geometry-and-trigonometry

[dark sparrow]

look at the flags of various countries around the world

take inspo from those

[wide nimbus]

how do i know which number goes first in an inverse trigonometric function?

[vernal pilot]

Wdym? Also w pfp

[frozen ocean]

depends on the problem

[regal phoenix]

@jaunty cradle hey, have you solved this question yet?

[jaunty cradle]

@regal phoenix no but i need help with another dm me

[regal phoenix]

i was thinking if these angle bisectors form two parallelograms

if they do, we already know the lenght of the two legs of the trapezoid

13 and 15

[jaunty cradle]

thats wrong

the answers dont match

[regal phoenix]

what are the answers?

[willow glacier]

Just saw this sorry you can send me an idea if you want

[vernal pilot]

If you’ve got a right angle triangle and two side lengths.

[vocal bison]

how do i use difference of squares when dealing with radicals

[dry lantern]

Just wondering, is that 73 and 75 or 13 and 15?

[raw anchor]

when writing the cos, sin, etc for a coterminal angle that is negative, do i write is as the negative or where the angle is on the unit circle

ex:

do i write

sin -pi/2 =
or
sin 3pi/2 =

[kindred spire]

yo can someone help me with math

[dry lantern]

idk if i can help but i can try

[dry lantern]

So 8x + 11 degrees should be 107.

[kindred spire]

it was 12

[dry lantern]

nice

[regal phoenix]

if you're refering to the image i sent, 13 and 15

[dry lantern]

Thx 🙂 ill try to figure it out

[novel vessel]

hello i need help with a proof

P1 = (x1, y1) and P2 = (x2, y2) such that y > 0 and x1 < x2. C is a circle that contains the points P1 and P2, and is tangent to the x-axis at the point (x3, 0). Prove that the condition x1< x3< x2 cannot always be fulfilled for a circle C given any two points (x1, y1) and (x2, y2).

anyone has some intuition as to where to start?

[glacial yacht]

In triangle ABC, ∠C is an acute angle m∠C = 2m∠B. AD ⊥ BC.
Prove that AC = BD − DC can we solve this question other than using trigono

[regal phoenix]

idk if that helps

[hearty forge]

In Fig is the centre of a circle of radius 5 cm, T is a point such that OT = 13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find the length of AB

can some one help me

[regal phoenix]

what have you done so far?

[dark sparrow]

this picture is discrimination against darkmode users

please upload one with a white background

[hearty forge]

my bad

wait

better?

ET=8 , let BE=y then BT=12-y

yeah my bad i got it

[lost drift]

I too solved a similar problem,

There are actually two ways of solving it.

1. Proving Similarity of triangles and then solving it

2. Deriving 2 equations using pythagoras theorem and then solving it

[lone panther]

split x sin³θ into x sin θ sin²θ
then replace the x sin θ with y cos θ
this allows you to simplify the first equation down to y = ||sin θ||
then with that, use the second equation to get x = ||cos θ||
from this you can see that x² + y² = 1

[tropic hare]

Can I ask a question real quick?

Is Ptutagoras Teorem Geometry

oh that was a stupid question

ofc its Geometry

srry

[regal phoenix]

don’t worry, every question is valid

no judgment here

[unkempt siren]

I’m having a really bad brain fart, how do I go from step 1 > step 2

[maiden brook]

first of all

no reason to factor like that

pull out an 5^x term

and u get 5^x(5 + 1)

the reason u can do this is because of exponent laws

[maiden brook]

do u remember what happens with exponents when u divide

[unkempt siren]

Subtract

[maiden brook]

so its 5^((x +1) - x)

or 5^1

= 5

[unkempt siren]

I’m still not understanding, when I pull out the the 5^x it divides with both and I am left with 5^x(1 + 1) what am I doing wrong?

OH I’m so stupid, ty sm now I see what you’re saying @maiden brook

Sorry another one, does anyone know how to simplify this?

[maiden brook]

factor

[unkempt siren]

I can take out a 2 from the top, and difference of squares on the bottom, but how do I proceed further

[maiden brook]

can u factor the top)

?

[unkempt siren]

Idk that’s where I’m stuck, Ifactored out a 2, I don’t know what’s next

[maiden brook]

do u know how to factor quadratics?

[unkempt siren]

Like simple and tricky tris?

[maiden brook]

huh?

[unkempt siren]

If you’re talking about factoring trinomials than yeah I do know how

[maiden brook]

quadratics

the top is 2(x^2 + 4x + 3)

factor x^2 + 4x + 3

[unkempt siren]

x^2 + 4x + 3 is a trinomial though?

[maiden brook]

and x^4 - 4x - 1 = 0 is also a trinomial on the left side

thats a lot harder to find the solutions to though

in general, its not usually helpful to think about how many terms there are

but anyway

[maiden brook]

do u know how to factor this?

[scenic creek]

factoring trinomials is part of ALGEBRA 1

as well as binomials and polynomials

[maiden brook]

that’s not true

Some trinomials of higher degree are much harder to factor

same with polynomials

[scenic creek]

the basic trinomials like the one he asked are from algebra 1

thats when you are introduced to them

[maiden brook]

yeah I asked him if u know how to factor but he didn’t respond

[eager warren]

does anyone know how to solve this

I cant figure it out

question c

[dark sparrow]

@eager warren can you show the entire question from part a onwards?

it looks like we're missing some context here about the rectangular tank.

[eager warren]

sure

[dark sparrow]

that... is categorically not what i asked for

i wanted the question from the start, from part a onwards.

[eager warren]

OHHH

I scrolled past that

I didnt even see this

lawd

[dark sparrow]

and part b?

[eager warren]

is the one with the cylinder

[dark sparrow]

so then you've done that, yes?

you know how much water is in the tank after we've poured one cylindrical bucketload in

@eager warren ?

[eager warren]

❌ imma do it rn

[eager warren]

I think its 56,400cm³

for the cylinder question

[dark sparrow]

correct answer, shoddy wording

but ok yeah

that is how much water is there now

are you now able to do part c?

