geometry-and-trigonometry

[formal raptor]

And the angle sum would have to be 90 I think

[silent plank]

yes the angle sum of the 2 acute angles is 90

and how would you determine whether the angle you have is the larger one or smaller one?
and how would you find the other angle

[formal raptor]

But I’m supposed to be finding the bigger one tho so like how the heck would I figure that out

Heck

[silent plank]

yes the angle sum of the 2 acute angles is 90

a + b = 90
and if you can find a, b = ?

(alternatively, use the other trig ratio)

[formal raptor]

So if I find the smaller one I just need to subtract it from 90 and that’s the answer?

[silent plank]

yes

though that can be avoided if you used arccos in this question

[upper karma]

Cosx -sinx =1 Find x. I got 0,2pi, but I am unable to get the rest of the solution

Please help

[umbral snow]

Square both sides @upper karma

[exotic pewter]

Are people allowed to asked questions here or is this a chat only to discuss geo/trig?

[devout shell]

you can ask

[exotic pewter]

o ok

I have a question on geometry, is the special segments for this triangle an altitude or a perpendicular bisector?

they seem like a perpendicular bisector, but I looked up examples and I haven't seen one go to the middle in a 90 degrees angle

I don't know if it's trying to fake it out with a trick question or I'm over-thinking it

<@&286206848099549185>

[void moat]

Perpendicular is a 90 degree angle

This seems to me as perpendicular because it has a 90

It’s not a right angled triangle though

[exotic pewter]

ye

but it also looks like a altitude as well

[void moat]

Here is a perpendicular bisector

[exotic pewter]

so it's a perpendicular bisector?

[void moat]

Let me look at an altitude one

[exotic pewter]

does it not matter if it goes straight to the middle

[exotic pewter]

hmmm...

[void moat]

Now look at this

[exotic pewter]

it's still a right at angle?

[void moat]

What

[exotic pewter]

nvm

so which is it?

[void moat]

Well on that altitude I just put that blue line

That blue line made another triangle

[exotic pewter]

so it'd beeeeeeeeee

angle bisector?

[void moat]

Well the line I drew was a median

It’s definitely an altitude btw

That’s the answer

Perpendicular would be like this

[exotic pewter]

it looks like a mash between perpendicular and altitude so it threw me off

thank you for your help : )

[void moat]

No problem

Perpendicular the line would be the other way

Like on the picture you showed me

[exotic pewter]

great

thank you so much again

and have a good day

[void moat]

No problem

[exotic pewter]

bye

[void moat]

And if you want a confirmation you can ask another person in this server

I am just giving you my answer but it could be incorrect!

[tender iron]

Would lines a and b be parallel

[weary drift]

how can you find out?

[tender iron]

This is the questions

I would assume it is drawn to scale

[long sapphire]

i mean even from just looking a and b don't look parallel to me

but i assume that you're not supposed to simply go by eye

[void moat]

You need to find slope or something

[tender iron]

I solved it already

I'm rushing my hw

[void moat]

Not good to do that

[tender iron]

cause its due in half an hour

[void moat]

What if you don’t get good

[tender iron]

and i really need help with this one

there are check answer buttons

and i get two tried

tries*

[tender iron]

I solved it after pondering the question. and getting help

[minor wren]

The markings between angle ABD and angle DBC make BD congruent to BD by the reflexive property, so there are 3 isosceles triangles here, right?

[dark sparrow]

BD is congruent to itself by the reflexive property anyway, regardless of any angle markings

can you name the three triangles in that picture that you think are isosceles?

[minor wren]

I know for sure that triangle ADC is an isosceles triangle, but I'm not certain if the angle markings make triangle ADB and DBC isosceles or not.

[dark sparrow]

in triangle ADB, are any two of the three angles known to be equal

[minor wren]

No, I see what you're saying

The triangles ABD and BDC are congruent to each other, but aren't isosceles.

[dark sparrow]

exactly

[minor wren]

Thanks

[queen python]

Proof unit eh

[wet ice]

Can anyone help me plug in something in a calculator that has sin cos and tan because I don’t have calculator 🙂

[dark sparrow]

use wolfram alpha

[wet ice]

NVM

And no

[copper moss]

Can someone tell me what the two lines mean? I'm currently reviewing vectors

[sullen ferry]

in terms of school math, it's length

magnitude

[copper moss]

oh ok, so the length of both is equal to the length of one times the length of the other? ._.

[sullen ferry]

it means that a magnitude of scalar times vector equals this scalar times magnitude of the vector

that might've sounded clunky but idk how to explain

[copper moss]

wat

[sullen ferry]

I believe λ here represents a number, not a vector

While u is vector.

[copper moss]

Oh I think I get it now. The lesson is basically saying that if the number is positive, the vector and the vector times the number are in the same direction.

[umbral snow]

It's exactly as written. You can always do this:
|λu| = λ|u|
For any scalar λ

How scalars interact with magnitudes

[copper moss]

There

oops, forgot to remove the vector on the scalar lol

[upper karma]

Anybody know how to do this?

[long sapphire]

|s|||u|| = ||su||

looks great

[twin yoke]

hey can i get help on an identifying trig question

[light tulip]

How do you solve to find one of each of the bases (b1 and b2) without knowing the area?

[upper karma]

@upper karma Hint: ADF, BDE, and EFC are all similar to ABC

And they are scaled by a factor of 1/2

From there the problem should be easier

I still don’t understand it

[junior frost]

anyone understand how they got that?

[sharp relic]

consider the pythagorean theorem @junior frost

[junior frost]

Which right triangle should I be looking at? @sharp relic

[sharp relic]

look at the answer given and try to see how the pythagorean theorem may be related

oh wait

lol

yea see how you can make right triangles in there

cut all of em up into right triangles n give some names to the sides

[undone mist]

Hi all!

