geometry-and-trigonometry

[inland mountain]

Okay so how would I find how far apart the deer from here?

[trim breach]

Have you learned trigonometric ratios yet?

[inland mountain]

Im pretty sure yes

I might have missed it tho when my teacher was talking about it

[trim breach]

You have two angles and a side in the right triangle from the furthest deer to the top of the hill, then straight down.

lol sorry, just finished an exam and I’m trying to eat while explaining, so I am not as my desk atm to draw it out.

[inland mountain]

Haha no problem im just thankful your explaining it at all

[trim breach]

But you have the adjacent side and the angle to use cosine.

For the hypotenuse to the furthest deer.

[inland mountain]

so use the adjacent side and the angle with cosine to figure out the hypotenuse of the furthest deer? Which would give me 46.6?

[trim breach]

Your hypotenuse should not be larger than a leg.

How did you use cosine?

[inland mountain]

acos(35/51)?

[trim breach]

Ah. Give me a sec again.

[inland mountain]

Okay np

[trim breach]

You want to use the cosine of this theta.

So cos(theta) = 51/y

And you solve for y.

[inland mountain]

Theta?

[trim breach]

Theta just means the angle you are using. In this case, it is 55 degrees since you subtract the angle of depression from 90 degrees.

Since we drew that imaginary perpendicular line from the top of the hill straight down.

[inland mountain]

Ah okay so id do cos 55= 51/y?

Okay I kinda understand but im still confused on how to get y somehow idk why tho

[trim breach]

That part is just algebra. Multiply both sides by y.

So now: y cos(55) = 51

Then divide both sides by cos(55).

So: y = 51/cos(55)

[inland mountain]

I got y = 88.91

[trim breach]

Yeah. I made a mistake earlier when I was explaining. That is just your hypotenuse for that triangle.

But you have two sides of that larger triangle now, so you can use Pythagorean Theorem to solve for that last side.

[inland mountain]

Okay I got 72.83 for the last side

[trim breach]

Yep.

[inland mountain]

Sorry I'm really bad at understanding stuff what would I do now to try and get x ?

[trim breach]

So now you have a strategy to find x.

x + z = 72.83

You find z by applying the steps we did in the first triangle to the new one.

[inland mountain]

I feel bad for making you explain this all but what are the steps before didnt we do like cos 55=51/y?

[trim breach]

Ignore the marker bleeding through.

Your hypotenuse is y.

No relation to what we solved for earlier.

[inland mountain]

Cos(43°)=51/88.19?

[trim breach]

Yeah, my bad…I should have used a different variable.

This y is not the y we solved for earlier.

Think of it as a new variable.

[inland mountain]

Ohhh

[trim breach]

Like a.

[inland mountain]

Okay so then I just multiple a on both sides then a cos(43°)= 51 a?

[trim breach]

Not quite.

When you multiply 51/a by a, the a’s cancel out.

But the a shows up on the other side.

a cos(43) = 51

So to isolate a, you divide both sides by cos(43).

Giving you: a = 51/cos(43)

[inland mountain]

So a=69.73?

[trim breach]

Yes!

So now you have two sides of the triangle.

Use Pythagorean Theorem to get z, the last leg.

[inland mountain]

So then z would be 47.55?

[trim breach]

Yes!

[inland mountain]

So then the distance between the deer is 25.28?

[trim breach]

Yep!

[inland mountain]

Yaya

[trim breach]

😃

[inland mountain]

Ay thanks so much

[trim breach]

Glad to help.

[vestal pendant]

Anyone know how to sketch this

[vestal pendant]

ty man ❤️

[eager plover]

how do i even solve this? my teacher just slapped these problems on me but i dont know how the heck i should do to solve this vector thing.

[prime linden]

Just draw

And from there you can also work on seeing how to solve without drawing aswell using algebra and some geometry

Recall how vector addition works purely by drawing

[obtuse tapir]

7,0

[inland mountain]

So I have done the math I just dont know if the x should be on the opposite or adjacent

<@&286206848099549185>

[twilit zenith]

The elevation angle is always created between the hypotenuse and the "shadow" of the hypotenuse on the ground, such as in the diagram.
In this case, you are being asked about how far Linus (the black circle) is from the point directly below the kite, which is the point where the right angle is.
This distance would be the golden side.
Since that side is one of the sides near the angle, then it would be the adjacent side, not the opposite.

[inland mountain]

Okay thank you so much

[twilit zenith]

566766992180298, normally, you should be told which is the leg with respect to the angle, but in this case, since it's an isosceles right triangle, it doesn't matter.
Are you familiar with how to use sine / cosine / tan in this case to find the measure of the side?
Moreover, are you familiar with sine / cosine / tan values of common angles?

Here's a diagram to help.

[dim cliff]

can someone help me with this? (find the area of triangle ABC)

[twilit zenith]

In this specific case, after drawing the diagram, you can see that you're given two sides, as well as the angle between them.
Are you familiar with the formula:
S = a · b · sin (C) / 2
for the area of a triangle given those pieces of data?

[foggy talon]

I only know 45-45-90, I’m not sure what sin cosine and tan is

[twilit zenith]

These are ratios between sides in a right-angled triangle.
The sine is the ratio between the opposite side to a specific acute angle and the hypotenuse.
The cosine is the ratio between the adjacent side to a specific acute angle and the hypotenuse.
The tangent is the ratio between the opposite side to a specific acute angle and the adjacent side to the same angle.

[foggy talon]

Im starting to understand, so in this problem it’s adjacent over hypotenuse

[twilit zenith]

Exactly. 👍
One more thing to note is that if you get one of these common angles, you may be expected to use these values as opposed to simply multiplying / dividing by sin (angle), cos (angle) or tan (angle) on a calculator.

The angles at the bottom of the top row are measured in degrees, as in the question you've shown.
The top row is in radians, which is another measurement of angles that you may encounter later on.

[dim cliff]

sorry! but yes i am familiar with that formula

[foggy talon]

@twilit zenith thank you so I have to follow a formula

[twilit zenith]

sorry! but yes i am familiar with that formula
In that case, since these are the exact pieces of data you're given (meaning, two sides and the angle between them), then you can use that formula to find out the area right away.

