geometry-and-trigonometry

[thorny forge]

oh waaaa

that’s weird

[dark sparrow]

wait hold on

...can you try calculating cos(100) on your calculator?

i have a sneaking suspicion

[thorny forge]

okii

it’s 0

[dark sparrow]

aha

thought so

[thorny forge]

0-0

[dark sparrow]

your calculator is in grad mode

[thorny forge]

ohhhhhh

wait

[dark sparrow]

grads are a rarely used unit of angle equal to 1/400 of a circle

you'll need to reset your calculator to degrees

[thorny forge]

that explains it i spent ages trying to find what i was doing wrong

changing to grad will not disturb anything will it

DEG*

[dark sparrow]

well changing to deg is kind of necessary

[thorny forge]

ohhh it’s not gonna change any answers that u know of

[dark sparrow]

??

[thorny forge]

ah nvm thank u so much for ur help :)

[lapis pine]

Length of PQ PR

[dark sparrow]

@lapis pine the last few problems i've seen you post, i haven't seen you demonstrate any attempts you made to solve them yourself

should this be taken as a sign that you are stuck not knowing how to begin?

[lapis pine]

Yes

Whenever I don't even know where to begin with. I post here. Cos I'm self studying

But this time I've made few attempts. Figure out length of PR

[dark sparrow]

you should be explicit in saying "i am stuck and don't know where to begin"

[lapis pine]

Figured out <P

[dark sparrow]

(and if you have progress, you can and should share all of it)

so you have calculated PR and angle P, have you?

[lapis pine]

Yes

[dark sparrow]

wait actually
how did you calculate angle P? it doesn't look like you could do that right away

[lapis pine]

Shit

[lapis pine]

Tan inverse (800/900)

We get total angle of pr with Y

[dark sparrow]

ah, so you just did arctan(800/900) - 30°

of course

[lapis pine]

Deduct 30

[dark sparrow]

based on the angles marked here in pencil, it will be possible to calculate all the angles in your triangle.

and then you can just use the law of sines

[lapis pine]

Subtract*

Yes but I don't know how to figure out even one angle from remaining two

[lapis pine]

How ?

Just give a lil hint

[dark sparrow]

well you can calculate angle R in a similar fashion as you did angle P

draw a ray RZ extending directly south from P, then angle QRZ will be 45° and angle PRZ is arctan(800/900)

[lapis pine]

Thanks ma'am

@dark sparrow 🙏

@dark sparrow what do you think how much dumb I'm? (Asking for self awareness)

[dark sparrow]

...what?

did you mean to ask "how dumb do you think i am?"

[lapis pine]

Yes

[dark sparrow]

i cannot answer that question

i have no value judgment of your intelligence in mind

[lapis pine]

😕

Np have a good day

[fast grove]

pretty stuck on this

can I have some help please?

[fast grove]

<@&286206848099549185>

[upper karma]

here you go

although I drew circles having almost the same diameter, the construction and the proof works for irregular diameters of S and T too.

[granite matrix]

Can someone help me with polygon rotation around a point

Im doing odds

[pearl lava]

Why is -cos60 equal to -sin30?

[wanton saddle]

can somebody solve this i have the right answer but im not sure if its right my answer: 9.82

[wanton saddle]

area

.

( my name is marwan 16 years old

)

[spiral plover]

Hello

Can someone help me with geometry

[lost patrol]

is -4sinxcosx equal to -2sin2x

[lone copper]

yes

[upper karma]

How do you get b and c

Data is... triangle is equilateral all vertex lie on the circle and each side is equal 8

Wait... that's too easy..... sorry guys no need to help now but i can give the answer if anyone is interested

[median cobalt]

hellooo, can anyone help me answer this? tia!

[scarlet plaza]

this was actually featured on a video that i watched

a few days ago when i was do ing quiz on youtube

let me try to find it

here it is! https://www.youtube.com/watch?v=cPNdvdYn05c&t

[uneven ferry]

can somebody help me?- not really much good at geometry

[uneven ferry]

just need the 3rd and 4th answer

[uneven ferry]

2 and 4th ^

[molten delta]

I identified some of the unknown angles but I’m stuck

[molten delta]

<@&286206848099549185>

[upper karma]

well A should be 20 not 100

180 - (80 + 80) = 20

[molten delta]

Oh right

Let me fix that real quick

Alright so mBAC (or whatever you’d like to call it) is 20

[molten delta]

So what do I do next?

[upper karma]

this is hard as hell lmao

anyways I managed to figure it out ... sorta

but I'd rather not act like I managed to solve it alone so here's the video with the exact same problem

https://www.youtube.com/watch?v=HQc-54hQ8kw

I hope this isn't ur homework 😭

[molten delta]

It actually is…..

But the fact that they just took the problem out of a YouTube video lmao

[upper karma]

I feel like ur teacher was trying to stump you all

but you can prove them wrong

😈

[molten delta]

Lol ty

[olive solstice]

ye no wonder it took so long to write
i thought i was missing the pt as well lmao

[dusky surge]

Wow! That's so hard

[dark thicket]

Ill think about it

[dark thicket]

Ok i realised something

The 100 on the right is wrong

Nvm someone already said that

Oh someone already helped you

ok

[dapper marlin]

does someone want to help me with a geometry paper it’s 10 questions

[wise hornet]

i've proved the equation holds via brute force

but i can't for the life of me relate it to the second half

and i feel like it's really obvious

i mean it says 'deduce'

and i'm just not seeing it

[solid mantle]

Can someone lmk what the equation would be?