[keen aspen]

hi

[ivory vortex]

how do i factor -2 from pi/3 to find the phase shift

i get -2pi/3 but the answer is -pi/6

[dark sparrow]

π/3 = (-2) * (π/3)/(-2)

[ivory vortex]

what i dont get it

doesnt that just simplify to pi/3

or what are you showing

[dark sparrow]

that's the point

but you wanted to factor out a (-2)

so i multiplied and divided by it

[ivory vortex]

how is the answer pi/6 then

[dark sparrow]

(π/3)/(-2) = -π/6

[ivory vortex]

doesnt that equal -2pi/3 ?

[dark sparrow]

no, it does not. dividing by -2 is not the same as multiplying by -2.

[ivory vortex]

ohh i get it

i was trying to do this

but i thought -2 was the d value

so i multiplied it wih the numerator

wait

isnt it like this??

[dark sparrow]

uh

okay so

first off

that notation on the lhs is HORRID

[ivory vortex]

what is lhs

[dark sparrow]

there's like 6 different things it could be interpreted as

left hand side

[ivory vortex]

well they're just variables

[dark sparrow]

i'm not talking about the variables

i'm talking about the fraction bars

the middlemost one should have been longer

$\frac{;\frac{a}{b};}{;\frac{c}{d};}$

[somber coyote]

Ann:

[dark sparrow]

and second, no, (a/b)/(c/d) is not (ac)/(bd)

it's (ad)/(bc), in fact.

but what i had isn't of the form (a/b)/(c/d).

[ivory vortex]

how so

isnt it (pi/3)/(-2/1)

[dark sparrow]

ok fine, but -2 isn't in the denominator of the inner fraction

[ivory vortex]

ok i think i get it now

but now this is confusing me

why are they multiplying a with c

shouldnt they multiply c with b

[dark sparrow]

why would they be multiplying c with b

this is $\frac{a}{b/c}$, not $\frac{a/b}{c}$.

[somber coyote]

Ann:

[woeful obsidian]

(a/b)/(c/d) = a/b ÷ c/d = a/b * d/c = ad/bc

And in your example a is actually a/1 to make it simple

So (2/1)/(30/4) = 2/1÷30/4 = 2/1 * 4/30 = (2 *4)/(1 *30)

Division is multiplying by the recriprocal ( I jacked up that spelling 😂)

And idk if this was even being asked, I just saw the end of the convo

[turbid rain]

hi

i have a line given by 2 endpoints

x1, y1, x2, y2

lets say i have a number=5

how can i get the coordinates of polygon around the line with given number margin?

[dark sparrow]

the coordinates of what?

[hasty flicker]

I'm trying to find angle theta.

This is the method I used using volumes:

[dark sparrow]

what am i looking at

[hasty flicker]

right

regular tetrahedron with sides x

and there is a centre point from which lines of length 1 go to each vertex of the tetrahedron

Here is how I found x using the equation for the volume of a tetrahedron

[dark sparrow]

well if you have x

then...

[hasty flicker]

yes ez

it seems

[dark sparrow]

law of cos

[hasty flicker]

but I got an angle of ~105.07 degrees

[dark sparrow]

cos(theta) = (1^2 + 1^2 - x^2)/(2*1*1)

[hasty flicker]

Whereas the bond angle in CH4(methane), which I am told is a perfect tetrahedron, is 109.5

why the discrepancy?

[dark sparrow]

maybe you calculated x wrong

4^(1/3) looks a bit weird

[turbid rain]

@dark sparrow let me draw what i mean

[dark sparrow]

yes please do

[turbid rain]

for the 2nd case it's pretty easy

but for the 1st one i can't figure out

[turbid rain]

sorry

i figured it out

it's just adding sqrt(2)*m to x and y

[junior frost]

What is the minimum value of f(x) = 3sin(2x) + 2cos(2x)?

[upper karma]

@junior frost first differentiate that function

then find the critical number by f'(x)=0

do you know how to do that?

[junior frost]

Yeah. I was wondering if there is any way to do this without calculus? This is from a competition that doesn't require calculus @upper karma

[upper karma]

i'm not sure

but if it is a quadratic equation it is possible

[junior frost]

Oh I guess cauchy Schwartz inequality works. Any other ways anyone can think of?

[flint osprey]

@junior frost you can convert A*cos(x) + B*sin(x) into the form C*cos(x - x_0)
the minimum will then be -|C|

[narrow storm]

is the volume formula of a parallelepiped the same as a rectangular prism?

[quiet mason]

netwonian mechanic motion

formula

,w parallelepiped

[somber coyote]

Display results online and refine query

Requested by Lionel#5497

[narrow storm]

what's this supposed to mean

its a cube?

[quiet mason]

a cube is a square prism if somwthing like that exists

but cuboid is the correct thing for rectangular prism

so yeah the volumes are same

as they are the same thing

considering a prism is a cuboid of glass

[junior frost]

@flint osprey how would you do that? I can't seem to combine that into one cosine

[flint osprey]

@junior frost one way is to expand cos(x - x_0) = cos(x)*cos(x_0) + sin(x)*sin(x_0)

[flint osprey]

another cute way is to note that y = C*cos(x - x_0) solves y'' + y = 0 but the general solution of this equation is y = A*cos x + B*sin x

[old bramble]

how to prove A'P/AA' + B'P/BB' + C'P/CC' =1

[quaint sinew]

is the sum of exterior angles of any polygon 360?

[wind heart]

Would y = 3/2x - 1 be the correct answer for this

“Through the given point that is parallel to the line” - the question continued

[twin prawn]

Does (2, 1) lay on that line?

[balmy roost]

How do i solve 4d?

[twin prawn]

I'd first try and simplify that equation

[upper karma]

@wind heart you can graph the function in geogebra and check if P(2, 1) lay on that line

[woven loom]

Easiest way to get angle pheta?

[dark sparrow]

pheta?