In more convoluted example, you may have to perform additional calculations before that, such as using the sine theorem in order to find out one of the sides you need.

thank you so I have to follow a formula
The general way this type of question is solved - yes.

The formulas you typically use are:
sin (angle) = opposite / hypotenuse
cos (angle) = adjacent / hypotenuse
tan (angle) = opposite / hypotenuse

So, for example, if the angle you're given is 30°, and you're told that the hypotenuse is 10 cm, and you want to find the opposite side, you use the sine formula:
sin (30°) = opposite / 10 cm
30° is a common angle, so you use its sine value: 0.5, and you get:
0.5 = opposite / 10 cm
and from there, you solve for the third side.
In some other occasions, you may have both sides, and then you need to calculate the sine, cosine or tangent of the angle, from which you can extract the angle.

[foggy talon]

Ah thank you so much for taking your time to explain it to me I’m very grateful @twilit zenith

[upper karma]

Anyone know how to do proofs?

[trim breach]

Yes. It is assumed someone will be able to help you here. Just post the question you have.

[compact jasper]

what is the best method to find cos(x) if you know sin(x)?

[pure cape]

use identities

there are plenty

for example, sin(theta) = cos(pi/2 - theta)

if you are using deg

its 90deg - theta

or pythag

[compact jasper]

not in all quadrants sin(theta) = cos(pi/2-theta) is it ?

but it general thats correct ...

[humble pulsar]

$\sin^{x}+\cos^2{x}=1$

[somber coyote]

83247839

[humble pulsar]

@compact jasper

[ancient oyster]

Can anyone explain how to do this please

[old fable]

shouldn't you just use pythagorean identities

for 90 45 45 triangle

[ancient oyster]

yea i’m just confused on this one now @old fable

[old fable]

the angle is 60 tight

right

and you can see from the triangle that the angle opposite of the 60 degr is 90

so it means that the triangle is a 30 60 90 one

[ancient oyster]

yea

[old fable]

you can use the same identities

[ancient oyster]

ohhh ok

[old fable]

you get it?

[ancient oyster]

yea ty

[old fable]

np

[ancient oyster]

i feel like i did something wrong @old fable

[old fable]

6/sqrt 3 is not 6sqrt3

only a small math error

[ancient oyster]

ohh my bad

thank u

[old fable]

np

[alpine shadow]

How do u do this

[lost hound]

if you're being given a single angle like 5pi/12 and you need to split that into two for a sum and difference are there any tricks to do it without a unit circle?

[elder dagger]

Do you know the Inscribed angle theorem? I'm not sure it is the only way to solve this but it should work...

[upper karma]

yo really need help with this

idk how to lay out geometry in english so hopefully it's clear

[lost hound]

i have an exam in 2 hours and theres a small concept i still don't understand in my trig class, can someone dm?

[old fable]

Ask here

[grizzled tundra]

trig identities is so confusing for me

i literally cannot do any hw i’m assigned 😭

[foggy talon]

Can someone help me with this one?

I got x is 5

Can someone tell me if I’m correct

[foggy talon]

<@&286206848099549185>

[silent plank]

that is incorrect

[foggy talon]

can you tell me where I went wrong

[silent plank]

i don't know how you got x=5

calculate a random side and hope for the best?
one of the legs has a length of 5 but it won't be the length of the longer leg

[foggy talon]

oh my i got the short side

instead of the long side

I misread

is it 5 sqrt 3

?

[silent plank]

yes

[foggy talon]

Thank you @silent plank

Is it 4?

[teal nebula]

can somebody help me with this one pls

[lucid tartan]

anyone knows a website that can graph vectors?

and find angle?

[versed river]

geogebra

is probably exactly what you're looking for

@lucid tartan

[lucid tartan]

oh didn't know geogebra can graph vector

[versed river]

yeah click on the line segment button and it should come up

[foggy talon]

can you help me with my problem too @versed river

[livid sand]

Can someone please explain to me how they figure out the letter order and what letters to use? (Probably not the best at explaining so message me if you need me to try and explain it another way)

[upper karma]

@naive tapir

[naive tapir]

I'm going to destroy you

[upper karma]

wdym letter order, if you want to name an angle the middle letter is the angle, the left and right dont really matter in the order of naming it, unless you’re gonna correspond it with another angle.

[viscid wave]

is anyone available to help explain some stuff to me??

[copper dove]

Could anyone help me out with some trigonometry?

[finite swan]

https://dontasktoask.com

[crystal quail]

For ellipses, does the square of the major axis mean anything in regards to its area? Trying to look at a geometric proof for a physics problem and it's buggin me hard. This is the proof, from the wiki page on "Shell theorem". I made it fine up to just after the ratios of the ellipses. Once it says "Therefore the two ellipses are similar, so their areas are as the squares of their major axes" that part kind of loses me.

I think the point trying to be made is that for each unit of distance from P, the area of the ellipse is proportional to the square of that distance

[icy dome]

is there any books on trigonometry that uses geometry to solve trig identities and build up on geometric intuition

And some books on problem solvinng for harder trig problems

[old fable]

use yt

it;s free

better than books

[tiny snow]

Well, I thoroughly enjoyed reading Euclid, but I know some who did not so I guess it's not for everyone. Worth checking out though: https://mathcs.clarku.edu/~djoyce/java/elements/toc.html

@icy dome ^

[dark sparrow]

let AB=1

then E is located at (1/sqrt(2), 1/sqrt(2)) and F is at (1, 2-sqrt(2))

[tiny snow]

Thanks @dark sparrow

[torn tundra]

Can someone solve 154 = x(x+15)?

[dark sparrow]

we don't give out answers here

[humble pulsar]

yes someone can solve that

I have the ability to solve that

[upper karma]

Please help me

[earnest echo]

You need to prove that the triangles ∆DEH and ∆DFG are congruent and the required result follows from that

[upper karma]

I fisnhed doing that one

Can u help me with another on3

[twilit zenith]

Since you're told that $\overline{VT}$ bisects $\measuredangle STU$, it often suggests that you should use the angle bisector theorem. The theorem states that the ratio between the two segments partitioned by the angle bisector is equal to the ratio between the lengths of the corresponding sides. In this case: $\$
$\frac{TS}{TU} = \frac{SV}{VU}$.

