geometry-and-trigonometry

[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.

[worthy eagle]

You asked the same question in help forum, did it get deleted somehow?

[wary edge]

Closed it because i found that i can put it here instead

[worthy eagle]

Oh

You don't need to find slope.

x/c + y/d = 1 is equation of the line in terms of x-intercept c and y-intercept d.

So in this case it becomes x/(-2) + y/4 = 1

y/4-x/2=1

Coordinates of points A and B are given, just put the value of x and y according to their coordinates and solve for a and b.

[wary edge]

It's okay i figured it out already thanks anyway

❤️

[fierce orchid]

just use the intercept form

[remote sigil]

Bro can someone help me with my geometry test

[fierce orchid]

uhh show

[remote sigil]

Look in “help-6”

[silent plank]

the test itself, no
study for it yes

[remote sigil]

Yeah I meant this

[twin escarp]

can anyone pls help with this

[dark sparrow]

do you still need help with this?

[twin escarp]

yes please i have no clue how to solve it

[dark sparrow]

ok let's see

can you tell me what kind of segment PQ is in relation to circle C on the diagram?

looking for a word here.

[twin escarp]

radius i think

[dark sparrow]

"you think"? why the uncertainty?

[twin escarp]

most likely radius

[dark sparrow]

...ok let's not

yes, it is the radius, and there is no reason whatsoever to be doubting that. ok?

anyway

now that that's out of the way,

what can we say about PQ and l?

a tangent to the circle and a radius to the point of contact are...?

(again, looking for a word here).

[fringe parcel]

Gradient of line PQ is 1/2 since the product of gradients of perpendicular line is -1.

[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]

i was gonna guide him to that.

and have him get to it himself, since @twin escarp said he had no clue about this problem.

[twin escarp]

oh yeah the gradient perpendicular of the line is 1/2

but the main issue is using the constant k to form a quadratic equation

[dark sparrow]

mixed there some up words

but ok, so we know that the gradient of line PQ is 1/2

[twin escarp]

yes

[dark sparrow]

can you write down the gradient of line PQ in terms of k, using the coordinates of P and Q? @twin escarp

[fringe parcel]

@dark sparrow Yes

[dark sparrow]

@fringe parcel may i politely ask you not to interrupt me here btw? i would like to continue helping OP myself.

[twin escarp]

yes ill try that

[dark sparrow]

ok, do it and ping me once you have a result.

do not simplify your equation until instructed!

[twin escarp]

This is what I got so far is it any good?

[dark sparrow]

the was you handwrite your k it almost doesn't look like a k

aside from that it is correct. but factorizing k^2 - 2k to k(k-2) was useless.

[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]

2nd time now

@twin escarp now simplify your equation.

[twin escarp]

Here

[dark sparrow]

ok good

[fringe parcel]

May I ask helping students by solving questions is it a crime?

[twin escarp]

so for b it says find a equation for c

[dark sparrow]

but do write your k's a bit neater so that you don't run into an issue with k being misread as 12.

[dark sparrow]

it is not a crime but it is against server policy.

and i have reminded you of it, now twice

[twin escarp]

ok

[dark sparrow]

yes

capital C though not lowercase c

[twin escarp]

yes captial C

[dark sparrow]

k is positive and k^2 - 3k - 10 = 0.

find the value of k.

do you see how to do that?

[twin escarp]

Value of k is 5

[dark sparrow]

ok good

do you see how to continue now? or not yet?

(both answers are okay.)

[twin escarp]

no not yet

[dark sparrow]

now that you know k = 5, you can work out the actual coordinates of P and Q.

do that.

[twin escarp]

Here’s what I got

[dark sparrow]

you meant Q(3,15) surely

[twin escarp]

no for y on p its actually 13

[dark sparrow]

that is a three?

you would benefit from working on your handwriting!

oh, nevermind, i was looking in the wrong place.

this is a very bad k though

anyway

P is (-1, 13) and Q is (3, 15).

[twin escarp]

yes

[dark sparrow]

P is the center of your circle and PQ is its radius.

do you see how to write down the equation of the circle?

[twin escarp]

x^2 y^2 ?

[dark sparrow]

$x^2y^2$ is not an equation.

[somber coyote]

Ann

[dark sparrow]

you're missing a lot from it, for sure.

[twin escarp]

x^2 + y^2 i meant

[dark sparrow]

x^2 + y^2 still isn't an equation.

and it is still missing a lot of info.

can you state the general form for the equation of a circle?

[twin escarp]

wait is it actually (x-a)^2 + (y-b)^2 = r^2

[upper karma]

Can anyone help me

[dark sparrow]

that's the general one, yes.

[dark sparrow]

you might wanna open a help channel or help-forum thread for this rn

[dark sparrow]

can you tell me what the letters a, b and r stand for here

[twin escarp]

a and b are the centre of the circle r is the radius

[dark sparrow]

a and b are the coordinates of the center, and r is the radius.

ok, so let's start with the last of these.

what is the radius of your circle?

[fringe parcel]

@upper karma First draw the line y=x-1

[twin escarp]

i dont really know?

[dark sparrow]

P is (-1, 13) and Q is (3, 15).

the radius is the distance between these two points.

do you know how to find the distance between two points?

yes or no

[twin escarp]

yes do you use y^2-y^1 over x^2-x^1

[dark sparrow]

no.

it should be _ not ^, but also what you wrote is the formula for the gradient.

while i want you to find the distance.

[twin escarp]

ok give me a sec

Is it 2 root 5?

[dark sparrow]

ok, yes, it is 2 sqrt(5).

that is your radius.

now,

the center of your circle is P.
the radius is 2 sqrt(5).

write down the equation of C by filling in this template:

(x-a)^2 + (y-b)^2 = r^2

where (a,b) are the coordinates of P
and r is the radius

[twin escarp]

That’s what I got

[dark sparrow]

$(2\sqrt{5})^2$ not just $2\sqrt{5}$.

[somber coyote]

Ann

[dark sparrow]

also remove the C=.

you can replace the equals sign in C= with a colon so it reads as C : if you want though.

[twin escarp]

oh right so the radius 2 root 5 needs to be 20 instead

thanks for your help

[upper karma]

What's that

[fringe parcel]

After drawing the line you draw perpendicular bisectors from each point of the triangle to obtain the new triangle.

[visual flume]

!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

[visual flume]

4

i got parallelogram and rhombus

and is this one all of the above

[warm tendon]

ignore the two 5m on the sides of the bottom triangle, the 5m should be in the middle perependicular to AC

can someone help me with this 3d word problem

[worthy eagle]

Use Pythagoras on ABG and BCG

@warm tendon

Ayo your figure also wrong G perpendicular to AC 5 m away, AG=5m
Using Pythagoras
5⁵=5⁵+x²
x=0
On left and on right, x=0
This contradicts that AB = 2x = 20m

[warm tendon]

the 5m on both sides is wrong

the 5m in the middle is the correct one

yeah the figure is wrong cause i took it from someone else

so i said ignore the two 5 metres on both sides

[fringe parcel]

Guys this is my thinking about the problem,I stand to be corrected.

[worthy eagle]

Oh sorry I didn't see the text with image

[worthy eagle]

This is wrong

h+5 ≠ perpendicular

It was a 3d figure not a 2d

@warm tendon

[past crag]

!noping

[lime crown]

Please do not ping individual helpers unprompted.

[pallid dragon]

ABCD is kite , |BC| = |DC| what is the value of x? I need help please

[obsidian harness]

#help-43 message - seems to be resolved

^

[pallid dragon]

maybe like this but i should ask this my teacher

[summer cradle]

how do you get those values?

[minor crow]

Thank you for the explanation, but how did you get from the line highlighted in green to the next, and from the line in red to the next?

[worthy eagle]

Green to Red: squared on both sides, on RHS, used the identity (A-B)²=A²+B²-2AB

[minor crow]

Oh yeah, I see that now. Thank you for the help. 🙂

And then the next line you just moved everything non x related to the other side

[worthy eagle]

Red to blue: "-x²" cancels on both sides,
875=400+1200-40√...
875-400-1200=-40√...
-725=-40√...
725/40 = √...

