sum to product

[meager acorn]

only i see is
sin a+sinb
sin a - sin b
cos a + cos b
cos a - cos b

which one do i need to use

i want to punch a wall beacuse frustration

[tropic shard]

Well sin(x+π/2)=cos(x)

12·sin(x)+16·sin(x+½·π)

Hm.. Now how to deal with the coefficients.

12·sin(x)+12·sin(x+½·π)+4·sin(x+½·π)

24·sin(2x+½·π)·cos(2x+½·π)+4·sin(x+½·π)

Ahh I thought I was unto something, sorry man.

[meager acorn]

<@&286206848099549185>

https://cdn.discordapp.com/attachments/326138757474680852/513576229379637268/unknown.png

@umbral snow

i understand i have to use sum to product

but which formulas i have to use

[river forge]

there's some different formula's

do you want to like derive them or just get hit with them

[meager acorn]

this is pre-calc

[umbral snow]

Start with the form you want:
Asin(x + θ)
= Asin(x)cos(θ) + Acos(x)sin(θ)

This shows you that
Acosθ = 12
Asinθ = 16

[meager acorn]

i dont know derive

[river forge]

derive != derivative

it means to show where they originate from

so take something you know, and turn it into the formula

im guessing that's what kaynex is gonna do so

ill step out

[umbral snow]

So you want to solve for both of those. Any guesses?

[meager acorn]

dont you need theta

since we know what A is

[umbral snow]

If you only want θ, then just divide bottom by top:
tanθ = 16/12

A divides out, and θ is findable

[meager acorn]

how did you or where did tan theta = 16/12 come from

[umbral snow]

Start with the form you want:
Asin(x + θ)
= Asin(x)cos(θ) + Acos(x)sin(θ)

This shows you that
Acosθ = 12
Asinθ = 16

Dividing one line by the other:
(Asinθ) / (Acosθ) = 16/12

tanθ = 16/12

[meager acorn]

where did = Asin(x)cos(θ) + Acos(x)sin(θ)
come from

[umbral snow]

I expanded Asin(x + θ)

Using a sum identity

I've never used a product identity for anything before, lel. Sums are worth remembering though

[meager acorn]

sumand difference identies

can i ask why did you choose tanθ

[umbral snow]

I could have divided the other way I guess? Got cotθ = 12/16

[meager acorn]

can i ask 1 more thing

why divided

Asin(x + θ)
= Asin(x)cos(θ) + Acos(x)sin(θ)

[umbral snow]

Do you understand how we got to
Acosθ = 12
Asinθ = 16

[meager acorn]

yes

from the formula

actuall no..

😦

[umbral snow]

Compare the form we WANT, vs the form we HAVE
WANT: Acosθsin(x) + Asinθcos(x)
HAVE: 12sin(x) + 16cos(x)

[meager acorn]

12sin x = A
16sin x = theta?

[umbral snow]

We can make those two things the same, if
Acosθ = 12
Asinθ = 16
So let them be the same thing

Note we have no control over x, and can't freely set that.

[meager acorn]

cos/sin = cot theta 16/12

okokokok

i see it now

[umbral snow]

There you go! And solving that:

== atan(16/12)

[charred spear]

atan(4/3) = 0.927295218001612

[umbral snow]

In radians. Maybe you want degrees?

[meager acorn]

theta = degree

3 decplaces

[umbral snow]

== atand(16/12)

[charred spear]

180×atan(4/3)/π = 53.130102354156

[umbral snow]

35.130 degrees

53*

[meager acorn]

whats atand

arc tan?

[umbral snow]

tan¯¹ in degrees

arctan in degrees

[meager acorn]

now the another questionis

why arctan

why cant i use just tan

[umbral snow]

tanθ = 16/12
θ = tan¯¹(16/12)

Or θ = arctan(16/12), since you are comfortable with the notation

[meager acorn]

i seeit

thank you!

[umbral snow]

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

[meager acorn]

been trying to solve for good 2hrs

t!rep @umbral snow

[loud cedar]

🆙 | Yupki has given @umbral snow a reputation point!

[umbral snow]

Yay.