[upper karma]

Theta I assume

[weary drift]

,rotate

[weary drift]

looks like phi to me

[upper karma]

Life begins where Pi meets Phi

[sudden locust]

I think most of you guys have heard of the Pythagorean Theorem. But do you know that it applies for different shapes? https://youtu.be/Ir3MUW_VkgQ In this video, I explain why it applies to semi-circles! 😊

[upper karma]

i mean, yes since the area of a circle is the area of a square of side r multiplied by a constant

[flint osprey]

if you do it with triangles you get a ridiculously short and cute proof of pythagoras

[woeful locust]

im searching for proofs of sin(alpha) = b/c (proof that the ratio is always same with same angle)

[queen python]

how do i prove that ASS doesnt work in words

@sudden locust do you know how i can say ASS doesnt work with words

do i say because we dont know fg=fe

[upper karma]

Could somebody explain why 8x+9 doesn’t not equal 5x-24

Like if the two lines are parallel, they would be intersecting at the same angle on the same line right

[silent plank]

can you clarify what you mean by

they would be intersecting at the same angle on the same line right?

what's the name of the parallel lines theorem you are trying to apply here?

[upper karma]

Oh wait these would not equal each other because these are not corresponding angles

These are consecutive interior angles correct?

Meaning that they add to 180?

[silent plank]

yes

[upper karma]

How do I do this problem? Segment BD is the shortest in one triangle but the longest in another

[summer spire]

doesn't matter which triangle it's in

absolute length

[upper karma]

@summer spire wdym absolute length

[summer spire]

"shortest in one triangle but the longest in another" doesn't impact the list

[upper karma]

How would I find absolute length

[summer spire]

do you know your trigonometric rules?

[upper karma]

I’m in geometry

[silent plank]

BD is the shortest in one triangle but the longest in another
so it'd be the 3rd longest or (3rd shortest) out of the 5 sides.

[upper karma]

But 80 degrees is the largest

Could you explain your reasoning behind that @silent plank

[silent plank]

BD is the longest side in triangle BCD right?

[summer spire]

that's the largest angle, sure. But large angle doesn't necessarily mean the longest side of both triangles

[upper karma]

Yes BD is the longest in BCD

[silent plank]

which means its greater than 2 sides : BC and CD right?

[upper karma]

Yes

[silent plank]

and as you said yourself,
its also the shortest side in triangle ABD. which means its less than AB and AD

[upper karma]

Ohhh

So should I just look at it from the point of view of the different triangles?

So do the shorter ones first

[umbral snow]

The smaller the angle, the larger the side across from it

In a triangle

[upper karma]

See that BD is the longest in that triangle but shortest in the other

@umbral snow yea but the angle was 80 degrees in one triangle

But that same side was 30 degrees in another

[umbral snow]

So that's the largest side of that triangle

[upper karma]

Ok I think I understand now

Thank you all so much

[chrome ice]

this is a basic question sorry but can a glide reflection have multiple reflections

or no

[upper karma]

How do I solve that?

Sorry that was the wrong picture

I need help with #10

[quiet mason]

1:triangle inequality

2:and if you have two angles in a triangle A and B such that A<B

the side opp to A is lesser in magnitude than the one opp. to B

now give it a shot

[upper karma]

Ok tysm

[upper karma]

@quiet mason I have a question

So when I do 15>20-x I get part of the answer

But when I do 15< 12+x I get something incorrect

Why is that?

Are we meant to go from least to greatest?

@silent plank could you help me with this please?

[silent plank]

you need to find the intersections of your inequalities
the first one will get you x>5
the second one will get you x>3
the common intersections of these is x>5

the next step is the apply the triangle inequality

[upper karma]

@silent plank sry for being a bother again but what is common intersections?

[silent plank]

values that satisfy both inequalities

[upper karma]

So is it just guess and check?

[silent plank]

no.

if you draw x>5 and x>3 on a number line,
you'll see that they overlap for x>5

[upper karma]

Oh so the one without overlap is the one we should use?

I also have a question regarding centroids

I’m having trouble calculating things

[silent plank]

wdym without the overlap?

[upper karma]

So you said if we drew x>3 and x>5, x>3 overlaps x>5 therefore the correct one is c>5

Is that right?

[silent plank]

they overlap each other

[upper karma]

So how do we decide which one to use?

[silent plank]

the values where they over overlap each other

(which is x>5)

[upper karma]

Ohhh

That makes sense now ty

[silent plank]

did you do the next step with the triangle inequality?

[upper karma]

Yea so I said x>5 + 15 < 20-x

And got the answer

[silent plank]

uh. can you make that less ambiguous?

what was your final answer?

[upper karma]

@silent plank 5<x<11.5

[silent plank]

mmmk. that looks ok.
dunno what you meant by
x>5 + 15 < 20-x
though

[upper karma]

Oh shoot

I mean 15+ 20-x < 12+X

[silent plank]

surely you mean > right?

[upper karma]

Triangle inequality theorem

Yea my ba

Bad

Do you mind answering a question about centroids for me?

[silent plank]

possibly

[upper karma]

So say for example we have this triangle:

And say for example that AE is 12

I know that AP is 8 because it’s 2/3 and PE is 1/3

But let’s say I only had the measure of AP which is 8

How would I find the measure of PE

[dark sparrow]

AP:PE = 2:1

the centroid divides all the medians in the ratio 2:1

[upper karma]

Ohhh

So I would divide AP by two to get PE?

And if I only have PE, I would multiply by two to get AP?

[dark sparrow]

sure

[upper karma]

Does it always work out like that?

@dark sparrow

[dark sparrow]

does what always work out like that

[upper karma]

Where I can divide the 2/3 by two to get the value of the 1/3 part

And vice versus

[dark sparrow]

i mean

(2/3)/2 is literally equal to 1/3

and (1/3)*2 is literally equal to 2/3

so i guess yes???

[upper karma]

Ok tysm

Is the centroid always on the inside of a triangle?

[dark sparrow]

yes

[thin tartan]

Can someone explain how this is done?

[dark sparrow]

,rotate -90

[thin tartan]

I know all the sides i just dont understand why it isnt 5/4

[dark sparrow]

well for one thing

5/4 is greater than 1

how can the cosine of anything be greater than 1?

(spoiler alert: it CAN'T)

[thin tartan]

but adjacent/hypo -> 5/4 how am i supposed to change that

[dark sparrow]

are you sure the adjacent leg is 5

how can a leg of a triangle be longer than its hypotenuse?