Yeah

[minor crow]

Yea thank you so much

[worthy eagle]

[minor crow]

[warm tendon]

Thanks for the explanation

[viral nimbus]

Can anyone help me w solving circle theroms

[fringe parcel]

To find Angle BAD we can use the fact that the angle subtended at the circumference is half the angle subtended at the center.

To find Angle DCB we can use the fact that opposite angles in a cyclic quadrilateral add up to 180°

[honest oxide]

Can someone help me with 3 and 4

3 and 5*

[cunning current]

whats the best way to memorize formulas? On my exam im going to have like half of them but I have to memorize alot which I only originally memorized cause I did it right before the test

[honest oxide]

think of words that go with it

[feral wind]

I was gonna make a cool looking shape and i been thwarted D:

from regular octahedron

[mental canopy]

im not sure how they got to the result 145.5mA

they're doing l1-l2 from something but im not sure what

[fringe parcel]

Add the 2 phasors as follows: (100cos(45) +50cos( 15)) +i(100sin(45) +50sin(15)) and finally find the magnitude and the phase.

[mental canopy]

we didn't learn imaginary numbers for this

[fringe parcel]

You resolve 100 <45°into x and y components as x component=100cos 45 and y component =100sin 45. You do the same to 50<15°. Then add x components and also add y components.To find phase use tan^-1(y/x), to find magnitude use √( x^2 +y^2).

[mental canopy]

oh i see

ok thank you

[green nest]

can anyone help me solve this? i have the entire height of the cone which is 6, theres a sircle cut inside which has an area of 225pi and the entire volume of the cone is 2800pi, i need to find the radius of the bottom and too of the cone, i tried a few ways and nothing seems to work.

[fringe parcel]

@green nest Use the formula V=1/3 πr^2 h

[green nest]

what does ^ mean?

times?

also mb i explained it badly

its a cut cone?

idk what the english word is but it has a radius on the top too

[fringe parcel]

@green nest r^2 means r raised to the power of 2

[pale sentinel]

exponentiation

[fringe parcel]

Does it mean it's a frustum of a cone.

[green nest]

yes yes

but theres another line exactly in the middle of the cone with a radius of 15

hold on i can draw it properly

[fringe parcel]

Paste the original question here please.

[green nest]

alright

The height of a truncated cone is 12cm, the area of ​​the median parallel section is 225Picm^2 and the volume is 2800Picm^3. Determine the radius of the bases.

[fringe parcel]

V=1/3 π(R^2 +R.r +r^2) h,use this to form the first equation ,where V is the volume of the frustum and R and r are the radii of lower and upper bases respectively.

[green nest]

I did, but then i got stuck with R and r, and i couldnt find any way to get those from that or anything else given

got to 700 = R^2 + R*r + r^2

then i got stuck and couldnt find any way to continue

[fringe parcel]

The radius of the median section is (R+r)/2,so the area of the median section is π( radius of the median section)^2,use this to find the second equation.

[green nest]

OOOH

thank you

[echo sigil]

Does anyone know how to solve this?
Bisectors AA1 and BB1 are drawn in triangle ABC.Prove that the distance from any point M of the segment A1B1 to the line AB is equal to the sum of the distances from M to the lines AC and BC.

[obsidian harness]

it's like an extension of Viviani's theorem it seems

[obsidian hornet]

Just use bisector definition

[honest oxide]

Yes

[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.

[reef narwhal]

Can anyone help me? I have to do a 10-minute oral in maths and I'm doing it on Gabriel's horn , but it's too short.

[grave pond]

Please don't crosspost the same question between all the pre-university channels.

[crisp folio]

hi
peles help me
I have a problem. I'm trying to solve time flexibility with math. We have a vision of a series of numbers that are repeating or a series of decimal numbers that are not divisible by anything at all.
p= 2X *9m =- 11
11 -= 0m + mp2 - 29 = 2.7580088765936458
OM = 0 -9 *1.2 + 0.4
mp = 8 + 7 * 78 - 3
and
x =?
m == X-2 * 82 ms

[grave pond]

Please don't crosspost the same question between all the pre-university channels. I cannot follow exactly what you're saying, but it looks like responses should probably be in #prealg-and-algebra.

[reef narwhal]

Could someone who knows something about geometry please explain to me how Toricelli managed to calculate the volume of Gabriel's trumpet using Cavalieri's principle?

I really need help please

[grave pond]

You've read the description at https://en.wikipedia.org/wiki/Gabriel's_horn#In_De_solido_hyperbolico_acuto?

[reef narwhal]

Its not about cavalieri principle

[grave pond]

I think it's more a kind of handwavy analogue -- but Cavalieri probably didn't use modern standards of rigor when stating the principle, so it may be a matter of interpretation whether it's the principle itself or an analogue in use.
Instead of parallel flat cross-sections, Toricelli cut the horn into cylindrical shells and then stacked flat circles of the same area as each shell to form a cylinder. The stacked circles directly form half of an application of Cavaleri -- and the cylindrical shells at least morally have the properties one expects to be able to use it: each shell is of uniform thickness (or, differently speaking, the perpendicular distance between a cylindrical cross-section and its neighbor cylinders is the same everywhere), and that uniform distance is also how closely spaced the circles are stacke to make the cylinder.

(An interesting observation one can make from a modern perspective is that one doesn't need to muck around with volumes of revolution in order to make a shape with infinite surface area and finite volume. Just consider, for example the volume of R^3 satisfying:
0 <= y <= 1
x >= 0
0 <= z <= 2^(-floor(x+1))
whose volume easily sums to 1, but even just its bottom surface is infinite).

[terse laurel]

can someone remind me how to convert between degress and radians

[compact lodge]

can someone mind explaining the basic of trig and how i do it ?

[grave pond]

pi radians is 180 degrees

[terse laurel]

ok

[sleek pendant]

#help-21｜아리스킨충1 pls

[flint nymph]

Is there a formula for this that I can know? I saw these type of questions would appear when studying. So, there might be a common solution for all of them. With the only difference being the actual numbers.

Example Question (The actual numbers could change):

A photo measuring 32 cm x 40 cm is attached to the cardboard, so that each on the left and right remains 4 cm and the remaining top = remaining bottom = p cm. If the photo and carton similarity, the p-value is...

[obsidian harness]

hopefully I'm interpreting this right

so the cardboard width is 32 + 4 + 4 = 40 cm

now if the height of the cardboard is h

by similarity (length ratios of similar shapes)

32 cm by 40 cm, and 40 cm by h cm

you must have 40/32 = h/40

solve for h

oh and then you have h = 40 + 2p, so you know p

[flint nymph]

Alright, thanks for the explanation. For clarification, I'm going to share the image that's usually included with these kind of questions.

Ah, I'm sorry. I misremembered. It turns out these kinds of questions usually doesn't have an image attached to them. I checked both of my worksheets as well as the ebook, and yep, there's no image on all those questions.

Okay, I'm just going to try to understand your explanation here. If I have any questions, may I ping you?

[obsidian harness]

pls don't ping me

just ask your follow-up questions in this channel and I or someone else will answer

[flint nymph]

Explanation Steps (Please correct me if I'm wrong here, thanks!):

1. Determine the width and height of the photo. Seeing as the height is usually the bigger value between the two, then the 40 cm is picked as the height.
2. Add the width of the photo with the remaining numerical value on both sides. For example, the 32 (the width of the photo) is added with the remains of 4 + 4 for the left and right.
3. Cross multiplication of Photo Height/Photo Width = Unknown Cardboard Height (i.e h)/Cardboard Width.
4. Do all the required multiplication and eventually division as well to get the unknown (i.e h) value.
5. Finally, calculate the Cardboard Height = Photo Height + 2 x p to get the unknown numerical value between the two Heights.

[worthy eagle]

(x+2dx)/x=(y+2y)/y
1+2dx/x=1+2dy/y
dx/x=dy/y

In your example, 4/32=p/40
p=160/32=5

x=length of photo
y=height of photo
dx=how far is the photo from left or right side of cardboard
dy=how far is the photo from top or bottom of cardboard

Just use dx/x=dy/y and get answer

This is the trick or shortcut to find the answer of this type of question @flint nymph

[grizzled lion]

if anyones here this one equations flipping me off and its basic geometry 💔

[worthy eagle]

!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/

[grizzled lion]

i have to basically find the parallel lines which ive found one i just dont know how to find the other one

[pale sentinel]

"The other one"?

