geometry-and-trigonometry

[grave thunder]

then you can change it to slope intercept form which it already is in

i didn't want to explain the whole answer

[foggy talon]

okay thank you

yes i have the answer

9=b

right

y=6x+9

?

[grave thunder]

yes

that's your final answer

wait

oh wait i made aa mistake sorry

[foggy talon]

okay which part was wrong

[grave thunder]

don't find b that's not the technique

its y-y1=m(x-x1)

how did i forget that

its the slope intercept form when one point is given

[foggy talon]

oh okay so i have to put the coordinattes into the equation

[grave thunder]

yes

in x1 and x2

sorry x1 and y1

[foggy talon]

and whats the next step

[grave thunder]

you will get y+3=6(x+2) right?

that's it

just solve it

[foggy talon]

yeah

so -3+3=6(-2+2)

i think i messed up

becuase i get 0=0

[grave thunder]

no don't do that

you have to get an equation not an answer

first you changed the GIVEN question to y=mx+c form right?

[foggy talon]

yes

[grave thunder]

then you found m=6 right?

[foggy talon]

yes

[grave thunder]

then since t is parallel to u both must have their m(slope) equal. right?

[foggy talon]

yes

[grave thunder]

so u hase m=6

when we are given slope of a line and one point its passing through we use formula y-y1=m(x-x1)

you have to put m , x1, x2as given inquestion

[foggy talon]

okay

y-y1=6x-3

-3=y1=6x-3

?

[grave thunder]

x-3 is in bracket

why are there 2 =s?

[foggy talon]

im not sure

[grave thunder]

y-y1=6(x-x1)

so y-(-3)=6(x-(-2))

ok?

[foggy talon]

yes

[grave thunder]

y+3=6(x+2)

y+3=6x+12

y=6x+12-3

y=6x+9

that's the answer

this time for sure

[foggy talon]

YES its right

[grave thunder]

i wanted you to understand but i made a mistake so here;s evrything

[foggy talon]

yes thank you for helping

[quick shale]

can someone please dm

me

i need help

with a 22 question

geo pretest

[dark sparrow]

are you trying to get someone to do the test for you

[foggy talon]

hey can someone check my answer

i got y=1/2x+15/2

my answer sounds a bit off

please check it

[grave thunder]

its correct

[foggy talon]

okay thank you

oh it was wrong

[silent plank]

check your signs

[kindred rapids]

Hi...can someone help with this please?
Just started trigonometry...but I'm not sure what to do here...

[grave thunder]

oh yes the signs i messed that

@kindred rapids try separating cos(a+b)

[kindred rapids]

Um how do I do that in this though..?

[grave thunder]

3/12+2/12

sorry i just told you directly

its just trial and errors

[kindred rapids]

But how'd you get that? > 3/12+2/12
@grave thunder

[grave thunder]

i told you trial and error

[kindred rapids]

Oh..

[grave thunder]

just try separating it into something that will finally give you easy answer

[kindred rapids]

I'm sorry but I'm a bit confused...here

[upper karma]

Remember your special angles?

[kindred rapids]

Yes

[upper karma]

Try to combine the special angles in such a way where their sum or difference is 5/12.

For example, 7π/12 = 3π/12 + 4π/12.

7π/12 = π/4 + π/3

After you find these two specific special angles, apply the addition or subtraction identity for cosine.

[kindred rapids]

Why are you saying addition or subtraction..? Could it be both that I can use..?

[upper karma]

Yup!

You can use either, depending on the special angles you found.

[kindred rapids]

Okay

Ty

[upper karma]

You're welcome.

[foggy talon]

can someone help with this one

[kindred rapids]

Guys I don't understand how I'm going to get one of the answers in the choices in the pic I sent jn....

[foggy talon]

huh

[kindred rapids]

@foggy talon line s is parallel to line r therefore they have the same gradient which is -8/3....all you have to do now is substitute the points given into the equation given to find c, which is y intercept for the line s .. (the eqn is already in the form y = mx + c so it's not like you have to change it around)
Then when you find c you plug that in to the y = mx+c eqn.....m is the same as they're parallel. That's it....you'll get something looking like that => y = -8/3 (+ or -) c for your answer.

-8/3 x **

[silent plank]

try to represent 5pi/12 as a sum or difference of two special angles

[kindred rapids]

So could it be pi/12 + 4pi/12...?

[silent plank]

well pi/12 isn't one of the common special angles so try using something more suited to you goal

consider 5 = 2 + 3

[kindred rapids]

Okay

Then what do I do?

[silent plank]

compound angle identities

[kindred rapids]

Put it into cosAcosB - sinAsinB...

?

[foggy talon]

@kindred rapids thank you

[silent plank]

cos(A+B) = that
yes

[kindred rapids]

Okay

..okay I've never done any question like that and idk how to work that out when I put it into the identity

[silent plank]

you should know the values of trig functions evaluated at special angles

[kindred rapids]

But its 3pi/2.....not 12

[silent plank]

can you show me what you have so far?

[foggy talon]

@kindred rapids so the answer is only y=-8/3x

theres no b?

[kindred rapids]

I didn't work it out but ...
When I said substitute the points given in the eqn given.....you had -6 = -8/3(1) + c ..right?

[foggy talon]

yes

[kindred rapids]

And when you work it out...what you got?

[foggy talon]

-6=-8/3+b

add -8/3 on both sides

10/3=b

?

[hollow raven]

remember lcd

wait thats right but theres something missing

[kindred rapids]

@foggy talon wait...

How about you carry -8/3 on the side with -6 and work it out...from there

So -6/1 + 8/3 = c

[foggy talon]

-26/3

[kindred rapids]

Yes and now put it into the eqn

y = mx+ c

This is for line s

m = -8/3

[foggy talon]

y=-8/3-26/3

?

