geometry-and-trigonometry

[vocal mauve]

But not a specific angle

16y - 20 = 60

[celest wave]

ohhhh

y=5

[lethal moss]

You have solved this earlier

[vocal mauve]

can someone help me for part b to this question:

[celest wave]

yes i dont understand

[vocal mauve]

the table asked you to find angle BAC and ACB so u just put in 60 i believe

[celest wave]

oh yes ur right

[celest wave]

can u explain in a easier way how u did that plz

[lethal moss]

The only important part is that you are trying to solve a single angle. If nothing else works for you just evaluate 16y-20 for y = 5.

[celest wave]

16(5)-20 = 60

[lethal moss]

And that is the measure you have been looking for

[vocal mauve]

because you know that its an equilateral triangle, (you know because all the sides are equal) then a rule for this type of triangle is that all the angles in the triangle will be equal. since 180/3 = 60, each angle in an equilateral triangle will aways be 60

[lethal moss]

a single angle

[vocal mauve]

themathboi can u help me with this

[celest wave]

and C is the same thing??

oohhh i get it

thank u

[lethal moss]

np

[lethal moss]

let me think for a moment

[vocal mauve]

ok thx

just so you know, my answer for one of the conditions was that BC = 12 because then the triangle is isoscles and there will be only 1 possible angle, I didnt get the mark for this answer

[lethal moss]

ohh, I was trying to do 8a

[vocal mauve]

oh no im only stuck with B

for a u just do sine rule and then use the sin graph to see two different positive values before 180

i just dont know how to do b

[lethal moss]

I think I can't help you this time ; /

[vocal mauve]

oof lol

[celest wave]

btw i need help with this llast one

[vocal mauve]

ill go ask in one of the other channels

thanks anyway

[lethal moss]

Knowing the side lengths, what can you say about the triangle overall? (Do not worry about specific features, I am asking mainly if the triangle is equilateral, isosceles, e.g.)

[celest wave]

isosoceles

[lethal moss]

What property do angles in isosceles triangles have?

[celest wave]

acute

[lethal moss]

Not always, this one is still isosceles.

any other ideas?

[red flax]

would it be 3x-4=65?

[lethal moss]

Let her figure it out herself

[red flax]

mb it just I am also learning this and wanted too see 😭

[celest wave]

yes because the bottom angles are congruent

[lethal moss]

by the way this was the feature I was asking for

But now you have an equation, you can solve it for x

sorry, my bad

[celest wave]

x=23

[lethal moss]

plug it into the table

[celest wave]

ok

how do i find y i get confused for this part

[lethal moss]

Now going for y,
You know the measures of 2 angles, and the identity that all 3 angles in a triangle sum up to 180 deg.
What do you do to find the missing angle?

[celest wave]

N and Q are both 65

5y+65=180 ?

[twin crag]

how do i find b

i already got a is 3

[twin crag]

wait i got a = 3 and b = 1/2

so would the equation be y=3sin(1/2x)

so a+b=7/2

?

[timber cargo]

https://tenor.com/view/obama-check-gun-point-gif-6281218

[sturdy rampart]

Any1 good at statement and proofs

[wary ibex]

yo can you subscribe to sjef entertaiment on yt

[vernal pilot]

No

[fickle rose]

how to calculate an angle of a non right triangle while only having one side and one angle given

[silent plank]

insufficient info

[upper karma]

it is not possible with just one angle and one side

unless it's a isoseles tringle

[vernal pilot]

With the info you are given it is not possible. However, if you have two sides and one angle you can find any side/angle with the law of cosines: c^2=a^2+b^2-2abcos(C).

[upper karma]

also known as the cosine rule or the cosine law

[somber coyote]

Mr. Macro

[stuck sleet]

what does it mean that "each face angle is 90 degrees to O"?

this is the answer:

how can I find line segment AB?

I found the line segments VA and VB and also try to solve the base

[stuck sleet]

What should I do next? Thanks!

[vernal pilot]

It means that that the angle in between OB and OV is 90 degrees, the angle between OV and OA is 90 degrees, and the angle between OA and OB is 90 degrees.

i think

might be wrong, but looking at the picture it makes sense, overall i think its worded pretty weirdly and they shoulu just annotate the diagram with what they mean.

[stuck sleet]

I get the answer now

The picture is wrong

[vernal pilot]

ok. what about the picture was wrong?

[stuck sleet]

you know a 3d drawing website?

I cant draw it properly

but I got the solution

[vernal pilot]

desmos.com/3d if you want to graph things in 3d

[dry bough]

do I need to be in radians or degrees when trying to find a length of a side using law of cosines

my tutor had it in radians now im confused since the angle is in degrees

[grave pond]

The angle is whichever unit you have measured the geometric angle with.

It's definitely wrong there to notate a length in degrees, though. Degrees are exclusively a measure of angles.

(also, according to my calculator, 34-30·cos(70°) should be 23.74 rather than 23.71).

[dry bough]

oh thanks for the look. out on that idk why I put 23.71 hmm.

[median pivot]

Help pls

[jolly nymph]

CALCULUS IS EVERYWHERE

[pliant tapir]

what grade r u guys in?

I'm a bit confused about our new topic

it's about circles

[vernal pilot]

q1 is impossible with the information given. we arent told whether all the angles or all the side lengths are the same (presuming it is an equilateral).

assuming it is an equilateral though, 2x=24, x=12 and 3y=24, y=8.

for q2, its an iscoleseles, so y must equal 65 degrees, and x = 180-2*65, so x=50 degrees.

for q3, its 2 isosceles triangles next to each other, connected with two lines. so we know x=180-2*59, or x=62 degrees, and we know that opposite angles are the same, so y=59 degrees.

as for q3, its an equilateral, so x+20=60, therefor x=40, and 4y=60, thus y=60/4=12 degrees.

