geometry-and-trigonometry

[gray minnow]

how is this true?

[dark sparrow]

$\cos^4(x) \leq \cos^2(x)$

[somber coyote]

Ann

[dark sparrow]

that's how

[gray minnow]

hmm

ok

i get it

equal when cos = 1

but fraction gets smaller when multiplied

[gray minnow]

is there any guide to use this bot

[dark sparrow]

in #resources there is a cheat sheet for common LaTeX commands

[gray minnow]

okok

thzz

thxx

[supple rose]

Anyone know this?

[slim path]

let x be short side
and y be long side
x1/x2 = y1/y2

i hope this helps

[slim path]

for here P(x) = x/total distance

or length

[tame quail]

hello i have a question about geometry

i just need to know what formula i should do for these questions

[silent plank]

formula for arc length

[tame quail]

arc length for all of them?

[silent plank]

same formula yes

[tame quail]

but what about stuff like circumference

or sector area

[silent plank]

you could consider the ratio of your arc to the circumference if you want

(or the angle to a full revolution)

[tame quail]

i see

so for number five how should i set it up

[silent plank]

do you know the formula for arc length?

[tame quail]

(arc length / Pi r 2)= (angle/360)

right?

[silent plank]

can you fix that formatting

[tame quail]

wdym

[silent plank]

wtf is TT

[tame quail]

oh its just pie lol

[silent plank]

pi not pie,
just write pi

[tame quail]

it always looked like two ts for me

my bad

but is that right

[silent plank]

also parentheses

$\frac{\text{arc length}}{2\pi r} = \frac{\text{angle}}{360\deg}$

[somber coyote]

ℝamonov

[tame quail]

alright good thanks

[silent plank]

yes, is something you can use

[tame quail]

156 would go in for arc length

but would go in for r?

[silent plank]

156 would go in for arc length
but would go in for r?
no

156° is your angle

r would represent the radius

[tame quail]

oh i see

[silent plank]

x represents the arc length you're trying to find

(in Q5)

[tame quail]

would the radius be 5

[silent plank]

5 ft yes

[tame quail]

alright

i got a big number

13.61356817

but i round to the nearest hundreth

13.61

[upper karma]

Can I get help with this?

<@&286206848099549185>

[steady shoal]

Do you know the formula for surface area of sphere

[upper karma]

4pir^2

[steady shoal]

Where are you getting stuck then

[upper karma]

Because I don’t understand what the question is asking

[steady shoal]

Do you know the relation bw diameter and radius ?

[upper karma]

Radius is half diameter

[upper karma]

Can u still help me?

[steady shoal]

Oh sorry I wasn’t active

[upper karma]

It’s alr

[steady shoal]

First find surface area of sphere 1

What is it

[upper karma]

Would the diameter be 2 for the first sphere?

[steady shoal]

Surface area 1 / surface area 2 what will it be

[upper karma]

3.5 and 153.94?

[steady shoal]

Surface area is 4pi r^2 which is 4pi (d/2)^2

[upper karma]

What would I plug in there tho

[steady shoal]

4pi is a constant when you take ratios ,constants get vanished or can say they cancel out

[upper karma]

I know but what’s the radius

[steady shoal]

TSA is total surface area

[upper karma]

So (2/7)^2?

[steady shoal]

Yes

[upper karma]

4/49

what’s that as a ratio?

Nvm

It’s 4:49

[south spade]

can someone help me to find the Surface Area

[dark sparrow]

do you know, in general, how to find the surface area of a polyhedron (i.e. a shape bounded by flat surfaces)?

@south spade

[south spade]

wdym

[dark sparrow]

if you're given a shape whose surface consists of a bunch of flat faces, how do you find its surface area?

the answer i expected here is "find the area of each face, then add them all up."

...

hello?

[south spade]

oh ok

[dark sparrow]

wait, so was that like... news to you?

did you not make the connection between the surface area of your shape and the areas of each face prior to posting here?

[south spade]

No I did but my teascher thought it a different way

and she just posted video notes

so didn't rlly understand it

[humble pulsar]

How do you find surface area a different way than finding the area of the surface

[south spade]

idk man ask my teacher she doesn't teach

[dark sparrow]

i mean ok like, im trying to probe how much knowledge you have about the concept of surface area in general

anyway ok

so can you tell me what faces your hexagonal prism has?

just their shapes.

[south spade]

rec

[dark sparrow]

"rec"?

[south spade]

yes rec

[dark sparrow]

do you mean rectangle?

also if you don't want me walking you through this please let me know

i would rather not pull info out of you like rotten teeth.

@south spade are you willing to continue & cooperate with my attempts to guide you towards the answer? Y/N

👻

[buoyant flower]

lost

[buoyant flower]

<@&286206848099549185>

[slim path]

draw DO

the measure of the arc is always the angle at the center

[high spindle]

<@&286206848099549185>

[weak lichen]

Do you guys know how to solve this

[unkempt crag]

use cosine

[slim path]

use cosθ = a/h

[unkempt crag]

in a right triangle the cos(x)=A/H, where A is the adjacent side and H is the hypotenuse. So in this case JK is adjacent and the hypotenuse is 9

cos(26)=JK/9
JK=9*cos(26)

[long ferry]

I am stuck at this

I got it done till here

[storm portal]

idk how you got here - but regardless, can you get the tan(x/2) in terms of sin and cos?