[kindred rapids]

Right.
So now you can multiple through my 3 to make it look better

Lol

By*

[foggy talon]

8.6?

but i cant leave it as decimial

[kindred rapids]

..lol no
Mutlolie through by 3 in the entire eqn
So... 3y = -8x - 26

That's the equation for line s

Understand?

[foggy talon]

yes thank you

[kindred rapids]

You're welcome:)

[kindred rapids]

Okay...I really need help for this question...please, I need more explanation because I'm confused by what yall are saying to me...

Hi...can someone help with this please?
Just started trigonometry...but I'm not sure what to do here...
@kindred rapids

[humble pulsar]

I dont see any question @kindred rapids

[grave thunder]

you are still stuck on that

[kindred rapids]

Yes because I have no idea what to do....this is like a new topic where I have no clue of anything about it.

@humble pulsar

Listen...I looked at a video but I do not know how he got what he got
From cospi/4 , he got root2/2, how?

[grave thunder]

i told you you just have to check for it one by one

there's no trick or formula for it

[humble pulsar]

$\frac{5\pi}{12} = \frac{\pi}{6} + \frac{\pi}{4}$

[kindred rapids]

@humble pulsar yes I understand that

[grave thunder]

you have yyour special angles check 1 by 1 to know what you can find

[somber coyote]

moshill1:

[humble pulsar]

cos(pi/4) = 1/sqrt(2) since its an exact value

[kindred rapids]

...how do you know that..? I'm just asking because I dont know

[grave thunder]

practice youcan say

[humble pulsar]

30-60-90 triangle

pi/4 = 45 degrees, cosine is adjacent on hypotenuse from sohcahtoa

so cos(pi/4) = x/(xsqrt(2)) = 1/sqrt(2)

[grave thunder]

@kindred rapids wait do you mean you don't know values of cospi/4 and stuff or your saying you don't know how 5pi/12 was divided to pi/6 and pi/4

[kindred rapids]

The values of cos pi/4

[grave thunder]

well he explainedit here^?

also for these special angles you just have to remember the table and that's all

[kindred rapids]

I dont have any table

[humble pulsar]

Just remember the triangles tbh, in my opinion it's easier

[kindred rapids]

How can I get this table..?

How is pi/4 ..45 degrees?

[humble pulsar]

pi radians = 180 degrees

so pi/4 = 180/4 = 45

[kindred rapids]

Okay

Makes sense

[humble pulsar]

but yeah, to answer your question: break 5pi/12 into the 2 exact value angles, use the compound angle identity, then evaluate the 4 trig values you get

[balmy meadow]

The radian is defined as the ratio of arclength to radius. If we choose a radius of 1, the circumference, or maximum arclength, is 2π (you may recall the circumference of a circle is C = 2πr ). Thus a 360° angle is equivalent to 2π radians.

2π/8 = π/4 just as 360°/8 = 45°

[kindred rapids]

@humble pulsar okay ty

[balmy meadow]

Just remember the triangles tbh, in my opinion it's easier
Seconding the triangles

[humble pulsar]

np

[kindred rapids]

I think I'd rather the table ....not the triangles

Could someone Send it or it cant be sent?

[humble pulsar]

just make one up tbh

[kindred rapids]

Lol alright then

[humble pulsar]

do sin, cos, tan for 0,30,45,60,90

that's all the table is

[kindred rapids]

@balmy meadow thank you

@humble pulsar yes i know

[upper karma]

There's a unit circle pinned at this channel too

[balmy meadow]

I should've checked there first. New to this server

[kindred rapids]

What's that...?@upper karma

[upper karma]

What's what? the unit circle?

[kindred rapids]

Yes lol

[upper karma]

Wow

[kindred rapids]

I'm also new...so

[upper karma]

Oh okay

[kindred rapids]

A little clueless

Sorry

[upper karma]

I just didn't have context

Don't be sorry

[balmy meadow]

The unit circle is that circle of radius 1 I was talking about. If we take any point on the circle and draw a line to the origin, we can obtain the coordinates of that point from the angle between the radius and the principle axis

[upper karma]

I have something i believe it's useful

,tex \unitcircle

Wait for the bot

[balmy meadow]

Noted

[kindred rapids]

@balmy meadow I see...

[upper karma]

You'll understand it easier with something texit's gonna send

[kindred rapids]

Okay

[upper karma]

well, not easier but like more in depth

[kindred rapids]

Alright

[somber coyote]

I have just restarted and am loading myself, please wait!

[upper karma]

Lol well that was unfortunate

[kindred rapids]

..lol

[upper karma]

Wait some more i guess

,tex \unitcircle

[kindred rapids]

Lol okay

[balmy meadow]

Is TeXit normally this slow?

[upper karma]

No, ig it's because it's restarting itself

Usually when you send it, it takes like 5 seconds

[kindred rapids]

Guys...what's tan 150° ?
Like how sin30° = 1/2 ....what's tan150°?

[somber coyote]

Al𝟛dium:

[upper karma]

There it is

I wrote this for beginners

[kindred rapids]

Okay ty

[umbral snow]

If you can get sin150 and cos150 then tan150 is immediate

[upper karma]

There's tears of hours spent on that pic

[kindred rapids]

Thank you...us beginners appreciate your effort to help us.

@umbral snow well I didnt get sin150 and cos 150 so...um

[balmy meadow]

There's tears of hours spent on that pic
I need to get more serious about LaTeX figures lol. Don't tell me that's using TikZ

[upper karma]

It is indeed using TikZ

[kindred rapids]

I'm sorry but how do I know what's tan 150° equal to?

[umbral snow]

Well, can you pinpoint 150 on the unit circle?

Use that to identify sin150 and cos150

[kindred rapids]

....um well I dont know where 150 is exactly...

[balmy meadow]

I like to think of it as 180° (on the opposite side of the circle) minus 30° (which we're familiar with from special triangles/that table from before)

We may want to explain quadrants and CAST since the unit circle was only just introduced

[umbral snow]

I suggest at least giving "unit circle" a quick Google.