[remote hill]

y=6tan(9x)

[storm portal]

You should try figuring it out yourself. You'll have to use the amplitude of the graph and its period to determine what A and w are

[solid mantle]

Yeah I actually got it it’s always a hit or misss with these

[ancient oyster]

can someone explain how this works please

[solid mantle]

No that wasn’t it

[remote hill]

:sadge:

[remote hill]

for 25 and 26 since both sides have same base

u can take the exponents of both sides and equate them then solve x

[ancient oyster]

so e = e^-2

?

[remote hill]

25 will be x=x^(2)-2

(-x) ^2+x+2=0

solve for x

[ancient oyster]

x = 2

[remote hill]

do u know how to solve quadratics

[ancient oyster]

not really

[chrome oak]

I need help finding the perpendicular of BC

<@&286206848099549185>

[tight frigate]

hey can I get some help with 4questions

please dm

[nocturne pebble]

Can someone help with problem 13?

[nocturne remnant]

q13 is a common configuration

[nocturne pebble]

C = 12.25 E = 6.8 D = 3.75 are the answers, but I don't know how they were found

[nocturne remnant]

and has tons of similar triangles

[nocturne remnant]

when you have an altitude cutting right through a right triangle, forming 2 other right triangles

then all 3 right triangle are similar

[nocturne pebble]

theyre similar?

[nocturne remnant]

yep!

its not hard to prove

[nocturne pebble]

I have a basic idea already, but I have no idea where to start

[nocturne remnant]

you are given 14 : 16

[nocturne pebble]

yes

[nocturne remnant]

so the corresponding ratios for other triangles are alos 14:16

[nocturne pebble]

I can't think of similarity equation to set up with those numbers only

[nocturne remnant]

consider the triangle with c, e, and 14

which two sides are in ratio 14:16?

[nocturne pebble]

14 is the hypotenuse and 16 is c + d?

[nocturne remnant]

yeh but

in triangle with c, e, and 14

c : 14 = 14 : 16

cuz of similar triangles

actually maybe someone help here im going off :/

[nocturne pebble]

Nvm, I figured it out. I just have really bad visualization skills

med/long correct?

[nocturne remnant]

after you find c, just pythag chase or sth to get to e and d

[nocturne pebble]

yeah thanks

[next pecan]

how does one do this? factoring out the sin2x to 2sinxcosx doesnt seem to help

[pure cape]

Assuming you are supposed to solve for x

Move the cos(x) to the left hand side and then factor out a cos(x)

Then just solve it normally

[upper merlin]

cosine rule?

[upper karma]

Without calculator?

[upper merlin]

i think cos 120 is standard

[upper karma]

But it said I only have to know up to 90... and the length will be in negative

[upper merlin]

uhh idt so cos120 = -0.5 and that won't give negative

[upper karma]

Yeah I know

[upper merlin]

other way could be to draw perpendicular from A

sorry i meant C

[upper karma]

Answer is wrong

[upper karma]

I did it and got the same result

I imagined that the triangle is a Bart of equilateral triangle where all sides are perpendicular

[upper karma]

Wait... that was wrong

I got it

[shut finch]

does anyone know graph function for golden spiral

[sonic fern]

Let CD = base(b), height = h

Area of ADP = h(DP)/2 and area of BPC = h(PC)

Sum = h(DP+PC)/2 = h(CD)/2 = bh/2

Area of ABP = bh/2

@upper karma

[rustic slate]

So I stumbled upon this strange property,
If you construct 2 circles on a big circle and then create a line like in the picture, then the purple segment has the same length as R (the radius of the big circle)
Does anyone know why that's the case?

[heady dove]

<@&286206848099549185>

I need help here please

I don't have much time to wait for 15 minutes

[upper karma]

uh

4x-7 = 3x+1

x = 8

m<ACB = 50

50 degrees i think

plug the x into the two equations

[heady dove]

Thanks a lot mate

How about this:

@upper karma

[upper karma]

c

[heady dove]

Thanks man

<@&286206848099549185>

[fallow edge]

bro

read #❓how-to-get-help

ffs

[heady dove]

lol

ok ok

I just need some help

calm down

[lime crown]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[heady dove]

okay understood, but can you help me with my question please. Since you are here

[tropic shard]

Read the image I've posted after the bot please.

[heady dove]

I did

[tropic shard]

Could you please do what it says?

[heady dove]

I did tried

[tropic shard]

Could you "show what you tried to solve the problem"?

[heady dove]

I am not good at math

[tropic shard]

Okay I might be too pedantic. Point is this is the third question you gave where you never showed your attempts so it seems you're ignoring what the underlined part says.

[heady dove]

because it on a computer?

[tropic shard]

Do you know what a kite is?

[heady dove]

no?

[tropic shard]

Okay so you see that we have two triangles, ACD and ABC

[heady dove]

mhm

[tropic shard]

Sides AD and AB must be the same length due to the angles and the side,

So the triangles can be said to be *blank*, you know what goes in the blank?

[heady dove]

No?

what next

[tropic shard]

How would you determine if two triangles are congruent?

[heady dove]

If two angles and the non included side of one triangle are equal to the corresponding angles and side of another triangle.

[tropic shard]

Sounds about right (I'm probably tired to not know what you mean by "not included side"),

[broken ingot]

does anyone here have any expeirence with trig?

[tropic shard]

But I think it's called ASA

for Angle-Side-Angle

Do you believe we can apply that in this case?

