geometry-and-trigonometry

[hearty meteor]

Spam practice problems from past regents exams

thats what ive been doing for like the last 4 days and its worked kind of

[vernal raptor]

what's the best way to learn geometry

[lofty spear]

as with anything else in math: practice!

[scarlet maple]

I keep having the same issue of not knowing what to use to prove the statement for this column proof, and everything I do put there just seems random

[plain scarab]

Oh ok

My teacher gave me 8/8 on it anyways xD

[harsh crow]

say straight angles supplementary

for 4

then do substitution for both 5 and 6

and algebra for 7 and 8 i guess

[scarlet maple]

algebra?

[austere sage]

How can I find angle BEG?

I got 27 degrees but im pretty sure that is incorrect

[grave pond]

Sounds right.

Angle BED is 72° and angle GED is 45°.

[austere sage]

How did you get 72 degrees for angle BED?

[upper karma]

W

[grave pond]

BE is parallel to CD.

[upper karma]

For 2+4 can't u say something about opposite angles or sum

I remember there was a theorem or something

Where if u had an outside angle u could get the opposite inside angle

For explanation 6 just say substitution

[austere sage]

ohh that makes sense, thanks

[upper karma]

Wow broski nice shape

4 triangles a square and a pentagon

[plain scarab]

yo I got a question

on another proof

am I allowed to put down definition of a straight line

[upper karma]

Uh

Depend

[plain scarab]

as a reason

Lemme send you the final

Once im done

[upper karma]

I know the definiyon of straight angle

If that's what u mean send that problem

[plain scarab]

how do I write this

m angle ZXY?

[upper karma]

Show the whole problem

[upper karma]

And yes

[upper karma]

If u were to convert it into words it would me

Measure angle zxy

[plain scarab]

Oh that’s what m stood for

Holy

I’m dumb

Shit

[upper karma]

Instead of straight line say straight

Angle

Since it adds up to 180

[plain scarab]

ok

this good?

I’ll fix line

[upper karma]

Yep

Very gud

How long have u been doing geo

[plain scarab]

Like a week ago

I’m just taking it over the summer

[upper karma]

Same bro

We r doomed for 2nd semester

[plain scarab]

Is it gonna be hard

I need to start paying attention

[trail tendon]

what about it

it can be used to know the results of certain inputs into trig functions...

[plain scarab]

how would I write a reason for 6

[trail tendon]

it's used for a lot more that i'm not thinking of off the top of my head lol

[plain scarab]

could I use the converse of the consecutive interior angles theorem?

or is there a simpler reason

[maiden brook]

wdym?

[harsh crow]

same side interior converse

[plain scarab]

write that as a reason?

[harsh crow]

yh

i think

[plain scarab]

how would I explain that

sorry I have no clue

[harsh crow]

i dont even know if it works

at the very least you can write same side interior angle supp. and draw an arrow to parallel lines

[upper karma]

Since 1 and 2 r something angles like 2 interior angles I briefly skimmed this but they add up to 180 and have a transverso they must be congruent therefore making the m n parallel

[plain scarab]

Ok

[plain scarab]

is this right?

[stuck sleet]

I cannot understand this phrase:

Find the locus of the centers of all circles passing through
two given points.

[nova herald]

No. Angles 1,3,5,7 are congruent and 2,4,6,8 are congruent

1,3 and 2,4 are vertical angles same with 6,8 and 7,5

Vertical angles are always congruent

[faint pasture]

Well let the centre be h,k
Then you can satisfy the points in the general equation
(X-h)^2+(y-k)^2=r^2

Youll get two equations

Subtract them to eliminate r

[gentle haven]

can someone explain how they did the step as i marked? this is the proof of the butterfly theorem on wikipedia. I don't understand how the intersecting chords theorem is used here.

[upper karma]

i think there are 2 values given for the MX/MY

so they just square mx/my and multiplied those values on the other side

as they are equivalent

[sand patrol]

Can anyone tell what does this means

[nocturne remnant]

i would assume perpendicular

[sand patrol]

K Tnx

[upper karma]

qr is perpendicular to ps

[gentle haven]

How do they get to this then? I get where the equation is coming from but I don't see how they are equivalent.

[plain scarab]

So all 3 angles are 60 degrees?

Or 120

[languid crater]

The equation for linear interpolation is:
n = a + (b-a) * t
And for lines, it is the same for each coordinate:
x = x1 + (x2-x1) * t
y = y1 + (y2-y1) * t

What I am trying to do is isolate for t on both, so I can find t from a point on said line

Just checking my algebra / geo here

[obsidian harness]

You have $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1}$ which is just the slope formula $\frac{x - x_1}{y - y_1} = \frac{x_2 - x_1}{y_2 - y_1}$ if you isolate for $t$

[somber coyote]

southy

[obsidian harness]

and yes your equation is correct, note that t ranges from 0 to 1 though

[languid crater]

yeah

[obsidian harness]

(when t = 0 you get a, and when t = 1 you get b)

[languid crater]

i know how the lerp works

im just isolating for t so I can get how far along the line a point is

without knowing t

[obsidian harness]

ah you can use similar triangles for that

so if I have say (1, 2) and (3, 4), then halfway along the line will have x-coordinate (1 + 3)/2

for example

so if I have a point with an x-coordinate of 2, that's (2 - 1)/(3 - 1)th of the way along the line

[languid crater]

im confused

where are you getting the numbers from

[obsidian harness]

it's just an example to show you stuff

it helps if you draw it out

this kind of thing applies to the x-coordinates (the base) and the y-coordinates (the height)

[languid crater]

Ohhh, similar triangles meaning two triangles with the same trig stuff

but different sidelengths

sorry, same angles

In this case I have the point representing the top of the man, the top of the street light, and the shade, and I'm trying to get t between 0 (street light) and 1 (shade) (or the other way around)

[obsidian harness]

yes

[upper karma]

ay nice

first hundo

thanks broski

i kinda forgot to do this test

(im on unit 5 this is unit 3)

so i js did it today

i'm surprised it was so short though

21 mins

anyways time to do unit 5 test

[sand epoch]

That's a lot of tabs

[nova herald]

No. What does congruent mean.

