geometry-and-trigonometry

[drifting parrot]

well we still haven't really learned how to do this yet:

[burnt arch]

Right

[drifting parrot]

i'm also wanting to get ahead of the game

[burnt arch]

Fair enough, well id recommend looking for some geometry textbook or use Khan academy or somm. If you have a specific question, just ask it here

[teal escarp]

Kinda encounter like this last year, let me just get my notes so lets see if i can help with that one

[drifting parrot]

ok thank you, i would really appreciate

[teal escarp]

if someone helps you rn, that would be great

[drifting parrot]

i'm tryna find a video

[toxic apex]

aren’t opposite angles equal or is this the wrong type of thing

[drifting parrot]

idk

[teal escarp]

Oh you can search up on youtube

[burnt arch]

Oh wait

[toxic apex]

if i’m not wrong it’s x+3+3x=171 if they’re equal

[teal escarp]

if you know "the organic chemistry tutor" channel is, you can find a video you need there

[burnt arch]

What type of angle does 171° and (x+3)° and (3x)° make?

[drifting parrot]

umm

[burnt arch]

In other words, what type of angle does PUQ and TUR make

[drifting parrot]

171

[burnt arch]

What kind of angle does it make though? You're on the right track

[drifting parrot]

either obtuse or acute

[teal escarp]

its obtuse

[drifting parrot]

ok

that's what i thought

but started second guessing myself

[teal escarp]

obtuse angles are always greater than 90 degree

[drifting parrot]

oh ok

[burnt arch]

Mai

[drifting parrot]

so it doesnt matter which direction they are facing

[burnt arch]

It makes a vertical angle

[toxic apex]

but like what are the two angles to each other i think is what he’s asking

yeah

[teal escarp]

it does

[burnt arch]

Guys I got this 😭

I've already studied geometry

So webdev, can we agree that the opposite sides of a vertical angle have the same angle?

[drifting parrot]

yes

[burnt arch]

A vertical angle being one that essentially makes an X

[drifting parrot]

so x = 171?

[burnt arch]

So we know that PUQ=171, what can we say about TUS and SUR?

[burnt arch]

Just follow along

[drifting parrot]

ok

[drifting parrot]

i'm trying to figure it out

[burnt arch]

Are TUS and SUR different angles or the same angle?

[drifting parrot]

different

[burnt arch]

But when you combine them, what is the name of that angle? (Hint: It's similar to PUQ)

[drifting parrot]

oh tur

[burnt arch]

What can you tell me about the angles PUQ and TUR?

[drifting parrot]

that they are the same

[burnt arch]

Correct, so what equation can we right that combines TUS and SUR to equal PUQ?

[drifting parrot]

add them

[burnt arch]

Well, write the equation then if you can

[drifting parrot]

wait how am i supposed to get the separate values for TUS and SUR

like yeah adding will get it to PUQ

but since they are different angles

I can't just do PUQ/2

[burnt arch]

Yeah I see, adding TUS and SUR gives us TUR which equals PUQ right?

[drifting parrot]

ywh

yeah*

[burnt arch]

Getting the exact values for TUS and SUR isn't important, we are just trying to solve for x. So try to write an equation for that.

I'll give you a slight hint, TUS+SUR=PUQ

[drifting parrot]

this part is obvious

but ok

so the next part of the equation

[burnt arch]

Okay let's break it down

Look at your picture

What is TUS?

[drifting parrot]

x+3

[burnt arch]

So TUS+SUR=PUQ, and since TUS=x+3, then we can just say x+3+SUR=PUQ right?

[drifting parrot]

Yeah

[burnt arch]

Do you see where I'm going with this?

[drifting parrot]

oh

turning this into algebra

[burnt arch]

Yeah, any time we're solving for a variable it's gonna be algebra (all of geometry does this)

[drifting parrot]

and then SUR=3x

[burnt arch]

Mhm

[drifting parrot]

so

it would be

x+3+3x=171

[burnt arch]

Correct

Now simply solve for x

[drifting parrot]

when we simplify it, it is:3+4x=171. Correct?

[burnt arch]

Right

[drifting parrot]

oh shoot

i'm not well versed in two step equations

sorry

[burnt arch]

Okay that's fine I'll walk you through this too

171 and 3 are both constants right? They're known values that are by themselves?

[drifting parrot]

yeah

[burnt arch]

Any way you can move 3 to the other side of the equals sign without making the equation incorrect?

[drifting parrot]

yeah

[burnt arch]

What do we do?

[drifting parrot]

171+3

since 3 is a positive value

[burnt arch]

Not quite, in order to move 3 over, we need to get rid of it on our original side while still moving it

[drifting parrot]

i already imagined that in my head lol

[burnt arch]

How would we get rid of a positive 3?

[drifting parrot]

forgot to mention it

subtract it by 3

oh wait

171-3

[burnt arch]

Correct

What's 171-3?

[drifting parrot]

168

[burnt arch]

So now we're left with 4x=168 right?

[drifting parrot]

do we then divide 4x by 4 and 168 by 4?

[burnt arch]

That's right

[drifting parrot]

x=42

[burnt arch]

That's it, you got it

[drifting parrot]

thank you so much for your time

and helping me

[burnt arch]

Mhm ofc, make sure to spend some time practicing algebra, it's a lot more useful than you'd thing

[drifting parrot]

ok

[burnt arch]

think*

[toxic apex]

i wanted to skip algebra I and go to honors geometry but they didn’t let me and i’m realizing that algebra is important to do things like that

[burnt arch]

You would have been so unimaginably fucked

[toxic apex]

still want to take ap classes though and i’m probably going to have to take 2 years of math next year

[burnt arch]

algebra is literally required to solve for a missing variable (you do this in literally every conceivable place in math)

[toxic apex]

bc i think alg II is required for ap calc so i need to take geometry and alg 2 if i want ap calc the year after

😔

and then ap phys req is getting raised to ap calc too

[drifting parrot]

man ixl assessements be hard sometimes

[toxic apex]

i hate ixl

[drifting parrot]

same

[toxic apex]

never make me use that again

[drifting parrot]

i have to

[toxic apex]

i got stuck on a limits by factoring khan academy lesson for like the longest time until i learned factoring in algebra 💀

[drifting parrot]

wth is factoring

[toxic apex]

and now i’m re stuck because idk what a conjugate is

[teal escarp]

"limits by factoring" sounds exciting for me to study even im still on 3rd year highschool

[toxic apex]

turning 4 + 4 into 2(2+2) is the simplest way to put it lol

[drifting parrot]

why tf is the formula for triangular prism so long

[drifting parrot]

oh tetration, pentation, and hexation?