[somber coyote]

uoɯpɐʞᴉoɹ

[tiny snow]

To anyone curious, the error is 0.3%, with the angle at approximately 15.04339 degrees.

[rough stone]

Anyone here to answer a quick question on a triangle

I'm assuming that from the right angle to the side is where you start from that is the hyp, and then I'm not sure how to determine from theta if "b" is the adja and "a" is opo.

I figured it out nvm I wrote it backwards.

Theta and the right angle is adj
and the other side "b" is opo

[crude patio]

question how do i find out the shaded region

[unreal beacon]

The radius of the big circle is equal to the diameter of a small circle

@crude patio

Can you solve the problem with this info ?

[sharp hull]

Hi, don't know if this should be posted in trig or in calculus, but I think it's more trig related so here :
Here, the person explaining this is trying to prove that the limit of sin(x)/x as it approaches 0 is 1, which, good on him and all. Is that actually necessary to know ? I mean I could literally just graph the function and I'd end up with the same result in a quicker way.
So the question is actually : Is that necessary to know as you're learning Calculus 1 ?

[wintry tundra]

if khan academy teaches it then its needed

[sharp hull]

But why ? In what part of calculus would I need that ?

[unreal beacon]

No you don`t need the proof of that to get through calc 1

[sharp hull]

But it is needed in calc 2 or 3 then ?

[unreal beacon]

Not even

It's extra learning material

[sharp hull]

Ah I see, well thank you! I was wondering, this felt like the formal definition of limits all over again

[unreal beacon]

Yeah... you just need to know what a limit is conceptually and how to find the limit of a function

if you want to get intuition behind something then look up a proof

Like I can know that the derivative of e^x is e^x. But why? Then I look up a proof

[sharp hull]

Yeah I see exactly what you mean, thank you ! It is great that Khan academy goes into details and proof like that though

[unreal beacon]

Yeah the number of available resources the internet offers is incredible for learning

[crude patio]

@unreal beacon thxxx i got the answer from before

[fast pulsar]

$\lim_{x\to 0} \frac{\sin(x)}{x}$ is quite the famous limit.

[somber coyote]

pᴉʞɔɐp

[fast pulsar]

There are multiple ways to go about it, but the way Khan does it is quite nice and geometrically clear.

Also, if you have learned about the definition of the derivative, this limit will appear when trying to find the derivative of sin(x).

[sharp hull]

I do indeed know about derivatives but not fully, our program had this weird math syllabus where we learned things that are in calc 1 without learning pre-calc beforehand, nor trigonometry, which is why I'm teaching myself all over again

But thanks! I didn't know that this was the derivative of sin(x) yet !
I'm not gonna lie I still prefer the graphical way of proving it but I'm sure the geometrical way has its benefits that I might just not be familiar with yet

[fast pulsar]

This isn't the derivative of sin(x), that is not what I meant.

Care if I show you what I mean in #calculus ?

[sharp hull]

Yes of course ^^

[fast pulsar]

Alright, I'll meet ya there

[nocturne thicket]

I know there's like a certain theorem/rule for this, but how do I solve this?

[potent sapphire]

The angle subtended at the centre is twice the angle at the circumference? If the angle at the centre is 90° (right angled), then ∅ is 45°... I think. Let someone else confirm

[dull reef]

can i get some help with this

[nocturne thicket]

x^2+x^2 = 9Rad2^2

that simplifies down to 2x^2 = 9Rad2^2

[dark sparrow]

not true. it's x^2 + x^2 = (9sqrt(2))^2

[tiny snow]

I am trying to inscribe an ellipse in a convex quadrilateral using five points. The four points where the ellipse intersects the sides are found using parallel projection. The fifth point is where some bisectors meet, as indicated by this sketch.

This first looks like it works, as seen here.

[nocturne thicket]

i forgot about the ^2 thanks for reminding me

[tiny snow]

However, for some arrangements it does not. Is this an error in my construction or an error in GeoGebra?

Fiddle here: https://www.geogebra.org/geometry/evpxxz6g

[tiny snow]

No ideas as to why that construction "mostly" works?

[gray minnow]

hello

why do we define trigonometric functions for angles larger than 90

[peak flower]

Because angles can be bigger than 90 degrees, @gray minnow ?

[gray minnow]

@peak flower not in triangle

isnt trigonometry about triangles

no wait

🤦‍♂️🤦‍♂️🤦‍♂️

[peak flower]

There are also "non right" triangles which have more than 90 degree angles

And also, trigonometry has many other uses.

[gray minnow]

sorry if this sounds stupid

but

why sin(pi/2 + a) = cos(a)

i saw unit circle definition on khan academy website

i did not understand why in 2nd quadrant triangle adjacent side to angle becomes sin

he said side OA becomes sin(theta + pi/2)

shouldn't it be cos(theta + pi/2) because base/hypotenuse = cosine??

[peak flower]

@gray minnow I don't really know how to explain it on the unit circle but look at the graph of sine and cosine, then it makes perfect sense.

,w plot sin(x) and cos(x)

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by veryhappyperson#1152

[gray minnow]

im starting to understand now

that unit circle definition doesnt really follow that ratio property outside 1st quadrant

in unit circle definition y coordinate is sine and x coordinate is cosine, the ratios doesnt reallly matter

correct me if im wrong

its because of this definition that sine and cosine graph follows that wave fashion

[hasty flare]

hi i need help with qn

[tiny snow]

@hasty flare Are those planes specified in hessian normal form?

Or is it the parametric form, but with only an r component?

[hasty flare]

Is the scalar product form

[tiny snow]

So, those two are parallel planes, given that the normal of the first is a multiple of the second. I don't know how to construct ABCE though, so I'll pass.