[kindred rapids]

Okay I did

So there is 5pi/6

But um...the answer has to contain root of something...that kind of answer

[wise hornet]

not always

[kindred rapids]

No...its a multiple choice lol

[wise hornet]

yeah

so you have sin150 which is 1/2

but cos150 has roots in

[balmy meadow]

Maybe it would be better to think of tan(150°) as
sin(30°)/(-cos(30°))

[wise hornet]

so when you divide them to get tan150 there's also a root

[balmy meadow]

The pinned unit circle is pretty helpful for this kind of thing

Just note that the coordinates are given as (cosθ, sinθ) and tanθ = sinθ/cosθ

[kindred rapids]

Cos 150 is what?

[balmy meadow]

cos(150°) = -cos(30°)

[umbral snow]

So you really need to be able to identify where 150 is on the unit circle.

[kindred rapids]

@balmy meadow okay

[umbral snow]

If you can't, I suggest googling how the unit circle works.

[alpine niche]

Hi, sorry for interrupting. Could anyone check if this is okay?

[balmy meadow]

It looks good to me. I would write the numerator as x•tanα so it doesn't look like x is a part of the argument though

I'm not super qualified. Unsure whether a helper has to check this sort of thing

Once you've found y, what do you plan to do to solve for h? (I assume that's what you're looking for, because of the = ? )

[alpine niche]

i got the formula in the first two steps

then i got h

[balmy meadow]

Well done (:

[kindred rapids]

So what's the answer for that question I just sent then..?

[balmy meadow]

Do you normally use this server to just obtain answers?

[kindred rapids]

No..

I normally do my own work but this is really confusing so I just asked for help here...

I also help people when i can.

Is that bad...? Should I stop..?

[balmy meadow]

So we've worked out that there's this thing called the unit circle that's a powerful tool for figuring out the values of different trig functions. We've reasoned out that

tan(150°) = sin(150°)/cos(150°)

We can obtain the exact value of the numerator and denominator since

sin(150°) = sin(30°) = 1/2
cos(150°) = -cos(30°) = -√3/2

So tan(150°) must be

1/2 ÷ -√3/2

[kindred rapids]

Yes

I'm working it out now...

[balmy meadow]

Best of luck

General question about trig for someone with more experience. Where do the "less appreciated" functions like versine and haversine show up in math? I've only used them with spherical triangles

[kindred rapids]

Wait how...is that root3/2 negative..?

[wise hornet]

it is

is there a problem

[kindred rapids]

I asked how..?

[wise hornet]

what's the problem

[balmy meadow]

For all intents and purposes here, the root of a negative number doesn't exist. This is a negative root

[wise hornet]

why can't it be negative

what?

[balmy meadow]

I think they're concerned by -√3 since it's being mistaken with √-3

[wise hornet]

debatable

[kindred rapids]

No...I dont understand how the - sign got there..

[wise hornet]

maybe they're just concerned about possible negative distance issues etc

it's like

it just is

it works

on the unit circle, 150 degrees is to the left of the origin

it's negative x

does that explain anything

[balmy meadow]

The negative got there because I considered the positive first quadrant with θ = 30°, and then reflected the point over the y-axis

[kindred rapids]

Cos150° = - cos30° ....how?

@balmy meadow ohh

[balmy meadow]

For all intents and purposes here, the root of a negative number doesn't exist. This is a negative root
I was misinterpreting your confusion. Sorry about that. If you're curious check out imaginary numbers (:

[kindred rapids]

It's okay...thank you for your help.

Is it okay if i ask for help again...? It's okay if I shouldn't
There are alot of questions we were assigned and most of them are not what i can understand and do..

[balmy meadow]

It's perfectly fine to ask for help! I'm not going to be around to answer every question (I have my own assignments to work on)

[kindred rapids]

Oh I know you have things to do.... I wasnt referring to you to help me alone. Lol I was just asking generally

[balmy meadow]

My main concern about directly addressing assignment questions is that you may not learn something as fully if you're simply presented with the answer. If you're not sure about whether something's allowed it's best you ask someone with more experience on this server

I only joined the server today

[kindred rapids]

Oh Alright then. Ty

[opal slate]

Count the Area of the "bright" in the cirkle, how do i do it?

[wise hornet]

the area of the bright is independent of the position of the hexagon so long as it's fully inside the circle

[opal slate]

I need to count the area of the hexagon, and then subtract it from the area of the circle, its the area of the hexagon that im not sure of how to count

[wise hornet]

ok

hexagons are polygons which means they can be split up into simpler shapes

try chopping them up a bit?

[opal slate]

i know i have to split it up by making triangles

[pearl lava]

is there a way to solve this without trig?

[wise hornet]

nrly, but it's not too hard

i think

[pearl lava]

I got the perimeter as 32.3 and for area I got 47.9. Is that right?

[silent plank]

you can apply properties and ratios of special triangles without explicitly using trig functions.

[wise hornet]

true, ig

just still feels like trig to me

[silent plank]

yeh

what were your exact answers?

[pearl lava]

my exact answer for perimeter is 32.31693247

and for area i got 47.91201442

[green tree]

https://i.imgur.com/5TLsZCl.png

[silent plank]

that isn't the exact answer

the exact answer would be irrational and consist of radicals

[pearl lava]

so would this be the perimeter?

and this is the area?

@silent plank

[silent plank]

yes

[alpine niche]

hi, can anyone help me to check one trigonometry problem? it's done, it's just to check if I messed up something

[distant thistle]

Anyone know how to do this one?

[humble pulsar]

@distant thistle isolate for sin(x), then get x using cast rule ideas

[wise hornet]

that won't be enough, the solution isn't nice unless i've messed something up

[humble pulsar]

you can still do cast rule

sin(x) = -8/15
Hence x is in Quad. 3 and 4

the principle angle is arcsin(-8/15) ~ -32 which means it's about 32 degrees

[opal slate]

How do i solve the area of the halfcircle and the penis

[humble pulsar]

"and the penis"

[cold siren]

does anyone know how do do this?