They're not all 3 congruent. Look at them they're clearly different angles

[plain scarab]

I see now

Is 5 sometimes?

idk

[cosmic plover]

I have a geometry regents tmr- any tips?

@everyone

[plain scarab]

How is this wrong

[upper karma]

first find the values of 2 and 4

according to my info angle 2 is congruent to angle u

due to the fact that they're alternate interior angles

then

2&3 are congruent to Q due to them being vertical angles

so i'm assuming if u were to add angle 2 and 3 you could get the value of q

that's just using logic never done these types of problems before

where u have multiple tranversos

you must use the given information to find angle 2 and 4

or maybe not since the u is exactly on the transverso line

[silent knoll]

Yes did you use any proofs to prove ur reasoning

Because I don’t see any

First one would be the parallel line theorem

[plain scarab]

Uh no

[quick zodiac]

I think the one I picked is right but I’m not 100% sure

[tall coral]

looks good on desmos 👍

[nova herald]

The answer is sometimes true. Alternate interior angles are congruent when they are formed by a transversal intersecting parallel lines.

[fickle rose]

there’s a theorem that will help u solve this and it has to do with the chord being perpendicular to the radius

[sudden shell]

I know that point P is the midpoint of AB

[fickle rose]

ok good

[sudden shell]

Which tells me Ap and BP equals 4

[fickle rose]

yes

try to make a triangle

[sudden shell]

I created a right triangle

[fickle rose]

yes

now u should be able to solve it

[fickle rose]

pythagorean theorem

[sudden shell]

Yep thank you

[fickle rose]

np

[sudden shell]

Got my answer it’s 3

[sudden shell]

Did I do this correctly

[lime dune]

,rccw

[lime dune]

your radius appears incorrect, check that again

[sudden shell]

It 0.84

I got it thank you

[silent knoll]

Anyone needs help with geometry

I can help

[grave marten]

What did your unit 2 go over? Just wondering because my summer course starts on Monday

[jovial sun]

In triangle ABC, G is the centroid, I is incircle that touches BC CA AB at D E F. It is given rhat AD BE CF intersect at X, the Gergonne point. Prove ABI, BCI, CAI have concurrent euler lines at Y, let GI and XY intersect at Q, prove 3GQ=2GI

[upper karma]

Basics of geometry what is a line ray segment

Segment addition postulate

Distance formula

Mid point formula

My unit 2 was about hmmm

I think it was just proofs in general

For angles

[grave marten]

Thanks

[upper karma]

Why so late?

[upper karma]

Ofc

[grave marten]

I have no idea lol some counties in Virginia start later

[upper karma]

Lucky

Bro gets a break

I've been doing this for almost 3 weeks

Finished 5 units of it out of 6

So we're getting there

[grave marten]

Bro no break sounds so bad 😨

[upper karma]

lmao

[plain scarab]

Is this 60?

[upper karma]

should be

[harsh crow]

yes bru assuming theres a degree symbol sumwhere

[upper karma]

alternate interior angles are always congruent

[upper karma]

u can also js add it

for the lower lvl geometry

[plain scarab]

Gotta make sure 😭😭

[upper karma]

lmao

don't doubt urself

[plain scarab]

You guys got 6 units?

we have 8 first semester i think

is this right?

just wanna make sure

before i submit

[upper karma]

yee

are they

??

[upper karma]

wait is that a 90 or 98

[plain scarab]

98

[upper karma]

oh ok then

you would be correct

if i can do math

98=98

x+98=180

x=180-98

x=82

[plain scarab]

Appreciate it

yo I got one more

i promise

[upper karma]

ofc

any works

why would u say its that answer?

if u truly want to learn u gotta answer questions to make ur mind work

[plain scarab]

4 and 8 are consecutive exterior angles

1 and 7 are consecutive exterior angles

2 and 5 are consecutive inferior angles

[upper karma]

mhm

why would u say its 2 and 5 though?

what is that rule

[plain scarab]

look like they add up to 180 degrees

supemtaey

supplementary

shit

[upper karma]

very gud

this just says that if its stated that 2 parallel lines are cut

i like to call it bisect

by a transverso then consecutive interior angles shall be supplementary

[plain scarab]

O ok

[upper karma]

as they have the same angle since

both lines are parralel

those r just thoughts though:)

you can also do process of elimination like you did

[plain scarab]

Preesh

[upper karma]

but better to understand

[maiden brook]

bisects is the wrong word

[upper karma]

ye expected

its used for triangles

but yk ykit looks like its bisecting parallel lines although it doesn't separate them

[maiden brook]

bisecting means to split something into two equal parts specifically

[upper karma]

yeah

[upper karma]

very good hockeydude85@maiden brook

[maiden brook]

ok can someone explain this to me including the motivation and thought process. thanks

Let ABC be a triangle with incenter I, and let D be the foot of the angle bisector from A to BC. Let Γ be the circumcircle of triangle BIC, and let PQ be a chord of Γ passing through D. Prove that AD bisects ∠PAQ.

[upper karma]

i'm assuming this is after mid segment of triangles

since incenter seems ahead of what i've done

isn't that like median or sum?

[maiden brook]

the incenter is the intersection of the angle bisectors in a triangle

[upper karma]

huh

[maiden brook]

what?

https://upload.wikimedia.org/wikipedia/commons/1/11/Incenter.svg

[upper karma]

awww nah

this is like geometry part 2

[sacred nimbus]

i got stuck on my regents 😔 when the pdf comes out im gonna ask some1 here

[maiden brook]

do u remember what the question was like?

[grave marten]

We have 11 units 💀

[stuck sleet]

Is this a valid form of angle bisector theorem?

wouldn't be the right relation is that ST/SV = TR/VR ????

[stuck sleet]

https://www.youtube.com/watch?v=7fR0QZnWT1w

what angle bisector theorem is this solution is talking about?

