geometry-and-trigonometry

[upper karma]

second video link

seriously, guys

[little osprey]

DSpider do you understand what im trying to do

[austere terrace]

This is probably really wrong

I got 14 - 256 - d^2

[little osprey]

If you substitute values right

what was on the LHS

[austere terrace]

I set up a meeting with my math teacher at 10:30 so I’ll be getting help

[little osprey]

when is that

[austere terrace]

45 mins

[little osprey]

I got 14^2 = (16-d)^2 + 14^2 - (16-d)^2

[austere terrace]

Uhh

[little osprey]

^

2 methods to do it

[upper karma]

Area = 1/2 * a * b
Area = 1/2 * base * height
(for triangles)

[little osprey]

yeahhhh

[runic peak]

Guys, I can't imagine what is radical position, what is the difference between radical and tangential?

[devout harbor]

do you mean radial

[runic peak]

Yeah

[pastel anchor]

hel p pls

[upper karma]

@austere terrace

@pastel anchor is angle CPB a right angle?

[pastel anchor]

i dont even know m8

[upper karma]

there should've been a little square marking it if it was

angle ACB is definitely a right angle

[pastel anchor]

im just gonna assume theres a right angle

actually i think ican use pythagoras and secant tangent

[upper karma]

well, you know AC = 8 and AB (which is the hypothenuse) = 10

[pastel anchor]

alright i figured it out it was 3.6

[fleet wolf]

you dont need to assume anythin since CPA is 90 (AC is a diameter and P a point on the circle) and APB are collinear

for the second one, call the centre O

then ABO are collinear

draw riht trianle OBC (or OBD doesnt matter by symmetry)

see what you get from there

[pastel anchor]

i got the equation CO^2 = 12^2+ r-4

but how do u do pythagoras with those numbers

[fleet wolf]

loo at CO

its special

its a line from the centre to a point on the circle

[pastel anchor]

what is special about it

[fleet wolf]

a circle is the set of points the same distance away from the centre

that distance is the radius

[pastel anchor]

so how do i find that distance

[fleet wolf]

the important thing is that CO is the radius

r

so substitute that in to ur formula and solve for r

do you see how its the radius?

[pastel anchor]

yea ok its the radius, but is the equation im kinda lost

[acoustic jungle]

baby pluto I thought we went over this 3 days ago

[pastel anchor]

actually i dipped on taht question and coming back into it

XD

[acoustic jungle]

The perpendicular bisector of a chord passes through the center of the circle

and you obtain the equation (r-4)^2+12^2=r^2

r=radius of the circle

[pastel anchor]

so how do you do that equation?

[acoustic jungle]

Pythagorean theorem

[pastel anchor]

im confused

[acoustic jungle]

You have to draw the picture on paper

You will not be able to understand it if you just look at what I am writing

[pastel anchor]

yea i did draw it on paper

[acoustic jungle]

just know a perpendicular bisector bisects the line into two equal pieces at a right angle

and the perpendicular bisector of any chord passes through the center of the circle

show me what you drew

[acoustic jungle]

CBC is a right triangle

so you can apply the pythagorean theorem

You did it correctly

[rose oracle]

Can i get help on ym math

[acoustic jungle]

put it in brackets though

[rose oracle]

my*

[acoustic jungle]

the r-4^2 should be (r-4)^2

Now you can solve for r

[rose oracle]

Find the length of the arc indicated in bold. Round your answer to the nearest tenth.

help

[pastel anchor]

@acoustic jungle so i put the thing into teh calculator, but it became r^2−8r+160=r^2

how did 12^2 go to 160

[austere terrace]

@upper karma Thanks! I just got off my meeting with my teacher!

[upper karma]

@austere terrace and was I right?

[runic peak]

@pastel anchor (r-4)^2 =r^2-8r+16,

[austere terrace]

Yeah

[runic peak]

16 + 12^2 =160

[pastel anchor]

where did u get the 16 @runic peak

[runic peak]

From the (r-4)^2

[pastel anchor]

oh

[runic peak]

@rose oracle product of the radius and radian of the angle =10yd×30degree×pi/180degree=5×pi/3=5,2yd

[upper karma]

need help with this

[silent plank]

all of them?

[upper karma]

@rose oracle I think the answer is 12 120, but I'm not 100% sure

[silent plank]

consider connecting some points and applying something like the inscribed angle theorem @upper karma

[rich wolf]

@rose oracle theta = s/r

[upper karma]

again, not 100% sure about this, but it's plausible

[gentle hollow]

Hey guys, if the radius of the circle is 5 and AB is 6, then the area of triangle is 25 right?

[quiet mason]

if AC is a diameter,yes

[gentle hollow]

Ok, just making sure!

[acoustic jungle]

No, it would be 24

[runic peak]

How?

[acoustic jungle]

@gentle hollow

[upper karma]

it is 24

the triangle is a 6,8,10 triangle

[acoustic jungle]

He probably thought BD is the altitude

BD is not the altitude

Unless it's a isosceles

which it isn't

[runic peak]

Wait a mỉnutes, guys, please explain to me what is racial position?

[upper karma]

racial?