[frank inlet]

I honesty

Looking for a geometry

Triangle circle exc questions asking angles

Where I can find it

Or like why I can t see advanced math

Server

Or math meme or discussed question learn new info

[finite jackal]

@gray minnow you are right in thinking that the adjacent side is related to cos, but the identity sin(pi/2+a) = cos(a) shows that they are equal. both things are saying pretty much the same thing. should it be cos(theta+pi/2)? well cos(theta+pi/2) = cos(theta) * cos(pi/2) - sin(theta) * sin(pi/2) = cos(theta) * 0 - 1 * sin(theta) = -sin(theta). does oa = -sin(theta)? cos(theta) = OA/hyp --> OA = hyp * cos(theta). In a unit circle the radius is 1, which is also the hypotenuse of this triangle. After you simplify you get OA = cos(theta) = sin(pi/2+a). Thus cos(theta+pi/2) does not equal OA. hopefully that helps

[gray minnow]

yes i can see that the trigonometry identity fits with the relation

[stone chasm]

can i have help please

[wintry tundra]

well whats the perimeter of a square

[finite swan]

Perimeter of a square is 4 times it's side length. If the perimeter is 100, then that means one side of the square is 25. If the side length is set to 4x-3, then 4x-3=25. Therefore 4x=28, and x=7.

[dull reef]

could i get some help on this

[torpid hearth]

use trig ratios:$\$ $\theta = 72$, adjacent = 8, opposite = x, hypotenuse = unkown
$$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$$
$$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$$
$$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$

[somber coyote]

keto11

[vapid stag]

@dull reef what do you think x is?^

opposite, hypotenuse or adjacent?

[dull reef]

adjacent

[wintry tundra]

whats the definition of adjacent @dull reef

[humble pulsar]

Im pretty sure they were helping Keane...

[upper karma]

hello
i have a problem
so if i have an arrow and its facing upwards, to move it i would decrease y
if its facing right, i would increase x
and etc
my question would be, how would i know how much x/y should i change by if the angle is unknown
for example if the object is facing upwards the angle is 0
if right then 90
and etc
but if the angle is for example 233, how would i then know by how much should i change x/y if i want to move it upwards

[torpid hearth]

I think you're talking about vectors, so you would just break it into components

[upper karma]

idk whats a vector

[torpid hearth]

uh

okay

does your problem look something like this?

[upper karma]

idk what this is

this is what it looks like

[torpid hearth]

okay

so you have a line with some angle

[upper karma]

well not really

[torpid hearth]

???

[upper karma]

can u join a call and i would screenshare (i wont talk)

this was just an example

its a bit different in reality

[upper karma]

can i screenshare

its impossible to explain over text, i would show in paint

[torpid hearth]

okay

but I can't talk

[fast hemlock]

im confused with the whole cosine and sine ting can some1 help?

[humble pulsar]

try posting a question youre confused about

[unique flower]

sure

there are many ways to explain it

this is the right triangle method of trig is the image below.

other things like include: cotangent = adjacent/opposite, secant = hypotenuse/adjacent, and cosecant = hypotenuse/opposite iirc

the full version goes for all angles, not just those from 0 to pi/2.

[light anchor]

notice how tan(theta) is a tangent line on the circle

[remote bronze]

is this correct

[wintry tundra]

hmm

i dont think so

[unique flower]

i think it is

actually

nvm

it isnt

wait

ok i made some miscalculations

false is the correct answer

but try to look for cases where this would be wrong

[rugged sage]

which ones are wrong if u guys can help me

my teacher takes homework seriously

#17 im working on still

[wintry tundra]

do you know the side ratios for a 30 60 90 triangle

[rugged sage]

i forgot

can you refresh me

[wintry tundra]

shortest side is a, longest leg is a root 3, and hypotenuse is 2a

[grand mango]

Can anyone please help with this question? I dont know where to start.

[upper karma]

@grand mango ez

if there is a gp with three terms

a ,b and c what is there relationship with each other?

||use that to throw 2 sin theta , 1 + cos theta and 4 sin theta and then multiply b and c and use (a+b)^2 = a^2 + 2ab + b^2 ||

|| then use sin^2 theta + cos^2 theta = 1 to get rid of the sin^2 theta ||

finally do a bit of manipulation and you have your result proven

the key to the question is to know the relationship of a geometric sequence's (first) three terms

[grand mango]

Thanks, but i dont understand what they want as a result

[upper karma]

who ?

[grand mango]

the question

[upper karma]

they want you to proove the identity

[grand mango]

This is too hard, lemme try

[upper karma]

i can send you the proof written down if you want

[grand mango]

Yes please. this is unfathomable

[upper karma]

@grand mango

[grand mango]

thanks a lot . respect

[upper karma]

man there is really no difference between my handwriting and a toddler's handwriting

[grand mango]

mines worse

[warm remnant]

The diameter of the barrel is 395 mm at each end, 477 mm long in quarter and three quarters high, 500 mm high in the middle. The total height is 581 mm. The volume of the barrel in liters is calculated according to the Simpson rule.

[rugged sage]

@fluid portal

hi

[fluid portal]

@rugged sage ayeeee

[dull pendant]

nah man

i really like it

[upper karma]

ty

[civic olive]

AYO IS ANYONE BIG BRAIN HERE

[silk patio]

How much money you got in yo pocket homie

[kind patio]

can anyone help me

[upper karma]

Does anyone know how to do this

[trim breach]

For part a, apply distance formula using the known point (5, 5).

Where a is obviously equal to the distance between R and Q or S.

Working on the proof atm.

[rocky orchid]

Can anyone explain this to me?

[storm portal]

Probably not with that small of a screenshot

[rocky orchid]

Can’t zoom in ⁉️

I already got it it’s fine tho thanks

[storm portal]

It's still small despite zooming in

:(

But nice on getting the problem :)

[sturdy grove]

Does anyone know how to do this?

[trim breach]

I think you should be able to solve that given that you can draw an apothem and have two sides of a right triangle.

Er, rather, you have one side of a 30-60-90 triangle, which you can use to find the apothem, which is also the radius of the circle.

[rough stone]

Anyone on to help?

@fast pulsar maybe you could help shed some light

?

[dark sparrow]

help with what

did you post a question earlier that got buried?

@rough stone could you (re)post your question please

[rough stone]

no it isn't

I have an issue computing a problem in a ti84

it is 4sin50degrees/8

[dark sparrow]

what's the issue

,w sin(50°)/2

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ann#0413

[rough stone]

so the answer comes out to 22.52 degrees

[dark sparrow]

(just to have a reference for the true value)

?

[rough stone]

but I'm getting 0.383

[dark sparrow]

excuse me what

can you show exactly what youre putting into it that gives you an answer of 22.52

[rough stone]

one second

sorry it was 4sin50/8

[dark sparrow]

?