[thin tartan]

i used pythag theorem and got 5 for that leg. idk how else i would go about that

[dark sparrow]

can you show your work

like, all of it

and exactly how you used the pythagorean theorem

[thin tartan]

one sec i think i saw my mistake

i think that leg is just sqrt(7)

idk what i was doing b4

[woeful locust]

are there proofs that the ratio of given triangle sides remain the same if a particular angle remains the same? (the basic principle of trigo functions)

[summer spire]

in a right triangle?

[subtle spear]

Guys, I have a quick question, how does one go about memorizing the unit circle?

[silent plank]

you don't really need to memorise the whole circle.
learn the special angles in quadrant 1
reference angles
memorise when each function is positive: ASTC

[tawdry pivot]

fun fact: if you memorise all the points on the circle, then your head will implode from all the information

[upper karma]

how do i rotate a point by theta?

[weary drift]

multiply by a rotation matrix

[upper karma]

that is what i am asking for

what is the rotation matrix

[weary drift]

http://polymathprogrammer.com/images/blog/200809/rotationmatrix2d.png

[upper karma]

thanks

[weary drift]

no prob

[velvet shard]

So idk if this is obvious, but I'm not sure how to find the area of this figure. I know the coords of A, B and length of |AB|. I also know the angel of v

[silent plank]

the diagram is a little unclear.
what is happening around BC. is the figure on the circumference?
(i'm assuming the red line is a tangent indicating that AB is the diameter?)

[velvet shard]

|AB| is diameter, and yes the figure is a little bit unclear put should just be seen as |AB| -> |BC| -> |CA| where |CB| is an arc part of the circle

I'm not native so I hope you understand what I'm trying to say lol

[velvet shard]

yes

[silent plank]

draw a line from the centre(O) to C

[tiny grail]

@velvet shard introduce center of the circle O, find triangle OAC and sector OCB separately

[silent plank]

^

[tiny grail]

LOL sorry i didnt see u were tryna guide him through it mb

[silent plank]

dw

[velvet shard]

how do i go about including the curved part of triangle OCB? can I cut it off and use this formula or is there a formula i can use that includes it

[tiny grail]

@velvet shard use area of the circle

[velvet shard]

oh ofc lol, tysm^^

[vague pagoda]

graph those

[maiden rain]

@vague pagoda I know tan = sin/cos

@vague pagoda what's arctan equals to?

sin/cos as well ?

[vague pagoda]

no

tan x = y

arctan y = x

[maiden rain]

So cos/sin ?

[vague pagoda]

no its not

[dark sparrow]

no

arctan(x) is not 1/tan(x)

[vague pagoda]

i mean cotangent is equal to adjacent/opposite

but not arctan

forget what i said

[maiden rain]

I'm not asking about cotangent... I want to know what's arctan equal to?

opposite/adjacent ?

or adjacent/opposite?

[vague pagoda]

NO

tan is equal to opposite/adjacent

neither

[maiden rain]

ok thanks

[olive glacier]

How is this done (boxed)

[sudden locust]

@queen python Hmm sorry for late answer, Its easy to show visually, much harder to describe in words

im afraid

"The last side can be rotated to make different size angles"

[upper karma]

Can someone help with the 9th question

Not sure what to do

[dark sparrow]

tan(C/2) = sin(C)/(1 + cos(C))

[upper karma]

How does that make use of A being 90°?

Let me try it tho

[dark sparrow]

it doesn't

but you now have an expression for tan(C/2) in terms of sin(C) and cos(C)

[upper karma]

Oh okay thanks solved it

Thanks for the help ann!

:D

[upper karma]

what are the proper rotation matricies for yaw, pitch, and roll?

[woeful obsidian]

Hey pre-cal is dead so I'm gonna ask here, can I have some guidance on the next step here? I was thinking of factoring, but I don't think that's it.

[umbral snow]

@woeful obsidian
For factoring to work EVER, you need one side to be equal to 0

Once that's done, factoring is the right move

[woeful obsidian]

That's what I did, pulled cos to the left then reacted l took out a cos

1-cos(cos-1)-1

I think that's what I did my notebook is in the room

No that's not what I had, sorry I'll have to look in a bit

I'm out with my kids right now

[woeful obsidian]

I'm pretty sure that's what I did before, but. I'm not sure it's right

I tried to do -cos^3+0cos^2 -cos+1 to factor, but I don't think that's correct

[dark sparrow]

$\sin^2(x) \cos^2(x) \neq 1 - \cos^3(x)$

[somber coyote]

Ann:

[woeful obsidian]

1-cos^4?

1-cos^2*cos^2?

[umbral snow]

There was no cos²(x) in the original question

[dark sparrow]

$\sin^2(x) \cos^2(x) \neq 1 - \cos^2(x) \cdot \cos^2(x)$

[somber coyote]

Ann:

[dark sparrow]

order of operations is a thing, yknow

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

[somber coyote]

Ann:

[fathom shard]

Given than AD =DB, AE =EC. prove that triangle AFG is isosceles

Any hints?

[dark sparrow]

angle-chasing

[turbid rain]

hi
i have a line segment
i.e. x1 y1 and x2 y2 coordinates
and a number m
m is for margin
i need to "enlarge" the area around the line segment with given margin
for example
if m=10, then

the math is easy here:

but if i have a line segment which is under some angle, e.g.

i'm not sure how to calculate the new polygon vertex coordinates

[mighty narwhal]

Well, if you want to expand it like in the picture:

You add (or subtract) m to the y, just like the first example

But for the x, you need to add cos(theta)m

Where theta is the slope of the line

If you can't see why:

Look at the left point of the line, and draw a right triangle there. M is the hyp, and the amount you add to x is the bottom leg.

Bottom leg=(cos theta)m, where theta is the angle that the bottom leg makes with m. It is also the angle of the line

[turbid rain]

okay i get it, but i do care about accuracy. is there some computational geometry technique to do this?

[mighty narwhal]

Im not sure what you mean by that.