[runic peak]

Radial

[upper karma]

i think its the angle made by two radii

@gentle hollow is angle ABC a right angle?

there's no marking to indicate that it is

@upper karma it has to be

how do you figure?

since it is the inscribed angle that intersects an arc with measure 180

and incribed angles have 1/2 the measure of the arc

[runic peak]

Or you can say: if a triangle is inscribed in a circle such that one side of that triangle is a diameter of the circle, then the angle of the triangle that is opposite the diameter is a right angle.

[acoustic jungle]

This wording

[upper karma]

I just (re)learned what inscribed angles are 🙂 thank you

@gentle hollow

sorry about the poorly drawn circle 😛

angle ABC is indeed a 90 degree angle

[little osprey]

Pretty good drawing

[upper karma]

@pastel anchor

[pastel anchor]

thanks

[acoustic jungle]

Why is the radius 4+5

what theorem is that

[little osprey]

yeah what theorem is that

[upper karma]

because it said BE = CD

[acoustic jungle]

ah

sorry I didn't see that part

[upper karma]

yeah, I cropped it out

[lyric summit]

im not sure how to start solving this problem

[little osprey]

Maybe

label

[lyric summit]

i have it printed out

but i cant send pictures rn

[upper karma]

where are the letters on the corners of the triangle? 🙂

[little osprey]

^Great start

[lyric summit]

?? i have it labeled on my paper already

[upper karma]

eh, even without the labels...

[little osprey]

Well what are you doing in class rn

[lyric summit]

like which unit this is from?

[little osprey]

^Important so we can get an idea of what we can use

Yeah ig

[little osprey]

Ah....

They teach this in school?

cool

[lyric summit]

my geometry teacher is scary

[little osprey]

well have you drawn everything

[lyric summit]

yes

[upper karma]

@upper karma

3. is ... 🙂

something, something, tangent

arc BA is twice angle CAB

arc BA?

oh?

doesn't look like an inscribed angle

well the thing is idk how to do it at all

my teacher never teached this

ty 4 the help

[silent plank]

ignore that work

[upper karma]

why is it wrong?

@silent plank my work?

[silent plank]

because it isn't using anything valid

how much circle geo have you been taught?

[upper karma]

arc BA is twice angle CAB
@upper karma
it's not called an inscribed angle, is it?

no

but its still true

i think i just started learning this

after learning the arcs and secant

I know that angle CAB is a supplementary angle to BAD (both form 180 degrees)

yes

draw the radii

connecting to B and A

[silent plank]

have you learned about:
inscribed angle theorem?

@upper karma

[upper karma]

180 - 74 = 106. So angle BAD is 106. And if you say that it's twice the angle... that means arch BEA = 2 * 106 = 212

correct?

yes

i have

but, i'm struggling

yes

@upper karma

[silent plank]

consider constructing RQ and/or PS

[upper karma]

there you go then, @upper karma answer to 3. is 212

[silent plank]

that value doesn't even make sense

wait

[upper karma]

but...
2 * 74 = 148
2 * 106 = 212
148 + 212 = 360 degrees

can u help me by step by step

[silent plank]

the question itself doesn't make sense

[upper karma]

@upper karma with 3. ?

with 1-4

ok, what didn't you get?

the first one

do you know what an inscribed angle is?
(I just re-learned this myself today)

[upper karma]

an inscribed angle is an angle with its vertex on the circle

https://youtu.be/-QBJtmEVg9s?t=18

"X" = "angle QTS" in your question 1.

"X" = half the measure of the arc

if you multiply something by 1/2 (or 0.5), you get half

i get it now

but how do u do numb 3

its different

from the rest

@upper karma said it's twice

scroll up

ok

BEA is twice of BAD so its 212

we got more questions

that needs helps

[acoustic jungle]

what is going on

[upper karma]

geometer question

@upper karma you don't expect us to do all your homework, do you?

no

just need adive

on 5-6

you learned what inscribed angles are, so give it a shot yourself, see where you get stuck, then ask for help

[acoustic jungle]

@upper karma your answer for 1 is incorrect

[upper karma]

oh?

how come?

.....

[tawdry mortar]

Yeah I thought so too

[silent plank]

why did you think that would work?

[acoustic jungle]

because you applied the wrong theorem

[silent plank]

^

[acoustic jungle]

there are no inscribed angles

the angles are not touching the circle

[upper karma]

its just arc and angles

[silent plank]

did you ignore that stuff and do it properly papi?

[acoustic jungle]

@upper karma https://www.youtube.com/watch?v=GppOSNTi5OA

[upper karma]

yeah it seemed little wrong

[acoustic jungle]

that being said all the answers from 1-4 are wrong

I guess except for 3 because he didn't do that one

for 4 you didn't even use the other angle provided by the question

It's okay though

[silent plank]

4 is a bit tricky

[upper karma]

@acoustic jungle thanks for the video

yeah, I messed up 🙂

my bad

[acoustic jungle]

🍉

[silent plank]

q3 had same bad notation.
minor and major aren't needed if you label your arc with 3 letters

and it doesn't make sense to call it minor especially since it takes up the majority of the circumference

[upper karma]

at 2. it says that arc GI = 76 degrees, but angle GKI = 55 degrees?