[neon hamlet]

How can you find x in this case?

[regal spruce]

7

@neon hamlet

[neon hamlet]

how'd you get it @regal spruce

[regal spruce]

Angle A + angle C = 180

[neon hamlet]

ah gotcha

[regal spruce]

Yo boi✌️

[fervent imp]

How in the world do I solve this?

Last question till I am done for the day. I got 270, and 540. I am not so sure if it is right, coz 540 is greater than 360.

[dark sparrow]

@fervent imp can you show your work? i might be able to pinpoint a mistake you're making

[neon hamlet]

Anyone know how to solve for this angle?

[dusky surge]

You can make use of this:
$$\frac{1-\tan{\frac{x}2}}{1+\tan{\frac{x}2}} = \pm \sqrt{\frac{1-\sin{x}}{1+\sin{x}}}$$
and this:
$$\cos{\alpha}=\frac{1-\tan^2{\frac{\alpha}2}}{1+\tan^2{\frac{\alpha}2}}$$

[somber coyote]

Biscuit

[dusky surge]

What do you know about tangents and alternate segments?

[neon hamlet]

aren't they congruent?

@dusky surge

[dusky surge]

Yea

[neon hamlet]

but how would that help in this case

or is YXZ (angle) the same as 38.4

[dusky surge]

the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment.

Hmm...

Alternate segment

Do you know which one is the alternate segment?

[neon hamlet]

im not sure

XZ?

[dusky surge]

Please take a look at this example 😁
https://brilliant.org/wiki/alternate-segment-theorem-2/

[neon hamlet]

oh yeah so would YXZ be congruent to MZX?

actually no.

[dusky surge]

" the angle in the alternate segment, i.e. the angle subtended by the chord in the opposite side of the previous angle."

[neon hamlet]

okay so XZ is touching the tangent MZ

[dusky surge]

Yea, then we find the alternate segment

The given angle is the MZX 38.5° side

So, MZX is congruent to XYZ

Just look at "the opposite side of the previous angle"

[neon hamlet]

ah so both are 38.5

[dusky surge]

Yep

[neon hamlet]

and uh

my last question

[dusky surge]

Yea?

[neon hamlet]

DWC: 43.5
and
SWY: 56?

[dusky surge]

Not sure at first glimpse

I think DWC is not 87/2

Like the 112° of SY means the angle at center is 112° right?

[neon hamlet]

yeah

[dusky surge]

So, since point W is on the circumference of the smaller circle, so SWY is 112/2

But W is not a point on the circumference of the bigger circle, so you cannot use angle at centre = 2 times angle at circumference

[neon hamlet]

true

[dusky surge]

Hint: adding a line passing through AD

[upper karma]

Anyone help with surface area questions?

[dusky surge]

Is it about geometry or trigonometry ?

[upper karma]

Geometry

[dusky surge]

Great! Then feel free to post it here 😁

Of course this channel is for pre-university level

[upper karma]

I’m stuck on this because wifi the apothem

Of*

[dusky surge]

Hmm... Did you try to draw the regular hexagon with apothem 8√3 out?

[upper karma]

I did but I have to find out the base and the problem is that I can’t use Pythagorean theorem since I only know the height

[dusky surge]

Note that a regular hexagon can be split into 6 congruent equilateral triangles. And then you can find the length of a side of the hexagon with the triangles

Are you not allowed to use Pythagorean Theorem?

[upper karma]

Nope

[dusky surge]

How about sine cosine tangent?

[upper karma]

Yeah I can

[dusky surge]

Then with the equilateral triangle, you can solve for the side

https://images.app.goo.gl/sT4yoLNmjvGfUyvJA

First split it like this

[upper karma]

Ok

[dusky surge]

Then you can find something similar to this

[upper karma]

Ah ok makes sense

[dusky surge]

Then you can find the length of one side of the equilateral triangle, and hence the area of it

[spice solar]

can anyone help me with this? ive been at it for a while now and im lost

[sharp plume]

u know what cos will be?

[upper karma]

wdym

[sharp plume]

which one will it be opposite/hypotenuse, adjacent/hypotenuse or opposite/adjacent

[abstract barn]

I have some homework to do, but my teacher never taught us how to solve it, can someone help me?

Its circle angles btw

[long garnet]

yeah sure

come at me

[abstract barn]

Its 20 questions

[long garnet]

you should get the gist of it after a few questions

[abstract barn]

Its not fluent

[long garnet]

alright

idc lmao

[abstract barn]

I already know I need to multiply it by 2, then try to make use of all 360 degrees

[long garnet]

oh so what do you need help wit then?

[abstract barn]

trying to figure it out

Im having trouble finding the missing parts'

[long garnet]

well a semicircle is 180 degrees, correct?

[abstract barn]

yes

[long garnet]

and BD is a semicircle

[abstract barn]

theres a bit of an angle on it

[long garnet]

hmm there's no other given information

[abstract barn]

So thats the main thing im stumped on for it

Yeah thats precisely why im not sure

theres a lack of information

[long garnet]

well you can't assume so i'll try and think of something else

[abstract barn]

ok

[long garnet]

yeah there's no other information given, you can only assume

[abstract barn]

And im bad at making assumptions

[long garnet]

...