[trim trellis]

i only need help with b(the second one), translation is as follows:

A kite with side lengths 10 and 24 was inscribed in a circle
b) Calculate the difference of the radiuses of the excircle and the incircle of the kite

[vernal pilot]

is that meant to be diagonal lengths? becuase if those are the sides then i dont think its possible.

[trim trellis]

sides

apparently its possible

idk how

[vernal pilot]

i might be able to do it with vectors but that just seems overkill.

[trim trellis]

im in year 8 i dont know no vectors

but i know some trig so that can be useful

[vernal pilot]

well you arent given any angles.

[trim trellis]

ye

[vernal pilot]

and unless i am forgetting a key part of kites, i dont think i can do anything.

[trim trellis]

idk

ur the undergrad im the y8

[vernal pilot]

wait, i have any idea

ill b back in 5 minutes

nvm, not possible

[trim trellis]

ok

wait

how do u think u can do that w vectors

i dont know no vectors

but i need the solution

the thing i tried is to use the equations for areas of quadrilaterals

idk how to say that in english

tangent

and the other ones

[vernal pilot]

nah my idea with vectors didnt work. i think they meant to say the "lengths" not side lengths.

because the "lengths" would just mean the diagonals.

because it is straight up impossible otherwise.

just say 120 units^2

idk man, i cant read russian and i dont know the terminology you learnt.

[trim trellis]

its not russian its serbia

n

[trim trellis]

tf does that mean

[vernal pilot]

sorry, i just say cyrrilic letters and assumed. and fuck i answered the wrong thing, but still its impossible.

[trim trellis]

oh its ok

[vernal pilot]

unless ofc they meant for the diagonals to equal 10 and 24

[trim trellis]

no no

with diagonals it would be too easy

[vernal pilot]

without diagonals its impossible.

[trim trellis]

ye

[trim trellis]

serbian has both cyrillic and latin

[vernal pilot]

just assume those are the diagonals and answer the question if you need an answer. otherwise ask your teacher (assuming this is homework or something), to clarify the question.

[trim trellis]

i kinda prefer latin, cause cursive cyrillic is my mortal enemy

[trim trellis]

its not hw

[vernal pilot]

well if you got this question off a teacher ask them to clarify it

[trim trellis]

there are some advanced math classes im taking, and it was one of the questions for competition prep

[vernal pilot]

otherwise see if there is an answer sheet

[trim trellis]

ok tyy

[trim trellis]

the competition is way easier lol

[vernal pilot]

because i completely do no believe this is possible with just the sides of the kite, so they probably just meant the diagonals.

[trim trellis]

they didnt

it would say dijagonale if it meant diagonals

[vernal pilot]

then they have left out a critical part of this question.

[trim trellis]

myb a typo i hope

[vernal pilot]

they may have used different terminology.

[vernal pilot]

dont worry about it, its just the internet so typos dont really matter.

[trim trellis]

nah there is no different terminology for sides and diagonals in serbian

sides have one word

diagonals have the other

[vernal pilot]

well, you can also just say lengths for the diagonals.

some textbooks (notably the inferior ones), say stuff like that.

[trim trellis]

this word means side lengths in serbian

[vernal pilot]

ahh, okay.

[trim trellis]

np bro

hard language to translate

[winter garden]

I was looking at a video the other day explaining why the formula for an ellipse π(a+b) is not accurate so I came up with this.

Is this more accurate or less and can you please tell me what I got wrong, this is based on 2 circle intersecting

The 2a under cos is actually a 180

I wrote that one wrong

[vernal pilot]

how did you derive the formula?

also is it in degrees or radians?

(assuming degrees because of the 180)

[winter garden]

Yes degrees

The formula if from the intersection of 2 circle

Is*

[vernal pilot]

its wrong for a circle with radius 1.

therefor, im going to go out on a limb, and say that it is wrong overall.

[smoky jetty]

any1 wanna try? I couldnt solve it mentally

[winter garden]

I checked and it does give 2pi so it isn't wrong for radius 1

[vernal pilot]

area of a circle radius 1 is pi

[winter garden]

Not are circumference

Area*

The formula is circumference

[vernal pilot]

oh, dam, nvm then

[winter garden]

The area of ellipse is accurately πab

[vernal pilot]

i was just a bit confused becase piab isnt how you calculate the circumference of an ellipse.

[smoky jetty]

nvm, this problem is actually crap

the given values are not coherent

[vernal pilot]

$c\approx\pi ab(1+\frac{3h}{10+\sqrt{4-h}})$

[winter garden]

Heh?

[vernal pilot]

latex isnt working

[somber coyote]

TheLord26

[vernal pilot]

im so good at latex

$h=\frac{(a-b)^{2}}{(a+b)^{2}$

[somber coyote]

TheLord26
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vernal pilot]

i hate latex

[winter garden]

Dude it just copied from the internet

Is an approximation

My goal is also to approximate

[smoky jetty]

can u calculate the circumference of ellipse by making a=b, then divide 2pi*r by the scale factor which makes a equal to b?

[vernal pilot]

we have good approximations already, if you want to see if yours is good just compare them

[winter garden]

But we don't have the right answer so how to compare

???

[vernal pilot]

just compare your approximation with one that is already out there

we have very good approximations.

theres, like at least 3 different ways to calculate it

[winter garden]

Mine is 9.30 they 9.6 correct answer unknown

A 1 b 2

[vernal pilot]

yours is probably wrong then.

[winter garden]

Probably

Still we can't just compare it this way

Because the formula they used is not 100%

[vernal pilot]

let me put it this way, we have formulas that, if they were to run forever, could accurately determine the circumference of an ellipse, but are limited by the fact that we cannot allow them to run forever. and that is how calculus works.