[silent plank]

yes

[upper karma]

but the video posted by Fishraider at 4:45 says otherwise

"angle ABC and arc AC have the same measure"

[silent plank]

K isnt the centre of that circle

[acoustic jungle]

it's good to prove the theorems so you don't accidentally misuse them

I think something is wrong with 5

I am not sure how to solve it unless it states those are tangent lines

[silent plank]

in circle geo its a reasonable assumption that things that look like tangents are tangents

also only 1 point is labelled at each contact

[acoustic jungle]

It could be a secant line that doesn't extend beyond that point

but I see what you mean

[upper karma]

the formula for 5. seems to be at 16:37 from the video above
if arc HF = 97, then arc HIF = 360 - 97

so angle FGH = 1/2 * ((360-97) - 97)

[acoustic jungle]

Yeah

That is still using the fact that they are tangent

and you could've also done it by drawing the right triangles and getting FGH+HF=180

[steep marsh]

Would somebody be able to explain how one would determine the angle of A for range of 180 < A < 360 where : sec A = 2

[silent plank]

do you know the definition of sec?

[little osprey]

^

[steep marsh]

I don't understand any of this honestly. Never did it anytime before now

[silent plank]

if you don't know definition, consider looking it up

[steep marsh]

Been looking up stuff related but now really sure what you mean by definition honestly

[silent plank]

tell me what you know about (the trig function) sec

[steep marsh]

It's something like the inverse of cosine where it's equal to 1/cos(theta) where 1 is the hypotenuse or something.
Most calculators and libraries don't have functions for it. Beyond that I'm not entirely sure about much

[silent plank]

be careful with the use of "inverse"

[steep marsh]

reciprocal

[silent plank]

it is the multiplicative inverse.
reciprocal function (of cos) works too

so for your question you can rewrite it as:

$\frac{1}{\cos(A)} = 2$

[somber coyote]

ramonov:

[steep marsh]

So, doesn't this just change the difficulty from sec to cos as A is unknown so I can't use the cos function on it

[silent plank]

perhaps if i then rearranged it to

[upper karma]

Geometry is awesome 👍

[silent plank]

$\cos(A) = \frac12$

[somber coyote]

ramonov:

[silent plank]

would you be able to determine the possible values of A

[steep marsh]

From what I know of math wouldn't it be 2*1 since whatever you do to one side you do to the other?

[silent plank]

wouldn't what be 2*1?

[steep marsh]

cos(A) = 2*1

[silent plank]

how are you getting that?

what exactly are you doing to both sides of the equation

[steep marsh]

I don't even know anymore. This stuff hurts my brain trying to figure out.

[silent plank]

i took the reciprocal of both sides to get cos(A) = 1/2

multiplying both sides by cos(A)/2 would've achieved the same result

[steep marsh]

I feel like there's information I'm completely missing or my brain has just melted. So sec A = 2 turns into the reciprocal of 1/cos(A) = 2, what the importance of moving the 1 over to the other side?

[silent plank]

wdym by moving 1 to the other side

[steep marsh]

moving from above cos to above the 2

[silent plank]

the idea is to rearrange it to a form that you should be familar with

that's cos(A) = 1/2
is something that you shouldn't have too much trouble solving

[steep marsh]

I'm familiar with exactly 0% of trigonometry. I apparently should have done it in high school but it was never covered.

[silent plank]

and if you can't then you pretty much have to start trig from the very beginning

try khan to get the basics

[steep marsh]

khan?

[silent plank]

khanacademy

[upper karma]

I love khan academy 👍

[steep marsh]

I'll have to look into that later, don't got enough time to properly learn this stuff at this time

[upper karma]

Good website! And it’s free 👍

I don't know if I'm putting this in the right chat, but i need someone to tell me if i did this right. I think this is supposed to calculate the size of the triangle or something like that... I'm not good at maths but the Points are A(-2,1) B(3,2) C(4,-5) thanks in advance

@upper karma it's not very clear what the problem is

[unborn jacinth]

logically speaking, why do they h and k also get flipped when the x and y do

it becomes y-k

[acoustic jungle]

@upper karma "Size of triangle" ? Do you mean area? Use the shoelace theorem

[upper karma]

it's fine i asked a classmate and he said it's correct so

[acoustic jungle]

@unborn jacinth well now you are looking at the graph in respective to x, so the values are flipped

It is not easy to explain

[upper karma]

also it wasn't clear what the problem was to me either since i missed out on 2 weeks since no one told me we were having online classes

[unborn jacinth]

ok

[acoustic jungle]

Maybe it's because 1/4p is unnecessary here

[unborn jacinth]

hmm

ok

[acoustic jungle]

like consider rotating the graph 180 degrees

then "subbing" y for x if that makes any sense

Wait a second, it doesn't make sense

@unborn jacinth those equations shouldn't be equal

[unborn jacinth]

they're not equal

theyre different equations

[acoustic jungle]

oh

what is your question

[unborn jacinth]

so normally, the vertex is h,k right

o wait

ooooohhh

nvm

I got it

[acoustic jungle]

also I just realized I meant to say 90 degrees not 180

[unborn jacinth]

oh ok

[wide fox]

can anyone help me with some of my geometry hw?