[wraith jackal]

hi everyone
theres this thing where you can rewrite
a sin x + b cos x
as R sin(x + alpha)

and theres other forms

but why can a singular sine wave model the sum of a sine and a cosine wave

ie are there any proofs that can show this

[nocturne remnant]

basically the idea is to find some angle theta such that cos theta : sin theta = a : b, then apply the compound angle formula

[wraith jackal]

ohh

since it holds for all x

you equate coefficients?

alright thanks

thats coolk

cool

[obsidian harness]

there's only one thing called the angle bisector theorem and it's the one in your YT video

[sacred nimbus]

it was like about a rectangle and they gave me one side BC and an angle of 37 degrees but asked me to find what diagonal AC was

[upper karma]

is AAS and ASA the same congruence condition only in euclidean geometry?

well i guess they're the same in euclidean geometry cuz AAS can always be translated to ASA

but not the case in spherical in non-euclidean geometry right? or spherical geometry?

[nocturne remnant]

I don’t think AAS always implies congruence on a sphere

[upper karma]

yep that's what i thought too was just making sure, thanks

btw it is the same in regular elementary geo right?

and not the same in hyperbolic geometry either i guess?

[nocturne remnant]

On the unit sphere, pick one vertex to be (0,0,1) and the other two vertices on the xy-plane

[nocturne remnant]

Well in euclidean AA always gives you the remaining A so they become the same

[twilit elk]

ah this one is pretty cool

[covert tinsel]

Can smn help me with this?

Question 6,7

My final is in a day, I need help like asap

[gritty surge]

(I may be wrong so someone correct me if I am) For question 6, it says they’re both congruent meaning they share the same side and angles. Therefore that would mean y is 48 degrees and the one to the right of it is 108 degrees.

Next to find the remaining angle, we use the idea that all triangles have 180 degrees. So add 108 and 48, which is 156. Subtract this from 180. Now we know that 2x-y = 24 so just solve from there

[covert tinsel]

Where did U get the 24 from ?

[gritty surge]

I did 180-156 to get 24

For number 7, use the same ideas. 2x + 4 matches up to 100 degrees to use algebra to find x

3y -3 = 12

[upper karma]

y=48

wait nvm

they're not right angles

js realized

but they r congruent

that means

u can substitute missing values no

and since both r equal to 180

u can substitute in equation

and then wrap it up with a kiss

[haughty arrow]

can anyone tell me what it is that i am dong wrong here

i used sin-1(3/7) as instructed in the help video

[gentle haven]

There are two solutions

It says find all angles between 0 and 180

[haughty arrow]

lol that was very vague but thanks

[plain scarab]

What am I missing?

[pine coral]

the angles must sum up 180 degrees

[plain scarab]

thanks

[upper karma]

leeme give u a few insights

[plain scarab]

?

[upper karma]

all triangles have a measurement of 180 degrees

all 90 right angles are congruent

all perpendicular bisectors create right angles

[plain scarab]

Kk

[upper karma]

a triangle has a base and a maximum point

basically the pinacle

a sphere has 41,253 square degrees

which can be represented by this

just a few insights for ur geo life

[plain scarab]

ok

[plain scarab]

looks like angle a and b have to be the same

[upper karma]

That’s kind of redundant

[upper karma]

That’s kind of vague

[upper karma]

?????

[trail tendon]

did you know all 89.8 degree angles are the same angle

[upper karma]

Wow!

They’re different angles but they’re congruent though

lmao

But they have the same measurements of degree

[upper karma]

what

[upper karma]

What are you talking about

[upper karma]

sphere

:>

[upper karma]

I mean where did the goofy ahh degrees come from

square degrees

Use steradian like a normal astronomer

[mild dagger]

Hello, 'm having trouble with this problem can someone help me I would really appreciate it

[lime crown]

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

[mild dagger]

im stuck after this part

[exotic yarrow]

,w sec(((45+(43/60)) deg)

[somber coyote]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[exotic yarrow]

,w sec(45.72 deg)

[somber coyote]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[exotic yarrow]

@mild dagger you put it in the calculator incorrectly

Although I would round like you did here anyway

[exotic yarrow]

Especially since you need to be so accurate

[exotic yarrow]

Doing this would be better instead

[mild dagger]

that was the first answer I had inputed but it was marked as incorrect

[exotic yarrow]

.

Read this

It rounds to 1.4322, but you got 1.4323 because you rounded in the middle, reducing your accuracy.

https://media.discordapp.net/attachments/326138757474680852/1254325972858306560/wolf.png?ex=66791598&is=6677c418&hm=378b71efd0ac5177461aa923f1340b9a477986f14fcf508c87d18fa0ecdd2a8b&

[mild dagger]

thank you that was the correct answer

[exotic yarrow]

[mild dagger]

yay

does anyone know any helpful youtube channels to help with Right Triangle Trigonometry. It would really help me.

[lime dune]

how about searching “right triangle trigonometry” on YouTube yourself instead of asking others to do what you can do yourself in 5 seconds??????

[upper karma]

Do u actually have to be this exact

[obsidian harness]

lmao just use steradians

1 steradian = 180^2 / pi^2 square degrees

https://en.wikipedia.org/wiki/Steradian#/media/File:BlankMap-World6_steradian.svg

this is pretty cool btw

[north kindle]

i feel like i am going crazy, i found the area to be 54 with herons formula, used that to find the inradius (3), reasoned to the fact thats the distance from the incenter to AC, so the distance from B to the incenter must be twice that so 6, which together with the distance from the incenter to AC gives the full line with length 9

but apparently that is not correct and i cant find why

[lime dune]

median != angle bisector

also there is one observation you can make here that greatly simplifies things

||ABC is a right triangle||

[north kindle]

oh wait shit

idk why i thought the incircle had something to do with medians

crossed wires

[lime dune]

lmao rip

[lime dune]

centroid is the intersection of the medians

incenter is the intersection of the (internal) angle bisectors

[north kindle]

i might have crammed a bit too much math into one day

[lime dune]

I remember when I would just spend hours upon hours on it in middle school

[faint pasture]

Just use coordinate instead of euclidean

(This one is actually not that bad with coordinate)

[golden copper]

Math sorecerer and professor Leonard

[maiden brook]

no this one is easy because it’s a right triangle

[faint pasture]

Is there some property of right angled triangle which i skipped over cz i dont really see how itll be easy

[maiden brook]

||where is the circumcenter of a right triangle||

[faint pasture]

Circumcentres the midpoint of hypotenuse

I dont see how thats related

[maiden brook]

oops one sec I misread

nvm I read again

@faint pasture

yeah that’s all u need

[covert tinsel]

Thank uu

[faint pasture]

Whats all i need?