[pearl wave]

Wdym?

[winter garden]

Also if you check a circle with radius 1 and 1 it gives π√1+6/10+√2

[vernal pilot]

i probably misremembered it.

[winter garden]

With their formula

Tho I do see that somewhere

[vernal pilot]

no its not.

$h=\frac{(a-b)^{2}}{(a+b)^{2}}$

[somber coyote]

TheLord26

[winter garden]

How do I use that latex thingy

[smoky jetty]

draw a line perpendicular to diameter through the center =8. However, by 60-30-90, that line must be = 4sqrt.3

[winter garden]

My writing sucks

And I am assuming H is the height

H*

h

[vernal pilot]

get gud, nah just type the cash signs, and then you use latex for the fancy functions, \frac{a}{b}, \sqrt{}, ^{}, _{}, \cdot (i dont think * works)

ect

also you got the wrong answer for r=1.

with the formula i remembered.

[winter garden]

Yeah I assumed h to be 2

Because I thought it was height

[vernal pilot]

h=0...

[winter garden]

Yes

I'll redo

Is 1π which is still wrong soo

I thinks that's area not circumference

Also the 9.3 9.6 is from Google formula not that one

[vernal pilot]

no its 2pi, cuz you do pi(a+b)(1+h/(10+sqrt(4-h))), the second half cancells out, leaving pi(a+b), which is 2pi if r=1

[winter garden]

You put ab not a +b

If it's a + b then yes it's 2π

[vernal pilot]

that was a mistake on my part.

[winter garden]

.

Okok

[pearl wave]

I got the ans as 8 for some reason

[vernal pilot]

but anyways, the point is, if you want your calculation to be considered accurate, first compare it to the other functions that have already be created, because most of them are highly optimised to get a VERY good approximation.

[winter garden]

I did and it's pretty close

Ill compare with yours now

[smoky jetty]

how come?

[pearl wave]

Just used the cosine rule

Twice

[smoky jetty]

hmm how?

[winter garden]

I'll keep the result here ellipse with short side = 1 and long side = 2 then

Mine, 9.27
Theirs, 10.11
Google, 9.6

[vernal pilot]

just trust google on this, okay?

[winter garden]

Yeah I should stop doing this shit

It ain't working out

[pearl wave]

If you connect the center to the other two vertices, you get two triangles in which one angle and two sides are known, so you can find the third side

[vernal pilot]

solve for y and z using trig then use cosine law to solve for x

[smoky jetty]

how'd u get z?

[vernal pilot]

wdym?

[smoky jetty]

could u lay out your solution?

[vernal pilot]

i havent solved it, but i have laid out how you solve it.

[smoky jetty]

coz to me, z would have 2 possible values, depending on the approach, if by cosine rule or 60-30-90 triangle sides

which makes the given values incoherent id say

[winter garden]

Cosine law need 3 input

So yeah it has not correct answer if using cosine law

However is a half circle so you need to take radius too

So z has base 4 and long side r=8 and 60 so you can find z as

[vernal pilot]

you can solve for y using the cosine law

$c^{2}=a^{2}+b^{2}+ab\cos(\theta)$

[somber coyote]

TheLord26

[vernal pilot]

screw this, this is too much using microsoft paint, im using desmos geometry

also i hate latex

[winter garden]

Cosine law calculated the line in between not the closing line

So you didn't fine y²

Also a and b are not the same

If they are it would be a equilateral triangle

I take that last sentence back tho

Is not equilateral

Nvm your right

I just realised you used 120

Not 60

[vernal pilot]

i would like to defend myself by stating my case, i hate latex.

too difficult to edit in discord chat.

[winter garden]

No no your right

I realised a solution was possible

Y = 9.211

[vernal pilot]

i dont think thats a viable solution as the radius is only 8. (4+12)/2=8

[winter garden]

Yeah but like I said cosine looks for the center line so you must use it twice

To find y the first answer is 8

Then you do it again and 8 got 9.211

[vernal pilot]

again, the radius is 8, so if you are saying that the length of y is longer than the radius i do not think that is the answer.

[winter garden]

Now to find z

To your do sound right how is y bigger than 8

Wait wait

It can only be 8

Sorry lower than 8

[vernal pilot]

the angle is 120

because you combine the two 60s

[winter garden]

Yes

[vernal pilot]

and y is definitely less than 8 units

[winter garden]

I did correct my self to that at the end

[winter garden]

.

Wait wait is a rombus

[vernal pilot]

sorry, just my lines

you are trying to find x

[winter garden]

Yes but we need y and z

[vernal pilot]

yes, i was just using the extra lines to get 60 degrees, dw about them really.

i got y=6

which is less than 8 and looks correct

now you solve for z, which i think equals 9.21

cuz i think you solved for the wrong side earlier

then you just plug the values into the cosine law

[winter garden]

Y is 6.92

Because is rhombus

[vernal pilot]

yeah i just realised i did +abcos instead of -2abcos

i hate this

math sucks

[winter garden]

It is +2abcos

Now we find these 2 I need to go to sleep

[vernal pilot]

nah its -2abcos

ty google, as always coming clutch

[winter garden]

Sorry yes it is -

[vernal pilot]

same bro, its 12:30am for me LOL

[winter garden]

Bye now this was fun

[vernal pilot]

gn

[vernal pilot]

$x=8$

[somber coyote]

TheLord26

[vernal pilot]

$64=y^{2}+4^{2}-2\cdot4y\cos(120) \newline
y\approx9.21$

[somber coyote]

TheLord26

[vernal pilot]

$64=z^{2}+4^{2}-2\cdot4\cos(60) \newline
z\approx5.21$

[somber coyote]

TheLord26

[vernal pilot]

$x=\sqrt{9.21^{2}+5.21^{2}-2\cdot4\cdot5.21\cos(60)} \newline
x\approx7.98 \newline \text{If you use more decimals the calculator} \newline \text{pretty much just evaluates it to 8,}\newline \text{so it probably is just 8.}$

[somber coyote]

TheLord26

[vernal pilot]

thats enough trig for the week.