[grizzled lantern]

@cold siren idk if that helps

[cold siren]

oo ty

[prime gull]

does anyone know how many johnson solids there are when all the polygons of the solid have to have the same edge length?

[wise hornet]

infinite

consider prisms

[prime gull]

huh, why?

[wise hornet]

whadja mean why

there's a prism for every 2d polygon

and prisms are johnson

[prime gull]

ohhhh

so like, except for a triangle there can be any shape with any number of sides

[wise hornet]

i mean even a triangle

triangular prisms are johnson if all the edges are equal

[prime gull]

oh ok thanks!

[foggy talon]

can someone help me with this pls

i cant fail this class

my parents will kick me out

[grizzled lantern]

@foggy talon no then won't

[foggy talon]

i have strict parents

my parents dont believe school is hard

but my dad dropped out

i tried to explain i was sad and they were like its becuase of that phone

and they took it away

[glacial dawn]

@foggy talon

Do u still need help

[foggy talon]

yes

[glacial dawn]

Alright

Are u familiar with the point-slope form of linear equations?

[foggy talon]

yes y=mx+c

[glacial dawn]

Nah, that’s slope-intercept form

[foggy talon]

oh then what is it?

[glacial dawn]

Do u understand what it is saying?

[foggy talon]

oh right yes

[gloomy mesa]

bro what does that mean jve been deprived of math for too long

due to covid

[glacial dawn]

Alright so

What can you do given the info provided in the problem as it relates to point-slope form?

Meaning

What can u substitute in

To get the new equation

[foggy talon]

y-3=1/5(x-7)

[glacial dawn]

Yup, so u got the x and y right

And the slope too I believe

Lol

[foggy talon]

yes

[glacial dawn]

So now just put it into slope-intercept form

How would you accomplish this?

[foggy talon]

im not sure of this part

[glacial dawn]

K so

What is slope intercept form?

[foggy talon]

y=mx+b

[glacial dawn]

Exactly

So what is on the left side

[foggy talon]

y=1/5x

[glacial dawn]

Yep, u gotta isolate the y

So try rewriting our equation like that

[foggy talon]

but how do i find b?

[glacial dawn]

Don’t worry abt that just yet, just isolate the y on the left side

Lmk what u get

[foggy talon]

3=1/5x

3=1/5x+b

[glacial dawn]

Try this

y-3=1/5(x-7)

How would u get the “y” by itself

[foggy talon]

adding 3 to 7

[glacial dawn]

Nono we cannot add the 3 to the 7 because the 7 is in the parentheses

But we could just add the 3 to the other side like u said

[foggy talon]

i would turn the eqution to y=1/5x-7

so 3 plus 1/5x

[upper karma]

https://prnt.sc/viaocg

is this correct?

[glacial dawn]

Nah

Yo bro this is being used

[upper karma]

oh my b

[glacial dawn]

All g

I’ll lyk after

So

Rosie

Let’s not worry abt the 1/5(x-7) yet

Just about the y and the 3

y-3=1/5(x-7)

To get the 3 to the other side, like u said, we gotta add it

What comes out on the right side when we just add the 3?

[foggy talon]

1/5?

[glacial dawn]

Nope, for the right side, u still gotta have the 1/5(x-7) but don’t even touch that yet

Just the 3

[foggy talon]

wait sorry do we just put it on the right

[glacial dawn]

Yep

So what does it come out to

[foggy talon]

y=1/5(x-7)+3

[glacial dawn]

Correct

[cold siren]

can anyone who knows how to do proofs dm me? thx!

[glacial dawn]

Now we can worry abt the 1/5(x-7)

So do u know how to distribute?

[foggy talon]

not really

but i think its 1/5x-7

[glacial dawn]

Close, u just forgot to multiply the 1/5 by the second term, which is -7

In distribution, u have to multiply both terms by the factor outside the parentheses

So what would be the result of the distribution?

[foggy talon]

7/5

?

[glacial dawn]

Bingo, so what’s the result

The first term and the second term together, I mean

[foggy talon]

y=1/5x+7/5

[glacial dawn]

U forgot one thing

What’s the sign given to the second term in the distribution?

[foggy talon]

-7/5

[glacial dawn]

Yup

[foggy talon]

y=1/5x-7/5

[glacial dawn]

What did we add to the right side at the beginning?

[foggy talon]

3

[glacial dawn]

So now what do we have?

[foggy talon]

y=-1/5x-7/5+3

[glacial dawn]

1 mistake

Can u spot it

[foggy talon]

-3?

[glacial dawn]

Nope, although it does have to do with signs

Revisit the distribution, what were the signs on the those terms?

[foggy talon]

plus

[glacial dawn]

For which?

[foggy talon]

7/5?

i mean 1/5

[glacial dawn]

Yea

So what’s our equation

[foggy talon]

y=1/5-7/5+3

[glacial dawn]

U forgot another thing this time xD

[foggy talon]

x

the x

[glacial dawn]

Yup

So what’s our equation

[foggy talon]

is that the complete answer

[glacial dawn]

Nope

[foggy talon]

y=1/5x-7/5+3

[glacial dawn]

Alright

Now we gotta add like terms

Do u want to do this in fractions or decimals?

[foggy talon]

fractions

[glacial dawn]

Alright so which terms could we combine together

[foggy talon]

-7/5 and 3

[glacial dawn]

U got it

What do u get?

[foggy talon]

-7/5+15/5

[glacial dawn]

Yup

Just add them together now

[foggy talon]

8/5

[glacial dawn]

Correct

So what’s the equation we have now

[foggy talon]

y=1/5x+8/5

[glacial dawn]

And what’s the slope intercept form again?

[foggy talon]

y=mx+b

[glacial dawn]

Does it match what we have?