180-38

[abstract barn]

so 142?

[long garnet]

yes

[abstract barn]

ok then

number 2, ive gotten an answer but wanna double check

[long garnet]

what'd you get?

oh

[abstract barn]

129

[long garnet]

?

WYZ is more than 180 degrees

[abstract barn]

the angle of it

I see it as a secant

[long garnet]

wyz is counterclockwise

and those are chords

[abstract barn]

Im not sure where this is going

[long garnet]

this is the angle you're trying to find

[abstract barn]

ok

[long garnet]

mm so try solving this

tell me what you get

unless you don't know how to solve it

[abstract barn]

I dont

I wrote down all the notes, but this just doesnt make sense anymore

[long garnet]

...

okay

so

you wanna add the semicircle's measure with XW

to get WYZ

understand?

[abstract barn]

ok

and for XW measure, I got 102

[long garnet]

how.

[abstract barn]

bc im pretty sure to find the measure I multiply the interior thing 51 by 2

[long garnet]

that's not the case for central angles

[abstract barn]

ok

[long garnet]

alright here

[leaden birch]

do you know about vertical angles? @abstract barn

[abstract barn]

yes

[long garnet]

this is what you're talking about

[abstract barn]

ok yeah

[long garnet]

central angles are congruent

[abstract barn]

gotcha

[leaden birch]

so going back to this, how would you go about solving for arc WYZ?

[abstract barn]

I dont have notes on how to figure out this situation

[leaden birch]

what is the measure of arc WXY?

[abstract barn]

Because everything I was shown had at least 2 things to go off of

[leaden birch]

@abstract barn can you figure this out?

[abstract barn]

no im really stupid

I hate this

[leaden birch]

well WXY is a semicircle

@abstract barn how many degrees are in a semicircle?

[abstract barn]

180

[leaden birch]

and WXY is a semicircle

right?

[abstract barn]

so its 180?

[leaden birch]

so how many degrees is WXY

yes

and which arc do you add to arc WXY to get arc YZ?

[abstract barn]

WX

[leaden birch]

sorry wrong question

which arc do you add to arc WXY to get arc WYZ?

[abstract barn]

YZ

?

[long garnet]

...

[leaden birch]

and what is the measure of arc YZ?

[abstract barn]

51 degrees

at least on the inside

Im honestly drawing a blank

[leaden birch]

in a circle, an arc that's drawn from the central angle has the same measure

[long garnet]

alright at this point it's logic tbh

[leaden birch]

can you vc?

[abstract barn]

yes

but cant talk or ill get yelled at

51

[spice solar]

ive been trying this for an hour on end? anyone able to help?

[sharp plume]

use the law of sines

[long garnet]

^

[abstract barn]

in relation to the whole circle, its a smaller piece?

[spice solar]

maybe my calculator is gone stoopid then

[long garnet]

download a calculator app

[abstract barn]

well since youre saying its 180 degrees, itd be 1/2?

[thin folio]

hey who can help me with 10 grade geometry

[abstract barn]

yes I see it

[thin folio]

51 i think its a vertical

[abstract barn]

180

51

231

[thin folio]

^

[abstract barn]

Im surprised my teacher didnt give us notes for this

[abstract barn]

listen, im REALLY dumb

this is another one I did but wanna double check

yes

47

I do not have an answer sheet no

so my math for this ended me with x=14

yes but its not correct

Its an example answer

That I have to try and get an answer for

I think I got this right as well

I will be honest, I need help with everything

ok so it is

ok

I dont understand this at all so I guessed

never saw an example about this

[leaden birch]

inscribed angles that intersect the same arc are congruent

[long garnet]

^

you set the two equations equal to each other to find x

[abstract barn]

x=19?

58

[leaden birch]

where are you learning geometry

[abstract barn]

in person

[leaden birch]

no like what school district

[abstract barn]

at an IB school

[leaden birch]

ib school?

[abstract barn]

International Baccaleureate

[leaden birch]

what country is that

[abstract barn]

America

[long garnet]

multiply 58 by 2

[abstract barn]

Its not a school district

[abstract barn]

116

[long garnet]

gj

that's the answer

[abstract barn]

I guessed 130 tbh

[leaden birch]

remember how an inscribed angle is one half of the intercepted arc?

[abstract barn]

yes?

[long garnet]

then half the semicircle's angle

[leaden birch]

so what is the measure of the arc that angle ABC is intercepting?

[abstract barn]

180 degrees?

[leaden birch]

yes

so since it's an inscribed angle...

what would the measure of ABC be?

[abstract barn]

90 degrees?

[leaden birch]

yes

[abstract barn]

ok

[long garnet]

gj

[leaden birch]

and REMEMBER THIS: any angle inscribed within a semicircle is 90 degrees

[abstract barn]

so this is also 90?

[long garnet]

no

[leaden birch]

which is 90?

[woven haven]

P is 90° right

[long garnet]

correct

okay so what's the measure of a triangle?

[abstract barn]

I set the variables to each other right?

180 degrees

x=16?

Ill be honest I feel really stupid right now

[woven haven]

P= 90°
∆QRP=
4x+3 + 90° + x+7=180°

[abstract barn]

so its 180 degrees?