[balmy carbon]

@vernal pilot

[vernal pilot]

Yes?

[balmy carbon]

can I ask you somth

[vernal pilot]

Ok

[balmy carbon]

so it wants the distance of AB yh

is it allowed if I make

a line from A to C

so it becomes a triangle for both?

like this

so I could get AC which is hypothenus

[vernal pilot]

Hmmm, I dunno. My brain is a bit fried rn. It’s 2am for me. But it looks to be on the right track.

[balmy carbon]

then do sin 98 = 10.59 /hypo

alr nws man

@pearl wave are you busy?

[pearl wave]

No wh5

Why

[balmy carbon]

here if you dont mind

I just wanna see if my way is correct

my apologies for the disturbance

I feel like my explaination is so shit lmk if u dont get it

[pearl wave]

If CD is perpendicular to AB then yeah you can do that

[balmy carbon]

oof it doesnt state that

[pearl wave]

No but BAC isn't a right triangle is it

[pearl wave]

Oh dear

[balmy carbon]

I can see its not now

[balmy carbon]

woow whats that

[vernal pilot]

10cos(89)

I think, it’s probably wrong though.

[pearl wave]

Oh I meant CD perpendicular to AD, sorry

[balmy carbon]

why 89 ?

[balmy carbon]

well AD is 3.5 cm and CD is 10

[vernal pilot]

180-98

I’m stupid

I’m tired.

[balmy carbon]

you said it yourself its 2 am bro

[vernal pilot]

82

[balmy carbon]

💀

[pearl wave]

So is that all the info given in the pic?

[balmy carbon]

Yes.

that's the full image

and the question is

[pearl wave]

Ok gimme a min

[vernal pilot]

I think my idea is correct tho.

[balmy carbon]

Given that Ad is 3.5 cm and DC is 10 cm.

Calculate

i) Length of AB

ii) Area of trapezium ABCD

[vernal pilot]

Just need to do the trig properly.

[balmy carbon]

im sorry thelord26 I couldn't catch what you are trying to do

where did this 10 come from

and why 180-98

sorry im actually a bit slow in this topic

or on this topic

[pearl wave]

Did it mention it's a trapezium? Coz according to part ii it is i think

[balmy carbon]

im not sure which is grammartically correct 🤓 ☝️

[balmy carbon]

yes the 2nd question states that

well it wants

Area of trapezium ABCD

so i think it must be a trapezium

[pearl wave]

But it's not given right?

If it isn't then ig we gotta show it first

[vernal pilot]

the bottom

[balmy carbon]

isnt that a trapezium

** why **

I knew I shouldn't have gone into the storage room last year 💀 🙏

[pearl wave]

That 10 is the same as the bottom, it's just an extra construction

[pearl wave]

That's right ig

[pearl wave]

Coz in the smaller triangle, angle B is 180-98

[vernal pilot]

Don’t mind the incorrect answer at the top. It is now 2:30am.

[pearl wave]

Since BC is a straight line

[vernal pilot]

AB=10.101…

[pearl wave]

Seems right

And area of trapezium would be area of the bigger rectangle- area of smaller triangle

[balmy carbon]

which triangle

[pearl wave]

This

[vernal pilot]

I may be severely sleep deprived, but my math is still perfect. (Don’t mind the absolute hot garbage math I was doing in this channel earlier nor confusing sin and cos).

[balmy carbon]

oh that

[pearl wave]

Interesting.

[balmy carbon]

hey about my technique earlier , when can I use it ?

or is it just not useable

[pearl wave]

The Pythagorean one?

[balmy carbon]

the line

[pearl wave]

Oh

[balmy carbon]

from a to c

yhh just incase

[pearl wave]

That could be used if they mentioned that the angle is 90⁰ or smth

[balmy carbon]

which angle ?

[pearl wave]

Between AD and DC

Then AC would be the hypotenuse

[balmy carbon]

oh alr I see

[pearl wave]

Did you get how we found AB now?

[balmy carbon]

yes

but isnt it just 2 points

[pearl wave]

Nice

[balmy carbon]

like

shouldn't it be 10.10 instead of 10.101

[balmy carbon]

another stupid question , just wondering

[pearl wave]

If you wanted to round off then yeah

10.10 is fine too

[balmy carbon]

alr and why is it allowed to add the 10 ?

[pearl wave]

Add?

[balmy carbon]

is it because its a rectangle

so its the same length?

sorry I mean't the

additional drawing there

aight nvm I think I got it

ignore my question 💀

[pearl wave]

Lol

[balmy carbon]

ion know how to get the area of trapezium tho

[pearl wave]

Just subtract the two areas

[balmy carbon]

what about 1/2 ( a+b) h

[pearl wave]

That works too but you'll need BC for that

[balmy carbon]

isn't its just 3.5 cm

since AD is 3.5 cm

[pearl wave]

But BC is smaller

[balmy carbon]

oh

couldnt I get it

with

the formula

[pearl wave]

If it was of the same length, then the construction wouldn't be possible

[balmy carbon]

root 10.1^2 - 10^2 = the left?

the leftover

• 5

like we know the O and H

we can just get the adjacent too can't we?

[pearl wave]

We can yeah

And then find the area of smaller triangle

[balmy carbon]

Why

sorry im a bit weak

[pearl wave]

So that

[balmy carbon]

about

insides of shapes

areas sorry

[pearl wave]

We can subtract it from the bigger rectangle(with sides 10 cm and 3.5 cm)
And also, it's alright

[balmy carbon]

oh yeah I see

so the bigger rectangle minus smaller triangle yeah?