[foggy talon]

yes

[glacial dawn]

That’s ur answer

[foggy talon]

okay thank you for helping me

[glacial dawn]

Yea np

[upper karma]

is this correct?
Yes.

[glacial dawn]

Thx ^^

For answering

[upper karma]

https://media.discordapp.net/attachments/477077539496656917/776599755697422346/unknown.png

anyone know how to do this

[umbral snow]

@upper karma
Start by squaring both sides

[upper karma]

yeah i did

[umbral snow]

Keep in mind that squaring will create false solutions

[upper karma]

i got like x+2sqrt(x)+1=x+9

[umbral snow]

Nice. Rearranging that:
√x = 4

[upper karma]

then i just square both sides again so x=16 and i just have to check?

[umbral snow]

Yeah, I suppose there's no false solutions here haha

x = 16 is the one you got.

[upper karma]

ty for your help

[foggy talon]

pls check my answer

[stuck dragon]

I think the slope is -2

the y intercept is good i think tho

[foggy talon]

are u sure

pls be sure

[stuck dragon]

yeah

ok im 100% sure

[foggy talon]

okay ty

pls check this one

[grizzled lantern]

@foggy talon you can check if a line goes through points by plugging them in

[foggy talon]

right but could you please check it i dont wanna mess up

[grizzled lantern]

it doesnt go through (-8,-6)

[foggy talon]

could u tell me where i went wrong

or the correct answer

pls

i got y=-1/4-4

does that work

[glacial dawn]

Rosie follow the steps i explained earlier

The purpose is to learn the content lol

Not just complete the assignment

[foggy talon]

yes i understand that but i really need to finsih quickly

i have alot of other work to do

its 8pm here

i dont want to stay til midnight

[small dock]

Can y’all help me

11 a. And b.

[vague forge]

when in doubt draw it out

[gloomy mesa]

yes draw

[stuck dragon]

yes

[rose tulip]

I've been taking Geo over the summer using Khan do you think it's advised to move up to Algebra 2?

Some of the math teachers are saying it might be hard once I get into Pre-Cal

What is there that i'm "missing"

[hoary trellis]

any help thanks

need to find area

[dense egret]

Complete the bigger square and minus the side pieces

[hoary trellis]

i got 1600 dunno if its right

[dense egret]

How u get that

[hoary trellis]

50x50 - 15x4

[dense egret]

Why is it 50x50

[hoary trellis]

isnt a square lxw

[dense egret]

But u got the corners missing

[hoary trellis]

ah

[dense egret]

So make it a square first

[hoary trellis]

3600

[dense egret]

And how u get that

[hoary trellis]

would u do 900+2500

[dense egret]

900?

[hoary trellis]

wait idid

15x4 = 60 then 60x60

[dense egret]

Wait I’ll show u the diagram and hopefully u can figure out

[main lintel]

Yes, focus on the big outer square, and from that remove the 4 minor edge squares.

[main lintel]

@upper karma Do you have a general idea what to do?

[urban knoll]

Stuck on this hw problem can I get some help?

I think it’s either triangle or square

But im not 100% sure

[dark sparrow]

,rccw

are the angles on the outside (exterior) or the inside (interior) of the corridor formed by the parallel lines?

[urban knoll]

Exterior

[dark sparrow]

well there you have it

[urban knoll]

Thanks

[dark sparrow]

what have you tried so far & where are you stuck

[hollow raven]

To add on to that, what defines an equilateral triangle?

[dark sparrow]

thank you wikipedia

do you know what the angles are in an equilateral triangle? @random stratus

[mighty wharf]

[slate kestrel]

@random stratus are you still here? because if not I will just ask my question already

[dark sparrow]

yamfe1 you can move to #help-8

[random stratus]

i have a question

[dark sparrow]

why did you delete your original post, cool332?

[obsidian hearth]

Hey guys

How to prove that M' is point of the radius ?

While M' = M1*M2

[upper karma]

@obsidian hearth what is the written construction

[solid flower]

Does anyone know SAS!

[jagged sapphire]

Can i get help

[fickle walrus]

angles on a straight line add to 180

@jagged sapphire

[jagged sapphire]

angles on a straight line add to 180
@fickle walrus sooo

that means

?

its 41

[fickle walrus]

yes

[odd blade]

Guys can I get hl I legit don’t understand this question

[cosmic pebble]

From

#❓how-to-get-help

Rule 1

The help channels are solely for help with math, so feel free to post your question. Asking whether you can ask a question or if anyone knows about some specific topic is unnecessary, so please try to avoid questions of that nature.

[jagged sapphire]

pls snipe

[grizzled lantern]

@jagged sapphire u need help?

wassup?

[odd blade]

How do you do this please help

[grizzled lantern]

alright, so they are similar triangles

[west basin]

if they are similar the sides have the same ratio

[jagged sapphire]

@jagged sapphire u need help?
@grizzled lantern na got a 80

[grizzled lantern]

so what we can do is setup a proportion

so we can do 18/24=x/8

when we do that, we can cross multiply and solve

[odd blade]

So 24x=

144

[grizzled lantern]

exaclty

[odd blade]

So the length of yx is x=6

[grizzled lantern]

yep

[odd blade]

Thanks

[grizzled lantern]

np

[pearl lava]

is the answer for #1 30?

[wise hornet]

30 is a solution

[pearl lava]

how would I find the other?

[wise hornet]

sketch the graph rq

[pearl lava]

would it look like this?

[wise hornet]

yes

[pearl lava]

so is the answer pi/6 and 5 pi/6?

[wise hornet]

yes

[pearl lava]

how would I do it without graphing?

[wise hornet]

you can see sin(x) is symmetrical about pi/2, 3pi/2, -pi/2, 5pi/2, etc.

so sin(pi/6) = sin(pi - pi/6) = sin(5pi/6)

[pearl lava]

oh ok

[cloud valve]

How do you prove this?

do I simplify them

[main lintel]

I'd maybe convert cos^2x to 1-sin^2x

And try that route @cloud valve

[cloud valve]

does that mean that I can write cos^4x

as 1-sin^4x?

nvm I got it

thank you

[main lintel]

No, but cos^4x is (cos^2x)^2

[cloud valve]

yes I saw it now, i solved it

it is indeed 0

[main lintel]

Cool

[untold peak]

I'm terrible with trig and I have to do this workbook. do all angles in a non right angle triangle still add up to 180?