[toxic apex]

what

no

i take out the greatest number i can multiply by all numbers in an equation to get back to the original

[drifting parrot]

oh wait simplifying it

[toxic apex]

that is calculus it is not exciting

[drifting parrot]

i'm quite literally gonna have to burn the triangular prism formula for surface area in my head

[burnt arch]

im taking my sweet ass time studying precalc cuz i am NOT doing calc w/o being ready

also lets move to #math-discussion pleaseeeee

[smoky jetty]

you in 10th grade? We're doing precalc as well

[bitter panther]

Hello i have a geometry problem i need urgent help

[burnt arch]

11th, im self studying (please go to #math-discussion please this is for geo and trig)

[smoky jetty]

sucks doing it knowing I've got a lot of gaps in algebra (hardly making any time to study 'em back)

[bitter panther]

Let ABC a non-isoscelis triangle with orthocenter H and the feet of altitudes A1, B1, C1. Let A2, B2, C2 be the projections of H on B1C1, C1A1, A1B1. a) Show that the circumscribed circles of HA1A2, HB1B2, HC1C2 have 2 points in common. b) Show that the circumscribed circles of HAA2, HBB2, HCC2 have 2 points in common.

I know i should make an inversion with center H for the circle (A2B2C2) right?

Or an inversion with center H for the incircle of A1B1C1

Please help

I have 30 min left for this problem and i am suck at geometry

[upper karma]

what does this mean

40°30'?

<@&286206848099549185>

[thick fable]

!15m

[lime crown]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[thick fable]

40 degree 30 minutes

[upper karma]

minutes?

[thick fable]

yes 1 degree = 60 minutes

[upper karma]

ok

[upper karma]

Degrees' minutes' seconds'

[full pivot]

the side of 7 is hyp?

[thick fable]

yes

[smoky jetty]

do u guys think is a 3d problem (involving volumes) or 2d?

coz idk how i would visualize the problem between the two

[nocturne remnant]

Amongus on the right

But does it make any difference? The heights are not given, so you can assume the height of the box can fit in the ellipse

[forest ocean]

guys there is a theo rem that states that tangents drawn from a point are equal
but,in this case it is not can someone please tell me why

[thick fable]

what do you mean in this case it is not

you can clearly see thier lengths are equal

[forest ocean]

no wait 1 min

[thick fable]

tangents drawn from a point to a circle are equal

[thick fable]

and PQRS are those points

[forest ocean]

look

[thick fable]

what?

oh are you thinking that PQ and RS must be equal?

[forest ocean]

yea

no

[thick fable]

no they are not

that is not what the theorem states

[forest ocean]

pq and qr should be

=

[thick fable]

according to theorem PA=PA, AQ=QB, BR=CR, DS=SC

[thick fable]

no

read the theorem again

[forest ocean]

ohhhhh so from a point p to the point of contact is the tangent huhu

i see

thanks

[smoky jetty]

exactly, the heights are not given. My classmate said there was some volume of the shapes involved

[nocturne remnant]

your classmate is trolling or sth

[smoky jetty]

acc to them, our teacher made them assume that since height wasnt given, height=width

tho Idk how that would help anyway

but the thing is, what steps can I do to find if it fits? clues would be fine

I tried creating a rectangle on a graph inside the ellipse, but idk how the edges of the rectangle would not be outside of it

like I dont know if the edges would actually go beyond the points on the ellipse unless I really plug the ellipse in desmos to get an accurate graph (which im not expected to do)

[upper karma]

AB = sqrt [(6+2)^2+5-1)^2] = sqrt(80)
BC = sqrt[(4-6)^2+(k-5)^2] = sqrt(4+(k-5^2)
CA = sqrt[(-2-4)^2+(1-k)^2] = sqrt(36+(1-k)^2
CA^2=AB^2+BC^2 by replacing we find k =9

[thick fable]

@upper karma i could be wrong tho but i found something on google related to volume of ellipse so maybe its related to volume

oh mb sorry pinged the wrong user

sorry

@smoky jetty

[upper karma]

yes

[smoky jetty]

mind to walk me through it?

[thick fable]

ok but im not sure tho

[smoky jetty]

it's fine, idk either so maybe I could get more ideas about it

[thick fable]

this is what google says

[smoky jetty]

yeah I saw that earlier

but what made me curious is how that's related to the "hole" the problem is refering to

[thick fable]

i mean the hole would have some volume

[smoky jetty]

how could that look like

i saw this one earlier as well

[thick fable]

idk its like you bury a rugby in the ground and take it out

oh

that would not be cylindtical

somethinng like

,w ellipsoid

[somber coyote]

Results provided by WolframAlpha

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

Requested by yajatjamal

[smoky jetty]

but like it was said to have been dug😭

[thick fable]

yea imgaine this inside the ground

[smoky jetty]

do u like dig sometihing like that

[thick fable]

idk if we can but its a question

[smoky jetty]

could it be only half of that ellipsoid's volume tho

so it looks like an elliptical hole on a ground

[thick fable]

also this question is pretty intresting as 3D ellipse would have a smaller cross section in the starting so at the time of putting that thing in this, it can be stuck too

just posting the question again so i dont have to scrol all the way

basically the length of the major axis is 0.25 and minor axis is 0.14

[smoky jetty]

yeah

im thinking it's either an elliptical cylinder or a 2d ellipse

the correct answer says it fits, not sure "how" tho

[grave pond]

The easiest way to work it out is probably to stretch the whole situation by a factor of 25/14 in the direction of the minor axis. Then we have a circular hole with a diameter of 25 cm, and a box measuring 18 cm by (8·25/14) cm.

(Oh, or better yet, scale by a factor of 25 in one direction and 14 in the other, so everything stays integers and the diameter of the hole becomes 350).

[smoky jetty]

what would the equation of ellipse be?

[grave pond]

In which coordinates -- before or after scaling?

[smoky jetty]

after scaling

[grave pond]

x² + y² = 175².

[smoky jetty]

when I graphed the ellipse, I assumed that center is at origin so initially it was: (x^2/12.5^2)+(y^2/7^2)=1

[smoky jetty]

oh so after the scale, the denominators cancel out?

[grave pond]

I'd write the equation of the unscaled ellipse as

(x/25)² + (y/14)² = 1/4
which is equivalent to yours. Multiplying that by (25·14)² gives
(14x)² + (25y)² = (25·14/2)² = 175²
The scaling then corresponds to switching to a new coordinate system (x',y') where x' = 14x and y' = 25y.