[eager warren]

Im gonna try

OH MY GA

I need to read questions fr, thanks for pointing it out

[dark sparrow]

you'd be surprised at how often this issue comes up

[desert hare]

any tips im started trig

[undone sapphire]

Hey everyone
Today was my last day in grade 10 math, I passed with a 98 and I really find an interest in the subject. I was wondering if anyone would be able to help teach me some things with respect to functions and/or trig to prepare myself for next year, as I know quite literally everyone here is more knowledgeable than I am lol

[upper karma]

become friends with the "unit circle".

[undone sapphire]

does (cos t, sin t) represent the coordinates of that point?

[upper karma]

yes!

assuming the origin is the center, of course

[undone sapphire]

Oh okok that's really helpful

The 1 there, is that the distance of the line?

[upper karma]

yes, the radius of the circle is 1... (the word "unit" sometimes means "1").

there are lots of ways to think about trig, but this is one I like very much

[undone sapphire]

Okk thank you so much

If the radius of the circle was 2, would this change the coordinate equation to (2cos t, 2sin t)

?

[upper karma]

I'm not sure off the top of my head, lets check!

well the equation of a circle is:
( x - h )^2 + ( y - k )^2 = r^2; where x and y are the point (x,y) on the circle, h and k are the x and y coordinates of the center of circle), and r is the radius

since we wisely put our circle on the origin, h=0 and k=0 leaving just:
x^2 + y^2 = r^2

I think we should plug your suggested coordinates in (2cos t, 2sin t)

(2cos(t))^2 + (2cos(t))^2 = (2)^2

the right hand side is just 4 (the square of the radius of 2), now I think we plug in the values of a well known angle on the left hand side to see if we get 4 (a circle with radius 2).

I'm going to choose pi/4 (45 degrees), the sine and cosine of pi/4 = 1/sqrt(2)

,ask (2/sqrt(2))^2 + (2/sqrt(2))^2 = (2)^2

[somber coyote]

Results provided by WolframAlpha

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

Requested by rocinantexyz

[upper karma]

seems like you nailed it

[undone sapphire]

hey thank you so much man, even just this concept of the unti circle is super helpful and you've explained it beautifully. I apologize for the late response also, I'm just coming home from vball practice
Do you mind if I add you if I have any other math-related questions?

[undone sapphire]

I remember this now actually from earlier in the semester, thank you for the refresher because I had totally forgotten

[upper karma]

sure np

[undone sapphire]

Thank you so much I really appreciate the help

[upper karma]

the subject can be confusing, practice helps

[ornate hemlock]

is anyone helping currently

😭

[lethal pendant]

just find the volume of cylinder using the formula pirsquareh and add that to the volume of cuboid , you get the total volume then

[ornate hemlock]

can you help with this?

[lethal pendant]

first u need to find the coordinates of b using midpoint formula

[ornate hemlock]

how if I don’t have Y

[lethal pendant]

but you have A and T

that is enough

it is given that A is the midpoint of seg TB

[ornate hemlock]

I got

(-0.5, 3.25)

[lethal pendant]

so u get coordinates of B as -8,-2

[ornate hemlock]

Then what

[lethal pendant]

then do midpoint formula again

for seg ty

B is the midpoint over here and we know the coordinates of T

Find y using that

[ornate hemlock]

-4.25,0.625

Omg it’s wrong

what am I doing wrong

[lethal pendant]

is the answer -18,-9

[ornate hemlock]

Let’s see

Omg it is

Thank you so much

[lethal pendant]

my pleasure

[glacial yacht]

I HAVE A HARD GEOMETRY TEST AFTER 2 HRS

[grave crypt]

gl

[glacial yacht]

Im sad

It scares me

In good at algebra but

The geometry.....

[grave crypt]

i used to be scared of geometry

geometry is kinda fun

[glacial yacht]

no its not

for me at least

[grave crypt]

i get your point

[glacial yacht]

i get f'ed

when i see

that

not

like co-ordinate

geometry

or trigono

but

[grave crypt]

smay if you find geometry to be amusing and intriguing you might need to change your mind 😭

its not fun i changed my mind

[glacial yacht]

XD

[light dock]

hello, i am going to ask y’all: as students who are taking geometry, can you comment about what you like/don’t like about two-column proofs? my friends are going to teach geometry next year and we had a big argument about this so we want real student perspective

[glacial yacht]

whats ur age cuh

[grave crypt]

i realized this was my homework

😭

[glacial yacht]

yeah

the cylic shit

and

[grave crypt]

no

[glacial yacht]

angle chasing

shiut

shit*

[light dock]

no as in you don’t like them? why?

[grave crypt]

geometry just sucks in general

for me at least

[glacial yacht]

agreed

[grave crypt]

ive been doing circle theorems for 3 months

[glacial yacht]

algebra is straight forward

[grave crypt]

i deeply regret

[glacial yacht]

tangent secant thm

did u study that

[grave crypt]

yes

[glacial yacht]

its fu'ed

f'ed

im fed up of

geometry

[grave crypt]

this aint nothing 🥶 💯

[glacial yacht]

ikr

did u learnm 9 point circle

[glacial yacht]

bro

actually

[grave crypt]

what is 9 point circle

[glacial yacht]

thois

is

an alien

from sun's core

[grave crypt]

we do this daily

😀

[glacial yacht]

any chance u could help with my exam XD

its in 15 mins

[grave crypt]

💀

nah

[glacial yacht]

nvm

im jk

[grave crypt]

bet

[glacial yacht]

?

like ok?

[grave crypt]

ye

[glacial yacht]

u'll help me?

[grave crypt]

no

[glacial yacht]

ye

[grave crypt]

💀

[vernal pilot]

It’s against rules to help someone cheat.

[glacial yacht]

ye

[vernal pilot]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[grave crypt]

omg itadori is here

hi

[glacial yacht]

its not

my school

[vernal pilot]

I forgot the other command

[vernal pilot]

It is I, itadori. Gimme your money.

This is a robbery.

[glacial yacht]

XD

[grave crypt]

shut up fam you aint NOTHING

🥶💀

im ur dad

ok that was cringe

nvm

[vernal pilot]

Very

[grave crypt]

😭

ik im unfunny

[glacial yacht]

can anybody help me with this

[vernal pilot]

start by finding all the angles. that will help a lot

[glacial yacht]

i did

just ac square

i cant get it

[vernal pilot]

well do you know the law of cosines?