Sorry if this sounds stupid but math is by far my worst subject

[white lichen]

yes

[untold peak]

okay

[subtle snow]

Can someone help me with this?

I’m so confused

[white lichen]

what is the question asking

do you need to find al the angles with letters

[subtle snow]

Yes

[white lichen]

Ok which one are you on

[subtle snow]

Well I attempted to it but I got everything wrong

This is my work

How should I do it or what did I do wrong?

@white lichen

[white lichen]

For the first one why did you think a and b had angle measures 79

[subtle snow]

Because it has to equal 180

That’s what think

[white lichen]

But why did you think they had to be the same

[subtle snow]

I’m not sure

[white lichen]

Have you learned about corresponding angles

[subtle snow]

I think so but I do t remember

[white lichen]

if the horizontal lines are parallel those two angles are equal

[subtle snow]

Can you go to Vic so we can talk?

*vc

[white lichen]

I dont rly wanna talk rn but i can listen and write here

[subtle snow]

Ok

[white lichen]

go mathematics

yes

Yes but that doesnt mean 2 angles need to be the same

Try using the picture I sent to figure out what a would be

wait ill type in #mathematics-voice

[untold peak]

Does this look right or did I get everything wrong?

[kindred rapids]

If tan (theta) = 4/3, what's tan2(theta) ?

[wise hornet]

i assume you mean tan^2(theta)?

generally, tan^2(theta) = (tan(theta))^2

and also sin^2(theta), cos^2(theta), sec^2(theta)... etc.

[kindred rapids]

@wise hornet no I meant tan2(theta)

[wise hornet]

tan(2theta)?

[kindred rapids]

tan2Ø <= it's like that.

[wise hornet]

ok

so tan(2theta)

[kindred rapids]

Ig

[wise hornet]

tan(2theta) = sin(2theta)/cos(2theta)

do you know those

[kindred rapids]

Yes

[wise hornet]

ok

so do it

and then if you need help expressing it in terms of tan(theta) again ask me

[kindred rapids]

What do I do? Tan(theta) is equal to 4/3 and they're asking what's tan2(theta) .....I'm not sure what to do with the sin2theta/cos2theta

[wise hornet]

so what's sin(2theta) in terms of sin(theta) and cos(theta)

oh sod this from now on sin(x) and cos(x)

[azure laurel]

How do I go about answering this

[upper karma]

@azure laurel consider the theorem of angle sum on a triangle

[azure laurel]

I got x = 34

But when I add up all of the sides using that, it equals out to 160

[upper karma]

How are you getting to 34? @azure laurel

[deft bear]

ive been stuck on this for like 20 minutes

[stuck dragon]

for y

since the purple lines are parallel

as shown with the arrows on that pair

and that you can view the green diagonal line as a transversal

you can say that measure of y=75

via alternate interior angles

for x

we could view it the same way but with the other transversal (the other line intersecting the 2 parallel lines)

this time we can say x=100

via corresponding angles

[pearl lava]

Can I use law of sines to figure out the side lengths?

[stuck dragon]

are u allowed to use calculators or no

[pearl lava]

yes

[stuck dragon]

oh ok

yeah u could find out the other 2 sides via law of sines

[pearl lava]

ok thanks

[stuck dragon]

np

[azure laurel]

I j want a rolly rolly rolly with a dab of ranch!

@limpid lagoon You are dogshit at aerials

[main lintel]

@deft bear you got your answer above

[upper karma]

please help!

[wary wind]

@upper karma what have you done so far? How have you approached it?

[upper karma]

I can't even think how to start

Its the hardest geometry question in my homework

I have no idea how to approach it

[wary wind]

Are you doing similar triangles rn?

[upper karma]

What you mean?

[wary wind]

What's the chapter you are doing in geometry right now?

[upper karma]

Im doing a weird cirriculum

chapter 2

[wary wind]

Probably review that chapter then.

To make the long story short - The idea is to use the property that base of the equilateral triangle and the straight line drawn from a mid-point of the base being parallel to each other.

[upper karma]

i read the chapter twice lol

I just couldnt put my finger where i could answer that question

[wary wind]

Okay, let's attack this another way then

How well versed are you with trigonometry?

[upper karma]

Yeah i know trig

btw this is the hardest qquestion in the hw

and it called a free question

so you could use anything

even outside the chapter

[wary wind]

Well, this is your homework - I would like you to think through this problem and apply your approach.

Also - stop saying "it's the hardest homework" - It's not as hard as you think it is.

Well anyways - what's angle of an equilateral triangle?

[upper karma]

60

[silk patio]

For alpha is less than 90 degrees can you work out the height of the triangle?

You know the base is always a/2 so you don’t need to worry about it

To work out the area you need the height

[upper karma]

hmmm

[wary wind]

I will just lay out the brief procedure that you can use to arrive at you result -

You angle alpha has three possibilities -

1. Less than 90°
2. Equal to 90°
3. More than 90°

In each of these cases - you'll form an interior triangle with your equilateral triangle.

Next, your goal should be to prove that the two triangles are similar (read up on similarity criterion)

using this - find out area ratio.

Euler has given the other details you can play around with.

[upper karma]

yeah he said you need to work out heihght

@wary wind how could i find the height?

of the small trianlge

without the area?

[wary wind]

You draw the altitude for the smaller triangle in the case where your alpha is less than 90°

[upper karma]

ok so from the highest point to the base

that creates 2 right angle triangles

[wary wind]

Yes, you have 2 more right triangles then but that's not what we want to use the altitude for.