[woven haven]

The entire triangle is 180° yes

[abstract barn]

@lapis snow I cant talk or ill be yelled at

[long garnet]

👍

[long garnet]

no

[long garnet]

this is correct

OR you can say

4x+3+x+7=180 just to simplify it a bit

[abstract barn]

I have to have it simplified to an angle

[leaden birch]

wouldn't it be 4x+3+x+7=90

[long garnet]

is that not what i just said.

[leaden birch]

180

[long garnet]

oh my bad

i meant 90 lmao

[abstract barn]

so is it 90 or 180?

[leaden birch]

90

[abstract barn]

ok

[long garnet]

90's not the answer

...

solve the inequality first

find x if 4x+3+x+7=90

[abstract barn]

x=16

[long garnet]

alright now plug it in

16+7

[abstract barn]

23

[abstract barn]

^

[long garnet]

gj

alright next question

[abstract barn]

so the answer is 23

[long garnet]

do you know the theorem for this question?

[abstract barn]

wait hold up

I just realized something

Every answer weve gotten has been in the previous answer thing on the nexdt question

[long garnet]

that's hwat i said.

[abstract barn]

Imma see if I cant get a 100 off that

[long garnet]

so do you understand everything

[leaden birch]

or you could try actually learning the applications of the maths

[abstract barn]

I technically have

Ive written down everything you guys have told me

[long garnet]

ouch

[long garnet]

alright tell us how to do this problem

https://cdn.discordapp.com/attachments/326138757474680852/842912139760500746/unknown.png

if you understand everything

[abstract barn]

That may take a while for me to work out on paper

[long garnet]

we can wait 🙂

[abstract barn]

I submitted and got a 95

Thanks for the help

have a good night/day

[long garnet]

alright so you were capping?

[abstract barn]

not capping, just very lazy

[long garnet]

then tell us how to solve this problem lmao

https://cdn.discordapp.com/attachments/326138757474680852/842912139760500746/unknown.png

just to make sure

you understand the application of mathematics

[abstract barn]

ok thats not a square, but im pretty sure counts as a quadrilateral

which means it needs to be 360 degrees

[long garnet]

yes

[abstract barn]

and since those angles seem to be congruent or at least similar, I believe I set the equations shown equal to each other to then find x

[long garnet]

no

[abstract barn]

ok

[long garnet]

yeah so like

you should probably study more math applications bc you need it imo 🙂

[abstract barn]

I am going to take math applications next year

[long garnet]

...

[abstract barn]

If I pass my math this year, I will be done with math for my graduation credits

[leaden birch]

why don't you put more effort in?

[abstract barn]

I try

Its kind of hard when I get yelled at every time my grade dips a little

[long garnet]

study theories and postulates that relate to geometry

[long garnet]

i wonder why your grades dip.

[abstract barn]

I actually try

its not due to lack of effort

[long garnet]

you should put more effort in learning theorems then, that's the whole concept of geometry

[abstract barn]

I was doing fine until now

I had an A- last quarter

[long garnet]

now that your grades are dipping you realize you should study more then, correct?

[abstract barn]

I studied before

If I hadnt studied, idve failed my tests

[long garnet]

how come you don't understand how to do this problem then

https://cdn.discordapp.com/attachments/326138757474680852/842912139760500746/unknown.png

[abstract barn]

Its just that I struggle with circle angles thats all

[leaden birch]

what part of it do you struggle with

[long garnet]

then go study circle angles? that's what studying is for

[abstract barn]

yeah

[long garnet]

why don't you actually put effort in

[leaden birch]

what are you struggling in

what specifically

[abstract barn]

I plan on asking my teacher if she has any recommended sites to study on

[abstract barn]

center points angles

[long garnet]

in the past hour, we can see he struggles in LOGIC

he doesn't understand addition and subtraction within circles

.

[abstract barn]

I understand that, I just overthink things

[long garnet]

we have some sources

[long garnet]

https://cdn.discordapp.com/attachments/326138757474680852/842912139760500746/unknown.png

then solve this

[abstract barn]

I overcomplicate it for myself for some unknown reason

[leaden birch]

"in a circle, the measure of a central angle is equal to the measure of the intercepted arc." do you know what this means?

[abstract barn]

Doesnt that mean the center is equal to the measure of the chords that intercept it?

[leaden birch]

what do you mean by the measure of the chords

[abstract barn]

The measure of the angles the chords make o the circle

*on

[leaden birch]

do you know what a chord is?

[abstract barn]

Im sorry to trouble you guys wih my idiocy

[abstract barn]

yes

[leaden birch]

can you give me a definition

or liek a general description

[abstract barn]

a line segment that is specifically straight with endpoints that are both on a circle arc

[leaden birch]

a circle arc?

[abstract barn]

circular arc

simplified, a segment whose endpoints are on a circle

Math is no my strongsuit

*not

English and History are

[long garnet]

so like this ?

[abstract barn]

no

because it is not touching another point in the circle

[long garnet]

😐

[clear sun]

like this

[long garnet]

it this correct?