Cos(x)m is the exact form, any inaccuracy is done by actualy calculating cos(x). But the precision most computers/calculators have with trig functions is more than enough for things like this

[turbid rain]

i had my first classes of computational geometry, and the lecturer said "we must avoid use trigonometric functions and division here" that's why i asked

[mighty narwhal]

:///
Weird

I dont think you can get the vertex without using trigonometry

Oooooor

Wait a minute

[long sapphire]

You can get two directions, the normalized form of a-b and the vector perpendicular to it

You can get all four corners via vector addition and subtraction

[turbid rain]

@long sapphire can you please elaborate?

I only know one of the vectors

[long sapphire]

Shitty drawing but w/e

You have the normalized form of a-b (as a and b are given to you), with that you can determine the positions of the yellow highlights

You can then get another direction, the vector perpendicular to a-b

You can then get the four corners by adding a subtracting that perpendicular vector to the yellow highlighted

[turbid rain]

what are the a and b?

[long sapphire]

The points given to you

It'd be the red points on the diagram

[turbid rain]

so a-b=(x1-x2, y1-y2)?

absolute values

[long sapphire]

Ah shit I'm stupid you said no division right

[turbid rain]

yes, if possible

i know i could use line equation which is with division

if that's what you were suggesting

[long sapphire]

Yeah I don't know how you're supposed to do this

It's impractical to circumvent tiny magnitudes of inaccuracies especially when dealing with rotations

[turbid rain]

i've heard about transformation matrix. can that help here?

[long sapphire]

No since construction of a transformation matrix requires trig

[turbid rain]

oh i see

how would i do that anyways?

[stark snow]

How can I prove that theta is equal to x?

The tangents originate from a point

So the angle where the tangents meet is 90 degrees since they're connected to a radius

[silent plank]

are you leaving out any information about x?

[olive glacier]

how is the conclusion of k made?

[stark snow]

x is just an angle and I have to prove theta is equal to it @silent plank

[silent plank]

as is, there is insufficient info

[stark snow]

Oh I see

[silent plank]

if it was like:

then they would be equal

[stark snow]

Ah yes I forgot to emtnion that was a right angle

How can I prove that they're equal?

Given that that is a right angle

[silent plank]

angle at the tangent, angle sum of triangle

[stark snow]

What do you mean?

[silent plank]

what's the angle the tangent line makes with the radius?

[stark snow]

90 degrees

[silent plank]

in the triangle i formed with that green line,
what's the 3rd angle? (at the center, in terms of theta)

[stark snow]

Excluding x? the one between x and the green line?

[silent plank]

to be more clear, what is y?

[stark snow]

180 - 90 - theta = y

[silent plank]

can you simplify that?

[quiet mason]

()

[stark snow]

y = 90 - theta

[silent plank]

and what is y + x?

[stark snow]

90

Ah so x has to be theta then right?

[silent plank]

sure

[stark snow]

THank you very much!!

[upper karma]

Hello

Hint?

[median crown]

How do you go from (1+cos(a)) to (1+cos(a))^2

[upper karma]

Squaring?

1+2cosa+cos^2a

[median crown]

Well

Yes

But we need to keep the fraction the same

if we multiply the bottom by (1+cos(a)) we must also multiply the top by what?

[upper karma]

1+cos(a)?

[median crown]

Yes

so do that

multiply (1+cos(a))(1-cos(a))

[upper karma]

Oh I get it since multiplying by (1+cosa)/(1+cosa) is still multiplying 1

Ty

[median crown]

Yes

[placid path]

@olive glacier u got mocks tomorrow?

was just doing that paper in preperation

[wheat briar]

Can someone explain me $\sin(\theta)=\sin(180-\theta)$ real quick

[somber coyote]

Lecko:

[wise pawn]

draw the unit circle and put two right triangles in it with theta and 180-theta as angles then draw the y components

https://prnt.sc/py88rz

if that's theta then on the left side we went 180 degrees but we go backwards by theta so it's 180-theta, both have the same sine

[wheat briar]

Ohhh that’s lit

Thank you

[wise pawn]

👌

[upper karma]

what does this mean?

i just expanded each side to

(sin^2a+cos^2a)(sina+cosa)(sina-cosa)

[dire rampart]

Why is the same q written twice lol

[upper karma]

6b has a equal sign under it

/=

[dire rampart]

Ah yes

Anyway yea

Factoring seems like a good approach

[upper karma]

So do I cancel factors from each side by dividing?

[steep temple]

@placid path what paper is that

[gentle dome]

Does arccosecant have any vertical asymptotes?

I’m pretty sure it doesn’t, right?

[dire rampart]

@upper karma yea u can cancel the sin^2+cos^2 since that's just 1

[torpid torrent]

guys

why does making two perpendicular bisectors work to find a center of rotation

<@&286206848099549185>

please

[fringe crater]

Hello. If you want to find the center of rotation, you're looking for the point where all the perpendicular bisectors meet. This image helps:

So when you rotate a shape, you connect the original and rotated points with a line segment (in red), find the midpoints of those line segments, and then construct perpendicular bisectors

Really you want the point where all the bisectors meet, but since lines only meet once, the point where two of them meet is all you need. That will be the point where they all meet.

And the reason is that you rotated this shape about a center of rotation, so the points of your shape as they rotate create concentric circles, varying in size but having their center at the center of rotation. What you're doing by finding the midpoint and then bisecting, is finding the circle whose chord is (Original Point, Rotated Point), and you're doing this for each set of points. The center of the circle whose chord is any two of these points could lie anywhere along a single perpendicular bisector, but when you have two perpendicular bisectors that meet, then you find the only point in space that can serve as a center for both concentric circles, the one containing (Original Point 1, Rotated Point 1) and the one containing (Original Point 2, Rotated Point 2) as chords

[torpid torrent]

thank you

[fringe crater]

You're welcome

[torpid torrent]

hm how do i simplify that tho

like that explanation

idk how

[fringe crater]

Hmm, I guess you could just explain what it's doing, like the second half is all you really need. More briefly, we could say that the point of intersection of the perpendicular bisectors serves as a center for all the circles traced by the rotation of the original points to their new positions.