[rough stone]

pic uploading

[dark sparrow]

you're trying to evaluate 4 sin(50°)/8, yes or no

[dark sparrow]

...

okay

[rough stone]

so I know the answer is incorrect

I'm lost at 4sin50/8

[dark sparrow]

so really you want the value of B given that sin(B) = 4sin(50°)/8

[rough stone]

yes

[dark sparrow]

,w arcsin(sin(50°)/2) in degrees

[rough stone]

I was trying to find the values of the angles to get the values of the sides

[somber coyote]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Ann#0413

[dark sparrow]

agh

,w arcsin(sin(50°)/2) * 180/pi

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ann#0413

[dark sparrow]

22.52

okay, so...

[rough stone]

so I take the inverse of sin

[dark sparrow]

you got 22.52 degrees from your calculator?

[rough stone]

so I've been putting 4sin50/8 and not the arc sign

hold on

I just don't know how to compute it in the calculator lol

I feel retarded

[brave zephyr]

What did I do wrong here?

Ignore the 4 extra row of problems, not sure why it’s doing that.

[dark sparrow]

show work?

[fast pulsar]

Sounds like Ann got you taken care of

[dark sparrow]

these solutions are a little hard to diagnose for faults

[fast pulsar]

Also, why is this equation showing up five times 🤔

Maybe it's 5 times as important

[rough stone]

I still don’t know how to input it in my calculator D:

I’m so confused

[fast pulsar]

This problem right?

[rough stone]

Yes

So I get 4sin(50)/8

[fast pulsar]

Okay, first, make sure your calculator is in degrees

[rough stone]

Yes it is

[fast pulsar]

Okay good.

[fast pulsar]

Now do $\sin^{-1}(ans)$

[somber coyote]

dackid

[fast pulsar]

Where ans stands for answer

[rough stone]

Wow how does that work o.0

I’ve been sitting here for 2 hours trying to figure out this problem

Lol

[fast pulsar]

The ans takes account of the last output given by the calculator

You can also store outputs as variables

[rough stone]

So let’s say we didn’t do it that way

Is there an easier way?

Or would that be the easiest to input

[fast pulsar]

1. answer is super easy to do, but we can do this in one go

[rough stone]

I tried in wolfram and get a different answer like 21.94

[fast pulsar]

[\sin^{-1}(\sin(50)/2)]

[somber coyote]

dackid

[rough stone]

So when inputting this I get 25

[fast pulsar]

That will get the job done

I got roughly 22.5°

[rough stone]

Unless I’m inputting it wrong again

[rough stone]

Oh

I didn’t close

[fast pulsar]

Yes you are. You are putting the 2 inside the sine, which is not where it should be

[rough stone]

So in terms of the half is that determined by the 4/8 reduced to 1/2?

[fast pulsar]

Yep

[rough stone]

Let’s say if it were to be 3/9 it would be arc sin(sin(50)/3

[fast pulsar]

You forgot to close the arcsine, but yes

[rough stone]

Yeah sorry was trying to type fast on my phone haha

Thank you both so much

[fast pulsar]

You betcha

[wanton storm]

Lol I used to think is is what people meant when they said algebraic geometry

[rough stone]

How can you determine whether a triangle has two triangles or one?

is that the arc sign of the answer?

[upper karma]

What does it mean for a triangle to "have two triangles or one"?

[tiny snow]

In what context did you see this wording?

[silent plank]

it was shiro's question above

related to the ambiguity when applying the sine rule

[tiny snow]

Ah, thanks

[silent plank]

find R using sine rule

using sine rule

or law of sines

by knowing that the sine rule or law of sines is or whatever

its called where you're from

[dark sparrow]

did the other person delete their question

[silent plank]

yes and then reposted

#help-8 message

[upper karma]

they banned me for sending cars on road

so my messages were deleted

[hearty hedge]

Anyone good at quadratics?

[storm portal]

@hearty hedge

https://dontasktoask.com/

[hearty hedge]

I’m asking Becuase I need help so I’m seeing if anyone is good at it smh

[storm portal]

https://dontasktoask.com/

The same idea

Just ask the question you have

It's not that hard.

Read the rules man

#❓how-to-get-help message

[hearty hedge]

You really had to link a dumb https for me like chill

[storm portal]

First rule

[storm portal]

Alright

[hearty hedge]

Sorry I didn’t read the rules clearly I guess but you didn’t have to be an ass and link some gyro coulda just said read the rules 😂

[storm portal]

Apparently you needed the second reminder

¯_(ツ)_/¯

[hearty hedge]

Go outside smh

[storm portal]

<@&268886789983436800> Someone is being rude to me

Guided them to the rules and is still being rude.

Told them what they had to do explicitly.

[hearty hedge]

Lmao naw u acting like one of the discord kids this is why discord is ruined by kids like u

[storm portal]

Okay?

Sent me a call in DMs

[hearty hedge]

Mia luck

[storm portal]

Don't feel comfortable with him/her here

[hearty hedge]

That was a misclick that’s why I hung up instantly meant to click reply

[storm portal]

[hearty hedge]

And I closed the dm to stop trying to get me in trouble for nothing

So*

[storm portal]

That's believeable

[hearty hedge]

Like man just leave me alone it was a mistake like your the type of kid to get bullied in school Becuase you look to get everyone in trouble

[weary drift]

@storm portal @hearty hedge please chill a bit. plant is right, do familiarize yourself with the server rules before using the help channels

[storm portal]

Sure?

[weary drift]

and mando knock it off with the insults

[storm portal]

I was never aggressive

I stated the truth.

But understood.

@weary drift ty

[hearty hedge]

Yeah I get that but then posting in the chat that I called u by accident my then you were just looking to get me in trouble I messed up on the rules part and I admit my mistake for that

[storm portal]

Okay, then we are good?

I mean I'm fine with you doing whatever but once insults start flying that's when I lose respect for people

[weary drift]

@hearty hedge none of this warranted using insults

[storm portal]

Because I have a hard time believing you DIDN'T mean those things.

[weary drift]

Lmao naw u acting like one of the discord kids this is why discord is ruined by kids like u|
your the type of kid to get bullied in school Becuase you look to get everyone in trouble
even if you think you've been wronged, stuff like this isn't ok in a response

[storm portal]

I agree!