You want the altitude so that you are able to express your area.

[upper karma]

ok so if i draw that

for the small triangle

i get 2 right angle triangles

with height h

and base

1/4 a for each triangle

[upper karma]

which question?

[kindred rapids]

Hi...so I just need help with two questions
I had 35 in total and I did 33 but can't seem to do these two...
First it's this one ...
I've never done questions like this one so far and I'm not sure how to start...

[earnest echo]

Try expanding sin(x+20)

[kindred rapids]

Okay

sinx + sin20°

[next jackal]

@kindred rapids What's the identity of Sin(A+B)?

[kindred rapids]

sinAcosB+sinBcosA

[next jackal]

got a hint?

[kindred rapids]

sinx cos20° + sin20°cosx

But I'm not sure what to do from there

[silent plank]

it should be simpler if you apply something like cos(a) = sin(90°-a)

[kindred rapids]

I'm not sure what that is..

[next jackal]

Oh

[silent plank]

complementary property of sine and cosine

[kindred rapids]

@silent plank I looked at a video and they said something like that but I've never seen anything like before...my teacher hasn't mentioned anything like that ...

So I'm not sure what to do with that
But is there a way to work it with the identity ..?

[next jackal]

Sin(70-x) = Cos(x)? is it right Ramonov?

[silent plank]

not quite

you didn't apply the identity properly

there's probably a different way you could do it, but may get very messy

[next jackal]

[silent plank]

i didn't say sin(a) = sin(90°-a)

[dark sparrow]

bruh

[silent plank]

sin is indeed a magical expression /s

[kindred rapids]

So expand it..instead?

[next jackal]

Ok, I got Sin(70)Cos(x)-Sin(x)Cos(70) = Cos(x)Cos(20)-Sin(x)Sin(20)

@silent plank Is there a simpler way to do this?

[kindred rapids]

Is it okay if I send the other question I need help with..? It's just two questions...this and the one I'm sending now.

[earnest echo]

it should be simpler if you apply something like cos(a) = sin(90°-a)

Use this only on one side

[kindred rapids]

@earnest echo yes but I don't understand what that is...

Could you explain?

[humble pulsar]

The one you just posted is double angle

[silent plank]

this is applying the complementary properties of sine and cosine

[earnest echo]

Could you explain?
e.g

sin(30°)=cos (90°-30°)

[silent plank]

which you could confirm using the compound angle identities

[next jackal]

So my expansion isn't helpful?

[silent plank]

not really, no

[next jackal]

Okayy

I got Cos(70-x) = Sin(x+20)

[kindred rapids]

So x is 35?

[silent plank]

applying that identity you'd reach
cos(70° - x) = cos(x°)
which allows you to easily identify one of the solutions

x=35 is one of the solutions,
there are actually infinitely many, but is probably not expected of you based on the phrasing / level you're at

[kindred rapids]

it should be simpler if you apply something like cos(a) = sin(90°-a)
@silent plank so this....is it always 90°...?,

[next jackal]

Ok I got Cos70Cos(x)+Sin70Sin(x)=Cos(x)

[silent plank]

its called a trig identity for a reason

why must you feel the need to apply the compound angle identity to expand cos(70°-x)

[humble pulsar]

There are like ~40-50 identities similar to cos(a) = sin(90-a)

all the related-corelated

[kindred rapids]

..ohh

[next jackal]

why must you feel the need to apply the compound angle identity to expand cos(70°-x)
@silent plank Because it's a compound angle?

[silent plank]

is it not clear that you could get one of the solutions from
equating the arguments

(and the rest if you introduce general solutions)

[next jackal]

I didn't get you

what else can you do about compound angles?

[silent plank]

cos(70° - x) = cos(x)
you can get one of the solutions simply equating
70° - x with x

[next jackal]

..........

[silent plank]

cosines of the same thing are equal

[next jackal]

Okay, thanks

[kindred rapids]

Guys...I sent the other question further up in this chat...can you all help with that please?
...is it double angle ?

[silent plank]

yes

[kindred rapids]

2tanA/ 1-tan^2A
?

[silent plank]

parentheses

pls

[brittle gyro]

I would suggest pyth

[kindred rapids]

I did that but ...how do I get tan2(theta) instead?
@brittle gyro

Would tan2(theta) be 8/3?

[silent plank]

stuff in inappropriate places

stuff like A and theta

[kindred rapids]

What are you referring to?@silent plank

[silent plank]

how do i get tan2(theta) instead

[kindred rapids]

...I don't get it

[silent plank]

$\tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)}$

[somber coyote]

ramonov:

[kindred rapids]

Yes..?

[silent plank]

then sub in your tan theta

[kindred rapids]

Alright then.

I got -24/7

[silent plank]

looks good

[kindred rapids]

Thank you all for the help

I'm grateful.

[next jackal]

@silent plank can you tell me how to derive tan(2x) identity?

I'm unable to figure it out

[silent plank]

from sin(2x)/cos(2x)

[next jackal]

Okay, I did that

[silent plank]

then apply double angle identities

[next jackal]

but I'm getting cos^2(x)-sin^2(x) as denominator

[silent plank]

and divide the numerator and denominator by cos^2(theta)

[upper karma]

but I'm getting cos^2(x)-sin^2(x) as denominator
There are more forms to that

[silent plank]

to get everything in terms of tan(theta)

[next jackal]

and divide the numerator and denominator by cos^2(theta)
@silent plank I got 2Sin(x)Cos(x)/[Cos^2(x)-Sin^2(x)]

[silent plank]

whoops i meant x

not theta

[next jackal]

okay, the denominator is cos(2x) right?

[silent plank]

divide the numerator and denominator by cos^2(x)

[next jackal]

okay#

Yeah got it. Thanks

Wait if I divide it by sin^2(x), I got 2cot(x)/(Cot^2(x)-1) Is that also one of the identities?