[glacial yacht]

yes

[wispy silo]

this shit hits different

[knotty pond]

You can easily get BP as BC given

Apply Sine law in Tr. APB

And find BP

Because we can see cosine at B would involve AC

And only thing stopping us from applying formula is BP

So we just find it

[knotty pond]

Then cosB=.../...

[ornate hemlock]

Urgent I need help Oj my math homework asap!!!

Someone please anyone

😭😭

[glacial yacht]

Ye

Ill

Do wait

Yes

[dark sparrow]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dark sparrow]

why urgent?

[maiden brook]

probably because it’s either not homework or they are asking last minute

[obtuse field]

(Please don't give out answers in this server)

[glacial yacht]

ok

[earnest valve]

How can you calculate the radius?

[upper karma]

I think we need a little more information

[earnest valve]

there are no other information
its a half-circle

[sturdy stone]

Is the 15 cm segment perpendicular to the diameter

[earnest valve]

yes

[sturdy stone]

Ok

Call the center of the circle O, call the intersection of the 15 cm segment and the diameter A and the intersection of the 15 cm segment and the circle B

Consider triangle AOB

[earnest valve]

took a bit long but I could get it thanks

[lyric hedge]

How could I solve where tanx = 1 on the interval 0 to 2pi?

Or in general how could solve where tanx equal anything over some interval

Confused on how I should approach it

[vernal pilot]

In which case find PB using pyhthag or law of cosines, then find PA using law of cosines, find PC using pythag or law of cosines, and AC you can find with law of cosines.

[silent plank]

<@&268886789983436800>

[vernal pilot]

Ty mods

[maiden brook]

yeah there was an aime problem pretty similar to this

[vernal pilot]

I’ve seen a few AIME problems, some of them look fine others look absolutely terrible.

[maiden brook]

absolutely terrible in terms of what

[vernal pilot]

I remember one that gave a complex number with some restrictions. I can’t remember what they exactly.

[vernal pilot]

2023 test 1 q15

[lethal pendant]

1. YES THEY ARE SIMILAR, ABC SIMILAR TO DEF by AA

2. NO THEY ARE NOT SIMILAR

3. THEY ARE SIMILAR BY SAS , TRIANGLE CBD SIMILAR TO EBA

[dark sparrow]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[stuck sleet]

what is the distinction between a bisector of an angle and an angle bisector of a triangle?

[grave crypt]

what?

[stuck sleet]

difference of:

bisector of an angle

and

bisector of a triangle

[grave crypt]

what even is a bisector of a triangle

you mean a line that splits a triangle into 2 other equal triangles?

[stuck sleet]

*angle bisector of a triangle

what I mean

[grave crypt]

both are angle bisectors

bisector of an angle splits that angle into 2 other equal angles

angle bisector of a triangle splits that angle into 2 other equal angles, and has a special property

they are the same tbh

[fickle rose]

same thing

[silent plank]

a triangle doesn't need to exist to have a bisector of an angle

[sonic willow]

Do you know formula for a circle in cartesian coordinates, If yes, then it's easy calculation

Or you can just use pythagorean theorem

Accualy two methods are equal

[maiden brook]

There’s like an IMO question about this

[twin crag]

how do i find this
lemme list my steps below

i converted the expression above into polar form and used de moivres theorem

then i just computed 2(cos(2011*pi/6))

which got me -sqrt(3)

which is marked as wrong

[dark sparrow]

the magnitude is 2^2011, not just 2

[snow plinth]

I don't know where else to post this, but consider a closed, non-empty set C that has a supporting hyperplane at each point of its boundary. Then the set is convex. I'm very new to this, and to prove this I believe the argument goes by contradiction. Assume C is not convex. Then a line segment between p,q in C must leave the set somewhere on the way. In particular, the line segment must go through a point where there's a hyperplane separating p and q. Now, apparently this is a contradiction, but I don't understand what it contradicts?

[twin crag]

oh shoot i forgot about that

[twin crag]

wait how do i evaluate 2^2011

the calculator wont do i

it

[dark sparrow]

yeah cause it's too large

you just leave it as is

except maybe for some simplifications further down the line

@twin crag can you show what the input box looks like

[desert hare]

anyone know how to get a and β?

[sturdy stone]

Inverse trigonometric functions

[somber coyote]

David K.

[somber coyote]

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

[icy pumice]

$\fbox{Improved}$ \
$x = \sqrt{9^2+5^2}$ \
$ x \approx 10.3$
$$$$
So
$$\alpha = sin^-1(\frac{9}{10.3})$$
$$\alpha \approx 60.9^{\circ}$$
And $\beta$ is;
$$\beta \approx 180^{\circ} - (90^{\circ} + 60.9^{\circ})$$
$$\beta \approx 29.9^{\circ}$$

[somber coyote]

David K.

[graceful apex]

I have one

[manic charm]

anyone know all the steps on finding the orthocenter on the graph and just from the points

[nocturne remnant]

what does "supporting hyperplane" mean? if it means the entire set lies on the same side of the boundary, then you have your contradiction as p and q are on different sides of the hyperplane

[shadow fable]

Have you seen this?

https://www.cuemath.com/geometry/orthocenter/

[snow plinth]

I believe that’s the contradiction they meant. Thanks!

[upper karma]

Anyone know ways to practice Postulates?

Like SAS, AAS, ASA, and H-L?

[echo heart]

I need help asp plz yall hear my cries pf help😭😭😭

From 8 to 11

I resend it bc the numbers changed

[dark sparrow]

,rccw

[echo heart]

Um so

No help

😕

[upper karma]

,rccw

Flip

[upper karma]

There

[echo heart]

,rccw

[echo heart]

I dont wanr to flip it i want a answer lamo

[dark sparrow]

!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.

[icy pumice]

!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.