[heady dove]

yes I think

[tropic shard]

Indeed,

So we know the two triangles (ACD and ABC) are the same. Are you able to visualize how you're able to move one to overlap the other? (that's how you can get the answers)

[formal pike]

8,2?

[upper karma]

sure?

[noble heath]

After some hard work I figured it out. Thing that leads to the solution: ||Prove that a rotation of 60 degrees clockwise about D maps triangle ACD to triangle GED in this diagram||

[surreal wolf]

How to find
Tan210-cot(-240)

[rustic slate]

that's really cool, thanks for the help and the insight!

[upper karma]

Guys I need help... help 23

[fallen sleet]

is this from euclidea?!

great game haha

[ivory path]

I'm currently studying the unit circle.. Is the co-ordinates 0,1 false in tangent?

[upper karma]

Hi guys... help 12 if anyone is free

[oblique hatch]

What's 12?

[upper karma]

Channel help 12

[oblique hatch]

Oh!

[upper karma]

Take a look please

[scarlet plaza]

@hard radish the guy didn't type .reopen

[hard radish]

lmao

[dusky surge]

#❓how-to-get-help

[wanton sparrow]

Thank you

[safe frigate]

why aint it 0

what did i do wrong

[dusky surge]

sin(x) ≠ x
sin(x) is just ≈ to x when x is "small enough"

[rustic slate]

yep it is lol

I still don't understand how all of those solutions in it work so yeah

[junior plinth]

anyone here a trig wiz?

[junior plinth]

i will pay someone to help tutor me

ive got a test next week and am in desperate help for someone thats knowledgable

<@&286206848099549185>

[ember heart]

You can ask some doubts that you've based on your understanding 😕

[loud shard]

can someone pls help

[sick sable]

remember the relation between tan, sin and cos

use the fundamental identity

then you can find either

after this, use the double angle formulas

turns into

[junior plinth]

@sick sable ill pay u for an hour tutor sesh?

[sick sable]

how are you planning to do it?

[junior plinth]

just over discord call?

[sick sable]

I'm not the best english speaker but we can try

[loud shard]

can u help me witht this one?

[sick sable]

also depending on the subject

[sick sable]

use the double angle formula in reverse, find cos, sen, then tan

the information about the interval is important to make sure you get the signal correct

[humble pulsar]

????

That's completely false

[upper karma]

Yes

I put wrong the equation

$3=\tan{a}=\frac{\sin{a}}{\cos{a}}$

[somber coyote]

Juan Caballo

[winter pumice]

Who are you responding to? Is this a new question? Use arctan. You will have a solution in Q1 and Q3 (why?).

This says you need a triangle whose side ratios are in y/x = 3/1 proportion

[feral charm]

uh can someone find the answer and explain it to me? i have 0 clue how to solve it after finding the circumcenters of both triangles

[winter pumice]

idk...8?

[feral charm]

yeah ur correct

i figured it out like 30 mins ago

tyvm

[winter pumice]

How though, b.c. I could only figure it for the right triangle case, which is very specific. I couldnt prove (to myself) it wouldn't matter, or why this works for all cases

@feral charm

[sonic fern]

a/sinA = 2R

Extended Sine Rule

[winter pumice]

ah! Thank you

[silk jacinth]

could someone help me troubleshoot a solution to a simple trigonometry problem? at the moment I get a result that's different from the textbook but I'm having a... let's say cognitively poor day

[dark sparrow]

ok, can you share the problem, your solution and the textbook's solution?

[silk jacinth]

the question is (I have to translate these since they're not in English):

"What solutions to the equation _sin 2α = 1/√2 _ are within [-π,2π]?"

My solution:

1. Determine what angle corresponds to the y-coordinate 1/√2 —> one such angle would be π/4

2. 2α = π/4, which means that α = π/8

3. Therefore, all angles with the same sine as α are:
   α + n *2π
   and π-α + n*2π (supplementary angle)

[silent plank]

you made the common mistake of isolating alpha before considering the period and supplementary identitiy

[silk jacinth]

so I'd need it to write the equation as

2α = π/4 + 2π*n instead ?

[silent plank]

for 2α in quad1, yes

[silk jacinth]

due to brain fog I'm having trouble understanding why this is

[silent plank]

why what is

[silk jacinth]

what I mean is that I don't understand how the period and supplementary identity are "considered"

[silent plank]

unit circle

[silk jacinth]

I don't understand. is that a hint or a prompt for me to draw one and see what the angles look like or something else?

[silent plank]

its a prompt to recall properties of the unit circle

something you should have a decent idea of

(and/or properties/itentities of trigonometric functions which are linked to that circle)

[silk jacinth]

clearly I've missed some crucial insight because I still don't understand even though I have the basic grasp of the unit circle and its relations to trig. functions

doh well I'll just hit the books

[livid moss]

sin(pi - t) = sin(t)

Using this gives you another solution in [0,2pi]

And then you have to add multiples of 2pi to both solutions to get the general solution

[silk jacinth]

I do know this. it's just that the 2α made it hard to comprehend

[livid moss]

t is now 2α

[silk jacinth]

[retried]

I understand that β = 2α is basically completely unnecessary

but it just made something in my head go click

[livid moss]

Sure, the last part is wrong

[silk jacinth]

oops

[livid moss]