The dimensions of the box in new coordinates become

w' = 14w = 14·18 = 252
h' = 24h = 25·8 = 200

[smoky jetty]

why does RHS = 1/4?

[grave pond]

Because I divided the x and y coordinates by the length of the entire minor and minor axes, so the LHS at the points of the end of those axes is (½)² + 0².

But if you're working with equations instead of just winging it like I did, you don't really need the scaling.
You can just plug x=9, y=4 into the equation and see that you get less than the constant on the RHS, so the corner point of a centered box is inside the ellipse.

[thick fable]

wow this is much more advanced that i thought it to be

[smoky jetty]

yeah same lmao

but if my classmates got it right and Im very sure they did not get into that process, I wonder how tf did they justify that it fits

[grave pond]

Sorry for muddying it with my scaling idea. That's a detour if you do it by the book with explicit equations.

[smoky jetty]

apparently, this was their justification sent too early

[grave pond]

🤦

[thick fable]

🤦‍♀️

[smoky jetty]

oh wait

they have more to share, one sec lmao

their solution had something to do with volume of rectangular prism and volume of ellipse

[smoky jetty]

acc to our teacher, if the length, width, and height of the ellipse measure greater compared tot he rectangular prism, then certainly it will fit

and they said that since height wasn't given, we can assume height=width

[smoky jetty]

mb for the misinformation (got confused with responses earlier), so there are no correct solutions so far in our class

[grave pond]

That would mean a square with side length 100 should fit into an an ellipse with minor and major axis 101 and 102.

[smoky jetty]

a friend tried to instead compare the volume of an elliptical cylinder to the volume of rectangular prism, (if V_Rec.Prism < V_Ell.Cyl. then it fits) based on this statement from our teacher

[smoky jetty]

would u say that the statement the teacher said was correct?

[smoky jetty]

that was my intuition as well

thats why I tried to find specific coordinates of points on the ellipse to see if a rectangle would fit

[grave pond]

No, it sounds tragically wrong. So either your teacher is incompetent, or you misunderstood them.

[smoky jetty]

dang I wish to talk to them about it tomorrow (if i got the confidence eek)

uhh, the only thing for sure was that it does fit

but yeah, no one got a correct solution tho

I got this idea ealier

so certainly the width of the rectangle would fit, and if the length of it is less than the maximum length of line from (x,4) to (x1,4), then the "box" fits inside the ellipse

but Im having a hard time finding the x coordinates of the two points

unless i refer to desmos XD

[smoky jetty]

basically, our teacher told them that if the dimensions of the ellipse > the dimensions of the box then the box fits. So my classmates were tasked to compare the volumes (or area in 2d?) of the ellipse and box, if V_box < V_ellipse, the box fits. Now, I (just) think that maybe she based these assumption on the fact that a box can be reshaped upon being fit inside to fit.

but i think it would be more of a physics problem, or idk

[lusty carbon]

Can someone explain how it went from 7pi/10 to 21pi over 30?

it's just that one step, I'll figure out the rest

[thick fable]

that is simple making the denominator common

[lusty carbon]

oh

lmao

lcd

[thick fable]

yes

[lusty carbon]

ok thank you

[bleak agate]

How do I find x

[frozen ocean]

using cosine and sine laws

[bleak agate]

Ohh

So would I jus use cosine rule

Or sine rule

[grave pond]

Start by finding x+1.6

That's just basic trig.

[smoky jetty]

regarding this problem, could we assume that it fits because the box (in 3d) can be fitted in certain orientations? Or we really just have to determine that no ends of the box goes outside of the hole

[slender hemlock]

can someone help me with Volume of cones & pyramids

[obtuse field]

Eclipse learns trig

[limber kraken]

fun fact, sin^2(x) - sin^4(x) = cos^2(x) - cos^4(X)

[surreal escarp]

I’m so lost on how to do this

[thick fable]

@surreal escarp is this the full question?

[surreal escarp]

pretty much

the matern furs and silver grivna is confusing me which is screwing the whole thing over

[thick fable]

its just written in a very weird way

[surreal escarp]

I think i t's asking what r all he things you can get with 100 girvnas??? maybe>?? but still idk how to solve it

[upper karma]

i would use Pythagoras theorem

first find the hyp

nvm i messed up

so if the angle of depression is 19 what would be the angle of elevation

then you could create an equation

$Tan(19)=1.4/x+1.6$

[somber coyote]

Mr. Macro

[upper karma]

if its not a right angle triagle then use the cosine law

[dark sparrow]

$\tan(19^{\circ}) = \frac{1.4}{x + 1.6}$

[somber coyote]

Ann

[dark sparrow]

if you're going to latex, then at the bare bare minimum you have to type fractions correctly.

otherwise nobody can tell $\frac{1.4}{x} + 1.6$ from $\frac{1.4}{x+1.6}$.

[somber coyote]

Ann

[upper karma]

idk how to use the bot (._.)

[dark sparrow]

#latex-help

also to tell you right now:

fractions are done with \frac{}{}

as you saw just now, three times

[upper karma]

yes

thx

wait

how do i get a multiply symbol

[nocturne remnant]

$\times$

[somber coyote]

elon mask

[upper karma]

Oh

[bleak agate]

Ohh okay

[bleak agate]

anyone know how I can solve this?

[thick fable]

@bleak agate what is ur question

[bleak agate]

nvm solved it

[lament tree]

anyone knows somewhere I can practice trigonometric identities?

[reef plank]

Probably Khan

[nova portal]

Can I get help with this please

[thick fable]

@nova portal use pythagoras theorem

[kind cairn]

Does anyone know where I can find the Three dimensional version of miquel's theorem?

[nova portal]

Thank you I figured it out

What would I use for this tho?

[thick fable]

@nova portal are u familiar with similarity rules?

[nova portal]

Nope I don’t remember unfortunately, maybe an example can tell me

[boreal juniper]

A point lies on a terminal arm at (-2,-7). Find the principal angle in radians.

anyone know how to solve this

I have the 3 trig ratios

but when i take arctan to find the angle in degrees. I get like 1 degree which isnt really possible for this

[cosmic dew]

i used arctan 8/6 to find the measure of angle I, then you can use tan(arctan(8/6)) = z/3. Multiply by 3 and you get 3tan(arctan(8/6)) which is || 4||

[upper karma]

is this right? have to find AK

[foggy parcel]

What's tg?
Do u mean tangent?

@upper karma

[upper karma]

tan

[granite harness]

Yes, it's correct

[upper karma]

but can u use sin cos tan for triangles without 90 angle?