[little osprey]

just

ask

[steep marsh]

Okay I don't even know if this is trig.
Convert 98 degrees 51 ' 16 " to an angle in decimal degrees.
What even is this?

[little osprey]

its a unit of measurement

@wide fox

post

[wide fox]

okay

[steep marsh]

How is that a unit of measurement? I can't even.

[wide fox]

@little osprey

[little osprey]

@steep marsh Google it perhaps?

[wide fox]

i did

[little osprey]

Well

What have you tried so far

[wide fox]

i have 3 other ones that far

i googled asked my friends and tried to solve it by plugging it in with these formulas

[little osprey]

wait

let me go get

paper and pencil

[wide fox]

okay thank you

[little osprey]

eh

ok

so basically

let x be the radius

therefore the slant height is 3x

do you follow so far

[wide fox]

yes

[little osprey]

ok tell me

surface area of a cone

[wide fox]

for the question?

[little osprey]

just in general

whats the formula

to finding the surface area of a cone

Im asking you so I can make sure you understand

[wide fox]

oh its the lateral plus the radius squared x pie

[little osprey]

hmm

surface area is all of the area

Do you recognise this

[wide fox]

yes

[little osprey]

ok thats how to find the surface area of a cone

everything clear so far?

[wide fox]

yes

[little osprey]

ok so

pi r^2 + pi r l = 144 pi

can you factorise the LHS please

[wide fox]

what is LHS?

[little osprey]

Left hand side

[wide fox]

oh

[little osprey]

HCF factoring

you got this

Do you know how to factorise?

[wide fox]

gimme a sec

[little osprey]

ok

[wide fox]

yes

so say that i got rid of the pies

say i did this

does this work too?

[little osprey]

uhm

ok lets go back to getting rid of the pis

[wide fox]

okay

[little osprey]

whats

2r^2 x 3r

wait no

ok so get rid of the pis

[wide fox]

yes

[little osprey]

pi (3r^2) + pi(r^2) = 144pi

3r^2 + r^2 = 144

[wide fox]

yes

[little osprey]

ok add

what do you get on the LHS

[wide fox]

4r^2

[little osprey]

good

now what do you do

4r^2 = 144

[wide fox]

square root

[little osprey]

give a guess even if youre not sure

[wide fox]

2=12

[little osprey]

nope

divide both sides by 4 first

you have to isolate the r

actually

you know

square root would work

but traditionally dividing by 4 comes first

ok lets do it your way

[wide fox]

okay

[little osprey]

2r = 12

[wide fox]

r=6

h=18

[little osprey]

nice nice

Good job

[wide fox]

we have more tho

[little osprey]

Did you get

the workng out

[wide fox]

yes

[little osprey]

Ok

just post it

[wide fox]

i will if i need to

[upper karma]

Yes 👍

[little osprey]

I need to head to bed soon

but uh

[wide fox]

thank you @little osprey

[little osprey]

no worries

[golden panther]

GUYS! can i turn r=2cosθ into rectangular coordinates

[simple slate]

How do I find the rest of d1?

[golden panther]

idk

[simple slate]

Fr?

[golden panther]

nah but if i did i would tell you

everybody must be asleep

[round anvil]

@simple slate problem is too blurry

[upper karma]

@keen nexus just do Pythagorean theorem

[round anvil]

pinged the wrong person there buddy

[upper karma]

@simple slate

Maybe idk

What’s that 54

Ya just do Pythagorean theorem I think since it’s a kite or rhombus

And diagonals of a kite are perpendicular

Your welcome

[golden panther]

GUYS! can i turn r=2cosθ into rectangular coordinates

[subtle gate]

....

Rectangular coordinates

i refuse that name

[golden panther]

thats what it says on the worksheet

please

[subtle gate]

Start by thinking how do you express r in terms of x and y?

and then how do you express θ in terms of x and y

then just plop that in and you should be good to go

[golden panther]

so its the same as a rectangular equation

[subtle gate]

?

what do you mean by that

[golden panther]

Forget i said that

[vocal hound]

I keep failing on this question can someone help me

[subtle gate]

is the base a regular hexagon?

[vocal hound]

yes i believe so

[subtle gate]

i don't get why they felt the need to specify the height to the middle you can calculate that

Do you know how to calculate the area of one of the triangles of the pyramid

[vocal hound]

I tried to view the example they give but I feel like the y skip a step

Wouldnt it just be the base x height /2

[subtle gate]

You see that all the triangles of the pyramid are identical

and that the base of them is 22m each and the height is also 22m

so what's the total area of the triangles?

[vocal hound]

1452?

[subtle gate]

that is correct

now you need to add the area of the bottom hexagon

[vocal hound]

how do i solve for the area of a hexagon

[subtle gate]

a hexagon is made up of 6 identical equilateral triangles

do you know that?

[vocal hound]

Yes I did

[subtle gate]

now do you know the height and the base of one of these triangles?

[vocal hound]

so the angle measure of each one would be 60

and the base is 22

[subtle gate]

the height is?

[vocal hound]

how do we figure out the height?