You need to apply the identity to the angle you are taking the sin of

β = pi - pi/4 + 2pi n

[silk jacinth]

okay, I'll try it once more 😄

sin 2α = 1/√2

let β = 2α

sin β = 1/√2
sin β = sin π/4

β = π/4 = π/4 + 2πn
or
β = π – π/4 = 3π/4 + 2πn

β = 2α:
2α = π/4 + 2πn —> α = π/8 + πn
or
2α = 3π/4 + 2πn —> α = 3π/8 + πn

[livid moss]

Yes

[silk jacinth]

I think I now understand why I need to apply the supplementary identity and "full circles" (not sure about these terms in English, I assume that's the "period" that was mentioned previously?) prior to any isolation

it was that β = 2α that helped me see this 😄

it was that β = 2α that helped me see this 😄

thanks

[livid moss]

👍

[silk jacinth]

I have another assignment, my answer is correct but the way I've reached it differs from the textbook's example and now I'm wondering if my method is valid:

I need to solve α here:

sin (α) * cos (α) + sin (α) = 0
sin α (cos (α) + 1) = 0
sin (α) = 0 or cos (α) + 1 = 0

1st half:
sin (α) = 0
sin (α) = sin 0 or sin (α) = sin π
—> α = 0 or α = π

2nd half:
cos (α) + 1 = 0
cos (α) = -1
cos (α) = cos π
—> α = π

—> α = 2πn or α = π-2πn

am I making any subtle mistakes here?

[lost fulcrum]

I am been stucked at this questions-

1. 16. In a ∆ ABC, ∠A = 35° and ∠B = 65°, then the measure of ∠C is:
       (a) 50°
       (b) 80°
       (c) 30°
       (d) 60°

2. An iron rod 5m long is placed against the wall in such a way that the foot of the rod is 3m away from the wall. Find how high the top of the iron rod reaches in the wall?

Can anyone help me out?

[winter pumice]

1. Do you remember what all triangles have in common?
2. This question says the rod is 5m long, and it's placed like this:

   /|
  / |
5/  |
/___|
  3

[upper karma]

Can you name a line with only one point?

[dusky surge]

If you are talking about straight lines, with one point, you can find a family of straight lines passing through that point

[upper karma]

I don't quite get what you mean, but I'll sketch it out for you

[dusky surge]

Hmmm...

[upper karma]

like it doesn't even tell you if it's a point on the line or the line itself

can you assume that "a" is a line then?

[dusky surge]

Usually, we will L for lines, and A,B,C… for points

And in case you are having more than 1 straight line on the same graph

We can use $L_1,L_2,…$

[somber coyote]

Biscuit

[upper karma]

bruh

I don't even know what you mean

[dusky surge]

That is

Let's say you have a point A.

Where a line is also on A.

It's hard to tell whether A is representing the point of the line.

So, we have some convention to use block letters other than L for points. And L for the line.

So, in this picture, A will be the point (0,2) and L is the line y=x+2

[winter pumice]

@dusky surge sometimes I think this server is an experiment designed for trolls to abuse mathematicians--like Westworld, except nobody's getting laid.

[ivory path]

Use pythagorean theorem, a^2 + b^2 = c^2

@lost fulcrum

Only if c^2 is the hypotenuse

for this we can do: c^2 - b^2 = a^2

5^2 - 3^2 = a^2

25 - 9 = a^2

16 = a^2

4 = a

[dusky surge]

lol

[ocean snow]

helpp problem 2

[dark sparrow]

@ocean snow is this a test?

[mystic karma]

1. 80 , 2)4 m

[wise pawn]

what have you tried

I think you mixed up the order, (1,-2) means x is 1 and y is -2

you're adding an extra 4 in there

y=-4x+b
-2=-4(1)+b

parallel means the same slope

the line you have is y=-4x+b so using that point (1,-2) you're solving for b

which will be different than the original line

what you're saying sounds partly wrong and partly right

what do you get for b

look at what you're saying

"-2=-4+-6"

think about it

this step is good

but you have to add 4 to both sides

4-2=4-4+b

the point is to get rid of -4 you have to cancel it with +4

what you're saying -2=-4+-6 makes no sense

cause -2 is not equal to -10

yeah, you have to because otherwise it won't be equal anymore

if both sides are equal, you have to do the same thing to both sides to keep it equal

you tell me

what is -4+2

nice

[nocturne pebble]

Can someone help with thid

this*

[mystic karma]

god usopp studiyng math

[robust tide]

Can someone help?

Problem:
Find the first three terms of the geometric sequence whose sum is 7/16 and product is 1/312

I don't seem to go anywhere with this problem, i've tried

x/y + x + xy = 7/16
x/y * x * xy = 1/312

But still I don't get the right answer

[ocean snow]

umm no, it's jus an assignment

[austere jewel]

hey guys, does anyone know if we can simplify this?

[dusky surge]

There are quite a few ways in my mind right now for your question.
But I would suggest this as it the results seems to look prettier.
https://image3.slideserve.com/6453499/slide41-l.jpg

Careful with the coefficients.

[broken ingot]

does anyone know how to write cos(x) in terms of csc(x)

[still mica]

Am I correct in assuming this is wrong?

[summer canopy]

yoo guys

I have a question on a simple matter

are those vertical?

so like are they congruent

[nocturne remnant]

vertically opposite
so yes, their measures are the same

[summer canopy]

one more question

is 6x-7 and 3x-29 = to 180?