[meager acorn]

do i need to find my deg to radian?

[tropic shard]

?

[meager acorn]

where do i start

[tropic shard]

So this is without a calculator?

[meager acorn]

yep

[tropic shard]

Okay well we know 165° is 15° from 180

[teal dragon]

i would go for 210 - 45

[meager acorn]

can we use 165 is 15

[tropic shard]

So sin(165°)/cos(165°)=sin(15°)/-cos(15°)

[meager acorn]

why -cos 15

[tropic shard]

Imagine the unit circle, if it passed the 90° point, the x-value is negative

But the y-value is still positive, because the angle is less than 180°

[graceful hull]

heya

Ive got a homework sheet i dont fully understand

Would anyone help me complete it?

[teal dragon]

yep

[graceful hull]

Ok let me go grab my phone real quick

Dm or here?

[tropic shard]

Here would be nice, so I can help too >:)

[graceful hull]

sure sure

4 tasks a side

covers all of the geometric rules etc

@teal dragon

[tropic shard]

Wait, so you don't know about translation via vectors?

[graceful hull]

Not really

I just find it so hard to make it accurate

And so I end up getting it wrong

[tropic shard]

I'm rusty, but I believe the -7 on top means move a point to the left by 7, and the -2 means move that same point down 2

Do that for all 4 points, then connect the dots and label 'B'

[graceful hull]

Right

[tropic shard]

Have you done that already?

[graceful hull]

doing now

done

[tropic shard]

Alright, now focusing back on image 'A'

[graceful hull]

Rotate your shape A 180 degrees about (1,0) and label the image C

[tropic shard]

The point of rotation is on one of the points,

The one at (1,0)

[graceful hull]

Is that the one on the bottom or left?

[tropic shard]

Very bottom.

[graceful hull]

yup

[tropic shard]

So that means that will be stationary throughout the rotation.

The other points will move the 180°, but in this case that simply means their y-value changes sign.

So for ex. move a point from (2,2) to (2,-2) in this case.

[graceful hull]

Uhh ok

ok

[tropic shard]

Wait

oh

[graceful hull]

So where do i move the bottom one to?

[tropic shard]

The one at (1,0) stays in place.

[graceful hull]

I am confusion

Ok

so start there and just replicate the shape?

[tropic shard]

The point at (2,2) goes to (0,-2)

[graceful hull]

i think ive got it

[tropic shard]

Just imagine rotating that shape to the other side, while holding the (0,1) point down with your thumb.

[graceful hull]

lemme check with you in a sec ill snap a pic when finished

[tropic shard]

Ye

[graceful hull]

I think thats right

Its reflection i find difficult

@tropic shard

[tropic shard]

That looks right.

[graceful hull]

great

[tropic shard]

Yea, you're good.

[graceful hull]

Reflect A in line M label the image D

[tropic shard]

So since it's like a secondary y-axis

since it goes up and down

The Y values will stay the same

[graceful hull]

So it stays as -2 for the Y?

[tropic shard]

Well I believe they want you to focus back on image 'A,' where none of the points are -2

so (1,0) will still be ( ,0) in the end

[graceful hull]

Dont forget line M

[tropic shard]

wait um

[graceful hull]

If you look up theres that bold line

[tropic shard]

There I fixed my coordinates lol.

Yeah I see lol.

[graceful hull]

I dont understand

Where abouts on the shape do I start?

and move where?

[tropic shard]

So pick a point, any point

[graceful hull]

Bottom one

[tropic shard]

Up, down, right, or left?

Ok

Count how far that is from the M-line

[graceful hull]

-6

0

[tropic shard]

So technically the boxes are ½, so it'd be -3 but you get it

[graceful hull]

The teacher says either or

[tropic shard]

Now, continue going that same direction the same amount of boxes you counted

[graceful hull]

You can use 1/2 or 1

Im using halfs because its easier to count, for me anyways

so another -6?

-12?

[tropic shard]

That's okay for these purposes, just make sure to stay consistent

Yea

So it'd be -12 away from it's original position

[graceful hull]

Yup

Now what

?