[subtle gate]

you can calculate it

but it's given to you

look onto the question's sketch again

[vocal hound]

Oh is it 11 rad 3

[subtle gate]

yeah so now do the same thing you did for the triangles before but with these measures

[vocal hound]

is it 726 rad 3

[subtle gate]

?

Not really

Oh wait shoot

That's for all 6

my bad

yes that is correct

[vocal hound]

ok so do we need to multiply that by rad 3

[subtle gate]

do you have a calculator on hand to approximate that

[vocal hound]

yes

[subtle gate]

what does it come out to?

[vocal hound]

1257.468888629....

[subtle gate]

so now add it to the original

which was 1452

and then round and that should be your answer

[vocal hound]

2709

[subtle gate]

now is that the correct answer?

[vocal hound]

yes thats right

thank you

[subtle gate]

good luck if there is more questions :)

[vocal hound]

thanks

[acoustic jungle]

are*

[golden panther]

naw bruh 😂

[compact spire]

Can someone help me with part A

[acoustic jungle]

is this geometry

[compact spire]

trig

learning vectors

but im just having trouble with $(-1-\sqrt{3})$

[somber coyote]

Pony Boy:

[compact spire]

how am I supposed to subtract with the square root?

@versed river you free?

[versed river]

hey mate try not to ping specific members, but yes

[compact spire]

Oh I'm sorry Sneaky

[versed river]

so what's the trouble you're having with this question?

[compact spire]

Okay part A is my problem with converting the W

[versed river]

so we have V=3i-j

and W=4i+sqrt(3)j

we want V-W

so we have 3i-j-(4i+sqrt(3)j)

do you agree so far?

[compact spire]

Yes

So (3-4)i+(-1-sqrt(3))j

[versed river]

we group and we have (3-4)i +(-1j-sqrt(3))j

ya

so thats -i-(1+sqrt(3))j

sometimes it may be appropriate to convert that to a decimal

[compact spire]

Okay so I can't do anything with that sqrt part

Are you talking about the sqrt part?

[versed river]

yes

i dont think its nessecary here

[compact spire]

Okay

[versed river]

and it will make your job easier on the questions finding the magnitude

[compact spire]

If I dont?

[versed river]

b, c, and d look the same as yesterday

if you don't what?

[compact spire]

If I dont turn to decimal

Itll be easier?

[versed river]

then you bracket 1+sqrt(3)

yes

because you're gonna have to square it later

is why i said you shouldnt turn to decimal

(in d when calculating the magnitude)

[compact spire]

Okay D is the other part I was stuck on

[versed river]

sure, what was your trouble?

[compact spire]

I was stuck due to the sqrt part

Let me write it out so I can show where

[versed river]

sure

[brittle ember]

heyy

[compact spire]

Part B is 5

Sorry I'm doing them all over again

[versed river]

5 degrees?

[compact spire]

Sorry I got 78 degrees

And part c I got 5

-78 degrees I believe

[versed river]

c is wrong

But i see what you did wrong

[compact spire]

Okay

[versed river]

You should have had sqrt(4^2+sqrt(3)^2)

Agreee?

[compact spire]

Okay

So it would be sqrt 19

[versed river]

Yes

[compact spire]

Okay

[versed river]

b sounds correct ill do it myself in a second

But tell me what you jad for d

[compact spire]

Okay

$(\sqrt{10}-\sqrt{19})$

[somber coyote]

Pony Boy:

[compact spire]

That's what I'm aiming for right?

[versed river]

Not nessecarily, no

[compact spire]

Oh

I'm trying to think where I'm messing up then.

Dont I use the answer from part a for part d?

[versed river]

You have the i and j components for v-w, so do the same thing you did to get magnitude for w

Nope

[compact spire]

Okay

$\sqrt{(-1)i^2-(1+\sqrt{3})j^2}$

[somber coyote]

Pony Boy:

[compact spire]

So something like this

[versed river]

Close

Hold on im not on pc ill write it up

[compact spire]

Yeah I'm not either, kinda hard on the phone lol

[versed river]

Also, we usually dont write the is and js under the sqrt

[compact spire]

Oh

[remote heart]

I don't think you'd square the i and the j

[versed river]

,rccw

[versed river]

What happens to a negative number when we square it?

Sir you're right, but i already corrected him on that

Also, we usually dont write the is and js under the sqrt
@versed river

[remote heart]

Sir you're right, but i already corrected him on that
sorry, I had a lot of lag and didn't see that

[versed river]

No worries mate

Can you make sense of what's there mr kappa pride?

[compact spire]

Square a negative?

Isnt that an imaginary number

[versed river]

Yes like (-2)^2

Imaginary numbers are an expression for the square root of negative numbers

[compact spire]

Oh if you square it itll cancwl

Cancel

[versed river]

Yes

[compact spire]

Sorry I was thinking sqrt

[versed river]

So what im getting at is you can rewrite that without negatives so you can solve it without a calculator

Although looking at it more you may still need a calculator

[upper karma]

uh how tf u calculate the linear measure of something

of an arc

[versed river]

Like the length of an arc of a circle?