[nocturne remnant]

ye

(6x-7)+(3x-29) = 180

[summer canopy]

cool!

and we call that format

consecutive interior angles?

did I do it right?

for this problem

[ivory path]

do 6x -7 = 8y + 17

[summer canopy]

No

Lol

[ivory path]

huh

[summer canopy]

I'd rather find what's x first

[ivory path]

then you need to find y first

you fool

[summer canopy]

Find x first

[ivory path]

y

bro

[summer canopy]

9x -36 = 180

[ivory path]

wait nvm-

nvm-

[summer canopy]

X= 24

[ivory path]

I thought it was 6x - 7 = 8x + 17

[summer canopy]

We plug it in with that

[ivory path]

thought they're the same variable-

[summer canopy]

Lol

[ivory path]

Ok then do 6x - 7 + 3x - 29 = 180

Combine like terms

get 24

then do:

6(24) - 7 = 8y + 17

[summer canopy]

.

I did it that way

Submitted already

:p

Thx tho

[ivory path]

oof, am sorry I didn't help lol

but yeah it seems like you did it correctly

[upper karma]

someone please help me in this ques....

[dark sparrow]

have you made any progress so far? @upper karma

[upper karma]

i tried but no conclusion

[upper karma]

i found the median by apollonius theorem and altitude also

[dark sparrow]

do you have a diagram or are you trying to do this blind?

[upper karma]

yes

[dark sparrow]

okay, then show your diagram please

[upper karma]

ok

this one

[dark sparrow]

,rccw

[dark sparrow]

okay, let's take a look

okay, the diagram checks out so far

[upper karma]

how to continue?

[dark sparrow]

i have a feeling that the area of AQPRB will be some rational multiple of the area of ABC

oh actually hold on

here's something

look at triangle CXD, for which CA and XP are medians

so XQ:QP = 2:1

you can find the ratio of area(QXA) to area(PXD) this way

find more area ratios in the same fashion and you should eventually arrive at the ratio of area(AQPRB) to area(ABC)

at which point it should be easy

[upper karma]

?

[dark sparrow]

hm?

is there something you don't understand in what i said?

[upper karma]

how PXD

[dark sparrow]

what do you mean?

[upper karma]

i mean how will i find area of pxd

[dark sparrow]

you don't need to

i'm telling you that it's possible to find the ratio area(QXA):area(PXD)

without necessarily knowing each individual area

[upper karma]

ok let me try

[upper karma]

unable to do please help....

[dark sparrow]

what are you unable to do?

[upper karma]

this area ratio

[dark sparrow]

QXA:PXD?

[upper karma]

yes

[dark sparrow]

do you understand this?

[upper karma]

yep since it is median of cxd

[dark sparrow]

please don't reply-ping me on every message. it's annoying.

[upper karma]

ok

[dark sparrow]

ok, and do you know the formula for the area of a triangle in terms of two sides and the angle between them?

[upper karma]

1/2 ab sin (angle between a and b)

ok i understood

[dark sparrow]

oh, you did?

[upper karma]

thanks

[dark sparrow]

does that mean you're able to continue all the way to the solution now?

[upper karma]

yes now i think i can

[dark sparrow]

ok, as you say.

[upper karma]

hey

please confirm my answer it is coming 56 sq. units

[dark sparrow]

i have the same answer as you.

[upper karma]

wow

thanks

i have one more question

[upper karma]

please help in this question also

[dark sparrow]

you're doing the reply pings again...

[upper karma]

o sorry

this time i will not

[dark sparrow]

and have you made any progress on this so far?

[upper karma]

i just know that this inequality holds when P0 is centroid and P is any other point in the triangular plane

[dark sparrow]

......

what did i say about reply-pings?

need i repeat it another thousand times?

[upper karma]

o sorry i forget it again and again

[dark sparrow]

also, i'm not so sure what you said there is true.

from what i googled just now, the point that minimizes AP+BP+CP is called the Fermat point, and i see no reason why it'd always coincide with the centroid.

[upper karma]

i have the proof

wait

[dark sparrow]

you do?

please share.

but this is about the minimizer of the sum of squared distances.

not the sum of the distances themselves.

[upper karma]

so u tell how can i solve that question

[dark sparrow]

....

https://en.wikipedia.org/wiki/Fermat_point here's what i found by googling.

maybe read that and take a look for yourself.

[upper karma]

ok

[upper karma]

i read that it means that P0 is the fermat point

[upper merlin]

does knowing that ABC is right-angled help?

[upper karma]

i know

but don't know how to use it

[upper merlin]

well u know the the point P0 is the centroid and the medians of a triangle intersect each other in the ratio 2:1, as this is a right-angled triangle its fairly easy to find the length of its medians

[upper karma]

but @dark sparrow said that fermat point concept will help

[upper merlin]

this looks like a tougher way of constructing the medians ig

cuz we still are joining the midpoint of that side to the vertex of the triangle

wait nvm its wrong

[upper karma]

yes

[upper merlin]

the only way I see to solve this is coordinate geometry

the fact that this is 30-60-90 triangle makes it a bit easy to get the coordinates

[upper karma]

yep

[upper merlin]

nicee u got the solution?

[austere jewel]

I just wanted to know if i did something wrong here i only worked out the left side and wanted to know if i did it correctly (trig identities)

[vagrant shale]

Is this right

[vague fox]

if you are doing a 180 degree rotation about the origin then both your new x and y values should become negative

so yes, you are correct

[compact wigeon]

can someone help me, i need to find the volume. (solid of revolution)

i got 26π but idk if that's right

[winter pumice]

I think you are just revolving the area under f(x) = -2x + 10 from x=2..5.