[torpid torrent]

hmm so like

the point of intersection between two perpendicular bisectors of pre image and image points yields the center of rotation because this point acts as the center of the circle of every point of rotation around the center

hmm idk

can i have some help

idk whwt to do

[fringe crater]

That sounds good, it depends on how in-depth they want you to go with it. If it's brief you can't really say everything about it. If I had to add anything I would say: the point of intersection between two perpendicular bisectors of the line segments formed by pre image and image points yields the center of rotation because this point acts as the center of the circle traced by every point's rotation around the center

[torpid torrent]

ok and ill provide an example

thank you sm

[fringe crater]

No problem. Good luck. If they let you draw a picture that might help, too

[torpid torrent]

ye its just rigid motion questions thhe one before was about how 4 vector translations makes a parallelogram

[fringe crater]

Ah ok

[torpid torrent]

like i have that much space

[fringe crater]

Ah I see, ok. Then a brief explanation like above plus an example would hopefully be alright

[torpid torrent]

im just gon print out the picture

wheres it from

the picture

so i can cite

@fringe crater

[fringe crater]

Oh, sorry

I was afk

The picture is from here: https://www.dummies.com/education/math/geometry/find-center-rotation/

But I made some marks on it

[median crown]

Where is the last question?

[torpid torrent]

yeah

@upper karma

lmfao what grade

[glacial basin]

been stuck with this for a whole hour

somebody help

[median crown]

Have you tried drawing it?

@glacial basin

[glacial basin]

yes

I drew a sphere

the middle diameter is 26 and radius is 13

[median crown]

ok

[glacial basin]

I drew a circle lower than that with a chord of diameter 20 and radius 10

[median crown]

nice

[glacial basin]

whatever it is

[median crown]

good

And?

what did you get?

[glacial basin]

then got the height of the triangle thing

which is like 8.3

[median crown]

ok

[glacial basin]

I don't know what the question is asking

[median crown]

Wdym

[glacial basin]

is it the depth of the water from bottom to surface

[median crown]

Yes

[glacial basin]

because that's 26-8.3

which is like

[median crown]

no

[glacial basin]

??

[median crown]

It’s a bowl

Not a sphere

How tf is the fish supposed to breath?

[glacial basin]

I can't see the big picture of it

[median crown]

The bowl isn’t actually a sphere

It just takes on the spherical shape

That’s why it says “spherical”

[glacial basin]

then how am I supposed to know the depth

[median crown]

wdym

,w circle

[somber coyote]

Display results online and refine query

Requested by maleb1964#6739

[glacial basin]

kinda like this?

[median crown]

except the bottom isn’t cut off

[glacial basin]

???? how is the bowl supposed to stand lmao

[median crown]

Something like this

[median crown]

yeah

you could have a stand for it

anyway, it looks like that

so now just find the depth

[median crown]

ok

Now find the the other depth

[glacial basin]

other depth?

[median crown]

two possible bowls

[glacial basin]

so if thats sideways

[median crown]

https://media.discordapp.net/attachments/326138757474680852/645861093423710209/2Q.png

That is also a possible bowl

[glacial basin]

ahhh

26-4.7

= 21.3

[median crown]

13+8.3 also equals 21.3

[glacial basin]

o

[median crown]

I guess it doesn’t matter tho

There you go

[glacial basin]

damn

[median crown]

put on labels and you’re done

[glacial basin]

my brain cells are not working until now

did 26-8.3 instead of those

thanks @median crown

I can finally sleep

[languid radish]

this is a basic question asked by someone in another server but I've forgotten some of the similarity stuff

can anyone help out?

[umbral snow]

These are important to recognize
AMB + AMC = 180
The sum of the angles of any triangle is 180

[languid radish]

yes

AB/AC= BM/MC=2

I thought of substituting amb=x and amc=180-x

[umbral snow]

It actually follows that, like you said, these triangles each have two matching angles and are similar

[languid radish]

but it's only one angle though

How would we know if ABM=ACM

if we do some ABM+60+X=ACM+60+180-X?

[umbral snow]

AMB + AMC = 180
The sum of the angles of any triangle is 180

You can't know what each angle is, but the opposites have two matching

[languid radish]

yes it's what I'm using,taking AMB as X and AMC as 180-X

[umbral snow]

Yeah that should work

[languid radish]

yeah but idk how I can go to the final answer I'm super slepp deprived and unable to think lol

I'll give it a shot and get back

to you I guess

using this, AMB+2X= ACM+180 but how do I go forward

nevermind

ABM+ACM would be 60 and the above eq. gives us ABM-ACM

2ABM=2(120+2X)
giving ABM= 120+2X and ACM=-(60+2X)

I d k

or x+60+(120+2x)=180 and

but

that'd give x=0 lol

or wait

60-60-2x=180 or -2x=180 giving 2x=-90?

that's wrong too

@umbral snow you there?

[fringe crater]

There's another way we could do it too, I'm not sure if it's the right way to do it, but it involves graphing a system of equations

And finding where they intersect

[languid radish]

oh wait the person I'm helping solve this for is only using some similarity stuff

[fringe crater]

Oh ok

[languid radish]

but,I'm interested,tell me your method

[fringe crater]

So let's call Angle ABM "Angle X", and Angle ACM "Angle Y"

Our first equation is X + Y = 60

because MAB + MAC = 120

[languid radish]

yes

yes

which is what I used above

[fringe crater]

And in a triangle, the ratio of two side lengths is the ratio of the sines of the angles opposite those sides

So Sin X / Sin Y = 1 / 2, since AB / 2AC

[languid radish]

sine rule?

[fringe crater]

Yep, law of sines

Graph those and find where they intersect, and that gives you the angles

[languid radish]

ok

thanks for telling me this

[fringe crater]

Yeah np. Once you got the angles you can use the law of sines and find the other side lengths

[languid radish]

yeah

I was trying to approach it from a similarity only pov

with some algebra

[fringe crater]

Ah ok

[languid radish]

because it's how the person who asked is supposed to solve

goodbye and thanks for the solution!

[fringe crater]

You're welcome 🙂

[silent plank]

i used
total area = area1 + area2
to immediately find x

and then apply cosine rule for BC

[placid path]

@steep temple its november 2018 edexcel paper 1 non calc

[steep temple]

thank you

[soft scarab]

what's the equation of a plane and a sphere?

aren't they the same?