[elder pivot]

why are these wrong

like 1/sin(43)=1.4662791856 which is 1.4663 rounded to four decimal places which is what i typed yet it is wrong

and for the second one we have x=-12 y=9 and r=15
so sine would be 9/15 or 3/5 and cos would be -12/15 or -4/5 and the inverses of those
it told me to simplify so what is wrong

[dark sparrow]

,calc 1/sin(43pi/180)

[somber coyote]

Result:

1.4662791856396

[dark sparrow]

yeah hm strange

all of the answers you put in seem correct to me

@elder pivot i'd imagine this is a bug with the question system

[elder pivot]

okay
i was assuming that because i followed the instructions exactly and compared my work to homework and could not find any errors but i didn’t want to just assume I was right

[tough viper]

How do I simplify this (prove the identify

[timber cargo]

Express everything in terms of sines and cosines

[tough viper]

For sec would cos multiply by the entire denominator

Figured it out

Tysm

[tough viper]

How do I do this

[cunning lion]

if you're trying to prove the identity, i would use the pythagorean identity

[tough viper]

How?

Oh

[jagged wyvern]

#1201677541254119474

Urgent help please

[vocal bison]

whats a calculator ready equation?

[remote lotus]

i think that means u can use a calculator on the problem?

[lethal pendant]

simplify LHS and RHS into Basic form

[tough viper]

Yep I got it already but thx

[lethal pendant]

when you simplify RHS also you get sin^4X/cos^2X

[lethal pendant]

fine

[lethal pendant]

I think you are supposed to use Exterior angle property

4x-4 = 60 + 2x

x = 32

is the answer 32?

[jagged wyvern]

It's 32 yeah i've went over it a while ago

#1201739938954235955

[jolly ravine]

Tell whether the information in the diagram allows you to conclude that point P
lies on the perpendicular bisector of LM
Explain your reasoning.

[gentle haven]

Do you know about the perpendicular bisector theorem and its converse

[obsidian flame]

help idk how to write x-5/x-1 = -3/x on deltamath

nvm

[vernal pilot]

Do you know the Pythagorean identities?

[wheat acorn]

If anyone has seen a question which looks like this (there also info about it above in the question.) Please tell me I think its 11marks with 3 parts.

[lone panther]

real

[vernal pilot]

do you have any more context?

like, what type of exam? does it have precalc on it (i assume so)? what else is going to be on it? circle theorems? triangle stuff?

also this isnt a place for solutions, although i dont think it breaks any rules

[wheat acorn]

I think that it's a-level. I think it was based on chords and tangents relating to circle's. I don't need the solution just the question.

Each of the lines were tangents of the circle but I'm not sure on much else just that there is a paragraph of info about it before the question but I'm not sure what info was giving. I think the center point of the circle was labeled A but can't be 100% certain

[potent fulcrum]

I apparently got this wrong can someone help ?

[noble lava]

Need help with this one

Nvm

[stiff moon]

I need help with this entire page. I dont understand

[wheat basin]

Use the basic proportionality theorem

[upper karma]

Does anyone have old geometry notes/hw they don't use anymore

I'm self studying geometry and just writing notes and khan academy doesn't feel like enough

[vocal bison]

what is angle of depression and angle of elevation? 😭😭 I’m confused

[dark sparrow]

it's the angle between the horizontal and your line of sight when you are looking at an object

if you're looking up, it's the angle of elevation; down, and it's the angle of depression

does that answer your question? @vocal bison

[vast spruce]

hi I have a simple qestion

Nvm

[vocal bison]

yes thanks 😄

[snow plinth]

I'm reading through a proof in convex analysis that every closed set in R^n with the nearest point property is convex. The proof goes by contradiction and assumes the set in question, which I denote A, to not be convex. They claim there exists a line segment between two points x,y whose intersection with A only contains the endpoints (see picture). Then they claim there exists a ball centered at (x+y)/2 with a radius small enough whose disjoint from A. I don't quite understand where (x+y)/2 in my sketch would be located. Depending on where I choose the origin, might not (x+y)/2 actually be in the set A?

[dark sparrow]

(x+y)/2 is halfway between x and y

[snow plinth]

yeah, you're right 😄 thank you

[grave tusk]

are you guys familiar w always sometimes or never questions

i need help on those w like quadrilaterals

[fickle rose]

yes

[fickle rose]

to get good at them, you have to know the properties of each type of quadrilateral

[grave tusk]

okay

how is a rectangle always a square but a square is not always a rectsngle

it it because a square has one characteristic of a rectangle but a rectangle doesnt have all the characteristics of a squsre

[grave tusk]

oh my god

[jagged wyvern]

Try using midsegment theorem for 24-26

[jagged wyvern]

Can someone find a worksheet that uses exterior angle theorem?

Cant seem to find any worksheet

[muted sorrel]

What is the exterior angle theorem

[jagged wyvern]

measure of an angle is equal to the sum of 2 interior angles

[muted sorrel]

ooooh

well what do u wanna know

abt those

like the angles related to what

[jagged wyvern]

The logic of the angles adding up to 180 degrees

Triangles

[muted sorrel]

hmmm

like u wanna know why all angles of the triangle add up to 180?

dayum i never asked myself that

[jagged wyvern]

No

Nvm

[muted sorrel]

WAIT THATS A GREAT QUESTION

ill go look it up

ooh this is interesting

[jagged wyvern]

Can I ask what grade youre in

[muted sorrel]

well

according to the American system

i think i'd be grade 11 rn

but I dont live in the US so we study some different thingies

[jagged wyvern]

you should know exterior angle theorem though...

some basic stuff before moving into trignometry

[muted sorrel]

well I alrdy did trigonometry

I still haven't rly dived into its relation with complex numbers

i think only the polar form covers that

[maiden brook]

it is commonly assumed that the converse of a statement must be true if the original is

but u needa prove the converse separately

for example

[muted sorrel]

my brain 😍

[maiden brook]

If i live in California, I live in America obviously not the same as If I live in America I live in California

[muted sorrel]

ooooh true

wait

but wouldnt that mean a square is always a rectangle but a rectangle isnt always a square?

[maiden brook]

yes and thats correct

[muted sorrel]

so it's basically sum of exterior angles adds up to 180?

[muted sorrel]

wait

both our statements are correct?