[maiden brook]

this

[faint pasture]

How? Any further hints

[maiden brook]

what is the circumcenter

[faint pasture]

Oh

Nvm

Just realised

[faint pasture]

Yea

Got it

[flint ingot]

if i have a ratio between the sines of two angles do i get a ratio between the angles themselves?

pls help

[kind rapids]

I think no

[nocturne remnant]

no

[kind rapids]

According to law of sines, you only know the ratio of two opposite sides.

[nocturne remnant]

but in computations the ratio of sines is often more useful than the ratio between the angles

[flint ingot]

Oh ok

[haughty arrow]

can someone help me figure out what i am doing wrong

i tried to subrtract and add 96.16 to 180 both answers were wrong

[heady violet]

yes

[exotic yarrow]

As a sanity test, an angle in a triangle can't be more than 180 degrees (since the angles in a triangle add to 180 degrees)

Anyway, since the angles in a triangle add to 180 degrees, it follows that $$\angle A+\angle B_2+\angle C_2=180^{\circ}$$

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

You have $\angle A$ and $\angle C_2$, so solving for $\angle B_2$ is straightforward

[somber coyote]

Civil Service Pigeon

[mild dagger]

thanks ill look into it

[stuck sleet]

If O is the center

How can I prove P-O-M as the diameter?

How can I prove that line MP passes through the center O?

[exotic yarrow]

find $MP$ using $\triangle MNP$ and $\triangle MQP$

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

(recall that a^2 + b^2 = c^2 + d^2)

[stuck sleet]

So the implication that the sum of squares of triangles legs are equal (sides of the quadrilateral) indicates that point O is in line MP? Which implies that the two triangles created are right triangles?

[exotic yarrow]

Yeah, it must be true that $\angle MNP=\angle MQP=90^{\circ}$, and hence $MP$ is a diameter

[somber coyote]

Civil Service Pigeon

[stuck sleet]

thanks

[soft pier]

.. I'm weak at geometry

🤕🤕

[lime dune]

!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/

[lime dune]

which topics specifically?

[soft pier]

.

[polar belfry]

Hints please

[upper karma]

AM-GM Inequality

[mystic oar]

can someone give me a very good level of trigonometry question for class 10 student

[tawny tide]

Dm me

[obsidian cosmos]

Does anyone know how to solve this i even asked chat gpt and the 4 solutions it gives me arent correct to the system

[trail tendon]

do you know the double angle formula thing

like the trig identity for sin(2a)

[obsidian cosmos]

ya ik the equations

but i wasnt sure how ot use it

to

[trail tendon]

sin(2a) = 2sin(a)cos(a)

right?

[obsidian cosmos]

ya thats the equation

[trail tendon]

yea

so then what do you get

[obsidian cosmos]

wait but its 5sin(2sin(a)cos(a)) + 4cos(a)

how do i make 5sin(2a) into 2sin(a)

[trail tendon]

you have 5sin(2a)

but

sin(2a) = 2sin(a)cos(a)

so 5sin(2a) = 5 * 2sin(a)cos(a)

right

[obsidian cosmos]

so then for just the a do i use half angle identity?

[trail tendon]

all you have to do is this identity

[trail tendon]

like this is close

but

;-;

[frozen jacinth]

You know herons formula, is there formulas similar to this in geometry?

[trail tendon]

if you have 5sin(2a) + 4cos(a) = 0 for 0<=a<=2pi
sin(2a) = 2sin(a)cos(a)
when you plulg in sin(2a) what do you get for the new equation

[trail tendon]

what do you mean similar?

[frozen jacinth]

oh ignore that please i didnt word it correctly.

is there a video that shows how you can integrate the area of a certain shape such as a triangle?

[trail tendon]

yeah sure

[obsidian cosmos]

this is what you get when you plug in sin(2a); 5 * 2sin(a)cos(a) + 4cos(a) = 0

[trail tendon]

if you can find the formulas of the lines and put them on a coordinate graph

you can integrate them

[obsidian cosmos]

so do i move the 4cos(a) to the other side so -4cos(a) and then divide cos of both side

[trail tendon]

nope

well i mean

you should just factor out 2cos(a)

because cos has the posibility of being 0 and you cannot divide by 0

[obsidian cosmos]

okay so does the 2 multiply to cos as well?

i thought it was just to sin

[trail tendon]

yes

5 * 2 * sin(a) * cos(a) you can move all these terms around

you can say cos(a) * 5 * sin(a) * 2 or 5 * cos(a) * sin(a) * 2, ect.

multipliation is communitive

[obsidian cosmos]

oh okay so i have to factor the values out first

[trail tendon]

communative?

idk how spell

[obsidian cosmos]

loll

[trail tendon]

yeah

XD

[obsidian cosmos]

okay bet thanks for the help

[hoary prism]

commutative

[trail tendon]

LOL

i had an n instead of t 😂

[meager skiff]

How do I start this js practicing before I start my quiz

[maiden brook]

what must be true about angles in a cyclic quadrilateral

[meager skiff]

Any two opposite angles must add up to 180 degrees

[maiden brook]

so

[meager skiff]

Wait ohhhhhh alright thank you:)

[steady vault]

Circle geo we've been there before

[upper karma]

guys don't concentrate on the hard questions concentrate on the stupid ones

HOW DOES THIS HAPPEN BROOOO

[lime dune]

😭

[upper karma]

I WAS FOCUSING ON THE CHALLENGING ONES

I GOT THE EASIEST ONE WRONG

https://tenor.com/view/crying-in-the-corner-cute-one-piece-mugiwara-straw-hats-gif-17513547

[steady vault]

omd

h

ans doesnt match

show me the process idk what im doing rong

[obsidian harness]

part b right?

then use part a, so you can either have sin 2x = 0, or that 2x = 0, 180, 360 or x = 0, 90, 180

or 1 - 5 cos 2x = 0, or cos 2x = 1/5

your domain is 0 <= 2x <= 360 so you get two solutions for cos 2x = 1/5 as well

2x = 78.5 or 360 - 78.5, cos(360 - u) = cos(-u) = cos(u)

[stuck sleet]

My answer is 36 sqrt(3) in. sq.