[foggy parcel]

Unfortunately not

[upper karma]

then what am i supposed to do

[foggy parcel]

AKB must be a right angle triangle

But its not

So let's just make one?

Draw a perpendicular from K to AB

Suppose it intersects AB at M

Then KM = CB

[granite harness]

Oh, really?
It must be 90 degrees?

[foggy parcel]

Yes

[upper karma]

what now?

[foggy parcel]

Those ratios are meant to be in right angle triangle

[foggy parcel]

KM = BC do u agree?

[upper karma]

well yeah

[foggy parcel]

Okay now apply sine ratio in triangle AKM

Sin60 = KM / AK

[granite harness]

Yes, you are right.
I just forgot that propriety.

[upper karma]

KM= 4√3?

[foggy parcel]

Yes

KCBM forms a ractangle
So basically KM = BC = 4√3

[upper karma]

what do i do next?

its giving me -22,7

[foggy parcel]

Dont break down the root too quickly

Sin60 = √3/2

So 4√3/(√3/2)
=> (4√3 × 2)/√3
=> 8

[upper karma]

oh

[upper karma]

by basic trigonometry

we can construct an equation'

[bleak agate]

Ah

Thanks, solved it though

[upper karma]

great!!

[bleak agate]

appreciate for trying to help tho

[upper karma]

np

[shut oxide]

Please help me finish it idk what I’m missing

[thick fable]

@shut oxide you're done with all three sides, you used 2 angle to prove it congruent, what do you think you're left with

[shut oxide]

I have no clue I literally selected all the reasons why segments are parallel but it’s still saying incorrect

[thick fable]

||you're missing an angle||

[shut oxide]

Bruhhh I’m stupid thanks I didn’t even notice

Thank u I finished it lol

[thick fable]

np

[leaden radish]

what do those triangles mean on the line

[grave pond]

Do you mean the arrowheads in the middle of HE and ID? They usually symbolize "these lines are parallel".

[leaden radish]

thank you

im pretty sure theres a formula for this but i cant remember or find one

[grave pond]

Consider an ant walking around the edge of the polygon. At each corner the direction the ant walks in changes by 180° - 156°. When it's gotten all the way around, the total amount it has turned by must be 360°.

[thick fable]

also i think you're talking about $\frac{(n-2)\times 180^{\circ}}{n}$

[leaden radish]

and then it would be 360/24

[thick fable]

this formula?

ah not again

[somber coyote]

🐱!Yajat! 【Catfan1398】🐱

[thick fable]

yea this one

[leaden radish]

probably, let me write it down

[upper oracle]

I made this up, is this possible? I came up with steps to find them all as I was making them. I’m failing geometry so I can’t guarantee that this makes sense at all lol

Angle A: find the angle FGE, and you get the vertical angle to that. B is a right angle that’s given, alt interior angles are congruent so you’ve got the part of angle D in the bottom right area, since squared add up to 360° u can do the rest from there to find <A

For angle C, B is right angle again, vertical angles are congruent so you have the part of angle G that’s in the bottom right area. To find the part of <E that’s in the bottom right area, use the given part of angle g to find the consecutive interior angle, then use that to find the supplementary angle. From there, since the angles add up to 360°, u know the rest

It’s given that FGD is isosceles, so you already have the first half of angle F. To find the other half, use the 2 angles from the top right area from earlier to find the other half, then add the 2 halves together to get angle F

[upper oracle]

If we ignore that line ED is useless as I just realized, would this actually work?

[upper karma]

whats the question, i dont like reading

what are we finding?

[upper oracle]

Find angles A, C, and F

[upper karma]

oh

[upper oracle]

That long wall of text is the solving method I came up with, but then I realized that one of the lines made that entire method completely useless

[upper karma]

ok so for this we use the cosine rule given the missing ones

are their measurements?

[upper oracle]

I haven’t learned trig yet, so I basically just went off if everything up to that

[upper karma]

what are the side lengths?

[upper oracle]

You aren’t given any

[upper karma]

💀

now i have to use my brain

[upper oracle]

I have no idea if this would actually work or not

[upper karma]

well we know opposite angles are equal right

[upper oracle]

Yeah

[upper karma]

so what makes angle g

[upper oracle]

The vertical angle to the isosceles triangle is 123°, also forgot to mention that I didn’t bother to calculate the angles of anything other than the vertical angle to the 123° and B

[upper karma]

wait i dont think its even possiable

is angle $90\circle%?

oops

is angle $90\circle$?

[somber coyote]

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

[upper oracle]

Which one?

[upper karma]

angle b

[upper oracle]

Given info says FB | ACF

so yeah it is

[upper karma]

🤦 this question is hurting my brain

[upper oracle]

This all basically says use ur vertical, supplementary, and consecutive interior angles a lot, and like 1 alt interior

[upper karma]

._.

[upper karma]

<@&286206848099549185> help, my brain hurts

[burnt arch]

bruhington that ping is for the help channel

!help

[lime crown]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[upper karma]

🤨

ive seen people use it in advanced

[burnt arch]

thats crazy

!help

[lime crown]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[upper karma]

ik right

[burnt arch]

if youre gonna ping helpers

[upper oracle]

Should I just paste all this in a help channel

[upper karma]

yes

[burnt arch]

yuh

[upper oracle]

Done

[upper karma]

i wanna see if someone can solve it, an i learn 😁

[upper oracle]

K I just realized that instead of all this, u can just find the consecutive interior of the vertical angle

[smoky jetty]

Given that the only property of the triangle is that FGD is isosceles and FB is perpendicular to ABC ( and not ACF?), this is all I could find

[junior flame]

what are we looking for here

[smoky jetty]

checked if finding the said angles was ppssible

[upper oracle]

To find <C and <F, find the consecutive interior angle on either the 57° or the 123° on the right half of ACF, use that to find the supplementary angle, then you have both halves of <E. From there just use all the degrees to find the missing degree like normal

Nvm the angles don’t even add up properly lol

[upper oracle]

c and f are impossible :(

At least with this method

[smoky jetty]

hmm, u may have confused the property of parallel lines cut by a transversal with vertical angles

[bright sundial]

doesnt this mean triangles ABC = BCF ?

which solves the entire problem

[smoky jetty]

abc?

afaik ABC is a line segment

[bright sundial]

oh sorry ABF = BCF

[smoky jetty]

hmm, Im not sure if those given properties are sufficient to say triangle ABF = triangle BCF.