[grizzled lion]

gotta graph 2 lines

they’re parallel

ive done the y = -4/3x + 4

[pale sentinel]

If you've got a straight line and a point not on that line, there exists exactly one parallel line that goes through that point

Or do you mean you have to draw them, and you can't draw the other one?

(because you already have the equations of both lines)

[grizzled lion]

it says writing the equation of the libe in slope intercept form passing through the point and parallel to the given line and to graph both lines on the same graph

[pale sentinel]

Right... and the straight line in the question?

[grizzled lion]

these are ones i finished

[pale sentinel]

You haven't drawn that one

,rccw

[pale sentinel]

Right - so you just haven't drawn the line $y + 3 = -\frac{4}{3} (x-1)$

[somber coyote]

Waes (Wires)

[grizzled lion]

how can i make that fit the y=mx+b formula

[pale sentinel]

Subtract 3 from both sides, expand the brackets and simplify

[grizzled lion]

the brackets are the distribution of multiplication

y+3=-4/3(x-1)

y+3=-4/3x+4/3

subtracting 3 by both sides (multiplying 3 on the right side to get a common denominator)

y=-4/1x+1

i think?

nvm

i see the solution but i dont know how to get to there

[somber coyote]

Waes (Wires)

[pale sentinel]

@grizzled lion

[grizzled lion]

-5/3?!?

…oh

i get it now

thank you 😭

[rich kite]

this is sus

Its for a cube whose circumscribed sphere has radius
and for a given point in its three-dimensional space with distances
from the cube's eight vertice

[visual dawn]

What do you mean "indian" scammer?

[safe pebble]

no racism dont worry

it just comes to ur mind the "your computer has virus" part

[visual dawn]

So you're assuming they're Indian because you don't like them?

I'll keep that in mind.

Thanks for the ping.

[safe pebble]

but why did you take it serious ? its a joke

jokes not allowed here?

& i actually like indians theyre friendly ive been to india bruh

[visual dawn]

You could show that by not making racist comments about Indians. No action will be taken for now.

[fading crater]

Can someone explain this to me?

The second one.

Nevermind, actually. I solved it.

I'm just braindead, took me time to get it.

[rotund hemlock]

🤓☝️

[dark sparrow]

takes a lot of balls to nerd-emoji a message from a moderator.

[regal kraken]

in a mathematics discord server as well ☠️

[shut solstice]

Hello guys, Can you help me with this interesting question
So in the picture we can see the cylinder with a radius of 6 √ 3 and with full height of 30 , and it's filled with water as you can see in the first picture
in the second picture, it is tilted by 30 degrees and the water inside is also tilted, but it does not spill
The task is to find the height of the water

[grave pond]

By symmetry, that must be the average of the largest and smallest side length wetted by the water in the tilted state.

[dark sparrow]

,rccw

[shut solstice]

To clarify, in the problem you need to find the height of the water in the first figure, but there may be an error in the problem where 6 √ 3 is diameter not a radius of cylinder

[grave pond]

As drawn here, it looks neither like a diameter nor like a radius. Best to let the explicit words decide.

[rancid pasture]

Hey guys, i got this problem : cos 2x - cos x = 2, i just used some formula, and switched place and i arrived at 2cos^2 x -cos x -3 = 0 here i just did delta which is 25, i arrived at 3/2 for cos x wich i deemed impossible and cos x= -1 , i arrived at x= pi +2kpi and -pi +2kpi (with k belonging to Z) and with that i just wrote that it is kpi. The teacher asked for the solutions between [0;2pi[ so i just wrote 0 and pi but my teacher said that 0 is not right somehow, can soemone enlighten me?

[grave pond]

If you plug x=0 into cos 2x - cos x, do you actually get 2?

[dark sparrow]

{pi + 2kpi | k ∈ Z} ∪ {-pi + 2kπ | k ∈ Z} = just {pi + 2kpi | k ∈ Z}, not all multiples of pi.

[rancid pasture]

Fair point, i don’t usually look at that after doing the formulas

[grave pond]

Instead of formula-crunching, I would note that cos 2x is at most 1 and cos x is at least -1.
So the only way their difference can be 2 is if cos2x = 1 and cosx=-1.
cosx=-1 happens only once between 0 and 2pi, so the only thing to check is whether that gives cos2x=1 or not.

[rancid pasture]

I cannot write that as kpi? Like isn’t pi + 2kpi the same thing beside from that 0 then

[grave pond]

pi+2kpi cannot be 0, but it also cannot be 2pi or 4pi or 6pi or ... hopefully you see a pattern.

[rancid pasture]

OH YEAH i am stoopid, thanks y’all

[dark sparrow]

nope, i think you are confusing it with if the other solution family was just kpi

[novel belfry]

Im failing geometry

[still jackal]

k

[worthy eagle]

,rccw

[worthy eagle]

Don't need to tell

[manic narwhal]

Height of water in first figure is 21

Tilted water height (smallest one ) is 12 in second figure

[manic narwhal]

U were doing wrong

Then again ur concept to come with volume of half cylinder was good
I immediately got ans then

[worthy eagle]

Yooo

I got it too

I was just taking the angle wrong

[worthy eagle]

Here angle between the base and diagonal is 60° and then you solve it and get h=21

[empty yew]

Does the cosine law reduce to the Pythagorean theorem if the angle is 90 degrees?

[cunning lion]

yes

[empty yew]

can the cosine law be applied to other angles of the right triangle? If yes, will it be consistent with the Pythagorean theorem of that triangle? @cunning lion

I tried that and it's working(just for the confirmation)

[cunning lion]

the cosine law can be applied to any angle of any triangle

any geometric law applied correctly should give you the same results

[empty yew]

ok and thank you!

[mystic umbra]

you can prove it for all three types of scalene triangles: acute, right angled, obtuse so it will hold true so obviously it will hold true for all triangles

[shut solstice]

Thank you very much bro
It would be great if you could show me all the calculation you did
I didn't understand it at all

[manic narwhal]

Somehow u should find blue indicating part volume

And then find whole cylinder volume

Then do subtract whole volume to blue one
The resultant volume will give u height

[manic narwhal]

To find blue part volume
Use this concept

[worthy eagle]

Volume of blue part = Half of volume of red cylinder

Volume of water in cylinder = Volume of entire cylinder - Volume of blue part

[torpid current]

,help

[somber coyote]

I can't DM you! Do you have DMs disabled?

May 24, 2025, 8:48 PM

[torpid current]

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[torpid current]

,list

[somber coyote]

My commands!

Use ,ls to obtain a briefer listing, and use ,help <cmd>to view detailed help for a particular command, or ,help to view general help.

If you still have questions, talk to our friendly support team here.

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[torpid current]

,calc 3/1 x m2 =-1

[somber coyote]

The following error occured while calculating:
Error: Invalid left hand side of assignment operator = (char 10)

[dark sparrow]

this bot's ,calc command is not an equation solver.

[torpid current]

Ohh okay

[torpid current]

How do we use this bot

[dark sparrow]

mostly for LaTeX rendering (writing math pretty) and for querying wolfram alpha with ,w

[shut solstice]

oh now i got it. thank you very much guys

[worthy eagle]

,w (3/1)*x = -1

[somber coyote]

Results provided by WolframAlpha

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

Requested by shubham1029

[worthy eagle]

Replace x with m2 and you get your answer

[still jackal]

,w (3/1)*x = -1

[somber coyote]

Results provided by WolframAlpha

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

Requested by cr7xhaaland._91903

[unkempt path]

how do you ask questions here

[grave pond]

#❓how-to-get-help

[unkempt path]

thc

thx

[still jackal]

just keep a comma and 'w' before your query

[cobalt scarab]

You drunk mate?