Am I right?

[obsidian harness]

let me check, so the blue shaded area can be broken up into 12 smaller equilateral triangles

there are also 6 white regions: split the blue and white triangles in half to get that the 1 blue triangle = 1 white triangle, so it's 12/18 * (area of regular hexagon)

yes, that answer is correct

[stuck sleet]

what I've done is that the I calculated first the area of the whole hexagon then I subtracted the area of the unshaded little triangles

54 sqrt(3) - 18 sqrt(3) = 36 sqrt(3)

[obsidian harness]

fair enough

[plain scarab]

Is this the last option?

[trail tendon]

am i trippin or is the third option exactly whats given

so its literally true

i guess the third and fourth

😂

[brazen monolith]

im looking for highschool geometry help. im taking an online course and have acouple days to finish 2 test

[vital pike]

is anyone doing the geometry course over the summer at the ut austin high school thing or does someoone know sm1 whos doing it

[vital pike]

me as well which place are you taking it

from

[brazen monolith]

lake zurich highschool

[fickle rose]

yea

i would guess the fourth one tho

if u only have one

[faint pasture]

Thats not by cpctc tho

[trail tendon]

🤔

i guess 😭

[faint pasture]

Is it?

[trail tendon]

its still true tho 😭

dum dum

[faint pasture]

But the problem specifically did mention it

Sooo i mean

XD

[trail tendon]

it should have said "which of these can we confirm using cpctc" or smthn

instead of saying "using cpctc which of these are true"

its jsutw orded badly imo :l

[faint pasture]

true

[vital pike]

does anyone know where to find the answers for the summative test

s

for the online ocurse

just to check

[plain scarab]

That’s what i was thinking lmao

[upper karma]

if sinθ+2cosθ=1, prove 2sinθ+cosθ=2

anyone know how to solve this?

ive been trying multiple ways but couldnt get an answer

[jagged rain]

https://gyazo.com/8bacfcec9c7af79d8da6fb13bdcedda4 Im struggling reading what quadrant this angle would be in my guess is quadrant 2 but could someone explain to me why?

[vivid vault]

it's in quadrant 1 since theta is less than or equal to 0 and less than or equal to pi/2

[sullen fern]

Hi what does this mean?

[sullen fern]

I have no idea how to solve it without the ^2

but maybe this can help?

[upper karma]

damn tysm

i tried to divide both sides in the first equation with cos to get tan+2=sec, then squared both sides

didnt really get me anywhere

[honest mountain]

I can ask help here?

Or still the same help channel?

[upper karma]

yea fr i need elp

[lime dune]

!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/

[lime dune]

feel free to ask your questions here lol

[upper karma]

oh

can u solve it or smthn

i js wanna know if it is correct

all i need rn

[lime dune]

what answer did you get

if possible show us your work

[faint pasture]

,w sinx + 2cosx =1

[somber coyote]

Results provided by WolframAlpha

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

Requested by enderdoesntmind

[upper karma]

this would be considered out of syllabus for us

its supposed to be solved using basic trigo identities

[faint pasture]

oh

nvm

im dumb

[faint pasture]

take the sin over to the other side

multiply by 1+sinx

youll see the rest

(in the first eqn)

[upper karma]

ahh i see

ty man

i appreciate it

[north kindle]

i know theres something super obvious here that im missing but i cant find it

OH ITS ANGLE BISECTOR THEOREM

[formal geyser]

Hello

The right top corner of the rectangle is splitted into 3 equal angles

What is the ratio of the shaded area to the whole rectangle's area?

Red triangle's area / rectangle's area?

I applied trigonometry to find the length of leg x

[formal geyser]

Then i found the area of the right triangle with a and x legs and substracted it from the area of the half of the rectangle

And i found the area of the shaded triangle

I dont know what is the right answer for this problem, but the ratio isn't just a number, there are also a and b variables, so i suppose i did something wrong

[dusky locust]

30-60-90 special triangles

[formal geyser]

Okay, but..

I think there is a mistake

[formal geyser]

Triangle abc seem to be 45-60-90, which is impossible, right?

[north kindle]

looks 30-60-90 to me

let the top right corner be D, angle DAB is 60, so BAC is 30, ACB is 90 and ABC is 60

[formal geyser]

Top right corner D? But it is B

[north kindle]

oh sorry i meant top left

[formal geyser]

Okay

There is something i do not understand about this problem. Is AB the diagonal of the whole rectangle?

It doesnt seem so to me

[formal geyser]

If so, angles dba and acb must be equal (45 degrees), but they are not

So in order abc to be 60° angle, the line AB can't be the diagonal. Moreover, it should not point toward bottom left corner

Oh

Sorry, i somehow thought that rectangles diagonal splits its angles in half like in a square

But it doesn't here

[robust jasper]

$\the answer is 70.6264 $

[plain ingot]

is there a way to simplify sin^-1(k * sin(phi))

i found that ksin(phi)is a good approximation

[plucky ledge]

i forgor how do this

[plucky ledge]

nvm

i fnshed

[upper karma]

LMAO

HAHA

[plain scarab]

could anyone please help me

im struggling

[bronze yacht]

ill find something zick

im know some sss

trigonometry

[plain scarab]

preciate bro idk wtf to do

this thing has been tripping me up

all unit

literally got everything else right besides that part

[bronze yacht]

the first one may be not enough info

[lime dune]

yea not enough info looks right

[grim flare]

🥺

[lime dune]

try drawing a picture like they suggest

[grim flare]

i did no comprendo

[static karma]

anyone want a random graph?