[wise hornet]

so tan(2x) = 2cot(x)/cot^2(x)-1

then cot(2x) = cot^2(x)-1/2cot(x)

which is nicer

[silent plank]

identities can have multiple forms

[high geode]

use a right triangle to write the expression as an algebratic expression. I'm a bit lost on this

[upper karma]

Yeah so

Can i see what you've tried

[high geode]

i ended up getting 5/x

[upper karma]

Like the whole process

Because maybe the triangle is well drawn

[high geode]

well sec is hyp over adj and tan is opp over adj

[upper karma]

...

Does it really hurt to send a picture of the whole thing

[high geode]

I ended up with 5/x

[upper karma]

Let me see

I'm guessing you want me to check your answer then

[high geode]

I was totally lost at first some of it came back to me a little

[upper karma]

Yeah 5/x is correct

Good job

[high geode]

oh thanks

[cosmic sun]

I made little visualization

the brightness is the ratio of a distance to point and line

and as you can see, a parabola emerges!

this is because the ratio on the parabola is going to 1, since by definition the distances between the point and line are equal

as such the brightness is going to be highest there

[upper karma]

got a cool unit circle acronym for y'all

AMERICAN STUDENTS TAKE COVER

[west basin]

lol ok

[earnest echo]

Wow

[unkempt venture]

we can j

Since p=IV

[devout echo]

So how do I find things like sss and sas for triangles, I didn’t understand it that much

[dark sparrow]

what do you mean by "find"

[devout echo]

Idk

My teacher says things so fast it’s hard for me to figure it out

[dark sparrow]

SSS and SAS (and ASA for that matter) are shorthands for triangle congruence tests

for example, if you can show two triangles have three pairs of matching sides (hence SSS, "side-side-side") they must necessarily be congruent

[devout echo]

Oh

[dark sparrow]

likewise if you find two pairs of matching sides and the angles in between also match (hence side-angle-side) that also gives you congruence

[devout echo]

Thanks

[valid dagger]

Can I ask for help regarding this question

I think ACB is 32° aswell but Im not sure how I can get ABC

[west basin]

ACB is 32

to get ABC subtract the two angle measurements in the triangle ABC from 180

since you know AB||OC you know AC bisects BAC and OAC

which means OAC = BAC

same reason ACB = ACO

[silent plank]

how are you getting ACB as 32

[west basin]

would that not be correct?

[silent plank]

doesn't sound right

there's no justification for it

[west basin]

is you are given that AB||OC you cant justify it?

idk geometry

[silent plank]

you aren't given that AO is parallel to BC,
instead you should apply stuff like angle sums / inscribed angle theorem to get <ABC first

[valid dagger]

so wait ACB isn't 32°?

So how am I supposed to find the other angle?

[dusky surge]

You can make use of OC and OA are the radii of the semicircle

[tender fable]

How do I sketch a graph that is domain 3-10 inclusive, and 1-4 inclusive? I get how to do it if there is only 2 brackets but how do i do it like this?

[silent plank]

do you have the exact question

are you plotting 2 different functions on the same plane?

[humble pulsar]

$x \in [3,10] \cup [1,4]$ ?

[tender fable]

it is one function, It does not give the function

[somber coyote]

moshill1:

[tender fable]

Do I just plug this in?

[humble pulsar]

no im confused what you're asking since the intervals you gave overlap

[tender fable]

I can just send the question if that makes it easier

[humble pulsar]

yeah that'd help

[dark sparrow]

how convenient of you to just skip over the word "range"

[tender fable]

my bad

[dark sparrow]

anyway you could probably just mark the points (3,1) and (10,4) and connect them with a straight line segment

[hybrid patrol]

https://i.imgur.com/RmwKWj9.png

whats the slope of this line

[upper karma]

Don't multipost

[cosmic pebble]

From

#❓how-to-get-help

Rule 3

Stick to one channel and don't post the same question in multiple channels. Please don't ask for help in other channels if no one is responding in the one you have posted your question in.

[steady bolt]

can i get help with this?

[round isle]

you know you can verify your answers by adding angles and see if they add up to 180. Like, add up angle 1, 2 and 3 and see if it equals 180 and also add angle 1 and 6 to see if the sum is 180

[steady bolt]

thank you

[final zenith]

I need to know the diagonal line in the middle and the one above that

[upper karma]

cos(25⁰) = 300/(length of diagonal line in middle)

[final zenith]

Yes i know that

But what is the answer

How do i get it

25 = cos^-1 (300/x)

What do i do now

[upper karma]

cos(25⁰) = 300/x

xcos(25⁰) = 300

x = 300/cos(25⁰)

[final zenith]

Does that work

[upper karma]

Yeah

[final zenith]

Wow i am stupid

[upper karma]

Practice makes perfect

usually

[final zenith]

I'll have to practice then

[flat viper]

help please?

and this
pi < x < 3pi/2 and 3sinx-5cosx = 0
cos^2.x - sin^2.x = ?

#geometry-and-trigonometry

[flat viper]

<@&286206848099549185>

[solar slate]

The lengths of two legs of an isosceles triangle are represented by the expressions (3x - 7) and (x + 5).

The perimeter of the triangle is 40cm.

Find the length of the base of the triangle.

Do NOT label your answer.

<@&286206848099549185>

[round isle]

the two sides will be equal since it is an isosceles triangle

so you can find x

use the perimeter to find the length of the base which is the remaining unknown side

[upper karma]

hi

[round isle]

hello

[warm rose]

can i get some help with this

just an efficient method would be great

[humble pulsar]

no real efficient way that im aware of

unless you were taught point-line distance formula

[warm rose]

i think the one most used

would be okay

if there is one

i know the distance formula though

[humble pulsar]

distance or point-line distance?

[warm rose]

a lot of steps i’ve seen are complicated to me

distance

i’m not sure what the point line one is

[humble pulsar]

Ok then dont use it lol

[warm rose]

ok

[humble pulsar]

but the typical way is you need to find the perpendicular to the base through the other vertex