[bright sundial]

does line BF spilt triangle ACF exactly through the middle?

[smoky jetty]

it didnt say so

[upper oracle]

I think the problem is that I’m trying to solve squares with a theorem meant for triangles

[smoky jetty]

interesting

[upper oracle]

It does, I’m just bad at drawing lol

[bright sundial]

Well if the line is perpendicular to A C and cuts through points B and F, then this means AB = BC

which then means AF = CF which means triangle ABF = BCF

which then means corner A = Corner C

[smoky jetty]

hmm, not necessarily

[bright sundial]

And corner F¹ = corner F²

[smoky jetty]

i mean if we'd really just base from the diagram itself, then ig we could make such assumptions

[bright sundial]

yeah true

[smoky jetty]

how come

[bright sundial]

nono i was agreeing

but i used a negative where i should have used a positive

[smoky jetty]

if it also stated that AF= FC i.e the biggest triangle formed is an isosceles, then we could say that line BF bisects angle F and the segment ABC

which makes your earlier statements true afaik

[bright sundial]

ahh okay

So A = C which is 61.5° and F = 28.5+28.5 = 57°

which adds up to

[smoky jetty]

ok durp says it does

[bright sundial]

61.5+61.5 + 57= 123+57 = 180

[smoky jetty]

so ig u should be right, Deem

[bright sundial]

so its right

@upper oracle solved it

[smoky jetty]

welp he made the problem, lol

[bright sundial]

yes, it wasnt really clear, even a small 0.1 degree difference could throw it off

[bright sundial]

yes i know, im wondering if im correct

[smoky jetty]

oh I thought u meant durp solved it

lmao mb (tried to insert a joke)

[bright sundial]

no worries, communication on discord is oftenly confusing

@smoky jetty you like space?

[smoky jetty]

definitely

[upper oracle]

That’s basically the entire triangle, I thought ED would complicate things, but it was completely useless and made it impossible

[bright sundial]

how interesting ? like do you want to study it?

[bright sundial]

wait how is it impossible?

[bright sundial]

calculate A C and F right?

[smoky jetty]

but hey, that's just some simple angle-chasing. Maybe other people much better in geometry can actually find those angles

[upper oracle]

The top right triangle angles add up to 180

For the square 360

Not enough room for another angle

[upper oracle]

Yeah

I probably incorrectly used consecutive interior angles

[bright sundial]

How can you say that E¹ and E² are 123° and 57°m

?

[smoky jetty]

yeah, but idk if i could actually study it due to several factors (off topic)

[upper oracle]

I tried using consecutive interior angles on both of the other angles

[bright sundial]

ahh okok i see, im going to study astronomy at a university in 2 years probably

[bright sundial]

but why would you try that? it doesnt really work here

in this situation

it's certainly not impossible

[upper oracle]

I thought it would for some reason

[bright sundial]

E¹ = 180- 57-28.5

[smoky jetty]

u sure you're not confusing that with this?

[bright sundial]

and E² = 180 - E¹

then you have correct angles for square BCEG

[upper oracle]

6 and 4 are CI, I also used supplementary angles to find them

[smoky jetty]

I wish u goodluck!!

[bright sundial]

thank you, its marked as a difficult study for a university so i need all the luck i can haha

[upper oracle]

This screenshot it just me adding on to someone else’s markings, I never actually calculated the angles even while making the problem

[smoky jetty]

note that in your diagram, lines BF and CF are not parallel

[bright sundial]

exactly

thats why im confused

[smoky jetty]

so angle FEG is not necessarily = 123

[bright sundial]

its not

it cant be

[upper oracle]

I forgot they had to be parallel

[bright sundial]

you can see that from the drawing

[smoky jetty]

make sure to get those maths geared up, you'll need 'em a lot XD

[bright sundial]

now i see what the problem is

[bright sundial]

yeah i know, thats why i am here haha

[smoky jetty]

if you're referring to properties of parallel lines cut by a line to find angle FEG, then yeah, FB must be parallel to CF

[bright sundial]

@upper oracle your assignment is very solveable, actually its a good practice for people who are just now starting geometry

but maybe not what you intended to make?

[upper oracle]

Main point of it is to use angles to find other angles to find the main 3 angles, but it didn’t work due to bad design

And me forgetting how stuff works

[bright sundial]

hahah doesnt matter, you learned right? Thats actually the most important part to success

[upper oracle]

Yeah

[bright sundial]

i say try to make another one ( if you want to)

[upper oracle]

What would you do for this? Here’s as far as I close get

Ik that the angles on EAB are wrong, I just didn’t fix them

Fixed angles

[bright sundial]

every angle in square ABCD is 90 so the angles from corner ABF and BCF are in total 90

[smoky jetty]

ohh this one again

[bright sundial]

you know that corner B is entirely 158°

[upper oracle]

I found it on a Reddit post and decided to go at it

[smoky jetty]

this problem honestly hasnt gotten off my mind

i tried angle chasing, but to no avail after 30 mins a few days ago

[bright sundial]

corner B vant be 158°

cant*

Every corner in square ABCD is 90

B is 124°

in total

[upper oracle]

Yeah, because I fixed one angle but not the other

[upper oracle]

Yeah

[bright sundial]

ohh

well then its easy

134-90

[smoky jetty]

your total interior angles for triangle ABE is incorrect

[bright sundial]

-24

20°

54°

[upper oracle]

112 + 34 (2) = 180

[smoky jetty]

yep

[upper oracle]

Then was that a typo

[upper karma]

Kinda don’t understand what you do to figure out x with this circle theorem any help would be appreciated

(The middle one)

[thick fable]

@upper karma does that x here represents the side length?

if yes, then diameter subtends 90 deg angles on the circumference of the circle, so u can use pythagoras to find x

[upper karma]

Yes it does

[upper karma]

Ok Tysm

I am sorry for being stupid but this is the last questions I will ask for help on but I have genuinely no idea how to solve them since I missed a lesson

[thick fable]

rules of similarity

[vague axle]

Sup

Guys how Can i find the equation of an hyperbola in the tridimensional Space?

[rocky marsh]

ik this question may seem stupid but i've been trying to solve it n i've gotten so many answers so i'd rly appreciate some help, tyyy

[upper karma]

remember opposite angles are equal

[rocky marsh]

i know, but i still don't understand how the answer can be anything besides 120

[upper karma]

thus might help

[rocky marsh]

is this correct?