[tropic shard]

Place the dot

[graceful hull]

done

[tropic shard]

And you're done for that one,

Pick another point any point

[graceful hull]

left

-12?

[tropic shard]

Nope

For each point you have to recount the distance from the M-Line

[graceful hull]

4

-4

to be exact

[tropic shard]

Yes

[graceful hull]

Then another -4?

[tropic shard]

Yeap

[graceful hull]

done

[tropic shard]

Nice, now another pont

point

[graceful hull]

finished it

[tropic shard]

no pick another on-

oh there goes my joke of the day, rats.

[graceful hull]

ill play along

rihgt

right

[tropic shard]

Alrighty, I guess you're done with that question

?

[graceful hull]

yeah

ty

ok

next is enlargement

Enlarge by sf 2 at point X

Centred at the point x*

Reflect in line M and Rotate 90* clockwise at the point x

[tropic shard]

Oh so for enlargement,

So basically count the dots x and y values from the X and double it

[graceful hull]

4,3?

I dunno if thats right?

the that becomes 8, 6 ?

How would I draw it?

[tropic shard]

It should look something like that.

[graceful hull]

oke

is that at point x?

@tropic shard

[tropic shard]

It's scaled in reference to reference point X, yes.

[graceful hull]

kk

Reflection i suck at

So pick a point

count the squares till the line?

[tropic shard]

So you have to count in two ways for each point

The verticle and horizontal,

[graceful hull]

I dont get it

so bottom left

start there

[tropic shard]

So like that for each point

(the lines are to be erased, leaving behind only the dot.

[graceful hull]

oke?

So ive dotted that part

But i dont get what i am doing

[teal dragon]

shouldn't it be more like

[tropic shard]

yes, yes it should

I was still scaling it by 2, mb

But same concept

Well

You see how horizontally you go to the right 2 squares?

Then go down by that amount.

[graceful hull]

so is that it

move all points right 2 and down 2?

[teal dragon]

no

for each point

find the horizontal distance from the line

[teal dragon]

^

[graceful hull]

uhm

ok

[tropic shard]

For each point it'd be different. Hopefully this picture could be worth a thousand words

[teal dragon]

the blue line is the horizontal distance

[graceful hull]

but what do i do to work it out

[teal dragon]

then the red line is that same distance downwarss

[graceful hull]

How do i know where to move it

[tropic shard]

Either way works

[teal dragon]

at the end of the red line

that's the final point

[tropic shard]

You can do all of them Horizontal to Vertical, or Vertical to Horizontal, you'll get the same result

That's Vertical to Horizontal, but again, no need to do both Vert. to Hori. and Hori. to Vert. for the same point.

(but it's a good technique to check your work)

[graceful hull]

Ok im going to restart it and draw in the lines

brb

failed miserably

wait

got it

I dont think it is meant to go over the line tho?

lol

[tropic shard]

What do you mean?

[graceful hull]

The top left of my shape goes over the line

[tropic shard]

Let's see

[tropic shard]

Almost,

[graceful hull]

oof

got it

Rotate 90* clockwise at point x

https://cdn.discordapp.com/attachments/326138757474680852/513604288677937173/image0.jpg

[tropic shard]

That one I use imagination, lol.

[graceful hull]

I can rotate it in my

head

but the whole point x

i dont know what that means

[serene field]

X is your center of rotation.

[graceful hull]

So i rotate off that and not around the shape?

[austere prairie]

Yep

[graceful hull]

Could someone draw how that would look? I just want to make sure im right

so would i draw the bottom 4 squares onto the x position?

[austere prairie]

The point right below x would be just to the left of it after the rotation

[graceful hull]

left of what?

[austere prairie]

Of x

[graceful hull]

The 4 squares on the bottom of the triangle

ok

[tropic shard]

I drew mine, show me how yours looks

[graceful hull]

im still working it out

hold up

I’d rule over it but it’s gonna get rubbed out Ik it xd

[serene field]

The distance from x to the closest point on the triangle should not change, as it's rotation.

[graceful hull]

ohhh

[serene field]

Castle, isn't that counterclockwise?