[sand patrol]

Can anyone help show ABP congruence CDQ

[faint pasture]

Is it given abcd is a parallelogram or smthn?

[sand patrol]

Yea

[faint pasture]

Then DC = AB

And RHS

[sand patrol]

Ohh Tnx

[sand patrol]

If BC^F=60 how to show ABC is a Equilateral triangle

[lime dune]

you have a pair of parallel lines in the figure, can you convert that to angle information?

[faint pasture]

Oh

Nvm

[sand patrol]

What do you mean

[faint pasture]

I was wrong

I mistaked bce for bcf because of the diagram marking

[lime dune]

are there any angles that you can show are equal?

[sand patrol]

Yea ABC = ACF

Z angels

[lime dune]

not quite

$\angle ABC=\angle FCB$

[somber coyote]

elrichardo1337

[lime dune]

the parallel lines mean that those two "alternate interior angles" are equal

[frozen jacinth]

am i allowed to write

^ above the b and c instead of putting that angle sign?

[obsidian harness]

no, that usually means the angle spanned by an arc in a circle

but check with your teacher or your books

[north kindle]

im having trouble with this, im trying to do something with the angle bisector theorem but i cant find a connection

[gilded thistle]

fellas im trying to learn geometry but i cant find a good source that has everything

do any of u know any ?

[upper karma]

like online sources?

all I know are like khan academy and delta math

[bronze owl]

Organic chemistry tutor pretty good

There's also channels like premath that i use for puzzles

[upper karma]

there’s also professor David explains and I think crash course?

[plain scarab]

Is this SAS?

[upper karma]

no

you need to have 2 congruent sides for it to be SAS

it would be not similar

[plain scarab]

gawd damn

do you know this by any chance

I have no clue

[trail tendon]

You see one side in common and one angle in common...

But hmm

[plain scarab]

sas?

U only need 2 to make a claim

Hm

[trail tendon]

Do you tho

[plain scarab]

That’s what the organic chemistry tutor told me

I was watching his video on this stuff

[trail tendon]

you need at least two, but sometimes more

[plain scarab]

ah ok

[trail tendon]

Like SAS

is three

Side angle side

But AA is two

Angle-angle

[obsidian harness]

basically this

[upper karma]

you would eliminate choice SSS since the 2 angles already share the same angle (vertical angles theorem)

[obsidian harness]

the key there is that AD and EC might not be parallel at all

[plain scarab]

Hm alright

thank you all

[obsidian harness]

for example

yeah so I would think of AA as a special case, as in if you know two angles you automatically know the third

[obsidian harness]

so usually you do need 3 pieces of information, not 2

np

[plain scarab]

So apparently reason 4 was incorrect?

I hate proofs

[upper karma]

me too man that was the worst unit in geometry for me

wait how would you use ASA congruence postulate if you only knew that angles 1 and 2 were congruent and the same side BD?

you don’t have the 2nd angle

i believe if you used angle 3 and 4 it wouldn’t work since that’s what you’re trying to prove

wait

is the problem correct

[obsidian harness]

yes, it's angle-angle-side

not angle-side-angle

[plain scarab]

ah

damn

[obsidian harness]

so you were close, like you can't assume that angle 3 = angle 4

[plain scarab]

Let’s go

More proofs

🔥🔥

Is 2 a good reason?

not entirely sure

[obsidian harness]

yes looks good

[plain scarab]

what reasons would I even write for 6 and 7?

for 6 could I write division property of equality

or substitute

[frozen jacinth]

is there a book pdf

where

it talks about trig basics

[faint pasture]

piracy?

[frozen jacinth]

whats ur opinion on khan academy?

[faint pasture]

Its fine

[frozen jacinth]

do u know any good yt playlists for prealgebra?

[faint pasture]

Not really

[frozen jacinth]

ah! ok

[harsh crow]

getting a textbook is always better than watching yt vids

its more painful but always pays off in the end

[harsh crow]

also if ur doing pre alg dont worry abt trig rn

[frozen jacinth]

i see

i got

a lot of gaps in my knowledge

and i would want a solid foundation

and build from there

[honest mountain]

Can somebody help me in geometry? #help-27

Line.

[plucky ledge]

I HATE TRIANGLES

im going le crazy

[trail tendon]

what did triangles ever do to u 😔

[plucky ledge]

my brain is frying

these videos arent helping me solve similar right triangles (geometric mean)

I DONT EVEN KNOW WUS HAPNING

HJASSHDJASd

[trail tendon]

Ah

Do u see the two right triangles

One is the part of the entire triangle with a right angle, one is the entire triangle

[plucky ledge]

yea

my brain is breaking

AHHHHHHHh

[trail tendon]

💀

[upper karma]

you could say the book is like an online course source of knowledge

since it brings you through everything from proofs to solutions theorems etc

i have 2 algebra books

one passed down from my grandpa one that i bought gotta say they're pretty useful (haven't had time to read them throughly just because i don't have much free time)

doing geometry to skip it next year

😛

[plucky ledge]

turns out i just needed to watch a short video then i got it

geometric mean gaming

[spiral oasis]

I got a test on some trig stuff tommrow

wish me luck

[faint pasture]

Geometry is just evil in general

[feral cloud]

heresy. geometry is beautifil

[mellow quest]

x = 34

but why

[gilded thistle]

fellas if im given an angle c and the length of AB how do i find the radius of the circumfrence around the triangle

[mellow quest]

https://en.wikipedia.org/wiki/Law_of_sines

[gilded thistle]

ty very much

[plain scarab]

How is this not congruent

[fallow marsh]

First apply Pythagoras in the lower triangle so that you get the base length and then apply sine law in the lower triangle to get the lower left angle and then you can also get the topmost angle,this way apply sine law in the whole triangle so that you get AC and then subtract AD from it.