[remote heart]

you can determine what fraction of the circle it is and then find the formula and multiply it by that fraction

[compact spire]

So it would just be $\sqrt{1+(1+\sqrt{3})^2}$

[remote heart]

binomial theorem

[upper karma]

like, I'm doing a math test online and it shows the measure of the arc is 135 but then it says calculate the linear measure and says "PQ = _" but somehow I have to have pi in the answer

[versed river]

The 1 +root3 would be squared but yeah

[compact spire]

But the negative wont cancel it?

[versed river]

Assuming the measure 135 means degrees find it as a fraction of the circumfrence

[upper karma]

how

[somber coyote]

Pony Boy:

[compact spire]

Can I simplify the 1 out of the sqrt?

[versed river]

You most defibitely cannot

[compact spire]

Oh

Because it's an equation

Not seperate

[versed river]

135 degrees is what fraction of a full circle

[compact spire]

So that would be the final answer? Seems like I did something wrong

When it comes to the squaring the negative

[remote heart]

or pascal's triangle

[versed river]

$\sqrt{(-1)^2+(-(1+\sqrt{3})^2})=\sqrt{1+(1+\sqrt{3})^2}$

[compact spire]

I'm confused now

[remote heart]

do you know the square of a binomial?

(a+b)^2

[compact spire]

Yes

[remote heart]

then just plug in a = -1, and b = -sqrt(3)

[compact spire]

a^2+ab+b^2

[remote heart]

indeed

[compact spire]

Right?

Okay

But I'm doing the positive

[somber coyote]

Sneaky:

[versed river]

what he's getting at is that theyre equal

[compact spire]

Okay so the answer is

[versed river]

and that you can expand with or without making the simplification that its positive and get the same answer

[compact spire]

$\sqrt{5+\sqrt{3}}$

[somber coyote]

Pony Boy:

[versed river]

close

i think he means what he means

[compact spire]

Hmm

[versed river]

i think 5 is pretty correct

[remote heart]

I forgot about the other 1

[compact spire]

Oh cool

[versed river]

but you're missing something, i believe

[compact spire]

The I and j

?

[remote heart]

it's 2ab

[versed river]

$\sqrt{1+(1+\sqrt{3})^2}=\sqrt{1+1+2\sqrt{3}+3}$

[compact spire]

Ohhh

[versed river]

where is my tex

[remote heart]

[versed river]

oh

extra dollar

[somber coyote]

Sneaky:

[compact spire]

2ab I forgot

[versed river]

so thats actually $\sqrt{5+2\sqrt{3}}$

[somber coyote]

Sneaky:

[versed river]

there we go

[compact spire]

Okay cool

[versed river]

that all?

[compact spire]

I have a lot more but I'm going to try and figure them out again

This is helping me a lot with understanding vectors

Thank you

[versed river]

sir i'll let you answer since you're typing

[compact spire]

Still answering one sec

[remote heart]

okay

[compact spire]

(-3i)+(-5j)

[versed river]

bruh you got it

[compact spire]

Yes!

[versed river]

oh wait

[remote heart]

wait, isn't it 3, and 5

[versed river]

no you didnt

yes

sorry

[compact spire]

Lol

[versed river]

i misread which one was which point

[compact spire]

Positive?

[remote heart]

yep

[high zephyr]

isnt it positive?

[versed river]

yes

yeah sorry

i saw the right numbers and just said correct

[compact spire]

Why is it positive?

[remote heart]

you subtract initial from terminal, I think you just got them switched up

[compact spire]

So its 3i+5j?

[remote heart]

indeed

[versed river]

yes. draw a diagram to conceptualise this if you aren't sure

[high zephyr]

it's positive becuz it's an increase of those values

[compact spire]

Ohhh

[high zephyr]

not decrease

[compact spire]

It goes ip

Up

Okay

[remote heart]

ye because the terminal values are greater than the initial

[compact spire]

Okay

[versed river]

show your diagram

consider which direction each force is in

[compact spire]

Something like this?

Okay idk if I'm doing this right

[upper karma]

@compact spire You can do it! I believe in you! We all do! 👍

[compact spire]

Thank you @xi

@upper karma

[carmine sundial]

lello?

[remote heart]

hello

[carmine sundial]

cani ask questions here?

[remote heart]

indeed you can

[carmine sundial]

trigo-related

[remote heart]

yessir

[carmine sundial]

#4

im stuck with 8cos(60)=x

[remote heart]

I'm sorry, I can't see the question

[carmine sundial]

i sent it through pic

[remote heart]

I can't see the question in the pic I mean

[carmine sundial]

4. In triangle RST, the measurment of angle RST is 60, and ST is 8, find RS

[remote heart]

thanks

so you said you're stuck with 8cos(60) = x?

[carmine sundial]

?

yeah

a guy sais the answer is 4

but i dont know how he did that

[remote heart]

ah

so, do you know the trig values?

[carmine sundial]

what trig values you mean?

sin cos? those?

[remote heart]

yes

[carmine sundial]

i know their formulas

only

[remote heart]

these:

[carmine sundial]

nope

hm wait

ok go in

on*

[remote heart]

well, you can use those to find cos(60) = 1/2

[silver radish]

https://gyazo.com/bbe47943913aff486d9fb5b4e7080b49

[remote heart]

then you just multiply 1/2*8

[carmine sundial]

oh its 1 half because the measurement is 60?