Thank you

;-;

[weary drift]

[upper karma]

yo any1 good w/ proofs

im so lost

[trim breach]

Post your question.

[inland salmon]

someone help please

[remote bronze]

is this correct

[glacial haven]

the area?

use law of sine

or

law of cosine

[sweet plume]

Its 1/2abSinC

[glacial haven]

uh just google it

find the third side and use herron’s formula

[sweet plume]

if you have an angle in between two sides you can use the trig area rule (1/2)absinC but you may need to manipulate the values based on which variables you have

[humble pulsar]

Ill never understand why people suggest heron's first and foremost >.>

[trim breach]

Those are tangents to a circle.

Do you know the formula to find the angle formed outside the circle by two tangents?

[tawdry quarry]

answers?

[valid sinew]

For the triagnle, use Pythagorus Theorem

That will then get you all of the polygon sides

[copper dove]

is the solution x = (7pi)/6 + (2pi)/6, n is an element of the integers equivalent to the ones in the screenshot?

[valid sinew]

x = (7pi)/6 + (2pi)/6, how come you divided 2pi?

[rugged sage]

"What name refers to an angle whose terminal ray is positioned on the x or y-axis"

is it standard position

[tough ember]

$$\cos{(a+b)}=\cos{a}\cos{b}-\sin{a}\sin{b}$$

[somber coyote]

ninnymonger

[tough ember]

What’s the standard name for this formula? (And its sister formula for sines.?)

It’s not the double angle formula.... that’s only true when a= b.

it’s not the law of cosines.

is it named after some famous dead persian math boi?

[unreal patrol]

how

how do

[tough ember]

You can prove it using complex algebra....dear lord I don’t know the name of its identity.

[silent plank]

compound angle identity

consider pythagoras and cosine law

[copper dove]

I was just trying to find a shorter solution to the answer, would (7pi)/6 + (2pi)/6n be an equivalent solution because adding (2pi)/6n can reach both 11pi/6 and 9pi/6 While replacing +(2pi)n if multiplied by 6?

[valid sinew]

You don't divide the 2pi. I'm trying to remember properly but from memory, 2pi is the period and the 7pi is when it occurs. n is how many times it goes around

[copper dove]

ok thank you

[remote bronze]

i need help with this

[tiny snow]

@remote bronze Marking the center and drawing radi to the corners will give you isosceles triangles. You are given the central angle for one of them. For the other two, use the central angle theorem together with the given inscribed angle.

[remote bronze]

ah ok

i got x = 65

[tiny snow]

That's correct.

[remote bronze]

ok ty

[compact dove]

Yo anyone know how to even start this problem I’ve been stuck for a while now

[tiny snow]

Find x using Pythagoras. Find the angle for which 6 is opposite and 15 the hyp. This will in turn give you the other angle in the bottom left corner. Use tan for that now known angle with y as opposite and x as adjacent (with x now known).

@compact dove ^

[compact dove]

thanks

[inland salmon]

someone help please!

i dont know what OYX is

[tiny snow]

From 10 you have <FOE. Think of what type of triangle OXY is.

What is the length of OY compared to OX?

@inland salmon ^

[outer mirage]

really need help

[silent plank]

where u stuck?

[outer mirage]

where to start

teachers never went over this

[silent plank]

pythag

it should be one of the first things they teach you when doing right triangle geometry

[torn tundra]

can someone help me with this problem plz

[outer mirage]

thats a^2 + b^2= c^2 right

[silent plank]

yes

[outer mirage]

so how do i know what values to subsitute it with

[silent plank]

google pythagorean theorem

[outer mirage]

no way

Thank you so much @silent plank

[torn tundra]

yes but i keep making a mistake while solving it

[storm portal]

6^2 = (x-7)(5 + x - 7)

i think?

[torn tundra]

yea

[storm portal]

ah

do you have work i can check?

[torn tundra]

yeah

not really lol

let me rewrite ikt

@storm portal

like this right

should i continue

[storm portal]

yup 🙂

6^2 = (x-7)(x-2)

[torn tundra]

continue?

@storm portal

[storm portal]

looks good to me

👍

idk if what i said is right

don't remember anything about geometry

[torn tundra]

well guess what

that formula u wrote

[storm portal]

okay so apparently i am right

[torn tundra]

was correct

yes

[storm portal]

that is great

[storm portal]

and solve for x

ideally by factoring

[torn tundra]

quadratic equation

[storm portal]

since that's easier than messing with the 9 and 22 in the QF

yup

[torn tundra]

how do i factor again

shit

@storm portal do u remember how

[storm portal]

oh yup

gotcha

basically

we want to factor x^2 - 9x - 22

thus, we are looking for 2 numbers that meet 2 requirements:

1. They add to be -9
2. They multiply to be -22

[torn tundra]

oh shit

yes

[storm portal]

remember?

[torn tundra]

i just went in a khan academy video

lol

[storm portal]

👍

[torn tundra]

much more efficient fr

[storm portal]

👍

The quadratic formula gets ugly

[storm portal]

there you go!

[torn tundra]

Thank you so much

[storm portal]

np

[torn tundra]

🤗

[storm portal]

🤗

you good with the rest of the problem?

if you think so - my only advice is to just be careful and think slowly through the rest of the problem

if you'd like some more help - i have no life

and i'll be here

just ping me

[torn tundra]

yes

11-9

@storm portal

what could 3 and 4 be

1 is 87 and 2 is 81

[storm portal]

there's probably some formula i don't remember

[silent plank]

apply properties related to opposite angles of cyclic quadrilaterals

[torn tundra]

@silent plank oh fuck i forgot all about those

[storm portal]

i don't remember those whatsoever

or i just never learned them or i just don't know them or a combination

¯_(ツ)_/¯

[torn tundra]

@silent plank this is my last wrong answer. do you know the answer to this?

or better question,

how to solve it?

[silent plank]

how'd you get 72.9

its awfully close to the correct answer. did you make a typo?