[upper karma]

👍

[rocky marsh]

tysm i've been struggling on this question forever

[upper karma]

np

[narrow crypt]

Hello so I was trying to help out and solve the same question as the last person (we’re friends) and I’m just wondering: wouldn’t it be x=152 degrees? I asked someone else and they got that as their answer and when they explained it, it made a lot of sense. This was their work.

They explained that by cutting the 88 in half and getting 44 you’re assuming the the way the 88 angle is rotated that when you bisect it you get it perfectly horizontal as to make the right angle

Sorry if that’s explained weird, I’m not good with terminology

[smoky jetty]

seems about right

[upper karma]

ah ha

makes sense when you approch it from a different angle

[fickle rose]

what the

ur answer is right but that’s a really weird thought process and i don’t get how that helps to solve

[waxen orchid]

I need help with something

[heady scaffold]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[waxen orchid]

do you know the answer

[heady scaffold]

well

[thick fable]

no

we dont tell answers

!status

[lime crown]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[heady scaffold]

ik

it looks like 3 to me

or 4

[waxen orchid]

1

[heady scaffold]

oh ok

do you know what these axioms mean?

[waxen orchid]

no

[heady scaffold]

so you dont know SSS, SAS, etc.?

[waxen orchid]

oh I know

but I’m not sure which one it is

[heady scaffold]

ok so

we do know that two sides are congruent right?

[waxen orchid]

ya

[heady scaffold]

and they share a third side, right?

[waxen orchid]

Right

[heady scaffold]

so

how is x related to x?

let's assume that side is x units

[waxen orchid]

Ok

[heady scaffold]

so

x = x, right?

[waxen orchid]

ya

[heady scaffold]

Therefore, the third side is also congruent

therefore, SSS

[waxen orchid]

Ty

[heady scaffold]

yw

[waxen orchid]

what about this

[heady scaffold]

so

[waxen orchid]

yes

[heady scaffold]

do you know VAT?

vertical angle theorem

[waxen orchid]

No

[heady scaffold]

basically vertical angles are congruent

that's all it says

[waxen orchid]

oh

[heady scaffold]

so

do you see any vertical angles here?

[waxen orchid]

2

[heady scaffold]

yes but that's 1 pair

right?

[waxen orchid]

oh so 1

ya

[heady scaffold]

so those angles are congruent

therefore

[waxen orchid]

Ya

[heady scaffold]

we can see this is congruent by SAS

[waxen orchid]

Ty again

[heady scaffold]

yw

🙂

[waxen orchid]

Last one

[heady scaffold]

. because i feel like no one will find this

[waxen orchid]

what about this

[heady scaffold]

are all three sides congruent?

[waxen orchid]

Yes

So it’s SSS?

@heady scaffold ?

[upper karma]

How do you know the 3rd side is congruent?

@broken cove dont directly give the answer

Help him getting to the right answer instead of just spoiling it

[broken cove]

My bad

[heady scaffold]

are they tho?

are the sides congruent?

[smoky jetty]

a pretty interesting problem, but i couldn't solve it after 30 mins of digging

[thick fable]

haha same, i think its missing something, i tried doing it all way, its just end up with 0=0 always🙄

[smoky jetty]

tried trigo bashing, area proportions, special triangles, and I got none lmao

[heady scaffold]

apparently

i looked at the solution

im supposed to use pythagorean

and

so

ya

idk

its fine tho

its just comp math

[thick fable]

pythagoras where i wanna know?

[heady scaffold]

uhh

well we're supposed to assume that r is radius of the unknown circle

and then

14^2 + r^2 = something i forgor :p

it's fine

just competition math like i said

[rocky marsh]

i honestly have no idea what rhe correct answer is because peoples answered ranged from 122, 134, 143, and 152. i'll have to wait and see and i'll ask him tmrw when i get to class, but ty i alr turned it in 😭

peop,e said they got 143 and got it right sooo

no idea

[true mauve]

forgor 😭 Kanye who

[heady scaffold]

https://knowyourmeme.com/photos/2146119-i-forgor-💀

[smoky jetty]

mind to share the source? just curious

im thinking it's 16 ft, but it says otherwise. who's wrong lmao

also, im particularly very interested to know the answer in the 2nd question. I'll appreciate yall's attempt at it

[heady scaffold]

https://artofproblemsolving.com/mathcounts_trainer/play

state division

[heady scaffold]

well i guess we can start off with establishing formulas

(x-h)^2/a^2 + (y-k)^2/b^2 = 1

if i remember correctly

and we want to find the point where the curve is 1.2 feet away from (h,k)

hmm

i actually dont know

[smoky jetty]

thats ight

[upper karma]

how can i prove those two?

[silent birch]

tanπ+tantheta/1-tanπtantheta

same for the π/2

[upper karma]

isnt this tan(π+theta)

[silent birch]

that's the formula for tan

[slate hearth]

Can any 1 say what is the Sum of first n AP

[upper karma]

im confused

[silent birch]

Sn=a/2(2a +(n-1) d)

[slate hearth]

ty

[silent birch]

I used the 6th formula

tan(A+B)

tanπ= 0

[upper karma]

ah ha

swell tnx

[upper karma]

this does the trick

[upper karma]

The tangent function is periodic, with its period being pi

so for all theta, the first one is true

[upper karma]

just use some geometry

[heady scaffold]

or Sn = n(a1 + an)/2

i like that one better

[slate hearth]

Ok thanks

[violet rose]

very dumb question

but like -pi is at wh at position??

[thick fable]

not sure what u r trying to ask, but anyways if your going the angle way its 180 degree the clockwise direction

[violet rose]

i thought so but thrn why is this inequality different if it is on [0;2π] than if it is on ]π;-π]?? sorry i am dumbass

[thick fable]

0 to 2 pi interval means you will take your answers that fall under this interval and same for pi to -pi

tho both interval have same length of 2pi

we dont provide such facilities nor do we appreciate such msgs sorry

[safe vapor]

mmk

[violet rose]

the difference is like in the place i start and end thrn only

sorry i have 0 logical thoughts and 0 mathematical knowledge

[thick fable]

no you dont

you can circle this circle (lmao) infinite times

ur only thinking like u can only cover it ones

its like you can cover an anlge of 1000 degrees here too

[violet rose]

i mean like where i start the notatoon of the

like solution

sorry if it doesnt make sense

wait i remembered a fancier way to say it

is the difference the place i start reading thr interval?

@thick fable ?

[thick fable]

i dont get what ur trying to say, please be clear

[violet rose]

is the difference between thode intervals therefore just on the place where you start reading the interval of solution?

[thick fable]

thode intervals?

[violet rose]

*those. why are you so pedantic, are you a matematician or what?