[pearl wave]

You could find BC and use the formula as you said too, its not wrong

But it would be longer

[pearl wave]

Yeah

[balmy carbon]

how do I get the area of this kind of triangle

I know about

the rectangle

3.5cm x 10 cm

[pearl wave]

Yes so

Once you find adjacent

You can use 1/2 bh

Since it's a right triangle the height is the adjacent

Basically one of the legs

[vernal pilot]

That’s not the point of my solution.

[balmy carbon]

right triangle ?

[pearl wave]

Ik

[vernal pilot]

Yes

[balmy carbon]

oh I thought the adjacent

was the base

💀

[pearl wave]

I was trying to say that the two radii aren't equal

[pearl wave]

I thought you meant adjacent to 82⁰

[balmy carbon]

yeah

that one

[pearl wave]

Yes

[balmy carbon]

this is correct yh?

[pearl wave]

Take the sqrt

Since it's side^2

Oh ok

I'm blind mb

[vernal pilot]

Formula for area of 4 sided polygon where two sides are the same is ab/2 * h, where a is small length and b is big length and h is constant length.

[balmy carbon]

Nah its the

lighting

alr so

[pearl wave]

Oh..

[balmy carbon]

which one is the height and which one is the base?!

[vernal pilot]

Don’t do base for it.

[pearl wave]

Doesnt matter, we multiply both together anyway

[balmy carbon]

oh

[pearl wave]

bh/2

[balmy carbon]

no because

oh yeah

so the hypo just

doesn't get used yh

?

[pearl wave]

Yeah

[balmy carbon]

so its 10x1.42/2

[pearl wave]

Ye

[balmy carbon]

7.1 cm

what now?!

now we have the triangle

so now we just need the rectangle yh?

and then rectangle areas minus triangle

[pearl wave]

cm^2**

[pearl wave]

Ye

[balmy carbon]

mb!

cm^3 for what ?

[pearl wave]

No I meant

Since it's the area

[balmy carbon]

no i meant

[pearl wave]

It would be cm^2

[balmy carbon]

cm^2 for area

what about ^3 lol

[pearl wave]

That's volume

[balmy carbon]

just wondering

💀

my bad that was random

[pearl wave]

It's alr

Happens.

[balmy carbon]

hey I got it

27.9 cm^2

[pearl wave]

That matches with the ans?

[balmy carbon]

theres no answer 💀

[pearl wave]

Oh lol

[balmy carbon]

,,,r

[somber coyote]

skrr

[balmy carbon]

,,r

[vernal pilot]

Area=24.85 units^2

[somber coyote]

skrr

[balmy carbon]

hey how do you flip the image

[vernal pilot]

Don’t

Trust

[balmy carbon]

,rccm

no I need to see it

[balmy carbon]

,rotate

[balmy carbon]

say what

?!

wait a minute

[pearl wave]

.

[balmy carbon]

how come you got

2.03 @vernal pilot

[balmy carbon]

I got 1.42 cm

[pearl wave]

That's y^2

[vernal pilot]

c^2=a^2+b^2

[pearl wave]

Not y

[balmy carbon]

my final answer is 27.9 cm^2 tho

[vernal pilot]

SHIT

It’s 3am for me.

[pearl wave]

Lol

Get back to sleep ig

Unless you don't have smth imp tom

[balmy carbon]

I have no idea if my answer is right or not

wtf

[vernal pilot]

The aforementioned curse word was not a reaction to my sleep schedule but the fact that I didn’t unsquare y.

[pearl wave]

I think we got that

Yeah?

[balmy carbon]

nah I thought

he cursed because

its 3 am

[pearl wave]

Oh

F

[balmy carbon]

so whos in the wrong here?!

🤔

🫵

[vernal pilot]

I was mainly clarifying for comedic effect.

[pearl wave]

We did go step by step tho
Make the construction, find AB and then subtract the areas to find area of ABCD

[vernal pilot]

You don’t need to.

[balmy carbon]

whats the other way thelord26

[vernal pilot]

ab/2 * h

[balmy carbon]

ab is which

sorry

I didnt really learn much last year

[pearl wave]

That directly should work too

[balmy carbon]

hey you guys should be my accountant

[pearl wave]

But wouldn't you need BC for that

[pearl wave]

Nah, I'm bad at math

[balmy carbon]

need some smartass in my company

future company

feel free to invest

[vernal pilot]

Long side and short side

[vernal pilot]

No

[balmy carbon]

im thinking about

making a product

bro it was so goated I can't remember what it is wtf

[pearl wave]

Oh

[balmy carbon]

you dont know

what's the short side tho

[pearl wave]

Yeah

[balmy carbon]

oh i tohught

[pearl wave]

That's what even skrr said

[balmy carbon]

it was the tall thing

💀 fuck is wrong with me

[vernal pilot]

h=10, b=3.5

And solve for a

[pearl wave]

Hmm

[balmy carbon]

wait how is h 10

shouldnt h be 3.5

[vernal pilot]

By using Pythagorean thereom

[pearl wave]

h is dist between parallel lines

[balmy carbon]

yh

[pearl wave]

Yh

[vernal pilot]

No. Because the shape we are calculating is on its side.

[balmy carbon]

oh parallel

parallel = equal right

[pearl wave]

Ye

[balmy carbon]

soo

[pearl wave]

Not necessary

[balmy carbon]

no thats

what it means in google

I memorised it few days back

[vernal pilot]

Luckily we have the assuming that AD and DC are perpendicular (I forgot what the letters were so I’m guessing).

[balmy carbon]

identical , equialevant , same , alike , consistent , equal , parallel means the word same

but AD and DC arent perpendicular ?