[storm portal]

,w calculate 2(27.4 - 5.9 + 5.9 + 12.2)

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Shen#1106

[slate pewter]

I have a intresting question it camed to my mind

If i have a jigsaw puzzle. 48cm X 34cm and it have 500 piece

How can i calculate each piece avrage possible size

[wintry tundra]

Hmm

Probably area over amount of pieces

[slate pewter]

Yeah maybe

[pure barn]

I need someone to explain how to find x pls and thank you

[silent plank]

apply properties of similar triangles

[wintry tundra]

^

also use substitution with pythagorean theorem after getting a certain side

if u dont wanna use similar triangles thing

ahh nvm

u end up getting two extraneous answers

[upper karma]

is it 79?
90*tan1/2 and move the decimal idk

[silent plank]

no, not even close

have you been introduced to inverse trig yet?

[upper karma]

ohh it’s 27

i forgot you had to inverse to find angles

thank you

[jolly cargo]

im so ridiculously confused

[lusty kraken]

wtf is $m$

[somber coyote]

spoon

[torpid hearth]

$\text{wtf is} : m$

[somber coyote]

keto11

[silent plank]

measure

[valid sinew]

Silly geometry question

Can I assume r is half the rectangle x length?

[trim breach]

Late reply, but ΔEFG and ΔHFJ are similar, so you can use that to set the angle measures equal to each other.

[trim breach]

I’m a bit confused by what you are asking…r is the radius of the circle and is not half the of the rectangle’s length. A radius touches the perimeter of the circle, so if you would draw that radius parallel to the rectangle’s longest pair of sides, it would be longer than half the rectangle’s length.

Since r can touch where the two sides of the rectangle meet (since it also touches the circle there), you can draw triangles if you would ever need to solve for things.

[valid sinew]

Well, I asked because the radius is counted at the corner of the rectangle so I thought maybe?

[trim breach]

The relationship with what I think your are talking about is a triangle where: r^2 = (b/2)^2 + (h/2)^2

Which would be what happens when you draw a line from the center of the circle to the midpoint of the shorter side of the rectangle.

[woeful tusk]

hey ive been trying to figure this out. can some of you help? you have to find the length of red and green lines. also the main lines are parallel

[silent plank]

apply properties of similar triangles

[woeful tusk]

yes i know i just dont know where to begin

[trim breach]

GA is similar to GX, and GB is similar to GY. Use a proportion since you know three of those lengths.

[woeful tusk]

yes thanks i just figured it out

why am i so dumb :((

thanks mr. vegatabel

[unreal patrol]

how do i do thi s

[dark sparrow]

one way is coordinate bashing i guess

[silent plank]

could be done with angle sums, properties of certain triangles

[upper karma]

what is the difference between arc sin of x and sin invere of x

[unreal patrol]

p sure arc sin literally means the inverse of sin

how 2 do this

[warped fjord]

Is 6 the answer?

[unreal patrol]

howd u get it

did u get the area of the rectangle

then like

got the square root

[warped fjord]

Yeah

[unreal patrol]

ooh

alright

thanks

[warped fjord]

Is 75 the answer?

[unreal patrol]

ye

i got it

[steady shoal]

Can anyone think of a quick approach to this

[dark sparrow]

N is the foot of the perpendicular to the tangent at what?

the last word(s) got cut off

[steady shoal]

Tangent at P

[dark sparrow]

ah.

hm

...which of these answer options even represent points on your ellipse in the first place?

it feels as if only option A fits just that alone

all the others don't give you 1 if you evaluate x^2/a^2 + y^2/b^2

[steady shoal]

I though of taking parametric coordinates ( a cosø ,b sinø) but the process is very tedious and long

Yes ik options can be put to get the answer but rn I’m more into the method without using brute calculus

[dark sparrow]

is this one of those "solve in under 10 seconds" questions

bc under THAT time constraint the only method is to test the answer options

okay so like

you can try letting $P = (x_0, y_0)$, try to write down the equation of the tangent, then find the coordinates of $N$ and maybe there's something clever to be done along the way that they expect you to just know

[somber coyote]

Ann

[steady shoal]

No definitely not a 10sec question but must be done within 10 min or so

[halcyon pelican]

I don't know what category of geometry this math goes under, but I'm building a function to make the 0-intersection of a hyperbola fit two spheres like one sphere is a light source and the other is the object casting a shadow. I have everything working except for the position of the 0-centered sphere.

Can't quite find where to insert the remaining aspect that takes Bx into account.

[remote bronze]

is this correct

[earnest echo]

Looks correct

[remote bronze]

ok

what about this

[earnest echo]

What test is this??

[remote bronze]

homework

[earnest echo]

,w (r-10)²+12²=r²

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by The Godfather#1908

[earnest echo]

,calc 61/5

[somber coyote]

Result:

12.2

[earnest echo]

Looks correct

[remote bronze]

ok ty

and this?

[upper karma]

In Right Triangle xyz, with integer side lengths x, y and z, P = 510, and A = kP for some prime number k. Determine all possible values of k

[upper karma]

<@&286206848099549185>

[unique robin]

holy shit lol

no idea how to even do a lot of this stuff

[wintry tundra]

That's how it be sometimes

[tawny cypress]

How do I solve this?

[wintry tundra]

Notice how c on dcf lies on the circle

A rule of circles is that if there's an angle with its midpoint lying on a circle the the arc it creates is twice the angle

Since a circumference is kinda measured in 360 degrees

Then If an arc created by an angle who's midpoint is in the center is measured in an angle then the angle is the same as the arc

So boom that should be what u need

[dark sparrow]

hmmm

maybe join M with A, B and C and consider the areas of the six triangles your ABC is split into?

just spitballing here, it may very well be useless to do that

[upper karma]

I think I've seen something similar

yeah try drawing || parallel to AC through M, parallel to AB through M and parallel to BC through M ||

bmo1 problem

[upper karma]

exactly

[upper karma]

yeah it's pretty weird

[nocturne junco]

hello. How did the 1-sintheta/costheta become 90-theta/2?

[tender prawn]

i may have copied it wrong

[tender prawn]

^

but i'm not sure if they're equal?

[honest jolt]

@tender prawn you have your angles in radians there you need to switch them to degrees in the settings (top right)

[tender prawn]

eek noob mistake my bad

[honest jolt]

xd

[tender prawn]

oh use =(1-cos(x))/sin(x) where x is 90-theta, and then cos(90-theta)=sin theta and vice versa @nocturne junco

hopefully that's right

[round delta]

hallo

plz

HALPP

[livid moss]

Did you draw a parallelogram?