[jagged wyvern]

pretty much

[maiden brook]

no ur statement is correctly

[muted sorrel]

yayy

i mean i knew that

but i didnt know it had a "theorem" for it

[maiden brook]

squares have four right angles, so it is always a rectangle, but a rectangle doesnt necessarily have four equal sides so its not necessarily a square

[muted sorrel]

oooooooh true

[opaque frigate]

Hello! I am working with Triangles.
I need to find length X,

Off the bat I'm seeing that $Tan(36) = 12/x$
Therefor $x = 12/tan(36)$

The website I'm working with doesn't like that answer.
are there special angles associated with a 36 degree angle that I am missing to find an exact answer?

Going through the long way and finding $Sin(36) =12/B$ is $12/sin(36) = B$
Then $Cos(36) = x/(12/sin(36)$ becoming $12Cos(36)/Sin(36)$

[somber coyote]

Jimmy'

[dark sparrow]

can you show what the answer box looks like? @opaque frigate

[dark sparrow]

maybe you're expected to give a decimal approximation instead...?

did it say "round your answer to 3 decimal places" or sth

[opaque frigate]

I don't see that anywhere, no.
I will try inputting the decimal

[dark sparrow]

like this is not about the math anymore

it's about input format requirements

[opaque frigate]

Okay, I found my issue.
I had tried inputting the decimal my calculator gave me, and it hadn't worked.

I googled a right triangle calculator, inputting my numbers, and it gave a different decimal. and that one work.

I had my Calc set to radians, not degrees 😆 😭

[dark sparrow]

cause 12/tan(36°) is correct mathematically

oh.

[opaque frigate]

I'm going to go hide in a hole now, thank you.

🫡

[jagged venture]

Let △ABC, M ∈ (AB),
AM/AB = 2/7
Knowing that MN ∥ BC, NP ∥ AB, P Q ∥ AC, QR ∥ BC,
RS ∥ AB, N, R ∈ (AC), P, S ∈ (BC), Q ∈ (AB).
a) Calculate AN/NC, CP/PB, BQ/AQ.
b) Show that AM = BQ.
c) Show that SM ∥ AC

[bold kindle]

hi can you take a picture a little more clear for that I understand better ?

[split pelican]

hello

[heady fulcrum]

somebody help me

@everybody

@peak fossil

@twin delta

@final portal

[sturdy stone]

Nobody can help you if you don't post the really hard maths problem

[heady fulcrum]

lol sorry

sin theta = m^2 + 2mn/m^2 +2mn + 2n^2

[twin delta]

can you not

[heady fulcrum]

and tan theta= m^2+2mn/2mn+2n^2

prove it

and the second one is, if cos theta - sin theta = underrot 2 sin theta

and cos theta +sin theta= underoot 2 cos theta

prove it

@sturdy stone

and dead

@twin delta

@peak fossil

[lime crown]

Please do not ping individual helpers unprompted.

[heady fulcrum]

got it

ciao

[ember grove]

damn i have an admission test for imperial and i have to revise a lot of stuff o.o panic

how many trig identity there are ???

[silent plank]

check the pinned image in the channel

[bronze wind]

how can I find #21

[flint ravine]

have you learned Pythagorean theorem yet

[bronze wind]

I thought that was for area

im struggling on perimeter

[flint ravine]

no its for the hypotenuse or the hypotenuse or the other side you don't have

[bronze wind]

oh ok

that makes sense

[flint ravine]

so your good to finnesh it?

[bronze wind]

yes

[flint ravine]

alright great

[grave tusk]

guys

does anyone have tips for geo like how to study

i genuinly think i got a 42 on my midterm so i acc need to lock in

but idk how

[flint ravine]

khan academy 100% just in general

[grave tusk]

Ok

is there anything else

[flint ravine]

if you jump strait to the unit test you can see what you need to learn still and go there

[bronze wind]

@flint ravine can you also help me with this real quick?:

A rectangle of 360 square yards is 10 times as long as it is wide. Find its leg and width

[flint ravine]

yeah one sec

360 yards a in area?

[bronze wind]

?

[upper karma]

Hello guys

Can anyone explain why cos(90-x) = sin x and cos(90 + x) = sin x? I'm really struggling to understand it.

[dark sparrow]

write out what sides' ratio is cos(90°-x)

observe that it is the ratio of the same sides as for sin(x)

cos(90°+x) = **-**sin(x) though, and that takes the unit circle to explain.

or angle-sum identities maybe.

[merry oar]

do uk cos(A+B)

[upper karma]

I got the first part about cos(90-x), i really appreciate your answer, but what about cos(90+x)? You said it takes unit circle to explain, i am familiar with unit circle and basic concepts, can you explain why cos(90+x) = -sin(x)

[dark sparrow]

are you familiar with cos(x+y) = cos(x)cos(y) - sin(x)sin(y)

[upper karma]

Yes, i am

[deep token]

made a chicken on a graphing calculator

oops the second image is zoomed in at 1st glance

[terse hull]

Big triangle is equilateral triangle. Little triangle is equilateral triangle. Area of little triangle is 1. Need to find the area of the grey triangle. Can someone give me an idea how to deal with it?

I thought about Pick's theorem, but as I remember it works only if your plane is cut into the squares

and here we have equilateral triangles

probably it still can work, but I dont know how to use it here

[dark sparrow]

you can calculate the areas of AXC, AYB and BCZ and subtract them from that of XYZ

[terse hull]

oh, I probably see

but

how to deat with this guys?

I mean

Maybe i should find the sides?

[terse hull]

I dont understand how should I do this

[dark sparrow]

AXC is a single small triangle in which one side has been stretched by 5 and the other by 3

[terse hull]

that makes some sence

[upper karma]

woah

[terse hull]

Now your explanation makes much more sense. I took a rest, went for a walk and now I solved it! Thank you sooo much!

[upper karma]

can someone please tell me what this means

idk how to set this bloody question up

i spent 3 hours trying this one question

because i can't wrap my head around how to draw it

could someone please help me draw it?

[merry mulch]

could you share with us how you did it?

[terse hull]

yeah

[merry mulch]

how did you find the area of those smaller triangles?

[terse hull]

they are given

[terse hull]

Area of little triangle is 1

[merry mulch]

i meant axc,aby and bcz

[terse hull]

ah

You need to find the side of small triangle using $$S=\frac{\sqrt{3}}{4}a^2$$

[somber coyote]

Umatriz

[terse hull]

The formula for the area of the triangle is $$S=\frac{1}{2}ab*sin60\degree$$

[somber coyote]

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

[merry mulch]

oh! i didn't knew this..