[abstract barn]

I memorize the terms by pairing them with visual qeues

[long garnet]

bc that's what u meant when u gave your definition

[clear sun]

that aint a circle

[long garnet]

exactly

your definition is mixed up

[abstract barn]

I reiterated it

a segment whose endpoints are on a circle

[long garnet]

soooooo

does this count as a chord?

[abstract barn]

that is not a circle

therefore it is not a chord

[long garnet]

so this is a chord

[clear sun]

jaoson are you trollin

[long garnet]

nope i'm trying to make him understand that definitions in mathematics are very important

[abstract barn]

a segment that has both endpoints inside a single circle

that better?

both endpoints inside a whole, single circle

[long garnet]

this is a chord?

[clear sun]

touching the circumference

called it

[abstract barn]

this makes me realize how much I hate about myself when it comes to learning

[clear sun]

is this correct?

[long garnet]

that's why it's important to study

[long garnet]

use specia lright triangesl

[leaden birch]

yea

[clear sun]

thx

which law do i use for this?

[long garnet]

90 60 30

it's another special right triagnle

do you know the formula?

not formula mb

[clear sun]

uh i forgot it

[abstract barn]

special right triangles I sortof remember

[abstract barn]

But imma get to studying so I dont get schooled like this again

[long garnet]

but you can't do this problem 😂 https://cdn.discordapp.com/attachments/326138757474680852/842912139760500746/unknown.png

yeah go study

[abstract barn]

time to hit the books, and sites

[clear sun]

oh so its x√3?

[long garnet]

well

aren't you trying to find what x equals

[clear sun]

50√3

x = 50 in this case

do i use sine cosine or tangent?

[leaden birch]

SOH CAH TOA

[long garnet]

do you need sources?

[clear sun]

so Toa?

[abstract barn]

yes please

dm them to me so I can reference them easier if possible

[leaden birch]

what would TOA be here?

[clear sun]

Tanget Opposite/Adjacent?

[leaden birch]

mhm

[clear sun]

so its COH?

[fickle sleet]

yes

[trim breach]

SOH*

[leaden birch]

okok so

[fickle sleet]

Now my eyes are playing tricks on me

[leaden birch]

@clear sun when using SOHCAHTOA, it's always looking from the perspective of the angle

so what's the angle in this question?

[clear sun]

67 degrees

[leaden birch]

yes

now what are the two sides that you're given?

[clear sun]

20/X

Hypo and Adj

[leaden birch]

which is the hypoteneuse and which is the adjacent?

[clear sun]

20 = H
X = A

[arctic isle]

X is opposite?

[clear sun]

i thought adj was at the bottom

[fickle sleet]

Common mistake

[leaden birch]

okok

[fickle sleet]

Adjacent is the side in contact with the angle

[leaden birch]

yes

[clear sun]

ohhh

[fickle sleet]

Opposite is the side not in contact with the angle

[clear sun]

i see

[arctic isle]

Nah adj is adj to the angle, but there are two, but one will be the opposite so which ever one is left over

X is opposite 67 and so the only other adj angle is the left side

[clear sun]

So its Op / Hypo?

SOH?

[arctic isle]

Yh

Sin(67) = x/20

So multiply both sides by 20 to get x

[clear sun]

so sin(67)x20?

[arctic isle]

I think anyway, trig is my least favourite topic

Yes

Then round to 1 d.p

[clear sun]

so 18.41?

[arctic isle]

Idk if that's what the calculator says then yes lol

[clear sun]

thank you

[arctic isle]

But round to nearest tenth as it asks

[fickle sleet]

The tenth place is 1 digit after the point

[clear sun]

ye i did

18.4

[arctic isle]

OK cool

[clear sun]

do i need to use a law for this question?

[arctic isle]

Probably sohcahtoa again

[fickle sleet]

sohcahtoa only works for 90 degree triangles. Non-applicable here for direct solving

[arctic isle]

Capital letters = angles and lowercase = sides, so it's probably best to draw a triangle with the given measurement

[clear sun]

oh ok

[arctic isle]

Actually true, I'm wrong

So sin or cos rule I think

[clear sun]

alr

[fickle sleet]

Since you're likely starting out, you're probably taught triangle congruences to find out whether to use cosine or sine law

[clear sun]

ye i think i was but i most likely forgot

[arctic isle]

I would add the labes ABC abc too just to make it easier

[clear sun]

oh ok

[arctic isle]

side a is opposite to angle A btw etc

[clear sun]

uh what does that mean?

[fickle sleet]

The side opposite to the angle of A is a

Which means 8 should be 'a'

[arctic isle]

So opposite of 40 will be a

So it should be a different side that is 8

[fickle sleet]

Actually yes, Flodz is right here

[clear sun]

so am i right?

[arctic isle]

No, bottom right side is b/8

As only other angle left is top left, which is B. Opposite of that is the side AC

[clear sun]

LIKE THIS?

oops caps

[arctic isle]

erm no

Sorry I might be explaining bad, trig isn't my thing

You're angles were correct, 'b' was just in the wrong place

[clear sun]

ohhh

[clear sun]

like this?