This is basically just asking you to list the properties of a parallelogram

[round delta]

nope

they just straight up

SEND DAT

AND SAID

HEY TURTLE ANSWER DIS

[livid moss]

wow

That's outrageous

[round delta]

so yeah..

idk

[livid moss]

So you don't know what a parallelogram is

?

[round delta]

yeah

[livid moss]

You know or you don't?

[round delta]

i dont...

[livid moss]

Do you have a textbook?

[round delta]

i dont.....

they dont

let me

[livid moss]

Where do you learn from then?

[round delta]

uh

google

and youtube

[livid moss]

Well, then google parallelogram

[tender prawn]

what did you learn before this

[round delta]

uh

algebra

and

dis new lesson

geometry

and i have no clue at all

[tender prawn]

so a parallelogram is a quadrilateral with opposite sides being parallel

did u learn any geometry before this

[round delta]

nope

previous lesson

is

algebra

then moved to geomtery

and dis paralelogram thing

[tender prawn]

hmm

[round delta]

tmx 4 try

[tender prawn]

do you know how to prove triangles congruent or anything like that

[round delta]

uh...

sadly no

cuz

yeah

[tender prawn]

yikes

[round delta]

yeah ik im so dumb

[tender prawn]

they probably just want you to google properties then

nah it's not rly ur fault

they kind of threw u into the deep end if u have no prior geo knowledge

[round delta]

even my friends asks me

da answers 4 dat

cuz

they dont know too

[tender prawn]

what grade/age?

[round delta]

9th grade

[tender prawn]

oh

hrm ok lemme try and draw some stuff

[round delta]

thank u

vewy

much

[tender prawn]

do you know about corresponding angles or alternate interior angles in parallel lines

i gotta do smt, here's a video

https://www.youtube.com/watch?v=X1klhqNHeJk

it has most of the stuff, if you have questions on why some of the stuff is true just ask

[round delta]

aight

tmx very muxh

imma look up to it

<333

tmx

alot

[tender prawn]

and this. p and q are parallel lines cut by transversal r, the numbers with the same color have the same measure, that should help with some of the angle stuff

[round delta]

currently looking up da video

imma try dat too

tmx alot

[obtuse tapir]

Type faster

[river meadow]

think the transversal is 't', not 'r' .

[fickle sleet]

Parallelograms in 9th grade right after algebra? That's a strange introduction to trig.

[gilded blaze]

if anyone can solve this ima be forever thankful

[valid sinew]

Do you know how to find angles?

[dark sparrow]

they left the server.

[obtuse tapir]

sad

[upper karma]

Hello, can anyone help me with this question <@&286206848099549185>

since the arcs are congruent i assumed that the central angles would also be congruent

[lime crown]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[upper karma]

oh sry

[upper karma]

but would the arc length be the same as well?

[supple onyx]

But what you have picked is correct

They tell you the arc length is the same

Since the radius is the same, so will the angle

[upper karma]

I was confused between a and c

[supple onyx]

Oh it is c sorry

I didnt even look at the other answers lmao

Yeah cause b is also correct for the same reasons so answer is def c

[upper karma]

Oh ok

Thank you

[velvet bronze]

can anyone help me on this pls

[sharp plume]

What's Sen

[earnest echo]

sin

[junior light]

@velvet bronze What have you tried so far?

[somber coyote]

Daksu
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[humble narwhal]

Can anyone help me?

[cerulean scarab]

"If sin A is 0.865 what is angle A"

guys is there a formula for how to do this question

?

oh okay thanks

[timid nebula]

Does anyone know how to do the law of sines

[humble pulsar]

yes

loads of people do in fact

[timid nebula]

Yeah im having hella trouble

[humble pulsar]

however, if you want help with using law of sines, ask the question rather than ask if anyone knows X

cause best bet is someone in a math discord knows that math

[timid nebula]

yeah my bad

solve the triangle a=65 b=45 a=30

[humble pulsar]

a cant both be 65 and 30

[timid nebula]

Thats how its showing on my thing and im confused

@humble pulsar

[humble pulsar]

yes, alpha is 65, not a

so $\frac{\sin{\alpha}}{a}=\frac{\sin{\beta}}{b}$

[somber coyote]

moshill1

[timid nebula]

ohh ok

[humble pulsar]

so then solve for b

then you can find the gamma angle by angle sum of a triangle, then use law of since by with gamma and c instead of beta and b

[timid nebula]

Thanks man

@humble pulsar its so much easier now

[thin skiff]

Find the straight line parallel with alpha1 : x +2y +4z = 8 and Alpha 2: x - y - 2z = 6 and trough p(2,1,0)

This is my solution. Is this correct?

[nocturne thicket]

Please help

[tiny snow]

@nocturne thicket https://en.wikipedia.org/wiki/Inscribed_angle

[nocturne thicket]

so 45 degrees then?

[tiny snow]

Yes

[nocturne thicket]

Thanks!

[quartz torrent]

please I dont know

[humble pulsar]

let GF=x, what's the length of EF?

[quartz torrent]

21-x

[humble pulsar]

right

so both GFH and EFH are right triangles

so you can determine the value of x since HF is a common side

[mortal dove]

Anybody can get the striped area

[quartz torrent]

I need help with this one

[night vigil]

They are teaching this in my Geometry CBE class

But for some reason he didnt talk about how to find JM

[silent plank]

apply the same formula for area

Area = base * (relative) altitude

[wintry tundra]

@quartz torrent wow, took a bit but I have the solution which I will lead u to

[night vigil]

Ok thanks!

[wintry tundra]

Notice how, when the medians are drawn, it creates 2 right triangles

[night vigil]

Yes, but how do I find the altitude?

the relative altitude?

[wintry tundra]

Relative altitude is the highest altitude throughout the whole quadrilateral I'm pretty sure

Which is given

[night vigil]

ok

[wintry tundra]

But anyways these are the 2 triangles created @quartz torrent

So now u can use substitution with the pythagorean theorem to find C

[mortal dove]

Can anybody get the area of the striped part of the sphere

[silent plank]

when i say relative altitude, i refer the the altitude relative to a certain base

@night vigil