[terse hull]

Its for this case

[merry mulch]

i see.

[terse hull]

https://en.wikipedia.org/wiki/Area_of_a_triangle#Using_trigonometry

and then we can find everything

[merry mulch]

thanks for your time.

[cerulean turtle]

Help

[dark sparrow]

ok i guess this is a better place

same question: have you made any progress so far? @cerulean turtle

!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

[dark sparrow]

ok so your handwriting just gets worse from line to line there

but anyway i think this is not a good route overall

better would probably be to write $\csc(\theta) + \cos(\theta)$ as $\frac{\cos(\theta)\sin(\theta) + 1}{\sin(\theta)}$, and then write out what $\frac{x^2+1}{x^2-1}$ will be based on that, and hope it simplifies down to $\frac{1}{\cos(\theta)}$

[somber coyote]

アンナ

[cerulean turtle]

I think i already tried this one

Its writing csc as 1/sin and taking lcm

[cerulean turtle]

[dark sparrow]

that is the first step of my suggestion

did you follow it through any further?

i mean, i can't force you to do it. if you really hate my suggestion, feel free to throw it into the trash.

[cerulean turtle]

Lemme try

[regal ferry]

Can someone help me with this: 2(tanx-1) + 3 = tanx + 1 + √3

This is probably easy for y'all but I suck at maths

[dark sparrow]

!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

[cerulean turtle]

1

[dark sparrow]

was asking sausageman.

you had gone off to try my suggestion, presumably.

and by the way, it looks like the problem asked you to prove a false statement.

(x^2 + 1)/(x^2 - 1) doesn't actually simplify down to 1/cos(θ)

@regal ferry you still there?

[regal ferry]

Ooo sorry I ended up opening a help channel 😭

[dark sparrow]

k

[regal ferry]

I've reached tanx=√3

[dark sparrow]

ok, and can you tell what special angle has tan equal to sqrt(3)?

yes, except it doesn't.

[regal ferry]

60, or π/3. So after that I draw the circle and notice that every 180º I get the same result of √3 so it means the final result is π/3 + kπ

[dark sparrow]

yes correct

@cerulean turtle i honestly do not understand why you are acting so hostile to me right now. it is as if i said something that is so deeply offensive it cuts to your v.ery soul...

[regal ferry]

Hip hip hurray! (I'm 17 and I've been doing trig for months why do I suck so bad )

[cerulean turtle]

Since when did I become hostile huhu🥹

[dark sparrow]

your barrage of in response and reaction to my messages.

that's how it comes across. as hostility.

[cerulean turtle]

Oh I'm sorry but it's a sad emoji right?

Tysm tho for helping

[dark sparrow]

is not a "sad" emoji

it's a "you just said some bullshit and i'm pointing that out to you" emoji.

[cerulean turtle]

I'm so sorry, I'm just too dumb for this world

[naive dagger]

How do I calculate the abc triangle, I figured out the bottom but I don’t k ow how to continue

[agile lintel]

do some angle chasing

[naive dagger]

whats that

i dont think ive learned that yet

[agile lintel]

its not that technical

you just label one angle x

and then try to figure out what the other angles are in terms of x

my tip is to label angleCAB x, then you can deduce what angle CBA is, and also the size of the two angles at C

then you will (hopefully) see something important about the triangles in the diagram

@naive dagger have you got it yet?

[naive dagger]

yeah ty

or i got 1745cm2 but the answer was 1733cm2 but thats probably just bcz of the way i rounded it up

[jagged wyvern]

Someone help please I need to review for a chapter test tmmrow 🙏 #1202775077729345586

[vocal bison]

ion get dis 😭

Ping pls

[covert cape]

how do i solve this?

[sudden shell]

How can I prove that RA is 6 root 2

[azure echo]

tan(kx) tan(kT) = 0, where T is the fundamental period.
why does this become
tan(kT) = 0, can't tan(kx) = 0 ?

[ocean lava]

Help

Im so confused as to how to proceed from here

[waxen moss]

Is it possible to easily find the full area when u know some of the area?

[dark sparrow]

,rccw

[dark sparrow]

@waxen moss i don't see what your question has to do w/ the problem

they give you the area of the full circle

[waxen moss]

Yea I just asked Someone to translate and realized

[dark sparrow]

translate into what language?

[cerulean turtle]

Help

[bright meteor]

angle over 360 eneh

pie D

[tough galleon]

I dont understand how to do this question

in this book before this ques there weren't really trig equations

and this is asking no. of rroots

[dark sparrow]

can you show part a of the same question?

maybe it sheds some light

@tough galleon

[muted sorrel]

Oooh I saw this on YouTube

I still remember one answer was the golden ratio

Idk if there were two or only one

[agile lintel]

the fraction a circular sector takes of its circle is the angle is creates/360

[rose karma]

hello, i will keep it short and simple my 10th grade starts in march and my finals end on 19th. My main goal this year is to complete grade 11th trigonometry, any suggestions how should I proceed with starting from some basics, and what all topics can I cover of grade 11th which are similar to grade 10th??

[tough galleon]

oh ye btw i didnt realize before

it asked to sketch both graphs

and then i realized the equation is basically the 2 functions to be sketched

so thats where we have to fund the intersections and tell the roots

[weak mauve]

why is mathway saying it isn't true?

theres my working

[sacred loom]

Because mathway isn't 100% perfect all the time

[weak mauve]

wait nvmd, the question asked me to prove that the 1-tan^2(x) thing is equal to 1-2sin^2(x) not what I have now

[pine bluff]

guys how to find sin cos and tan and why do i need it to calculate something

[sacred loom]

Althought that doesn't seem to be correct

[weak mauve]

yeah my bad

put 2 in front of sin

[sacred loom]

Ah

Then yeah that works

[weak mauve]

how do you prove this?