[vernal pilot]

Parallel means they never touch, even if you extend the line to go on forever.

[pearl wave]

Well they haven't mentioned it

So even if it is

We can't go with the assumption

[balmy carbon]

alr bro no offence your technique is too smart for my brain to comprehend

[vernal pilot]

ITS LITERALLY GRADE 6 MATH

[pearl wave]

What technique

[vernal pilot]

Area=(a*b)/2 *h

[balmy carbon]

bro

[pearl wave]

Ig it's (a+b) in place of ab no?

[balmy carbon]

what is this astute way?

[vernal pilot]

Yes

I’m very tired

[pearl wave]

Happens.

[balmy carbon]

nah no way I spent

2 hours

for this question ?

[pearl wave]

But for the next one you won't spend as much

[balmy carbon]

wow

thats true

[vernal pilot]

Idk man, I gave you a method nearly 10 second after you asked the question.

[balmy carbon]

I literally gained alot of knowledge

[balmy carbon]

🤨

[vernal pilot]

I drew the diagram.

[balmy carbon]

yeah man thank you

[vernal pilot]

For finding AB

[pearl wave]

In the middle of the night

[balmy carbon]

now I know that you could do that

before I didnt know

[vernal pilot]

Yes

[balmy carbon]

you could

thanks for sacrificing your hours lord26

even tho its 3:20 am or something

[pearl wave]

Lol

[balmy carbon]

for you

🫡

and you too lorenzo

[vernal pilot]

Geometry is just you drawing lines, so if you are ever doing a difficult question just draw more lines.

[pearl wave]

Np

[pearl wave]

Hmm

[smoky jetty]

ight i see

[full estuary]

can anybody help me with understanding the basics of transformations, translations, and dilations of a linear function?

[worldly scroll]

I'm struggling with this.
Treating it as a unit circle of r=1 I'm saying the distance t from the point (0,1) is 2pi/3, but for calculations from point (1,0) is 3pi/2.
Then I do cos(3pi/2) = x, sin(3pi/2) = y

[shut oxide]

Please help me understand what I’m missing I have no clue

[broken crescent]

@here what would be the next statement and reasoning?

[fossil mica]

Anyone here can help me polish my proof of pythagorean theorem, going to include it in a video.

Just to make sure it follows notations and what not

[lean tartan]

Send it

[vocal bison]

help

[limber peak]

Why are there so many trig functions. We pretty much only learned reciprocal and pythagorean yet when I get super stumped on questions lo and behold it is a trig identity we never learned like double angle identities and half angle identities. What are some of the most common ones that I’ll need to know? Or should I just work on memorizing all of them? Where do they come from? They honestly all look super random. Why does sin^2 x=(1-cos2x)/2? Pythagorean says sin^2 x=1-cos^2 x so in the half angle identity why does the exponent on the cosine turn into not an exponent and the whole side gets divided by it? Is there even a why? Or is that just the way of the world and I’ll just have to memorize everything

[obsidian harness]

do you understand why $\cos 2x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$ (the last equality comes from $\cos^2 x = 1 - \sin^2 x$ btw)

then it's just using algebra to rearrange for cos^2 x and sin^2 x

[somber coyote]

south

[obsidian harness]

also double angle identities are special cases of when A = B

you just need to remember sin(A + B) and cos(A + B) actually

cause for sin(A - B) just substitute $B \to -B$

[somber coyote]

south

[obsidian harness]

in the formula

[limber peak]

I don’t understand why but I see the algebra part

The last bit is a rearranged Pythagorean identity sin^2 x+cos^2 x=1 rewritten to cos^2 x=1-sin^2 x then plug that for cos^2 x in 2cos^2 x-1 to get 1-2sin^2 x. What I got lost on was why cos2x=2cos^2 x-1

[steep pendant]

$cos(2x)=cos^2x-sin^2x$

[somber coyote]

A2

[steep pendant]

and you can also use $cos^2x+sin^2x=1$

[somber coyote]

A2

[limber peak]

Okay, I think it’s a double angle formula. But how do I know when to set cos2x=2cos^2 x-1 because it also seems to equal several other things like cos2x=1-2sin^2 x and cos2x=2cos^2 x-sin^2 x

[obsidian harness]

depends on the situation

[obsidian harness]

oh right this follows directly from cos(A + B), where A = B = x

one less thing to remember

[steep pendant]

Just like this

most of the time the best thing to do is to find as many connections as possible

[obsidian harness]

exactly

the need to memorise in maths definitely isn't a lot compared to other subjects you might have in school

[steep pendant]

For most of the time trig identities come from pythagorean and some algebra, and they are very useful. They do, however, seem to be totally random. My suggestion is that u might try to be exposed to more of these (while still keeping the mindset of finding connections with what you know)
You would naturally be able to apply the right knowledge on the problems u encounter.
This sounds like a random "work hard" thing, but I think there is a point where you get really used to these and acutally see through bunch of trigs without putting special effort to simply "memorize" them

[limber peak]

Okay, I understand what you mean by this I think

[steep pendant]

Fantastic 😄

[limber peak]

So pretty much just practice questions and prepare to face trig until I start recognizing patterns

[steep pendant]

Well most of the people who were good at those mathy stuff basically did a lot of math (of course).
I would refrain from telling you a specfic way to study math (like "solve 1000 questions per day" or sth), since I am honestly not an expert at how to learn/teach stuff, so I would not want to misguide you ||(oh no I'm causing more ambiguity)||
My point was that those random equations and identities would sort of look familiar to you at some point if you stick on to it ||(yep yep basiclly a "never give up")||.