[compact wigeon]

and what's the answer

[winter pumice]

gj

[steady shoal]

Volume of a cone =1/3 pi r^2 h
Revolving a right angled triangle about its length won’t change its radius or height
So r=3
h=6
V=1/3 pi 9*6=18pi

[upper karma]

i dont think this is a simple cne

they explicitly stated the distance between the line and the right triangle

if we rottate the right triangle along the line

we would get a frustrum

[sturdy trail]

Can someone help me how to do this?

[lapis pine]

Can anyone explain me
How cot² (45 + ∅/2)
Becomes
(1 - sin∅)/(1 + sin∅)

[dark sparrow]

don't use the empty set symbol for the letter phi

@lapis pine do i understand correctly that 45 is supposed to be 45°, i.e. pi/4?

[lapis pine]

Yes

[lapis pine]

Ok

[dark sparrow]

so you're asking how to prove $\cot^2(\pi/4 + \varphi/2) = \frac{1 - \sin(\varphi)}{1 + \sin(\varphi)}$

[somber coyote]

Ann

[lapis pine]

Yes

[dark sparrow]

have you made any progress so far?

[lapis pine]

Asking coz no idea

No

[dark sparrow]

may i suggest writing cot as cos over sin?

[lapis pine]

Yes

[dark sparrow]

well then do it

and see what you end up with

maybe it'll help you progress

[lapis pine]

But 45 + X/2 ??

I'm really noob at trigo

Cos²/sin² (45 + x/2)

[dark sparrow]

$\frac{\cos^2(\pi/4 + x/2)}{\sin^2(\pi/4 + x/2)}$

[somber coyote]

Ann

[dark sparrow]

consider now that cos^2(t) = (1 + cos(2t))/2 and sin^2(t) = (1 - cos(2t))/2

[lapis pine]

Thank you @dark sparrow

[stray pollen]

can someone help me with this

no call

just want an explanation with answers

please

[obsidian robin]

For 13, AB=AC
So, you can use the property that of opposite sides are equal then the corresponding opposite angles are equal
So, mAngleB=mAngleC= x
Hence, x+x+96=180
=>2x=84
=>x=42
Now, you may know Vertically opposite angles are equal
So, mAngleACB=mAngleECD= 42 degrees
By the same property that we used first, mAngleE=mAngleD=y
y+y+42=180
=>2y=138
=>y=69
Hence, mAngleE=69 degrees

[obsidian robin]

For 15th, you can use the same property of mAngleD=mAngleE
=>x^2=3x+10
you got a quadratic equation, so on solving(https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:quadratics-solve-factoring/v/example-1-solving-a-quadratic-equation-by-factoring) this, you'll get your x value, then using the property that the sum f all angles in a triangle is 180degrees, you can find the value of mAngle F.

[upper karma]

no! unable to do

[upper merlin]

after you have got this, you just have to find the intersection of the lines CE and BD which will be P0

[glossy charm]

can i ask for help here?

anyways

[upper karma]

ok i will try

[marsh quiver]

I need to find the "d" side, also A and C angles (they are equal since the triangle is isoceles)

does anyone have the answer?

[dusky surge]

We don't give away answers.

And as for how to find it

Have you learnt cosine rule?

[marsh quiver]

i just want to know if this is possible

[dusky surge]

It's possible.

[marsh quiver]

How.

[dusky surge]

And d is unique.

[marsh quiver]

wdym by that?

[dusky surge]

Nothing special, sometimes we may have more than 1 answer

[marsh quiver]

can you guide me to the answer? make me understand

[dusky surge]

Hint: draw the height of the triangle with base d that passes through B

[marsh quiver]

i did this, so, if i divide the triangle, would angle B be 75° or 60°? after all, the isoceles triangle became a rectangular triangle

[graceful halo]

75

[dusky surge]

As it is 150°, and we bisect it, it's just 150/2

[dusky surge]

Yes

[marsh quiver]

The only thing I know is the hypotenuse of these rectangular triangles and the angles

[dusky surge]

Looks good

Then now we can use sine cosine

[marsh quiver]

Hmm

but how

[dusky surge]

Have you learnt sine cosine tangent?

[marsh quiver]

A little

[dusky surge]

https://www.aplustopper.com/wp-content/uploads/2016/12/quick-review-working-sine-cosine-tangent-1.jpg

In this case, if you use 15°, then you can use cos15°

[marsh quiver]

cos15° = 0,9659 right?

[dusky surge]

Yea

And now find adjacent and hypotenuse

[marsh quiver]

in this case, adjacent would be d/2

and hypotenuse = 1cm

d/2/1?

i dont know

[dusky surge]

Yes

Then $\frac{\frac{d}2}1$

[somber coyote]

Biscuit

[dusky surge]

Is just d/2

But

[marsh quiver]

and why do i need it?

[dusky surge]

d/2 is not the adjacent

So

Be careful😁

The hypotenuse is the line on the other side of the right angle

[marsh quiver]

I divide the triangle, the base was d, so it became d/2, right?

[dusky surge]

Wait....

I was wrong!

[marsh quiver]

😱

[dusky surge]

Sorry man, I must be tried

You are correct

d/2 /1

Right?

[marsh quiver]

yes i think

[dusky surge]

So we have

cos(15°)=d/2

Then 2*0.9659=d

Then we have d

Done :D

[marsh quiver]

omg.

ty

[dusky surge]

That's how we work with sine cosine tangent, they are useful!

[marsh quiver]

i see

[lofty frigate]

how u do this

[oblique hatch]

Did you have a question about this?