[plucky ledge]

i just did the opposite of geometric mean and got the answer

it was 20

[fallow marsh]

Good one,multiple ways..mine was longer ig

Brain ain't braining

[plain scarab]

Is it similar?

my teacher is tripping

[fallow marsh]

What?

[plain scarab]

the problem

[fallow marsh]

The triangle one?

You can't comment by just looking,they are both right angled so if one more angle turn outs to be equal between both then they would be similar

[trail tendon]

I don't know why it says "the angle measurements are unknown" I feel like they can only be one thing :l

like if you have all the side lengths of a shape I thought you could find the angles xd

[plucky ledge]

quadrilaterals cant have side side side side congruence i think

cuz like

they can be like

angled slightly differently

and the angles can be different measures

only triangles have SSS congruence iirc

[plucky abyss]

Yes like a square of 3 cm and a rhombus of 3 cm

I think their sides angles diagonals and fb profile should be same to be congruent

[hard moon]

one way is to apply the law of sines;

(1) you can obtain the ratio between AB and AD using triangle ABD

(2) you can obtain the ratio between AB and AC (in terms of x) using triangle ABC

(3) you can obtain the ratio between AD and AC (in terms of x) using triangle ACD

(4) taking the equations from (1), (2), and (3), you can obtain an equation where only the variable x remains, and you can solve for x

[fresh meteor]

I don’t even know how to start

[hard moon]

interestingly though, if you change one of the angles just a little bit (say you change 44° to 45°), then x is no longer a nice angle; rather it's:

so it is possible there is a more elegant way to do it than law of sines since x happens to be nice when the given angles are "just right"

but idk

[hardy salmon]

https://math.stackexchange.com/questions/4939366/a-question-about-general-equations-of-hyperbolas?noredirect=1#comment10558067_4939366

[formal geyser]

Hello

There are 3 squares in the right triangle

What is the sum of areas of two shaded squares

[grave pond]

First prove that the corner of the unshaded square splits the length labeled 6 into two equal pieces.

[grave pond]

Hmm, interesting. I get 34° by coordinate-bashing with a calculator too, but it is not at all clear why the result would even be a nice rational angle.
I notice that 34 is half of (104+26)-(18+44), but that coincidence certainly doesn't stay true for other random choices of the four known angles ...

[formal geyser]

How do i do this?

[grave pond]

I used similarity to compare the ratios BC/AE and CD/AE.

E.g. BC/AE = BC/AC · AC/AE = CE/AE · AC/AE.

[formal geyser]

Oh, man, that's difficult to me, but thanks

[formal geyser]

I know what similarity is but, i dont understand this

[grave pond]

I used similarity to see BC/AC = CE/AE; the rest is just algebra.

(In fact, what I did on my scratch paper was more something like:
Suppose there's a number f such that AB = f·AC and AC=f·AE and CD = f·DE.
Then which fraction of AE does the 6 represent?
It turned out this was f(1-f) + (1-f)f, where each of the two equal terms represent BC and CD.)

[dull pilot]

how can i denote the major arc

[grave pond]

Name a point on it and write something like "arc APC"?

[dull pilot]

i cant do that with only 2 points?

yea

no

i cant

why did i even ask this question in the first place

[faint pasture]

or just call major ac

[dull pilot]

cant you do this in math syntax

[faint pasture]

nah then you need three point

[mellow quest]

how to solve this

[faint pasture]

,w calc sin(x degrees)*sin(70degrees) sin(18 degrees) = sin(44 degrees)sin(50-x degrees ) sin (58 degrees)

[somber coyote]

Results provided by WolframAlpha

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

Requested by enderdoesntmind

[faint pasture]

yea no that doesnt have a pretty solution at least

oh nvm i guess i see something
180/(pi/2) ( pi*n + 1.4818)

cant see the 1.4818 though

[plain scarab]

Do reasons 2 and 3 make sense?

[hard moon]

you can use the identity for sin(a - b), and from there, you can rewrite the equation in the form tanx = #

[plain scarab]

anybody know?

[plucky ledge]

makes sense yea cuz it says it's the midpoint

[void lynx]

Hey I am weak at inverse trigonometry any solutions

?

[plucky abyss]

Is inverse trigonometry under trigonometry

[obsidian harness]

ye

[plucky abyss]

Okay

[harsh crow]

wdym inverse trigonometry

[upper karma]

trymenogotri

[trail tendon]

arcsin n stuff prob

[upper karma]

Mine was better

😔😔😔

[trail tendon]

Fair

[upper karma]

wanna learn trymenogotri with me\

[lime dune]

triggerednometry

[upper karma]

riggedonometry

[lime dune]

trigonomeWHY

[upper karma]

Ygonometry

[upper karma]

Simplify this expression into one single expression ((t²-1)/(t²+1) + (cos(x/2))×cot(4x))×sec(9x/2) where t=tan((x-pi)/4))

[tall coral]

can we get that in latex?

[upper karma]

Idk how to latex that

[hard moon]

what's a single expression

[upper karma]

Like cosec4x sec4x Cosec2x sec2x

[tall coral]

$t=tan\left ( \frac x-\pi \right))$
$(\frac{t^2+1}{t^2-1}+\left(\cos\left(\fracx2\right)\right)\cot(4x))\sec \left( \frac{9x}{2} \right)$

[somber coyote]

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

[upper karma]

Nah

[tall coral]

oh that went really wrong lol

[upper karma]

Soo any clue how to solve this @tall coral

[tall coral]

no

[hard moon]

try simplifying (t^2 - 1)/(t^2 + 1) first; it might be useful to know cos(2a) = cos^2(a) - sin^2(a); also, remember sine and cosine are horizontal shifts of each other

after that, simplify the whole thing; try to recognize the identity cos(a + b) when simplifying

[manic junco]

who wants to help me with my online school geometry work

[plucky abyss]

Press the power button to get instant help

[trail tendon]

alt F4 works as well

[plucky abyss]

Yes

[stuck sleet]

A goat is tethered to a barn by a 12-ft chain. If the chain is connected to the barn at a point 6ft from one end of the barn, what is the area of the pasture that the goat is able to graze?