[still jackal]

better lad

[signal vapor]

just type ,w

why complicate it

[knotty quiver]

that's a command for a wolfram alpha query

[empty yew]

do using arccosine in an inequality reverse the inequality?

like

,, 0 \le x <1

[somber coyote]

Monkey•D•Luffy

[empty yew]

on using arcos in it

,, \frac\pi2 \ge \cos^{-1}x > 0

[somber coyote]

Monkey•D•Luffy

[cunning lion]

arccos is a decreasing function, applying any decreasing function to an inequality reverses it

[empty yew]

ohh i remember it thank you

[empty yew]

Is reversing not standard for non-monotonic functions?

[cunning lion]

increasing functions preserve an inequality, decreasing functions reverse it, non-monotonic functions just don't play nice with inequalities at all

[empty yew]

ohh thank you so much

do monotonicity have anything to do with continuity?

[cunning lion]

well it's known that monotonic functions can only have jump discontinuities, and they can have at most countably many

[empty yew]

Is it possible for a function to be both monotonic and periodic?(I am asking this because every trig functions are periodic and non-monotonic but every inverse trig are monotonic and aperiodic)

[cunning lion]

any nonstrictly monotonic and periodic function would have to be constant. i don't think there's such a thing as a strictly monotonic and periodic function at all

[empty yew]

what do you mean by nonstrictly here?

[cunning lion]

a nonstrictly increasing function is one such that x < y implies f(x) ≤ f(y)
a strictly increasing function is one such that x < y implies f(x) < f(y)
apply analogous definitions for decreasing

[empty yew]

ohh i see

Thank you so much, cloud!

maybe good night, cloud!

[deft sail]

I have x=135 . Am I right?

[worthy eagle]

Yes

Would you mind telling how you got 2z=180?

[deft sail]

Transversal Angle Relationships Î think

[bold bear]

Hai

I gave myself a real challenging question earlier that I can’t divine the answer to

Suppose a square and circle, of the same area, share centres. The section in yellow (as per the diagram) has area 1.2. Determine the area of the square.

[upper echo]

r=2a/sqrt(pi)
cos(theta)=sqrt(pi)/2
the area of the yellow segment = theta * r^2 - a^2 * tan(theta)
then just solve it with respect to a^2 (you can write it in terms of arccos(sqrt(pi)/2).
So the area of the square is 530.117471...

[bold bear]

530????

[upper echo]

approximately

[bold bear]

Why’s it so big lmao

[upper echo]

bc the yellow part (which is 12) is small as compared to the square

[bold bear]

I mean it’s not really right to approach these things visually im aware but you’re sure 530 of those tiny little parts can fit inside the square?

[upper echo]

Yep, I'm sure. The side of the square is just around 23

Oh, you say the area 1.2 not 12

[bold bear]

Ye

So it’s 53

[upper echo]

then the area of the square is 39.75881...

[upper echo]

nope

[bold bear]

Oh

[bold bear]

Thanks for the answer

[upper echo]

wait a bit Ill check my computation

[upper echo]

yes, of course it is 53.0117471...
sorry, that 39 was incorrect

[bold bear]

No prob

That’s an interesting lil problem I posited

And a very nice answer you gave

[upper echo]

This is exact value for the area if you are interested

[bold bear]

Walk me through this a lil?

[upper echo]

this #geometry-and-trigonometry message

[worthy eagle]

Your answer is correct but I want to confirm that you used correct method to get to that answer.

Do you mean that you think sum of vertically opposite angles is 180° and you used this to find z?

[stoic ocean]

I have a quick question . The Volume of the whole cone is 128 pi cm3 and I know the raport of the whole volume and the volume of the trunk that I sectioned is 1/7 . How can I find the volume of the sectioned part ?

[worthy eagle]

Cross multiply

[stoic ocean]

Yes but it does ‘t make sense the smaller one get’s a bigger volume

I tried checkinf for any mistakes

but i don’t see any

[worthy eagle]

Then i think your ratio is wrong, maybe you meant 7:1

[stoic ocean]

Nope my question says 1:7

[worthy eagle]

Send the original question

[stoic ocean]

If you can translate it from Romanian to English

[worthy eagle]

Yes np

Just send

[worthy eagle]

It says ratio of volume of the bodies obtained is 1:7. It means the ratio of volume of the cone obtained to the volume of frustum is 1:7.

[stoic ocean]

Oh , so the small cone basically not the whole volume?

[worthy eagle]

What is the issue the 2nd cone obtained can have less volume than the frustum

[worthy eagle]

Yee

There can be 2 answers

If you consider ratio of cone to frustom = 1:7 or frustom to cone 1:7

[stoic ocean]

I solved it , I got the volume of the frustom of the cone 112

[worthy eagle]

Both are possible, in english translation it said ratio of bodies, so both are the cases

[stoic ocean]

Ye

My teacher will correct me anyways if it's wrong

[worthy eagle]

Ok gl

[stoic ocean]

Thx

[deft sail]

Yeah

[worthy eagle]

That's wrong. Sum of vertically opposite angles is not always 180°

It was in this case particularly, but it's not in general.

You can't use that to get 2z=180

[knotty quiver]

,rccw

[somber coyote]

Couldn't find an attached image in the last 10 messages.

May 27, 2025, 10:43 AM

[knotty quiver]

last 10 messages.
I see

[zealous pike]

Does anyone has any tips to solve 67? I'm clueless about how to approach it

[obsidian harness]

well, if O is the origin

then angle QOP = pi/6 + pi/6 = pi/3

R has an angle of pi/2, so angle POR = pi/2 - .... ?

yeah and then QOP and POQ are both isosceles triangles, so that means they are congruent triangles from this

[zealous pike]

Is the triangle important to solve this or can I solve it without considering triangles?

[obsidian harness]

that explains why PQ = PR

if you can figure that out then the rest of the reasoning is unrelated

just distance formula PQ = PR then, well for PQ you don't need the formula

[zealous pike]

I see. I was able to see that 2y = sqrt(x² + (y-1)²)

[obsidian harness]

then it's a bit more algebra

$2y = \sqrt{(x^2 + y^2) - 2y + 1} \implies 4y^2 = (2 - 2y)^2$

[somber coyote]

south

[obsidian harness]

also don't forget y > 0

so actually you can do difference of two squares

$(2y)^2 = (2 - 2y)^2 \implies (2y + 2 - 2y)(2y - 2 + 2y) = 0$

[somber coyote]

south

[zealous pike]

With that we get to y = 1/2?

[obsidian harness]

mhm

so yes, sin(pi/6) = 1/2

and you can use the relation x^2 + y^2 = 1 to find cos(pi/6)

[zealous pike]

From here it is easy to see that x = sqrt(3)/2

[obsidian harness]

mhm

that's a fairly creative approach

I've seen something new, other than "half an equilateral triangle" and boom

[zealous pike]

Indeed. I've learned how to prove the sin and cosine for those angles with the right triangle, then I bumped into this one

And got curious

[obsidian harness]

q68 is also nice

[zealous pike]

I'll try it right away

Thx for the help man

[obsidian harness]

no worries!

[dusty torrent]

hey guys I have a question about trig proofs so I've been struggling a lot lately, and I just dont know what to do onto the next step
any tips?

[worthy eagle]

!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/

[dusty torrent]

my bad im used to ti

[pure knoll]

ill pay someone 20 dollars to do my geometry work

all if it

[dusty torrent]

bruh

what is it even about

[pure knoll]

circles and areas

[worthy eagle]

!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.

[solid spire]

is the answer 12 degrees?

[worthy eagle]

6 days

Bro was checking 6 days old messages for what reasons

[solid spire]

i was just looking through preuni channels out of curiosity

i normally have these channels removed from my list

[oak laurel]

Does anyone have some good websites to study for Geometry finals?

[obsidian harness]

Khan Academy

[trim herald]

Ayuda / Help please

[dark sparrow]

@trim herald what's that blob next to point B? is it meant to be a right angle marker?

[lone plinth]

Can you translate the question in English "exactly" as it is written?

[lone plinth]

These are questions, solutions and explanations on Mathematics standardized tests/exams:
https://successexams.appspot.com/
Look for the topic: Geometry and solve each topic under the exam.
Check back often as contents are updated frequently.
If you have any question regarding any solution/explanation, please ask.
Feel free to share with others.