[grand dune]

Hi can someone help me with ratios

[upper merlin]

what do you want to find?

[still laurel]

How is the ratio used to find the side values of 2 and -root 3?

[wintry tundra]

wdym the ratio

you mean the hyp/adj?

[still laurel]

yeah, this isnt my solution, im just trying to understand it

like how does secant 5pi/6 give -2/root 3

my main question is how do we find the side values of the triangle

[wintry tundra]

because the hypotenuse is 2 high and the adjacent is root 3 to the left

it says on the triangle

[still laurel]

right but that was skecthed later on, what if i just have the ratio, how do i find the side values then

for example like what if the ratio was sin 2pi/3, how do i find the value

[wintry tundra]

i mean the unit circle always has set values with the angles

bc the hypotenuse has to be a length of 1 constantly

looks like this circle has a hypotenuse of length 2 because itll make the ratio be less weird

[still laurel]

so what if i only had the ratio, what exactly is the process to find the value of it?

[wintry tundra]

so if you only had sec 5pi/6?

or only knew what sec meant

[still laurel]

nono ik what secant means

im saying whats the process

like for example heres the textbook question

[wintry tundra]

ok

so just the process

should i do it with the secant or one of these ones

[still laurel]

uhh actually would you mind doing it with the sin cuz then i think i can understand the rest after i see an example

[wintry tundra]

eh sure

so 2pi is all the way around so a third of that is 120 degrees

since 360/3

[still laurel]

yup

[wintry tundra]

so 120 is gonna have the same sin as 60

and its gonna have the negative cos that 60 has

bc its in the positive y but now its in the negative x

but we only care about the sin so we could really just use sin(60) to find it

or sin pi/3

[still laurel]

so how do we determine what other sin value to use

[wintry tundra]

?

what do you mean the other sin value

theres only one sin

a lot of these just have the same value since its on a circle

[still laurel]

120 has the same sin as 60, as you said, so what role does that play in the question

im just confused

[wintry tundra]

well that is relevant because a lot of these you just have to memorize and know what they are, so like sin 90 is one and sin 45 is rt2/2

[still laurel]

i have the solution from a textbook, sorry if im not understanding, maybe this could help explain it better?

[wintry tundra]

yeah

well theyre doing something a bit different with the circle by making it with a radius of 2

but sin is defined as opposite over hypotenuse so you just take the root 3 and the 2

since root 3 is the opposite of the angle

[still laurel]

ok but like where did the root 3 over 2 come from

[wintry tundra]

the opposite side is rt3 and the the hypotenuse is 2

because of how the triangle works

[still laurel]

no i understand that but couldnt also have been a diff triangle like 1,1 rt 2

how is the side value for this determined

[wintry tundra]

at a precalc level its kind of just the way it is

some of these you can logic out

[still laurel]

but like how am i supposed to know that

if its sin 2pi/3 is that triangle just something i should know automatically

as in memorize it

[wintry tundra]

i mean you can know the basic sin and cos values for angles

like pi/2 and pi/3

then you use logic from there to find out harder ones

[still laurel]

right but how would that help with finding the value for this triangle

like how would i apply that

[wintry tundra]

because if you know the height for pi/3, thats equal to 60 degrees

which you might notice is the same as 120 degrees but reflected over the y axis

so its the same triangle but with a reflected x or adjacent value

so here you should know that the opposite side from pi/3 is always root 3

then the hypotenuse is always 2

because unit circles have the same hypotenuse length

so then you plug opp/hyp for sin

and get rt3/2

@still laurel if you wanna know more about how to actually find these sin and cosine values with algebra i think you would be interested in the taylor series, but its kinda complicated

the taylor series is how ur calculator can find any number in you input sin cos or tan into it

[still laurel]

so really it just involves memorizing or knowing, to a certain extent, most of the values on the unit circle?

[wintry tundra]

yeah

which i can relate to that misfortune i hate memorizing stuff

its kind of like how 2 means 1+1 because it just is

theres not much more to it besides thats just what it happens to be

it is pretty cool to know that you can build the values for sin or cos with a series of powers of just one variable

so if you wanna get into the more algebraic stuff that relates to this then the taylor series would be pretty cool

hopefully i could be of help here, though its kind of difficult to explain intrinsic things

[normal roost]

help pls T^T

[upper merlin]

which question are you stuck at?

[lofty frigate]

I need angle dce

[normal roost]

everything

[upper merlin]

can u send a zoomed in picture of that question?

[nocturne remnant]

Omg begula is here

[upper karma]

Find locus of middle points of chords of a circle which pass through a fixed point.

Anyone want to try this out

[marsh quiver]

do I use cosine(75°)?

i need to find PQ

i named it "d"

it's ok

ty

ok

[marsh quiver]

@upper karma hey

sorry for ping

[upper merlin]

you can either use cos(75) or the cosine rule

[marsh quiver]

yeah i discorvered i think

sin15 = 0,2588 = d/2/8
sin15 = 0,2588 . 8 = d/2
d = 2,0704 . 2
d = 4,1
@upper merlin

i think is correct

[upper merlin]

yeap the answer is correct

[marsh quiver]

gg

[upper karma]

Cos75 is cos(30+45)

Use the identity

@nocturne remnant

[upper karma]

Try this

It’s pretty cool

[nocturne remnant]

It’s a circle right?