[obsidian harness]

yeah also trig and geometry are super connected

you can get identities like this for example

like sin(-x) = -sin(x)

cos(-x) = cos(x) all follow from the unit circle

so don't worry if it takes you time to process everything

it's going to pay off later, again and again

[limber peak]

Okay I found another cos(2x) through Pythagorean identities and plugging. So can I repeat random plugging and identities to eventually find all the trig identities? Or does this only work on double angle identities?

[obsidian harness]

yep

so that works with all kinds of trig identities

[limber peak]

Oh, wow

[obsidian harness]

like for instance, starting from sin^2 x + cos^2 x = 1

divide this by sin^2 x

now start again and divide by cos^2 x

[limber peak]

Okay the trig suddenly seems a lot less random

[steep pendant]

Wow

[limber peak]

Thank you, things definitely make more sense now

[steep pendant]

||forgive me torturing you|| If this came up in your class or it does in some future moment

You can start from that cos(2x) = 2cos^2(x)-1 = 1 - 2sin^2(x)
you derived earlier

and write cos^2(x) and sin^2(x) in terms of cos(2x)

(the tan^2(x) just comes from dividing sin^2(x) by cos^2(x) )

[obsidian harness]

exactly

tan^2 x identity is underrated

[obsidian harness]

npnp!

[stuck sleet]

why 100%/90%

why multiply it

[obsidian harness]

cause you need more, if some is going to be wasted

so if the original is 90

you would need 100, cause 100 * (1 - 10 percent) = 90

[stuck sleet]

so I can also go to the right answer if I multiply the lateral area with 110%???

[obsidian harness]

no

that's different

90 * 1.1 = 99

you start off with x, and you take away x * 0.10

that equals 494.599

so x - x * 0.10 = 494.599

[stuck sleet]

(Needed Surface Area) (1-0.1) = (Surface Area Computed)

[obsidian harness]

yeah, that explains why it's 1 -

[stuck sleet]

Thanks for explaining

[opaque swan]

In hope this solves it

[obsidian harness]

this is how we get there

no worries

[stuck sleet]

I'm sorry for being not to get it quickly

[obsidian harness]

no worries

[opaque swan]

Np

Sometimes our brain also stops braining

[stuck sleet]

braining

[opaque swan]

Yeah

[stuck sleet]

yeah

[stuck sleet]

That's a lit word 🔥 🔥 🔥

[opaque swan]

Yeah very underrated word

I think your math wasn't mathing there

[stuck sleet]

yeah man

I passed calc and diff eq and still I suck at algebra

[opaque swan]

Oh shi-

Great respect man u passe calc and diff eq

I am still in 10th grade

[stuck sleet]

Because it is online (I take the calculus class and differential equations class in pandemic)

all are applications to physics and engineering

[opaque swan]

Ohh

[stuck sleet]

like vibrating springs and electrical circuits

also on population

but I forgot it somehow

[opaque swan]

U r in college i suppose then?

[stuck sleet]

yeah and I can't get out of here

[opaque swan]

I'm still schooling

[stuck sleet]

I stopped right now, didn't enroll at this term because my basics sucks

appreciate the help though @opaque swan @obsidian harness

[opaque swan]

Arigatou

[obsidian harness]

dozo dozo

ningun problema

[somber coyote]

Heckin' Beluga

Heckin' Beluga

[opaque swan]

@obsidian harness

U also watch anime

?

[stuck sleet]

heck yea

[shy shoal]

HELPP

stuck on extension question

abc is icosceles

[vernal pilot]

Are E and D the midpoints?

[trim trellis]

yo i found the solution for the kites

so basically because kites have an excircle, means that the sums of 2 opposite angle pairs are equal

and bc the angle between the sides is 90 degrees means the diagonal is 26(pythagorean theorem)

[shy shoal]

no info given

[frosty rose]

Can someone send me an effective youtube guide to 10th grade trigonometry ? Ts has me bamboozled I'm ngl

[vernal pilot]

Is this the original question, or is there a text book with where this question came from?

[shy shoal]

original

i saw it on a sophomore's textbook and it was an extension

(im a freshman)

[vernal pilot]

Organic Chemistry Teacher and Khan Academy probably have a few videos.

[mystic charm]

Is everything correct in the picture I sent?

[shy shoal]

yes

[mystic charm]

Alright

Do you agree that PM = DG

[vernal pilot]

He said it was a rectangle, so it would be.

[mystic charm]

?

Yes

[mystic charm]

And we know that AM = 12

And PM = y

Therefore AP = 12 - y

Clear enough?

[vernal pilot]

You put F instead of P in your drawing.

[mystic charm]

It doesn’t matter

The solution

Is important

[shy shoal]

Thanks!!

man i gotta work on geo

[mystic charm]

You solved it?

Cuz it’s not solved yet💀

[winter garden]

Since AED is equivalent to ABC then I am going to assume EDM is also equivalent of this is true then X = 9 and y = 6

if*

[shy shoal]

workin

[mystic charm]

Alright

[mystic charm]

the question doesn’t want you to find the answers for x and y

(It’s not possible to find x and y tho)

[winter garden]

Exactly so how I going to find area

You*

[mystic charm]

You will find the area based on x

[winter garden]

Xy

[mystic charm]

Like x^2 or something

[dusty field]

can someone tell me what is a website that could help you do math

[winter garden]

X(1.5X)

[mystic charm]

Khan academy is a good website

[winter garden]

1.5X^2

I think

[dusty field]

is a course?

[winter garden]

Not 0.6x^2

Still wrong

[mystic charm]

I haven’t used it very much, but as far as I know it’s a good site for math

[mystic charm]

Idk

[winter garden]

π for a 4 sided shape is 4

[mystic charm]

I haven’t solved the question

What

[winter garden]

So 41/2x1/2y

Maybe

[mystic charm]

Lemme send a picture rq

[winter garden]

4(1/2)^2(X)^2

[mystic charm]

Have you ever heard about this thing?