[fickle sleet]

^

[arctic isle]

Bit unclear as the letters aren't that thick but as I've done

[clear sun]

ohhhhhh i see now

[fickle sleet]

It's really important to be able to recognize opposite sides to angles

[clear sun]

ye i think i understand

[fickle sleet]

Practice more and you'll be able to get it really quickly

[clear sun]

ok

[arctic isle]

It shouldn't matter which angles you choose to label as AB or C just make sure the opposites are correct and the measurements are correct too

[clear sun]

I see

[arctic isle]

But now with those you can use the sin/cos rule

[leaden birch]

is that the law of sines?

[arctic isle]

Yeah

[vapid stag]

hey umm. If tan(x) = -1/4 and x is in Quadrant IV
doesn't sin = -1
cos = 4?

[arctic isle]

I'm not sure, someone else can help with that

[fickle sleet]

Not possible. No values of sin and cos in the unit circle can go beyond 1,-1

[vapid stag]

the person used some of the Pythagorean identities. I'm just confused on why I can't just use tan(x) as tan(x) = sin(x)/cos(x)

i must of missed some info on the logic behind tan(x) = sin(x)/cos(x) and assuming tan(x) = y/x

[clear sun]

so a is 12.2?

[fickle sleet]

Use theta instead of x for future reference when working with unit circle

let me think for a sec

[vapid stag]

yes of course. i just typed it here for the sake without having to type sin(theta)

[arctic isle]

@clear sun I'm not sure as I'm in bed and can't work it out, but it'll be sin A/a = Sin B/b

[clear sun]

oh i see

[arctic isle]

Which is Sin(40)/a = Sin(25)/8

[clear sun]

so i just have to solve Sin(25)/8

[arctic isle]

Yes then divide by Sin(40)

Wait

[clear sun]

i feel like the number would be to small

[arctic isle]

No I'm wrong

[vapid stag]

😢 @fickle sleet pls save me

[arctic isle]

Erm wait how do I rearrange sin(25)/8 = Sin(40)/a
To get a on its own?

Sin (40)/ (Sin(25)/8)

[clear sun]

seems confusing but ill try

[arctic isle]

First do Sin 25/8

Just so you know that

[fickle sleet]

Ok so like I remember trying this before to get tan(theta) = y/x and it was definitely wrong. But I don't exactly remember why

[clear sun]

aight done

[arctic isle]

What's the answer for that?

[clear sun]

i got 0.42261826174069943618697848964773

[arctic isle]

So sin(40) = a x 0.422...
So sin (40) / 0.422... should be a

[clear sun]

oh ok

[arctic isle]

I have a strong feeling I've done this wrong

If you have the time, I'd go to the maths help part of this server

There are better people there who could help

[clear sun]

oh ok

[leaden birch]

i'm pretty sure i got the cos(theta) part right, but how would i go about rationalizing the denominator for tan(theta)?

[fickle sleet]

Well here's the thing. tan(theta) is actually y/x on the unit circle, so you're right about that. But -1/4 doesn't necessarily imply that y is -1 and x is 4 or that y is 1 and x is -4. All it really says is that the ratio between y and x is -1/4. In the 4th quadrant, cosine is positive and sine is negative. This is the only thing you can be 100% sure of. So by this logic, it should be y = -1 and x should be 4 correct? But then that doesn't make sense considering that x cannot go above 1. Wait you must be thinking, so that means that you just have to reduce their ratios until one of them is 1 or -1 right? Not exactly. In this case if you did that, y = -1/4 and x = 1 since you divided them both by 4. Now you're close, but the problem is that the coordinate doesn't touch the unit circle. Have a look for yourself.

So judging by the unit circle, both values have to be smaller, but still keep the -1/4 ratio. But you can't determine this normally

Here's something interesting. If you draw a slope of -1/4, you'll actually be able to find the exact point at which this slope intersects with the unit circle. That point is what the values of the cos(theta) and sin(theta) are

and correlating with the answer sheet your teacher gave, those numbers matches up with this

[clear sun]

Did I do this right?

[fickle sleet]

Your drawing is correct. Just make sure to use lowercase letters for sides, and uppercase letters for Angles

[clear sun]

oh ok

[clear sun]

uh what do i do after this step? 😅

[leaden birch]

law of cosines

do you know that?

[clear sun]

ill search it up

[fickle sleet]

Anyways if the above graphs I sent were a bit unclear, I apologize for that. This is a more cleaned up version of the graph. Here's a brief overview of it. Red circle is the unit circle. The red point represents the y and x values of what you think they are. The blue line is the -1/4 ratio, visualized as a slope. The green dot is the correct coordinates for the values of sin(theta) and cos(theta) as your teacher gave on the sheet you sent. You can see that the ratios of both points are on the blue line. However, notice that the only correct coordinate is the one actually touching the unit circle. Hopefully this helped you realize as to why you can't directly find sin(theta) and cos(theta) given tan(theta), and need to solve for it using other methods.

[tawny dragon]

Does anyone know how to find missing angles and arcs in a circle?

[vapid stag]

@fickle sleet From my understanding one of my friend told me the concept i understood is correct. but the combination is infinite. so assuming tan(theta) = sin(theta)/cos(theta). The theta for sin and cos can be 1/1, 2/2, 3/3, and so on and so forth. so it is not a way to find the exact value that is the solution.

I think i understand a bit more based on that.

[long garnet]

which one do u need?