My answer is ( 48pi + 18 sqrt(3) ) sq. ft.

Am I correct or wrong?

[obsidian harness]

we need the picture as there's not enough info here

Depends where exactly the chain is connected

[obsidian harness]

Cool

[stuck sleet]

I saw the answer in the internet but I think it is wrong

[obsidian harness]

Should be pi * 12^2 * 1/2 + pi * 6^2 * 1/4 right

[stuck sleet]

the answer in the net is 63pi ft^2

[obsidian harness]

That's less than the semicircle so impossible

[stuck sleet]

yeah

a quarter circle + (quarter circle - the half of a segment)

This segment:

[obsidian harness]

The goat can't go inside the rectangle

[stuck sleet]

this number plus the quarter circle

[stuck sleet]

I am sorry for confusion, English is not my mothertongue is it asking for the area in the red or is the area on the blue?

[obsidian harness]

The red area

[stuck sleet]

Ah..

So the area must be 72pi + (area half of the segment)

I get it now

[obsidian harness]

Yes this

Cool

[stuck sleet]

A triangle has sides of lengths 6 cm, 8 cm, and 10 cm.
Find the distance between the center of the inscribed circle
and the center of the circumscribed circle for this triangle.
Give the answer to the nearest tenth of a centimeter.

help

I tried to solve it but

I Just cant

the answer is 2.2 cm

though there is on the internet telling something about "Euler Triangle Formula"

[grave pond]

It's a 3-4-5 triangle, so if you draw it in a sufficiently large scale on graph paper, all the relevant point can be grid points.

[stuck sleet]

yeah that was my first step

then the the radius of the incircle is 2 cm

then I utilized two tangent principles

one is the the two tangents are equal to each other

then the other is the tangent line is perpendicular to the incircle's radius

[grave pond]

If you have the inradius, then you know the incenter is at (2,2) and the circumcenter is at (3,4) in suitable coordinates...

(Since the circumcenter of a right triangle is always the midpoint of the hypotenuse).

[stuck sleet]

the origin is the point of right angle right?

[grave pond]

Yes.

[stuck sleet]

But without using analytic geometry

how would you analyze it?

just pure synthetic euclidean geometry

[stuck sleet]

Thanks!

[grave pond]

There are trigonometric formulas that apply to general triangles -- in particular the law of sines and the law of cosines -- but the basic relations that hold in right triangles cannot just be applied as written to general triangles.

[grave pond]

Hmm, I suppose you could use your two-tangent principle to show that the point where the incircle touches the hypotenuse divides it in parts of 4 and 6 cm (namely 6-2 and 8-2), whereas the circumcenter divides it in parts of 5 and 5 cm. So you have a small right triangle with legs 2 (the radius) and 1 (along the hypotenuse).

[maiden brook]

if O is the circumcenter of this triangle and R is the circumradius, and I is the incenter and r the inradius, then (what we desire) OI^2 = R(R - 2r)

The circumradius is equal to the length of the hypotenuse over 2, so 10/2 = 5

And the inradius is 24/12 = 2

OI^2 = 5(5 - 2(2))

so OI^2 = 5 and OI = sqrt(5)

which, rounded to the nearest tenth, is 2.2 cm, your answer

[upper karma]

Hey guys

struggling with trihedral angle exercises

this is the exercise
If in a trihedral angle ∠𝑂 − 𝐴𝐵𝐶 the dihedral angle ∠𝐶 − 𝑂𝐴 − 𝐵 measures 90° and the angles of the faces 𝐶 = 𝐵 = 60°. Find the measure of face 𝐴

[rich grove]

I understand that the triangle is the half of the length and width but how is 5 related to the area?

Doesnt it make more sense if the area of the triangle is 1/2 times 2 times 7 instead of 1/2 times 2 times 5

[harsh crow]

i got 81 pi

[faint pasture]

The formula is 1/2 bh

[plucky abyss]

Height means it should be perpendicular

Little kitty

Read the book cathematics to revise

[rich grove]

Ahhhh yeesss I see. Thanks y'all

[meager skiff]

😭

[sturdy stone]

Half the difference of the intercepted arcs formula

[limpid onyx]

Let O be the circumcenter of ABC, let N be Nagel point. The circle Gamma1 is tangent to BA, CA's extensions at E,D such that AD=AE=BC, similarily define Gamma2 and Gamma 3. Let circle U be the circle externally tangent to Gamma 1,2,3. Prove N, O, center of U are collinear.

[nova kraken]

I need help

Do you guys have any recommendations on learning basic Geometry?

[obsidian harness]

Cavalieri's principle btw

Just move all the slices left to right

[tranquil bolt]

im 13 and i completed algebra 1 & 2, and advice what to do next?

[trail tendon]

geometry?

[tranquil bolt]

ok

[trail tendon]

bro asks in the geometry channel 😂

[tranquil bolt]

its under prealg & alg lol

[ashen grove]

precalc if ur geometry is solid enough

get familiar w calc concepts if u have time

you dont have to be proficient but it helps to have an idea

[tranquil bolt]

ok

Il do after i learn geometry nicely

[ashen grove]

usually in geometry first i think

[maiden brook]

combinatorics

[plucky abyss]

It's very hard

[obsidian harness]

deranged /j

[maiden brook]

I mean since he’s already learned both alg I and ii by 13 it actually does make sense to learn that

[obsidian harness]

I guess so

[formal geyser]

Hi

How do i solve it?

[lime dune]

can we do something with the angles in the center?

[round lake]

the ans is 20 degree for x

[faint pasture]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[vapid hinge]

maybe try labelling some random angle y and try to get a simultaneous equation

[gritty topaz]

trigonometric ceva theorem

[faint pasture]

💀 thats such an overkill