[remote heart]

ye, because we want cos of 60

[silver radish]

anyone know how to do my question?

[remote heart]

that looks like a quiz on canvas

[carmine sundial]

cos(60)=1/2*8 = RS?

[remote heart]

indeed

[carmine sundial]

nice

[silver radish]

its called online school

[remote heart]

yes

[compact spire]

I think I figured out my problem 🙂

[remote heart]

nice!

[compact spire]

yeah I had to redraw my diagram

[remote heart]

what was your final diagram?

[compact spire]

ill post a picture of part a and b, need me to repost the question too?

[remote heart]

I've got the question

[compact spire]

Okay

[carmine sundial]

@remote heart what do i use to find ST if ever i want to?

this shows as false in photomath

[remote heart]

uhm, cos(60) = 1/2

only

[carmine sundial]

no 8?

[remote heart]

but you said you had 8*cos(60)

[remote heart]

nice!

[compact spire]

I got it good right!?

[remote heart]

I believe so

[compact spire]

Cool

Okay im having trouble with this next problem because I dont remember how to calculate it properly

Okay so I have this so far which I believe is correct

not sure if you can read it

[remote heart]

ye

[compact spire]

I think I might know how to calculate it if its the same as part d from earlier

where I just sqrt it all and square the v1 and v2

[remote heart]

sorry

I had to go eat

[compact spire]

Okay

thats all good

[remote heart]

is there a target?

[compact spire]

one sec

going to write where im at on this

[remote heart]

okay

[compact spire]

$\sqrt{(2509.4)^2+(1312.1)^2$

[somber coyote]

Pony Boy:
Compile Error! Click the reaction for details. (You may edit your message)

[compact spire]

I was at $ <1000(cos(0)+2000(cos(41)),1000(sin(0))+2000(sin(41))>$

[somber coyote]

Pony Boy:

[compact spire]

and I narrowed it down to $<2509.4,1312.1>$

[somber coyote]

Pony Boy:

[compact spire]

$||v1+v2|| = \sqrt{(2509.4)^2+(1312.1)^2$

[somber coyote]

Pony Boy:
Compile Error! Click the reaction for details. (You may edit your message)

[remote heart]

just a sec

[compact spire]

Okay

[remote heart]

isn't it asking for the horizontal force on the stump?

[compact spire]

yes

my final answer is $2831lbs$

[somber coyote]

Pony Boy:

[rich wolf]

$\mathrm{you, didn't, really, need, to, type, that, out, in, latex}$

[compact spire]

lol thanks @rich wolf

once I get in the groove I can't stop sometimes

[remote heart]

lul

[somber coyote]

AMD:

[rich wolf]

$💩$

[somber coyote]

AMD:

[remote heart]

oh god

what happened

to

his mouth

[little osprey]

either the black hole is his moutn and the brown part is his beard

or

[remote heart]

@compact spire I think to solve that problem you would find the i components of each one and then add them together, no?

[compact spire]

how do you mean?

[remote heart]

you know how you can visualize vectors using graphs?

[compact spire]

Okay

[remote heart]

the i component would just be the change in x

[compact spire]

Can you run me through it?

I'm not quite picking up

[remote heart]

imagine it like a right triangle where 2000 is the hypotenuse

we have two angles

[compact spire]

Okay

[remote heart]

41 and 90

[compact spire]

like in my drawing

okay

[remote heart]

yesyes

I'm not too sure how to describe this, but we want to find the leg that coincides with the x axis

the leg adjacent to the 41 degree angle

[compact spire]

so the opposite

[remote heart]

yes

[silver radish]

The volume of a pyramid is 50 cubic units. The base is a square with sides of length 5. What is the height?

how do i solve

[remote heart]

what are you stuck on?

[compact spire]

can I use the form Side-Angle-Side?

[silver radish]

idk how to start it should i use volume equation for pyramid?

[remote heart]

side-angle-angle, I believe

[little osprey]

yes

[remote heart]

yes, yk

[little osprey]

recall the volume of a pyramid

[compact spire]

2000lbs(hyp)-41deg(theta)-90deg

[remote heart]

ye

[silver radish]

A cone and a cylinder have the same radius and height. The volume of the cone is
100 ∏
cubic feet. What is the volume of the cylinder? Round your answer to the nearest tenth.

how do i start this

[remote heart]

do you know the relation between a cone's volume and a cylinder's?

[silver radish]

v=pier2h ?

[high zephyr]

no, their relationship

[remote heart]

^

[silver radish]

nope

[high zephyr]

go on mister

[remote heart]

do you know the formula for the volume of a cone, or the area for the volume of a cylinder?

[silver radish]

yeah

[remote heart]

ah

1/3(pi*r^2*h)

do you know the volume of a cylinder formula?

[silver radish]

v= pie r2 h

[high zephyr]

compare the two formulas, what relationship do u see?

[silver radish]

pie r2 h

[versed river]

pie

[high zephyr]

no, i meant relationship

[remote heart]

yes, but in the cone it's being multiplied by what?