[austere prairie]

Clockwise, not counterclockwise castle

[tropic shard]

Oh

I didn't read that part, I just read 90°, so I was thinking like the unit circle xD

[serene field]

It happens to the best of us 👍

[austere prairie]

I thought the same thing at first as well lol

[serene field]

Check all vertices of the triangle.

[graceful hull]

wait

[serene field]

Make sure the angle subtended by the arc from initial to final is 90 degrees.

[graceful hull]

did i do it backwards?

[tropic shard]

Not quite there yet, you seem to have 2 points correct.

[graceful hull]

Is it backwards?

[austere prairie]

Not really

[graceful hull]

Whats to correct? I dont get where i went wrong

[austere prairie]

That top-most point needs to be below the horizontal line

[tropic shard]

Think of the rotation in your head

[austere prairie]

Directly below the leftmost vertex

To better visualize it, take your paper, hold your finger on x, and spin the paper 90 degrees

Such that the top of the paper is facing to the right

[tropic shard]

Yea, just don't lose your original position lol.

(have it visualized the whole time)

[graceful hull]

Uh

I keep failing

lemme take another crack

[tropic shard]

@serene field Sorry about that random FR, I won't feel bad if you declined that xD

[serene field]

¯_(ツ)_/¯

More friends to drag into mathematical hell

send help

[tropic shard]

Oh, I like that kind of help.

Hell*

rip

[serene field]

A hint: Check the right-most vertex of the original.

[graceful hull]

i brought that under the line

under x*

the 4 square line

[tropic shard]

Let's see

[graceful hull]

I dont think thats right tho

[tropic shard]

rip

[tropic shard]

You see Jythro's rotation?

[graceful hull]

Yeah

thats the one i did previously

[tropic shard]

You did, but it was misplaced.

[serene field]

Try imagining the triangle were a flag of sorts attached to X

And you were moving the flag around X

[austere prairie]

Literally take your paper, hold the tip of your pencil on X, and rotate your paper such that the top of the paper faces 3 o'clock

See where the original triangle moves to relative to it's original position

[austere prairie]

Stop and do exactly what I said, then take a picture of what you see

[serene field]

Rotation should not change the distance between points to the axis of rotation.

[graceful hull]

i feel so incredibly dumb rn

[serene field]

For example, if you were to paint a dot on a car tire

And you were to spin the tire 90 degrees

The distance to that dot from the axis of rotation, the center, is still the same.

[hazy granite]

the closest point to X is one line away rite

shouldnt it keep that distance

[tropic shard]

In rotations, the points must reserve the same distance away from the point of rotation.

[austere prairie]

What you need to do is stop and take a step back

Literally take the time to rotate something irl

[serene field]

Get a piece of cheese from your fridge, cut it into a triangle shape

And spin it around

*a triangle piece of paper works too.

[hazy granite]

or just cut out something the same shape

and put it on the figire

add a circle to it

[sharp fox]

Lol

[hazy granite]

radius is 1 gridline

[graceful hull]

I’m so confused

[austere prairie]

Yes

[serene field]

👍

[austere prairie]

That's it

[graceful hull]

Finallyyyy

[austere prairie]

Do you see why?

[hazy granite]

take the point closest to X

[graceful hull]

Gotcha

[hazy granite]

keep that distance

its 1 gridline

[graceful hull]

uhh ok

[hazy granite]

now, move the point 90° while still keeping the distance of 1 gridline

[graceful hull]

How’s this so far?

Yeah i get what you mean big

Thanks for the help on a quick note

Means so much

[serene field]

A is transformed to B by a [ ] through [ ]

Yeah, that'd be right. It's one in both the x and y directions.

A to C looks right as well.

[graceful hull]

Is it a vector I’m looking for on the 2nd part of B?

[austere prairie]

Not the grammar of the prompting sentence tho lol (ignore this, side comment)

[serene field]

That first sentence threw me way off lmao

Doesn't sound right, but it works

What do you need to do to A to get it to B?

Try looking at one vertex of A and its corresponding vertex in B.

[graceful hull]

D is a reflection of A draw the mirror line to reflect the diagram

[serene field]

Think of it as a mirror line.