[thick fable]

no im not, english is not my first language also i have no knowledge of schooling systems outside my country, there are many words we dont use here but are said in other countries for similar concepts, i just asked to be clear

also we dont use ] ] these kind of intervals

[violet rose]

sorry english not my first language either snd so that might be why i was unclear in the firsy place and might have been just slightly frustrated about not being understood. sorry for being slightly rude

[heady scaffold]

🙂

[thick fable]

no problem

coming back to ur problem

[heady scaffold]

those mean > or equal to

so

in a domain

[5, infinity]

would be all values between 5 and infinity, inclusive

[thick fable]

bruh chill ik

[heady scaffold]

ok lol

[smoky jetty]

uhh not sure if that geo prob was this

[heady scaffold]

NOOOOO

I WAS ON STATE ROUND

also

it's randomized

ya something tells me thats not the same problem

💀

[thick fable]

@violet rose do you know how to find the principle solutions of any trig function?

[violet rose]

what's pronciple solution

shit

[thick fable]

general solutions i meant

like this

[smoky jetty]

oof

[violet rose]

i dont think we ddi this

[thick fable]

soosh wants to help u i think...

[opal carbon]

@finite cairn I was looking something else up in a book and happened to see a proof along the lines of the identity you were asking about earlier. sent you a link to stack exchange but this one uses only very simple concepts you would definitely know even if it looks a bit long at first, in case you wanna have a look

[thick fable]

ok nvm

okay so tell me at value of x is cos x =-1/2

[finite cairn]

ok will def have a look

ty

[violet rose]

wait what sorry i dont understand

[thick fable]

yea like if sin x = 1/2, u say x = pi/6

similarly im asking for this

[violet rose]

at 2pi/3 and -2pi/3

[thick fable]

,w cos(-2pi/3)

[somber coyote]

Results provided by WolframAlpha

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

Requested by yajatjamal

[thick fable]

yea

that means you have this that x< 2pi/3 and x > -2pi/3

[violet rose]

yeah and how would that make an difference in solution with two identical intervals that are 0 to 2pi and -pi to pi??

[thick fable]

this answer is for -pi to pi

if we had 0 to 2pi

we could have taken 2pi/3 and 4pi/3

[violet rose]

i think i officially give up

and accept a c or d or whatever

[wheat meteor]

what is the side of the orange rhombus if the yellow rhombus has sides 120x, both are regular

could you please help

i meant hexagon

sorry

[hard coral]

Can someone help with finding this F function?

For context, it's finding the angle between the reflection vector and 2pi

For light reflection

[lusty carbon]

can someone explain how it went from 5pi/12+pi/12 to pi/3?

*pi/2

[heady scaffold]

5pi + pi = 6pi

6pi/12 = pi/2

[lusty carbon]

lmao

ok thank you

[heady scaffold]

🙂

[lusty carbon]

what i did was 5pi+pi=6 and also sum the denominator. No wonder why my answer was wrong lol

[heady scaffold]

lol

glad to be of help

[sudden glacier]

Guys how do I find theta in this

im new cus im not taking trig rn im just learning outside of school

[misty tusk]

basically since we're looking for theta we can use inverse trig functions here

so we're given the adjacent and the opposite right

[sudden glacier]

cos-1?

[misty tusk]

if you remember tangent relates the opposite to the adjacent

so if we use tangent we get something that looks a bit like this

now in order to solve this for theta we need to get theta on it's own right

[sudden glacier]

yes

[misty tusk]

so we can take the inverse tangent of both sides
this cancels out on the theta side and gives us an answer on the 3/4 side
it basically looks like this

[sudden glacier]

36.9?

[misty tusk]

yep

[sudden glacier]

ok thanks a lot

[misty tusk]

the inverse trig functions also work in other cases

if you were given the hypotenuse and the opposite you may have used inverse sine instead

same goes for the adjacent and hypotenuse and inverse cosine

[sudden glacier]

you can get hypotenuse with c=a2+b2(in square root) right?

[misty tusk]

yep

[upper karma]

Do you know how to solve equations?

If not then learning it will make things easier for you. Trigonometry and geometry requires solving equations and creating solutions

[vernal pilot]

hey man, if you aren't taking trig in class i assume you are still doing very basic prealg, like exponents. So instead of learning geometry and trig i think if you learnt the basics of prealg and algebra first it will make things so much easier. But if you are intending to continue learning trig here are a few things you should learn:

Pythagorean Theorem and its applications in geometry/trigonomety
SOH CAH TOA - sin=opposite/hypotenuse, cos=adjacent/hypotenuse, tan=opposite/adjacent.
tan=sin/cos
Exact Value Triangles - they are basically just common angles
Radians - a different, and preferred, unit of measuring angles. they are measured in terms of pi

AND please do not get into trigonometric proofs right now, you are not ready to learn trig identities. Anyways, good luck.

[weary haven]

Does anyone know what is the name of the formula $\cos \alpha =\frac{x-x_0}{d}$?

[somber coyote]

Herr Person

[haughty lichen]

solution ??

[heady scaffold]

well is E a midpoint??

[haughty lichen]

no it is not.

[heady scaffold]

ahh

so that makes this harder

[haughty lichen]

do you think something is missing here?

[heady scaffold]

let me think about it

[haughty lichen]

sure take your time

[heady scaffold]

we might be able to solve it using this formula

but it doesnt look like it

because only one vertex is touching the chord

[haughty lichen]

to use this we need to have chord and CD is not a chord so how are we gonna do this?

[heady scaffold]

yes but CD is half of a chord

i think that might be important

[haughty lichen]

hmm ill consider it sure. lets see what others would say.

[heady scaffold]

yk what

if we find OD it will be smooth sailing from there

so let's set OD to a variable y

sqrt(y^2 + 36) = OE

so sqrt(y^2 + 36)^2 + 64 = FO^2

so y^2 + 36 + 64 = FO^2

so y^2 + 100 = FO^2

so

OB^2 is also y^2 + 100

[haughty lichen]

mhmm, i'm following you.

[heady scaffold]

ya im just writing this down on pen and paper

[haughty lichen]

okay, try it further

[heady scaffold]

are the colors just for aesthetic?

or are they congruence?

[haughty lichen]

they have nothing to do with additional info about problem

FO and CD are parallel , would that help ?

[heady scaffold]

alr

well ik that

where did you find this problem?

[haughty lichen]

oh okay , it wasn't given though

[haughty lichen]

my cousin asked me , their teacher gave them at school. and he is in 9th grade

[heady scaffold]

u can tell cuz they're both 90 degree angles

[heady scaffold]

is this everything?