[weary drift]

,w plane equation

[somber coyote]

Display results online and refine query

Requested by RokettoJanpu#6002

[weary drift]

,w sphere equation

[somber coyote]

Display results online and refine query

Requested by RokettoJanpu#6002

[soft scarab]

thanks

[weary drift]

Np

[soft scarab]

and for 2D functions it's usually y=x format, how do you represent 3D function?

z = x+y?

[weary drift]

First bot search result

[soft scarab]

and if i take whatever point in 3d and insert it into the plane equation, the sign of the result should tell me if it's on one side or the other, correct?

[tropic bolt]

can i break this triangle into two right angled triangles?

i need to solve for <AB

[normal harness]

Does anyone know a formula to find the centroid of a triangle that exists in 3D space?

By that I mean there are 3 points that exist in 3D space with X Y Z coordinates. Connecting these points forms some triangle that exists in 3D space. Assuming we know the components for points A, B, and C, is there a formula to find the coordinates of the centroid which we can label as point D?

Nvm figured it out

Just gotta expand this into 3D

[tropic bolt]

,calc cos(pi)

[somber coyote]

Result:

-1

[tropic bolt]

,calc cos(pi-2)

[somber coyote]

Result:

0.41614683654714

[chrome ice]

Is my solution correct

And this is how I solved it

Ah whoops it’s rotated wrong

Here’s the problem

[vagrant steeple]

do i just plug this in and it'd be
x = 22^2 + 27^2 -2(22x27)cos(25) ?

[dark sparrow]

don't use x for multiplication especially when you have a variable called x

but yes that's what it would be

[languid radish]

yes plug

[idle bloom]

@tropic bolt I mean you know the angles have to add up to pi radiens right? So use that sum of angles in a triangle property to find theta

[iron kite]

Embedded assessment for geometry. @ me

I need help with the rigid motion questions

[proven rapids]

Can someone help me with this question?

[smoky mountain]

Since the tangent of an angle is the opposite side length / adjacent side length, you can take the inverse tangent of that to find the angle measure.

For A, you'd do tan-1 (16 / 30) and for C, you'd do tan-1 (30 / 16)

[coral zealot]

ok

[upper karma]

all the interior angles of a triangle equal 180

and you know that two of the three interior angles are 81 and 31

that means 81+31+v = 180

to find out what the interior angles of a polygon add up to, the formula is: (s-2)180

S is the number of sides you have

[dark sparrow]

all the interior angles of a triangle equal 180

no, they ADD UP to 180°

no one of them can ever be EQUAL to 180°

@upper karma

[upper karma]

Yea my bad that’s what I meant

all the interior angles of a triangle

[dark sparrow]

you may note that you just copied precisely the part i did not correct

[upper karma]

Yea I’m saying that by saying the copied part I mean to say that <1 + <2 + ❤️ in a triangle = 180

< 3*

But my mistake I understand where you are coming from

What I said could have been misinterpreted

[dark sparrow]

yeah just know that "all the angles" does not automatically mean their sum

[sinful marten]

Hello people!
I need help with my homework and I'm really struggling to find alpha in this triangle. I could do the first exercise without any issues but I'm really struggling with this because I'm only allowed to use sin and tan.
https://cdn.discordapp.com/attachments/172645867570987008/646661443361439757/56dea9373372b_353bb3239d41c9f6097b1c7dc10daf0fd7d76215.png

b = 9.7cm
c = 12.5cm
gamma = 90°
I need to find a, alpha and beta

I don't want the exact solution but only help on how do go on about it

According to a friend we're allowed to use the Pythagorean theorem

Which isn't stated in the exercise and I couldn't ask the teacher

So I just calculate a and then alpha and beta using arctan right?

If anyone could tell me if I did it correctly that would be greatly appreciated https://btw.i-use-ar.ch/i/ssfaj9g6.png

[silent plank]

beta is wrong
you used the wrong variable
you had arctan(b/a) but you used c/a

[sinful marten]

in the second exercise?

Ok ye I found it and fixed it

Is the rest right?

[silent plank]

a and alpha look alright

[sinful marten]

Noice

Now I'm happy

I finally understand something lol

[silent plank]

your question a) looks weird

[sinful marten]

why?

To get the hypotenuse I used a formula off the internet that seemed right

It's basically the same reformation of how I got b in exercise c)

tan(alpha) = a/b -> b = a/tan(alpha

[silent plank]

recalculate alpha

these are all right triangle questions right?
alpha + beta should be 90°

[sinful marten]

still the same result for me

yea they are

oh nvm

I had 49 and not 45

Was a typo rip

[silent plank]

which would also affect c

[sinful marten]

in c) alpha and beta look right

~45.6 + ~44.4

yea

[silent plank]

i meant c in q a)

[sinful marten]

Ye I noticed after the edit

Fixed it now

Thanks a ton for the help

[silk crown]

If I pick a point on the asymptote of an hyperbola, where will the two tangents that I draw from there touch the hyperbola?

At two different branches or two points at a single branch

[fringe crater]

Hello CoolShot, if I'm reading your question correctly, a point on the asymptote of a hyperbola is never on the hyperbola, because the hyperbola will never reach its asymptote

[silk crown]

Yeah

But I can draw tangents to the hyperbola from it, right

Here, I have drawn two tangents from the point (1,4) which are tangent to the curve at the two different branches

Here, I do it from (1,0) and both are tangent to the curve at the same branch

And I know that I can divide the plane into four "sections" based on the asymptotes which will determine if the tangents touch the curve at both branches or just one

But what happens when my point is on the asymptote

[fringe crater]

Hmm, that's an interesting question. Thinking about it, I think your tangents should touch the curve at only one branch, and that you'll only be able to draw one tangent line, unless your point lies at the intersection of the asymptotes, where you won't be able to draw any tangent lines. I could be wrong here though, so we'll see if someone else can weigh in on it too.