[serene field]

👍

[graceful hull]

right

?

[hazy granite]

its dilation

or scaling

whatever you want to call it

ok to find it

take any line in the smaller figure

and take the corresponding line in the larger figure

[graceful hull]

For B?

[hazy granite]

yeah

[graceful hull]

Is it by sf 2?

[hazy granite]

yeah

[graceful hull]

how would i write it as a answer?

Enlargement by sf 2?

[hazy granite]

yeah probably

which one do you want to find

[graceful hull]

C now

[hazy granite]

so keep the basic figure of c in your head

now rotate figure a

actually nevermind

it's reflection

[graceful hull]

uhh ok

Should I draw a line for it?

[hazy granite]

yeah

it states that in the question

[graceful hull]

that would go facing what way?

And in the halfway of both on a 45* angle?

[hazy granite]

yeah

[graceful hull]

What way does it face?

[serene field]

Could be answered as reflection or rotation + reflection 🤔

Try imagining a mirror between A and C

[hazy granite]

the closest point to each other is the right angle

[serene field]

It's diagonal, yeah, but that does look slightly off

[hazy granite]

yep

[graceful hull]

Great

so answered as

[serene field]

Your professor might dock points if they're stingy

[hazy granite]

it should be directly in between the 2 right angle points

[graceful hull]

Reflection (Mirrored line drawn above)?

Ill redraw

done

I’ve done the next part, and semi completed the part after

Then just geometric reasoning after

[hazy granite]

so just draw the mirrors first

coordinates are given

[graceful hull]

Would that be on a diagonal?

Left to right?

[hazy granite]

no

it would be a straight line

perpendicular to 0

[graceful hull]

Ok

[serene field]

x = -2 for all y

[hazy granite]

on the opposite axis

[serene field]

Likewise, y = 1 for all x

[graceful hull]

im confusion

drawn the line where?

[hazy granite]

take point -2 on the x axis

[graceful hull]

-2?

on the black line

?

[hazy granite]

yeah

now draw a line perpendicular to the black line

connecting x -2 and the dges

[graceful hull]

Like so?

[hazy granite]

yea

[serene field]

The horizontal line is y = 0.

[hazy granite]

wait

[serene field]

Once you redraw it, you may want to extend it to the right.

[hazy granite]

yeah that should be

[graceful hull]

extend what to the right?

[serene field]

The line supposedly extends infinitely in both directions

You'll be reflecting the image, which will likely end up on the right

[graceful hull]

uhm ok

so what do i do now?

[serene field]

Reflect the triangle across the axes

Remember that the distance from the original's vertices and the mirror is the same as the distance from the mirror to the image's vertices.

[graceful hull]

so to the right of the line?

[hazy granite]

yes

[graceful hull]

ok

[hazy granite]

you just have to reflect all 3 points

then you can construct the triangle from there

[graceful hull]

Slightly off the bottom, but my teacher isn’t picky

I’ll redraw the bottom anyways

Ok so now do the same thing underneath the X axis?

[hazy granite]

yeah

draw a line which y=1

[graceful hull]

How’s this

[serene field]

It asks to reflect the image

Not the original

[hazy granite]

perfect except you have to reflect the image

actually

the mirror is not placed correctly

[serene field]

The second mirror is still along y = 0

[graceful hull]

Correct?

So what am I doing with the image

[hazy granite]

yeah the mirror is placed correctly

extend the mirror though

[serene field]

Uh.

That's y = 2

[hazy granite]

no that's y = 1

the gridlines = 0.5

[graceful hull]

^

[serene field]

Woo

Scale is fun

[graceful hull]

So I’m just redrawing it again?

[hazy granite]

extend the line

no don't redraw it

[serene field]

Extend it to the right

[hazy granite]

extend the mirror

[graceful hull]

ok

and the Y axis as well

bring that down?

[hazy granite]

yeah

[graceful hull]

oke

so what next?

[hazy granite]

reflect the image

[graceful hull]

was i doing it right before?

[hazy granite]

yeah

[graceful hull]

kk

done