[haughty lichen]

he says so, but if you think there is something missing. I may ask him for enquire to teacher about this.

[heady scaffold]

dont tell him to ask his teacher. i suck at geometry

i specialize in algebra and precalc

[haughty lichen]

we should wait for other. they may have seen or done such prob

Its reposted,, don't scroll up

[hard coral]

Can someone help me get the function F? For context, it's finding the angle between the reflection vector and 2pi. This whole process is for calculating light reflection angles

[smoky jetty]

if FE was extended to the lower right, would it pass through B?

and is ED = OD?

[haughty lichen]

Both aren't given , nor can i prove such claims.

[smoky jetty]

oof

[upper karma]

Hi there I am trying to get into topology

I am looking at the poincare conjecture

if a line cannot be included

as it has two fixed endpoints

by the same token

isn't a circle a line with two points? that has just be joined together

so by the same token the circle is not allowed?

Can i get a clarification

[thick fable]

i think this question belongs in #point-set-topology

[upper karma]

it says i have to be an undergradute course

[thick fable]

got no idea about topology tho

[upper karma]

i've just finished high school lol

[thick fable]

there's no prolem in it

get the undergraduate role

[upper karma]

but i don't have an undergraduate in topology?

[thick fable]

ok ok, but then i dont think anyone might help u here

[upper karma]

gg

[smoky jetty]

i tried to apply geometric mean of the triangles, but to no avail

i honestly think that FE when extended passes through B

but i dont got any proof

[haughty lichen]

Thanks for your work , I'll let you know if any progress would happen ✌️

[grizzled violet]

How does tan(2x) equal 2tan(x)/1-tan^^2(x)

i found this but i dont know what to do with it

[gentle agate]

Can anyone help me to solve this...

[upper karma]

I dont know is it right

Or wrong answer

[smoky jetty]

ight, it's actually easier than I expected. So, if we create a line segment that joins each center of the semi-circles, then we can create a right triangle with base and height equal to the radius of semi circle AQ, and BQ minus the radius of the smaller semicircle respectively. Hence, the pythagorean formula would be (14-r)^2 + 7^2 = (7+r)^2; where r= = radius of smaller semi-circle

got a help from an olympiad server

finally, r = ||14/3||

[heady scaffold]

that was...simple

here's another wun i cant do

not exactly geometry

but...

[smoky jetty]

oof, there goes my fcked up algebra

[heady scaffold]

wait where does 7 + r come from??

how is the hypoteneuse 7 + r??

[smoky jetty]

the dark green tangent line intersects both at one point (point of tangency), so their radii equals to the hypotenuse

[heady scaffold]

ahhh

that makes sense

thx

[smoky jetty]

i forgot mention that it'll only work "if" the the red line passes through their point of tangency, and sure enough the problem states that the semi circles are tangential to each other

[smoky jetty]

np, cant do it without the nerds who helped

[nocturne remnant]

If the semicircles are not tangent, then there is simply not enough info to solve the problem

[smoky jetty]

ooh, thx for that!

[heady scaffold]

nvm solved it

it was surprisingly simple

[smoky jetty]

how about this aforementioned problem earlier, do u think it's solvable?

[heady scaffold]

i tried that one

couldnt solve it]

but yk

im dumb

[smoky jetty]

nah, I doubt

[heady scaffold]

🙂

so close to national......

[nocturne remnant]

Let FE and CD meet the circle again at H and Z again respectively, then FE=EH=8, CE=x, EZ=x+12; by the interesting chords theorem we have
8*8=x(x+12),
So we get x=4

[heady scaffold]

thats interesting

[smoky jetty]

ohh rightt the chords theoremmm

gosh I overlooked that

[smoky jetty]

but how come does H lie on the circle again such that EH = 8?

or that E becomes a midpoint of line FH when FE is extended again to meet the circle at H. i.e how are u sure that the extended line to meet H has length = FE = 8?

[keen osprey]

double angle formula

[twin crag]

how do i solve this

[keen osprey]

you can use the formula for an infinite geometric series, and then the double angle formula for cos(theta)

[heady scaffold]

arithmetic*

[keen osprey]

no, its a geometric series

the common ratio is cos^2(theta) and the first term is 1

[twin crag]

huh

when i use the formula a/1-r

i get 5sec(x) when simplifying

cos^2(x) is the first tierm (a) right

[keen osprey]

no its 1

[twin crag]

oh

crap

i didnt see that

[keen osprey]

its 1/[1-cos^2(theta)]

[twin crag]

so then cosx would be (2sqrt5)/5

right?

then i just plug that in to the double formua

oh i need to find sinx

wha

[keen osprey]

we are trying to find cos(2x), so we dont need to find cos(x). we can use that 1-cos^2(x) = sin^2(x) by the pythagorean theorem.

so 1/sin^2(x) = 5, i.e. sin^2(x) = 1/5. then note that cos(2x) = 1 - 2sin^2(x)

[twin crag]

oh i see

wait what

gimme a second to understand this sorry

ok i see how you got sin^2x=1/5

ah okay i see

ty

[twin crag]

How do I isolate cos(theta)

i tried expanding the left side into the double angle formula and stuff but nothing has worked

[frozen ocean]

!show

[lime crown]

Show your work, and if possible, explain where you are stuck.

[frozen ocean]

,tex .double angle

[somber coyote]

𝔄𝔨𝔦𝔯𝔞 🍇

[twin crag]

sorry i forgot there was more than 1 double angle formula for cosine

showing work in a second

i got this far
and then realized that you can't factor the expression on the left by 2cos(x)
so now im stuck

[frozen ocean]

yup but do you know why?

it's because 2cos(x) is not a common factor of all the terms in the expression

[frozen ocean]

look here

[twin crag]

yep

so now idk how to isolate cosx now

[keen osprey]

so u have the equation 2cos^2(x) - 2cos(x) -1 =0 right?

[twin crag]

yep

[keen osprey]

notice that if we substitute lets say z = cos(x), we have 2z^2-2z-1 = 0, which is just a quadratic

i.e. we just have a quadratic in terms of cos(x)

[twin crag]

so now we just have to solve for z in that quadratic?

or do we just have to isolate z

oh i see wha tto do now

[twin crag]

ok so i got this for z

so that is what cos(x) is?

and i just take arccos of that

[keen osprey]

be careful here, you have gotten an additional solution - this is because cos(x) is only defined between -1 and 1

[1+sqrt(3)]/2 is greater than 1, so we disregard that value of z

so