I got up to this stage

[sacred loom]

you may be able to split the 1-tan(x)^2 into (1-tan(x))(1+tan(x)) by factoring

but idk if that will do anything

[weak mauve]

oh yeah then it will cancel out

[weak mauve]

nope still does the same thing

[dark sparrow]

$\frac{1-\tan^2(x)}{1/\cos^2(x)} = \paren{1 - \tan^2(x)} \cdot \cos^2(x) \neq 1 - \tan^2(x) \cos^2(x)$

[somber coyote]

アンナ

[dark sparrow]

you didn't rewrite the fraction properly

[weak mauve]

wdym?

isn't sec^2(x) same as (1/cos^2(x))??

[dark sparrow]

i mean exactly what i said. i just told you where you went wrong and how to fix it.

you rewrote $\frac{1 - \tan^2(x)}{1/\cos^2(x)}$ as $1 - \paren{\tan^2(x) \cos^2(x)}$, which is incorrect.

[somber coyote]

アンナ

[sacred loom]

Dividing by secant is just like multiplying by cosine

[weak mauve]

oh yeah I didn't multiply 1 by cos

[dark sparrow]

yes that was your mistake

as i said twice now :p

[weak mauve]

but then how would I proceed to turn this:

into

that

[sacred loom]

use a trig identity on the cosine squared

[weak mauve]

omg

[sacred loom]

it'll turn into 1 - sin squared

[weak mauve]

thank you!!!

[sacred loom]

yuh

[dark sparrow]

you can and should use the symbol ^ for exponents.

so sin^2(x)

etc.

[sacred loom]

ok.

but I got my idea across without rude about it

[hollow barn]

What am I meant to do with tan(x) = 4/3

[waxen moss]

Indonesia

[dark sparrow]

good question, what are you meant to do with it?

!xy

[lime crown]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[hollow barn]

Find x using unit circle theorem

[dark sparrow]

using what

[hollow barn]

Radians

[dark sparrow]

...

what does "unit circle theorem" mean?

[dark sparrow]

!xy

[lime crown]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[silent plank]

!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

[twin crag]

well

i dont really know where to begin

my only idea is

arcsin(-0.7)=theta

[shrewd kiln]

well it says find the sum of all the possible angle values right

so arcsin of -.7 is uh

-0.775397497 radians

which is -44.427004 degrees

and since sin can only be negative in quadrants 3 and 4

we hv to find angles in those quadrants with reference angles of 44.427 degrees

so i got 315.57

and 224.43

but it says in radians so that would be

5.50773552 + 3.91704244 = sum of values of theta

um so i got 9.42477796 radians lmk if that works

[proper ice]

Hello

idk if this have somthing in relation with geometry

but i think yes

It´s about phi

can smebody help me with this?

[silent plank]

what does this have anything to do with phi

what are you being asked to do

[proper ice]

If i remember right, this have relation with the "Divine propotion"

(srry if my english isn't good, i speak spanish)

[silent plank]

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

[obsidian harness]

So (L - 2) / L = 2 / (L - 2)

[dark sparrow]

so you do not even remember your own problem.

[twin crag]

ty

it worked

[stuck ore]

Guysss

I need help

[dark sparrow]

with what

[stuck ore]

How to do this

[dark sparrow]

have you made a diagram?

[stuck ore]

I have but i dont sure it’s right or not

[dark sparrow]

well you will have to show your diagram to us

[stuck ore]

Oh kekw

[twin crag]

i'm supposed to find the equation for this graph in the form y=acos(bx-c)
i got y=3cos(4x-c)

and i'm supposed to find the smallest value of c, how do i do this?

[halcyon crane]

I suppose you can start by taking some points of the graph and resolving the equation y=3cos(4x-c) to find some values of c?

not sure tho

[twin crag]

oh yeah that works ty
c=pi/2

wait do u know how to convert this into an integer

i dont think i can put pi/2 for c

[halcyon crane]

but try for different x

[halcyon crane]

why couldn't you?

[twin crag]

i still get pi/2 for each value of x where y=0

[twin crag]

oh nvm you can
i was just assuming c needs to be an integer

[twin crag]

wait but the amplitude and period of the function are both correct right

[halcyon crane]

that I don't know

the amplitude yes

the period I don't know

but you can just verify on your calculator and see if you're right or not

[twin crag]

wait wdym

oh i graph it

yeah its correct

[proper ice]

Yeah, ,ty i solved it yesterday in the night

it´s L=3±√ 5

haha

[vocal bison]

Help pls idk how to set this up

[gentle haven]

Basically it's two different triangles, from the boats, to the bottom of the lighthouse, and to the top of the lighthouse.

You know one angle is 90 degrees

And you know the angles from the top of the lighthouse to the boats for each triangle

And you know the height of the skyscraper

Basically you know two angles and the included side for both triangles

Use trigonometry to find the distance from the boats to the light house.

From there you should be able to solve it.

[vocal bison]

oh okay thanks

[hybrid cosmos]

can someone please help me derive those formulas?

from the basic trig identities
just to make sure I don't forget them on the exam

[halcyon crane]

you want to derivate the quotients with tan?

and see if they're equal to -sin and cos?

[hybrid cosmos]

well, I want to be able to derive them from scratch

[halcyon crane]

so

[hybrid cosmos]

and not just check if those equalities are equal or not (which they obv are)

[halcyon crane]

what's the formula for the derivative of u/v?

let's start by here

[hybrid cosmos]

= (u • 1/v)'

xD

[halcyon crane]

but what's the formula haha

[hybrid cosmos]

(vu' - uv')/v^2
probably this way

[halcyon crane]

yep

so now

what are u and v here?

[hybrid cosmos]

hm, tbh I don't see where u are heading

why would u want to differentiate those fractions

or u want to prove them with calc? 🙂

[halcyon crane]

to derivate the quotient

isn't that what you wanted to do?

[hybrid cosmos]

like to show the derivative is zero?

[halcyon crane]

no

to show the derivative is -sin and the other cos

I thought that's what you wanted to do

to show the quotients were equal to cos and sin

[hybrid cosmos]

yes, but why would u need derivatives for that 🙂

[halcyon crane]

hmmm it wouldn't even be a good way of thinking

[halcyon crane]

what did you want to do?

[hybrid cosmos]

I want to derive those formulas, not differentiate them 🙂

[halcyon crane]

what's the difference?

I don't know it in english lol

[hybrid cosmos]

rediscover two of those formulas only using trigonometry

[halcyon crane]

only using trig???

is it even possible🤣

[vernal pilot]

Use trig identities.