I seen it before

[upper karma]

Damn

U figured it out

[nocturne remnant]

Or an arc whatever lol

[upper karma]

It’s a circle

How u do it

[nocturne remnant]

I seen it before

[upper karma]

Oof

Big brain

[nocturne remnant]

If point is outside circle I think it would be arc only

[upper karma]

No it would still be circle

Afaik

Might be wrong

It will be two intersecting circles

OG and one from locus of points

[loud shard]

can someone please help me

[vagrant bolt]

anyone know about barycentric coordinates who can give clear intuition on the diagram shown here

i think it get it but ehh

what do they exactly mean by can be computed with two additions

rather than full area

[slender mist]

anyone here ?

[slender mist]

oh

ty

ahh

i got it

ty

can you help me

with this one though

haha its ok. I appreciate you

ahh

its asking for the

tan a

in simplest fractipon

fraction

tan a in simplest fraction form

[wintry tundra]

yeah so you take the opposite of a and then the adjacent

and do opp/adj

which as pebble said simplifies to 24/7

[slender mist]

and thats the simplest fraction form !?

[wintry tundra]

actually

its 7/24

[slender mist]

:hmm:

[wintry tundra]

cot a is 24/7

[wintry tundra]

yes because 14/48 simplifies to 7/24

[slender mist]

ah

[wintry tundra]

do you see why the tangent is why it is tho?

like why opposite and adjacent are those sides

oh you have math dyslexia?

i forgot what its called

my alg 2 teacher had it tho

so big respect for continuing in math despite it if you do

ah i see

[slender mist]

i be focusing all the time

and still have a d-

in maths

everybody thinks im dumb lol

even my tutor

hnmmmm

yes

🗿

What is cosz

wait i think

i got the answer

is it

28/35

i got a lil more to go!!!

let me

find how high the ladder reaches up to the building

to the nearest meter

ye

it

doesnt

find how high the ladder reaches up to the building!

[somber coyote]

Pebble

[slender mist]

ok!

i put it in my calculator

and it pops up as

error

[wintry tundra]

the thing about trig that frustrates me is all the memorization with it

yeah

its also annoying to have to say the reason the sin or cos or tan of a number is the way it is is because it just is

[slender mist]

oh.

[wintry tundra]

also pebble id try to explain it more

than just give answer

the whole teach a man to fish thing

[slender mist]

i got -3.048

[wintry tundra]

ur probably in radians

[slender mist]

i got

8.66

oh

[slender mist]

haha ty both

[wintry tundra]

np

[slender mist]

oh.

OH

thats what i needed to do

[wintry tundra]

ill explain it from a beginning standpoint

so the reason why you even bother to do sine is because notice how the height is opposite of the angle given

@slender mist do you see that on the triangle?

how side AB is opposite of the 60 degrees

[slender mist]

yeah

[wintry tundra]

and remember how sine is opp/hyp

bc SOHcahtoa

[slender mist]

opposite

over hypot

[wintry tundra]

yeah

just opposite of the angle

[slender mist]

ahh'

all i gotta do is

memorize

sohcahto

[wintry tundra]

yeah

sohcahtoa

[slender mist]

a

[slender mist]

and then i'll be set

right ?!

hmm

[wintry tundra]

you still gotta use some logic tho

[slender mist]

does it contain

[wintry tundra]

so ill keep explaining

[slender mist]

elements

from other topics

?

[wintry tundra]

AB is opposite of 60, so since ur given the hypotenuse and you need to find the opposite side you use sine

[slender mist]

let me

write this down

[wintry tundra]

since sine uses opposite and hypotenuse

@ztco

[slender mist]

i see

[wintry tundra]

so yeah

you have sin(60) = AB/10

[slender mist]

let me

see if i can do this

question

on my own then

with the knowledge you just gave me

[wintry tundra]

do you need me to explain any more? or do u wanna try it on ur own

also sin(60) isnt a variable, its has a set value so you can solve for AB

[slender mist]

i wanna

try this on my own

and then see if im right

[wintry tundra]

if ur wondering $sin(60) = \frac{\sqrt{3}}{2}$

[somber coyote]

hiidostuff

[wintry tundra]

so you can just plug that in

if you really want to

[slender mist]

using it rn

tan a

would be

[wintry tundra]

yeah

alright find the ratio first

sohcahTOA

[slender mist]

tan a would be 12/13

?

[wintry tundra]

no

because 12 is adjacent

and 13 is hypotenuse

[slender mist]

12 / 5

[wintry tundra]

super close

you put the hypotenuse on top

[slender mist]

:pain:

[slender mist]

oh

[wintry tundra]

cotangent is the reciprocal of tangent

[slender mist]

5/13

[wintry tundra]

so close

[slender mist]

:confusion:

[wintry tundra]

wait

yeah no thats the sine

because the longest side is the hypotenuse

remember the hypotenuse is always the longest side

[slender mist]

yea

slet me bring this down here

tan would be 5/12

then

oh

ty

@wintry tundra ty too

[wintry tundra]

do you understand why that fraction is the way it is tho zyco?

just to make sure you dont have to guess

[slender mist]

because of the way sohcahtoa is

[wintry tundra]

yeah

but do you know what sohcahtoa means

[slender mist]

opp of hypo

[wintry tundra]

yeah

over

[slender mist]

adj over hypo

and oppo of adja

[wintry tundra]

like how the SOH part means S = sine, O = opposite and H = hypotenuse

so the last 2 you put as a fraction

S = O/H

C = A / H

T = O/A

[slender mist]

so tan c

would be

5/12

[wintry tundra]

so so close

thats the cotangent

[slender mist]

tan c is 5/13

?

i mean 13