[dark sparrow]

Find the area of the shaded region if AOB and COD are circular sectors, where theta = 2pi/9 and BC = 3 m.

@lone plinth why the quote marks around the word exactly?

[trim herald]

please

[trim herald]

yes

right angle

[lone plinth]

By "exactly", I mean that there should be no error in the translation.

[dark sparrow]

ok but what was the purpose of enclosing the word in quote marks

@trim herald ok let's see

[trim herald]

can you help me, please?

[dark sparrow]

do you know how to find the area of one circular sector

[trim herald]

Yes

If you are busy, could you tell me the steps to solve it?

[dark sparrow]

well step 1 is to recall how to find the area of a sector

like the formula

bc your region is clearly sector COD minus sector AOB

[trim herald]

Ok, I got it

[trim herald]

yes

could you do it on paper :/

[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.

[trim herald]

Ok, so what else is next?

in the steps

[dark sparrow]

it's not going to make sense without having done step 1 first tbh

[trim herald]

I already did step 1

[dark sparrow]

ok can you show me

[trim herald]

the drawing or what? So far you've only told me that the shaded area is the difference between the total and the smaller part.

[dark sparrow]

no, the area formula.

preferably in general.

[trim herald]

if the formulas for the area of ​​a circular sector are known

[lone plinth]

Do you know the formula for the area of a sector if the angle is in RAD?

[narrow tendon]

Is this the math III section?

[obsidian harness]

what is math III?

the channel description here is: "euclidean geometry, coordinate geometry, trigonometry"

[narrow tendon]

Oh thanks, I have to study algebra 2 for next year

[obsidian harness]

#prealg-and-algebra then

[stoic ocean]

In this cube I know that AB = 30 cm . And that the highlighted part down has a volume of 9liters . How much liters do I need to add to fill the whole cube ?

I guess it is double but idk how to prove

Nvm I figured it out

[fringe parcel]

Find the Volume of the cube first in litres the subtract 9l.

I need help on this guys.

[dark sparrow]

#calculus

[idle goblet]

https://cdn.discordapp.com/attachments/326138757474680852/1320617825144016926/Circle-trig6.svg.png?ex=68389371&is=683741f1&hm=7b43e11c4917afdf9671f8debc924745c4006087d2bf213b8b38f31582b41c72&

@chrome token

haver functions were used in ancient navigation apparently and are still used in spherical geometry

[arctic hatch]

Yo guys I have a question, pls help

How can we calculate the surd form answer easily in these type of question?

@everyone

please someone pls help

How can we calculate the surd form answer easily in these type of question?

[mystic umbra]

!noping especially everyone

[lime crown]

Please do not ping individual helpers unprompted.

[mystic umbra]

erm anyways

in triangle ADC, tan 60 = AC/DC

same with sine ratio

[empty yew]

Is $\sqrt{1-\sin(x)}=\sin(\frac{x}{2})-\cos(\frac{x}{2})$?If yes, how?

[somber coyote]

Monkey•D•Luffy

[void inlet]

you gotta put a modulus around the expression on the right hand side

[empty yew]

yea but how?

[void inlet]

are you asking about how there is a modulus or how the expression has come?

[cunning lion]

,align \ab(\sin \frac x2 - \cos\frac x2)^2 &= \sin^2 \frac x2 + \cos^2 \frac x2 - 2\sin\frac x2 \cos\frac x2 \
&= 1 - \sin x

[somber coyote]

cloud

[cunning lion]

so we have [ \abs*{\sin \frac x2 - \cos\frac x2} = \sqrt{1 - \sin x} ]

[somber coyote]

cloud

[empty yew]

oh thank you!

[arctic hatch]

How to clculte surd form in this question??

[obsidian harness]

which part?

[dark sparrow]

is your A key broken

[knotty quiver]

-# LaTeX would be helpful in that situation

-# ,tex How to c^^61lcul^^61te surd form in this question??

[empty yew]

,tex c^^6lcul^^6te @knotty quiver

[somber coyote]

Monkey•D•Luffy

[obsidian harness]

cul in French....

[empty yew]

(Anyway I don't convey that in french)

[delicate topaz]

I read through everyone else's responses and i genuinely feel like they forgot about special triangles
Part A here is specifically focusing on triangle ADC
And its a 30-60 right triangle and has specific proportions ||1:sqrt3:2 where each part cooresponds to short leg, long leg, hypotenuse||
I took a quick google on what the hell surd form was as i mever heard of it, and its basically simplify your radicals

Ex) $\sqrt8$ would be $2\sqrt2$

[somber coyote]

Pai

[delicate topaz]

This whole problem uses special right triangles
Even the trig problem in part C requires the use of special right triangles when setting up the proportion of sin15
Then from there you need to numerically rationalize the denominator (as in make the denominator be an integer by multiplying numerator and denominator with a specific squareroot so that the denominator becomes the squareroot of a perfect square)

[arctic hatch]

Can dm pls??

[delicate topaz]

Absolutely 😁

[knotty quiver]

what's this. it's 61, not just 6

[empty yew]

Is dm really required?😂

[delicate topaz]

I dont mind
A private one on one can help so that its easier to understand if youre the type of person who cant listen to several people at once
Sometimes a person cant function when they have to listen to multiple outputs 🤣

[empty yew]

If you've ever get the helpful role, you can't function as there are many dms lol

[lone plinth]

Are you good with this question, or do you still need help with it?

[timber storm]

Can someone please help me out with this problem

[delicate topaz]

He is currently asleep. But i did give him a very general walkthrough for it. Inwas about to post here so other people can view it too 🧡

[lone plinth]

Looks like you did several questions for him.

For Number (3.), is it okay if I post my solution? May I have your permission?

I understand he wanted you to work with him. Hence, I ask for your permission.

[delicate topaz]

Oh the first bit of it was some examples bc he had never used the special right triangle ratios and needed some extra practice with radicals
But the actual problems just had a step by step guide on what to look at and what to use without providing the actual work

[delicate topaz]

I'm down for seeing the solution
I dont mind you dming me so that unicorn can still work it out himself

[halcyon vale]

Hello, I just joined this server. As I recall, since BH is perpendicular to AC, 3x+5x must be equal to 90°, that is, 8x=90, which results in x being 11.25°, meaning ABC is a right triangle.

[lone plinth]

What theorem did you use to justify your statement? Prove that Angle ABC is a right angle.

[halcyon vale]

Complementary angles formed by perpendicular lines, I was also surprised, it must be taken into account that the graph is only illustrative, the demonstration is mathematical

[lone plinth]

What are the complementary angles in context? What are the perpendicular lines in which those complementary angles are formed?

[halcyon vale]

It doesn't have a name like "theorem of...", But you could call it: "Property of complementary angles formed by a perpendicular altitude."

It's something that I didn't even know, but when using the value of 11.25 in X, I added all the sides of the different triangles that are formed in the exercise, and indeed they all give me 180°

[lone plinth]

May you show your work? Please specify the triangle (s)

[halcyon vale]

I just analyzed and started to investigate more because I remember hearing this and well I'm not using a relationship that has nothing to do with it, the original exercise is a scalene triangle and I'm using the theorem of the height of the leg to solve this and well I realize that it is not the correct method, that is to say the value in x works for me because I'm literally changing the figure, that's why the values match but I don't realize that my theorem is only valid as long as it is a right triangle but not in my solution it is wrong

Mb

[worthy eagle]

tan5x/AH = tan3x/HC
tan5x/tan3x = AH/HC

AH×tanx = HC×tan2x
AH/HC = tan2x/tanx

→ tan5x/tan3x = tan2x/tanx
→ tan5x tanx = tan2x tan3x

Now, use formula to convert this mess in terms of tanx and get your answer.

I am not even touching this, who gives tan 5x in paper bruh

Could there be an easier way to this?