[silver radish]

1/3

[remote heart]

indeed

[silver radish]

so 1/3 x 100?

[remote heart]

100 is the volume of the cone

[high zephyr]

you're trying to find the volume of cylinder

[silver radish]

substract 1/3 from 100?

[remote heart]

why subtract?

[silver radish]

he cone's volume is exactly one third (
1
3
) of a cylinder's volume.

so divide?

[high zephyr]

@compact spire taken atm

[versed river]

use a #❓how-to-get-help channel

[remote heart]

indeed, 100/(1/3)

[silver radish]

o

300

ty

[prime jewel]

hey guys i have a question, so i have this example and i can't really find the way to get the answer with my calculator

I'm trying to get that -0.893

Idk what i'm doing wrong

I'm following the formula.

The main question is about 2 forces that lie on 1 point A

[wary bone]

what are you trying to solve for in the end?

[prime jewel]

I'm trying to find the angle between two forces

[wary bone]

so angle abc

[prime jewel]

It looks like 2 triangles

With angle A B C and D

[wary bone]

y i know 😄

if you search angle ABC all you need to du is solve for it, right?

[prime jewel]

I get an error on my calculator when i put the whole thing in

Yes.

[wary bone]

no.

[haughty summit]

set AC=1012, AB to 567...

[prime jewel]

That is what i did, but i get an error.

[somber coyote]

Result:

-0.89288231148696

[wary bone]

cos is a function. you cannot get rid of that by division.

[haughty summit]

watch me

[wary bone]

@prime jewel so you got cos (abc) = somehting. you want only (abc) in the end. remember what you do then?

[silver radish]

A pyramid has a height of 4 inches and a volume of 40 cubic inches. Circle all figures that could be the base for this pyramid.
a 5 inch by 2 inch rectangle

a 3 inch by 10 inch rectangle

a heart with area 30 square inches

a right triangle with one side 5 inches and the hypotenuse 13 inches

a triangle with height 10 inches and base 3 inches

[haughty summit]

arccosine lul

[remote heart]

taken, post in a questions channel

[prime jewel]

How would you go into the problem, you replace the Terms, and then you solve, you square it and you move all terms to the left

and then arccosine.

But when i do it i get an error

[silver radish]

oof

[remote heart]

what's the error?

[prime jewel]

Could you pop into a channel so i can explain?

Anyone?

I'm on Mathematics voice channel rn.

[wary bone]

no voice here 😦

[upper karma]

Does anyone know what the period for this graph is?

[little osprey]

Youre backkkkkk

[upper karma]

I know that it’s 2pi/b

[remote heart]

the period is just when it returns to the starting value

[upper karma]

Yeah man. Thanks to sneaky I passed my test with a 98/100

[little osprey]

Amp is not -3

[remote heart]

da

[upper karma]

But the graph starts from 0 and goes down

Which makes it -3sin?

[remote heart]

where does it go down to?

like what's the minimum value

[upper karma]

5

I’m dumb

Good catch

[little osprey]

yeah no we all make mistakes

[remote heart]

just a happy accident

[upper karma]

Thanks g

What’s is b in the graph

[little osprey]

frequency

[upper karma]

Is it pi/4

[little osprey]

Imma let Mistersir handle this

it'll be confusing if both of us type over each other

[remote heart]

it's when it y returns to original value after y > 0 and y < 0

so a full cycle

[upper karma]

So is B pi/4?

[remote heart]

y has not yet been > 0 at pi/4

[upper karma]

Oh

So it’s 3pi/8

[wary bone]

full circle 🤔

so a full cycle
@remote heart

[remote heart]

at 3pi/8 y isn't 0

[upper karma]

Wth

[remote heart]

full circle 🤔
@remote heart
@wary bone

[wary bone]

guess both work 😄

[little osprey]

it says full cycle?

[upper karma]

bro i cant figure out B in the graph

then is it pi/2

[remote heart]

yes!

[upper karma]

hmm, so why at pi/2 it is considered to be a full cycle

[remote heart]

because at pi/2 y has already been > 0 and < 0 and at pi/2 it is 0

so it has returned to it's original value

[upper karma]

I see

[remote heart]

and been greater and less than zero

[little osprey]

Its had one maximum and minimum value done and came back to the principle axis

[upper karma]

wow

i get it

[remote heart]

yes

big bean explanation is jc

[upper karma]

gonna screen shot this so i dont forget

[wary bone]

you can look at the distance between two maximums aswell. or two minimums.

[remote heart]

oh yea

[upper karma]

also to get the period, we do : 2pi/(pi/2)

[wary bone]

yeah because the default period would be 2pi, have to adapt to that

[remote heart]

da

[little osprey]

are you Russian by any chance mister

[remote heart]

nyet

[little osprey]

Ukrainian?

[wary bone]

can you help me FIND an exercise?
i need a real life example of any circle, where you know the surface area but not the radius. (i only find ideas where you know/measure the radius first)

[remote heart]

a circular kiddie pool covered in water?

but just the surface

no, that's a bad example

[upper karma]

So guys is the graphs equation come out to f(x)= -5sin(4x)