[tawny dragon]

@long garnet arc CD

[mint turret]

I know none of this ;c

[rugged sage]

how does one convert from cos to tan

ik cos to sin is sin theta = cos90-theta

[dark sparrow]

why exactly do you need such a conversion?

[edgy scarab]

does anyone here know how to find volume of sphere

[mint turret]

does anyone here know how to find volume of sphere
@edgy scarab it’s like 4/3 *pi * r^2

[edgy scarab]

ok

@dark sparrow do you know this?

[dark sparrow]

you should not ping random people like this

[edgy scarab]

sorry

i am new to the server and didn’t know that

but do you know how to find lateral and surface area of this?

[dark sparrow]

this is a cylinder with a hemisphere stuck on top of it

act accordingly

[edgy scarab]

i’m confused

[dark sparrow]

do you know how to find the surface areas of spheres and cylinders?

[edgy scarab]

i know cylinder

but not sphere

?

[dark sparrow]

then look up "surface area of a sphere formula"

[edgy scarab]

ok

[topaz slate]

Did you look it up ?

It is 4πr²

[edgy scarab]

yeah i did

[upper karma]

Given an isosceles triangle, write a function that shows the dependency betwen the angle at the base (isosceles angle)
and the angle between two heights coming out of these angles(srry for bad terminology i have no idea how math or any of that stuff works in english)

who the heck came up with this

[unique condor]

ABCDE is a regular pentagon

What is measurement of angle alpha by degrees

Very golden ratio question

[mental crystal]

Is geometry hard for many people ?

I get some tasks telling about some points moving

I just cannot cannot do anything with it, I just cannot answer things correctly

It makes no sense to me

[civic zenith]

i don't know i enjoy it but i would not say that i am that good at it

[mental crystal]

I just cannot visualize transformations

Someway

in my head

[prisma glade]

geometry is easier than algebra for me

[mental crystal]

I find algebra easier

Geometry crashes my brain

[prisma glade]

at first, i thought algebra was easier tbh

but then i got used to geometry

i caught a grip

[mental crystal]

I hope I will ever understand it

It really anoys and demotivates the crap out of me

[prisma glade]

i was same

i think when you get the hang of it

itll be easier

[mental crystal]

Like I saw something like tan(40)

1 / tan (50)

or something like that

that those were equal

I just..

How

Someone please explain why tan 40 = 1 / tan 50

[civic zenith]

?

if i put them in my calculator i don't get the same thin

g

[mental crystal]

I do get the same thing

tan (40) = 1 / tan (50)

results in ~ 0.839

[civic zenith]

oh yeah sorry

okay

i'll explain

1/tan(50)= cot(50)

tan a= cot(90-a) so tan 40= cot (90-40)=1/ tan 50

[mental crystal]

What's cot ?

[civic zenith]

oh yeah well in my country we use cot for 1/ tan

[mental crystal]

Ah okay

[civic zenith]

oh well it is also right in english

cotangent

but this has nothing to do with visualisations of transformations

[mental crystal]

Thank you for explaining

I guess i'm going to do pre-geometry on khanacadamy

And then do high-school version

[stoic plank]

Good afternoon, People

I was assigned Khan Academy by my teacher, but the thing is. . . I don't understand it

I am in 7th grade

Does anyone here know how to solve or can help me out with formulas of how to solve for Surface Area?

[silent plank]

khan should be providing you with the relevant information in text and videos

what have you been viewing and what don't you understand?

[fervent imp]

You can search up the formulas for surface area.

[stoic plank]

Oh my

Sorry to respond to your advice so late

I was searching up formulas

[fervent imp]

What is the problem?

[stoic plank]

Thanks @silent plank and @fervent imp

[stoic plank]

Well kinda just the four questions I was assigned

Do you want me to take screen shots and do them together?

[fervent imp]

1 problem at a time.

[stoic plank]

Yeah

Np

1 problem is already a lot of help

I think to find one part is to do base * height and then cut that into half, but from there I am confused

[fervent imp]

Give me a second to work it out.

[stoic plank]

Okay

Thanks

[fervent imp]

What formulas are given in Khan Academy?

What formulas did you learn so far?

[stoic plank]

So

This is the video I learned from

https://www.youtube.com/watch?v=918qWRjICbs

I looked up how to solve for the surface area of a triangular prism

And that came up

[fervent imp]

just verify if it is correct.

Do you have the answers?

[stoic plank]

Nope

It is khan

[fervent imp]

type the answer and let me know if it is right.

What is your next question?

[stoic plank]

Okay

It was correct

[fervent imp]

Find the height of the pyramid first.

Then use the formula and you will get the answer

[stoic plank]

Sure

As in how you solve these type of problems?

[fervent imp]

What is the height?

Use Pythagorean theorem

[stoic plank]

?

The height is 6

[fervent imp]

Use the Pythagorean theorem to find the height.

[fervent imp]

No.

[stoic plank]

It think one step is to do 6 x 8 = 48

And then cut that in half

28

So then you would do 28 + ( 8 x 8 ) = 88

So SA = 88?

[fervent imp]

Is this for the first question?

I am confused.

[stoic plank]

For that question

[fervent imp]

This is what I did :

Pythagorean theorem states that: a^2 + b^2 = c^2

[stoic plank]

Oh

[fervent imp]

So, x^2 + 4^2 = 6^2

Solve for x.. which is the square root of 20.