[empty yew]

,, \arctan(1-x^2 - 1/x^2)+\arcsin(x^2 + 1/x^2 -1)

[somber coyote]

Monkey•D•Luffy

[edgy axle]

how can you find the angle in trig

[hoary totem]

in what context

whats the problem

if thats ur sole question the only answer i can give is "use the arctrig functions"

for example if sin(a) = 0.4 then
a = arcsin(0.4) + 2nπ = 0.4115 + 2nπ rad
or
a = π - arcsin(0.4) + 2nπ = 2.7301 + 2nπ rad

[upper echo]

that expression in arcsin is always greater than 1 in absolute value. So, if x is real then your formula makes no sense.

[upper echo]

x^2+1/x^2-1>=2-1=1. So arcsin is undefined.

oh wait

x=1

and x=-1

so they are the only possible values

then it is arctan(-1)+arcsin(1)=-pi/4+pi/2=pi/4

[empty yew]

thank you i can carry on from here

but how do you find that this expression is always greater than or equal to one?

[upper echo]

you can use AG inequality x^2+1/x^2>=2sqrt(x^2/x^2)=2.

[empty yew]

oh i see

[lone plinth]

I think "easier" is a relative term. I'm not sure if I would say "easier", but would you like to see the solution?

[worthy eagle]

Yes

[worthy eagle]

I am really waiting sir

Abra Kadabara Maths Guy summon

[lone plinth]

Here you go, Sir.
This is the detailed solution:
Number (10.) on
https://examssuccess.appspot.com/MATHEMATICS/Trigonometry.html#one-twenty

Please let me know if you have any questions on the solution.
I'm not here 24/7 but will answer questions when I come here.

[sinful cradle]

Hi! Can someone help me out with these two problems?

[slow pasture]

its true tho lol

you answered correctly

[unkempt topaz]

yeah thats right

if the planes are at right angles theyre perpendicular

[sinful cradle]

Ohhh okay haha lol thank you so much though!

One more question, I just wanted to clarify if I got these correct too, I’m not rlly sure

[delicate topaz]

Yeah, youre correct
Aside from actually knowing the definition of skew, you can use process of elimination as the indicated segments are not perpendicular nor are they coplanar
Hope that helps

[sinful cradle]

Got it! Thanks!

[weary drift]

$\trig$

[somber coyote]

ロケットジャンプ
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[weary drift]

enjoy my trig sheet v2

[delicate topaz]

Thats extremely helpful. Thanks rocket 🧡

[worthy eagle]

@lone plinth did you make that website?

[empty yew]

Isn't the q1 the two lines are parallel?

they two lines AB and GH are coplanar and not intersect

[empty yew]

Is that command only used by you?

[weary drift]

everyone can make their own texit commands via ,preamble

[empty yew]

Oh i see but why does it show compile error(If it is very technical nvm)

[weary drift]

idk but idc as long as it renders fine

[delicate topaz]

Theres no plane connecting the two segments, at least not one drawn in. Unless i'm looking at the prism incorrectly, AB would be the front and top edge of a box while GH would be at the bottom and back side
Does that make sense?

[empty yew]

the plane in which they are parallel is not shown but it lies in that space

can you see it?

[delicate topaz]

Yes, thats true and i can see it, but i'm under the assumption we only need to look at whats given to us by the figure

[empty yew]

Do they tell that the plane of the rectangles are the only planes in that space?

I mean in that question

If it is then you're correct

[delicate topaz]

It doesnt say. Theres usually directions above the box that contains the problems
But i'm assuming it might be high school or general math college (basically high school from what my friends had shown me in the past)

So i dont think itd make the student add in extra junk unless it says to

[empty yew]

I don't think someone will learn parallel lines with only some planes provided!

but anyways I don't wanna argue

[delicate topaz]

Theres 6 different planes provided though 😅
Nah thats fair though
I guess whenever OP gets bavk to check your reply they can answer, right?

Thats still good insight to have though
And i did consider that when i first glanced the problem
But i have learned that theres room for overthinking if you dont look at the multiple choice options closely

[empty yew]

I have seen the options, and we only need to know whether there is a limited number of planes or not.

[worthy eagle]

Question 1 , option 1st should be correct. They are coplanar, parallel, and aren't skew.

But let's leave it, you can just check the answer key after that.

[lone plinth]

Yes Sir.

Hosted by Google (because of the subdomain: appspot.com)

But, it is not free. Google charges me for it.

[stuck nest]

Sinx^2+cosx^2=3

[delicate topaz]

Is that sin(x²) or (sinx)²? And same for cosine

[stuck nest]

Sinx.sinx

[worthy eagle]

Isn't it 1

[stuck nest]

Yes its 1

[delicate topaz]

Cool. I was thinking why it wasnt =1 but i had to clarify where the squared was first

[stuck nest]

Bro i just randonly writed something

dont take it serious pls

[delicate topaz]

Theres other channels if you want to mess around though. I dont think this and the other two channels where you said geom was better was really appropriate for randomness tbvh
Like you could have gone to #chill to do that 🤷‍♀️

[dark sparrow]

#bots or #latex-testing or #chill

[noble urchin]

Hi guys, i wanted to know how to solve this problem, can anyone explain

[delicate topaz]

I could be wrong, but because there is a square inscribed the circle, the square's vertices evenly cuts the circle and the measure of the arcs are equal

[hoary totem]

chords of equal length subtend equal angles at the center

[obsidian harness]

By SAS the four triangles JNK, KNL, LNM, NMJ are congruent

The radii are all equal

Hence mLM = angle LNM = 360/4 = 90

Due to sum of angles around a point

[upper karma]

what is the best book that is comprehnsive for highschool geometry

[maiden brook]

aops intro to geometry

[dark atlas]

honestly its pretty easy to memorize mosty of it its mostly just memorize distance & midpoint formulas, and memorize Surface area, area and volume formulas, that's pretty much it besides common sense like angle ab+angle bc= angle ac

[frigid spruce]

can someone teach me the basic approach whenever you see a question that has trig 2 and asks you something like

$0 \leq x \leq k\pi$ \ \
$\cos (4x) \cdot \sin (x) + 2 \sin ^2 (2x) = 1$ \

How many roots of x are there?

ah i cant even

use

[somber coyote]

Red.

[frigid spruce]

should i ask this in some help channel or

cuz this feels like

im asking for

sources or what to study

more than help with a specific question

i mean questions LIKE that

not just that question

thats just one that I couldn't solve and I was stumped

the answer was like 9

[signal vapor]

Simplify by taking sin² term to th other side

[frigid spruce]

i mean what'd that achieve

surely

okay so

$\cos (4x) \cdot \sin (x) = 1 - 2 \sin ^2 (2x)$

[somber coyote]

Red.

[frigid spruce]

1-2sin^2 sounds familiar

from something

i dont got my trig 2 down very well

idk what am i supposed to know here

i know cos(2x) / cos(a+b)

and sin(2x) / sin (a+b)

therefore also tan(2x) or tan(a+b)

sure actually

ill ask this in help ig

[tender fulcrum]

cos2x identity

so essentially the RHS is equal to cos4x

[tribal rose]

Did I do something wrong here?

[worthy eagle]

Nothing wrong @tribal rose

[tribal rose]

Thank you!

[main spindle]

Requesting help on this quadrilateral problem

[worthy eagle]

What other information do you get from the figure other than that it is a parallelogram

[main spindle]

m<1 = 3y - 6

[worthy eagle]

From the figure only

There are symbols used

[main spindle]

Two congruent sides, bisected right angle

The right angle is bisected by diagonals

[worthy eagle]

*two adjacent congruent sides

[main spindle]

Oh yeah they're adjacent

[worthy eagle]

And what does a parallelogram with right angle mean

What kind of shape it means

[main spindle]

A rectangle?

[worthy eagle]

Yea

[main spindle]

Oh alright thanks

That fills the first blank

[worthy eagle]

A rectangle with congruent adjacent side is called

[worthy eagle]

That's not complete

[main spindle]

Just pointing it out

[main spindle]

A square

[worthy eagle]

That's the answer

[main spindle]

Got it

[worthy eagle]

You could do this one more way.

1. Parallelogram with adjacent congruent sides means it's a rhombus.
2. Rhombus with 1 right angle means it's a square.

[main spindle]

Yes that method works as well

[worthy eagle]

You know x,y,z?