That's my thinking ^

[wind heart]

I’m really confused on 5. Can someone please help? Thanks

[dark sparrow]

what exactly is confusing you @wind heart

and do you still need help with it

[wind heart]

I found C, which is .99 but I don’t know how to implement it into an equation

[dark sparrow]

what's C

[wind heart]

The rate of change

[dark sparrow]

of what

[wind heart]

Of the minute of call thingy

[dark sparrow]

,,,

,calc 10.86 - 5.92

[somber coyote]

Result:

4.94

[dark sparrow]

,calc 4.94/5

[somber coyote]

Result:

0.988

[dark sparrow]

the \textbf{cost per minute} is $0.988, not $0.99

[somber coyote]

Ann:

[upper karma]

Just got a 35 on a trig test

It was so fucked

[dark sparrow]

also @wind heart this isn't really the right channel for this

[wind heart]

Oh

[dark sparrow]

maybe #prealg-and-algebra

@upper karma 35 out of how much

[wind heart]

Yeah I’m doing geometry but I guess were doing a different thing

[upper karma]

I got 10/35

[dark sparrow]

,calc 10/35

[somber coyote]

Result:

0.28571428571429

[upper karma]

10/30

Sorry

L

Lol

[dark sparrow]

33%, then

[upper karma]

Yea

So fucked

[dark sparrow]

still a pretty yikes result

[upper karma]

The whole class did shit

[dark sparrow]

what was the top grade

[upper karma]

So the teacher said don’t worry cuz he’s dropping the test

The top was like a 70 percent

[dark sparrow]

and what was the test on

[upper karma]

Trig identities

I hate identities

Some guy got like a 6 percent

[dark sparrow]

m

yeah there's like a handful that are useful to know and find uses in stuff outside trig questions but those "prove this identity: (jumble of trig) = (jumble of trig)" questions can be a bit of a mess to untangle

just wondering, do you go to the gym often

there's an analogy i wanna make but i wanna make sure it gets my point across

basically these "prove this identity" questions are like the math equivalent of weight lifting in a way

[upper karma]

@dark sparrow I play soccer

And I have a job

Like when tf am I gonna use trig identities in life

So useless

[dark sparrow]

,,, right, are you in college or sth

[upper karma]

High school

@dark sparrow

[dark sparrow]

i'm gonna guess retail or food

[upper karma]

?

[dark sparrow]

aight nvm

[upper karma]

Lol wut

[dark sparrow]

forget it

anyway

hhf

kinda tired of people doing the "when am i gonna use X in real life"

[upper karma]

Like functions are actually useful like graphs

Identities are not useful

[dark sparrow]

do you make the same sort of complaint in English class with Shakespeare

[upper karma]

No

That’s useful

[dark sparrow]

how

[upper karma]

Because English is the language we speak

We must speak it properly it’s common sense

In the real world

Identities you’ll never use in life guaranteed

Functions and graphs yes

@dark sparrow

[dark sparrow]

,,,

okay no i don't have the energy to deal with this rn

[upper karma]

Not retarded stupid identities lol

So dumb

I’m correct

[upper karma]

@upper karma life is useless

[odd rock]

Yes Mr trump

[storm crown]

Plenty of people can communicate well in English without ever having read Shakespeare, so the purpose of reading literature isn’t to teach you English.

It’s the analysis and critical thinking skills you learn by analyzing literature, and the cultural experience you get

Similar with teaching math

[stone carbon]

i don't know how "advanced" my question is because it's not related to a class at all. i need a function to do something for me in 3d space to scale bones for a video game's character creation system. it's a bit of a doozy to explain the premise so i've drawn a picture if anyone wants to tackle it... basically, the rotation affects the axis and makes scaling incorrect. i need a formula that takes an angle of rotation and gives me the scaling on both x and y. should be using sin and cos https://gyazo.com/7e4435ff825c56f1f9b68fb7326acd82

[upper karma]

Whoa

[stone carbon]

i missed a rather important detail here. with bones they're in a hierarchy so the first scale affects both bones... the blue portion only affects the 2nd bone

[upper karma]

lol it’s twice as fat

[stone carbon]

the point is to undo the scaling from the first when you get to the 2nd... so the first would be 2x but the 2nd needs to be .5 to change it back again

right, because the scale happens in the parent's local space, it makes something bent at a 90 degree angle twice as fat instead of twice as tall

[upper karma]

It ate too much McDonald’s

[stone carbon]

then when you go to scale the new one, the rotation angle needs to be considered, scaling the x .5 would undo this change, scaling y .5 just makes it even worse

[upper karma]

Too much Big Macs haha

Poor guy

How much Mickey d’s did u feed that thing lmaoo

@stone carbon

[stone carbon]

0%

[upper karma]

Wow

So he just was born like that I guess

Low metabolism for the poor guy

@stone carbon

[gritty sail]

$2$

[somber coyote]

Jbao:

[wanton valve]

can someone please do these questions for me

the ones with answers in them as well so I can double check it all

[silent plank]

what's the angle sum of consecutive interior angles on parallel lines? @wanton valve

[wanton valve]

adds to 180 degrees @silent plank

@silent plank i mainly do not understand number 11

[final veldt]

wait isnt that giving you AC nvm thats 13

[silent plank]

did you fix the errors in 2 and 5

[wanton valve]

yes @silent plank

@vivid wraith helped me with those

[vivid wraith]

so the two diagonals of a rhombus are perpendicular, giving you 12y = 90. we get y = 15/2. because we know that the diagonals of a rhombus bisect the opposite angle, we get that angle BCD is 2(4y-1) or 58. We also know that opposite angles of a rhombus are equal so we get angle BAD is 58

[wanton valve]

@vivid wraith thanks

[upper karma]

i do not get the geometric intution of why cos180 = -1

or cos pi = -1

[smoky mountain]

180 degrees = pi radians

[upper karma]

ye i know

i am asking why it equals -1

or why they both equal -1

i get why cos 0 = 1

[smoky mountain]

180 degrees just faces the opposite direction as 0

so if you think of an xy coordinate plane, where the center is at (0, 0), 180 degrees is in the negative x half

[upper karma]

i was looking at the wrong angle the whole time