Then, use the formula.

[stoic plank]

But does it ask for the Pythagorean theorem?

[fervent imp]

PS: Don't use the formula if it is not givern to you. In my school we have a data booklet with formula.

[stoic plank]

I haven't learned that

[fervent imp]

Okay. What formulas have you learnt

[stoic plank]

None for pyramid

[fervent imp]

We could find the SA for all 4 triangles and the square then add it up.

[stoic plank]

I was taught rectangular and triangular prism

But not pyramid

[fervent imp]

Okay, find the surface area for 1 triangle then times it by 4. Then, find the surface area for the square. Then add all surface areas up.

Got it?

[stoic plank]

This is what it said

I got it wrong

[fervent imp]

Okay, ya, that is like what I said.

[fervent imp]

^^^

Sorry I am not much use of help. I am still in highschool

[stoic plank]

No no

I am in middle school

[stoic plank]

You taking your time to try to help somebody else is already so much help

[fervent imp]

Thanks.

What is your third question?

[stoic plank]

That is my 3rd question

[fervent imp]

I am solving it now.

Just a second

[stoic plank]

Ok

Thx

[stoic plank]

I am trying too

[stoic plank]

Oh

God that was fast

[fervent imp]

Look at how I solved it and try solving it on your own without looking at the solution.

@wet loom is correct.

[fervent imp]

Lol, I thought I wasn't fast enough.

[stoic plank]

I wish I could try to do what you did for this problem

[fervent imp]

Did you get the answer correct?

[stoic plank]

Yeah

Thank you so much

So 4 * 2 1/2?

More like

4 (2.25)

Oh right

Sorry I got it mixed up for a second

[fervent imp]

A=37.5. Did you get that?

[stoic plank]

So SA = 37.5?

Because I did 6 * 2.5 * 2

[fervent imp]

Yes.

Use the formula.

How did you write that in fancy handwriting?

[stoic plank]

So 37.5 is correct?

[fervent imp]

yes

[stoic plank]

Okeh, lemme try it

[fervent imp]

Grave marks?

[stoic plank]

Yay it is correct

Except I got a 75% percent on it since one was wrong

Oh whelp

At least it is something

Better than 25%

[fervent imp]

I don't have that.

[stoic plank]

Do you guys wanna try again and speed run through it?

Like I would do it and you guys would correct me?

[fervent imp]

Yeah sure.

[stoic plank]

okah

I think you would do

5 x 8

Then split that in half

so 20

From there I believe you do 8 x 4

32 + 20 = 52

I think the Surface area is 52 squared meters

[fervent imp]

@stoic plank Sorry dude, I got to go. It is midnight. Good luck with math

[stoic plank]

Midnight?

For me it is 3:55 PM

Interesting time zone

[fervent imp]

It is going to be the next day in 4 minutes.

[stoic plank]

Oh god

Okay

It was nice meeting you, @fervent imp

IS the surface area 52 squared meters?

Oh

[fervent imp]

Did you get the same?

[stoic plank]

Oh no

Now I got fractions

As a decimal

0.75?

right?

[fervent imp]

6(2/3)^2

i got 2.666??

So, the formula is 6a^2

= 8/3

[stoic plank]

I am so confused

I get the 6 * (2/3), but not the = 6 * 4 / 9 = 24 / 9?

[fervent imp]

The formula for SA for square is 6a^2. Over here "a" represents the length of the side. So, substitute in the values. 6 (2/3)^2 and solve.

[stoic plank]

What is " ^ "?

Like what does it represent

[fervent imp]

i was just emphasizing it.

[stoic plank]

Oh as in

3^5 is the same as 3 * 3 * 3 * 3 * 3

Oh

Okay

But I am still confused on how to solve this problem

Like if I make it into a decimal

do I multiply it by 6 and then by 4?

I believe it is "Length * Width"

All sides have the same value, right? ( Or at least for this problem )

so 6 (2/3)

But idk how to make that 2/3 into a decimal

Oh

So when you multiply fractions the denominator stays the same, right?

So

6 x 2 = 12

so 12/3

right?

This is just so confusing

Mhm

Each square is 2/3

Mhm

Oh

This one

If I am correct

Here lemme find your message

"Ok do its the same concept. realize that a pyramid with a square base is 4 triangles and one square.

the triangles on each side are just two right triangles joined together. remember the area of a right triangle is just (l * w) / 2.
so the the area of one of those obtuse triangles is 2 * (4 * 5) / 2 since we have two of them which is 4 * 5 which is 20 so the total area of all the triangles is 4 * 20 which is 80.

the base of the pyramid is a square (thats why its called a square pyramid). we know that the area of a square is just l^2. 8^2 = 8 * 8 = 64.

now we the total area of the triangles and the area of the square and we get 80 + 64 which is 144"

Wait no

This is for triangle

Right?

This one

15^4?

Oh

So I believe the answer for the problem with the present is 1350 squared cm

Ogeh

I can NOT thank you enough

Thank you so much @wet loom and @fervent imp

It was really nice of the both of you to take your time and help me out

I am really appreciative for the both of you

Thank you

[jolly dove]

oh sorry wrong channel

[upper karma]

hey