[main spindle]

Yeah, I figured how to solve them

[worthy eagle]

Ok

[rugged silo]

can someone help me

I rlly don't understand geometry,I've asked her if she could dumb it down a little for me but she refuses

[delicate topaz]

Happy to help
This first part here you're showing is pretty straight forward
Go ahead click the dots on the ruler to place it over segment WY and share another SS

[maiden brook]

who’s “she”

[delicate topaz]

Probably the teacher

[charred berry]

can someone help with this pls

[delicate topaz]

start with with drawing a coordinate plane and then draw a triangle to match that cosine
think unit circle

[mystic umbra]

ern just use the relation

sin²x+cos²x=1

[empty yew]

also check for the quadrants if you use this relation.

[dark sparrow]

do you still need help with this

[empty yew]

With telescopic summation I got $\tan^{-1}((n+1)^2)+\tan^{-1}(n^2)+\tan^{-1}(1)+\tan^{-1}(0)$

[somber coyote]

Monkey•D•Luffy

[empty yew]

Here I get $\pi - \frac{\pi}{4}$

[somber coyote]

Monkey•D•Luffy

[empty yew]

since arctan x approaches pi/2 as x approaches infinity

but can we use the identity $\arctan(x)+arctan(y)=arctan\left(\frac{x+y}{1-xy}\right)$?

[somber coyote]

Monkey•D•Luffy

[frigid spruce]

this is insulting

nanananah

i aint gonna accept this

trig 0/4

im insulted

im deaddd

[daring geyser]

trig

[untold void]

$$ \calc

Awww man

[worthy eagle]

Split it up into trig1 and trig2

[empty yew]

Ping him not the bot

[cunning lion]

<@&268886789983436800>

[empty yew]

What do you mean really?

Do you mean like two images?

or

trig levels

??

[unborn sparrow]

anyone here?

[worthy eagle]

Into two images

[empty yew]

I see I too want that

[weary drift]

but i want everything in one place

,tex \trig

[somber coyote]

ロケットジャンプ

[graceful talon]

two pins?

idk how much they differ but i think they're similar

[empty yew]

This is a more comprehensive list of trig identities than the previous one lol!

[graceful talon]

I see

[weary drift]

it also has better formatting

[ancient sky]

I'm not sure if this would be the correct channel for this sort of question, but I'm curious how they cooked up the red underlined relation.
Is this just a fact about angles of a triangle (that I am unaware of)? I'm not sure what to google to find this idea either

[worthy eagle]

It's saying if you draw a line AD perpendicular to BC, then angle ABC = 90°±BAD

In case of acute triangle, ADB=90°, so ABC = 180°-90°-BAD = 90°-BAD.

In case of obtuse angle, the altitude AD will lie outside the triangle as angle B is obtuse.
So, BAD + ADB + ABD = 180°
BAD + 90° + (180°-ABC) = 180°
BAD + 90° - ABC = 0
ABC = 90° + BAD

[worthy eagle]

Relate with this figure what I wrote for case of obtuse triangle ABC.

[ancient sky]

goated, thankyou so much

[empty kayak]

Can someone help me with b please

[obsidian harness]

$\cos x \cot x - \cos x \cot(3x - 50) = 0$, so if you now factor, either $\cos x = 0$, or $\cot x - \cot(3x - 50) = 0$

[somber coyote]

south

[empty kayak]

What should I do after that?

I get cos-1(0)=90

And then how do I solve to the cot part?

[mellow pollen]

u got a level tmr too?

[worthy eagle]

cot x = cot 3x-50
When is this true? Think and apply the condition for 0<x<180

[dark sparrow]

$\cot(x)=\cot(3x)-50$

[somber coyote]

Ann

[dark sparrow]

is this what you wrote?

[worthy eagle]

no cot(3x-50)

[worthy eagle]

i thought it would be clear from here 🥶

[empty kayak]

Yah 😭

Gonna kms

[mellow pollen]

bro dww itll be fine, i gotta deal w further maths next year too 😭

[empty kayak]

😭😭good lord

Good luck Brodie

[mellow pollen]

gonna do last minute revision tmr too so we should be good

[mellow pollen]

ahaha thanks

and good luck tmr as well

i just hope we dont get some bs questions

[viral nimbus]

Can anyone give me a full explanation n not js the workin?

[plucky crag]

the triangle inequality states: the sides sum of a and b must be greater than or equal to the hypotenuse. being equal to creats a degenerate triangle, where the two sides fall in line with the hypotenuse. this fits all the trig laws and is therefore a triangle. does it have any uses? or any cool properties that i would be able to use in high school maths

just anything to help or like that i should know about degenerate triangles?

[plucky crag]

for the last one, if you use rules for parrallel lines such as Vertically opposite angles are the same, and corresponding angles are the same, angles along a straight line add to 180 degrees. therefore angle ECD is 37. angle CED is 76. therefore CDE is 67 (angles add to 180 degrees) as vertically opposite angles are the same x is 67 degrees. in problems like this just look for common rules such as these

[mellow pollen]

@empty kayak how did ur exam go

[empty kayak]

It was great

Honestly was pretty easy imo

I couldn’t do the integration question tho 😭

I tried u sub and by parts and it didn’t work out for me

I was honestly tweaking out

@mellow pollen what about you?

[mellow pollen]

yeah it was rlly easy

which integration question do u mean

the last one?

bc u sub worked for me

[empty kayak]

Yah the x/(2x+3)^2

I tried u sub and I ended up with u^4/4 +u^3/4

[urban hamlet]

Idk if this is the right channel to ask this, but does anyone have any book recommendations for geometry? I am teaching geometry over the summer and I thought it would be fun to get some historical background

[dusty current]

@urban hamlet

teach me please

im gonna have a exam 30 june

[obsidian hornet]

Euclid's Elements) Ok, for serious: Geometry revisited (Coxeter Greitzer), Geometry in figures (A.Akopyan), lemmas in olympiad geometry(there’s no need to introduce), some additional historical content: Blez Pascal thoughts, Viktor Tebo problem series in American Monthly magazine, should be enough.

[whole thorn]

im struggling in that sine cosine and tan

and exam is in 3 days

[obsidian harness]

Khan Academy and org chem tutor on YT

[zealous pike]

I wonder if it wants me to come to a conclusion through the equation. My answer for a is that the equation allows to say that the areas of the light blue shaded parts is equal to the dark blue shaded triangle.

I + II + IV + V = II + III + IV
I + V = III + (IV + II) - (IV + II)
I + V = III

is this the only conclusion there is for that?

[olive agate]

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

[grave pond]

You've concluded (correctly) that they are equal. That's the strongest imaginable conclusion about "how they compare".

[zealous pike]

Then I don't understand question 24. Why? Because the equation says so.

[grave pond]

Just give your algebraic argument there.

[zealous pike]

So far it has only showed simple axioms on angles so I dunno anything about how that's related to the link the guy above sent or how they are equal. But if the author says so then thats it for now

[grave pond]

I think the Wikipedia link was just meant to show as an aside that the diagram is a somewhat famous configuration and suggest an argument for why the problem is right about I+II+IV+V = II+III+IV in the first place (which you were not asked to answer).

[olive agate]

The link was just to inform you that this is a historically important exercise. An early version of what became in Euclid the areas of similar shapes on the catheti of a right triangle summing to the area of the similar shape on the hypotenuse was used in the first conserved proof (extant) in history and your Alhazen version has that as a special case

[grave pond]

It's a shame that Hippocrates's proof has not survived. Today we can say relatively quickly that each of the semicircles (I+II), (II+III+IV), and (IV+V) has an area that is the same fraction of the square of its diameter, and then apply the Pythagorean theorem to the right triangle III -- but "same fraction" would not have been easy to say in the style of that period.

[olive agate]

Ok, what's known is from a commentary to Aristotle's Physics by Simplicius of Cilicia with reports of Eudemos of Rhodes reporting of Hippocrates of Chios. The screenshots are of the video https://www.youtube.com/watch?v=nNtCv25kQIU?t=690 where a likely proof of the times is sketched

[grave pond]

Unfortunately those screenshots just seem to assert that "the segment on the base equals the sum of those on the sides". Which is true, but doesn't really tell us how Hippocrates argued for that fact.