First decide which side you want to manipulate.

(I would chose the RHS).

[upper karma]

do you guys think csc(x) and sec(x) are unnecessary to know and should just be written as 1/sin(x) or 1/cos(x)

Knowing that sin alpha = - (4/5) , 3pi / 2 < a < 2 pi
a) represent the angle and the values of sin alpha, cos alpha, tan alpha and the circumpherence
b) calculate the incognito values from the other 2 functions

??? I need help with second b question

[upper karma]

i would say it is necessary if you're talking about the fundamental ratios in a triangle

[upper karma]

is incognito here meaning undefinied values?

[upper karma]

no im just talking about notation

[upper karma]

in terms of notation, then no, it doesn't matter

[upper karma]

I don't understand b question

[upper karma]

$\sin{\alpha} = - \frac{4}{5}$\$\frac{3\pi}{2}< \alpha < 2 \pi$

[somber coyote]

ursmomsfavnotification

[upper karma]

let me help, one sec

[upper karma]

Knowing that sin alpha = - (4/5) , 3pi / 2 < a(lpha) < 2 pi
a) represent the angle and the values of sin alpha, cos alpha, tan alpha and the circumpherence
b) calculate the incognito values from the other 2 functions

[upper karma]

im sort of uncertain with what it means by incognito values though

[upper karma]

I think incognito mean undefined values now that I think about it

that's what i was assuming

[upper karma]

do you think you can help me?

[upper karma]

sorry im back

[upper karma]

okayy

but i believe it might involve finding the range of alpha values

hii @spare haven, do you think you can help me?

[upper karma]

ohh okay, any more tips or no

[upper karma]

not sure because the question is a little vague

[upper karma]

okayy, if you think of anything new that might help me, it would be really appreciated!

[upper karma]

just ping me if you think of any new tips yk

[upper karma]

i will, sorry about that

[spare haven]

here's an example involving sine

you can do the same thing with cosine, best to transform it using cos 2x = 2cos² x - 1

[upper karma]

it's okayy, thanks for trying to help anyways

[spare haven]

sorry, nope, i don't know what b) means either

[upper karma]

lol it's okay

[spare haven]

fyi it's seen as bad form in this server to ping people for help

outside of rules in #❓how-to-get-help i mean

[upper karma]

ohh my bad

[wispy mantle]

Show that the parabola y²=4x and the circle 4x²+4y²-25+y+3=0 have a common point of contact. Find the tangent at this point

Help....

[grave tusk]

guys help plesse idk ehere to start

[lime dune]

wrong channel

[grave tusk]

this is literally geo dummy

[lime dune]

how is that geo

oh truth and logic stuff i guesssss?

but it’s not explicitly geo, might be better to put it in #discrete-math

[grave tusk]

ok

[dim plank]

!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

[random fern]

analytic geometry

[lime dune]

that is not analytic geo 😭😭😭 that is just propositional logic

“analytic” geometry is coordinate geometry, I don’t see any coordinates in use there

[cunning lion]

it's not geometry, but geometry class is when formal logic is often taught in high schools

[lime dune]

well said high schools seem way better than mine lmao

we only discussed the most basic case of single direction conditionals

[random fern]

its like the equations

on a graph

oh nvm

[torpid egret]

Hey y’all can anyone help me with this problem? I need to calculate the volume. But i don’t understand the ratio stuff

[vernal pilot]

i dont understand the drawing

[lethal needle]

is that even 2d or 3d

[runic pasture]

3d it's about a ramp

[lethal needle]

oof

[runic pasture]

it's basically asking for the volume (cubic meters) of the ramp

i am not pretty sure about the ratio but i assume 1:4 refers to the lengths of the rectangle for example

the rectangle's width is 4.20 meters them the length should be four times as much

[naive geyser]

I was doing pair of straight lines from analytic geometry.
Is it possible to be abc+2fgh-af^2-bg^2-ch^2 =0
And h^2 -ab<0 in General equation of second degree?

[toxic crescent]

hi

if a math questions states that vectors C = A + B, and that vector C is in the -y direction with a magnitude of 12. Does that mean that vector C is pointing somewhere in quadrants III / IV ? Or does it mean that vector C is pointing straight down -y and x = 0?

[plush garden]

Hi

Can someone help?

I need to find the value of Y

is for tomorrow and I just don't undestand

[pliant nest]

Is there a way to solve this without graphing? A shortcut like how

• If the Line of Reflection = x-axis, just change the sign of the Point's y-value
• If the Line of Reflection = y-axis, just change the sign of the Point's x-value

I feel like there is, I have a strong feeling that there is, I just can't figure it out rn

[lime dune]

translate everything so the line of reflection is one of the axes

flip sign

then translate back

so for the above you translate everything up 1

so now you're reflecting R'(-3,-4), N'(-4,1), V'(-2,0), E'(0,-3) over the line y=0

reflect

then translate the points back down 1

in fact, you can sorta generalize this process to reflecting over any line, not just the horizontal/vertical ones

[cinder oak]

can someone help me with geometry proofs pls

i have the stesps i just need the correct order

[maiden brook]

it is not geometry but ig

[lime dune]

it might instead help to write the whole thing as a “flowchart”

(two column proofs are fucking stupid, nobody actually writes proofs like that)

[upper karma]

welcome to geometry and trigonometry ( 💀 )

[frozen ocean]

<@&268886789983436800>

[obtuse field]

Sorry, Discord only allows users who are at least 13

[bronze iron]

not me in greometry

ughhh

[lucid thicket]

Please help me guys

[lime dune]

can you make another right triangle that’ll be useful?

[lucid thicket]

Wdym?

[lime dune]

look at the diagram

are there any lines you can draw in to form a useful right triangle?

[obsidian harness]

Instead of a radius of 4 cm, try thinking of the diameter of 8 cm

As it touches the ground

[lucid thicket]

15 + 8 to get a hypotenuse of 17

But I’m not too sure what do do after that

[lime dune]

you already have all the information you need

draw a right triangle in the diagram that includes the lengths you know

[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

[dark sparrow]

well you status'd yourself

so what is your status?

[hybrid shoal]

hi

am lil concerned about trignometery