[winter garden]

Yes 8th grade

[mystic charm]

Yes

Now

For the question above

[frosty rose]

Not to be rude and cut u off but what was this relation called

[mystic charm]

The same thing applies for PM/AM=ED/BC

[winter garden]

Not really we don't know if it's the center

[mystic charm]

Honestly I don’t know its English name

[frosty rose]

oh

[mystic charm]

M is the midpoint

And

The triangle is isosceles

Ion ok

Oh ok

[mystic charm]

.

[winter garden]

Ooo

I understand now

X/18 = y/12

[mystic charm]

Well

It’s something like that but a little different

We said

PM= y

AP= 12- y

So

[winter garden]

Now we can find the second first to y interms of X then multiply by x

[mystic charm]

x/18=(12-y)/12

[mystic charm]

Yes

Exactly

The numbers are messed up

What is bro typing 💀

[winter garden]

Y = -1(12x - 216)/18

[mystic charm]

Exactly

You know the solution

[winter garden]

NOW I DO

[mystic charm]

Just to finish it off

[winter garden]

Area

=

(-12x^2 + 216x)/18

[mystic charm]

Yes exactly

[winter garden]

Damn that was fun

[mystic charm]

Yeah ikr

[winter garden]

Imma go to sleep now bye maybe see you Tomorrow

[mystic charm]

Good night and bye

[glad orbit]

you are looking at a 40ft building with a 20 degree angle of elevation. how far away from the building are you

i need help

[deep bluff]

information is insufficient i guess

[mystic charm]

Are you familiar with sin and cos and tan and cot?

[mystic charm]

This is generally your question

[frosty rose]

Can someone help me solve these

In a ik that I should use a trigonometric identity but idk how I should get to cos(18°)

[hybrid citrus]

ach

tell me

what is the relationship between cos(pi+theta) and cos(theta) and cos(2pi+theta)?

@frosty rose

ach wait I think u gotta open a channel 4 help

[frosty rose]

I think it relates to quadrants I'm not sure

[frosty rose]

wym

[hybrid citrus]

find a help channel and paste ur question into it

not discuss channell

[frosty rose]

Oh

Mb

[hybrid citrus]

Ill help u there

[frosty rose]

I gotta get off tho so I'll check when I can

Thanks

[hybrid citrus]

K, so one thing u hav 2 know is that cos(2pi+theta) = cos(theta-2pi) = cos(theta), so does sin

and cos(pi+theta) cos(theta-pi) = -cos(theta)

[vernal pilot]

You don’t need to know sin18 or cos18

It’s asking you to simplify, not get exact values.

[wheat tangle]

Does anyone know how to do triangle proofs? I was absent for 10 days and my teacher made me do a test which I failed. so now I need help on the test corrections

[vernal pilot]

Do you mean trig proofs?

[wheat tangle]

No, triangle proofs

[vernal pilot]

Give an example if you don’t mind.

[vernal pilot]

Nah I’m absolutely garbage at these.

[maiden brook]

it’s the definition

of hl

[maiden brook]

dw this one is rlly easy

[wheat tangle]

I wish I knew how to do it. Idek why he made me take the test I wasn’t even there for 10 days

[wheat tangle]

Yeah but there should be like more than one statement/reasoning other than the given and the final statement

[maiden brook]

not necessary

[vernal pilot]

i think im going to sit this one out, ill stick to anything thats not geometry related.

[wheat tangle]

😂

[maiden brook]

wait nvm

[wheat tangle]

Yeah there should be

[maiden brook]

u have AB = AB

[wheat tangle]

I got reflexive property

[wheat tangle]

Yes^

[vernal pilot]

ill have a guess though, cant you say that because of SSS congruency they are equal?

its been so long since ive done them, so even if im wrong, thats the most ill try.

[wheat tangle]

That’s all good. Thank you tho

It’s so dumb he’s making it apart of my grad

Grade*

My math grade dropped from a 95 to a F because of this

[vernal pilot]

why not talk to your principle about it if you think its so unfair (meaning you didnt have a chance to apply for an extension).

or even if you did have a chance, just ask your principle about it.

or like, a head of department.

[wheat tangle]

I can’t even ask for an extension because it was a quiz we took and we already passed that unit of proving triangles congruency

If I get a 90+ on the test corrections then my math grade is saved. I just don’t know how to do them

It’s 4 questions to😂

[vernal pilot]

Uhh, must be a difference in country. Are you in college or highschool?

or uni

[wheat tangle]

Highschool

[maiden brook]

there’s no way this is a uni geometry course

yeah

[wheat tangle]

In America

[maiden brook]

u have test corrections also?

[vernal pilot]

just ask your head of department about it. should work.

[wheat tangle]

Yes, very fortunate of it

[maiden brook]

my geo tests were hard and no test corrections

[wheat tangle]

Dang, that’s rough

[wheat tangle]

This like the first test I failed

So going back on def. Of HL

Guessing HL is hypotenuse leg?

What would the statement be?

[maiden brook]

don’t memorize

but

Just if two triangles are right and have two given sides to be equal to each other, then the triangles are congruent

[wheat tangle]

Yeah I’m pretty sure I think

[vernal pilot]

SAS congruency

but cant you do SSS congruency?

[wheat tangle]

I think so

[pseudo nebula]

You can do any congruence

[vernal pilot]

i feel like SSS

is easiest

because you are explicitly told each side is equal

[maiden brook]

u cannot rlly for the problem above

[vernal pilot]

Yeah I looked at it again. I’m way too tired to be doing math. Pulled an all nighter to get something done so I’m a bit out of it rn. Pair that with my absolute horrendous geometry skills and I definitely wasn’t gonna get this right.