[olive agate]

No, but do watch the video, the relevant five minutes are interesting

[zealous pike]

I'm seeking to learn whatever is needed to be able to understand that stuff

[grave pond]

I'm not going to watch an 2-hour video just to figure out what "the relevant five minutes" are.

[olive agate]

I gave the 690 second mark to you...

[grave pond]

When I click on the video it starts at 0:00 / 1:51:04.

[olive agate]

read the link

[grave pond]

If you have a point to make that you would like me to know about, I invite you to make that point here. ¯_(ツ)_/¯

[olive agate]

ok, the crux is the isosceles case is so much easier Hippocrates could do the argument we use today

[grave pond]

The only thing that looks easier to me is how to establish the a²+b²=c² relation for the triangle. The problems with speaking about the relation between the circular segments and the corresponding squares seem to be the same.

[olive agate]

The inscription in a semicircle in the screenshot doesn't entice you?

[grave pond]

This one? No, it doesn't seem so show anything about how to reason about the ratios without having numbers for them (which the Greeks didn't).

[olive agate]

In this one

[simple vigil]

guys i am in 10th i struggle with maths can you guys help

?

[olive agate]

Ok, trying to keep things in a thread is futile it seems, so I'll yield @grave pond

[grave pond]

At exactly 13:55 the lecturer says: "he doesn't say why".

[olive agate]

ok, I see you don't subscribe to the Zeuthen view

[grave pond]

And then he goes on to speak about the Pythagorean theorem, but the Pythagorean theorem speaks about squares summing up to each other, not about circular segments summing up to each other.

I don't know what "the Zeuthen view" is.

[zealous pike]

Is the only way to develop good geometry intuition doing thousands of exercises?

[grave pond]

Maybe non thousands, but at least scores. That's the only way in any kind of mathematics.

[olive agate]

But maybe you do https://link.springer.com/article/10.1007/s00407-015-0169-5

[grave pond]

A paywalled article, really? Again, if you have a point to make, make it here.

[urban hamlet]

Thank you! I’ll check those out!

[olive agate]

Privilege can’t see that kind of thing, sorry. Zeuthen on Wikipedia then. He was both a mathematician and a math historian

[grave pond]

Are you claiming that https://en.wikipedia.org/wiki/Hieronymus_Georg_Zeuthen contains anything of relevance for Hippocrates?

[olive agate]

The point I am making is that you could ascribe intelligence to the ancients

[grave pond]

I think you're just being bellingerent for conflict's sake here. If you have point to make, make it rather than hiding behind links to obscure sources outside the chat.

[olive agate]

There’a an aspect of that on both sides I think. But to me you seem to discount the possibility that there was a proof we’d accept today at the time

[grave pond]

But to me you seem to discount the possibility that there was a proof we’d accept today at the time
How on earth do you form that conclusion?

[olive agate]

Ok. As I said above, this is not the forum, especially since you take things out of thread all the time. For the sake of the forum, I yield

[grave pond]

Speaking of "yield" definitely makes you sound aggressive, like talking in a chat is a game that aims at finding a winner ...

And this is the channel the server has for talking about geometry, even ancient geometry.

[olive agate]

Ok if this really is the forum, what makes you say Hippocrates couldn’t talk about other similar figures than squares on the right triangle?

[grave pond]

Nothing makes me say that, and I didn't say that.

[olive agate]

What did you say?

There’s where I derailed

[grave pond]

I said that we don't know how Hippocrates got from a statement about sum of squares to a statement about sum of circular segments because his writings where he made that connection have not survived.

[olive agate]

He might not even have gone from squares

[grave pond]

The more to lament that we don't know how he did it, right?

[olive agate]

We agree at last!

[arctic hatch]

@meager raft This guy is very good at maths

[meager raft]

What

No

[arctic hatch]

Teach trigonometry

[zealous pike]

Can't we assume that it was by similar methods as those used by Aristotle to approximate the area of a parabola?

Wait, was it Aristotle who did that or was it Erasthotenes

Neither

It was Archimedes

Exhaustive approximation of curves with squares and rectangles seems to have been very common back then, idk

[grave pond]

Wikipedia states that the method of exhaustion started with Antiphon, who was a contemporary of Hippocrates, "although it is not entirely clear how well he understood it". If Hippocrates had explained that involved, it feels like Simplicius would probably have chosen that to gush about instead of the comparatively simple adding/subtraction of areas that comes after.

[zealous pike]

interesting

If we use the knowledge on mathematics by the context they had back then, wouldn't it be possible to simulate or approximate his idea?

Most demonstrations about that theorem uses notation and knowledge that were polished much later in history, which may make room for confusion.

[grave pond]

Hmm, perhaps I was overly skeptical about how easy it would be to speak about equal proportions in the language of antiquity. Euclid, a century or two later, does prove (via exhaustion) that all circles have the same ratio to the squares of their diameters (proposition XII.2) and has shown earlier how to add or subtract such proportions (e.g. proposition V.19).

[zealous pike]

Although it may be considerably hard to come up with something as smart as that

[zealous pike]

[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/

[past crag]

,iamnotstudying

[somber coyote]

Removed the studying! role from you.

[fading crater]

How do i find where on the trigononometric circle the intended value? For example:

[zealous pike]

Are you using radians?

[fading crater]

No.

It's just equations now.

[zealous pike]

3x = π/6

Isn't it x = π/18?

3π/18 = π/6

sin(π/6) = 1/2

[fading crater]

I mean, it makes sense to me, not that i would know.

[zealous pike]

For that, it is good to know at least the standard terminal points and their values for sine or cosine. If you know a value x for sine to become 1/2, then u are half way done

Like, for sine 5x = 1/2

We know that sine x is 1/2 when x = π/6. can we find a value x so that 5x =π/6

?

[fading crater]

But there is no 5x.

It's only 3x.

[zealous pike]

It was just an example

[fading crater]

Oh, okay.

[zealous pike]

The logic is the same though. Which value x so that 3x = π/6?

[fading crater]

Well, it comes down to being sin3x = 1/2.

[zealous pike]

Isn't that what you are trying to solve?

[fading crater]

Yes, but i don't know how?

[zealous pike]

Hmm

Do you already understand that sin(π/6) = 1/2?

[fading crater]

No.

[zealous pike]

Oke

[zealous pike]

Do you already know that Sin = opposite side / hypotenuse?

If you do, it may be simpler to understand this. Trigonometry of right triangles is also sort of possible to apply here

[fading crater]

Kind of.

[zealous pike]

Can you already convert radians to degrees?

[fading crater]

Like pi=180° for example?

[zealous pike]

Yes

[fading crater]

Yes, i do.

[zealous pike]

Let's say we have an equilateral triangle whose sides are 2 units. What happens if we divide that triangle in 2 with a perpendicular line from its tip to the base? What kinds of triangle do we end up with

[fading crater]

Two right angled triangles?

[zealous pike]

That's right. Let's call the base = b, height = h and hypotenuse = a. What are their units in the new right triangle

?

[fading crater]

Sides?

[zealous pike]

Yes

Their , length, measures… are?

[fading crater]

Hypotenuse is the longest.

So sin is h/a, i think.

cos is b/a

[zealous pike]

No

It depends actually

Let's start this by finding the lengths of the sides. Before we've split the triangle, the sides were 2 units each.

With it split, what length is the base now?

[fading crater]

Half of what it once was?

Two equal halves?

[zealous pike]

Right

The base is 2× 1/2, half of what it was before. Base = 1

What about the hypotenuse?

[upper karma]

help

[fading crater]

Hypotenuse is the same length as 2 new bases

[zealous pike]

That's right

Can you tell what angles that triangle has? You can do so either in radians or in degrees

[fading crater]

60° - 90° - 30°?

[zealous pike]

That's right. You got this

Therefore, if we know that sin is the ratio of the opposite side by the hypotenuse, do you understand how sin(30°) = 1/2?

[fading crater]

No, that confuses me.

Unless.

[zealous pike]

You don't understand now, but you are in the right way. You can do this